[
  {
    "solution": "\\boxed{28^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, rectangle $ABCD$ is folded along the diagonal $BD$ such that point $C$ falls on point $F$, and $BF$ intersects $AD$ at point $E$. If $\\angle BDC = 62^\\circ$, what is the measure of $\\angle DBF$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral $ABCD$ is a rectangle,\\\\\nit follows that $AD\\parallel BC$, and $\\angle ADC=90^\\circ$,\\\\\nAlso, since $\\angle BDC=62^\\circ$,\\\\\nit then follows that $\\angle BDE=90^\\circ-\\angle BDC=90^\\circ-62^\\circ=28^\\circ$,\\\\\nthus, $\\angle CBD=\\angle BDE=28^\\circ$,\\\\\nSince the rectangle $ABCD$ is folded along the diagonal $BD$,\\\\\nit then follows that $\\angle FBD=\\angle CBD=\\boxed{28^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52076087.png"
  },
  {
    "solution": "\\boxed{x < 0}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y=kx+b$ is presented. What is the range of $x$ when $kx+b>3$?\n",
    "completion": "\\textbf{Solution:} Given that $\\because kx + b > 3$ and $y = kx + b$,\\\\\n$\\therefore y > 3$,\\\\\n$\\because$ when $y > 3$, by analyzing the graph, it is determined that the graph meeting the requirements intersects the y-axis above the given point,\\\\\n$\\therefore x < 0$, thus, the answer is: $\\boxed{x < 0}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52354786.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the length of line segment $AB$ is $12$, $C$ is a point on line segment $AB$, $AC=4$, $M$ is the midpoint of $AB$, $N$ is the midpoint of $AC$. Find the length of line segment $MN$.\n",
    "completion": "\\textbf{Solution:} Given: $AM = \\frac{1}{2}AB = 6$, $AN = \\frac{1}{2}AC = 2$\\\\\n$\\therefore MN = AM - AN = \\boxed{4}$",
    "image_path": "dataset/mm_math/images/MM_Math/53010381.png"
  },
  {
    "solution": "\\boxed{52.5^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, two triangular boards OAB and OCD are positioned with $\\angle ABO = 30^\\circ$ and $\\angle ODC = 45^\\circ$. If OM and ON are the angle bisectors of $\\angle AOD$ and $\\angle BOC$ respectively, what is the degree measure of $\\angle MON$?\n",
    "completion": "\\textbf{Solution:}\\\\\n$\\because \\angle ABO=30^\\circ$ and $\\angle ODC=45^\\circ$\\\\\n$\\therefore \\angle COB=90^\\circ-45^\\circ=45^\\circ$, $\\angle AOB=90^\\circ-30^\\circ=60^\\circ$\\\\\n$\\because$ OM and ON bisect $\\angle AOD$ and $\\angle BOC$ respectively\\\\\n$\\therefore \\angle NOB= \\frac{1}{2}\\angle COB= \\frac{1}{2}\\times45^\\circ=22.5^\\circ$\\\\\n$\\angle BOM= \\frac{1}{2}\\angle AOB= \\frac{1}{2}\\times60^\\circ=30^\\circ$\\\\\n$\\therefore \\angle MON=\\angle NOB+\\angle BOM=\\boxed{52.5^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/8928654_71.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{2}{3}x+\\frac{26}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, in the Cartesian coordinate system, point $O$ is the origin, and quadrilateral $ABCO$ is a rhombus. The coordinates of point $A$ are $(-5, 12)$, and point $C$ is located on the positive half of the x-axis. Line $AC$ intersects the y-axis at point $M$, and side $AB$ intersects the y-axis at $H$. Find the analytical expression of line $AC$.\n",
    "completion": "\\textbf{Solution:} Given that the side length of the rhombus $ABCO$ is $13$,\\\\\nhence $OC=13$,\\\\\nthus $C(13,0)$,\\\\\nlet the equation of line $AC$ be: $y=kx+b$ (where $k\\neq 0$),\\\\\nsubstituting $A(-5,12)$ and $C(13,0)$ into the equation yields: $\\begin{cases}-5k+b=12\\\\ 13k+b=0\\end{cases}$,\\\\\nsolving this system gives: $\\begin{cases}k=-\\frac{2}{3}\\\\ b=\\frac{26}{3}\\end{cases}$,\\\\\ntherefore, the equation of line $AC$ is: \\boxed{y=-\\frac{2}{3}x+\\frac{26}{3}}.",
    "image_path": "dataset/mm_math/images/MM_Math/52395385_99.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $\\angle BOC = 120^\\circ$, $BD = 12$, where $P$ is a moving point on side $AD$. What is the minimum value of $OP$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, if we draw OP$\\perp$AD from point O, then the length of OP is the smallest at this time.\\\\\nSince quadrilateral ABCD is a rectangle,\\\\\nit follows that AC=BD,\\\\\nSince $AO=\\frac{1}{2}AC$ and $DO=\\frac{1}{2}DB$,\\\\\nit follows that AO=DO=$\\frac{1}{2}DB=6$,\\\\\nSince $\\angle$AOD=$\\angle$BOC=120$^\\circ$,\\\\\nit follows that $\\angle$OAD=30$^\\circ$,\\\\\nSince $\\angle$OPA=90$^\\circ$,\\\\\nit follows that OP=$\\frac{1}{2}AO=\\boxed{3}$",
    "image_path": "dataset/mm_math/images/MM_Math/52422701.png"
  },
  {
    "solution": "\\boxed{(3,5)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: \\textit{As shown in the figure, the intersection point of the lines $y=mx-3m$ and $y=-\\frac{1}{2}x+n$ has a coordinate of 5. What is the solution set for the system of inequalities in terms of $x$ $\\begin{cases} -\\frac{1}{2}x+n>mx-3m \\\\ mx-3m>0 \\end{cases}$?}\n",
    "completion": "\\textbf{Solution:} Given that the intersection point of the lines $y=mx-3m$ and $y=-\\frac{1}{2}x+n$ has a coordinate of 5,\\\\\nit follows from the graph that when $-\\frac{1}{2}x+n>mx-3m$, it is solved that $x<5$;\\\\\nSince the graph shows that $y=mx-3m$ increases as $x$ increases,\\\\\nhence $m>0$\\\\\nTherefore, when $mx-3m>0$, it is solved that $x>3$;\\\\\nThus, $3<x<5$.\\\\\nSo, the answer is: $x\\in\\boxed{(3,5)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50482001.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{3}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within the quadrilateral $ABCD$, points E and F are the midpoints of sides AD and BC, respectively. Extend CD to point G such that DG=CD. Construct quadrilateral $DGME$ outside of quadrilateral $ABCD$ using DG and DE as sides. Connect BM to intersect AD at point N, and connect FN. If DG=DE=1 and $\\angle ADC=60^\\circ$, what is the length of FN?\n",
    "completion": "\\textbf{Solution:} As shown in the figure: connect EF, AF\\\\\n$\\because$ quadrilateral $DGME$, $DG=CD=1$,\\\\\n$\\therefore$ $DC\\parallel EF$,\\\\\n$\\because$ $\\angle ADC=60^\\circ$\\\\\n$\\therefore$ $\\angle AEF=60^\\circ$\\\\\n$\\because$ points E, F are the midpoints of sides AD, BC respectively\\\\\n$\\therefore$ $AB\\parallel EF$, $AE=DE$\\\\\n$\\therefore$ Quadrilateral ABFE is a parallelogram, $\\angle BAN=\\angle MEN$\\\\\n$\\because$ $DG=DC$, $DG=DE$\\\\\n$\\therefore$ $AE=EF=AB=ME=1$\\\\\n$\\because$ $\\angle AEF=60^\\circ$\\\\\n$\\therefore$ $\\triangle AEF$ is an equilateral triangle\\\\\nIn $\\triangle ABN$ and $\\triangle EMN$\\\\\n$\\left\\{\\begin{array}{l}\nME=AB \\\\\n\\angle MEN=\\angle BAN \\\\\n\\angle MNE=\\angle ANB\n\\end{array}\\right.$\\\\\n$\\therefore$ $\\triangle ABN \\cong \\triangle EMN$ (by SAA)\\\\\n$\\therefore$ $AN=NE$\\\\\n$\\therefore$ $NE=\\frac{1}{2}AE=\\frac{1}{2}$\\\\\n$\\therefore$ $FN\\perp AE$\\\\\n$\\therefore$ $FN=\\sqrt{1^2-\\left(\\frac{1}{2}\\right)^2}=\\boxed{\\frac{\\sqrt{3}}{2}}$",
    "image_path": "dataset/mm_math/images/MM_Math/53498591.png"
  },
  {
    "solution": "\\boxed{(0, -6)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the graph, in the Cartesian coordinate system $xOy$, the direct proportion function $y_{1}=-\\frac{3}{2}x$ and the inverse proportion function $y_{2}=\\frac{k}{x}$ intersect at point A in the second quadrant, and the x-coordinate of point A is $-2$. The coordinates of point B are $(-4, 0)$. If point P is on the y-axis, and the area of $\\triangle AOP$ is equal to the area of $\\triangle AOB$, find the coordinates of point P.\n",
    "completion": "\\textbf{Solution:} Since ${S}_{\\triangle AOB}=\\frac{1}{2}\\times 4\\times 3=6$,\\\\\nit follows that ${S}_{\\triangle APO}=\\frac{1}{2}\\times 2OP=OP$,\\\\\ntherefore $OP=6$,\\\\\nthus, the coordinates of point P are $\\boxed{(0, 6)}$ or $\\boxed{(0, -6)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52602886_85.png"
  },
  {
    "solution": "\\boxed{30}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle ABCO, point D is on side BC. Connect AD and fold $\\triangle ABD$ along AD, making point B exactly coincide with point E on side OC. Given B $(10, 8)$ and that the inverse proportion function $y= \\frac{k}{x}$ ($k\\neq 0$) passes through point D, what is the value of $k$?\n",
    "completion": "\\textbf{Solution:}\n\nGiven a rectangle ABCO with point B at (10, 8),\n\n\\(\\therefore \\) AB=OC=10, BC=OA=8,\n\n\\(\\because \\) triangle ABD is folded along AD such that point B exactly coincides with point E on side OC,\n\n\\(\\therefore \\) AE=AB=10, DE=BD,\n\n\\(\\therefore \\) OE\\(=\\sqrt{AE^{2}-OA^{2}}=\\sqrt{10^{2}-8^{2}}=6\\),\n\n\\(\\therefore \\) EC=OC-OE=10-6=4,\n\nLet CD=\\(b\\), and BD=DE=8-\\(b\\),\n\n\\(\\because \\) \\(CD^{2}+CE^{2}=DE^{2}\\),\n\n\\(\\therefore \\) \\(b^{2}+4^{2}=(8-b)^{2}\\),\n\nSolving for \\(b\\), we find \\(b=3\\),\n\n\\(\\therefore \\) The coordinates of point D are (10, 3),\n\n\\(\\because \\) the graph of the inverse proportion function passes through point D,\n\n\\(\\therefore \\) \\(k=10\\times 3=\\boxed{30}\\),",
    "image_path": "dataset/mm_math/images/MM_Math/52160918.png"
  },
  {
    "solution": "\\boxed{56^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, rectangle $ABCD$ is folded along the diagonal $BD$, with point $C$ falling on point $E$. $BE$ intersects $AD$ at point $F$. Given that $\\angle BDC=62^\\circ$, what is the measure of $\\angle DFE$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral ABCD is a rectangle, \\\\\n$\\therefore$ $AD\\parallel BC$ and $\\angle ADC=90^\\circ$, \\\\\nGiven $\\angle BDC=62^\\circ$, \\\\\n$\\therefore$ $\\angle FDB=90^\\circ-\\angle BDC=90^\\circ-62^\\circ=28^\\circ$, \\\\\nSince $AD\\parallel BC$, \\\\\n$\\therefore$ $\\angle CBD=\\angle FDB=28^\\circ$, \\\\\nSince rectangle ABCD is folded along the diagonal BD, \\\\\n$\\therefore$ $\\angle FBD=\\angle CBD=28^\\circ$, \\\\\n$\\therefore$ $\\angle DFE=\\angle FBD+\\angle FDB=28^\\circ+28^\\circ=\\boxed{56^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52354910.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, there is a right-angled triangular scrap ABC, where $\\angle BAC=90^\\circ$. Now, a rectangular strip DEFG is cut from it, where E and F are on BC, G is on AB, and D is on AC. If $BF=4.5\\text{cm}$, and $CE=2\\text{cm}$, then what is the length of the strip GF in centimeters?\n",
    "completion": "\\textbf{Solution:} Given $\\because$ $\\angle BAC=90^\\circ$,\\\\\n$\\therefore$ $\\angle AGD+\\angle ADG=90^\\circ$,\\\\\n$\\because$ rectangle GFED,\\\\\n$\\therefore$ $\\angle BFG=\\angle DEC=\\angle GFE=\\angle DEF=\\angle GDE=90^\\circ$, DG$\\parallel$BC, FG=DE,\\\\\n$\\therefore$ $\\angle ADG+\\angle CDE=90^\\circ$, $\\angle AGD=\\angle B$,\\\\\n$\\therefore$ $\\angle AGD=\\angle CDE$,\\\\\n$\\therefore$ $\\angle B=\\angle CDE$,\\\\\n$\\therefore$ $\\triangle BFG\\sim \\triangle DEC$,\\\\\n$\\therefore$ $\\frac{FG}{CE}=\\frac{BF}{DE}$\\\\\n$\\therefore$ FG$^{2}=BF\\cdot CE=4.5\\times 2=9$\\\\\nSolving for FG yields $\\boxed{3}$ (taking the positive value)",
    "image_path": "dataset/mm_math/images/MM_Math/52861189.png"
  },
  {
    "solution": "\\boxed{\\frac{2\\pi}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the radius of the circle $ \\odot O$ is 2 cm, and a regular hexagon is inscribed in $ \\odot O$. What is the area of the shaded region in the figure?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, connect BO, CO, and OA.\\\\\nFrom the given information, both $\\triangle OBC$ and $\\triangle AOB$ are equilateral triangles,\\\\\n$\\therefore \\angle AOB = \\angle OBC = 60^\\circ$,\\\\\n$\\therefore OA \\parallel BC$,\\\\\n$\\therefore$ The area of $\\triangle OBC$ is equal to the area of $\\triangle ABC$,\\\\\n$\\therefore$ The area of the shaded region is equal to the area of the sector OBC $= \\frac{60^\\circ \\pi \\times 2^2}{360^\\circ} = \\boxed{\\frac{2\\pi}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52766428.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{1}{2}x^{2}-\\frac{3}{2}x+2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, within the Cartesian coordinate system, the line $y=\\frac{1}{2}x+2$ intersects the x-axis at point A and the y-axis at point C. The parabola $y=ax^2+bx+c$ has its axis of symmetry at $x=-\\frac{3}{2}$ and passes through points A and C, with another intersection point with the x-axis being point B. Determine the equation of the parabola.\n",
    "completion": "Solution: The parabola $y=ax^{2}+bx+c$ passes through the points $A(-4, 0)$, $B(1, 0)$, so we can set the equation of the parabola to $y=a(x+4)(x-1)$. Moreover, the parabola passes through the point $C(0, 2)$, $\\therefore$ $-4a=2$, $\\therefore$ $a=-0.5$, \\\\\n$\\therefore$ \\boxed{y=-\\frac{1}{2}x^{2}-\\frac{3}{2}x+2}.",
    "image_path": "dataset/mm_math/images/MM_Math/51457966_269.png"
  },
  {
    "solution": "\\boxed{20−10\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the top of a building is equipped with a billboard CD. Xiao Ming measures the angle of elevation to the bottom of the billboard D to be $60^\\circ$ at point A at the foot of the hill. While walking up the hill to point B, he measures the angle of elevation to the top of the billboard C to be $45^\\circ$. Given that the slope angle of AB is $30^\\circ$, $AB=10$ meters, and $AE=15$ meters, what is the height of the billboard CD? (in meters)\n",
    "completion": "\\textbf{Solution:} Draw perpendicular lines from point B to AE and DE, with the perpendicular feet being G and F, respectively, as shown in the diagram. \\\\\nIn right triangle $\\triangle ABG$, $\\angle BAG=30^\\circ$ \\\\\n$\\therefore BG=\\frac{1}{2}AB=5$ meters, $AG=AB\\cdot \\cos 30^\\circ=10\\times \\frac{\\sqrt{3}}{2}=5\\sqrt{3}$ (meters) \\\\\n$\\therefore EG=AG+AE=(5\\sqrt{3}+15)$ meters \\\\\n$\\because BG\\perp AE$, $BF\\perp ED$, $AE\\perp ED$ \\\\\n$\\therefore$ Quadrilateral BGEF is a rectangle. \\\\\n$\\therefore EF=BG=5$ meters, $BF=EG=(5\\sqrt{3}+15)$ meters \\\\\n$\\because \\angle CBF=45^\\circ$, $BF\\perp ED$ \\\\\n$\\therefore \\angle BCF=\\angle CBF=45^\\circ$ \\\\\n$\\therefore CF=BF=(5\\sqrt{3}+15)$ meters \\\\\nIn right triangle $\\triangle DAE$, $\\angle DAE=60^\\circ$, AE=15 meters \\\\\n$\\therefore DE=AE\\cdot \\tan\\angle DAE=15\\sqrt{3}$(meters) \\\\\n$\\therefore CD=CF+EF−DE=5\\sqrt{3}+15+5−15\\sqrt{3}=\\boxed{20−10\\sqrt{3}}$ meters",
    "image_path": "dataset/mm_math/images/MM_Math/51446383.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a billboard CD is erected at the top of a building. Xiaoming and his classmates measured the elevation angle of the bottom of the billboard D to be $53^\\circ$ from point A at the foot of the hill. Walking up the slope AB to point B, they measured the elevation angle of the top of the billboard C to be $45^\\circ$. It is known that the slope $i$ of hill AB is $1\\colon \\sqrt{3}$, $AB=12$ meters, $AE=24$ meters. (The height of the theodolite is negligible, and the result should be accurate to 0.1 meters. Reference data: $\\sqrt{2}\\approx 1.41$, $\\sqrt{3}\\approx 1.73$, $\\sin 53^\\circ\\approx \\frac{4}{5}$, $\\cos 53^\\circ\\approx \\frac{3}{5}$, $\\tan 53^\\circ\\approx \\frac{4}{3}$). How many meters is point B above the horizontal ground AE?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, draw $BM\\perp AE$ and $BN\\perp CE$ through point $B$, with the perpendiculars touching points $M$ and $N$ respectively,\\\\\nit is known from the problem that $\\angle CBN=45^\\circ$, $\\angle DAE=53^\\circ$, $i=1\\colon \\sqrt{3}$, $AB=12$ meters, $AE=24$ meters,\\\\\nsince $i=1\\colon \\sqrt{3}=\\frac{BM}{AM}=\\tan\\angle BAM$,\\\\\nit follows that $\\angle BAM=30^\\circ$,\\\\\nthus $BM=\\frac{1}{2}AB=6$ (meters),\\\\\nwhich means the height of point $B$ from the horizontal ground $AE$ is $\\boxed{6}$ meters;",
    "image_path": "dataset/mm_math/images/MM_Math/51445934_97.png"
  },
  {
    "solution": "\\boxed{y=\\frac{1}{2}x}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, it is known that in parallelogram $OABC$, point $O$ is the coordinate vertex, point $A(3, 0)$, $C(1, 2)$, and the graph of the function $y=\\frac{k}{x} \\ (k\\ne 0)$ passes through point $C$. Determine the value of $k$ and the equation of line $OB$.\n",
    "completion": "\\textbf{Solution:} According to the problem, point $C(1, 2)$ is on the graph of the inverse proportion function $y=\\frac{k}{x}(k\\ne 0)$, \\\\\nhence $k=xy=2$, \\\\\nsince $A(3, 0)$, \\\\\nhence $CB=OA=3$, \\\\\nalso, $CB$ is parallel to the x-axis, \\\\\nhence $B(4, 2)$, \\\\\nlet the equation of line $OB$ be $y=ax$, \\\\\nhence $2=4a$, \\\\\nhence $a=\\frac{1}{2}$, \\\\\ntherefore, the equation of line $OB$ is $\\boxed{y=\\frac{1}{2}x}$;",
    "image_path": "dataset/mm_math/images/MM_Math/50677341_143.png"
  },
  {
    "solution": "\\boxed{1\\colon \\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the quadrilateral $ABCD$ is a rectangle, point $M$ is on $BC$, $BM=CD$, point $N$ is on $CD$, and $DN=CM$, $DM$ intersects with $BN$ at point $P$, then what is the ratio of $DM\\colon BN$?\n",
    "completion": "\\textbf{Solution:} Let $BM=CD=a$, $DN=CM=b$,\\\\\n$\\therefore BC=a+b$, $NC=a-b$,\\\\\n$\\because$ Quadrilateral $ABCD$ is a rectangle,\\\\\n$\\therefore \\angle DCB=90^\\circ$,\\\\\nIn right triangles $\\triangle DCM$ and $\\triangle BCN$, by the Pythagorean theorem, we have\\\\\n$DM=\\sqrt{DC^2+MC^2}=\\sqrt{a^2+b^2}$,\\\\\n$BN=\\sqrt{BC^2+CN^2}=\\sqrt{(a+b)^2+(a-b)^2}=\\sqrt{2}\\cdot\\sqrt{a^2+b^2}$,\\\\\n$\\therefore DM\\colon BN=1\\colon \\sqrt{2}$,\\\\\nTherefore, the answer is: $DM\\colon BN=\\boxed{1\\colon \\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50481553.png"
  },
  {
    "solution": "\\boxed{-x^{2}-\\frac{9}{2}x}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, in the Cartesian coordinate system $xOy$, the parabola $y=ax^2+bx$ passes through the point $(-2, 5)$, and intersects with the line $y=-\\frac{1}{2}x$ at point A in the second quadrant. Through point A, draw $AB\\perp x$ axis, with the foot of the perpendicular at point $B\\left(-4, 0\\right)$. Suppose P is a point on the parabola above the line OA, through point P draw $PC\\perp x$ axis at point C, intersecting OA at point D. Connect $OP$ and $PA$. What is the analytical expression of the parabola?\n",
    "completion": "\\textbf{Solution:} Let $AB$ be perpendicular to the $x$-axis, with the point $B(-4, 0)$,\\\\\n$\\therefore y=-\\frac{1}{2}\\times 4=-2$,\\\\\n$\\therefore A(-4, 2)$. Furthermore, since the parabola passes through $(-2, 5)$,\\\\\n$\\therefore \\left\\{\\begin{array}{l}\n16a-4b=2\\\\\n4a-2b=5\n\\end{array}\\right.$, solving this gives: $\\left\\{\\begin{array}{l}\na=-1\\\\\nb=-\\frac{9}{2}\n\\end{array}\\right.$,\\\\\n$\\therefore$ the equation of the parabola is $y=-x^{2}-\\frac{9}{2}x=\\boxed{-x^{2}-\\frac{9}{2}x}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51323941_196.png"
  },
  {
    "solution": "\\boxed{y=x+1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that in the Cartesian coordinate system $xOy$, the graph of the linear function $y=kx+b$ ($k\\neq 0$) passes through points A and $B(-1,0)$, and the graph of the inverse proportion function $y=\\frac{6}{x}$ also passes through point A, whose abscissa is 2. Find the analytical expression of the linear function.\n",
    "completion": "\\textbf{Solution:} Since the graph of the inverse proportion function $y=\\frac{6}{x}$ passes through point A, and the abscissa of point A is 2,\\\\\nhence $y_{A}=\\frac{6}{2}=3$, that is, $A(2,3)$.\\\\\nSince the graph of the linear function $y=kx+b(k\\ne 0)$ passes through points A, $B(-1,0)$,\\\\\nthus $\\left\\{\\begin{array}{l}3=2k+b \\\\ 0=-k+b \\end{array}\\right.$,\\\\\nsolving this, we get: $\\left\\{\\begin{array}{l}k=1 \\\\ b=1 \\end{array}\\right.$,\\\\\ntherefore the explicit formula of the linear function is $\\boxed{y=x+1}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52917576_249.png"
  },
  {
    "solution": "\\boxed{x>2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the graphs of the linear functions $y=kx+3$ and $y=-x+b$ intersect at point $P(2,4)$. What is the solution set of the linear inequality in $x$, $kx+3>-x+b$?\n",
    "completion": "\\textbf{Solution:} Solve: When $x>2$, we have $kx+3>-x+b$,\\\\\nwhich means the solution set of the inequality $kx+3>-x+b$ is $x>2$.\\\\\nTherefore, the answer is: $\\boxed{x>2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53706737.png"
  },
  {
    "solution": "\\boxed{12\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the rhombus $ABCD$, $CE \\perp AB$ at point $E$. If $CE=4$ and $AE=2BE$, what is the perimeter of the rhombus $ABCD$?\n",
    "completion": "\\textbf{Solution:} Given that $AE=2BE$, \\\\\nit follows that $AB=3BE$. \\\\\nSince quadrilateral $ABCD$ is a rhombus, \\\\\nwe have $AB=BC=CD=AD=3BE$. \\\\\nIn $\\triangle BCE$, we have $BE^2+CE^2=BC^2$, \\\\\ntherefore $BE^2+4^2=(3BE)^2$, solving this gives $BE=\\sqrt{2}$ (neglect the negative value), \\\\\nthus $AB=BC=CD=AD=3BE=3\\sqrt{2}$, \\\\\nhence the perimeter of rhombus $ABCD$ is $3\\sqrt{2}\\times 4=\\boxed{12\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51510416_127.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, $\\angle A=30^\\circ$. Points D and E are the midpoints of sides AB and AC, respectively. Extend BC to point F so that CF = $\\frac{1}{2}$BC. If EF = 4, what is the length of DE?\n",
    "completion": "\\textbf{Solution:} \nDraw $DC$,\\\\\nsince points D and E are respectively the midpoints of sides AB and AC,\\\\\nit follows that $DE\\parallel BC$ and $DE=\\frac{1}{2}BC$\\\\\nExtending BC to point F such that $CF=\\frac{1}{2}BC$,\\\\\nit follows that $CF=DE$ and $CF\\parallel DE$\\\\\nTherefore, quadrilateral $CDEF$ is a parallelogram\\\\\nThus $CD=EF=4$\\\\\nTherefore $AB=2CD=8$\\\\\nSince $\\angle ACB=90^\\circ$ and $\\angle A=30^\\circ$,\\\\\nit follows that $BC=\\frac{1}{2}AB=4$\\\\\nTherefore $DE=\\frac{1}{2}BC=\\boxed{2}$",
    "image_path": "dataset/mm_math/images/MM_Math/52344788.png"
  },
  {
    "solution": "\\boxed{18S}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, $P$ is a point inside rectangle $ABCD$. Lines are drawn connecting $P$ to each vertex of rectangle $ABCD$. The vertices of rectangle $EFGH$ lie respectively on $AP$, $BP$, $CP$, and $DP$. It is known that $AE = 2EP$ and $EF \\parallel AB$. The sum of the areas of the two shaded regions in the diagram is $S$. What is the area of rectangle $ABCD$?\n",
    "completion": "\\textbf{Solution:} Given that $AE=2EP$,\\\\\nit follows that $\\frac{PE}{PA}=\\frac{1}{3}$,\\\\\nsince quadrilaterals $ABCD$ and $EFGH$ are rectangles,\\\\\nwe have $\\angle DAB=\\angle HEF=90^\\circ$,\\\\\nsince $EF\\parallel AB$,\\\\\nit results in $\\angle PEF=\\angle PAB$,\\\\\nwhich leads to $\\angle PEH=\\angle PAD$,\\\\\nhence $EH\\parallel AD$,\\\\\nsimilarly, $FG\\parallel BC$,\\\\\nbecause $EH\\parallel AD$,\\\\\nit implies $\\triangle PEH\\sim \\triangle PAD$ with a similarity ratio of $\\frac{PE}{PA}=\\frac{1}{3}$,\\\\\nthus $\\frac{S_{\\triangle PEH}}{S_{\\triangle PAD}}=\\left(\\frac{PE}{PA}\\right)^2=\\frac{1}{9}$,\\\\\nsimilarly, $\\frac{S_{\\triangle PFG}}{S_{\\triangle PBC}}=\\frac{1}{9}$,\\\\\nsince $S_{\\triangle PAD}+S_{\\triangle PBC}=\\frac{1}{2}S_{\\text{rectangle}ABCD}$,\\\\\nit follows that $S=\\frac{1}{9}\\left(S_{\\triangle PAD}+S_{\\triangle PBC}\\right)=\\frac{1}{9}\\times \\frac{1}{2}S_{\\text{rectangle}ABCD}$,\\\\\ntherefore, the area of rectangle $ABCD$ is $\\boxed{18S}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53080285.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{5}}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, $OB=5$, $\\sin\\angle AOB= \\frac{3}{5}$, and the coordinates of point $A$ are $(10,0)$. What is the value of $\\sin\\angle OAB$?\n",
    "completion": "\\textbf{Solution:} Given the information in the problem, the coordinates of point A are (10, 0),\\\\\n$\\therefore OA=10$,\\\\\n$\\because OC=4$,\\\\\n$\\therefore AC=6$,\\\\\nIn $\\triangle ABC$, by Pythagoras' theorem we have\\\\\n$AB=\\sqrt{AC^2+BC^2}=\\sqrt{6^2+3^2}=3\\sqrt{5}$,\\\\\n$\\therefore \\sin\\angle OAB=\\frac{BC}{AB}=\\frac{3}{3\\sqrt{5}}=\\boxed{\\frac{\\sqrt{5}}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51083577_100.png"
  },
  {
    "solution": "\\boxed{8\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rhombus $ABCD$, $AB=4$, $E$ is the midpoint of $BC$, $AE \\perp BC$ at point $E$, $AF \\perp CD$ at point $F$, $CG \\parallel AE$, $CG$ intersects $AF$ at point $H$, and intersects $AD$ at point $G$. What is the area of the rhombus $ABCD$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, connect $AC$. Since $E$ is the midpoint of $BC$ and $AE\\perp BC$, it follows that $AB=AC$. Furthermore, since quadrilateral $ABCD$ is a rhombus, we have $AB=BC$. Therefore, $\\triangle ABC$ is an equilateral triangle. Hence, $BE=\\frac{1}{2}BC=2$, and $AE=\\sqrt{AB^{2}-BE^{2}}=\\sqrt{4^{2}-2^{2}}=2\\sqrt{3}$. Consequently, the area of rhombus $ABCD$ is given by $BC\\cdot AE=4\\times 2\\sqrt{3}=\\boxed{8\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51614476_115.png"
  },
  {
    "solution": "\\boxed{20}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AD$ is the altitude to side $BC$. Within this triangle, there is an inscribed rectangle $PQMN$, with one side lying on $BC$, and the other two vertices respectively on $AB$, $AC$. If $BC=60$ and $AD=40$, find the lengths of the sides of rectangle $PQMN$ when the area of $PQMN$ is maximized.\n",
    "completion": "\\textbf{Solution:} Let $PN=x$, \\\\\nthen $DE=PN=x$, $AE=AD-DE=40-x$, \\\\\nsince $\\triangle APQ \\sim \\triangle ABC$, \\\\\nwe have $\\frac{AE}{AD}=\\frac{PQ}{BC}$, i.e., $\\frac{40-x}{40}=\\frac{PQ}{60}$, \\\\\nthus $PQ=60-\\frac{3}{2}x$, \\\\\ntherefore, the area of rectangle $PQMN$ is $PQ \\cdot PN= \\left(60-\\frac{3}{2}x\\right)x$ \\\\\n$= -\\frac{3}{2}x^{2}+60x$ \\\\\n$= -\\frac{3}{2}(x-20)^{2}+600$, \\\\\nwhen $x=20$, the area of rectangle $PQMN$ reaches its maximum value of $600$, \\\\\nat this point, $PN=20, PQ=60-\\frac{3}{2}x=60-\\frac{3}{2}\\times 20=30$. \\\\\nTherefore, the dimensions of the rectangle are $\\boxed{30}$ and $\\boxed{20}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52602086_189.png"
  },
  {
    "solution": "\\boxed{D(-3, 2)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, it is known that the graph of the linear function $y=\\frac{1}{2}x+1$ intersects the $x$ axis and the $y$ axis at points $A$ and $B$, respectively. A square $ABCD$ is constructed in the second quadrant with $AB$ as its side. What are the coordinates of points $C$ and $D$?\n",
    "completion": "\\textbf{Solution:} Construct $CE \\perp y$-axis and $DF \\perp x$-axis, we can get $\\angle CEB = \\angle AFD = \\angle AOB = 90^\\circ$,\\\\\nsince square $ABCD$,\\\\\nthus $BC = AB = AD$, $\\angle DAB = \\angle ABC = 90^\\circ$,\\\\\nthus $\\angle DAF + \\angle BAO = 90^\\circ$, $\\angle ABO + \\angle CBE = 90^\\circ$,\\\\\nsince $\\angle DAF + \\angle ADF = 90^\\circ$, $\\angle BAO + \\angle ABO = 90^\\circ$,\\\\\nthus $\\angle BAO = \\angle ADF = \\angle CBE$,\\\\\nthus $\\triangle BCE \\cong \\triangle DAF \\cong \\triangle ABO$,\\\\\nthus $BE = DF = OA = 2$, $CE = AF = OB = 1$,\\\\\nthus $OE = OB + BE = 2 + 1 = 3$, $OF = OA + AF = 2 + 1 = 3$,\\\\\nthus $\\boxed{C(-1, 3)}$, $\\boxed{D(-3, 2)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52386761_124.png"
  },
  {
    "solution": "\\boxed{66^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle A=90^\\circ$, $\\angle B=38^\\circ$, points E and F are located on sides BC and AC respectively. Fold $\\triangle CEF$ along the line containing EF, so that the corresponding point $C'$ of C falls on AB, and $CE=BC$, what is the measurement of $\\angle AFC'$?\n",
    "completion": "\\textbf{Solution:} Given $\\because$ $\\angle A=90^\\circ$,\\\\\n$\\therefore$ $\\angle B+\\angle C=90^\\circ$,\\\\\n$\\therefore$ $\\angle C=90^\\circ-38^\\circ=52^\\circ$,\\\\\n$\\because$ folding $\\triangle CEF$ along the line containing $EF$, such that the corresponding point of $C$, denoted $C'$, falls on $AB$,\\\\\n$\\therefore$ $C'E=CE=BC'$\\\\\n$\\therefore$ $\\angle CEF=\\angle C'EF=\\frac{1}{2}(180^\\circ-38^\\circ)=71^\\circ$, $\\angle CFE=\\angle C'FE$,\\\\\n$\\therefore$ $\\angle CFE=\\angle C'FE=180^\\circ-\\angle C-\\angle CEF=180^\\circ-52^\\circ-71^\\circ=57^\\circ$,\\\\\n$\\therefore$ $\\angle AFC' = 180^\\circ-\\angle CFE-\\angle C'FE=180^\\circ-57^\\circ-57^\\circ=\\boxed{66^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52815748.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{1}{4}x^2+\\frac{1}{2}x+\\frac{15}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the parabola $C_1\\colon y=-\\frac{1}{4}x^2+bx+c$ passes through the points $(-3,0)$ and $(3,3)$, and intersects the y-axis at point C. What is the equation of the parabola $C_1$?\n",
    "completion": "\\textbf{Solution:} Since the graph of $y=-\\frac{1}{4}x^2+bx+c$ passes through the points $(-3,0)$ and $(3,3)$,\\\\\nwe have \n\\[\n\\begin{cases}\n-\\frac{1}{4}(-3)^2+3b+c=0\\\\\n-\\frac{1}{4}3^2+3b+c=3\n\\end{cases},\n\\]\nfrom which we deduce\n\\[\n\\begin{cases}\nb=\\frac{1}{2}\\\\\nc=\\frac{15}{4}\n\\end{cases},\n\\]\nTherefore, the equation of the parabola $C_1$ is $\\boxed{y=-\\frac{1}{4}x^2+\\frac{1}{2}x+\\frac{15}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52916776_257.png"
  },
  {
    "solution": "\\boxed{70^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the direction of ray $OA$ is $15^\\circ$ east of north, and the direction of ray $OB$ is $40^\\circ$ west of north. $OA$ is the angle bisector of $\\angle BOC$. $OF$ is the straight extension in the opposite direction of $OB$. What is the degree measure of $\\angle COF$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle BOC = \\angle AOB + \\angle AOC$\\\\\n$=55^\\circ \\times 2 = 110^\\circ$,\\\\\nthen $\\angle COF = 180^\\circ - \\angle BOC$\\\\\n$=180^\\circ - 110^\\circ$\\\\\n$=70^\\circ$.\\\\\nTherefore, the measure of $\\angle COF$ is $\\boxed{70^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53404231_76.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC \\cong \\triangle DEF$, with points B, E, C, F lying on the same straight line. If $BC=7$ and $EC=4$, then what is the length of $CF$?\n",
    "completion": "\\textbf{Solution:} Given that $\\triangle ABC \\cong \\triangle DEF$,\\\\\nhence $BC = EF$.\\\\\nSince points B, E, C, and F are on the same line, and $BC = 7$, $EC = 4$,\\\\\ntherefore $CF = EF - EC = BC - EC = 7 - 4 = \\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51368820.png"
  },
  {
    "solution": "\\boxed{35^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, point $P$ is the midpoint of diagonal $BD$, points $E$ and $F$ are the midpoints of $AB$ and $CD$, respectively, $AD=BC$, $\\angle CBD=30^\\circ$, and $\\angle ADB=100^\\circ$, then what is the measure of $\\angle PFE$?\n",
    "completion": "\\textbf{Solution:} Since point P is the midpoint of BD, and point E is the midpoint of AB,\\\\\nit follows that PE is the midline of $\\triangle ABD$,\\\\\ntherefore PE=$\\frac{1}{2}$AD, PE$\\parallel$AD,\\\\\nthus $\\angle$EPD=$180^\\circ-\\angle$ADB=$80^\\circ$,\\\\\nSimilarly, PF=$\\frac{1}{2}$BC, PF$\\parallel$BC,\\\\\nthus $\\angle$FPD=$\\angle$CBD=$30^\\circ$,\\\\\nsince AD=BC,\\\\\nit follows that PE=PF,\\\\\nthus $\\angle$PFE=$\\frac{1}{2}\\times(180^\\circ-110^\\circ)=\\boxed{35^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52416424.png"
  },
  {
    "solution": "\\boxed{\\sqrt{34}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the side length of the square $ABCD$ is $8$. On each side, we sequentially take $AE=BF=CG=DH=5$. What is the area of the quadrilateral $EFGH$?\n",
    "completion": "\\textbf{Solution:}\\\\\nSince quadrilateral ABCD is a square,\\\\\n$\\therefore$ $\\angle A=\\angle B=\\angle C=\\angle D=90^\\circ$, AB=BC=CD=DA,\\\\\nSince AE=BF=CG=DH,\\\\\n$\\therefore$ AH=BE=CF=DG.\\\\\nIn $\\triangle AEH$, $\\triangle BFE$, $\\triangle CGF$, and $\\triangle DHG$,\\\\\n$\\left\\{\\begin{array}{l}\nAE=BF=CG=DH \\\\\n\\angle A=\\angle B=\\angle C=\\angle D \\\\\nAH=BE=CF=DG\n\\end{array}\\right.$\\\\\n$\\therefore$ $\\triangle AEH \\cong \\triangle BFE \\cong \\triangle CGF \\cong \\triangle DHG$ (SAS),\\\\\n$\\therefore$ EH=FE=GF=GH, $\\angle AEH=\\angle BFE$,\\\\\n$\\therefore$ Quadrilateral EFGH is a rhombus,\\\\\nSince $\\angle BEF+\\angle BFE=90^\\circ$,\\\\\n$\\therefore$ $\\angle BEF+\\angle AEH=90^\\circ$,\\\\\n$\\therefore$ $\\angle HEF=90^\\circ$,\\\\\n$\\therefore$ Quadrilateral EFGH is a square,\\\\\nSince AB=BC=CD=DA=8, AE=BF=CG=DH=5,\\\\\n$\\therefore$ EH=FE=GF=GH=$\\sqrt{5^2+3^2}=\\sqrt{34}$\\\\\nThus, the area of the square EFGH is $\\boxed{\\sqrt{34}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52250669.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, within the Cartesian coordinate system, the coordinates of point $B$ are $(0, 5)$. After $\\triangle OAB$ is translated to the right along the $x$-axis, it becomes $\\triangle O^{\\prime}A^{\\prime}B^{\\prime}$. The corresponding point of $B$, point $B^{\\prime}$, is on the line $y=\\frac{5}{6}x$. What is the distance between point $A$ and its corresponding point $A^{\\prime}$?\n",
    "completion": "\\textbf{Solution:} Given that point B has coordinates (0,5), and triangle $OAB$ is translated to the right along the x-axis to form triangle $O'A'B'$, where the corresponding point of B, denoted as B', lies on the line $y=\\frac{5}{6}x$,\\\\\n$\\therefore$ the y-coordinate of point B' is: 5,\\\\\nSo, we have $5=\\frac{5}{6}x$,\\\\\nSolving this gives: $x=6$,\\\\\nimplying the distance from B to B' is 6,\\\\\nthus, the distance between point A and its corresponding point A' is $\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52425091.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points C and D lie on a semicircle with AB as the diameter. AD bisects $\\angle BAC$, and the line segment OD intersects BC at point E. OD $\\parallel$ AC. If AB $=10$ and BC $=8$, and we connect BD, find the length of BD.\n",
    "completion": "\\textbf{Solution:} Let AD intersect BC at M,\\\\\nbecause AB is the diameter of circle O,\\\\\ntherefore, $\\angle C=\\angle ADB=90^\\circ$,\\\\\nbecause OD$\\parallel$AC,\\\\\ntherefore, $\\angle DEM=90^\\circ$,\\\\\nby the Pythagorean theorem, we have $AC=\\sqrt{AB^{2}-BC^{2}}=\\sqrt{10^{2}-8^{2}}=6$,\\\\\nbecause AO=BO and OD$\\parallel$AC,\\\\\ntherefore, CE=BE=$\\frac{1}{2}\\times 8=4$,\\\\\nbecause AC=6,\\\\\ntherefore, OE=$\\frac{1}{2}$AC=3,\\\\\nbecause OD=$\\frac{1}{2}$AB=$\\frac{1}{2}\\times 10=5$,\\\\\ntherefore, DE=5-3=2,\\\\\nbecause OD$\\parallel$AC,\\\\\ntherefore, $\\triangle DEM\\sim \\triangle ACM$,\\\\\ntherefore, $\\frac{AC}{DE}=\\frac{CM}{EM}$,\\\\\ntherefore, $\\frac{6}{2}=\\frac{4-EM}{EM}$,\\\\\nsolving for EM gives: EM=1, thus, BM=4+1=5,\\\\\nin $\\triangle DEM$, by the Pythagorean theorem, we have $DM=\\sqrt{DE^{2}+ME^{2}}=\\sqrt{2^{2}+1^{2}}=\\sqrt{5}$,\\\\\nin $\\triangle MDB$, by the Pythagorean theorem, we have $BD=\\sqrt{BM^{2}-DM^{2}}=\\sqrt{5^{2}-(\\sqrt{5})^{2}}=\\boxed{2\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53009966_90.png"
  },
  {
    "solution": "\\boxed{(4, 1)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, it is known that the line $y=kx+b$ passes through points $A\\left(6, 0\\right)$ and $B\\left(2, 2\\right)$. If the line $y=x-3$ intersects with line $AB$ at point $C$, find the coordinates of point $C$.\n",
    "completion": "\\textbf{Solution}: Given the problem, if the line $y=x-3$ intersects with line $AB$ at point $C$,\\\\\nthen we have the system of equations $\\left\\{\\begin{array}{l}y=-\\frac{1}{2}x+3 \\\\ y=x-3\\end{array}\\right.$, solving this gives $\\left\\{\\begin{array}{l}x=4\\\\ y=1\\end{array}\\right.$\\\\\n$\\therefore$ The coordinates of point $C$ are $\\boxed{(4, 1)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52056013_216.png"
  },
  {
    "solution": "\\boxed{31^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $O$ is on the line $AB$, $OD$ bisects $\\angle COB$, and $\\angle AOC=118^\\circ$. What is the degree measure of $\\angle COD$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle AOC$ and $\\angle BOC$ are adjacent supplementary angles,\\\\\nwe have $\\angle AOC+\\angle BOC=180^\\circ$,\\\\\nGiven that $\\angle AOC=118^\\circ$,\\\\\nthus $\\angle BOC=180^\\circ-118^\\circ=62^\\circ$,\\\\\nSince $OD$ bisects $\\angle BOC$,\\\\\ntherefore $\\angle COD=\\frac{1}{2}\\angle BOC=\\boxed{31^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53404318.png"
  },
  {
    "solution": "\\boxed{\\frac{5}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, both $\\triangle ADB$ and $\\triangle BCD$ are equilateral triangles. Extend $AD$ to $E$ such that $\\angle AEC=90^\\circ$, and $AD=5$. A moving point $M$ starts from point $B$ and moves towards $D$ at a speed of $1$ unit per second, while point $N$ moves from $D$ towards $C$ at a speed of $2$ units per second. The movement stops when one of them reaches their endpoint. Connect $AM$, $CM$, $MN$, $NE$. Let the movement time be $t$ ($0\\leq t\\leq 2.5$). At what value of $t$ is $MN\\parallel BC$?\n",
    "completion": "\\textbf{Solution:} Given that $\\triangle ADB$ and $\\triangle BCD$ are both equilateral triangles, and $AD=5$,\\\\\nthus $BD=DC=AD=5$,\\\\\ntherefore $BM=t$, $DN=2t$,\\\\\ngiven $MN\\parallel BC$;\\\\\nthus $\\angle NMD=\\angle DBC=60^\\circ=\\angle MDN$,\\\\\ntherefore $\\triangle MDN$ is an equilateral triangle,\\\\\nthus $DM=DN$, i.e., $5-t=2t$,\\\\\nsolving gives $t=\\boxed{\\frac{5}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51403181_162.png"
  },
  {
    "solution": "\\boxed{5\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of a semicircle, and $BC$ is a chord of the semicircle. The arc $\\overset{\\frown}{BC}$ is folded along the chord $BC$ to intersect the diameter $AB$ at point D. When $AD=BD=5$, what is the length of $BC$?\n",
    "completion": "\\textbf{Solution:} We have $BC = \\boxed{5\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51458277_54.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $OA \\perp OB$, $\\angle COD = 60^\\circ$. If $OC$ bisects $\\angle AOD$, what is the degree measure of $\\angle BOC$?\n",
    "completion": "\\textbf{Solution:} Since $OC$ bisects $\\angle AOD$, $\\angle COD=60^\\circ$,\\\\\nthus $\\angle AOC=\\angle COD=60^\\circ$,\\\\\nsince $OA\\perp OB$,\\\\\nthus $\\angle AOB=90^\\circ$,\\\\\ntherefore $\\angle BOC=\\angle AOB-\\angle AOC=90^\\circ-60^\\circ=\\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51482433_174.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, $ \\triangle ABC$ is inscribed in circle $\\odot O$, with D and F respectively on arcs $\\overset{\\frown}{AC}$ and $\\overset{\\frown}{AB}$, and $\\overset{\\frown}{BF} = \\overset{\\frown}{DA}$. Extend AF to intersect the extension of CB at point E, and connect AD, CD. $\\triangle AEB \\sim \\triangle CAD$; if $\\triangle ABC$ is an equilateral triangle with side length 6, and $AD = 3, AE = 8$. Find $\\tan\\angle DAC$?\n",
    "completion": "\\textbf{Solution:} Given that $\\because \\triangle AEB\\sim \\triangle CAD$,\\\\\n$\\therefore \\frac{AD}{EB}=\\frac{AC}{EA}$, $\\angle DAC=\\angle AEC$,\\\\\n$\\because AD=3, AE=8, AB=AC=BC=6$,\\\\\n$\\therefore \\frac{3}{EB}=\\frac{6}{8}$,\\\\\n$\\therefore EB=4$,\\\\\n$\\therefore EC=EB+BC=4+6=10$,\\\\\n$\\therefore AE^2+AC^2=8^2+6^2=10^2=EC^2$,\\\\\n$\\therefore \\triangle AEC$ is a right-angled triangle,\\\\\n$\\therefore \\tan\\angle DAC=\\tan\\angle AEC=\\frac{AC}{AE}=\\frac{6}{8}=\\boxed{\\frac{3}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53043623_131.png"
  },
  {
    "solution": "\\boxed{40^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle B=46^\\circ$, $\\angle C=54^\\circ$, $AD$ bisects $\\angle BAC$, intersects $BC$ at $D$, and $DE\\parallel AB$, intersecting $AC$ at $E$. What is the measure of $\\angle ADE$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle B=46^\\circ$ and $\\angle C=54^\\circ$, \\\\\nit follows that $\\angle BAC=180^\\circ-\\angle B-\\angle C=180^\\circ-46^\\circ-54^\\circ=80^\\circ$, \\\\\nSince AD bisects $\\angle BAC$, \\\\\nit follows that $\\angle BAD=\\frac{1}{2}\\angle BAC=40^\\circ$, \\\\\nSince DE $\\parallel$ AB, \\\\\nit follows that $\\angle ADE=\\angle BAD=\\boxed{40^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51212034.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, $AD$ is the median, $E$ is the midpoint of $AD$, and $F$ is the midpoint of $BE$. Connect $DF$. If $AC=4\\sqrt{3}$ and $DF\\perp BE$, then what is the length of $DF$?\n",
    "completion": "\\textbf{Solution:} Connect CE. \\\\\nSince $AD$ is the median of side $BC$, and $F$ is the midpoint of $BE$, \\\\\ntherefore $DF$ is the median of $\\triangle BCE$, \\\\\nthus $CE = 2DF$, and $DF \\parallel CE$, \\\\\ntherefore $\\angle BDF = \\angle DCE$, $\\angle EDF = \\angle DEC$, \\\\\nalso since $DF \\perp BE$, \\\\\ntherefore $\\angle EDF = \\angle BDF$, \\\\\ntherefore $\\angle DEC = \\angle DCE$, \\\\\ntherefore $CD = ED$, \\\\\nsince $E$ is the midpoint of $AD$, and $\\angle ACB = 90^\\circ$, \\\\\ntherefore $CE = ED = CD = \\frac{1}{2}AD$, \\\\\ntherefore $AD = 4DF$, \\\\\nsince $AC = 4\\sqrt{3}$, \\\\\ntherefore $AD^{2} - CD^{2} = AD^{2} - (\\frac{1}{2}AD)^{2} = AC^{2} = 48$, \\\\\ntherefore $AD = 8$, \\\\\ntherefore $DF = \\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52302537.png"
  },
  {
    "solution": "\\boxed{(-4, -6)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the parabola $y=ax^2+bx-4$ passes through the points $A(-8, 0)$ and $B(2, 0)$, and the line $x=-4$ intersects the x-axis at point $C$ and intersects the parabola at point $D$. Point $P$ is on the parabola, and point $E$ is on the line $x=-4$. If the quadrilateral with vertices $A$, $O$, $E$, $P$ is a parallelogram, find the coordinates of point $P$.\n",
    "completion": "\\textbf{Solution:} Let point $P(m, \\frac{1}{4}m^{2}+\\frac{3}{2}m-4)$. When $AO$ is a side,\\\\\n$\\because$ the quadrilateral formed by vertices $A$, $O$, $E$, $P$ is a parallelogram, with point $A(0, -8)$,\\\\\n$\\therefore AO=PE=8$,\\\\\n$\\therefore |m+4|=8$,\\\\\nSolving gives: $m_{1}=-12$, $m_{2}=4$,\\\\\nWhen $m=4$, $\\frac{1}{4}m^{2}+\\frac{3}{2}m-4=\\frac{1}{4}\\times 16+\\frac{3}{2}\\times 4-4=6$;\\\\\nWhen $m=-12$, $\\frac{1}{4}m^{2}+\\frac{3}{2}m-4=\\frac{1}{4}\\times 144+\\frac{3}{2}\\times (-12)-4=14$,\\\\\n$\\therefore$ the coordinates of point $P$ are $(-12, 14)$ or $(4, 6)$.\\\\\nWhen $AO$ is a diagonal,\\\\\n$\\because$ point $A(-8, 0)$, point $C(-4, )$,\\\\\n$\\therefore$ points $A$, $O$ are symmetric about the line $x=-4$, the intersection points of line $x=-4$ with the parabola are $P$,\\\\\nWhen $x=-4$, $\\frac{1}{4}x^{2}+\\frac{3}{2}x-4=\\frac{1}{4}\\times 16-\\frac{3}{2}\\times 4-4=-6$,\\\\\n$\\therefore$ the coordinates of point $P$ are $(-4, -6)$.\\\\\n$\\therefore$ the coordinates of point $P$ are $\\boxed{(-12, 14)}$ or $\\boxed{(4, 6)}$ or $\\boxed{(-4, -6)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52936367_231.png"
  },
  {
    "solution": "\\boxed{\\frac{12}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB=4$, $CD=6$, point $F$ is on $BD$, $BC$ and $AD$ intersect at point $E$, and $AB \\parallel CD \\parallel EF$. What is the length of $EF$?\n",
    "completion": "\\textbf{Solution:} Since $CD\\parallel EF$, \\\\\n$\\therefore \\triangle BEF\\sim \\triangle BCD$, \\\\\n$\\therefore \\frac{EF}{CD}=\\frac{BF}{BD}$, \\\\\nSimilarly: $\\frac{EF}{AB}=\\frac{DF}{BD}$, \\\\\n$\\therefore \\frac{EF}{CD}+\\frac{EF}{AB}=\\frac{BF}{BD}+\\frac{DF}{BD}=1$, \\\\\n$\\therefore \\frac{EF}{6}+\\frac{EF}{4}=1$, \\\\\nSolving for $EF$, we get $EF=\\boxed{\\frac{12}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51445794_167.png"
  },
  {
    "solution": "\\boxed{5\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the angle of elevation to the top of the flagpole A at point C is measured to be $30^\\circ$. Moving forward 10 meters to point D, the angle of elevation to point A is measured to be $60^\\circ$. What is the height of the flagpole? Meters.\n",
    "completion": "\\textbf{Solution:} Given from the problem, $\\angle C=30^\\circ$ and $\\angle ADB=60^\\circ$,\\\\\n$\\therefore \\angle DAC=\\angle ADB-\\angle C=30^\\circ$,\\\\\n$\\therefore \\angle DAC=\\angle C$,\\\\\n$\\therefore AD=DC=10$ meters,\\\\\nIn $\\triangle ADB$, $\\sin \\angle ADB= \\frac{AB}{AD}$,\\\\\nThen $AB=AD \\cdot \\sin \\angle ADB=10 \\times \\frac{\\sqrt{3}}{2} = 5\\sqrt{3}$ (meters),\\\\\nTherefore, the height of the flagpole is $AB=\\boxed{5\\sqrt{3}}$ (meters).",
    "image_path": "dataset/mm_math/images/MM_Math/51174881.png"
  },
  {
    "solution": "\\boxed{18}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, four triangular paper pieces $\\triangle ABC$, $\\triangle AB_{1}C_{1}$, $\\triangle AB_{2}C_{2}$, and $\\triangle AB_{3}C_{3}$ are perfectly superimposed and arranged as depicted. Given $BC=\\sqrt{10}$ and $AB=1$, what is the area of quadrilateral $CC_{1}C_{2}C_{3}$?\n",
    "completion": "\\textbf{Solution:} From the given problem, we have Rt$\\triangle ABC \\cong$ Rt$\\triangle AB_{1}C_{1} \\cong$ Rt$\\triangle AB_{2}C_{2} \\cong$ Rt$\\triangle AB_{3}C_{3}$,\\\\\n$\\therefore AC = AC_{1} = AC_{2} = AC_{3}$, and AB$ = AB_{1} = 1$.\\\\\nIn Rt$\\triangle ABC$, by the Pythagorean theorem, we have $AC = \\sqrt{BC^{2} - AB^{2}} = 3$,\\\\\n$\\therefore Area_{\\text{quadrilateral} CC_{1}C_{2}C_{3}} = 4 \\times \\frac{3 \\times 3}{2} = \\boxed{18}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52836476.png"
  },
  {
    "solution": "\\boxed{(-3, 6)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given the parabola $y=ax^2+bx+3$ intersects the x-axis at points A and $B(1,0)$, and the y-axis at point C. Connect $AC$. A moving point D is on the line segment $AC$. Draw a perpendicular line from D to the x-axis, which intersects the parabola at point E and the x-axis at point F. Given $AB=4$, let the x-coordinate of point D be $m$. When $m=-2$, is there a point Q in the plane such that the quadrilateral formed by vertices B, C, E, Q is a parallelogram? If such a point exists, please provide the coordinates of point Q directly; if not, please explain why.\n",
    "completion": "\\textbf{Solution:} When point $Q$ is at $\\boxed{(3, 0)}$, $\\boxed{(-1, 0)}$, or $\\boxed{(-3, 6)}$, the quadrilateral formed by vertices B, C, E, and Q is a parallelogram.",
    "image_path": "dataset/mm_math/images/MM_Math/53107133_273.png"
  },
  {
    "solution": "\\boxed{(2, 1)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola passes through points $A(4, 0)$, $B(1, 0)$, and $C(0, -2)$. There exists a point $D$ on the parabola above line $AC$ for which the area of $\\triangle DCA$ is maximized. Determine the coordinates of point $D$.\n",
    "completion": "\\textbf{Solution:} Let the $x$-coordinate of point $D$ be $t(0<t<4)$, then the $y$-coordinate of point $D$ is $\\left|-\\frac{1}{2}t^{2}+\\frac{5}{2}t-2\\right|$.\\\\\nDraw a line parallel to the $y$-axis through point $D$ that intersects $AC$ at point $E$.\\\\\nFrom the given information, the equation of line $AC$ is $y=\\frac{1}{2}x-2$.\\\\\n$\\therefore$ The coordinates of point $E$ are $\\left(t, \\frac{1}{2}t-2\\right)$.\\\\\n$\\therefore DE=-\\frac{1}{2}t^{2}+\\frac{5}{2}t-2-\\left(\\frac{1}{2}t-2\\right)=-\\frac{1}{2}t^{2}+2t$,\\\\\n$\\therefore S_{\\triangle DAC}=S_{\\triangle DCE}+S_{\\triangle DEA}=\\frac{1}{2}DE\\cdot h+\\frac{1}{2}DE\\cdot (4-h)=\\frac{1}{2}DE\\cdot 4$,\\\\\n$\\therefore S_{\\triangle DAC}=\\frac{1}{2}\\times \\left(-\\frac{1}{2}t^{2}+2t\\right)\\times 4=-t^{2}+4t=-(t-2)^{2}+4$,\\\\\n$\\therefore$ When $t=2$, the area of $\\triangle DAC$ is maximized,\\\\\n$\\therefore D\\left(2, 1\\right)$. Hence, the coordinates of point $D$ are $\\boxed{(2, 1)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52513841_163.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within a circle $\\odot O$ with a radius of $50\\,mm$, the chord $AB$ has a length of $50\\,mm$. Find the measure of $\\angle AOB$ in degrees.\n",
    "completion": "\\textbf{Solution:} Since OA and OB are radii of the circle O, \\\\\nthus OA = OB = 50mm, \\\\\nand since AB = 50mm, \\\\\ntherefore OA = OB = AB, \\\\\ntherefore $\\triangle$ AOB is an equilateral triangle, \\\\\ntherefore $\\angle$AOB = \\boxed{60^\\circ}",
    "image_path": "dataset/mm_math/images/MM_Math/52978029_150.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point C is on the line segment $AD$, and point B is the midpoint of $CD$, with $AD=9\\,cm$, $BC=2\\,cm$. What is the length of $AC$ in cm?\n",
    "completion": "Solution: Since $B$ is the midpoint of $CD$, and $BC=2\\,cm$, \\\\\nit follows that $CD=2BC=4\\,cm$, \\\\\nGiven $AD=9\\,cm$, \\\\\nthus $AC=AD-CD=\\boxed{5}\\,cm$.",
    "image_path": "dataset/mm_math/images/MM_Math/53066707_136.png"
  },
  {
    "solution": "\\boxed{72^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated, with point C of the regular pentagon ABCDE fixed, and rotating in the clockwise direction, to have the vertex \\(D^{\\prime}\\) of the new pentagon \\(A^{\\prime}B^{\\prime}CD^{\\prime}E^{\\prime}\\) fall onto the line BC, what is the angle of rotation?\n",
    "completion": "\\textbf{Solution:} Since the external angle of a regular pentagon is $360^\\circ \\div 5 = 72^\\circ$,\\\\\nby fixing point C of the regular pentagon ABCDE and rotating it clockwise, a rotation of $72^\\circ$ will result in the vertices of the new pentagon A$^{\\prime}$B$^{\\prime}$C$^{\\prime}$D$^{\\prime}$E$^{\\prime}$ with vertex D$^{\\prime}$ falling on the line BC.\\\\\nTherefore, the answer is: \\boxed{72^\\circ}.",
    "image_path": "dataset/mm_math/images/MM_Math/50481439.png"
  },
  {
    "solution": "\\boxed{15^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB \\parallel CE$, $\\angle ABC = 30^\\circ$, $\\angle BDE = 45^\\circ$, what is the measure of $\\angle DBC$ in degrees?\n",
    "completion": "\\textbf{Solution:} Since $AB\\parallel CE$, and $\\angle ABC=30^\\circ$,\\\\\nit follows that $\\angle ABC=\\angle BCE=30^\\circ$,\\\\\nsince $\\angle BDE=45^\\circ$,\\\\\nit implies that $\\angle DBC=\\angle BDE-\\angle BCE=45^\\circ-30^\\circ=\\boxed{15^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51434732.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{2}a}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the length of the line segment $AB$ is $a$. Extend the line segment $AB$ to $C$ so that $BC=b$. Let points $M$ and $N$ be the midpoints of $AC$ and $BC$, respectively. Find the length of $MN$.\n",
    "completion": "\\textbf{Solution:} Given that $AB=a$ and $BC=b$,\\\\\nthus $AC=AB+BC=a+b$,\\\\\nand since points $M$ and $N$ are the midpoints of $AC$ and $BC$ respectively,\\\\\nthus $CM=\\frac{1}{2}AC=\\frac{1}{2}(a+b)$ and $CN=\\frac{1}{2}BC=\\frac{1}{2}b$,\\\\\nthus $MN=CM-CN=\\frac{1}{2}(a+b)-\\frac{1}{2}b=\\boxed{\\frac{1}{2}a}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51407373_76.png"
  },
  {
    "solution": "\\boxed{25^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB = 90^\\circ$, $\\angle A = 40^\\circ$, the bisector of the external angle $\\angle CBD$ of $\\triangle ABC$ intersects the extension line of $AC$ at point $E$. Through point $D$, draw $DF \\parallel BE$, which intersects the extension line of $AC$ at point $F$. What is the measure of $\\angle F$ in degrees?\n",
    "completion": "\\textbf{Solution:} Since $\\angle ACB= 90^\\circ$ and $\\angle CBE=65^\\circ$, \\\\\nit follows that $\\angle CEB=90^\\circ- 65^\\circ= 25^\\circ$. \\\\\nSince $DF \\parallel BE$, it follows that $\\angle F=\\boxed{25^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50710244_210.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, an equilateral triangle $\\triangle DCE$ is constructed on the exterior of the rhombus $ABCD$. The segments $AE$ and $DE$ are connected. If the diagonal $AC$ is equal to the side $AB$, then what is the measure of $\\angle DEA$ in degrees?\n",
    "completion": "\\textbf{Solution:} Draw AC, \\\\\nsince quadrilateral ABCD is a rhombus, AC=AB,\\\\\ntherefore, AD=CD=AB=AC,\\\\\nsince $\\triangle DCE$ is an equilateral triangle,\\\\\nthus DE=CD=CE, $\\angle$CED=60$^\\circ$,\\\\\ntherefore, AD=AC=CE=DE,\\\\\nthus quadrilateral ACED is a rhombus,\\\\\nhence $\\angle$DEA= $\\frac{1}{2}$ $\\angle$CED=30$^\\circ$.\\\\\nTherefore, the answer is: $\\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52252067.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{5}}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, the vertices of $\\triangle ABC$ are all on the lattice points of a square grid. What is the value of $\\sin\\angle ACB$?\n",
    "completion": "\\textbf{Solution:} Draw BD $\\perp$ AC, \\\\\nsince BC $= \\sqrt{1^{2}+3^{2}} = \\sqrt{10}$, and BD $= \\sqrt{1^{2}+1^{2}} = \\sqrt{2}$, \\\\\ntherefore, $\\sin\\angle ACB = \\frac{\\sqrt{2}}{\\sqrt{10}} = \\boxed{\\frac{\\sqrt{5}}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52585578.png"
  },
  {
    "solution": "\\boxed{\\sqrt{13}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the regular hexagon $ABCDEF$ is inscribed in circle $\\odot O$ with a radius of 2 cm. How many centimeters is $CD$? $G$ is the midpoint of $CD$. Connect $AG$. How many centimeters is $AG$?\n",
    "completion": "\\textbf{Solution:} Let us connect $AD$, $AC$, $OC$, as depicted:\\\\\nSince the regular hexagon $ABCDEF$ is inscribed in $\\odot O$,\\\\\nit follows that $AD$ is the diameter of $\\odot O$, and $\\angle COD=60^\\circ$,\\\\\nwhich implies $\\angle ACD=90^\\circ$,\\\\\nGiven that $OC=OD=2\\text{cm}$,\\\\\nit follows that $\\triangle COD$ is an equilateral triangle,\\\\\nthus, $CD=OC=2\\text{cm}$, and $\\angle ADC=60^\\circ$,\\\\\ntherefore, $AC=CD\\cdot \\tan\\angle ADC=2\\sqrt{3}\\text{cm}$,\\\\\nSince $G$ is the midpoint of $CD$,\\\\\nwe have $CG=\\frac{1}{2}CD=1\\text{cm}$,\\\\\ntherefore, $AG=\\sqrt{AC^2+CG^2}=\\boxed{\\sqrt{13}}\\text{cm}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53149296.png"
  },
  {
    "solution": "\\boxed{15^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: A right-angled triangle board is placed as shown in the figure, with $D$, $F$ on side $AB$, point $C$ on side $DE$, and $EF$, $AC$ intersecting at point $G$, $\\angle A=30^\\circ$, $\\angle E=45^\\circ$, then what is the degree measure of $\\angle FGA$?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle DEF$ is a right-angled triangle, and $\\angle E=45^\\circ$, $\\angle A=30^\\circ$,\\\\\nthen, $\\angle DFE=45^\\circ$.\\\\\nSince $\\angle DFE=\\angle A+\\angle FGA$,\\\\\nit follows that $\\angle FGA=\\angle DFE-\\angle A=45^\\circ-30^\\circ=\\boxed{15^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52914195.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, during the final physical examination, the relationship between the flight height $y$ (meters) and the horizontal distance $x$ (meters) of the shot put thrown by Xiao Zhu approximately satisfies the quadratic function $y=\\frac{1}{10}x^{2}+\\frac{3}{5}x+\\frac{8}{5}$. What was Xiao Zhu's score in this shot put event?\n",
    "completion": "\\textbf{Solution:} Solve: In $y=\\frac{1}{10}x^{2}+\\frac{3}{5}x+\\frac{8}{5}$, let $y=0$ to get:\\\\\n$-\\frac{1}{10}x^{2}-\\frac{3}{5}x-\\frac{8}{5}=0$,\\\\\nwe find $x=-2$ (discard) or $x=8$,\\\\\n$\\therefore$ Therefore, Xiao Zhu's shot put result is $\\boxed{8}$ meters.",
    "image_path": "dataset/mm_math/images/MM_Math/52913784.png"
  },
  {
    "solution": "\\boxed{72^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, quadrilateral $ABCD$ is inscribed in circle $\\odot O$. If $\\angle B=108^\\circ$, what is the measure of $\\angle D$?\n",
    "completion": "\\textbf{Solution:} Given that the quadrilateral $ABCD$ is inscribed in the circle $\\odot O$, and $\\angle B=108^\\circ$,\\\\\n$\\therefore \\angle D=180^\\circ-\\angle B=180^\\circ-108^\\circ=\\boxed{72^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51262084.png"
  },
  {
    "solution": "\\boxed{55^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, the straight line $DE$ intersects with $BC$ at point $O$, $\\angle COE$ and $\\angle AOE$ are complementary, and $\\angle BOD=35^\\circ$. What is the measure of $\\angle AOE$ in degrees?\n",
    "completion": "Solution: Since $\\angle COE = \\angle BOD$,\\\\\n$\\angle BOD = 35^\\circ$,\\\\\ntherefore $\\angle COE = 35^\\circ$,\\\\\nsince $\\angle COE$ and $\\angle AOE$ are complementary,\\\\\nthus $\\angle COE + \\angle AOE = 90^\\circ$,\\\\\ntherefore $\\angle AOE = 90^\\circ - \\angle COE = \\boxed{55^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51482445.png"
  },
  {
    "solution": "\\boxed{50^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\angle AOC = 30^\\circ$, $\\angle BOC = 80^\\circ$, and $OC$ bisects $\\angle AOD$. What is the measure of $\\angle BOD$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle AOC=30^\\circ$ and OC bisects $\\angle AOD$,\\\\\nit follows that $\\angle COD=\\angle AOC=30^\\circ$,\\\\\nSince $\\angle BOC=80^\\circ$,\\\\\nit follows that $\\angle BOD=\\angle BOC-\\angle COD=\\boxed{50^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55190284.png"
  },
  {
    "solution": "\\boxed{31.5^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Please provide the text you wish to have translated into English in a format suitable for LaTeX. Without the original text, I'm unable to assist with your request.\n",
    "completion": "\\textbf{Solution:} Given that $AB=AC$ and $\\angle A=39^\\circ$, \\\\\nit follows that $\\angle ABC=\\angle ACB=70.5^\\circ$, \\\\\nsince $DM$ is the perpendicular bisector of $AB$, \\\\\nit implies $DA=DB$ and thus $\\angle A=\\angle DBA=39^\\circ$, \\\\\ntherefore, $\\angle DBC=\\angle ABC-\\angle DBA=\\boxed{31.5^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51368553_107.png"
  },
  {
    "solution": "\\boxed{18}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given point B on line segment AC and point D on line segment AB. As shown in Figure 2, if $BD=\\frac{1}{4}AB=\\frac{1}{3}CD$, and E is the midpoint of line segment AB, with $EC=12\\,\\text{cm}$, find the length of line segment AC.\n",
    "completion": "\\textbf{Solution:} Let $BD=x$ cm,\\\\\nsince $BD= \\frac{1}{4}AB= \\frac{1}{3}CD$,\\\\\nthus $AB=4BD=4x$ cm, $CD=3BD=3x$ cm,\\\\\nmoreover, since $DC=DB+BC$,\\\\\nthus $BC=3x-x=2x$,\\\\\nand since $AC=AB+BC$,\\\\\nthus $AC=4x+2x=6x$ cm,\\\\\nsince $E$ is the midpoint of segment $AB$,\\\\\nthus $BE= \\frac{1}{2}AB= \\frac{1}{2} \\times 4x=2x$ cm,\\\\\nmoreover, since $EC=BE+BC$,\\\\\nthus $EC=2x+2x=4x$ cm,\\\\\nand since $EC=12$ cm,\\\\\nthus $4x=12$,\\\\\nsolving gives: $x=3$,\\\\\nthus $AC=6x=6 \\times 3=\\boxed{18}$ cm.",
    "image_path": "dataset/mm_math/images/MM_Math/51438092_90.png"
  },
  {
    "solution": "\\boxed{(2, 2)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, two balls are thrown from point A on the slope, passing exactly through the basket at point B, and landing at point O at the bottom of the slope. Taking O as the origin, a Cartesian coordinate system is established. The coordinates of B are $\\left(\\frac{1}{2}, \\frac{7}{8}\\right)$, the slope ratio of the slope is $1\\colon 2$, the height of point A from the ground is 1.5 meters, and the trajectory of the ball can be described by a quadratic function. What are the coordinates of the highest point the ball reaches?\n",
    "completion": "\\textbf{Solution:} Since $y=-\\frac{1}{2}x^{2}+2x=-\\frac{1}{2}(x-2)^{2}+2$,\\\\\ntherefore, when $x=2$, $y$ reaches its maximum value, which is $2$,\\\\\nthus, the highest point the ball reaches has the coordinates $\\boxed{(2, 2)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52796984_40.png"
  },
  {
    "solution": "\\boxed{70^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, fold $\\triangle ABC$ along the segment $DE$ so that point $B$ falls on point $F$; if $AC \\parallel DE$, $\\angle A = 70^\\circ$, $AB = AC$, then what is the degree measure of $\\angle CEF$?\n",
    "completion": "\\textbf{Solution}: We fold triangle $ABC$ along the line segment $DE$, such that point $B$ falls on point $F$,\\\\\nthus, triangle $BDE \\cong FDE$,\\\\\nhence, $\\angle DEB = \\angle DEF$,\\\\\nsince $\\angle A=70^\\circ$, and $AB=AC$,\\\\\nthus, $\\angle B = \\angle C = \\frac{1}{2} \\times (180^\\circ - 70^\\circ) = 55^\\circ$,\\\\\nsince $AC \\parallel DE$,\\\\\nhence, $\\angle DEB = \\angle C = 55^\\circ = \\angle DEF$,\\\\\nthus, $\\angle FEC = 180^\\circ - \\angle DEB - \\angle DEF = \\boxed{70^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53106961.png"
  },
  {
    "solution": "\\boxed{\\frac{16}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in rectangle $ABCD$, $AB=4$, $BC=3$. The triangle $\\triangle BCD$ is translated along ray $BD$ by a distance $a$ ($a>0$) to obtain $\\triangle B'C'D'$. When line segments $AB'$ and $AD'$ are connected, what is the value of $a$ when $\\triangle AB'D'$ forms a right-angled triangle?\n",
    "completion": "\\textbf{Solution:} Consider two cases:\\\\\n(1) As shown in Figure 1, $\\angle D'AB' = 90^\\circ$, extend C'B' to meet AB at G, and draw D'H$\\perp$AB, intersecting the extension of BA at H,\\\\\n$\\therefore \\angle H = \\angle AGB' = \\angle BGB' = 90^\\circ$,\\\\\n$\\because$ quadrilateral ABCD is a rectangle,\\\\\n$\\therefore \\angle BAD = \\angle C = 90^\\circ$, AD = BC = 3,\\\\\n$\\because \\tan \\angle ABD = \\frac{AD}{AB} = \\frac{B'G}{BG}$, thus $\\frac{B'G}{BG} = \\frac{3}{4}$,\\\\\nLet B'G = 3x, BG = 4x,\\\\\n$\\therefore BB' = a = 5x$,\\\\\nBy translation: DD' = BB' = 5x,\\\\\n$\\therefore D'H = 3 + 3x$, AH = BG = 4x,\\\\\n$\\therefore AG = AB = BG = 4 - 4x$,\\\\\n$\\because \\angle D'AB' = \\angle HAD' + \\angle BAB' = 90^\\circ$,\\\\\n$\\angle AD'H + \\angle HAD' = 90^\\circ$,\\\\\n$\\therefore \\angle AD'H = \\angle GAB'$,\\\\\n$\\because \\angle H = \\angle AGB' = 90^\\circ$,\\\\\n$\\therefore \\triangle D'HA \\sim \\triangle AGB'$,\\\\\n$\\therefore \\frac{D'H}{AG} = \\frac{AH}{B'G}$, i.e.,\\\\\n$\\frac{3 + 3x}{4 - 4x} = \\frac{4x}{3x}$,\\\\\n$\\therefore x = \\frac{7}{25}$,\\\\\n$\\therefore a = 5 \\times \\frac{7}{25} = \\boxed{\\frac{7}{5}}$;\\\\\n(2) As shown in Figure 2, $\\angle AB'D' = 90^\\circ$, extend C'B' to meet AB at M, then C'M$\\perp$AB,\\\\\n$\\therefore \\angle AMB' = 90^\\circ$,\\\\\nBy translation: B'C' = BC = 3,\\\\\nSimilarly, let B'M = 3m, BM = 4m, then BB' = a = 5m,\\\\\n$\\therefore AM = 4 - 4m$,\\\\\n$\\because \\angle AB'M + \\angle D'B'C' = 90^\\circ$, $\\angle MAB' + \\angle AB'M = 90^\\circ$,\\\\\n$\\therefore \\angle D'B'C' = \\angle MAB'$,\\\\\n$\\because \\angle C' = \\angle AMB' = 90^\\circ$,\\\\\n$\\therefore \\triangle D'C'B' \\sim \\triangle B'MA$,\\\\\n$\\therefore \\frac{C'D'}{MB'} = \\frac{B'C'}{AM}$, i.e.,\\\\\n$\\frac{4}{3m} = \\frac{3}{4 - 4m}$,\\\\\n$\\therefore m = \\frac{16}{25}$,\\\\\n$\\therefore a = 5m = 5 \\times \\frac{16}{25} = \\boxed{\\frac{16}{5}}$;\\\\\nIn summary, the value of a is $\\frac{7}{5}$ or $\\frac{16}{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51406974.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Line AB intersects line CD at point O, and OE bisects $\\angle BOD$. As shown in Figure 2, ray OF is inside $\\angle AOD$.\n",
    "completion": "\\textbf{Solution:}\n\n(1) OF is the bisector of $\\angle$AOD, for the following reason:\n\nSince OF $\\perp$ OE, therefore $\\angle$EOF $=90^\\circ$,\n\nthus $\\angle$BOE + $\\angle$AOF = $90^\\circ$.\n\nSince OE bisects $\\angle$BOD, therefore $\\angle$BOE = $\\angle$DOE,\n\nthus $\\angle$DOE + $\\angle$AOF = $90^\\circ$.\n\nTherefore, $\\angle$DOE + $\\angle$DOF = $90^\\circ$,\n\ntherefore, $\\angle$AOF = $\\angle$DOF,\n\ntherefore, OF is the bisector of $\\angle$AOD.\n\n(2) Since $\\angle AOF = \\frac{5}{3}\\angle DOF$,\n\nlet $\\angle$DOF = 3x, then $\\angle$AOF = 5x.\n\nSince OF bisects $\\angle$AOE,\n\nthus $\\angle$AOF = $\\angle$EOF = 5x, hence $\\angle$DOE = 2x.\n\nSince OE bisects $\\angle$BOD, therefore $\\angle$BOD = 4x,\n\ntherefore 5x + 3x + 4x = $180^\\circ$, therefore x = $15^\\circ$, therefore $\\angle$BOD = 4x = $\\boxed{60^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/54962530_120.png"
  },
  {
    "solution": "\\boxed{20}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the slope ratio of the water-facing slope $AB$ of the river embankment is $1\\colon \\sqrt{3}$. Given that the height of the embankment $BC=10m$, what is the length of the slope $AB$ in meters?\n",
    "completion": "\\textbf{Solution:} Given that the slope ratio of the upstream slope $AB$ is $1\\colon \\sqrt{3}$,\\\\\nit follows that $\\frac{BC}{AC}=\\frac{1}{\\sqrt{3}}=\\frac{\\sqrt{3}}{3}$\\\\\nTherefore, $\\tan A=\\frac{BC}{AC}=\\frac{\\sqrt{3}}{3}$\\\\\nThus, $\\angle A=30^\\circ$\\\\\nHence, $AB=2BC=\\boxed{20}\\text{m}$",
    "image_path": "dataset/mm_math/images/MM_Math/52537147.png"
  },
  {
    "solution": "\\boxed{B\\left(4, 0\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=ax^2-3ax-4a$ passes through point $C(0, 2)$, intersects the $x$ axis at points $A$ and $B$ (with point $A$ to the left of point $B$), connects $BC$, and the line $y=kx+1$ ($k>0$) intersects the $y$ axis at point $D$, intersects the parabola above $BC$ at point $E$, and intersects $BC$ at point $F$. Find the analytical expression of the parabola and the coordinates of points $A$ and $B$.\n",
    "completion": "\\textbf{Solution:} \n\nSubstitute $C\\left(0, 2\\right)$ into $y=a{x}^{2}-3ax-4a$, which gives $-4a=2$. Solving this, we find $a=-\\frac{1}{2}$.\n\n$\\therefore$ The equation of the parabola is $y=-\\frac{1}{2}{x}^{2}+\\frac{3}{2}x+2$.\n\nSetting $-\\frac{1}{2}{x}^{2}+\\frac{3}{2}x+2=0$,\n\nwe find:${x}_{1}=-1, {x}_{2}=4$.\n\n$\\therefore \\boxed{A\\left(-1, 0\\right)}, \\boxed{B\\left(4, 0\\right)}$",
    "image_path": "dataset/mm_math/images/MM_Math/52977788_220.png"
  },
  {
    "solution": "\\boxed{18^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $ \\angle DAC = 2x$, $ \\angle ACB = 4x$, $ \\angle ABC = 3x$, and $AD = BC$, then what is the value of $ \\angle BAD$?\n",
    "completion": "\\textbf{Solution:} Extend BC to point E such that CE=BD,\\\\\n$\\therefore$BD+CD=CE+CD,\\\\\n$\\therefore$BC=DE,\\\\\n$\\because$AD=BC,\\\\\n$\\therefore$AD=DE,\\\\\n$\\therefore \\angle E=\\angle DAE$,\\\\\n$\\because \\angle ACD=4x$,\\\\\n$\\therefore \\angle CAE+\\angle E=4x$,\\\\\n$\\therefore \\angle E=4x-\\angle CAE=\\angle DAE$,\\\\\n$\\therefore 4x-\\angle CAE =2x+\\angle CAE$,\\\\\n$\\therefore \\angle CAE =x$, $\\angle DAE =\\angle E =3x$,\\\\\n$\\because \\angle ABC =3x$,\\\\\n$\\therefore \\angle ABC =\\angle E$,\\\\\n$\\therefore AB=AE$,\\\\\n$\\therefore \\triangle ABD \\cong \\triangle AEC$,\\\\\n$\\therefore \\angle ADB=\\angle ACE$,\\\\\n$\\therefore \\angle ADC=\\angle ACD=4x$,\\\\\n$\\because \\angle ADC+\\angle ACD+\\angle DAC =180^\\circ$,\\\\\n$\\therefore 4x+4x+2x=180^\\circ$,\\\\\nSolving for: $x=18^\\circ$,\\\\\n$\\therefore \\angle BAD= \\boxed{18^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52837435.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, line AE intersects with CD at point B, and BF $\\perp$ AE. If $\\angle DBE=60^\\circ$, what is the degree measure of $\\angle FBD$?\n",
    "completion": "\\textbf{Solution:} Since BF$\\perp$AE,\\\\\nit follows that $\\angle$DBF + $\\angle$DBE = $90^\\circ$,\\\\\nsince $\\angle$DBE = $60^\\circ$,\\\\\nthus $\\angle$DBF = $90^\\circ$ - $\\angle$DBE = $30^\\circ$.\\\\\nHence, the measure of $\\angle FBD$ is $\\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52837576_97.png"
  },
  {
    "solution": "\\boxed{2.5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within the rhombus $ABCD$, the diagonals $AC$ and $BD$ intersect at point $O$, $E$ is the midpoint of side $AD$, and $AC=6$, $BD=8$, then what is the length of the line segment $OE$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rhombus with AC=6 and BD=8,\\\\\nit follows that OA=OC=3 and OB=OD=4, with AO$\\perp$BO,\\\\\nIn the right triangle AOD, by the Pythagorean theorem, we get: AD=$\\sqrt{OA^{2}+OD^{2}}$=5,\\\\\nSince E is the midpoint of AD,\\\\\nit follows that OE=$\\frac{1}{2}$AD=$\\boxed{2.5}$",
    "image_path": "dataset/mm_math/images/MM_Math/52188830.png"
  },
  {
    "solution": "\\boxed{}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the rhombus $ABCD$, $AD\\parallel x$-axis, the coordinates of point $A$ are $(0, 4)$, and the coordinates of point $B$ are $(3, 0)$. The line $y_{1}=mx+n$ containing side $CD$ intersects the $x$-axis at point $C$ and intersects the hyperbola $y_{2}=\\frac{k}{x} (x<0)$ at point $D$. How many units should the rhombus $ABCD$ be translated in the positive direction of the $y$-axis so that point $C$ lies on the hyperbola $y_{2}=\\frac{k}{x} (x<0)$?\n",
    "completion": "\\textbf{Solution:} Given $C(-2,0)$,\\\\\nsubstitute $x=-2$ into ${y}_{2}=-\\frac{20}{x}$ $(x<0)$ to find, $y=-\\frac{20}{-2}=10$,\\\\\ntherefore, by translating the rhombus $ABCD$ $10$ units in the positive direction of the $y$ axis, point $C$ will lie on the hyperbola $y_{2}=\\frac{k}{x}\\left(x<0\\right)$.",
    "image_path": "dataset/mm_math/images/MM_Math/52250758_124.png"
  },
  {
    "solution": "\\boxed{44^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, it is known that $\\angle AOB=160^\\circ$, $\\angle COE$ is a right angle, and $OF$ bisects $\\angle AOE$. If $\\angle COF=32^\\circ$, then what is the degree measure of $\\angle BOE$?\n",
    "completion": "\\textbf{Solution:} We have $\\angle BOE = \\boxed{44^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51385668_79.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the diagonals AC and BD of rectangle ABCD intersect at point O. Points E and F are the midpoints of AO and AD, respectively. If EF = 4 and AB = 8, what is the measure of $\\angle ACB$?\n",
    "completion": "\\textbf{Solution:} Given rectangle ABCD,\\\\\nthus OD=OB=OA=$\\frac{1}{2}$BD, $\\angle$ABC=$90^\\circ$,\\\\\nsince points E and F are the midpoints of AO and AD respectively,\\\\\nthus EF is the midline of $\\triangle$AOD,\\\\\nthus EF=$\\frac{1}{2}$OD=4,\\\\\nthus OD=8,\\\\\nthus AB=OB=OA,\\\\\nthus $\\triangle$AOB is an equilateral triangle\\\\\nthus $\\angle$BAC=$60^\\circ$,\\\\\nthus $\\angle$ACB=$90^\\circ$−$\\angle$BAC=$90^\\circ$−$60^\\circ$=$\\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52160915.png"
  },
  {
    "solution": "\\boxed{\\sqrt{6}+\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ is inscribed in $\\odot O$. If $\\angle A=60^\\circ$ and $\\angle C=45^\\circ$, what is the length of $AC$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, connect BO and extend it to intersect the circle at point D. Draw AD, BD, CD, and construct DH $\\perp$ AC, \\\\\n$\\because$ BD is the diameter, \\\\\n$\\therefore$ $\\angle$BAD=$\\angle$BCD=$90^\\circ$, \\\\\n$\\because$ $\\angle$ADB=$\\angle$ACB=$45^\\circ$, $\\angle$BDC=$\\angle$BAC=$60^\\circ$, \\\\\n$\\therefore$ AD=$\\frac{\\sqrt{2}}{2}$BD=$2\\sqrt{2}$, CD=$\\frac{1}{2}$BD=2, \\\\\n$\\because$ $\\angle$DAH=$\\angle$CBD=$30^\\circ$, $\\angle$DCH=$\\angle$ABD=$45^\\circ$, \\\\\nCH=$\\frac{\\sqrt{2}}{2}$CD=$\\sqrt{2}$, AH=$\\frac{\\sqrt{3}}{2}$AD=$\\sqrt{6}$, \\\\\n$\\therefore$ AC=AH+HC=$\\sqrt{6}+\\sqrt{2}$. \\\\\nThus, the answer is: $AC=\\boxed{\\sqrt{6}+\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51061530.png"
  },
  {
    "solution": "\\boxed{126^\\circ 43'}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $O$ is a point on the straight line $AB$, $\\angle AOC = 53^\\circ 17'$, then what is the degree measure of $\\angle BOC$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle AOC=53^\\circ 17'$, \\\\\nthus $\\angle BOC=180^\\circ-53^\\circ 17'$ \\\\\n$=179^\\circ 60'-53^\\circ 17'$ $=\\boxed{126^\\circ 43'}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53119829.png"
  },
  {
    "solution": "\\boxed{49}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, inside $\\triangle ABC$, points $D$, $E$, and $F$ lie on $AB$, $BC$, and $AC$ respectively, with $DE\\parallel AC$, and $EF\\parallel AB$. $\\triangle BDE\\sim \\triangle EFC$. Given that $\\frac{AF}{FC}=\\frac{4}{3}$ and the area of $\\triangle EFC$ is 9, find the area of $\\triangle ABC$.\n",
    "completion": "\\textbf{Solution:} Given that $FE\\parallel AB$,\\\\\n$\\therefore \\angle A=\\angle EFC, \\angle B=\\angle FEC$.\\\\\n$\\therefore \\triangle ABC\\sim \\triangle FEC$.\\\\\nGiven that $\\frac{AF}{FC}=\\frac{4}{3}$,\\\\\n$\\therefore AF=\\frac{4}{3}FC$.\\\\\n$\\therefore \\frac{FC}{AC}=\\frac{FC}{AF+FC}=\\frac{FC}{\\frac{4}{3}FC+FC}=\\frac{3}{7}$.\\\\\n$\\therefore \\frac{S_{\\triangle EFC}}{S_{\\triangle ABC}}=\\left(\\frac{3}{7}\\right)^2=\\frac{9}{49}$.\\\\\nSince the area of $\\triangle EFC$ is 9,\\\\\n$\\therefore S_{\\triangle ABC}=\\boxed{49}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52872420_160.png"
  },
  {
    "solution": "\\boxed{9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the length of segment $AB=4$. Extend $AB$ to $C$ such that $BC=\\frac{1}{2}AB$. Extend $AB$ in the opposite direction to $D$ such that $AD=\\frac{1}{2}AC$. What is the length of segment $CD$?\n",
    "completion": "\\textbf{Solution:} Given the problem,\\\\\n$\\because AB=4$, $BC=\\frac{1}{2}AB$,\\\\\n$\\therefore BC=\\frac{1}{2}\\times 4=2$.\\\\\n$\\therefore AC=AB+BC=4+2=6$.\\\\\n$\\because AD=\\frac{1}{2}AC$,\\\\\n$\\therefore AD=\\frac{1}{2}\\times 6=3$.\\\\\n$\\therefore CD=AD+AC=3+6=\\boxed{9}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55000649_113.png"
  },
  {
    "solution": "\\boxed{\\frac{48}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, a cup filled with water is placed tilted on a horizontal table surface. Its cross-section can be considered a rectangle with width $BC=6\\, \\text{cm}$ and length $CD=16\\, \\text{cm}$. When the water surface touches the edge of the cup, half of side $CD$ is above the water surface. What is the height of the water surface in centimeters at this moment?\n",
    "completion": "\\textbf{Solution:} Construct $BF\\perp AF$ at $F$, as shown in the diagram:\\\\\nSince $BC=6$ cm, $CD=16$ cm, and $CE=\\frac{1}{2}CD$,\\\\\nTherefore $CE=8$ cm,\\\\\nSince $\\angle C=90^\\circ$,\\\\\nBy the Pythagorean theorem, we have $BE=\\sqrt{BC^{2}+CE^{2}}=\\sqrt{6^{2}+8^{2}}=10$,\\\\\nSince $\\angle BCE=\\angle FBE=90^\\circ$,\\\\\nTherefore $\\angle EBC=\\angle ABF$,\\\\\nSince $\\angle BCE=\\angle BFA=90^\\circ$,\\\\\nTherefore $\\triangle CBE\\sim \\triangle FBA$,\\\\\nTherefore $\\frac{BE}{AB}=\\frac{BC}{BF}$,\\\\\ni.e., $\\frac{10}{16}=\\frac{6}{BF}$,\\\\\nTherefore $BF=\\boxed{\\frac{48}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/41274449.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{6}-8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, there is a flask with a spherical bottom, where the radius of the sphere is $5\\,cm$. Originally, the maximum depth of the liquid inside the flask, denoted as $CD$, is $4\\,cm$. After some of the liquid evaporated, the maximum depth of the liquid inside the flask decreased to $2\\,cm$. How much does the length of the chord $AB$ in the cross-sectional circle decrease by (keep the result in square root form)?\n",
    "completion": "\\textbf{Solution:} Let $A^{\\prime}B^{\\prime}$ intersect $OD$ at $E$, \\\\\nAccording to the problem: $OA=O{A}^{\\prime}=OD=5\\text{cm}$, $OD\\perp AB$, $OD\\perp A^{\\prime}B^{\\prime}$, \\\\\n$\\therefore AC=BC$, $A^{\\prime}E=B^{\\prime}E$, \\\\\n$\\because CD=4\\text{cm}$, \\\\\n$\\therefore OC=OD-CD=1\\text{cm}$, \\\\\n$\\therefore AC=\\sqrt{OA^{2}-OC^{2}}=\\sqrt{5^{2}-1^{2}}=2\\sqrt{6}\\text{cm}$, \\\\\n$\\therefore AB=2AC=4\\sqrt{6}\\text{cm}$, \\\\\n$\\because DE=2\\text{cm}$, \\\\\n$\\therefore OE=OD-DE=3\\text{cm}$, \\\\\n$\\therefore A^{\\prime}D=\\sqrt{O{A^{\\prime}}^{2}-OE^{2}}=\\sqrt{5^{2}-3^{2}}=4\\text{cm}$, \\\\\n$\\therefore A^{\\prime}B^{\\prime}=2A^{\\prime}D=8\\text{cm}$, \\\\\n$\\therefore AB-A^{\\prime}B^{\\prime}=(4\\sqrt{6}-8)\\text{cm}$, \\\\\nThus, the length of the chord $AB$ in the circular section decreased by $\\boxed{4\\sqrt{6}-8}\\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52872109.png"
  },
  {
    "solution": "\\boxed{35^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the right angle of the set square $DEF$ is placed inside $\\triangle ABC$, where the two right-angle sides of the set square pass through points B and C, respectively. If $\\angle A=55^\\circ$, what is the sum of $\\angle ABD+\\angle ACD$?\n",
    "completion": "\\textbf{Solution:} In $\\triangle ABC$, $\\angle A=55^\\circ$, \\\\\n$\\therefore \\angle ABC+\\angle ACB=180^\\circ-\\angle A=125^\\circ$, \\\\\nIn $\\triangle DBC$, $\\because \\angle BDC=90^\\circ$, \\\\\n$\\therefore \\angle DBC+\\angle DCB=180^\\circ-\\angle BDC=90^\\circ$, \\\\\n$\\angle ABD+\\angle ACD=\\angle ABC+\\angle ACB-(\\angle DBC+\\angle DCB)=125^\\circ-90^\\circ=\\boxed{35^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52827533.png"
  },
  {
    "solution": "\\boxed{\\frac{7}{4}\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the figure, in $ \\triangle ABC$, $ \\angle C=90^\\circ$, $AC=BC=3$. Points $D$ and $E$ are on sides $AC$ and $AB$ respectively. Folding $ \\triangle ADE$ along the line $DE$ makes point $A$ land on side $BC$, denoted as point $F$. If $CF=1$, what is the value of $BE$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $FG\\perp AB$ at point G through point F,\n\n$\\because$ in $\\triangle ABC$, $\\angle C=90^\\circ$, $AC=BC=3$,\n\n$\\therefore AB=\\sqrt{AC^2+BC^2}=3\\sqrt{2}$, $\\tan B=\\frac{AC}{BC}=1$,\n\n$\\therefore \\angle A=\\angle B=45^\\circ$,\n\n$\\therefore \\triangle FGB$ is an isosceles right triangle,\n\n$\\therefore BG=FG=FB\\cdot \\sin B=\\sqrt{2}$,\n\nLet $BE=x$, then $AE=3\\sqrt{2}-x$, $EG=BE-BG=x-\\sqrt{2}$,\n\n$\\because$ folding triangle $ADE$ along line $DE$, point A lands on the side $BC$, denoted as point F,\n\n$\\therefore EA=EF=3\\sqrt{2}-x$,\n\nIn $\\triangle EFG$, $EF^2=EG^2+FG^2$,\n\ni.e., $\\left(3\\sqrt{2}-x\\right)^2=\\left(x-\\sqrt{2}\\right)^2+\\left(\\sqrt{2}\\right)^2$,\n\nsolving for $x$, we get $x=\\boxed{\\frac{7}{4}\\sqrt{2}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51318991.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the side length of square $ABCD$ is 10. Points $E$ and $F$ are inside the square where $AE=CF=8$, and $BE=DF=6$. What is the length of line segment $EF$?\n",
    "completion": "\\textbf{Solution:} Extend DF to intersect AE at G, as shown in the figure,\\\\\n$\\because$ AB=10, AE=6, BE=8,\\\\\n$\\therefore$ AB$^{2}$=AE$^{2}$+BE$^{2}$,\\\\\n$\\therefore$ $\\triangle$ABE is a right-angled triangle,\\\\\nSimilarly, $\\triangle$DFC is a right-angled triangle,\\\\\n$\\because$ Quadrilateral ABCD is a square,\\\\\n$\\therefore$ $\\angle$BAD=$\\angle$ABC=$\\angle$BCD=$\\angle$ADC=$90^\\circ$, AB= BC=CD=AD,\\\\\n$\\therefore$ $\\angle$BAE+$\\angle$DAG=$90^\\circ$,\\\\\n$\\because$ In $\\triangle$ABE and $\\triangle$CDF,\\\\\n$\\begin{cases} AB=CD \\\\ AE=CF \\\\ BE=DF \\end{cases}$,\\\\\n$\\therefore$ $\\triangle$ABE$\\cong$ $\\triangle$CDF,\\\\\n$\\therefore$ $\\angle$BAE=$\\angle$DCF,\\\\\n$\\because$ $\\angle$DCF+$\\angle$CDF=$\\angle$ADF+$\\angle$CDF=$90^\\circ$,\\\\\n$\\therefore$ $\\angle$DCF=$\\angle$ADG,\\\\\n$\\therefore$ $\\angle$BAE=$\\angle$ADG,\\\\\n$\\because$ $\\angle$BAE+$\\angle$DAG=$90^\\circ$,\\\\\n$\\therefore$ $\\angle$ADG+$\\angle$DAG=$90^\\circ$,\\\\\n$\\therefore$ $\\angle$DGA=$90^\\circ$, i.e., $\\triangle$AGD is a right-angled triangle,\\\\\n$\\because$ In $\\triangle$AGD and $\\triangle$BAE,\\\\\n$\\begin{cases} \\angle AGD=\\angle BEA \\\\ \\angle ADG=\\angle BAE \\\\ AD=BA \\end{cases}$,\\\\\n$\\therefore$ $\\triangle$AGD$\\cong$$\\triangle$BAE,\\\\\n$\\therefore$ AG=BE=6, DG=AE=8,\\\\\n$\\therefore$ EG=8−6=2, GF=8−6=2,\\\\\n$\\because$ $\\angle$DGA=$90^\\circ$,\\\\\n$\\therefore$ $\\angle$EGF=$180^\\circ−90^\\circ=90^\\circ$,\\\\\n$\\therefore$ In $\\triangle$EFG, $EF=\\sqrt{2^2+2^2}=\\boxed{2\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52380808.png"
  },
  {
    "solution": "\\boxed{4.5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $AM$ and $DN$ are the medians of $\\triangle ABC$ and $\\triangle DCE$ respectively, $B$, $C$, $D$ are on the same straight line, $AB\\parallel CD$, $AC\\parallel DE$. If $BC=6$, what is the length of $MN$?\n",
    "completion": "\\textbf{Solution:} Given that $AB\\parallel CD$, $AC\\parallel DE$,\\\\\nit follows that $\\angle B = \\angle DCE$, $\\angle E = \\angle ACB$,\\\\\nthus, $\\triangle ABC \\sim \\triangle DCE$,\\\\\nsince $AM$ and $DN$ are the medians of $\\triangle ABC$ and $\\triangle DCE$ respectively,\\\\\nit implies that $\\frac{BC}{CE} = \\frac{AM}{DN} = 2$, $CM = \\frac{1}{2}BC = 3$, $CN = \\frac{1}{2}CE$,\\\\\nthus, $\\frac{6}{CE} = 2$,\\\\\ntherefore $CE = 3$,\\\\\nhence $CN = 1.5$,\\\\\nconsequently $MN = CM + CN = 3 + 1.5 = \\boxed{4.5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53059515.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{1}{2}x+6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the Cartesian coordinate system, the graph of line $L_1$: $y=kx+b$ passes through point $C(-2, 7)$, and intersects the x-axis at point $B$, and intersects line $L_2$: $y=\\frac{1}{2}x$ at point $A$, where the x-coordinate of point $A$ is $6$. What is the equation of line $AB$?\n",
    "completion": "\\textbf{Solution:} Given the problem's statement, point $A$ lies on the line $y=\\frac{1}{2}x$ and the abscissa of point $A$ is $6$,\\\\\n$\\therefore$ the ordinate of point $A$ is $3$.\\\\\n$\\because$ line $L_1$ passes through point $C(-2, 7)$, we can establish the system of equations $\\left\\{\\begin{array}{l}-2k+b=7, \\\\ 6k+b=3,\\end{array}\\right.$ which yields $\\left\\{\\begin{array}{l}k=-\\frac{1}{2}, \\\\ b=6.\\end{array}\\right.$\\\\\n$\\therefore$ the equation of line $AB$ is $y=-\\frac{1}{2}x+6$; thus, the answer is $\\boxed{y=-\\frac{1}{2}x+6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50520352_179.png"
  },
  {
    "solution": "\\boxed{38^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, it is known that $\\angle AOB = 90^\\circ$, OE and OD bisect $\\angle AOB$ and $\\angle BOC$ respectively. If $\\angle EOD = 64^\\circ$, what is the degree measure of $\\angle BOC$?\n",
    "completion": "\\textbf{Solution:} It is known from the problem that $\\angle BOC = \\boxed{38^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51439127_175.png"
  },
  {
    "solution": "\\boxed{\\frac{7}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AM$ is the bisector of $\\angle CAB$, and $CN$ is the bisector of the exterior angle $\\angle GCB$. $BE \\perp AM$ at point E, and $BD \\perp CN$ at point D, with $DE$ connected. If $AB=4$, $BC=5$, and $AC=6$, what is the length of $DE$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, \\\\\nextend $BE$ to meet $AC$ at point $F$, and extend $AC$ and $BD$ to meet at point $G$,\\\\\nsince $AM$ bisects $\\angle CAB$ and $BE\\perp AM$,\\\\\nit follows that $\\angle CAM=\\angle BAM$, $\\angle AEF=\\angle AEB=90^\\circ$,\\\\\nthus $\\angle AFE=90^\\circ-\\angle CAM$, $\\angle ABE=90^\\circ-\\angle BAM$,\\\\\nhence $\\angle AFE=\\angle ABE$,\\\\\nwhich implies $AF=AB$,\\\\\ngiven that $AB=4$, $AC=6$, and $BC=5$,\\\\\nit follows that $FC=AC-AF=AC-AB=6-4=2$,\\\\\nsince $CN$ bisects $\\angle GCB$ and $BD\\perp CN$,\\\\\nit follows that $\\angle GCN=\\angle BCN$, $\\angle CDB=\\angle CDG=90^\\circ$,\\\\\nthus $\\angle CGD=90^\\circ-\\angle GCN$, $\\angle CBD=90^\\circ-\\angle BCN$,\\\\\nhence $\\angle CGD=\\angle CBD$,\\\\\nwhich implies $CG=CB=5$,\\\\\ntherefore $FG=FC+CG=2+5=7$,\\\\\nsince $AF=AB$ and $BE\\perp AM$,\\\\\nit follows that $AE$ is the median to side $BF$, that is, point $E$ is the midpoint of $BF$,\\\\\nsince $CG=CB$ and $BD\\perp CN$,\\\\\nit follows that $CD$ is the median to side $BG$, that is, point $D$ is the midpoint of $BG$,\\\\\nthus $DE$ is the midsegment of $\\triangle BFG$,\\\\\ntherefore $DE=\\frac{1}{2}FG=\\boxed{\\frac{7}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52395428.png"
  },
  {
    "solution": "\\boxed{\\frac{7}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, where $AB=4$ and $BC=5$, $P$ is a moving point on ray $BC$. Construct $PE\\perp AP$, with $PE$ intersecting ray $DC$ at point $E$, and ray $AE$ intersecting ray $BC$ at point $F$. Let $BP=x$ and $CF=y$. When $\\frac{PE}{AP}=\\frac{1}{2}$, what is the length of $CF$?\n",
    "completion": "\\textbf{Solution:}\n\n(1) When point $P$ is on segment $BC$, and $E$ is on segment $CD$,\n\n$\\because$ $\\triangle PCE\\sim \\triangle ABP$,\n\n$\\therefore$ $\\frac{PC}{AB}=\\frac{CE}{BP}=\\frac{PE}{AP}=\\frac{1}{2}$,\n\n$\\therefore$ $PC=2$, $PB=3$,\n\n$\\therefore$ $CE=\\frac{3}{2}$,\n\n$\\because$ $\\triangle ECF\\sim \\triangle ABF$,\n\n$\\therefore$ $\\frac{CF}{BF}=\\frac{CE}{AB}$, that is $\\frac{CF}{5+CF}=\\frac{\\frac{3}{2}}{4}$,\n\nSolving, we get $CF=\\boxed{3}$;\n\n(2) When $P$ is to the right of $C$, and $E$ is on the extension of segment $DC$,\n\n$\\because$ $\\angle EPC+\\angle BPA=90^\\circ$, $\\angle BPA+\\angle BAP=90^\\circ$,\n\n$\\therefore$ $\\angle EPC=\\angle BAP$,\n\n$\\because$ $\\angle B=\\angle PCE$,\n\n$\\therefore$ $\\triangle ABP\\sim \\triangle PCE$,\n\n$\\therefore$ $\\frac{PE}{AP}=\\frac{CE}{BP}=\\frac{PC}{AB}=\\frac{1}{2}$,\n\n$\\therefore$ $CP=2$,\n\n$\\therefore$ $CE=\\frac{7}{2}$,\n\n$\\because$ $\\angle AFB=\\angle CFE$, $\\angle B=\\angle FCE$,\n\n$\\therefore$ $\\triangle ABF\\sim \\triangle ECF$,\n\n$\\therefore$ $\\frac{AB}{EC}=\\frac{BF}{CF}$, that is $\\frac{4}{\\frac{7}{2}}=\\frac{5-CF}{CF}$,\n\n$\\therefore$ $CF=\\boxed{\\frac{7}{3}}$,\n\nIn summary, the length of $CF$ could be \\boxed{3} or \\boxed{\\frac{7}{3}}.",
    "image_path": "dataset/mm_math/images/MM_Math/52977987_136.png"
  },
  {
    "solution": "\\boxed{y = x + 1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the graph of the linear function $y=kx+b$ intersects the $x$-axis and the $y$-axis at points $A$ and $B$, respectively, and intersects the graph of the inverse proportion function $y= \\frac{m}{x}$ at point $C$ in the first quadrant. The line $CD$ is perpendicular to the $x$-axis, with $D$ being the foot of the perpendicular. If $OA=OB=OD=1$, find the expression for the linear function.\n",
    "completion": "\\textbf{Solution:} From equation (1), we find points A(-1, 0) and B(0, 1), therefore the equation of line AB is $y = x + 1$, \\\\\nthus, the equation of the linear function is: \\boxed{y = x + 1}.",
    "image_path": "dataset/mm_math/images/MM_Math/51546388_69.png"
  },
  {
    "solution": "\\boxed{70^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $OC$ is the bisector of $\\angle AOD$, and $OE$ is the bisector of $\\angle BOD$. If $\\angle AOB=140^\\circ$, then what is the degree measure of $\\angle COE$?\n",
    "completion": "\\textbf{Solution:} Since $OC$ is the bisector of $\\angle AOD$ and $OE$ is the bisector of $\\angle BOD$,\\\\\n$\\therefore \\angle COD=\\frac{1}{2}\\angle AOD$, $\\angle DOE=\\frac{1}{2}\\angle BOD$,\\\\\nGiven that $\\angle AOB=140^\\circ$,\\\\\n$\\therefore \\angle AOD+\\angle BOD=140^\\circ$,\\\\\n$\\therefore \\angle COE=\\angle COD+\\angle DOE=\\frac{1}{2}\\left(\\angle AOD+\\angle BOD\\right)=\\frac{1}{2}\\times 140^\\circ=\\boxed{70^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51508981_134.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, there are two squares with side length of 1 forming a $2\\times 1$ lattice graph. Points $A$, $B$, $C$, $D$ are all on the lattice points. Lines $AB$ and $CD$ intersect at point $P$. What is the value of $\\tan\\angle BPD$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $DH \\perp BP$ at point $H$,\\\\\nsince the side length of the square is 1,\\\\\ntherefore $AD=2$,\\\\\nin $\\triangle BCD$, $CD=\\sqrt{1^2+1^2}=\\sqrt{2}$,\\\\\nin $\\triangle ABD$, $AB=\\sqrt{2^2+1^2}=\\sqrt{5}$,\\\\\nsince $\\frac{1}{2}AB\\cdot DH=\\frac{1}{2}AD\\cdot BD$,\\\\\ntherefore $DH=\\frac{2\\sqrt{5}}{5}$,\\\\\nsince BC$\\parallel$AD,\\\\\ntherefore $\\angle DAB=\\angle CBP$, $\\angle ADP=\\angle BCP$,\\\\\ntherefore, $\\triangle APD\\sim \\triangle BPC$,\\\\\ntherefore $\\frac{DP}{CP}=\\frac{AD}{BC}=\\frac{2}{1}$,\\\\\ntherefore DP=2PC,\\\\\ntherefore $PD=\\frac{2}{3}CD=\\frac{2\\sqrt{2}}{3}$,\\\\\nin $\\triangle PHD$,\\\\\n$PH=\\sqrt{PD^2-DH^2}=\\frac{2\\sqrt{5}}{15}$,\\\\\ntherefore $\\tan\\angle BPD=\\frac{DH}{PH}=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53148889.png"
  },
  {
    "solution": "\\boxed{N(5, 1)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the Cartesian coordinate system, the line $y=\\frac{1}{2}x+m$ intersects the $x$-axis and the $y$-axis at points $A$ and $B$ respectively, with point $A$ having the coordinates $\\left(-2, 0\\right)$, and point $C$ having the coordinates $\\left(3, 0\\right)$. Point $M$ is the intersection of the line $y=x+m$ and line $l$, and point $N$ is in the first quadrant. When $\\triangle BMN$ forms an isosceles right triangle, directly write out the coordinates of point $N$.\n",
    "completion": "\\textbf{Solution:} From (1) we know $m=1$. Solving the system of equations $\\begin{cases}y=x+1\\\\ y=3x-4\\end{cases}$ gives $x=\\frac{5}{2}$ and $y=\\frac{7}{2}$, that is $M\\left(\\frac{5}{2}, \\frac{7}{2}\\right)$,\\\\\n$\\therefore BM=\\sqrt{\\left(\\frac{5}{2}-0\\right)^{2}+\\left(\\frac{7}{2}-1\\right)^{2}}=\\frac{5\\sqrt{2}}{2}$. Point $N$ is in the first quadrant. When $\\triangle BMN$ is an isosceles right triangle, the following cases are considered;\\\\\n(1) When $BM$ is the hypotenuse, it is evident from the diagram that when $\\triangle BMN$ is an isosceles right triangle, $N\\left(\\frac{5}{2}, 1\\right)$, fulfilling: $BM^{2}=BN^{2}+MN^{2}$ and $BN=MN=\\frac{5}{2}$, thus $\\triangle BMN$ is an isosceles right triangle;\\\\\n(2) When $BM$ is a leg, based on symmetry, the point $N$ at this instance, with $M$ being symmetrical to the point $N$ mentioned in (1), $\\therefore N\\left(\\frac{5}{2}, -\\frac{3}{2}\\right)$, which contradicts with point $N$ being in the first quadrant and is thus discarded,\\\\\n(3) When $BM$ is a leg, based on symmetry, the point $N$ at this instance, with $B$ being symmetrical to the point $N$ mentioned in (1), $\\therefore N(5, 1)$, fulfilling $BN^{2}=MB^{2}+MN^{2}$ and $MN=MB$. When $N$ is on the upper left side of $BM$, there exists a point $N$ that meets the conditions,\\\\\nIn summary: $N\\left(\\frac{5}{2}, 1\\right)$ and $N(5, 1)$ are the coordinates of point $N$, directly presented as \\boxed{N\\left(\\frac{5}{2}, 1\\right)} and \\boxed{N(5, 1)}.",
    "image_path": "dataset/mm_math/images/MM_Math/50553714_109.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{13}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, $\\angle BAC=30^\\circ$, and $BC=1$. Equilateral triangles $ABD$ and $ACE$ are constructed on sides $AB$ and $AC$, respectively. Line $DE$ is drawn and intersects $AB$ at point $F$. What is the length of $DF$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, \\\\\ndraw DG$\\perp$AB at G, and EH$\\perp$DA intersecting the extension of DA at H, \\\\\nsince $\\angle$EAC＝$60^\\circ$ and $\\angle$BAC＝$30^\\circ$, \\\\\nthus $\\angle$EAG＝$\\angle$AGD＝$90^\\circ$, \\\\\nsince BC＝1, \\\\\nthus in $\\triangle ABC$, AC＝$\\sqrt{3}$ and AB＝2, \\\\\nalso since $\\triangle ABD$ and $\\triangle ACE$ are equilateral triangles, \\\\\nthus AE＝$\\sqrt{3}$ and DG＝$\\sqrt{3}$, \\\\\nthus DG＝AE, \\\\\nalso since $\\angle$DFG＝$\\angle$EFA, \\\\\nthus $\\triangle AEF \\cong \\triangle GDF$ (AAS), \\\\\nthus DF＝$\\frac{1}{2}$DE, \\\\\nalso since in $\\triangle AEH$, $\\angle$EAH＝$30^\\circ$, \\\\\nthus HE＝$\\frac{1}{2}$AE＝$\\frac{1}{2}\\sqrt{3}$ and AH＝$\\frac{3}{2}$, \\\\\nthus DH＝DA+AH＝2+$\\frac{3}{2}$＝$\\frac{7}{2}$, \\\\\nthus in $\\triangle DEH$, DE＝$\\sqrt{HE^2+DH^2}$＝$\\sqrt{\\left(\\frac{1}{2}\\sqrt{3}\\right)^2+\\left(\\frac{7}{2}\\right)^2}$＝$\\sqrt{13}$, \\\\\nthus the length of DF is $\\boxed{\\frac{\\sqrt{13}}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52699778.png"
  },
  {
    "solution": "\\boxed{x < -3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the graph of the linear function $y=kx+b$ passes through point $A(-2, 4)$, and intersects with the graph of the direct proportion function $y=-\\frac{2}{3}x$ at point $B(a, 2)$. Write down directly the solution set of the inequality $-\\frac{2}{3}x>kx+b$ with respect to $x$.\n",
    "completion": "\\textbf{Solution:} From the given problem, it is known based on the graph that the solution set for $-\\frac{2}{3}x > kx + b$ is: $\\boxed{x < -3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50553639_112.png"
  },
  {
    "solution": "\\boxed{38}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $D$, $E$, and $F$ are the midpoints of $AB$, $BC$, and $CA$, respectively, with $AC=20$ and $BC=18$. What is the perimeter of the quadrilateral $DECF$?\n",
    "completion": "\\textbf{Solution:} Since D, E, F are the midpoints of AB, BC, and CA, respectively, and given that $AC=20$, $BC=18$,\\\\\nit follows that $DE=CF=\\frac{1}{2}AC=10$, $DF=CE=\\frac{1}{2}BC=9$,\\\\\ntherefore, the perimeter of the quadrilateral DECF $=DE+CF+DF+CE=10+10+9+9=\\boxed{38}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52416340.png"
  },
  {
    "solution": "\\boxed{-6\\le x<-4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y=kx+b$ intersects with the graph of the inverse proportion function $y=\\frac{m}{x}$ at points A and B, and intersects with the x-axis at point C. The coordinate of point A is $(2, 3)$, and the coordinate of point B is $(-6, n)$. Based on the graph, directly write down the solution set of the system of inequalities $\\frac{m}{x}\\le kx+b<0$.\n",
    "completion": "\\textbf{Solution:} From the graph, it is known that the graph of the inverse proportion function $y=\\frac{m}{x}$ is below that of the linear function $y=kx+b$. The range of the variable $x$, which is less than 0, is: $-6\\le x<-4$. Therefore, the solution set is $\\boxed{-6\\le x<-4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50496315_70.png"
  },
  {
    "solution": "\\boxed{\\frac{15}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In $\\triangle ABC$, $AC=BC=10$, $\\tan A = \\frac{3}{4}$, points $D$ and $E$ are movable points on sides $AB$ and $AC$ respectively, and segment $DE$ is connected. Construct $\\triangle ADE$ symmetric to $\\triangle A'DE$ about $DE$. As shown in figure 1, when point $A'$ coincides with point $C$, what is the length of $DE$?\n",
    "completion": "\\textbf{Solution:} Given that $AE=CE$, $\\angle AED=90^\\circ$,\\\\\nsince $AC=10$,\\\\\nit follows that $AE=5$,\\\\\nsince $\\tan A=\\frac{3}{4}=\\frac{DE}{AE}$,\\\\\nthen $DE=5\\times \\frac{3}{4}=\\boxed{\\frac{15}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53080186_104.png"
  },
  {
    "solution": "\\boxed{\\sqrt{2}-1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that in $\\triangle ABC$, $DE\\parallel BC$ intersects sides $AB$ and $AC$ at points D and E, respectively, and $DE$ divides $\\triangle ABC$ into two parts of equal area. Fold $\\triangle ADE$ along line $DE$ so that point A falls onto point F, $DF$ intersects $BC$ at point G, connect $CF$. If $CF\\parallel AB$, what is the value of the ratio $CF\\colon AB$?\n",
    "completion": "\\textbf{Solution:} Draw line $AF$, intersecting $DE$ at $M$, and $BC$ at $N$,\\\\\n$\\because$ folding triangle $ADE$ along the line $DE$, point $A$ lands on point $F$,\\\\\n$\\therefore$ $AF\\perp BC$, $AM=FM$,\\\\\n$\\because$ $DE\\parallel BC$,\\\\\n$\\therefore$ $\\triangle ADE\\sim \\triangle ABC$,\\\\\n$\\because$ $DE$ divides $\\triangle ABC$ into two parts of equal area,\\\\\n$\\therefore$ $\\frac{S_{\\triangle ADE}}{S_{\\triangle ABC}}=\\frac{1}{2}$,\\\\\n$\\therefore$ $\\frac{AM}{AN}=\\frac{1}{\\sqrt{2}}$,\\\\\n$\\therefore$ $\\frac{AM}{MN}=\\frac{1}{\\sqrt{2}-1}=\\sqrt{2}+1$,\\\\\n$\\therefore$ $AM=(\\sqrt{2}+1)MN$,\\\\\n$\\therefore$ $MF=AM=(\\sqrt{2}+1)MN$, $AN=AM+MN=(\\sqrt{2}+1)MN+MN=(\\sqrt{2}+2)MN$,\\\\\n$\\therefore$ $NF=MF-MN=(\\sqrt{2}+1)MN-MN=\\sqrt{2}MN$,\\\\\n$\\because$ $CF\\parallel AB$,\\\\\n$\\therefore$ $\\frac{CF}{AB}=\\frac{NF}{AN}=\\frac{\\sqrt{2}MN}{(\\sqrt{2}+2)MN}=\\boxed{\\sqrt{2}-1}$,",
    "image_path": "dataset/mm_math/images/MM_Math/52917636.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $ \\triangle ABC \\cong \\triangle DEF $, with B, E, C, F lying on the same straight line. If $ BF = 8\\text{cm} $ and $ BE = 2\\text{cm} $, then what is the length of $ CE $ in $\\text{cm}$?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABC \\cong \\triangle DEF$,\\\\\nit follows that $BC = EF$,\\\\\ntherefore $BC - CE = EF - CE$,\\\\\nhence $BE = CF$,\\\\\ngiven that $BE = 2\\text{cm}$,\\\\\nit implies $CF = BE = 2\\text{cm}$,\\\\\nsince $BF = 8\\text{cm}$,\\\\\nthus $CE = BF - BE - CF = 8 - 2 - 2 = \\boxed{4}\\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52836462.png"
  },
  {
    "solution": "\\boxed{75^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\triangle ABC$, point $D$ is on side $BC$, $BD=AD=AC$, $E$ is the midpoint of $CD$, and $ \\angle CAE=20^\\circ$. What is the measure of $ \\angle BAC$?\n",
    "completion": "\\textbf{Solution:} Because $AD=AC$, and point E is the midpoint of $CD$,\n\\[\n\\therefore AE\\perp CD,\n\\]\n\\[\n\\therefore \\angle AEC=90^\\circ,\n\\]\nBecause $\\angle CAE=20^\\circ$,\n\\[\n\\therefore \\angle C=90^\\circ-\\angle CAE=70^\\circ, \\quad \\angle DAC=2\\angle CAE=40^\\circ,\n\\]\nBecause $AD=AC$,\n\\[\n\\therefore \\angle ADC=\\angle C=70^\\circ,\n\\]\nBecause $AD=BD$,\n\\[\n\\therefore \\angle B=\\angle BAD,\n\\]\nBecause $\\angle B+\\angle BAD=\\angle ADC=70^\\circ$,\n\\[\n\\therefore \\angle B=\\angle BAD=\\frac{1}{2}\\angle ADC=35^\\circ,\n\\]\n\\[\n\\therefore \\angle BAC=\\angle DAC+\\angle BAD=40^\\circ+35^\\circ=\\boxed{75^\\circ}.\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/52978730.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $AD \\parallel BC$, $S_{\\triangle AOD} = 4$, $S_{\\triangle BOC} = 9$, then what is the area of $\\triangle ABO$?\n",
    "completion": "\\textbf{Solution:} Since $AD\\parallel BC$,\\\\\nit follows that $\\triangle AOD\\sim \\triangle COB$,\\\\\nhence $\\frac{{S}_{\\triangle AOD}}{{S}_{\\triangle COB}}=\\left(\\frac{OA}{OC}\\right)^{2}$, that is, $\\left(\\frac{OA}{OC}\\right)^{2}=\\frac{4}{9}$,\\\\\nfrom which we find $\\frac{OA}{OC}=\\frac{2}{3}$,\\\\\nthus $\\frac{{S}_{\\triangle AOB}}{{S}_{\\triangle BOC}}=\\frac{OA}{OC}=\\frac{2}{3}$, that is, $\\frac{{S}_{\\triangle ABO}}{9}=\\frac{2}{3}$,\\\\\nsolving this, we get ${S}_{\\triangle ABO}=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52917672.png"
  },
  {
    "solution": "\\boxed{59^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points $A$, $O$, $B$ lie on the same line, and $OC$ bisects $\\angle AOB$. If $\\angle COD = 28^\\circ$, and $OE$ bisects $\\angle BOD$, find the degree measure of $\\angle COE$.\n",
    "completion": "\\textbf{Solution:} Since $OE$ bisects $\\angle BOD$,\\\\\nwe have $\\angle DOE = \\frac{1}{2}\\angle BOD = 31^\\circ$,\\\\\nhence, $\\angle COE = \\angle DOE + \\angle COD = 31^\\circ + 28^\\circ = \\boxed{59^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53211413_118.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{2}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $AB=5$, $AD=3$, point $E$ is on $AB$, and point $F$ is on $BC$. If $AE=2$, $CF=1$, what is the value of $\\sin(\\angle 1+\\angle 2)$?\n",
    "completion": "\\textbf{Solution:} Connect EF, \\\\\n$\\because$ Quadrilateral ABCD is a rectangle, \\\\\n$\\therefore$ $\\angle A=\\angle B=\\angle C=\\angle ADC=90^\\circ$, BC=AD=3, CD=AB=5, \\\\\nIn $\\triangle ADE$, AD=3, AE=2, \\\\\n$\\therefore$ $DE^2=AD^2+AE^2=3^2+2^2=13$, \\\\\n$\\because$ AB=5, \\\\\n$\\therefore$ BE=AB−AE=3, \\\\\n$\\because$ CF=1, \\\\\n$\\therefore$ BF=BC−CF=2, \\\\\nIn $\\triangle EBF$, \\\\\n$\\therefore$ $EF^2=BE^2+BF^2=3^2+2^2=13$, \\\\\n$\\therefore$ EF=DE \\\\\nIn $\\triangle CDF$, \\\\\n$\\therefore$ $DF^2=DC^2+CF^2=5^2+1^2=26$, \\\\\n$\\because$ 26=13+13, i.e., $DF^2=DE^2+EF^2$, \\\\\n$\\therefore$ $\\angle DEF=90^\\circ$, \\\\\n$\\therefore$ $\\angle EDF=\\angle DFE=45^\\circ$, \\\\\n$\\therefore$ $\\angle 1+\\angle 2=\\angle ADC-\\angle EDF=45^\\circ$, \\\\\n$\\therefore$ $\\sin(\\angle 1+\\angle 2)=\\sin45^\\circ=\\boxed{\\frac{\\sqrt{2}}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52719763.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, there are three points $D$, $B$, and $E$ successively on the line segment $AC$, with $AD=\\frac{1}{2}DB$, $E$ is the midpoint of $BC$, and $BE=\\frac{1}{5}AC=2$. Find the length of the line segment $AB$.\n",
    "completion": "\\textbf{Solution:} Since $BE=\\frac{1}{5}AC=2$, and $E$ is the midpoint of $BC$, \\\\\n$\\therefore BC=2BE=4$, $AC=10$, \\\\\n$\\therefore AB=\\frac{3}{5}AC=\\boxed{6}$,",
    "image_path": "dataset/mm_math/images/MM_Math/53119836_134.png"
  },
  {
    "solution": "\\boxed{110^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, both $\\triangle ABC$ and $\\triangle CDE$ are equilateral triangles, with point E inside $\\triangle ABC$. Line segments AE, BE, and BD are connected. If $\\angle EBD=50^\\circ$, what is the measure of $\\angle AEB$?\n",
    "completion": "\\textbf{Solution:}\n\nSince $\\triangle ABC$ is an equilateral triangle,\n\nit follows that AC = BC, $\\angle ACB=\\angle ABC=\\angle BAC=60^\\circ$,\n\nSince $\\triangle CDE$ is an equilateral triangle,\n\nit follows that CD = CE, $\\angle DCE=60^\\circ$\n\nTherefore, $\\angle ACB=\\angle DCE$,\n\nTherefore, $\\angle ACB-\\angle BCE=\\angle DCE-\\angle BCE$,\n\nthat is $\\angle BCD=\\angle ACE$,\n\nIn $\\triangle ACE$ and $\\triangle BCD$,\n\n$\\left\\{\\begin{array}{l}AC=BC\\\\ \\angle ACE=\\angle BCD\\\\ EC=CD\\end{array}\\right.$,\n\nTherefore, $\\triangle ACE \\cong \\triangle BCD$ (SAS),\n\nTherefore, $\\angle CAE=\\angle CBD$,\n\nSince $\\angle EBD=\\angle EBC+\\angle CBD=50^\\circ$,\n\nTherefore, $\\angle CAE+\\angle EBC=50^\\circ$,\n\nSince $\\angle CAB+\\angle ABC=60^\\circ+60^\\circ=120^\\circ$,\n\nTherefore, $\\angle ABE+\\angle BAE=120^\\circ-(\\angle CAE+\\angle EBC)=120^\\circ-50^\\circ=70^\\circ$,\n\nTherefore, $\\angle AEB=180^\\circ-(\\angle ABE+\\angle BAE)=180^\\circ-70^\\circ=\\boxed{110^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52720676.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $OE$ is the angle bisector of $\\angle AOC$, and $OD$ is the angle bisector of $\\angle BOC$. If $\\angle AOC=120^\\circ$ and $\\angle BOC=30^\\circ$, what is the degree measure of $\\angle DOE$?\n",
    "completion": "\\textbf{Solution.} Since OE is the bisector of $\\angle AOC$ and OD is the bisector of $\\angle BOC$, with $\\angle AOC=120^\\circ$ and $\\angle BOC=30^\\circ$,\ntherefore, $\\angle EOC=60^\\circ$ and $\\angle DOC=15^\\circ$,\nhence $\\angle DOE=\\angle EOC-\\angle DOC=60^\\circ-15^\\circ=\\boxed{45^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51355508_172.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{21}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In right-angled triangle $ABC$, with $\\angle BAC=90^\\circ$, E and F are the midpoints of BC and AC, respectively. Extend BA to point D such that $AB = 2AD$. Connect DE, DF, AE, EF, and the intersection of AF and DE is point O. If $AB=8$ and $BC=12$, what is the length of DE?\n",
    "completion": "\\textbf{Solution:} Given that in $\\triangle ABC$, $\\angle BAC=90^\\circ$, $AB=8$, $BC=12$,\\\\\ntherefore, by the Pythagorean theorem, $AC=\\sqrt{BC^2-AB^2}=\\sqrt{12^2-8^2}=4\\sqrt{5}$\\\\\nAdditionally, given that $OA=OF$ and $AF=CF$,\\\\\nthus $OA=\\frac{1}{2}AF=\\frac{1}{4}AC=\\sqrt{5}$,\\\\\ntherefore, in $\\triangle AOD$, $\\angle DAO=90^\\circ$, $AD=\\frac{1}{2}AB=4$, $OA=\\sqrt{5}$,\\\\\nthus by the Pythagorean theorem, $DO=\\sqrt{DA^2+OA^2}=\\sqrt{4^2+(\\sqrt{5})^2}=\\sqrt{21}$,\\\\\nhence $DE=2DO=2\\sqrt{21}=\\boxed{2\\sqrt{21}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53498569_70.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, an athlete is shot putting. The shot is released at point A and lands at point B. It is known that the trajectory followed by the shot is a parabola described by the equation \\(y =-\\frac{1}{12}(x-4)^{2}+3\\). What is the horizontal distance from the athlete to the landing point B of the shot in meters?\n",
    "completion": "\\textbf{Solution:} Let $y=0$, \\\\\nthus $-\\frac{1}{12}(x-4)^2+3=0$, \\\\\nwe obtain: $x=-2$ or $x=10$, \\\\\n$\\therefore B(10,0)$, \\\\\n$\\therefore$ the horizontal distance from the shot put landing point B to the athlete is $\\boxed{10}$ meters.",
    "image_path": "dataset/mm_math/images/MM_Math/52660903.png"
  },
  {
    "solution": "\\boxed{38^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: A ruler and a right-angled triangle ruler containing a $30^\\circ$ angle are placed as shown in the figure. If $\\angle 1 = 22^\\circ$, what is the degree measure of $\\angle 2$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw line CE through point C such that CE$\\parallel$AB, \\\\\n$\\because$AB$\\parallel$GH, $\\therefore$AB$\\parallel$GH$\\parallel$CE, $\\therefore$$\\angle$1=$\\angle$DCE=$22^\\circ$, $\\angle$2=$\\angle$ECF, $\\therefore$$\\angle$2=$\\angle$ECF=$\\angle$DCF−$\\angle$DCE=$60^\\circ−22^\\circ=\\boxed{38^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55452494.png"
  },
  {
    "solution": "\\boxed{-4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph represents the inverse proportion function $y=\\frac{k}{x}$ in the second quadrant. If the area of the rectangle $OABC$ in the graph is $4$, what is the value of $k$?\n",
    "completion": "\\textbf{Solution:} Given the inverse proportion function $y=\\frac{k}{x}$, and the area of rectangle OABC is 4,\\\\\n$\\therefore |k|=4$, which means $k=\\pm 4$,\\\\\nSince the graph of the inverse proportion function $y=\\frac{k}{x}$ is in the second quadrant, $k<0$,\\\\\n$\\therefore k=\\boxed{-4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52905636.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{10}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of a linear function intersects the x-axis at point A and the y-axis at point B. What is the length of $AB$?\n",
    "completion": "\\textbf{Solution:} Given $A(-6, 0)$ and $B(0, 2)$, \\\\\n$\\therefore OA=6$ and $OB=2$, \\\\\nIn the right-angled triangle $AOB$, $AB=\\sqrt{OA^{2}+OB^{2}}=\\sqrt{6^{2}+2^{2}}=\\boxed{2\\sqrt{10}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50555422_89.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in parallelogram $ABCD$, diagonal $AC$ is perpendicular to side $CD$, $\\frac{AB}{AC}=\\frac{3}{4}$, the perimeter of quadrilateral $ABCD$ is $16$, point $E$ is on the extension of $AD$, point $F$ is on ray $AB$, $\\angle CED=\\angle CDF$. If point $F$ coincides with point $B$, find the cotangent of $\\angle AFD$?\n",
    "completion": "\\textbf{Solution:} Given the problem, \\\\\nsince $\\frac{AB}{AC}=\\frac{3}{4}$, \\\\\nlet AB=3k, AC=4k, and the intersection of AC and BD be O, \\\\\nsince quadrilateral ABCD is a parallelogram, \\\\\nthus OA=OC=$\\frac{1}{2}AC=2k$, and CD$\\parallel$AB, \\\\\nsince AC$\\perp$CD, \\\\\nthus AC$\\perp$AB, \\\\\nthus BC=$\\sqrt{AC^2+AB^2}=5k$, \\\\\nsince the perimeter of quadrilateral ABCD is 16, \\\\\nthus $5k+5k+3k+3k=16$, \\\\\nsolving gives $k=1$, \\\\\nthus AB=3, AC=4, BC=5, OA=2, \\\\\nthus $\\cot \\angle AFD = \\frac{AB}{AO} = \\boxed{\\frac{3}{2}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51379461_93.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, $\\angle A=30^\\circ$, and the perpendicular bisector of $AB$ intersects $AB$ at $D$ and $AC$ at $E$. If $AC=9$ cm, what is the length of $AE$ in centimeters?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect $BE$,\\\\\n$\\because$ $\\angle C=90^\\circ$ and $\\angle A=30^\\circ$, the perpendicular bisector of $AB$ intersects $AB$ at $D$,\\\\\n$\\therefore \\angle ABC=60^\\circ$, $BE=AE$, $\\angle A=\\angle EBA=30^\\circ$,\\\\\n$\\therefore \\angle CBE=30^\\circ$,\\\\\n$\\therefore BE=2CE=AE$,\\\\\n$\\therefore AC=CE+AE=9$,\\\\\n$\\therefore 3CE=9$, hence $CE=3$,\\\\\n$\\therefore AE=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52653549.png"
  },
  {
    "solution": "\\boxed{48m}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a hobby group intends to determine the distance between two points, $A$ and $B$, which are separated by a pond. They select a point $C$ outside the line $AB$, connect $AC$ and $BC$, and identify their midpoints $D$ and $E$, respectively, then connect $DE$. Now, it's measured that $AC=30m$, $BC=40m$, and $DE=24m$. What is the distance between points $A$ and $B$ in meters?\n",
    "completion": "\\textbf{Solution:} It is given that since $D$ and $E$ are the midpoints of $AC$ and $BC$ respectively,\\\\\ntherefore $DE$ is the midline of $\\triangle ABC$,\\\\\ntherefore $AB=2DE$\\\\\nsince $DE=24m$\\\\\ntherefore $AB=\\boxed{48m}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52386573.png"
  },
  {
    "solution": "\\boxed{48}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system, the side OC of the rhombus OABC lies on the x-axis, with B(18,6). The graph of the inverse proportion function $y=\\frac{k}{x}$ ($k \\ne 0$) passes through point A and intersects OB at point E. Find the value of k.\n",
    "completion": "\\textbf{Solution:} From equation (1), we find that the coordinate of point A is (8, 6). Substituting point A (8, 6) into $y= \\frac{k}{x}$, we solve for $k=\\boxed{48}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51433106_156.png"
  },
  {
    "solution": "\\boxed{\\frac{36}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $l_{1}\\colon y=ax+b$ (with constants $a<0$, $b>0$) intersects the x-axis and the y-axis at points A and B, respectively, while the line $l_{2}\\colon y=ca+d$ (with constants $c>0$, $d>0$) intersects the x-axis and the y-axis at points C and D, respectively. The lines $l_{1}$ and $l_{2}$ intersect at point E, and $\\triangle AOB\\cong \\triangle COD$. $AB\\perp CD$. If $a=-2$ and $b=4$, what is the area of $\\triangle ADE$?\n",
    "completion": "\\textbf{Solution}: From the given information, we have $a=-2$, $b=4$, \\\\\n$\\therefore$ the line $l_1\\colon y=-2x+4$, \\\\\nLet $x=0$, we get $y=4$; let $y=0$, we get $-2x+4=0$, solving this gives $x=2$, \\\\\n$\\therefore$ point A is $(2, 0)$, point B is $(0, 4)$, \\\\\n$\\therefore OA=2$, $OB=4$, \\\\\n$\\because \\triangle AOB \\cong \\triangle COD$, \\\\\n$\\therefore OC=OA=2$, $OD=OB=4$, \\\\\n$\\therefore$ points C and D are $(0, 2)$ and $(-4, 0)$ respectively, \\\\\nSolving gives: $\\begin{cases} c=\\frac{1}{2} \\\\ d=2 \\end{cases}$, \\\\\n$\\therefore$ the line $l_2\\colon y=\\frac{1}{2}x+2$, \\\\\nBy solving the system of equations $\\begin{cases} y=\\frac{1}{2}x+2 \\\\ y=-2x+4 \\end{cases}$, we get: $\\begin{cases} x=\\frac{4}{5} \\\\ y=\\frac{12}{5} \\end{cases}$, \\\\\n$\\therefore$ the coordinates of point E are $\\left(\\frac{4}{5}, \\frac{12}{5}\\right)$, \\\\\n$\\therefore$ the area of $\\triangle ADE$ is $\\frac{1}{2}AD\\cdot y_E=\\frac{1}{2}\\times (2+4)\\times \\frac{12}{5}=\\boxed{\\frac{36}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53066328_128.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $C$ is a moving point on line segment $BD$, with perpendiculars $AB \\perp BD$ and $ED \\perp BD$ drawn through points $B$ and $D$, respectively. The lines $AC$ and $EC$ are also drawn. Given that $AB=5$, $DE=1$, and $BD=8$, and letting $CD=x$, what is the minimum value of $AC+CE$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram: \\\\\nWhen points A, C, and E are collinear, AC + CE is minimized;\\\\\nWhen points A, C, and E are on the same line,\\\\\nextend AB and construct line EF $\\perp$ AB at point F,\\\\\n$\\because$AB = 5, DE = 1,\\\\\n$\\therefore$AF = 6,\\\\\n$\\because$$\\angle$ABD = $90^\\circ$,\\\\\n$\\therefore$$\\angle$FBD = $90^\\circ$,\\\\\n$\\because$$\\angle$BDE = $\\angle$BFE = $90^\\circ$,\\\\\n$\\therefore$Quadrilateral BFED is a rectangle,\\\\\n$\\therefore$BD = EF = 8,\\\\\n$\\therefore$AE = $ \\sqrt{AF^{2} + EF^{2}}$ = $ \\sqrt{6^{2} + 8^{2}}$ = \\boxed{10}.",
    "image_path": "dataset/mm_math/images/MM_Math/53320060.png"
  },
  {
    "solution": "\\boxed{x<2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=kx+b$ ($k<0$) passes through the point $P(2,0)$. What is the range of $x$ when $kx+b>0$?\n",
    "completion": "\\textbf{Solution:} The line $y=kx+b$ ($k<0$) passes through the point $P(2,0)$.\\\\\nWhen $kx+b>0$, the range of $x$ is $x<2$,\\\\\nTherefore, the answer is: $\\boxed{x<2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52342563.png"
  },
  {
    "solution": "\\boxed{\\frac{25}{11}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $O$ is the circumcircle of $\\triangle ABC$, $AB$ is the diameter of $O$, $AB \\perp CD$ at point $E$, and $P$ is a point on the extension line of $AB$, with $\\angle BCP = \\angle BCD$. Extend $DO$ to intersect $AC$ at point $F$ and intersect $O$ at point $G$, and connect $GC$. Given that the radius of $O$ is $5$ and $OE=3$, find the lengths of $GC$ and $OF$.\n",
    "completion": "\\textbf{Solution:} Since $AB\\perp CD$ at point E, \\\\\nhence E is the midpoint of CD, \\\\\nSince O is the midpoint of GD, \\\\\nhence OE is the median of $\\triangle DCG$, \\\\\nhence $GC=2OE=6$, $OE\\parallel GC$, \\\\\nSince $AO\\parallel GC$, \\\\\nhence $\\triangle GCF\\sim \\triangle OAF$, \\\\\nhence $\\frac{GC}{OA}=\\frac{GF}{OF}$, \\\\\nthat is $\\frac{6}{5}=\\frac{GF}{OF}$, \\\\\nSince $GF+OF=5$, \\\\\nhence $OF=\\boxed{\\frac{25}{11}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51350909_110.png"
  },
  {
    "solution": "\\boxed{(-2, 0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, for rectangle $ABOC$ with side $BO$ and $CO$ on the x-axis and y-axis respectively, the coordinate of point $A$ is $(-12, 8)$. Points $D$ and $E$ are the midpoints of $AC$ and $OC$ respectively. Point $P$ is a moving point on $OB$. What are the coordinates of point $P$ when $PD+PE$ is at its minimum?\n",
    "completion": "\\textbf{Solution:} Consider point E's symmetric point with respect to the x-axis, denoted as E', and connect DE' to intersect OB at point P. \\\\\n$\\because$ The sides BO and CO of rectangle ABOC are respectively on the x-axis and y-axis, and the coordinates of point A are (-12, 8),\\\\\n$\\therefore$ AC = 12, OC= 8,\\\\\n$\\because$ Points D and E are the midpoints of AC and OC, respectively,\\\\\n$\\therefore$ D (-6, 8), E (0, 4),\\\\\n$\\therefore$ E' (0, -4),\\\\\nLet the equation of line DE' be $y= kx + b$,\\\\\nFrom the problem statement, we have:\\\\\n$\\left\\{\\begin{array}{l}\n-6k+b=8\\\\\nb=4\n\\end{array}\\right.$,\\\\\n$\\therefore$ $\\left\\{\\begin{array}{l}\nk=-2\\\\\nb=-4\n\\end{array}\\right.$,\\\\\n$\\therefore$ The equation of line DE' is $y= -2x - 4$\\\\\nWhen y=0, $-2x - 4=0$,\\\\\n$\\therefore$ x = -2,\\\\\n$\\therefore$ The coordinates of P are $\\boxed{(-2, 0)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50677495.png"
  },
  {
    "solution": "\\boxed{\\frac{5}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the figure, within $\\triangle ABC$ and $\\triangle ADE$, $\\angle ACB=\\angle AED=90^\\circ$, $\\angle ABC=\\angle ADE$. Connecting BD and CE, if $AC:BC=3:4$, what is $BD:CE$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle ACB=90^\\circ$ and $AC\\colon BC=3\\colon 4$, we have $\\frac{AB}{AC}=\\frac{5}{3}$.\\\\\nSince $\\angle ACB=\\angle AED=90^\\circ$ and $\\angle ABC=\\angle ADE$, we conclude that $\\triangle ABC \\sim \\triangle ADE$, leading to $\\frac{AB}{AD}=\\frac{AC}{AE}$ and $\\angle DAE=\\angle BAC$. Therefore, $\\angle DAB=\\angle EAC$, implying that $\\triangle DAB\\sim \\triangle EAC$, and thus $\\frac{BD}{CE}=\\frac{AB}{AC}=\\boxed{\\frac{5}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51155308.png"
  },
  {
    "solution": "\\boxed{3.4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that in $\\triangle ABC$, $\\angle C=90^\\circ$, $BC=3$, and $AC=5$. The triangle is folded along $DE$ such that point $A$ coincides with point $B$. What is the length of $AE$?\n",
    "completion": "\\textbf{Solution:} By the properties of the folding, we have: $BE=AE$,\\\\\nLet $BE=AE=x$,\\\\\nIn the right triangle $\\triangle BEC$, by the Pythagorean theorem, we have: $BE^2=BC^2+EC^2$,\\\\\nthus $x^2=3^2+(5-x)^2$,\\\\\nthus $x=\\boxed{3.4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52096657.png"
  },
  {
    "solution": "\\boxed{7}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB \\perp CD$, and $AB = CD$, $CE \\perp AD$ at $E$, $BF \\perp AD$ at $F$. Given $CE = 6$, $BF = 3$, $EF = 2$, what is the length of $AD$?\n",
    "completion": "\\textbf{Solution:} Since $AB\\perp CD$, $CE\\perp AD$, and $BF\\perp AD$,\\\\\nit follows that $\\angle A+\\angle D=90^\\circ$, $\\angle C+\\angle D=90^\\circ$, and $\\angle CED=\\angle AFB=90^\\circ$,\\\\\nthus, $\\angle A=\\angle C$,\\\\\nin $\\triangle ABF$ and $\\triangle CDE$,\\\\\n$\\begin{cases}\n\\angle A=\\angle C \\\\\n\\angle AFB=\\angle CED \\\\\nAB=CD\n\\end{cases}$,\\\\\ntherefore, $\\triangle ABF\\cong \\triangle CDE$ (AAS),\\\\\nhence, $AF=CE=6$, $BF=DE=3$,\\\\\ntherefore, $AD=AF-EF+DE=6-2+3=\\boxed{7}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53033151.png"
  },
  {
    "solution": "\\boxed{70^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the quadrilateral $ABCD$ is inscribed in a circle $\\odot O$, where $\\overset{\\frown}{AD}=\\overset{\\frown}{CD}$, $\\angle ABD=33^\\circ$, and $\\angle ACB=44^\\circ$. What is the degree measure of $\\angle BAC$?\n",
    "completion": "\\textbf{Solution:} Given $\\because \\overset{\\frown}{AD}=\\overset{\\frown}{CD}$,\\\\\n$\\therefore \\angle ABD=\\angle DBC=\\angle DAC=33^\\circ$,\\\\\n$\\therefore \\angle ABC=\\angle ABD+\\angle DBC=33^\\circ+33^\\circ=66^\\circ$,\\\\\n$\\therefore \\angle BAC=180^\\circ-\\angle ABC-\\angle ACB=180^\\circ-66^\\circ-44^\\circ=\\boxed{70^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52774665_127.png"
  },
  {
    "solution": "\\boxed{(1,1)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, $A(0,4)$, $B(2,4)$, and $C(4,2)$. What are the coordinates of the center $M$ of the circle containing the arc passing through points $A$, $B$, and $C$?\n",
    "completion": "\\textbf{Solution:} Let $M$ be the center of the circle that passes through points $A$, $B$, and $C$. The coordinates of $M$ are $\\boxed{(1,1)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52623631_55.png"
  },
  {
    "solution": "\\boxed{(4,11)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the Cartesian coordinate system, the point $O$ is the origin. The parabola $y=-x^2+6x+3$ intersects the $y$-axis at point $A$. A line segment $AB$ is drawn parallel to the $x$-axis, intersecting the parabola at point $B$, and $OB$ is connected. Point $P$ lies above $AB$ on the parabola, and $PA$ is connected. A perpendicular $PQ$ is drawn to $AB$ at the foot point $H$, intersecting $OB$ at point $Q$. When $\\angle APQ = \\angle B$, what are the coordinates of point $P$?\n",
    "completion": "\\textbf{Solution:} Let $P(m, -m^{2}+6m+3)$,\\\\\nsince $\\angle P = \\angle B$ and $\\angle AHP = \\angle OAB = 90^\\circ$,\\\\\nit follows that $\\triangle ABO \\sim \\triangle HPA$, hence $\\frac{HP}{AB}=\\frac{AH}{AO}$,\\\\\nthus $\\frac{-m^{2}+6m}{6}=\\frac{m}{3}$,\\\\\nsolving for $m$ gives $m=4$.\\\\\nTherefore, $P(4,11)$. Thus, the coordinates of point $P$ are $\\boxed{(4,11)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51317435_147.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=ax^2+bx-4$ ($a\\neq 0$) intersects the $x$-axis at points $A(-2, 0)$ and $B(4, 0)$, and intersects the $y$-axis at point $C$. Line segment $BC$ is drawn. Point $P$ is a moving point on the parabola below line segment $BC$ (not coinciding with points $B$, $C$). A line parallel to the $y$-axis is drawn through point $P$, intersecting line segment $BC$ at $M$ and the $x$-axis at $N$. Let the $x$-coordinate of point $P$ be $t$. Represent the length of segment $PM$ with an algebraic expression in terms of $t$, and find the maximum value of $PM$ and the coordinates of point $M$ at this time.\n",
    "completion": "\\textbf{Solution:} Let $x=0$, we get $y=-4$,\\\\\n$\\therefore C(0, -4)$,\\\\\nAssume the equation of line BC is $y=kx+b$,\\\\\nthen $\\begin{cases}b=-4\\\\ 4k+b=0\\end{cases}$,\\\\\n$\\therefore \\begin{cases}k=1\\\\ b=-4\\end{cases}$,\\\\\n$\\therefore$ the equation of line BC is: $y=x-4$,\\\\\nLet the x-coordinate of P be t, and $PM\\parallel y$ axis,\\\\\n$\\therefore P\\left(t, \\frac{1}{2}t^{2}-t-4\\right)$, $M\\left(t, t-4\\right)$,\\\\\n$\\therefore PM=t-4-\\left(\\frac{1}{2}t^{2}-t-4\\right)=-\\frac{1}{2}t^{2}+2t$,\\\\\n$\\therefore$ when $t=2$, the maximum value of $PM=\\boxed{2}$, at this time $M\\left(2, -2\\right)$.",
    "image_path": "dataset/mm_math/images/MM_Math/52774825_54.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{7}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In quadrilateral $ABCD$, $AB\\parallel DC$, $AB=AD$, diagonals $AC$, $BD$ intersect at point $O$, $AC$ bisects $\\angle BAD$, and through point $C$, draw $CE\\parallel BD$ intersecting the extension of $AB$ at point $E$, connect $OE$. Quadrilateral $ABCD$ is a rhombus: If $\\angle DAB=60^\\circ$ and $AB=4$, what is the length of $OE$?\n",
    "completion": "\\textbf{Solution:} Given that (1) informs us that quadrilateral $ABCD$ is a rhombus, with $\\angle DAB=60^\\circ$, hence $\\angle OAB=30^\\circ$, and $\\angle AOB=90^\\circ$. Since $AB=4$, it follows that $OB=2$, and $AO=OC=2\\sqrt{3}$. Since $CE\\parallel DB$ and $DC\\parallel BE$, quadrilateral $DBEC$ is a parallelogram. Thus, $CE=DB=4$, and $\\angle ACE=90^\\circ$. Therefore, $OE=\\sqrt{OC^{2}+CE^{2}}=\\sqrt{(2\\sqrt{3})^{2}+4^{2}}=\\boxed{2\\sqrt{7}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50496217_116.png"
  },
  {
    "solution": "\\boxed{\\left(\\frac{3}{2}t, 2t\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the vertices of $\\triangle OAB$ are located at $O(0, 0)$, $A(3, 4)$, and $B(6, 0)$, respectively. Moving points $P$ and $Q$ start simultaneously from point $O$, moving in the positive directions of the x-axis and y-axis at speeds of 3 units per second and 2 units per second, respectively. They stop moving when point $P$ reaches point $B$. Through point $Q$, draw $MN \\parallel OB$, intersecting $AO$ and $AB$ at points $M$ and $N$, respectively, and connect $PM$ and $PN$. Let the movement time be $t$ (seconds). What are the coordinates of point $M$ (expressed in terms of $t$)?\n",
    "completion": "\\textbf{Solution:} Let the equation of the line $OA$ be $y=ax$. Substituting the coordinates of point $A$, we get $4=3a$, which yields $a=\\frac{4}{3}$.\\\\\n$\\therefore$ The equation of the line $OA$ is $y=\\frac{4}{3}x$.\\\\\n$\\because$ $MN\\parallel OB$ and $OQ=2t$,\\\\\n$\\therefore$ The ordinate of $M$ is $2t$.\\\\\nSubstituting $y=2t$ into $y=\\frac{4}{3}x$ gives $x=\\frac{3}{2}t$,\\\\\n$\\therefore$ The abscissa of $M$ is $\\boxed{\\left(\\frac{3}{2}t, 2t\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51434142_203.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{1}{2}x+6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the Cartesian coordinate system, the line $y=-\\frac{1}{2}x$ intersects with the inverse proportion function $y=\\frac{k}{x}$ at two points, $A$ and $B$, which are symmetrical about the origin. It is known that the ordinate of point $A$ is 3. After the line $y=-\\frac{1}{2}x$ is translated upwards, it intersects with the inverse proportion function in the second quadrant at point $C$. If the area of $\\triangle ABC$ is 36, what is the equation of the translated line?\n",
    "completion": "\\textbf{Solution:} Set the line after translation intersects the $y$-axis at point $F$, and connect $AF$, $BF$ as shown in the figure. Assume the equation of the line after translation is $y=-\\frac{1}{2}x+b$,\\\\\n$\\because$ this line is parallel to line $AB$,\\\\\n$\\therefore S_{\\triangle ABC}=S_{\\triangle ABF}$,\\\\\n$\\because$ the area of $\\triangle ABC$ is 36,\\\\\n$\\therefore S_{\\triangle ABF}=\\frac{1}{2}OF\\cdot (x_{B}-x_{A})=36$,\\\\\nBy symmetry, it is known: $x_{B}=-x_{A}$,\\\\\n$\\because x_{A}=-6$,\\\\\n$\\therefore x_{B}=6$,\\\\\n$\\therefore \\frac{1}{2}b\\times 12=36$,\\\\\n$\\therefore b=6$.\\\\\n$\\therefore$ The equation of the line after translation is $\\boxed{y=-\\frac{1}{2}x+6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50496356_66.png"
  },
  {
    "solution": "\\boxed{(-2,1)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the Cartesian coordinate system, $OA=OB=\\sqrt{5}$, and $AB=\\sqrt{10}$. If the coordinates of point $A$ are (1, 2), what are the coordinates of point $B$?\n",
    "completion": "\\textbf{Solution:} Construct line BN$\\perp$x-axis, AM$\\perp$x-axis,\\\\\n$\\because$OA=OB=$\\sqrt{5}$, AB=$\\sqrt{10}$,\\\\\n$\\therefore$AO$^{2}$+OB$^{2}$=AB$^{2}$,\\\\\n$\\therefore$$\\angle$BOA=$90^\\circ$,\\\\\n$\\therefore$$\\angle$BON+$\\angle$AOM=$90^\\circ$,\\\\\n$\\because$$\\angle$BON+$\\angle$NBO=$90^\\circ$,\\\\\n$\\therefore$$\\angle$AOM=$\\angle$NBO,\\\\\n$\\because$$\\angle$AOM=$\\angle$NBO, $\\angle$BNO=$\\angle$AMO, BO=OA,\\\\\n$\\therefore$$\\triangle$BNO$\\cong$$\\triangle$OMA,\\\\\n$\\therefore$NB=OM, NO=AM,\\\\\n$\\because$the coordinates of point A are (1,2),\\\\\n$\\therefore$the coordinates of point B are $\\boxed{(-2,1)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51367276.png"
  },
  {
    "solution": "\\boxed{AECF}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in parallelogram \\(ABCD\\), \\(\\angle B = 60^\\circ\\), \\(BC = 6\\), and point \\(E\\) is the midpoint of side \\(BC\\). Triangle \\(ABE\\) is folded to the right along \\(AE\\), with point \\(B\\) falling on \\(B'\\). Connect \\(CB'\\) and extend it to intersect \\(AD\\) at point \\(F\\). Prove that quadrilateral \\(AECF\\) is a parallelogram.\n",
    "completion": "Proof:\n\nBecause $\\triangle ABE$ is folded along $AE$ with point $B$ landing on $B'$,\n\nit follows that $\\triangle ABE \\cong \\triangle AB'E$,\n\nthus $\\angle AEB = \\angle AEB' = \\frac{1}{2}\\angle BEB'$, and $EB = EB'$,\n\nsince $E$ is the midpoint of side $BC$,\n\nwe have $EB = EC = EB'$,\n\ntherefore, $\\angle EB'C = \\angle ECB' = \\frac{1}{2}\\angle BEB'$,\n\nhence, $\\angle BEA = \\angle ECB'$,\n\nwhich implies $AE \\parallel CF$,\n\ngiven that quadrilateral $ABCD$ is a parallelogram,\n\nwe have $AD \\parallel BC$,\n\nthus, quadrilateral \\boxed{AECF} is a parallelogram.\n\nThis completes the proof.",
    "image_path": "dataset/mm_math/images/MM_Math/51825231_76.png"
  },
  {
    "solution": "\\boxed{0\\le m\\le 2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, in the Cartesian coordinate system xOy, the coordinates of point A are (2, 1). OA is perpendicular to OB, and the length of OA is equal to OB. BC is parallel to the x-axis and intersects the y-axis at point D. The angle $\\angle BCA$ equals $135^\\circ$. Point P(0, m) is on segment OD. A line parallel to the axis is drawn through point P (as shown in Figure 2), and this line intersects the sides of quadrilateral AOBC, resulting in a segment of length $d$. Find the function expression of $d$ in terms of $m$, and directly state the range of the independent variable $m$.\n",
    "completion": "\\textbf{Solution:} From the problem, we have $d=\\begin{cases}\n\\frac{5}{2}m, & 0\\le m<1, \\\\\n-\\frac{m}{2}+3, & 1\\le m\\le 2\n\\end{cases}$.\n\nTherefore, the range of the independent variable $m$ is $\\boxed{0\\le m\\le 2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51698836_167.png"
  },
  {
    "solution": "\\boxed{x>1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the graphs of the functions $y=x+b$ and $y=ax+4$ intersect at point $P$. What is the solution set of the inequality $x+b>ax+4$?\n",
    "completion": "\\textbf{Solution:} Since the intersection point of the graphs of the function $y=x+b$ and $y=ax+4$ has an x-coordinate of $1$, \\\\\nit follows that the solution set for the inequality $x+b>ax+4$ is $\\boxed{x>1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50560169.png"
  },
  {
    "solution": "\\boxed{\\frac{5}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=-x+3$ intersects with the inverse proportion function $y=\\frac{2}{x}\\ (x>0)$ at points A and B. As shown in Figure 1, point E is a point on segment AC. Connect OE and OA. If $\\angle AOE=45^\\circ$, what is the value of $\\frac{AE}{EC}$?\n",
    "completion": "\\textbf{Solution:} Since $y=-x+3$ intersects the $x$-axis at point $C$, $C(3, 0)$. Since $\\angle OCA = \\angle AOE = 45^\\circ$ and $\\angle OAE = \\angle CAO$, $\\triangle AOE \\sim \\triangle ACO$. Therefore, $\\frac{AO}{AE} = \\frac{AC}{AO}$, and thus $AO^2 = AE \\cdot AC$. Given $A(1, 2)$ and $C(3, 0)$, we find $AO = \\sqrt{2^2 + 1^2} = \\sqrt{5}$ and $AC = \\sqrt{2^2 + 2^2} = \\sqrt{8} = 2\\sqrt{2}$. Therefore, $AE = \\frac{AO^2}{AC} = \\frac{5}{2\\sqrt{2}}$ and $EC = \\frac{3}{2\\sqrt{2}}$. Hence, $\\frac{AE}{EC} = \\boxed{\\frac{5}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51407111_116.png"
  },
  {
    "solution": "\\boxed{x>1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the function $y=kx+b$ ($k\\neq 0$) passes through the point $(1,2)$. What is the solution set of the inequality $kx+b>2$?\n",
    "completion": "\\textbf{Solution:} From the graph, it can be inferred that when $x>1$, $kx+b>2$,\\\\\nthus, the solution set of the inequality $kx+b>2$ is $x>1$,\\\\\ntherefore, the answer is $\\boxed{x>1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50468935.png"
  },
  {
    "solution": "\\boxed{96^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, the perpendicular bisector of BC intersects with the angle bisector of $\\angle BAC$ at point D, with the foot of the perpendicular being point P. If $\\angle BAC=84^\\circ$, then what is the degree measure of $\\angle BDC$?\n",
    "completion": "\\textbf{Solution:} Construct DE $\\perp$ AB at point E, extending through AB, and DF $\\perp$ AC at F,\\\\\nsince AD is the bisector of $\\angle$BAC,\\\\\ntherefore DE = DF,\\\\\nsince DP is the perpendicular bisector of BC,\\\\\ntherefore BD = CD,\\\\\nIn right triangles $\\triangle$DEB and $\\triangle$DFC,\\\\\n$\\left\\{\\begin{array}{l}DB=DC\\\\ DE=DF\\end{array}\\right.$,\\\\\ntherefore right $\\triangle$DEB $\\cong$ right $\\triangle$DFC (HL).\\\\\ntherefore $\\angle$BDE = $\\angle$CDF,\\\\\ntherefore $\\angle$BDC = $\\angle$EDF,\\\\\nsince $\\angle$DEB = $\\angle$DFC = $90^\\circ$,\\\\\ntherefore $\\angle$EAF + $\\angle$EDF = $180^\\circ$,\\\\\nsince $\\angle$BAC = $84^\\circ$,\\\\\ntherefore $\\angle$BDC = $\\angle$EDF = \\boxed{96^\\circ},",
    "image_path": "dataset/mm_math/images/MM_Math/53080967.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the quadrilateral ABCD is a parallelogram, with circle $\\odot O$ centered at O and passing through AB at point E. Line segments DE and DA are connected, with $DA = 2\\sqrt{2}$, $DE = \\sqrt{7}$, and $DC = 5$. A straight line $l$ is drawn through point E. Through point C, draw $CH \\perp l$ with the foot of the perpendicular at H. Through point A, draw $AM \\perp l$ with the foot of the perpendicular at M. As the straight line $l$ rotates about point E, what is the maximum value of $CH - 4AM$?\n",
    "completion": "\\textbf{Solution:} Draw BN $\\perp l$ at N through point B and draw BT $\\parallel l$ intersecting CH at T, \\\\\n$\\because$BN $\\perp l$, CH $\\perp l$, \\\\\n$\\therefore$BN $\\parallel$ CH, \\\\\n$\\therefore$ Quadrilateral BCHN is a parallelogram, \\\\\n$\\therefore$HT=BN, \\\\\nSimilarly, it can be proved that AM $\\parallel$ BN, \\\\\n$\\therefore$ $\\triangle AME \\sim \\triangle BNE$, \\\\\n$\\therefore$ $\\frac{BN}{AM}=\\frac{BE}{AE}=4$, \\\\\n$\\therefore$BN=4AM, \\\\\n$\\therefore$HT=4AM, \\\\\n$\\therefore$CH-4AM=CH-HT=CT, \\\\\nAlso, $\\because$ $CT \\le BC$ \\\\\n$\\therefore$ when the line $l$ is perpendicular to the line BC, $CT=BC$, i.e., at this time, the maximum value of CH-4AM is BC, \\\\\n$\\because$ Quadrilateral ABCD is a parallelogram, \\\\\n$\\therefore$ $BC=AD=2\\sqrt{2}$, \\\\\n$\\therefore$ The maximum value of CH-4AM is $\\boxed{2\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51351656_171.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{2}x + 3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the line $l_1\\colon y_1=kx+b$ that passes through point $A(-6, 0)$ intersects with the line $l_2\\colon y_2=2x$ at point $B(m, 4)$. What is the analytical expression of line $l_1$?\n",
    "completion": "\\[ \\textbf{Solution:} \\]\nSince point B $(m, 4)$ lies on the line $y = 2x$,\\\\\nwe have $2m = 4$. Solving this gives $m = 2$.\\\\\nTherefore, point B is $(2, 4)$;\\\\\nSince the line $l_1$ passes through point A $(-6, 0)$ and also goes through point B $(2, 4)$,\\\\\nwe get \n\\[\\left\\{\\begin{aligned}\n-6k + b &= 0\\\\ \n2k + b &= 4\n\\end{aligned}\\right.\\]\nSolving these equations, we find\n\\[\\left\\{\\begin{aligned}\nk &= \\frac{1}{2}\\\\ \nb &= 3\n\\end{aligned}\\right.,\\]\nTherefore, $y_1 = \\frac{1}{2}x + 3 = \\boxed{\\frac{1}{2}x + 3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52188279_130.png"
  },
  {
    "solution": "\\boxed{55^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $OB$ is the bisector of $\\angle AOC$, and $OD$ is the bisector of $\\angle COE$. If $\\angle AOE=160^\\circ$ and $\\angle COD=25^\\circ$, then what is the measure of $\\angle AOB$ in degrees?\n",
    "completion": "\\textbf{Solution}: Since $OD$ is the bisector of $\\angle COE$,\\\\\nit follows that $\\angle COE=2\\angle COD=2\\times 25^\\circ=50^\\circ$,\\\\\nthus $\\angle AOC=\\angle AOE-\\angle COE=160^\\circ-50^\\circ=110^\\circ$,\\\\\nalso, since $OB$ is the bisector of $\\angle AOC$,\\\\\nit follows that $\\angle AOB=\\frac{1}{2}\\angle AOC=\\frac{1}{2}\\times 110^\\circ=\\boxed{55^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53119839_134.png"
  },
  {
    "solution": "\\boxed{65^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\angle AOB$ is a right angle, $\\angle AOC = 40^\\circ$, and $OD$ bisects $\\angle BOC$. What is the measure of $\\angle AOD$ in degrees?\n",
    "completion": "\\textbf{Solution:} Because $\\angle BOC = \\angle AOB - \\angle AOC$, \\\\\ntherefore $\\angle BOC = 90^\\circ - 40^\\circ = 50^\\circ$, \\\\\nBecause OD bisects $\\angle BOC$, \\\\\ntherefore $\\angle COD = \\frac{1}{2}\\angle BOC = \\frac{1}{2} \\times 50^\\circ = 25^\\circ$, \\\\\ntherefore $\\angle AOD = \\angle AOC + \\angle COD = 40^\\circ + 25^\\circ = \\boxed{65^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53207134.png"
  },
  {
    "solution": "\\boxed{\\frac{25}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC$, points $P$ and $D$ are on sides $BC$ and $AC$ respectively, and $\\angle APD = \\angle B$. $AC\\cdot CD = CP\\cdot BP$; If $AB=10$, $BC=12$, when $PD \\parallel AB$, find the length of $BP$.\n",
    "completion": "\\textbf{Solution:} Since $PD\\parallel AB$,\\\\\n$\\therefore \\angle APD=\\angle BAP$.\\\\\nAlso, since $\\angle APD=\\angle C$,\\\\\n$\\therefore \\angle BAP=\\angle C$.\\\\\nAlso, since $\\angle B=\\angle B$,\\\\\n$\\therefore \\triangle BAP \\sim \\triangle BCA$,\\\\\n$\\therefore \\frac{BA}{BC}=\\frac{BP}{BA}$.\\\\\nAlso, since $AB=10$, $BC=12$,\\\\\n$\\therefore \\frac{10}{12}=\\frac{BP}{10}$,\\\\\n$\\therefore BP=\\boxed{\\frac{25}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52513835_121.png"
  },
  {
    "solution": "\\boxed{\\sqrt{13}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, points $E$ and $F$ are on sides $AD$ and $DC$, respectively, $BE\\perp EF$. Given $AB=6$, $AE=9$, $DE=2$, what is the length of $EF$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral $ABCD$ is a rectangle, \\\\\n$\\therefore \\angle A=\\angle D=90^\\circ$, \\\\\n$\\therefore \\angle ABE+\\angle AEB=90^\\circ$, \\\\\n$\\because BE\\perp EF$, \\\\\n$\\therefore \\angle DEF+\\angle AEB=90^\\circ$, \\\\\n$\\therefore \\angle DEF=\\angle ABE$, \\\\\nIn $\\triangle DEF$ and $\\triangle ABE$, $\\left\\{\\begin{array}{l}\\angle D=\\angle A=90^\\circ \\\\ \\angle DEF=\\angle ABE\\end{array}\\right.$, \\\\\n$\\therefore \\triangle DEF\\sim \\triangle ABE$, \\\\\n$\\therefore \\frac{DF}{AE}=\\frac{DE}{AB}$, \\\\\n$\\because AB=6, AE=9, DE=2$, \\\\\n$\\therefore \\frac{DF}{9}=\\frac{2}{6}$, \\\\\nSolving gives $DF=3$, \\\\\nIn $\\triangle DEF$, $EF=\\sqrt{DE^2+DF^2}=\\boxed{\\sqrt{13}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52977597.png"
  },
  {
    "solution": "\\boxed{3\\sqrt{7}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $ \\triangle ABC$ is an equilateral triangle with a side length of 6. $CE$ bisects the exterior angle $\\angle ACF$. Point $D$ is on $AC$, line $BD$ is extended to intersect $CE$ at point $E$. If $CD= \\frac{1}{2} AD$, then what is the value of $ \\frac{CE}{BC}$? What is the value of $BE$?\n",
    "completion": "\\textbf{Solution:} Given that $ \\triangle ABC$ is an equilateral triangle with a side length of 6,\\\\\n$ \\therefore AB=BC=6$, $\\angle A=\\angle ACB=60^\\circ$,\\\\\nsince $CE$ bisects the exterior angle $\\angle ACF$,\\\\\n$ \\therefore \\angle FCE=\\angle DCE=\\frac{1}{2}\\angle ACF=\\frac{1}{2}(180^\\circ-\\angle ACB)=60^\\circ$,\\\\\nin $\\triangle ABD$ and $\\triangle CED$, $\\left\\{\\begin{array}{l}\\angle ADB=\\angle CDE \\\\ \\angle A=\\angle DCE=60^\\circ \\end{array}\\right.$,\\\\\n$ \\therefore \\triangle ABD\\sim \\triangle CED$,\\\\\n$ \\therefore \\frac{CE}{AB}=\\frac{CD}{AD}=\\frac{1}{2}$, that is $\\frac{CE}{6}=\\frac{1}{2}$,\\\\\nsolving this gives $CE=3$,\\\\\n$ \\therefore \\frac{CE}{BC}=\\frac{3}{6}=\\boxed{\\frac{1}{2}}$,\\\\\nas shown, draw $EG\\perp BF$ at point $G$,\\\\\nin $Rt\\triangle CEG$, $\\angle CEG=90^\\circ-\\angle FCE=30^\\circ$,\\\\\n$ \\therefore CG=\\frac{1}{2}CE=\\frac{3}{2}$, $EG=\\sqrt{CE^2-CG^2}=\\frac{3}{2}\\sqrt{3}$,\\\\\n$ \\therefore BG=BC+CG=\\frac{15}{2}$,\\\\\nin $Rt\\triangle BEG$, $BE=\\sqrt{BG^2+EG^2}=\\sqrt{\\left(\\frac{15}{2}\\right)^2+\\left(\\frac{3}{2}\\sqrt{3}\\right)^2}=\\boxed{3\\sqrt{7}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51155352.png"
  },
  {
    "solution": "\\boxed{48}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, line segment $BD = \\frac{1}{3}AB = \\frac{1}{4}CD$, points $M$ and $N$ are the midpoints of line segments $AB$ and $CD$, respectively, and $MN = 20\\,cm$, then what is the length of $AC$ in $cm$?\n",
    "completion": "\\textbf{Solution:} Since BD $= \\frac{1}{3}$AB $= \\frac{1}{4}$CD, we have AB=3BD and CD=4BD.\\\\\nBecause points M and N are the midpoints of segments AB and CD, respectively,\\\\\nit follows that AM=BM $= \\frac{1}{2}AB = \\frac{3}{2}$BD and DN=CN $= \\frac{1}{2}CD = 2$BD.\\\\\nUsing the property of segment addition and subtraction, we find BN=DN−BD=2BD−BD=BD and BC=CD−BD=4BD−BD=3BD,\\\\\ntherefore, MN=MB+BN $= \\frac{3}{2}$BD+BD=20.\\\\\nSolving gives BD=8cm.\\\\\nThus, AC=AB+BC=3BD+3BD=6BD=6×8=$\\boxed{48}$,",
    "image_path": "dataset/mm_math/images/MM_Math/53404004.png"
  },
  {
    "solution": "\\boxed{150^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the right-angle vertex P of the right-angled triangle \\( ABP \\) lies on the line \\( CD \\), with points A and B on the same side of line \\( CD \\). Rotate the right-angled triangle \\( ABP \\) about point P so that points A and B are on opposite sides of line \\( CD \\), as shown in figure (3). When \\( \\angle APC = 4\\angle BPD \\), find the degree measure of \\( \\angle BPC \\).\n",
    "completion": "\\textbf{Solution:} Let $\\angle BPD=x$, then $\\angle APC=4x$,\\\\\nsince $\\angle APB=90^\\circ$,\\\\\ntherefore $\\angle APD=90^\\circ-x$,\\\\\naccording to the problem, we have $4x+(90^\\circ-x)=180^\\circ$,\\\\\nyielding $3x=90^\\circ$,\\\\\nsolving for $x$, we get $x=30^\\circ$,\\\\\nthus $\\angle BPC=180^\\circ-\\angle BPD=180^\\circ-30^\\circ=\\boxed{150^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53413592_118.png"
  },
  {
    "solution": "\\boxed{\\left(\\frac{4}{3},\\frac{7}{3}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, it is known that in the Cartesian coordinate system $xOy$, the graph of the linear function $y=kx+b(k\\ne 0)$ passes through points A and $B(-1,0)$, and the graph of the hyperbolic function $y=\\frac{6}{x}$ also passes through point A, where the abscissa of point A is 2. Under this condition, when connecting $AD$, is there a point F on the graph of the linear function $y=kx+b(k\\ne 0)$ such that triangles $\\triangle EAD$ and $\\triangle ECF$ are similar? If such a point exists, please write down the coordinates of point F directly; if not, please explain why.\n",
    "completion": "\\textbf{Solution:} Exist, the coordinates of point F are $\\boxed{\\left(\\frac{4}{3},\\frac{7}{3}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52917576_251.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=-x^2+2x+1$ intersects the x-axis at points A and B, and intersects the y-axis at point C. Point D is the vertex of the parabola, and the point symmetrical to C with respect to the axis of symmetry of the parabola is point E. Points G and F are located on the x-axis and the y-axis, respectively. What is the minimum perimeter of the quadrilateral EDFG?\n",
    "completion": "Solution: Let $x=0$, then $y=1$,\\\\\n$\\therefore$ $C(0,1)$,\\\\\n$\\because$ $y=-x^2+2x+1=-{(x-1)}^2+2$,\\\\\n$\\therefore$ $D(1,2)$, the axis of symmetry of the parabola is the line $x=1$,\\\\\n$\\because$ the symmetric point of C with respect to the axis of symmetry of the parabola is point E,\\\\\n$\\therefore$ $E(2,1)$,\\\\\n$\\therefore$ $DE=\\sqrt{1^2+1^2}=\\sqrt{2}$,\\\\\nConstruct the symmetric point ${D}^{\\prime }$ of D with respect to the y-axis, and the symmetric point ${E}^{\\prime }$ of E with respect to the x-axis, connect ${D}^{\\prime }{E}^{\\prime }$, intersecting the x-axis at point G and the y-axis at point F, as shown in the figure,\\\\\nThen ${D}^{\\prime }(-1,2)$, ${E}^{\\prime }(2,-1)$, $FD=F{D}^{\\prime }$, $GE=G{E}^{\\prime }$,\\\\\nAt this time, $DF+FG+EG=F{D}^{\\prime }+FG+G{E}^{\\prime }={D}^{\\prime }{E}^{\\prime }$,\\\\\n$\\therefore$ At this time, the perimeter of the quadrilateral EDFG is minimal,\\\\\nExtend $D{D}^{\\prime }$ and $E{E}^{\\prime }$, they intersect at point H, as shown in the figure,\\\\\nThen ${D}^{\\prime }H=3$, $H{E}^{\\prime }=3$,\\\\\n$\\therefore$ ${D}^{\\prime }{E}^{\\prime }=\\sqrt{3^2+3^2}=3\\sqrt{2}$,\\\\\n$\\therefore$ The minimum perimeter of the quadrilateral EDFG is $DE+DF+FG+EG=DE+{D}^{\\prime }{E}^{\\prime }=\\sqrt{2}+3\\sqrt{2}=\\boxed{4\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52731640.png"
  },
  {
    "solution": "\\boxed{35^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, line AB and line CD intersect at point O, and OB bisects $\\angle EOD$. If $\\angle EOC = 110^\\circ$, what is the degree measure of $\\angle BOD$?\n",
    "completion": "\\textbf{Solution:} Given $\\because \\angle DOC = 180^\\circ$ and $\\angle EOC = 110^\\circ$, \\\\\n$\\therefore \\angle DOE = 180^\\circ - \\angle EOC = 180^\\circ - 110^\\circ = 70^\\circ$, \\\\\nSince $OB$ bisects $\\angle EOD$, \\\\\n$\\therefore \\angle BOD = \\frac{1}{2} \\angle DOE = \\frac{1}{2} \\times 70^\\circ = \\boxed{35^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53206081_104.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the waterfront slope $AB$ of the dam cross-section has a slope ratio of $1\\colon \\sqrt{3}$. If the dam height $BC$ is 4m, what is the length of $AC$?\n",
    "completion": "\\textbf{Solution:} According to the given information, we have $\\frac{BC}{AC}=\\frac{1}{\\sqrt{3}}$,\\\\\n$\\therefore$ $\\frac{4}{AC}=\\frac{1}{\\sqrt{3}}$,\\\\\nSolving this, we get: $AC=\\boxed{4\\sqrt{3}}$m.",
    "image_path": "dataset/mm_math/images/MM_Math/51446412.png"
  },
  {
    "solution": "\\boxed{4.8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the rectangle $ABCD$, where $AB=6$, $BC=8$, and $AE\\perp BD$ at point $E$, what is the length of $AE$?\n",
    "completion": "\\textbf{Solution:}\\\\\nSince quadrilateral ABCD is a rectangle,\\\\\nthus AO$=\\frac{1}{2}$AC.\\\\\nTherefore, S$\\triangle$AOB$=\\frac{1}{2}$S$\\triangle$ABC.\\\\\nSince in right-angled triangle ABC, AB=6, BC=8.\\\\\nTherefore, S$\\triangle$ABC$=\\frac{1}{2}$AB$\\cdot$BC=24\\\\\nTherefore, S$\\triangle$AOB=12.\\\\\nSince in right-angled triangle ABC,\\\\\nAC$=\\sqrt{AB^{2}+BC^{2}}=\\sqrt{6^{2}+8^{2}}=10$. \\\\\nTherefore, BO$=\\frac{1}{2}$AC=5.\\\\\nSince AE$\\perp$BD.\\\\\nTherefore, S$\\triangle$AOB$=\\frac{1}{2}$OB$\\cdot$AE=12\\\\\nTherefore, AE$=\\frac{24}{OB}=\\frac{24}{5}=\\boxed{4.8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52302787.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{5} + 4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AD$ is the altitude on side $BC$, $AE$ is the median on side $BC$, $\\angle C=45^\\circ$, $\\sin B=\\frac{2}{3}$, and $AD=4$. What is the length of $BC$?\n",
    "completion": "\\textbf{Solution:} Given $\\triangle ABC$, since AD is the height from A to side BC,\\\\\nthus $\\angle ADB = \\angle ADC = 90^\\circ$.\\\\\nIn $\\triangle ADC$, since $\\angle ADC = 90^\\circ$, $\\angle C = 45^\\circ$, and AD = 4,\\\\\nthus DC = AD = 4.\\\\\nIn $\\triangle ADB$, since $\\angle ADB = 90^\\circ$, $\\sin B = \\frac{2}{3}$, and AD = 4,\\\\\nthus AB = $\\frac{AD}{\\sin B} = 6$\\\\\nthus BD = $\\sqrt{AB^2 - AD^2} = 2\\sqrt{5}$,\\\\\nthus BC = BD + DC = $2\\sqrt{5} + 4 = \\boxed{2\\sqrt{5} + 4}$",
    "image_path": "dataset/mm_math/images/MM_Math/51318176_96.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the inverse proportion function $y=\\frac{k}{x} (k \\neq 0)$ passes through the point $(-2, 4)$ and the point $A(a, -2)$. Find the analytical expression of this inverse proportion function and the value of $a$.\n",
    "completion": "\\textbf{Solution}: Since the graph of the inverse proportion function $y=\\frac{k}{x} (k\\ne 0)$ passes through the points $(-2, 4)$ and point $A(a, -2)$, \\\\\n$\\therefore k=-2\\times 4=-8$, \\\\\n$\\therefore$ the inverse proportion function is $y=-\\frac{8}{x}$, \\\\\n$\\therefore -2a=-8$, \\\\\nSolving for $a$ gives: $a=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52720153_115.png"
  },
  {
    "solution": "\\boxed{y=\\frac{1}{2}x+1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the line \\(l\\) passes through points \\(A(0, 1)\\) and \\(B(-2, 0)\\). Find the equation of the line \\(l\\).\n",
    "completion": "\\textbf{Solution:} Set the expression for line $l$ as: $y=kx+b$, passing through $A(0,1)$, $B(-2,0)$,\\\\\n$\\therefore \\left\\{\\begin{array}{l}b=1 \\\\ -2k+b=0 \\end{array}\\right.$,\\\\\n$\\therefore \\left\\{\\begin{array}{l}k=\\frac{1}{2} \\\\ b=1 \\end{array}\\right.$,\\\\\n$\\therefore$ the expression for line $l$ is: $\\boxed{y=\\frac{1}{2}x+1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51662963_202.png"
  },
  {
    "solution": "\\boxed{A(1, 3)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the Cartesian coordinate system, point O is the origin, and the line $BC$ intersects the y-axis at point B and the x-axis at point C. The coordinates of point B are $\\left(0,4\\right)$, and $OB=OC$. If the area of $\\triangle ABO$ is 2, find the coordinates of point A.\n",
    "completion": "\\textbf{Solution:} Construct $AD \\perp y$-axis at point D, \\\\\n$\\therefore {S}_{\\triangle ABO} = \\frac{1}{2}BO \\times AD = \\frac{1}{2} \\times 4AD = 2AD$, \\\\\n$\\because {S}_{\\triangle ABO} = 2$, \\\\\n$\\therefore 2AD = 2$, \\\\\n$\\therefore AD = 1$, \\\\\n$\\therefore$ The x-coordinate of point A is 1, \\\\\nWhen $x=1$, $y = -1 + 4 = 3$, \\\\\n$\\therefore A\\left(1, 3\\right)$. Therefore, the coordinates of point A are $\\boxed{A(1, 3)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52189291_145.png"
  },
  {
    "solution": "\\boxed{17^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the given figure, within $\\triangle ABC$, $AD \\perp BC$, $AE$ is the angle bisector of $\\angle BAC$. If $\\angle B = 72^\\circ$ and $\\angle C = 38^\\circ$, what is the measure of $\\angle DAE$?\n",
    "completion": "\\textbf{Solution:} Given that in $\\triangle ABC$, $\\angle B=72^\\circ$ and $\\angle C=38^\\circ$,\\\\\nthus $\\angle BAC=180^\\circ-\\angle B-\\angle C=180^\\circ-72^\\circ-38^\\circ=70^\\circ$,\\\\\nsince $AD\\perp BC$,\\\\\nthus $\\angle ADC=90^\\circ$,\\\\\nhence $\\angle DAC=90^\\circ-\\angle C=90^\\circ-38^\\circ=52^\\circ$;\\\\\nsince $AE$ bisects $\\angle BAC$,\\\\\nthus $\\angle EAC= \\frac{1}{2}\\angle BAC=35^\\circ$;\\\\\ntherefore $\\angle DAE=\\angle DAC-\\angle EAC=52^\\circ-35^\\circ=\\boxed{17^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52661181.png"
  },
  {
    "solution": "\\boxed{\\frac{5}{6}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, it is known that the graph of the function $y=x+1$ intersects the $y$-axis at point $A$. The graph of the linear function $y=kx+b$ passes through point $B(0, -1)$ and intersects the $x$-axis and the graph of $y=x+1$ at points $C$ and $D$, respectively. Find the area of quadrilateral $AOCD$ (i.e., the area of the shaded region in the diagram).\n",
    "completion": "\\textbf{Solution:} From equation (1), substituting $x=0$ into $y=x+1$, we get $y=1$,\\\\\n$\\therefore A(0, 1)$. Substituting $y=0$ into $y=3x-1$ gives $x=\\frac{1}{3}$,\\\\\n$\\therefore C\\left(\\frac{1}{3}, 0\\right)$,\\\\\n$\\therefore$ the area of quadrilateral $AOCD=S_{\\triangle AOD}+S_{\\triangle COD}$\\\\\n$=\\frac{1}{2}\\times 1\\times 1+\\frac{1}{2}\\times \\frac{1}{3}\\times 2=\\boxed{\\frac{5}{6}}$",
    "image_path": "dataset/mm_math/images/MM_Math/50482144_77.png"
  },
  {
    "solution": "\\boxed{-\\frac{1}{2}x^2+\\frac{3}{2}x+2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the parabola $y=-\\frac{1}{2}x^2+bx+c$ intersects the x-axis at points $A(-1, 0)$ and $B(4, 0)$, and intersects the y-axis at point C. What is the expression of this parabola?\n",
    "completion": "\\textbf{Solution}: Let us consider the parabola $y=-\\frac{1}{2}x^2+bx+c$ intersects the x-axis at points $A(-1, 0)$, and $B(4, 0)$. Thus,\n\\[\n\\begin{cases}\n-\\frac{1}{2}(-1)^2+b(-1)+c=0\\\\\n-\\frac{1}{2}(4)^2+4b+c=0\n\\end{cases}\n\\]\nSolving the system of equations, we find:\n\\[\n\\begin{cases}\nb=\\frac{3}{2}\\\\\nc=2\n\\end{cases}\n\\]\nTherefore, the equation of the parabola is $y=-\\frac{1}{2}x^2+\\frac{3}{2}x+2=\\boxed{-\\frac{1}{2}x^2+\\frac{3}{2}x+2}$",
    "image_path": "dataset/mm_math/images/MM_Math/53065871_91.png"
  },
  {
    "solution": "\\boxed{5\\pi}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ is inscribed in $\\odot O$, and $AB=AC$. Let $P$ be a point on the arc $\\overset{\\frown}{AC}$, and $\\angle BAC=30^\\circ$. If the radius of $\\odot O$ is 6, find the length of the arc $\\overset{\\frown}{APC}$ (keep $\\pi$ in the result).\n",
    "completion": "\\textbf{Solution:} Draw the lines OA and OC.\n\n$\\because \\angle ABC=75^\\circ$,\n\n$\\therefore \\angle AOC=2\\angle ABC=2\\times 75^\\circ=150^\\circ$,\n\n$\\therefore L_{\\overset{\\frown}{APC}} = \\frac{150\\pi R}{180}=\\frac{5}{6}\\pi \\times 6=\\boxed{5\\pi}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51458286_107.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, $\\angle BAD=120^\\circ$. Points E and F are on the extensions of CD and BC, respectively, with AE $\\parallel$ BD and EF $\\perp$ BC. Given that EF $= 5\\sqrt{3}$, what is the length of AB?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a parallelogram, and $\\angle BAD=120^\\circ$, it follows that AB$\\parallel$DC and AB=CD, and $\\angle BAD=\\angle BCD=120^\\circ$. Hence, $\\angle FCE=180^\\circ-120^\\circ=60^\\circ$. As AE$\\parallel$BD, quadrilateral ABDE is also a parallelogram. Therefore, AB=DE, and hence AB=DE=CD, making D the midpoint of CE. Given that EF$\\perp$BC, we have $\\angle EFC=90^\\circ$. In $\\triangle CEF$, $\\angle CEF=30^\\circ$. Let CF=x, and CE=2x, thus EF$^{2}$+CF$^{2}$=CE$^{2}$ leads to ($5\\sqrt{3}$)$^{2}$+x$^{2}$=4x$^{2}$. Solving this, we find x=5 (taking the positive value). Therefore, CE=5×2=10, which makes CD=AB=$\\frac{1}{2}$CE=5. Thus, the answer is: $\\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51825012.png"
  },
  {
    "solution": "\\boxed{\\frac{9}{2}cm}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known: the length of line segment $AB=20\\text{cm}$, points $C$ and $D$ are two points on line segment $AB$, and $BC=\\frac{1}{5}AB$, $AD=\\frac{1}{4}AB$. Points $M$ and $N$ are the midpoints of line segments $AC$ and $BD$, respectively. Question: what is the length of line segment $MN$?\n",
    "completion": "\\textbf{Solution:} Since $AB=20cm$, $BC=\\frac{1}{5}AB$, $AD=\\frac{1}{4}AB$,\\\\\nhence $BC=4cm$, $AD=5cm$,\\\\\nthus $AC=AB-BC=16cm$, $BD=AB-AD=15cm$,\\\\\nas points $M$ and $N$ are the midpoints of line segments $AC$ and $BD$ respectively,\\\\\nthus $AM=\\frac{1}{2}AC=8cm$, $BN=\\frac{1}{2}BD=\\frac{15}{2}cm$,\\\\\ntherefore $MN=AB-AM-BN=20-8-\\frac{15}{2}=\\boxed{\\frac{9}{2}cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53211136.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point D is a point on side AC of $\\triangle ABC$. Points B and C are symmetric with respect to DE. If $AC=6$ and $AD=2$, what is the length of segment BD?\n",
    "completion": "\\textbf{Solution:} \\quad Since $AC=6$ and $AD=2$, \\\\\n$\\therefore CD=AC-AD=6-2=4$, \\\\\nSince $B$ and $C$ are symmetrical about $DE$, \\\\\n$\\therefore DB=DC=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52653234.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ is inscribed in $\\odot O$, $\\angle ABC>90^\\circ$, the bisector of its exterior angle $\\angle EAC$ intersects $\\odot O$ at point D, connect $DB, DC$, and $DB$ intersects $AC$ at point F. If $\\angle EAD=75^\\circ$, what is the degree measure of $\\overset{\\frown}{BC}$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle EAD=75^\\circ$ and $AD$ bisects $\\angle EAC$,\\\\\nit follows that $\\angle EAC=2\\angle DAE=150^\\circ$,\\\\\nthus $\\angle BAC=180^\\circ-\\angle EAC=30^\\circ$,\\\\\ntherefore, the degree measure of $\\overset{\\frown}{BC}$ is $30^\\circ \\times 2=\\boxed{60^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/55497770_66.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{13}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $O$ is the intersection point of the diagonals of rectangle $ABCD$, $DE \\parallel AC$, $CE \\parallel BD$, and $DE$ intersects $CE$ at point $E$. Given that $AB=4$ and $AD=6$, what is the perimeter of quadrilateral $OCED$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rectangle,\\\\\nit follows that $\\angle BAD=90^\\circ$, OB=OD=OC=OA,\\\\\nthus BD= $\\sqrt{AB^{2}+AD^{2}}= 2\\sqrt{13}$,\\\\\nsince OD= $\\frac{1}{2}$BD= $\\sqrt{13}$,\\\\\nand DE$\\parallel$ AC, CE$\\parallel$ BD,\\\\\nit follows that quadrilateral OCED is a parallelogram,\\\\\nsince OC=OD,\\\\\nthus quadrilateral OCED is a rhombus,\\\\\nhence the perimeter of quadrilateral OCED $=4$OD$= 4\\sqrt{13}=\\boxed{4\\sqrt{13}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51614485.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in right-angled triangle $\\triangle ABC$, with $\\angle ACB=90^\\circ$, and $CD\\perp AB$ at $D$, if $AD=4$ and $BD=9$, what is the length of $CD$?\n",
    "completion": "\\textbf{Solution:} Given: $\\because$ $\\angle ACD + \\angle BCD = \\angle BCD + \\angle B$,\\\\\n$\\therefore$ $\\angle ACD = \\angle B$,\\\\\nAlso, $\\angle ADC = \\angle BDC = 90^\\circ$,\\\\\n$\\therefore$ $\\triangle ADC \\sim \\triangle BDC$,\\\\\n$\\therefore$ $\\frac{AD}{CD} = \\frac{CD}{BD}$,\\\\\n$\\therefore$ $CD^2 = AD \\cdot BD = 36$,\\\\\n$\\therefore$ $CD = \\boxed{6}$ or $-6$ (discard).",
    "image_path": "dataset/mm_math/images/MM_Math/51299758.png"
  },
  {
    "solution": "\\boxed{-2\\leq y<3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: The graph of the function $y=|2x+1|-2$ is shown in the figure. What is the range of the function values $y$ when $-2<x<2$?\n",
    "completion": "\\textbf{Solution:} When $x=2$, we have $y=\\left|2x+1\\right|-2=3$,\\\\\nsince $\\left|2x+1\\right|\\ge 0$,\\\\\ntherefore, the minimum value of the function is $-2$,\\\\\naccording to the graph, when $-2<x<2$, the range of the function value $y$ is $-2\\leq y<3$.\\\\\nThus, the answer is: $\\boxed{-2\\leq y<3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50560886.png"
  },
  {
    "solution": "\\boxed{\\frac{24}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in rectangle $ABCD$, point $M$ lies on diagonal $BD$. A circle passing through points $A$, $B$, and $M$ intersects $BC$ at point $E$. If $AM=4$, $EB=EM=3$, find the length of $BM$.\n",
    "completion": "\\textbf{Solution:} Let $AE$ intersect $BD$ at point $G$. Consider the circle passing through points $A$, $B$, and $M$, denoted as $\\odot O$. In rectangle $ABCD$, we have $\\angle ABC=\\angle BAD=90^\\circ$, \\\\\nsince $\\angle ABC=90^\\circ$, \\\\\nthus, chord $AE$ is the diameter of $\\odot O$,\\\\\ntherefore, $\\angle AME=90^\\circ=\\angle ABC$,\\\\\nsince $EB=EM=3$, $AE=AE$,\\\\\ntherefore, $Rt\\triangle AEB\\cong Rt\\triangle AEM$,\\\\\ntherefore, $AB=AM$,\\\\\nsince $EB=EM=3$,\\\\\ntherefore, $\\angle EAB=\\angle EAM$, that is, $AG$ bisects $\\angle BAM$,\\\\\ntherefore, by the \"angle bisector theorem\", $AG\\perp BM$, $BG=GM$,\\\\\nin $Rt\\triangle AEM$, $EM=3$, $AM=4$,\\\\\ntherefore, $AE=\\sqrt{AM^{2}+EM^{2}}=5$,\\\\\nsince $\\text{Area}_{\\triangle AME}=\\frac{1}{2}\\times AM\\times EM=\\frac{1}{2}\\times AE\\times GM$,\\\\\ntherefore, $GM=\\frac{3\\times 4}{5}=\\frac{12}{5}$,\\\\\nsince $BG=GM$,\\\\\ntherefore, $BM=BG+GM=\\boxed{\\frac{24}{5}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53102171_23.png"
  },
  {
    "solution": "\\boxed{y=-2x+8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=kx+b$ intersects the x-axis at point $A(4,0)$ and the y-axis at point $B(0,8)$. What is the equation of line $AB$?\n",
    "completion": "\\textbf{Solution:} Let the equation of line AB be $y=kx+b$. Substituting A(4,0) and B(0,8) into it gives $\\begin{cases}4k+b=0\\\\ b=8\\end{cases}$, solving this system yields $\\begin{cases}k=-2\\\\ b=8\\end{cases}$, \\\\\n$\\therefore$ the equation of line AB is $\\boxed{y=-2x+8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52342732_119.png"
  },
  {
    "solution": "\\boxed{\\frac{16}{15}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $E$ is a point on $BC$, $DF\\perp AE$ at point $F$, and let $\\frac{AD}{AE}=\\lambda$ ($\\lambda>0$). If $AB=3$, $AD=4$, and $D$, $B$, $F$ are on the same straight line, find the value of $\\lambda$.\n",
    "completion": "\\textbf{Solution:} When points D, B, and F are on the same line as depicted:\\\\\nSince quadrilateral $ABCD$ is a rectangle,\\\\\nit follows that $\\angle BAD=\\angle ABC=90^\\circ$,\\\\\nIn $\\triangle ADB$, where $AB=3$ and $AD=4$,\\\\\nit follows that $\\tan\\angle ABD=\\frac{AD}{AB}=\\frac{4}{3}$,\\\\\nSince $DF\\perp AE$,\\\\\nit follows that $\\angle BFE=90^\\circ$,\\\\\nSince $\\angle ABD+\\angle DBC=90^\\circ$ and $\\angle DBC+\\angle FEB=90^\\circ$,\\\\\nit follows that $\\angle FEB=\\angle ABD$,\\\\\nHence $\\frac{AB}{BE}=\\tan\\angle FEB=\\tan\\angle ABD=\\frac{4}{3}$,\\\\\nSince $AB=3$,\\\\\nit follows that $BE=\\frac{9}{4}$,\\\\\nIn $\\triangle ABE$, by the Pythagorean theorem, $AE=\\sqrt{3^2+\\left(\\frac{9}{4}\\right)^2}=\\frac{15}{4}$,\\\\\nThus, $\\lambda=\\frac{AD}{AE}=\\frac{4}{\\frac{15}{4}}=\\boxed{\\frac{16}{15}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53080796_10.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, $DE$ is the angle bisector of the exterior angle $\\angle FDC$ of $\\triangle DBC$, intersecting the extension of $BC$ at point $E$. The extension of $DE$ intersects the circumscribed circle of $\\triangle DBC$ at point $A$. $AB=AC$; if $\\angle DCB=90^\\circ$, $\\sin E=\\frac{\\sqrt{5}}{5}$, and $AD=4$, find the length of $BD$.\n",
    "completion": "\\textbf{Solution:} Given the problem, \\\\\n$\\because \\angle DCB=90^\\circ$, \\\\\n$\\therefore \\angle DCE=\\angle BAD=90^\\circ$, \\\\\n$\\therefore \\angle E+\\angle CDE=\\angle ABD+\\angle ADB=90^\\circ$, \\\\\n$\\because \\angle ADB=\\angle FDE=\\angle CDE$, \\\\\n$\\therefore \\angle ABD=\\angle E$, \\\\\n$\\because \\sin E=\\frac{\\sqrt{5}}{5}$, \\\\\n$\\therefore \\sin\\angle ABD=\\frac{AD}{BD}=\\frac{\\sqrt{5}}{5}$, \\\\\n$\\because AD=4$, \\\\\n$\\therefore BD=\\boxed{4\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52514026_89.png"
  },
  {
    "solution": "\\boxed{14}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given the diagram, $P$ is a point inside the right-angled triangle $ABC$ where $\\angle BAC=90^\\circ$, and $PA=3$, $PB=7$, $PC=9$. What is the maximum value of $BC$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw PD$\\perp$AB at D, PF$\\perp$AC at F, then draw BG$\\perp$PF at G through B, and CE$\\perp$PD at E through C, intersecting BG at H, connect AH, PH\\\\\n$\\because$ $\\angle$BAC$=90^\\circ$,\\\\\n$\\therefore$ the quadrilaterals ADPF, ADEC, ABGF, PGHE, and ABHC are all rectangles\\\\\n$\\therefore$ PD=BG, PF=AD=EC, PG=HE, BC=AH\\\\\n$\\therefore$ $PA^{2}=AD^{2}+PD^{2}=PF^{2}+PD^{2}$\\\\\n$PC^{2}=EC^{2}+PE^{2}=PF^{2}+PE^{2}$\\\\\n$PH^{2}=PE^{2}+HE^{2}=PE^{2}+PG^{2}$\\\\\n$PB^{2}=PG^{2}+BG^{2}=PG^{2}+PD^{2}$\\\\\n$\\therefore$ $PB^{2}+PC^{2}=PA^{2}+PH^{2}=PF^{2}+PE^{2}+PG^{2}+PD^{2}$\\\\\n$\\because$PA=3, PB=7, PC=9\\\\\n$\\therefore$ $7^{2}+9^{2}=3^{2}+PH^{2}$,\\\\\nSolving gives $PH=11$\\\\\n$\\because$ in $\\triangle PAH$, PA=3, $PH=11$\\\\\n$\\therefore$ $8<AH<14$\\\\\n$\\because$ when points A, P, H are collinear, $AH=14$\\\\\n$\\therefore$ when points A, P, H are collinear, $BC=AH=14$ is the maximum\\\\\nHence, the answer is: $BC=\\boxed{14}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51722347.png"
  },
  {
    "solution": "\\boxed{16^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AD$ is the altitude of $\\triangle ABC$ on side $BC$, $AE$ bisects $\\angle BAC$. Given that $\\angle B=40^\\circ$, $\\angle C=72^\\circ$, find the measures of $\\angle AEC$ and $\\angle DAE$.\n",
    "completion": "\\textbf{Solution:} Given $\\because \\angle BAC + \\angle B + \\angle C = 180^\\circ$, $\\angle B = 40^\\circ$, $\\angle C = 72^\\circ$,\\\\\n$\\therefore \\angle BAC = 68^\\circ$,\\\\\n$\\because$ AE bisects $\\angle BAC$,\\\\\n$\\therefore \\angle BAE = \\angle CAE = \\frac{1}{2} \\angle BAC = 34^\\circ$,\\\\\n$\\therefore \\angle AEC = \\angle B + \\angle BAE = 74^\\circ$,\\\\\n$\\because$ AD$\\perp$BC,\\\\\n$\\therefore \\angle ADE = 90^\\circ$,\\\\\n$\\therefore \\angle DAE = 90^\\circ - \\angle AEC = \\boxed{16^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52978931.png"
  },
  {
    "solution": "\\boxed{9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $DE \\parallel BC$, and $AE:AC = 2:3$, $AD=6$. What is the length of $AB$?\n",
    "completion": "\\textbf{Solution:} Since $DE\\parallel BC$,\\\\\nit follows that $\\frac{AE}{AC}=\\frac{AD}{AB}$,\\\\\nGiven $AE\\colon AC=2\\colon 3$ and $AD=6$,\\\\\nit follows that $\\frac{2}{3}=\\frac{6}{AB}$,\\\\\nSolving yields: $AB=\\boxed{9}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53102067.png"
  },
  {
    "solution": "\\boxed{115^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, $\\angle ABC=70^\\circ$. Point D is the midpoint of AB, and point E is a moving point on side AC. $\\triangle ADE$ is folded along DE, placing point A at point A'. When $A'E \\parallel BC$, then what is the degree measure of $\\angle ADE$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, when $A'E\\parallel BC$,\\\\\n$\\therefore \\angle A'EA=\\angle C=90^\\circ$,\\\\\n$\\because \\angle ABC=70^\\circ$,\\\\\n$\\therefore \\angle A=90^\\circ-70^\\circ=20^\\circ$,\\\\\nBy folding, we know: $\\angle A'ED=\\angle AED=\\frac{1}{2}\\angle A'EA=45^\\circ$,\\\\\n$\\therefore \\angle ADE=180^\\circ-\\angle A-\\angle AED=180^\\circ-20^\\circ-45^\\circ=\\boxed{115^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52634474.png"
  },
  {
    "solution": "\\boxed{34^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, $\\triangle ABC$ is inscribed in circle $\\odot O$, $CD$ is the diameter of $\\odot O$, $\\angle BCD=56^\\circ$, what is the degree measure of $\\angle A$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect $BD$, \\\\\nsince $CD$ is the diameter of $\\odot O$, \\\\\nthus $\\angle DBC=90^\\circ$, \\\\\nsince $\\angle BCD=56^\\circ$, \\\\\ntherefore $\\angle BDC=90^\\circ-56^\\circ=34^\\circ$, \\\\\nsince $\\overset{\\frown}{BC}=\\overset{\\frown}{BC}$, \\\\\nthus $\\angle A=\\boxed{34^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51446258.png"
  },
  {
    "solution": "\\boxed{40^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, OA and OB are two radii of the circle $O$, and point C is on the circle $O$. If $\\angle AOB = 80^\\circ$, what is the measure of $\\angle ACB$?\n",
    "completion": "\\textbf{Solution:} Since OA and OB are two radii of the circle O, with point C on the circle O, and $\\angle$AOB=80°, \\\\\nit follows that $\\angle$ACB=$\\frac{1}{2}\\angle$AOB=$\\boxed{40^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/52814897.png"
  },
  {
    "solution": "\\boxed{\\sqrt{6}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, points $D$ and $E$ are on sides $AB$ and $AC$ respectively, with $DE \\parallel BC$. If $AD = 2EC$, $BD = 3$, and $AE = 4$, then what is the value of $CE$?\n",
    "completion": "\\textbf{Solution:} Since $DE\\parallel BC$,\\\\\nit follows that $\\frac{AD}{BD}=\\frac{AE}{EC}$, which means $AD\\cdot EC=AE\\cdot BD$,\\\\\ngiven $AD=2EC, BD=3, AE=4$,\\\\\nwe have $2EC^{2}=4\\times 3=12$,\\\\\ntherefore, $EC=\\boxed{\\sqrt{6}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52872100.png"
  },
  {
    "solution": "\\boxed{y=\\frac{12}{x} \\left(x>0\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system, the coordinates of point B are $(6, 0)$, and $AO=AB=5$, $AH \\perp x$-axis at point H. A reciprocal function $y=\\frac{k}{x}\\ (x>0)$ passes through point A and intersects the $x$-axis at point B, creating line $BC \\perp x$-axis. Line $OC$ intersects line $AB$ at point D and line $AH$ at point M. What is the expression of this reciprocal function?\n",
    "completion": "Since point B has the coordinates $(6, 0)$, and $AO=AB=5$, $AH \\perp x$-axis, \\\\\nthus $OB=6$, $OH=HB=3$,\\\\\nthus $AH= \\sqrt{AO^{2}-OH^{2}}=\\sqrt{5^{2}-3^{2}}=4$,\\\\\nthus $A(3,4)$,\\\\\nthus $k=3\\times 4=12$,\\\\\nthus, the expression of the inverse proportion function is: \\boxed{y=\\frac{12}{x} \\left(x>0\\right)}.",
    "image_path": "dataset/mm_math/images/MM_Math/51458261_152.png"
  },
  {
    "solution": "\\boxed{(1,-1)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system $xOy$, line $l_{1}$ passes through points $A(0, 1)$ and $B(2, 2)$. Line $l_{1}$ is translated downwards by $m$ units to obtain line $l_{2}$. It is known that line $l_{2}$ passes through point $(-1, -2)$ and intersects the x-axis at point $C$. Point $D$ is a point on line $l_{2}$. If points $A$, $B$, $C$, and $D$ can form a parallelogram, find the coordinates of point $D$.\n",
    "completion": "Solution: From the problem, we know that AB$\\parallel$CD. When the four points A, B, C, and D form the parallelogram ABDC, \\\\\n$\\because$A (0,1), B (2,2), C (3,0), \\\\\n$\\therefore$point A moves 3 units to the right and then 1 unit down to coincide with point C, \\\\\n$\\therefore$point B moves 3 units to the right and then 1 unit down to coincide with point D, at this time the coordinates of D are $\\boxed{(5,1)}$; \\\\\n$\\because$AB$\\parallel$CD, when points A, B, C, and D form the parallelogram ABCD, \\\\\n$\\because$A (0,1), B (2,2), C (3,0), \\\\\n$\\therefore$point B moves 1 unit to the right and then 2 units down to coincide with point C, \\\\\n$\\therefore$point A moves 1 unit to the right and then 2 units down to coincide with point D, at this time the coordinates of D are $\\boxed{(1,-1)}$; \\\\\nOverall, if points A, B, C, and D can form a parallelogram, the coordinates of point D are either \\boxed{(5,1)} or \\boxed{(1,-1)}.",
    "image_path": "dataset/mm_math/images/MM_Math/50446864_231.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $O$ is the midpoint of the diagonal $BD$ of rectangle $ABCD$, point $E$ is on $CD$, and $OE \\parallel AD$. If $AB = 8$ and $OE = 3$, what is the length of $OC$?\n",
    "completion": "\\textbf{Solution:} Given that point O is the midpoint of the diagonal BD of rectangle ABCD, and OE $\\parallel$ AD,\\\\\nit follows that OE $\\parallel$ CB, and OE is the midline of $\\triangle$BCD;\\\\\nGiven that OE = 3,\\\\\nit follows that AD = BC = 2OE = 6,\\\\\nGiven that AB = 8, and $\\angle$BAD = 90$^\\circ$,\\\\\nit follows that BD = 10,\\\\\nIn $\\triangle$BCD, OC = $\\frac{1}{2}$BD = \\boxed{5};",
    "image_path": "dataset/mm_math/images/MM_Math/51968606.png"
  },
  {
    "solution": "\\boxed{18\\text{cm}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, there are three points C, D, and E sequentially on the line segment AB, dividing the line segment AB into four parts with the ratios $2: 3: 4: 5$. Given that $AB=28$, what is the length of the line segment AE?\n",
    "completion": "Solution: Let $AC=2x$, then $CD$, $DE$, and $EB$ are $3x$, $4x$, and $5x$ respectively,\\\\\nGiven that, $2x+3x+4x+5x=28$\\\\\nSolving for $x$, we get $x=2$\\\\\nThen, $AC$, $CD$, $DE$, and $EB$ are $4\\text{cm}$, $6\\text{cm}$, $8\\text{cm}$, and $10\\text{cm}$ respectively,\\\\\nTherefore, $AE=AC+CD+DE=\\boxed{18\\text{cm}}$",
    "image_path": "dataset/mm_math/images/MM_Math/52794225_128.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the direct proportion function $y_1=x$ intersects with that of the quadratic function $y_2=x^2-bx$ at points $O(0,0)$ and $A(4,4)$. What is the value of $b$?\n",
    "completion": "\\textbf{Solution:} Substitute point A (4,4) into $y_2=x^2-bx$, we get $16-4b=4$, $4b=12$,\\\\\nsolving for $b$, we find $b=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51458362_258.png"
  },
  {
    "solution": "\\boxed{75^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given that O is a point on line AB, $\\angle COD$ is a right angle, and OE bisects $\\angle BOC$. As shown in Figure 1, if $\\angle AOC= 30^\\circ$, find the measures of $\\angle COE$ and $\\angle BOD$.\n",
    "completion": "\\textbf{Solution:} Given that $\\angle AOC=30^\\circ$ and $\\angle COD$ is a right angle,\\\\\nit follows that $\\angle BOC= 180^\\circ-\\angle AOC=150^\\circ$ and $\\angle COD= 90^\\circ$,\\\\\nthus, $\\angle BOD= 180^\\circ-\\angle AOC-\\angle COD=60^\\circ$.\\\\\nSince OE bisects $\\angle BOC$,\\\\\nit implies that $\\angle COE=\\frac{1}{2}\\angle BOC=75^\\circ$;\\\\\ntherefore, $\\angle BOD=\\boxed{60^\\circ}$ and $\\angle COE=\\boxed{75^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/54927325_11.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, $BD$ is the bisector of $\\angle ABC$, $DE \\perp AB$, and the foot of the perpendicular is E. If $AC=5$ and $DE=2$, what is the length of $AD$?\n",
    "completion": "\\textbf{Solution:} Since BD is the bisector of $\\angle ABC$, $DE \\perp AB$, and $\\angle C = 90^{\\circ}$,\\\\\nit follows that $DE = CD = 2$,\\\\\nSince $AC = 5$,\\\\\nit follows that $AD = AC - CD = 5 - 2 = \\boxed{3}$",
    "image_path": "dataset/mm_math/images/MM_Math/51212156.png"
  },
  {
    "solution": "\\boxed{52^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $CD$ bisects $\\angle ACB$, $\\angle A=68^\\circ$, $\\angle ACD=30^\\circ$. Find: the degree measure of $\\angle B$?\n",
    "completion": "\\textbf{Solution:} Given that $CD$ bisects $\\angle ACB$ and $\\angle ACD=30^\\circ$, \\\\\n$\\therefore \\angle ACB=2\\angle ACD=60^\\circ$, \\\\\n$\\therefore \\angle B=180^\\circ-\\angle A-\\angle ACB=180^\\circ-68^\\circ-60^\\circ=\\boxed{52^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53106969_148.png"
  },
  {
    "solution": "\\boxed{7}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle A=90^\\circ$, $AB=3$, and $AC=4$. D, E, and F are the midpoints of sides AB, BC, and AC, respectively. Lines DE, DF, and EF are connected, and the measure of $\\angle ADF$ is $53^\\circ$. What is the perimeter of quadrilateral ADEF?\n",
    "completion": "\\textbf{Solution:} Given that D, E, F are the midpoints of sides AB, BC, AC respectively, \\\\\n$\\therefore$ DE$\\parallel$ AC and $DE=\\frac{1}{2}AC$, EF$\\parallel$ AB and $EF=\\frac{1}{2}AB$, AF$=\\frac{1}{2}$AC, and AD$=\\frac{1}{2}$AB.\\\\\nSince $AB=3$ and $AC=4$, \\\\\n$\\therefore$ AF=DE=2 and $AD=EF=\\frac{3}{2}$.\\\\\nTherefore, the perimeter of quadrilateral ADEF $=\\left(2+\\frac{3}{2}\\right)\\times 2=\\boxed{7}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53359538_148.png"
  },
  {
    "solution": "\\boxed{135^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: After folding a circular paper (with center at point O) along diameter $MN$, and then folding it into six equal parts as shown in figure 1 to obtain figure 2, if we cut along the dashed line $AB$ and then unfold $\\triangle AOB$ to get a hexagon star as shown in figure 3. If $\\angle CDE = 75^\\circ$, what is the degree measure of $\\angle OBA$?\n",
    "completion": "\\textbf{Solution:} Connect OC, EO\\\\\nDue to the folding property, we have: $\\angle$EOC=$ \\frac{360^\\circ}{12}=30^\\circ$, EC=DC, OC bisects $\\angle$ECD\\\\\n$\\therefore$ $\\angle$ECO=$ \\frac{1}{2}\\angle ECD=\\frac{1}{2}(180^\\circ−2\\times 75^\\circ)=15^\\circ$\\\\\n$\\therefore$ $\\angle$OEC=$180^\\circ$−$\\angle$ECO−$\\angle$EOC=$135^\\circ$\\\\\nThus, the measure of $\\angle OBA$ is $\\boxed{135^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/53296951.png"
  },
  {
    "solution": "\\boxed{-2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the parabola $y=\\frac{1}{2}x^{2}+bx+c$ (where $b$ and $c$ are constants) has its vertex at the coordinates $\\left(\\frac{3}{2}, -\\frac{25}{8}\\right)$, intersects the $x$-axis at points $A$ and $B$ (with point $A$ to the left of point $B$), and intersects the $y$-axis at point $C$. Point $C$ and point $D$ are symmetric about the $x$-axis. Connect $AD$ and draw line $BD$. What are the values of $b$ and $c$?\n",
    "completion": "\\textbf{Solution:} Given the parabola $y=\\frac{1}{2}x^{2}+bx+c$ has its vertex at $\\left(\\frac{3}{2}, -\\frac{25}{8}\\right)$,\\\\\nthus, the equation of the parabola is $y=\\frac{1}{2}\\left(x-\\frac{3}{2}\\right)^{2}-\\frac{25}{8}$\\\\\nSimplifying, we obtain $y=\\frac{1}{2}x^{2}-\\frac{3}{2}x-2$,\\\\\nthus, $b=\\boxed{-\\frac{3}{2}}$ and $c=\\boxed{-2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53079018_48.png"
  },
  {
    "solution": "\\boxed{120^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, for the quadrilateral $ABCD$ inscribed in a circle, if $\\angle A=60^\\circ$, then what is the measure of $\\angle C$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral $ABCD$ is inscribed in a circle, \\\\\nit follows that $\\angle A + \\angle C = 180^\\circ$, \\\\\nsince $\\angle A = 60^\\circ$, \\\\\nit follows that $\\angle C = 180^\\circ - 60^\\circ = \\boxed{120^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53106905.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the rhombus $ABCD$, $AC=4$ and $BD=2$, what is the perimeter of this rhombus?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral ABCD is a rhombus,\\\\\nit follows that $OA=\\frac{1}{2}AC=2$, $OB=\\frac{1}{2}BD=1$, and $AC\\perp BD$, with AB=BC=CD=DA,\\\\\ntherefore, in $\\triangle OAB$, $AB=\\sqrt{OA^{2}+OB^{2}}=\\sqrt{2^{2}+1^{2}}=\\sqrt{5}$,\\\\\nthus, the perimeter of this rhombus is: $4\\sqrt{5}$,\\\\\nhence, the answer is: $\\boxed{4\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52189321.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the right triangle $\\triangle ABC$ where $\\angle ACB=90^\\circ$ and $AC=BC=8$. $F$ is the midpoint of $AC$, and $D$ is a moving point on segment $AB$. Connect $CD$ and rotate segment $CD$ $90^\\circ$ counterclockwise around point $C$ to obtain segment $CE$. Connect $EF$. During the movement of point $D$, what are the maximum and minimum values of $EF$ respectively?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, let G be the midpoint of BC, and connect DG,\\\\\nby rotation, we have DC=EC, $\\angle$DCE$=90^\\circ$,\\\\\nalso, since $\\angle$ACB$=90^\\circ$, and AC=BC=8, with F being the midpoint of AC,\\\\\ntherefore, CG=CF, $\\angle$DCG+$\\angle$ACD$=\\angle$ECF+$\\angle$ACD$=90^\\circ$,\\\\\ntherefore, $\\angle$DCG=$\\angle$ECF,\\\\\ntherefore, $\\triangle DCG \\cong \\triangle ECF$ by SAS,\\\\\ntherefore, EF=DG,\\\\\n(1) As shown in Figure 1,\\\\\nwhen GD$\\perp$AB, DG is the shortest, at this time $\\triangle BDG$ is an isosceles right triangle,\\\\\ntherefore, DG=BG$\\times \\sin 45^\\circ =4\\times \\frac{\\sqrt{2}}{2}=2\\sqrt{2}$,\\\\\nthus, the minimum value of EF is $2\\sqrt{2}$;\\\\\n(2) When D coincides with B, DG=BG=4;\\\\\n(3) As shown in Figure 2,\\\\\nWhen D coincides with A, DG=$\\sqrt{CG^{2}+AC^{2}}=\\sqrt{4^{2}+8^{2}}=4\\sqrt{5}>4$,\\\\\nthus, the maximum value of EF is $\\boxed{4\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52511902.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=x^2+bx+c$ (where $b$ and $c$ are constants) has the vertex $C$, intersects the x-axis at points $A$ and $B$, with $A(1, 0)$, and $AB=4$. Let point $P$ be a moving point on segment $AB$, and draw $PQ\\parallel BC$ intersecting $AC$ at point $Q$. Determine the maximum area of $\\triangle CPQ$ and the coordinates of point $P$ at that maximum.\n",
    "completion": "\\textbf{Solution:} From equation (1), we get the parabola equation as $y=x^{2}+2x-3$,\\\\\nthe vertex form is: $y=(x+1)^{2}-4$,\\\\\nthus the coordinates of point C are: $(-1,-4)$,\\\\\nfrom B $(-3,0)$ and C $(-1,-4)$, we can find the equation of line BC as: $y=-2x-6$,\\\\\nfrom A $(1,0)$ and C $(-1,-4)$, we can find the equation of line AC as: $y=2x-2$,\\\\\nsince $PQ\\parallel BC$,\\\\\nlet the equation of line PQ be: $y=-2x+n$, intersecting the x-axis at P$\\left(\\frac{n}{2}, 0\\right)$,\\\\\nfrom $\\begin{cases}y=-2x+n\\\\ y=2x-2\\end{cases}$, we find: $Q\\left(\\frac{n+2}{4}, \\frac{n-2}{2}\\right)$,\\\\\nsince $P$ is on the segment AB,\\\\\ntherefore $-3<\\frac{n}{2}<1$,\\\\\ntherefore the range of n is $-6<n<2$,\\\\\nthen the area of $\\triangle CPQ$ equals the area of $\\triangle CPA$ minus the area of $\\triangle APQ$\\\\\n$=\\frac{1}{2}\\times \\left(1-\\frac{n}{2}\\right)\\times 4-\\frac{1}{2}\\times \\left(1-\\frac{n}{2}\\right)\\times \\left(-\\frac{n-2}{2}\\right)$\\\\\n$=-\\frac{1}{8}(n+2)^{2}+2$\\\\\ntherefore, when $n=-2$, i.e., P $(-1,0)$, the area of $\\triangle CPQ$ is maximized, with the maximum value being $\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53043661_108.png"
  },
  {
    "solution": "\\boxed{12\\sqrt{10}\\text{cm}^2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, if two small squares with areas of $15\\text{cm}^2$ and $24\\text{cm}^2$ are cut from a large square, find the area of the remaining part.\n",
    "completion": "\\textbf{Solution:} Because the areas of the two small squares are $15\\text{cm}^2$ and $24\\text{cm}^2$ respectively,\\\\\nhence their side lengths are $\\sqrt{15}$cm and $\\sqrt{24}=2\\sqrt{6}$cm respectively,\\\\\ntherefore, the side length of the large square is $\\sqrt{15}+2\\sqrt{6}$,\\\\\ntherefore, the area of the remaining part (i.e., the shaded area) is $(\\sqrt{15}+2\\sqrt{6})^2−15−24$\\\\\n$=15+2\\times \\sqrt{15}\\times 2\\sqrt{6}+24−15−24$\\\\\n$=4\\sqrt{9\\times 10}$\\\\\n$=12\\sqrt{10}\\text{cm}^2$,\\\\\nthus, the area of the remaining part is $\\boxed{12\\sqrt{10}\\text{cm}^2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53183097.png"
  },
  {
    "solution": "\\boxed{5 \\sqrt{6}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the four corners of the rectangular piece of paper $ABCD$ are folded inward to form a quadrilateral $EFGH$ without gaps or overlaps. The quadrilateral $EFCH$ is a rectangle. If $AH=2$ and $HD=3$, what is the area of the rectangle $EFCH$?\n",
    "completion": "\\textbf{Solution:} Method 1: Knowing that the quadrilateral EFGH is a rectangle implies that EF=GH, and $\\angle B=\\angle D$, $\\angle BFE=\\angle CGF=\\angle DHG$,\\\\\n$\\therefore \\triangle BFE\\cong \\triangle DHG$, $\\therefore BF=DH$, $\\therefore JF=BF=HD=HK=3$, $\\therefore HF=HJ+JF=2+3=5$.\\\\\nFrom folding, it is known that AE=EJ=BE, let AE=x,\\\\\n$\\because \\angle HEF=90^\\circ$, $\\therefore EH^2+ EF^2= HF^2$,\\\\\n$\\therefore 2^2+x^2+3^2+x^2=(2+3)^2$, $\\therefore x= \\sqrt{6}$,\\\\\n$S_{\\text{Rectangle} EFGH}=2S_{\\triangle EFH}=2\\times \\frac{1}{2} \\times5\\times \\sqrt{6} = \\boxed{5 \\sqrt{6}}$.\\\\\nMethod 2: Similarly, we obtain JF=BF=HD=HK=3, HF=5, thus HO=FO=2.5, JO=KO=0.5,\\\\\n$\\because$ quadrilateral EFGH is a rectangle, EG=FH=5, EO=GO=2.5,\\\\\n$\\therefore$, knowing that EJ$\\perp$HF, $\\therefore HJ^2+EJ^2=EH^2, FJ^2+EJ^2= EF^2$,\\\\\n$\\therefore \\angle EJO=90^\\circ$, $\\therefore EJ^2= EO^2−JO^2 =2.5^2−0.5^2= 6$,\\\\\n$\\therefore EJ= \\sqrt{6}$,\\\\\n$\\therefore S_{\\text{Rectangle} EFGH}=2S_{\\triangle EFH}=2\\times \\frac{1}{2} \\times5\\times \\sqrt{6} = \\boxed{5 \\sqrt{6}}$.\\\\\nMethod 3: As above, we obtain HF=5, BF=3, drawing a line HP$\\perp$BC at point P, then PF=l, HP= $\\sqrt{6}$,\\\\\n$\\therefore S_{\\text{Rectangle} EFGH}= \\frac{1}{2} S_{\\triangle EFH}= \\frac{1}{2} \\times5\\times2 \\sqrt{6} = \\boxed{5 \\sqrt{6}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50475228_145.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the perimeter of $\\triangle ABC$ is $20\\text{cm}$, $BC=6\\text{cm}$. Circle $O$ is the inscribed circle of $\\triangle ABC$. The tangent line $MN$ of circle $O$ intersects $AB$ and $CA$ at points $M$ and $N$, respectively. What is the perimeter of $\\triangle AMN$?\n",
    "completion": "\\textbf{Solution:} Given that circle O is the inscribed circle of $\\triangle ABC$, and the tangent of circle O, MN, intersects AB and AC at points M and N, respectively,\n\n$\\therefore$BF=BE, CF=CD, DN=NG, EM=GM, AD=AE,\n\n$\\because$ the perimeter of $\\triangle ABC$ is 20cm, and BC=6cm,\n\n$\\therefore$AE=AD=$\\frac{AB+AC-BC}{2}$=$\\frac{20-BC-BC}{2}$=$\\frac{20-12}{2}$=4(cm),\n\n$\\therefore$ the perimeter of $\\triangle AMN$ is AM+MG+NG+AN=AM+ME+AN+ND=AE+AD=4+4=\\boxed{8}(cm).",
    "image_path": "dataset/mm_math/images/MM_Math/51402984.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that lines AB $\\parallel$ CD $\\parallel$ EF, BD $= 2$, and DF $= 4$. What is the value of $ \\frac{AC}{AE}$?\n",
    "completion": "\\textbf{Solution:} Given that $BD=2$ and $DF=4$,\\\\\nthus $BF=BD+DF=2+4=6$;\\\\\nsince $AB\\parallel CD\\parallel EF$,\\\\\nit follows that $\\frac{AC}{AE}=\\frac{BD}{BF}$,\\\\\nhence $\\frac{AC}{CE}=\\frac{2}{6}=\\boxed{\\frac{1}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52694081.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\square ABCD$, $AD=8$, points $E$ and $F$ are the midpoints of $BD$ and $CD$ respectively, what is the length of $EF$?\n",
    "completion": "\\textbf{Solution}: Since quadrilateral ABCD is a parallelogram, and $AD=8$, \\\\\nthen $BC=AD=8$.\\\\\nSince points E and F are midpoints of BD and CD, respectively, \\\\\nthen $EF=\\frac{1}{2}BC=\\frac{1}{2}\\times 8=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53232715.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points B, E, C, and F lie on the same straight line. Given that triangles $ABC \\cong DEF$, with $BC = 8$ and $BF = 13$, what is the length of BE?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABC \\cong \\triangle DEF$, BC=8, BF=13,\\\\\ntherefore, EF=BC=8,\\\\\ntherefore, BE=BF-EF=13-8=\\boxed{5}",
    "image_path": "dataset/mm_math/images/MM_Math/53044201.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points A, B, and C are all on the circle $\\odot O$. Connect OA, OB, AC, and BC. If OA $\\perp$ OB, what is the degree measure of $\\angle C$?\n",
    "completion": "\\textbf{Solution:} Since $OA\\perp OB$, \\\\\n$\\therefore$ $\\angle AOB=90^\\circ$, \\\\\n$\\therefore$ $\\angle C=\\frac{1}{2}\\angle AOB=45^\\circ$, \\\\\nThus, the answer is: $\\boxed{45^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51351223.png"
  },
  {
    "solution": "\\boxed{3.5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, we have $AB=AC$, and $AD$ is the altitude on $BC$. $E$ is a point on side $AB$, and $EF\\perp BC$ at point $F$, intersecting the extension of $CA$ at point $G$. It is given that $EF=2$ and $EG=3$. What is the length of $AD$?\n",
    "completion": "\\textbf{Solution:} Draw $AH \\perp FG$ at point H,\\\\\nsince AB=AC, and AD is an altitude,  \\\\\ntherefore, $\\angle DAC = \\angle DAB$,\\\\\nsince $EF \\perp BC$,\\\\\ntherefore, $\\angle ADC = \\angle GFC = 90^\\circ$,\\\\\ntherefore, $AD \\parallel GF$,\\\\\ntherefore, $\\angle G = \\angle DAC$, $\\angle AEG = \\angle BAD$,\\\\\ntherefore, $\\angle G = \\angle GEA$,\\\\\ntherefore, $AG = AE$,\\\\\nsince $AH \\perp EG$,\\\\\ntherefore, $EH= \\frac{1}{2}EG=1.5$;\\\\\nsince $\\angle AHF = \\angle ADF = \\angle HFD = 90^\\circ$,\\\\\ntherefore, quadrilateral ADFH is a rectangle,\\\\\ntherefore, $AD = HF = HE + EF = 1.5 + 2 = \\boxed{3.5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52807802.png"
  },
  {
    "solution": "\\boxed{4-\\sqrt{10}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, given rectangle $ABCD$, with $AB=4\\sqrt{3}$ and $BC=3$, point $P$ is a moving point on the right side of line $AB$, and $\\angle APB=60^\\circ$, $CE=2\\sqrt{3}-3$. What is the maximum distance of point $P$ to line $AB$? What is the minimum value of $PE$?\n",
    "completion": "\\textbf{Solution:} Given the problem, point $P$ is on the major arc $AB$ of the circumcircle of the equilateral triangle with side $AB$. If we draw $OF \\perp AB$ at point $F$, then $OF$ is the perpendicular bisector of $AB$. Therefore, when point $P$ is on $OF$, the distance from point $P$ to line $AB$ is maximized, with the maximum value being the length of $PF$. At this time, $\\triangle PAB$ is an equilateral triangle, therefore $PB=PA=AB=4\\sqrt{3}$, $\\angle PBA=60^\\circ$, therefore $PF=\\frac{\\sqrt{3}}{2}PB=6$. When points $P$, $E$, and $O$ are collinear, the value of $PE$ is minimized. As shown in the figure, connect $OB$. In $\\triangle BFO$, $\\angle BOF=60^\\circ$, $\\angle FBO=30^\\circ$, $BF=CG=\\frac{1}{2}AB=2\\sqrt{3}$, therefore $OF=\\frac{\\sqrt{3}}{3}BF=2$, therefore $OG=3-2=1$, $GE=CG-CE=3$, therefore $OE=\\sqrt{1^2+3^2}=\\sqrt{10}$, therefore $PE_{\\min}=OP-OE=4-\\sqrt{10}$. Thus, the answers are: $PF=\\boxed{6}$, $PE_{\\min}=\\boxed{4-\\sqrt{10}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55138396.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{3}{4}x+6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system $xOy$, points A and B are the intersections of the line $y=\\frac{3}{4}x+3$ with the x-axis and y-axis, respectively. Given the condition in (1), point D is on the x-axis. If the quadrilateral $ABCD$ is an isosceles trapezoid, find the equation of line $CD$.\n",
    "completion": "Solution: As shown in the figure, draw CM $\\perp$ x-axis at M, then $\\angle AOB=\\angle DMC=90^\\circ$, M(4,0),\\\\\n$\\because$ quadrilateral ABCD is an isosceles trapezoid,\\\\\n$\\therefore \\angle BAO=\\angle CDM$, AB=DC; in $\\triangle ABO$ and $\\triangle DCM$,\\\\\n$\\left\\{\\begin{array}{l}\n\\angle AOB=\\angle DMC \\\\\n\\angle BAO=\\angle CDM \\\\\nAB=DC\n\\end{array}\\right.$,\\\\\n$\\therefore \\triangle ABO \\cong \\triangle DCM$ (AAS),\\\\\n$\\therefore AO=DM=4$,\\\\\n$\\therefore OD=OM+MD=8$,\\\\\n$\\therefore D(8,0)$,\\\\\nLet the equation of line CD be $y=k(x-8)$, substituting point C(4,3),\\\\\nthen $k= -\\frac{3}{4}$,\\\\\n$\\therefore$ the equation of line CD is $y=-\\frac{3}{4}x+6$. Therefore, the equation of line CD is $\\boxed{y=-\\frac{3}{4}x+6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50446991_227.png"
  },
  {
    "solution": "\\boxed{(0, \\frac{3}{5})}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, the graph of the linear function $y=-\\frac{1}{2}x+4$ intersects the $x$-axis and $y$-axis at points $A$ and $B$ respectively, and intersects the graph of the direct proportion function $y=\\frac{3}{2}x$ at point $C$. By moving point $C$ 1 unit to the right and then 6 units downward, point $D$ is obtained. As shown in Figure 2, point $P$ is a moving point on the $y$-axis. When the sum of $CP+PD$ is minimized, what are the coordinates of point $P$?\n",
    "completion": "\\textbf{Solution:} Let point $D'$ be the symmetric point of point $D$ about the $y$-axis, then $D'(-3, -3)$. Connect $CD'$ and let it intersect the $y$-axis at point $P'$. Connect $P'D$. At this time, $CP'+PD$ is minimal. Let the equation of line $CD'$ be $y=kx+b$. Substituting points $C(2,3)$ and $D'(-3,-3)$, we obtain:\n\\[\n\\left\\{\n\\begin{array}{l}\n2k+b=3 \\\\\n-3k+b=-3\n\\end{array}\n\\right.\n\\]\nSolving this yields $k=\\frac{6}{5}$ and $b=\\frac{3}{5}$. Therefore, the equation of line $CD'$ is $y=\\frac{6}{5}x+\\frac{3}{5}$. When $x=0$, we have $y=\\frac{3}{5}$. Therefore, the coordinates of point $P'$ are $\\boxed{(0, \\frac{3}{5})}$, which represent the coordinates of point $P$ when $CP+PD$ is minimized.",
    "image_path": "dataset/mm_math/images/MM_Math/50481662_83.png"
  },
  {
    "solution": "\\boxed{40^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB=AC$, point D is a point inside $\\triangle ABC$, $\\angle D=110^\\circ$, $\\angle 1=\\angle 2$, then what is the measure of $\\angle A$ in degrees?\n",
    "completion": "\\textbf{Solution:}\\\\\nSince $\\angle D=110^\\circ$ and $\\angle 1=\\angle 2$,\\\\\nit follows that $\\angle D=180^\\circ-\\angle 1-\\angle DCB=110^\\circ$,\\\\\nhence $\\angle 1+\\angle DCB=70^\\circ$,\\\\\nSince $AB=AC$,\\\\\nit follows that $\\angle ABC=\\angle ACB$,\\\\\nthus $\\angle ABD=\\angle BCD$,\\\\\nSince $\\angle 1+\\angle DCB=70^\\circ$ and $\\angle 1=\\angle 2$,\\\\\nit follows that $\\angle ACB=\\angle 2+\\angle DCB=70^\\circ$,\\\\\nhence $\\angle ABC+\\angle ACB=140^\\circ$,\\\\\nthus $\\angle A=180^\\circ-140^\\circ=\\boxed{40^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52827540.png"
  },
  {
    "solution": "\\boxed{50^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, straight lines AB and CD intersect at point O, OE bisects $\\angle AOD$, $OF\\perp CD$, and  $\\angle AOE=70^\\circ$, find the degree measure of $\\angle BOF$.\n",
    "completion": "\\textbf{Solution:} Given that OE bisects $\\angle AOD$ and $\\angle AOE = 70^\\circ$, \\\\\nthus $\\angle AOD = 2\\angle AOE = 140^\\circ$, \\\\\nsince A, O, B are collinear, \\\\\nthus $\\angle BOD = \\angle AOC = 180^\\circ - \\angle AOD = 180^\\circ - 140^\\circ = 40^\\circ$, \\\\\ngiven that $OF \\perp CD$, \\\\\nthus $\\angle FOD = 90^\\circ$, \\\\\nthus $\\angle BOF = \\angle FOD - \\angle BOD = 90^\\circ - 40^\\circ = \\boxed{50^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/53205972.png"
  },
  {
    "solution": "\\boxed{24}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the diagonals of rhombus ABCD intersect at point O, E is the midpoint of CD, and OE = 3. What is the perimeter of the rhombus?\n",
    "completion": "\\textbf{Solution:} Since the diagonals of the rhombus ABCD intersect at point O,\\\\\nit follows that OB=OD,\\\\\ngiven that point E is the midpoint of CD,\\\\\nit then follows that DE=CE,\\\\\nthus, OE is the median of $\\triangle BCD$,\\\\\nhence, BC=2OE=6,\\\\\ntherefore, the perimeter of the rhombus is: $4BC=4\\times 6=\\boxed{24}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52387322.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is a centrally symmetric figure with $A$ as the center of symmetry. If $\\angle C=90^\\circ$, $\\angle B=30^\\circ$, and $AC=3$, what is the length of $BB'$?\n",
    "completion": "\\textbf{Solution:} In the right triangle $ABC$, $\\angle B=30^\\circ$, $AC=3$,\\\\ \n$\\therefore AB=2AC=6$,\\\\\n$\\because$ B and B' are centrally symmetric about A,\\\\\n$\\therefore BB'=2AB=\\boxed{12}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51920359.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{26}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in square $ABCD$, $AB=2$. Extend $AD$ to point $E$ so that $DE=1$, $EF \\perp AE$, and $EF=AE$. Connect $AF$ and $CF$ respectively. Let $M$ be the midpoint of $CF$. What is the length of $AM$?\n",
    "completion": "\\textbf{Solution:} Extend FE and BC to meet at H, and connect $AC$,\\\\\nIn square ABCD: $\\angle ADC=\\angle BCD=90^\\circ$, $\\angle DAC=45^\\circ$, $AB=CD=AD=2$,\\\\\n$\\therefore \\angle EDC=\\angle HCD=90^\\circ$,\\\\\n$\\because EF\\perp AE$,\\\\\n$\\therefore \\angle AEF=\\angle AEH=90^\\circ$,\\\\\n$\\therefore \\angle EDC=\\angle HCD=\\angle AEH=90^\\circ$,\\\\\n$\\therefore$ quadrilateral $DCHE$ is a rectangle,\\\\\n$\\therefore CD=EH=2$, $DE=CH=1$,\\\\\n$\\because EF=AE$, $\\angle AEF=90^\\circ$,\\\\\n$\\therefore \\angle FAE=45^\\circ$,\\\\\n$\\because DE=1$, $AD=2$,\\\\\n$\\therefore AE=EF=3$,\\\\\n$\\therefore FH=5$,\\\\\nAccording to the Pythagorean theorem: $CF=\\sqrt{CH^2+FH^2}=\\sqrt{1^2+5^2}=\\sqrt{26}$,\\\\\n$\\because \\angle CFA=\\angle FAE+\\angle EAF=90^\\circ$,\\\\\n$\\therefore \\triangle CAF$ is a right triangle,\\\\\n$\\because M$ is the midpoint of $CF$,\\\\\nAccording to the fact that the median to the hypotenuse of a right triangle is half the length of the hypotenuse: $AM=\\frac{1}{2}CF=\\boxed{\\frac{\\sqrt{26}}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53572100.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{2} \\alpha}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, it is known that $O$ is a point on line $AB$, $\\angle COD=90^\\circ$, and $OE$ bisects $\\angle BOC$. If $\\angle AOC=\\alpha$, find the measure of $\\angle DOE$ (expressed as an algebraic expression containing $\\alpha$).\n",
    "completion": "\\textbf{Solution:} Given that $\\angle AOC = \\alpha$ \\\\\nTherefore, $\\angle BOC = 180^\\circ - \\alpha$ \\\\\nSince OE bisects $\\angle BOC$ \\\\\nThen, $\\angle COE = \\frac{1}{2} \\angle BOC = 90^\\circ - \\frac{1}{2} \\alpha$ \\\\\nGiven that $\\angle COD$ is a right angle \\\\\nTherefore, $\\angle COD = 90^\\circ$ \\\\\nHence, $\\angle DOE = 90^\\circ - (90^\\circ - \\frac{1}{2} \\alpha) = \\boxed{\\frac{1}{2} \\alpha}$",
    "image_path": "dataset/mm_math/images/MM_Math/51317331_61.png"
  },
  {
    "solution": "\\boxed{9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $D$, $E$, $F$ are points on $AB$, $AC$, and $BC$, respectively, and $DE \\parallel BC$, $EF \\parallel AB$. If $BF:FC=2:3$ and $AB=15$, then what is the value of $BD$?\n",
    "completion": "\\[\n\\text{{Solution:}} \\quad \\because \\text{{ DE }} \\parallel \\text{{ BC}}, \\text{{ EF }} \\parallel \\text{{ AB}}, \\\\\n\\therefore \\text{{ quadrilateral BFED is a parallelogram, }} \\triangle \\text{{ EFC }} \\sim \\triangle \\text{{ ABC}}, \\\\\n\\therefore \\frac{{EF}}{{AB}} = \\frac{{CF}}{{CB}}, \\text{{ EF }} = \\text{{ BD }}, \\\\\n\\because \\text{{ BF:FC }} = 2:3, \\text{{ AB }} = 15, \\\\\n\\therefore \\text{{ EF }} = \\frac{3}{5}\\text{{ AB }} = 9, \\\\\n\\therefore \\text{{ BD }} = \\boxed{9};\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/52977300.png"
  },
  {
    "solution": "\\boxed{\\frac{5}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the line $l_1$ passing through point $B(1,0)$ intersects with line $l_2\\colon y=2x+4$ at point $P(-1,a)$. Find the area of the quadrilateral $PAOC$.\n",
    "completion": "\\textbf{Solution:} \n\nSince the line $l_1\\colon y=-x+1$ intersects with the y-axis at point C,\n\nthe coordinates of C are (0, 1),\n\nAnd since the line $l_2\\colon y=2x+4$ intersects with the x-axis at point A,\n\nthe coordinates of A are (-2, 0),\n\nBecause $S_{\\text{quadrilateral} PAOC}=S_{\\triangle PAB}-S_{\\triangle BOC}$,\n\ntherefore, $S_{\\text{quadrilateral} PAOC}=\\frac{1}{2}\\times3\\times2-\\frac{1}{2}\\times1\\times1=\\boxed{\\frac{5}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50431079_190.png"
  },
  {
    "solution": "\\boxed{5.4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated in the diagram, a street lamp (point O) is 6.4m above the ground. Xiao Ming, who is 1.6m tall, walks from point A, which is 9m away from the base of the street lamp (point P), along a straight line in the direction of AP to point D. When he stops, the length of his shadow under the street lamp has reduced by 1.8m. Calculate the distance Xiao Ming has walked.\n",
    "completion": "\\textbf{Solution:} From the given information, we have $OP=64m$, $AB=DE=1.6m$, $AP=9m$, $DF=AC-1.8$,\\\\\n$\\because$ $\\angle OPC=\\angle BAC$, $\\angle BCA=\\angle OCP$,\\\\\n$\\therefore$ $\\triangle BCA\\sim\\triangle OCP$,\\\\\n$\\therefore$ $\\frac{BA}{OP}=\\frac{AC}{PC}$, that is $\\frac{1.6}{6.4}=\\frac{AC}{AC+9}$,\\\\\n$\\therefore$ $AC=3$,\\\\\n$\\therefore$ $DF=AC-1.8=3-1.8=1.2\\text{(m)}$,\\\\\n$\\because$ $\\angle OPC=\\angle EDF=90^\\circ$, $\\angle EFD=\\angle OFP$,\\\\\n$\\therefore$ $\\triangle EFD\\sim\\triangle OFP$,\\\\\n$\\therefore$ $\\frac{DE}{OP}=\\frac{DF}{PF}$, that is $\\frac{1.6}{6.4}=\\frac{1.2}{PF}$,\\\\\n$\\therefore$ $PF=4.8$,\\\\\n$\\therefore$ $AD=AP-PF+DF=9-4.8+1.2=\\boxed{5.4}\\text{(m)}$,\\\\",
    "image_path": "dataset/mm_math/images/MM_Math/51402923.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{3}+4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, with $DE \\parallel BC$, $AD=EC$, $BD=4$, and $AE=3$, what is the length of $AB$?\n",
    "completion": "\\textbf{Solution:} Since $DE\\parallel BC$,\\\\\nit follows that $\\frac{AD}{BD}=\\frac{AE}{EC}$,\\\\\nsince $AD=EC$,\\\\\nthen $AD^2=AE\\cdot BD=3\\times 4=12$,\\\\\nthus $AD=2\\sqrt{3}$,\\\\\ntherefore $AB=AD+BD=2\\sqrt{3}+4$,\\\\\nhence the answer is: $AB=\\boxed{2\\sqrt{3}+4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52513747.png"
  },
  {
    "solution": "\\boxed{135^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the quadrilateral $ABCD$ is inscribed in a circle $\\odot O$, $OC=2$, and $AC=2\\sqrt{2}$. What is the measure of $\\angle ADC$ in degrees?\n",
    "completion": "\\textbf{Solution:} From the given condition, we have\\\\\n$\\because \\angle AOC=90^\\circ,$\\\\\n$\\therefore \\angle B= \\frac{1}{2}\\angle AOC=45^\\circ,$\\\\\n$\\because$ quadrilateral ABCD is inscribed in circle $O$,\\\\\n$\\therefore \\angle ADC=180^\\circ-45^\\circ=\\boxed{135^\\circ}.$",
    "image_path": "dataset/mm_math/images/MM_Math/53078807_106.png"
  },
  {
    "solution": "\\boxed{\\frac{14}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In rectangle $ABCD$, where $AB = 6$ and $AD = 8$, $P$ is a point on segment $AC$. If $\\triangle PCD$ is an isosceles triangle, then what is the length of $AP$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral ABCD is a rectangle,\\\\\n$\\therefore$ $\\angle$ADC=$90^\\circ$, DC=AB=6, AD=8,\\\\\n$\\therefore$ AC=$\\sqrt{AD^{2}+DC^{2}}=\\sqrt{8^{2}+6^{2}}=10$,\\\\\nConsider three cases:\\\\\n(1) When CP=CD=6, AP=AC−CP=10−6=4;\\\\\n(2) When PD=PC, $\\angle$PDC=$\\angle$PCD,\\\\\n$\\because$ $\\angle$PCD+$\\angle$PAD=$\\angle$PDC+$\\angle$PDA=$90^\\circ$,\\\\\n$\\therefore$ $\\angle$PAD=$\\angle$PDA,\\\\\n$\\therefore$ PD=PA,\\\\\n$\\therefore$ PA=PC,\\\\\n$\\therefore$ AP=$\\frac{1}{2}$AC=5;\\\\\n(3) When DP=DC, as illustrated, draw DQ$\\perp$AC at Q,\\\\\nthen PQ=CQ,\\\\\n$\\because$ Area$_{\\triangle ADC}$=$\\frac{1}{2}$AD$\\cdot$DC=$\\frac{1}{2}$AC$\\cdot$DQ,\\\\\n$\\therefore$ DQ=$\\frac{AD\\cdot DC}{AC}=\\frac{8\\times 6}{10}=\\frac{24}{5}$,\\\\\n$\\therefore$ CQ=$\\sqrt{DC^{2}-DQ^{2}}=\\sqrt{6^{2}-\\left(\\frac{24}{5}\\right)^{2}}=\\frac{18}{5}$,\\\\\n$\\therefore$ PC=2CQ=$\\frac{36}{5}$,\\\\\n$\\therefore$ AP=AC−PC=10−$\\frac{36}{5}=\\boxed{\\frac{14}{5}}$;\\\\\nIn summary, if $\\triangle$PCD is an isosceles triangle, the length of AP could be 4, 5, or $\\frac{14}{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53436917.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rhombus $ABCD$, we have $AE=1$, and $AF=BE=2$. If $P$ is a moving point on diagonal $BD$, what is the minimum value of $EP+FP$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, draw $EE'\\perp BC$, intersecting $BD$ at $E'$, then draw line $E'F$ intersecting $BD$ at $P'$, and connect $EP'$,\\\\\nsince quadrilateral $ABCD$ is a rhombus,\\\\\ntherefore $\\angle ABD = \\angle CBD$,\\\\\ntherefore $BD$ is the perpendicular bisector of $EE'$,\\\\\ntherefore $P'E = P'E'$,\\\\\ntherefore $PE + PF > P'E' + P'F' = E'F$,\\\\\ntherefore the minimum value of $EP + FP$ is $E'F$,\\\\\nsince $B'E = BE = AF$, $BE \\parallel AF$,\\\\\ntherefore quadrilateral $ABE'F$ is a parallelogram,\\\\\ntherefore $E'F = AB = AE + BE = \\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51614470.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the ratio of similarity between $\\triangle ADE$ and $\\triangle ABC$ is $1:2$, and the area of $\\triangle ADE$ is $1$. What is the area of the quadrilateral $DBCE$?\n",
    "completion": "\\textbf{Solution:} Since the similarity ratio between $\\triangle ADE$ and $\\triangle ABC$ is $1\\colon 2$,\\\\\nit follows that $\\frac{S_{\\triangle ADE}}{S_{\\triangle ABC}}=\\left(\\frac{1}{2}\\right)^2=\\frac{1}{4}$,\\\\\nthus $S_{\\triangle ABC}=4S_{\\triangle ADE}=4$,\\\\\ntherefore, the area of the quadrilateral $DBCE=4-1=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51299525.png"
  },
  {
    "solution": "\\boxed{\\left(\\frac{8}{3},\\frac{4\\sqrt{5}}{3}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, place the isosceles $\\triangle ABC$ as in Figure 1, so that the base $BC$ coincides with the $x$-axis, at this moment the coordinates of point $A$ are $\\left(2, \\sqrt{5}\\right)$. If this triangle is placed as in Figure 2, such that the side length $AB$ coincides with the $x$-axis, what are the coordinates of point $C$ then?\n",
    "completion": "\\textbf{Solution:} As shown in Figure 1, draw $AD\\perp OC$ through point A intersecting the x-axis at D,\\\\\nsince the coordinates of point A are $(2,\\sqrt{5})$,\\\\\nthus $BD=2$, $AD=\\sqrt{5}$,\\\\\naccording to the Pythagorean theorem, $BA=\\sqrt{BD^{2}+AD^{2}}=\\sqrt{2^{2}+(\\sqrt{5})^{2}}=3$,\\\\\nsince $\\triangle ABC$ is an isosceles triangle,\\\\\nthus $AB=AC=3$, $BC=2BD=4$,\\\\\nAs shown in Figure 2, draw $CE\\perp AB$ through point C intersecting the x-axis at E,\\\\\nlet $BE=x$, then $AE=3-x$,\\\\\nin $\\triangle CBE$, $CE^{2}=BC^{2}-BE^{2}=16-x^{2}$,\\\\\nin $\\triangle CEA$, according to the Pythagorean theorem,\\\\\n$CE^{2}+AE^{2}=CA^{2}$,\\\\\n$16-x^{2}+(3-x)^{2}=3^{2}$,\\\\\n$16+9-6x=0$,\\\\\n$x=\\frac{8}{3}$,\\\\\nthus $CE=\\sqrt{BC^{2}-BE^{2}}=\\sqrt{4^{2}-\\left(\\frac{8}{3}\\right)^{2}}=\\frac{4\\sqrt{5}}{3}$,\\\\\nthus the coordinates of point C are $\\left(\\frac{8}{3},\\frac{4\\sqrt{5}}{3}\\right)=\\boxed{\\left(\\frac{8}{3},\\frac{4\\sqrt{5}}{3}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52836352.png"
  },
  {
    "solution": "\\boxed{3+10\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the archaeological team measured the angle of elevation of the top $C$ of the ancient tower $BC$ at point $A$ to be $60^\\circ$. The length of the slope $AD$ is 5 meters, with a gradient of $i=3\\colon 4$, and the length of $BD$ is 6 meters. What is the height of the ancient tower $BC$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, draw $AE\\perp BC$ with $E$ as the foot of the perpendicular, and draw $AF\\perp BD$ with $F$ as the foot of the perpendicular, then quadrilateral $AEBF$ forms a rectangle, hence $AE=FB$,\\\\\nsince the slope of the ramp $AD$ has a gradient of $i=3\\colon 4$, let $AF=3x$, $DF=4x$,\\\\\ntherefore $AD=5x$,\\\\\nsince $AD=5$, then $x=1$,\\\\\nthus, $AF=3$, $DF=4$,\\\\\nsince the length of $BD$ is $6$,\\\\\ntherefore, $AE=FB=DF+BD=4+6=10$,\\\\\nsince $\\angle CAE=60^\\circ$,\\\\\ntherefore, $CE=AE\\times \\tan\\angle CAE=10\\sqrt{3}$,\\\\\nthus, $BC=BE+CE=10\\sqrt{3}+3$,\\\\\nwhich means the height of the ancient tower $BC$ is $\\boxed{3+10\\sqrt{3}}$ meters.",
    "image_path": "dataset/mm_math/images/MM_Math/52977728.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ is an equilateral triangle, $\\angle CBD = 90^\\circ$, $BD = BC$, and $AD$ is connected and intersects $BC$ at point $E$. What is the measure of $\\angle CAD$?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABC$ is an equilateral triangle, \\\\\ntherefore AB＝BC, and $\\angle ABC = \\angle CAB = 60^\\circ$, \\\\\nsince BD＝BC, \\\\\ntherefore AB＝BD, \\\\\ntherefore $\\angle BAD = \\angle BDA$, \\\\\nsince $\\angle CBD = 90^\\circ$, \\\\\ntherefore $\\angle ABD = 90^\\circ + 60^\\circ = 150^\\circ$, \\\\\ntherefore $\\angle BDA = 15^\\circ$, \\\\\ntherefore $\\angle CAD = \\angle CAB - \\angle BDA = 60^\\circ - 15^\\circ = \\boxed{45^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/53105794.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ABC=60^\\circ$, $BC=16$, point D lies on the extension of BA, $CA=CD$, and $BD=10$. What is the length of $AD$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw CE $\\perp$ AD at point E,\\\\\n$\\therefore$ $\\angle$BEC = $90^\\circ$,\\\\\n$\\because$ $\\angle$ABC = $60^\\circ$,\\\\\n$\\therefore$ $\\angle$BEC = $30^\\circ$,\\\\\n$\\therefore$ BE = $\\frac{1}{2}$BC = 8,\\\\\n$\\therefore$ DE = BD - BE = 10 - 8 = 2,\\\\\n$\\because$ CA = CD,\\\\\n$\\therefore$ AE = DE = 2,\\\\\n$\\therefore$ AD = \\boxed{4}.",
    "image_path": "dataset/mm_math/images/MM_Math/52907826.png"
  },
  {
    "solution": "\\boxed{16}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\triangle ABC$, $ \\angle ABC=90^\\circ$, $ \\angle A=60^\\circ$. Point D is on side $AC$, $ \\angle DBC=30^\\circ$, $ AC=12\\, \\text{cm}$. What is the perimeter of $ \\triangle ABD$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a parallelogram, \\\\\n$\\therefore$OB=OD, AB=CD,\\\\\nSince E is the midpoint of BC,\\\\\n$\\therefore$OE is the median of $\\triangle BCD$,\\\\\n$\\therefore$CD=2EO,\\\\\nSince EO=8,\\\\\n$\\therefore$CD=2EO=16,\\\\\n$\\therefore$AB=CD=\\boxed{16}.",
    "image_path": "dataset/mm_math/images/MM_Math/52019362.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle B=90^\\circ$, $AB=4$, $BC>AB$, and point $D$ is on side $BC$. Among all the parallelograms $ADCE$ with $AC$ as the diagonal, what is the minimum value of $DE$?\n",
    "completion": "Solution: The intersection point of the diagonals of parallelogram ADCE is the midpoint O of AC. When $OD \\perp BC$, OD is minimal, which means DE is minimal.\\\\\nSince $AB \\perp BC$,\\\\\nTherefore, $OD \\parallel AB$,\\\\\nTherefore, at this time, OD is the median of $\\triangle ABC$,\\\\\nTherefore, $OD = \\frac{1}{2}AB = 2$,\\\\\nSince parallelogram ADCE is a parallelogram,\\\\\nTherefore, $DE = 2OD = \\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52345523.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB \\parallel CD$, $AC$ and $BD$ intersect at point P. E and F are the midpoints of $AC$ and $BD$, respectively. If $AB = 10$ and $CD = 6$, what is the length of $EF$?\n",
    "completion": "\\textbf{Solution:} Connect CF and extend it to meet AB at G, \\\\\n$\\because$AB$\\parallel$DC, \\\\\n$\\therefore$$\\angle$D＝$\\angle$B, \\\\\n$\\because$F is the midpoint of BD, \\\\\n$\\therefore$DF＝BF, \\\\\nIn $\\triangle DFC$ and $\\triangle BFG$, $\\left\\{\\begin{array}{l}\\angle D=\\angle B \\\\ DF=BF \\\\ \\angle DFC=\\angle BFG\\end{array}\\right.$, \\\\\n$\\therefore$$\\triangle DFC\\cong \\triangle BFG$ by ASA, \\\\\n$\\therefore$BG＝CD＝6, CF＝FG, \\\\\n$\\therefore$AG＝AB−BG＝4, \\\\\n$\\because$CF＝FG, CE＝EA, \\\\\n$\\therefore$EF＝$\\frac{1}{2}$AG＝$\\frac{1}{2}\\times 4＝\\boxed{2}$",
    "image_path": "dataset/mm_math/images/MM_Math/53333629.png"
  },
  {
    "solution": "\\boxed{\\left(\\sqrt{2}, 1\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a right-angled triangle paper $ABO$ is placed in the Cartesian coordinate system, with point $A(\\sqrt{3}, 0)$, point $B(0, 1)$, and point $O(0, 0)$. $P$ is a moving point on side $AB$ (point $P$ does not coincide with points $A$ or $B$), and the paper is folded along $OP$ to obtain the corresponding point $A'$ of point $A$. As shown in Figure 1, when point $A'$ is in the first quadrant and satisfies $A'B \\perp OB$, what are the coordinates of point $A'$?\n",
    "completion": "\\textbf{Solution:} Let point $A\\left(\\sqrt{3}, 0\\right)$ and point $B\\left(0, 1\\right)$. Hence, $OA=\\sqrt{3}$ and $OB=1$. According to the property of folding, we have $OA'=OA=\\sqrt{3}$. Since $A'B\\perp OB$, it follows that $\\angle A'BO=90^\\circ$. In right triangle $\\triangle A'OB$, we find that $A'B=\\sqrt{OA'^{2}-OB^{2}}=\\sqrt{2}$,\\\\\nthus the coordinates of point $A'$ are $\\boxed{\\left(\\sqrt{2}, 1\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53572668_144.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AD$ bisects $\\angle BAC$. Given $AB=AC=3$ and $AD=2$, find the length of $BC$.\n",
    "completion": "\\textbf{Solution:} Since $AB=AC$ and $AD$ bisects $\\angle BAC$, \\\\\n$\\therefore AD \\perp BC$, and $BD=DC$, \\\\\n$\\therefore BD= \\sqrt{AB^{2}-AD^{2}}= \\sqrt{5}$, \\\\\n$\\therefore BC=2BD=2\\sqrt{5}=\\boxed{2\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51385808.png"
  },
  {
    "solution": "\\boxed{9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points D and E are on sides AC and AB of $\\triangle ABC$, respectively, with $AE=6$, $AC=10$, $BC=15$, and $\\frac{AD}{AB}=\\frac{3}{5}$. Find the length of DE.\n",
    "completion": "\\textbf{Solution:} Since $AE:AC = 6:10 = 3:5$, and $AD:AB = 3:5$,\\\\\nit follows that $\\frac{AD}{AB} = \\frac{AE}{AC}$.\\\\\nAlso, $\\angle A = \\angle A$,\\\\\ntherefore $\\triangle ADE \\sim \\triangle ABC$,\\\\\nthus $\\frac{DE}{BC} = \\frac{AE}{AC} = \\frac{3}{5}$.\\\\\nGiven that $BC = 15$,\\\\\nit follows that $DE = \\frac{3}{5} \\times 15 = \\boxed{9}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51367526.png"
  },
  {
    "solution": "\\boxed{10-4\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, points $D$ and $E$ are on $BC$ and $AC$ respectively. $\\triangle CDE$ is folded along $DE$ to result in $\\triangle FDE$. Lines $BF$ and $CF$ are drawn, with $\\angle BFC=90^\\circ$. If $EF\\parallel AB$, $AB=4\\sqrt{3}$, and $EF=10$, what is the length of $AE$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, extend $ED$ to meet $CF$ at point $G$,\\\\\nlet $\\angle BFE=\\alpha$\\\\\nBy folding, it is known that $DG\\perp CF$,\\\\\nsince $BF\\perp CF$,\\\\\nthus $ED\\parallel BF$,\\\\\ntherefore $\\angle FED=\\angle BFE=\\alpha$,\\\\\nextend $EA$, $FB$ to meet at point $M$,\\\\\nsince $AB\\parallel EF$,\\\\\ntherefore $\\angle BAC=\\angle FEC=2\\alpha$, $\\angle ABM=\\angle BFE=\\alpha$,\\\\\ntherefore $\\angle M=\\angle BAC-\\angle ABM=\\alpha$,\\\\\nsince $\\angle M=\\angle BFE=\\alpha$, $\\angle M=\\angle ABM=\\alpha$,\\\\\nthus $EM=EF=10$, $AM=AB=4\\sqrt{3}$,\\\\\ntherefore $AE=EM-AM=10-4\\sqrt{3}=\\boxed{10-4\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53232972.png"
  },
  {
    "solution": "\\boxed{\\frac{2}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: If a dart is thrown randomly onto the dartboard shown in the figure, where each point on the board has an equal chance of being hit, what is the probability that the dart will land in the white region?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, let the area of each small triangle be $a$,\\\\\nthen the area of the shaded region is $6a$, the area of the hexagon is $18a$, and the area of the white region is $12a$,\\\\\n$\\therefore$ throwing a dart at the dartboard randomly, the probability of it landing in the white area is $\\frac{12a}{18a}=\\boxed{\\frac{2}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52991321.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the diagonals of square $ABCD$ intersect at point $O$. $E$ is any point on $BC$, $EG \\perp BD$ at point $G$, and $EF \\perp AC$ at point $F$. If $AC=10$, what is the value of $EG+EF$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral ABCD is a square,\\\\\n$\\therefore \\angle OBC = \\angle OCB = 45^\\circ$, AC$\\perp$BD, and OC=$\\frac{1}{2}$AC=5,\\\\\n$\\because EG \\perp$BD and EF$\\perp$AC,\\\\\n$\\therefore quadrilateral EFOG is a rectangle, and  \\triangle BEG$, $\\triangle CEF$ are isosceles right triangles,\\\\\n$\\therefore EG=FO, CF=EF,\\\\\n\\therefore EG+EF=FO+CF=CO=\\boxed{5}$",
    "image_path": "dataset/mm_math/images/MM_Math/51614618.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{2}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In a square grid, the position of $\\triangle ABC$ is shown as in the figure, where points A, B, and C are all on the grid points. What is the value of $\\cos B$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw AD $\\perp$ BC and extend BC to meet point D,\\\\\n$\\because$ AD=BD=4, $\\angle$ADB=90$^\\circ$,\\\\\n$\\therefore$ $\\triangle$ABD is an isosceles right triangle,\\\\\n$\\therefore$ $\\angle$B=45$^\\circ$\\\\\n$\\therefore$ $\\cos B=\\frac{\\sqrt{2}}{2}$\\\\\nHence, the answer is: $\\boxed{\\frac{\\sqrt{2}}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51285474.png"
  },
  {
    "solution": "\\boxed{2.5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In $\\triangle ABC$, $\\angle C=90^\\circ$, $\\angle B=15^\\circ$, and $DE$ is the perpendicular bisector of $AB$, with $BE=5$. Find the length of $AC$.\n",
    "completion": "\\textbf{Solution:} Connect AE, \\\\\n$\\because$ DE is the perpendicular bisector of AB, \\\\\n$\\therefore$ BE = AE, \\\\\n$\\therefore$ $\\angle B = \\angle EAB = 15^\\circ$, \\\\\n$\\therefore$ $\\angle AEC = 30^\\circ$, \\\\\n$\\because$ $\\angle C = 90^\\circ$, \\\\\n$\\therefore$ AC = $\\frac{1}{2}$ AE = $\\frac{1}{2}$ BE = \\boxed{2.5}",
    "image_path": "dataset/mm_math/images/MM_Math/9100176.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{1}{2}x+10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the Cartesian coordinate system, the sides $OA$ and $OC$ of the rectangle $OABC$ lie on the positive half of the x-axis and the positive half of the y-axis, respectively, and the lengths of $OA$ and $OC$ are the two roots of the equation ${x}^{2}-20x+100=0$. As shown in the figure, under the condition when $FG: GO = 3:5$, find the equation of line $AD$.\n",
    "completion": "\\textbf{Solution:}\n\nLet $\\because FG:GO=3:5$, \\\\\nlet $FG=3a(a>0)$, \\\\\n$\\therefore GO=5a$, \\\\\nIt is known from (2) that $\\triangle EGH\\cong \\triangle CGO$, \\\\\n$\\therefore HG=GO=5a$, \\\\\n$\\therefore FH=2a$, \\\\\n$\\therefore AD=OH=10a$, \\\\\nConnect $OD$, \\\\\n$S_{\\triangle AOD}=\\frac{1}{2}AO\\cdot |x_D|=\\frac{1}{2}AD\\cdot OF$, \\\\\n$10a\\cdot 8a=10\\times 10$, \\\\\n$\\therefore$ solving gives $a=\\pm \\frac{\\sqrt{5}}{2}$, \\\\\n$\\because a>0$, \\\\\n$\\therefore a=\\frac{\\sqrt{5}}{2}$, \\\\\n$\\therefore AD=10a=5\\sqrt{5}$, \\\\\nIn $Rt\\triangle ABD$, $BD=\\sqrt{AD^2-AB^2}=5$, \\\\\n$\\therefore CD=BC-BD=5$, \\\\\n$\\therefore D(10, 5)$; \\\\\nLet the equation of line $AD$ be $y=kx+b(k\\ne 0)$, \\\\\n$\\because A(0, 10), D(10, 5)$, \\\\\n$\\therefore \\left\\{\\begin{array}{l}b=10 \\\\ 10k+b=5\\end{array}\\right.$, \\\\\n$\\therefore \\left\\{\\begin{array}{l}k=-\\frac{1}{2} \\\\ b=10\\end{array}\\right.$, \\\\\n$\\therefore$ the equation of line $AD$ is: \\boxed{y=-\\frac{1}{2}x+10}.",
    "image_path": "dataset/mm_math/images/MM_Math/53728996_97.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{10}}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of $\\odot O$, with $C$ and $D$ being two points on the circle. Connect $AC$ and $CD$, and let $AC=CD$. Extend $DC$ to intersect with the extension of $BA$ at point $E$. $\\triangle EAC\\sim \\triangle ECO$; if $\\tan\\angle EOC=\\frac{3}{4}$, find the value of $\\frac{EC}{EO}$.\n",
    "completion": "\\textbf{Solution:} Draw $CF\\perp EO$ through point C, \\\\\nsince $\\tan\\angle EOC=\\frac{3}{4}$, \\\\\ntherefore, $\\frac{CF}{OF}=\\frac{3}{4}$. Let $CF=3x$, then $OF=4x$, \\\\\nthus, $OC= \\sqrt{(3x)^{2}+(4x)^{2}}=5x=OA$, \\\\\ntherefore, $AF=5x-4x=x$, \\\\\nthus, $AC= \\sqrt{(3x)^{2}+x^{2}}=\\sqrt{10}x$, \\\\\ntherefore, $\\frac{AC}{OC}=\\frac{\\sqrt{10}x}{5x}=\\frac{\\sqrt{10}}{5}$. From (1), we get $\\triangle EAC\\sim \\triangle ECO$, \\\\\nthus, $\\frac{EC}{EO}=\\frac{AC}{OC}$, \\\\\ntherefore, $\\frac{EC}{EO}=\\boxed{\\frac{\\sqrt{10}}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51433210_146.png"
  },
  {
    "solution": "\\boxed{65^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, from a point $P$ outside the circle $\\odot O$, rays $PA$ and $PB$ are drawn to tangentially touch the circle $\\odot O$ at points $A$ and $B$, respectively, with $\\angle P = 50^\\circ$. Point $C$ is on the minor arc $AB$. A tangent to circle $\\odot O$ at point $C$ intersects $PA$ and $PB$ at points $D$ and $E$, respectively. What is the measure of $\\angle DOE$ in degrees?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, connect OA, OC, OB,\n\nsince PA, PB, DE are tangential to the circle at points A, B, and E, respectively,\n\ntherefore, $OA \\perp PA$,\n\n$OB \\perp PB$,\n\n$OC \\perp DE$,\n\nsince $\\angle P = 50^\\circ$,\n\ntherefore, $\\angle AOB = 180^\\circ - \\angle P = 130^\\circ$,\n\nsince $OA = OB = OC$,\n\ntherefore, DO bisects $\\angle ADC$, EO bisects $\\angle BEC$,\n\ntherefore, $\\angle ADO = \\angle CDO$,\n\n$\\angle CEO = \\angle BEO$,\n\ntherefore, $\\angle AOD = \\angle COD$,\n\n$\\angle COE = \\angle BOE$,\n\ntherefore, $\\angle DOE = \\angle COD + \\angle COE = \\frac{1}{2}\\angle AOC + \\frac{1}{2}\\angle BOC = \\frac{1}{2}\\angle AOB = \\boxed{65^\\circ}$,",
    "image_path": "dataset/mm_math/images/MM_Math/51433712.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Place the right-angled vertex $O$ of the right-angled triangle $OMN$ on the line $AB$, and let the ray $OC$ bisect $\\angle AON$. As shown in the figure, if $\\angle BON=60^\\circ$, what is the degree measure of $\\angle COM$?\n",
    "completion": "\\textbf{Solution:}\\\\\n$\\because \\angle BON=60^\\circ$,\\\\\n$\\therefore \\angle AON=120^\\circ$,\\\\\n$\\because OC$ bisects $\\angle AON$,\\\\\n$\\therefore \\angle CON=60^\\circ$,\\\\\n$\\because \\angle MON=90^\\circ$,\\\\\n$\\therefore \\angle COM=90^\\circ-60^\\circ=\\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51385746_142.png"
  },
  {
    "solution": "\\boxed{\\left(4, 1\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the parabola $y=ax^2+bx+c$ has its vertex at $A(4,3)$ and intersects the $y$-axis at point $B(0,-5)$. The axis of symmetry is the line $l$. Let point $M$ be the midpoint of segment $AB$. Assume that $P$ and $Q$ are moving points on the parabola and the axis of symmetry $l$ respectively. When the quadrilateral with vertices $A$, $P$, $Q$, $M$ forms a parallelogram, determine the coordinates of points $P$ and $Q$.\n",
    "completion": "\\textbf{Solution:} Let point $Q(4, s)$ and point $P(m, -\\frac{1}{2}m^2+4m-5)$,\\\\\n(1) When $AM$ is one side of the parallelogram,\\\\\nWhen point Q is below A,\\\\\nPoint A is moved 2 units to the left and 4 units down to get M,\\\\\nSimilarly, point $P(m, -\\frac{1}{2}m^2+4m-5)$ moved 2 units to the left and 4 units down to get $Q(4, s)$\\\\\ni.e., $m-2=4, -\\frac{1}{2}m^2+4m-5-4=s$\\\\\nSolving yields: $m=6, s=-3,$\\\\\nTherefore, when point Q is above point A, $AQ=MP=2,$\\\\\nSimilarly, the coordinates of point Q are found to be $(4, 5),$\\\\\n(2) When $AM$ is the diagonal of the parallelogram,\\\\\nBy the midpoint theorem, $4+2=m+4, 3-1=-\\frac{1}{2}m^2+4m-5+s$\\\\\nSolving yields: $m=2, s=1$\\\\\nTherefore, the coordinates of points $P, Q$ are respectively $\\left(2, 1\\right), \\left(4, 1\\right)$\\\\\nIn summary, the coordinates of $P, Q$ are respectively $\\left(6, -3\\right), \\left(4, 5\\right)$ or $\\boxed{\\left(2, 1\\right)}, \\boxed{\\left(4, 1\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53042870_134.png"
  },
  {
    "solution": "\\boxed{1\\,cm}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: The line segments $AB$ and $CD$ are on the same straight line, with $M$ and $N$ being the midpoints of line segments $AB$ and $CD$, respectively. It is known that $AB=6\\,cm$ and $CD=8\\,cm$. When the points $A$ and $C$ coincide, what is the length of $MN$ in $cm$?\n",
    "completion": "Solution: Since $M$ and $N$ are the midpoints of line segments $AB$ and $CD$ respectively, and $AB=6\\,cm$, $CD=8\\,cm$, with points $A$ and $C$ coinciding,\\\\\ntherefore, $AM=3\\,cm$, $AN=4\\,cm$,\\\\\nhence $MN=AN-AM=\\boxed{1\\,cm}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52916287_137.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=ax^2+bx+c$ passes through points $A(-1,0)$, $B(3,0)$, and $C(0,3)$. Point $D$ is the vertex, and the axis of symmetry is the line $DE$. If point $M$ is a point on the parabola to the top right of $BD$, and the x-coordinate of $M$ is denoted as $m$, and the area of $\\triangle MBD$ is $S$. Does $S$ have a maximum value? If yes, please find the maximum value and the coordinates of $M$. If no, please explain why.\n",
    "completion": "\\textbf{Solution:}\nGiven that $y=-x^2+2x+3=-\\left(x-1\\right)^2+4$,\\\\\nthus, the coordinates of point D are $(1,4)$, and the axis of symmetry is the line $x=1$,\\\\\nlet the equation of line BD be $y=kx+f$,\\\\\nthus, $\\left\\{\\begin{array}{l}\n3k+f=0 \\\\\nk+f=4\n\\end{array}\\right.$,\\\\\nsolving this yields $\\left\\{\\begin{array}{l}\nk=-2 \\\\\nf=6\n\\end{array}\\right.$,\\\\\nthus, $y=-2x+6$,\\\\\ndrawing line $MG\\parallel y$ axis to intersect BD at point G from M,\\\\\nsince $M\\left(m,-m^2+2m+3\\right)$,\\\\\nthus, $G\\left(m,-2m+6\\right)$,\\\\\nthus, $MG=-m^2+4m-3$,\\\\\nthus, $S=\\frac{1}{2}\\times(3-1)\\times(-m^2+4m-3)=-m^2+4m-3=-\\left(m-2\\right)^2+1$,\\\\\nthus, when $m=2$, $S$ reaches its maximum value $\\boxed{1}$, at this time $M(2,3)$;",
    "image_path": "dataset/mm_math/images/MM_Math/52699702_87.png"
  },
  {
    "solution": "\\boxed{2.4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle BAC=90^\\circ$, $AB=3$, $AC=4$. Let P be a moving point on side $BC$, with $PE \\perp AB$ at E and $PF \\perp AC$ at F. What is the minimum value of $EF$?\n",
    "completion": "\\textbf{Solution:} Connect AP, as shown in the diagram: \\\\\nSince $PE\\perp AB$ and $PF\\perp AC$, \\\\\nTherefore, $\\angle AEP=\\angle AFP=90^\\circ$, \\\\\nSince $\\angle BAC=90^\\circ$, \\\\\nTherefore, the quadrilateral AFPE is a rectangle, \\\\\nTherefore, $EF=AP$, \\\\\nTo minimize EF, we only need to minimize AP, \\\\\nWhen $AP\\perp BC$, AP is the shortest, \\\\\nSince $\\angle BAC=90^\\circ$ and $AB=3$ and $AC=4$, \\\\\nTherefore, $BC=\\sqrt{AB^{2}+AC^{2}}=\\sqrt{3^{2}+4^{2}}=5$, \\\\\nSince the area of $\\triangle ABC$ $=\\frac{1}{2}\\times 4\\times 3=\\frac{1}{2}\\times 5\\times AP$, \\\\\nTherefore, $AP=2.4$, \\\\\nThat is, $EF=\\boxed{2.4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53572608.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rhombus $ABCD$, $\\angle D=120^\\circ$, then what is the degree measure of $\\angle 1$?\n",
    "completion": "\\textbf{Solution:} Given rhombus $ABCD$,\\\\\nit follows that $AB\\parallel CD$ and $\\angle 1=\\frac{1}{2}\\angle DAB$,\\\\\nsince $\\angle D=120^\\circ$,\\\\\nit follows that $\\angle DAB=60^\\circ$,\\\\\nhence $\\angle 1=\\frac{1}{2}\\angle DAB=\\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52190011.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: The lines $y_1 = -x + 3$ and $y_2 = kx - 2$ intersect the y-axis at points $A$ and $B$, respectively, and intersect at point $C(2, m)$. What is the area of $\\triangle ABC$?\n",
    "completion": "\\textbf{Solution:} When $x=0$, $y_1=3$,\\\\\nthen $A(0,3)$;\\\\\nWhen $x=0$, $y_2=-2$,\\\\\nthen $B(0,-2)$,\\\\\n$\\therefore$ The area of $\\triangle ABC = \\frac{1}{2} \\times (3+2) \\times 2 = \\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51968557_185.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $D$ and $E$ are points on $AC$ and $AB$ respectively, $\\triangle ADE \\sim \\triangle ABC$, and $DE=8$, $BC=24$, $CD=18$, $AD=6$. Find the lengths of $AE$ and $BE$.\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ADE \\sim \\triangle ABC$,\\\\\nwe have $\\frac{AE}{AC}=\\frac{AD}{AB}=\\frac{DE}{BC}$,\\\\\nSince $DE=8$, $BC=24$, $CD=18$, $AD=6$,\\\\\nthus $AC=AD+CD=24$,\\\\\nthus $AE=\\boxed{8}$, $AB=18$,\\\\\nthus $BE=AB-AE=\\boxed{10}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51322393.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, AB and CD are two chords of $\\odot O$ that intersect at point P. AD and BD are connected, and it is known that AD = BD = 4 and PC = 6. What is the length of CD?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect AC,\\\\\naccording to the inscribed angle theorem, we know that $\\angle C = \\angle B$,\\\\\nsince $AD=BD$,\\\\\ntherefore $\\angle B = \\angle DAB$,\\\\\nthus $\\angle DAP = \\angle C$,\\\\\nhence $\\triangle DAP \\sim \\triangle ACA$,\\\\\ntherefore $AD \\colon CD = DP \\colon AD$,\\\\\nit follows that $AD^2 = DP \\cdot CD = CD(CD - PC)$,\\\\\nsubstituting $AD = 4$, $PC = 6$ into the equation yields,\\\\\n$CD = \\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51545190.png"
  },
  {
    "solution": "\\boxed{-3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, the line BC intersects the x-axis at point C and the y-axis at point B, and intersects the line $y=ax$ at point A. The x-coordinate of point A is 2, $\\angle ACO = 45^\\circ$, and the area of $\\triangle ABO$ is 1. If the line $y=ax+m$ intersects the y-axis at point D, when the area of $\\triangle ABD$ is 4, what is the value of $m$?\n",
    "completion": "\\textbf{Solution:} Since the line $y=ax+m$ intersects the y-axis at point D, \\\\ \ntherefore $D(0, m)$, \\\\ \nsince the area of $\\triangle ABD=\\frac{1}{2}BD\\cdot x_A=\\frac{1}{2}BD\\times 2=4$, \\\\ \nthus $BD=4$, \\\\ \nsince $B(0, 1)$, \\\\ \ntherefore the coordinates of point D are $(0, 5)$ or $(0, -3)$, \\\\ \nthus $m=\\boxed{5}$ or \\boxed{-3};",
    "image_path": "dataset/mm_math/images/MM_Math/52056412_201.png"
  },
  {
    "solution": "\\boxed{95^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, $ \\triangle ABC \\cong \\triangle ADE$, and $\\angle EAB = 120^\\circ$, $\\angle B = 30^\\circ$, $\\angle CAD = 10^\\circ$, what is the degree measure of $\\angle CFD$?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABC \\cong \\triangle ADE$, \\\\\nit follows that $\\angle EAD = \\angle CAB$, \\\\\nand since $\\angle EAB = 120^\\circ$ and $\\angle CAD = 10^\\circ$, \\\\\nit follows that $\\angle EAD = \\angle CAB = 55^\\circ$, \\\\\ntherefore, $\\angle CFD = \\angle FAB + \\angle B = 10^\\circ + 55^\\circ + 30^\\circ = \\boxed{95^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/53019700.png"
  },
  {
    "solution": "\\boxed{56^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given the diagram, place the right-angled vertex \\(O\\) of the right-angled triangle \\(MON\\) on the line \\(AB\\), and let the ray \\(OC\\) bisect \\(\\angle AON\\). As shown in Figure 1, if \\(\\angle MOC = 28^\\circ\\), what is the degree measure of \\(\\angle BON\\)?\n",
    "completion": "\\textbf{Solution:} As shown in Figure 1, since $\\angle MOC=28^\\circ$ and $\\angle MON=90^\\circ$,\\\\\nit follows that $\\angle NOC=90^\\circ-28^\\circ=62^\\circ$,\\\\\nand since $OC$ bisects $\\angle AON$,\\\\\nit follows that $\\angle AOC=\\angle NOC=62^\\circ$,\\\\\ntherefore $\\angle BON=180^\\circ-2\\angle NOC=180^\\circ-62^\\circ\\times 2=\\boxed{56^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51481976_180.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $O$ is any point inside the equilateral triangle $ABC$. Lines $OD$, $OE$, and $OF$ are drawn through point $O$ parallel to $AB$, $AC$, and $BC$, respectively, intersecting $AC$, $BC$, and $AB$ at points $G$, $H$, and $I$. Given that the perimeter of the equilateral triangle $ABC$ is 18, what is the sum of $OD + OE + OF$?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABC$ is an equilateral triangle with a perimeter of 18, \\\\\n$\\therefore$ $\\angle B=\\angle C=\\angle A=60^\\circ$, AC=6\\\\\nSince OD$\\parallel$AB, OE$\\parallel$AC, and OF$\\parallel$BC, \\\\\n$\\therefore$ Quadrilateral OHCF is a parallelogram, Quadrilateral AEOG is a parallelogram, $\\angle C=\\angle OHD=\\angle GFO=60^\\circ$, $\\angle ODH=\\angle B=60^\\circ$, $\\angle OGF=\\angle A=60^\\circ$\\\\\n$\\therefore$ $\\triangle ODH$ and $\\triangle OGF$ are equilateral triangles, OH=CF\\\\\n$\\therefore$ OD=OH, OF=FG, OE=AG\\\\\n$\\therefore$ OD=CF,\\\\\nSince AC=AG+GF+CF\\\\\n$\\therefore$ AC=OE+OF+OE=6.\\\\\nThus, the answer is: $\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50475020.png"
  },
  {
    "solution": "\\boxed{x<\\frac{3}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=kx+b\\ (k<0)$ intersects the y-axis at point A and intersects the x-axis at point B, and $(AB+OA)(AB-OA)=\\frac{9}{4}$. What is the solution set of the inequality $kx+b>0$?\n",
    "completion": "\\textbf{Solution:}\\\\\nSince $(AB+OA)(AB−OA)=\\frac{9}{4}$,\\\\\nit follows that $AB^{2}−OA^{2}=\\frac{9}{4}$,\\\\\ntherefore $OB^{2}=\\frac{9}{4}$,\\\\\nhence $OB=\\frac{3}{2}$,\\\\\ntherefore, the solution set of the inequality $kx+b>0$ is $\\boxed{x<\\frac{3}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52355189.png"
  },
  {
    "solution": "\\boxed{(4,0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, the parabola $y=-x^2+bx+c$ passes through points $A(-1,0)$ and $B(4,0)$, and intersects with the y-axis at point $C$. Connect $BC$, and let point $P$ be a moving point on the parabola. Perpendicular line $l$ is drawn from point $P$ to the x-axis, intersecting line $BC$ at point $G$ and the x-axis at point $E$. When $P$ moves on the section of the parabola to the right of the y-axis, draw $CF\\perp l$ through point $C$, where $F$ is the foot of the perpendicular. At what position of $P$ will the triangle formed by $P$, $C$, and $F$ be similar to $\\triangle OBC$? Also, find the coordinates of point $P$ at this time.\n",
    "completion": "\\textbf{Solution:} Set $x=0$, then we get $y=4$, and the point $C(0,4)$,\\\\\n\\(\\therefore OC=4\\).\\\\\n\\(\\therefore OC=OB\\).\\\\\n\\(\\therefore \\triangle OBC\\) is an isosceles right triangle,\\\\\n\\(\\because \\angle CFP=\\angle COB=90^\\circ\\),\\\\\n\\(\\therefore\\) when $FC=PF$, \\(\\triangle PFC\\sim \\triangle OBC\\).\\\\\nLet the coordinates of point $P$ be \\((a, -a^{2}+3a+4)\\) (\\(a>0\\)).\\\\\nThen \\(CF=a\\), \\(PF= |-a^{2}+3a+4-4|=|a^{2}-3a|\\).\\\\\n\\(\\therefore |a^{2}-3a| =a\\).\\\\\n\\(\\therefore a^{2}-3a=a\\),\\\\\nSolving this gives $a=4$, $a=0$ (discard the latter),\\\\\nWhen $a=4$, $y=-4^{2}+3\\times 4+4=0$,\\\\\n\\(\\therefore\\) Point $P(4,0)$,\\\\\nor $a^{2}-3a=-a$,\\\\\nSolving this gives $a=2$, $a=0$ (discard the latter),\\\\\nWhen $a=2$ then, $y=-2^{2}+3\\times 2+4=6$.\\\\\n\\(\\therefore\\) Point $P(2,6)$,\\\\\n\\(\\therefore\\) In synthesis, the coordinates of point $P$ can be \\boxed{(2,6)} or \\boxed{(4,0)}.",
    "image_path": "dataset/mm_math/images/MM_Math/51458091_167.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{3}{8}x^2-\\frac{9}{4}x+6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, in the Cartesian coordinate system, $\\odot P$ is the inscribed circle of $\\triangle AOC$, with points D, E, and F being the points of tangency. Line CD intersects $\\odot P$ at point G. The radius of $\\odot P$ is 2, $EG \\parallel$ to the x-axis, and $AB=AC$. A parabola passes through points A, B, and C. $\\triangle ADP\\cong \\triangle COD$; what is the equation of the parabola?\n",
    "completion": "\\textbf{Solution:} From (1), we know $\\triangle APD\\cong \\triangle CDO$, $\\therefore OD=PD=2$, $AD=OC=AE$, $\\because$ circle $P$ is the inscribed circle of $\\triangle AOC$, $\\therefore CE=CF$, let $CE=CF=n$, then $AE=AD=OC=n+2$, $OA=n+4$, $AC=2n+2$. In right triangle $\\triangle ACO$, $OA^2+OC^2=AC^2$, $\\therefore (n+4)^2+(n+2)^2=(2n+2)^2$, solving this gives: $n_1=4$, $n_2=-2$ (discard), $\\therefore OA=8$, $OC=6$, $AC=10$, $\\because AB=AC$, $\\therefore AB=10$, $\\therefore OB=AB-OA=2$, $\\therefore A(-8, 0)$, $B(2, 0)$, $C(0, 6)$. The equation of the parabola is $y=a(x+8)(x-2)$. Substituting $C(0, 6)$ into it gives: $6=a(0+8)(0-2)$, solving for $a$ gives: $a=-\\frac{3}{8}$, $\\therefore y=-\\frac{3}{8}(x+8)(x-2)=-\\frac{3}{8}x^2-\\frac{9}{4}x+6$, $\\therefore$ the equation of the parabola is $y=-\\frac{3}{8}x^2-\\frac{9}{4}x+6$; thus, the equation of the parabola is $\\boxed{y=-\\frac{3}{8}x^2-\\frac{9}{4}x+6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52949747_132.png"
  },
  {
    "solution": "\\boxed{\\frac{4}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in square $ABCD$ with a side length of $4$, point $E$ is a point on side $CD$, and point $F$ is symmetric to point $D$ with respect to line $AE$. Connect $AF$ and $BF$. If $\\tan\\angle ABF=2$, what is the length of $DE$?\n",
    "completion": "\\textbf{Solution:} Construct FN$\\perp$AB at point N, and extend NF to intersect CD at point M,\\\\\nsince AB$\\parallel$CD,\\\\\ntherefore MN$\\perp$CD,\\\\\ntherefore $\\angle$FME＝$90^\\circ$,\\\\\nsince $\\tan\\angle$ABF＝2,\\\\\ntherefore $\\frac{FN}{BN} = 2$,\\\\\nLet BN＝x, then FN＝2x,\\\\\ntherefore AN＝4−x,\\\\\nsince point F is the symmetric point of point D with respect to line AE,\\\\\ntherefore DE＝EF, DA＝AF＝4,\\\\\nsince AE＝AE,\\\\\ntherefore $\\triangle$ADE$\\cong\\triangle$AFE (SSS),\\\\\ntherefore $\\angle$D＝$\\angle$AFE＝$90^\\circ$,\\\\\nsince AN$^{2}$＋NF$^{2}$＝AF$^{2}$,\\\\\ntherefore $(4−x)^{2}＋(2x)^{2}=4^{2}$,\\\\\ntherefore $x_{1}=0$ (discard), $x_{2}=\\frac{8}{5}$,\\\\\ntherefore AN＝4−x＝4−$\\frac{8}{5}$＝$\\frac{12}{5}$, MF＝4−2x＝4−$\\frac{16}{5}$＝$\\frac{4}{5}$,\\\\\nsince $\\angle$EFM+$\\angle$AFN＝$\\angle$AFN+$\\angle$FAN＝$90^\\circ$,\\\\\ntherefore $\\angle$EFM＝$\\angle$FAN,\\\\\ntherefore $\\cos\\angle$EFM＝$\\cos\\angle$FAN,\\\\\ntherefore $\\frac{FM}{EF} = \\frac{AN}{AF}$, that is $\\frac{\\frac{4}{5}}{EF}=\\frac{\\frac{12}{5}}{4}$,\\\\\ntherefore EF＝$\\boxed{\\frac{4}{3}}$,\\\\\ntherefore DE＝EF＝$\\frac{4}{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50987121.png"
  },
  {
    "solution": "\\boxed{\\frac{2}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, $AC=2$, and $AB=3$, what is the value of $\\sin B$?\n",
    "completion": "\\textbf{Solution:} Given that in the right-angled triangle $ABC$, $\\angle C=90^\\circ$, $AC=2$, $AB=3$,\\\\\ntherefore, $\\sin B=\\frac{AC}{AB}=\\boxed{\\frac{2}{3}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51446419.png"
  },
  {
    "solution": "\\boxed{65^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, AE is the diameter of the circumscribed circle $\\odot O$ of the quadrilateral ABCD, $AD=CD$, $\\angle B=50^\\circ$, then what is the degree measure of $\\angle DAE$?\n",
    "completion": "\\textbf{Solution:} Draw OC and OD, \\\\\n$\\because$ $\\angle B=50^\\circ$, \\\\\n$\\therefore$ $\\angle AOC=2\\angle B=100^\\circ$, \\\\\n$\\because$ AD=CD, \\\\\n$\\therefore$ $\\overset{\\frown}{AD}=\\overset{\\frown}{CD}$, \\\\\n$\\therefore$ $\\angle AOD=\\angle COD= \\frac{1}{2}\\angle AOC=50^\\circ$, \\\\\n$\\because$ OA=OD, \\\\\n$\\therefore$ $\\angle OAD=\\angle ODA$, \\\\\n$\\therefore$ $\\angle DAE=(180^\\circ-50^\\circ)\\div2=\\boxed{65^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51402794.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $CE \\perp BD$ with point $E$ being the foot of the perpendicular. Connect $AE$. If $\\angle DCE : \\angle ECB = 3 : 1$, then what is the measure of $\\angle ACE$?\n",
    "completion": "\\textbf{Solution:} Given that rectangle $ABCD$, with $CE \\perp BD$, \\\\\nhence, $\\angle DCB = \\angle CEB = \\angle CED = 90^\\circ$, and $OC = OA = OD = OB$, \\\\\ntherefore, $\\angle OCB = \\angle OBC$, \\\\\nsince $\\angle DCE : \\angle ECB = 3 : 1$ and $\\angle ECB + \\angle DCE = 90^\\circ$, \\\\\nthus, $4\\angle ECB = 90^\\circ$, \\\\\ntherefore, $\\angle ECB = 22.5^\\circ$, \\\\\nhence, $\\angle OCB = \\angle OBC = 90^\\circ - \\angle ECB = 90^\\circ - 22.5^\\circ = 67.5^\\circ$, \\\\\nhence, $\\angle OCE = \\angle OCB - \\angle ECB = 67.5^\\circ - 22.5^\\circ = 45^\\circ$, \\\\\ntherefore, $\\angle ACE = \\boxed{45^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51435952.png"
  },
  {
    "solution": "\\boxed{\\frac{60}{17}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AC=BC=10\\text{cm}$, $CD\\perp AB$, $CD=8\\text{cm}$. Point P moves from B to A at a speed of $1\\text{cm/s}$, simultaneously, points Q and R move from C to B and A to C, respectively, at a speed of $2\\text{cm/s}$. The movement of all points stops when one of them stops. The movement time is $t$ (unit: $s$, $0<t<5$). At what value of $t$ is $PR\\parallel BC$?\n",
    "completion": "\\textbf{Solution:} Given that $AC=BC=10\\text{cm}$ and $CD\\perp AB$,\\\\\nit follows that $AD=BD=\\sqrt{10^2-8^2}=6\\text{cm}$,\\\\\nthus $AB=12\\text{cm}$.\\\\\nAccording to the problem, let $BP=t\\text{cm}$ and $AR=2t\\text{cm}$,\\\\\nthus $AP=AB-BP=(12-t)\\text{cm}$.\\\\\nSince $PR\\parallel BC$,\\\\\nit follows that $\\frac{AR}{AC}=\\frac{AP}{AB}$, i.e., $\\frac{2t}{10}=\\frac{12-t}{12}$,\\\\\nsolving this yields: $t=\\boxed{\\frac{60}{17}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53043392_175.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in right triangle \\( \\triangle ABC \\), where \\( \\angle ACB=90^\\circ \\), \\( CD \\) is the median to the hypotenuse \\( AB \\), and it is given that \\( CD=2 \\) and \\( AC=3 \\), what is the value of \\( \\cos A \\)?\n",
    "completion": "\\textbf{Solution:} Since $D$ is the midpoint of $AB$, and $CD=2$, \\\\\n$\\therefore$ $AC=2CD=4$, \\\\\n$\\cos A= \\frac{AC}{AB}=\\boxed{\\frac{3}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51299761.png"
  },
  {
    "solution": "\\boxed{20^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $ \\triangle ABC \\cong \\triangle DCB$, $ \\angle A = 80^\\circ$, and $ \\angle ACB = 40^\\circ$. What is the measure of $ \\angle ABD$?\n",
    "completion": "\\textbf{Solution:} Given that $\\because$ $\\triangle ABC\\cong \\triangle DCB$\\\\\n$\\therefore$ $\\angle DBC=\\angle ACB=40^\\circ$\\\\\nIn $\\triangle ABC$\\\\\n$\\angle ABC=180^\\circ-\\angle A-\\angle ACB=60^\\circ$\\\\\n$\\therefore$ $\\angle ABD=\\angle ABC-\\angle DBC=20^\\circ$\\\\\nHence, the answer is: $\\boxed{20^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53066418.png"
  },
  {
    "solution": "\\boxed{\\frac{27}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points $P$, $Q$, and $R$ are located on the graph of the inverse proportion function $y=\\frac{k}{x}$ (where the constant $k>0$ and $x>0$), respectively. Lines parallel to the $x$-axis and the $y$-axis through these three points are drawn, forming shaded regions in the figure with areas denoted as ${S}_{1}$, ${S}_{2}$, and ${S}_{3}$ from left to right. If $OE=ED=DC$ and ${S}_{1}{S}_{3}=27$, what is the value of ${S}_{2}$?\n",
    "completion": "\\textbf{Solution:} Let us assume $CD=DE=OE=a$,\\\\\nHence, $P\\left(\\frac{k}{3a}, 3a\\right)$, $Q\\left(\\frac{k}{2a}, 2a\\right)$, $R\\left(\\frac{k}{a}, a\\right)$,\\\\\nThus, $CP=\\frac{k}{3a}$, $DQ=\\frac{k}{2a}$, $ER=\\frac{k}{a}$,\\\\\nTherefore, $OG=AG$, $OF=2FG$, $OF=\\frac{2}{3}GA$,\\\\\nConsequently, $S_{1}=2S_{2}$, $S_{3}=3S_{2}$,\\\\\nGiven $S_{1}+S_{3}=27$,\\\\\nIt implies $2S_{2}+3S_{2}=27$,\\\\\nHence, $S_{2}=\\boxed{\\frac{27}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51615154.png"
  },
  {
    "solution": "\\boxed{20}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In rhombus $ABCD$, if $AC=6$ and $BD=8$, then what is the perimeter of rhombus $ABCD$?\n",
    "completion": "\\textbf{Solution:} Given that the quadrilateral $ABCD$ is a rhombus,\\\\\n$\\therefore$ $AO=\\frac{1}{2}AC=\\frac{1}{2}\\times 6=3$,\\\\\n$BO=\\frac{1}{2}BD=\\frac{1}{2}\\times 8=4$,\\\\\n$AC\\perp BD$,\\\\\n$\\therefore$ $\\triangle ABO$ is a right triangle,\\\\\nBy Pythagorean theorem, we have $AB=\\sqrt{3^2+4^2}=5$,\\\\\n$\\therefore$ The perimeter of the rhombus $ABCD$ is: $4\\times 5=\\boxed{20}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50497049.png"
  },
  {
    "solution": "\\boxed{50^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the right-angled vertex $P$ of the triangle ruler $ABP$ lies on the line $CD$, with points $A$, $B$ on the same side of line $CD$. As shown in fig. (1), if $\\angle APC=40^\\circ$, find the degree measure of $\\angle BPD$.\n",
    "completion": "\\textbf{Solution:} By the given conditions,\\\\\n$ \\because \\angle APB=90^\\circ$, $ \\angle APC=40^\\circ$\\\\\n$ \\therefore \\angle BPD=180^\\circ-\\angle APB-\\angle APC=180^\\circ-90^\\circ-40^\\circ=\\boxed{50^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/53413592_116.png"
  },
  {
    "solution": "\\boxed{(1,3)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y_{1}=x+2$ is the line $l_{1}$, and the graph of the linear function $y_{2}=kx+b$ is the line $l_{2}$. The two lines intersect at point $A(1,a)$. It is known that the points where lines $l_{1}$ and $l_{2}$ intersect the $x$-axis are points $B$ and $C$, respectively, and line $l_{2}$ intersects the $y$-axis at point $E(0,4)$. What are the coordinates of point $A$? What are the coordinates of point $B$?\n",
    "completion": "\\textbf{Solution:} The coordinates of point $A$ are $\\boxed{(1,3)}$, while the coordinate information for point $B$ is not provided.",
    "image_path": "dataset/mm_math/images/MM_Math/52395370_108.png"
  },
  {
    "solution": "\\boxed{\\frac{3+\\sqrt{5}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, five congruent isosceles triangles are pieced together to form two regular pentagons of different sizes, inside and outside. If the side length of the smaller regular pentagon is 1, what is the side length of the larger regular pentagon?\n",
    "completion": "\\textbf{Solution:} Extend CF to intersect BA at point H, \\\\\nsince assembling 5 congruent isosceles triangles forms two differently sized regular pentagon patterns, \\\\\ntherefore $\\angle$BAE=$\\angle$CBA=108$^\\circ$, \\\\\ntherefore $\\angle$ABG+$\\angle$BAG=108$^\\circ$, \\\\\n$\\angle$ABG+$\\angle$BAG+$\\angle$AGB=180$^\\circ$, \\\\\ntherefore $\\angle$AGB=$\\angle$BAG=180$^\\circ$−108$^\\circ$=72$^\\circ$, \\\\\ntherefore $\\angle$ABG=180$^\\circ$−72$^\\circ \\times 2$=$36^\\circ$, \\\\\nsince AB=BC, \\\\\ntherefore $\\angle$BHC=(180$^\\circ$−108$^\\circ$)÷2=$36^\\circ$, \\\\\ntherefore $\\angle$CBF=$\\angle$CFB=108$^\\circ$−36$^\\circ$=72$^\\circ$=$\\angle$AFG, \\\\\ntherefore $\\angle$FHG=180$^\\circ$−72$^\\circ \\times 2$=36$^\\circ$=$\\angle$ABG, \\\\\ntherefore points A and H coincide, \\\\\ntherefore $\\angle$FAG=72$^\\circ$−36$^\\circ$=36$^\\circ$=$\\angle$ABG, \\\\\ntherefore $\\triangle$BAG$\\sim$$\\triangle$FGA, \\\\\ntherefore $\\frac{BA}{AG}=\\frac{AG}{FG}$ \\\\\nLet BF=AF=AG=x, then BG=AB=x+1, \\\\\ntherefore $\\frac{1+x}{x}=\\frac{x}{1}$ \\\\\nSolving it: $x_{1}=\\frac{1+\\sqrt{5}}{2}$, $x_{2}=\\frac{1−\\sqrt{5}}{2}$ (discard since it does not meet the conditions), \\\\\ntherefore AB=1+$\\frac{1+\\sqrt{5}}{2}$=\\boxed{\\frac{3+\\sqrt{5}}{2}}.",
    "image_path": "dataset/mm_math/images/MM_Math/51369883.png"
  },
  {
    "solution": "\\boxed{(0,1.5)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=\\frac{4}{3}x+4$ intersects the x-axis and y-axis at points A and B, respectively. Point M is a point on OB, and if $\\triangle ABM$ is folded along AM, point B exactly falls on point $B_{1}$ on the x-axis. Find the coordinates of point M.\n",
    "completion": "\\textbf{Solution:} Given the coordinates of points $A$ and $B$, we know that $OA=3$, $BO=4$. Therefore, $AB=\\sqrt{OA^2+BO^2}=5=AB_1$,\\\\\nthus the coordinates of point $B_1$ are $(2,0)$,\\\\\nlet the coordinates of point $M$ be $(0,m)$,\\\\\naccording to the problem, $MB=MB_1$, which means $m^2+4=(m-4)^2$. Solving this, we find $m=1.5$,\\\\\ntherefore, the coordinates of point $M$ are $\\boxed{(0,1.5)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52978850_131.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the rhombus $ABCD$, as shown in the figure, $\\angle ADC = 120^\\circ$. Point $E$ is on the diagonal $AC$, and $AE = AB$. When connecting $DE$, if $CE = 4$, what is the length of $DE$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, draw $EF \\perp CD$ at F,\\\\\nthen $\\angle EFC = \\angle EFD = 90^\\circ$,\\\\\nsince quadrilateral ABCD is a rhombus,\\\\\ntherefore $AB = AD = CD$,\\\\\ntherefore $\\angle DCA = \\angle DAC$,\\\\\nsince $\\angle ADC = 120^\\circ$,\\\\\ntherefore $\\angle DCA = \\angle DAC = 30^\\circ$,\\\\\ntherefore $EF = \\frac{1}{2}CE = 2$,\\\\\nsince $AE = AB$,\\\\\ntherefore $AE = AE$,\\\\\ntherefore $\\angle ADE = \\angle AED = \\frac{1}{2}(180^\\circ - \\angle DAC) = 75^\\circ$,\\\\\ntherefore $\\angle EDF = \\angle ADC - \\angle ADE = 120^\\circ - 75^\\circ = 45^\\circ$,\\\\\ntherefore $\\triangle DEF$ is an isosceles right-angled triangle,\\\\\ntherefore $DE = \\sqrt{2}EF = 2\\sqrt{2}$,\\\\\nthus the answer is: $DE = \\boxed{2\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53706925.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{5}}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, AB is the diameter of semicircle O, and point C is a point on the semicircle. CD$\\perp$AB at point D. If AB=10 and CD=4, what is the value of $\\sin\\angle BCD$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect OC, \\\\\nsince AB is the diameter of the semicircle O with AB=10, \\\\\nthus OC=OB=5, \\\\\nsince CD$\\perp$AB at point D, and CD=4, \\\\\nthus OD=$\\sqrt{OC^{2}-CD^{2}}=\\sqrt{5^{2}-4^{2}}=3$, \\\\\nthus BD=OB−OD=5−3=2, \\\\\nthus BC=$\\sqrt{CD^{2}+BD^{2}}=\\sqrt{4^{2}+2^{2}}=2\\sqrt{5}$, \\\\\nthus $\\sin\\angle BCD=\\frac{BD}{BC}=\\frac{2}{2\\sqrt{5}}=\\boxed{\\frac{\\sqrt{5}}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51261932.png"
  },
  {
    "solution": "\\boxed{50^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AD$ is the altitude of $\\triangle ABC$, and $CE$ is the angle bisector of $\\triangle ADC$. If $\\angle BAD = \\angle ECD$ and $\\angle B = 70^\\circ$, what is the degree measure of $\\angle CAD$?\n",
    "completion": "\\textbf{Solution:}\n\nSince $AD$ is the altitude of $\\triangle ABC$,\n\\[\\therefore \\angle ADB=\\angle ADC=90^\\circ\\]\n\nSince $\\angle B=70^\\circ$,\n\\[\\therefore \\angle BAD=20^\\circ\\]\n\nSince CE is the bisector of $\\triangle ADC$,\n\\[\\therefore \\angle ECD=\\frac{1}{2}\\angle ACD\\]\n\nSince $\\angle BAD=\\angle ECD=20^\\circ$,\n\\[\\therefore \\angle ACD=40^\\circ\\]\n\nTherefore, in $\\triangle ACD$, $\\angle CAD=90^\\circ-40^\\circ=\\boxed{50^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51378752.png"
  },
  {
    "solution": "\\boxed{y=3x-4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the Cartesian coordinate system, the line $y=\\frac{1}{2}x+m$ intersects the $x$-axis and the $y$-axis at points $A$ and $B$, respectively. The coordinates of point $A$ are $\\left(-2, 0\\right)$, and the coordinates of point $C$ are $\\left(3, 0\\right)$. If points $B$ and $C$ are symmetrical about line $l$, draw line $l$ in the auxiliary figure and then find the analytical expression of line $l$.\n",
    "completion": "\\textbf{Solution:} As shown in Figure 1, let the intersection of line $l$ with the x-axis be $E$, and it intersects the y-axis at $F$. Let $OE=x$, and connect $BE$, then $BE=CE=3-x$. In $\\triangle BOE$, we have $BE^2=OB^2+OE^2$, $\\therefore (3-x)^2=1+x^2$. Solving this, we find $x=\\frac{4}{3}$, $\\therefore E\\left(\\frac{4}{3}, 0\\right)$. Let $H$ be the intersection of $BC$ and $l$, then $H\\left(\\frac{3}{2}, \\frac{1}{2}\\right)$. Let the equation of line $l$ be $y=kx+b$, then $\\begin{cases}\\frac{3}{2}k+b=\\frac{1}{2}\\\\ \\frac{4}{3}k+b=0\\end{cases}$, solving this gives $\\begin{cases}k=3\\\\ b=-4\\end{cases}$, $\\therefore$ the equation of line $l$ is \\boxed{y=3x-4}.",
    "image_path": "dataset/mm_math/images/MM_Math/50553714_108.png"
  },
  {
    "solution": "\\boxed{\\frac{10}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ is inscribed in circle $O$, with $AB$ being the diameter of circle $O$. $\\triangle ABC$ is rotated about point $C$ to $\\triangle EDC$, where point $E$ is on the circle. Given that $AE=2$ and $\\tan D=3$, what is the length of $AB$?\n",
    "completion": "\\textbf{Solution:} Since AB is the diameter,\\\\\nit follows that $\\angle ACB=90^\\circ$\\\\\nBecause rotating $\\triangle ABC$ around point C to $\\triangle EDC$,\\\\\nit follows that $\\angle ABC=\\angle D$, $\\angle ACB=\\angle ECD=90^\\circ$, ED=AB, BC=CD, AC=CE,\\\\\ntherefore, $\\angle ACE=\\angle BCD=\\angle ABE$,\\\\\nthus, $\\angle ABE+\\angle ABC+\\angle CBD=\\angle BCD+\\angle D+\\angle CBD=180^\\circ$,\\\\\ntherefore, points E, B, D lie on the same line,\\\\\nin $\\triangle ECD$, $\\tan D= \\frac{EC}{CD}=3$,\\\\\nLet CD=x, then EC=3x,\\\\\ntherefore, $ED=\\sqrt{CD^2+EC^2}=\\sqrt{x^2+9x^2}=\\sqrt{10}x$;\\\\\nBecause $\\overset{\\frown}{AC}=\\overset{\\frown}{AC}$\\\\\nit follows that $\\angle ACE=\\angle BCD$, $\\angle D=\\angle ABC=\\angle AEC$,\\\\\ntherefore, $\\triangle ACE\\sim \\triangle BCD$,\\\\\ntherefore, $\\frac{CE}{CD}=\\frac{AE}{BD}=3$,\\\\\n$\\angle CBD=\\angle CAE$,\\\\\nthus, $\\frac{2}{BD}=3$ Solving this yields: $BD=\\frac{2}{3}$;\\\\\ntherefore, $BE=DE−BD=\\sqrt{10}x-\\frac{2}{3}$\\\\\nIn $\\triangle AEB$, $AE^2+BE^2=AB^2$\\\\\nthus, solving $2^2+\\left(\\sqrt{10}x-\\frac{2}{3}\\right)^2=\\left(\\sqrt{10}x\\right)^2$ yields: $x=\\frac{\\sqrt{10}}{3}$\\\\\ntherefore, $AB=DE=\\frac{\\sqrt{10}}{3}\\times \\sqrt{10}=\\boxed{\\frac{10}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53059869.png"
  },
  {
    "solution": "\\boxed{x<1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, the graph of the linear function $y=kx+b$ intersects the $x$-axis at point $A$, where $OA=4$, and intersects the graph of the direct proportion function $y=3x$ at point $B$, whose abscissa is 1. Please directly specify the range of $x$ for which $kx+b>3x$.\n",
    "completion": "\\textbf{Solution:} From the result of (1), we know the inequality is: $-x+4>3x$. Given $B(1, 3)$, when the graph of the linear function $y=-x+4$ is above the graph of the direct proportionality function $y=3x$, the range of the variable $x$ is: $\\boxed{x<1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52381215_154.png"
  },
  {
    "solution": "\\boxed{105^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $\\angle BAC=130^\\circ$, $AB=AC$, and the perpendicular bisector of $AC$ intersects $BC$ at point $D$. What is the degree measure of $\\angle BAD$?\n",
    "completion": "\\textbf{Solution:} Given that DE is the perpendicular bisector of AC,\\\\\nit follows that AD=CD,\\\\\nthus, $\\angle C=\\angle CAD$,\\\\\nsince AB=AC,\\\\\nit also follows that $\\angle B=\\angle C$,\\\\\nin $\\triangle ABC$, $\\angle ABC+\\angle B+\\angle C=180^\\circ$,\\\\\ntherefore, $130^\\circ+2\\angle C=180^\\circ$,\\\\\nsolving this, we find $\\angle C=25^\\circ$,\\\\\nthus, $\\angle ADB=\\angle CAD+\\angle C=25^\\circ+25^\\circ=50^\\circ$.\\\\\nTherefore, $\\angle BAD=180^\\circ-\\angle B-\\angle ADB=\\boxed{105^\\circ}$,",
    "image_path": "dataset/mm_math/images/MM_Math/52096691.png"
  },
  {
    "solution": "\\boxed{(1, 2)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system \\(xOy\\), the line \\( l_{1}: y_{1}=2x \\) intersects with the line \\( l_{2}: y_{2}=-x+3 \\) at point A. What are the coordinates of point A?\n",
    "completion": "\\textbf{Solution:} From the given, we have\n\\[\n\\begin{cases}\ny=2x \\\\\ny=-x+3\n\\end{cases}\n\\]\nSolving these equations, we get:\n\\[\n\\begin{cases}\nx=1 \\\\\ny=2\n\\end{cases}\n\\]\n$\\therefore$ The coordinates of point A are $\\boxed{(1, 2)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52211015_116.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the diameter $AB$ of circle $\\odot O$ is perpendicular to chord $CD$ at point $E$, and $BD$ is connected. If $CD=8$ and $OE=3$, what is the length of $BD$?\n",
    "completion": "\\textbf{Solution:} Draw line $OD$, \\\\\nsince $AB \\perp CD$, $AB$ passes through the center $O$, $CD=8$, \\\\\ntherefore $CE=DE=4$, $\\angle OED=\\angle DEB=90^\\circ$, \\\\\nsince $OE=3$, \\\\\ntherefore $OD=\\sqrt{OE^2+DE^2}=\\sqrt{3^2+4^2}=5$, \\\\\ntherefore $OB=OD=5$, \\\\\ntherefore $BE=OB-OE=5-3=2$, \\\\\nby the Pythagorean theorem, we get \\\\\ntherefore $BD=\\sqrt{BE^2+DE^2}=\\sqrt{2^2+4^2}=\\boxed{2\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52949314.png"
  },
  {
    "solution": "\\boxed{(2, 3)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, in the Cartesian coordinate system, let point $A(0,6)$ and let point $B$ be a moving point on the positive half of the $x$-axis. Connect $AB$ and take the midpoint $M$ of $AB$. Rotate the line segment $MB$ around point $B$ in a clockwise direction by $90^\\circ$ to obtain the line segment $BC$. Construct a perpendicular line to the $x$-axis through point $B$, intersecting line $AC$ at point $D$. Suppose the coordinates of point $B$ are $(2t,0)$. When $t=2$, what are the coordinates of point $M$?\n",
    "completion": "\\textbf{Solution:} When $t=2$, the coordinates of point $M$ are $\\boxed{(2, 3)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53490616_90.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, fold the rectangular paper ABCD so that edge DC falls on the diagonal line AC, with the fold line being CE, and point D falling on point F on the diagonal line. If AB=6 and AD=8, what is the length of ED?\n",
    "completion": "Solution: In rectangle $ABCD$, $AB=6$, $AD=8$, \\\\\n$\\therefore$ $DC=6$, \\\\\n$\\therefore$ $AC=\\sqrt{AD^{2}+DC^{2}}=10$, \\\\\nFrom the folding, it follows that $\\triangle DEC \\cong \\triangle FEC$, \\\\\n$\\therefore$ $FC=DC=6$, $DE=FE$, \\\\\nLet $ED=x$, then $FE=x$, $AF=AC−CF=4$, $AE=8−x$, \\\\\nIn $\\triangle AEF$: $AF^{2}+EF^{2}=AE^{2}$, \\\\\n$4^{2}+x^{2}=(8−x)^{2}$, \\\\\nSolving this yields: $x=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52225520.png"
  },
  {
    "solution": "\\boxed{2.5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, point D is on $BC$, and $DE\\perp AB$ at point E, $AE=AC$. If $BC=4$ and $DE=1.5$, what is the value of $BD$?\n",
    "completion": "\\textbf{Solution:} Since $DE\\perp AB$,\\\\\nit follows that $\\angle AED=\\angle C=90^\\circ$,\\\\\nsince $AE=AC$ and $AD=AD$,\\\\\nit follows that $\\triangle ACD\\cong \\triangle AED\\left(HL\\right)$,\\\\\nhence $DE=CD=1.5$,\\\\\nthus $BD=BC-CD=4-1.5=\\boxed{2.5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52867066.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in parallelogram $ABCD$, $DE\\perp AB$ at point $E$, and $DF\\perp BC$ at point $F$. The perimeter of parallelogram $ABCD$ is $28$, and its area is $40$. The ratio of $AB:AD$ is $4:3$. Find: What is the length of $DE$?\n",
    "completion": "\\textbf{Solution:} In parallelogram $ABCD$, where $AB=CD$ and $AD=BC$, \\\\\nsince the perimeter of the parallelogram $ABCD$ is 28, \\\\\nit follows that $AB+AD=14$, \\\\\nand given $AB\\colon AD=4\\colon 3$, \\\\\nthus, $AB=8$ and $AD=6$, \\\\\nfurthermore, the area of parallelogram $ABCD$ is given by $AB \\times DE = 40$, \\\\\ntherefore, $DE=\\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52486708_91.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, there is an advertising board CD at the top of a building. Xiaoming and his classmates measured the elevation angle of the bottom of the advertising board D to be $53^\\circ$ at the foot of a hill at point A. Walking up the slope AB to point B, they measured the elevation angle of the top of the advertising board C to be $45^\\circ$. Given that the gradient of hill AB is $i=1\\colon \\sqrt{3}$, $AB=10$ m, and $AE=21$ m. (Ignoring the height of the protractor, and the results are accurate to 0.1 meter. Reference data: $\\sqrt{2}\\approx1.41$, $\\sqrt{3}\\approx1.73$, $\\sin53^\\circ\\approx\\frac{4}{5}$, $\\cos53^\\circ\\approx\\frac{3}{5}$, $\\tan53^\\circ\\approx\\frac{4}{3}$). Calculate the height of point B above the horizontal ground AE in meters.\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, draw BM$\\perp$AE and BN$\\perp$CE through point B, where M and N are the feet of the perpendiculars, respectively. \\\\\n$\\because \\angle CBN=45^\\circ$, $\\angle DAE=53^\\circ$, $i=1\\colon \\sqrt{3}=\\frac{BM}{AM}=\\tan\\angle BAM$, \\\\\n$\\therefore \\angle BAM=30^\\circ$, \\\\\n$\\therefore BM=\\frac{1}{2}AB=5\\text{ (meters)}$, \\\\\nthus, the height of point B from the horizontal ground AE is \\boxed{5}\\text{ meters};",
    "image_path": "dataset/mm_math/images/MM_Math/52601742_132.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\angle AOE = \\angle BOE = 15^\\circ$, $EF \\parallel OB$, $EC \\perp OB$. If $EC = 2$, what is the length of $EF$?\n",
    "completion": "\\textbf{Solution:} Construct $EG \\perp OA$ at $G$, as shown in the figure:\\\\\nSince $EF \\parallel OB$, $\\angle AOE = \\angle BOE = 15^\\circ$, and $EC \\perp OB$,\\\\\nit follows that $\\angle OEF = \\angle COE = 15^\\circ$ and $EG = CE = 2$,\\\\\nSince $\\angle AOE = 15^\\circ$,\\\\\nit follows that $\\angle EFG = 15^\\circ + 15^\\circ = 30^\\circ$,\\\\\nTherefore, $EF = 2EG = \\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51379084.png"
  },
  {
    "solution": "\\boxed{7}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in an equilateral triangle $ABC$ with side length $9$, where $BD=3$, and $\\angle ADE=60^\\circ$, what is the length of $AE$?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABC$ is an equilateral triangle, we have $\\angle B=\\angle C=60^\\circ$, and $AB=BC$.\\\\\nTherefore, $CD=BC-BD=9-3=6$, and $\\angle BAD+\\angle ADB=120^\\circ$.\\\\\nSince $\\angle ADE=60^\\circ$, $\\angle ADB+\\angle EDC=120^\\circ$. Thus, $\\angle DAB=\\angle EDC$.\\\\\nMoreover, since $\\angle B=\\angle C=60^\\circ$, we conclude that $\\triangle ABD\\sim \\triangle DCE$.\\\\\nThus, $\\frac{AB}{BD}=\\frac{DC}{CE}$, which gives $\\frac{9}{3}=\\frac{6}{CE}\\Rightarrow CE=2$.\\\\\nTherefore, $AE=AC-CE=9-2=\\boxed{7}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52623552.png"
  },
  {
    "solution": "\\boxed{\\frac{2}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, D is a point on the extension of side BC of $\\triangle ABC$, and $CD=BC$, with line DE intersecting AB and AC at points E and F, respectively. If $EA = EB$, then $\\frac{AF}{AC}=?$.\n",
    "completion": "\\textbf{Solution:} Construct line $EG \\parallel BD$ through point $E$, intersecting $AC$ at point $G$,\\\\\nsince $AE=BE$ and $BC=CD$,\\\\\ntherefore $AB=2AE$,\\\\\ntherefore $\\triangle AEG \\sim \\triangle ABC$,\\\\\ntherefore $\\frac{AE}{AB}=\\frac{AG}{AC}=\\frac{EG}{BC}=\\frac{EG}{CD}=\\frac{1}{2}$,\\\\\nsince $EG\\parallel BD$,\\\\\ntherefore $\\triangle EFG \\sim \\triangle DFC$,\\\\\ntherefore $\\frac{EG}{CD}=\\frac{GF}{CF}=\\frac{1}{2}$,\\\\\ntherefore $CF=2GF$,\\\\\ntherefore $AG=CG=3GF$,\\\\\ntherefore $AF=AG+CF=4GF$,\\\\\ntherefore $AC=2CG=6GF$,\\\\\ntherefore $\\frac{AF}{AC}=\\frac{4GF}{6GF}=\\boxed{\\frac{2}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52936222.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $D$ is a point inside $\\angle MAN$, $DE\\perp AM$ at $E$, $DF\\perp AN$ at $F$, and $DE=DF$. Point $B$ is on ray $AM$ with $AB=6$, $BE=2$. A point $C$ is taken on ray $AN$ such that $DC=DB$. What is the length of $AC$?\n",
    "completion": "Solution:\n(1) As shown in Figure 1, when point C is on line segment $AF$, connect $AD$, \\\\\n$\\because$ $DE\\perp AM$ at E, $DF\\perp AN$ at F, \\\\\n$\\therefore$ $\\angle DEB=\\angle DFC=90^\\circ$, \\\\\nIn $Rt\\triangle DEB$ and $Rt\\triangle DFC$, $\\left\\{\\begin{array}{l}DC=DB\\\\ DF=DE\\end{array}\\right.$, \\\\\n$\\therefore$ $Rt\\triangle DEB\\cong Rt\\triangle DFC\\left(HL\\right)$, \\\\\n$\\therefore$ $CF=BE=2$, \\\\\nAlso, $\\because$ in $Rt\\triangle DEA$ and $Rt\\triangle DFA$, $\\left\\{\\begin{array}{l}DA=DA\\\\ DF=DE\\end{array}\\right.$, \\\\\n$\\therefore$ $Rt\\triangle DEA\\cong Rt\\triangle DFA\\left(HL\\right)$, \\\\\n$\\therefore$ $AF=AE=AB+BE=6+2=8$, \\\\\n$\\therefore$ $AC=AF−CF=8−2=\\boxed{6}$; \\\\\n(2) As shown in Figure 2, when point C is on the extension of line segment $AF$, \\\\\nSimilarly, it is obtained that $AF=AE=8$, $CF=BE=2$, \\\\\n$\\therefore$ $AC=AF+CF=8+2=10$.",
    "image_path": "dataset/mm_math/images/MM_Math/52793335.png"
  },
  {
    "solution": "\\boxed{75^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the incircle $\\odot O$ of $\\triangle ABC$ is tangent to side $BC$ at point $D$, $\\angle A=70^\\circ$, $\\angle C=80^\\circ$. Connect $OB$ and $OD$. What is the degree measure of $\\angle BOD$?\n",
    "completion": "\\textbf{Solution:} Given $\\because$ $\\angle A=70^\\circ$ and $\\angle C=80^\\circ$, \\\\\n$\\therefore$ $\\angle ABC=180^\\circ-70^\\circ-80^\\circ=30^\\circ$, \\\\\n$\\because$ the incircle $\\odot$O of $\\triangle ABC$ touches side $BC$ at point D, \\\\\n$\\therefore$ $OB$ bisects $\\angle ABC$, and $OD\\perp BC$, \\\\\n$\\therefore$ $\\angle OBD=\\frac{1}{2}\\angle ABC=\\frac{1}{2}\\times 30^\\circ=15^\\circ$, \\\\\n$\\therefore$ $\\angle BOD=90^\\circ-\\angle OBD=\\boxed{75^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52977736.png"
  },
  {
    "solution": "\\boxed{\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the right triangle $ABC$, $\\angle ACB=90^\\circ$, $\\angle A>\\angle B$. $CD$ is the altitude on the hypotenuse, and $CE$ is the angle bisector of $\\triangle ABC$. $FG$ is the perpendicular bisector of side $AB$, and $FG$ intersects sides $BC$ and $AB$ at points $F$ and $G$, respectively. If $\\angle DCE=\\angle B$, what is the value of $\\frac{BF}{FC}$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect $AF$,\n\n$\\because$ $CD$ is the altitude on the hypotenuse of right $\\triangle ABC$,\n\n$\\therefore$ $\\angle ACD + \\angle CAB = 90^\\circ$.\n\n$\\because$ $\\angle ACB = 90^\\circ$,\n\n$\\therefore$ $\\angle CAB + \\angle B = 90^\\circ$,\n\n$\\therefore$ $\\angle ACD = \\angle B$,\n\n$\\therefore$ $\\angle ACD = \\angle DCE = \\angle B$.\n\n$\\because$ $CE$ is the bisector of $\\angle ACB$,\n\n$\\therefore$ $\\angle ACE = \\frac{1}{2}\\angle ACB = 45^\\circ$,\n\n$\\therefore$ $\\angle ACD + \\angle DCE = 45^\\circ$,\n\n$\\therefore$ $\\angle B = \\angle ACD = \\angle DCE = \\frac{1}{2} \\times 45^\\circ = 22.5^\\circ$.\n\n$\\because$ $FG$ is the perpendicular bisector of side $AB$,\n\n$\\therefore$ $AF = BF$,\n\n$\\therefore$ $\\angle FAB = \\angle B = 22.5^\\circ$,\n\n$\\therefore$ $\\angle AFC = \\angle B + \\angle FAB = 45^\\circ$,\n\n$\\therefore$ $\\triangle ACF$ is an isosceles right triangle,\n\n$\\therefore$ $\\frac{AF}{FC} = \\sqrt{2}$, that is, $\\frac{BF}{FC} = \\boxed{\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52837902.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that in $\\triangle ABC$, $\\angle C=60^\\circ$, BD bisects $\\angle ABC$, and AD bisects the external angle $\\angle CAE$. What is the degree measure of $\\angle D$?\n",
    "completion": "\\textbf{Solution:} Since BD bisects $\\angle ABC$ and AD bisects the exterior angle $\\angle CAE$,\\\\\nit follows that $\\angle DBA= \\frac{1}{2}\\angle ABC$ and $\\angle DAE= \\frac{1}{2}\\angle CAE$,\\\\\nGiven that $\\angle CAE=\\angle ABC+\\angle C$,\\\\\nand $\\angle DAE=\\angle ABD+\\angle D$,\\\\\nit can be derived that $\\angle D= \\frac{1}{2}\\angle CAE- \\frac{1}{2} \\angle ABD= \\frac{1}{2} (\\angle CAE-\\angle ABD)= \\frac{1}{2} \\angle C=\\frac{1}{2}\\times 60^\\circ=\\boxed{30^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51434865.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{2}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within a grid composed of small squares with a side length of 1, points A, B, and C are located on the grid points, and point D is on the circumcircle of $ \\triangle ABC$. What is the value of $\\sin\\angle ADC$?\n",
    "completion": "\\textbf{Solution:} Given $\\because \\overset{\\frown}{AC}=\\overset{\\frown}{AC}$,\\\\\n$\\therefore \\angle ABC=\\angle ADC$,\\\\\n$\\because AB^{2}=3^{2}+1^{2}=10, AC^{2}=3^{2}+1^{2}=10, BC^{2}=2^{2}+4^{2}=20$,\\\\\n$\\therefore AB^{2}+AC^{2}=BC^{2}$, which means $\\triangle ABC$ is an isosceles right triangle,\\\\\n$\\therefore \\angle ABC=\\angle ACB=\\angle ADC=45^\\circ$,\\\\\n$\\therefore \\sin\\angle ADC=\\sin45^\\circ=\\boxed{\\frac{\\sqrt{2}}{2}}$",
    "image_path": "dataset/mm_math/images/MM_Math/53043249.png"
  },
  {
    "solution": "\\boxed{10.1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, it is known that the distance from the rooftop to the ground $GE$ is $18.5$ meters. When Xiaoliang stands at point $B$ in front of the building, measuring from point $A$, which is directly above point $B$, the angle of elevation to the top of the banner $G$ is $37^\\circ$. Then, moving $15$ meters towards the direction of the teaching building to point $D$ (points $E$, $B$, and $D$ are on a straight line and at the base of the building), and from point $C$, which is directly above point $D$, the angle of elevation to the bottom end $F$ of the banner is $42^\\circ$. If $AB$ and $CD$ are both $1.7$ meters (meaning quadrilateral $ABCD$ is a rectangle), please help Xiaoliang calculate the length of the banner $GF$. (The result should be accurate to $0.1m$, reference data: $\\sin37^\\circ\\approx 0.60$, $\\cos37^\\circ\\approx 0.80$, $\\tan37^\\circ\\approx 0.75$, $\\sin42^\\circ\\approx 0.67$, $\\cos42^\\circ\\approx 0.74$, $\\tan42^\\circ\\approx 0.90$)\n",
    "completion": "\\textbf{Solution:} Given from (1) that \\(CH = 7.4\\) meters, in the right triangle \\(\\triangle FCH\\), since \\(\\angle FCH = 42^\\circ\\),\\\\\ntherefore, \\(\\tan 42^\\circ = \\frac{FH}{CH} = \\frac{FH}{7.4} \\approx 0.90\\),\\\\\nhence, \\(FH = 6.66\\),\\\\\nthus, \\(FG = GH - FH = 16.8 - 6.66 \\approx \\boxed{10.1}\\) (meters).",
    "image_path": "dataset/mm_math/images/MM_Math/53080171_106.png"
  },
  {
    "solution": "\\boxed{16^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure (1), given point $O$ is the intersection point of lines $AB$ and $CD$, $\\angle BOD=32^\\circ$, $OE$ bisects $\\angle AOD$, $\\angle EOF=90^\\circ$, find the degree measure of $\\angle COF$.\n",
    "completion": "Solution: $\\because$ $\\angle BOD=32^\\circ$,\\\\\n$\\therefore$ $\\angle AOD=180^\\circ-\\angle BOD=180^\\circ-32^\\circ=148^\\circ$.\\\\\n$\\because$ $OE$ bisects $\\angle AOD$,\\\\\n$\\therefore$ $\\angle DOE=\\frac{1}{2}\\angle AOD=74^\\circ$.\\\\\n$\\because$ $\\angle EOF=90^\\circ$,\\\\\n$\\therefore$ $\\angle COF=180^\\circ-\\angle DOE-\\angle EOF=180^\\circ-74^\\circ-90^\\circ=\\boxed{16^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/52732940_84.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the area of rectangle $OABC$ is $18$, and its diagonal $OB$ intersects with the hyperbola $y= \\frac{k}{x}$ at point $D$, where $OD: DB=2:1$. What is the value of $k$?\n",
    "completion": "\\textbf{Solution:} Given the problem statement, let the coordinates of point D be $(x_D, y_D)$. Therefore, the coordinates of point B are $\\left(\\frac{3}{2}x_D, \\frac{3}{2}y_D\\right)$,\\\\\n$\\therefore$ the area of rectangle OABC $=\\left|\\frac{3}{2}x_D \\times \\frac{3}{2}y_D\\right|=18$,\\\\\n$\\because$ the figure lies in the first quadrant,\\\\\n$\\therefore k=x_D \\cdot y_D=\\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51368180.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{15}}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given: As shown in the figure, in the circle $\\odot O$ with a radius of $4$, $AB$ and $CD$ are two diameters, $M$ is the midpoint of $OB$, and the extension line of $CM$ intersects $\\odot O$ at point $E$, with $EM > MC$. Connect $DE$, and $DE = \\sqrt{15}$. Find the value of $\\sin\\angle EOB$.\n",
    "completion": "\\textbf{Solution:} Draw $EF \\perp AB$ through point $E$, with the foot of the perpendicular at point $F$.\n\nSince $OE=4$ and $EM=4$,\n\nit follows that $OE=EM$,\n\nthus $OF=FM=1$,\n\nso $EF=\\sqrt{4^2-1^2}=\\sqrt{15}$,\n\ntherefore, $\\sin\\angle EOB=\\boxed{\\frac{\\sqrt{15}}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53066251_86.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within rectangle $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$, $AC=10$, and $\\angle ACB = 30^\\circ$. What is the length of $CD$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rectangle,\\\\\nit follows that $\\angle ABC=90^\\circ$.\\\\\nSince AC=10 and $\\angle ACB=30^\\circ$,\\\\\nit follows that AB=5.\\\\\nSince quadrilateral ABCD is a rectangle,\\\\\nit follows that CD=AB=5.\\\\\nTherefore, the length of CD is $\\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50475076.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within the rhombus $ABCD$, the diagonals $AC$ and $BD$ intersect at point $O$, and their lengths are $6\\text{cm}$ and $8\\text{cm}$, respectively. A line passing through point $O$ intersects $AD$ and $BC$ at points $E$ and $F$, respectively. What is the sum of the areas of the shaded regions in $\\text{cm}^2$?\n",
    "completion": "\\textbf{Solution:} Since $AC$ and $BD$ are the diagonals of the rhombus $ABCD$,\\\\\nit follows that $AC \\perp BD$, and $OA=OC$, $AD\\parallel BC$,\\\\\ntherefore, $\\angle OAE=\\angle OCF$, $\\angle AEO=\\angle CFO$,\\\\\nthus, $\\triangle AOE\\cong \\triangle COF$,\\\\\nhence, the areas are $S_{\\triangle AOE}=S_{\\triangle COF}$,\\\\\ntherefore, $S_{\\text{shaded}}=S_{\\triangle BCD}=\\frac{1}{2}S_{\\text{rhombus} ABCD}$,\\\\\nGiven the lengths of the diagonals of the rhombus are 6cm and 8cm,\\\\\nwhich means: $S_{\\text{shaded}}=S_{\\triangle BCD}=\\frac{1}{2}S_{\\text{rhombus} ABCD}=\\frac{1}{2}\\times (\\frac{1}{2}\\times 6\\times 8)=\\boxed{12}\\text{cm}^2$",
    "image_path": "dataset/mm_math/images/MM_Math/52344189.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y=kx+b$ intersects with the graph of the inverse proportion function $y=\\frac{4}{x}$ ($x>0$) at two points $A(m, 4)$ and $B(2, n)$. It also intersects with the coordinate axes at points $M$ and $N$ respectively. What is the area of $\\triangle AOB$?\n",
    "completion": "\\textbf{Solution:} For $y=-2x+6$, let $y=0$, then $-2x+6=0$,\\\\\nsolving this yields $x=3$,\\\\\n$\\therefore$ the coordinates of point N are $(3,0)$,\\\\\n$\\therefore$ ${S}_{\\triangle AOB}={S}_{\\triangle AON}-{S}_{\\triangle BON}=\\frac{1}{2}ON\\cdot y_A-\\frac{1}{2}ON\\cdot y_B=\\frac{1}{2}\\times 3\\times (4-2)=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52601975_178.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ABC=90^\\circ$, circle $\\odot O$ with diameter $AB$ intersects $AC$ at point $D$, $AE$ is perpendicular to the tangent at point $D$, with the foot of the perpendicular being $E$. $AD$ bisects $\\angle BAE$; If $CD=\\frac{9}{5}$ and $BC=3$, find the length of segment $AB$.\n",
    "completion": "\\textbf{Solution:} Connect BD, as shown in the figure,\n\n$\\because$ AB is a diameter,\n\n$\\therefore$ $\\angle$ADB$=90^\\circ$,\n\n$\\because$ $\\angle 2+\\angle$ABD$=90^\\circ$, $\\angle 3+\\angle$ABD$=90^\\circ$,\n\n$\\therefore$ $\\angle 2=\\angle 3$,\n\n$\\because$ $\\angle$DCB$=\\angle$BCA,\n\n$\\therefore$ $\\triangle CDB\\sim \\triangle CBA$,\n\n$\\therefore$ $\\frac{CD}{CB}=\\frac{CB}{AC}$,\n\n$\\therefore$ $\\frac{\\frac{9}{5}}{3}=\\frac{3}{AC}$,\n\n$\\therefore$ AC$=5$,\n\n$\\therefore$ AB$=\\sqrt{AC^2-BC^2}=\\sqrt{5^2-3^2}=\\boxed{4}$",
    "image_path": "dataset/mm_math/images/MM_Math/51403136_120.png"
  },
  {
    "solution": "\\boxed{\\frac{25}{8}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the diameter $AB$ of the circle $\\odot O$ and chord $CD$ are perpendicular to each other at point $E$. Also, $AD=5$ and $\\cos\\angle BCD=\\frac{4}{5}$. Find the radius of $\\odot O$.\n",
    "completion": "\\textbf{Solution:} Let us connect $OD$, with the radius being $r$, thus $OE = 4 - r$, $OD = r$.\\\\\nIn $\\triangle BCE$, we have $OE^{2} + DE^{2} = OD^{2}$,\\\\\n$\\therefore$ $(4-r)^{2} + 9 = r^{2}$,\\\\\n$\\therefore$ $r = \\boxed{\\frac{25}{8}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51322362_126.png"
  },
  {
    "solution": "\\boxed{\\frac{3\\sqrt{10}}{7}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: The large square ABCD is composed of four congruent right-angled triangles and one small square, as shown in the diagram. Draw a perpendicular line from point D to F, intersecting the extension of the diagonal EF of the small square at point G. Connect CG and extend BE to intersect CG at point H. If $AE=2BE$, what is the value of $\\frac{CG}{BH}$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw a line GM$\\perp$DG at point G, intersecting the extended line of CF at point M,\\\\\n$\\because$DF$\\perp$DG,\\\\\n$\\therefore$$\\angle$GDF=$\\angle$DFM=$\\angle$MGD=$90^\\circ$,\\\\\n$\\therefore$Quadrilateral DFMG is a rectangle,\\\\\n$\\therefore$DF=MG;\\\\\n$\\because$The large square ABCD is composed of four congruent right-angled triangles and a small square,\\\\\n$\\therefore$let BE=AT=CN=x,\\\\\n$\\because$AE=2BE,\\\\\n$\\therefore$BH=CF=2x, TF=DF=x,\\\\\n$\\because$Square ETFH,\\\\\n$\\therefore$$\\angle$TFE=$\\angle$DFG=$45^\\circ$,\\\\\nIn $\\triangle$DFG, $\\angle$GDF=$90^\\circ$,\\\\\n$\\therefore$$\\triangle$DFG is an isosceles right-angled triangle,\\\\\n$\\therefore$DF=DG,\\\\\n$\\therefore$Quadrilateral DFMG is a square,\\\\\n$\\therefore$MF=DF=x,\\\\\n$\\therefore$CM=2x+x=3x\\\\\nIn Rt$\\triangle$CMG,\\\\\n$CG=\\sqrt{M{G}^{2}+C{M}^{2}}=\\sqrt{{x}^{2}+{\\left(3x\\right)}^{2}}=\\sqrt{10}x$;\\\\\n$\\because$NH$\\parallel$MG,\\\\\n$\\therefore$$\\triangle$CHN$\\sim$$\\triangle$CMG\\\\\n$\\therefore$$\\frac{NH}{MG}=\\frac{CN}{CM}$, thus $\\frac{NH}{x}=\\frac{x}{3x}$\\\\\n$\\therefore$$NH=\\frac{1}{3}x$,\\\\\n$\\therefore$$BH=BN+NH=2x+\\frac{1}{3}x=\\frac{7}{3}x$\\\\\n$\\therefore$$\\frac{CG}{BH}=\\frac{\\sqrt{10}x}{\\frac{7}{3}x}=\\boxed{\\frac{3\\sqrt{10}}{7}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52694089.png"
  },
  {
    "solution": "\\boxed{y=x+1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the direct proportion function $y=2x$ intersects with the graph of the linear function $y=kx+b$ at point $A(m, 2)$. The graph of the linear function passes through point $B(-2, -1)$ and intersects with the $x$-axis at point $C$. What is the analytical expression of the linear function?\n",
    "completion": "\\textbf{Solution:} Since the graph of the direct proportion function $y=2x$ intersects with the graph of the linear function $y=kx+b$ at point $A(m, 2)$,\\\\\ntherefore $2m=2$, that is, $m=1$.\\\\\nSubstituting $(1, 2)$ and $(-2, -1)$ into $y=kx+b$, we get\\\\\n$\\begin{cases}\nk+b=2\\\\\n-2k+b=-1\n\\end{cases}$, which solves to $\\begin{cases}\nk=1\\\\\nb=1\n\\end{cases}$,\\\\\ntherefore the equation of the linear function is $\\boxed{y=x+1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53183208_207.png"
  },
  {
    "solution": "\\boxed{4+2\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given: As shown in the figure, in rectangle $ABCD$, $AB=3$, $AD=4$. Point $P$ is a point inside rectangle $ABCD$, and satisfies ${S}_{\\triangle PBC}=\\frac{1}{3}{S}_{\\text{rectangle} ABCD}$, what is the minimum perimeter of $\\triangle ADP$?\n",
    "completion": "\\textbf{Solution:} Draw $MN\\perp AD$ through point $P$, intersecting $AD$ at point $M$ and $BC$ at point $N$, \\\\\nsince $S_{\\triangle PBC}=\\frac{1}{3}S_{\\text{rectangle }ABCD}$, \\\\\ntherefore $\\frac{1}{2}\\times BC\\times PN=\\frac{1}{3}\\times BC\\times MN$, \\\\\nhence $PN=\\frac{2}{3}MN$, \\\\\nsince $AB=3$, \\\\\nthus $MP=1$, \\\\\nDraw $GH\\parallel AD$ through point $P$, intersecting $AB$ at point $G$ and $CD$ at point $H$, construct the symmetric point $A'$ of point $A$ about $GH$, and the intersection of $A'D$ with $GH$ will be the desired point $P$, \\\\\nsince $AP=A'P$, \\\\\nthus $AP+PD=A'D$, \\\\\nsince $AG=1$, \\\\\nthus $AA'=2$, \\\\\nIn $\\triangle AA'D$, with $AD=4$ and $AA'=2$, \\\\\nthus $A'D=2\\sqrt{5}$, \\\\\ntherefore, the minimum perimeter of $\\triangle ADP$ is $2\\sqrt{5}+4$, \\\\\nhence the answer is $\\boxed{4+2\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50481557.png"
  },
  {
    "solution": "\\boxed{15^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $\\triangle ABC$ is an equilateral triangle, and points $B$, $C$, $D$, $E$ are on the same line, with $CG=CD$, $DF=DE$. What is the measure of $\\angle EFD$?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABC$ is an equilateral triangle,\\\\\nit follows that $\\angle ACB=60^\\circ$.\\\\\nSince $\\angle ACB=\\angle CGD+\\angle CDG$,\\\\\nit follows that $\\angle CGD+\\angle CDG=60^\\circ$.\\\\\nSince $CG=CD$,\\\\\nit follows that $\\angle CGD=\\angle CDG=30^\\circ$.\\\\\nSince $\\angle CDG=\\angle DFE+\\angle E$,\\\\\nit follows that $\\angle DFE+\\angle E=30^\\circ$.\\\\\nSince $DF=DE$,\\\\\nit follows that $\\angle DFE=\\angle E=15^\\circ$.\\\\\nTherefore, $\\angle EFD=\\boxed{15^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52828908.png"
  },
  {
    "solution": "\\boxed{B(1, 0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the parabola $y=ax^2+bx+c$ ($a\\ne 0$) has its vertex at $Q(2, -1)$, and intersects the $y$-axis at point $C(0, 3)$, and intersects the $x$-axis at points $A$ and $B$ (point $A$ is to the right of point $B$). Point $P$ is a moving point on the parabola, moving from point $C$ towards point $A$ along the parabola (point $P$ does not coincide with $A$). A perpendicular line is drawn from point $P$ to line $AC$ at point $D$. What is the equation of this parabola, and the coordinates of points $A$ and $B$?\n",
    "completion": "\\textbf{Solution:} Given that the vertex of the parabola is $Q(2, -1)$,\\\\\nthus, the equation of the parabola can be assumed as $y=a(x-2)^2-1$.\\\\\nSubstituting $C(0, 3)$ into $y=a(x-2)^2-1$,\\\\\nwe get: $3=a(0-2)^2-1$,\\\\\nfrom which it follows that $a=\\boxed{1}$.\\\\\nTherefore, the equation of the parabola is $y=(x-2)^2-1=x^2-4x+3$,\\\\\nSetting $y=0$, gives $x^2-4x+3=0$,\\\\\nfrom which we find: $x_1=1$, $x_2=3$,\\\\\nsince point $A$ is to the right of point $B$,\\\\\ntherefore, $\\boxed{A(3, 0)}$, $\\boxed{B(1, 0)}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52977854_223.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the Cartesian coordinate system, the graph of the quadratic function $y=x^2+bx+c$ passes through points $A(3, 0)$ and $B(2, -3)$. By translating the graph of this quadratic function to the right by a certain number of units, the resulting graph can be made to pass through the origin.\n",
    "completion": "\\textbf{Solution:} When $y=0$, we have $x^2-2x-3=0$,\\\\\nSolving this yields: $x_1=-1$, $x_2=3$,\\\\\n$\\therefore$ the coordinates of the intersection points of the parabola $y=x^2-2x-3$ with the x-axis are $C(-1,0)$ and $A(3,0)$,\\\\\n$\\therefore$ by shifting the graph of the quadratic function one unit to the right, the resultant graph can pass through the origin.\\\\\nThus, the answer is: by shifting the graph of the quadratic function $\\boxed{1}$ unit to the right, the shifted graph can pass through the origin.",
    "image_path": "dataset/mm_math/images/MM_Math/52731589_77.png"
  },
  {
    "solution": "\\boxed{165^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given, as shown in the figure, line $AB$ intersects with line $CD$ at point $O$. $\\angle BOE = 90^\\circ$, $\\angle AOD = 30^\\circ$, and $OF$ is the angle bisector of $\\angle BOD$. Find the measure of $\\angle EOF$.\n",
    "completion": "\\textbf{Solution:} Given that $\\angle BOC=30^\\circ$,\\\\\nit follows that $\\angle BOD=180^\\circ-30^\\circ=150^\\circ$.\\\\\nSince OF is the angle bisector of $\\angle BOD$,\\\\\nwe have $\\angle BOF= \\frac{1}{2} \\angle BOD= \\frac{1}{2} \\times 150^\\circ=75^\\circ$.\\\\\nTherefore, $\\angle EOF=\\angle EOC+\\angle BOC+\\angle BOF=60^\\circ+30^\\circ+75^\\circ=\\boxed{165^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/46785011_223.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that in $\\triangle ABC$, $\\angle ACB=90^\\circ$, $AC=6$, and $BC=4$. Rotate $\\triangle ABC$ $90^\\circ$ clockwise around the right-angle vertex $C$ to get $\\triangle DEC$. If point $F$ is the midpoint of $DE$, and connect $AF$, then what is the length of $AF$?\n",
    "completion": "\\textbf{Solution:} Draw $FH \\perp AC$ at H, as shown in the figure,\\\\\nsince $\\triangle ABC$ is rotated clockwise by $90^\\circ$ around the right angle vertex C to obtain $\\triangle DEC$,\\\\\ntherefore, $CE=BC=4$, $CD=CA=6$, and $\\angle ECD=\\angle ACB=90^\\circ$,\\\\\nthus, $AE=AC-CE=2$, and $FH \\parallel CD$,\\\\\ntherefore, $\\triangle EHF \\sim \\triangle ECD$,\\\\\nsince point F is the midpoint of DE,\\\\\nthus, $\\frac{EH}{EC}=\\frac{HF}{CD}=\\frac{EF}{ED}=\\frac{1}{2}$,\\\\\ntherefore, $EH=\\frac{1}{2}EC=2$, $HF=\\frac{1}{2}CD=3$,\\\\\nthus, $AH=AE+EH=4$,\\\\\nin right-angled $\\triangle AFH$, $AF=\\sqrt{AH^2+HF^2}=\\sqrt{4^2+3^2}=\\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53078938.png"
  },
  {
    "solution": "\\boxed{\\sqrt{37}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in rectangle $ABCD$, rectangle $ABCD$ is folded along $EF$ with point $A$ landing on point $A'$, and point $B$ landing on point $B'$ on side $CD$. Line $BB'$ intersects $EF$ at point $G$, and point $M$ is on $A'B'$ with $A'M=2B'M$. Given that $CD=3$ and $AD=6$, during the folding process, when point $B'$ is at different positions on side $CD$, what is the minimum value of $MG+ B'G$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect GC, and choose a point Q on the segment AB such that AQ=2BQ. Connect QG and QC,\\\\\n$\\because$AB=CD=3,\\\\\n$\\therefore$BQ=1,\\\\\n$\\because$rectangle ABCD is folded along EF, point A falls on point A', and point B falls on point B' on side CD, with A'M=2B'M,\\\\\n$\\therefore$Q and M are symmetrical about EF,\\\\\n$\\therefore$QG=MG,\\\\\nAlso, $\\because$G is the midpoint of BB', and $\\angle$BCB'=$90^\\circ$,\\\\\n$\\therefore$GC=B'G,\\\\\n$\\therefore$MG+ B'G=QG+GC, $\\therefore$when points Q, G, and C are collinear, CQ=QG+GC, at this time, QG+GC is the shortest,\\\\\n$\\therefore$The minimum value of MG+ B'G is the length of CQ,\\\\\nIn right $\\triangle$BCQ, BQ=1, BC=6,\\\\\n$\\therefore$CQ=$\\sqrt{BQ^2+BC^2}$=$\\sqrt{1^2+6^2}$=$\\sqrt{37}$.\\\\\nTherefore, the answer is: $\\boxed{\\sqrt{37}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52082370.png"
  },
  {
    "solution": "\\boxed{9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, $O$ is the origin, and point $A(3, 3)$ is on the graph of the inverse proportional function $y = \\frac{k}{x}$. What is the value of $k$?\n",
    "completion": "\\textbf{Solution}: Since the point $A(3, 3)$ lies on the graph of the inverse proportion function $y=\\frac{k}{x}$, \\\\\nit follows that $k=3\\times 3=\\boxed{9}$",
    "image_path": "dataset/mm_math/images/MM_Math/53066069_86.png"
  },
  {
    "solution": "\\boxed{M(-1, 0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system, it's known that the graph of the linear function $y=\\frac{1}{2}x+1$ intersects the $x$-axis and $y$-axis at points $A$ and $B$, respectively. A square $ABCD$ is constructed inside the second quadrant with $AB$ as its side. Does there exist a point $M$ on the $x$-axis such that the perimeter of $\\triangle MDB$ is minimized? Determine the coordinates of point $M$.\n",
    "completion": "\\textbf{Solution:} To solve this, find the point $B'$ which is symmetrical to $B$ with respect to the $x$-axis, and connect $BD$, intersecting the $x$-axis at point $M$. At this time, the perimeter of $\\triangle BMD$ is minimized,\\\\\n$\\because B(0, 1)$,\\\\\n$\\therefore B'(0, -1)$\\\\\nLet the equation of line $B'D$ be $y=kx+b$,\\\\\nSubstituting the coordinates of $B'$ and $D$ yields: $\\begin{cases}b=-1\\\\ -3k+b=2\\end{cases}$,\\\\\nSolving this, we get: $\\begin{cases}k=-1\\\\ b=-1\\end{cases}$, that is, the equation of line $B'D$ is $y=-x-1$,\\\\\nSetting $y=0$, we find $x=-1$, i.e., \\boxed{M(-1, 0)}.",
    "image_path": "dataset/mm_math/images/MM_Math/52386761_125.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the side length of square $ABCD$ is $4$, and point $E$ is the midpoint of $AD$. When $\\triangle ABE$ is folded along the line containing $BE$ to obtain $\\triangle FBE$, and $BF$ is extended to intersect $CD$ at point $G$, what is the length of $DG$?\n",
    "completion": "\\textbf{Solution:}\nAs shown in the figure, connect EG,\n\n$\\because$ $\\triangle FBE$ is obtained by folding $\\triangle ABE$,\n\n$\\therefore$ AE=FE, $\\angle EFB=\\angle EAB=90^\\circ$,\n\n$\\therefore$ $\\angle EFG=\\angle EDG=90^\\circ$.\n\n$\\because$ AE=DE,\n\n$\\therefore$ FE=DE.\n\n$\\because$ EG=EG,\n\n$\\therefore$ Rt$\\triangle EFG \\cong$ Rt$\\triangle EDG$ (HL).\n\n$\\therefore$ DG=FG,\n\nLet DG=FG=x,\n\n$\\because$ AB=BF=4,\n\nAccording to the problem: BG=4+x, CG=4-x,\n\nBy the Pythagorean theorem: BG$^{2}$=BC$^{2}$+CG$^{2}$,\n\nWe get: $(4+x)^{2}=4^{2}+(4-x)^{2}$,\n\nSolve to get: $x=\\boxed{1}$.\n\n$\\therefore$ DG=1,",
    "image_path": "dataset/mm_math/images/MM_Math/52188952.png"
  },
  {
    "solution": "\\boxed{8\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the radius of the circumcircle of $\\triangle ABC$ is $8$, and $\\angle ACB=60^\\circ$, then what is the length of $AB$?\n",
    "completion": "\\textbf{Solution:} Connect OA and OB, and draw OH $\\perp$ AB at H, \\\\\n$\\because$ $\\angle$ACB=$60^\\circ$, \\\\\n$\\therefore$ $\\angle$AOB=$2\\angle$ACB=$120^\\circ$, \\\\\n$\\because$ OB=OA=8, \\\\\n$\\therefore$ $\\angle$AOH=$\\angle$BOH=$60^\\circ$, \\\\\n$\\therefore$ $\\angle$OAB=$30^\\circ$, \\\\\n$\\therefore$ OH=$\\frac{1}{2}$OA=4, \\\\\n$\\therefore$ AH=$\\sqrt{OA^{2}-OH^{2}}=\\sqrt{8^{2}-4^{2}}=4\\sqrt{3}$, \\\\\n$\\therefore$ AB=2AH=$\\boxed{8\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51445882.png"
  },
  {
    "solution": "\\boxed{20^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $\\angle AOB=90^\\circ$, $\\angle AOC$ is an acute angle, $ON$ bisects $\\angle AOC$, and ray $OM$ is inside $\\angle AOB$. If $\\angle AOC=50^\\circ$ and $\\angle MON=45^\\circ$, find the measure of $\\angle AOM$.\n",
    "completion": "\\textbf{Solution:} Since $ON$ bisects $\\angle AOC$ and $\\angle AOC=50^\\circ$,\\\\\nit follows that $\\angle AON=\\angle CON=\\frac{1}{2}\\angle AOC=25^\\circ$,\\\\\nSince $\\angle MON=45^\\circ$,\\\\\ntherefore, $\\angle AOM=\\angle MON-\\angle AON=45^\\circ-25^\\circ=\\boxed{20^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53403787_54.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $D$ is the midpoint of $AC$, $CB = 4\\,cm$, $DB = 7\\,cm$, what is the length of $AB$ in $cm$?\n",
    "completion": "Solution: Since $DC=DB-CB=7-4=3\\,cm$,\\\\\nand since point D is the midpoint of AC,\\\\\nit follows that $AD=DC=3\\,cm$\\\\\nTherefore, $AB=AD+DB = 3+7=\\boxed{10}\\,cm$",
    "image_path": "dataset/mm_math/images/MM_Math/51438522.png"
  },
  {
    "solution": "\\boxed{\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $E$ is the midpoint of $AD$, and $AE = 1\\,\\text{cm}$. The perpendicular bisector of $BE$, $MN$, exactly passes through point $C$. What is the length of $AB$ in centimeters?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect EC, and BE intersects MN at point F, \\\\\nsince rectangle ABCD, E is the midpoint of AD, and AE=1cm, \\\\\ntherefore AB=CD, AD=BC=2AE=2cm, ED=1cm, \\\\\nsince FC is perpendicular bisector of BE, \\\\\ntherefore EC=BC=2cm, $\\angle$BFC=$\\angle$EFC=$90^\\circ$, \\\\\ntherefore in Rt$\\triangle$EFC, CD= $\\sqrt{EF^{2}-ED^{2}}=\\sqrt{2^{2}-1^{2}}=\\sqrt{3}$, \\\\\ntherefore AB=CD= $\\boxed{\\sqrt{3}}$cm.",
    "image_path": "dataset/mm_math/images/MM_Math/51614400.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, $\\angle BAC=30^\\circ$, and $BC=2$. The line segment $BC$ rotates around point $B$ to $BD$. Connect $AD$, where $E$ is the midpoint of $AD$. Connect $CE$. What is the maximum value of $CE$?\n",
    "completion": "\\textbf{Solution:} Construct the midpoint O of AB and connect OE as shown in the diagram: \\\\\nAccording to the problem, $BD=BC=2$, \\\\\nSince point E is the midpoint of AD, and point O is the midpoint of AB, \\\\\nit follows that $OE=\\frac{1}{2}BD=1$, \\\\\nthus, the trajectory of point E is a circle centered at O with a radius of 1, \\\\\nhence, when point E is on the extension line of CO, CE is maximum, \\\\\nsince $\\angle ACB=90^\\circ$, $\\angle BAC=30^\\circ$, $BC=2$ \\\\\nit follows that $AB=4$, \\\\\nsince point O is the midpoint of AB, \\\\\nit follows that $OC=\\frac{1}{2}AB=2$, \\\\\nthus, the maximum value of $CE$ is $OC+OE=2+1=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52457427.png"
  },
  {
    "solution": "\\boxed{(2, 2)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system, the parabola $y=\\frac{1}{2}x^{2}+bx+c$ passes through points $A\\left(0, 2\\right)$ and $B\\left(1, \\frac{3}{2}\\right)$. Given that point C is symmetric to point A with respect to the axis of symmetry of this parabola, what are the coordinates of point C?\n",
    "completion": "\\textbf{Solution:} Since $y=\\frac{1}{2}x^{2}-x+2=\\frac{1}{2}(x-1)^{2}+\\frac{3}{2}$,\\\\\nthe axis of symmetry of the parabola is the line $x=1$,\\\\\nand since point C is symmetric to point A with respect to this axis of the parabola,\\\\\nthe coordinates of point C are $\\boxed{(2, 2)}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53033041_157.png"
  },
  {
    "solution": "\\boxed{20}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the square $ABCD$, $E$ is the midpoint of $BC$, $F$ is a point on side $CD$, and $CD=4CF$. Extend line $EF$ to intersect the extension of line $AD$ at point $G$. $\\triangle ABE\\sim \\triangle ECF$. If the side length of the square $ABCD$ is 8, find the length of $AG$.\n",
    "completion": "\\textbf{Solution:} Given that the side length of square ABCD is 8, it follows that BC=CD=AD=8, and BC is parallel to AD, leading to $\\angle CEF=\\angle G$. Since $\\angle CFE=\\angle DFG$, triangles CEF and DGF are similar, implying $\\frac{CF}{DF}=\\frac{CE}{DG}$. Given that E is the midpoint of BC and CD equals 4 times CF, we find CF=2, DF=6, and CE=4. Thus, $\\frac{2}{6}=\\frac{4}{DG}$, which gives DG=12. Therefore, AG=DG+AD=\\boxed{20}.",
    "image_path": "dataset/mm_math/images/MM_Math/51458200_164.png"
  },
  {
    "solution": "\\boxed{y=\\frac{2}{x}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y=k_1x+b$ intersects with the graph of the inverse proportion function $y=\\frac{k_2}{x}$ at points $A(1, 2)$ and $B(-2, n)$. What are the analytical expressions of the linear function and the inverse proportion function?\n",
    "completion": "\\textbf{Solution:} From the problem, we know that the inverse proportion function $y=\\frac{k_2}{x}$ passes through $A(1, 2)$,\n\\[ \\therefore k_2=1\\times 2=2, \\]\n\\[ \\therefore \\] the inverse proportion function is $y=\\frac{2}{x}$,\nMoreover, since $B(-2, n)$ lies on the graph of the inverse proportion function $y=\\frac{2}{x}$,\n\\[ \\therefore n=\\frac{2}{-2}=-1, \\]\n\\[ \\therefore B(-2, -1), \\]\nAlso, since the line $y=k_1x+b$ passes through $A(1, 2)$, $B(-2, -1)$,\n\\[ \\therefore \\left\\{\\begin{aligned} k_1+b&=2 \\\\ -2k_1+b&=-1 \\end{aligned}\\right., \\] solving this yields $\\left\\{\\begin{aligned} k_1&=1 \\\\ b&=1 \\end{aligned}\\right.$,\n\\[ \\therefore \\] the linear function is described by $y=x+1$,\nThe inverse proportion function is described by $y=\\frac{2}{x}$.\nThus, the formula for the linear function is $\\boxed{y=x+1}$, and the formula for the inverse proportion function is $\\boxed{y=\\frac{2}{x}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51402407_30.png"
  },
  {
    "solution": "\\boxed{-4<x<2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $l_1: y=kx+b \\, (k \\neq 0)$ and $l_2: y=mx+n \\, (m \\neq 0)$ intersect the x-axis at points $A(-4, 0)$ and $B(2, 0)$, respectively. What is the solution set of the system of inequalities $\\begin{cases} kx+b>0 \\\\ mx+n>0 \\end{cases}$?\n",
    "completion": "Solution: From the graph, we know\n\n$\\because$ $kx+b>0$\n\n$\\therefore$ $x>-4$\n\n$\\because$ $mx+n>0$\n\n$\\therefore$ $x<2$\n\n$\\therefore$ the solution set of the system of inequalities $\\begin{cases}kx+b>0\\\\ mx+n>0\\end{cases}$ is $-4<x<2$\n\nTherefore, the answer is: $\\boxed{-4<x<2}$",
    "image_path": "dataset/mm_math/images/MM_Math/50493304.png"
  },
  {
    "solution": "\\boxed{\\frac{16\\sqrt{3}}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of $\\odot O$, $PA$ and $PC$ are tangents to $\\odot O$, with $A$ and $C$ being the points of tangency. Connect $AC$ and $PO$, intersecting at point $D$. $\\angle BAC=\\angle OPC$; extend $PO$ to intersect $\\odot O$ at point $E$, and connect $BE$ and $CE$. If $\\angle BEC=30^\\circ$ and $PA=8$, what is the length of $AB$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, since $\\angle BEC=30^\\circ,$\\\\\nthus $\\angle BAC=30^\\circ,$ and $AC\\perp OP,$\\\\\nthus $\\angle AOP=60^\\circ,$\\\\\nsince $\\angle OAP=90^\\circ, PA=8,$\\\\\nthus $\\tan\\angle AOP=\\frac{PA}{AO}=\\tan60^\\circ=\\sqrt{3},$\\\\\nthus $AO=\\frac{8}{\\sqrt{3}}=\\frac{8\\sqrt{3}}{3},$\\\\\nthus $AB=2AO=\\boxed{\\frac{16\\sqrt{3}}{3}}.$",
    "image_path": "dataset/mm_math/images/MM_Math/51402246_139.png"
  },
  {
    "solution": "\\boxed{80^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle BAC=60^\\circ$, $\\angle BCE=40^\\circ$, $AD$ bisects $\\angle BAC$, $CE \\perp AB$ at point E, then what is the degree measure of $\\angle ADC$?\n",
    "completion": "\\textbf{Solution:} Since $CE\\perp AB$,\n\\[\\therefore \\angle BEC=90^\\circ,\\]\nSince \\(\\angle BCE=40^\\circ\\),\n\\[\\therefore \\angle B=50^\\circ,\\]\nSince \\(\\angle BAC=60^\\circ\\) and $AD$ bisects \\(\\angle BAC\\),\n\\[\\therefore \\angle BAD=\\frac{1}{2}\\angle BAC=30^\\circ,\\]\n\\[\\therefore \\angle ADC=\\angle B+\\angle BAD\\]\n\\[=50^\\circ+30^\\circ\\]\n\\[=\\boxed{80^\\circ}.\\]",
    "image_path": "dataset/mm_math/images/MM_Math/53102441.png"
  },
  {
    "solution": "\\boxed{\\begin{cases}x=1\\\\ y=2\\end{cases}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the Cartesian coordinate system $xOy$, the graphs of linear functions $y=kx$ and $y=-x+3$ are shown as in the figure. Then, what is the solution to the system of linear equations $\\begin{cases}y=kx\\\\ y=-x+3\\end{cases}$?\n",
    "completion": "\\textbf{Solution}: Since the graph of the linear function $y=kx$ intersects with the graph of $y=-x+3$ at point $(1, 2)$,\ntherefore the solution to the system of linear equations $\\begin{cases}y=kx\\\\ y=-x+3\\end{cases}$ is $\\boxed{\\begin{cases}x=1\\\\ y=2\\end{cases}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50590774.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $\\angle BOC = 60^\\circ$, $OE$ bisects $\\angle AOC$, and $OF$ bisects $\\angle BOC$. If $AO \\perp BO$, then what is the measure of $\\angle EOF$ in degrees?\n",
    "completion": "\\textbf{Solution:} Given $\\because \\angle BOC=60^\\circ$, and OF bisects $\\angle BOC$, $\\therefore \\angle COF= \\frac{1}{2}\\angle BOC=30^\\circ$, $\\because AO\\perp BO$, $\\therefore \\angle AOB=90^\\circ$, hence $\\angle AOC=90^\\circ+60^\\circ=150^\\circ$, $\\because OE$ bisects $\\angle AOC$, $\\therefore \\angle COE= \\frac{1}{2}\\angle AOC=75^\\circ$, thus $\\angle EOF=\\angle COE-\\angle COF=\\boxed{45^\\circ};$",
    "image_path": "dataset/mm_math/images/MM_Math/55190308_104.png"
  },
  {
    "solution": "\\boxed{\\frac{10}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line segment $CD \\parallel AB$, $AD$ and $BC$ intersect at point $E$. $AE \\cdot CE = DE \\cdot BE$; through point $E$, draw $EF \\parallel AB$, intersecting $AC$ at point $F$. If $AB=5$ and $EF=2$, what is the length of $CD$?\n",
    "completion": "\\textbf{Solution:} Since EF$\\parallel$AB,\\\\\nit follows that $\\angle$EFC$=\\angle$BAC, $\\angle$CEF$=\\angle$B,\\\\\ntherefore, $\\triangle$ CEF$\\sim$ $\\triangle$ CBA,\\\\\nthus, $\\frac{CF}{CA}=\\frac{EF}{AB}=\\frac{2}{5}$,\\\\\nso, $\\frac{AF}{CA}=\\frac{3}{5}$,\\\\\nSince CD$\\parallel$AB and EF$\\parallel$AB,\\\\\nit follows that EF$\\parallel$CD,\\\\\ntherefore, $\\angle$EAF$=\\angle$DAC, $\\angle$EFA$=\\angle$DCA,\\\\\nthus, $\\triangle$ AEF$\\sim$ $\\triangle$ ADC,\\\\\nit follows that $\\frac{EF}{CD}=\\frac{AF}{CA}$,\\\\\nhence, $\\frac{2}{CD}=\\frac{3}{5}$,\\\\\ntherefore, CD$=\\boxed{\\frac{10}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51446069_103.png"
  },
  {
    "solution": "\\boxed{65^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rhombus $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$, and $OE \\perp AB$ at point $E$. If $\\angle ADC = 130^\\circ$, what is the measure of $\\angle AOE$?\n",
    "completion": "\\textbf{Solution:} In rhombus $ABCD$, $\\angle ADC=130^\\circ$, \\\\\n$\\therefore \\angle BAD=180^\\circ-130^\\circ=50^\\circ$, \\\\\n$\\therefore \\angle BAO= \\frac{1}{2}\\angle BAD= \\frac{1}{2}\\times50^\\circ=25^\\circ$, \\\\\n$\\because OE\\perp AB$, \\\\\n$\\therefore \\angle AOE=90^\\circ-\\angle BAO=90^\\circ-25^\\circ=\\boxed{65^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52344787.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, $AM=AC$, $BN=BC$. When the degree of $\\angle A$ changes, does the degree of $\\angle MCN$ change? If it does not change, find the degree of $\\angle MCN$.\n",
    "completion": "\\textbf{Solution:}\\\\\nSince $\\angle AMC=\\frac{1}{2}\\left(180^\\circ-\\angle A\\right), \\angle CNB=\\frac{1}{2}\\left(180^\\circ-\\angle B\\right)$,\\\\\nit follows that $\\angle MCN=180^\\circ-\\angle AMC-\\angle BNC=\\frac{1}{2}\\left(\\angle A+\\angle B\\right)=45^\\circ$,\\\\\ntherefore, as the measure of $\\angle A$ changes, the measure of $\\angle MCN$ remains constant, $\\angle MCN=\\boxed{45^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52867075_158.png"
  },
  {
    "solution": "\\boxed{\\frac{1000\\pi}{9}m}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a turning point of a road is a section of a circular arc (the arc $ \\overset{\\frown}{AB}$ in the figure), where point O is the center of this arc, $ \\angle AOB=80^\\circ$. Point C is on segment AB, and OC is perpendicular to AB at D. Given that $AB=300m$ and $CD=50m$, what is the length of the curved road AB in meters?\n",
    "completion": "\\textbf{Solution:} Let the radius be $r$,\\\\\nthen $OD=r-CD=(r-50)m$,\\\\\n$\\because$ $OC\\perp AB$,\\\\\n$\\therefore$ $AD=BD=\\frac{1}{2}AB=\\frac{1}{2}\\times 300=150m$,\\\\\nIn $\\triangle AOD$, $AO^2=AD^2+OD^2$,\\\\\ni.e., $r^2=150^2+(r-50)^2$,\\\\\nsolving this yields $r=250m$,\\\\\nthe radius of this curved segment is $250m$.\\\\\n$\\because$ $\\angle AOB=80^\\circ$,\\\\\n$\\therefore$ $\\overset{\\frown}{AB}=\\frac{80\\pi \\times 250}{180}=\\frac{1000\\pi}{9}m$.\\\\\nThus, the length of this curved segment AB is $\\boxed{\\frac{1000\\pi}{9}m}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53101887.png"
  },
  {
    "solution": "\\boxed{7}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the equilateral triangle $\\triangle ABC$ has a side length of 8. $P$ and $Q$ are points on sides $AC$ and $BC$, respectively. Line segments $AQ$ and $BP$ intersect at point $O$, with $AP = CQ$. What is the measure of $\\angle AOB$ in degrees? If $BQ = 5$, what is the length of $AQ$?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABC$ is an equilateral triangle, \\\\\nwe have $AB=AC$, $\\angle BAC=\\angle C=60^\\circ$, \\\\\nIn triangles $\\triangle ABP$ and $\\triangle ACQ$, \\\\\n$\\left\\{\\begin{array}{l}\nAB=AC\\\\\n\\angle BAP=\\angle CAQ\\\\\nAP=CQ\n\\end{array}\\right.$, \\\\\nthus $\\triangle ABP\\cong \\triangle ACQ$ (SAS), \\\\\ntherefore, $\\angle ABP=\\angle CAQ$, \\\\\nthus $\\angle BOQ=\\angle ABO+\\angle BAQ=\\angle CAQ+\\angle BAQ=\\angle BAC=60^\\circ$, \\\\\nhence, $\\angle AOB=120^\\circ$; \\\\\nDraw line AD $\\perp$ BC at D through point A, \\\\\nSince BC=8 and $\\triangle ABC$ is an equilateral triangle, \\\\\nthen BD=CD=4, $\\angle ABD=60^\\circ$, \\\\\nthus AD=$\\sqrt{AB^2-BD^2}=\\sqrt{8^2-4^2}=4\\sqrt{3}$ \\\\\nGiven BQ=5, \\\\\nthus DQ=1, \\\\\ntherefore, AQ=$\\sqrt{AD^2+DQ^2}=\\sqrt{(4\\sqrt{3})^2+1^2}=\\boxed{7}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52913897.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated in the square $ABCD$, where $E$ is a moving point on side $CD$ (not coinciding with points $C$ or $D$), connect $AE$. By folding $\\triangle ADE$ along $AE$, $\\triangle AEF$ is obtained. Extend $EF$ to intersect $BC$ at point $G$, and connect $AG$. Draw $EH\\perp AE$ passing through point $E$ and extending to intersect the extended line of $AG$ at point $H$, and then connect $CH$. What is the measure of $\\angle EAH$ in degrees?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral $ABCD$ is a square, we have $AD=DC=AB$ and $\\angle BAD=\\angle B=\\angle D=90^\\circ$. By the property of folding, we get $AF=AD$ and $\\angle AFE=\\angle D=90^\\circ$, $\\angle EAF=\\angle EAD$. Thus, $AB=AF$ and $\\angle AFG=90^\\circ$. In right triangles $Rt\\triangle ABG$ and $Rt\\triangle AFG$, we have $\\left\\{\\begin{array}{l}AG=AG \\\\ AB=AF\\end{array}\\right.$, hence $Rt\\triangle ABG \\cong Rt\\triangle AFG$ (by HL congruence), implying $\\angle BAG=\\angle FAG$. Consequently, $\\angle EAH=\\angle FAG+\\angle EAF=\\angle BAG+\\angle EAD$. Also, since $\\angle FAG+\\angle EAF+\\angle BAG+\\angle EAD=\\angle BAD=90^\\circ$, we thus have $\\angle EAH+\\angle EAH=90^\\circ$. Solving this, we find $\\angle EAH=\\boxed{45^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50482868_118.png"
  },
  {
    "solution": "\\boxed{\\frac{13}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, P is a moving point on side AD of a square $ABCD$ with side length 6 (P does not coincide with A or D). Connect $CP$, and draw $BH\\perp CP$ from point B. Fold $\\triangle DPC$ along the straight line containing $CP$ to obtain $\\triangle CPD^{\\prime}$. Extend $PD^{\\prime}$ to meet the extension of $CB$ at point G. When $BH = \\frac{18\\sqrt{13}}{13}$, what is the length of $PG$?\n",
    "completion": "\\textbf{Solution:}\n\nIn right triangle $BCH$, $\\angle BHC=90^\\circ$, $BH=\\frac{18\\sqrt{13}}{13}$, $BC=6$,\n\naccording to the Pythagorean theorem, $CH=\\sqrt{BC^2-BH^2}=\\sqrt{6^2-\\left(\\frac{18\\sqrt{13}}{13}\\right)^2}=\\frac{12\\sqrt{13}}{13}$,\n\nsince square $ABCD$, $DC=6$,\n\nthus $AD\\parallel BC$,\n\ntherefore $\\angle DPC=\\angle PCB$,\n\ntherefore $\\triangle PDC\\sim \\triangle CHB$,\n\ntherefore $\\frac{PD}{CH}=\\frac{DC}{HB}$,\n\ntherefore $\\frac{PD}{\\frac{12\\sqrt{13}}{13}}=\\frac{6}{\\frac{18\\sqrt{13}}{13}}$,\n\nsolving yields $PD=4$,\n\nsince $\\triangle DPC$ is reflected along line $CP$ to get $\\triangle CPD'$,\n\ntherefore $PD'=PD=4$, $D'C=DC=6$, $\\angle D'PC=\\angle DPC=\\angle PCB$,\n\nthus $GP=GC$,\n\nlet $GP=GC=x$, then $GD'=GP-PD'=x-4$,\n\nin right triangle $CD'G$, $\\angle CD'G=90^\\circ$,\n\naccording to the Pythagorean theorem, $GC^2=GD'^2+CD'^2$,\n\nthus $x^2=(x-4)^2+6^2$,\n\nwhich gives $8x=52$,\n\nsolving yields $x=\\boxed{\\frac{13}{2}}$,\n\nthus, $PG=\\frac{13}{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52866267.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, quadrilateral $ABCD$ is inscribed in circle $\\odot O$, with $\\angle C = 2\\angle A$. $DE$ is the diameter of $\\odot O$, and $BD$ is connected. What is the degree measure of $\\angle A$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral $ABCD$ is inscribed in circle $\\odot O$,\\\\\nit follows that $\\angle C+\\angle A=180^\\circ$,\\\\\nand since $\\angle C=2\\angle A$,\\\\\nit results in $\\angle A=\\boxed{60^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53019741_126.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within the circle $\\odot O$, the radius $OC$ is perpendicular to the chord $AB$ at point $D$, and given $AB=8$ and $OC=5$, what is the length of $OD$?\n",
    "completion": "\\textbf{Solution:} Given that radius $OC$ is perpendicular to chord $AB$ at point $D$, and $AB=8$,\\\\\nhence, $AD=\\frac{1}{2}AB=4$, and $\\angle ADO=90^\\circ$,\\\\\nsince $OA=OC=5$,\\\\\nthus, $OD=\\sqrt{OA^{2}-AD^{2}}=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53080808.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $D$ is inside $\\triangle ABC$, where $CD$ bisects $\\angle ACB$, $BD\\perp CD$, and $\\angle A = \\angle ABD$. If $AC = 7$ and $BC = 4$, what is the length of $BD$?\n",
    "completion": "\\textbf{Solution:} Extend BD to meet AC at point E,\\\\\n$\\because$ $\\angle A = \\angle ABD$,\\\\\n$\\therefore$ BE = AE,\\\\\n$\\because$ BD $\\perp$ CD,\\\\\n$\\therefore$ BE $\\perp$ CD,\\\\\n$\\because$ CD bisects $\\angle ACB$,\\\\\n$\\therefore$ $\\angle BCD = \\angle ECD$,\\\\\n$\\therefore$ $\\angle EBC = \\angle BEC$,\\\\\n$\\therefore$ $\\triangle BEC$ is an isosceles triangle,\\\\\n$\\therefore$ BC = CE,\\\\\n$\\because$ BE $\\perp$ CD,\\\\\n$\\therefore$ 2BD = BE,\\\\\n$\\because$ AC = 7, BC = 4,\\\\\n$\\therefore$ CE = 4,\\\\\n$\\therefore$ AE = AC - EC = 7 - 4 = 3,\\\\\n$\\therefore$ BE = 3,\\\\\n$\\therefore$ BD = $\\boxed{\\frac{3}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52991478.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a regular hexagon $ABCDEF$ is inscribed in circle $\\odot O$. Connecting $BD$, what is the degree measure of $\\angle CDB$?\n",
    "completion": "\\textbf{Solution}: As shown in the figure, connect $OB$, $OC$,\\\\\nsince regular hexagon $ABCDEF$ is inscribed in $\\odot O$,\\\\\ntherefore $\\angle BOC=\\frac{360^\\circ}{6}=60^\\circ$,\\\\\ntherefore $\\angle CBD=\\frac{1}{2}\\angle COD=\\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53033414.png"
  },
  {
    "solution": "\\boxed{(4,0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated in the diagram, the line $l_{1}: y=x+2$ intersects the $x$ axis at point $A$ and the $y$ axis at point $B$, while the line $l_{2}: y=4x-4$ intersects the $y$ axis at point $C$ and the $x$ axis at point $D$. The lines $l_{1}$ and $l_{2}$ intersect at point $P$. If there exists a point $Q$ on the $x$ axis such that the quadrilateral formed by vertices $A$, $C$, $P$, $Q$ is a parallelogram, then what are the coordinates of point $Q$?\n",
    "completion": "\\textbf{Solution:} Given $y=x+2$, when $y=0$, we have $x+2=0$,\\\\\nSolving yields: $x=-2$,\\\\\n$\\therefore$ The coordinates of point A are $(-2,0)$,\\\\\nIn $y=4x-4$, when $x=0$, we have $y=-4$,\\\\\n$\\therefore$ The coordinates of point C are $(0,-4)$,\\\\\nBy solving the system of equations $\\begin{cases} y=x+2 \\\\ y=4x-4 \\end{cases}$,\\\\\nWe find: $\\begin{cases} x=2 \\\\ y=4 \\end{cases}$,\\\\\n$\\therefore$ The coordinates of point P are $(2,4)$,\\\\\nLet the coordinates of point Q be $(x,0)$,\\\\\n$\\because$ Point Q is on the x-axis,\\\\\n$\\therefore$ when the quadrilateral with vertices A, C, P, Q is a parallelogram, AQ and PC are diagonals,\\\\\n$\\therefore$ $\\frac{-2+x}{2}=\\frac{2+0}{2}$,\\\\\nSolving yields: $x=4$,\\\\\n$\\therefore$ The coordinates of point Q are $\\boxed{(4,0)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50545595.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given: In quadrilateral $ABCD$, $DE\\perp AC$ and $BF\\perp AC$ with perpendicular feet $E$ and $F$ respectively. Extend $DE$ and $BF$ to intersect $AB$ at point $H$ and $BC$ at point $G$, if $AD\\parallel BC$ and $AE=CF$. If $\\angle DAH=\\angle GBA$, $GF=2$, and $CF=4$, find the length of $AD$.\n",
    "completion": "\\textbf{Solution:} Since $DE\\perp AC$ and $BF\\perp AC$, it follows that $DH\\parallel BG$. Consequently, $\\angle GBA=\\angle DHA$. Given that $\\angle DAH=\\angle GBA$, we have $\\angle DAH=\\angle DHA$, which implies $AD=DH$. In $\\triangle GCF$, where $GF=2$ and $CF=4$, we find that $CG=\\sqrt{CF^2+GF^2}=2\\sqrt{5}$. Because quadrilateral $ABCD$ is a parallelogram, we have $DC\\parallel AB$ and $AB=CD$, which means $DG\\parallel HB$. Given that $DH\\parallel GB$, quadrilateral $DHBG$ forms a parallelogram, thus $DG=HB$. Therefore, $AH=CG=2\\sqrt{5}$. In $\\triangle AEH$, where $AE=CF=4$ and $AH=2\\sqrt{5}$, it follows that $EH=2$. In $\\triangle ADE$, we have $AD^2=DE^2+AE^2$, which gives us $AD^2=(AD-2)^2+4^2$. Solving this, we find $AD=\\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52354589_103.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Fold the rectangular paper ABCD as shown in the figure, and you get the rhombus AECF exactly. If \\(AD = \\sqrt{3}\\), what is the area of the rhombus AECF?\n",
    "completion": "\\textbf{Solution:} We know from the properties of folding that $\\angle DAF = \\angle OAF$ and $OA = AD = \\sqrt{3}$,\\\\\nIn rhombus AECF, $\\angle OAF = \\angle OAE$,\\\\\n$\\therefore \\angle OAE = \\frac{1}{3} \\times 90^\\circ = 30^\\circ$,\\\\\n$\\therefore AE = \\frac{AO}{\\cos 30^\\circ} = \\frac{\\sqrt{3}}{\\frac{\\sqrt{3}}{2}} = 2$,\\\\\n$\\therefore$ the area of rhombus AECF $= AE \\cdot AD = 2\\sqrt{3}$.\\\\\nThus, the answer is: \\boxed{2\\sqrt{3}}.",
    "image_path": "dataset/mm_math/images/MM_Math/53101881.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a rectangular piece of paper ABCD (with AD > DC) is folded along a straight line passing through point D, so that point A lands on side BC, with the landing point being E, and the crease intersects side AB at point F. Given that \\(BE:EC = 1:4\\) and \\(CD = 9\\), find the length of \\(BF\\).\n",
    "completion": "\\textbf{Solution:} Let $\\because BE:EC=1:4$, assume $BE=x$, $EC=4x$, by the properties of the rectangle, we have: $AD=BC=BE+EC=5x$. From the property of folding, we get: $DE=AD=5x$. In $\\triangle DEC$, we have $CD^2+CE^2=DE^2$, that is $9^2+(4x)^2=(5x)^2$. Solving this gives $x=3$ (neglect the negative value), hence $BE=3$. Let $BF=a$, then $EF=AF=AB-BF=9-a$. By the Pythagorean theorem, we have $BE^2+BF^2=EF^2$, that is $3^2+a^2=(9-a)^2$. Solving this yields $a=\\boxed{4}$, so $BF=4$.",
    "image_path": "dataset/mm_math/images/MM_Math/51155328_177.png"
  },
  {
    "solution": "\\boxed{y=\\frac{2}{3}x+2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the line $l_{1}$ intersects the x-axis at point A $(-3,0)$, intersects the y-axis at point B, and intersects with line $l_{2}: y=\\frac{4}{3}x$ at point C $(a,4)$. What is the equation of line $l_{1}$?\n",
    "completion": "\\textbf{Solution:} Given point C(3, 4), let the equation of line $l_{1}$ be $y=kx+b$, \\\\\n$\\left\\{\\begin{array}{l}\n3k+b=4\\\\\n-3k+b=0\n\\end{array}\\right.$ Solving this system, we get $\\left\\{\\begin{array}{l}\nk=\\frac{2}{3}\\\\\nb=2\n\\end{array}\\right.$\\\\\n$\\therefore$ The equation of line $l_{1}$ is: $\\boxed{y=\\frac{2}{3}x+2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52226206_105.png"
  },
  {
    "solution": "\\boxed{\\frac{5\\sqrt{3}}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, given three equilateral triangles arranged from left to right with side lengths of 1, 2, and 3 respectively, what is the area of the shaded region?\n",
    "completion": "\\textbf{Solution:} Let points M, N, H be the intersections of AC with BE, BF, and CF respectively,\n\n$\\because$ AB=1, BC=2, CD=3,\n\n$\\therefore$ AC=CG,\n\n$\\therefore\\angle CAG=\\angle CGA=30^\\circ$,\n\n$\\therefore$ AG $\\perp$ BE,\n\nAlso $\\because\\angle EBA=\\angle FCA=60^\\circ$,\n\n$\\therefore$ EB $\\parallel$ CF, AG $\\perp$ CF,\n\n$\\therefore$ AH=HG, AM=MN,\n\n$\\therefore$ Area of $\\triangle CHG$ = Area of $\\triangle CHA$,\n\nSimilarly, Area of $\\triangle BMN$ = Area of $\\triangle BMA$,\n\nIn $\\triangle ABM$, AB=1, BM= $\\frac{1}{2}$ AB= $\\frac{1}{2}$, AM= $\\frac{\\sqrt{3}}{2}$,\n\n$\\therefore$ Area of $\\triangle ABM$= $\\frac{1}{2} \\times BM \\times AM= \\frac{1}{2}\\times \\frac{1}{2}\\times \\frac{\\sqrt{3}}{2} = \\frac{\\sqrt{3}}{8}$,\n\nSimilarly, it can be shown that Area of $\\triangle ACH$= $\\frac{1}{2} \\times CH \\times AH= \\frac{1}{2}\\times \\frac{3}{2}\\times \\frac{3\\sqrt{3}}{2} = \\frac{9\\sqrt{3}}{8}$,\n\n$\\therefore$ The area of the shaded part= Area of $\\triangle ABM$ + Area of $\\triangle ACH$= $\\frac{\\sqrt{3}}{8}+\\frac{9\\sqrt{3}}{8} = \\boxed{\\frac{5\\sqrt{3}}{4}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51213181.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{13}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\cos\\angle BAC=\\frac{12}{13}$, $AB=26$, $AC=15$, $BD\\perp AC$ with the foot being D, E is the midpoint of BD, and AE intersects BC at point F. Find the value of $\\frac{EF}{AE}$?\n",
    "completion": "\\textbf{Solution:} Construct line EM $\\parallel$ DC intersecting BC at M.\\\\\nSince E is the midpoint of BD,\\\\\nThus EM$= \\frac{1}{2}$DC=4.5\\\\\nSince EM$\\parallel$CA,\\\\\nThus $\\triangle$EMF$\\sim$$\\triangle$ACF,\\\\\nThus $\\frac{EF}{AF}=\\frac{EM}{AC}=\\frac{4.5}{15}=\\frac{3}{10}$,\\\\\nThus $\\frac{EF}{AE}=\\boxed{\\frac{3}{13}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52602500_139.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $BD$ is the angle bisector of $\\triangle ABC$, and $E$ is a point on the extension of $BD$ with $AE=AB$. $\\triangle ADE \\sim \\triangle CDB$. If $AB=4$, $\\frac{DC}{AD}=\\frac{1}{2}$, find the length of $BC$.\n",
    "completion": "Solution: Since $AE=AB$, and $AB=4$,\\\\\nit follows that $AE=4$.\\\\\nSince $\\triangle ADE\\sim \\triangle CDB$,\\\\\nit follows that $\\frac{BC}{AE}=\\frac{DC}{AD}=\\frac{1}{2}$.\\\\\nTherefore, $BC=\\frac{1}{2}AE=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52871853_110.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB = 90^\\circ$, CE is the angle bisector of $\\triangle ABC$, and $\\angle AEC = 105^\\circ$. What is the measure of $\\angle B$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle ACB=90^\\circ$ and CE is the angle bisector of $\\triangle ABC$, \\\\\nit follows that $\\angle ACE=\\angle BCE=45^\\circ$,\\\\\nFurthermore, since $\\angle AEC=105^\\circ$,\\\\\nit results in $\\angle B=\\angle AEC-\\angle BCE=105^\\circ-45^\\circ=\\boxed{60^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52807551.png"
  },
  {
    "solution": "\\boxed{\\frac{35}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in rhombus $ABCD$, $\\angle ABC=120^\\circ$, $AB=10$, and points $M$ and $N$ are on sides $AD$ and $AB$, respectively. If $\\triangle AMN$ is folded along $MN$ to obtain $\\triangle PMN$, with point $P$ exactly at the midpoint of side $CD$, then what is the length of $AN$?\n",
    "completion": "\\textbf{Solution:} The length of $AN$ is \\boxed{\\frac{35}{4}}.",
    "image_path": "dataset/mm_math/images/MM_Math/53151061_72.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the figure, within $ \\triangle ABC$, where $ \\angle ABC=90^\\circ$, and the median $CD$ on the hypotenuse $AB$ measures $5$cm, what is the length of $AB$ in cm?\n",
    "completion": "\\textbf{Solution:} \\\\\nGiven that in the right-angled triangle $ABC$, with the median $CD$ on the hypotenuse $AB$ equals $5$,\\\\\nit follows that $AB = 2CD = \\boxed{10}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52874608.png"
  },
  {
    "solution": "\\boxed{125^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AD$ is the altitude, $AE$ and $BF$ are the angle bisectors, and they intersect at point $O$. Given $\\angle ABC=50^\\circ$ and $\\angle C=70^\\circ$, find the measures of $\\angle DAE$ and $\\angle BOA$.\n",
    "completion": "\\textbf{Solution:} Since $AD \\perp BC$,\\\\\nit follows that $\\angle ADC = 90^\\circ$,\\\\\nSince $\\angle C = 70^\\circ$,\\\\\nit follows that $\\angle CAD = 180^\\circ - 90^\\circ - 70^\\circ = 20^\\circ$,\\\\\nSince $\\angle ABC = 50^\\circ$, \\\\\nit follows that $\\angle BAC = 180^\\circ - \\angle ABC - \\angle C = 60^\\circ$,\\\\\nSince AE is the bisector of $\\angle BAC$,\\\\\nit follows that $\\angle EAC = \\angle BAE = 30^\\circ$,\\\\\ntherefore $\\angle EAD = \\angle EAC - \\angle CAD = 30^\\circ - 20^\\circ = \\boxed{10^\\circ}$,\\\\\nSince BF is the bisector of $\\angle ABC$,\\\\\nit follows that $\\angle ABO = 25^\\circ$,\\\\\ntherefore $\\angle BOA = 180^\\circ - \\angle BAO - \\angle ABO = 180^\\circ - 30^\\circ - 25^\\circ = \\boxed{125^\\circ}$,\\\\\nHence, the measures of $\\angle DAE$ and $\\angle BOA$ are $10^\\circ$ and $125^\\circ$, respectively.",
    "image_path": "dataset/mm_math/images/MM_Math/52950688.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the quadrilateral $ABCD$, $\\angle D = 90^\\circ$, $\\angle B = 60^\\circ$, $AD=6$, $AB^2=\\frac{100}{3}$, $AB \\perp AC$. A point E is chosen on $CD$, and $AE$ is connected. The triangle $ADE$ is folded along $AE$ such that point D falls on point F on $AC$. Find: What is the length of $DE$?\n",
    "completion": "\\textbf{Solution:} Let $ED=x$, thus $EF=x, CE=8-x$.\\\\\nFrom the folding, we have: $AF=AD=6, CF=10-6=4$.\\\\\nIn $\\triangle CFE$ right-angled at $F$, we have $CF^2+FE^2=CE^2$,\\\\\n$\\therefore 4^2+x^2=(8-x)^2$,\\\\\nSolving this gives $x=\\boxed{3}$.\\\\\nTherefore, $DE=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52731085_119.png"
  },
  {
    "solution": "\\boxed{x<\\frac{16}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the graph, in the cartesian coordinate system, the line $L_1\\colon y=-\\frac{1}{2}x+8$ intersects the $x$-axis and $y$-axis at points $B$ and $C$, respectively, and intersects the line $L_2\\colon y=x$ at point $A$. There is a moving point $P$ on line $L_1$. A perpendicular line to the $x$-axis passing through point $P$ intersects line $L_2$ at point $N$. Find the range of values for $x$ when point $P$ is above point $N$.\n",
    "completion": "\\textbf{Solution:} Solve: The line $L_{1}\\colon y=-\\frac{1}{2}x+8$ intersects with the line $L_{2}\\colon y=x$ at point $A$,\\\\\nA perpendicular line from point $P$ to the $x$-axis intersects line $L_{2}$ at point $N$. When point $P$ is above point $N$,\\\\\nThat is: $-\\frac{1}{2}x+8>x$,\\\\\nWe find: $\\boxed{x<\\frac{16}{3}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52019348_97.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in rectangle $ABCD$, $BC>CD$, and $BC$ and $CD$ are the two roots of the quadratic equation $x^2-14x+48=0$ respectively. Connect $BD$. A moving point $P$ starts from $B$ and moves towards $D$ along $BD$ at a speed of 1 unit per second. At the same time, another moving point $Q$ starts from $D$ and moves along the ray $DA$ at the same speed. When point $P$ reaches point $D$, point $Q$ stops moving. Let the duration of the motion be $t$ seconds. Construct $\\text{Rt}\\triangle PQM$ with $PQ$ as the hypotenuse, such that point $M$ falls on segment $BD$. What is the length of segment $BD$?\n",
    "completion": "\\textbf{Solution:} Given $x^{2}-14x+48=0$,\\\\\n$\\therefore$ $(x-6)(x-8)=0$. Solving this gives $x_{1}=6$, $x_{2}=8$. \\\\\n$\\therefore$ $BC=8$, $CD=6$,\\\\\n$\\because$ Rectangle ABCD,\\\\\n$\\therefore$ $\\angle C=90^\\circ$,\\\\\n$\\therefore$ $BD=\\sqrt{BC^{2}+CD^{2}}=\\sqrt{8^{2}+6^{2}}=\\boxed{10}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52774682_57.png"
  },
  {
    "solution": "\\boxed{(-1,\\frac{84}{3})}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Comprehensive and Practice\n\nAs shown in the figure, the parabola $y=\\frac{4}{3}x^{2}+bx+c$ intersects the x-axis at points $A(-1,0)$ and $B$ (with point $B$ to the right of point $A$) and intersects the y-axis at point $C(0,-4)$. Point $D$ is on the parabola. Line $ED$ intersects line $BC$ at point $G$. When $DG=2GE$, directly write down the coordinates of point $D$.\n",
    "completion": "\\textbf{Solution:} From the construction, it is found that the parabola intersects the x-axis at $A(-1,0)$, and the vertex $E$ has coordinates $(1,-\\frac{16}{3})$,\n\n$\\therefore$ the other intersection point of the parabola with the x-axis is $B(3,0)$,\n\nAssume the equation of line $BC$ is $y=kx-4$,\n\nthus $0=3k-4$,\n\nSolving gives $k=\\frac{4}{3}$,\n\n$\\therefore$ the equation of line $BC$ is $y=\\frac{4}{3}x-4$,\n\n$\\because$ the coordinates of vertex $E$ are $(1,-\\frac{16}{3})$,\n\n$\\therefore$ the coordinates of point $M$ are $(-1,-\\frac{16}{3})$,\n\nAssume the coordinates of point $D$ are $(a, \\frac{4}{3}a^{2}-\\frac{8}{3}a-4)$, then the coordinates of point $N$ are $(a^{2}-2a, \\frac{4}{3}a^{2}-\\frac{8}{3}a-4)$,\n\nFrom the construction, it is known $ME\\parallel DN$ and $DG=2GE$,\n\n$\\therefore\\triangle MEG\\cong \\triangle NDG$,\n\nFrom the properties of similar triangles, it follows that $DN=2ME$,\n\n$ME=1-(-1)=2$, $DN= a^{2}-2a-a=a^{2}-3a$,\n\n$\\therefore a^{2}-3a=4$, i.e., $a^{2}-3a-4=0$,\n\nSolving gives: $a=4$ or $a=-1$,\n\n$\\therefore$ the coordinates of point $D$ are $\\boxed{(4,\\frac{20}{3})}$ or $\\boxed{(-1,\\frac{84}{3})}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52719833_113.png"
  },
  {
    "solution": "\\boxed{N\\left(-4, -\\frac{15}{8}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the parabola $y=-\\frac{1}{4}x^2+bx+c$ intersects the line $KB$ at point $B(2, 0)$ and $K(-12, -14)$, and intersects the y-axis at point C. Under the given conditions, the parabola is translated downwards by $\\frac{3}{8}$ units and to the right by 3 units, so that point P corresponds to point $P^{\\prime}$. Point S is a point on the axis of symmetry of the new parabola. Is there a point N in the plane such that the quadrilateral with vertices $P^{\\prime}$, S, A, N is a rhombus? If it exists, please write down the coordinates of point N directly; if it does not exist, please explain why.\n",
    "completion": "\\textbf{Solution:} Exist, $\\boxed{N\\left(-12, \\frac{949}{80}\\right)}$ or $\\boxed{N\\left(-4, -\\frac{15}{8}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53080553_31.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated, the line $y=k_1x+b$ intersects the hyperbola $y=\\frac{k_2}{x}$ at points $A$ and $B$. The line $AB$ intersects the $x$-axis at point $C$ and the $y$-axis at point $D(0, -5)$. Given that $OA=5$ and $\\tan\\angle AOC=\\frac{3}{4}$, and the $y$-coordinate of point $B$ is $-8$, what is the area of $\\triangle AOD$?\n",
    "completion": "\\textbf{Solution:} Substituting $A(-4, 3)$ and $D(0, -5)$ into $y=k_1x+b$, we obtain $b=-5$ and $-4k_1+b=3$. Solving these equations yields $k_1=-2$ and $b=-5$, \\\\\n$\\therefore$ the equation of line $AB$ is $y=-2x-5$, \\\\\nSubstituting $y=0$ into $y=-2x-5$, we get $x=-\\frac{5}{2}$, \\\\\n$\\therefore$ the coordinates of point $C$ are $\\left(-\\frac{5}{2}, 0\\right)$, and $OC=\\frac{5}{2}$, \\\\\n$\\therefore$ the area of $\\triangle AOD$ is $\\frac{1}{2}\\times \\frac{5}{2}\\times 3+\\frac{1}{2}\\times \\frac{5}{2}\\times 5=\\boxed{10}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51446358_248.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y=x+3$ denoted as ${l}_{1}$ intersects the $x$-axis at point $B$ and intersects with the graph of another linear function that passes through point $A(3, 0)$, denoted as ${l}_{2}$, at point $C(1, m)$. What is the area of $\\triangle ABC$?\n",
    "completion": "\\textbf{Solution:} Let $CD\\perp x$-axis at point $D$ passing through point $C$,\\\\\n$\\therefore B(-3, 0)$,\\\\\n$\\therefore AB=6$, $CD=4$,\\\\\n$\\therefore$ The area of $\\triangle ABC = \\frac{1}{2} \\cdot AB \\cdot CD = \\frac{1}{2} \\times 6 \\times 4 = \\boxed{12}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52250956_86.png"
  },
  {
    "solution": "\\boxed{x<1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, given that the graphs of the functions $y=x+b$ and $y=ax+3$ intersect at point $P$, what is the solution set of the inequality $x+b<ax+3$?\n",
    "completion": "\\textbf{Solution:} Since the x-coordinate of the intersection point of the two lines is: $x=1$,\\\\\nby examining the graph, it is evident that when $x<1$, $x+b<ax+3$;\\\\\ntherefore, the answer is $\\boxed{x<1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52381419.png"
  },
  {
    "solution": "\\boxed{\\frac{7}{4}\\text{cm}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, there is a rectangular piece of paper $ABCD$, where $AD=8\\text{cm}$ and $AB=6\\text{cm}$. First, fold along the diagonal $BD$, and point $C$ falls onto the position of point $C'$, with $BC'$ intersecting $AD$ at point $G$. $BG=DG$; what is the length of $C'G$ in centimeters?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle GAB \\cong \\triangle GC'D$,\\\\\n$\\therefore AG=C'G$,\\\\\nLet $C'G=x$, then $GD=BG=8-x$,\\\\\n$\\therefore x^2+6^2=(8-x)^2$,\\\\\nSolving this gives: $x=\\frac{7}{4}$,\\\\\n$\\therefore C'G=\\boxed{\\frac{7}{4}\\text{cm}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51920381_133.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC$, the perpendicular bisector of $AB$ intersects side $AB$ at point $D$ and side $AC$ at point $E$. If the perimeters of $\\triangle ABC$ and $\\triangle EBC$ are 15 and 9, respectively, what is the length of $BC$?\n",
    "completion": "\\textbf{Solution:} Since $AB=AC$, the perimeter of $\\triangle ABC$ is 15. \\\\\nTherefore, $2AC+BC=15$. \\\\\nSince $E$ is on the perpendicular bisector of $AB$, \\\\\nwe have $AE=BE$. \\\\\nSince the perimeter of $\\triangle BCE$ is $BC+CE+BE=BC+CE+AE=BC+AC=9$, \\\\\nwe find $AC=15-9=6$, \\\\\nthus $BC=9-6=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53066575.png"
  },
  {
    "solution": "\\boxed{42^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB=AD$, $AC=AE$, $BC=DE$, with point $E$ on $BC$. If $\\angle EAC=\\angle BAD$ and $\\angle EAC=42^\\circ$, what is the degree measure of $\\angle DEB$?\n",
    "completion": "\\textbf{Solution:} Since $AC=AE$, and $\\angle EAC=42^\\circ$, \\\\\ntherefore $\\angle AEC=\\angle C= \\frac{1}{2} \\times(180^\\circ-42^\\circ)=69^\\circ$. \\\\\nSince $\\triangle ABC \\cong \\triangle ADE$, \\\\\ntherefore $\\angle AED=\\angle C=69^\\circ$, \\\\\nthus $\\angle DEB=180^\\circ-\\angle AED-\\angle C=180^\\circ-69^\\circ-69^\\circ=\\boxed{42^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51368838_105.png"
  },
  {
    "solution": "\\boxed{35}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points C and D are two points on line segment AB, with $AC:BC = 3:2$, and point D is the midpoint of AB. As shown in Figure 2, if E is the midpoint of AC and $ED = 7$, find the length of line segment AB.\n",
    "completion": "\\textbf{Solution:} Let $AC=3x, BC=2x$, thus $AB=5x$, \\\\\nsince point D is the midpoint of AB.\\\\\nTherefore, $AD=\\frac{1}{2}AB=\\frac{5}{2}x$, \\\\\nsince E is the midpoint of AC, \\\\\nTherefore, $AE=\\frac{1}{2}AC=\\frac{3}{2}x$, \\\\\nTherefore, $DE=AD−AE=\\frac{5}{2}x−\\frac{3}{2}x=x$, \\\\\nsince $ED=7$, \\\\\nTherefore, $x=7$, \\\\\nTherefore, $AB=5x=\\boxed{35}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51354427_194.png"
  },
  {
    "solution": "\\boxed{70^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=CB$, $\\angle ABC=90^\\circ$, $F$ is a point on the extension of $AB$, and point $E$ is on $BC$, with $BE=BF$. $\\triangle ABE \\cong \\triangle CBF$. If $\\angle CAE = 20^\\circ$, what is the measure of $\\angle ACF$?\n",
    "completion": "\\textbf{Solution:} Given that $AB=BC$ and $\\angle ABC=90^\\circ$, it follows that $\\angle CAB=\\angle ACB=45^\\circ$. Also, given that $\\angle BAE=\\angle CAB-\\angle CAE=45^\\circ-20^\\circ=25^\\circ$ and according to (1), $\\triangle ABE\\cong \\triangle CBF$. Hence, $\\angle BCF=\\angle BAE=25^\\circ$, which means $\\angle ACF=\\angle BCF+\\angle ACB=45^\\circ+25^\\circ=\\boxed{70^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53059814_131.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, fold the rectangular paper ABCD along AE so that point D falls on point F on side BC. If $\\angle AEF = 75^\\circ$, what is the degree measure of $\\angle BAF$?\n",
    "completion": "\\textbf{Solution:} Given the problem, we have: $\\angle AFE = \\angle D = 90^\\circ$, $\\angle EAF = \\angle DAE$, $\\angle BAD = 90^\\circ$,\\\\\n$\\because \\angle AEF = 75^\\circ$,\\\\\n$\\therefore \\angle EAF = 90^\\circ - \\angle AEF = 15^\\circ$,\\\\\n$\\therefore \\angle DAE = 15^\\circ$,\\\\\n$\\therefore \\angle DAF = \\angle DAE + \\angle EAF = 30^\\circ$,\\\\\n$\\therefore \\angle BAF = \\angle BAD - \\angle DAF = \\boxed{60^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51545363.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the sidelength of square $ABCD$ is $4$. Let the point $E$ be a moving point on side $AD$, and the point $F$ be on side $CD$ with the constraint that the length of segment $EF$ is $4$. Let the point $G$ be the midpoint of segment $EF$. Connecting $BG$ and $CG$, what is the minimum value of $BG+\\frac{1}{2}CG$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, \\\\\nin right triangle $\\triangle DEF$, where $G$ is the midpoint of $EF$, \\\\\n$\\therefore DG = \\frac{1}{2}EF = 2$, \\\\\n$\\therefore$ point $G$ moves on a circle with $D$ as the center and 2 as the radius, \\\\\nCut $DI = 1$ on $CD$, and connect $GI$, \\\\\n$\\therefore \\frac{DI}{DG} = \\frac{DG}{CD} = \\frac{1}{2}$, \\\\\n$\\therefore \\angle GDI = \\angle CDG$, \\\\\n$\\therefore \\triangle GDI \\sim \\triangle CDG$, \\\\\n$\\therefore \\frac{IG}{CG} = \\frac{DI}{DG} = \\frac{1}{2}$, \\\\\n$\\therefore IG = \\frac{1}{2}CG$, \\\\\n$\\therefore BG + \\frac{1}{2}CG = BG + IG \\geq BI$, \\\\\n$\\therefore$ when $B$, $G$, and $I$ are collinear, $BG + \\frac{1}{2}CG$ is minimized $=$ $BI$, \\\\\nIn right $\\triangle BCI$, where $CI = 3$ and $BC = 4$, \\\\\n$\\therefore BI = \\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51322498.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: If point $E$ is the midpoint of $BC$, and $CD=4$, by folding the square $ABCD$ along $FH$ such that point $D$ precisely falls on the midpoint $E$ of side $BC$, and the corresponding point of point $A$ is point $P$, then what is the length of the crease $HF$?\n",
    "completion": "\\textbf{Solution:}\n\nDraw $HG \\perp CD$ at point G, and connect $DE$, as shown in the diagram:\\\\\nThen, $\\angle DGH=\\angle FGH=90^\\circ$,\\\\\nIn the square $ABCD$, $\\angle A=\\angle D=\\angle C=90^\\circ$, $AD=CD=BC=4$,\\\\\n$\\therefore$ Quadrilateral $ADGH$ is a rectangle,\\\\\n$\\therefore GH=AD=CD$,\\\\\nAccording to the folding, $FH$ is a perpendicular bisector of $DE$,\\\\\n$\\therefore \\angle FDE+\\angle DFH=90^\\circ$,\\\\\n$\\because \\angle FDE+\\angle DEC=90^\\circ$,\\\\\n$\\therefore \\angle DFH=\\angle DEC$,\\\\\nIn $\\triangle HGF$ and $\\triangle DCE$,\\\\\n$\\left\\{\\begin{array}{l}\n\\angle DFH=\\angle DEC\\\\\n\\angle FGH=\\angle ECD\\\\\nHG=CD\n\\end{array}\\right.$\\\\\n$\\therefore \\triangle HGF \\cong \\triangle DCE \\left(AAS\\right)$,\\\\\n$\\therefore FH=DE$,\\\\\n$\\because E$ is the midpoint of $BC$,\\\\\n$\\therefore EC=2$,\\\\\nIn $Rt\\triangle CDE$, according to the Pythagorean theorem, $DE=\\sqrt{4^2+2^2}=2\\sqrt{5}$,\\\\\n$\\therefore FH=\\boxed{2\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53706819.png"
  },
  {
    "solution": "\\boxed{50^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $\\angle BOC=60^\\circ$, $OE$ bisects $\\angle AOC$, and $OF$ bisects $\\angle BOC$. If $\\angle AOC + \\angle EOF = 210^\\circ$, then how many degrees is $\\angle EOF$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle BOC=60^\\circ$ and OF bisects $\\angle BOC$,\\\\\nit follows that $\\angle COF= \\frac{1}{2}\\angle BOC=30^\\circ$,\\\\\nGiven that OE bisects $\\angle AOC$,\\\\\nit follows that $\\angle AOC=2\\angle COE$,\\\\\nGiven that $\\angle AOC+\\angle EOF=210^\\circ$,\\\\\nit follows that $2\\angle COE+\\angle COE-\\angle COF=210^\\circ$,\\\\\nwhich simplifies to $3\\angle COE-30^\\circ=210^\\circ$,\\\\\nand therefore $\\angle COE=80^\\circ$,\\\\\nconsequently, $\\angle EOF=\\angle COE-\\angle COF=\\boxed{50^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55190308_105.png"
  },
  {
    "solution": "\\boxed{25^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $D$ is a point on side $AC$ of $\\triangle ABC$. An arc is drawn with $A$ as the center and $AD$ as the radius, intersecting the extension line of $BA$ at point $E$, and $ED$ is connected. If $\\angle B=60^\\circ$ and $\\angle C=70^\\circ$, what is the degree measure of $\\angle ADE$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle B=60^\\circ$ and $\\angle C=70^\\circ$,\\\\\nit follows that $\\angle CAB=180^\\circ-\\angle B-\\angle C=50^\\circ$,\\\\\nsince an arc is drawn with A as the center and AD as the radius, intersecting the extension of BA at point E,\\\\\nit implies AE=AD,\\\\\nthus $\\angle AED=\\angle ADE$,\\\\\nsince $\\angle AED+\\angle ADE=\\angle CAB$,\\\\\ntherefore $\\angle ADE+\\angle ADE=50^\\circ$,\\\\\nwhich gives $\\angle ADE=\\boxed{25^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52603249.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{1}{2}(x-2)^2+8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, water is sprayed out from window A of a building, forming a parabola (the plane of the parabola is perpendicular to the wall). The height of point A from the ground is 6 meters, the vertical distance of the highest point P from the wall is 2 meters, and its vertical distance from the ground is 8 meters. A Cartesian coordinate system is established as shown in the figure. What is the analytical equation of the parabola?\n",
    "completion": "Solution: From the given information, we know that the vertex of the parabola is at $(2,8)$, and it passes through point $A(0,6)$,\\\\\n$\\therefore$let the equation of the parabola be $y=a(x-2)^2+8$,\\\\\nSubstituting $A(0,6)$ into the equation, we get $4a+8=6$,\\\\\nSolving for $a$, we find $a=-\\frac{1}{2}$,\\\\\n$\\therefore$the equation of the parabola is $y=-\\frac{1}{2}(x-2)^2+8$,\\\\\nHence, the equation of the parabola is $\\boxed{y=-\\frac{1}{2}(x-2)^2+8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51446099_264.png"
  },
  {
    "solution": "\\boxed{(6,7)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that in the Cartesian coordinate system $xOy$, the graph of the linear function $y=kx+b(k\\ne 0)$ passes through points A and $B(-1,0)$, the graph of the inverse proportion function $y=\\frac{6}{x}$ also passes through point A, and the abscissa of point A is 2. Point C is a point on the positive half of the x-axis, connect $AC$, $\\tan\\angle ACB=\\frac{3}{4}$. A line through point C is drawn perpendicular to the x-axis and intersects the graphs of the inverse proportion function $y=\\frac{6}{x}$ and the linear function $y=kx+b(k\\ne 0)$ at points D and E, respectively. Find the coordinates of points D and E.\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $AH \\perp x$-axis at point H.\n\n$\\therefore \\tan\\angle ACB = \\frac{AH}{CH} = \\frac{3}{4}$.\n\n$\\because A(2,3)$,\n\n$\\therefore AH=3$, $OH=2$\n\n$\\therefore CH=4$,\n\n$\\therefore OC=OH+CH=6$,\n\n$\\therefore C(6,0)$.\n\n$\\because x_C=x_D=x_E=6$.\n\n$\\therefore y_D=\\frac{6}{x_D}=1$, $y_E=x_E+1=7$,\n\n$\\therefore D(6,1)$, $E(6,7)$;\n\nHence, the coordinates of points D and E are $D=\\boxed{(6,1)}$, $E=\\boxed{(6,7)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52917576_250.png"
  },
  {
    "solution": "\\boxed{y=-2x+1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the hyperbola $y=\\frac{k}{x}$ ($k\\ne 0$) and the line $y=mx+n$ intersect at points $A$ and $B$, with $B$ having coordinates $(2, -3)$. The line $AC$ is perpendicular to the $y$-axis at point $C$, and $AC=\\frac{3}{2}$. What are the analytical equations of the hyperbola and the line?\n",
    "completion": "\\textbf{Solution:}\n\nSince $B$ lies on the hyperbola $y=\\frac{k}{x}$ ($k\\ne 0$), and the coordinates of point $B$ are $(2, -3)$,\\\\\nhence $k=-6$,\\\\\ntherefore the equation of the hyperbola is: $y=-\\frac{6}{x}$.\\\\\nSince $AC$ is perpendicular to the y-axis at point $C$, and $AC=\\frac{3}{2}$,\\\\\nthus the x-coordinate of point $A$ is $-\\frac{3}{2}$,\\\\\nsubstituting into $y=-\\frac{6}{x}$ gives $y=-\\frac{6}{-\\frac{3}{2}}=4$,\\\\\nhence $A\\left(-\\frac{3}{2}, 4\\right)$,\\\\\nsubstituting the coordinates of A and B into $y=mx+n$, we obtain $\\begin{cases}-\\frac{3}{2}m+n=4\\\\ 2m+n=-3\\end{cases}$,\\\\\nsolving this yields $\\begin{cases}m=-2\\\\ n=1\\end{cases}$,\\\\\ntherefore, the equation of line $AB$ is $\\boxed{y=-2x+1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52622054_224.png"
  },
  {
    "solution": "\\boxed{2.4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, there is a triangular paper ABC, with point D lying on side BC. Connect AD and fold $\\triangle ABD$ along AD to obtain $\\triangle AED$, where DE intersects AC at point F. If point F is the midpoint of DE, $AD=9$, $EF=2.5$, and the area of $\\triangle AEF$ is 9, then what is the distance from point F to BC?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect BE, intersecting AD at point O. Through point E, draw $EH\\perp BC$ at point H, and from point F, draw $FG\\perp BC$ at point G,\\\\\nfrom the folding, it is known that AB=AE, $\\angle BAO=\\angle EAO$, BD=DE,\\\\\nfurthermore $\\because$AO=AO,\\\\\n$\\therefore$ $\\triangle BAO\\cong \\triangle EAO(SAS)$,\\\\\n$\\therefore$BO=EO, $\\angle BOA=\\angle EOA$,\\\\\n$\\therefore$ $AD\\perp BE$.\\\\\n$\\because$ point F is the midpoint of DE, EF=2.5,\\\\\n$\\therefore$DF=EF=2.5, BD=DE=5,\\\\\n$\\therefore$ $\\triangle ADF$ and $\\triangle AEF$ have the same base and height,\\\\\n$\\therefore$ ${S}_{\\triangle ADE}={S}_{\\triangle AEF}+{S}_{\\triangle ADF}=18$.\\\\\n$\\because$ ${S}_{\\triangle ADE}=\\frac{1}{2}AD\\cdot OE$,\\\\\n$\\therefore$ $\\frac{1}{2}\\times 9\\times OE=18$,\\\\\nsolving gives: $EO=4$.\\\\\n$\\therefore$ in $Rt\\triangle ODE$, $OD=\\sqrt{DE^{2}-OE^{2}}=\\sqrt{5^{2}-4^{2}}=3$,\\\\\n$\\because$ $BE=OB+OE=8$.\\\\\n$\\therefore$ ${S}_{\\triangle BDE}=\\frac{1}{2}BE\\cdot OD=\\frac{1}{2}\\times 8\\times 3=12$.\\\\\nMoreover, since ${S}_{\\triangle BDE}=\\frac{1}{2}BD\\cdot EH=\\frac{1}{2}\\times 5\\times EH$,\\\\\n$\\therefore$ $\\frac{1}{2}\\times 5\\times EH=12$,\\\\\nsolving gives: $EH=4.8$.\\\\\n$\\because$ point F is the midpoint of DE, $EH\\perp BC$, $FG\\perp BC$,\\\\\n$\\therefore$FG is the midline of $\\triangle DEH$,\\\\\n$\\therefore$ $FG=\\frac{1}{2}EH=\\boxed{2.4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51546401.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ is folded along the line $AD$ so that point $B$ coincides with point $E$ on side $AC$. If $AB=AD=4$ and $AC=6$, what is the length of $DC$?\n",
    "completion": "\\textbf{Solution:} Due to the property of folding, we have $\\angle BAD=\\angle DAE$, $\\angle ADB-\\angle ADE$, $AB=AE=4$,\\\\\nsince $AB=AD=4$,\\\\\nthus $\\angle B=\\angle ADB$,\\\\\nlet $\\angle B=\\angle ADB=x$,\\\\\nin $\\triangle ABD$, we have $\\angle BAD=180^\\circ-2x$,\\\\\nsince $\\angle EDC=180^\\circ-\\angle ADB-\\angle ADE=180^\\circ-2x$,\\\\\nthus $\\angle BAD=\\angle EDC$,\\\\\nsince $\\angle BAD=\\angle DAE$,\\\\\nthus $\\angle DAE=\\angle EDC$,\\\\\nsince $\\angle C=\\angle C$,\\\\\nthus $\\triangle DCE\\sim \\triangle ACD$,\\\\\ntherefore, $\\frac{EC}{DC}=\\frac{DC}{AC}=\\frac{6-4}{DC}=\\frac{DC}{6}$,\\\\\nhence, $DC=\\boxed{2\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51433090.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{5}}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In rectangle $ABCD$, where $E$ is a point on side $AB$, fold $\\triangle ADE$ along $DE$ such that the corresponding point of $A$, point $F$, falls exactly on side $BC$. Connect $AF$ and intersect $DE$ at point $N$, then connect $BN$. Given $AD=3$ and $BF=\\frac{1}{3}BC$, find the value of $\\sin\\angle BNF$.\n",
    "completion": "\\textbf{Solution:} We have $\\sin\\angle BNF = \\boxed{\\frac{\\sqrt{5}}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51318132_41.png"
  },
  {
    "solution": "\\boxed{\\frac{8}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the side length of rhombus $ABCD$ is 4, $E$ is the midpoint of $BC$, $AF$ bisects $\\angle EAD$ and intersects $CD$ at point $F$, $FG \\parallel AD$ and intersects $AE$ at point $G$. If $AB=AE$, what is the length of $FG$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, we draw $AH$ perpendicular to $BC$ at H, extend $AE$ and $DC$ to meet at point M,\\\\\nsince $AB=AE$,\\\\\ntherefore $BH=EH$, $\\angle ABE=\\angle AEB$,\\\\\nsince the side length of the rhombus $ABCD$ is 4,\\\\\ntherefore $AB=AE=AD=BC=4$, $AB\\parallel DC$,\\\\\nsince $E$ is the midpoint of $BC$,\\\\\ntherefore $BE=CE=2$,\\\\\ntherefore $BH=EH=\\frac{1}{2}BE=1$,\\\\\nsince $AB\\parallel DC$,\\\\\ntherefore $\\angle B=\\angle MCE$,\\\\\nsince $\\angle AEB=\\angle MEC$,\\\\\ntherefore $\\angle MEC=\\angle MCE$,\\\\\ntherefore $AE=AB=EM=CM=4$,\\\\\nsince $FG\\parallel AD$,\\\\\ntherefore $\\angle DAF=\\angle AFG$,\\\\\nsince $AF$ bisects $\\angle EAD$,\\\\\ntherefore $\\angle GAF=\\angle DAF$,\\\\\ntherefore $\\angle GAF=\\angle AFG$,\\\\\ntherefore $AG=GF$,\\\\\nlet $GF=x$,\\\\\nthen $AG=x$, $GE=4-x$, $MG=GE+EM=8-x$,\\\\\nsince $GF\\parallel AD\\parallel BC$,\\\\\ntherefore $\\triangle MGF\\sim \\triangle MEC$,\\\\\ntherefore $\\frac{EC}{FG}=\\frac{EM}{MG}$,\\\\\ntherefore $\\frac{2}{x}=\\frac{4}{8-x}$,\\\\\nsolving for $x$ yields $x=\\boxed{\\frac{8}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52866966.png"
  },
  {
    "solution": "\\boxed{116^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, quadrilateral $ABCD$ is inscribed in $\\circ O$, and $\\angle CBE$ is one of its exterior angles. If $\\angle CBE=58^\\circ$, what is the degree measure of $\\angle AOC$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral $ABCD$ is inscribed in circle $O$,\\\\\nit follows that $\\angle ADC+\\angle ABC=180^\\circ$,\\\\\nsince $\\angle CBE+\\angle ABC=180^\\circ$ and $\\angle CBE=58^\\circ$,\\\\\nthen $\\angle ADC=\\angle CBE=58^\\circ$,\\\\\nBy the Inscribed Angle Theorem, $\\angle AOC=2\\angle ADC=\\boxed{116^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52916763.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, D, E, and F are the midpoints of BC, AC, and AB, respectively. $AH \\perp BC$ at H, and $HE=8$. What is the length of DF?\n",
    "completion": "\\textbf{Solution:} Given that $AH \\perp BC$ and E is the midpoint of side AC, \\\\\n$\\therefore$ AC = 2HE = 16, \\\\\nGiven that D and F are the midpoints of sides BC and AB respectively, \\\\\n$\\therefore$ DF = $\\frac{1}{2}$ AC = \\boxed{8}",
    "image_path": "dataset/mm_math/images/MM_Math/50496860.png"
  },
  {
    "solution": "\\boxed{(0,2.5)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the quadrilateral $OABC$ is a rectangular sheet placed in the Cartesian coordinate system, where $O$ is the origin, point $A$ is on the positive semi-axis of the $x$-axis, and point $C$ is on the positive semi-axis of the $y$-axis, with $OA=5$; $OC=4$. A point $D$ is taken on side $OC$, and the sheet is folded along $AD$ so that point $O$ falls on point $E$ on side $BC$. What are the coordinates of point $D$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral OABC is a rectangle, with OA=5 and OC=4.\n\nHence, $OA=BC=5$, $OC=AB=4$,\n\n$\\angle AOC=\\angle B=\\angle OCB=90^\\circ$,\n\nBy the property of folding, we get: OA=AE=5, OD=DE,\n\nIn $\\triangle ABE$, $BE=\\sqrt{AE^2-AB^2}=3$,\n\nTherefore, $CE=BC-BE=2$,\n\nLet D$(0,x)$, then OD=DE=$x$, CD=$4-x$,\n\nTherefore, in $\\triangle DCE$, by the Pythagorean theorem, we get: $(4-x)^2+2^2=x^2$,\n\nSolving this gives: $x=\\frac{5}{2}$,\n\nHence, $D\\boxed{(0,2.5)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51604337.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rhombus $ABCD$, the diagonals intersect at point $O$, with $DE \\parallel AC$ and $CE \\parallel BD$. Quadrilateral $OCED$ is a rectangle; if $AD = 5$ and $BD = 8$, calculate the value of $DE$.\n",
    "completion": "Solution: Since quadrilateral ABCD is a rhombus, and $BD=8$, \\\\\ntherefore, $OD=\\frac{1}{2}BD=4$, $OC=OA$, and $AD=CD$, \\\\\nsince $AD=5$, \\\\\nthus, $OC=AO=\\sqrt{AD^{2}-OD^{2}}=3$, \\\\\nsince quadrilateral OCED is a rectangle, \\\\\ntherefore, $DE=OC=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52049906_111.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, AB is a straight line, and OC is the bisector of $\\angle AOD$. As shown in Figure (2), if the ratio $\\angle DOE : \\angle BOD = 2 : 5$ and $\\angle COE = 80^\\circ$, find the degree measure of $\\angle BOE$.\n",
    "completion": "\\textbf{Solution:} Let $\\angle COD=x^\\circ$, $\\angle DOE=y^\\circ$,\\\\\n$\\therefore x+y=80$,\\\\\n$\\because$ OC is the bisector of $\\angle AOD$,\\\\\n$\\therefore \\angle AOD=2\\angle COD=2x^\\circ$,\\\\\n$\\because \\angle DOE\\colon \\angle BOD=2\\colon 5$,\\\\\n$\\therefore \\angle BOD= \\frac{5}{2}\\angle DOE=\\frac{5}{2}y^\\circ$,\\\\\n$\\because \\angle AOD+\\angle BOD=180^\\circ$,\\\\\n$\\therefore 2x+\\frac{5}{2}y=180$,\\\\\n$\\therefore$ solving the above equations we get $y=40$,\\\\\n$\\therefore x=80-40=40$,\\\\\n$\\therefore \\angle AOD=2x^\\circ=80^\\circ$,\\\\\n$\\therefore \\angle BOD=180^\\circ-80^\\circ=100^\\circ$,\\\\\n$\\therefore \\angle BOE=\\angle BOD-\\angle DOE=100^\\circ-40^\\circ=\\boxed{60^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51508770_137.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ and $\\triangle A^{\\prime}B^{\\prime}C^{\\prime}$ are similar, with the center of similarity at point O, and $\\frac{A^{\\prime}C^{\\prime}}{AC}=\\frac{2}{3}$. If the area of $\\triangle ABC$ is 9, what is the area of $\\triangle A^{\\prime}B^{\\prime}C^{\\prime}$?\n",
    "completion": "\\textbf{Solution}: Since $\\triangle ABC$ and $\\triangle A^{\\prime}B^{\\prime}C^{\\prime}$ are similar, with the center of similarity being point O, and $\\frac{A'C'}{AC}=\\frac{2}{3}$,\\\\\nit follows that $\\frac{S_{\\triangle A^{\\prime}B^{\\prime}C^{\\prime}}}{S_{\\triangle ABC}}=\\left(\\frac{A^{\\prime}C^{\\prime}}{AC}\\right)^2=\\left(\\frac{2}{3}\\right)^2=\\frac{4}{9}$,\\\\\nSince the area of $\\triangle ABC$ is 9,\\\\\ntherefore, the area of $\\triangle A^{\\prime}B^{\\prime}C^{\\prime}$ is $\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51918244.png"
  },
  {
    "solution": "\\boxed{\\frac{21}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the side length of the square $ABCD$ is $8$, $E$ is a point on side $BC$, and $BE=2.5$. $F$ is a moving point on side $AB$ connected with $EF$. An equilateral triangle $\\triangle EFG$ is constructed to the right side with $EF$ as one of its sides. Connect $CG$. What is the minimum value of $CG$?\n",
    "completion": "\\textbf{Solution:} It is known from the problem that point $F$ is the driving point, and point $G$ is the driven point. Point $F$ moves on a line segment, thus point $G$ must also move on a straight line trajectory.\\\\\nRotate $\\triangle EFB$ about point $E$ by $60^\\circ$ so that $EF$ coincides with $EG$, obtaining $\\triangle EFB\\cong \\triangle EHG$,\\\\\nTherefore, it can be inferred that $\\triangle EBH$ is an equilateral triangle, and point $G$ is on line $HN$, which is perpendicular to $HE$,\\\\\nDraw $CM\\perp HN$ through point $C$, then $CM$ is the minimum value of $CG$,\\\\\nDraw $EP\\perp CM$ through point $E$, it can be seen that quadrilateral $HEPM$ forms a rectangle,\\\\\nThus $CM=MP+CP=HE+\\frac{1}{2}EC=2.5+\\frac{11}{4}=\\boxed{\\frac{21}{4}}$,",
    "image_path": "dataset/mm_math/images/MM_Math/50496337.png"
  },
  {
    "solution": "\\boxed{\\frac{4}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, in $\\triangle ABC$, $\\angle BCA=90^\\circ$, $AC=3$, and $BC=4$. Point $P$ is located on side $AB$, and a line is drawn through point $P$ such that $PD\\perp AC$, intersecting $AC$ and $BC$ at points $D$ and $E$, respectively. If $P$ is the midpoint of $AB$, with $PD\\perp AC$ and $PE\\perp BC$, what is the value of $\\frac{PD}{PE}$?\n",
    "completion": "\\textbf{Solution:} According to the given conditions, we have $ \\frac{PD}{PE}=\\boxed{\\frac{4}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51446440_75.png"
  },
  {
    "solution": "\\boxed{80^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $P$ is a point inside $\\angle AOB$. The symmetric point of $P$ with respect to $OA$ is $C$, and the symmetric point of $P$ with respect to $OB$ is $D$. Connect $CD$ intersecting $OA$, $OB$ at points $M$ and $N$ respectively, and then connect $PM$, $PN$. If $\\angle AOB=50^\\circ$, what is the measure of $\\angle MPN$ in degrees?\n",
    "completion": "\\textbf{Solution:} Connect PC, PM, PD, PN. \\\\\n$\\because$ Point P is symmetrical to point C with respect to line OA, and to point D with respect to line OB, \\\\\n$\\therefore OC=OP=OD$, $\\angle AOP=\\angle AOC$, $\\angle POB=\\angle BOD$, \\\\\n$\\because \\angle AOB=50^\\circ$, \\\\\n$\\therefore \\angle COD=2\\angle AOB=100^\\circ$. $\\because OC=OP$, $\\angle AOP=\\angle AOC$, $OM=OM$ $\\therefore \\triangle OCM\\cong \\triangle OPM$, $\\therefore \\angle OCM=\\angle OPM$, similarly $\\angle ODN=\\angle OPN$ \\\\\n$\\therefore \\angle MPN=\\angle OPM+\\angle OPN=\\angle OCM+\\angle ODN=180^\\circ−2\\angle AOB=\\boxed{80^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52457275.png"
  },
  {
    "solution": "\\boxed{(-\\infty, -4)\\cup(0, 2)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y=kx+2$ intersects the x-axis and the y-axis at points $A$ and $B$, respectively. It also intersects the graph of the inverse proportion function $y=\\frac{n}{x} (n\\ne 0)$ at point $C$ in the first quadrant. If $OA=OB$ and $B$ is the midpoint of the line segment $AC$. Directly write down the solution set of the inequality $kx+2<\\frac{n}{x}$.\n",
    "completion": "\\textbf{Solution:} The solution set of $x<-4$ or $0<x<2$ is $x\\in(-\\infty, -4)\\cup(0, 2)$. Therefore, the solution set for the inequality $kx+2<\\frac{n}{x}$ is $x\\in\\boxed{(-\\infty, -4)\\cup(0, 2)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53065570_75.png"
  },
  {
    "solution": "\\boxed{\\frac{27}{8}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the Cartesian coordinate system, the graph of the quadratic function $y=ax^2+bx+c\\ (a\\ne 0)$ intersects the x-axis at points $A(-3,0)$ and $B(1,0)$, and intersects the y-axis at point $C(0, -3)$. The vertex is denoted as $D$, and its axis of symmetry intersects the x-axis at point $E$. Let point $P$ be a point on the parabola in the third quadrant. The area of $\\triangle APC$ is denoted as $S$. Find the maximum value of $S$ and the coordinates of point $P$ at this time?\n",
    "completion": "\\textbf{Solution:} Suppose the equation of line $AC$ is: $y=kx+m$,\\\\\nsince $A(-3,0)$, $B(0, -3)$,\\\\\nthen $\\left\\{\\begin{array}{l}\n-3k+m=0\\\\\nm=-3\n\\end{array}\\right.$,\\\\\nthus $k=-1$, $m=-3$,\\\\\ntherefore $y=-x-3$.\\\\\nDraw a perpendicular line from point $P$ to the x-axis intersecting $AC$ at $G$, let $P\\left(x, -x-3\\right)$,\\\\\nthen $G\\left(x, -x-3\\right)$.\\\\\nSince point $P$ is in the third quadrant,\\\\\ntherefore $PG=-x-3-\\left(-x^2+2x-3\\right)$\\\\\n$=-x-3-x^2-2x+3$\\\\\n$=-x^2-3x$,\\\\\nthus the area of $\\triangle APC=\\frac{1}{2}PG\\cdot OA$\\\\\n$=\\frac{1}{2}\\left(-x^2-3x\\right)\\times 3$\\\\\n$=-\\frac{3}{2}x^2-\\frac{9}{2}x$\\\\\n$=-\\frac{3}{2}\\left(x+\\frac{3}{2}\\right)^2+\\frac{27}{8}$.\\\\\nTherefore, when $x=-\\frac{3}{2}$, the maximum area $S_{\\text{max}}=\\boxed{\\frac{27}{8}}$, at this point $P\\left(-\\frac{3}{2}, -\\frac{15}{4}\\right)$.",
    "image_path": "dataset/mm_math/images/MM_Math/53147831_65.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, two lines are intersected by three parallel lines, with $DE = 4$, $EF = 6$, and $AB = 2$. What is the value of $AC$?\n",
    "completion": "\\textbf{Solution:} Given: $l_1 \\parallel l_2$,\\\\\nTherefore, $\\frac{DE}{EF}=\\frac{AB}{BC}$,\\\\\nTherefore, $\\frac{4}{6}=\\frac{2}{BC}$,\\\\\nTherefore, $BC=3$,\\\\\nTherefore, $AC=AB+BC=2+3=\\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51368178.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\angle AOB = 30^\\circ$, point $P$ is located inside $\\angle AOB$ with $OP = 3$. Points $M$ and $N$ are moving points on the rays $OA$ and $OB$, respectively. What is the minimum perimeter of $\\triangle PMN$ when its perimeter is at its minimum?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, construct the symmetric points $E$ and $D$ of point $P$ with respect to $OA$ and $OB$, respectively, and connect $DN$, $EM$, $DE$, $OD$, $OE$,\\\\\nthen $OA$ perpendicularly bisects $PE$ and $OB$ perpendicularly bisects $PD$,\\\\\n$\\therefore EM=PM$, $DN=PN$, $OD=OP=OE=3$,\\\\\n$\\therefore$ the perimeter of $\\triangle PMN$ is $PM+MN+PN=EM+MN+DN$,\\\\\nKnowing that the length of the segment between two points is the shortest, when points $E$, $M$, $N$, $D$ are collinear, the value of $EM+MN+DN$ is minimal, and the minimum value is equal to the length of $DE$,\\\\\n$\\because OE=OP$, $OA\\perp PE$,\\\\\n$\\therefore \\angle AOE=\\angle AOP$ (the three lines of an isosceles triangle coincide),\\\\\nSimilarly, it can be derived: $\\angle BOD=\\angle BOP$,\\\\\n$\\because \\angle AOB=30^\\circ$,\\\\\n$\\therefore \\angle BOP+\\angle AOP=30^\\circ$,\\\\\n$\\therefore \\angle DOE=\\angle BOD+\\angle BOP+\\angle AOP+\\angle AOE=2\\left(\\angle BOP+\\angle AOP\\right)=60^\\circ$,\\\\\nAlso, $\\because OD=OE=3$,\\\\\n$\\therefore \\triangle DOE$ is an equilateral triangle,\\\\\n$\\therefore DE=OD=3$,\\\\\n$\\therefore$ the minimum value of $EM+MN+DN$ is 3,\\\\\n$\\therefore$ the minimum perimeter of $\\triangle PMN$ is $\\boxed{3}$,\\\\",
    "image_path": "dataset/mm_math/images/MM_Math/53058551.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, if $AB$ and $AC$ are respectively one side of an inscribed equilateral triangle and one side of an inscribed square of circle $O$, and if $BC$ is one side of an inscribed regular $n$-sided polygon of circle $O$, then what is the value of $n$?\n",
    "completion": "\\textbf{Solution:} Draw lines OA, OC, and OB. \\\\\nSince AB and AC are respectively one side of a regular triangle and a regular square inscribed in circle O, \\\\\ntherefore, $\\angle$AOC = $360^\\circ \\div 4 = 90^\\circ$, $\\angle$AOB = $360^\\circ \\div 3 = 120^\\circ$, \\\\\ntherefore, $\\angle$BOC = $\\angle$AOB - $\\angle$AOC = $120^\\circ - 90^\\circ = 30^\\circ$, \\\\\ntherefore, n = $360^\\circ \\div 30^\\circ = \\boxed{12}$, \\\\\ntherefore, BC is definitely one side of a regular dodecagon inscribed in circle O.",
    "image_path": "dataset/mm_math/images/MM_Math/52916624.png"
  },
  {
    "solution": "\\boxed{10\\,\\text{cm}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $C$ is the midpoint of line segment $AB$, and point $D$ divides line segment $AB$ into two parts, $AD$ and $DB$, in the ratio of $3\\colon 2$. If $CD=1\\text{cm}$, what is the length of line segment $AB$?\n",
    "completion": "\\textbf{Solution:} Let $AD=3x\\,(\\text{cm})$, and $BD=2x\\,(\\text{cm})$, thus $AB=5x\\,(\\text{cm})$ \\\\\nSince $C$ is the midpoint of $AB$, \\\\\nThen $AC=BC=\\frac{5}{2}x\\,(\\text{cm})$ \\\\\nTherefore $CD=BC-BD=\\frac{1}{2}x\\,(\\text{cm})$ \\\\\nSince $CD=1\\,\\text{cm}$, \\\\\nThen $\\frac{1}{2}x=1$, \\\\\nTherefore $x=2\\,\\text{cm}$, \\\\\nTherefore $AB=5x=\\boxed{10\\,\\text{cm}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/54565727_211.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the inverse proportion function $y=\\frac{k}{x}$ $(x>0)$ passes through points A and B. Draw $AC\\perp y$ axis at point C through A, and draw $BD\\perp y$ axis at point D through B. Also, draw $BE\\perp x$ axis at point E through B, and connect AD. It is known that $AC=1$, $BE=1$, and the area of rectangle $BDOE$ is $4$. Then, what is the area of $\\triangle ACD$?\n",
    "completion": "\\textbf{Solution:} Construct AH $\\perp$ x-axis at point H, intersecting BD at point F, \\\\\n$\\therefore$ Quadrilateral ACOH, quadrilateral ACDF, quadrilateral ODBE are all rectangles, \\\\\n$\\because$ The area of rectangle BDOE is 4, and the inverse proportion function $y=\\frac{k}{x}$ (x＞0) passes through point B, with k＞0, \\\\\n$\\therefore$k＝4 \\\\\n$\\therefore$The area of rectangle ACOH is 4, \\\\\n$\\because$AC＝1 \\\\\n$\\therefore$OC＝4÷1＝4, \\\\\n$\\therefore$CD＝OC−OD＝OC−BE＝4−1＝3, \\\\\n$\\therefore$The area of rectangle ACDF＝1×3＝3, \\\\\n$\\therefore$${S}_{\\triangle ACD}=\\frac{1}{2}{S}_{\\text{rectangle }CDFA}=\\boxed{\\frac{3}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51615108.png"
  },
  {
    "solution": "\\boxed{26^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, it is known that $AB$ is the diameter of the circle $\\odot O$, point $C$ lies on circle $\\odot O$, and the tangent line to $\\odot O$ passing through point $C$ intersects the extended line of $AB$ at point $P$. Point $D$ is a point on arc $AC$. Connect $BD$, $BC$, and $DC$. If $\\angle D = 26^\\circ$, what is the measure of $\\angle PCB$?\n",
    "completion": "\\textbf{Solution:} Let's connect CO,\\\\\n$\\because$ $\\angle D=26^\\circ$,\\\\\n$\\therefore$ $\\angle COB=52^\\circ$,\\\\\n$\\because$ $OB=OC$,\\\\\n$\\therefore$ $\\angle BCO=\\frac{180^\\circ-\\angle COB}{2}=64^\\circ$,\\\\\n$\\because$ PC is tangent to $\\odot O$,\\\\\n$\\therefore$ $\\angle PCO=90^\\circ$,\\\\\n$\\therefore$ $\\angle PCB=\\angle PCO-\\angle BCO=26^\\circ$;\\\\\nHence, the measure of $\\angle PCB$ is $\\boxed{26^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51319037_127.png"
  },
  {
    "solution": "\\boxed{\\frac{25}{8}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In $\\triangle ABC$, $AC=BC=10$, and $\\tan A = \\frac{3}{4}$. Points $D$ and $E$ are moving points on sides $AB$ and $AC$, respectively. Connect $DE$ and construct the triangle $\\triangle A'DE$ symmetric to $\\triangle ADE$ with respect to $DE$. As shown in figure 2, $E$ is the midpoint of side $AC$. When $\\triangle A'DE$ is an isosceles triangle, what is the length of $AD$?\n",
    "completion": "\\textbf{Solution:} Given that $\\triangle ADE$ and $\\triangle A'DE$ are symmetric about $DE$,\\\\\nthen to find the length of $AD$ when $\\triangle A'DE$ is an isosceles triangle is equivalent to finding the length of $AD$ when $\\triangle ADE$ is an isosceles triangle,\\\\\nsince $E$ is the midpoint of side $AC$,\\\\\nthen $AE=5$.\\\\\n(1) When $AE=AD$,\\\\\n$AD=AE=5$;\\\\\n(2) When $AE=ED$,\\\\\nas shown in the diagram, draw $EM \\perp AD$ intersecting $AD$ at $M$,\\\\\nsince $AE=ED$ and $EM \\perp AD$,\\\\\nthen $EM$ is the perpendicular bisector of $AD$,\\\\\nsince $\\tan A=\\frac{3}{4}$ and $AE=5$,\\\\\nthen $AM=4$,\\\\\nthus $AD=8$;\\\\\n(3) When $AD=ED$,\\\\\nas shown in the diagram, draw $DN \\perp AE$ intersecting $AE$ at $N$,\\\\\nsince $DN \\perp AE$ and $AD=ED$,\\\\\nthen $DN$ is the perpendicular bisector of $AE$,\\\\\nthus $AN=\\frac{5}{2}$,\\\\\nsince $\\tan A=\\frac{3}{4}$,\\\\\nthen $AD=\\frac{25}{8}$;\\\\\ntherefore, the length of $AD$ is $5$, $8$, or \\boxed{\\frac{25}{8}}.",
    "image_path": "dataset/mm_math/images/MM_Math/53080186_105.png"
  },
  {
    "solution": "\\boxed{{F}_{1}\\left(-3, 0\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, quadrilateral $ABCD$ is a parallelogram with $AD=6$. If the lengths of $OA$ and $OB$ are the two roots of the quadratic equation in $x$, ${x}^{2}-7x+12=0$, and $OA>OB$, and if point M is within the Cartesian coordinate system, then is there a point F on line $AB$ such that the quadrilateral with vertices A, C, F, and M, with $AC$ and $AF$ as adjacent sides, forms a rhombus? If such a point exists, provide the coordinates of point F. If it does not exist, explain why.\n",
    "completion": "\\textbf{Solution:} According to the calculated data, $OB=OC=3$, $\\because$ $AO\\perp BC$, $\\therefore$ $AO$ bisects $\\angle BAC$. Consider two scenarios:\n- (1) If $AC$ and $AF$ are adjacent sides, and point F is on the ray $AB$, then $AF=AC=5$, $\\therefore$ point F coincides with B, that is $F\\left(-3, 0\\right)$;\n- (2) If $AC$ and $AF$ are adjacent sides, and point F is on the ray $BA$, then M should be on line $AD$, and $FC$ perpendicularly bisects $AM$. Given $A\\left(0, 4\\right)$, $D\\left(6, 4\\right)$, $C\\left(3, 0\\right)$, $FC=8$, ${x}_{C}={x}_{F}=3$, $\\therefore$ in this case, the coordinates of point F are $\\left(3, 8\\right)$.\n\nIn conclusion, there are two points that meet the conditions: \\boxed{{F}_{1}\\left(-3, 0\\right)}; \\boxed{{F}_{2}\\left(3, 8\\right)}.",
    "image_path": "dataset/mm_math/images/MM_Math/51458229_161.png"
  },
  {
    "solution": "\\boxed{75^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a right-angle triangle plate is stacked in the manner shown. Where $ \\angle A=45^\\circ$ and $ \\angle D=30^\\circ$. If $ DF\\parallel BC$, then what is the degree measure of $ \\angle AGE$?\n",
    "completion": "\\textbf{Solution}: Since $\\triangle ABC$ and $\\triangle DFE$ are both right-angled triangles, with $\\angle A=45^\\circ$ and $\\angle D=30^\\circ$,\\\\\nthus $\\angle B=45^\\circ$ and $\\angle DFE=90^\\circ-30^\\circ=60^\\circ$,\\\\\nsince $DF\\parallel BC$ and $\\angle D=30^\\circ$,\\\\\nthus $\\angle FEC=\\angle DFE=60^\\circ$,\\\\\nthus $\\angle GEB=90^\\circ-\\angle DFE=90^\\circ-60^\\circ=30^\\circ$,\\\\\nthus $\\angle AGE=\\angle GEB+\\angle B=30^\\circ+45^\\circ=\\boxed{75^\\circ}$,",
    "image_path": "dataset/mm_math/images/MM_Math/51546405.png"
  },
  {
    "solution": "\\boxed{(2, 3)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=-\\frac{1}{2}x^2+bx+c$ intersects the x-axis at points $A$ and $B(4, 0)$, and intersects the y-axis at point $C$. The line $y=-\\frac{1}{2}x+n$ passes through points $B$ and $C$. Point $D$ is a moving point on line $BC$. A perpendicular line $DQ$ is drawn from point $D$ to the x-axis with $Q$ as the foot of the perpendicular, and it intersects the parabola at point $P$. When point $P$ is located above line $BC$ and the area of $\\triangle PBC$ is the maximum, what are the coordinates of $P$?\n",
    "completion": "\\textbf{Solution:} Let $P\\left(t, -\\frac{1}{2}t^{2}+\\frac{3}{2}t+2\\right)$. Then $D\\left(t, -\\frac{1}{2}+2\\right)$, \\\\\n$\\therefore PD=-\\frac{1}{2}t^{2}+\\frac{3}{2}t+2+\\frac{1}{2}t-2=-\\frac{1}{2}t^{2}+2t$, \\\\\n$\\therefore S_{\\triangle PBC}=\\frac{1}{2}\\times 4\\times \\left(-\\frac{1}{2}t^{2}+2t\\right)=-t^{2}+4t=-(t-2)^{2}+4$, \\\\\n$\\therefore$ when $t=2$, the area of $\\triangle PBC$ is maximized, \\\\\nAt this time $P(2, 3)$, hence the coordinates of $P$ are $\\boxed{(2, 3)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52634644_174.png"
  },
  {
    "solution": "\\boxed{\\frac{3\\sqrt{3}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the diameter $AB$ of the circle $\\odot O$ is $2$. Point $C$ is on the circumference, and $\\angle CAB=30^\\circ$. Point $D$ is a moving point on the circle, with $DE\\parallel AB$ intersecting the extension of $CA$ at point $E$, and $CD$ is connected intersecting $AB$ at point $F$. As shown in figure 2, when point $F$ is the midpoint of $CD$, find the area of $\\triangle CDE$.\n",
    "completion": "\\textbf{Solution:} As shown in the figure: connect OC\\\\\n$\\because$ AB is the diameter, and point F is the midpoint of CD\\\\\n$\\therefore$ AB$\\perp$CD, CF=DF,\\\\\n$\\because$ $\\angle$COF$=2\\angle$CAB$=60^\\circ$,\\\\\n$\\therefore$ OF$=\\frac{1}{2}$OC$=\\frac{1}{2}$, CF$=\\sqrt{3}$ OF$=\\frac{\\sqrt{3}}{2}$,\\\\\n$\\therefore$ CD$=2$CF$=\\sqrt{3}$, AF$=$OA$+$OF$=\\frac{3}{2}$,\\\\\n$\\because$ AF$\\parallel$AD, and point F is the midpoint of CD,\\\\\n$\\therefore$ DE$\\perp$CD, AF is the midline of $\\triangle CDE$,\\\\\n$\\therefore$ DE$=2$AF$=3$,\\\\\n$\\therefore$ $S_{\\triangle CED}=\\frac{1}{2}\\times 3\\times \\sqrt{3}=\\boxed{\\frac{3\\sqrt{3}}{2}}$",
    "image_path": "dataset/mm_math/images/MM_Math/52766163_166.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the length of line segment $AB=8$, point $C$ is the midpoint of line segment $AB$, and point $D$ is the midpoint of line segment $BC$. What is the length of line segment $AD$?\n",
    "completion": "\\textbf{Solution:} Since $AB=8$ and C is the midpoint of AB, it follows that $AC=BC=4$. Since D is the midpoint of BC, it follows $CD=DB=\\frac{1}{2}BC=2$. Therefore, $AD=AC+CD=4+2=\\boxed{6};$",
    "image_path": "dataset/mm_math/images/MM_Math/52579842_71.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, the straight lines $l_{1} \\parallel l_{2} \\parallel l_{3}$ intersect the straight lines $m$ and $n$ at points A, B, C, D, E, F respectively. Given that $AB=4$, $BC=6$, $DE=2$, what is the length of $EF$?\n",
    "completion": "\\textbf{Solution:} Since $l_1\\parallel l_2\\parallel l_3$,\\\\\nit follows that $\\frac{AB}{BC}=\\frac{DE}{EF}$,\\\\\ngiven $AB=4, BC=6, DE=2$,\\\\\nthus $\\frac{4}{6}=\\frac{2}{EF}$,\\\\\nsolving gives $EF=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51403188.png"
  },
  {
    "solution": "\\boxed{\\frac{13}{6}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=-2x+7$ intersects the x-axis and y-axis at points C and B, respectively, and intersects the line $y=\\frac{3}{2}x$ at point A. If there exists a point P on the y-axis such that $\\triangle OAP$ is an isosceles triangle with $OA$ as its base, what are the coordinates of point P?\n",
    "completion": "\\textbf{Solution:} Let the coordinates of point P be $(0, y)$,\\\\\nsince $\\triangle OAP$ is an isosceles triangle with $OA$ as its base,\\\\\ntherefore $OP=PA$,\\\\\nthus $2^2+(3-y)^2=y^2$,\\\\\nsolving this yields $y=\\frac{13}{6}$,\\\\\ntherefore, the coordinates of point P are $\\left(0, \\boxed{\\frac{13}{6}}\\right)$;",
    "image_path": "dataset/mm_math/images/MM_Math/52342217_127.png"
  },
  {
    "solution": "\\boxed{50^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AC=BC$, $\\angle A=40^\\circ$. Observing the traces of the construction with compass and straightedge in the figure, what is the measure of $\\angle BCG$?\n",
    "completion": "\\textbf{Solution:} Since $AC=BC$ \\\\\nit follows that $\\angle B=\\angle A=40^\\circ$ \\\\\nThus, $\\angle ACB=180^\\circ-2\\angle A=100^\\circ$ \\\\\nFrom compass and straightedge construction, we know that $CF$ is perpendicular to $AB$, and according to the congruence of medians, altitudes, and angle bisectors in an isosceles triangle, we get $\\angle BCG= \\frac{1}{2}\\angle ACB=50^\\circ$ \\\\\nTherefore, $\\angle BCG=\\boxed{50^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/51434796.png"
  },
  {
    "solution": "\\boxed{2\\,\\text{cm}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that point $M$ is the midpoint of line segment $AB$, point $N$ is on line segment $MB$, $AM=\\frac{5}{3}MN$, if $MN=3\\,cm$, find the lengths of line segments $AB$ and $NB$.\n",
    "completion": "\\textbf{Solution:} Given that $AM= \\frac{5}{3}MN$, and $MN=3\\,\\text{cm}$\\\\\n$\\therefore AM=5\\,\\text{cm}$\\\\\nFurthermore, since point $M$ is the midpoint of line segment $AB$\\\\\n$\\therefore AM=BM= \\frac{1}{2}AB$,\\\\\n$\\therefore AB=\\boxed{10\\,\\text{cm}}$\\\\\nAlso, since $NB=AB-AM-MN$,\\\\\n$\\therefore NB=\\boxed{2\\,\\text{cm}}$",
    "image_path": "dataset/mm_math/images/MM_Math/52580413.png"
  },
  {
    "solution": "\\boxed{1\\colon 3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, points D and E are the midpoints of AB and AC, respectively. What is the ratio of the area of $\\triangle ADE$ to the area of the quadrilateral DBCE?\n",
    "completion": "\\textbf{Solution:} Given that in $\\triangle ABC$, points $D$ and $E$ are the midpoints of $AB$ and $AC$ respectively,\\\\\nthus $\\frac{AD}{AB}=\\frac{1}{2}$,\\\\\ntherefore $\\frac{{S}_{\\triangle ADE}}{{S}_{\\triangle ABC}}={\\left(\\frac{AD}{AB}\\right)}^{2}=\\frac{1}{4}$,\\\\\nthus $\\frac{{S}_{\\triangle ADE}}{{S}_{\\text{quadrilateral }DBCE}}=\\frac{1}{3}$,\\\\\nmeaning the ratio of the area of $\\triangle ADE$ to that of quadrilateral DBCE is $\\boxed{1\\colon 3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52966970.png"
  },
  {
    "solution": "\\boxed{\\frac{25}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, point $M$ is on $BC$, and $EM \\perp AM$ meets the extension of $AD$ at point $E$. $\\triangle ABM \\sim \\triangle EMA$; if $AB=4$ and $BM=3$, find the value of $AE$.\n",
    "completion": "Solution: Since $AB=4$ and $BM=3$, \\\\\nwe have $AM= \\sqrt{AB^{2}+BM^{2}}=\\sqrt{4^{2}+3^{2}}=5$, \\\\\nBecause $\\triangle ABM \\sim \\triangle EMA$, \\\\\nit follows that $\\frac{AM}{AE}=\\frac{AB}{AM}$, \\\\\nThus, $AE= \\frac{AM^{2}}{AB}=\\boxed{\\frac{25}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51433170_149.png"
  },
  {
    "solution": "\\boxed{3 + \\sqrt{73}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In figure, in $\\triangle ABC$, $AD$ is the height from $A$ to side $BC$, $\\cos C = \\frac{1}{2}$, $AB=10$, and $AC=6$, then what is the length of $BC$?\n",
    "completion": "\\textbf{Solution:} In right triangle $\\triangle ADC$, we have $CD = \\cos C \\times AC = \\frac{1}{2}AC = \\frac{1}{2} \\times 6 = 3$,\\\\\ntherefore $AD = \\sqrt{AC^{2} - CD^{2}} = \\sqrt{6^{2} - 3^{2}} = 3\\sqrt{3}$,\\\\\nIn right triangle $\\triangle ADB$, $BD = \\sqrt{AB^{2} - AD^{2}} = \\sqrt{10^{2} - (3\\sqrt{3})^{2}} = \\sqrt{73}$,\\\\\ntherefore $BC = CD + BD = 3 + \\sqrt{73}$,\\\\\nThus, the answer is: $BC = \\boxed{3 + \\sqrt{73}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51262133.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{2}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, we have $AB=AC$, and $BD$ bisects $\\angle ABC$ and intersects $AC$ at point $D$. Point $E$ is on the extension line of $BC$, and $BD=DE$. Point $G$ is the midpoint of the line segment $BC$. We draw line $AG$ and it intersects $BD$ at point $F$. A line is drawn through point $D$ such that $DH\\perp BC$ with $H$ as the foot of the perpendicular. $\\triangle DCE$ is an isosceles triangle; if $\\angle CDE=22.5^\\circ$ and $DC=\\sqrt{2}$, find the length of $GH$.\n",
    "completion": "\\textbf{Solution:}\\\\\nSince $\\angle CDE=22.5^\\circ$ and $CD=CE=\\sqrt{2}$,\\\\\nit follows that $\\angle DCH=45^\\circ$ and $DH\\perp BC$,\\\\\nthus $\\angle HDC=\\angle DCH=45^\\circ$,\\\\\nimplying $DH=CH$,\\\\\nsince $DH^2+CH^2=DC^2=2$,\\\\\nwe have $DH=CH=1$,\\\\\nsince $\\angle ABC=\\angle DCH=45^\\circ$,\\\\\n$\\triangle ABC$ is an isosceles right triangle.\\\\\nFurthermore, since point G is the midpoint of BC,\\\\\n$AG\\perp BC$ and $AG=GC=BG$,\\\\\nand since $BD=DE$ and $DH\\perp BC$,\\\\\nit follows $BH=HE=\\sqrt{2}+1$,\\\\\nsince $BH=BG+GH=CG+GH=CH+GH+GH=\\sqrt{2}+1$,\\\\\nthus $1+2GH=\\sqrt{2}+1$,\\\\\nyielding $GH=\\boxed{\\frac{\\sqrt{2}}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52913917_89.png"
  },
  {
    "solution": "\\boxed{\\frac{5}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $BD$ bisects $\\angle ABC$, and $DE\\parallel BC$ intersects $BA$ at point $E$. If $DE=\\frac{5}{2}$, then what is the value of $EB$?\n",
    "completion": "\\textbf{Solution:} Since BD bisects $\\angle$ABC,\\\\\nit follows that $\\angle$CBD=$\\angle$ABD,\\\\\nSince DE$\\parallel$BC,\\\\\nit follows that $\\angle$BDE=$\\angle$CBD,\\\\\nthus $\\angle$BDE=$\\angle$DBE,\\\\\nwhich means BE=DE,\\\\\nSince DE= $\\frac{5}{2}$,\\\\\nit follows that EB= $\\boxed{\\frac{5}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51435242.png"
  },
  {
    "solution": "\\boxed{75^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, the diameter $AB$ of the circle $\\odot O$ is $2$, point $C$ is on the circumference of the circle, and $\\angle CAB=30^\\circ$. Point $D$ is a moving point on the circle, $DE\\parallel AB$ intersects the extension line of $CA$ at point $E$, and $CD$ is connected, intersecting $AB$ at point $F$. As shown in Figure 1, when $DE$ is tangent to $\\odot O$, what is the degree measure of $\\angle CFB$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure: Connect OD\\\\\n$\\because$ DE is tangent to $\\odot$O\\\\\n$\\therefore \\angle$ODE$=90^\\circ$\\\\\n$\\because$ AB $\\parallel$ DE\\\\\n$\\therefore \\angle$AOD$+\\angle$ODE$=180^\\circ$\\\\\n$\\therefore \\angle$AOD$=90^\\circ$\\\\\n$\\because \\angle$AOD$=2\\angle$C\\\\\n$\\therefore \\angle$C$=45^\\circ$\\\\\n$\\because \\angle$CFB$=\\angle$CAB$+\\angle$C\\\\\n$\\therefore \\angle$CFB$=\\boxed{75^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/52766163_165.png"
  },
  {
    "solution": "\\boxed{D\\left(0, 4\\right)$ or $D\\left(0, -4\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the line $l_1$ intersects the $x$-axis at point $A$ and the $y$-axis at point $B$. The coordinates of point $B$ are $(0, 3)$. The line $l_2: y=2x$ intersects line $l_1$ at point $C$, and the x-coordinate of point $C$ is 1. If point $D$ is on the $y$-axis and the area of $\\triangle OCD$ is $\\frac{2}{3}$ of the area of $\\triangle AOC$, what are the coordinates of point $D$?\n",
    "completion": "\\textbf{Solution:} From (1), we can get the equation of line $l_1$ as $y=-x+3$, and point $C(1, 2)$,\\\\\n$\\therefore$ when $y=0$, then $-x+3=0$, solving this we get: $x=3$,\\\\\n$\\therefore$ point $A(3, 0)$,\\\\\n$\\therefore$ the area of $\\triangle AOC$ with OA as the base and the distance from point C to the x-axis as the height, i.e., $S_{\\triangle AOC}=\\frac{1}{2}\\times 3\\times 2=3$,\\\\\n$\\therefore$ $S_{\\triangle OCD}=\\frac{2}{3}S_{\\triangle AOC}=2$,\\\\\nLet point $D(0, a)$, then the area of $\\triangle OCD$ has OD as the base and the distance from point C to the y-axis as the height,\\\\\n$\\therefore$ $S_{\\triangle OCD}=\\frac{1}{2}\\times 1\\cdot |a|=2$,\\\\\nSolving this, we find: $a=\\pm 4$,\\\\\n$\\therefore$ $\\boxed{D\\left(0, 4\\right)$ or $D\\left(0, -4\\right)}$;",
    "image_path": "dataset/mm_math/images/MM_Math/50481348_87.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $AB\\parallel CD\\parallel EF$, $AD:DF=2:3$, $BE=15$, what is the length of $BC$?\n",
    "completion": "\\textbf{Solution:} Since $AB \\parallel CD \\parallel EF$, \\\\\nit follows that $\\frac{BC}{BE} = \\frac{AD}{AF}$, \\\\\nwhich means $\\frac{BC}{15} = \\frac{2}{2+3}$, \\\\\nhence, BC$=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53124716.png"
  },
  {
    "solution": "\\boxed{84^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, given: $\\angle COD$ is inside $\\angle AOB$, and $\\angle AOB=160^\\circ$, $\\angle COD=\\frac{1}{4}\\angle AOB$, ray $OE$ bisects $\\angle AOD$, $\\angle COE=18^\\circ$. Find: What is the degree measure of $\\angle BOC$?\n",
    "completion": "\\textbf{Solution:} From equation (1), we have $\\angle AOE=\\angle EOD=58^\\circ$,\\\\\nsince $\\angle COE=18^\\circ$ and $\\angle AOB=160^\\circ$,\\\\\ntherefore $\\angle AOC=\\angle AOE+\\angle COE=58^\\circ+18^\\circ=76^\\circ$,\\\\\nthus $\\angle BOC=\\angle AOB-\\angle AOC=160^\\circ-76^\\circ=\\boxed{84^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53211137_115.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point \\(A\\) lies on the extension line of diameter \\(CD\\) of circle \\(\\odot O\\), and point \\(B\\) is on circle \\(\\odot O\\), with \\(\\angle A=\\angle C=30^\\circ\\). Line \\(AB\\) is a tangent to circle \\(\\odot O\\); if \\(BC=2\\sqrt{3}\\), what is the length of \\(AC\\)?\n",
    "completion": "\\textbf{Solution:} Construct $BE\\perp AC$ at point $E$\\\\\nSince $\\angle C=30^\\circ$ and $BC=2\\sqrt{3}$\\\\\nIt follows that $BE=\\sqrt{3}$\\\\\nThus, $CE=\\sqrt{BC^{2}-BE^{2}}=\\sqrt{(2\\sqrt{3})^{2}-(\\sqrt{3})^{2}}=3$\\\\\nSince $\\angle A=\\angle C=30^\\circ$\\\\\nIt implies $AB=BC$\\\\\nSince $BE\\perp AC$\\\\\nIt concludes $AC=2EC=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53110769_67.png"
  },
  {
    "solution": "\\boxed{\\frac{2\\pi }{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of $\\odot O$, and the chord $CD \\perp AB$ at point $E$, $\\angle CDB=30^\\circ$, $CD=2\\sqrt{3}$. Find the area of the shaded figure.\n",
    "completion": "Solution: As shown in the figure, connect $BC$\\\\\n$\\because \\angle CDB=30^\\circ$\\\\\n$\\therefore \\angle COB=60^\\circ$\\\\\n$\\because OC=OB$\\\\\n$\\therefore \\triangle OBC$ is an equilateral triangle\\\\\n$\\because CE\\perp OB$\\\\\n$\\therefore OE=BE$\\\\\n$\\because CE=DE$, $\\angle OEC=\\angle BED$\\\\\n$\\therefore \\triangle OCE\\cong \\triangle BDE$\\\\\n$\\therefore {S}_{\\text{shaded}}={S}_{\\text{sector} OBC}=\\frac{n\\pi {r}^{2}}{360}=\\frac{60^\\circ \\cdot \\pi \\cdot {2}^{2}}{360}=\\boxed{\\frac{2\\pi }{3}}$",
    "image_path": "dataset/mm_math/images/MM_Math/52872308_104.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Known in $\\triangle ABC$, $\\angle C=90^\\circ$, $AB=4$, $AC=\\sqrt{7}$. Find $\\sin A$.\n",
    "completion": "\\textbf{Solution:} We have $\\sin A = \\frac{BC}{AB} = \\boxed{\\frac{3}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51262135_109.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, given an equilateral $\\triangle ABC$ with a side length of 4, where points $D$ and $E$ are the midpoints of $AC$ and $BC$ respectively. Draw $DF\\perp DE$, intersecting the extension of $BC$ at point $F$. Find the length of $DF$.\n",
    "completion": "\\textbf{Solution:} Given that $\\triangle ABC$ is an equilateral triangle with side length 4,\\\\\n$\\therefore AC=BC=AB=4, \\angle ACB=60^\\circ=\\angle B=\\angle A,$\\\\\nGiven that points D and E are the midpoints of $AC$ and $BC$, respectively,\\\\\n$\\therefore DE\\parallel AB$ and $DE=\\frac{1}{2}AB=2,$\\\\\n$\\therefore \\angle DEC=\\angle B=60^\\circ,$\\\\\nGiven that $DE\\perp DF,$\\\\\n$\\therefore \\angle F=30^\\circ$, and $DE\\colon DF\\colon EF=1\\colon \\sqrt{3}\\colon 2,$\\\\\n$\\therefore DF=\\sqrt{3}DE=2\\sqrt{3}=\\boxed{2\\sqrt{3}}.$",
    "image_path": "dataset/mm_math/images/MM_Math/52250885.png"
  },
  {
    "solution": "\\boxed{110^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, points D and E are located on sides AC and BC respectively, and $AD=DE$, $AB=BE$, $\\angle A=70^\\circ$, what is the measure of $\\angle CED$ in degrees?\n",
    "completion": "\\textbf{Solution:} Since $AD=DE$, and $AB=BE$\\\\\nAlso $BD= BD$\\\\\nTherefore, $\\triangle ABD \\cong \\triangle EBD$ (SSS)\\\\\nThus, $\\angle BED = \\angle A = 70^\\circ$\\\\\nTherefore, $\\angle CED = 180^\\circ - \\angle BED = 180^\\circ - 70^\\circ = \\boxed{110^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/52653585.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the vertices of $\\triangle ABC$ are all located on the grid points of a square grid. What is the value of $\\tan C$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, in the lattice $\\triangle ADC$, we have $AD=2$, $DC=4$, and $\\tan C= \\frac{AD}{DC} = \\frac{2}{4}=\\boxed{\\frac{1}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52536703.png"
  },
  {
    "solution": "\\boxed{\\frac{72}{7}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the Cartesian coordinate system $xOy$, the linear function $y=-4x-2$ intersects the $y$-axis at point $C$ and intersects the line $AB: y=kx+2$ at point $D$, and the area of $\\triangle ACD$ is $\\frac{16}{7}$. Connect $BC$. What is the area of $\\triangle BCD$?\n",
    "completion": "\\textbf{Solution:} Substituting the coordinates of point D into the equation of line $AB: y=kx+2$,\\\\\nwe get $-\\frac{8}{7}k+2=\\frac{18}{7}$,\\\\\n$\\therefore k=-\\frac{1}{2}$,\\\\\n$\\therefore$ the equation of line $AB$ is $y=-\\frac{1}{2}x+2$;\\\\\n$\\because$ B is the intersection point of line AB and the x-axis,\\\\\n$\\therefore$ the coordinates of point B are (4, 0),\\\\\n$\\therefore S_{\\triangle ABC}=\\frac{1}{2}AC\\cdot OB=\\frac{1}{2}\\times4\\times4=8$,\\\\\n$\\therefore$ the area of $\\triangle BCD$ $=S_{\\triangle ACD}+S_{\\triangle ABC}=\\frac{16}{7}+8=\\boxed{\\frac{72}{7}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52056193_212.png"
  },
  {
    "solution": "\\boxed{14^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, fold the rectangular paper ABCD upwards along AE, so that point B falls at point F on side DC. If $\\angle FEC = 28^\\circ$, what is the value of $\\angle EAB$ in degrees?\n",
    "completion": "\\textbf{Solution:} By folding, we obtain $\\angle BEA = \\angle FEA$,\\\\\n$\\because \\angle FEC = 28^\\circ$,\\\\\n$\\therefore \\angle AEB = (180^\\circ - 28^\\circ) \\div 2 = 76^\\circ$,\\\\\n$\\therefore \\angle EAB = 90^\\circ - \\angle AEB = \\boxed{14^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51213212.png"
  },
  {
    "solution": "\\boxed{x-3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y_{1}=kx+b$ intersects the graph of the inverse proportion function $y_{2}=\\frac{m}{x}$ at points $A(-2, -5)$ and $C(5, n)$, intersects the y-axis at point B, and intersects the x-axis at point D. What are the expressions for the inverse proportion function and the linear function?\n",
    "completion": "\\textbf{Solution:} Solution: Since substituting point A ($-2$, $-5$) into $y_{2}=\\frac{m}{x}$ gives: $-5=\\frac{m}{-2}$,\\\\\nthus $m=10$,\\\\\ntherefore, the expression for the inverse proportion function is $y_{2}=\\frac{10}{x}$,\\\\\nsince substituting point C ($5$, $n$) into the inverse proportion function $y_{2}=\\frac{10}{x}$ yields: $n=\\frac{10}{5}=2$,\\\\\ntherefore, the coordinates of point C are ($5$, $2$),\\\\\nsince substituting the coordinates of A and C into $y_{1}=kx+b$ gives: $\\left\\{\\begin{array}{l} -5=-2k+b \\\\ 2=5k+b \\end{array}\\right.$,\\\\\nthus $\\left\\{\\begin{array}{l} k=1 \\\\ b=-3 \\end{array}\\right.$\\\\\ntherefore, the expression for the linear function is $y_{1}=x-3$;\\\\\nHence, the expression for the inverse proportion function is $y_{2}=\\boxed{\\frac{10}{x}}$, and the expression for the linear function is $y_{1}=\\boxed{x-3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52601939_121.png"
  },
  {
    "solution": "\\boxed{40^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $PA$, $PB$ are tangent to the circle $\\odot O$ at points $A$, $B$ respectively, and point $C$ is on the circle $\\odot O$. If $\\angle C = 70^\\circ$, what is the degree measure of $\\angle P$?\n",
    "completion": "\\textbf{Solution:} Connect OA and OB, \\\\\n$\\because PA$ and $PB$ are tangents to $\\odot O$, \\\\\n$\\therefore PA\\perp OA$, $PB\\perp OB$, \\\\\n$\\therefore \\angle PAO=\\angle PBO=90^\\circ$, \\\\\n$\\because \\angle P+\\angle PAO+\\angle AOB+\\angle PBO=360^\\circ$, \\\\\n$\\therefore \\angle P=180^\\circ-\\angle AOB$, \\\\\n$\\because \\angle C=70^\\circ$, \\\\\n$\\therefore \\angle AOB=2\\angle C=140^\\circ$, \\\\\n$\\therefore \\angle P=180^\\circ-140^\\circ=\\boxed{40^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52457096.png"
  },
  {
    "solution": "\\boxed{3\\pi}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within the right-angled triangle $ABC$ with a $30^\\circ$ angle, point $M$ is the midpoint on the hypotenuse $AB$. After rotating the triangle clockwise by $90^\\circ$ around the right-angle vertex $C$, we obtain $\\triangle A^\\prime B^\\prime C$. If $AC=6$, what is the length of the path traveled by point $M$?\n",
    "completion": "\\textbf{Solution:} As illustrated, connect CM and CM',\\\\\nsince in the $30^\\circ$ right-angled triangle ABC, point M is the midpoint on the hypotenuse AB,\\\\\ntherefore CM = $\\frac{1}{2}AB$,\\\\\nsince $\\angle B = 30^\\circ$,\\\\\ntherefore AB = 2AC = 12,\\\\\ntherefore CM = 6,\\\\\nAccording to the problem, the path of point M is an arc with CM as the radius and point C as the center,\\\\\nsince the triangle is rotated $90^\\circ$ clockwise around the right-angle vertex C to form $\\triangle A^{\\prime}B^{\\prime}C$,\\\\\ntherefore $\\angle MCM' = 90^\\circ$,\\\\\nthus, the length of the path traveled by point M is\\\\\n$ \\frac{90^\\circ\\pi \\times 6}{180} = \\boxed{3\\pi} $.",
    "image_path": "dataset/mm_math/images/MM_Math/51569291.png"
  },
  {
    "solution": "\\boxed{\\frac{25}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points B and C divide the line segment AD into three parts in the ratio 2:3:4, and $CD=20$. If point P is the midpoint of AD, and point Q is the midpoint of CD, what is the length of line segment PQ?\n",
    "completion": "\\textbf{Solution:} Since point P is the midpoint of AD,\\\\\nwe have $PD=\\frac{1}{2}AD=\\frac{45}{2}$.\\\\\nSince point Q is the midpoint of CD,\\\\\nwe have $QD=\\frac{1}{2}CD=10$.\\\\\nTherefore, $PQ=PD−QD=\\frac{45}{2}−10=\\boxed{\\frac{25}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53206048_102.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{7}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, in $\\triangle ABC$, $\\angle ACB = 90^\\circ$, $\\angle CAB= 30^\\circ$, and $\\triangle ABD$ is an equilateral triangle. As shown in Figure 2, when quadrilateral $ACBD$ is folded to make $D$ coincide with $C$, and $EF$ is the crease, what is the sine value of $\\angle ACE$?\n",
    "completion": "\\textbf{Solution:} Given that in $\\triangle ABC$, $\\angle ACB=90^\\circ$, $\\angle BAC=30^\\circ$, let AB=2a,\\\\\nthen it follows AC= $\\sqrt{3}a$, BC=a;\\\\\nGiven that $\\triangle ABD$ is an equilateral triangle,\\\\\nthus AD=AB=2a;\\\\\nLet DE=EC=x, then AE=2a−x;\\\\\nIn $\\triangle AEC$, by the Pythagorean theorem, we get: $(2a−x)^{2}+3a^{2}=x^{2}$,\\\\\nsolving for x yields $x= \\frac{7}{4}a$;\\\\\nthus AE= $\\frac{1}{4}a$, EC= $\\frac{7}{4}a$,\\\\\nthus $\\sin\\angle ACE= \\frac{AE}{CE}=\\boxed{\\frac{1}{7}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50126432.png"
  },
  {
    "solution": "\\boxed{50^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\angle B = \\angle D = 90^\\circ$, $BC = CD$, $\\angle 1 = 40^\\circ$, then what is the value of $\\angle 2$?\n",
    "completion": "\\textbf{Solution:} Given $\\because$ $\\angle B = \\angle D = 90^\\circ$,\\\\\n$\\therefore$ $\\triangle ABC$ and $\\triangle ADC$ are both right-angled triangles,\\\\\nIn $\\triangle ABC$ and $\\triangle ADC$,\\\\\n$\\left\\{\\begin{array}{c} BC = DC \\\\ AC = AC \\end{array}\\right.$,\\\\\n$\\therefore$ $\\triangle ABC \\cong \\triangle ADC$ (HL),\\\\\n$\\therefore$ $\\angle 1 = \\angle CAD = 40^\\circ$,\\\\\n$\\therefore$ $\\angle 2 = 90^\\circ - \\angle CAD = \\boxed{50^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51367862.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, $AC=4$, and $BC=3$, with $CD$ being the altitude of $\\triangle ABC$. What is the value of $\\tan\\angle BCD$?\n",
    "completion": "\\textbf{Solution:} In $\\triangle ABC$ and $\\triangle BCD$, $\\angle A + \\angle B = 90^\\circ$, $\\angle BCD + \\angle B = 90^\\circ$.\\\\\n$\\therefore \\angle A = \\angle BCD$.\\\\\n$\\therefore \\tan \\angle BCD = \\tan A = \\frac{BC}{AC} = \\boxed{\\frac{3}{4}}$",
    "image_path": "dataset/mm_math/images/MM_Math/52601757.png"
  },
  {
    "solution": "\\boxed{8.5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a mathematics interest group measures the height of an ancient tree $BH$ and a teaching building $CG$. First, at point $A$, they use an altimeter with a height of $1.5$ meters to measure the angle of elevation $\\angle HDE=45^\\circ$ of the top of the ancient tree $H$. At this moment, the top of the teaching building $G$ is exactly in the line of sight $DH$. Then, they move forward $7$ meters to point $B$ and measure the angle of elevation $\\angle GEF=60^\\circ$ of the top of the teaching building $G$. Points $A$, $B$, $C$ are on the same horizontal line. Calculate the height of the ancient tree $BH$ in meters.\n",
    "completion": "\\textbf{Solution:} Solution: According to the problem statement: Quadrilateral ABED is a rectangle, so we get $DE=AB=7$ meters.\\\\\nIn $\\triangle DEH$, $\\because \\angle EDH=45^\\circ$,\\\\\n$\\therefore HE=DE=7$ meters,\\\\\n$\\therefore BH=EH+BE=\\boxed{8.5}$ meters.",
    "image_path": "dataset/mm_math/images/MM_Math/51103651_86.png"
  },
  {
    "solution": "\\boxed{x<-2$ or $x>8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the graph, the quadratic function $y_{1}=ax^{2}+bx+c$ ($a>0$) intersects with the linear function $y_{2}=kx+m$ ($k\\neq 0$) at points $A(-2,4)$ and $B(8,2)$. What is the range of values of $x$ for which $y_{1}>y_{2}$?\n",
    "completion": "\\textbf{Solution:} From the given conditions, we have: when $x<-2$ or $x>8$, $y_{1}>y_{2}$,\\\\\n$\\therefore$ the range of $x$ for which $y_{1}>y_{2}$ holds true is $x<-2$ or $x>8$.\\\\\nTherefore, the solution is: $\\boxed{x<-2$ or $x>8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52574130.png"
  },
  {
    "solution": "\\boxed{19}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=10$, $AC=7$, $BC=9$, and points $D$, $E$, and $F$ are the midpoints of $AB$, $AC$, and $BC$, respectively. What is the perimeter of the quadrilateral $DBFE$?\n",
    "completion": "\\textbf{Solution:} Since points $D$, $E$, and $F$ are the midpoints of $AB$, $AC$, and $BC$, respectively,\\\\\n$\\therefore DE$ is a midline of $\\triangle ABC$, $EF$ is also a midline of $\\triangle ABC$,\\\\\n$\\therefore DE=BF=\\frac{1}{2}BC=\\frac{1}{2}\\times 9=\\frac{9}{2}$, $EF=BD=\\frac{1}{2}AB=\\frac{1}{2}\\times 10=5$,\\\\\n$\\therefore$ the perimeter of the quadrilateral $DBFE$ is $DE+BF+EF+BD=9+10=\\boxed{19}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52422738.png"
  },
  {
    "solution": "\\boxed{40^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $AD$ and $CE$ are two altitudes of $\\triangle ABC$, with $AD=CE$ and $\\angle CAD=25^\\circ$. What is the measure of $\\angle OCD$ in degrees?\n",
    "completion": "\\textbf{Solution:} Since AD and CE are the two altitudes of $\\triangle ABC$,\\\\\nit follows that AD$\\perp$CB and CE$\\perp$AD,\\\\\ntherefore $\\angle ADC=\\angle CEA=90^\\circ$,\\\\\nIn Rt$\\triangle ADC$ and Rt$\\triangle CEA$,\\\\\n$\\left\\{\\begin{array}{l}AC=CA\\\\ AD=CE\\end{array}\\right.$,\\\\\nthus, Rt$\\triangle ADC\\cong$Rt$\\triangle CEA$ (HL),\\\\\ntherefore, $\\angle CAD=\\angle ACE=25^\\circ$,\\\\\nsince $\\angle ACD=90^\\circ-\\angle CAD=90^\\circ-25^\\circ=65^\\circ$,\\\\\nthus, $\\angle OCD=\\angle ACD-\\angle ACE=65^\\circ-25^\\circ=\\boxed{40^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52914131.png"
  },
  {
    "solution": "\\boxed{\\left(-\\frac{8}{7}, \\frac{18}{7}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the Cartesian coordinate system $xOy$, the linear function $y=-4x-2$ intersects with the $y$-axis at point $C$ and intersects with the line $AB: y=kx+2$ at point $D$, and the area of $\\triangle ACD$ is $\\frac{16}{7}$. What are the coordinates of point $D$?\n",
    "completion": "\\textbf{Solution:} Start by considering the line $y=-4x-2$ which intersects the y-axis at point C, \\\\\nthus the coordinates of point C are $(0, -2)$, \\\\\nFrom the line $AB: y=kx+2$, \\\\\nwe get the coordinates of point A as $(0, 2)$, \\\\\n$\\therefore AC=4$, \\\\\n$\\because$ the area of $\\triangle ACD=\\frac{16}{7}$, \\\\\n$\\therefore$ the height of $\\triangle ACD$ from side $AC=\\frac{8}{7}$, \\\\\n$\\therefore$ the x-coordinate of point D is $-\\frac{8}{7}$, \\\\\nWhen $x=-\\frac{8}{7}$, $y=-4x-2=\\frac{18}{7}$ \\\\\n$\\therefore$ the coordinates of point D are $\\boxed{\\left(-\\frac{8}{7}, \\frac{18}{7}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52056193_211.png"
  },
  {
    "solution": "\\boxed{15^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle BAC=100^\\circ$, $AD\\perp BC$ at point D, and $AE$ bisects $\\angle BAC$ and intersects $BC$ at point E. If $\\angle C=25^\\circ$, find the degree measure of $\\angle DAE$.\n",
    "completion": "\\textbf{Solution:} Since AE bisects $\\angle BAC$, \\\\\nit follows that $\\angle EAC = \\frac{1}{2}\\angle BAC = \\frac{1}{2}\\times 100^\\circ = 50^\\circ$, \\\\\nSince $\\angle C = 25^\\circ$, \\\\\nit follows that $\\angle AED = \\angle C + \\angle EAC = 25^\\circ + 50^\\circ = 75^\\circ$, \\\\\nSince AD is perpendicular to BC, \\\\\nit follows that $\\angle ADE = 90^\\circ$, \\\\\ntherefore, $\\angle DAE = 90^\\circ - 75^\\circ = \\boxed{15^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53009855.png"
  },
  {
    "solution": "\\boxed{(0,4)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the parabola $y=\\frac{1}{2}x^{2}+bx+c$ passes through point $A(-4, 0)$. Point $M$ is the vertex of the parabola, and point $B$ is on the $y$-axis, with $OA=OB$. The line $AB$ intersects the parabola at point $C(2, 6)$ in the first quadrant. Find the equation of line $AB$ and the value of $\\sin\\angle ABO$; connect $OC$. If a line through point $O$ intersects segment $AC$ at point $P$, dividing the area of triangle $AOC$ into two parts in the ratio $1:2$, find the coordinates of point $P$.\n",
    "completion": "\\textbf{Solution:} Given that $A(-4,0)$, \\\\\ntherefore $OA=4$, \\\\\nsince $OA=OB$, \\\\\nthus $OB=4$, \\\\\ntherefore, $B(0,4)$, \\\\\nLet the equation of line AB be $y=kx+b$, \\\\\nSubstituting $A(-4,0)$ and $B(0,4)$ into the equation yields: $\\left\\{\\begin{array}{l}0=-4k+b \\\\ 4=b\\end{array}\\right.$, \\\\\nSolving this, we get $\\left\\{\\begin{array}{l}k=1 \\\\ b=4\\end{array}\\right.$, \\\\\nTherefore, the equation of line AB is $y=x+4$, \\\\\nIn $\\triangle AOB$, $AB=\\sqrt{OA^2+OB^2}=4\\sqrt{2}$, \\\\\nthus $\\sin\\angle ABO=\\frac{OA}{AB}=\\frac{4}{4\\sqrt{2}}=\\frac{\\sqrt{2}}{2}$. \\\\\nA line through the point O intersects segment AC at point P, dividing the area of triangle AOC into two parts with a ratio of 1:2. Draw $PQ\\perp x$-axis at Q, and draw $CH\\perp x$-axis at H, \\\\\nWe consider two cases: \\\\\n(1) If $S_{\\triangle AOP}:S_{\\triangle COP}=1:2$, \\\\\nsince $S_{\\triangle AOP}:S_{\\triangle AOC}=1:3$, \\\\\ntherefore $PQ:CH=1:3$, \\\\\ngiven that $C(2,6)$, \\\\\nthus $CH=6$, \\\\\ntherefore $PQ=2$, i.e., $y_P=2$, \\\\\nIn $y=x+4$, setting $y=2$ gives, $2=x+4$, \\\\\nthus $x=-2$, \\\\\ntherefore $P(-2,2)$; \\\\\n(2) If $S_{\\triangle COP}:S_{\\triangle AOP}=1:2$, \\\\\nsince $S_{\\triangle COP}:S_{\\triangle AOP}=1:2$, \\\\\ntherefore $S_{\\triangle AOP}:S_{\\triangle AOC}=2:3$, \\\\\nthus $PQ:CH=2:3$, \\\\\ngiven that $CH=6$, \\\\\ntherefore $PQ=4$, i.e., $y_P=4$, \\\\\nIn $y=x+4$, setting $y=4$ gives, $4=x+4$, \\\\\nthus $x=0$, \\\\\ntherefore $P(0,4)$; \\\\\nIn summary, a line through the point O intersects segment AC at point P, dividing the area of triangle AOC into two parts with a ratio of 1:2, then the coordinates of P are \\boxed{(-2,2)} or \\boxed{(0,4)}.",
    "image_path": "dataset/mm_math/images/MM_Math/53043807_115.png"
  },
  {
    "solution": "\\boxed{1+\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, it is known that the radius of $\\odot O$ is 2, points A and B are on $\\odot O$, $\\angle BAC=90^\\circ$, and $\\tan\\angle ACB=2$. What is the maximum value of line segment OC?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect OA, OB, and draw AD$\\perp$AO such that $\\angle ADO=\\angle ABC$,\\\\\n$\\because$ in $\\triangle ABC$, $\\angle BAC=90^\\circ$, $\\tan \\angle ACB=2$,\\\\\n$\\therefore$ $\\frac{AB}{AC}=2$, i.e., $AB=2AC$,\\\\\n$\\because$ $\\angle ADO=\\angle ABC$, $\\angle OAD=\\angle CAB=90^\\circ$,\\\\\n$\\therefore$ $\\triangle ADO\\sim \\triangle ABC$,\\\\\n$\\therefore$ $\\frac{AO}{AC}=\\frac{AD}{AB}$, $\\angle AOD=\\angle ACB$,\\\\\n$\\because$ $\\angle DAO=\\angle BAC$,\\\\\n$\\therefore$ $\\angle DAB=\\angle OAC$,\\\\\n$\\therefore$ $\\triangle DAB\\sim \\triangle OAC$,\\\\\n$\\therefore$ $\\frac{BD}{OC}=\\frac{AB}{AC}=2$,\\\\\n$\\therefore$ $OC=\\frac{1}{2}BD$,\\\\\nIn $\\triangle AOD$, $\\tan \\angle AOD=\\tan \\angle ACB=\\frac{AD}{OA}=2$,\\\\\n$\\therefore$ $AD=2AO=4$,\\\\\n$\\therefore$ $OD=\\sqrt{OA^2+AD^2}=2\\sqrt{5}$,\\\\\n$\\because$ $BD\\le OD+OB=2+2\\sqrt{5}$,\\\\\n$\\therefore$ the maximum value of BD is $2+2\\sqrt{5}$,\\\\\n$\\therefore$ the maximum value of OC is $\\boxed{1+\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53148597.png"
  },
  {
    "solution": "\\boxed{\\frac{9}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB=4$, $CD=6$, point $F$ is on $BD$, lines $BC$ and $AD$ intersect at point $E$, and $AB \\parallel CD \\parallel EF$. If $AE=3$, what is the length of $ED$?\n",
    "completion": "\\textbf{Solution:} Given $AB\\parallel CD$,\\\\\n$\\therefore \\triangle AEB\\sim \\triangle DEC$,\\\\\n$\\therefore \\frac{AE}{DE}=\\frac{AB}{CD}$,\\\\\nGiven $AB=4$, $CD=6$, $AE=3$,\\\\\n$\\therefore \\frac{3}{DE}=\\frac{4}{6}$,\\\\\nSolving gives: $DE=\\boxed{\\frac{9}{2}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51445794_166.png"
  },
  {
    "solution": "\\boxed{\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the radius of $\\odot O$ is 1, and the point $P$ is a point outside $\\odot O$, with $OP=2$. If $PT$ is a tangent to $\\odot O$ with $T$ as the point of tangency, and $OT$ is connected, then what is the length of $PT$?\n",
    "completion": "\\textbf{Solution:} Since $PT$ is a tangent to the circle $\\odot O$ with $T$ as the point of tangency, \\\\\nit follows that $\\angle OTP=90^\\circ$. \\\\\nThus, $PT=\\sqrt{OP^{2}-OT^{2}}$. \\\\\nGiven that the radius of $\\odot O$ is 1, \\\\\nit follows that $OT=1$. \\\\\nTherefore, $PT=\\sqrt{OP^{2}-OT^{2}}=\\sqrt{2^{2}-1}=\\boxed{\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53078950.png"
  },
  {
    "solution": "\\boxed{70^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $O$ is a point on the line $AB$, with $\\angle BOC=40^\\circ$, and $OD$ bisects $\\angle AOC$. What is the degree measure of $\\angle AOD$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle BOC=40^\\circ$,\\\\\nit follows that $\\angle AOC=180^\\circ-\\angle BOC=140^\\circ$,\\\\\nsince OD bisects $\\angle AOC$,\\\\\nthus $\\angle AOD= \\frac{1}{2} \\angle AOC= \\boxed{70^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51494981_148.png"
  },
  {
    "solution": "\\boxed{(2,0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, square ABCD and square EFOG are similar figures, where point A corresponds to point E. The coordinate of point A is $(-4, 2)$, and the coordinate of point E is $(-1, 1)$. What are the coordinates of the center of similitude of these two squares?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, extend lines AE and DG until they intersect the x-axis at point N. \\\\\nAssume the equation of line $AE$ is $y=kx+b$. Substituting points $A(-4,2)$ and $E(-1,1)$ into the equation, we get $\\left\\{\\begin{array}{l}-4k+b=2 \\\\ -k+b=1\\end{array}\\right.$. \\\\\nSolving these equations gives $\\left\\{\\begin{array}{l}k=-\\frac{1}{3} \\\\ b=\\frac{2}{3}\\end{array}\\right.$. \\\\\nThus, $y=-\\frac{1}{3}x+\\frac{2}{3}$. \\\\\nSetting $y=0$ yields $x=2$, \\\\\n$\\therefore$ the coordinates of N are $(2,0)$; \\\\\nHence, the answer is: the coordinates of $N$ are $\\boxed{(2,0)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50322915.png"
  },
  {
    "solution": "\\boxed{\\frac{85}{52}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $AB=2BC$, point $M$ is the midpoint of side $CD$, points $E$ and $F$ are on sides $AB$ and $BC$ respectively, and $AF \\perp ME$, with $G$ being the foot of the perpendicular. If $EB=2$ and $BF=1$, what is the area of quadrilateral $BFGE$?\n",
    "completion": "\\textbf{Solution:} Set $BC=a$, then $AB=2a$, $DM=MC=a$.\\\\\nConstruct $MH\\perp AB$ at $H$,\\\\\nthen $\\angle EMH=90^\\circ-\\angle MEA=\\angle FAB$.\\\\\nHence $Rt\\triangle EMH\\backsim Rt\\triangle FAB$.\\\\\nTherefore, $\\frac{MH}{HE}=\\frac{AB}{BF}$,\\\\\nthat is, $\\frac{a}{a-2}=\\frac{2a}{1}$,\\\\\nsolving this gives $a=\\frac{5}{2}$.\\\\\nThus, $BC=\\frac{5}{2}$, $AB=5$.\\\\\nTherefore, $AF=\\sqrt{AB^2+BF^2}=\\sqrt{5^2+1^2}=\\sqrt{26}$,\\\\\n$S_{\\triangle ABF}=\\frac{1}{2}AB\\times BF=\\frac{1}{2}\\times 5\\times 1=\\frac{5}{2}$.\\\\\nAlso, $Rt\\triangle AEG\\backsim Rt\\triangle AFB$,\\\\\nTherefore, $\\frac{S_{\\triangle AEG}}{S_{\\triangle AFB}}=\\left(\\frac{AE}{AF}\\right)^2=\\left(\\frac{5-2}{\\sqrt{26}}\\right)^2=\\frac{9}{26}$.\\\\\nThus, $S_{\\triangle AEG}=\\frac{9}{26}S_{\\triangle AFB}=\\frac{9}{26}\\times \\frac{5}{2}=\\frac{45}{52}$.\\\\\nThus, $S_{\\text{quadrilateral} BFGE}=S_{\\triangle AFB}-S_{\\triangle AEC}=\\frac{5}{2}-\\frac{45}{52}=\\boxed{\\frac{85}{52}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51722326.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\triangle ABC$, $ AB=AC$, points E, F are on side BC, and $ \\angle EAF=\\angle C$. $ \\triangle ABF\\sim \\triangle ECA$; if $ BF=8$ and $ CE=6$, find the length of AB.\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABF\\sim \\triangle ECA$, \\\\\nwe have $\\frac{AB}{EC}=\\frac{BF}{AC}$, \\\\\nSince $BF=8$, $CE=6$, $AB=AC$, \\\\\nit follows that $\\frac{AB}{6}=\\frac{8}{AB}$, that is $AB^{2}=6\\times 8=48$, \\\\\nHence, $AB=\\boxed{4\\sqrt{3}}$ (negative value discarded).",
    "image_path": "dataset/mm_math/images/MM_Math/52977747_132.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $A$, $B$, $C$ are vertices of a small square, and the side length of each small square is 1. What is the value of $\\sin\\angle ACB$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw AD $\\perp$ BC at point D.\\\\\nIn $\\triangle BCE$, $BC = \\sqrt{BE^2 + CE^2} = \\sqrt{3^2 + 1^2} = \\sqrt{10}$.\\\\\n$\\therefore AC = BC = \\sqrt{10}$.\\\\\n$\\therefore CH \\perp AB$, and $CH = 3$.\\\\\n$\\because \\frac{1}{2}BC \\times AD = \\frac{1}{2}AB \\times CH$,\\\\\n$\\therefore AD = \\frac{AB \\times CH}{BC} = \\frac{2 \\times 3}{\\sqrt{10}} = \\frac{3\\sqrt{10}}{5}$.\\\\\nIn $\\triangle ADC$, $\\sin\\angle ACB = \\frac{AD}{AC} = \\frac{3\\sqrt{10}}{5} \\div \\sqrt{10} = \\boxed{\\frac{3}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52537075.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle ABCD, AB = 2, $\\angle DAC=30^\\circ$, the point M is the midpoint of side BC, point P is a moving point on the diagonal AC ($0<CP<1.5$). Triangle CPM is folded along PM, such that point C falls on point C$^{\\prime}$, segment MC$^{\\prime}$ intersects AC at point N. Segment AC$^{\\prime}$ is drawn. When $\\triangle ANC^{\\prime}$ forms a right triangle, what is the length of segment AC$^{\\prime}$?\n",
    "completion": "\\textbf{Solution:} When $\\angle ANC'$ is a right angle, $\\triangle ANC'$ is a right-angled triangle, then $\\angle CNM=90^\\circ$,\\\\\n$\\because$ in the rectangle ABCD, AB=2, $\\angle DAC=30^\\circ$,\\\\\n$\\therefore \\angle ACB=30^\\circ$, AC=2AB=4, BC=$\\sqrt{AC^2-AB^2}=2\\sqrt{3}$,\\\\\n$\\because$ point M is the midpoint of side BC, $\\therefore MC=\\frac{1}{2}BC=\\sqrt{3}$,\\\\\nIn $\\triangle CNM$, $\\angle CNM=90^\\circ$, $\\angle MCN=30^\\circ$,\\\\\n$\\therefore MN=\\frac{1}{2}MC=\\frac{\\sqrt{3}}{2}$, NC=$\\sqrt{MC^2-MN^2}=\\frac{3}{2}$,\\\\\n$\\therefore AN=AC-NC=\\frac{5}{2}$,\\\\\nAs folded $MC'=MC=\\sqrt{3}$,\\\\\n$\\therefore NC'=MC'-MN=\\frac{\\sqrt{3}}{2}$,\\\\\nIn $\\triangle ANC'$, $\\angle ANC'=90^\\circ$,\\\\\n$\\therefore AC'=\\sqrt{\\left(\\frac{5}{2}\\right)^2+\\left(\\frac{\\sqrt{3}}{2}\\right)^2}=\\sqrt{7}$;\\\\\nWhen $\\angle AC'N=90^\\circ$, $\\triangle ANC'$ is a right-angled triangle, then $\\angle AC'M=90^\\circ$,\\\\\nConnecting AM,\\\\\nIn $\\triangle ABM$, AB=2, MC=BM=$\\sqrt{3}$,\\\\\nAM=$\\sqrt{2^2+(\\sqrt{3})^2}=\\sqrt{7}$,\\\\\nAs folded $MC'=MC=\\sqrt{2}$,\\\\\n$\\therefore AC'=\\sqrt{(\\sqrt{7})^2-(\\sqrt{3})^2}=2$;\\\\\nIn conclusion, the length of the line segment AC' is $\\boxed{2}$ or $\\sqrt{7}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52188838.png"
  },
  {
    "solution": "\\boxed{14\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of $\\odot O$, and point $P$ is on $\\odot O$ with $PA=PB$. Point $M$ is outside $\\odot O$, and $MB$ is tangent to $\\odot O$ at point $B$. Connect $OM$, and draw $AC\\parallel OM$ through point $A$ to intersect with $\\odot O$ at point $C$, and connect $BC$ to intersect $OM$ at point $D$. $MC$ is a tangent to $\\odot O$. Given that $AB=20$ and $BC=16$, connect $PC$. What is the length of $PC$?\n",
    "completion": "\\textbf{Solution:} Since $AB$ is the diameter of $\\odot O$,\\\\\nit follows that $\\angle ACB=\\angle APB=90^\\circ$,\\\\\nSince $AB=20$,\\\\\nwe have $PA=PB=\\frac{\\sqrt{2}}{2}\\times 20=10\\sqrt{2}$,\\\\\nSince $BC=16$,\\\\\nwe get $AC=\\sqrt{AB^{2}-BC^{2}}=\\sqrt{20^{2}-16^{2}}=12$,\\\\\nDraw $AH\\perp PC$ at point H through point A,\\\\\nthus $\\angle AHC=\\angle AHP=90^\\circ$,\\\\\nSince $\\angle ACH=\\angle ABP=45^\\circ$,\\\\\nwe have $AH=CH=AC\\times \\sin 45^\\circ=12\\times \\frac{\\sqrt{2}}{2}=6\\sqrt{2}$,\\\\\nTherefore, $PH=\\sqrt{PA^{2}-AH^{2}}=\\sqrt{(10\\sqrt{2})^{2}-(6\\sqrt{2})^{2}}=8\\sqrt{2}$,\\\\\nHence, $PC=PH+CH=\\boxed{14\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53101898_76.png"
  },
  {
    "solution": "\\boxed{(4,\\frac{9}{4})}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, place a square paper in the Cartesian coordinate system. The side length of the square paper ABCD is 3, with points E and F located on BC and CD, respectively. Fold AB and AD along AE and AF such that points B and D both precisely fall on point G. Given that $3BE=BC$, there exists a point M on line EF, such that the area of $\\triangle AFM$ is half the area of $\\triangle AEF$. What are the coordinates of point M?\n",
    "completion": "\\textbf{Solution:} Suppose there exists a point $M$ on the line $EF$ such that the area of $\\triangle AFM$ is half of the area of $\\triangle AEF$, \\\\\n$\\because$ the heights of $\\triangle AEF$ and $\\triangle AFM$ are equal, \\\\\n$\\therefore \\frac{S_{\\triangle AFM}}{S_{\\triangle AEF}}=\\frac{FM}{EF}=\\frac{1}{2}$, \\\\\n$\\therefore FM=\\frac{1}{2}EF$, \\\\\nWhen $M$ is on the segment $EF$, specifically, when $M$ is the midpoint of $EF$, we denote this point as $M_{1}$, \\\\\n$\\because$ the coordinate of $E$ is $(1,0)$, and the coordinate of $F$ is $(3,\\frac{3}{2})$, \\\\\n$\\therefore \\left\\{\\begin{array}{l}x_{M_{1}}=\\frac{1+3}{2}=2 \\\\ y_{M_{1}}=\\frac{0+\\frac{3}{2}}{2}=\\frac{3}{4}\\end{array}\\right.$, \\\\\n$\\therefore$ the coordinates of $M_{1}$ are $\\boxed{(2,\\frac{3}{4})}$; \\\\\nWhen $M$ lies on the extension of $EF$, we denote this point as $M_{2}$, then $FM_{1}=FM_{2}$, which means $F$ is the midpoint of $M_{1}M_{2}$, \\\\\n$\\therefore \\left\\{\\begin{array}{l}3=\\frac{2+x_{M_{2}}}{2} \\\\ \\frac{3}{2}=\\frac{y_{M_{2}}+\\frac{3}{4}}{2}\\end{array}\\right.$, \\\\\n$\\therefore \\left\\{\\begin{array}{l}x_{M_{2}}=4 \\\\ y_{M_{2}}=\\frac{9}{4}\\end{array}\\right.$, \\\\\n$\\therefore$ the coordinates of $M_{2}$ are $\\boxed{(4,\\frac{9}{4})}$; \\\\\n$\\therefore$ In summary, when the coordinates of $M$ are $\\boxed{(2,\\frac{3}{4})}$ or $\\boxed{(4,\\frac{9}{4})}$, the area of $\\triangle AFM$ is half of the area of $\\triangle AEF$.",
    "image_path": "dataset/mm_math/images/MM_Math/53182535_151.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $AB$ is the diameter of $\\odot O$, $\\angle ACD$ is the angle subtended by arc $AD$ on the circumference, $\\angle ACD=30^\\circ$. What is the measure of $\\angle DAB$ in degrees?\n",
    "completion": "\\textbf{Solution:} Connect BD, \\\\\n$\\because$ AB is the diameter of the circle, \\\\\n$\\therefore \\angle ADB=90^\\circ$, \\\\\n$\\because \\angle ACD=\\angle B=30^\\circ$, \\\\\n$\\therefore \\angle DAB=90^\\circ-\\angle B=90^\\circ-30^\\circ=\\boxed{60^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52874717_95.png"
  },
  {
    "solution": "\\boxed{\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $BD$ is a diagonal of rectangle $ABCD$. Extend $BC$ to point $E$ such that $CE=BD$. Connect $AE$. Given $AB=1$ and $\\angle AEB=15^\\circ$, find the length of $AD$.\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect AC and intersect BD at point O. Since quadrilateral ABCD is a rectangle, it follows that AD$\\parallel$BE and AC=BD, thus $OA=\\frac{1}{2}AC$ and $OD=\\frac{1}{2}BD$, therefore $OA=OD$. Consequently, $\\angle ADB=\\angle CAD$ and $\\angle E=\\angle DAE$. Also, since $BD=CE$, it follows that $CE=CA$ and thus $\\angle E=\\angle CAE$. Since $\\angle CAD=\\angle CAE+\\angle DAE=30^\\circ$, it follows that $\\angle ADB=30^\\circ$ and hence $BD=2AB=2$, leading to $AD=\\sqrt{BD^2−AB^2}=\\boxed{\\sqrt{3}}.$",
    "image_path": "dataset/mm_math/images/MM_Math/51614402.png"
  },
  {
    "solution": "\\boxed{(2, 6)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in a Cartesian coordinate system, the graph of the linear function $y=kx+b$ intersects the x-axis and the y-axis at points A and B, respectively, and passes through point $D(-2, 6)$. It intersects the graph of the direct proportion function $y=3x$ at point C, where the x-coordinate of point C is 1. If point E is a moving point on line $OC$, and the area of $\\triangle ADE$ is 6 times the area of $\\triangle BOC$, please directly write down the coordinates of point E.\n",
    "completion": "\\textbf{Solution:}\n\nFrom given, we have: $6S_{\\triangle BOC}=6\\times 2=12$,\\\\\nAs per the diagram: $S_{\\triangle AED}=\\frac{1}{2}\\times 6\\times AE=\\frac{1}{2}\\times 6\\times 4=12$,\\\\\n$\\therefore E(0, 0)$,\\\\\nPoint E is a variable point on line $OC$. According to symmetry, when point $E$ at the origin is symmetrical about point $C$ and falls on $y=3x$, it also satisfies the condition,\\\\\nBased on symmetry about a point, we obtain: $E(2, 6)$,\\\\\nTherefore, the coordinates that satisfy the condition are: $E=\\boxed{(0, 0)}$, $E=\\boxed{(2, 6)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50520008_185.png"
  },
  {
    "solution": "\\boxed{(1, -4)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the parabola $y=x^2+bx+c$ passes through points $A(-1,0)$ and $B(3,0)$. What are the equation of the parabola and the coordinates of its vertex?\n",
    "completion": "\\textbf{Solution:} Substituting the points $A(-1,0)$ and $B(3,0)$ into $y=x^{2}+bx+c$, \\\\\nwe get: $\\begin{cases}1-b+c=0\\\\ 9+3b+c=0\\end{cases}$, \\\\\nSolving, we find: $\\begin{cases}b=-2\\\\ c=-3\\end{cases}$, \\\\\n$\\therefore$ The equation of the parabola is $y=x^{2}-2x-3$, \\\\\n$\\because y=x^{2}-2x-3=(x-1)^{2}-4$, \\\\\n$\\therefore$ The vertex coordinates are $\\boxed{(1, -4)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52978190_213.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given a right-angled $\\triangle ABC$, where $\\angle ACB=90^\\circ$, $AC=BC$, and $AB=4\\sqrt{2}$, let $D$ be the midpoint of $BC$, and $E$ be a point on line segment $AB$. Connect $CE$ and $DE$. What is the minimum value of $CE+DE$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, construct the symmetric point \\(C'\\) of point \\(C\\) with respect to \\(AB\\), and connect \\(C'D\\), which intersects \\(AB\\) at point \\(E\\). Construct \\(DF \\perp CC'\\) at point \\(F\\),\\\\\nthen \\(CE = C'E\\), \\(CE + DE = C'E + DE\\), and the line segment \\(C'D\\) is the minimum value of \\(CE + DE\\).\\\\\nSince \\(\\angle ACB = 90^\\circ\\) and \\(AC = BC\\), \\(AB = 4 \\sqrt{2}\\),\\\\\nthus \\(AC = BC = 4\\).\\\\\nSince \\(D\\) is the midpoint of \\(BC\\),\\\\\nthus \\(CD = BD = \\frac{1}{2} BC = \\frac{1}{2} \\times 4 = 2\\).\\\\\nthus \\(CF = DF = \\sqrt{2}\\), \\(CC' = 2CG = 2 \\times 2\\sqrt{2} = 4\\sqrt{2}\\)\\\\\nthus \\(C'F = CC' - CF = 4\\sqrt{2} - \\sqrt{2} = 3\\sqrt{2}\\)\\\\\nthus \\(C'D = \\sqrt{C'F^2 + FD^2} = \\sqrt{(3\\sqrt{2})^2 + (\\sqrt{2})^2} = \\boxed{2\\sqrt{5}}\\)",
    "image_path": "dataset/mm_math/images/MM_Math/51435470.png"
  },
  {
    "solution": "\\boxed{\\left(-4, 4\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the line $l_{1}: y=k_{1}x+b$ ($k_{1}\\neq 0$) passes through points $A(4,0)$ and $B(0,2)$, and intersects with line $l_2: y=k_{2}x$ ($k_{2}\\neq 0$) at point $P(a,1)$. Point $C$ is on line $l_{1}$. A line $m$ is drawn through $C$ and is perpendicular to the x-axis at point $E$, and line $m$ intersects $l_2$ at point $D$. When $CD=3ED$, find the coordinates of point $C$.\n",
    "completion": "\\textbf{Solution:} Let the point $C(t, -\\frac{1}{2}t+2)$, then the point $D(t, \\frac{1}{2}t)$, and the point $E(t, 0)$, \\\\\n$\\therefore CD=\\left|-\\frac{1}{2}t+2-\\frac{1}{2}t\\right|=\\left|-t+2\\right|$, $DE=\\left|\\frac{1}{2}t\\right|$, \\\\\n$\\because CD=3DE$, \\\\\n$\\therefore \\left|-t+2\\right|=3\\left|\\frac{1}{2}t\\right|$, \\\\\n$\\therefore -t+2=\\frac{3}{2}t$ or $-t+2-\\frac{3}{2}t=0$, \\\\\n$\\therefore t=\\boxed{\\frac{4}{5}}$ or $t=-4$, \\\\\n$\\therefore$ the point $C\\boxed{\\left(\\frac{4}{5}, \\frac{8}{5}\\right)}$ or $\\boxed{\\left(-4, 4\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51699549_170.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the length of line segment $AB=20\\,\\text{cm}$, point C is on line $AB$, and $AC=4\\,\\text{cm}$. M and N are the midpoints of $AC$ and $BC$, respectively. What is the length of $MN$? in cm.\n",
    "completion": "\\textbf{Solution:} Given that $AB=20\\,cm$ and $AC=4\\,cm$,\\\\\nthus, $BC=AB-AC=20-4=16\\,cm$,\\\\\nsince M and N are the midpoints of $AC$ and $BC$, respectively,\\\\\nit follows that $MC=\\frac{1}{2}AC$ and $CN=\\frac{1}{2}BC$,\\\\\nthus, $MN=\\frac{1}{2}(AC+BC)=\\frac{1}{2}(4+16)=\\boxed{10}\\,cm$.",
    "image_path": "dataset/mm_math/images/MM_Math/53404526.png"
  },
  {
    "solution": "\\boxed{D\\left(-6, -\\frac{3}{2}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, $O$ is the origin. The point $A(3, 3)$ lies on the graph of the inverse proportion function $y = \\frac{k}{x}$. Through point $A$, draw $AB \\perp x$-axis at point $B$ and $AC \\perp y$-axis at point $C$. Point $D$ is on the graph of the function in the fourth quadrant. Connect $OD$ and $CD$. If the area of $\\triangle OCD$ is equal to the area of quadrilateral $ABOC$, find the coordinates of point $D$.\n",
    "completion": "\\textbf{Solution:} Since point $A(3, 3)$ lies on the graph of the inverse proportion function $y=\\frac{9}{x}$,\\\\\nit follows that ${S}_{\\text{quadrilateral} ABOC}=3\\times 3=9$,\\\\\nSince $D$ is on the graph of the inverse proportion function $y=\\frac{9}{x}$,\\\\\nlet $D\\left(x, \\frac{9}{x}\\right)$,\\\\\nthus ${S}_{\\triangle OCD}=\\frac{1}{2}OC\\cdot\\left|x_{D}\\right|=\\frac{1}{2}\\times 3\\times \\left(-x_{D}\\right)$,\\\\\nleading to $-\\frac{3}{2}x_{D}=9$,\\\\\nsolving this gives $x_{D}=-6$,\\\\\ntherefore \\boxed{D\\left(-6, -\\frac{3}{2}\\right)}.",
    "image_path": "dataset/mm_math/images/MM_Math/53066069_87.png"
  },
  {
    "solution": "\\boxed{11}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, D is a point on the bisector of $\\angle MAN$. Point B lies on ray AM, with $DE \\perp AM$ at point E, and $DF \\perp AN$ at point F. Connect AD. Take a point C on ray AN such that $DC=DB$. If $AB=7$ and $BE=2$, what is the length of $AC$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, since $D$ is a point on the bisector of $\\angle MAN$, and $DE \\perp AM$ at point $E$, $DF \\perp AN$ at point $F$, therefore $\\angle DEB = \\angle DFC = 90^\\circ$, $DE=DF$, and $\\angle BAD=\\angle CAD$. In right triangles $\\triangle DEB$ and $\\triangle DFC$, we have $\\left\\{\\begin{array}{l}BD=DC\\\\ DE=DF\\end{array}\\right.$, therefore right $\\triangle DEB \\cong$ right $\\triangle DFC$ (HL), hence $\\angle DBE=\\angle DCF$, and therefore $\\angle ABD=\\angle ACD$. In $\\triangle ABD$ and $\\triangle ACD$, we have $\\left\\{\\begin{array}{l}\\angle BAD=\\angle CAD\\\\ \\angle ABD=\\angle ACD\\\\ DB=DC\\end{array}\\right.$, therefore $\\triangle ABD \\cong \\triangle ACD$ (AAS), hence $AC=AB=7$. In right triangles $\\triangle DEA$ and $\\triangle DFA$, we have $\\left\\{\\begin{array}{l}AD=DA\\\\ DE=DF\\end{array}\\right.$, therefore right $\\triangle DEA \\cong$ right $\\triangle DFA$ (HL), hence $AF=AE=2+7=9$, therefore $AC=AF+CF=9+2=11$. In summary, the length of $AC$ is $\\boxed{7}$ or $\\boxed{11}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52845727.png"
  },
  {
    "solution": "\\boxed{\\frac{10}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, M is the midpoint of CD, EM$\\perp$CD. If CD$=4$ and EM$=6$, what is the radius of the circle containing arc $\\overset{\\frown}{CED}$?\n",
    "completion": "\\textbf{Solution:} Connect OC, \\\\\n$\\because$ M is the midpoint of CD, and EM $\\perp$ CD, \\\\\n$\\therefore$ EM passes through the center point O of $\\odot$O, CM = $\\frac{1}{2}$CD = 2, \\\\\nLet the radius be x, \\\\\n$\\because$ EM = 6, \\\\\n$\\therefore$ OM = EM - OE = 6 - x, \\\\\nIn Rt$\\triangle$OCM, we have OM$^{2}$ + CM$^{2}$ = OC$^{2}$, which gives (6-x)$^{2}$ + $2^{2}$ = x$^{2}$, \\\\\nSolving for x, we get x = $\\frac{10}{3}$. \\\\\n$\\therefore$ The radius of the circle that contains the arc $CED$ is $\\boxed{\\frac{10}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51351843.png"
  },
  {
    "solution": "\\boxed{\\frac{6\\sqrt{10}}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $E$ is the midpoint of $BC$, $DF \\perp AE$ at $F$, with $AB=6$ and $BC=4$. What are the lengths of $AE$ and $DF$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral $ABCD$ is a rectangle,\\\\\nwe have $\\angle BAE+\\angle DAE=90^\\circ$, and $\\angle ADF+\\angle DAF=90^\\circ$,\\\\\nthus, $\\angle BAE=\\angle ADF$,\\\\\nand since $\\angle AFD=\\angle B=90^\\circ$,\\\\\nwe obtain $\\triangle ADF\\sim \\triangle EAB$,\\\\\nwhich implies $\\frac{AD}{AE}=\\frac{DF}{AB}$,\\\\\nknowing that $E$ is the midpoint of $BC$ and $BC=4$,\\\\\nwe find $BE=2$,\\\\\nthus, $AE=\\sqrt{6^2+2^2}=2\\sqrt{10}$,\\\\\nleading to $\\frac{4}{2\\sqrt{10}}=\\frac{DF}{6}$,\\\\\nsolving this, we find: $DF=\\boxed{\\frac{6\\sqrt{10}}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51379170.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Steel balls are commonly used in engineering to measure the width of small circular holes in parts. Suppose the diameter of the steel ball is 10mm, and the distance from the top of the steel ball to the surface of the part is measured to be 8mm, as shown in the figure. What is the length of the width $AB$ of this small circular hole?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, point O is the center of the circle, and through point O, draw OC$\\perp$AB,\\\\\nAccording to the perpendicular diameter theorem, it follows that: AC=BC,\\\\\n$\\because$The diameter is 10mm,\\\\\n$\\therefore$OA=5mm, OC=8−5=3mm,\\\\\nIn $\\triangle AOC$, $\\angle OCA=90^\\circ$,\\\\\n$\\therefore$ $AC=\\sqrt{OA^{2}-OC^{2}}=\\sqrt{5^{2}-3^{2}}=4$,\\\\\n$\\therefore$AB=2AC=8mm,\\\\\nThus, the answer is: $AB=\\boxed{8}$mm.",
    "image_path": "dataset/mm_math/images/MM_Math/52719996.png"
  },
  {
    "solution": "\\boxed{20}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the Cartesian coordinate system xOy, the graph of the linear function $y = -\\frac{1}{2}x + 5$, denoted by $l_1$, intersects the x and y axes at points A and B, respectively. The graph of the direct proportionality function, denoted by $l_2$, intersects $l_1$ at point C$(m, 4)$. What is the area $S_{\\triangle AOC}$ of $\\triangle AOC$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $CD\\perp AO$ at D, then $CD=4$. Since points A and B are on the linear function $y=-\\frac{1}{2}x+5$, let $y=0$, then $x=0$, $x=10$, thus $O(0,0)$, $A(10,0)$, hence $AO=10$. Therefore, the area of $\\triangle AOC$ is $\\frac{1}{2}\\times 10\\times 4=\\boxed{20}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50676403_168.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, the line BC intersects the x-axis at point C and the y-axis at point B, and it intersects the line $y=ax$ at point A. The x-coordinate of point A is 2, $\\angle ACO = 45^\\circ$, and the area of $\\triangle ABO$ is 1. What is the value of $a$ and the equation of the line BC?\n",
    "completion": "\\textbf{Solution:} Let the coordinates of point A be $(x_A, y_A)$,\n\n$\\because$ $S_{\\triangle ABO} = \\frac{1}{2} BO \\cdot x_A = \\frac{1}{2} \\times BO \\times 2 = 1$,\n\n$\\therefore$ $BO = 1$,\n\n$\\therefore$ $B(0, 1)$,\n\n$\\because$ $\\angle ACO = 45^\\circ$, $\\angle BOC = 90^\\circ$,\n\n$\\therefore$ $\\angle CBO = \\angle ACO = 45^\\circ$,\n\n$\\therefore$ OC = BO = 1\n\n$\\therefore$ C$(-1,0)$\n\nLet the analytic expression for BC be $y=kx+b$, where $k\\neq 0$\n\nSubstituting the coordinates of points B and C into $y=kx+b$, we get: $\\left\\{ \\begin{array}{l} b=1 \\\\ -k+b=0 \\end{array} \\right.$\n\n$\\therefore$ $\\left\\{ \\begin{array}{l} k=1 \\\\ b=1 \\end{array} \\right.$\n\n$\\therefore$ The analytic expression for BC is $y=x+1$,\n\n$\\because$ when $x=2$, $y=3$,\n\n$\\therefore$ $A(2, 3)$,\n\nAlso $\\because$ A is on the line $y=ax$,\n\n$\\therefore$ $a=\\boxed{\\frac{3}{2}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52056412_200.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, within $ \\triangle ABC$, $DE \\parallel BC$, $AD=9$, $DB=3$, and $DE=6$. What is the length of $BC$?\n",
    "completion": "\\textbf{Solution:} Given that $DE \\parallel BC$, \\\\\nthen $\\triangle ADE \\sim \\triangle ABC$, \\\\\nhence $\\frac{AD}{AB} = \\frac{DE}{BC}$, \\\\\nsince $AD = 9$, and $DB = 3$, \\\\\nthen AB = AD + BD = 12, \\\\\nalso, because $DE = 6$, \\\\\ntherefore, $\\frac{9}{12} = \\frac{6}{BC}$, \\\\\nthus, BC = \\boxed{8}.",
    "image_path": "dataset/mm_math/images/MM_Math/53106639.png"
  },
  {
    "solution": "\\boxed{5^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AD$ is the angle bisector, and $AE$ is the height. Given that $\\angle B=50^\\circ$ and $\\angle C=60^\\circ$, then what is the degree measure of $\\angle EAD$?\n",
    "completion": "\\[ \\textbf{Solution:} \\]\n\\[ \\because \\angle BAC = 180^\\circ - \\angle B - \\angle C, \\]\n\\[ \\therefore \\angle BAC = 180^\\circ - 50^\\circ - 60^\\circ = 70^\\circ; \\]\n\\[ \\because \\text{AD is the bisector of the angle}, \\]\n\\[ \\therefore \\angle DAC = \\frac{1}{2} \\angle BAC = 35^\\circ; \\]\n\\[ \\because \\text{AE is the altitude}, \\]\n\\[ \\therefore \\angle AEC = 90^\\circ, \\]\n\\[ \\therefore \\angle EAC = 90^\\circ - \\angle C = 90^\\circ - 60^\\circ = 30^\\circ, \\]\n\\[ \\therefore \\angle EAD = \\angle DAC - \\angle EAC = 35^\\circ - 30^\\circ = \\boxed{5^\\circ}. \\]",
    "image_path": "dataset/mm_math/images/MM_Math/52807793.png"
  },
  {
    "solution": "\\boxed{12\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, it is known that in the equilateral $\\triangle ABC$, $BE$ and $CD$ are medians on sides $AC$ and $AB$, respectively, and $BE$ and $CD$ intersect at point $O$. Points $M$ and $N$ are the midpoints of line segments $OB$ and $OC$, respectively. Quadrilateral $DENM$ is a rectangle; if the side length of the equilateral $\\triangle ABC$ is $12$, find the area of rectangle $DENM$.\n",
    "completion": "\\textbf{Solution:} Given $\\triangle ABC$ with side length 12, and BE as the median,\\\\\nthus $BC=AC=12$, $\\angle BEC=90^\\circ$, CE=6,\\\\\nthus $DE=MN=\\frac{1}{2}BC=6$,\\\\\nusing the Pythagorean theorem we get $BE=\\sqrt{BC^{2}-EC^{2}}=6\\sqrt{3}$,\\\\\nfrom (1) we know, OM=OE, with M being the midpoint of OB,\\\\\nthus $BM=OM=OE$,\\\\\nthus $OM=OE=\\frac{1}{3}BE=2\\sqrt{3}$,\\\\\nsince quadrilateral DENM is a rectangle, and $\\angle DEO=30^\\circ$,\\\\\nthus $\\angle OEN=60^\\circ$, implying $\\triangle OEN$ is an equilateral triangle, hence EN=OE=2$\\sqrt{3}$,\\\\\nthus the area of rectangle DENM $=EN \\times DE=2\\sqrt{3} \\times 6=\\boxed{12\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53313103_108.png"
  },
  {
    "solution": "\\boxed{50^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given $\\angle AOB = 100^\\circ$, rays $OE$ and $OF$ are the angle bisectors of $\\angle AOC$ and $\\angle COB$, respectively. If ray $OC$ rotates around point $O$ within $\\angle AOB$, what is the degree measure of $\\angle EOF$?\n",
    "completion": "\\textbf{Solution:} Given that,\n\\[\n\\because \\angle AOB=100^\\circ,\n\\]\n\\[\n\\therefore \\angle AOC+\\angle COB=100^\\circ,\n\\]\n\\[\n\\because OE \\text{ is the bisector of } \\angle AOC, OF \\text{ is the bisector of }\\angle COB,\n\\]\n\\[\n\\therefore \\angle COE=\\frac{1}{2}\\angle AOC, \\angle COF=\\frac{1}{2}\\angle COB,\n\\]\n\\[\n\\therefore \\angle EOF=\\angle COE+\\angle COF=\\frac{1}{2}\\left(\\angle AOC+\\angle COB\\right)=\\boxed{50^\\circ}.\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/50518431_213.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in rhombus $ABCD$, $O$ is the intersection point of the two diagonals. Three lines passing through point $O$ divide the rhombus into shaded and unshaded areas. When the side length of the rhombus is $5$ and one of the diagonals is $8$, what is the area of the shaded region?\n",
    "completion": "\\textbf{Solution:} Draw lines AC and BD as shown in the figure: \\\\\n$\\because$ Quadrilateral ABCD is a rhombus, \\\\\n$\\therefore$ AB=5, OA=OC=$\\frac{1}{2}$AC=4, OB=OD, and AC$\\perp$BD, \\\\\n$\\therefore$ OB=$\\sqrt{AB^{2}-OA^{2}}=\\sqrt{5^{2}-4^{2}}=3$, \\\\\n$\\therefore$ BD=2OB=6, \\\\\n$\\therefore$ The area of rhombus ABCD= $\\frac{1}{2}$AC$\\times$BD= $\\frac{1}{2}\\times6\\times8=24$, \\\\\n$\\because$ O is the intersection point of the diagonals of the rhombus, and a rhombus is a centrally symmetric figure, \\\\\n$\\therefore$ The area of the shaded part=$\\frac{1}{2}\\times24=\\boxed{12}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53313404.png"
  },
  {
    "solution": "\\boxed{12cm}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, $AD \\parallel BC$, $\\angle B = 70^\\circ$, $\\angle C = 40^\\circ$, $DE \\parallel AB$ intersects $BC$ at point $E$. If $AD = 5\\,cm$, $BC = 17\\,cm$, then how many $cm$ is $CD$?\n",
    "completion": "\\textbf{Solution:} Given that $AD\\parallel BC$, $DE\\parallel AB$,\\\\\ntherefore, quadrilateral ABED is a parallelogram,\\\\\nthus, BE=AD=5,\\\\\ntherefore, EC=BC-BE=17cm-5cm=12cm,\\\\\nsince $DE\\parallel AB$,\\\\\nwe have $\\angle DEC=\\angle B=70^\\circ$,\\\\\ngiven that $\\angle C=40^\\circ$,\\\\\nthus, $\\angle EDC=180^\\circ-\\angle DEC-\\angle C=70^\\circ$,\\\\\ntherefore, $\\angle EDC=\\angle DEC$,\\\\\nthus, CD=CE=\\boxed{12cm}.",
    "image_path": "dataset/mm_math/images/MM_Math/52019513.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated, $\\triangle PAC \\cong \\triangle PBD$, given that $\\angle A=40^\\circ$ and $\\angle BPD=20^\\circ$, what is the measure of $\\angle PCD$?\n",
    "completion": "Solution: Since $\\triangle PAC \\cong \\triangle PBD$,\\\\\nit follows that $\\angle A=\\angle B, \\angle APC=\\angle BPD, \\angle PCA=\\angle PDB$\\\\\nSince $\\angle A=40^\\circ, \\angle BPD=20^\\circ$\\\\\nit follows that $\\angle APC=20^\\circ$\\\\\nTherefore, $\\angle PCD=\\angle A+\\angle APC=40^\\circ+20^\\circ=\\boxed{60^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52933303.png"
  },
  {
    "solution": "\\boxed{-\\frac{25}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the vertex $O$ of the rhombus $ABOC$ is at the origin, the side $BO$ is on the negative half axis of the $x$ axis, and the coordinates of vertex $C$ are $(-3, 4)$. The graph of the inverse proportion function $y = \\frac{k}{x}$ intersects the diagonal $AO$ of the rhombus at point $D$. Line $BD$ is drawn. When $BD \\perp x$ axis, what is the value of $k$?\n",
    "completion": "\\textbf{Solution:} Given that $C(-3,4)$,\\\\\nhence $OC= \\sqrt{3^{2}+4^{2}}=5$,\\\\\nsince the quadrilateral OBAC is a rhombus,\\\\\nit follows that $AC=OB=OC=5$, and $AC\\parallel OB$,\\\\\nthus $B(-5,0)$, $A(-8,4)$,\\\\\nlet the equation of line OA be $y=mx$,\\\\\nsubstituting $A(-8,4)$ into it gives $-8m=4$, leading to $m= -\\frac{1}{2}$,\\\\\nhence, the equation of line OA is $y= -\\frac{1}{2}x$,\\\\\nfor $x=-5$, we have $y= -\\frac{1}{2}x = \\frac{5}{2}$, therefore $D(-5,\\frac{5}{2})$,\\\\\nsubstituting $D(-5,\\frac{5}{2})$ into $y= \\frac{k}{x}$,\\\\\nthus $k= -\\frac{5\\times 5}{2} = -\\frac{25}{2}$,\\\\\nso, the answer is: $k=\\boxed{-\\frac{25}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50496331.png"
  },
  {
    "solution": "\\boxed{\\text{Indeterminate}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $l_{1}: y=2x+2$ intersects the $x$-axis at point $A$ and the $y$-axis at point $C$; the line $l_{2}: y=kx+b$ intersects the $x$-axis at point $B(3, 0)$ and intersects line $l_{1}$ at point $D$, with the ordinate of point $D$ being 4. What is the solution set of the inequality $kx+b>2x+2$?\n",
    "completion": "\\textbf{Solution:} Given the inequality $kx+b>2x+2$.\n\nFirst, convert the inequality into standard form, that is, move all terms to one side to obtain $kx - 2x + b - 2 > 0$.\n\nTo solve the inequality, we need to determine the values of $k$ and $b$. However, these values are not provided in the question, hence we cannot directly find the solution set for this inequality.\n\nHowever, the given information is $x<1$, which does not directly assist us in resolving the original problem, as there is insufficient information to determine the solution set of the inequality $kx+b>2x+2$.\n\n\\boxed{\\text{Indeterminate}}, due to the lack of specific information about $k$ and $b$.",
    "image_path": "dataset/mm_math/images/MM_Math/50555324_94.png"
  },
  {
    "solution": "\\boxed{(0,3)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, two lines $l_1$ and $l_2$ passing through point $A(2, 0)$ intersect the $y$-axis at points $B$ and $C$, respectively, where point $B$ is above the origin and point $C$ is below the origin. It is known that $AB = \\sqrt{13}$. What are the coordinates of point $B$?\n",
    "completion": "\\textbf{Solution:} Given that $\\because A(2,0)$, \\\\\n$\\therefore OA=2$, \\\\\nIn the right-angled triangle $Rt\\triangle AOB$, $\\angle AOB=90^\\circ$, \\\\\n$OB=\\sqrt{AB^{2}-OA^{2}}=\\sqrt{13-4}=3$, \\\\\n$\\therefore$ the coordinates of point B are $\\boxed{(0,3)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52056156_204.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the perpendiculars to the x-axis and y-axis through the point $M(-3, 2)$ intersect the graph of the inverse proportion function $y=\\frac{4}{x}$ at points A and B, respectively. What is the area of the quadrilateral MAOB?\n",
    "completion": "\\textbf{Solution}: As shown in the figure\\\\\nSubstitute $y=2$ into $y=\\frac{4}{x}$ to find $x=2$\\\\\n$\\therefore B\\left(2,2\\right)$, $C\\left(0,2\\right)$\\\\\nSubstitute $x=-3$ into $y=\\frac{4}{x}$ to find $y=-\\frac{4}{3}$\\\\\n$\\therefore A\\left(-3,-\\frac{4}{3}\\right)$, $D\\left(-3,0\\right)$\\\\\n$\\therefore {S}_{\\text{quadrilateral }MAOB}={S}_{\\triangle BOC}+{S}_{\\triangle AOD}+{S}_{\\text{rectangle }MDOC}$\\\\\n$=\\frac{1}{2}OC\\times BC+\\frac{1}{2}OD\\times AD+OD\\times OC$\\\\\n$=\\frac{1}{2}\\times 2\\times 2+\\frac{1}{2}\\times 3\\times \\frac{4}{3}+3\\times 2$\\\\\n$=\\boxed{10}$",
    "image_path": "dataset/mm_math/images/MM_Math/51446463.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle CEF$, $\\angle E=80^\\circ$, $\\angle F=55^\\circ$, $AB\\parallel CF$, $AD\\parallel CE$, connect BC and CD, then what is the degree measure of $\\angle A$?\n",
    "completion": "\\textbf{Solution}: Connect AC and extend it to intersect EF at point M,\\\\\nsince AB$\\parallel$CF, and AD$\\parallel$CE,\\\\\ntherefore, $\\angle$3=$\\angle$1, $\\angle$2=$\\angle$4,\\\\\ntherefore, $\\angle$BAD=$\\angle$3+$\\angle$4=$\\angle$1+$\\angle$2=$\\angle$FCE,\\\\\nsince $\\angle$FCE=$180^\\circ$−$\\angle$E−$\\angle$F=$180^\\circ−80^\\circ−55^\\circ=45^\\circ$,\\\\\ntherefore, $\\angle$BAD=\\boxed{45^\\circ}.",
    "image_path": "dataset/mm_math/images/MM_Math/52693977.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the figure, in $\\triangle ABC$, $AB=CB=6$, $BD\\perp AC$ at point D, $F$ is on $BC$ and $BF=2$. Connect AF, with E being the midpoint of AF. Connect DE. What is the length of DE?\n",
    "completion": "\\textbf{Solution:} Given that $CB=6, BF=2$, \\\\\nthus $CF=CB-BF=4$, \\\\\nsince $AB=CB=6, BD\\perp AC$, \\\\\nit follows that $AD=CD$ (the three sides of an isosceles triangle are equal), meaning point $D$ is the midpoint of $AC$,\\\\\nsince $E$ is the midpoint of $AF$,\\\\\nthus $DE$ is the midsegment of $\\triangle ACF$,\\\\\ntherefore $DE=\\frac{1}{2}CF=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52354868.png"
  },
  {
    "solution": "\\boxed{x<2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the function $y=mx+\\frac{4}{3}$ is a straight line $l_{1}$, and the graph of the function $y=kx+b$ is a straight line $l_{2}$. The lines $l_{1}$ and  $l_{2}$ intersect the x-axis at point $B$ and point $C$ ($3$, $0$) respectively, and intersect the y-axis at point $D$ and $E$ respectively. Lines $l_{1}$ and $l_{2}$ intersect at point $A$ ($2$, $2$). Directly write down the solution set of the inequality $mx+\\frac{4}{3}<kx+b$.\n",
    "completion": "\\textbf{Solution:} Since lines $l_1$ and $l_2$ intersect at point $A(2, 2)$,\\\\\nit is known from the graph that when $x<2$, $mx+\\frac{4}{3}<kx+b$,\\\\\ntherefore, the solution set of the inequality $mx+\\frac{4}{3}<kx+b$ is $x<2$;\\\\\nHence, the answer is: $\\boxed{x<2}$;",
    "image_path": "dataset/mm_math/images/MM_Math/50589493_155.png"
  },
  {
    "solution": "\\boxed{1:2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, $E$ is the midpoint of $AB$, and $EC$ intersects $BD$ at point $F$. What is the ratio of the area of $\\triangle BEF$ to the area of $\\triangle BCF$?\n",
    "completion": "\\textbf{Solution:} Given that $ABCD$ is a parallelogram,\\\\\nit follows that $BE\\parallel CD$, \\\\\nthus $\\angle BEC=\\angle DCF$, $\\angle EBF=\\angle CDF$, \\\\\nwhich means $\\triangle EBF \\sim \\triangle CDF$,\\\\\nhence $\\frac{EF}{CF}=\\frac{BE}{CD}=\\frac{1}{2}$,\\\\\nFor triangles $BEF$ and $BCF$ taking $EF$ and $CF$ as the base respectively, the heights are the same, \\\\\ntherefore $S_{\\triangle BEF} : S_{\\triangle BCF} = EF : CF = 1:2$, \\\\\nThus, the answer is: $\\boxed{1:2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52602101.png"
  },
  {
    "solution": "\\boxed{\\left(\\frac{12}{5}, \\frac{22}{5}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, line $ {l}_{1}$ passes through points A($-2$, $0$) and B($0$, $2$), and line $ {l}_{2}: y=-4x+14$ intersects $ {l}_{1}$ at point C. Find the coordinates of point C.\n",
    "completion": "\\textbf{Solution}: Since line $l_1$ intersects with line $l_2$ at point C, we have $\\begin{cases}y=x+2\\\\ y=-4x+14\\end{cases}$. Solving these equations yields $\\begin{cases}x=\\frac{12}{5}\\\\ y=\\frac{22}{5}\\end{cases}$. Therefore, the coordinates of point C are $\\boxed{\\left(\\frac{12}{5}, \\frac{22}{5}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52252074_72.png"
  },
  {
    "solution": "\\boxed{\\frac{8\\sqrt{5}+12}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, it illustrates an ancient trebuchet capable of long-range attacks, while Figure 2 depicts a schematic diagram of the trebuchet in action at a given moment. Here, $GP$ stands for the lever, with the sling attached at point $G$ and the counterweight hanging at point $P$. Point $A$ serves as the fulcrum, and point $D$ is located on the horizontal base $BC$, where $AD=AC=3$ meters, and $CD=3.6$ meters. During the operation of the trebuchet, the extension line of $AP$ intersects segment $DC$ at point $E$. Assuming the ratio $DE:CE=5:1$, how many meters above the ground is point $G$?\n",
    "completion": "\\textbf{Solution}: The point $G$ is $\\boxed{\\frac{8\\sqrt{5}+12}{5}}$ meters away from the ground.",
    "image_path": "dataset/mm_math/images/MM_Math/52808005_37.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{58}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ABC=45^\\circ$, $AB=6\\sqrt{2}$, $AC=8$, $BC>6$. Points $E$ and $F$ are located on sides $BC$ and $AC$, respectively, and $AF=CE$. What is the minimum value of $AE+BF$?\n",
    "completion": "\\textbf{Solution:} Construct line $AG \\parallel BC$ through point $A$ such that $AG=AC$, and connect $FG$, $BG$. \\\\\nSince $AG \\parallel BC$, \\\\\nthen $\\angle GAF=\\angle ACE$. \\\\\nSince $AF=CE$, $AG=AC$, \\\\\nthen $\\triangle ACE$ is congruent to $\\triangle GAF$, \\\\\nthus $AE=GF$, \\\\\nhence $AE+BF=GF+BF$. \\\\\nWhen points $B$, $F$, $G$ are collinear, $BF+GF$ is minimized. \\\\\nConstruct $AI \\perp BC$ through point $A$, intersecting $BC$ at point $I$, and construct $GH \\perp BC$ through point $G$, intersecting the extended line of $BC$ at point $H$. \\\\\nSince $\\angle ABC=45^\\circ$, \\\\\nthen $AI=BI=6$. \\\\\nIt is easy to see that quadrilateral $AIHG$ is a rectangle, thus $GH=AI=6$, $IH=AG=8$, \\\\\ntherefore $BH=BI+IH=6+8=14$. \\\\\nIn $\\triangle BGH$, $BG= \\sqrt{BH^2+GH^2}=\\sqrt{14^2+6^2}=2\\sqrt{58}$. \\\\\nThus, the minimum value of $AE+BF$ is $\\boxed{2\\sqrt{58}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52874507.png"
  },
  {
    "solution": "\\boxed{rectangle}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\square ABCD$, $\\triangle ACE$ is constructed with $AC$ as the hypotenuse, and $\\angle BED$ is a right angle. Prove that: $\\square ABCD$ is a rectangle.\n",
    "completion": "\\textbf{Solution.} Proof: Connect OE\\\\\n$\\because$ quadrilateral ABCD is a parallelogram,\\\\\n$\\therefore$ AO=CO, BO=DO.\\\\\n$\\because$ $\\angle$AEC=$\\angle$BED=90$^\\circ$,\\\\\n$\\therefore$ AC=2OE, BD=2OE,\\\\\n$\\therefore$ AC=BD,\\\\\n$\\therefore$ $\\square ABCD$ is a \\boxed{rectangle}.",
    "image_path": "dataset/mm_math/images/MM_Math/51614760.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{1}{2}x+2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system, the parabola $y=\\frac{1}{2}x^{2}+bx+c$ (where $b$ and $c$ are constants) has its vertex at $\\left(\\frac{3}{2}, -\\frac{25}{8}\\right)$ and intersects the $x$-axis at points $A$ and $B$ (with point $A$ to the left of point $B$) and intersects the $y$-axis at point $C$. Point $C$ and point $D$ are symmetric about the $x$-axis. Connect points $A$ and $D$, and draw line $BD$. What is the equation of line $BD$?\n",
    "completion": "\\textbf{Solution:} Let $x=0$, then $y=-2$, hence $C(0, -2)$,\\\\\nsince point C and point D are symmetric with respect to the x-axis,\\\\\ntherefore $D(0, 2)$,\\\\\nlet the equation of the line $BD$ be $y=mx+n$, by the problem statement, we have\\\\\n\\[\n\\begin{cases}\nn=2\\\\\n4m+n=0\n\\end{cases}\n\\], solving this we get \\[\n\\begin{cases}\nm=-\\frac{1}{2}\\\\\nn=2\n\\end{cases}\n\\],\\\\\nthus, the equation of the line $BD$ is $\\boxed{y=-\\frac{1}{2}x+2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53079018_50.png"
  },
  {
    "solution": "\\boxed{70^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $AD \\parallel BC$, $\\angle B=35^\\circ$, and $DB$ bisects $\\angle ADE$, what is the measure of $\\angle DEC$?\n",
    "completion": "\\textbf{Solution:} Since $AD\\parallel BC$ and $\\angle B=35^\\circ$, \\\\\nit follows that $\\angle ADB=\\angle B=35^\\circ$, \\\\\nSince DB bisects $\\angle ADE$, \\\\\nit follows that $\\angle ADE=2\\angle ADB=70^\\circ$, \\\\\nSince $AD\\parallel BC$, \\\\\nit follows that $\\angle DEC =\\angle ADE =\\boxed{70^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52732599.png"
  },
  {
    "solution": "\\boxed{\\frac{2}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a rectangular paper ABCD (AD > DC) is folded along a straight line passing through point D, so that point A falls on side BC at point E, and the crease intersects side AB at point F. If BE = 1, EC = 2, then what is the value of $\\sin\\angle EDC$?\n",
    "completion": "\\textbf{Solution}: We have $\\sin\\angle EDC = \\boxed{\\frac{2}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51155328_176.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $G$ is the centroid of $\\triangle ABC$, extending $BG$ to intersect $AC$ at point $D$, and extending $CG$ to intersect $AB$ at point $E$. $P$ and $Q$ are respectively the centroids of $\\triangle BCE$ and $\\triangle BCD$, the length of $BC$ is $12$, then what is the length of $PQ$?\n",
    "completion": "\\textbf{Solution:} Draw line EP and extend EP to intersect BC at point F, connect DE,\\\\\nsince point G is the centroid of $\\triangle ABC$,\\\\\ntherefore, BE and CE are the medians of $\\triangle ABC$,\\\\\ntherefore, DE is the midsegment of $\\triangle ABC$,\\\\\ntherefore, DE=$\\frac{1}{2}BC=\\frac{1}{2}\\times 12=6$,\\\\\nsince P and Q are the centroids of $\\triangle BCE$ and $\\triangle BCD$ respectively,\\\\\ntherefore, the extensions of EP and DQ intersect at point F, and point F is on BC,\\\\\ntherefore, $\\frac{PF}{EF}=\\frac{FQ}{DF}=\\frac{1}{3}$,\\\\\nsince $\\angle PFQ=\\angle DFE$,\\\\\ntherefore, $\\triangle PFQ\\sim \\triangle DFE$,\\\\\ntherefore, $\\frac{PF}{EF}=\\frac{PQ}{DE}=\\frac{1}{3}=\\frac{PQ}{6}$,\\\\\nSolving gives: $PQ=\\boxed{2}$",
    "image_path": "dataset/mm_math/images/MM_Math/52861184.png"
  },
  {
    "solution": "\\boxed{85^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $OB$ is the bisector of $\\angle AOC$, $\\angle COD = \\frac{1}{3}\\angle BOD$, $\\angle COD = 17^\\circ$. What is the measure of $\\angle AOD$?\n",
    "completion": "\\textbf{Solution:} Given $\\because$ $\\angle COD = \\frac{1}{3} \\angle BOD$, and $\\angle COD = 17^\\circ$,\\\\\n$\\therefore$ $\\angle BOC = 2\\angle COD = 2 \\times 17^\\circ = 34^\\circ$,\\\\\nsince OB is the bisector of $\\angle AOC$,\\\\\n$\\therefore$ $\\angle AOC = 2\\angle BOC = 2 \\times 34^\\circ = 68^\\circ$,\\\\\n$\\therefore$ $\\angle AOD = \\angle AOC + \\angle COD = 68^\\circ + 17^\\circ = \\boxed{85^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/51284781.png"
  },
  {
    "solution": "\\boxed{9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $DE$ is the perpendicular bisector of side $AC$ of $\\triangle ABC$, with the foot of the perpendicular at point $E$, intersecting $AB$ at point $D$. Connect $CD$, $\\angle B=30^\\circ$, $CD\\perp BC$, $CD=3$, then what is the length of $AB$?\n",
    "completion": "\\textbf{Solution:} Since $DE$ is the perpendicular bisector of side $AC$ of $\\triangle ABC$, and $CD=3$, \\\\\nit follows that $AD=CD=3$, \\\\\nBecause $\\angle B=30^\\circ$ and $CD\\perp BC$, \\\\\nit follows that $BD=2CD=6$, \\\\\nHence, $AB=AD+BD=\\boxed{9}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53066383.png"
  },
  {
    "solution": "\\boxed{10cm}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, if $CB=4\\,\\text{cm}$, $DB=7\\,\\text{cm}$, and $D$ is the midpoint of $AC$, find the length of $AB$.\n",
    "completion": "\\textbf{Solution:} Since $CB = 4$cm, $DB = 7$cm,\\\\\nthus $DC = DB - CB = 3$cm.\\\\\nAlso, since $D$ is the midpoint of $AC$,\\\\\nthus $AD = DC = 3$cm,\\\\\ntherefore $AB = AD + DB = \\boxed{10cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51366778.png"
  },
  {
    "solution": "\\boxed{\\frac{27}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, $AC=6$, $BC=9$, and point D is the midpoint of side $BC$. The $\\triangle ACD$ is folded along $AD$ so that point C falls on point $C'$ in the same plane. Connect $BC'$. What is the length of $BC'$?\n",
    "completion": "\\textbf{Solution:} Suppose the intersection point of CC' and AD is E, as shown in the figure: \\\\\nSince point D is the midpoint on side BC, and BC = 9, \\\\\nThus, CD = BD = 4.5, \\\\\nSince $\\triangle ACD$ is folded along AD, making point C coincide with point C' in the same plane, \\\\\nThus, CD = C'D, $CC'\\perp AD$, CE = C'E, \\\\\nThus, CD = BD = C'D, \\\\\nThus, $\\angle DCC'=\\angle DC'C$, $\\angle DBC'=\\angle DC'B$, \\\\\nSince $\\angle DCC'+\\angle DC'C+\\angle DBC'+\\angle DC'B=180^\\circ$, \\\\\nThus, $\\angle CC'B=\\angle CC'D+\\angle DC'B= \\frac{1}{2} (\\angle DCC'+\\angle DC'C+\\angle DBC'+\\angle DC'B)=90^\\circ$, \\\\\nSince $\\angle C=90^\\circ$ and AC = 6, \\\\\nThus, AD = $\\sqrt{AC^{2}+CD^{2}}$ = 7.5, \\\\\nSince $S_{\\triangle ACD}$ = $\\frac{1}{2}$ AC$\\cdot$CD = $\\frac{1}{2}$ AD$\\cdot$CE, \\\\\nThus, CE = $\\frac{AC\\cdot CD}{AD}$ = 3.6, \\\\\nThus, CC' = 7.2, \\\\\nIn Rt$\\triangle BC'C$, BC' = $\\sqrt{BC^{2}-CC'^{2}}= \\boxed{\\frac{27}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50584300.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $BD$ is the diameter of the circle $\\odot O$. A tangent to $\\odot O$ at point A intersects the extension of line $BD$ at point C. Lines $AB$ and $AD$ are drawn. If $\\angle ABC=30^\\circ$, what is the measure of $\\angle C$?\n",
    "completion": "\\textbf{Solution:} Connect $OA$,\n\nsince $OA=OB$,\n\nit follows that $\\angle OAB=\\angle OBA=30^\\circ$,\n\ntherefore $\\angle AOC=\\angle OAB+\\angle OBA=60^\\circ$,\n\nsince $OA$ is the tangent to the circle $O$,\n\nit follows that $\\angle OAC=90^\\circ$,\n\ntherefore $\\angle C=180^\\circ-\\angle OAC-\\angle AOC=180^\\circ-90^\\circ-60^\\circ=\\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53019902.png"
  },
  {
    "solution": "\\boxed{\\frac{4}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the direct proportion function $y=kx$ intersects with the graph of the inverse proportion function $y=\\frac{m}{x}$ ($x>0$) at point $A(2, 2)$. By translating the line $y=kx$ downwards, we obtain line $l$. If line $l$ intersects with the graph of the inverse proportion function at point $B(3, n)$. What are the values of $m$ and $n$?\n",
    "completion": "\\textbf{Solution}: From the given information, substituting point A$(2,2)$ into the inverse proportion function $y=\\frac{m}{x}$, we get: $m=2\\times 2=4$,\\\\\n$\\therefore y=\\frac{4}{x}$. Then, substituting B$(3,n)$ into $y=\\frac{4}{x}$, we find: $n=\\frac{4}{3}$;\\\\\nHence, $m=\\boxed{4}$, $n=\\boxed{\\frac{4}{3}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52204038_127.png"
  },
  {
    "solution": "\\boxed{4.8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, quadrilateral ABCD is a rhombus with $AC=8\\,\\text{cm}$ and $DB=6\\,\\text{cm}$. If $DH \\perp AB$ at H, then what is the length of DH?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, let AC and BD intersect at point O, \\\\\n$\\because$ Quadrilateral ABCD is a rhombus, AC$=8\\text{cm}$, DB$=6\\text{cm}$,\\\\\n$\\therefore$AC$\\perp$BD, OA$=$OC$=4\\text{cm}$, OB$=$OD$=3\\text{cm}$,\\\\\n$\\therefore$ $AB=\\sqrt{OA^{2}+OB^{2}}=\\sqrt{4^{2}+3^{2}}=5\\text{cm}$,\\\\\n$\\because$DH$\\perp$AB,\\\\\n$\\therefore$the area of the rhombus $=AB\\cdot DH=\\frac{1}{2}AC\\cdot BD=\\frac{1}{2}\\times 8\\times 6=24\\text{cm}^{2}$,\\\\\n$\\therefore$ $DH=\\boxed{4.8}\\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52212478.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the Cartesian coordinate system $xOy$, points $A$ and $B$ lie on the line $y=\\frac{1}{2}x$, as shown in the figure. The graph of the quadratic function $y=ax^2+bx-2$ also passes through points $A$ and $B$, and intersects the $y$-axis at point $C$. If line $BC$ is parallel to the $x$-axis, and the $x$-coordinate of point $A$ is $2$. Let the vertex of the graph of this quadratic function be $M$, calculate the value of $\\tan\\angle AMC$?\n",
    "completion": "\\textbf{Solution:} Draw $CH\\perp AM$ at point $C$, with the foot of the perpendicular being $H$.\\\\\nLet the equation of line $AM$ be $y=kx+b$,\\\\\nthen $\\left\\{\\begin{array}{l}2k+b=1 \\\\ -2k+b=-3\\end{array}\\right.$,\\\\\nwhich gives us $k=1$ and $b=-1$.\\\\\nTherefore, the equation of line $AM$ is $y=x-1$.\\\\\nLet the intersections of line $AM$ with the $x$-axis and $y$-axis be points $P$ and $Q$ respectively,\\\\\nthen the coordinates of point $P$ are $(1, 0)$, and the coordinates of point $Q$ are $(0, -1)$,\\\\\ntherefore, $\\triangle OPQ$ is an isosceles right triangle, $\\angle OQP=45^\\circ$, $OQ=OP=1$,\\\\\nsince $\\angle OQP=\\angle HQC$,\\\\\nit follows that $\\angle HQC=45^\\circ$,\\\\\ngiven that the coordinates of point $C$ are $(0, -2)$,\\\\\nit follows that $CQ=1$,\\\\\ntherefore, $HC=HQ=\\frac{\\sqrt{2}}{2}$,\\\\\nand $MQ=\\sqrt{(-2-0)^2+(-3-(-1))^2}=2\\sqrt{2}$,\\\\\ntherefore, $MH=MQ-HQ=\\frac{3\\sqrt{2}}{2}$,\\\\\nthus, $\\tan\\angle AMC=\\frac{HC}{MH}=\\boxed{\\frac{1}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51379457_209.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, right-angled triangles $AOB$ and $COD$ are positioned as depicted (with the right-angled vertices coinciding). Given that $\\angle AOC=45^\\circ$, what is the measure of $\\angle DEB$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle AOC=45^\\circ$ and $\\angle C=45^\\circ$,\\\\\nit follows that $\\angle AFE=\\angle AOC+\\angle C=90^\\circ$.\\\\\nSince $\\angle A=30^\\circ$,\\\\\nit follows that $\\angle AEF=60^\\circ$,\\\\\ntherefore, $\\angle DEB=\\angle AEF=\\boxed{60^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53147637.png"
  },
  {
    "solution": "\\boxed{\\frac{15}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the inverse proportion function $y = \\frac{k_1}{x}$ intersects with the graph of the direct proportion function $y = k_2x$ at points $A(a, 1)$ and $B$. The point $M(a-3, a)$ lies on the graph of the inverse proportion function. Connect $OM$ and extend $BM$ to intersect the y-axis at point $N$. What is the area of $\\triangle BOM$?\n",
    "completion": "\\textbf{Solution:} Since the graph of the inverse proportion function $y=\\frac{k_1}{x}$ intersects with the graph of the direct proportion function $y=k_2x$ at points A and B, and given A(4, 1),\\\\\nit follows that the coordinates of point B are $(-4, -1)$,\\\\\nAssuming the equation of line $BM$ is $y=mx+b$,\\\\\nSubstituting points $B(-4, -1)$ and $M(1, 4)$ we have\\\\\n$\\left\\{\n\\begin{array}{ll}\n-4m+b=-1\\\\\nm+b=4\n\\end{array}\n\\right.$,\\\\\nSolving this gives $\\left\\{\n\\begin{array}{ll}\nm=1\\\\\nb=3\n\\end{array}\n\\right.$,\\\\\ntherefore, the equation of line $BM$ is $y=x+3$,\\\\\nwhen $x=0$, $y=3$,\\\\\nthus, the coordinates of point N are (0, 3),\\\\\nAs shown in the figure, vertical lines to the y-axis are drawn through points M and B, with the feet of the perpendiculars being points P and Q, respectively,\\\\\nthus, $PM=1$, $BQ=4$,\\\\\nTherefore, $S_{\\triangle BOM}=S_{\\triangle BON}+S_{\\triangle MON}=\\frac{1}{2}\\times 3\\times 4+\\frac{1}{2}\\times 3\\times 1=\\boxed{\\frac{15}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53043068_130.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the side length of the square $ABCD$ is $4$, and the point $P$ is on the diagonal $AC$ (not including endpoints). With $PA$ and $PB$ as adjacent sides, a square $PAQB$ is formed. What is the minimum value of the diagonal length $PQ$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, let AB and PQ intersect at point O, and draw $OE\\perp AC$ through point $O$, \\\\\nsince quadrilateral $AQBP$ is a parallelogram, \\\\\ntherefore $AO=OB=\\frac{1}{2}AB=2$, \\\\\nsince quadrilateral $ABCD$ is a square, \\\\\ntherefore $\\angle AOE=\\angle OAE=45^\\circ$, \\\\\nthus, OE=AE, \\\\\ntherefore $OE^{2}+AE^{2}=AO^{2}$, \\\\\nthus, OE=$\\frac{\\sqrt{2}}{2}AO$, \\\\\naccording to the problem, $P$ is a moving point on $AC$, when point P coincides with point E, PO reaches its minimum value. \\\\\nAt this time, $PQ=2OE=2\\times \\frac{\\sqrt{2}}{2}AO=\\boxed{2\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52354643.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, two flat strips of paper, each with a width of 1, are overlaid in a cross fashion. The angle between the edges of the two strips is $\\alpha =30^\\circ$. What is the area of their overlapping part (the shaded area in the figure)?\n",
    "completion": "\\textbf{Solution:} As shown in the figure: Draw AE$\\perp$BC and AF$\\perp$CD at F, with the feet of the perpendiculars being E and F,\\\\\n$\\therefore$ $\\angle$AEB=$\\angle$AFD=$90^\\circ$,\\\\\n$\\because$AD$\\parallel$CB and AB$\\parallel$CD,\\\\\n$\\therefore$quadrilateral ABCD is a parallelogram,\\\\\n$\\because$the width of the strip is 1,\\\\\n$\\therefore$AE=AF=1,\\\\\nIn $\\triangle$ABE and $\\triangle$ADF\\\\\n$\\left\\{\\begin{array}{l}\n\\angle ABE=\\angle ADF=30^\\circ \\\\\n\\angle AEB=\\angle AFD=90^\\circ \\\\\nAE=AF\n\\end{array}\\right.$,\\\\\n$\\therefore$ $\\triangle$ABE$\\cong$ $\\triangle$ADF (AAS),\\\\\n$\\therefore$AB=AD,\\\\\n$\\therefore$quadrilateral ABCD is a rhombus.\\\\\n$\\therefore$BC=AB,\\\\\n$\\because$ $\\frac{AE}{AB}=\\sin 30^\\circ$,\\\\\n$\\therefore$BC=AB=$\\frac{1}{\\sin 30^\\circ}=2$,\\\\\n$\\therefore$the area of the overlapping part (the shaded area in the figure) is: BC×AE=2×1=\\boxed{2},",
    "image_path": "dataset/mm_math/images/MM_Math/51212700.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the diameter of $\\odot O$ is 4, and $\\angle ACB = 45^\\circ$. What is the length of AB?\n",
    "completion": "\\textbf{Solution:} Connect OA and OB, as shown in the figure,\\\\\n$\\because$ $\\angle AOB = 2\\angle ACB = 2\\times 45^\\circ = 90^\\circ$,\\\\\n$\\therefore$ $\\triangle AOB$ is an isosceles right triangle,\\\\\n$\\therefore$ AB = $\\sqrt{2}$OA = $2\\sqrt{2}$.\\\\\nTherefore, AB = \\boxed{2\\sqrt{2}}.",
    "image_path": "dataset/mm_math/images/MM_Math/53080774.png"
  },
  {
    "solution": "\\boxed{2\\alpha}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In triangles $AOB$ and $COD$, $\\angle AOB = \\angle COD$. When $\\angle AOB = \\alpha$, the two triangles are combined as shown in the diagram. Express $\\angle AOC + \\angle BOD$ using an algebraic formula involving $\\alpha$.\n",
    "completion": "\\textbf{Solution:} Given:\n\\[\n\\because \\angle AOC = \\angle AOB + \\angle BOC = \\alpha + \\angle BOC,\n\\]\n\\[\n\\angle DOB = \\angle DOC - \\angle BOC = \\alpha - \\angle BOC,\n\\]\n\\[\n\\therefore \\angle AOC + \\angle DOB = \\alpha + \\angle BOC + \\alpha - \\angle BOC = \\boxed{2\\alpha}.\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/17008253_59.png"
  },
  {
    "solution": "\\boxed{3\\sqrt{15}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that quadrilateral $ABCD$ is inscribed in circle $\\odot O$, with radius $AO=6$. The diagonals $AC$ and $BD$ intersect at point $E$, and $AB=BD$, $EC=2$. Then what is the length of $AD$?\n",
    "completion": "\\textbf{Solution:} Connect BO and extend it to meet AD at point F, draw OD, and CD, \\\\\n$\\because$BA=BD, OA=OD, \\\\\n$\\therefore$BF is the perpendicular bisector of line segment AD, \\\\\n$\\therefore$BF$\\perp$AD, \\\\\n$\\because$AC is the diameter of $\\odot$O, \\\\\n$\\therefore$$\\angle$ADC$=90^\\circ$, \\\\\nthat is, AD$\\perp$DC, \\\\\n$\\therefore$BF$\\parallel$CD, \\\\\n$\\therefore$$\\triangle $BOE$\\sim $$\\triangle $DCE, \\\\\n$\\therefore$$\\frac{OB}{CD}=\\frac{EO}{EC}$, \\\\\n$\\because$AO=6, EC=2, \\\\\n$\\therefore$OB=6, OC=6, \\\\\n$\\therefore$OE=4, \\\\\n$\\therefore$$\\frac{6}{CD}=\\frac{4}{2}$, \\\\\nSolving, we get CD=3, \\\\\nIn the right triangle ADC, $\\angle$ADC$=90^\\circ$, AC=12, CD=3, \\\\\n$\\therefore$AD=$\\sqrt{AC^2−CD^2}=\\sqrt{12^2−3^2}=\\boxed{3\\sqrt{15}}$, \\\\",
    "image_path": "dataset/mm_math/images/MM_Math/51322738.png"
  },
  {
    "solution": "\\boxed{y=-x-2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given in the diagram, points $A(-4, 2)$ and $B(n, -4)$ are two intersection points of the graph of the linear function $y=kx+b$ and the graph of the inverse proportion function $y=\\frac{m}{x}$. Find the equations of the linear function and the inverse proportion function?\n",
    "completion": "\\textbf{Solution:} Given that $A(-4, 2)$ lies on the curve of $y=\\frac{m}{x}$,\\\\\n$\\therefore m=-8$.\\\\\n$\\therefore$ The equation of the inverse proportion function is $y=-\\frac{8}{x}$.\\\\\nGiven that point $B(2, -4)$ is on $y=-\\frac{8}{x}$,\\\\\n$\\therefore n=2$.\\\\\n$\\therefore B(2, -4)$.\\\\\nGiven $y=kx+b$ passes through $A(-4, 2)$ and $B(2, -4)$,\\\\\n$\\therefore \\left\\{\\begin{array}{l} -4k+b=2 \\\\ 2k+b=-4 \\end{array}\\right.$.\\\\\nSolving yields: $\\left\\{\\begin{array}{l} k=-1 \\\\ b=-2 \\end{array}\\right.$.\\\\\n$\\therefore$ The linear function equation is $\\boxed{y=-x-2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52622329_234.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{2} + 1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In $\\triangle ABC$, $AD$ is the altitude from $A$ to side $BC$, $\\angle C=45^\\circ$, $\\sin B=\\frac{1}{3}$, and $AD=1$. Find the length of $BC$?\n",
    "completion": "\\textbf{Solution:} Given $\\tan C = \\frac{AD}{DC}$, it follows that $\\tan 45^\\circ = \\frac{1}{DC} = 1$,\\\\\nthus, DC = 1.\\\\\nSince $\\sin B = \\frac{AD}{AB} = \\frac{1}{3}$, this implies $\\frac{1}{AB} = \\frac{1}{3}$,\\\\\nthus, AB = 3.\\\\\nIn right triangle $ABD$, we have $AD^2 + BD^2 = AB^2$,\\\\\ntherefore, $BD = \\sqrt{AB^2 - AD^2} = \\sqrt{3^2 - 1^2} = \\sqrt{8} = 2\\sqrt{2}$.\\\\\nHence, BC = BD + DC = $2\\sqrt{2} + 1 = \\boxed{2\\sqrt{2} + 1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51323853.png"
  },
  {
    "solution": "\\boxed{120^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, point $O$ is on line $AB$, $\\angle COD$ is a right angle, and $OE$ bisects $\\angle BOC$. As shown in figure 2, if $\\angle COE = \\angle DOB$, find the measure of $\\angle AOC$.\n",
    "completion": "\\textbf{Solution:} Given that $OE$ bisects $\\angle BOC$,\\\\\nthus $\\angle COE = \\angle BOE$,\\\\\nand since $\\angle COE = \\angle DOB$ and $\\angle COD = 90^\\circ$,\\\\\nit follows that $\\angle COE = \\angle DOB = \\angle BOE = 30^\\circ$,\\\\\nthus $\\angle BOC = 60^\\circ$,\\\\\ntherefore $\\angle AOC = 180^\\circ - 60^\\circ = \\boxed{120^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50938639_220.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC$, point $P$ is a moving point on the bisector $CD$ of $\\angle ACB$, $CE=\\frac{1}{2}BC=8$, and the area of $\\triangle ABC$ is $48$. What is the minimum value of $PA+PE$?\n",
    "completion": "\\textbf{Solution:} Since CD is the bisector of $\\angle ACB$,\\\\\nwe construct the symmetric point $E'$ of point E with respect to line CD on BC, and draw line AE', where its intersection with line CD is point P.\\\\\nSince CD perpendicularly bisects EF,\\\\\nwe have PE=PE', and CE=CE',\\\\\nthus PA+PE=PA+PE'=AE'.\\\\\nSince $CE=\\frac{1}{2}BC=8$,\\\\\nwe get $BC=16$,\\\\\nwhich means point E' is the midpoint of BC.\\\\\nMoreover, since AB=AC,\\\\\nwe have AE'$\\perp$BC.\\\\\nGiven that the area of $\\triangle ABC$ is 48,\\\\\nwe deduce ${S}_{\\triangle ABC}=\\frac{1}{2}BC\\cdot AE'=\\frac{1}{2}\\times 16\\times AE'=48$,\\\\\nthus AE'=6,\\\\\ntherefore, the minimum value of PA+PE is $\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51367328.png"
  },
  {
    "solution": "\\boxed{3\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the quadrilateral $ABCD$ is inscribed in the circle $\\odot O$, $\\angle ABC=135^\\circ$, $OE\\perp AC$, and $\\angle AOE=\\angle D$; if $AC=6$, what is the radius of $\\odot O$?\n",
    "completion": "\\textbf{Solution}: From (1), we know that $\\triangle AOC$ is an isosceles right triangle, which means $\\angle OAC=\\angle OCA=45^\\circ$, \\\\\nBecause $AC=6$, \\\\\nTherefore, $OA=OC=6\\times \\sin 45^\\circ=3\\sqrt{2}$, \\\\\nHence, the radius of circle $O$ is $\\boxed{3\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53080615_96.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC$, $AD$ is the angle bisector of $\\triangle ABC$, and $E$ is the midpoint of $AC$. Connect $DE$. If $DE = 3$, then what is the value of $AB$?\n",
    "completion": "\\textbf{Solution:} Since $AB=AC$ and $AD$ bisects $\\angle BAC$, \\\\\nit follows that $D$ is the midpoint of $BC$, \\\\\nand since $E$ is the midpoint of $AC$, \\\\\nit follows that $AB=2DE=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51368926.png"
  },
  {
    "solution": "\\boxed{y=\\frac{2}{x}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, the graph of the inverse proportion function $y = \\frac{k}{x}$ passes through points $A(1, m)$ and $B(2, 1)$. The coordinates of point $P$ are $(6, 1)$. What is the expression of the inverse proportion function?\n",
    "completion": "\\textbf{Solution:} Since the graph of the inverse proportion function $y=\\frac{k}{x}$ passes through point $B(2,1)$,\\\\\nhence $k=2\\times 1=2$,\\\\\ntherefore, the equation of the inverse proportion function is $\\boxed{y=\\frac{2}{x}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51445690_116.png"
  },
  {
    "solution": "\\boxed{70^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral ABCD, connect AC and BD. If $\\triangle ABC$ is an equilateral triangle, with $AB=BD$ and $\\angle ABD=20^\\circ$, then what is the measure of $\\angle BDC$?\n",
    "completion": "\\textbf{Solution:} Given an equilateral $\\triangle ABC$ where $AB=BC$, $\\angle ABC=60^\\circ$, \\\\\nsince $AB=BD$, $\\angle ABD=20^\\circ$, \\\\\nit follows that $BD=BC$, $\\angle CBD=60^\\circ-20^\\circ=40^\\circ$, \\\\\nthus, $\\angle BDC=(180^\\circ-40^\\circ)\\div 2=\\boxed{70^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52991670.png"
  },
  {
    "solution": "\\boxed{20}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $BC=120$, the height $AD=60$, and square $EFGH$ has one side on $BC$, with points $E$, $F$ respectively on $AB$, $AC$, and $AD$ intersects $EF$ at point $N$. What is the length of $AN$?\n",
    "completion": "\\textbf{Solution:} Let the side length EF=EH=x of the square EFGH, \\\\\n$\\because$ EFGH is a square, \\\\\n$\\therefore$ $\\angle$HEF=$\\angle$EHG=90$^\\circ$, EF$\\parallel$BC, \\\\\n$\\therefore$ $\\triangle$AEF$\\sim$ $\\triangle$ABC, \\\\\n$\\because$ AD is the altitude of $\\triangle$ABC, \\\\\n$\\therefore$ $\\angle$HDN=90$^\\circ$, \\\\\n$\\therefore$ the quadrilateral EHDN is a rectangle, \\\\\n$\\therefore$ DN=EH=x, \\\\\n$\\because$ $\\triangle$AEF$\\sim$ $\\triangle$ABC, \\\\\n$\\therefore$ $\\frac{AN}{AD}=\\frac{EF}{BC}$, \\\\\n$\\because$ BC=120, AD=60, \\\\\n$\\therefore$ AN=60−x, \\\\\n$\\therefore$ $\\frac{60−x}{60}=\\frac{x}{120}$, \\\\\nSolving for x, we get: x=40, \\\\\n$\\therefore$ AN=60−x=60−40=\\boxed{20}.",
    "image_path": "dataset/mm_math/images/MM_Math/51322868.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AD \\perp BC$ at point $D$. Point $E$ is on $AC$, and $BE$ intersects $AD$ at point $F$. If $BF=AC$ and $DF=DC$, what is the sum of $\\angle 1$ and $\\angle 2$?\n",
    "completion": "\\textbf{Solution:} Since $AD\\perp BC$ at point $D$,\\\\\nit follows that $\\angle BDF=\\angle ADC=90^\\circ$,\\\\\nIn $Rt\\triangle BDF$ and $Rt\\triangle ADC$,\\\\\n$\\left\\{\\begin{array}{l}BF=AC\\\\ DF=DC\\end{array}\\right.$,\\\\\ntherefore $Rt\\triangle BDF \\cong Rt\\triangle ADC\\left(HL\\right)$,\\\\\nthus $\\angle DBF=\\angle 2$, $BD=AD$,\\\\\nhence $\\angle DBA=\\angle DAB=45^\\circ$,\\\\\nthus $\\angle 1+\\angle 2=\\angle 1+\\angle DBF=\\angle DBA=45^\\circ$,\\\\\ntherefore the sum of $\\angle 1$ and $\\angle 2$ is $\\boxed{45^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53009747.png"
  },
  {
    "solution": "\\boxed{-\\frac{1}{2}x + \\frac{5}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, within a Cartesian coordinate system, points $E$, $F$, and $G$ are located on the sides of rectangle $ABCO$. When triangle $EFO$ is folded along $EF$, point $O$ coincides exactly with point $G$. $GH$ is perpendicular to the x-axis at point $H$, and point $M$ is the intersection of $GH$ and $EF$. Given that $CG=2$ and $B(6,4)$, determine the analytical expression of line $EF$.\n",
    "completion": "\\textbf{Solution:} Since $GH \\perp x$-axis, and $OC \\perp x$-axis,\\\\\ntherefore $GH \\parallel OC$, thus $\\angle OFE = \\angle GMF$.\\\\\nSince $\\angle OFE = \\angle GFE$, therefore $\\angle GFM = \\angle GMF$,\\\\\nthus $GF = GM = \\frac{5}{2}$, and $HM = 4 - \\frac{5}{2} = \\frac{3}{2}$.\\\\\nTherefore $M(2, \\frac{3}{2})$.\\\\\nLet the equation of line $EF$ be $y = kx + b$.\\\\\nTherefore $\\left\\{ \\begin{array}{l} b = \\frac{5}{2}, \\\\ 2k + b = \\frac{3}{2}, \\end{array} \\right.$ thus $\\left\\{ \\begin{array}{l} k = -\\frac{1}{2}, \\\\ b = \\frac{5}{2}. \\end{array} \\right.$\\\\\nThus, the equation of line $EF$ is $y = -\\frac{1}{2}x + \\frac{5}{2} = \\boxed{-\\frac{1}{2}x + \\frac{5}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52019464_108.png"
  },
  {
    "solution": "\\boxed{28}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, it is known that $AB$ is the diameter of $\\odot O$, $\\triangle ABC$ is inscribed in $\\odot O$, $AB=10$, $BC=6$, and point $D$ is a moving point on $\\odot O$ (point $D$ does not coincide with points $A$ or $B$).\n\nAs shown in Figure 2, if $CD$ bisects $\\angle ACB$, and $AD$ is connected, determine the area of $\\triangle ACD$.\n",
    "completion": "\\textbf{Solution:} Draw line BD and let AG $\\perp$ CD at point G,\\\\\nsince AB is the diameter,\\\\\nit follows that $\\angle ACB=\\angle ADB=90^\\circ$.\\\\\nSince CD bisects $\\angle ACB$,\\\\\nwe have $\\angle ACD=\\angle BCD=45^\\circ$,\\\\\nwhich implies $\\overset{\\frown}{AD}=\\overset{\\frown}{BD}$,\\\\\nhence $AD=BD=\\frac{\\sqrt{2}}{2}AB=5\\sqrt{2}$, $AC=\\sqrt{AB^{2}-BC^{2}}=8$,\\\\\nthus $AG=CG=\\frac{\\sqrt{2}}{2}AC=4\\sqrt{2}$,\\\\\nand therefore $DG=\\sqrt{AD^{2}-AG^{2}}=3\\sqrt{2}$,\\\\\nwhich means $CD=DG+CG=7\\sqrt{2}$,\\\\\nso $S_{\\triangle ACD}=\\frac{1}{2}CD\\cdot AG=\\frac{1}{2}\\times 7\\sqrt{2}\\times 4\\sqrt{2}=\\boxed{28}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52793804_149.png"
  },
  {
    "solution": "\\boxed{\\frac{40}{7}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\mathrm{Rt}\\triangle ABC$ where $\\angle ACB=90^\\circ$, $AB=25$, and $AC=15$, point $P$ starts from point $B$ and moves toward $A$ at a speed of $2$ units per second. At the same time, point $Q$ starts from $C$ and moves toward $B$ at a speed of $1$ unit per second. When one point reaches its destination, the other point stops moving. Connect $PQ$. On ray $PB$, take $PN=PQ$ and construct $\\square PQMN$ using $PN$ and $PQ$ as sides. Let the movement time be $t$ seconds ($t>0$). Find the value of $t$ when $\\square PQMN$ is a square.\n",
    "completion": "\\textbf{Solution:} As shown in the figure,\\\\\n$\\because$ Quadrilateral PQMN is a square,\\\\\n$\\therefore$ $\\angle$BPQ$=90^\\circ$,\\\\\n$\\therefore$ $\\triangle BPQ \\sim \\triangle BCA$,\\\\\n$\\therefore$ $\\frac{PB}{BQ}=\\frac{BC}{AB}=\\frac{20}{25}=\\frac{4}{5}$,\\\\\n$\\therefore$ BP$=\\frac{4}{5}$BQ,\\\\\n$\\because$ BP$=2t$, CQ$=t$,\\\\\n$\\therefore$ $2t=\\frac{4}{5}(20-t)$,\\\\\n$\\therefore$ $t=\\boxed{\\frac{40}{7}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51445941_163.png"
  },
  {
    "solution": "\\boxed{y=\\frac{4}{x}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the linear function $y_{1}=x+b$ and the inverse proportion function $y_{2}=\\frac{k}{x}$ intersect at points $A\\left(1, a\\right)$ and $D\\left(-4, -1\\right)$, and intersect the $y$-axis and $x$-axis at points $B$ and $C$, respectively. What is the expression of the inverse proportion function?\n",
    "completion": "\\textbf{Solution:} Since point $D(-4, -1)$ lies on the graph of the inverse proportion function $y=\\frac{k}{x}$,\\\\\ntherefore $k=-4\\times (-1)=4$,\\\\\nthus, the expression for the inverse proportion function is $\\boxed{y=\\frac{4}{x}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51379335_196.png"
  },
  {
    "solution": "\\boxed{9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, points D and E are on segments AB and AC, respectively, with $DE\\parallel BC$, and $AD: AB = 1: 3$. If the area of $\\triangle ADE$ is 1, what is the area of $\\triangle ABC$?\n",
    "completion": "\\textbf{Solution:} Given that $DE \\parallel BC$,\n\n$\\therefore$ $\\triangle ADE \\sim \\triangle ABC$,\n\n$\\therefore$ $\\frac{S_{\\triangle ADE}}{S_{\\triangle ABC}}=\\left(\\frac{AD}{AB}\\right)^2=\\left(\\frac{1}{3}\\right)^2$,\n\n$\\therefore$ $\\frac{1}{S_{\\triangle ABC}}=\\frac{1}{9}$,\n\n$\\therefore$ $S_{\\triangle ABC}=\\boxed{9}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52807991.png"
  },
  {
    "solution": "\\boxed{-\\frac{1}{2}x+5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, the line $y=kx+b$ and the inverse proportion function $y=\\frac{8}{x}$ (where $x>0$) intersect at points $A(m, 4)$ and $B(8, n)$, and intersect the coordinate axes at points $C$ and $D$ respectively. What is the analytical expression of line $AB$?\n",
    "completion": "\\textbf{Solution:} From the given information, point $A\\left(2, 4\\right)$ and point $B\\left(8, 1\\right)$ lie on the graph of $y=\\frac{8}{x}$,\\\\\n$\\therefore m=\\frac{8}{4}=2$, $n=\\frac{8}{8}=1$,\\\\\nthat is, $A\\left(2, 4\\right)$, $B\\left(8, 1\\right)$.\\\\\nSubstituting points $A\\left(2, 4\\right)$ and $B\\left(8, 1\\right)$ into $y=kx+b$ yields\\\\\n$\\left\\{\\begin{array}{l}\n4=2k+b \\\\\n1=8k+b\n\\end{array}\\right.$, solving for which we get: $\\left\\{\\begin{array}{l}\nk=-\\frac{1}{2} \\\\\nb=5\n\\end{array}\\right.$,\\\\\nTherefore, the equation of line $AB$ is: $y=-\\frac{1}{2}x+5=\\boxed{-\\frac{1}{2}x+5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51403084_60.png"
  },
  {
    "solution": "\\boxed{22.5^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within square $ABCD$, point $E$ is located on diagonal $BD$, and $BE=AB$. Connect $CE$ and $AE$. Then, what is the measure of $\\angle DAE$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral ABCD is a square,\\\\\nit follows that $\\angle ABD=45^\\circ$, $\\angle BAD=90^\\circ$,\\\\\nsince BE=AB,\\\\\nit follows that $\\angle BAE=\\angle BEA= \\frac{1}{2} \\times (180^\\circ-45^\\circ)=67.5^\\circ$,\\\\\ntherefore, $\\angle DAE=\\angle BAD-\\angle BAE=90^\\circ-67.5^\\circ=\\boxed{22.5^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53729175.png"
  },
  {
    "solution": "\\boxed{\\frac{50}{9}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that in rhombus $ABCD$, $AB=5$, $\\cos B=\\frac{3}{5}$. Points E and F are on sides $BC$ and $CD$, respectively. The extension of line $AF$ intersects the extension of line $BC$ at point G, and $\\angle EAF=\\frac{1}{2}\\angle BAD$. It is given that $AE^{2}=EC\\cdot EG$; if point F is the midpoint of side $CD$, what is the value of $S_{\\triangle ABE}$?\n",
    "completion": "\\textbf{Solution:} Draw $AH\\perp BC$ through point A, with H as the foot of the perpendicular.\\\\\nSince quadrilateral $ABCD$ is a rhombus, $\\therefore$ $AD=AB=BC=CD=5$. $\\therefore$ $\\frac{AD}{CG}=\\frac{DF}{CF}$.\\\\\nSince point F is the midpoint of side $CD$, $\\therefore$ $DF=CF$. $\\therefore$ $CG=AD=5$.\\\\\nLet $BE=x$, then $EC=5-x$.\\\\\n$\\therefore$ $EG=EC+CG=5-x+5=10-x$.\\\\\nSince $AE^2=EC\\cdot EG$, $AE^2=(5-x)(10-x)$.\\\\\nIn $\\triangle ABH$ with a right angle, $\\cos B=\\frac{BH}{AB}=\\frac{3}{5}$,\\\\\nand since $AB=5$, $\\therefore$ $BH=3$. $\\therefore$ $AH=4$, $EH=3-x$.\\\\\nIn $\\triangle AEH$ with a right angle, $AE^2=AH^2+EH^2$, $\\therefore$ $AE^2=(3-x)^2+4^2$.\\\\\n$\\therefore$ $(5-x)(10-x)=(3-x)^2+4^2$. Solving this, we get $x=\\frac{25}{9}$. $\\therefore$ the area of $\\triangle ABE$ is $\\frac{1}{2}BE\\cdot AH=\\frac{1}{2}\\times \\frac{25}{9}\\times 4=\\boxed{\\frac{50}{9}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52942187_155.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of the circle $O$, and the chord $CD\\perp AB$ with the foot of the perpendicular being $E$. Connect $AC$ and $BC$. If $\\angle BAC=30^\\circ$ and $CD=6\\,\\text{cm}$. What is the degree measure of $\\angle BCD$?\n",
    "completion": "\\textbf{Solution:} Since diameter $AB \\perp CD$, \\\\\nit follows that $\\overset{\\frown}{BC} = \\overset{\\frown}{BD}$,\\\\\ntherefore, $\\angle BCD = \\angle CAB = 30^\\circ$,\\\\\nhence, $\\angle BCD = \\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53009955_43.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, quadrilateral $ABCD$ is an inscribed quadrilateral of circle $\\odot O$, the radius of $\\odot O$ is 2, and $\\angle B=135^\\circ$. What is the measure of $\\angle ADC$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral $ABCD$ is inscribed in circle $O$, and $\\angle B=135^\\circ$,\n\n$\\therefore$ $\\angle ADC=180^\\circ-\\angle B=45^\\circ$,\n\nthus $\\angle ADC=\\boxed{45^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/52871840.png"
  },
  {
    "solution": "\\boxed{20}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, if $\\tan A = \\frac{1}{2}$ and $AB = 10$, then what is the area of $\\triangle ABC$?\n",
    "completion": "\\textbf{Solution:} We have $\\because \\tan A=\\frac{1}{2}=\\frac{BC}{AC}$\\\\\n$\\therefore AC=2BC$\\\\\nBy the Pythagorean theorem, we have $AB^{2}=AC^{2}+BC^{2}$, which gives $100=(2BC)^{2}+BC^{2}$\\\\\nSolving this, we find $BC^{2}=20$\\\\\n$\\therefore {S}_{\\triangle ABC}=\\frac{1}{2}AC\\times BC=BC^{2}=\\boxed{20}$",
    "image_path": "dataset/mm_math/images/MM_Math/52601718.png"
  },
  {
    "solution": "\\boxed{(-1,6)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y= -x^{2} + bx + c$ intersects the x-axis at points $A(2,0)$ and $B(-4,0)$. If the parabola intersects the y-axis at point $C$, is there a point $Q$ on the axis of symmetry of the parabola such that the perimeter of $\\triangle QAC$ is minimized? If such a point exists, find the coordinates of point $Q$; if it does not exist, please explain why.\n",
    "completion": "\\textbf{Solution:} Exist, Reason:\\\\\n$\\because$ the perimeter of $\\triangle QAC = AC + QA + QC$,\\\\\n$\\therefore$ to minimize the perimeter of $\\triangle QAC$, we only need to minimize $QA + QC$,\\\\\nBased on the given information, points A and C are symmetric with respect to the axis $x = -1$,\\\\\n$\\therefore$ the intersection of line $AC$ and line $x = -1$ is point $Q$, at this time $QA + QC$ is minimized, i.e., the perimeter of $\\triangle AQC$ is minimized,\\\\\nFor $y = -x^2 - 2x + 8$, let $x = 0$, then $y = 8$,\\\\\n$\\therefore$ C$(0,8)$,\\\\\nLet the equation of line $AC$ be $y = kx + 8$ ($k \\neq 0$),\\\\\nSubstituting point C$(0,8)$, we get: $-4k + 8 = 0$, solving gives: $k = 2$,\\\\\n$\\therefore$ the equation of line $AC$ is $y = 2x + 8$,\\\\\nWhen $x = -1$, $y = 2 \\times (-1) + 8 = 6$,\\\\\n$\\therefore$ the coordinates of point Q are $\\boxed{(-1,6)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52602758_131.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $E$ is the midpoint of side $BC$ of square $ABCD$. Connect $AE$ and extend it to intersect the extension line of $DC$ at point $F$. Connect $AC$, $BF$, and $\\angle AEC=2\\angle ABC$. Quadrilateral $ABFC$ is a rectangle; if $\\triangle AFD$ is an equilateral triangle with a side length of $4$, what is the area of quadrilateral $ABFC$?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle AFD$ is an equilateral triangle, \\\\\n$\\therefore$ $\\angle AFC=60^\\circ$, AF=DF=4, \\\\\n$\\therefore$ CF=CD=2, \\\\\nSince quadrilateral ABFC is a rectangle, \\\\\n$\\therefore$ $\\angle ACF=90^\\circ$, $\\angle AFC=60^\\circ$, $\\angle FAC=30^\\circ$, \\\\\nBy the Pythagorean theorem, AC=$2\\sqrt{3}$, \\\\\n$\\therefore$ The area of quadrilateral ABFC=AC$\\cdot$CF=\\boxed{4\\sqrt{3}}.",
    "image_path": "dataset/mm_math/images/MM_Math/53359575_114.png"
  },
  {
    "solution": "\\boxed{\\frac{2}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the coordinate system shown, let $O(0,0)$, $A(3,3\\sqrt{3})$, and $B(6,0)$. If $\\triangle OAB$ is folded along line $CD$ such that point $A$ precisely falls on point $E$ on segment $OB$ and given that $OE=\\frac{6}{5}$, what is the value of $AC: AD$?\n",
    "completion": "\\textbf{Solution:} Draw AF$\\perp$OB at F, as shown in the figure: \\\\\nSince A(3, 3$\\sqrt{3}$), B(6, 0),\\\\\ntherefore AF $=$ 3$\\sqrt{3}$, OF $=$ 3, OB $=$ 6,\\\\\ntherefore BF $=$ 3,\\\\\ntherefore OF $=$ BF,\\\\\ntherefore AO $=$ AB,\\\\\nsince $\\tan\\angle AOB = \\frac{AF}{OF} = \\sqrt{3}$,\\\\\ntherefore $\\angle AOB = 60^\\circ$,\\\\\ntherefore $\\triangle AOB$ is an equilateral triangle,\\\\\ntherefore $\\angle AOB = \\angle ABO = 60^\\circ$,\\\\\nsince folding $\\triangle OAB$ along the line CD such that point A precisely falls on the point E on segment OB,\\\\\ntherefore $\\angle CED = \\angle OAB = 60^\\circ$,\\\\\ntherefore $\\angle OCE = \\angle DEB$,\\\\\ntherefore $\\triangle CEO \\sim \\triangle EDB$,\\\\\ntherefore $\\frac{OE}{BD} = \\frac{CE}{ED} = \\frac{CO}{BE}$,\\\\\nsince OE $= \\frac{6}{5}$,\\\\\ntherefore BE $=$ OB - OE $= 6 - \\frac{6}{5} = \\frac{24}{5}$,\\\\\nLet CE $= a$, then CA $= a$, CO $= 6-a$, ED $= b$, then AD $= b$, DB $= 6-b$,\\\\\nthen $\\frac{\\frac{6}{5}}{6-b} = \\frac{a}{b}$, $\\frac{6-a}{\\frac{24}{5}} = \\frac{a}{b}$,\\\\\ntherefore $6b = 30a - 5ab$ (1), $24a = 30b - 5ab$ (2),\\\\\n(2) - (1) gives: $24a - 6b = 30b - 30a$,\\\\\ntherefore $\\frac{a}{b} = \\frac{2}{3}$,\\\\\nthat is AC$\\colon$AD $=$ $\\boxed{\\frac{2}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51131123.png"
  },
  {
    "solution": "\\boxed{x>1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=kx+b$ intersects the line $y=mx+n$ at point $C(1, 5)$. What is the solution set for the inequality $mx+n<kx+b$?\n",
    "completion": "\\textbf{Solution:} From the graph, we can see that to the left of point C, the line $y=mx+n$ is above the line $y=kx+b$, and to the right of point C, the line $y=mx+n$ is below the line $y=kx+b$,\\\\\n$\\because$ The line $y=kx+b$ intersects the line $y=mx+n$ at point C(1, 5),\\\\\n$\\therefore$ The solution set of the inequality $mx+n<kx+b$ is $\\boxed{x>1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51919592.png"
  },
  {
    "solution": "\\boxed{-2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that point A lies on the graph of the inverse proportion function $y=\\frac{k}{x}$ ($x<0$). A line is drawn through point A perpendicular to the y-axis, and the foot of the perpendicular is B. If the area of $\\triangle OAB$ is 1, what is the value of $k$?\n",
    "completion": "\\textbf{Solution:} Since $AB \\perp$ y-axis, \\\\\ntherefore, we have $S_{\\triangle OAB} = \\frac{1}{2}|k| = 1$, \\\\\nand given $k < 0$, \\\\\ntherefore, we find $k = \\boxed{-2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52602900.png"
  },
  {
    "solution": "\\boxed{50^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a rectangular sheet of paper is folded along EF, with points D and C corresponding to positions D' and C', respectively. If $\\angle EFC'=115^\\circ$, what is the measure of $\\angle AED'$?\n",
    "completion": "\\textbf{Solution:} Given $AD \\parallel BC$ and $\\angle EFC' = 115^\\circ$,\\\\\n$\\therefore \\angle DEF = 180^\\circ - 115^\\circ = 65^\\circ$.\\\\\nSince $\\angle D'EF$ is formed by folding $\\angle DEF$,\\\\\n$\\therefore \\angle DED' = 2\\angle DEF = 130^\\circ$,\\\\\n$\\therefore \\angle AED' = 180^\\circ - 130^\\circ = \\boxed{50^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53151168.png"
  },
  {
    "solution": "\\boxed{36}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, point $E$ is the midpoint of side $AD$, and $EC$ intersects the diagonal $BD$ at point $F$. If the area of $\\triangle DEF$ is $3$, what is the area of quadrilateral $ABCD$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral $ABCD$ is a parallelogram,\n\n$\\therefore$ $AD\\parallel BC$.\n\n$\\therefore$ $\\angle EDB=\\angle DBC$, $\\angle DEC=\\angle BCE$\n\n$\\therefore$ $\\triangle EFD\\sim \\triangle CFB$.\n\nAdditionally, since $E$ is the midpoint of $AD$,\n\n$\\therefore$ $DE=\\frac{1}{2}CB$.\n\n$\\therefore$ ${S}_{\\triangle BCF}=4{S}_{\\triangle EDF}=12$.\n\nLet ${S}_{\\triangle DFC}=x$, then $3+x=\\frac{1}{2}(12+x)$,\n\nSolving for $x$ gives: $x=6$.\n\n$\\therefore$ ${S}_{\\triangle BCD}=12+6=18$.\n\n$\\therefore$ ${S}_{\\square ABCD}=2{S}_{\\triangle BCD}=\\boxed{36}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52977891.png"
  },
  {
    "solution": "\\boxed{y=-x^{2}+4x}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, within the Cartesian coordinate system, the line AB passing through point A (4, 0) intersects with the y-axis at point B (0, 4). The parabola $y=-x^2+bx+c$ that passes through the origin O intersects the line AB at points A and C. The vertex of the parabola is D. What is the expression of the parabola $y=-x^2+bx+c$?\n",
    "completion": "Solution: From the problem statement, we have\n\\[\n\\left\\{\n\\begin{array}{l}\nc=0\\\\\n-16+4b+c=0\n\\end{array}\n\\right.\n\\]\nSolving this, we get\n\\[\n\\left\\{\n\\begin{array}{l}\nb=4\\\\\nc=0\n\\end{array}\n\\right.\n\\]\n$\\therefore$ The quadratic function is: \\boxed{y=-x^{2}+4x}.",
    "image_path": "dataset/mm_math/images/MM_Math/52936243_164.png"
  },
  {
    "solution": "\\boxed{35^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, line AB and CD intersect at point O, with OA bisecting $\\angle EOC$. If $\\angle EOC = 70^\\circ$, what is the degree measure of $\\angle BOD$?\n",
    "completion": "\\textbf{Solution:} \\\\\n$\\because$ OA bisects $\\angle EOC$, \\\\\n$\\therefore \\angle AOC= \\frac{1}{2}\\angle EOC$, \\\\\n$\\because \\angle EOC=70^\\circ$, \\\\\n$\\therefore \\angle AOC= \\frac{1}{2}\\times 70^\\circ=35^\\circ$, \\\\\n$\\because$ Straight lines AB and CD intersect at point O, \\\\\n$\\therefore \\angle BOD=\\angle AOC=\\boxed{35^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53223463_89.png"
  },
  {
    "solution": "\\boxed{x<4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given the graph of the linear function $y=kx+b$ ($k\\ne 0$) as shown, what is the solution set of the inequality $kx+b>0$ with respect to $x$?\n",
    "completion": "\\textbf{Solution:} The graph of the function $y=kx+b$ passes through the point $(4,0)$, and the function value $y$ decreases as $x$ increases,\\\\\ntherefore, when $x<4$, the function value is greater than $0$, meaning the solution set for the inequality $kx+b>0$ with respect to $x$ is $\\boxed{x<4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50591574.png"
  },
  {
    "solution": "\\boxed{x>2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y=kx+b$ (where $k\\neq 0$) is shown. Then, what is the solution set of the inequality $kx+b>3$?\n",
    "completion": "\\textbf{Solution:} As can be seen from the graph, the solution set of the inequality $kx+b>3$ is $\\boxed{x>2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50561376.png"
  },
  {
    "solution": "\\boxed{y=\\frac{1}{2}x}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the line $l_{1}: y=k_{1}x+b$ ($k_{1}\\neq 0$) passes through points $A(4,0)$ and $B(0,2)$, and intersects with the line $l_2: y=k_{2}x$ ($k_{2}\\neq 0$) at point $P(a,1)$. What are the equations of lines $l_{1}$ and $l_{2}$?\n",
    "completion": "\\textbf{Solution:} Let $\\because$ the line ${l}_{1}\\colon y={k}_{1}x+b\\left({k}_{1}\\ne 0\\right)$ passes through point A (4, 0) and B (0, 2),\\\\\n$\\therefore$ $\\begin{cases}b=2\\\\ 4{k}_{1}+b=0\\end{cases}$,\\\\\n$\\therefore$ $\\begin{cases}{k}_{1}=-\\frac{1}{2}\\\\ b=2\\end{cases}$,\\\\\n$\\therefore$ the equation of line ${l}_{1}$ is $y=-\\frac{1}{2}x+2$,\\\\\nWhen $y=1$, then $-\\frac{1}{2}x+2=1$,\\\\\n$\\therefore x=2$,\\\\\n$\\therefore$ point P (2, 1),\\\\\n$\\therefore$ $1=2{k}_{2}$,\\\\\n$\\therefore$ ${k}_{2}=\\frac{1}{2}$,\\\\\n$\\therefore$ the equation of line ${l}_{2}$ is $\\boxed{y=\\frac{1}{2}x}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51699549_169.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{6}}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle A = 90^\\circ$, $AD \\perp BC$ at point D. If $BD = 3$ and $CD = 2$, then what is the value of $\\tan B$?\n",
    "completion": "\\textbf{Solution:} Given $\\because$ $\\angle BAC=90^\\circ$, $AD\\perp BC$,\\\\\n$\\therefore$ $\\angle BAC=\\angle ADC=\\angle ADB=90^\\circ$,\\\\\n$\\therefore$ $\\angle B+\\angle BAD=\\angle BAD+\\angle CAD=90^\\circ$,\\\\\n$\\therefore$ $\\angle B=\\angle CAD$,\\\\\n$\\therefore$ $\\triangle ADB\\sim \\triangle CAD$,\\\\\n$\\therefore$ $\\frac{AD}{CD}=\\frac{BD}{AD}$,\\\\\n$\\therefore$ $AD^2=BD\\cdot CD$,\\\\\n$\\because$ $BD=3$, $CD=2$,\\\\\n$\\therefore$ $AD=\\sqrt{6}$,\\\\\n$\\therefore$ $\\tan B=\\frac{AD}{BD}=\\boxed{\\frac{\\sqrt{6}}{3}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/50913440.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in parallelogram $ABCD$, $AE \\perp BC$, $AF \\perp CD$ with perpendicular feet at $E$ and $F$ respectively, $\\angle EAF = 60^\\circ$, and $DF = 3\\,\\text{cm}$. What is the length of $AD$ in cm?\n",
    "completion": "\\textbf{Solution:} Since $AE \\perp BC$ and $AF \\perp CD$, and $\\angle EAF = 60^\\circ$, \\\\\nit follows that $\\angle AEC = \\angle AFC = \\angle AFD = 90^\\circ$,\\\\\nMoreover, since the sum of interior angles of quadrilateral AECF is $360^\\circ$, we get\\\\\n$\\angle C + \\angle AEC + \\angle AFC + \\angle EAF = 360^\\circ$,\\\\\nthus $\\angle C + 90^\\circ + 90^\\circ + 60^\\circ = 360^\\circ$,\\\\\ntherefore $\\angle C = 120^\\circ$,\\\\\nSince quadrilateral ABCD is a parallelogram,\\\\\nit follows that $AD \\parallel BC$,\\\\\nthus $\\angle D = 180^\\circ - \\angle C = 60^\\circ$,\\\\\ntherefore $\\angle DAF = 180^\\circ - \\angle D - \\angle AFD = 30^\\circ$.\\\\\nthus $AD = 2DF = \\boxed{6} \\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53728977.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the right triangle $\\triangle ABC$, the height $AD=4$ on the hypotenuse $BC$, and $\\cos B =\\frac{4}{5}$, what is the length of $AC$?\n",
    "completion": "\\textbf{Solution:} Given in right triangle $ABC$, $AD \\perp BC$,\\\\\n$\\therefore \\angle BAD + \\angle B = \\angle BAD + \\angle DAC = 90^\\circ$,\\\\\n$\\therefore \\angle B = \\angle DAC$,\\\\\nSince $\\cos B = \\frac{4}{5}$,\\\\\n$\\therefore \\cos \\angle DAC = \\frac{AD}{AC} = \\frac{4}{5}$,\\\\\nGiven that $AD = 4$,\\\\\n$\\therefore AC = \\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51285481.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Fig. 1, $Rt\\triangle BAC$ and $Rt\\triangle DAE$ are both isosceles right-angled triangles, and $AB=AC=2\\sqrt{5}$, $AD=AE=2\\sqrt{2}$, $\\angle BAC=\\angle DAE=90^\\circ$. $\\triangle DAE$ rotates counterclockwise around point A; connect CD. When $\\tan\\angle EAC=\\frac{1}{3}$, find the length of CD.\n",
    "completion": "\\textbf{Solution:} Draw $DW \\perp AC$ at $W$,\\\\\nsince $\\angle DAE=90^\\circ$ and $\\tan\\angle EAC=\\frac{1}{3}$,\\\\\nthus $\\angle ADW+\\angle DAW=\\angle DAW+\\angle EAC$,\\\\\nhence $\\angle EAC=\\angle ADW$,\\\\\nin right triangle $\\triangle AWD$ with $\\angle AWD=90^\\circ$,\\\\\nthus $\\tan\\angle ADW=\\frac{AW}{DW}=\\frac{1}{3}$,\\\\\nassume $DW=3k$ and $AW=k$, then $(3k)^2+k^2=(2\\sqrt{2})^2$,\\\\\nhence $AW=\\frac{2\\sqrt{5}}{5}$ and $DW=\\frac{6\\sqrt{5}}{5}$,\\\\\nthus $WC=AC-AW=\\frac{8\\sqrt{5}}{5}$,\\\\\nin right triangle $\\triangle DWC$ with $\\angle DWC=90^\\circ$,\\\\\nthus $DW^2+WC^2=DC^2$,\\\\\ntherefore, $DC=\\boxed{2\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51496741_135.png"
  },
  {
    "solution": "\\boxed{\\frac{16\\pi}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, AB is the diameter of the circle $\\odot O$. The chord CD is perpendicular to AB at point E. G is a moving point on the arc AC and does not coincide with points A and C. The extensions of AG and DC intersect at point F. Connect BC. The length of CD is $4\\sqrt{3}$ and BE is 2. What is the area of the sector DOC?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect OC, \\\\\n$\\because$ in $\\triangle ODE$, $DO=4$, $OE=2$, \\\\\n$\\therefore \\angle ODE=30^\\circ$, \\\\\n$\\because$ $OD=OC$, \\\\\n$\\therefore \\angle OCD=30^\\circ$, \\\\\n$\\therefore \\angle DOC=120^\\circ$, \\\\\n$\\therefore$ the area of sector DOC $=\\frac{120^\\circ \\cdot \\pi \\cdot 16}{360^\\circ}=\\boxed{\\frac{16\\pi}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52733204_137.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of $\\odot O$, with points $C$ and $D$ located on opposite sides of diameter $AB$ on $\\odot O$. Connecting $AC$, $AD$, $BD$, $CD$, if the radius of $\\odot O$ is $5$ and $BD = 8$, then what is the value of $\\sin\\angle ACD$?\n",
    "completion": "\\textbf{Solution:} Given that $AB$ is the diameter of circle $O$,\\\\\nit follows that $\\angle ADB=90^\\circ$\\\\\ntherefore, $\\angle ACD=\\angle ABD$\\\\\nGiven that the radius of circle $O$ is $5$ and $BD=8$,\\\\\nit follows that $\\sin\\angle ACD=\\sin\\angle ABD=\\frac{6}{10}=\\boxed{\\frac{3}{5}}$",
    "image_path": "dataset/mm_math/images/MM_Math/50891003.png"
  },
  {
    "solution": "\\boxed{n}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the inverse proportion function $y=\\frac{1}{x}$ intersects the sides AB and BC of the rectangle ABCO at points E and F, respectively, where points A and C are on the coordinate axes. If $AB=(n+1)AE$, what is the area of the quadrilateral OEBF?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, connect OB.\\\\\n$\\because$ AB=$(n+1)$AE\\\\\n$\\because$ BE＝nAE,\\\\\n$\\therefore$ S$_{\\triangle OBE}$＝n$\\cdot$S$_{\\triangle OAE}$,\\\\\n$\\because$ E, F lie on y＝$\\frac{1}{x}$ and quadrilateral AOCB is a rectangle,\\\\\n$\\therefore$ S$_{\\triangle AEO}$＝S$_{\\triangle OCF}$＝$\\frac{1}{2}$, S$_{\\triangle OBC}$＝S$_{\\triangle OBA}$,\\\\\n$\\therefore$ S$_{\\triangle OBE}$＝S$_{\\triangle OBF}$＝$\\frac{1}{2}$n,\\\\\n$\\therefore$ S$_{\\text{quadrilateral}OEBF}$＝\\boxed{n}.",
    "image_path": "dataset/mm_math/images/MM_Math/53490696.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $AB=6$, $AD=10$. If point $E$ is on side $BC$ and $AE$ bisects $\\angle DEB$, what is the length of $BE$?\n",
    "completion": "\\textbf{Solution}: In rectangle $ABCD$, \\\\\n$\\because$ $AD\\parallel BC$, \\\\\n$\\therefore$ $\\angle AEB=\\angle EAD$, \\\\\nAlso, $\\because$ $AE$ bisects $\\angle DEB$, \\\\\n$\\therefore$ $\\angle AEB=\\angle AED$, \\\\\n$\\therefore$ $\\angle EAD=\\angle AED$, \\\\\n$\\therefore$ $AD=DE=10$, \\\\\nIn right triangle $\\triangle CDE$, \\\\\n$\\because$ $DE=10$, $DC=AB=6$, \\\\\nAccording to the Pythagorean theorem, we have: $CE=\\sqrt{DE^{2}-DC^{2}}=\\sqrt{10^{2}-6^{2}}=8$, \\\\\nAlso, $\\because$ $BC=AD=10$, \\\\\n$\\therefore$ $BE=BC-CE=10-8=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53182907.png"
  },
  {
    "solution": "\\boxed{\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, within the square $ABCD$, $E$ is a point on the side $BC$. Connect $AE$ and draw $DF\\parallel AE$ through point $D$ to intersect the extension of $BC$ at point $F$. $AG$ bisects $\\angle DAE$ and intersects $DF$ at point $G$, and intersects $DC$ at point $M$. If $\\angle DAM=30^\\circ$ and $CF=1$, what is the length of $AB$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a square,\\\\\nit follows that AD $ \\parallel $ EF, AB=DC, $\\angle$ABE=$\\angle$DCB=$\\angle$DCF=$\\angle$DAB=$90^\\circ$,\\\\\nSince AE $ \\parallel $ DF,\\\\\nit implies that quadrilateral AEFD is a parallelogram,\\\\\nhence, AE=DF,\\\\\nIn right triangles $\\triangle$ABE and $\\triangle$DCF, $\\left\\{\\begin{array}{l}AE=DF\\\\ AB=DC\\end{array}\\right.$,\\\\\nit follows that right triangle $\\triangle$ABE$\\cong$$\\triangle$DCF (HL),\\\\\ntherefore $\\angle$BAE=$\\angle$CDF,\\\\\nSince AM bisects $\\angle$DAE, and $\\angle$DAM=$30^\\circ$,\\\\\nit follows that $\\angle$EAM=$\\angle$DAM=$30^\\circ$,\\\\\nthus $\\angle$BAE=$90^\\circ-30^\\circ-30^\\circ=30^\\circ$,\\\\\ntherefore $\\angle$CDF=$\\angle$BAE=$30^\\circ$,\\\\\nhence, DF=2CF,\\\\\nSince CF=1,\\\\\nit follows that DF=2,\\\\\nthus, DC=$\\sqrt{DF^2-CF^2}$=$\\sqrt{2^2-1^2}=\\sqrt{3}$,\\\\\ntherefore AB=DC=$\\boxed{\\sqrt{3}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53728604_84.png"
  },
  {
    "solution": "\\boxed{8-\\pi}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, for the square $ABCD$ with a side length of 4, and a semicircle $O$ with diameter $BC$ intersecting the diagonal $AC$ at point $E$, what is the area of the shaded region?\n",
    "completion": "\\textbf{Solution:} As shown in the figure,\\\\\n$\\because$ $BC$ is the diameter of the circle.\\\\\n$\\therefore$ $BE\\perp AC$\\\\\nAlso, $\\because$ $BC=AB$\\\\\n$\\therefore$ $BE=EA=EC$,\\\\\n$\\because$ $O$ is the midpoint of $BC$\\\\\n$\\therefore$ $EO$ is the median of $\\triangle ABC$,\\\\\n$\\therefore$ $EO\\parallel AB$,\\\\\n$\\therefore$ $\\angle BOE=90^\\circ$,\\\\\n$\\therefore$ the area of the shaded part $=S_{\\triangle ABC}-S_{\\text{sector} BOE}$\\\\\n$=\\frac{1}{2}AB\\times BC-\\frac{90^\\circ\\times \\pi \\times OB^{2}}{360}$\\\\\n$=\\frac{1}{2}\\times 4\\times 4-\\frac{90^\\circ\\times \\pi \\times 2^{2}}{360}$\\\\\n$=8-\\pi$.\\\\\nTherefore, the answer is: \\boxed{8-\\pi}.",
    "image_path": "dataset/mm_math/images/MM_Math/52871844.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{3}+1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle B=30^\\circ$, $AB=4$, $AD\\perp BC$ at point $D$ and $\\tan \\angle CAD=\\frac{1}{2}$, what is the length of $BC$?\n",
    "completion": "\\textbf{Solution:} Since $AD \\perp BC$ at point D, \\\\\ntherefore, $\\triangle ADB$ and $\\triangle ADC$ are right-angled triangles, \\\\\nin $\\triangle ADB$, $\\angle B=30^\\circ$, $AB=4$, \\\\\ntherefore, $AD=\\frac{1}{2}AB=2$, $BD=\\sqrt{AB^{2}-AD^{2}}=\\sqrt{4^{2}-2^{2}}=2\\sqrt{3}$, \\\\\nin $\\triangle ADC$, $\\tan\\angle CAD=\\frac{1}{2}$, $AD=2$, \\\\\ntherefore, $\\tan\\angle CAD=\\frac{CD}{AD}=\\frac{1}{2}$, \\\\\ntherefore, $CD=1$, \\\\\ntherefore, $BC=BD+CD=2\\sqrt{3}+1=\\boxed{2\\sqrt{3}+1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51350901.png"
  },
  {
    "solution": "\\boxed{3+\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle A=30^\\circ$, $\\angle B=45^\\circ$, and $AC=2\\sqrt{3}$. What is the length of $AB$?\n",
    "completion": "\\textbf{Solution:} Draw $CD \\perp AB$ at $D$, \\\\\n$\\therefore$ $\\angle ADC=\\angle BDC=90^\\circ$.\\\\\n$\\because$ $\\angle B=45^\\circ$, \\\\\n$\\therefore$ $\\angle BCD=\\angle B=45^\\circ$, \\\\\n$\\therefore$ $CD=BD$. \\\\\n$\\because$ $\\angle A=30^\\circ$, $AC=2\\sqrt{3}$, \\\\\n$\\therefore$ $CD=\\sqrt{3}$, \\\\\n$\\therefore$ $BD=CD=\\sqrt{3}$. \\\\\nBy the Pythagorean theorem, $AD=\\sqrt{AC^2-CD^2}=3$, \\\\\n$\\therefore$ $AB=AD+BD=3+\\sqrt{3}$. \\\\\nHence, the answer is: $AB=\\boxed{3+\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51032242.png"
  },
  {
    "solution": "\\boxed{(2,6)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=ax^2+bx-4a$ (where $a \\neq 0$) passes through points $A(-1,0)$ and $C(0,4)$, and intersects the x-axis at another point $B$. Lines $AC$ and $BC$ are drawn. Point $P$ is on the line segment $OB$ (not coinciding with points $B$ or $O$). A line through $P$ perpendicular to the x-axis intersects the parabola at point $M$, and lines $CM$ and $BM$ are drawn. Find the maximum area of $\\triangle BCM$ and the coordinates of point $M$ at this condition.\n",
    "completion": "\\textbf{Solution:} Let $y=0$, then we have $-x^2+3x+4=0$, which simplifies to $x^2-3x-4=0$.\\\\\nSolving this, we get $x=-1$ or $x=4$,\\\\\n$\\therefore$ the coordinates of point B are $(4,0)$;\\\\\nLet the equation of line BC be $y=kx+n$,\\\\\n$\\begin{cases}\nb=4 \\\\\n4k+b=0\n\\end{cases}$,\\\\\n$\\therefore$\n$\\begin{cases}\nk=-1 \\\\\nb=4\n\\end{cases}$,\\\\\n$\\therefore$ the equation of line BC is $y=-x+4$,\\\\\nLet PM intersect segment BC at point E, with M$(m, -m^2+3m+4)$, then E$(m, -m+4)$\\\\\nME$=-m^2+4m$,\\\\\n$\\therefore$ the area of $\\triangle BMC=S_{\\triangle BME}+S_{\\triangle CME}$\\\\\n$=\\frac{1}{2}\\times 4(-m^2+4m)$\\\\\n$=-2m^2+8m$\\\\\n$=-2(m^2-4m+4)+8$\\\\\n$=-2(m-2)^2+8$ $(0<m<4)$\\\\\nSince $-2<0$,\\\\\n$\\therefore$ when $m=2$, $S_{\\triangle BMC}$ reaches its maximum value of $\\boxed{8}$,\\\\\n$\\therefore$ the coordinates of M are $\\boxed{(2,6)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52536660_154.png"
  },
  {
    "solution": "\\boxed{100^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points A, B, and C are three points on the circle $\\odot O$, with $\\angle C = 50^\\circ$, what is the degree measure of $\\angle AOB$?\n",
    "completion": "\\textbf{Solution:} Since arc AB = arc AB,\\\\\nit follows that $\\angle AOB = 2\\angle C = 2\\times50^\\circ = \\boxed{100^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51545215.png"
  },
  {
    "solution": "\\boxed{\\frac{29}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $CD$ is the diameter of the circle $\\odot O$, $AB$ is a chord of $\\odot O$, $AB=10$, $AB\\perp CD$ at the foot M, and $CM=2$. What is the length of the radius?\n",
    "completion": "\\textbf{Solution:} Connect OA, let the radius be $r$, as shown in the following figure:\\\\\nThen, we have $OA=r$, $OM=r-2$\\\\\nAccording to the perpendicular bisector theorem, we obtain $AM=\\frac{1}{2}AB=5$,\\\\\nBy the Pythagorean theorem, we find: $OM^2+AM^2=OA^2$, that is, $(r-2)^2+5^2=r^2$\\\\\nSolving this, we find $r=\\boxed{\\frac{29}{4}}$",
    "image_path": "dataset/mm_math/images/MM_Math/52991558.png"
  },
  {
    "solution": "\\boxed{\\frac{16\\sqrt{7}}{7}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given that line segment $AD$ is translated down by $m$ units and then translated to the right by $n$ units to line segment $BC$, with corresponding points of $A$, $D$ being points $B$ and $C$ respectively, and connecting $AB$, $CD$, $AC$, $BD$, where $AC$ intersects $BD$ at point $O$. Under the given conditions, as $H$ moves on $BC$, when $\\triangle CDH$ is an isosceles triangle, write down the length of $HC$.\n",
    "completion": "\\textbf{Solution:}\\\\\n(1) When $HC=DC=8$, as shown in the figure, an isosceles triangle is formed.\\\\\n(2) When $HD=HC$, as shown in the figure, an isosceles triangle is formed.\\\\\nAt this time, let $HM$ be $x$,\\\\\n$\\because H{M}^{2}+M{D}^{2}=H{D}^{2}=M{C}^{2}$,\\\\\n$\\therefore {x}^{2}+{6}^{2}={\\left(x+2\\sqrt{7}\\right)}^{2}$,\\\\\nSolving this, we get: $x=\\frac{2\\sqrt{7}}{7}$,\\\\\n$\\therefore HC=x+MC= \\frac{2\\sqrt{7}}{7}+2\\sqrt{7}=\\frac{16\\sqrt{7}}{7}$.\\\\\n(3) When $DH=DC$, as shown in the figure, an isosceles triangle is formed:\\\\\nIn this case $HC=2MC= 4\\sqrt{7}$.\\\\\nTherefore, the length of $HC$ can be: $8$ or \\boxed{\\frac{16\\sqrt{7}}{7}} or $4\\sqrt{7}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52386348_128.png"
  },
  {
    "solution": "\\boxed{\\left(\\frac{4}{3}, 3\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: It is known that the rectangle $OCBA$ is located as shown in the diagram within the Cartesian coordinate system. Point $C$ is on the positive semi-axis of $x$, and point $A$ is on the positive semi-axis of $y$. It is known that point $B$ has coordinates $(4,2)$, and the graph of the inverse proportion function $y=\\frac{k}{x}$ passes through the midpoint $D$ of $AB$ and intersects $BC$ at point $E$. Suppose the equation of line $DE$ is $y=mx+n$. Connect $OD$ and $OE$. Let point $P$ be a point on the $x$-axis, and point $Q$ be a point on the graph of the inverse proportion function $y=\\frac{k}{x}$. Are there points $P$ and $Q$ such that the quadrilateral with vertices $P$, $Q$, $D$, and $E$ forms a parallelogram? If such points exist, provide the coordinates of point $Q$ directly; if not, explain why.\n",
    "completion": "\\textbf{Solution:} Yes, there exists. The coordinates of point $Q$ are $\\boxed{(-4, -1)}$ or $\\boxed{\\left(\\frac{4}{3}, 3\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52536619_159.png"
  },
  {
    "solution": "\\boxed{(-3, -2)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, for the linear function $y=kx+1$, there are points A and B on its graph, with point A in the first quadrant and point B on the x-axis. Point D is on the positive half of the x-axis, and the coordinates of point C are $(1, -2)$. The quadrilateral OADC is a rhombus. Let point P be a moving point on line AB, and $S_{\\triangle AOP}=\\frac{1}{2}S_{\\text{rhombus} OADC}$. Find the coordinates of point P.\n",
    "completion": "\\textbf{Solution:} Given by the problem, $A(1, 2), D(0, 2)$,\\\\\n$\\therefore S_{\\text{rhombus} AOCD}=2\\times \\frac{1}{2}\\times 2\\times 2=4$,\\\\\nAs shown in the diagram, let the intersection of AB with the y-axis be Q,\\\\\nwhen $x=0$, then $y=1$, i.e., Q$(0, 1)$,\\\\\nLet point P be $(x, x+1)$,\\\\\n$\\therefore S_{\\triangle AOP}=S_{\\triangle AOQ}+S_{\\triangle POQ}=\\frac{1}{2}\\times OQ\\times |x-1|=\\frac{1}{2}|x-1|$,\\\\\nSince $S_{\\triangle AOP}= \\frac{1}{2}S_{\\text{rhombus} AOCD}$,\\\\\n$\\therefore \\frac{1}{2}|x-1|=2$,\\\\\nSolving gives $x=5$ or $x=-3$,\\\\\nWhen $x=5$, $y=6$, when $x=-3$, $y=-2$,\\\\\n$\\therefore$ The coordinates of point P are $\\boxed{(5, 6)}$ or $\\boxed{(-3, -2)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52211814_82.png"
  },
  {
    "solution": "\\boxed{64^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, two identically sized set squares, ACF and CFB, containing $45^\\circ$ angles, are placed as depicted. Point D is on side AC, and point E is on side BC, with $\\angle CFE=13^\\circ$, and $\\angle CFD=32^\\circ$. What is the measure of $\\angle DEC$?\n",
    "completion": "\\textbf{Solution:} Construct $FH$ perpendicular to $FE$, intersecting $AC$ at point $H$,\n\\[\n\\because \\angle AFC=\\angle EFH=90^\\circ\n\\]\nAlso,\n\\[\n\\because \\angle AFC=\\angle AFH+\\angle CFH,\\ \\angle HFE=\\angle CFE+\\angle CFH\n\\]\n\\[\n\\therefore \\angle AFH=\\angle CFE=13^\\circ\n\\]\n\\[\n\\because \\angle A=\\angle FCE=45^\\circ,\\ FA=CF\n\\]\n\\[\n\\therefore \\triangle FAH\\cong \\triangle FCE\\ (\\text{ASA})\n\\]\n\\[\n\\therefore FH=FE\n\\]\n\\[\n\\because \\angle DFE=\\angle DFC+\\angle EFC=32^\\circ+13^\\circ=45^\\circ\n\\]\n\\[\n\\because \\angle DFH=\\angle HFE−\\angle DFE=90^\\circ−45^\\circ=45^\\circ\n\\]\n\\[\n\\therefore \\angle DFE=\\angle DFH\n\\]\nAlso,\n\\[\n\\because DF=DF\n\\]\n\\[\n\\therefore \\triangle HDF\\cong \\triangle EDF\\ (\\text{SAS})\n\\]\n\\[\n\\therefore \\angle DHF=\\angle DEF\n\\]\n\\[\n\\because \\angle DHF=\\angle A+\\angle HFA=45^\\circ+13^\\circ=58^\\circ\n\\]\n\\[\n\\therefore \\angle DEF=58^\\circ\n\\]\n\\[\n\\because \\angle CFE+\\angle CEF+\\angle FCE=180^\\circ\n\\]\n\\[\n\\therefore \\angle CEF=180^\\circ−\\angle CFE−\\angle FCE=180^\\circ−13^\\circ−45^\\circ=122^\\circ\n\\]\n\\[\n\\therefore \\angle DEC=\\angle CEF−\\angle DEF=122^\\circ−58^\\circ=\\boxed{64^\\circ}\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/52699726.png"
  },
  {
    "solution": "\\boxed{{B}^{\\prime}(3, 1)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within the Cartesian coordinate system, the vertices of rectangle OABC are $A\\left(0, 3\\right)$, and $C\\left(-1, 0\\right)$. Now, rectangle OABC is rotated clockwise by $90^\\circ$ around the origin O to obtain rectangle $O{A}^{\\prime}{B}^{\\prime}{C}^{\\prime}$. The line $B{B}^{\\prime}$ intersects the x-axis at point M and the y-axis at point N. The graph of the parabola $y=ax^{2}+bx+c$ passes through points C, M, and N. Please directly write down the coordinates of points B and ${B}^{\\prime}$.\n",
    "completion": "\\textbf{Solution:} Point $B$ and point ${B}^{\\prime}$ have coordinates $\\boxed{B(-1, 3)}$ and $\\boxed{{B}^{\\prime}(3, 1)}$, respectively.",
    "image_path": "dataset/mm_math/images/MM_Math/52602539_162.png"
  },
  {
    "solution": "\\boxed{70^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ is rotated clockwise around point C to obtain $\\triangle EDC$. If point A is exactly on the extension of line ED and $\\angle ABC=110^\\circ$, then what is the degree measure of $\\angle ADC$?\n",
    "completion": "Solution: It is known from the rotation that $\\triangle ABC \\cong \\triangle EDC$,\\\\\n$\\therefore \\angle EDC = \\angle ABC = 110^\\circ$,\\\\\n$\\because$ Points A, D, E are collinear,\\\\\n$\\therefore \\angle EDC + \\angle ADC = 180^\\circ$,\\\\\n$\\therefore \\angle ADC = 180^\\circ - \\angle EDC = 180^\\circ - 110^\\circ = \\boxed{70^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55377877.png"
  },
  {
    "solution": "\\boxed{32^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that in circle $\\odot O$, CD is the diameter of $\\odot O$, and points A and B are on $\\odot O$ with $AC=AB$. If $\\angle BCD=26^\\circ$, what is the degree measure of $\\angle ABC$?\n",
    "completion": "\\textbf{Solution:} Since CD is the diameter,\\\\\n$\\therefore\\angle CAD=90^\\circ$,\\\\\n$\\therefore\\angle ACD+\\angle ADC=90^\\circ$,\\\\\nSince AC$=$AB,\\\\\n$\\therefore\\angle ACB=\\angle B$,\\\\\nSince $\\angle D=\\angle B$,\\\\\n$\\therefore\\angle ACB=\\angle D$,\\\\\n$\\therefore\\angle ACB+26^\\circ+\\angle D=90^\\circ$,\\\\\n$\\therefore\\angle ACB=32^\\circ$,\\\\\n$\\therefore\\angle ABC=\\angle ACB=\\boxed{32^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/51368365.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In square $ABCD$, $AB=4$, point $P$ is on line $AC$, draw $PM\\perp AB$ at point $M$, and $PN\\perp BC$ at point $N$. What is the minimum value of $MN$?\n",
    "completion": "\\textbf{Solution:} Connect PB, and through point B draw $BP' \\perp AC$ at $P'$, as shown in the figure: \\\\\n$\\because$ Quadrilateral ABCD is a square,\\\\\n$\\therefore \\angle DCB=\\angle ABC=90^\\circ$, and $DC=BC=AD=AB=4$,\\\\\n$\\therefore AC=\\sqrt{AD^2+DC^2}=4\\sqrt{2}$,\\\\\n$\\because PM\\perp AB$ and $PN\\perp BC$,\\\\\n$\\therefore$ Quadrilateral BNPM is a rectangle,\\\\\n$\\therefore PB=MN$,\\\\\nWhen $BP\\perp AC$, BP is minimal, that is, as shown in the figure, the length of $BP'$,\\\\\n$\\because$ The area of $\\triangle ABC = \\frac{1}{2}AB\\times CB = \\frac{1}{2}AC\\times BP'$,\\\\\n$\\therefore BP'=\\frac{AB\\cdot BC}{AC}=\\frac{4\\times 4}{4\\sqrt{2}}=2\\sqrt{2}$,\\\\\n$\\therefore$ The minimum value of MN is $\\boxed{2\\sqrt{2}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52049854.png"
  },
  {
    "solution": "\\boxed{(1,2)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $l: y=k_{1}x+b$ intersects with the hyperbola $y=\\frac{k_{2}}{x}(x>0)$ at points A and B, and intersects with the x-axis at point C. If the x-coordinates of points A and B are 1 and 2 respectively, please directly provide the solution set of $k_{1}x+b>\\frac{k_{2}}{x}$.\n",
    "completion": "\\textbf{Solution:} From the given problem, we have $k_{1}x+b>\\frac{k_{2}}{x}$,\\\\\nthe solution set is $x\\in(1,2)$,\\\\\n$\\therefore$ the solution set is $x\\in\\boxed{(1,2)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52601705_136.png"
  },
  {
    "solution": "\\boxed{x<2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given the figure, the line $y=x+1$ intersects the line $y=ax+b$ at point $A(m,3)$. What is the solution set of the inequality $x+1<ax+b$ with respect to $x$?\n",
    "completion": "\\textbf{Solution:}\n\nGiven that the line $y=x+1$ intersects the line $y=ax+b$ at point $A(2,3)$,\\\\\n$\\therefore m+1=3$,\\\\\nSolving this gives: $m=2$,\\\\\n$\\therefore A(2,3)$,\\\\\n$\\therefore$ the solution set of the inequality $x+1<ax+b$ with respect to $x$ is $\\boxed{x<2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52189468.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, $AD$ is the angle bisector of $\\triangle ABC$, $DE\\perp AB$ with the foot at $E$, the area of $\\triangle ABC$ is $8$, $DE=2$, and $AB=5$. What is the length of $AC$?\n",
    "completion": "\\textbf{Solution:} Construct $DF\\perp AC$ at $F$, \\\\\nsince $AD$ is the angle bisector of $\\triangle ABC$, $DE\\perp AB$, and $DF\\perp AC$, \\\\\nthus $DE=DF=2$, \\\\\ntherefore the area of $\\triangle ADB=\\frac{1}{2}AB\\cdot DE=\\frac{1}{2}\\times 5\\times 2=5$, \\\\\nsince the area of $\\triangle ABC$ is 8, \\\\\nthus the area of $\\triangle ADC$ is $8-5=3$, \\\\\nthus $\\frac{1}{2}AC\\cdot DF=3$, \\\\\nthus $\\frac{1}{2}AC\\times 2=3$, \\\\\nthus $AC=\\boxed{3}$,",
    "image_path": "dataset/mm_math/images/MM_Math/53102451.png"
  },
  {
    "solution": "\\boxed{110^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the lines $AB$ and $CD$ intersect at point O, and $OE \\perp CD$ at O. $OD$ bisects $\\angle AOF$, and $\\angle AOE = 55^\\circ$. What is the degree measure of $\\angle BOF$?\n",
    "completion": "\\textbf{Solution:} According to the construction,\\\\\nsince $OE \\perp CD$ and $\\angle AOE = 55^\\circ$,\\\\\nit follows that $\\angle AOD = 90^\\circ - \\angle AOE = 35^\\circ$,\\\\\nsince $OD$ bisects $\\angle AOF$,\\\\\nit yields $\\angle DOF = \\angle AOD = 35^\\circ$,\\\\\nsince lines $AB$ and $CD$ intersect at point O,\\\\\nthus $\\angle BOC = \\angle AOD = 35^\\circ$,\\\\\ntherefore $\\angle BOF = 180^\\circ - \\angle BOC - \\angle DOF = \\boxed{110^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53206228.png"
  },
  {
    "solution": "\\boxed{\\frac{5}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, D and E are points on sides BC and AC, respectively, and AD intersects BE at point F. If E is the midpoint of AC, and the ratio of BD to DC is 2:3, what is the value of AF: FD?\n",
    "completion": "\\textbf{Solution:} Let line DH be drawn through point D such that DH $\\parallel$ AC, intersecting BE at H.\\\\\n$\\therefore \\triangle DHF\\sim \\triangle AEF$, $\\triangle BDH\\sim \\triangle BCE$,\\\\\n$\\therefore \\frac{DH}{AE}=\\frac{DF}{AF}$, $\\frac{DH}{CE}=\\frac{BD}{BC}$,\\\\\n$\\because$ If E is the midpoint of AC,\\\\\n$\\therefore$ CE＝AE,\\\\\n$\\therefore \\frac{DF}{AF}=\\frac{BD}{BC}$,\\\\\n$\\because$ The ratio BD: DC＝2: 3,\\\\\n$\\therefore$ The ratio BD: BC＝2: 5,\\\\\n$\\therefore$ The ratio DF: AF＝2: 5,\\\\\n$\\therefore$ The ratio AF: FD＝$\\boxed{\\frac{5}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51322496.png"
  },
  {
    "solution": "\\boxed{55^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, $AB$ is a straight line, and $OC$ is the bisector of $\\angle AOD$. As illustrated in figure (1), if $\\angle COE$ is a right angle, and $\\angle AOD=70^\\circ$, what is the measure of $\\angle BOE$?\n",
    "completion": "\\textbf{Solution:} Since $OC$ is the bisector of $\\angle AOD$ and $\\angle AOD = 70^\\circ$,\\\\\nit follows that $\\angle AOC = \\angle DOC = \\frac{1}{2}\\angle AOD = \\frac{1}{2}\\times 70^\\circ = 35^\\circ$,\\\\\nSince $\\angle COE is a right angle,\\\\\nwe have $\\angle COE = 90^\\circ$,\\\\\nHence, $\\angle BOE = 180^\\circ - \\angle AOC - \\angle COE = 180^\\circ - 35^\\circ - 90^\\circ = \\boxed{55^\\circ};$",
    "image_path": "dataset/mm_math/images/MM_Math/51508770_136.png"
  },
  {
    "solution": "\\boxed{(\\sqrt{3}, 2)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, it is known that the coordinate of point $C$ is ($\\sqrt{3}$, $-2$), and the line $AB$ intersects the x-axis and y-axis at point $A$ and point $B$, respectively, with the coordinate of point $B$ being (0, 3), and $\\angle BAO=30^\\circ$. If point $D$ is a moving point on the y-axis, and point $E$ ($\\sqrt{3}$, $m$) is on line $AB$, find the coordinates of points $D$ and $E$ when $CD+DE$ is minimized?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, construct the symmetric point $C'$ of point $C$ with respect to the $y$-axis, and connect $DC'$ and $EC'$. By the property of axial symmetry, we have $C'(-\\sqrt{3}, -2)$, $DC'=CD$, $\\therefore CD+DE=DC'+DE$. According to the principle that the shortest distance between two points is a straight line, when points $C'$, $D$, and $E$ are collinear, meaning point $D$ is the intersection of line $EC'$ and the $y$-axis, $DC'+DE$ achieves its minimum value. \n\nSubstituting point $E(\\sqrt{3}, m)$ into $y=-\\frac{\\sqrt{3}}{3}x+3$ yields: \n\\[ m=-\\frac{\\sqrt{3}}{3}\\times \\sqrt{3}+3=2 \\]\nhence $E(\\sqrt{3}, 2)$. \n\nAssume the equation of line $EC'$ is $y=k_0x+b_0$. Substituting points $E(\\sqrt{3}, 2)$ and $C'(-\\sqrt{3}, -2)$ into it yields: \n\\[ \\left\\{\\begin{array}{l}\\sqrt{3}k_0+b_0=2 \\\\ -\\sqrt{3}k_0+b_0=-2\\end{array}\\right. \\]\nsolving this system of equations gives \n\\[ \\left\\{\\begin{array}{l}k_0=\\frac{2\\sqrt{3}}{3} \\\\ b_0=0\\end{array}\\right. \\]\nTherefore, the equation of line $EC'$ is $y=\\frac{2\\sqrt{3}}{3}x$. When $x=0$, $y=0$, meaning when $CD+DE$ is minimized, the coordinates of point $D$ are $D(0, 0)$. \n\nThus, the answer is $D=\\boxed{(0, 0)}$, $E=\\boxed{(\\sqrt{3}, 2)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50482871_122.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $a\\parallel b\\parallel c$, $\\frac{AC}{CE}=\\frac{1}{2}$, and $DF=12$, what is the length of $BD$?\n",
    "completion": "\\textbf{Solution:} Since $a\\parallel b\\parallel c$,\\\\\nit follows that $\\frac{AC}{CE}=\\frac{BD}{DF}=\\frac{1}{2}$,\\\\\nsince $DF=12$,\\\\\nit follows that $\\frac{BD}{12}=\\frac{1}{2}$,\\\\\ntherefore, $BD=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51402904.png"
  },
  {
    "solution": "\\boxed{(2+\\sqrt{3}, 1)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the vertices $C$ and $A$ of the square $ABCO$ are on the $x$-axis and $y$-axis, respectively. $BC$ is the diagonal of the rhombus $BDCE$. If $\\angle D = 60^\\circ$ and $BC = 2$, what are the coordinates of point $D$?\n",
    "completion": "\\textbf{Solution:} As illustrated, draw line DF perpendicular to BC at point F from point D,\\\\\n$\\because$ rhombus BDCE,\\\\\n$\\therefore$ BD = CD,\\\\\nAlso $\\because$ $\\angle$D = 60$^\\circ$,\\\\\n$\\therefore$ $\\triangle$BDC is an equilateral triangle,\\\\\n$\\therefore$ $\\angle$DFC = 90$^\\circ$, BF = CF = 1, so the y-coordinate of point D is: 1,\\\\\n$\\therefore$ in $\\triangle$DFC, DF = $\\sqrt{CD^{2}-CF^{2}} = \\sqrt{2^{2}-1^{2}} = \\sqrt{3}$,\\\\\n$\\because$ square ABCD, BC = 2,\\\\\n$\\therefore$ OC = 2,\\\\\n$\\therefore$ the x-coordinate of point D is: OC + DF = 2 + $\\sqrt{3}$,\\\\\nThus, the answer is: $(2+\\sqrt{3}, 1) = \\boxed{(2+\\sqrt{3}, 1)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51435951.png"
  },
  {
    "solution": "\\boxed{35^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that straight lines AB and CD intersect at point O. OA bisects $\\angle EOC$, and $\\angle EOC = 70^\\circ$. What is the degree measure of $\\angle BOD$?\n",
    "completion": "\\textbf{Solution:} Since OA bisects $\\angle EOC$ and $\\angle EOC=70^\\circ$,\n\\[\n\\therefore \\angle AOC=\\frac{1}{2}\\angle EOC=\\frac{1}{2}\\times 70^\\circ=35^\\circ,\n\\]\n\\[\n\\therefore \\angle BOD=\\angle AOC=\\boxed{35^\\circ}\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/55426353.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{10}-4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the central angle $\\angle AOB$ corresponding to arc AB is $90^\\circ$, and the radius is 4. Point C is the midpoint of OB, and point D is a point on arc AB. CD rotates $90^\\circ$ counterclockwise around point C to get CE. What is the minimum value of AE?\n",
    "completion": "\\textbf{Solution:}\n\nConstruct $MC\\perp OB$ passing through point C such that $CM=OC$. Connect EM and OD. Then, $\\angle OCM=90^\\circ$,\n\nSince point C is the midpoint of OB,\n\n$\\therefore OC=BC=\\frac{1}{2}OB=2$,\n\n$\\therefore CM=OC=2$,\n\nSince CD rotates $90^\\circ$ counterclockwise around point C to get CE,\n\n$\\therefore CD=CE$, $\\angle DCE=90^\\circ$,\n\n$\\therefore \\angle OCM=\\angle DCE$,\n\n$\\therefore \\angle OCM+\\angle OCE=\\angle DCE+\\angle OCE$,\n\n$\\therefore \\angle ECM=\\angle DCO$,\n\nIn $\\triangle ECM$ and $\\triangle DCO$,\n\n$\\left\\{\\begin{array}{l}\nEC=DC \\\\\n\\angle ECM=\\angle DCO \\\\\nCM=CO\n\\end{array}\\right.$,\n\n$\\therefore \\triangle ECM\\cong\\triangle DCO$ (SAS),\n\n$\\therefore EM=OD=4$,\n\n$\\therefore$ point E is on the circle with center M and radius 4,\n\n$\\therefore$ when points A, E, M are collinear, AE takes the minimum value,\n\nConstruct $MN\\perp AO$ extending AO and intersecting at point N,\n\n$\\therefore \\angle MNO=\\angle MCO=\\angle CON=90^\\circ$,\n\n$\\therefore$ quadrilateral COMN is a rectangle,\n\nSince $CM=OC$,\n\n$\\therefore$ quadrilateral COMN is a square,\n\n$\\therefore MN=OC=ON=2$,\n\n$\\therefore AN=AO+ON=6$,\n\n$\\therefore AM=\\sqrt{AN^2+MN^2}=2\\sqrt{10}$,\n\n$\\therefore$ the minimum value of AE is $AM-EM=2\\sqrt{10}-4$,\n\nThus, the answer is: $AE=\\boxed{2\\sqrt{10}-4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52160124.png"
  },
  {
    "solution": "\\boxed{12\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the inverse proportion function $y = \\frac{k}{x}$ ($k \\neq 0$) passes through the vertices $A$ and $B$ of an equilateral triangle $\\triangle ABC$, and the origin $O$ is exactly on $AB$. Given that the coordinates of point $C$ are (3, 3), a line passing through point $C$ and parallel to $AB$ intersects the graph of the inverse proportion function $y = \\frac{k}{x}$ ($k \\neq 0$) at points $M$ and $N$. What is the area of $\\triangle MON$?\n",
    "completion": "Solution: Given point C $(3,3)$, \\\\\ntherefore the equation for line OC is $y=kx$, substituting the coordinates of point C, \\\\\nthus, $3=3k$, \\\\\ntherefore, $k=1$, \\\\\nthus, the equation for OC becomes $y=x$, \\\\\nsince points A and B lie on the graph of an inverse proportional function, \\\\\nthus, points A and B are symmetric about the origin O, with O being the midpoint of AB, \\\\\nsince $\\triangle ABC$ is an equilateral triangle, \\\\\nthus, OC $\\perp$ AB, \\\\\nthus, $\\angle ACO = \\frac{1}{2}\\angle ACB=\\frac{1}{2}\\times 60^\\circ=30^\\circ$, \\\\\ntherefore, according to the Pythagorean theorem, OC $=\\sqrt{3^2+3^2}=3\\sqrt{2}$, \\\\\nthus, AO $=OC\\tan 30^\\circ=3\\sqrt{2}\\times \\frac{\\sqrt{3}}{3}=\\sqrt{6}$, \\\\\nsince AB $\\perp$ OC, with the equation for OC being $y=x$, \\\\\nthus, $\\angle COy=45^\\circ$, \\\\\nthus, $\\angle AOy=45^\\circ$, \\\\\nthus, the y-coordinate of point A is $\\sqrt{6}\\sin 45^\\circ=\\sqrt{6}\\times \\frac{\\sqrt{2}}{2}=\\sqrt{3}$, the x-coordinate of point A is $-\\sqrt{6}\\cos 45^\\circ=-\\sqrt{6}\\times \\frac{\\sqrt{2}}{2}=-\\sqrt{3}$, \\\\\nthus, point A $(-\\sqrt{3}, \\sqrt{3})$, \\\\\nsince point A lies on the graph of the inverse proportion function $y=\\frac{k}{x}$, \\\\\nthus, $k=-\\sqrt{3}\\times \\sqrt{3}=-3$, \\\\\nthus, the inverse proportion function is $y=-\\frac{3}{x}$, \\\\\nLet the equation for OA be $y=k_1x$, by substituting the coordinates of point A, \\\\\n$\\sqrt{3}=-\\sqrt{3}k_1$, \\\\\nsolving yields: $k_1=-1$, \\\\\nthus, the equation for OA is $y=-x$, \\\\\nsince MN $\\parallel$ AB, \\\\\nlet the equation for MN be $y=-x+b$, passing through point C, \\\\\nthus, $3=-3+b$, \\\\\nsolving gives $b=6$, \\\\\nthus, the equation for MN is $y=-x+6$, \\\\\npoints M, N also lie on the graph of the inverse proportion function $y=-\\frac{3}{x}$, \\\\\nthus, $\\left\\{\\begin{array}{l}y=-x+6 \\\\ y=-\\frac{3}{x}\\end{array}\\right.$, \\\\\neliminating $y$ yields $x^2-6x-3=0$, \\\\\nsolving gives $x=3\\pm 2\\sqrt{3}$, \\\\\nthus, $x_1=3+2\\sqrt{3}, x_2=3-2\\sqrt{3}$, \\\\\nthus, $y_1=3-2\\sqrt{3}, y_2=3+2\\sqrt{3}$, \\\\\nPoint M $(3-2\\sqrt{3}, 3+2\\sqrt{3})$, point N $(3+2\\sqrt{3}, 3-2\\sqrt{3})$, \\\\\ntherefore, NM $=\\sqrt{((3-2\\sqrt{3}-3-2\\sqrt{3})^2+((3+2\\sqrt{3}-3+2\\sqrt{3})^2}=4\\sqrt{6}$, \\\\\nthus, $S_{\\triangle MON}=\\frac{1}{2}OC\\cdot MN=\\frac{1}{2}\\times 3\\sqrt{2}\\times 4\\sqrt{6}=\\boxed{12\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52550173.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $P$ is the centroid of $\\triangle ABC$. Connect $AP$ and extend it to intersect $BC$ at point $D$. If the area of $\\triangle ABC$ is $20$, what is the area of $\\triangle ADC$?\n",
    "completion": "\\textbf{Solution:} Since $P$ is the centroid of $\\triangle ABC$,\n\nit follows that $AD$ is the median of $\\triangle ABC$,\n\nhence, the area of $\\triangle ADC$ is equal to half the area of $\\triangle ABC$,\n\nand since the area of $\\triangle ABC$ is 20,\n\nthus, the area of $\\triangle ADC$ is $\\boxed{10}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52978950.png"
  },
  {
    "solution": "\\boxed{20^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle B=\\angle C$, point $D$ is on side $BC$, and point $E$ is on $AC$, $\\angle ADE=\\angle AED$. If $\\angle BAD=40^\\circ$, what is the measure of $\\angle CDE$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle ADC$ is an exterior angle of $\\triangle ABD$, \\\\\nwe have $\\angle ADC=\\angle BAD+\\angle B=40^\\circ+\\angle B$, \\\\\nSince $\\angle AED$ is an exterior angle of $\\triangle DEC$, \\\\\nwe have $\\angle AED=\\angle C+\\angle EDC$, \\\\\nSince $\\angle ADE=\\angle AED$, \\\\\nwe have $\\angle ADE=\\angle C+\\angle EDC$, \\\\\nSince $\\angle ADC=\\angle ADE+\\angle EDC=\\angle C+2\\angle EDC$, \\\\\nand since $\\angle B=\\angle C$, \\\\\nwe have $\\angle B+40^\\circ=\\angle B+2\\angle EDC$, \\\\\nTherefore, $\\angle EDC=\\boxed{20^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52634676.png"
  },
  {
    "solution": "\\boxed{x<-1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the direct proportion function $y_1=-2x$ intersects with the graph of the linear function $y_2=ax+3$ at point $A(-1, m)$. What is the solution set of the inequality $-2x>ax+3$ with respect to $x$?\n",
    "completion": "\\textbf{Solution:} From the graph, it is known that when $x<-1$, $-2x>ax+3$. \\\\\nTherefore, the answer is $\\boxed{x<-1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50545751.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, given a rectangular piece of paper ABCD, with \\(AB=6\\), \\(BC=20\\), and \\(M\\) being the midpoint of \\(BC\\), point \\(P\\) moves along the polyline \\(B-A-D\\). If the paper is folded to the right along the crease \\(MP\\), making point \\(B\\) fall on the edge of the rectangle, what is the length of the crease \\(MP\\)?\n",
    "completion": "Solution: \n(1) As shown in the figure, when point P is on side AB, draw ME $\\perp$ AD at E, so quadrilateral ABME is a rectangle,\\\\\n$\\therefore$ EM = AB = 6, AE = BM,\\\\\nAlso, $\\because$ BC = 20, and M is the midpoint of BC,\\\\\n$\\therefore$ by folding: B$^{\\prime }$M = BM = AE = 10,\\\\\nIn Rt$\\triangle$EMB$^{\\prime}$, according to the Pythagorean theorem: B$^{\\prime}$E = $\\sqrt{10^{2}-6^{2}} = 8$,\\\\\n$\\therefore$ AB$^{\\prime} = $ AE - B$^{\\prime }$E = 2,\\\\\nLet BP = x, then AP = 6 - x, PB$^{\\prime }$ = x,\\\\\nIn Rt$\\triangle$PAB$^{\\prime}$, according to the Pythagorean theorem: PB$^{\\prime }$$^{2} =$ AP$^{2}+$AB$^{\\prime }$$^{2}$,\\\\\nThus x$^{2} = $($6 - x)^{2} + 2^{2}$,\\\\\nSolving for x, we find x = $\\frac{10}{3}$,\\\\\n$\\therefore$ PB = $\\frac{10}{3}$,\\\\\nIn Rt$\\triangle$BMP, according to the Pythagorean theorem: PM = $\\sqrt{\\left(\\frac{10}{3}\\right)^{2} + 10^{2}} = \\frac{10}{3}\\sqrt{10}$;\\\\\n(2) As shown in the figure, when point P is on side AD, draw ME $\\perp$ AD at E, so quadrilateral ABME is a rectangle,\\\\\n$\\therefore$ EM = AB = 6,\\\\\nAlso, $\\because$ BC = 20, and M is the midpoint of BC,\\\\\n$\\therefore$ by folding: B$^{\\prime}$M = BM = 10,\\\\\nIn Rt$\\triangle$EMB$^{\\prime}$, according to the Pythagorean theorem: B$^{\\prime}$E = $\\sqrt{10^{2} - 6^{2}} = 8$,\\\\\nSince AD $\\parallel$ BC, $\\angle$DPM = $\\angle$BMP,\\\\\nBy folding, $\\angle$PMB$^{\\prime}$ = $\\angle$BMP,\\\\\n$\\therefore$$\\angle$DPM = $\\angle$PMB$^{\\prime}$,\\\\\n$\\therefore$ B$^{\\prime}$M = B$^{\\prime}$P = 10,\\\\\n$\\therefore$ PE = 10 - 8 = 2,\\\\\nIn Rt$\\triangle$PEM, according to the Pythagorean theorem: PM = $\\sqrt{2^{2} + 6^{2}} = 2\\sqrt{10}$;\\\\\n(3) As shown in the figure, when point B$^{\\prime}$ coincides with point C, because $\\angle$A = $\\angle$B = $\\angle$BMP = $90^\\circ$, quadrilateral ABMP is a rectangle,\\\\\nAt this point, PM = AB = 6.\\\\\nIn summary, the length of the crease MP is: $\\frac{5}{3}\\sqrt{13}$ or $2\\sqrt{10}$ or \\boxed{6}.",
    "image_path": "dataset/mm_math/images/MM_Math/51807104.png"
  },
  {
    "solution": "\\boxed{\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in rectangle $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$. If $\\angle AOD = 60^\\circ$ and $AD = 1$, what is the length of $AB$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral $ABCD$ is a rectangle,\\\\\nit follows that $AO=OC$, $BO=DO$, and $AC=BD$,\\\\\ntherefore $AO=DO$,\\\\\nsince $\\angle AOD=60^\\circ$,\\\\\nit follows that $\\triangle AOD$ is an equilateral triangle,\\\\\nthus $AO=OD=OB=AD=1$,\\\\\ntherefore $BD=2$,\\\\\nIn the right triangle $ABD$, by the Pythagorean theorem, we can find: $AB=\\sqrt{BD^{2}-AD^{2}}=\\boxed{\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52076092.png"
  },
  {
    "solution": "\\boxed{90^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points $A, B, C$ are on the circle $\\odot O$ with $\\angle A=45^\\circ$. What is the degree measure of $\\angle BOC$?\n",
    "completion": "\\textbf{Solution:}\n\nSince, as shown in the figure: $\\angle A = 45^\\circ$, \\\\\ntherefore, $\\angle BOC = 2\\angle A = 90^\\circ$, \\\\\nthus, the answer is: $\\boxed{90^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52747341.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{3}-4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In $\\triangle ABC$, $\\angle CAB=90^\\circ$, and $AC=AB$. Let point $D$ be a point on $AC$, and connect $BD$. Rotate $BD$ $90^\\circ$ clockwise around point $B$ to get $BE$, and connect $CE$ to intersect $AB$ at point $F$. As shown in figure 3, if $AB=4$ and $D$ is the midpoint of $AC$, rotate $\\triangle ABD$ around point $B$ to get $\\triangle A'BD'$. Connect $A'C$ and $A'D$. When $A'D+ \\frac{\\sqrt{2}}{2} A'C$ is minimized, find $S_{\\triangle A'BC}$.\n",
    "completion": "\\textbf{Solution:} Let $AB=a$, then $BC=\\sqrt{2}a$. Take the midpoint $N$ of $BC$, connect $A'D$, $A'C$, and $A'N$, also connect $DN$. By rotation, we know $A'B=AB=a$, \\\\\n$\\because \\frac{A'B}{BN}=\\frac{a}{\\frac{\\sqrt{2}}{2}a}=\\sqrt{2}$, $\\frac{BC}{A'B}=\\frac{\\sqrt{2}a}{a}=\\sqrt{2}$, \\\\\n$\\therefore \\frac{A'B}{BN}=\\frac{BC}{A'B}=\\sqrt{2}$. Also, $\\angle A'BN=\\angle CBA'$, \\\\\n$\\therefore \\triangle A'BN\\sim \\triangle CBA'$, \\\\\n$\\therefore \\frac{A'N}{A'C}=\\frac{A'B}{BC}=\\frac{\\sqrt{2}}{2}$, \\\\\n$\\therefore A'N=\\frac{\\sqrt{2}}{2}A'C$. According to rotation and the principle that the shortest distance between two points is a straight line, $A'D+\\frac{\\sqrt{2}}{2}A'C$ is minimized, which means $A'D+A'N$ is minimized. At this point, $D$, $A'$, $N$ are collinear, that is, $A'$ is on the line segment $DN$. Assume at this moment $A'$ falls at the position $A''$. Draw $A''F\\perp AB$ at $F$, connect $AA''$, \\\\\n$\\because D$, $N$ are the midpoints of $AC$, $BC$ respectively, \\\\\n$\\therefore DN$ is the median of $\\triangle ABC$, \\\\\n$\\therefore DN\\parallel AB$, \\\\\n$\\because AB\\perp AC$, \\\\\n$\\therefore DN\\perp AC$, \\\\\n$\\because \\angle A=\\angle A''FA=\\angle A''DA=90^\\circ$, \\\\\n$\\therefore$ quadrilateral $A''FAD$ is a rectangle, \\\\\n$\\therefore AF=A''D$, $A''F=AD=2$, also $A''B=AB=4$. Let $AF=x$, in the right triangle $A''FB$, $A''B^2=A''F^2+BF^2$, \\\\\n$\\therefore 16=4+(4-x)^2$, solving this gives $x=4-2\\sqrt{3}$. \\\\\n$\\therefore$ at this moment, $S_{\\triangle A''BC}=S_{\\triangle ABC}-S_{\\triangle AA''B}-S_{\\triangle A''AC}=\\frac{1}{2}AB\\cdot AC-\\frac{1}{2}AB\\cdot A''F-\\frac{1}{2}AC\\cdot A''D=4\\sqrt{3}-4$. \\\\\nHence, $S_{\\triangle A'BC}=\\boxed{4\\sqrt{3}-4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51368402_95.png"
  },
  {
    "solution": "\\boxed{50^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in $\\triangle ABC$, $\\angle BAC=115^\\circ$, the perpendicular bisectors of $AB$ and $CD$ intersect $BC$ at points $E$ and $F$, respectively. What is the measure of $\\angle EAF$?\n",
    "completion": "\\textbf{Solution:} Since the perpendicular bisectors of AB and AC intersect BC at points E and F, respectively,\\\\\nwe have EA = EB, thus $\\angle B = \\angle EAG$,\\\\\nLet $\\angle B = \\angle EAG = x$ degrees,\\\\\nSince FA = FC, then $\\angle C = \\angle FAH$,\\\\\nLet $\\angle C = \\angle FAH = y$,\\\\\nSince $\\angle BAC = 115^\\circ$,\\\\\nTherefore, $x + y + \\angle EAF = 115^\\circ$,\\\\\nAccording to the triangle angle sum theorem, $x + y + x + y + \\angle EAF = 180^\\circ$,\\\\\nSolving this gives $\\angle EAF = \\boxed{50^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53066348.png"
  },
  {
    "solution": "\\boxed{E_2(-1, 8)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system, the line $y=2x+6$ intersects the x-axis at point A and intersects the y-axis at point C. The parabola $y=-2x^2+bx+c$ passes through points A and C and intersects the x-axis at another point B. There is a moving point E on the parabola above line $AC$. Connect $BE$ and let it intersect line $AC$ at point F. When $EF=\\dfrac{1}{2}BF$, what are the coordinates of point E?\n",
    "completion": "\\textbf{Solution:} Let $-2x^2-4x+6=0$, we find $x_1=-3$, $x_2=1$, therefore $B(1, 0)$,\n\nSuppose the x-coordinate of point E is t, then $E\\left(t, -2t^2-4t+6\\right)$,\n\nAs shown in the diagram, draw $EH \\perp x$ axis at point H, and draw $FG \\perp x$ axis at point G, then $EH \\parallel FG$, therefore $\\triangle BFG \\sim \\triangle BEH$,\n\nSince $EF=\\frac{1}{2}BF$,\n\nTherefore $\\frac{BF}{BE}=\\frac{BG}{BH}=\\frac{FG}{EH}=\\frac{2}{3}$,\n\nSince $BH=1-t$,\n\nTherefore $BG=\\frac{2}{3}BH=\\frac{2}{3}-\\frac{2}{3}t$,\n\nTherefore the x-coordinate of point F is $\\frac{1}{3}+\\frac{2}{3}t$,\n\nTherefore $F\\left(\\frac{1}{3}+\\frac{2}{3}t, \\frac{20}{3}+\\frac{4}{3}t\\right)$,\n\nTherefore $-2t^2-4t+6=\\frac{3}{2}\\left(\\frac{20}{3}+\\frac{4}{3}t\\right)$,\n\nTherefore $t^2+3t+2=0$,\n\nWe find $t_1=-2$, $t_2=-1$,\n\nWhen $t_1=-2$, $-2t^2-4t+6=6$,\n\nWhen $t_2=-1$, $-2t^2-4t+6=8$,\n\nTherefore $E_1(-2, 6)$, $E_2(-1, 8)$,\n\nHence, the coordinates of point E are $\\boxed{E_1(-2, 6)}$ or $\\boxed{E_2(-1, 8)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53066295_94.png"
  },
  {
    "solution": "\\boxed{25}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in square $ABCD$, point $E$ is on $AD$, and point $F$ is on $AC$, with $\\angle BFE=90^\\circ$. Line $BE$ intersects $AC$ at point $G$. If $AG=24$ and $CG=32$, what is the length of $GF$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect $DF$, and rotate $\\triangle BCF$ $90^\\circ$ counterclockwise about point $B$ to get $\\triangle BAM$, connect $MG$,\n\n$\\because$ Quadrilateral $ABCD$ is a square,\n\n$\\therefore \\angle BAC = \\angle DAC = \\angle ACB = 45^\\circ$, $AB=AD$, $\\angle BAD = 90^\\circ$,\n\nIn $\\triangle ABF$ and $\\triangle ADF$,\n\n$\\begin{cases}\nAB=AD\\\\\n\\angle BAC=\\angle DAC\\\\\nAF=AF\n\\end{cases}$,\n\n$\\therefore \\triangle ABF \\cong \\triangle ADF$ (SAS),\n\n$\\therefore BF=DF$, $\\angle ABF = \\angle ADF$;\n\n$\\because \\angle BFE = 90^\\circ$,\n\n$\\therefore \\angle BFE = \\angle BAD = 90^\\circ$,\n\n$\\because \\angle ABF + \\angle AEF + \\angle BFE + \\angle BAD = 360^\\circ$,\n\n$\\therefore \\angle ABF + \\angle AEF = 180^\\circ$,\n\nAlso, $\\because \\angle DEF + \\angle AEF = 180^\\circ$,\n\n$\\therefore \\angle DEF = \\angle ABF$,\n\n$\\therefore \\angle DEF = \\angle ADF$,\n\n$\\therefore EF = DF$,\n\n$\\therefore EF = BF$,\n\n$\\because \\angle BFE = 90^\\circ$,\n\n$\\therefore \\angle EBF = \\angle BEF = 45^\\circ$;\n\nFrom the rotation, we know: $\\angle BAM = \\angle BCF = 45^\\circ$, $AM=CF$, $BM=BF$, $\\angle MBF = \\angle ABC = 90^\\circ$,\n\n$\\therefore \\angle MAG = \\angle BAM + \\angle BAC = 45^\\circ + 45^\\circ = 90^\\circ$,\n\nIn $\\triangle AMG$, according to the Pythagorean theorem: $AM^2 + AG^2 = MG^2$;\n\n$\\because AM = CF = CG - GF = 32 - GF$, $AG=24$,\n\n$\\therefore (32 - GF)^2 + 24^2 = MG^2$;\n\n$\\because \\angle MBF = 90^\\circ$, $\\angle EBF = 45^\\circ$,\n\n$\\therefore \\angle MBG = 90^\\circ - \\angle EBF = 45^\\circ$,\n\n$\\therefore \\angle MBG = \\angle FBG = 45^\\circ$,\n\nIn $\\triangle MBG$ and $\\triangle FBG$,\n\n$\\begin{cases}\nBM=BF\\\\\n\\angle MBG=\\angle FBG\\\\\nBG=BG\n\\end{cases}$,\n\n$\\therefore \\triangle MBG \\cong \\triangle FBG$,\n\n$\\therefore MG = FG$,\n\n$\\therefore (32 - GF)^2 + 24^2 = GF^2$,\n\nRearranging gives: $64GF = 1600$,\n\nSolving gives $GF = \\boxed{25}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53728981.png"
  },
  {
    "solution": "\\boxed{85}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In a right-angled triangle $ABC$, with $\\angle ABC=90^\\circ$, $AC=13$, and $AB=12$, what is the sum of the perimeters of the five small right-angled triangles in the figure?\n",
    "completion": "\\textbf{Solution:} In $\\triangle ABC$, $\\angle ABC=90^\\circ$, $AC=13$, and $AB=12$. By the Pythagorean theorem, $BC= \\sqrt{AC^2-AB^2}=\\sqrt{13^2-12^2}=5$. Based on the properties of translation, we know that the sum of the two legs of the five small right-angled triangles in the figure equals $AB+BC=12+5=17$,\n\n$\\therefore$ The perimeter of the five small right-angled triangles in the figure is $5 \\times (AB+BC) = 5 \\times 17 = \\boxed{85}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52755383.png"
  },
  {
    "solution": "\\boxed{\\left[\\frac{(n+1)^2}{2},\\frac{n+1}{2}\\right]}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system, the points are \\(A_{1}(1,0)\\), \\(A_{2}(3,0)\\), \\(A_{3}(6,0)\\), \\(A_{4}(10,0)\\), and so forth. The first square \\(A_{1}C_{1}A_{2}B_{1}\\) is constructed with \\(A_{1}A_{2}\\) as its diagonal, the second square \\(A_{2}C_{2}A_{3}B_{2}\\) with \\(A_{2}A_{3}\\) as its diagonal, the third square \\(A_{3}C_{3}A_{4}B_{3}\\) with \\(A_{3}A_{4}\\) as its diagonal, and so on, with vertices \\(B_{1}\\), \\(B_{2}\\), \\(B_{3}\\), etc., all in the first quadrant. Following this pattern, what are the coordinates of the point \\(B_n\\)?\n",
    "completion": "\\textbf{Solution}: Draw perpendiculars from points $B_1$, $B_2$, $B_3$ to the x-axis at points $D$, $E$, $F$ respectively, such that $B_1D\\perp x$-axis, $B_2E\\perp x$-axis, $B_3F\\perp x$-axis, Since $A_1(1,0)$,\\\\\nTherefore $A_1A_2=3-1=2$, $A_1D=1$, $OD=2$,\\\\\n$B_1D=A_1D=1$,\\\\\nWe obtain $B_1(2,1)$, Since $A_2(3,0)$,\\\\\nTherefore $A_3A_2=6-3=3$,\\\\\n$EB_2=\\frac{3}{2}$,\\\\\n$B_2E=EA_2=\\frac{3}{2}$,\\\\\n$OE=6-\\frac{3}{2}=\\frac{9}{2}$,\\\\\nWe obtain $B_2\\left(\\frac{9}{2},\\frac{3}{2}\\right)$,\\\\\nSimilarly, we find: $B_3(8,2)$, $B_4\\left(\\frac{25}{2},\\frac{5}{2}\\right)$, $\\dots$,\\\\\nSince the x-coordinates of $B_1$, $B_2$, $B_3$, $\\dots$ are respectively: $\\frac{4}{2}$, $\\frac{9}{2}$, $\\frac{16}{2}$, $\\frac{25}{2}\\dots$,\\\\\nTherefore, the x-coordinate of point $B_n$ is: $\\frac{(n+1)^2}{2}$,\\\\\nSince the y-coordinates of $B_1$, $B_2$, $B_3$, $\\dots$ are respectively: 1, $\\frac{3}{2}$, 2, $\\frac{5}{2}$, $\\dots$,\\\\\nTherefore, the y-coordinate of point $B_n$ is: $\\frac{n+1}{2}$,\\\\\nTherefore, the coordinates of point $B_n$ are: $\\left[\\frac{(n+1)^2}{2},\\frac{n+1}{2}\\right]$.\\\\\nHence, the answer is: $\\boxed{\\left[\\frac{(n+1)^2}{2},\\frac{n+1}{2}\\right]}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51604338.png"
  },
  {
    "solution": "\\boxed{22^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the lines AB and CD intersect at point O, $\\angle EOC = 90^\\circ$, OF is the angle bisector of $\\angle AOE$, $\\angle COF = 34^\\circ$. What is the degree measure of $\\angle BOD$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle EOC=90^\\circ$ and $\\angle COF=34^\\circ$, \\\\\nit follows that $\\angle EOF=56^\\circ$,\\\\\nsince OF is the angle bisector of $\\angle AOE$,\\\\\nit follows that $\\angle AOF=\\angle EOF=56^\\circ$,\\\\\nthus $\\angle AOC=\\angle AOF-\\angle COF=22^\\circ$,\\\\\nhence $\\angle BOD=\\angle AOC=\\boxed{22^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51251865.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the quadrilateral $ABCD$ is a parallelogram with $AD=6$. If the lengths of $OA$ and $OB$ are the two roots of the quadratic equation in $x$, ${x}^{2}-7x+12=0$, and $OA>OB$, find the lengths of $OA$ and $OB$.\n",
    "completion": "\\textbf{Solution:} To solve the equation $x^2-7x+12=0$, factorize it to get: $\\left(x-3\\right)\\left(x-4\\right)=0$. So, we have: $x-3=0$ and $x-4=0$. Therefore, the solutions are: $x_1=3$ and $x_2=4$. Since $OA>OB$, \\\\\nit follows that $OA=\\boxed{4}$ and $OB=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51458229_159.png"
  },
  {
    "solution": "\\boxed{36^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, line AB and line CD intersect at point O, and OB bisects $\\angle EOD$. If $\\angle DOE : \\angle EOC = 2 : 3$, find the degree measure of $\\angle AOC$.\n",
    "completion": "Solution:\\\\\nSince $\\angle DOE \\colon \\angle EOC = 2 \\colon 3$\\\\\nTherefore, $\\angle DOE = 180^\\circ \\times \\frac{2}{5} = 72^\\circ$\\\\\nSince $OB$ bisects $\\angle EOD$\\\\\nTherefore, $\\angle DOB = \\frac{1}{2} \\times 72^\\circ = 36^\\circ$\\\\\nTherefore, $\\angle AOC = \\angle DOB = \\boxed{36^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53206081_105.png"
  },
  {
    "solution": "\\boxed{\\frac{25\\sqrt{2}}{8}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the parabola $y=x^{2}-2x+c$ intersects with the line $y=x+1$ at points $A$ and $C$, and the coordinates of point $A$ are $(-1,0)$. If point $P$ is a moving point on the parabola below line $AC$, what is the maximum distance from point $P$ to line $AC$?\n",
    "completion": "\\textbf{Solution}: Draw $PM\\perp x$-axis from point $P$ to meet $AC$ at point $M$, as shown in the diagram: Let $P(m,m^{2}-2m-3)$ ($-1<m<5$), then $M(m,m+1)$,\\\\\n$\\therefore PM=m+1-(m^{2}-2m-3)=-m^{2}+3m+4$,\\\\\n$\\therefore S_{\\triangle ACP}=\\frac{1}{2}\\times 5(-m^{2}+3m+4)=-\\frac{5}{2}(m-\\frac{3}{2})^{2}+\\frac{125}{8}$,\\\\\n$\\therefore$when $m=\\frac{3}{2}$, $S_{\\triangle ACP}$ is maximized to $\\frac{125}{8}$,\\\\\n$\\because$point $A(-1,0)$, point $C(4,5)$,\\\\\n$\\therefore AC=\\sqrt{(4+1)^{2}+5^{2}}=5\\sqrt{2}$,\\\\\nLet the distance from point $P$ to line $AC$ be $h$,\\\\\n$\\therefore S_{\\triangle ACP}=\\frac{1}{2}\\times 5\\sqrt{2}\\times h=\\frac{125}{8}$,\\\\\n$\\therefore h=\\frac{25\\sqrt{2}}{8}$,\\\\\n$\\therefore$the maximum distance from point $P$ to line $AC$ is $\\boxed{\\frac{25\\sqrt{2}}{8}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52605441_117.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given: As shown in the figure, $AB$ is the diameter of $\\odot O$, the chord $CD$ intersects $AB$ at point E, $BE=1$, $AE=5$, and $\\angle AEC=30^\\circ$. What is the length of $CD$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $OF\\perp CD$ at F, and connect OD, \\\\\n$\\therefore$ $\\angle OFE=90^\\circ$, $CD=2FD$, \\\\\n$\\because$ $\\angle AEC=30^\\circ$, \\\\\n$\\therefore$ $OF=\\frac{1}{2}OE$, \\\\\n$\\because$ BE=1, AE=5, \\\\\n$\\therefore$ AB=BE+AE=6, \\\\\n$\\because$ AB is the diameter, \\\\\n$\\therefore$ $OB=OD=\\frac{1}{2}AB=3$, \\\\\n$\\therefore$ $OE=OB-BE=2$, \\\\\n$\\therefore$ $OF=\\frac{1}{2}OE=1$, \\\\\n$\\therefore$ $DF=\\sqrt{OD^2-OF^2}=2\\sqrt{2}$, \\\\\n$\\therefore$ $CD=2DF=\\boxed{4\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51213419.png"
  },
  {
    "solution": "\\boxed{x\\geq 1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $ l_{1}: y = x + 1 $ intersects with the line $ l_{2}: y = mx + n $ at the point $ P(a, 2) $. What is the solution set of the inequality $ x + 1 \\ge mx + n $ with respect to $x$?\n",
    "completion": "\\textbf{Solution:} Since the line $l_{1}\\colon y=x+1$ intersects with the line $l_{2}\\colon y=mx+n$ at point P$(1,2)$,\\\\\n$\\therefore a+1=2$,\\\\\nSolving this gives: $a=1$,\\\\\nObserving the graph, we know that the solution set of the inequality $x+1\\geq mx+n$ with respect to $x$ is $\\boxed{x\\geq 1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51698708.png"
  },
  {
    "solution": "\\boxed{-\\frac{1}{2}x^2+x+4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y = ax^{2} + bx + c \\left(a \\ne 0\\right)$ intersects the y-axis at point $C(0, 4)$ and intersects the x-axis at points A and B, where the coordinates of point A are $(-2, 0)$. The axis of symmetry of the parabola, $x = 1$, intersects the parabola at point D and intersects line $BC$ at point E. What is the equation of the parabola?\n",
    "completion": "\\textbf{Solution:} Given the problem, we know the parabola $y=ax^2+bx+c$ passes through points $C(0, 4)$ and $A(-2, 0)$, and its axis of symmetry is the line $x=1$,\n\\[\n\\therefore \\left\\{\\begin{array}{l}\nc=4 \\\\\n4a-2b+c=0 \\\\\n-\\frac{b}{2a}=1\n\\end{array}\\right.,\n\\]\nsolving this system yields\n\\[\n\\left\\{\\begin{array}{l}\na=-\\frac{1}{2} \\\\\nb=1 \\\\\nc=4\n\\end{array}\\right.,\n\\]\n\\[\n\\therefore \\text{the equation of the parabola is } y=-\\frac{1}{2}x^2+x+4=\\boxed{-\\frac{1}{2}x^2+x+4}.\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/53080245_25.png"
  },
  {
    "solution": "\\boxed{242^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, if $\\angle A = 31^\\circ$, then what is the value of $\\angle A + \\angle B + \\angle C + \\angle D + \\angle E$?\n",
    "completion": "\\textbf{Solution:} According to the given conditions,\\\\\n\\[ \\because \\angle AFG+\\angle AGF=180^\\circ-\\angle A, \\angle CFG+\\angle AFG=180^\\circ, \\angle EGF+\\angle AGF=180^\\circ\\]\n\\[ \\therefore \\angle CFG+\\angle EGF=360^\\circ-\\left(\\angle AFG+\\angle AGF\\right)=360^\\circ-\\left(180^\\circ-\\angle A\\right)=180^\\circ+\\angle A\\]\nAlso \\[ \\because \\angle CFG=\\angle B+\\angle C, \\angle EGF=\\angle D+\\angle E,\\]\n\\[ \\therefore \\angle A+\\angle B+\\angle C+\\angle D+\\angle E\\]\n\\[ =\\angle A+\\angle CFG+\\angle EGF\\]\n\\[ =180^\\circ+2\\angle A\\]\n\\[ =180^\\circ+2\\times 31^\\circ\\]\n\\[ =\\boxed{242^\\circ}\\]",
    "image_path": "dataset/mm_math/images/MM_Math/53107186.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in $\\triangle ABC$, $D$ and $E$ are the midpoints of $AB$ and $AC$, respectively. A line through point $C$ is drawn such that $CF \\parallel BD$, intersecting the extension of $DE$ at point $F$. Quadrilateral $BCFD$ is a parallelogram; if $BC = 6$, find the length of $EF$.\n",
    "completion": "\\textbf{Solution:} Given that $DE$ is the mid-segment of $\\triangle ABC$,\\\\\nhence $DE = \\frac{1}{2}BC = 3$,\\\\\nsince quadrilateral BCFD is a parallelogram,\\\\\nthus $DF = BC = 6$,\\\\\ntherefore $EF = DF - DE = 6 - 3 = \\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52416493_102.png"
  },
  {
    "solution": "\\boxed{9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in triangle $ABC$, $DE \\parallel BC$, $AD = 2$, $AB = 6$, $AE = 3$, then what is the length of $AC$?\n",
    "completion": "\\textbf{Solution:} \\\\\n\\[\n\\because DE\\parallel BC, \\\\\n\\therefore \\triangle ADE\\sim \\triangle ABC, \\\\\n\\therefore \\frac{AD}{AB}=\\frac{AE}{AC}, \\\\\n\\therefore \\frac{2}{6}=\\frac{3}{AC}, \\\\\n\\therefore AC=\\boxed{9}\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/52866129.png"
  },
  {
    "solution": "\\boxed{119^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $PM$ and $PN$ are tangent to $\\odot O$ at points $A$ and $B$ respectively, and $C$ is a point on $\\odot O$ that is different from $A$ and $B$. Connect $AC$ and $BC$. If $\\angle P=58^\\circ$, what is the measure of $\\angle ACB$ in degrees?\n",
    "completion": "\\textbf{Solution:} Connect $OA$ and $OB$ as shown in the figure:\\\\\nSince $PM$ and $PN$ are tangent to circle $O$ at points A and B, respectively,\\\\\nit follows that $OA\\perp PA$ and $OB\\perp PB$,\\\\\nhence $\\angle OAP=\\angle OBP=90^\\circ$,\\\\\nthus $\\angle AOB=360^\\circ-90^\\circ-90^\\circ-58^\\circ=122^\\circ$,\\\\\nWhen point C is on the major arc $AB$, $\\angle ACB=\\frac{1}{2}\\angle AOB=\\frac{1}{2}\\times 122^\\circ=\\boxed{61^\\circ}$,\\\\\nWhen point C is on the minor arc $AB$, $\\angle ACB=180^\\circ-61^\\circ=\\boxed{119^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53043223.png"
  },
  {
    "solution": "\\boxed{y=\\frac{-8}{x}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y_1=k_1x+b$ intersects with the graph of the inverse proportion function $y_2=\\frac{k_2}{x}\\ (x<0)$ at points $A$ and $B$. The intersections with the coordinate axes are $\\left(-6, 0\\right)$ and $\\left(0, 6\\right)$, respectively, and the x-coordinate of point $B$ is $-4$. Determine the equation of the inverse proportion function?\n",
    "completion": "\\textbf{Solution:} Let the equation of the linear function be $y=kx+b$,\\\\\n$\\because$ The intersections of the linear function with the coordinate axes are $\\left(-6, 0\\right)$ and $\\left(0, 6\\right)$,\\\\\n$\\therefore$\n$\\begin{cases}\n-6k+b=0, \\\\\nb=6,\n\\end{cases}$\\\\\n$\\therefore$\n$\\begin{cases}\nk=1, \\\\\nb=6,\n\\end{cases}$\\\\\n$\\therefore$ The equation of the linear function is: $y=x+6$,\\\\\n$\\therefore$ For point $B\\left(-4, 2\\right)$,\\\\\n$\\therefore$ The equation of the inverse proportional function is: $\\boxed{y=\\frac{-8}{x}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52622495_248.png"
  },
  {
    "solution": "\\boxed{\\frac{2}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $P$ is a point inside the square $ABCD$, with lines $AP$, $BP$, $DP$ connected, and $\\angle 1 = \\angle 2$. If $AB = 2$, and the tangent of one of the acute angles in $\\triangle PAB$ is 2, then what is the area of $\\triangle PAD$?\n",
    "completion": "\\textbf{Solution:} Therefore, the area of $\\triangle PAD$ is $\\boxed{\\frac{2}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53065849_138.png"
  },
  {
    "solution": "\\boxed{120^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in triangle $ABC$, the angle bisectors of $\\angle ABC$ and $\\angle ACB$ intersect at point $P$. When $\\angle A=60^\\circ$, what is the measure of $\\angle BPC$ in degrees?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, since BP is the bisector of $\\angle ABC$,\\\\\nthen $\\angle PBC = \\frac{1}{2}\\angle ABC$. (Definition of an angle bisector)\\\\\nSince CP is the bisector of $\\angle ACB$,\\\\\nthen $\\angle PCB = \\frac{1}{2}\\angle ACB$,\\\\\nthus $\\angle PBC + \\angle PCB = \\frac{1}{2}(\\angle ACB + \\angle ABC)$,\\\\\nSince $\\angle A = 60^\\circ$,\\\\\nthen $\\angle ACB + \\angle ABC = 120^\\circ$,\\\\\nthus $\\angle PBC + \\angle PCB = \\frac{1}{2}\\left(\\angle ACB + \\angle ABC\\right) = 60^\\circ$,\\\\\ntherefore $\\angle BPC = 180^\\circ - \\angle PBC - \\angle PCB = 180^\\circ - (\\angle PBC + \\angle PCB) = 180^\\circ - 60^\\circ = \\boxed{120^\\circ}.$",
    "image_path": "dataset/mm_math/images/MM_Math/52732451_78.png"
  },
  {
    "solution": "\\boxed{x<-2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $l$ is the graph of the linear function $y=kx+b$. Based on the graph of the function, directly write down the range of $x$ when $y<0$.\n",
    "completion": "\\textbf{Solution:} According to the graph of the function, we can directly state that when $y<0$, the range of $x$ is $\\boxed{x<-2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52423839_227.png"
  },
  {
    "solution": "\\boxed{30}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, the perimeter of $\\triangle DEF$ formed by the three medians is $15\\text{cm}$. What is the perimeter of $\\triangle ABC$ in $\\text{cm}$?\n",
    "completion": "\\textbf{Solution:} Since DE is the median of $\\triangle ABC$,\\\\\nit follows that DE=$\\frac{1}{2}$BC, similarly EF=$\\frac{1}{2}$AB, and DF=$\\frac{1}{2}$AC,\\\\\nthus $\\frac{DE}{BC}=\\frac{EF}{AB}=\\frac{DF}{AC}=\\frac{1}{2}$,\\\\\ntherefore $\\triangle ABC \\sim \\triangle DEF$,\\\\\nhence the perimeter of $\\triangle ABC$: the perimeter of $\\triangle DEF$=2:1,\\\\\nthus the perimeter of $\\triangle ABC$=2×15=\\boxed{30}\\text{cm}.",
    "image_path": "dataset/mm_math/images/MM_Math/51491488.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, given $\\triangle ABC\\sim \\triangle ADE$, $AD=6$, $BD=3$, $DE=4$, what is the length of $BC$?\n",
    "completion": "\\textbf{Solution}: Since $AD=6$ and $BD=3$,\\\\\nthen $AB=9$,\\\\\nSince $\\triangle ABC\\sim \\triangle ADE$ and $DE=4$,\\\\\nthen $AD: AB = DE: BC$,\\\\\nthus $BC=\\frac{AB\\cdot DE}{AD}$,\\\\\nSince $DE=4$,\\\\\nthus $BC=\\frac{9\\times 4}{6}=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52874674.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, one of the vertices $A$ of a triangular plate containing a $45^\\circ$ angle coincides with the vertex of the rectangle $ABCD$, the right-angled vertex $E$ is located on side $BC$, and another vertex $F$ falls exactly at the midpoint of side $CD$. If $BC = 12$, what is the length of $AB$?\n",
    "completion": "\\textbf{Solution:} Given that isosceles right triangle AEF,\n\\[\n\\therefore AE=EF, \\angle AEF=90^\\circ,\n\\]\n\\[\n\\therefore \\angle AEB+\\angle CEF=90^\\circ,\n\\]\nGiven rectangle ABCD,\n\\[\n\\therefore \\angle B=\\angle C=90^\\circ,\n\\]\n\\[\n\\therefore \\angle AEB+\\angle BAE=90^\\circ,\n\\]\n\\[\n\\therefore \\angle BAE=\\angle CEF,\n\\]\n\\[\n\\therefore \\triangle ABE \\cong \\triangle CEF \\text{ (AAS)},\n\\]\n\\[\n\\therefore AB=EC, BE=CF,\n\\]\nGiven F is the midpoint of CD and AB=CD,\n\\[\n\\therefore BE=\\frac{1}{2}AB,\n\\]\nGiven BC=12,\n\\[\n\\therefore \\frac{1}{2}AB + AB = 12,\n\\]\n\\[\n\\therefore AB=\\boxed{8}.\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/52076993.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, quadrilateral $ABCD$ is inscribed in circle $\\odot O$, with $BD$ bisecting $\\angle ABC$. Through point $D$, draw $DE\\parallel AB$, intersecting $BC$ at point $E$. Connect $AE$ and intersect $BD$ at point $F$. Given $\\angle AFD = \\angle ADB + \\angle CDE$, and if $CE = 2$, $AB = 8$, what is the length of $DE$?\n",
    "completion": "\\textbf{Solution:} From equation (2), it is known that $\\triangle ABE \\cong \\triangle BAD$ (SAS), $CE=2$, $AB=8$, \\\\\n$\\therefore BE=BC-CE=4-2=2$, \\\\\n$\\therefore DE=BE=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52991542_120.png"
  },
  {
    "solution": "\\boxed{\\left(-\\frac{5}{2}, \\frac{3}{2}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the parabola $y=x^{2}+5x+4$ intersects the $x$-axis at point $A$ (point $A$ is to the left of the axis of symmetry of this parabola), and intersects the $y$-axis at point $B$. The line segment $AB$ intersects the axis of symmetry of the parabola at point $P$. $OA=OB$. Find the coordinates of point $P$.\n",
    "completion": "\\[ \\textbf{Solution:} \\text{ Let the equation of line AB be } y=kx+b, \\]\n\\[ \\text{then } \\begin{cases} b=4 \\\\ 4k+b=0 \\end{cases} \\]\n\\[ \\text{Solving yields } \\begin{cases} k=-1 \\\\ b=4 \\end{cases} \\]\n\\[ \\therefore \\text{ the equation of line AB is } y=-x+4, \\]\n\\[ \\because \\text{ the axis of symmetry of the parabola is the line } x=-\\frac{5}{2}, \\]\n\\[ \\therefore \\text{ when } x=-\\frac{5}{2}, y=-\\frac{5}{2}+4=\\frac{3}{2}, \\]\n\\[ \\therefore \\text{ the coordinates of point } P \\text{ are } \\boxed{\\left(-\\frac{5}{2}, \\frac{3}{2}\\right)}. \\]",
    "image_path": "dataset/mm_math/images/MM_Math/52914172_247.png"
  },
  {
    "solution": "\\boxed{\\frac{16}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the rectangle $AOBC$, taking the lines containing $OB$ and $OA$ as the x-axis and y-axis respectively, a Cartesian coordinate system is established as shown. The coordinates of point A are $(0, 3)$ and the coordinates of point B are $(4, 0)$. Let F be a moving point on $BC$ (not coinciding with B or C). The graph of the inverse proportion function $y=\\frac{4}{x}$ ($x>0$) passing through point F intersects side $AC$ at point E. Connect $OE$ and $OF$, and draw line EF. Given $CF=2$, calculate the area of $\\triangle EOF$.\n",
    "completion": "\\textbf{Solution:}\n\nSubstituting $y=3$ into $y=\\frac{4}{x}$, we get $3=\\frac{4}{x}$.\n\nSolving, we find $x=\\frac{4}{3}$.\n\n$\\therefore$ The coordinates of point E are $\\left(\\frac{4}{3}, 3\\right)$.\n\n$\\therefore AE=\\frac{4}{3}$, and $CE=AC-AE=\\frac{8}{3}$.\n\nSince $BF=1$,\n\n$\\therefore CF=2$.\n\nGiven that $S_{\\triangle EOF}=S_{\\text{rectangle} AOBC}-S_{\\triangle CEF}-S_{\\triangle AOE}-S_{\\triangle BOF}$,\n\n$=OA \\times OB - \\frac{1}{2} \\times CF \\times CE - \\frac{1}{2} \\times OA \\times AE - \\frac{1}{2} \\times OB \\times BF$,\n\n$=3 \\times 4 - \\frac{1}{2} \\times 2 \\times \\frac{8}{3} - \\frac{1}{2} \\times 3 \\times \\frac{4}{3} - \\frac{1}{2} \\times 4 \\times 1$,\n\n$=12 - \\frac{8}{3} - 2 - 2$,\n\n$=\\boxed{\\frac{16}{3}}$.\n\n$\\therefore$ The area of $\\triangle EOF$ is $\\frac{16}{3}\\text{cm}^2$.",
    "image_path": "dataset/mm_math/images/MM_Math/51446287_231.png"
  },
  {
    "solution": "\\boxed{\\frac{50}{9}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in right triangle $ABC$, $\\angle C=90^\\circ$, $AC=8\\,\\text{cm}$, $BC=6\\,\\text{cm}$. Now, a moving point $P$ starts from point $B$, moving towards point $A$ along segment $BA$, and another moving point $Q$ starts from point $A$, moving towards the end point along the polyline $AC-CB$. If the speed of point $P$ is $1\\,\\text{cm/s}$ and the speed of point $Q$ is $1\\,\\text{cm/s}$. They start at the same time, and when one point reaches the end point, the other point also stops moving. Let the time of motion be $t$ seconds. As shown in Figure 1, $Q$ is on $AC$, at what time $t$ in seconds, will the triangle formed by points $A$, $P$, $Q$ be similar to $\\triangle ABC$?\n",
    "completion": "Solution: As shown in Figure 1, when $\\angle AQP=90^\\circ$, $\\triangle AQP\\sim \\triangle ABC$,\n\\[\\therefore \\frac{AQ}{AC}=\\frac{AP}{AB}.\\]\nIn $\\triangle ABC$, by the Pythagorean theorem, we have\n\\[AB= \\sqrt{AC^2+BC^2}=\\sqrt{8^2+6^2}=10 \\text{ (cm)}.\\]\n\\[\\because BP=t,\\ AQ=t,\\]\n\\[\\therefore PA=10-t,\\]\n\\[\\therefore \\frac{t}{8}=\\frac{10-t}{10},\\]\n\\[\\therefore t= \\boxed{\\frac{40}{9}};\\]\nAs shown in Figure 2, when $\\angle APQ=90^\\circ$, $\\triangle APQ\\sim \\triangle ABC$,\n\\[\\therefore \\frac{AQ}{AB}=\\frac{AP}{AC},\\]\n\\[\\therefore \\frac{t}{10}=\\frac{10-t}{8},\\]\n\\[t= \\boxed{\\frac{50}{9}}.\\]\nIn conclusion, when $t= \\boxed{\\frac{40}{9}} or t= \\boxed{\\frac{50}{9}}, the triangle with vertices A, P, and Q is similar to $\\triangle ABC$;",
    "image_path": "dataset/mm_math/images/MM_Math/51368352_97.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given: As shown in Figure 1, point $A(4, n)$ lies on the graph of the inverse proportion function $y=\\frac{8}{x}$ ($x>0$). As shown in Figure 2, after rotating the graph of the inverse proportion function $y=\\frac{8}{x}$ ($x>0$) counterclockwise around the origin $O$ by $45^\\circ$, it intersects the $y$-axis at point $M$. Find the length of segment $OM$.\n",
    "completion": "\\textbf{Solution:} As shown in Figure 1, rotate the y-axis clockwise by $45^\\circ$, intersecting the graph of $y=\\frac{8}{x}(x>0)$ at point N,\\\\\nthen $OM=ON$,\\\\\nthe equation of line ON is $y = x$,\\\\\nfrom $\\left\\{\\begin{array}{l}y=\\frac{8}{x}\\\\ y=x\\end{array}\\right.$, we get: $\\left\\{\\begin{array}{l}x=2\\sqrt{2}\\\\ y=2\\sqrt{2}\\end{array}\\right.$ or $\\left\\{\\begin{array}{l}x=-2\\sqrt{2}\\\\ y=-2\\sqrt{2}\\end{array}\\right.$ (discard)\\\\\n$\\therefore$Point N ($2\\sqrt{2}, 2\\sqrt{2}$)\\\\\n$\\therefore OM=ON=\\sqrt{(2\\sqrt{2})^{2}+(2\\sqrt{2})^{2}}=\\boxed{4}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52250480_92.png"
  },
  {
    "solution": "\\boxed{\\left(\\frac{11}{8}, 0\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a linear function $y=x+2$ intersects the x-axis and y-axis at points A and B, respectively. Point $M(1, m)$ is a point on line segment AB, and line segment MC intersects the x-axis at point $C\\left(\\frac{5}{2}, 0\\right)$. If point P is a moving point on line segment AC, and BP, MP are connected, and if the area of $\\triangle ABP$ is twice the area of $\\triangle MPC$, find the coordinates of point P.\n",
    "completion": "\\textbf{Solution:} For $y=x+2$, when $x=0$, then $y=2$; when $y=0$, then $x=-2$. Therefore, points $A(-2, 0)$ and $B(0, 2)$ exist, and therefore $OA=OB=2$. Let point $P(a, 0)$, then $AP=a+2$, $PC=\\frac{5}{2}-a$. Since the area of $\\triangle ABP$ is twice the area of $\\triangle MPC$, it follows that $\\frac{1}{2}(a+2)\\times 2=2\\times \\frac{1}{2}\\left(\\frac{5}{2}-a\\right)\\times 3$. Solving this, we find $a=\\frac{11}{8}$. Therefore, the coordinates of point $P$ are $\\boxed{\\left(\\frac{11}{8}, 0\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52381160_168.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, if the leg $AF=6$ and the chord $AD=10$, then what is the area of the small square $EFGH$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, the right-angled triangles in the \"Zhao Shuang Chord Diagram\" are all congruent triangles, that is $AE=BF$:\\\\\n$\\because$ leg $AF=6$, chord $AD=AB=10$,\\\\\n$\\therefore$ leg $BF=\\sqrt{AB^{2}-AF^{2}}=\\sqrt{100-36}=\\sqrt{64}=8$,\\\\\n$\\therefore$ the side length of the small square $EF=AF-AE=AF-BF=8-6=2$,\\\\\n$\\therefore$ the area of the small square $=2^{2}=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53273921.png"
  },
  {
    "solution": "\\boxed{34}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the quadrilateral $ABCD$ is a square, and the lines $a$, $b$, $c$ pass through points $A$, $D$, and $C$ respectively, with $a \\parallel b \\parallel c$. If the distance between $a$ and $b$ is $3$, and the distance between $b$ and $c$ is $5$, what is the area of square $ABCD$?\n",
    "completion": "\\textbf{Solution:} Draw AE$\\perp$b at point E through point A, and draw CF$\\perp$b at point F through point C, \\\\\n$\\therefore \\angle AED=\\angle CFD=90^\\circ$, $\\angle CDF+\\angle DCF=90^\\circ$, $\\because$ square ABCD, $\\therefore \\angle ADC=\\angle ADE+\\angle CDF=90^\\circ$, AD=CD, $\\therefore \\angle DCF=\\angle ADE$;\\\\\n$\\because a\\parallel b\\parallel c$. If the distance between a and b is 3, and the distance between b and c is 5, $\\therefore AE=3, CF=5$;\\\\\nIn $\\triangle ADE$ and $\\triangle DCF$\n$\\left\\{\\begin{array}{l}\n\\angle AED=\\angle CFD\\\\\n\\angle ADE=\\angle DCF\\\\\nAD=DC\n\\end{array}\\right.$\\\\\n$\\therefore \\triangle ADE\\cong \\triangle DCF$ (AAS), thus $AE=DF=3$,\\\\\nIn $\\triangle DCF$, $DC^{2}=DF^{2}+CF^{2}=3^{2}+5^{2}=34$.\\\\\n$\\therefore$ The area of the square ABCD is $\\boxed{34}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52251663.png"
  },
  {
    "solution": "\\boxed{\\left(\\frac{7}{2}, -\\frac{3}{2}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the line $y =\\frac{3}{4}x +\\frac{3}{2}$ intersects the x-axis at point A, and passes through point B (2, a). There is a moving point P on the y-axis, and a moving point M on the line BC, with C (3, 0). If $\\triangle APM$ is an isosceles right triangle with segment AM as its hypotenuse, what are the coordinates of point M?\n",
    "completion": "\\textbf{Solution:} The coordinates of point M are $\\boxed{\\left(\\frac{7}{2}, -\\frac{3}{2}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52837765_68.png"
  },
  {
    "solution": "\\boxed{y=2x}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle AOB$, $\\angle ABO=90^\\circ$, $OB=4$, $AB=8$. The line $y=-x+b$ intersects $OA$ and $AB$ at points $C$ and $D$, respectively, and the area of $\\triangle BOD$ is $4$. What is the equation of line $AO$?\n",
    "completion": "\\textbf{Solution:} Given that $OB=4$, $AB=8$, and $\\angle ABO=90^\\circ$,\\\\\nthus, the coordinates of point A are (4, 8).\\\\\nAssume the equation of line $AO$ is $y=kx$, then $4k=8$,\\\\\nsolving this gives $k=2$, thus the equation of line $AO$ is $\\boxed{y=2x}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52076885_174.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, rhombus \\(ABCD\\) has a side length of 10, \\(E\\) is the midpoint of \\(AD\\), \\(O\\) is the intersection point of the diagonals, and rectangular \\(OEFG\\) has one side on \\(AB\\), with \\(EF = 4\\). What is the length of \\(BG\\)?\n",
    "completion": "\\textbf{Solution:} Since $ABCD$ is a rhombus, and $E$ is the midpoint of side $AD$,\\\\\ntherefore $AD=AB=10$, $AE=\\frac{1}{2}AD=5$, $AC\\perp BD$,\\\\\nthus $EO=\\frac{1}{2}AD=5$. Since $OEFG$ is a rectangle,\\\\\ntherefore $FG=EO=5$, $\\angle EFA=90^\\circ$,\\\\\nin $\\triangle EFA$, where $AE=5$, $EF=4$,\\\\\nthus $AF=\\sqrt{5^2-4^2}=3$,\\\\\ntherefore $BG=AB-AF-FG=10-3-5=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51942331.png"
  },
  {
    "solution": "\\boxed{15^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the rhombus $ABCD$, $\\angle ABC=30^\\circ$. Connect $AC$. Follow the steps below to draw: Take $C$ and $B$ as centers, and with the length of $BC$ as the radius, draw arcs. The arcs intersect at point $P$. Connect $BP$ and $CP$. What is the measure of $\\angle ACP$?\n",
    "completion": "\\textbf{Solution:} According to the steps given in the question, the diagram is completed as follows:\\\\\nSince rhombus ABCD, $\\angle$ABC=30$^\\circ$,\\\\\ntherefore, $\\angle$BCD=150$^\\circ$,\\\\\nthus, $\\angle$ACB=75$^\\circ$,\\\\\n(1) When point P is above BC,\\\\\nsince BP=PC=BC,\\\\\nthus, $\\triangle BPC$ is an equilateral triangle,\\\\\ntherefore, $\\angle$PCB=60$^\\circ$,\\\\\nhence, $\\angle$ACP=$\\angle$ACB−$\\angle$PCB=75$^\\circ$−60$^\\circ$=\\boxed{15^\\circ};\\\\\n(2) When point P' is below BC,\\\\\nsince BP'=P'C=BC,\\\\\nthus, $\\triangle BP'C$ is an equilateral triangle,\\\\\ntherefore, $\\angle$P'CB=60$^\\circ$,\\\\\nhence, $\\angle$ACP'=$\\angle$ACB+$\\angle$P'CB=75$^\\circ$+60$^\\circ$=135$^\\circ$,\\\\\nIn conclusion, the degree of $\\angle$ACP is either $\\boxed{15^\\circ}$ or 135$^\\circ$.",
    "image_path": "dataset/mm_math/images/MM_Math/52205063.png"
  },
  {
    "solution": "\\boxed{(1, -\\frac{4\\sqrt{3}}{3})}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated, it is known in $\\triangle ABC$ that $\\angle BAC=30^\\circ$ and $\\angle C=90^\\circ$. The coordinates of point A are $(-1,0)$ and those of point B are $(3,0)$. The parabola $y_1$ has its vertex labeled as Q and passes through the three vertices A, B, and C of $\\triangle ABC$ (with point A to the left of B and point C below the x-axis). Another parabola $y_2$ also intersects the x-axis at points A and B, with its vertex labeled as P. The task is to find the coordinates of point C and the coordinates of the vertex Q of the parabola $y_1$.\n",
    "completion": "Solution: Since $\\angle BAC=30^\\circ$ and $\\angle C=90^\\circ$, with A located at $(-1,0)$ and B at $(3,0)$, it follows that $AC=2\\sqrt{3}$. Consequently, $y_C=-\\sqrt{3}$. Therefore, the coordinates of point C are $(2, -\\sqrt{3})$. Given that the parabola $y_1$ passes through points A, B, and C, let the equation of the parabola $y_1$ be $y_1=a(x+1)(x-3)$. It follows that $-\\sqrt{3}=-3a$, hence $a=\\frac{\\sqrt{3}}{3}$. Therefore, $y_1=\\frac{\\sqrt{3}}{3}(x+1)(x-3)=\\frac{\\sqrt{3}}{3}(x-1)^2-\\frac{4\\sqrt{3}}{3}$. Hence, the coordinates of the vertex Q are $\\boxed{(1, -\\frac{4\\sqrt{3}}{3})}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51317409_150.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, AB is the diameter of the circle with center O, and the chord CD is perpendicular to AB at point H. If AB = 10 and CD = 8, what is the length of OH?\n",
    "completion": "\\textbf{Solution:} Given that AB is the diameter of $\\odot$O, and chord CD is perpendicular to AB at point H, if AB=10, and CD=8, \\\\\n$\\therefore CH=\\frac{1}{2}CD=4$, $OC=\\frac{1}{2}AB=5$\\\\\nIn $\\triangle OHC$, $ OH=\\sqrt{OC^{2}-CH^{2}}=\\sqrt{5^{2}-4^{2}}=\\boxed{3}$",
    "image_path": "dataset/mm_math/images/MM_Math/51350892.png"
  },
  {
    "solution": "\\boxed{\\frac{2\\sqrt{5}}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within a square grid with side length 1, points A, B, and C are on grid points, and lines AB and CD intersect at point D. What is the value of $\\sin \\angle ADC$?\n",
    "completion": "\\textbf{Solution:} Extend CD to meet another vertex of the square at $E$, and connect BE, as shown in the diagram:\\\\\nFrom the problem statement, it is known that: $\\angle BED=90^\\circ$, $\\angle ADC=\\angle BDE$,\\\\\nBased on the side length of the small square grid and the Pythagorean theorem, it can be derived that: $BE=2\\sqrt{2}$, $BD=\\sqrt{10}$,\\\\\n$\\therefore$ In $\\triangle BDE$, $\\sin\\angle BDE=\\frac{BE}{BD}=\\frac{2\\sqrt{5}}{5}$,\\\\\n$\\therefore \\sin\\angle ADC=\\sin\\angle BDE=\\boxed{\\frac{2\\sqrt{5}}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55551954.png"
  },
  {
    "solution": "\\boxed{\\left(1, 4\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=-x^2+bx+c$ passes through point $A(3, 0)$ and point $B(0, 3)$. Find the expression of the parabola and the coordinates of vertex $C$.\n",
    "completion": "\\textbf{Solution:} Substitute $A\\left(3, 0\\right)$ and $B\\left(0, 3\\right)$ into the parabola equation $y=-{x}^{2}+bx+c$ and get:\n\\[\n\\begin{cases}\n3b+c-9=0\\\\\nc=3\n\\end{cases}\n\\]\nSolving these, we find $b=2$ and $c=3$;\\\\\nTherefore, the equation of the parabola is: $y=-{x}^{2}+2x+3$\\\\\nSince $y=-{x}^{2}+2x+3=-{\\left(x-1\\right)}^{2}+4$,\\\\\nTherefore, the coordinates of the vertex $C$ of the parabola are: $\\boxed{\\left(1, 4\\right)}$",
    "image_path": "dataset/mm_math/images/MM_Math/53043922_118.png"
  },
  {
    "solution": "\\boxed{2\\colon 1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the parallelogram $ABCD$, $AC$ and $BD$ intersect at point $O$, $E$ is the midpoint of $OA$. Connect $BE$ and extend it to intersect $AD$ at point $F$. What is the ratio of $FD:AF$?\n",
    "completion": "\\textbf{Solution:} Given quadrilateral ABCD,\\\\\nit follows that $AD=BC$, $AD\\parallel BC$, and OA=OC,\\\\\nfurthermore, since point E is the midpoint of OA,\\\\\nit follows that $AE=OE=\\frac{1}{2}OA=\\frac{1}{4}AC$, and $CE=\\frac{3}{4}AC$\\\\\nSince $AD\\parallel BC$\\\\\nit follows that $\\triangle AEF\\sim \\triangle CEB$\\\\\nTherefore, $\\frac{AF}{BC}=\\frac{AE}{CE}=\\frac{\\frac{1}{4}AC}{\\frac{3}{4}AC}=\\frac{1}{3}$\\\\\nThus, $AF=\\frac{1}{3}BC$\\\\\nSince $AF+FD=AD=BC$\\\\\nit follows $FD=\\frac{2}{3}BC$\\\\\nTherefore, $FD\\colon AF=2\\colon 1$\\\\\nHence, the answer is: $FD\\colon AF=\\boxed{2\\colon 1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51445486.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{1}{2}x^{2}+\\frac{5}{2}x-2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the parabola passes through three points $A(4,0)$, $B(1,0)$, and $C(0,-2)$. Find the equation of the parabola.\n",
    "completion": "\\textbf{Solution:} Let the equation of the parabola be $y=a(x-4)(x-1)$. Substituting the coordinates of point $C(0, -2)$ into the equation gives: $-2=a(0-4)(0-1)$, solving for $a$ yields $a=-\\frac{1}{2}$, $\\therefore y=-\\frac{1}{2}(x-4)(x-1)=-\\frac{1}{2}x^{2}+\\frac{5}{2}x-2$. Hence, the equation of the parabola is $\\boxed{y=-\\frac{1}{2}x^{2}+\\frac{5}{2}x-2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52874656_286.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y=mx+2$ passes through points $A(2, 4)$ and $B(n, -1)$. Connect $OA$ and $OB$. Find the area of $\\triangle OAB$.\n",
    "completion": "\\textbf{Solution:} As shown in the figure, let the intersection of line $AB$ with the x-axis be $C$. Let $y=0$, then $x=-2$, $\\therefore$ the coordinates of point $C$ are $(-2, 0)$, $\\therefore$ $OC=2$, $\\therefore$ the area of $\\triangle OAB$ equals the area of $\\triangle AOC$ plus the area of $\\triangle BOC$\\\\\n$=\\frac{1}{2}\\times OC\\times y_A+\\frac{1}{2}\\times OC\\times |y_B|$\\\\\n$=\\frac{1}{2}\\times 2\\times 4+\\frac{1}{2}\\times 2\\times 1=\\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50590307_153.png"
  },
  {
    "solution": "\\boxed{\\frac{2\\sqrt{3}}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle BAC=90^\\circ$, and $AD$ is drawn perpendicular to $BC$ at point $D$. If $\\frac{BD}{CD} = \\frac{4}{3}$, what is the value of $\\tan C$?\n",
    "completion": "\\textbf{Solution:}\n\nGiven $\\frac{BD}{CD} = \\frac{4}{3}$,\\\\\nlet $BD=4x$, and $CD=3x$.\\\\\nSince $AD \\perp BC$,\\\\\nwe have $\\angle ADB = \\angle ADC = \\angle BAC = 90^\\circ$,\\\\\nwhich implies $\\angle BAD + \\angle CAD = \\angle C + \\angle CAD = 90^\\circ$,\\\\\nhence $\\angle C = \\angle BAD$,\\\\\nleading to $\\triangle BAD \\sim \\triangle ACD$,\\\\\nthus $\\frac{AD}{BD} = \\frac{CD}{AD}$, i.e., $AD^2 = BD \\cdot CD = 3x \\cdot 4x = 12x^2$,\\\\\ntherefore, $AD = 2\\sqrt{3}x$,\\\\\nfrom which it follows $\\tan C = \\frac{AD}{CD} = \\frac{2\\sqrt{3}x}{3x} = \\boxed{\\frac{2\\sqrt{3}}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51083564.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point P $(x, y)$ is a moving point in the first quadrant, and it lies on the line $y=-2x+8$. The line intersects the x-axis at point A. What is the area of $\\triangle APO$ when the x-coordinate of point P is 3?\n",
    "completion": "\\textbf{Solution:} Let $y=0$, then $-2x+8=0$, solving this we find $x=4$,\\\\\n$\\therefore OA=4$.\\\\\n$\\because$ Point P$(x,y)$ is a moving point in the first quadrant, and lies on the line $y=-2x+8$,\\\\\n$\\therefore$ when $x=3$, $y=-2\\times 3+8=2$.\\\\\n$\\therefore S_{\\triangle APO}=\\frac{1}{2}\\times 4\\times 2=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52422771_83.png"
  },
  {
    "solution": "\\boxed{y=(x-2)^{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the vertex $C$ of the parabola passing through points $A\\left(0, 4\\right)$ and $B\\left(5, 9\\right)$ is located on the positive half of the $x$-axis. What is the analytic equation of the parabola?\n",
    "completion": "\\textbf{Solution:} Since the vertex $C$ of the parabola is on the positive half of the x-axis,\\\\\nlet the equation of the parabola be $y=a(x-h)^{2}$.\\\\\nSubstituting points $A(0, 4)$ and $B(5, 9)$ into $y=a(x-h)^{2}$, we have:\\\\\n$\\begin{cases}\na(0-h)^{2}=4\\\\\na(5-h)^{2}=9\n\\end{cases}$,\\\\\nSolving this gives $h=-10$ (discard) or $h=2$,\\\\\nthus $a=1$,\\\\\ntherefore, the equation of the parabola is: \\boxed{y=(x-2)^{2}}.",
    "image_path": "dataset/mm_math/images/MM_Math/52551573_131.png"
  },
  {
    "solution": "\\boxed{76^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the incircle $\\odot O$ of $\\triangle ABC$ is tangent to $AB$, $BC$, $CA$ at points $D$, $E$, and $F$, respectively. If $\\angle DEF = 52^\\circ$, then what is the degree measure of $\\angle A$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect OD, and OF, \\\\\n$\\because$ The incircle $\\odot O$ of $\\triangle ABC$ is tangent to the sides AB, BC, CA at points D, E, F, respectively,\\\\\n$\\therefore$ $\\angle ADO=\\angle AFO=90^\\circ$；\\\\\n$\\therefore$ $\\angle A+\\angle DOF=180^\\circ$，\\\\\n$\\because$ $\\angle DEF=52^\\circ$, and $\\angle DOF$ and $\\angle DEF$ are the central angle and the inscribed angle subtended by the same arc $\\overset{\\frown}{DF}$, respectively,\\\\\n$\\therefore$ $\\angle DOF=2\\angle DEF=104^\\circ$；\\\\\n$\\therefore$ $\\angle A=180^\\circ-\\angle DOF=\\boxed{76^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52622404.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y=kx+b$ intersects with the graph of the inverse proportion function $y=\\frac{m}{x}$ at points A and B, and intersects with the x-axis at point C. The coordinates of point A are $(2, 3)$, and the coordinates of point B are $(-6, n)$. Connect AO and OB, then find the area of $\\triangle AOB$.\n",
    "completion": "\\textbf{Solution:} When $y=0$, $x=4$, \\\\\n$\\therefore C(-4,0)$, \\\\\n$\\therefore OC=4$. Construct $AD\\perp x$ axis at point $D$, and $BE\\perp x$ axis at point $E$. \\\\\nThen, $AD=3$, $BE=1$, \\\\\n$\\therefore$ the area of $\\triangle AOB$ is $S_{\\triangle AOB} = S_{\\triangle BOC} + S_{\\triangle AOC} = \\frac{1}{2} \\times 4 \\times 1 + \\frac{1}{2} \\times 4 \\times 3 = \\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50496315_69.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y=-\\frac{1}{2}x+3$ intersects the x-axis at point A and the y-axis at point B, and intersects the graph of the direct proportionality function $y=x$ at point C. What is the ratio of the areas of $\\triangle BOC$ to $\\triangle AOC$?\n",
    "completion": "\\textbf{Solution:} Given that the graph of the linear function $y=-\\frac{1}{2}x+3$ intersects the x-axis at point A and the y-axis at point B, \\\\\nthus $A\\left(6,0\\right)$, and $B\\left(0,3\\right)$, \\\\\nGiven that the linear function $y=-\\frac{1}{2}x+3$ intersects the graph of the direct proportionality function $y=x$ at point C, \\\\\nthus solving $\\left\\{\\begin{array}{l}y=-\\frac{1}{2}x+3 \\\\ y=x\\end{array}\\right.$ gives $C\\left(2,2\\right)$, \\\\\nthus, the area of $\\triangle BOC$ is $\\frac{1}{2}\\times 3\\times 2=3$, and the area of $\\triangle AOC$ is $\\frac{1}{2}\\times 6\\times 2=6$, \\\\\nthus, the ratio of the areas of $\\triangle BOC$ and $\\triangle AOC$ is $\\boxed{\\frac{1}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52342410.png"
  },
  {
    "solution": "\\boxed{20^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points $A$, $B$, $C$, $D$ are all on the circle $\\odot O$, $\\angle BAC=15^\\circ$, $\\angle BOD=70^\\circ$, and $DE$ is tangent to $\\odot O$ at $D$. What is the measure of $\\angle CDE$?\n",
    "completion": "\\textbf{Solution:} Draw line $OC$, \\\\\nsince $\\angle BAC=15^\\circ$, \\\\\nthus $\\angle BOC=2\\angle BAC=30^\\circ$, \\\\\nsince $\\angle BOD=70^\\circ$, \\\\\nthus $\\angle COD=70^\\circ-30^\\circ=40^\\circ$, \\\\\nsince $OC=OD$, \\\\\nthus $\\angle ODC=\\angle OCD=\\frac{1}{2}\\left(180^\\circ-40^\\circ\\right)=70^\\circ$, \\\\\nsince $DE$ is tangent to circle $O$ at $D$, \\\\\nthus $OD\\perp DE$, \\\\\nthus $\\angle CDE=90^\\circ-70^\\circ=\\boxed{20^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/52977065.png"
  },
  {
    "solution": "\\boxed{-\\frac{1}{4}(x-4)^2+6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the cross-section of a tunnel is composed of a parabola and a rectangle. The rectangle has a length of 8m and a width of 2m. The highest point of the tunnel, point P, is located at the center of AB and is 6m above the ground. Establish the coordinate system as shown in the figure. Find the expression of the parabola.\n",
    "completion": "\\textbf{Solution:} Given the problem, we know that the vertex of the parabola is at $(4,6)$. Assume the equation of the parabola is $y=a(x-4)^2+6$. Since point A$(0,2)$ lies on the parabola, it follows that $2=a(0-4)^2+6$. Solving this gives $a=-\\frac{1}{4}$. Therefore, the equation of the parabola is: $y=-\\frac{1}{4}(x-4)^2+6=\\boxed{-\\frac{1}{4}(x-4)^2+6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51446670_261.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, taking points A and B as the centers respectively, arcs are drawn with radii greater than $\\frac{1}{2}AB$ and intersecting at points M and N. Line $MN$ is drawn, intersecting $BC$ at point D. $AD$ is connected. If the perimeter of $\\triangle ABC$ is $17$ and $AB=7$, what is the perimeter of $\\triangle ADC$?\n",
    "completion": "Solution: From the method, we know: MN is the perpendicular bisector of AB,\\\\\n$\\therefore$AD=BD,\\\\\n$\\because$ The perimeter of $\\triangle ABC$ is 17 and AB=7,\\\\\n$\\therefore$AC+BC=10,\\\\\n$\\therefore$ The perimeter of $\\triangle ADC$ is:\\\\\nAC+CD+AD=AC+CD+BD=AC+BC=\\boxed{10}",
    "image_path": "dataset/mm_math/images/MM_Math/51367859.png"
  },
  {
    "solution": "\\boxed{16}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown, the area of the paper triangle $\\triangle ABC$ is $12\\,\\text{cm}^2$, where one side $BC$ has a length of $6\\,\\text{cm}$. After two cuts, it is assembled into a seamless, non-overlapping rectangle $BCDE$. What is the perimeter of the rectangle in $\\,\\text{cm}$?\n",
    "completion": "\\textbf{Solution:} Extend AT to intersect BC at point P,\\\\\n$\\because$AP$\\perp$BC,\\\\\n$\\therefore$ $\\frac{1}{2}$•BC•AP=12,\\\\\n$\\therefore$ $\\frac{1}{2}$×6×AP=12,\\\\\n$\\therefore$AP=4 (cm),\\\\\nAccording to the problem, AT=PT=2 (cm),\\\\\n$\\therefore$BE=CD=PT=2 (cm),\\\\\n$\\because$DE=BC=6cm,\\\\\n$\\therefore$The perimeter of rectangle BCDE is 6+6+2+2=\\boxed{16} (cm).",
    "image_path": "dataset/mm_math/images/MM_Math/51635596.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that point $M$ is the centroid of $\\triangle ABC$, and $AB=18$. If $MN\\parallel AB$, what is the value of $MN$?\n",
    "completion": "\\textbf{Solution:} Given that point $M$ is the centroid of $\\triangle ABC$ and $AB=18$,\\\\\nit follows that $AD=DB=\\frac{1}{2}AB=9$, and $\\frac{CM}{CD}=\\frac{2}{3}$.\\\\\nSince $MN\\parallel AB$,\\\\\nit implies that $\\triangle CMN\\sim \\triangle CDB$,\\\\\ntherefore, $\\frac{MN}{BD}=\\frac{CM}{CD}=\\frac{2}{3}$, which means $\\frac{MN}{9}=\\frac{2}{3}$,\\\\\nSolving this gives: $MN=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51322831.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, $\\angle B=15^\\circ$, the perpendicular bisector of AB intersects AB at E and intersects BC at D. Given that BD=8, what is the length of AC?\n",
    "completion": "\\textbf{Solution:} As shown in the figure: \\\\\nSince in $\\triangle ABC$, $\\angle C=90^\\circ$, $\\angle B=15^\\circ$, \\\\\ntherefore $\\angle BAC=180^\\circ-\\angle C-\\angle B=180^\\circ-90^\\circ-15^\\circ=75^\\circ$. \\\\\nDraw AD. \\\\\nSince ED is the perpendicular bisector of AB, therefore AD=BD=8, $\\angle B=\\angle 1=15^\\circ$, \\\\\ntherefore $\\angle 2=\\angle BAC-\\angle 1=75^\\circ-15^\\circ=60^\\circ$. \\\\\nIn $\\triangle ACD$, $\\angle 2=60^\\circ$, $\\angle C=90^\\circ$, \\\\\ntherefore $\\angle 3=180^\\circ-\\angle C-\\angle 2=180^\\circ-90^\\circ-60^\\circ=30^\\circ$. \\\\\nTherefore, $AC=\\frac{1}{2}AD=\\frac{1}{2}BD=\\frac{1}{2}\\times 8=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52655068.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown, \\(O\\) is the intersection point of the diagonals of rhombus \\(ABCD\\), \\(DE \\parallel AC\\), \\(CE \\parallel BD\\). If \\(AC = 6\\), \\(BD = 8\\), then what is the length of line segment \\(OE\\)?\n",
    "completion": "\\textbf{Solution:} Given $\\because$DE$\\parallel$AC, CE$\\parallel$BD,\\\\\n$\\therefore$ quadrilateral OCED is a parallelogram,\\\\\n$\\because$ quadrilateral ABCD is a rhombus,\\\\\n$\\therefore$ $\\angle$COD$=90^\\circ$,\\\\\n$\\therefore$ quadrilateral OCED is a rectangle,\\\\\nFurther $\\because$AC$=6$, BD$=8$,\\\\\n$\\therefore$OC$=3$, OD$=4$,\\\\\n$\\therefore$CD$=\\sqrt{OC^{2}+OD^{2}}=\\sqrt{3^{2}+4^{2}}=\\boxed{5}$,\\\\\nIn rectangle OCED, OE$=CD=5$.",
    "image_path": "dataset/mm_math/images/MM_Math/52386862.png"
  },
  {
    "solution": "\\boxed{55^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, quadrilateral $ABCD$ is inscribed in $\\odot O$, with $E$ being a point on the extension of $BC$, and point $C$ being the midpoint of the arc $\\overset{\\frown}{BD}$. If $\\angle DCE=110^\\circ$, what is the degree measure of $\\angle BAC$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral ABCD is inscribed in circle O,\\\\\nit follows that $\\angle DCE=\\angle BAD$ \\\\\nSince $\\angle DCE=110^\\circ$ \\\\\nthen $\\angle BAD=110^\\circ$ \\\\\nGiven that point C is the midpoint of the arc $BD$\\\\\nit follows that arc $BC=$ arc $DC$ \\\\\nTherefore, $\\angle BAC=\\angle DAC=\\frac{1}{2}\\angle BAD=\\frac{1}{2}\\times 110^\\circ=55^\\circ$ \\\\\nHence, the measure of $\\angle BAC$ is $\\boxed{55^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/51402325.png"
  },
  {
    "solution": "\\boxed{(-1-2\\sqrt{2}, -4)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the parabola $y=-(x+1)^{2}+4$ intersects the x-axis at points A and B, and intersects the y-axis at point C. Is there a point P on the parabola such that the area of $\\triangle PAB$ is 8? If it exists, write down the coordinates of point P directly; if not, explain the reason.\n",
    "completion": "\\textbf{Solution:} Given that, the coordinates of point P are \\boxed{(-1, 4)}, \\boxed{(-1+2\\sqrt{2}, -4)}, or \\boxed{(-1-2\\sqrt{2}, -4)}.",
    "image_path": "dataset/mm_math/images/MM_Math/53066218_99.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the graph of the linear function $y=2x+4$ intersects with the graph of the inverse proportion function $y=\\frac{k}{x}$ at points $A$ and $B$, where the x-coordinate of point $B$ is $1$. A line is drawn from point $A$ such that $AC \\perp y$ axis at point $C$, and line $BC$ is connected. What is the area of $\\triangle ABC$?\n",
    "completion": "\\textbf{Solution:}\n\nSince the graph of the linear function $y=2x+4$ intersects the graph of the inverse proportionality function $y=\\frac{6}{x}$ at points A and B, and the x-coordinate of point B is 1,\n\nSubstituting $x=1$ into $y=2x+4$ gives $y=6$,\n\nTherefore, point B is (1, 6),\n\nTherefore, $6=\\frac{6}{1}$, solving for $k$ gives $k=6$,\n\nTherefore, the equation of the inverse proportionality function is $y=\\frac{6}{x}$,\n\nSolving $\\left\\{\\begin{array}{l}y=\\frac{6}{x}\\\\ y=2x+4\\end{array}\\right.$ yields: $\\left\\{\\begin{array}{l}x=1\\\\ y=6\\end{array}\\right.$ or $\\left\\{\\begin{array}{l}x=-3\\\\ y=-2\\end{array}\\right.$,\n\nTherefore, point A is (-3, -2),\n\nSince AC is perpendicular to the y-axis at point C,\n\nTherefore, AC equals 3,\n\nTherefore, the area of $\\triangle ABC=\\frac{1}{2}\\times3\\times(6+2)=\\boxed{12}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51402479.png"
  },
  {
    "solution": "\\boxed{84}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle ABCD, the diagonals AC and BD intersect at point O. A line parallel to BD is drawn through point C, and a line parallel to AC is drawn through point D, with the two lines intersecting at point P. Quadrilateral CODP is a rhombus. Given that $AD=7$, and $AC=25$, what is the area of quadrilateral CODP?\n",
    "completion": "\\textbf{Solution:} Given that $AD=7$ and $AC=25$, \\\\\ntherefore $DC=\\sqrt{AC^2-AD^2}=24$, \\\\\nsince $AO=CO$, \\\\\nthus the area of $\\triangle COD$ is half of that of $\\triangle ADC$, which equals $\\frac{1}{2}\\times \\frac{1}{2}AD\\cdot CD=42$, \\\\\nsince the quadrilateral CODP is a rhombus, \\\\\ntherefore the area of the rhombus CODP is twice the area of $\\triangle COD$, which equals $\\boxed{84}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53359542_111.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{10}}{10}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the vertices of $\\triangle ABC$ are all on the grid points of the graph paper. What is the value of $\\sin A$?\n",
    "completion": "\\textbf{Solution:} Given the diagram, we have $AC=\\sqrt{1^2+3^2}=\\sqrt{10}$, \\\\\n$\\therefore \\sin A=\\frac{1}{AC}=\\frac{1}{\\sqrt{10}}=\\boxed{\\frac{\\sqrt{10}}{10}}$",
    "image_path": "dataset/mm_math/images/MM_Math/50934145.png"
  },
  {
    "solution": "\\boxed{\\frac{2\\sqrt{5}}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the vertices of $\\triangle ABC$ are all lattice points in a square grid. What is the value of $\\cos\\angle ABC$?\n",
    "completion": "\\textbf{Solution:} Let $\\triangle ABE$ be a right-angled triangle, where $CE=2, BE=4$, \\\\\n$\\therefore$ $BC= \\sqrt{CE^2+BE^2} = \\sqrt{2^2+4^2} = 2\\sqrt{5}$, \\\\\n$\\therefore$ $\\cos\\angle ABC = \\frac{BE}{BC} = \\frac{4}{2\\sqrt{5}} = \\boxed{\\frac{2\\sqrt{5}}{5}}$",
    "image_path": "dataset/mm_math/images/MM_Math/55574671.png"
  },
  {
    "solution": "\\boxed{(-2,-4)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=kx+b$ intersects the $x$-axis at point $B$ and the $y$-axis at point $C(0,3)$, and it intersects $y=2x$ at point $A(a,2a)$. Point $M$ lies on the line $OA$. Is there a point $M$ such that the area of $\\triangle OMC$ is equal to the area of $\\triangle OAB$? Find the coordinates of point $M$.\n",
    "completion": "\\textbf{Solution:} Let there be a point $M$ such that the area of $\\triangle OMC$ is equal to the area of $\\triangle OAB$, for the following reasons:\\\\\n$\\because$ Point $C(0,3)$,\\\\\n$\\therefore OC=3$, $\\therefore OB=OC=3$.\\\\\n$\\because$ The area of $\\triangle OMC$ is equal to the area of $\\triangle OAB$,\\\\\n$\\therefore$ The distance from $M$ to the $y$-axis is equal to the $y$-coordinate of point $A$, which is $2$,\\\\\n$\\therefore$ The $x$-coordinate of point $M$ is $2$ or $-2$;\\\\\nWhen the $x$-coordinate of $M$ is $2$, in $y=2x$,\\\\\nwhen $x=2$, $y=4$, then the coordinates of $M$ are $(2,4)$;\\\\\nWhen the $x$-coordinate of $M$ is $-2$, in $y=2x$,\\\\\nwhen $x=-2$, $y=-4$, then the coordinates of $M$ are $(-2,-4)$.\\\\\nIn summary: The coordinates of point $M$ are either $\\boxed{(2,4)}$ or $\\boxed{(-2,-4)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52354898_184.png"
  },
  {
    "solution": "\\boxed{y=2x+2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the graph of the linear function $y=kx+b$ passes through points $A(-2, -2)$ and $B(1, 4)$, and intersects the x-axis at point C and the y-axis at point D. What is the analytical expression of this linear function?\n",
    "completion": "\\textbf{Solution}: Substituting $A(-2, -2)$ and $B(1, 4)$ into $y=kx+b$, we get \n\\[\n\\begin{cases} \n-2k+b=-2 \\\\ \nk+b=4 \n\\end{cases},\n\\]\nsolving this yields \n\\[\n\\begin{cases} \nk=2 \\\\ \nb=2 \n\\end{cases},\n\\]\n\\[\n\\therefore\n\\] the equation of the linear function is $\\boxed{y=2x+2}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51663193_196.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the triangular paper $ABC$, $\\angle C=90^\\circ$, $\\angle A=30^\\circ$, and $AC=9$. Fold the paper so that point $C$ falls on point $D$ on side $AB$, and the fold line $BE$ intersects $AC$ at point $E$. What is the length of the fold line $BE$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle C=90^\\circ$ and $\\angle A=30^\\circ$,\\\\\nit follows that $\\angle ABC=60^\\circ$.\\\\\nSince the paper is folded such that point C lies on point D on side AB,\\\\\nit follows that $\\angle CBE=\\angle ABE=30^\\circ$, CE=DE, and $\\angle ADE=90^\\circ$,\\\\\nwhich implies that $\\angle A=\\angle ABE$ and AE=2CE\\\\\nTherefore, AE=BE=\\boxed{6}.",
    "image_path": "dataset/mm_math/images/MM_Math/51435245.png"
  },
  {
    "solution": "\\boxed{x>6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Synthesis and Inquiry: As shown in the figure, in the Cartesian coordinate system, the line $l_1\\colon y_1 = -\\frac{1}{2}x + 6$ intersects the x-axis and y-axis at points B and C, respectively, and intersects with the line $l_2\\colon y_2 = \\frac{1}{2}x$ at point A. When $y_1 < y_2$, write down directly the range of values for x.\n",
    "completion": "\\textbf{Solution:} Given point A$(6,3)$ and the graph, when $y_1<y_2$, then $\\boxed{x>6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50677413_140.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y=kx+b$ (where $k\\ne 0$) intersects with the graph of the inverse proportion function $y=\\frac{6}{x}$ (where $x>0$) at points $A(m, 6)$ and $B(3, n)$. Find the area of triangle $\\triangle AOB$.\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, lines are drawn from point A and point B perpendicular to the x-axis, meeting the x-axis at points E and C respectively. The line AB intersects the x-axis at point D.\\\\\nLet $-2x+8=0$, we get $x=4$, that is, D(4,0).\\\\\nSince A(1,6) and B(3,2),\\\\\nit follows that AE=6 and BC=2,\\\\\ntherefore, the area of triangle AOB is calculated as $S_{\\triangle AOB}=S_{\\triangle AOD}-S_{\\triangle BOD}=\\frac{1}{2}\\times4\\times6-\\frac{1}{2}\\times4\\times2=\\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53043725_112.png"
  },
  {
    "solution": "\\boxed{}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, $AB=5$, and $BC=3$. Point P starts from point A and moves along $AC-CB-BA$ at a speed of 2 units per second until it stops at point A. When point P does not coincide with the vertices of $\\triangle ABC$, a perpendicular line is drawn from point P to the side it lies on, intersecting another side of $\\triangle ABC$ at point Q. Let the movement time of point P be t seconds. What is the length of side $AC$?\n",
    "completion": "\\textbf{Solution:} Due to the insufficient information provided by the problem, it is impossible to directly determine the length of side $AC$. Therefore, specific solution steps and results cannot be provided.",
    "image_path": "dataset/mm_math/images/MM_Math/51318739_99.png"
  },
  {
    "solution": "\\boxed{\\frac{8}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, $AB\\parallel CD$, $AB=BC=2CD$, $E$ is the midpoint of diagonal $AC$, and $F$ is the midpoint of side $BC$, with segments $DE$ and $EF$ connected. Quadrilateral $CDEF$ is a rhombus; segment $DF$ intersects $AC$ at point $G$. Given that $DF=2$ and $CD=\\frac{5}{3}$, find the area of quadrilateral $CDEF$.\n",
    "completion": "\\textbf{Solution:} As shown in the figure, $DF$ intersects $EC$ at point $G$\\\\\nSince quadrilateral $CDEF$ is a rhombus and $DF=2$,\\\\\nit follows that $DG=1$, $DF\\perp CE$, and $EG=GC$,\\\\\nSince $CD=\\frac{5}{3}$,\\\\\nwe have $CG=\\sqrt{CD^{2}-DG^{2}}=\\sqrt{\\left(\\frac{5}{3}\\right)^{2}-1^{2}}=\\frac{4}{3}$\\\\\nTherefore, $CE=2CG=\\frac{8}{3}$\\\\\nHence, the area of quadrilateral $CDEF$ is $S_{\\text{quadrilateral}CDEF}=\\frac{1}{2}DF\\cdot CE=\\frac{1}{2}\\times \\frac{8}{3}\\times 2=\\boxed{\\frac{8}{3}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51863903_105.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C = 90^\\circ$, $AB = 4$, and $AC = \\sqrt{7}$. What is the length of $BC$?\n",
    "completion": "\\textbf{Solution:} In $\\triangle ABC$, $\\angle C=90^\\circ$, $AB=4$, $AC=\\sqrt{7}$.\\\\\n$BC=\\sqrt{AB^2-AC^2}=\\sqrt{4^2-(\\sqrt{7})^2}=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51103676_83.png"
  },
  {
    "solution": "\\boxed{41}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $M$ is the midpoint of $BC$, $AN$ bisects $\\angle BAC$, $BN\\perp AN$ at point $N$, and $MN$ is connected. If $AB=10$, $BC=15$, and $MN=3$, what is the perimeter of $\\triangle ABC$?\n",
    "completion": "\\textbf{Solution:} Extend $BN$ to meet $AC$ at point $D$,\\\\\n$\\because AN$ bisects $\\angle BAC$, that is $\\angle BAN = \\angle DAN$,\\\\\n$BN \\perp AN$, that is $\\angle BNA = \\angle DNA = 90^\\circ$, and $AN$ is the common side of $\\triangle ABN$ and $\\triangle ADN$\\\\\n$\\therefore \\triangle ABN \\cong \\triangle ADN$, $AD = AB = 10$\\\\\n$BN = DN$, that is, $N$ is the midpoint of $BD$\\\\\nAlso, $M$ is the midpoint of $BC$, $\\therefore CD = 2MN = 6$\\\\\n$\\therefore$ The perimeter is: $AB + BC + CD + AD = 10 + 15 + 6 + 10 = \\boxed{41}$",
    "image_path": "dataset/mm_math/images/MM_Math/52284901.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the cross-section of a drainpipe is a circle with a radius of $5$. If the width of the water surface $AB = 8$, then what is the depth of the water $CD$?\n",
    "completion": "\\textbf{Solution:} Connect OB. \\\\\nAccording to the problem, OD $\\perp$ AB, intersecting AB at point C, \\\\\nsince AB = 8, \\\\\nthus BC = $\\frac{1}{2}$ AB = $\\frac{1}{2} \\times 8$ = 4, \\\\\nIn $\\triangle OBC$, \\\\\nsince OB = 5, BC = 4, \\\\\nthus OC = $\\sqrt{OB^{2} - BC^{2}} = \\sqrt{5^{2} - 4^{2}}$ = 3, \\\\\ntherefore CD = OD - OC = 5 - 3 = \\boxed{2}.",
    "image_path": "dataset/mm_math/images/MM_Math/52861883.png"
  },
  {
    "solution": "\\boxed{(-3, -2)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the intersection point of the lines $y=-x+a$ and $y=x+b$ has an abscissa of $-2$, and the abscissas of the points where the two lines intersect the $x$-axis are $-1$ and $-3$, respectively. What is the solution set of the inequality $-x+a > x+b > 0$ with respect to $x$?\n",
    "completion": "\\textbf{Solution:} Let us assume that the intersection point of the line $y = -x + a$ and the line $y = x + b$ has an abscissa of $-2$. The abscissas where the two lines intersect the x-axis are $-1$ and $-3$, respectively.\\\\\nTherefore, the solution set of the inequality in $x$, $-x + a > x + b > 0$, corresponds to the range of $x$ values for which the line $y = -x + a$ is above the line $y = x + b$, i.e., $-3 < x < -2$.\\\\\nHence, the solution set is $x \\in \\boxed{(-3, -2)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50482062.png"
  },
  {
    "solution": "\\boxed{40^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, the perpendicular bisector DM of side AC intersects AC at D, and the perpendicular bisector EN of side CB intersects BC at E. DM and EN intersect at point F. If $\\angle MFN=70^\\circ$, what is the measure of $\\angle MCN$?\n",
    "completion": "\\textbf{Solution:} Given that MD$\\perp$AC and NE$\\perp$BC, \\\\\nit follows that $\\angle ACB=180^\\circ-\\angle MFN=110^\\circ$, \\\\\nwhich leads to $\\angle A+\\angle B=70^\\circ$, \\\\\nGiven MA=MC and NB=NC, \\\\\nit follows that $\\angle MCA=\\angle A$ and $\\angle NCB=\\angle B$, \\\\\nHence, $\\angle MCN=\\angle ACB -\\angle MCA-\\angle NCB =\\angle ACB -(\\angle A+\\angle B)=\\boxed{40^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51367169_130.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the straight lines $a$ and $b$ intersect perpendicularly at point $O$. The curve $C$ is centrally symmetric with respect to point $O$. The symmetric point of point $A$ is $A'$, $AB \\perp a$ at point $B$, and $A'D \\perp b$ at point $D$. If $OB=4$ and $OD=3$, what is the total area of the shaded regions?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, \\\\\nsince line a and b are perpendicular and intersect at point O, and curve C is centrally symmetric about point O, with point A's symmetric point being point A', AB $\\perp$ a at point B, and A'D $\\perp$ b at point D, OB = 4, OD = 3, \\\\\ntherefore, AB = 3, \\\\\nthus, the areas of shape (1) and shape (2) are equal, \\\\\nhence, the sum of the areas of the shaded parts equals the area of rectangle ABOE, which is $4 \\times 3 = \\boxed{12}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51212138.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Consider point $P$ on the parabola $y=-x^2+x+2$ in the first quadrant. Perpendicular lines are drawn from point $P$ to the $x$-axis and the $y$-axis, with the feet of the perpendiculars being points $A$ and $B$, respectively. What is the maximum perimeter of the quadrilateral $OAPB$?\n",
    "completion": "\\textbf{Solution:} Solve when $y=0$, we have $-x^{2}+x+2=0$\\\\\nSolving this, we get $x_{1}=-1$ and $x_{2}=2$,\\\\\nTherefore, when $0<x<2$ then $y>0$\\\\\nLet the point $P(x, -x^{2}+x+2)$,\\\\\nTherefore, the perimeter of the rectangle $OAPB$ is $2(x-x^{2}+x+2)=-2(x-1)^{2}+6$,\\\\\nSince the parabola opens downwards,\\\\\nTherefore, when $x=1$ the maximum perimeter of the rectangle $OAPB$ is $\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52694176.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points A, D, C, E are on the same straight line, $AB \\parallel EF$, $AB=EF$, $\\angle B = \\angle F$, $AE=10$, $AC=6$, then what is the length of CD?\n",
    "completion": "\\textbf{Solution:} Since $AB\\parallel EF$,\\\\\n$\\therefore \\angle A = \\angle E$,\\\\\nIn $\\triangle ABC$ and $\\triangle EFD$,\\\\\n$\\angle A = \\angle E$, $AB = EF$, $\\angle B = \\angle F$,\\\\\n$\\therefore \\triangle ABC \\cong \\triangle EFD$ (ASA),\\\\\n$\\therefore AC = ED = 6$,\\\\\n$\\therefore AD = AE - ED = 10 - 6 = 4$,\\\\\n$\\therefore CD = AC - AD = 6 - 4 = \\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53080193.png"
  },
  {
    "solution": "\\boxed{400\\text{cm}^{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, quadrilateral AEFH and quadrilateral ABCD are similar figures, with a similarity ratio of $\\frac{2}{3}$. If the area of quadrilateral ABCD is 900 cm$^2$, then what is the area of quadrilateral AEFH in cm$^2$?\n",
    "completion": "\\textbf{Solution:} Because quadrilateral AEFH and quadrilateral ABCD are similar figures with a similarity ratio of $\\frac{2}{3}$,\\\\\ntherefore, the ratio of the area of quadrilateral AEFH to the area of quadrilateral ABCD is $4\\colon 9$,\\\\\nfurthermore, since the area of quadrilateral ABCD is $900\\text{cm}^{2}$,\\\\\ntherefore, the area of quadrilateral AEFH is $\\boxed{400\\text{cm}^{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51638146.png"
  },
  {
    "solution": "\\boxed{120^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is given that $\\angle AOC = 2\\angle BOC$, $OD$ bisects $\\angle AOB$, and $\\angle COD = 20^\\circ$. Find the measure of $\\angle AOB$.\n",
    "completion": "\\textbf{Solution:} Given that $\\angle AOC = 2\\angle BOC$,\\\\\nand $\\angle AOC + \\angle BOC = \\angle AOB$,\\\\\nit follows that $\\angle AOB = 3\\angle BOC$,\\\\\nSince $OD$ bisects $\\angle AOB$,\\\\\nwe have $\\angle AOB = 2\\angle BOD$,\\\\\nthus, $\\angle BOC = \\frac{2}{3}\\angle BOD$,\\\\\nGiven $\\angle BOD = \\angle BOC + \\angle COD$,\\\\\nit follows that $\\angle COD = \\frac{1}{3}\\angle BOD$,\\\\\nGiven $\\angle COD = 20^\\circ$,\\\\\nit follows that $\\angle BOD = 60^\\circ$,\\\\\ntherefore, $\\angle AOB = \\boxed{120^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51457634.png"
  },
  {
    "solution": "\\boxed{14\\text{cm}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC=9\\text{cm}$, $DE$ is the perpendicular bisector of $AB$, intersecting $AB$ and $AC$ at points $D$ and $E$, respectively. If $BC=5\\text{cm}$, what is the perimeter of $\\triangle BCE$ in $\\text{cm}$?\n",
    "completion": "\\textbf{Solution:} Given that $DE$ is the perpendicular bisector of $AB$,\\\\\nit follows that $AE=BE$,\\\\\ngiven $AC=9\\text{cm}$, $BC=5\\text{cm}$, and $AE=BE$,\\\\\nthus, the perimeter of $\\triangle BCE$ is $BC+BE+CE=BC+AE+CE=BC+AC=9\\text{cm}+5\\text{cm}=\\boxed{14\\text{cm}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53033459.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{3}{4}(x-2)^{2}+1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the parabola $C_1$ is centrally symmetric about the origin with the parabola $C_2$. If the equation of the parabola $C_1$ is $y=\\frac{3}{4}(x+2)^2-1$, what are the coordinates of the vertex of the parabola $C_2$, and what is its equation?\n",
    "completion": "\\textbf{Solution:} Given that the equation of parabola $C_1$ is $y=\\frac{3}{4}(x+2)^{2}-1$,\\\\\n$\\therefore$ the parabola $C_1$ opens upwards, and its vertex is at: $\\left(-2,-1\\right)$ \\\\\n$\\because$ Parabola $C_1$ and parabola $C_2$ are centrally symmetric about the origin.\\\\\n$\\therefore$ Parabola $C_2$ opens downwards, and its vertex is at: $\\boxed{\\left(2,1\\right)}$ \\\\\n$\\therefore$ The equation of parabola $C_2$ is $\\boxed{y=-\\frac{3}{4}(x-2)^{2}+1}$",
    "image_path": "dataset/mm_math/images/MM_Math/53019808.png"
  },
  {
    "solution": "\\boxed{4.8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the diagonals of rhombus $ABCD$ intersect at point $O$, with $AC=12$ and $BD=16$. Point $P$ lies on side $BC$ and does not coincide with points $B$ or $C$. Through $P$, construct $PE \\perp AC$ at $E$ and $PF \\perp BD$ at $F$, and connect $EF$. What is the minimum value of $EF$?\n",
    "completion": "\\textbf{Solution:} Connect $OP$,\\\\\n\\textit{Since} quadrilateral $ABCD$ is a rhombus, $AC=12$, $BD=16$\\\\\n\\textit{Therefore} $AC\\perp BD$,\\\\\n$BO=\\frac{1}{2}BD=8$, $OC=\\frac{1}{2}AC=6$\\\\\n\\textit{Therefore} $BC=\\sqrt{OB^2+OC^2}=\\sqrt{64+36}=10$\\\\\n\\textit{Since} $PE\\perp AC$, $PF\\perp BD$, $AC\\perp BD$\\\\\n\\textit{Therefore} quadrilateral $OEPF$ is a rectangle,\\\\\n\\textit{Therefore} $FE=OP$\\\\\nWhen $OP\\perp BC$, $OP$ is at its minimum,\\\\\nThen $S_{\\triangle OBC}=\\frac{1}{2}OB\\cdot OC=\\frac{1}{2}BC\\cdot OP$\\\\\n\\textit{Therefore} $OP=\\frac{6\\times 8}{10}=4.8$\\\\\n\\textit{Therefore} the minimum value of $EF$ is $\\boxed{4.8}$",
    "image_path": "dataset/mm_math/images/MM_Math/52386935.png"
  },
  {
    "solution": "\\boxed{20^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the point $O$ is a point on the line $AB$, $OD$ bisects $\\angle BOC$, $\\angle COE=90^\\circ$, if $\\angle AOC=40^\\circ$, what is the measure of $\\angle DOE$ in degrees?\n",
    "completion": "\\textbf{Solution:} Given that $O$ is a point on the line $AB$ and $\\angle AOC=40^\\circ$, \\\\\nit follows that $\\angle BOC=180^\\circ-\\angle AOC=140^\\circ$.\\\\\nSince $OD$ bisects $\\angle BOC$, \\\\\nwe have $\\angle COD=\\frac{1}{2}\\angle BOC=70^\\circ$.\\\\\nGiven $\\angle COE=90^\\circ$, \\\\\nit follows that $\\angle DOE=\\angle COE-\\angle COD=\\boxed{20^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53119262.png"
  },
  {
    "solution": "\\boxed{85^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $CE$ is the bisector of the exterior angle $\\angle ACD$ of $\\triangle ABC$. If $\\angle B = 45^\\circ$ and $\\angle ACE = 65^\\circ$, then what is the measure of $\\angle A$?\n",
    "completion": "\\textbf{Solution:} Since $CE$ is the bisector of $\\angle ACD$ and $\\angle ACE=65^\\circ$,\\\\\nit follows that $\\angle ACD=130^\\circ$.\\\\\nSince $\\angle ACD$ is an exterior angle of $\\triangle ABC$,\\\\\nwe have $\\angle ACD=\\angle A + \\angle B$.\\\\\nGiven that $\\angle B=45^\\circ$,\\\\\nit follows that $\\angle A=\\boxed{85^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52826806.png"
  },
  {
    "solution": "\\boxed{\\left(\\frac{7}{3},-\\frac{20}{9}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a parabola intersects the $x$-axis at points $A$ and $B(3, 0)$, and intersects the $y$-axis at point $C(0, -3)$. The vertex of the parabola is $D$. Consider a point $P$ on the parabola below the line $BC$, a perpendicular line $PQ$ is drawn to $BC$ at point $Q$, and line $CP$ is connected. When one of the angles in $\\triangle CPQ$ equals $\\angle ACO$, find the coordinates of point $P$.\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, let there be a point P on the parabola below the line BC. Draw PQ perpendicular to BC at point Q, connect CP, and draw PH $\\perp$ y-axis at point H, let P $(x,x^{2}-2x-3)$, \\\\\n$\\therefore$PH $=x$, CH $=|x^{2}-2x-3-(-3)|=|x^{2}-2x|$, \\\\\n$\\therefore \\angle PQC = \\angle CHP = 90^\\circ$, \\\\\n$\\therefore$ Points P, Q, H, C are concyclic, \\\\\n$\\therefore \\angle PCQ = \\angle CPH$, \\\\\nWhen $\\angle PCQ = \\angle ACO$, \\\\\n$\\therefore \\angle HPC = \\angle ACO$, \\\\\n$\\therefore \\tan \\angle HPC = \\tan \\angle ACO$, \\\\\n$\\therefore \\frac{HC}{PH} = \\frac{AO}{CO} = \\frac{1}{3}$, \\\\\n$\\therefore$PH $= 3$HC, \\\\\n$\\therefore x = 3|x^{2}-2x|$, \\\\\nSolving, we get $x_{1}=\\frac{5}{3}, x_{2}=0$ (discard), or $x_{3}=\\frac{7}{3}, x_{4}=0$ (discard), \\\\\n$\\therefore$ The coordinates of point P are $\\boxed{\\left(\\frac{5}{3},-\\frac{32}{9}\\right)}$ or $\\boxed{\\left(\\frac{7}{3},-\\frac{20}{9}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52733221_73.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, rectangle $ABCD$ is folded along the diagonal $AC$, making point $D$ fall on point $M$, with $AM$ intersecting $BC$ at point $E$, and $AB=4$, $AD=8$. What is the length of $BE$?\n",
    "completion": "\\textbf{Solution:} Given the properties of the rectangle and the fold, we have: $AB=CD=MC=4$, $AD=BC=AM=8$, $\\angle AMC=90^\\circ$,\\\\\nIn $\\triangle ABE$ and $\\triangle CME$,\\\\\n$\\left\\{\\begin{array}{l}\n\\angle ABC=\\angle CME\\\\\n\\angle AEB=\\angle CEM\\\\\nAB=CM\n\\end{array}\\right.$\\\\\n$\\therefore \\triangle ABE\\cong \\triangle CME$ (AAS),\\\\\n$\\therefore BE=ME$,\\\\\nLet $BE=x$, then $ME=x$, $AE=8-x$,\\\\\nIn $\\triangle ABE$, by the Pythagorean theorem: $AB^2+BE^2=AE^2$, i.e., $4^2+x^2=(8-x)^2$,\\\\\nSolving this, we get: $x=\\boxed{3}$, hence $BE=3$.",
    "image_path": "dataset/mm_math/images/MM_Math/52699614.png"
  },
  {
    "solution": "\\boxed{120^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $O$ is a point on the line segment $AB$, with $ \\angle BOC=40^\\circ$, and $OD$ bisects $ \\angle AOC$. Construct ray $OE$ such that $ \\angle BOE=\\frac{2}{3}\\angle COE$. Find the degree measure of $ \\angle COE$.\n",
    "completion": "\\textbf{Solution:}\n\n(1) As shown in Figure 1, when ray OE is above AB, $\\angle BOE = \\frac{2}{3} \\angle COE$,\n\n$\\because \\angle BOE + \\angle COE = \\angle BOC$,\n\n$\\therefore \\frac{2}{3} \\angle COE + \\angle COE = 40^\\circ$,\n\n$\\therefore \\angle COE = \\boxed{24^\\circ}$;\n\n(2) As shown in Figure 2, when ray OE is below AB, $\\angle BOE = \\frac{2}{3} \\angle COE$,\n\n$\\because \\angle COE - \\angle BOE = \\angle BOC$,\n\n$\\therefore \\angle COE - \\frac{2}{3} \\angle COE = 40^\\circ$,\n\n$\\therefore \\angle COE = \\boxed{120^\\circ}$;\n\nIn conclusion: The degree of $\\angle COE$ can be either \\boxed{24^\\circ} or \\boxed{120^\\circ}.",
    "image_path": "dataset/mm_math/images/MM_Math/51494981_149.png"
  },
  {
    "solution": "\\boxed{\\frac{24}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rhombus $ABCD$, the diagonals $AC$ and $BD$ intersect at point $O$. $BE \\parallel AC$ and $AE \\parallel BD$, where $OE$ intersects $AB$ at point $F$. If $OE = 5$ and $AC = 8$, what is the height of rhombus $ABCD$?\n",
    "completion": "\\textbf{Solution:} Given that $BE\\parallel AC$ and $AE\\parallel BD$,\\\\\nit follows that quadrilateral AEBO is a parallelogram,\\\\\nand since the diagonals of rhombus ABCD intersect at point O,\\\\\nthen $OA=\\frac{1}{2}AC=4$, $OB=OD$, and $AC\\perp BD$,\\\\\nwhich means $\\angle AOB=90^\\circ$.\\\\\nThus, parallelogram AOBE is a rectangle,\\\\\nand so $AB=OE=5$,\\\\\nwhich leads to $OB=\\sqrt{AB^{2}-OA^{2}}=\\sqrt{5^{2}-4^{2}}=3$,\\\\\nand therefore $BD=2OB=6$.\\\\\nLet the height of rhombus ABCD be $h$,\\\\\nthen the area $S_{\\text{rhombus }ABCD}=\\frac{1}{2}AC\\cdot BD=AB\\cdot h$,\\\\\nyielding $h=\\frac{\\frac{1}{2}\\times 8\\times 6}{5}=\\boxed{\\frac{24}{5}}$,\\\\\nmeaning the height of rhombus ABCD is $\\frac{24}{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52345212.png"
  },
  {
    "solution": "\\boxed{25^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in circle $O$, $OA \\perp BC$, $\\angle AOC = 50^\\circ$, then what is the degree measure of $\\angle ADB$?\n",
    "completion": "\\textbf{Solution:} Connect OB\\\\\nSince $OA\\perp BC$, and $\\angle AOC=50^\\circ$,\\\\\nthen $\\angle AOB=50^\\circ$,\\\\\nthus $\\angle ADB=\\frac{1}{2}\\angle AOB=\\boxed{25^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/52720185.png"
  },
  {
    "solution": "\\boxed{23\\text{m}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the basketball player Xiao Ming throws the ball from point O. The ball is thrown from point A, which is 1 meter above the ground. The ball reaches its highest point, B, 6 meters away from point O, with the highest point C being 4 meters above the ground. It bounces again and lands at point E, with the distance between E and F being 2 meters. According to experiments, the parabola described by the basketball on its second bounce on the court has the same shape as the first, but the maximum height is reduced to half of the original maximum height. Taking the standing position of Xiao Ming at point O as the origin, a Cartesian coordinate system is established as shown in the figure. (All calculated results are to be rounded to the nearest integer, with $ \\sqrt{3}\\approx1.75$ and $ \\sqrt{6}\\approx2.5$). What is the distance from point E, where the basketball lands for the second time, to point O?\n",
    "completion": "\\textbf{Solution:} Let $y=0$, $-\\frac{1}{12}(x-6)^2+4=0$,\\\\\n$\\therefore (x-6)^2=48$,\\\\\nWe find: $x_1=4\\sqrt{3}+6\\approx 13$, $x_2=-4\\sqrt{3}+6$,\\\\\n$\\therefore$ the coordinates of point D are $(13,0)$.\\\\\n$\\therefore OD=13\\text{m}$\\\\\nAs shown in the diagram, draw $MG\\parallel x$ axis from the vertex M of the parabola DE, intersecting the parabola ACD at points G and H.\\\\\n$\\because$ The trajectory of the basketball after the second bounce on the court forms a parabola identical in shape to the first, but its maximum height is reduced to half of the first\\\\\n$\\therefore$ The parabola DME is equivalent to the parabola ACD moved two units downward and then moved to the right.\\\\\n$\\therefore GH=DE$, the $y$-coordinates of points G, H, and M are 2.\\\\\nWhen $y=2$, $-\\frac{1}{12}(x-6)^2+4=2$\\\\\n$x_1=6-2\\sqrt{6}$, $x_2=6+2\\sqrt{6}$\\\\\n$\\therefore DE=GH=x_2-x_1=4\\sqrt{6}\\approx 10$\\\\\n$OE=OD+DE$\\\\\n$=13+10$\\\\\n$=\\boxed{23\\text{m}}$\\\\\n$\\therefore$ The distance from point E, the basketball's second landing point, to point O is $23\\text{m}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51403162_47.png"
  },
  {
    "solution": "\\boxed{125^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the central angle of sector OAB is $110^\\circ$, and C is a point on $\\overset{\\frown}{AB}$. What is the degree measure of $\\angle C$?\n",
    "completion": "\\textbf{Solution:} Construct $\\overset{\\frown}{AB}$'s corresponding central angle $\\angle ADB$, as shown in the diagram,\\\\\n$\\therefore \\angle ADB = \\frac{1}{2}\\angle AOB = \\frac{1}{2} \\times 110^\\circ = 55^\\circ$,\\\\\n$\\because \\angle ADB + \\angle C = 180^\\circ$,\\\\\n$\\therefore \\angle C = 180^\\circ - 55^\\circ = \\boxed{125^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53148594.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of $\\odot O$, and chord $CD \\perp AB$ at $P$. If $CD = AP = 8$, what is the radius of $\\odot O$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, draw line $OC$.\\\\\nSince $AP=8$,\\\\\nit follows that $AP=OA+OP=OC+OP=8$.\\\\\nLet $OP=x$, then $OC=8-x$.\\\\\nSince $AB$ is the diameter of the circle $\\odot O$, and chord $CD$ is perpendicular to $AB$ at point $P$,\\\\\nit follows that $CP=\\frac{1}{2}CD=4$.\\\\\nIn $\\triangle OCP$, $\\angle OPC=90^\\circ$,\\\\\nhence $OP^2+CP^2=OC^2$.\\\\\nTherefore, $(8-x)^2=x^2+4^2$.\\\\\nTherefore, $x=3$.\\\\\nTherefore, $OC=8-x=5$.\\\\\nTherefore, the radius of the circle $\\odot O$ is $\\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52861797.png"
  },
  {
    "solution": "\\boxed{2.5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, the diagonals $AC$ and $BD$ of the rhombus $ABCD$ intersect at point $O$, with $BD = 6$ and $AC = 8$. If $E$ is the midpoint of $CD$, then what is the length of $OE$?\n",
    "completion": "Solution: Since quadrilateral ABCD is a rhombus,\\\\\nit follows that OC=OA=$\\frac{1}{2}AC=4$, OB=OD=$\\frac{1}{2}BD=3$, and $\\angle$DOC=$90^\\circ$,\\\\\nthus CD=$\\sqrt{OC^2+OA^2}=5$,\\\\\ngiven that E is the midpoint of AB, and O is the midpoint of AC,\\\\\nthus OE is the median to the hypotenuse of $\\triangle COD$,\\\\\ntherefore OE=$\\frac{1}{2}CD=\\boxed{2.5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52423571.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rhombus $ABCD$, $DE \\perp AB$, $\\cos A = \\frac{3}{5}$, and $AE = 6$, then what is the length of $AB$?\n",
    "completion": "\\textbf{Solution:} Since $DE\\perp AB$,\\\\\nit follows that $\\triangle ADE$ is a right-angled triangle,\\\\\nhence $DA=\\frac{AE}{\\cos A}=6\\times \\frac{5}{3}=10$,\\\\\nsince quadrilateral ABCD is a rhombus,\\\\\nit follows that AB=AD=10,\\\\\nthus, the answer is: AB=\\boxed{10}.",
    "image_path": "dataset/mm_math/images/MM_Math/51457944.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $D$ and $E$ are points on $AC$ and $BC$, respectively. If $\\triangle ADB \\cong \\triangle EDB \\cong \\triangle EDC$, what is the measure of $\\angle C$?\n",
    "completion": "\\textbf{Solution:} Given that $\\triangle ADB \\cong \\triangle EDB \\cong \\triangle EDC$,\\\\\nit follows that $\\angle A = \\angle BED = \\angle CED$, and $\\angle ABD = \\angle EBD = \\angle C$,\\\\\nsince $\\angle BED + \\angle CED = 180^\\circ$,\\\\\nit implies that $\\angle A = \\angle BED = \\angle CED = 90^\\circ$,\\\\\nthus $\\angle ABD + \\angle EBD + \\angle C = 90^\\circ$,\\\\\nhence $\\angle C = \\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52661291.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given that in $\\triangle ABC$, $\\angle C = 90^\\circ$, $AB=4$, and $AC=\\sqrt{7}$. Find $BC$?\n",
    "completion": "\\textbf{Solution:} In triangle $\\triangle ABC$, $\\angle C=90^\\circ$, $AB=4$, $AC=\\sqrt{7}$.\\\\\n$BC=\\sqrt{AB^2-AC^2}=\\sqrt{4^2-(\\sqrt{7})^2}=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51403070_126.png"
  },
  {
    "solution": "\\boxed{x>-1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y_1=3x+6$ intersects the x-axis and y-axis at points A and B respectively. The line $y_2=kx+b$ passes through point D (3, 0) and intersects with the line $y_1=3x+6$ at point C (m, 3). Based on the diagram, directly write the solution set for the inequality $3x+6>kx+b$ in terms of x.\n",
    "completion": "\\textbf{Solution:} From equation (1), we have point C at $(-1,3)$,\\\\\n$3x+6>kx+b$ implies $y_1>y_2$,\\\\\nmeaning the graph of $y_1=3x+6$ is above that of $y_2=-\\frac{3}{4}x+\\frac{9}{4}$,\\\\\nBased on the graph above, the solution set is $\\boxed{x>-1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52159946_164.png"
  },
  {
    "solution": "\\boxed{2:5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in quadrilateral $ABCD$, $AE:EB=2:3$, and $DE$ intersects $AC$ at $F$. Calculate the ratio of the perimeters of $\\triangle AEF$ and $\\triangle CDF$.\n",
    "completion": "\\textbf{Solution:} Given that in quadrilateral $ABCD$, \\\\\n$\\therefore AB=CD, AB\\parallel CD$, \\\\\n$\\therefore \\triangle AEF\\sim \\triangle CDF$, \\\\\n$\\therefore$ The ratio of the perimeters of $\\triangle AEF$ and $\\triangle CDF$ is equal to the similarity ratio $AE:CD$, \\\\\n$\\because$ $AE:EB=2:3$, $AB=CD$, \\\\\n$\\therefore AE:AB=AE:CD=2:5$, \\\\\n$\\therefore$ The ratio of the perimeters of $\\triangle AEF$ and $\\triangle CDF$ is $ \\boxed{2:5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52917377_69.png"
  },
  {
    "solution": "\\boxed{15cm}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the side length of square $ABCD$ is $6\\,cm$, and $E$ is a point on side $AB$, with the length of $AE$ being $1\\,cm$. A moving point $P$ starts from point $B$ and moves in the direction of ray $BC$ at a speed of $1\\,cm$ per second. Triangle $EBP$ is folded along $EP$, and point $B$ falls at point ${B}^{\\prime}$. Let the moving time be $t$ seconds. If the distance from point ${B}^{\\prime}$ to line $AD$ is $3\\,cm$, what is the length of $BP$ in $cm$?\n",
    "completion": "\\textbf{Solution:} We have the length of $BP$ as $\\boxed{\\frac{5}{7}\\sqrt{21}}$ or $\\boxed{15cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53359333_49.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, the perimeter of $\\triangle ABC$ is 19. Points $D$ and $E$ are on side $BC$. The bisector of $\\angle ABC$ is perpendicular to $AE$ at point $N$. The bisector of $\\angle ACB$ is perpendicular to $AD$ at point $M$. If $BC = 7$, what is the length of $DE$?\n",
    "completion": "\\textbf{Solution:} Given that the angle bisector of $\\angle ABC$ is perpendicular to $AE$, with the foot of the perpendicular being point $N$, \\\\\nit follows that $\\angle ABN=\\angle EBN$, and $\\angle BNA=\\angle BNE=90^\\circ$, \\\\\nIn $\\triangle BNA$ and $\\triangle BNE$, \\\\\nsince $\\left\\{\\begin{array}{l}\\angle ABN=\\angle EBN \\\\ BN=BN \\\\ \\angle BNA=\\angle BNE\\end{array}\\right.$, \\\\\nit implies $\\triangle BNA\\cong \\triangle BNE\\left(ASA\\right)$, \\\\\nthus $BA=BE$, \\\\\nSimilarly, it can be proved that $\\triangle CAM\\cong \\triangle CDM\\left(ASA\\right)$, \\\\\nhence, $AC=DC$.\\\\\nGiven that the perimeter of $\\triangle ABC$ is 19, \\\\\nit follows that $AB+AC+BC=19$, \\\\\nSince $BA=BE$ and $AC=DC$, \\\\\nit follows that $BE+DC+BC=19$, \\\\\nthus $BD+DE+DE+EC+BC=19$, \\\\\nGiven $BC=7$, \\\\\nit follows that $BD+DE+DE+EC=19-BC=12$, \\\\\nthus $BD+DE+EC+DE=BC+DE=12$, \\\\\nwhich means $DE=12-BC=12-7=\\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53033284.png"
  },
  {
    "solution": "\\boxed{P\\left(-\\frac{5}{3}, \\frac{32}{9}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the Cartesian coordinate system, point O is the origin. The parabola $y=a{x}^{2}+bx+3$ passes through point $B(-3, 0)$ and point $C(1, 0)$, and intersects the y-axis at point A. As shown in the figure, under the given conditions, connect $CD$, draw $AE\\perp CD$ at point F where $AE=CD$, draw $EG\\perp AB$ at point G, draw $GH\\parallel DP$ intersecting the x-axis at point H, and connect $FH$ and $FO$. If $\\angle HFO=90^\\circ$, find the coordinates of point P.\n",
    "completion": "\\textbf{Solution:}\n\n(1) Construct $EL \\perp OA$ at L, extend $PD$ to intersect the x-axis at T.\\\\\n$\\therefore \\angle AFC = \\angle AOC = 90^\\circ$\\\\\n$\\because \\angle 1 = \\angle 2$\\\\\n$\\therefore \\angle EAL = \\angle DCT$\\\\\n$\\because AE = CD$\\\\\n$\\therefore \\triangle AEL \\cong \\triangle CDT$\\\\\n$\\therefore EL = DT$, $AL = CT$\\\\\n$\\because D(t, t+3)$\\\\\n$\\therefore EL = t+3$, $AL = CT = 1-t$\\\\\n$\\because OL = OA - AL = 3 - (1 - t) = 2 + t$\\\\\n$\\therefore E(-t-3, 2+t)$\\\\\n(2) Construct $EW \\parallel y$-axis, intersecting $AB$ at W, and $OB$ at V, extend $GE$ to intersect the x-axis at R.\\\\\nThus, $W(-t-3, -t)$, $V(-t-3, 0)$, $WE = -2t-2$, $WV = -t$\\\\\nFrom the coordinates of points A and B, we get: $OA = OB$\\\\\n$\\therefore \\angle ABO = 45^\\circ$,\\\\\n$\\therefore \\angle BWV = \\angle WEG = \\angle GRB = \\angle ABO = 45^\\circ$\\\\\n$\\therefore WG = GE$, $WV = BV$, $GR = BG$\\\\\n$\\therefore$ By Pythagorean theorem, $GW = \\frac{\\sqrt{2}}{2}WE = -\\sqrt{2}(t+1)$, $BW = \\sqrt{2}WV = -\\sqrt{2}t$,\\\\\n$\\therefore BG = BW - GW = \\sqrt{2}$\\\\\nBy Pythagorean theorem, $BR = \\sqrt{2}BG = 2$\\\\\nAlso, point H is the midpoint of $BR$, $OB = 3$\\\\\n$\\therefore G(-2, 1)$, $R(-1, 0)$, $H(-2, 0)$\\\\\n(3) Extend $GE$ to intersect the x-axis at R, connect $DH$ and $OE$.\\\\\n$\\because D(t, t+3), G(-2, 1), E(-t-3, 2+t), R(-1, 0)$\\\\\n$\\therefore DG = ER$, $OR = GH = 1$, $\\angle DGH = \\angle FRO = 135^\\circ$\\\\\n$\\therefore \\triangle DGH \\cong \\triangle ERO$\\\\\n$\\therefore DH = EO$, $\\angle GDH = \\angle REO$\\\\\nConnect $AC$\\\\\n$\\because \\angle HFO = \\angle DFE = 90^\\circ$,\\\\\n$\\therefore \\angle DFH = \\angle EFO$\\\\\n$\\because EG \\perp AB$, $AF \\perp CD$, $\\angle DAF = \\angle EAG$\\\\\n$\\therefore \\angle ADF = \\angle AEG$,\\\\\n$\\therefore \\angle GDF = \\angle FER$\\\\\n$\\because \\angle GDH = \\angle REO$\\\\\n$\\therefore \\angle FEO = \\angle FDH$,\\\\\n$\\because DH = EO$\\\\\n$\\therefore \\triangle DHF \\cong \\triangle EOF$\\\\\n$\\therefore DF = EF$\\\\\n$\\because AE = CD$\\\\\n$\\therefore AF = CF$\\\\\n$\\therefore \\angle FAC = 45^\\circ$\\\\\nIn right triangle $AOC$, $\\tan\\angle OAC = \\frac{1}{3}$\\\\\n$\\because \\angle BAO = \\angle FAC = 45^\\circ$\\\\\n$\\therefore \\angle GAE = \\angle OAC$\\\\\n$\\therefore \\tan\\angle GAE = \\frac{EG}{AG} = \\frac{1}{3}$\\\\\nThat is, $\\frac{\\sqrt{2}}{2}(-2t-2) = \\frac{2}{3}\\sqrt{2}$\\\\\nSolving for $t$, we get: $t = -\\frac{5}{3}$\\\\\n$\\therefore \\boxed{P\\left(-\\frac{5}{3}, \\frac{32}{9}\\right)}$",
    "image_path": "dataset/mm_math/images/MM_Math/53019865_162.png"
  },
  {
    "solution": "\\boxed{D\\text{ is the midpoint of }BD}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, $CB=CA$, $CE\\perp AB$ at point $E$, point $F$ is on $CE$, connect $AF$ and extend it to intersect $BC$ at point $D$, $CG\\perp AD$ at point $G$, connect $EG$. $CD^2=DG\\cdot DA$; as shown in Figure 1, if $CF=2EF$, prove that point $D$ is the midpoint of $BC$.\n",
    "completion": "\\textbf{Solution:} As shown in Figure 1, draw EH$\\parallel$AD through point E and intersecting BC at point H,\\\\\nsince HE$\\parallel$AD,\\\\\ntherefore, BH:HD = BE:EA, and CD:HD = CF:EF,\\\\\nsince CB=CA, and $\\angle$ACB$=90^\\circ$, with CE$\\perp$AB,\\\\\ntherefore, E is the midpoint of AB,\\\\\ntherefore, BE:EA=1,\\\\\ntherefore, BH:HD=BE:EA=1,\\\\\nsince CF=2EF,\\\\\ntherefore, CD:HD=CF:EF=2,\\\\\ntherefore, BH=HD, and CD=2HD,\\\\\ntherefore, BD=BH+HD=2HD,\\\\\ntherefore, BD=CD,\\\\\ntherefore, \\boxed{D\\text{ is the midpoint of }BD}.",
    "image_path": "dataset/mm_math/images/MM_Math/51212645_94.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB = 90^\\circ$, $CD \\perp AB$ at D, $AF$ bisects $\\angle CAB$, intersects $CD$ at point E, and intersects $CB$ at point F. Given $AC=6$, $BC=8$. Find the length of $CF$ and the value of $\\tan\\angle BAF$.\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $FG\\perp AB$ at point G through point F.\\\\\n$\\because$AF bisects $\\angle CAB$,\\\\\n$\\therefore \\angle BAF=\\angle CAF$, $CF=FG$.\\\\\n$\\because$ ${S}_{\\triangle ABC}={S}_{\\triangle ACF}+{S}_{\\triangle ABF}=\\frac{1}{2}AC\\cdot CF+\\frac{1}{2}AB\\cdot FG$,\\\\\n$\\therefore \\frac{1}{2}AC\\cdot CF+\\frac{1}{2}AB\\cdot CF=\\frac{1}{2}AC\\cdot BC$, i.e., $6CF+10CF=6\\times 8$,\\\\\nSolving yields: $CF=\\boxed{3}$.\\\\\n$\\therefore \\tan\\angle BAF=\\tan\\angle CAF=\\frac{CF}{AC}=\\frac{3}{6}=\\boxed{\\frac{1}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52765791_142.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, it is known that points $D$, $E$, and $F$ are respectively the midpoints of $BC$, $AD$, and $EC$, and the area of $\\triangle ABC$, ${S}_{\\triangle ABC}=8$. What is the area $S$ of the shaded part?\n",
    "completion": "\\textbf{Solution:}\\\\\nSince point D is the midpoint of BC,\\\\\ntherefore, the area of $\\triangle ABD$ equals the area of $\\triangle ACD$ equals $\\frac{1}{2}$ the area of $\\triangle ABC$,\\\\\nSince point E is the midpoint of AD,\\\\\ntherefore, the area of $\\triangle BDE$ equals $\\frac{1}{2}$ the area of $\\triangle ABD$ equals $\\frac{1}{4}$ the area of $\\triangle ABC$, and the area of $\\triangle CDE$ equals $\\frac{1}{2}$ the area of $\\triangle ACD$ equals $\\frac{1}{4}$ the area of $\\triangle ABC$,\\\\\ntherefore, the area of $\\triangle BCE$ equals the area of $\\triangle BDE$ plus the area of $\\triangle CDE$, which equals $\\frac{1}{2}$ the area of $\\triangle ABC$,\\\\\nSince point F is the midpoint of EC,\\\\\ntherefore, the area of $\\triangle BEF$ equals $\\frac{1}{2}$ the area of $\\triangle BCE$ equals $\\frac{1}{2} \\times \\frac{1}{2} \\times 8 = \\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52905802.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{2}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the vertices of $\\triangle ABC$ are lattice points of a square grid. What is the value of $\\sin B$?\n",
    "completion": "\\textbf{Solution:} By the Pythagorean theorem, we have $AB=\\sqrt{3^2+3^2}=3\\sqrt{2}$,\\\\\ntherefore, $\\sin B=\\frac{3}{3\\sqrt{2}}=\\boxed{\\frac{\\sqrt{2}}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53043702.png"
  },
  {
    "solution": "\\boxed{-1\\leq k\\leq\\frac{1}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the Cartesian coordinate system, the coordinates of point $A$ are $(6, 4)$. The lines $AB$ and $AC$ are drawn from point $A$ perpendicular to the x-axis and y-axis at points $B$ and $C$ respectively. The graph of the linear function $y=kx+b$ passes through the point $P(2, 3)$. When $1\\le x\\le 5$, the function value $y$ satisfies $-1\\le y\\le 4$. Determine the range of values for $k$.\n",
    "completion": "\\textbf{Solution:} Given the problem statement, when $k>0$, $y$ increases as $x$ increases. When $x=1$, we have ${y}_{\\min}=k-2k+3=3-k$. When $x=5$, we have ${y}_{\\max}=5k-2k+3=3k+3$. From the given conditions, we have\n\\[\n\\left\\{\n\\begin{array}{l}\n3-k\\geq-1, \\\\\n3k+3\\leq4.\n\\end{array}\n\\right.\n\\]\nSolving this, we find $k\\leq\\frac{1}{3}$. Therefore, $0<k\\leq\\frac{1}{3}$.\n\nWhen $k<0$, $y$ decreases as $x$ increases. When $x=1$, we have ${y}_{\\max}=k-2k+3=3-k$. When $x=5$, we have ${y}_{\\min}=5k-2k+3=3k+3$. From the given conditions, we have\n\\[\n\\left\\{\n\\begin{array}{l}\n3k+3\\geq-1, \\\\\n3-k\\leq4.\n\\end{array}\n\\right.\n\\]\nSolving this, we find $k\\geq-1$. Therefore, $-1\\leq k<0$.\n\nTherefore, combining the two cases, the range of values for $k$ is: $-1\\leq k<0$, $0<k\\leq\\frac{1}{3}$. Hence, the solution is the range of values for $k$: \\boxed{-1\\leq k\\leq\\frac{1}{3}}.",
    "image_path": "dataset/mm_math/images/MM_Math/52342947_116.png"
  },
  {
    "solution": "\\boxed{9−4\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, there is a billboard AB on top of the building. From point D, which is 12 meters away from the building BC, the angle of elevation to the top A of the billboard is $37^\\circ$, and the angle of elevation to the bottom B of the billboard is $30^\\circ$. Calculate the height of the billboard AB. (Leave the result in square root form, reference data: $\\sin 37^\\circ \\approx 0.60$, $\\cos 37^\\circ \\approx 0.80$, $\\tan 37^\\circ \\approx 0.75$)\n",
    "completion": "\\textbf{Solution:} According to the problem, we know\\\\\n$\\angle ADC=37^\\circ$, $\\angle BDC=30^\\circ$, $CD=12$ meters, $AC\\perp DC$.\\\\\nIn $\\triangle DCA$, $\\tan\\angle ADC=\\frac{AC}{CD}$, thus\\\\\n$AC=CD\\cdot \\tan\\angle ADC=12\\times \\tan37^\\circ\\approx 12\\times 0.75=9$ meters.\\\\\nIn $\\triangle DBC$, $\\tan\\angle BDC=\\frac{BC}{CD}$, thus\\\\\n$BC=CD\\cdot \\tan\\angle BDC=12\\times \\tan30^\\circ=12\\times \\frac{\\sqrt{3}}{3}=4\\sqrt{3}$ meters.\\\\\nTherefore, $AB=AC−BC=9−4\\sqrt{3}$.\\\\\nAnswer: The height of the billboard $AB$ is $\\boxed{9−4\\sqrt{3}}$ meters.",
    "image_path": "dataset/mm_math/images/MM_Math/52478922.png"
  },
  {
    "solution": "\\boxed{}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, we have $AB=AC$. A circle $\\odot O$ with diameter $AC$ intersects $BC$ at point $D$. A line through point $D$ and perpendicular to $AB$ is drawn, with the foot of the perpendicular being point $E$. Extend $BA$ to intersect $\\odot O$ at point $F$. $DE$ is a tangent to $\\odot O$. Given $DE=2$ and $AF=3$, find the length of $AE$.\n",
    "completion": "Given that the provided information is only the number \"1\", there is not enough context or specific LaTeX code to perform any cleaning or modification. Please provide the complete LaTeX code or a more detailed description of the problem, so that I can offer accurate assistance.",
    "image_path": "dataset/mm_math/images/MM_Math/52814924_88.png"
  },
  {
    "solution": "\\boxed{80^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points $A, B, C$ lie on circle $\\odot O$, and $\\angle ACB = 40^\\circ$. What is the degree measure of arc $AB$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle AOB$ and $\\angle ACB$ both subtend $\\overset{\\frown}{AB}$,\\\\\nit follows that $\\angle AOB=2\\angle ACB=2\\times 40^\\circ=80^\\circ$,\\\\\nthus, the measure of arc $AB$ is $\\boxed{80^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53009646.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in rectangle $ABCD$, points $E$ and $F$ are located on sides $BC$ and $AD$ respectively. The rectangle is folded along $EF$ such that point $B$ falls exactly on point $H$ on side $AD$. Given that the area of rectangle $ABCD$ is $16\\sqrt{3}$ and $FH = 2HD$, what is the length of the crease $EF$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram,\\\\\nsince folding the rectangle along EF makes point B precisely overlap with point H on edge AD, \\\\\nthus, $AF=FG$, $\\angle EHG=\\angle B=\\angle A=\\angle G=90^\\circ$, $HG=AB$, $\\angle BEF=\\angle FEH$,\\\\\nsince $\\angle BEF+\\angle FEH+\\angle HEC=180^\\circ$, $\\angle HEC=60^\\circ$,\\\\\nthus, $\\angle BEF=\\angle FEH=60^\\circ$,\\\\\nsince $AD\\parallel BC$,\\\\\nthus, $\\angle FHE=\\angle HEC=60^\\circ$,\\\\\nthus, $\\triangle EFH$ is an equilateral triangle,\\\\\nthus, $\\angle FHE=60^\\circ$,\\\\\nthus, $\\angle GHF=30^\\circ$,\\\\\nthus, $FG=\\frac{1}{2}FH$,\\\\\nsince $FH=2HD$,\\\\\nthus, $DH=FG=AF$,\\\\\nlet $DH=FG=AF=x$,\\\\\nthus, $AB=HG=\\sqrt{3}x$, $AD=4x$,\\\\\nsince the area of rectangle ABCD is $16\\sqrt{3}$,\\\\\nthus, $\\sqrt{3}x \\cdot 4x = 16\\sqrt{3}$,\\\\\nthus, $x=2$ (negative value is discarded),\\\\\nthus, $AB=2\\sqrt{3}$,\\\\\nDraw $FM\\perp BC$ at $M$ through F,\\\\\nthen, $FM=AB=2\\sqrt{3}$,\\\\\nthus, $EF=\\frac{2\\sqrt{3}}{3}FM=\\boxed{4}$,\\\\",
    "image_path": "dataset/mm_math/images/MM_Math/53572616.png"
  },
  {
    "solution": "\\boxed{\\frac{4}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $I$ is the incenter of $\\triangle ABC$, $O$ is a point on side $AB$, and $\\odot O$ passes through point $B$ and is tangent to $AI$ at point $I$. If $\\tan\\angle BAC = \\frac{24}{7}$, then what is the value of $\\sin\\angle ACB$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, extend line OI to intersect AC at D and connect BI.\\\\\n$\\because$ AI is tangent to the circle with center O,\\\\\n$\\therefore$ AI $\\perp$ OD,\\\\\n$\\therefore$ $\\angle$AIO = $\\angle$AID = $90^\\circ$,\\\\\n$\\because$ I is the incenter of $\\triangle ABC$,\\\\\n$\\therefore$ $\\angle$OAI = $\\angle$DAI, $\\angle$ABI = $\\angle$CBI,\\\\\n$\\because$ AI = AI,\\\\\n$\\therefore$ $\\triangle AOI \\cong \\triangle ADI$ (ASA),\\\\\n$\\therefore$ AO = AD,\\\\\n$\\because$ OB = OI,\\\\\n$\\therefore$ $\\angle$OBI = $\\angle$OIB,\\\\\n$\\therefore$ $\\angle$OIB = $\\angle$CBI,\\\\\n$\\therefore$ OD $\\parallel$ BC,\\\\\n$\\therefore$ $\\angle$ADO = $\\angle$C,\\\\\nConstruct OE $\\perp$ AC at E,\\\\\n$\\because$ $\\tan\\angle$BAC = $\\frac{OE}{AE}$ = $\\frac{24}{7}$,\\\\\n$\\therefore$ assume OE = 24k, AE = 7k without loss of generality,\\\\\n$\\therefore$ OA = AD = 25k,\\\\\n$\\therefore$ DE = AD - AE = 18k,\\\\\n$\\therefore$ OD = $\\sqrt{OE^2 + DE^2}$ = 30k,\\\\\n$\\therefore$ $\\sin\\angle$ACB = $\\frac{OE}{OD}$ = $\\frac{24k}{30k}$ = $\\boxed{\\frac{4}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51322497.png"
  },
  {
    "solution": "\\boxed{C\\left(-\\sqrt{10}+1, 0\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the graph of the linear function $y=3x-3$ intersects the $x$-axis at point $A$, and intersects the graph of the direct proportion function $y=kx$ at point $B(a, 3)$. Taking point $A$ as the center and the length of $AB$ as the radius, an arc is drawn that intersects the $x$-axis at point $C$. Please directly write down the coordinates of point $C$.\n",
    "completion": "\\textbf{Solution:} From the problem statement, we have $y=3x-3$. When $y=0$, it follows that $3x-3=0$, which solves to: $x=1$. Therefore, $A(1, 0)$ and hence $\\left|AB\\right|=\\sqrt{(3-0)^2+(2-1)^2}=\\sqrt{10}$. Draw an arc with point $A$ as the center and the length of $AB$ as the radius, intersecting the x-axis at point $C$. Therefore, $C\\left(\\sqrt{10}+1, 0\\right)$ or $\\left(-\\sqrt{10}+1, 0\\right)$. Hence, the coordinates of point $C$ are $\\boxed{C\\left(\\sqrt{10}+1, 0\\right)}$ or $\\boxed{C\\left(-\\sqrt{10}+1, 0\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50555435_92.png"
  },
  {
    "solution": "\\boxed{\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point E is on the side BC of the square ABCD. If $CE=1$ and $DE=2$, what is the length of AD?\n",
    "completion": "\\textbf{Solution}: Since quadrilateral $ABCD$ is a square,\\\\\nit follows that $AD=CD$ and $\\angle C=90^\\circ$,\\\\\nsince $CD=\\sqrt{DE^{2}-CE^{2}}=\\sqrt{4-1}=\\sqrt{3}$,\\\\\ntherefore $AD=CD=\\boxed{\\sqrt{3}}$",
    "image_path": "dataset/mm_math/images/MM_Math/53706920.png"
  },
  {
    "solution": "\\boxed{\\frac{8\\sqrt{10}}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate plane, point $A(2, m)$ and point $B(6, 2)$ are both on the graph of an inverse proportion function. Find the area of $\\triangle AOB$ and the distance $AH$ from point $A$ to line $OB$.\n",
    "completion": "\\textbf{Solution:} Let the point of intersection between line $AB$ and the $x$-axis be $C$. Within the line equation $AB$ as $y=-x+8$, by setting $y=0$ we get $x=8$. Therefore, $C(8, 0)$. Thus, the area of $\\triangle AOB$ can be calculated as ${S}_{\\triangle AOB}={S}_{\\triangle AOC}-{S}_{\\triangle BOC}=\\frac{1}{2}\\times 8\\times 6-\\frac{1}{2}\\times 8\\times 2=\\boxed{16}$. Given $B(6, 2)$, it follows that $OB=\\sqrt{6^2+2^2}=2\\sqrt{10}$. Since ${S}_{\\triangle AOB}=\\frac{1}{2}OB\\cdot AH=16$, it concludes that $AH=\\frac{16\\times 2}{2\\sqrt{10}}=\\boxed{\\frac{8\\sqrt{10}}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50447041_224.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points $D$, $E$, and $F$ are respectively the midpoints of the sides of $\\triangle ABC$. If the perimeter of $\\triangle ABC$ is $12$, what is the perimeter of $\\triangle DEF$?\n",
    "completion": "\\textbf{Solution:} Given that D, E, F are the midpoints of the sides of $\\triangle ABC$ respectively,\\\\\nthus DE, DF, EF are the midsegments of $\\triangle ABC$,\\\\\ntherefore, $BC=2DF, AC=2DE, AB=2EF,$\\\\\nhence, the perimeter of $\\triangle ABC$ is $AB+BC+AC=2(DF+FE+DE)=12,$\\\\\nwhich implies $DF+FE+DE=12\\times\\frac{1}{2}=6.$\\\\\nTherefore, the perimeter of $\\triangle DEF$ is $\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52354642.png"
  },
  {
    "solution": "\\boxed{\\frac{4\\sqrt{3}}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y= -\\frac{\\sqrt{3}}{3}x+b$ intersects the $x$-axis at point $A$ and the $y$-axis at point $B$. Taking the line segment $AB$ as a side, an equilateral triangle $ABC$ is constructed inside the first quadrant. The graph of the hyperbolic function $y= \\frac{k}{x}$ (where $k\\neq 0$) just passes through the midpoint $D$ of side $BC$ and intersects side $AC$ at point $E$. If the area of $\\triangle ODE$ is $\\sqrt{3}$, what is the value of $k$?\n",
    "completion": "\\textbf{Solution:} Given the linear function $y= -\\frac{\\sqrt{3}}{3}x+b$ intersects with the x-axis and y-axis at points $A$ and $B$, respectively.\\\\\nTherefore, when $y=0$, it holds that $-\\frac{\\sqrt{3}}{3}x+b=0$,\\\\\nsolving this gives $x= \\sqrt{3}b$,\\\\\nthus, the coordinate of point $A$ is ($\\sqrt{3}b$, $0$), and $OA=\\sqrt{3}b$,\\\\\nin the linear function $y= -\\frac{\\sqrt{3}}{3}x+b$, when $x=0$, we have $y=b$,\\\\\nthus, point $B$ is at $(0, b)$, and $OB=b$,\\\\\nhence, $\\tan\\angle BAO= \\frac{OB}{OA}=\\frac{b}{\\sqrt{3}b}=\\frac{\\sqrt{3}}{3}$, and $AB= \\sqrt{(\\sqrt{3}b)^{2}+b^{2}}=2b$,\\\\\ntherefore, $\\angle BAO=30^\\circ$,\\\\\ngiven that $\\triangle ABC$ is an equilateral triangle,\\\\\nit follows that $\\angle CAB=60^\\circ$, and $AC=AB=2b$,\\\\\nthus, $\\angle CAB+\\angle BAO=\\angle CAO=90^\\circ$, which implies $AC\\perp$ the x-axis,\\\\\ntherefore, point $C$ is located at ($\\sqrt{3}b$, $2b$), and point $E$ at ($\\sqrt{3}b$, $\\frac{\\sqrt{3}k}{3b}$), with the area of $\\triangle OAE$ being $\\frac{1}{2}k$,\\\\\ngiven that point $D$ is the midpoint of $BC$,\\\\\nthus, the coordinates of point $D$ are ($\\frac{\\sqrt{3}}{2}b$, $\\frac{3}{2}b$),\\\\\ngiven that the inverse proportion function $y= \\frac{k}{x}$ (with $k\\neq 0$) exactly passes through the midpoint $D$ of $BC$,\\\\\nit follows that $\\frac{3}{2}b\\times \\frac{\\sqrt{3}}{2}b=k$ (1),\\\\\ndraw $DF\\perp$ x-axis at point $F$,\\\\\nthus, $OF=\\frac{\\sqrt{3}}{2}b$ and $DF=\\frac{3}{2}b$, with the area of $\\triangle ODF$ being $\\frac{1}{2}k$,\\\\\ngiven that the area of $\\triangle ODE$ equals the area of trapezoid $AEDF$ plus the area of $\\triangle ODF$ minus the area of $\\triangle AOE$, which is equal to the area of trapezoid $AEDF$,\\\\\nthus, $\\frac{1}{2}(DF+AE)\\times AF=\\sqrt{3}$, i.e., $\\frac{1}{2}\\left(\\frac{3}{2}b+\\frac{\\sqrt{3}k}{3b}\\right)\\times \\left(\\sqrt{3}b-\\frac{\\sqrt{3}}{2}b\\right)=\\sqrt{3}$,\\\\\nwhich implies $\\frac{1}{2}\\left(\\frac{3}{2}b+\\frac{\\sqrt{3}k}{3b}\\right)\\times \\frac{\\sqrt{3}}{2}b=\\sqrt{3}$,\\\\\nthus, $\\frac{1}{2}\\times \\frac{3}{2}b\\times \\frac{\\sqrt{3}}{2}b+\\frac{1}{2}\\times \\frac{\\sqrt{3}k}{3b}\\times \\frac{\\sqrt{3}}{2}b=\\sqrt{3}$,\\\\\ntherefore, $\\frac{1}{2}k+\\frac{1}{4}k=\\sqrt{3}$,\\\\\nhence, $k=\\boxed{\\frac{4\\sqrt{3}}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/54967449.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, with AB as the diameter, construct $\\odot O$. Take a point C on $\\odot O$, extend AB to point D, and connect DC, where $\\angle DCB=\\angle DAC$. Through point A, draw $AE\\perp AD$ to meet the extension of DC at point E. CD is the tangent to $\\odot O$; if $CD=4$ and $DB=2$, find the length of AE.\n",
    "completion": "\\textbf{Solution:} Since $\\angle DCO = 90^\\circ$ and $OC = OB$, it follows that $OC^2 + CD^2 = OD^2$, which implies $OB^2 + 4^2 = (OB + 2)^2$. Hence, $OB = 3$, leading to $AB = 6$ and $AD = 8$. \\\\\nSince $AE \\perp AD$ and $AB$ is the diameter of the circle $O$, it follows that $AE$ is the tangent of circle $O$. \\\\\nSince $CD$ is the tangent of circle $O$, it implies $AE = CE$. In $\\triangle ADE$, we have $AD^2 + AE^2 = DE^2$, so $8^2 + AE^2 = (4+AE)^2$, resulting in $AE = \\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52602536_178.png"
  },
  {
    "solution": "\\boxed{y=\\frac{8}{x}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the direct proportion function $y_{1}=kx$ intersects with the graph of the inverse proportion function $y_{2}=\\frac{m}{x}$ at points A$(2,4)$ and point B. The coordinates of point C are $(4,0)$, and point D lies on the graph of the inverse proportion function $y_{2}=\\frac{m}{x}$. What is the expression of the inverse proportion function?\n",
    "completion": "\\textbf{Solution:} Since the graph of the inverse proportion function $y=\\frac{m}{x}$ passes through point A$(2,4)$,\\\\\nthen $m=2\\times 4=8$,\\\\\nthus, the expression of the inverse proportion function is $\\boxed{y=\\frac{8}{x}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51403218_42.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system, the rhombus OABC has its side OA on the positive half of the $x$-axis, and the graph of the inverse proportional function $y=\\frac{k}{x}(x>0)$ passes through the midpoint $D$ of the diagonal OB and the vertex $C$. If the area of the rhombus OABC is 12, what is the value of $k$?\n",
    "completion": "\\textbf{Solution:} Let the coordinates of point A be $(a, 0)$, and the coordinates of point C be $\\left(c, \\frac{k}{c}\\right)$, \\\\\n$\\because$ the area of the rhombus is 12, \\\\\n$\\therefore a \\cdot \\frac{k}{c}=12$, \\\\\n$\\because$ D is the midpoint of the diagonal OB, \\\\\n$\\therefore$ the coordinates of point D are: $\\left(\\frac{a+c}{2}, \\frac{k}{2c}\\right)$, \\\\\n$\\therefore \\frac{a+c}{2} \\cdot \\frac{k}{2c}=k$, \\\\\nthus $a=3c$, \\\\\n$\\therefore 3c \\cdot \\frac{k}{c}=12$, \\\\\nsolving this gives $k=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51510317.png"
  },
  {
    "solution": "\\boxed{100}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, a point A is selected on the opposite side of the river. Then, on this side of the river, points B and C are chosen such that $AB \\perp BC$. Another point E is selected such that $EC \\perp BC$, and the line of sight determines that BC and AE intersect at point D. At this moment, it is measured that $BD=120\\,m$, $DC=60\\,m$, and $EC=50\\,m$. What is the distance between the two sides of the river, that is, the length of AB?\n",
    "completion": "\\textbf{Solution:} Given that $AB\\perp BC$, $EC\\perp BC$,\\\\\n$\\therefore AB\\parallel CE$\\\\\n$\\therefore \\triangle ABD\\sim \\triangle ECD$\\\\\n$\\therefore \\frac{BD}{DC}=\\frac{AB}{EC}$\\\\\nGiven $BD=120m$, $DC=60m$, $EC=50m$,\\\\\n$\\therefore AB=\\frac{BD\\cdot EC}{DC}=\\frac{120\\times 50}{60}=\\boxed{100}m$",
    "image_path": "dataset/mm_math/images/MM_Math/52719807.png"
  },
  {
    "solution": "\\boxed{\\frac{15}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$. Point $D$ is on $AC$, and $\\angle DBC=\\angle A$. If $AC=4$ and $AB=5$, what is the length of $BD$?\n",
    "completion": "\\textbf{Solution:} In the right triangle $ABC$, by the Pythagorean theorem, we have $BC=\\sqrt{AB^{2}-AC^{2}}=3$,\\\\\nsince $\\angle DBC = \\angle A$, and $\\angle C = \\angle C$,\\\\\nthus $\\triangle ABC \\sim \\triangle BDC$,\\\\\ntherefore $\\frac{AC}{BC}=\\frac{AB}{BD}$, that is, $\\frac{4}{3}=\\frac{5}{BD}$,\\\\\nhence $BD=\\boxed{\\frac{15}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51322771.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a set of triangles (ignoring thickness) is placed on the table, with the right angle $\\angle DAE$ and the $60^\\circ$ angle $\\angle BAC$ vertices coinciding. The triangle $ADE$ can rotate around point A. When $AD$ bisects $\\angle BAC$, find the degree of $\\angle 1$.\n",
    "completion": "\\textbf{Solution:} Since AD bisects $\\angle BAC$, and $\\angle BAC=60^\\circ$,\\\\\nit follows that $\\angle DAC=\\frac{1}{2}\\angle BAC=30^\\circ$.\\\\\nSince $\\angle DAE=90^\\circ$,\\\\\nit follows that $\\angle 1=\\angle DAE-\\angle DAC=\\boxed{60^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51321764_135.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, D, E, and F are the midpoints of BC, AD, and CE, respectively. If $S_{\\triangle ABC}=8\\text{cm}^2$, what is $S_{\\triangle DEF}$?\n",
    "completion": "\\textbf{Solution:} Given that $S_{\\triangle ABC}=8\\,\\text{cm}^2$ and D is the midpoint of BC,\\\\\nit follows that $S_{\\triangle ABD}=S_{\\triangle ADC}=\\frac{1}{2}S_{\\triangle ABC}=\\frac{1}{2}\\times 8=4\\,\\text{cm}^2$,\\\\\nsince E is the midpoint of AD,\\\\\nthen $S_{\\triangle DEC}=\\frac{1}{2}S_{\\triangle ADC}=\\frac{1}{2}\\times 4=2\\,\\text{cm}^2$,\\\\\nsince F is the midpoint of EC,\\\\\nthen $S_{\\triangle DEF}=\\frac{1}{2}S_{\\triangle DEC}=\\frac{1}{2}\\times 2=\\boxed{1}\\,\\text{cm}^2$,",
    "image_path": "dataset/mm_math/images/MM_Math/52604074.png"
  },
  {
    "solution": "\\boxed{\\sqrt{10}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=2\\angle B$, and $CD$ bisects $\\angle ACB$. If $AD=2$, $BD=3$, what is the length of $AC$?\n",
    "completion": "\\textbf{Solution}:\\\\\nSince $CD$ bisects $\\angle ACB$,\\\\\nit follows that $\\angle ACD=\\angle BCD=\\frac{1}{2}\\angle ACB$.\\\\\nGiven $\\angle ACB=2\\angle B$,\\\\\nit follows that $\\angle ACD=\\angle B$.\\\\\nConsidering $\\triangle ADC$ and $\\triangle ACB$, we have\\\\\n$\\left\\{\\begin{array}{l}\n\\angle ACD=\\angle B \\\\\n\\angle A=\\angle A\n\\end{array}\\right.$\\\\\nHence, $\\triangle ADC\\sim \\triangle ACB$\\\\\nwhich implies $\\frac{AD}{AC}=\\frac{AC}{AB}$, i.e.,\\\\\n$AC^{2}=AD\\cdot AB$.\\\\\nGiven $AD=2, BD=3$,\\\\\nit follows that $AB=AD+BD=5$\\\\\nand thus $AC^{2}=AD\\cdot AB=10$.\\\\\nSince the length of a segment is a positive number, $AC=\\sqrt{10}$. Therefore, the solution is $AC=\\boxed{\\sqrt{10}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51433574.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, D and E are the midpoints of BC and AC respectively, the area of $\\triangle CDE$ is 2, then what is the area of $\\triangle ABC$?\n",
    "completion": "\\textbf{Solution:} Since $D$ and $E$ are the midpoints of $BC$ and $AC$ respectively, \\\\\ntherefore, $AD$ and $DE$ are the medians of $\\triangle ABC$ and $\\triangle ADC$ respectively, \\\\\ntherefore, ${S}_{\\triangle ADC}=2{S}_{\\triangle CDE}=2\\times 2=4$, \\\\\ntherefore, ${S}_{\\triangle ABC}=2{S}_{\\triangle ADC}=2\\times 4=\\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53059011.png"
  },
  {
    "solution": "\\boxed{y = -x + 2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the vertex A of the right triangle $\\triangle ABO$ is the intersection point in the second quadrant between the hyperbola $y=\\frac{k}{x}$ and the line $y=-x-(k+1)$. $\\overline{AB} \\perp x$-axis at B, and the area of $\\triangle ABO$ is $\\frac{3}{2}$. What are the equations of these two functions?\n",
    "completion": "\\textbf{Solution:} Since $AB \\perp x$-axis at B, and the area of $\\triangle ABO = \\frac{3}{2}$,\\\\\nit follows that $\\frac{1}{2}|k| = \\frac{3}{2}$. Solving this, we find $k = \\pm 3$.\\\\\nSince the graph of the inverse proportionality function lies in the second and fourth quadrants,\\\\\nwe have $k < 0$,\\\\\nthus $k = -3$,\\\\\nwhich means the equation of the inverse proportionality function is $\\boxed{y = -\\frac{3}{x}}$, and the equation of the linear function is $\\boxed{y = -x + 2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51458226_261.png"
  },
  {
    "solution": "\\boxed{15^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: When a set of triangular plates are stacked together as shown in the figure, what is the degree measure of $\\angle \\alpha$ in the diagram?\n",
    "completion": "\\textbf{Solution:} As shown in the figure,\\\\\naccording to the special nature of the angles on the drafting triangle, it is known that $\\angle 1=45^\\circ,\\ \\angle 2=60^\\circ$. By applying the property of the external angle of a triangle, it follows that $\\angle \\alpha =\\angle 2-\\angle 1=60^\\circ-45^\\circ=\\boxed{15^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53044250.png"
  },
  {
    "solution": "\\boxed{120^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given the diagram, $OC\\perp AB$ at point $O$, and $\\angle COD= \\frac{1}{4}\\angle BOD$, with $OE$ bisecting $\\angle BOD$. Find the measures of $\\angle COE$ and $\\angle AOE$?\n",
    "completion": "\\textbf{Solution:} Since OC $\\perp$ AB, it follows that $\\angle BOC = \\angle AOC = 90^\\circ$. Since $\\angle COD = \\frac{1}{3}\\angle BOC = 30^\\circ$, it follows that $\\angle BOD = 120^\\circ$. Since OE bisects $\\angle BOD$, it follows that $\\angle BOE = \\angle DOE = 60^\\circ$. Therefore, $\\angle COE = \\angle BOC - \\angle BOE = 90^\\circ - 60^\\circ = \\boxed{30^\\circ}$, $\\angle AOE = 180^\\circ - \\angle BOE = 180^\\circ - 60^\\circ = \\boxed{120^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52774109_158.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle A=90^\\circ$, $BD$ bisects $\\angle ABC$ and intersects $AC$ at point $D$, with $AB=4$, $BC=12$, $AD=3$. If point $P$ moves along $BC$, what is the minimum value of the length of segment $DP$?\n",
    "completion": "\\textbf{Solution:} When $DP \\perp BC$, the length of the line segment $DP$ is minimized, \\\\\n$\\because BD$ bisects $\\angle ABC$ and $\\angle A = 90^\\circ$, \\\\\nwhen $DP \\perp BC$, $DP = AD$, \\\\\n$\\because AD = 3$, \\\\\n$\\therefore$ the minimal value of $DP$ is $\\boxed{3}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52095800_113.png"
  },
  {
    "solution": "\\boxed{\\frac{2}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\triangle ABC$, $ DE\\parallel BC$. If $ \\frac{AD}{DB}=\\frac{2}{3}$, then what is the value of $ \\frac{DE}{BC}$?\n",
    "completion": "\\textbf{Solution:} Given that $\\because \\frac{AD}{DB}=\\frac{2}{3}$,\\\\\n$\\therefore \\frac{AD}{AB}=\\frac{2}{5}$,\\\\\nsince $DE\\parallel BC$,\\\\\n$\\therefore \\frac{DE}{BC}=\\frac{AD}{AB}=\\boxed{\\frac{2}{5}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51403120.png"
  },
  {
    "solution": "\\boxed{\\frac{4}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that in $\\triangle ABC$, $\\angle C=90^\\circ$, $AB=10$, and $AC=8$. What is the value of $\\tan B$?\n",
    "completion": "\\textbf{Solution:} In the right-angled $\\triangle ABC$, by the Pythagorean theorem, we have $BC=\\sqrt{AB^{2}-AC^{2}}=\\sqrt{10^{2}-8^{2}}=6$, \\\\\n$\\therefore \\tan B=\\frac{AC}{BC}=\\frac{8}{6}=\\boxed{\\frac{4}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51367507.png"
  },
  {
    "solution": "\\boxed{x>0}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, if the graph of the function $y=ax+b$ is as depicted, then what is the solution set of the inequality $ax+b<2$ with respect to $x$?\n",
    "completion": "\\textbf{Solution:} From the graph, it is known that the coordinates of the intersection point of the function $y=ax+b$ with the $y$-axis are $(0,2)$,\\\\\nhence, the solution set of the inequality $ax+b<2$ is $x>0$,\\\\\ntherefore, the answer is $\\boxed{x>0}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50590397.png"
  },
  {
    "solution": "\\boxed{\\left(-3,6\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=kx-3$ intersects the x-axis and y-axis at points B and A, respectively, with $OB=\\frac{1}{3}OA$. Point C is a point on line AB, located in the second quadrant. When the area of $\\triangle OBC$ is 3, what are the coordinates of point C?\n",
    "completion": "\\textbf{Solution:} Draw $CH\\perp x$ axis at point H, as shown in the figure:\\\\\nSince the line $y=kx-3$ intersects the x-axis and y-axis at points B and A respectively,\\\\\nthus, by setting $x=0$, we get $y=-3$, i.e., $OA=3$,\\\\\nsince $OB=\\frac{1}{3}OA$,\\\\\nthus, $OB=1$, i.e., $B(-1,0)$, substituting into the line equation yields: $0=-k-3$, solving for $k$ yields: $k=-3$;\\\\\nthus, the equation of line AB is $y=-3x-3$,\\\\\nsince the area of $\\triangle OBC$ is 3,\\\\\nthus, $\\frac{1}{2}OB\\cdot CH=3$,\\\\\ntherefore, $CH=6$, i.e., the y-coordinate of point C is 6,\\\\\ntherefore, $-3x-3=6$, solving for: $x=-3$,\\\\\ntherefore, $C\\left(-3,6\\right)$;\\\\\nThus, the answer is $\\boxed{\\left(-3,6\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52386546.png"
  },
  {
    "solution": "\\boxed{\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\Delta DBE$ is inscribed in $\\odot O$, with $BD$ as the diameter and $DE=EB$. Point $C$ is on $\\odot O$ (not coinciding with $D$, $B$, $E$), $\\angle A=45^\\circ$, point $A$ is on line $CD$, and line $AB$ is drawn. $\\triangle ABD \\sim \\triangle CBE$, $DC=6$, $DB=10$. What is the length of segment $CE$?\n",
    "completion": "\\textbf{Solution:} By the given problem, \\\\\nsince $\\angle BCD=90^\\circ$, \\\\\ntherefore $BC= \\sqrt{BD^{2}-CD^{2}}=8$, \\\\\nsince $\\triangle ABC$ and $\\triangle BED$ are isosceles right triangles,\\\\\ntherefore $AB= \\sqrt{2}BC= 8\\sqrt{2}$, $BE= \\frac{\\sqrt{2}}{2}BD= 4\\sqrt{2}$,\\\\\ntherefore $AD=AC-CD=8-6=2$,\\\\\nby similarity $\\triangle ADB\\sim \\triangle CEB$,\\\\\ntherefore $\\frac{BE}{BC}=\\frac{CE}{AD}$, that is $\\frac{4\\sqrt{2}}{8}=\\frac{CE}{2}$,\\\\\nsolving for $CE$ yields $CE= \\boxed{\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51369960_130.png"
  },
  {
    "solution": "\\boxed{\\sqrt{2}+\\sqrt{6}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In $\\triangle ABC$, $\\angle CAB=90^\\circ$, $AC=AB$. If point $D$ lies on $AC$, and $BD$ is connected and rotated $90^\\circ$ clockwise around point $B$ to obtain $BE$, and $CE$ is connected and intersects $AB$ at point $F$, as shown in Figure 1. If $\\angle ABE=75^\\circ$ and $BD=4$, what is the length of $AC$?\n",
    "completion": "\\textbf{Solution:} Draw DG$\\perp$BC from D with the foot as G, as shown in Fig. 1:\\\\\n$\\because$ Rotating BD $90^\\circ$ clockwise around point B yields BE,\\\\\n$\\therefore$ $\\angle$EBD$=90^\\circ$,\\\\\n$\\because$ $\\angle$ABE$=75^\\circ$,\\\\\n$\\therefore$ $\\angle$ABD$=15^\\circ$,\\\\\n$\\because$ $\\angle$ABC$=45^\\circ$,\\\\\n$\\therefore$ $\\angle$DBC$=30^\\circ$,\\\\\n$\\therefore$ in right $\\triangle$BDG, we have $DG=\\frac{1}{2}BD=2$, $BG=\\sqrt{3}DG=2\\sqrt{3}$,\\\\\n$\\because$ $\\angle$ACB$=45^\\circ$,\\\\\n$\\therefore$ in right $\\triangle$DCG, CG$=DG=2$,\\\\\n$\\therefore$ BC$=BG+CG=2+2\\sqrt{3}$,\\\\\n$\\therefore$ AC$=\\frac{\\sqrt{2}}{2}BC=\\sqrt{2}+\\sqrt{6}$;\\\\\nThus, the length of AC is $\\boxed{\\sqrt{2}+\\sqrt{6}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51368402_93.png"
  },
  {
    "solution": "\\boxed{52^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $ \\angle POQ$ is an acute angle, and point A is a point on side $ OP$, with ray $ AB\\parallel OQ$. It is known that the two acute angle vertices of the triangular plate $ CDE$ are located on $ AB$ and $ OQ$ respectively, $ DE\\parallel OP$, $ \\angle E=90^\\circ$, and $ \\angle CDE=60^\\circ$. As shown in Figure 1, if $ \\angle ACD=112^\\circ$, what is the degree measure of $ \\angle POQ$?\n",
    "completion": "\\textbf{Solution:} Given $\\because AB\\parallel OQ, \\angle ACD=112^\\circ$,\\\\\n$\\therefore \\angle CDQ=\\angle ACD=112^\\circ$.\\\\\n$\\because \\angle CDE=60^\\circ$,\\\\\n$\\therefore \\angle EDQ=\\angle CDQ-\\angle CDE=112^\\circ-60^\\circ=52^\\circ$.\\\\\nFurthermore, $\\because DE\\parallel OP$,\\\\\n$\\therefore \\angle POQ=\\boxed{52^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/54474420_113.png"
  },
  {
    "solution": "\\boxed{55^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, straight lines AB and CD intersect at point O, EO $\\perp$ AB with the foot of the perpendicular at O, and $\\angle EOC = 35^\\circ$. What is the measure of $\\angle BOD$?\n",
    "completion": "\\textbf{Solution:} Given that $EO\\perp AB$,\\\\\nit follows that $\\angle AOE=90^\\circ$,\\\\\nand given $\\angle EOC=35^\\circ$,\\\\\nit results in $\\angle AOC=\\angle AOE-\\angle EOC=55^\\circ$,\\\\\ntherefore, $\\angle BOD=\\angle AOC=\\boxed{55^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55377894.png"
  },
  {
    "solution": "\\boxed{96}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, $AC$ and $BD$ intersect at point $O$, and $AO=CO$. Point $E$ is on line segment $BO$, and $\\angle CEO=\\angle ADO$. If $AB=BC$, $CD=10$, and $AC=16$, find the area of quadrilateral $AECD$.\n",
    "completion": "\\textbf{Solution:} Given $AB=BC$ and $OA=OC$, it follows that $BO \\perp AC$. Within $\\triangle COD$, we have $OC=\\frac{1}{2}AC=8$, therefore $OD=\\sqrt{CD^{2}-OC^{2}}=\\sqrt{10^{2}-8^{2}}=6$. Consequently, $DE=2OD=12$. Thus, the area of quadrilateral AECD $=\\frac{1}{2}\\times DE\\times AC=\\frac{1}{2}\\times 12\\times 16=\\boxed{96}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52424185_126.png"
  },
  {
    "solution": "\\boxed{230^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the three vertices of $\\triangle ABC$ are on the circle $\\odot O$, $D$ is a point on $\\overset{\\frown}{AB}$, and $E$ is a point on $\\overset{\\frown}{AC}$. If $\\angle BAC=50^\\circ$, what is the sum of $\\angle D+\\angle E$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect OA, OB, and OC.\\\\\nAccording to the problem: $\\angle BOC=2\\angle BAC=100^\\circ$,\\\\\n$\\therefore \\angle AOB+\\angle AOC=260^\\circ$,\\\\\nAlso, $\\because \\angle D= \\frac{1}{2}(\\angle BOC+\\angle AOC)$, $\\angle E= \\frac{1}{2}(\\angle BOC+\\angle AOB)$,\\\\\n$\\therefore \\angle D+\\angle E= \\frac{1}{2}(\\angle BOC+\\angle AOC+\\angle BOC+\\angle AOB)= \\frac{1}{2}(260^\\circ+100^\\circ+100^\\circ)=\\boxed{230^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52991553.png"
  },
  {
    "solution": "\\boxed{20\\text{m}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, to estimate the width of the fish pond $AB$, Xiao Kai took points $C$, $D$, and $E$ on the land such that $A$, $C$, $D$ are on the same straight line and $B$, $C$, $E$ are on the same straight line. It was measured that $CD = \\frac{1}{2}AC$ and $CE = \\frac{1}{2}BC$. Xiao Kai measured the length of $DE$ to be 10 meters. What is the length of the fish pond width $AB$ in meters?\n",
    "completion": "\\textbf{Solution:} Given that $CD=\\frac{1}{2}AC, CE=\\frac{1}{2}BC$,\\\\\nit follows that $\\frac{CD}{AC}=\\frac{CE}{BC}=\\frac{1}{2}$,\\\\\nand since $\\angle DCE=\\angle ACB$,\\\\\nit implies $\\triangle DCE\\sim \\triangle ACB$,\\\\\nhence $\\frac{ED}{AB}=\\frac{CD}{AC}=\\frac{1}{2}$.\\\\\nGiven $ED=10\\text{m}$,\\\\\ntherefore $AB=\\boxed{20\\text{m}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52916840.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the diagonals of square $ABCD$ intersect at point $O$. An arbitrary line passing through point $O$ intersects $AD$ and $BC$ at points $E$ and $F$, respectively. If the length of the diagonal of the square is $\\sqrt{2}$, what is the area of the shaded region in the figure?\n",
    "completion": "\\textbf{Solution:} In the square $ABCD$,\n\n$AO=CO=DO$, $\\angle OAE=\\angle OCF=45^\\circ$, $\\angle AOD=90^\\circ$,\n\nAlso, since $\\angle AOE=\\angle COF$,\n\n$\\therefore$ $\\triangle AOE \\cong \\triangle COF$,\n\n$\\therefore$ $S_{\\triangle COF}=S_{\\triangle AOE}$,\n\n$\\therefore$ the area of the shaded region is equal to $S_{\\triangle COF}+S_{\\triangle DOE}=S_{\\triangle AOE}+S_{\\triangle DOE}=S_{\\triangle AOD}$,\n\nSince the diagonal of a square is $\\sqrt{2}$,\n\n$\\therefore$ $AO=DO=\\frac{\\sqrt{2}}{2}$,\n\n$S_{\\triangle AOD}=\\frac{1}{2}AO\\cdot DO=\\frac{1}{2}\\times \\frac{\\sqrt{2}}{2}\\times \\frac{\\sqrt{2}}{2}=\\frac{1}{4}$,\n\n$\\therefore$ the area of the shaded region is equal to $\\boxed{\\frac{1}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52344118.png"
  },
  {
    "solution": "\\boxed{y=\\frac{1}{2}x-1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, two lines $l_{1}$ and $l_{2}$ passing through point A intersect the y-axis at points B and C, respectively, where point B is above the origin and point C is below the origin. It is known that $AB = \\sqrt{13}$ and $B(0,3)$. If the area of $\\triangle ABC$ is 4, what is the equation of the line $l_{2}$?\n",
    "completion": "\\textbf{Solution:} Let $\\because S_{\\triangle ABC}=\\frac{1}{2}BC \\cdot OA$,\\\\\n$\\therefore 4=\\frac{1}{2} \\cdot BC \\times 2$, solving this gives $BC=4$,\\\\\n$\\therefore OC=BC-OB=4-3=1$,\\\\\n$\\therefore C(0, -1)$,\\\\\nLet the equation of the line $l_{2}$ be $y=kx+b$,\\\\\nSubstituting $A(2, 0)$, $C(0, -1)$ into $y=kx+b$, we get:\\\\\n$\\begin{cases}\n0=2k+b\\\\\n-1=b\n\\end{cases}$, solving this gives $\\begin{cases}\nk=\\frac{1}{2}\\\\\nb=-1\n\\end{cases}$,\\\\\n$\\therefore$ the equation of line $l_{2}$ is $\\boxed{y=\\frac{1}{2}x-1}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52096283_101.png"
  },
  {
    "solution": "\\boxed{100^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, OC is the bisector of $\\angle AOB$, and $\\angle COD=20^\\circ$. If $\\angle AOD=30^\\circ$, what is the measure of $\\angle AOB$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle COD=20^\\circ$ and $\\angle AOD=30^\\circ$,\\\\\n$\\therefore \\angle AOC=\\angle COD+\\angle AOD=20^\\circ+30^\\circ=50^\\circ$,\\\\\nSince OC is the bisector of $\\angle AOB$,\\\\\n$\\therefore \\angle AOB=2\\angle AOC=\\boxed{100^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52837816_94.png"
  },
  {
    "solution": "\\boxed{20}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $O$ is the midpoint of the diagonal $AC$ of rectangle $ABCD$, and $M$ is the midpoint of $AD$. If $AB=5$ and $AD=12$, what is the perimeter of quadrilateral $ABOM$?\n",
    "completion": "\\textbf{Solution:} Since $O$ is the midpoint of the diagonal $AC$ of rectangle $ABCD$, and $M$ is the midpoint of $AD$,\\\\\ntherefore $OM = \\frac{1}{2}CD = \\frac{1}{2}AB = 2.5$,\\\\\nand since $AB=5$, $AD=12$,\\\\\ntherefore $AC = \\sqrt{5^{2}+12^{2}}=13$,\\\\\nSince $O$ is the midpoint of the diagonal $AC$ of rectangle $ABCD$,\\\\\ntherefore $BO = \\frac{1}{2}AC = 6.5$,\\\\\ntherefore, the perimeter of quadrilateral $ABOM$ is $AB+AM+BO+OM = 5+6+6.5+2.5 = \\boxed{20}$",
    "image_path": "dataset/mm_math/images/MM_Math/52380847.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{2}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, in the intelligent triangle $\\triangle ABC$, where $AB=2$, $BC=2\\sqrt{2}$, and $AD$ is the median on side $BC$, find the value of $\\frac{AD}{AC}$.\n",
    "completion": "\\textbf{Solution:} Let: $AD$ be the median of $BC$, $AB=2$, $BC=2\\sqrt{2}$,\\\\\nThus, $BD=\\frac{1}{2}BC=\\sqrt{2}$,\\\\\nTherefore, $\\frac{BD}{AB}=\\frac{AB}{BC}=\\frac{\\sqrt{2}}{2}$,\\\\\nGiven that, $\\angle B=\\angle B$,\\\\\nIt follows that, $\\triangle ABD\\sim \\triangle CBA$,\\\\\nTherefore, $\\frac{AD}{AC}=\\frac{BD}{AB}=\\boxed{\\frac{\\sqrt{2}}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53081503_92.png"
  },
  {
    "solution": "\\boxed{265^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that in $\\triangle ABC$, $\\angle A=65^\\circ$. When $\\angle B$ and $\\angle C$ are folded as shown, if $\\angle ADB'=35^\\circ$, then what is the sum of $\\angle 1+\\angle 2+\\angle 3$?\n",
    "completion": "\\textbf{Solution:}\\\\\nFrom the folding, we know: $\\angle B=\\angle B'$, $\\angle C=\\angle C'$.\\\\\n$\\because \\angle 3=\\angle B+\\angle 4$, $\\angle 4=\\angle ADB'+\\angle B'$,\\\\\n$\\therefore \\angle 3=\\angle B+\\angle ADB'+\\angle B'$\\\\\n$=2\\angle B+35^\\circ$.\\\\\n$\\because \\angle 1+\\angle 2=180^\\circ-\\angle C'GC+180^\\circ-\\angle C'FC$\\\\\n$=360^\\circ-(\\angle C'FC+\\angle C'GC)$,\\\\\n$\\angle C'FC+\\angle C'GC=360^\\circ-\\angle C-\\angle C'$\\\\\n$=360^\\circ-2\\angle C$,\\\\\n$\\therefore \\angle 1+\\angle 2=360^\\circ-(\\angle C'FC+\\angle C'GC)$\\\\\n$=360^\\circ-(360^\\circ-2\\angle C)$\\\\\n$=2\\angle C$.\\\\\n$\\therefore \\angle 1+\\angle 2+\\angle 3$\\\\\n$=2\\angle C+2\\angle B+35^\\circ$\\\\\n$=2(\\angle C+\\angle B)+35^\\circ$\\\\\n$=2(180^\\circ-\\angle A)+35^\\circ$\\\\\n$=2(180^\\circ-65^\\circ)+35^\\circ$\\\\\n$=\\boxed{265^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53106968.png"
  },
  {
    "solution": "\\boxed{\\frac{9\\sqrt{7}}{7}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, in rhombus $ABCD$, $AC$ is the diagonal, and $AB=AC=6$. Points $E$ and $F$ are points on edges $AB$ and $BC$, respectively, satisfying $AE=BF$. Lines $AF$ and $CE$ intersect at point $G$. As shown in Figure 2, draw $DH\\perp CE$ meeting $CE$ at point $H$. If $CF=4$ and $AF=2\\sqrt{7}$, determine the value of $GH$.\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, extend GA to point K so that $AK = CG$. \\\\\n$\\because \\angle BAF = \\angle ACE$, \\\\\n$\\therefore \\angle FAC = \\angle GCF$, \\\\\n$\\because \\angle GFC = \\angle AFC$, \\\\\n$\\therefore \\triangle AFC \\sim \\triangle CFG$, \\\\\n$\\therefore \\frac{FG}{FC} = \\frac{FC}{FA} = \\frac{CG}{AC}$, \\\\\n$\\because CF = 4$, $AF = 2\\sqrt{7}$, $AC = 6$, \\\\\n$\\therefore GF = \\frac{8\\sqrt{7}}{7}$, $AG = \\frac{6\\sqrt{7}}{7}$, $CG = \\frac{12\\sqrt{7}}{7}$, \\\\\n$\\because \\angle FGC = 60^\\circ$, $\\angle ADC = 60^\\circ$, \\\\\n$\\therefore \\angle AGC + \\angle ADC = 180^\\circ$, \\\\\n$\\therefore \\angle GAD + \\angle GCD = 180^\\circ$, \\\\\n$\\because \\angle KAD + \\angle GAD = 180^\\circ$, \\\\\n$\\therefore \\angle KAD = \\angle GCD$, \\\\\nMoreover, $\\because DC = DA$, \\\\\n$\\therefore \\triangle ADK \\cong \\triangle CDG$ (SAS), \\\\\n$\\therefore DK = DG$, $\\angle KDA = \\angle GDC$, \\\\\n$\\therefore \\angle KDG = \\angle ADC = 60^\\circ$, \\\\\n$\\therefore \\triangle DKG$ is an equilateral triangle, \\\\\n$\\therefore \\angle AGD = \\angle DGH = 60^\\circ$, $DG = KG = AK + AG = AG + CG = \\frac{18\\sqrt{7}}{7}$, \\\\\n$\\because DH \\perp CE$, $\\angle DGH = 60^\\circ$, \\\\\n$\\therefore GH = \\frac{1}{2}DG = \\boxed{\\frac{9\\sqrt{7}}{7}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51152250_120.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the side length of the square is 1. Take a point E on ray AB, connect DE, and rotate $\\triangle ADE$ $90^\\circ$ around point D. Point E falls on point F, connect EF, which intersects with the line containing diagonal BD at point M, and intersects with ray DC at point N. When $AE=\\frac{1}{3}$, what is the value of $\\tan\\angle EDB$?\n",
    "completion": "\\textbf{Solution:} Draw $EH \\perp BD$ at H,\n\nsince the side length of the square is 1, $AE=\\frac{1}{3}$,\n\nthus $EB=1-AE=1-\\frac{1}{3}=\\frac{2}{3}$,\n\nsince BD is the diagonal of the square,\n\nthus BD bisects $\\angle ABC$,\n\nthus $\\angle ABD=45^\\circ$,\n\nsince $EH \\perp BD$,\n\nthus $\\angle BEH=180^\\circ-\\angle EBH-\\angle EHB=180^\\circ-45^\\circ-90^\\circ=45^\\circ$,\n\nthus $EH=BH$,\n\nthus $EH=BH=BE\\sin45^\\circ= \\frac{2}{3}\\times \\frac{\\sqrt{2}}{2}=\\frac{\\sqrt{2}}{3}$,\n\nthus $BD=1\\div\\frac{\\sqrt{2}}{2}=\\sqrt{2}$,\n\nthus $DH=DB-BH= \\sqrt{2}-\\frac{\\sqrt{2}}{3}=\\frac{2\\sqrt{2}}{3}$,\n\nthus $\\tan\\angle EDB=\\frac{EH}{HD}=\\frac{\\frac{\\sqrt{2}}{3}}{\\frac{2\\sqrt{2}}{3}}=\\boxed{\\frac{1}{2}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51403292_158.png"
  },
  {
    "solution": "\\boxed{40^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, it is known that $AB=AC$, $BC$ bisects $\\angle ABD$, and $AC\\parallel BD$; if $\\angle A=100^\\circ$, find the measure of $\\angle 1$.\n",
    "completion": "\\textbf{Solution}: Since $\\angle A=100^\\circ$,\\\\\nit follows that $\\angle ABC=\\angle ACB=\\frac{180^\\circ-100^\\circ}{2}=40^\\circ$,\\\\\ntherefore $\\angle 1=\\boxed{40^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53033465_96.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the area of $\\triangle ABC$ is 48, with $AB=AC=8$. Point D is on side $BC$, and lines $DE \\perp AB$ at E, and $DF \\perp AC$ at F are drawn. If $DF=2DE$, then what is the length of $DF$?\n",
    "completion": "\\textbf{Solution:} Construct line segment AD, \\\\\nsince $S_{\\triangle ABD} + S_{\\triangle ACD} = S_{\\triangle ABC}$, AB = AC = 8, \\\\\ntherefore $\\frac{1}{2} \\times 8 \\cdot DF + \\frac{1}{2} \\times 8 \\cdot DE = 48$, \\\\\ntherefore DE + DF = 12, \\\\\nsince DF = 2DE, \\\\\ntherefore DF = \\boxed{8}.",
    "image_path": "dataset/mm_math/images/MM_Math/52693972.png"
  },
  {
    "solution": "\\boxed{8\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the square $ABCD$, the diagonal $AC=4$. What is the perimeter of the square $ABCD$?\n",
    "completion": "\\textbf{Solution:} Given that the diagonal $AC=4$,\\\\\nhence $\\sqrt{AB^{2}+BC^{2}}=4$,\\\\\nsince in square ABCD $AB=BC$,\\\\\nthus $AB=BC=2\\sqrt{2}$,\\\\\ntherefore, the perimeter of square ABCD is: $2\\sqrt{2}\\times 4=\\boxed{8\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53359985.png"
  },
  {
    "solution": "\\boxed{y=2x-5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, given the parabola $y = ax^2 + bx + c$ with its vertex at $A(4, 3)$ and intersecting the y-axis at point $B(0, -5)$, with the line of symmetry being the line $l$, and the point $M$ being the midpoint of segment $AB$. Find the coordinates of point $M$ and the equation of line $AB$.\n",
    "completion": "\\textbf{Solution:} Let the coordinates of midpoint $M$ be $(x, y)$,\\\\\nsince $A(4, 3), B(0, -5)$,\\\\\nthus $x-0=4-x$, i.e., $x=\\frac{4+0}{2}=2$,\\\\\nSimilarly, we get $y=\\frac{-5+3}{2}=-1$,\\\\\ntherefore the midpoint $\\boxed{M(2, -1)}$,\\\\\nLet the equation of the line $AB$ be: $y=kx-5$,\\\\\nSubstituting the coordinates of point $A$ into the equation yields: $3=4k-5$, solving gives: $k=2$,\\\\\nHence, the equation of the line $AB$ is: $\\boxed{y=2x-5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53042870_133.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In a rectangular piece of paper $ABCD$, where $AD=10$ and $AB=4\\sqrt{3}$, $M$ and $N$ are the midpoints of $AB$ and $CD$, respectively. Now, fold the paper as shown in the figure so that point $B$ falls on point $F$ on $MN$. What is the length of $EF$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, draw $FH\\perp BC$ at point H, obtaining rectangle $BMFH$.\\\\\nSince M is the midpoint of $AB$,\\\\\nit follows that $AM=2\\sqrt{3}$, $\\angle AMF=90^\\circ$,\\\\\ntherefore, $\\angle AFM=30^\\circ$,\\\\\nthus, $MF=AF\\cdot \\cos 30^\\circ=6$.\\\\\nSince quadrilateral $BMFH$ is a rectangle,\\\\\nit follows that $BH=MF=6$.\\\\\nLet $EF=x$, then $EH=6-x$, $FH=2\\sqrt{3}$.\\\\\nIn $\\triangle EFH$, $EF^2=EH^2+FH^2$,\\\\\nwhich gives $x^2=(6-x)^2+(2\\sqrt{3})^2$,\\\\\nsolving this yields $x=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53019951.png"
  },
  {
    "solution": "\\boxed{50^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AD$ is the altitude of $\\triangle ABC$, $CE$ is the angle bisector of $\\angle ACB$ in $\\triangle ACB$, and $F$ is the midpoint of $AC$, with $\\angle ACB=50^\\circ$ and $\\angle BAD=65^\\circ$. What is the degree measure of $\\angle AEC$?\n",
    "completion": "\\textbf{Solution:}\n\nSince $AD$ is the altitude of $\\triangle ABC$,\\\\\ntherefore $\\angle ADB=90^\\circ$,\\\\\nSince $\\angle BAD=65^\\circ$,\\\\\ntherefore $\\angle ABD=90^\\circ-65^\\circ=25^\\circ$,\\\\\nSince $CE$ is the bisector of $\\triangle ACB$, and $\\angle ACB=50^\\circ$,\\\\\ntherefore $\\angle ECB=\\frac{1}{2} \\angle ACB=25^\\circ$,\\\\\ntherefore $\\angle AEC=\\angle ABD+\\angle ECB=25^\\circ+25^\\circ=\\boxed{50^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52815585_121.png"
  },
  {
    "solution": "\\boxed{20}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB = AC = 10$. Point $P$ is on $BC$. Parallel lines to $AC$ and $AB$ are drawn through point $P$, intersecting $AB$ and $AC$ at points $M$ and $N$ respectively. What is the perimeter of the quadrilateral $AMPN$?\n",
    "completion": "\\textbf{Solution:} Since $AB=AC$, \\\\\nit follows that $\\angle B=\\angle C$, \\\\\nSince $PM\\parallel AC$ and $PN\\parallel AB$, \\\\\nit implies that quadrilateral AMPN is a parallelogram, and $\\angle MPB=\\angle C$, \\\\\nHence, $AM=PN$, $MP=AN$, and $\\angle B=\\angle MPB$, \\\\\nTherefore, $MB=MP$, \\\\\nConsequently, the perimeter of quadrilateral AMPN is $AM+MP+PN+AN=2(AM+MP)=2(AM+MB)=2AB=\\boxed{20}$",
    "image_path": "dataset/mm_math/images/MM_Math/51968621.png"
  },
  {
    "solution": "\\boxed{y=-x+10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system $xOy$, point $A$ is on the positive half of the $y$-axis, and point $B$ is on the positive half of the $x$-axis, with $OA = OB = 10$. What is the equation of line $AB$?\n",
    "completion": "\\textbf{Solution:} Since point $A$ is on the positive half of the y-axis, and point $B$ is on the positive half of the x-axis, with $OA=OB=10$, \\\\\nthus $A\\left(0, 10\\right)$, $B\\left(10, 0\\right)$. \\\\\nLet the equation of line $AB$ be $y=kx+b$, \\\\\nhence $\\begin{cases}10k+b=0\\\\ b=10\\end{cases}$, solving gives: \\\\\n$\\begin{cases}k=-1\\\\ b=10\\end{cases}$, \\\\\nthus the equation of line $AB$ is $\\boxed{y=-x+10}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50493321_219.png"
  },
  {
    "solution": "\\boxed{36^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the figure, A, B, and C are three points on the circle $\\odot O$, with $\\angle AOB = 72^\\circ$. What is the measure of $\\angle ACB$?\n",
    "completion": "\\textbf{Solution:} We have $\\angle ACB=\\frac{1}{2}\\angle AOC=\\frac{1}{2}\\times 72^\\circ=\\boxed{36^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52837271.png"
  },
  {
    "solution": "\\boxed{1\\colon 3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle A'B'C'$ is obtained from $\\triangle ABC$ through a similarity transformation centered at point O. If $OA':AA = 1:2$, what is the ratio of the perimeter of $\\triangle A'B'C'$ to that of $\\triangle ABC$?\n",
    "completion": "\\textbf{Solution:} Given that $OA'\\colon AA=1\\colon 2$,\\\\\nit follows that $OA'\\colon OA=1\\colon 3$,\\\\\nsince $\\triangle A'B'C'$ is obtained from $\\triangle ABC$ through a similarity transformation centered at point O,\\\\\nit implies $A'B' \\parallel AB$ and $\\triangle A'B'C' \\sim \\triangle ABC$,\\\\\nthus $\\frac{A'B'}{AB}=\\frac{OA'}{OA}=\\frac{1}{3}$,\\\\\ntherefore, the ratio of the perimeter of $\\triangle A'B'C'$ to the perimeter of $\\triangle ABC$ is $\\boxed{1\\colon 3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51367505.png"
  },
  {
    "solution": "\\boxed{\\left(\\frac{3}{2}, -\\frac{15}{2}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Synthesis and Practice\\\\\nAs shown in the figure, the parabola $y=2x^2-4x-6$ intersects the x-axis at points $A$ and $B$, where point $A$ is to the left of point $B$, and intersects the y-axis at point $C$. Point $D$ is a moving point on the parabola. As shown in Figure 2, when point $D$ is in the fourth quadrant, connect $BD$, $CD$, and $BC$ to form $\\triangle BCD$. When the area of $\\triangle BCD$ is maximized, what are the coordinates of point $D$?\n",
    "completion": "Solution: Let the coordinates of point D be $\\left(m, 2m^{2}-4m-6\\right)$.\n\nAs shown in the diagram, draw $DH \\perp x$-axis at point H, and draw $DG \\perp y$-axis at point G, then connect $OD$.\n\n$\\therefore DG=m$, $DH=-2m^{2}+4m+6$.\n\n$\\because$ The coordinates of point B are $\\left(3, 0\\right)$, and the coordinates of point C are $\\left(0, -6\\right)$.\n\n$\\therefore OB=3$, $OC=6$.\n\n$\\because S_{\\triangle BCD}=S_{\\triangle OCD}+S_{\\triangle OBD}-S_{\\triangle OBC}$,\n\n$\\therefore S_{\\triangle BCD}=\\frac{OC\\cdot DG}{2}+\\frac{OB\\cdot DH}{2}-\\frac{OC\\cdot OB}{2}$\n\n$S_{\\triangle BCD}=\\frac{6m}{2}+\\frac{3(-2m^{2}+4m+6)}{2}-\\frac{6\\times 3}{2}$\n\nSimplifying, we get $S_{\\triangle BCD}=-3\\left(m-\\frac{3}{2}\\right)^{2}+\\frac{27}{4}$.\n\n$\\because -3<0$,\n\n$\\therefore$ When $m=\\frac{3}{2}$, the maximum area of $\\triangle BCD$ is $\\frac{27}{4}$.\n\n$\\therefore 2m^{2}-4m-6=2\\times \\left(\\frac{3}{2}\\right)^{2}-4\\times \\frac{3}{2}-6=2\\times \\frac{9}{4}-6-6=-\\frac{15}{2}$.\n\n$\\therefore$ The coordinates of point D are $\\boxed{\\left(\\frac{3}{2}, -\\frac{15}{2}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52977819_206.png"
  },
  {
    "solution": "\\boxed{\\frac{5}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the grid shown, each small square has a side length of 1, and each vertex of the small squares is called a grid point. Points $O$, $A$, and $B$ are all grid points. If the sector $AOB$ in the figure is the lateral surface development of a cone, what is the radius of the base circle of the cone?\n",
    "completion": "\\textbf{Solution:} Let the radius of the circular base of the cone be $r$. \\\\\nSince each small square has a side length of 1, \\\\\nthus $OA=OB=\\sqrt{3^2+4^2}=5$, $AB=\\sqrt{1^2+7^2}=5\\sqrt{2}$ \\\\\ntherefore $OA^2+OB^2=AB^2$ \\\\\nthus $\\angle AOB=90^\\circ$, \\\\\nsince the sector AOB is a lateral surface of the cone unfolded, \\\\\nthus the arc $AB$ is equal to the circumference of the base circle. \\\\\nLet the radius of the base circle be $r$ \\\\\nthus $\\frac{90^\\circ \\pi \\times 5}{180}=2\\pi r$, \\\\\ntherefore $r=\\boxed{\\frac{5}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51366872.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points $B$ and $C$ divide the line segment $AD$ into three parts in the ratio $2:5:3$, with $M$ being the midpoint of $AD$. Given that $BM=6\\,\\text{cm}$, what is the length of $CM$ in centimeters?\n",
    "completion": "Solution: Let AB=2x cm, BC=5x cm, CD=3x cm,\\\\\n$\\therefore$AD=AB+BC+CD=10x cm,\\\\\n$\\because$ M is the midpoint of AD,\\\\\n$\\therefore$AM=MD= $\\frac{1}{2}$ AD=5x cm,\\\\\n$\\therefore$BM=AM−AB=5x−2x=3x cm,\\\\\n$\\because$BM=6 cm,\\\\\n$\\therefore$x=2 cm,\\\\\nhence CM=BC−BM=5×2−6=\\boxed{4}cm.",
    "image_path": "dataset/mm_math/images/MM_Math/52948987.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $AB$: $y=kx+b$ intersects with the linear function $y=2x$ at point $B$. What is the value of $k+b$?\n",
    "completion": "\\textbf{Solution:} Substitute $x=1$ into $y=2x$, to get $y=2\\times1=2$,\\\\\n$\\therefore B(1,2)$,\\\\\nSubstitute $x=1$, $y=2$ into $y=kx+b$, we get: $k+b=\\boxed{2}$",
    "image_path": "dataset/mm_math/images/MM_Math/50493369.png"
  },
  {
    "solution": "\\boxed{145^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\angle AOC$ and $\\angle BOD$ are both right angles. If $\\angle DOC = 35^\\circ$, then what is the degree measure of $\\angle AOB$?\n",
    "completion": "\\textbf{Solution:} We have $\\angle AOB = \\boxed{145^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51438537_64.png"
  },
  {
    "solution": "\\boxed{(0, 4)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the graph of the linear function $y=-2x$ intersects with the graph of the inverse proportion function $y=\\frac{k}{x}$ at point $A(-1, n)$. If $P$ is a point on the coordinate axis and satisfies $PA=OA$, write down the coordinates of point $P$ directly.\n",
    "completion": "\\textbf{Solution:} The coordinates of point $P$ are \\boxed{(-2, 0)} or \\boxed{(0, 4)}.",
    "image_path": "dataset/mm_math/images/MM_Math/51402287_58.png"
  },
  {
    "solution": "\\boxed{26}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, D and E are the midpoints of AB and AC, respectively. Extend DE to the point F such that EF=DE. Given AB=10 and BC=8, what is the perimeter of the quadrilateral BCFD?\n",
    "completion": "\\textbf{Solution:} Since $D$ and $E$ are the midpoints of $AB$ and $AC$ respectively,\\\\\nit follows that $DE \\parallel BC$ and $DE = \\frac{1}{2} BC$, $BD = \\frac{1}{2} AB = 5$,\\\\\nsince $DE = EF$,\\\\\nit follows that $DE + EF = DF = BC$,\\\\\ntherefore, quadrilateral $DBCF$ is a parallelogram,\\\\\nhence, the perimeter of quadrilateral $BCFD$ is $2(BD + DF) = 2(5 + 8) = \\boxed{26}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51510247.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, water is sprayed outwards from the window $A$ of a building, forming a parabolic trajectory (the plane of the parabola is perpendicular to the wall). The height of point $A$ from the ground is $6$ meters. The highest point of the parabola, $P$, is $2$ meters away from the wall in the perpendicular direction, and $8$ meters away from the ground in the vertical direction. A Cartesian coordinate system is established as shown in the figure. Determine the maximum distance from the wall where the water hits the ground, in meters.\n",
    "completion": "\\textbf{Solution:} Let $y=0$, we get $-\\frac{1}{2}(x-2)^2+8=0$,\\\\\nSolving this gives: $x_1=6$, $x_2=-2$ (discard),\\\\\n$\\therefore$ Hence, the furthest distance the water falls from the wall is $\\boxed{6}$ meters.",
    "image_path": "dataset/mm_math/images/MM_Math/51446099_265.png"
  },
  {
    "solution": "\\boxed{0<x\\leq 1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line passing through the point $(3,0)$: $y=-x+b$ intersects with the line: $y=ax$ at point $P(n,2)$. What is the solution set of the system of inequalities $0<ax\\leq -x+b$?\n",
    "completion": "\\textbf{Solution:} Given the line passing through the point $(3,0)$ can be expressed as $y=-x+b$ and intersects with the line $y=ax$ at point $P(n,2)$,\\\\\n$\\therefore$ $-3+b=0$,\\\\\n$\\therefore$ $b=3$,\\\\\n$\\therefore$ $y=-x+3$,\\\\\n$\\therefore$ $2=-n+3$,\\\\\n$\\therefore$ $n=1$,\\\\\n$\\therefore$ $P(1,2)$,\\\\\nFrom the graph, the solution set of the system of inequalities $0<ax\\leq -x+b$ is $0<x\\leq 1$.\\\\\nHence, the answer is $\\boxed{0<x\\leq 1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52019291.png"
  },
  {
    "solution": "\\boxed{\\frac{2\\pi}{3}-\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, $\\odot O$ is the circumcircle of the right-angled triangle $ABC$, with the diameter $AC=4$. A tangent line is drawn through point $C$ to $\\odot O$, intersecting the extended line of $AB$ at point $D$. $M$ is the midpoint of $CD$, and lines $BM$ and $OM$ are drawn, with $BC$ and $OM$ intersecting at point $N$. $BM$ is tangent to $\\odot O$; when $\\angle BAC=60^\\circ$, what is the area of the shape enclosed by chord $AB$ and arc $AB$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle BAC=60^\\circ$ and $OA=OB$, \\\\\nit follows that $\\triangle ABO$ is an equilateral triangle, \\\\\nwhich implies that $\\angle AOB=60^\\circ$.\\\\\nGiven that $AC=4$, it follows that $OA=2$,\\\\\ntherefore, the area of the region bounded by chord $AB$ and arc $AB$ $=S_{\\text{sector}}AOB-S_{\\triangle}AOB$\\\\\n$=\\frac{60^\\circ\\pi \\times 2^2}{360^\\circ}-\\frac{\\sqrt{3}}{4}\\times 2^2=\\boxed{\\frac{2\\pi}{3}-\\sqrt{3}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51402523_135.png"
  },
  {
    "solution": "\\boxed{x>\\frac{4}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, line $l_1$: $y=2x-1$ intersects the x-axis and y-axis at points $A$ and $B$, respectively. Line $l_2$: $y=-\\frac{1}{2}x+1$ intersects the x-axis and y-axis at points $P$ and $C$, respectively. Connect $AC$, and lines $l_1$ and $l_2$ intersect at point $D$. Find the coordinates of point $D$, and directly write the solution set of the inequality $2x-1>-\\frac{1}{2}x+1$.\n",
    "completion": "\\textbf{Solution:} Solve the system of equations\n\\[\n\\begin{cases}\ny=2x-1\\\\\ny=-\\frac{1}{2}x+1\n\\end{cases}\n\\]\nto find\n\\[\n\\begin{cases}\nx=\\frac{4}{5}\\\\\ny=\\frac{3}{5}\n\\end{cases}\n\\]\n$\\therefore D\\left(\\frac{4}{5}, \\frac{3}{5}\\right)$\n\nFrom the graph of the functions, it is known that the solution set for $2x-1>-\\frac{1}{2}x+1$ is $\\boxed{x>\\frac{4}{5}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52386314_130.png"
  },
  {
    "solution": "\\boxed{(-1,9)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=-x^{2}+bx+c$ intersects the x-axis at points $A(2, 0)$ and $B(-4, 0)$. Determine the axis of symmetry and the coordinates of the vertex of the parabola.\n",
    "completion": "\\textbf{Solution:} Since $y=-x^2-2x+8=-\\left(x+1\\right)^2+9$,\\\\\ntherefore the axis of symmetry is the line $x=-1$, and the coordinates of the vertex are $\\boxed{(-1,9)}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52602758_130.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the four corners of the rectangular paper $ABCD$ are folded inward to precisely form a quadrilateral $EFGH$ with no gaps and no overlaps. If $EH=3$ and $EF=4$, what is the length of side $BC$?\n",
    "completion": "\\textbf{Solution:} We have $\\because \\angle HEM=\\angle AEH$ and $\\angle BEF=\\angle FEM$, \\\\\n$\\therefore \\angle HEF=\\angle HEM+\\angle FEM=\\frac{1}{2}\\times 180^\\circ=90^\\circ$, \\\\\nSimilarly, we find: $\\angle EHG=\\angle HGF=\\angle EFG=90^\\circ$, \\\\\n$\\therefore$ Quadrilateral $EFGH$ is a rectangle, \\\\\n$\\therefore GH=EF$ and $GH\\parallel EF$, \\\\\n$\\therefore \\angle GHN=\\angle EFM$, \\\\\nIn $\\triangle GHN$ and $\\triangle EFM$ we have\\\\\n$\\left\\lbrace\\begin{array}{l}\n\\angle GNH=\\angle EMF\\\\\n\\angle NHG=\\angle MFE\\\\\nHG=EF\n\\end{array}\\right.$, \\\\\n$\\therefore \\triangle GHN\\cong \\triangle EFM(AAS)$, \\\\\n$\\therefore HN=MF=HD$, \\\\\n$\\therefore AD=AH+HD=HM+MF=HF$, \\\\\n$HF=\\sqrt{EH^2+EF^2}=\\sqrt{3^2+4^2}=5$, \\\\\n$\\therefore BC=AD=\\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53308051.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in rectangle $ABCD$, it is known that $AB=8\\,cm$, $BC=10\\,cm$. Fold $AD$ along the straight line $AE$, making point D land on point F which is on side $BC$. What is the length of $CF$ in $cm$?\n",
    "completion": "\\textbf{Solution:} In the rectangle $ABCD$, \\\\\n$\\because$ $AB=8\\text{cm}$, $BC=10\\text{cm}$, let $CF=x\\text{cm}$, \\\\\n$\\therefore$ $AD=BC=10\\text{cm}$, $BF=(10-x)\\text{cm}$, $\\angle B=90^\\circ$, \\\\\n$\\because$ folding $AD$ along the line $AE$, such that point $D$ falls on point $F$ on side $BC$, \\\\\n$\\therefore$ $\\triangle ADE \\cong \\triangle AFE$, \\\\\n$\\therefore$ $AD=AF=10\\text{cm}$, \\\\\nIn $\\triangle ABF$, by the Pythagorean theorem, $AB^2+BF^2=AF^2$, \\\\\n$\\therefore$ $8^2+(10-x)^2=10^2$, solving for $x$, we find $x=\\boxed{4}$, \\\\\n$\\therefore$ $CF=4\\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52731253.png"
  },
  {
    "solution": "\\boxed{x > -1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $l_{1}: y_{1}=ax+b$ and the line $l_{2}: y_{2}=mx$ are depicted in the same Cartesian coordinate system. What is the solution set of the inequality $mx<ax+b$ with respect to $x$?\n",
    "completion": "\\textbf{Solution:} From the graph, it is known that the two lines intersect at the point $(-1,2)$, \\\\\nwhen $x > -1$, $mx < ax+b$,\\\\\n$\\therefore$ the solution set is $\\boxed{x > -1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52212195.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $ \\triangle ABC \\cong \\triangle DEF$. If $ DF = 6$, $ AB = 3$, $ EF = 5$, and $ DC = 4$, what is the length of $ AD$?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABC \\cong \\triangle DEF$,\\\\\ntherefore $AC = DF = 6$,\\\\\nthus $AD = AC - DC = 6 - 4 = \\boxed{2}$",
    "image_path": "dataset/mm_math/images/MM_Math/52978387.png"
  },
  {
    "solution": "\\boxed{\\left(0, \\frac{12\\sqrt{5}}{5}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that O is the origin. The coordinate of point A is $(6, 0)$. Point B is a moving point on the positive half-axis of the y-axis. Rectangle OBCA is formed with OB and OA as sides. OC is the diagonal of rectangle OBCA. OE bisects $\\angle BOC$ and intersects BC at point E. CF bisects $\\angle ACO$ and intersects OA at point F. Through point E, draw $EG \\perp OC$, with G as the foot of the perpendicular. Through point F, draw $FH \\perp OC$, with H as the foot of the perpendicular. When points G and H trisect the diagonal OC, find the coordinates of point B.\n",
    "completion": "\\textbf{Solution:} Given that $\\because$OE bisects $\\angle BOC$, $EG\\perp OC$, $EB\\perp OB$,\\\\\n$\\therefore \\angle BOE=\\angle GOE$, $\\angle OBE=\\angle OGE=90^\\circ$,\\\\\nAlso $\\because OE=OE$,\\\\\n$\\therefore \\triangle OEB\\cong \\triangle OEG$,\\\\\n$\\therefore OG=OB$. Similarly, $CH=CA$. Since $OB=CA$,\\\\\n$\\therefore OG=CH$,\\\\\nWhen point G is between points O and H, as shown in figure 5:\\\\\n$\\because$Points G, H trisect the diagonal OC,\\\\\n$\\therefore AC=OG=GH=HC$. Let $AC=m$, then $OC=3m$,\\\\\nIn $\\triangle OAC$ with a right angle at $A$, $OA=6$,\\\\\n$\\because AC^2+OA^2=OC^2$,\\\\\n$\\therefore m^2+6^2=(3m)^2$, solving this gives $m=\\frac{3\\sqrt{2}}{2}$,\\\\\n$\\therefore OB=AC=\\frac{3\\sqrt{2}}{2}$,\\\\\n$\\therefore$The coordinates of point B are $\\left(0, \\frac{3\\sqrt{2}}{2}\\right)$;\\\\\nWhen point H is between points O and G, as shown in figure 6,\\\\\nSimilarly, it follows that $OG=CH=AC$. Let $OH=n$, then $AC=HC=2n$,\\\\\nIn $\\triangle OAC$ with a right angle at $A$, $OA=6$,\\\\\n$\\because AC^2+OA^2=OC^2$,\\\\\n$\\therefore (2n)^2+6^2=(3n)^2$, solving this gives $n=\\frac{6\\sqrt{5}}{5}$,\\\\\n$\\therefore AC=OB=\\frac{12\\sqrt{5}}{5}$,\\\\\n$\\therefore$The coordinates of point B are $\\left(0, \\frac{12\\sqrt{5}}{5}\\right)$,\\\\\n$\\therefore$The coordinates of point B that satisfy the conditions are $\\boxed{\\left(0, \\frac{3\\sqrt{2}}{2}\\right)}$ or $\\boxed{\\left(0, \\frac{12\\sqrt{5}}{5}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52212635_94.png"
  },
  {
    "solution": "\\boxed{8\\pi}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, AB is the diameter of the circle $\\odot O$, and the quadrilateral ABCD is inscribed in $\\odot O$. If $BC=CD=DA=4\\,cm$, then what is the circumference of $\\odot O$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect OD and OC.\\\\\nSince AB is the diameter of $\\odot O$, and quadrilateral ABCD is inscribed in $\\odot O$, if BC = CD = DA = 4cm,\\\\\nthen $\\overset{\\frown}{AD}$ = $\\overset{\\frown}{CD}$ = $\\overset{\\frown}{BC}$,\\\\\nthus $\\angle AOD = \\angle DOC = \\angle BOC = 60^\\circ$.\\\\\nAlso, since OA = OD,\\\\\nthus $\\triangle AOD$ is an equilateral triangle,\\\\\ntherefore OA = AD = 4cm,\\\\\nthus the circumference of $\\odot O$ = 2×4$\\pi$ = $\\boxed{8\\pi}$ (cm).",
    "image_path": "dataset/mm_math/images/MM_Math/53009943.png"
  },
  {
    "solution": "\\boxed{x<-1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given the diagram, in the Cartesian coordinate system, the graph of the function $y=-3x$ intersects with the graph of $y=kx+b$ $(k>0)$ at point $P(m, 3)$. What is the solution set of the inequality $kx+b<-3x$?\n",
    "completion": "\\textbf{Solution:} Solve: when $y=3$, we have $-3x=3$,\\\\\nSolving this, we get $x=-1$,\\\\\nFrom the graph, the solution set of the inequality $kx+b<-3$ is: $\\boxed{x<-1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52211719.png"
  },
  {
    "solution": "\\boxed{\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=kx+b$ intersects the $x$-axis at point $A(4,0)$ and intersects the $y$-axis at point $B(0,8)$. If point $P(2,m)$ and point $Q(n,2)$ are two points on line $AB$, what is the length of the segment $\\overline{PQ}$?\n",
    "completion": "\\textbf{Solution:} Given the problem statement,\\\\\n$\\because$ the point P(2, m) lies on the line $y=-2x+8$,\\\\\n$\\therefore m=-2\\times 2+8=4$,\\\\\n$\\therefore$ the coordinates of point P are (2, 4);\\\\\n$\\because$ the point Q(n, 2) lies on the line $y=-2x+8$,\\\\\n$\\therefore -2n+8=2$, solving for $n$ yields $n=3$,\\\\\n$\\therefore$ the coordinates of point Q are (3, 2),\\\\\n$\\therefore PQ=\\sqrt{{(2-3)}^2+{(4-2)}^2}=\\boxed{\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52342732_120.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{10}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, it is known that the radius of $\\odot O$ is 5, and the chord $AB = 4\\sqrt{5}$. Points C and D are on the major arc $\\overset{\\frown}{BCA}$ (points B, C, D are arranged clockwise), $\\overset{\\frown}{CD} = \\overset{\\frown}{AB}$. Point F is a moving point on $\\overset{\\frown}{CD}$. Connect CF and extend it to intersect the extension line of chord AD at point E. $BC \\parallel AD$. As shown in Figure 3, when $\\angle B = 90^\\circ$ and $CE = 5\\sqrt{5}$, connect AC, OE, and let OE intersect $\\odot O$ at point K. Calculate the ratio of the area of $\\triangle KDE$ to the area of $\\triangle EOC$.\n",
    "completion": "\\textbf{Solution:} In Figure 3, connect $CD$,\\\\\n$\\because$ $\\angle B=90^\\circ$,\\\\\n$\\therefore$ $AC$ is the diameter of $\\odot O$,\\\\\n$\\therefore$ $\\angle ADC=90^\\circ$, $AC=2AO=10$,\\\\\nIn $Rt\\triangle CDE$, $CE=5\\sqrt{5}$, $CD=AB=4\\sqrt{5}$,\\\\\n$\\therefore$ $DE=\\sqrt{CE^{2}-CD^{2}}=3\\sqrt{5}$,\\\\\nIn $Rt\\triangle ADC$, $AD=\\sqrt{AC^{2}-CD^{2}}=2\\sqrt{5}$,\\\\\n$\\therefore$ $AE=AD+DE=5\\sqrt{5}$,\\\\\n$\\therefore$ $AE=CE$, thus $\\triangle AEC$ is an isosceles triangle,\\\\\n$\\because$ $AO=OC$,\\\\\n$\\therefore$ $OE\\perp AC$,\\\\\nIn $Rt\\triangle AOE$, $OE=\\sqrt{AE^{2}-OA^{2}}=\\sqrt{(5\\sqrt{5})^{2}-5^{2}}=10$,\\\\\n$\\therefore$ $KE=OE-OK=5$,\\\\\nDraw $KM\\perp DE$ at M through point K, then $\\angle KME=\\angle AOE=90^\\circ$, and also $\\angle KEM=\\angle AEO$,\\\\\n$\\therefore$ $\\triangle KME\\sim \\triangle AOE$,\\\\\n$\\therefore$ $\\frac{KM}{AO}=\\frac{KE}{AE}$, that is $\\frac{KM}{5}=\\frac{5}{5\\sqrt{5}}$,\\\\\nSolving for $KM$, we get: $KM=\\sqrt{5}$,\\\\\n$\\therefore$ $\\frac{S_{\\triangle KDE}}{S_{\\triangle EOC}}=\\frac{\\frac{1}{2}DE\\cdot KM}{\\frac{1}{2}OC\\cdot OE}=\\frac{3\\sqrt{5}\\times \\sqrt{5}}{5\\times 10}=\\boxed{\\frac{3}{10}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52933588_183.png"
  },
  {
    "solution": "\\boxed{14}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, where $AB=AC=10$ and $BC=8$, the line $AD$ bisects $\\angle BAC$ and intersects $BC$ at point $D$. Point $E$ is the midpoint of $AC$. After connecting $DE$, what is the perimeter of $\\triangle CDE$?\n",
    "completion": "\\textbf{Solution:} Given that $AB=AC$ and $AD$ bisects $\\angle BAC$, with $BC=8$, \\\\\nit follows that $AD\\perp BC$, and $CD=BD= \\frac{1}{2} BC=4$, \\\\\nsince point E is the midpoint of AC, \\\\\nthen $DE=CE= \\frac{1}{2} AC=5$, \\\\\nhence, the perimeter of $\\triangle CDE$ is $CD+DE+CE=4+5+5=\\boxed{14}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52807500.png"
  },
  {
    "solution": "\\boxed{50^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC \\cong \\triangle ADE$, the extension of $BC$ intersects $DE$ at point $F$, $\\angle B=30^\\circ$, $\\angle AED=110^\\circ$, $\\angle DAC=10^\\circ$, what is the measure of $\\angle DFB$?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABC \\cong \\triangle ADE$,\\\\\nit follows that $\\angle AED = \\angle ACB = 110^\\circ$, and $\\angle D = \\angle B = 30^\\circ$,\\\\\ntherefore $\\angle ACF = 180^\\circ - 110^\\circ = 70^\\circ$,\\\\\nGiven the sum of angles in a triangle is $180^\\circ$, we have: $\\angle DAC + \\angle ACF = \\angle D + \\angle DFC$,\\\\\nthus $\\angle DFC = 70^\\circ + 10^\\circ - 30^\\circ = \\boxed{50^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/53033279.png"
  },
  {
    "solution": "\\boxed{62^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points $A$, $O$, and $B$ are on the same line, $OC$ bisects $\\angle AOB$. If $\\angle COD = 28^\\circ$, what is the degree measure of $\\angle BOD$?\n",
    "completion": "\\textbf{Solution:} Since points A, O, and B lie on the same line, and $OC$ bisects $\\angle AOB$,\\\\\n$\\therefore$ $\\angle AOC=\\angle BOC=90^\\circ$,\\\\\n$\\therefore$ $\\angle BOD=\\angle BOC-\\angle COD=90^\\circ-28^\\circ=\\boxed{62^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53211413_117.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\square ABCD$, $EF\\parallel AB$, $DE:EA=2:3$, and $EF=4$, what is the length of $CD$?\n",
    "completion": "\\textbf{Solution:} Given $EF \\parallel AB$,\\\\\nit follows that $\\angle DEF = \\angle A$, $\\angle DFE = \\angle DBA$,\\\\\nwhich means $\\triangle DEF \\sim \\triangle DAB$.\\\\\nGiven $DE : EA = 2 : 3$,\\\\\nit implies $EF : AB = DE : DA = 2 : 5$,\\\\\nand since $EF = 4$,\\\\\nwe find that $AB = 10$.\\\\\nTherefore, in quadrilateral $ABCD$, $CD = AB = \\boxed{10}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51379199.png"
  },
  {
    "solution": "\\boxed{B(2,3)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the Cartesian coordinate system, the parabola $y=x^2+(k-1)x-k$ intersects the line $y=kx+1$ at points $A$ and $B$, where point $A$ is to the left of point $B$. As shown in Fig. 1, when $k=1$, directly write down the coordinates of points $A$ and $B$.\n",
    "completion": "\\textbf{Solution}: Let $\\boxed{A(-1,0)}$, $\\boxed{B(2,3)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53009969_92.png"
  },
  {
    "solution": "\\boxed{70\\,cm^{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, the ratio $AE:EB=2:3$, and $DE$ intersects $AC$ at $F$. If the area of $\\triangle CDF$ is $25\\,cm^{2}$, what is the area of quadrilateral $ABCD$?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle AEF\\sim \\triangle CDF$, \\\\\nit follows that $\\frac{AF}{CF}=\\frac{AE}{CD}=\\frac{2}{5}$, \\\\\nthus, $\\frac{{S}_{\\triangle AFD}}{{S}_{\\triangle CDF}}=\\frac{AF}{CF}=\\frac{2}{5}$, \\\\\nGiven that the area of $\\triangle CDF$ is $25\\,cm^{2}$, \\\\\nhence, ${S}_{\\triangle AFD}=\\frac{2}{5}{S}_{\\triangle CDF}=\\frac{2}{5}\\times 25=10\\,cm^{2}$; \\\\\nTherefore, ${S}_{\\triangle ACD}=10+25=35\\,cm^{2}$, \\\\\nConsequently, ${S}_{\\square ABCD}=2{S}_{\\triangle ACD}=2\\times 35=70\\,cm^{2}$. \\\\\nHence, the area of quadrilateral $ABCD$ is $\\boxed{70\\,cm^{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52917377_70.png"
  },
  {
    "solution": "\\boxed{5\\text{cm}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In right triangle $ABC$, $\\angle ACB=90^\\circ$, $BC=2\\,\\text{cm}$, $CD\\perp AB$. Point $E$ is taken on $AC$ such that $EC=2\\,\\text{cm}$. A line through point $E$ and perpendicular to $AC$ intersects the extension of $CD$ at point $F$. If $AE=3\\,\\text{cm}$, what is the length of $EF$ in centimeters?\n",
    "completion": "\\textbf{Solution:}\\\\\n$\\because \\angle ACB=90^\\circ$\\\\\n$\\therefore \\angle ECF+\\angle BCD=90^\\circ$\\\\\n$\\because CD\\perp AB$\\\\\n$\\therefore \\angle BCD+\\angle B=90^\\circ$\\\\\n$\\therefore \\angle ECF=\\angle B$\\\\\nIn $\\triangle ABC$ and $\\triangle FEC$\\\\\n$\\because \\angle ECF=\\angle B$, $EC=BC$, $\\angle ACB=\\angle FEC=90^\\circ$\\\\\n$\\therefore \\triangle ABC \\cong \\triangle FCE$ (ASA)\\\\\n$\\therefore AC=EF$\\\\\n$\\because AC=AE+CE=3+2=5\\text{cm}$,\\\\\n$\\therefore EF=\\boxed{5\\text{cm}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51366532.png"
  },
  {
    "solution": "\\boxed{4\\sin47^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the barrier at the entrance of a parking lot rotates from the horizontal position $AB$ around point $O$ to the position of $AB$. Given that $AO=4$ meters and the rotation angle $\\angle AOA=47^\\circ$, what is the vertical distance $AH$ that the endpoint $A$ rises?\n",
    "completion": "\\textbf{Solution:} Refer to the diagram, draw $A^{\\prime}H\\perp AB$ at $H$ through point $A^{\\prime}$,\\\\\naccording to the problem, we get $OA^{\\prime}=OA=4$ meters,\\\\\nin $\\triangle OA^{\\prime}H$, $\\angle A^{\\prime}OH=47^\\circ$, $\\sin\\angle AOH=\\frac{AH}{OA}$,\\\\\n$\\therefore$ the vertical distance the railing's endpoint $A$ rises is $AH=\\sin\\angle AOH\\cdot OA=4\\sin47^\\circ$ meters,\\\\\nthus, $AH=\\boxed{4\\sin47^\\circ}$ meters.",
    "image_path": "dataset/mm_math/images/MM_Math/51351226.png"
  },
  {
    "solution": "\\boxed{2.5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in square $ABCD$, $G$ is a point on diagonal $BD$, $GE \\perp CD$, $GF \\perp BC$, with $E$ and $F$ being the respective foot of the perpendiculars. Connect $EF$. Let $M$ and $N$ be the midpoints of $AB$ and $BG$ respectively. If $EF=5$, what is the length of $MN$?\n",
    "completion": "\\textbf{Solution:} Connect AG and CG, \\\\\n$\\because$ In the square ABCD, $\\angle$BCD$=90^\\circ$, \\\\\n$\\because$ GE$\\perp$CD, GF$\\perp$BC, \\\\\n$\\therefore$ Quadrilateral CFGE is a rectangle, \\\\\n$\\therefore$ CG$=EF=5$, \\\\\n$\\because$ AB$=BC$, $\\angle$ABD$=\\angle$CBD$=45^\\circ$, \\\\\n$\\because$ BG$=BG$, \\\\\n$\\therefore$ $\\triangle ABG\\cong \\triangle CBG$ (SAS), \\\\\n$\\therefore$ AG$=CG=5$, \\\\\n$\\because$ M and N are respectively the midpoints of AB and BG, \\\\\n$\\therefore$ MN$=\\frac{1}{2}AG=\\boxed{2.5}$, \\\\",
    "image_path": "dataset/mm_math/images/MM_Math/53729285.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\angle AOB=90^\\circ$, $OM$ bisects $\\angle AOC$, and $ON$ bisects $\\angle BOC$. If $\\angle BOC=30^\\circ$, what is the measure of $\\angle MON$?\n",
    "completion": "\\textbf{Solution}: Given that, \\\\\n$\\because \\angle AOB=90^\\circ, \\angle BOC=30^\\circ$,\\\\\n$\\therefore \\angle AOC=\\angle AOB+\\angle BOC=120^\\circ$.\\\\\n$\\because OM$ bisects $\\angle AOC, ON$ bisects $\\angle BOC$,\\\\\n$\\therefore \\angle MOC=\\frac{1}{2}\\angle AOC=120^\\circ \\times \\frac{1}{2}=60^\\circ, \\angle NOC=\\frac{1}{2}\\angle BOC=\\frac{1}{2}\\times 30^\\circ=15^\\circ$.\\\\\n$\\therefore \\angle MON=\\angle MOC-\\angle NOC=\\boxed{45^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53365775_78.png"
  },
  {
    "solution": "\\boxed{\\frac{4\\sqrt{3}}{7}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, we have $AB=5$, $AC=8$, and $\\angle A=60^\\circ$. Find $\\sin B$.\n",
    "completion": "\\textbf{Solution:} In $\\triangle BCD$,\\\\\n$\\because$ from (1) we know that $CD=4\\sqrt{3}$ and $BC=7$,\\\\\n$\\therefore$ $\\sin B=\\frac{CD}{BC}=\\frac{4\\sqrt{3}}{7}=\\boxed{\\frac{4\\sqrt{3}}{7}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52602114_148.png"
  },
  {
    "solution": "\\boxed{\\frac{4\\sqrt{5}}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the right-angled triangle $\\mathrm{Rt}\\triangle ABO$, $\\angle O=90^\\circ$. A circle with point $O$ as its center and $OB$ as its radius intersects $AB$ at point $C$ and intersects $OA$ at point $D$. If $BO=2$ and $AO=4$, find the length of $BC$.\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $OQ\\perp AB$ at Q, and $\\angle AOB=90^\\circ$,\\\\\n$\\therefore \\angle BOQ+\\angle B=90^\\circ=\\angle B+\\angle A=90^\\circ$,\\\\\n$\\therefore \\angle BOQ=\\angle A$,\\\\\n$\\because OB=2$, $AO=4$,\\\\\n$\\therefore AB=\\sqrt{2^2+4^2}=2\\sqrt{5}$,\\\\\n$\\therefore \\sin A=\\frac{OB}{AB}=\\frac{2}{2\\sqrt{5}}=\\frac{\\sqrt{5}}{5}$,\\\\\n$\\therefore \\sin\\angle BOQ=\\frac{BQ}{OB}=\\frac{BQ}{2}=\\frac{\\sqrt{5}}{5}$,\\\\\n$\\therefore BQ=\\frac{2\\sqrt{5}}{5}$,\\\\\n$\\because OQ\\perp AB$,\\\\\n$\\therefore BC=2BQ=\\boxed{\\frac{4\\sqrt{5}}{5}}.$",
    "image_path": "dataset/mm_math/images/MM_Math/51029323_110.png"
  },
  {
    "solution": "\\boxed{2.5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the rhombus $ABCD$, the diagonals $AC$ and $BD$ intersect at point $O$, and $E$ is the midpoint of $AB$, with $AC=8$ and $BD=6$. What is the length of $OE$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rhombus, \\\\\n$\\therefore$AC$\\perp$BD, AO=CO, DO=BO, \\\\\nSince AC$=8$, BD$=6$, \\\\\n$\\therefore$AO$= \\frac{1}{2}AC=\\frac{1}{2}\\times 8=4$, OD$= \\frac{1}{2}BD=\\frac{1}{2}\\times 6=3$, \\\\\nIn $\\triangle AOD$, AD$= \\sqrt{OD^{2}+OA^{2}}=\\sqrt{3^{2}+4^{2}}=5$, \\\\\nAlso, since point E is the midpoint of side AB, and O is the midpoint of BD, \\\\\n$\\therefore$OE$= \\frac{1}{2}AB=\\frac{1}{2}\\times 5=\\frac{5}{2}=\\boxed{2.5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53183009.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, an equilateral triangle $ABC$ with a side length of 4 is translated along ray $BC$ to obtain triangle $DEF$. Points $M$ and $N$ are the midpoints of $AC$ and $DF$, respectively. Point $P$ is the midpoint of segment $MN$. When connecting $PA$ and $PC$, if triangle $APC$ forms a right-angled triangle, what is the length of $BE$?\n",
    "completion": "\\textbf{Solution:} As shown in Figure 1, when $\\angle APC=90^\\circ$, \\\\\n$\\because$AM=MC, AC=4,\\\\\n$\\therefore$PM is the median to the hypotenuse of $\\triangle APC$,\\\\\n$\\therefore$AM=CM=PM=2,\\\\\n$\\therefore$PN=2,\\\\\n$\\therefore$MN=4,\\\\\nThus, translating $\\triangle ABC$ 4 units to the right is sufficient,\\\\\n$\\therefore$BE=\\boxed{4};\\\\\nAs shown in Figure 2, when $\\angle ACP=90^\\circ$,\\\\\n$\\because$$\\triangle ABC$ is an equilateral triangle, AM=MC,\\\\\n$\\therefore$$\\angle BMC=90^\\circ$,\\\\\n$\\therefore$$\\angle BMC=\\angle ACP$,\\\\\n$\\therefore$BM$\\parallel$CP,\\\\\n$\\because$$\\triangle ABC$ is an equilateral triangle, $\\triangle DEF$ is an equilateral triangle, M and N are the midpoints of AC and DF respectively,\\\\\n$\\therefore$$\\angle ACB=\\angle DFE= 60^\\circ$, CM=NF,\\\\\n$\\therefore$MC$\\parallel$NF,\\\\\n$\\therefore$Quadrilateral MCFN is a parallelogram,\\\\\n$\\therefore$MP$\\parallel$BC,\\\\\n$\\therefore$Quadrilateral BCPM is a parallelogram,\\\\\n$\\therefore$PM=4,\\\\\n$\\therefore$PN=4,\\\\\n$\\therefore$MN=8,\\\\\nThus, translating $\\triangle ABC$ 8 units to the right is sufficient,\\\\\n$\\therefore$BE=\\boxed{8};",
    "image_path": "dataset/mm_math/images/MM_Math/52091240.png"
  },
  {
    "solution": "\\boxed{-\\frac{3}{2}\\le m \\le 3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the line $l_{1}$ intersects the x-axis at point $A(-3,0)$ and intersects the y-axis at point $B$. Additionally, it intersects line $l_{2}\\colon y=\\frac{4}{3}x$ at point $C(a,4)$. The line $l_{1}$ is translated 3 units downward along the y-axis to obtain line $l_{3}$. Point $P(m,n)$ is a moving point on line $l_{2}$, from which a perpendicular to the x-axis is drawn, intersecting lines $l_{1}$ and $l_{3}$ at points $M$ and $N$ respectively. When point $P$ is on the segment $MN$, determine the range of values for $m$.\n",
    "completion": "\\textbf{Solution:} Since the line $l_{1}$ is translated 3 units downwards along the y-axis to obtain the line $l_{3}$,\\\\\ntherefore, the equation of the line $l_{3}$ is $y=\\frac{2}{3}x+2-3=\\frac{2}{3}x-1$\\\\\nSolving for: $\\left\\{\\begin{array}{l}\ny=\\frac{2}{3}x-1\\\\ \ny=\\frac{4}{3}x\n\\end{array}\\right.$\\\\\nYields: $\\left\\{\\begin{array}{l}\nx=-\\frac{3}{2}\\\\ \ny=-2\n\\end{array}\\right.$\\\\\nTherefore, the intersection point of line $l_{2}$ and line $l_{3}$ is $\\left(-\\frac{3}{2}, -2\\right)$;\\\\\nSince the intersection point of line $l_{2}$ and line $l_{3}$ is given as $(3, 4)$,\\\\\ntherefore, the range of values for $m$ is: \\boxed{-\\frac{3}{2}\\le m \\le 3}",
    "image_path": "dataset/mm_math/images/MM_Math/52226206_107.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $BC=20$, and $D$, $E$ are the midpoints of $AB$, $AC$, respectively. What is the length of $DE$?\n",
    "completion": "\\textbf{Solution:} Since $D$ and $E$ are the midpoints of $AB$ and $AC$ respectively,\\\\\ntherefore, $DE$ is the mid-segment of $\\triangle ABC$,\\\\\ntherefore, $DE = \\frac{1}{2}BC$,\\\\\nsince $BC = 20$,\\\\\ntherefore, $DE = \\frac{1}{2} \\times 20 = \\boxed{10}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52204217.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, within a grid formed by small squares, the vertices of $\\triangle ABC$ are all lattice points (intersections of the grid lines). What is the value of $\\tan\\angle ABC$?\n",
    "completion": "\\textbf{Solution}: As shown in the figure, draw AD$\\perp$BC at D. \\\\\nIn $\\triangle ABD$, $\\tan\\angle ABC=\\frac{AD}{BD}=\\frac{1}{2}$, \\\\\nthus, $\\tan\\angle ABC=\\boxed{\\frac{1}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53102359.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $O$ is a point on line $AB$, $OM$ and $ON$ are on opposite sides of line $AB$, and $\\angle MON=90^\\circ$, $OE$ bisects $\\angle MOB$, $OF$ bisects $\\angle AON$. Given $\\angle BOM=150^\\circ$, what are the degrees of $\\angle BOE$ and $\\angle NOF$?\n",
    "completion": "\\textbf{Solution:} \\\\\nSince $OE$ bisects $\\angle MOB$ and $\\angle BOM=150^\\circ$,\\\\\nwe have $\\angle BOE=\\frac{1}{2}\\angle BOM=\\frac{1}{2}\\times 150^\\circ=\\boxed{75^\\circ}$;\\\\\nSince $\\angle AOM=180^\\circ-\\angle BOM=180^\\circ-150^\\circ=30^\\circ$,\\\\\nand since $\\angle MON=90^\\circ$,\\\\\nwe get $\\angle AON=\\angle MON-\\angle AOM=90^\\circ-30^\\circ=60^\\circ$,\\\\\nSince $OF$ bisects $\\angle AON$,\\\\\ntherefore, $\\angle NOF=\\frac{1}{2}\\angle AON=\\frac{1}{2}\\times 60^\\circ=\\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52948847_171.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, segment $AB=16\\, \\text{cm}$. A point $C$ is taken on $AB$, where $M$ is the midpoint of $AB$ and $N$ is the midpoint of $AC$. If $MN=3\\, \\text{cm}$, what is the length of segment $AC$?\n",
    "completion": "\\textbf{Solution:} Let $CM = a$, and $CN = CM + MN = a + 3$, \\\\\nSince $M$ is the midpoint of $AB$, and $N$ is the midpoint of $AC$, \\\\\nTherefore, $AM = \\frac{1}{2}AB = \\frac{1}{2} \\times 16 = 8$, and $AN = CN = a + 3$, \\\\\nSince $AM = AN + MN = 8$, this implies $a + 3 + 3 = 8$, \\\\\nTherefore, $a = 2$, \\\\\nTherefore, $AC = AM + CM = 8 + 2 = \\boxed{10}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53210764.png"
  },
  {
    "solution": "\\boxed{(-2, 0) \\cup (3, +\\infty)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the Cartesian coordinate system, the line $y_1=k_1x+b$ intersects the hyperbola $y_2=\\frac{k_2}{x}$ at points $A(-2, 3)$ and $B(m, -2)$. Based on the graph of the function, directly write the solution set of the inequality $k_1x+b<\\frac{k_2}{x}$ with respect to $x$.\n",
    "completion": "Solution:\nAccording to the graph of the function, we can directly write the solution set of the inequality $k_1x + b < \\frac{k_2}{x}$ with respect to $x$ as $-2 < x < 0$ or $x > 3$.\n\nHence, the answer is $x \\in \\boxed{(-2, 0) \\cup (3, +\\infty)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53043125_138.png"
  },
  {
    "solution": "\\boxed{20}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points $E$, $F$, $G$, and $H$ are the midpoints of the sides of square $ABCD$, and lines $BE$, $DG$, $CF$, $AH$ are connected. If $AB=10$, what is the area of quadrilateral $MNPQ$?\n",
    "completion": "\\textbf{Solution:} Given that points $E$, $F$, $G$, $H$ are midpoints of the sides of the square $ABCD$,\\\\\n$\\therefore DE=BG=DH=CG=\\frac{1}{2}AB$, and $DE\\parallel BG$,\\\\\n$\\therefore$ quadrilateral $DEBG$ is a parallelogram,\\\\\n$\\therefore BE\\parallel DG$,\\\\\n$\\therefore AM=QM$,\\\\\nSimilarly, $DQ=PQ$ can be deduced,\\\\\nIn $\\triangle ADH$ and $\\triangle DCG$,\\\\\n$\\left\\{\\begin{array}{l}AD=DC\\\\ \\angle ADH=\\angle DCG\\\\ DH=CG\\end{array}\\right.$,\\\\\n$\\therefore \\triangle ADH \\cong \\triangle DCG\\ (SAS)$,\\\\\n$\\therefore \\angle DAH=\\angle CDG$,\\\\\n$\\because \\angle CDG+\\angle ADQ=90^\\circ$,\\\\\n$\\therefore \\angle DAH+\\angle ADQ=90^\\circ$,\\\\\n$\\therefore \\angle AQD=90^\\circ$,\\\\\nSimilarly, it can be deduced that $\\angle BMA=\\angle CNB=\\angle DPC=90^\\circ$,\\\\\n$\\therefore$ quadrilateral $MNPQ$ is a rectangle,\\\\\nIn $\\triangle ADQ$ and $\\triangle DCP$,\\\\\n$\\left\\{\\begin{array}{l}\\angle AQD=\\angle DPC\\\\ \\angle DAQ=\\angle CDP\\\\ AD=DC\\end{array}\\right.$,\\\\\n$\\therefore \\triangle ADQ \\cong \\triangle DCP\\ (AAS)$,\\\\\n$\\therefore AQ=DP$,\\\\\n$\\therefore AM=MQ=DQ=PQ$,\\\\\n$\\therefore$ quadrilateral $MNPQ$ is a square,\\\\\nLet $MQ=x$, then $AQ=2x$, $DQ=x$,\\\\\nIn right triangle $\\triangle ADQ$, $x^2+(2x)^2=100$,\\\\\n$\\therefore x^2=20$,\\\\\n$\\therefore$ the area of quadrilateral $MNPQ$ is $\\boxed{20}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52204555.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $C$ is the midpoint of line segment $AB$, point $D$ is the midpoint of line segment $AC$, and $BE=\\frac{1}{3}BC$. If $BE=2\\,\\text{cm}$, what is the length of $AB$ in cm?\n",
    "completion": "\\textbf{Solution:} Since $BE=\\frac{1}{3}BC$ and $BE=2\\,\\text{cm}$, \\\\\nthen $BC=3BE=6\\,\\text{cm}$. \\\\\nGiven that point $C$ is the midpoint of line segment $AB$, \\\\\nit follows that $AB=2BC=\\boxed{12}\\,\\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53205976_50.png"
  },
  {
    "solution": "\\boxed{80^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $ \\angle 1 : \\angle 2 : \\angle 3 = 1 : 3 : 6$, what is the value of $ \\angle 4$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle 2\\colon \\angle 3=3\\colon 6$ and $\\angle 2+\\angle 3=180^\\circ$,\\\\\ntherefore, $\\angle 2=\\frac{3}{9}\\times 180^\\circ=60^\\circ$,\\\\\nhence, $\\angle 1=60^\\circ\\times \\frac{1}{3}=20^\\circ$,\\\\\nthus, $\\angle 4=180^\\circ-20^\\circ-80^\\circ=\\boxed{80^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/53147641.png"
  },
  {
    "solution": "\\boxed{\\sqrt{13}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In $\\triangle ABC$, $\\angle ABM = 45^\\circ$, $AM \\perp BM$ with the foot of the perpendicular at $M$, and point $C$ lies on the extension of $BM$. Connect $AC$. If $AB = \\sqrt{18}$ and $BC = 5$, what is the length of $AC$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle ABM=45^\\circ$ and $AM\\perp BM$,\\\\\nit follows that $AM=BM=\\frac{\\sqrt{2}}{2}AB=\\frac{\\sqrt{2}}{2}\\times 3\\sqrt{2}=3$,\\\\\nthus $CM=BC-BM=5-3=2$,\\\\\ntherefore $AC=\\sqrt{AM^2+CM^2}=\\sqrt{3^2+2^2}=\\sqrt{13}$;\\\\\nhence, the length of $AC$ is $\\boxed{\\sqrt{13}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52991780_114.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, an equilateral triangle $ABC$ is folded such that point $A$ falls on point $D$ on side $BC$ (not coinciding with $B$ or $C$), with the crease being $EF$. We have $\\triangle BDE \\sim \\triangle CFD$; if $BD=6$ and $DC=2$, calculate the perimeters of $\\triangle BDE$ and $\\triangle CFD$ respectively.\n",
    "completion": "\\textbf{Solution:} Given the problem, \\\\\n$\\because BD=6$ and $DC=2$, \\\\\n$\\therefore BC=BD+DC=8$, \\\\\n$\\because$ triangle $ABC$ is equilateral, \\\\\n$\\therefore AB=AC=BC=8$, \\\\\n$\\because$ triangle $ABC$ is folded such that point $A$ falls on point $D$ on side $BC$, \\\\\n$\\therefore AE=ED$ and $AF=FD$, \\\\\n$\\therefore$ the perimeter of $\\triangle BDE$ is: $BD+DE+BE=BD+AE+BE=BD+AB=6+8=\\boxed{14}$, \\\\\nthe perimeter of $\\triangle CFD$ is: $CD+DF+CF=CD+AF+FC=CD+AC=2+8=\\boxed{10}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53110772_169.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, there are points A and B on the graph of the inverse proportion function $y=\\frac{k}{x} (k>0, x>0)$. Their $x$-coordinates are 2 and 4, respectively. If the area of $\\triangle ABO$ is 6, what is the value of $k$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $AC \\perp x$-axis at point $C$, and draw $BD \\perp x$-axis at point $D$,\\\\\nsince the $x$-coordinates of points $A$ and $B$ are 2 and 4, respectively,\\\\\ntherefore $A\\left(2, \\frac{k}{2}\\right)$, $B\\left(4, \\frac{k}{4}\\right)$, and the area of $\\triangle AOC$ and $\\triangle BOD$ are both $\\frac{1}{2}|k|$,\\\\\nsince the area of $\\triangle AOB$ equals the area of $\\triangle AOC$ plus the area of trapezoid $ACDB$ minus the area of $\\triangle BOD$,\\\\\nthus the area of $\\triangle AOB$ equals the area of trapezoid $ACDB$ which is 6,\\\\\nthus $\\frac{1}{2}\\left(\\frac{k}{2}+\\frac{k}{4}\\right)\\times (4-2)=6$,\\\\\nsolving for $k$ gives $k=\\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51457867.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AD \\parallel BE \\parallel CF$, where points $B, E$ are on $AC, DF$ respectively, $\\frac{AB}{BC} = \\frac{2}{3}, EF = 6$. What is the length of $DE$?\n",
    "completion": "\\textbf{Solution:} Since $AD\\parallel BE\\parallel CF$, \\\\\nit follows that $\\frac{AB}{BC}=\\frac{DE}{FE}$, \\\\\nthus $DE=\\boxed{4}$",
    "image_path": "dataset/mm_math/images/MM_Math/55604325.png"
  },
  {
    "solution": "\\boxed{1:6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in parallelogram $ABCD$, $AC$ and $BD$ intersect at point $O$. $E$ is the midpoint of $OD$. When $AE$ is extended to intersect $DC$ at point $F$, what is the ratio of the area of $\\triangle DEF$ to the area of $\\triangle OAB$?\n",
    "completion": "\\textbf{Solution:} Let $O$ be the intersection point of the diagonals of the parallelogram $ABCD$,\n\n$\\therefore$ $DO=BO$, $AB\\parallel DC$,\n\nAlso, since $E$ is the midpoint of $OD$,\n\n$\\therefore$ $DE=\\frac{1}{2}OD=\\frac{1}{4}DB$,\n\n$\\therefore$ $DE:EB=1:3$,\n\nAlso, since $AB\\parallel DC$,\n\n$\\therefore$ $\\triangle DEF\\sim \\triangle BAE$,\n\n$\\therefore$ $\\frac{S_{\\triangle DEF}}{S_{\\triangle BEA}}=\\left(\\frac{1}{3}\\right)^2=\\frac{1}{9}$,\n\n$\\therefore$ $S_{\\triangle DEF}=\\frac{1}{9}S_{\\triangle BEA}$,\n\nSince $\\frac{S_{\\triangle AOB}}{S_{\\triangle BEA}}=\\frac{OB}{EB}=\\frac{2}{3}$,\n\n$\\therefore$ $S_{\\triangle AOB}=\\frac{2}{3}S_{\\triangle BEA}$,\n\n$\\therefore$ $S_{\\triangle DEF}:S_{\\triangle OAB}=\\frac{\\frac{1}{9}S_{\\triangle BEA}}{\\frac{2}{3}S_{\\triangle BEA}}=\\boxed{1:6}$",
    "image_path": "dataset/mm_math/images/MM_Math/52536707.png"
  },
  {
    "solution": "\\boxed{59^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in diamond ABCD, M and N are on AB and CD, respectively, and $AM = CN$, MN intersects AC at point O, connect $BO$, if $\\angle DAC = 31^\\circ$, what is the degree measure of $\\angle OBC$?\n",
    "completion": "\\textbf{Solution}: Since quadrilateral ABCD is a rhombus,\\\\\nit follows that $AB\\parallel CD$ and $AB=BC$,\\\\\ntherefore $\\angle MAO=\\angle NCO$ and $\\angle AMO=\\angle CNO$,\\\\\nin $\\triangle AMO$ and $\\triangle CNO$,\\\\\nsince $\\left\\{\\begin{array}{l}\\angle MAO=\\angle NCO\\hfill \\\\ AM=CN\\hfill \\\\ \\angle AMO=\\angle CNO\\hfill \\end{array}\\right.$,\\\\\nit follows that $\\triangle AMO \\cong \\triangle CNO\\left(ASA\\right)$,\\\\\ntherefore $AO=CO$,\\\\\nsince $AB=BC$,\\\\\nit follows that $BO\\perp AC$,\\\\\ntherefore $\\angle BOC=90^\\circ$,\\\\\nsince $\\angle DAC=31^\\circ$,\\\\\nit follows that $\\angle BCA=\\angle DAC=31^\\circ$,\\\\\ntherefore $\\angle OBC=90^\\circ-31^\\circ=\\boxed{59^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52395005.png"
  },
  {
    "solution": "\\boxed{26}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, \\(ABCD\\) is a rectangular ground, with length \\(AB = 20m\\) and width \\(AD = 10m\\). In the middle, there stands a brick wall with height \\(MN = 2m\\). A grasshopper needs to crawl from point \\(A\\) to point \\(C\\), and it must climb over the wall in the middle. What is the minimum distance it has to travel?\n",
    "completion": "Solution: As shown in the figure, unfolding the figure increases the length by 2MN,\\\\\nif the original length increases by 4 meters, then AB=20+4=24(m),\\\\\nconnect AC,\\\\\n$\\because$ quadrilateral ABCD is a rectangle, with AB=24m and width AD=10m,\\\\\n$\\therefore$AC=$\\sqrt{AB^{2}+BC^{2}}=\\sqrt{24^{2}+10^{2}}=\\boxed{26}$(m),\\\\\n$\\therefore$, for the grasshopper to crawl from point A to point C, it needs to travel at least 26m.",
    "image_path": "dataset/mm_math/images/MM_Math/52603284.png"
  },
  {
    "solution": "\\boxed{2\\frac{2}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, isosceles right triangle $ABC$ has $\\angle C=90^\\circ$, with $AC=BC=4$. $M$ is the midpoint of $AB$, and $\\angle PMQ=45^\\circ$, where the two sides of $\\angle PMQ$ intersect $BC$ at point $P$ and $AC$ at point $Q$ respectively. If $BP=3$, what is the length of $AQ$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, connect CM, draw $PF\\perp CM$ at point F through point P, and draw $MD\\perp AC$ at point D through point M.\n\nIn $\\triangle ABC$, $AB=\\sqrt{AC^2+BC^2}=\\sqrt{4^2+4^2}=4\\sqrt{2}$,\n\nSince M is the midpoint of AB,\n\nThen $BM=AM=\\frac{1}{2}AB=2\\sqrt{2}$,\n\nSince $AC=BC$,\n\nThen $CM\\perp AB$, $\\angle 1=\\angle 2=\\frac{1}{2}\\angle ACB=45^\\circ$,\n\nSince in $\\triangle BMC$, $\\tan\\angle 1=\\frac{BM}{CM}=1$,\n\nThen $CM=BM=2\\sqrt{2}$,\n\nSince $BP=3$,\n\nThen $PC=BC-BP=4-3=1$,\n\nIn $\\triangle PFC$, $\\sin\\angle 1=\\frac{PF}{PC}=\\frac{\\sqrt{2}}{2}$, $\\cos\\angle 1=\\frac{CF}{PC}=\\frac{\\sqrt{2}}{2}$,\n\nTherefore, $PF=\\frac{\\sqrt{2}}{2}$, $CF=\\frac{\\sqrt{2}}{2}$,\n\nTherefore, $FM=\\frac{3\\sqrt{2}}{2}$,\n\nIn $\\triangle CMD$, $\\sin\\angle 2=\\frac{DM}{CM}=\\frac{\\sqrt{2}}{2}$, $\\cos\\angle 2=\\frac{CD}{CM}=\\frac{\\sqrt{2}}{2}$,\n\nTherefore, $DM=2$, $CD=2$,\n\nTherefore, $AD=AC-CD=4-2=2$,\n\nSince $\\angle PMQ=45^\\circ$,\n\nThen $\\angle 3+\\angle 4=45^\\circ$,\n\nSince $\\angle CMD=90^\\circ-\\angle 2=45^\\circ$,\n\nThen $\\angle 4+\\angle 5=45^\\circ$,\n\nTherefore, $\\angle 3=\\angle 5$,\n\nSince in $\\triangle MPF$, $\\tan\\angle 3=\\frac{PF}{FM}=\\frac{\\frac{\\sqrt{2}}{2}}{\\frac{3\\sqrt{2}}{2}}=\\frac{1}{3}$,\n\nThen in $\\triangle MDQ$, $\\tan\\angle 5=\\frac{DQ}{MD}=\\frac{1}{3}$,\n\nTherefore, $DQ=\\frac{1}{3}\\times MD=\\frac{1}{3}\\times 2=\\frac{2}{3}$,\n\nTherefore, $AQ=AD+DQ=2+\\frac{2}{3}=\\boxed{2\\frac{2}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51319026.png"
  },
  {
    "solution": "\\boxed{x\\leq 1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, if the graph of the linear function $y_{1}=x+a$ intersects with the graph of the linear function $y_{2}=kx+b$ at point P(1,3), what is the solution set of the inequality $x+a\\le kx+b$ with respect to $x$?\n",
    "completion": "\\textbf{Solution:} According to the graph, when $x\\leq 1$, $x+a\\leq kx+b$,\\\\\nthat is, the solution set of the inequality $x+a\\leq kx+b$ with respect to $x$ is $\\boxed{x\\leq 1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51983394.png"
  },
  {
    "solution": "\\boxed{65^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, $\\angle A=40^\\circ$, and the bisector of the exterior angle $\\angle CBD$ of $\\triangle ABC$, labeled as $BE$, intersects the extension of $AC$ at point $E$. Point $F$ is located on the extension of $AC$, and $DF$ is connected. What is the measure of $\\angle CBE$ in degrees?\n",
    "completion": "\\textbf{Solution:} Since $\\angle ACB=90^\\circ$ and $\\angle A=40^\\circ$,\n\\[\\therefore \\angle CBD=\\angle A+\\angle ACB=130^\\circ,\\]\nSince $BE$ bisects $\\angle CBD$,\n\\[\\therefore \\angle CBE=\\frac{1}{2}\\angle CBD=\\boxed{65^\\circ}.\\]",
    "image_path": "dataset/mm_math/images/MM_Math/53206635_95.png"
  },
  {
    "solution": "\\boxed{(0, 2)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y=kx+2$ intersects with the graph of the inverse proportion function $y=\\frac{m}{x}$ (where $x>0$) at point $P$. Point $P$ is located in the first quadrant, with $PA\\perp x$-axis at point A. The graph of the linear function intersects the x-axis and y-axis at points $C$ and $D$, respectively, and ${S}_{\\triangle COD}=1$, $\\frac{CO}{OA}=\\frac{1}{2}$. What are the coordinates of point $D$?\n",
    "completion": "\\textbf{Solution:} In the equation $y=kx+2$, let $x=0$, then $y=2$. \\\\\n$\\therefore$ The coordinates of point D are $\\boxed{(0, 2)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53043682_121.png"
  },
  {
    "solution": "\\boxed{\\frac{9}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, OA is the radius of the circle $\\odot O$, with chord BC $\\perp$ OA at point D. AC is connected. If BC = $4\\sqrt{2}$ and AC = 3, then what is the length of the radius of $\\odot O$?\n",
    "completion": "\\textbf{Solution:} As illustrated in the diagram, connect OC.\\\\\nSince BC$\\perp$OA,\\\\\nit follows that $\\angle ADC=\\angle ODC=90^\\circ$, $CD=BD=\\frac{1}{2}BC=2\\sqrt{2}$,\\\\\ntherefore $AD=\\sqrt{AC^{2}-CD^{2}}=1$.\\\\\nLet $OA=OC=r$, then $OD=OA-AD=r-1$,\\\\\nsince $OD^{2}+CD^{2}=OC^{2}$,\\\\\nit can be derived that $(r-1)^{2}+\\left(2\\sqrt{2}\\right)^{2}=r^{2}$,\\\\\nsolving this gives $r=\\boxed{\\frac{9}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52720284.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the Cartesian coordinate system shown in the figure, $O$ is the origin. The parabola $y=-x^{2}+6x+3$ intersects the $y$-axis at point $A$. A line $AB$ is drawn parallel to the $x$-axis from point $A$, and it intersects the parabola at point $B$. The line segment $OB$ is connected. A point $P$ is located above segment $AB$ on the parabola, and a line $PA$ is drawn. Perpendicular $PQ$ is dropped from $P$ to $AB$, with the foot of the perpendicular being $H$, and it intersects $OB$ at point $Q$. Find the length of $AB$.\n",
    "completion": "\\textbf{Solution:} Given $y=-x^{2}+6x+3$, let $x=0$, then we have $y=3$. Thus, point $A(0,3)$, \\\\\nlet $y=-x^{2}+6x+3=3$, solving this gives $x=0$ or $x=6$. Hence, point $B(6,3)$, \\\\\n$\\therefore AB=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51317435_146.png"
  },
  {
    "solution": "\\boxed{\\frac{m}{\\cos\\alpha}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a village plans to plant trees on a hillside with a slope angle of $ \\alpha $. It is required that the horizontal distance between two adjacent trees be $m$ (m). What is the distance AB between these two trees on the slope?\n",
    "completion": "\\textbf{Solution:} From the given information, we have $\\cos\\alpha =\\frac{m}{AB}$,\\\\\nthus $AB=\\frac{m}{\\cos\\alpha}=\\boxed{\\frac{m}{\\cos\\alpha}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51322892.png"
  },
  {
    "solution": "\\boxed{\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the vertices A, B, C of the rhombus OABC are on the circle $\\odot O$. A tangent to $\\odot O$ passing through point B intersects the extension of OA at point D. If the radius of $\\odot O$ is 1, what is the length of BD?\n",
    "completion": "\\textbf{Solution:} Draw line OB, then OB=OA=1, since BD is the tangent of circle O, thus $\\angle$OBD=90$^\\circ$, since quadrilateral OABC is a rhombus, thus OA=OB, therefore $\\triangle$OAB is an equilateral triangle, hence $\\angle$AOB=60$^\\circ$, therefore $\\angle$ADB=30$^\\circ$, hence OD=2OB=2, therefore BD=$ \\sqrt{OA^{2}-OB^{2}}$=$ \\boxed{\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55138394.png"
  },
  {
    "solution": "\\boxed{y=x-2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system, line $l_1: y = -2x + 1$ intersects the y-axis at point A, and line $l_2$ intersects the y-axis at point B $(0, -2)$ and intersects line $l_1$ at point C. The ordinate of point C is -1. Point D is an arbitrary point on line $l_2$. A perpendicular line to the x-axis is drawn from point D and intersects line $l_1$ at point E. What is the equation of line $l_2$?\n",
    "completion": "\\textbf{Solution:} Since the ordinate of point $C$ is $-1$ and point $C$ lies on line $l_1$,\\\\\nwe have $-1=-2x+1$, solving this gives $x=1$,\\\\\nthus the coordinates of point $C$ are $(1, -1)$,\\\\\nLet the equation of line $l_2$ be $y=kx+b$,\\\\\nsince line $l_2$ intersects the $y$-axis at point $B(0, -2)$, and intersects line $l_1$ at point $C$,\\\\\nwe obtain $\\begin{cases}k+b=-1\\\\ b=-2\\end{cases}$ solving this gives $\\begin{cases}k=1\\\\ b=-2\\end{cases}$\\\\\nTherefore, the equation of line $l_2$ is $y=x-2$, thus the equation of line $l_2$ is $\\boxed{y=x-2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50555108_101.png"
  },
  {
    "solution": "\\boxed{\\left(\\frac{1}{2}, -\\frac{3}{4}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the parabola $y=x^{2}+(k-1)x-k$ intersects with the line $y=kx+1$ at points $A$ and $B$, where point $A$ is to the left of point $B$. Given $A(-1, 0)$ and $B(2, 3)$, let point $P$ be a moving point on the parabola, situated below the line $AB$. Determine the maximum area of $\\triangle ABP$ and the coordinates of point $P$ under this condition.\n",
    "completion": "\\textbf{Solution}: Let $P(x, x^{2}-1)$. Draw $PF\\parallel y$-axis through point P, intersecting line $AB$ at point $F$, hence $F(x, x+1)$.\\\\\n$\\therefore PF=y_{F}-y_{P}=(x+1)-(x^{2}-1)=-x^{2}+x+2$.\\\\\n$S_{\\triangle ABP}=\\frac{1}{2}PF(x_{F}-x_{A})+\\frac{1}{2}PF(x_{B}-x_{F})=\\frac{1}{2}PF(x_{B}-x_{A})=\\frac{3}{2}PF$\\\\\n$\\therefore S_{\\triangle ABP}=\\frac{3}{2}(-x^{2}+x+2)=-\\frac{3}{2}(x-\\frac{1}{2})^{2}+\\frac{27}{8}$\\\\\nWhen $x=\\frac{1}{2}$, $y_{P}=x^{2}-1=-\\frac{3}{4}$.\\\\\n$\\therefore$ The maximum area of $\\triangle ABP$ is $\\boxed{\\frac{27}{8}}$, with point $P$ coordinates at $\\boxed{\\left(\\frac{1}{2}, -\\frac{3}{4}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53009969_93.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the vertex $C$ of the parabola passing through points $A(0, 4)$ and $B(5, 9)$ is on the positive $x$-axis. Let $P(x, y)$ be a point on line segment $AB$, where $1 \\le x \\le 4$. Draw $PM$ parallel to the $y$-axis to intersect the parabola at point $M$. What are the maximum and minimum values of $PM$?\n",
    "completion": "\\textbf{Solution:} Let the equation of $AB$ be: $y=kx+b$,\\\\\nSubstituting points $A\\left(0, 4\\right)$ and $B\\left(5, 9\\right)$ into $y=kx+b$ gives:\\\\\n$\\begin{cases}\nb=4\\\\\n5k+b=9\n\\end{cases}$,\\\\\nSolving yields: $\\begin{cases}\nk=1\\\\\nb=4\n\\end{cases}$,\\\\\n$\\therefore$ The equation of $AB$ is: $y=x+4$,\\\\\n$\\because$ Point $P$ is a point on segment $AB$, and point $M$ is a point on the parabola $y=(x−2)^{2}$, with $1\\le x\\le 4$, and $PM\\parallel$ to the $y$ axis,\\\\\n$\\therefore$ When $x=1$, $P\\left(1, 5\\right)$, $M\\left(1, 5\\right)$,\\\\\n$\\therefore PM=5−1=4$,\\\\\nWhen $x=4$, $P\\left(4, 8\\right)$, $M\\left(4, 4\\right)$,\\\\\n$\\therefore PM=8−4=4$,\\\\\nWhen $x=2$, $P\\left(2, 6\\right)$, $M\\left(2, 0\\right)$,\\\\\n$\\therefore PM=6−0=6$,\\\\\nLet $P\\left(n, n+4\\right)$, $M\\left(n, n^{2}−4n+4\\right)$,\\\\\n$\\therefore PM=n+4−\\left(n^{2}−4n+4\\right)$\\\\\n$=−n^{2}+5n$\\\\\n$=−\\left(n−\\frac{5}{2}\\right)^{2}+\\frac{25}{4}$,\\\\\n$\\therefore$ When $n=\\frac{5}{2}$, the maximum value of $PM$ is: $\\boxed{\\frac{25}{4}}$,\\\\\n$\\therefore$ The maximum value of $PM$ is $\\frac{25}{4}$, and the minimum value is $\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52551573_133.png"
  },
  {
    "solution": "\\boxed{\\sqrt{31}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $ \\triangle ABC$ is an equilateral triangle with $O$ as the center of its circumcircle, and point $D$ is on the arc $ABC$. Connect $AD$ and $BD$. $AE$ is a chord on the right side of $AD$, and $\\angle DAE=30^\\circ$. Connect $CE$. What is the angle between the lines containing $BD$ and $CE$? When $BD=\\sqrt{3}$, and $CE=4$, what is the radius of circle $O$?\n",
    "completion": "\\textbf{Solution:} Extend BD and CE to intersect at point F, connect OB, OC, CD, and BE,\\\\\nsince $\\triangle ABC$ is an equilateral triangle,\\\\\ntherefore $\\angle BAC=60^\\circ$,\\\\\nsince $\\angle DBC=\\angle DAC$, $\\angle ECB=\\angle BAE$,\\\\\ntherefore $\\angle BFC=180^\\circ-\\angle FBC-\\angle FCB$\\\\\n$=180^\\circ-\\angle DAC-\\angle BAE$\\\\\n$=180^\\circ-(\\angle DAC+\\angle BAD+\\angle DAE)$\\\\\n$=180^\\circ-(\\angle BAC+\\angle DAE)$\\\\\n$=180^\\circ-(60^\\circ+30^\\circ)$\\\\\n$=90^\\circ$,\\\\\ntherefore the angle between the lines containing BD and CE is $\\boxed{90^\\circ}$;\\\\\nsince $\\angle DAE=30^\\circ$,\\\\\ntherefore $\\angle DBE=\\angle ECD=30^\\circ$,\\\\\nin triangle $\\triangle BEF$, let DF$=m$, EF$=n$,\\\\\ntherefore BF$=BD+DF=\\sqrt{3}+m$,\\\\\nsince $\\angle FBE=30^\\circ$,\\\\\ntherefore $\\tan 30^\\circ=\\frac{EF}{BF}$,\\\\\ntherefore $\\frac{\\sqrt{3}}{3}=\\frac{n}{\\sqrt{3}+m}$,\\\\\ntherefore $n=1+\\frac{\\sqrt{3}}{3}m$,\\\\\nin triangle $\\triangle CDF$, CE$=4$,\\\\\ntherefore CF$=CE+EF=4+n$,\\\\\nsince $\\angle FCD=30^\\circ$,\\\\\ntherefore $\\tan 30^\\circ=\\frac{DF}{CF}$,\\\\\ntherefore $\\frac{\\sqrt{3}}{3}=\\frac{m}{4+n}$,\\\\\ntherefore $m=\\frac{\\sqrt{3}}{3}n+\\frac{4\\sqrt{3}}{3}$,\\\\\ntherefore $m=\\frac{\\sqrt{3}}{3}\\left(1+\\frac{\\sqrt{3}}{3}m\\right)+\\frac{4\\sqrt{3}}{3}$,\\\\\nsolving for $m$ yields: $m=\\frac{5\\sqrt{3}}{2}$,\\\\\ntherefore $n=\\frac{7}{2}$,\\\\\ntherefore $BF=\\sqrt{3}+m=\\sqrt{3}+\\frac{5\\sqrt{3}}{2}=\\frac{7\\sqrt{3}}{2}$,\\\\\n$CF=4+n=4+\\frac{7}{2}=\\frac{15}{2}$,\\\\\ntherefore $BC=\\sqrt{BF^2+CF^2}=\\sqrt{\\left(\\frac{7\\sqrt{3}}{2}\\right)^2+\\left(\\frac{15}{2}\\right)^2}=\\sqrt{93}$,\\\\\ndraw line OH perpendicular to BC at point H through point O,\\\\\nsince OB$=$OC,\\\\\ntherefore $BH=\\frac{1}{2}BC=\\frac{\\sqrt{93}}{2}$,\\\\\nsince $\\angle BOC=2\\angle BAC=120^\\circ$,\\\\\ntherefore $\\angle BOH=60^\\circ$,\\\\\ntherefore $\\sin 60^\\circ=\\frac{BH}{OB}$,\\\\\ntherefore $OB=\\frac{BH}{\\frac{\\sqrt{3}}{2}}=\\boxed{\\sqrt{31}}$,\\\\\nthus, the radius of circle $O$ is $\\boxed{\\sqrt{31}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52872415.png"
  },
  {
    "solution": "\\boxed{\\frac{21}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that in $\\triangle ABC$, point $F$ is the centroid of the triangle. Line $DE \\parallel BC$ passes through point $F$ and intersects $AB$ at point $D$ and $AC$ at point $E$. If $DE=7$, what is the value of $BC$?\n",
    "completion": "\\textbf{Solution:} Draw line $CF$ and extend it to intersect $AB$ at point H, and connect $GH$, as shown in the figure,\\\\\nSince point $F$ is the centroid of $\\triangle ABC$,\\\\\nTherefore, $AG$ and $CH$ are the medians of $\\triangle ABC$,\\\\\nTherefore, points $G$ and $H$ are respectively the midpoints of $BC$ and $AB$,\\\\\nTherefore, $GH=\\frac{1}{2}AC$ and $GH\\parallel AC$,\\\\\nTherefore, $\\frac{AF}{FG}=\\frac{AC}{HG}=2$,\\\\\nTherefore, $\\frac{AF}{AG}=\\frac{2}{3}$,\\\\\nSince $DE\\parallel BC$,\\\\\nTherefore, $\\frac{DE}{BC}=\\frac{AF}{AG}=\\frac{2}{3}$,\\\\\nSince $DE=7$,\\\\\nTherefore, $BC=\\frac{3DE}{2}=\\frac{3\\times 7}{2}=\\boxed{\\frac{21}{2}}$,",
    "image_path": "dataset/mm_math/images/MM_Math/53102179.png"
  },
  {
    "solution": "\\boxed{P\\left(3,12\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the Cartesian coordinate system, let $O$ be the origin. The parabola $y=-x^2+6x+3$ intersects the $y$-axis at point $A$. A line segment $AB$ is drawn parallel to the $x$-axis from $A$ and intersects the parabola at point $B$. The segment $OB$ is then connected. Let $P$ be a point on the parabola above segment $AB$, and draw $PA$ and $PQ \\perp AB$ with the foot of the perpendicular on $AB$ at $H$ and intersecting $OB$ at point $Q$. When the area of $\\triangle APH$ is twice the area of the quadrilateral $AOQH$, find the coordinates of point $P$.\n",
    "completion": "\\textbf{Solution:} When the area of $\\triangle APH$ is twice the area of quadrilateral $AOQH$,\\\\\nthen $2(AO+HQ)=PH$,\\\\\n$\\therefore 2\\left(3+ \\frac{6-m}{2}\\right)=-m^2+6m$,\\\\\nsolving this gives: $m_1=4$, $m_2=3$,\\\\\n$\\therefore \\boxed{P(4,11)} or \\boxed{P\\left(3,12\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51317435_148.png"
  },
  {
    "solution": "\\boxed{\\frac{9}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in parallelogram $ABCD$, point $E$ is a point on side $BC$, and $AE$ and $BD$ intersect at point $F$. If $\\angle DAF = \\angle ABD$, $\\frac{AF}{FE} = \\frac{3}{2}$, and $BE = \\sqrt{15}$, what is the length of $DF$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral $ABCD$ is a parallelogram, and $AD\\parallel BC$, with $\\frac{AF}{FE}=\\frac{3}{2}$,\\\\\nit follows that $\\triangle ADF\\sim \\triangle EBF$,\\\\\nthus $\\frac{AF}{EF}=\\frac{AD}{EB}=\\frac{DF}{BF}=\\frac{3}{2}$,\\\\\nalso, since $BE=\\sqrt{15}$,\\\\\nthen $AD=\\frac{3}{2}BE=\\frac{3\\sqrt{15}}{2}$, and $BF=\\frac{2}{3}DF$\\\\\nGiven $\\angle DAF=\\angle ABD=\\angle BEA$,\\\\\nit implies that $\\triangle ABD\\sim \\triangle FEB$,\\\\\nthus $\\frac{AD}{BF}=\\frac{BD}{BE}$,\\\\\nwhich means $\\frac{AD}{\\frac{2}{3}DF}=\\frac{BF+DF}{BE}=\\frac{\\frac{2}{3}DF+DF}{BE}$,\\\\\nhence $\\frac{\\frac{3\\sqrt{15}}{2}}{\\frac{2}{3}DF}=\\frac{\\frac{5}{3}DF}{\\sqrt{15}}$,\\\\\nsolving gives $DF=\\boxed{\\frac{9}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52991515.png"
  },
  {
    "solution": "\\boxed{12\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the slope of a river embankment $BC=12\\,m$. In order to reinforce the embankment, it is necessary to thicken the dam. After completion, the slope ratio changes from the original $1\\colon 2$ to $1\\colon 3$. What is the length of the reinforced slope $AD$ in meters?\n",
    "completion": "\\textbf{Solution:} Construct $CE\\perp AB$ through point C, and $DF\\perp AB$ through point D, with the respective perpendicular feet being E and F.\\\\\nLet $CE=x$, then $BE=2x$, $DF=CE=x$, $AF=3x$,\\\\\nsince in $\\triangle CEB$, $\\angle BEC=90^\\circ$, and $BC=12$,\\\\\nthus $x^2+(2x)^2=12^2$, yielding: $x=\\frac{12\\sqrt{5}}{5}$.\\\\\nSince in $\\triangle ADF$, $\\angle AFD=90^\\circ$,\\\\\nthus $AD=\\sqrt{AF^2+DF^2}=\\sqrt{10}x=\\sqrt{10}\\times \\frac{12\\sqrt{5}}{5}=12\\sqrt{2}$ m.\\\\\nAnswer: The reinforced slope $AD$ is $\\boxed{12\\sqrt{2}}$ meters.",
    "image_path": "dataset/mm_math/images/MM_Math/51174836.png"
  },
  {
    "solution": "\\boxed{144}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle BAC=90^\\circ$, squares are constructed externally on sides $AB$, $CA$, $BC$. The area of square $ABIH$ is $25$, and the area of square $BDEC$ is $169$. What is the area of square $ACFG$?\n",
    "completion": "\\textbf{Solution:} In the right triangle $ABC$, $\\angle BAC=90^\\circ$, \\\\\n$\\therefore AB^2+AC^2=BC^2$, \\\\\n$\\because$ the area of the square $ABIH$ is $25$, and the area of the square $BDEC$ is $169$, \\\\\n$\\therefore AB^2=25, BC^2=169$, \\\\\n$\\therefore AC^2=BC^2-AB^2=169-25=144$, \\\\\n$\\therefore$ the area of the square $ACFG$ is $AC^2=\\boxed{144}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52836463.png"
  },
  {
    "solution": "\\boxed{65^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line AB intersects with line MN at point O, with OC $\\perp$ AB, and OA bisects $\\angle MOD$. If $\\angle BON = 25^\\circ$, what is the degree measure of $\\angle COD$?\n",
    "completion": "\\textbf{Solution:} Given $\\angle BON = 25^\\circ$, \\\\\nthen $\\angle AOM = 25^\\circ$, \\\\\nsince OA bisects $\\angle MOD$, \\\\\nthen $\\angle AOD = \\angle MOA = 25^\\circ$, \\\\\nsince OC $\\perp$ AB, \\\\\nthen $\\angle AOC = 90^\\circ$, \\\\\ntherefore $\\angle COD = 90^\\circ - 25^\\circ = \\boxed{65^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53206132.png"
  },
  {
    "solution": "\\boxed{\\frac{13\\sqrt{3}}{4}-\\frac{3\\pi}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a circle $\\odot O$ with a radius of 2 is tangent to the regular hexagon ABCDEF at points C and F. What is the area of the shaded region in the figure?\n",
    "completion": "\\textbf{Solution:} Connect $OF$, $OC$, $CF$, from point $O$ draw $OH\\perp ED$ at point H, it intersects $FC$ at point P,\\\\\nin the quadrilateral $OCDH$, $OC\\perp CD$, $OH\\perp HD$, $\\angle CDH=120^\\circ$,\\\\\n$\\therefore$ $\\angle OCD=90^\\circ$, $\\angle DHO=90^\\circ$,\\\\\n$\\therefore$ $\\angle COH=360^\\circ-\\angle OCD-\\angle CDH-\\angle DHO=360^\\circ-90^\\circ-120^\\circ-90^\\circ=60^\\circ$,\\\\\nSimilarly, $\\angle FOH=60^\\circ$,\\\\\nSince $OC=OF$,\\\\\n$\\therefore$ $OP$ is the perpendicular bisector of $FC$,\\\\\nIn $\\triangle OPC$, $\\angle OPC=90^\\circ$, $\\angle COP=60^\\circ$, $OC=2$,\\\\\n$\\therefore$ $\\angle OCP=180^\\circ-\\angle OPC-\\angle COP=180^\\circ-90^\\circ-60^\\circ=30^\\circ$,\\\\\n$\\therefore$ $OP=\\frac{1}{2}\\times OC=\\frac{1}{2}\\times 2=1$,\\\\\n$PC=\\sqrt{OC^2-OP^2}=\\sqrt{2^2-1^2}=\\sqrt{3}$,\\\\\n$\\therefore$ $FC=2PC=2\\times \\sqrt{3}=2\\sqrt{3}$,\\\\\nDraw $DM\\perp FC$ from point D, draw $EN\\perp FC$ from point E,\\\\\n$\\therefore$ $\\angle DMC=\\angle ENF=90^\\circ$,\\\\\nBecause $\\angle OCD=90^\\circ$, $\\angle OCP=30^\\circ$,\\\\\n$\\therefore$ $\\angle PCD=\\angle OCD-\\angle OCP=90^\\circ-30^\\circ=60^\\circ$,\\\\\nSimilarly, $\\angle PFE=60^\\circ$,\\\\\nIn $\\triangle DMC$, $\\angle MDC=180^\\circ-\\angle DMC-\\angle MCD=180^\\circ-90^\\circ-60^\\circ=30^\\circ$,\\\\\n$\\therefore$ $MC=\\frac{1}{2}DC$,\\\\\nIn $\\triangle EFN$, $\\angle FEN=180^\\circ-\\angle EFN-\\angle ENF=180^\\circ-90^\\circ-60^\\circ=30^\\circ$,\\\\\n$\\therefore$ $NF=\\frac{1}{2}EF$,\\\\\n$\\therefore$ $FC=FN+NM+MC=2ED=2\\sqrt{3}$,\\\\\nSince $EF=DE=CD=NM$,\\\\\n$\\therefore$ $ED=CD=EF=NM=\\sqrt{3}$,\\\\\n$MC=NF=\\frac{1}{2}(FC-NM)=\\frac{1}{2}\\times (2\\sqrt{3}-\\sqrt{3})=\\frac{\\sqrt{3}}{2}$,\\\\\n$\\therefore$ $M=\\sqrt{CD^2-MC^2}=\\sqrt{\\sqrt{3}^2-\\left(\\frac{\\sqrt{3}}{2}\\right)^2}=\\frac{3}{2}$,\\\\\nThen $PH=DM=\\frac{3}{2}$,\\\\\n$\\therefore$ ${S}_{OFEDC}={S}_{\\triangle OFC}+{S}_{FEDC}=\\frac{1}{2}\\times 2\\sqrt{3}\\times 1+\\frac{1}{2}\\times (\\sqrt{3}+2\\sqrt{3})\\times \\frac{3}{2}=\\frac{13\\sqrt{3}}{4}$,\\\\\n$S_{\\stackrel{\\frown}{OFC}}=\\frac{\\frac{3\\pi}{2}}{2\\pi}\\cdot\\pi\\cdot 2^2=\\frac{3\\pi}{2}$,\\\\\n$\\therefore$ The area of the shaded part = ${S}_{OFEDC}-{S}_{\\stackrel{\\frown}{OFC}}=\\frac{13\\sqrt{3}}{4}-\\frac{3\\pi}{2}=\\boxed{\\frac{13\\sqrt{3}}{4}-\\frac{3\\pi}{2}}$,",
    "image_path": "dataset/mm_math/images/MM_Math/51458354.png"
  },
  {
    "solution": "\\boxed{100^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, for the inscribed quadrilateral $ABCD$ in the circle $\\odot O$, if $\\angle BCD=130^\\circ$, then what is the degree measure of $\\angle BOD$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral $ABCD$ is inscribed in circle $O$, \\\\\nthus $\\angle A = 180^\\circ - \\angle C = 50^\\circ$, \\\\\ntherefore $\\angle BOD = 2\\angle A = \\boxed{100^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/51318244.png"
  },
  {
    "solution": "\\boxed{\\left(-\\frac{3}{2}, -\\frac{9}{4}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, in the Cartesian coordinate system, the parabola $y = ax^2 - 2ax - 3a$ ($a < 0$) intersects the x-axis at points A and B (with A to the left of B), and intersects the y-axis at point C. Point D is the vertex of the parabola, and the axis of symmetry intersects the x-axis at point E, with $OB = OC$. Connect BD. Is there a point F on the parabola such that $\\angle FBO = \\angle BDE$? If such a point exists, find the coordinates of point F; if not, explain why.\n",
    "completion": "\\textbf{Solution:} From the given information, we know that $A(-1, 0)$, $B(3, 0)$, $D(1, 4)$,\\\\\n$\\therefore DE=4$, $BE=2$,\\\\\nLet $F(x, -x^2+2x+3)$,\\\\\nWhen $F$ is at ${F}_{1}$, draw ${F}_{1}H\\perp$ to the x-axis,\\\\\n$\\because \\angle BDE=\\angle {F}_{1}BH$, $\\angle {F}_{1}HB=\\angle BED=90^\\circ$,\\\\\n$\\therefore \\triangle BDE\\sim \\triangle {F}_{1}BH$,\\\\\n$\\therefore \\frac{BE}{DE}=\\frac{{F}_{1}H}{BH}$, which means $\\frac{1}{2}=\\frac{-x^2+2x+3}{3-x}$,\\\\\n$\\therefore 2x^2-5x-3=0$,\\\\\nSolve to get: $x=3$ or $x=-\\frac{1}{2}$,\\\\\n$\\because x<0$,\\\\\n$\\therefore x=-\\frac{1}{2}$, at this time $y=\\frac{7}{4}$,\\\\\nThus, ${F}_{1}\\left(-\\frac{1}{2}, \\frac{7}{4}\\right)$;\\\\\nWhen $F$ is at ${F}_{2}$, draw ${F}_{2}K\\perp$ to the x-axis,\\\\\n$\\because \\angle BDE=\\angle {F}_{2}BK$, $\\angle {F}_{2}KB=\\angle BED=90^\\circ$,\\\\\n$\\therefore \\triangle BDE\\sim \\triangle {F}_{2}BK$,\\\\\n$\\therefore \\frac{BE}{DE}=\\frac{{F}_{2}K}{BK}$, which means $\\frac{1}{2}=\\frac{-(-x^2+2x+3)}{3-x}$,\\\\\n$\\therefore 2x^2-3x-9=0$,\\\\\nSolve to get: $x=3$ or $x=-\\frac{3}{2}$,\\\\\n$\\because x<0$,\\\\\n$\\therefore x=-\\frac{3}{2}$, at this time $y=-\\frac{9}{4}$,\\\\\n$\\therefore$ the coordinates of point $F$ are $\\boxed{\\left(-\\frac{3}{2}, -\\frac{9}{4}\\right)}$ or $\\left(-\\frac{1}{2}, \\frac{7}{4}\\right)$;",
    "image_path": "dataset/mm_math/images/MM_Math/51367443_124.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $A$ is on the line segment $DE$, $AB\\perp AC$ with the foot of the perpendicular at $A$, and $AB = AC$, $BD\\perp DE$, $CE\\perp DE$ with the perpendicular feet at $D$ and $E$ respectively. If $ED = 12$ and $BD = 8$, what is the length of $CE$?\n",
    "completion": "\\textbf{Solution:} Given that $BD\\perp DE$ and $CE\\perp DE$,\\\\\nit follows that $\\angle D=\\angle E=90^\\circ$ and $\\angle ABD+\\angle BAD=90^\\circ$,\\\\\nsince $AB\\perp AC$,\\\\\nit also follows that $\\angle BAD+\\angle EAC=90^\\circ$,\\\\\ntherefore, $\\angle ABD=\\angle EAC$,\\\\\nin $\\triangle ABD$ and $\\triangle CAE$, we have $\\left\\{\\begin{array}{l}\n\\angle D=\\angle E \\\\\nAB=CA \\\\\n\\angle ABD=\\angle EAC\n\\end{array}\\right.$,\\\\\nthus, $\\triangle ABD \\cong \\triangle CAE$ by ASA,\\\\\nresulting in $BD=AE=8$, and $AD=CE$,\\\\\ntherefore, $AD=ED-AE=12-8=4$,\\\\\nhence, $CE=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52603508.png"
  },
  {
    "solution": "\\boxed{20^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, $ \\angle AOC=90^\\circ$, $OD$ bisects $ \\angle AOB$, $ \\angle COD=35^\\circ$, what is the degree measure of $ \\angle BOC$?\n",
    "completion": "\\textbf{Solution:} Because $OD$ bisects $\\angle AOB$, \\\\\ntherefore $\\angle AOD = \\angle BOD = \\frac{1}{2} \\angle AOB$, \\\\\nbecause $\\angle AOC = 90^\\circ$, $\\angle COD = 35^\\circ$, \\\\\ntherefore $\\angle BOD = \\angle AOD = 90^\\circ - 35^\\circ = 55^\\circ$, \\\\\ntherefore $\\angle BOC = \\angle BOD - \\angle COD = 55^\\circ - 35^\\circ = \\boxed{20^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53195260.png"
  },
  {
    "solution": "\\boxed{48^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $CA=CB$, $D$ is the midpoint of $AB$, $\\angle B=42^\\circ$, what is the measure of $\\angle ACD$ in degrees?\n",
    "completion": "\\textbf{Solution:} Given that $CA=CB$, \\\\\nit follows that $\\triangle ABC$ is an isosceles triangle, \\\\\nsince $\\angle B=42^\\circ$, \\\\\nthen $\\angle A=\\angle B=42^\\circ$, \\\\\nthus $\\angle ACB=96^\\circ$, \\\\\nfurthermore, since D is the midpoint of AB, meaning CD bisects the base AB, \\\\\nit implies that CD bisects $\\angle ACB$, \\\\\ntherefore $\\angle ACD= \\frac{1}{2}\\angle ACB=\\boxed{48^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52653199.png"
  },
  {
    "solution": "\\boxed{120^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $OC$ is a ray inside $\\angle AOB$, $OD$ is the angle bisector of $\\angle AOB$, the ratio of the degrees of $\\angle AOC$ to $\\angle BOC$ is $3:2$, and $\\angle DOC=12^\\circ$. What is the degree measure of $\\angle AOB$?\n",
    "completion": "\\textbf{Solution:} Solution 1,\\\\\n$\\because \\angle AOC : \\angle BOC = 3 : 2$,\\\\\n$\\therefore$ let $\\angle AOC = (3x)^\\circ$, $\\angle BOC = (2x)^\\circ$,\\\\\n$\\therefore \\angle AOB = \\angle AOC + \\angle BOC = (5x)^\\circ$,\\\\\n$\\therefore OD$ bisects $\\angle AOB$,\\\\\n$\\therefore \\angle AOD = (2.5x)^\\circ$,\\\\\n$\\because \\angle COD = 12^\\circ$ and $\\angle COD = \\angle AOC - \\angle AOD$,\\\\\n$\\therefore 3x - 2.5x = 12$,\\\\\nSolving for $x$ gives $x = 24$,\\\\\n$\\therefore \\angle AOB = 5 \\times 24^\\circ = \\boxed{120^\\circ}$.\\\\\nSolution 2,\\\\\n$\\because \\angle AOC : \\angle BOC = 3 : 2$,\\\\\n$\\therefore \\angle AOC = 2\\angle BOC$,\\\\\n$\\because \\angle AOB = \\angle AOC + \\angle BOC$,\\\\\n$\\therefore \\angle AOB = \\angle BOC$,\\\\\n$\\because OD$ bisects $\\angle AOB$,\\\\\n$\\therefore \\angle AOD = \\angle AOB = \\angle BOC$,\\\\\n$\\because \\angle COD = \\angle AOC - \\angle AOD$,\\\\\n$\\therefore \\angle COD = \\angle BOC - \\angle BOC = \\angle BOC$,\\\\\ni.e., $\\angle BOC = 4\\angle COD$,\\\\\n$\\because \\angle COD = 12^\\circ$,\\\\\n$\\therefore \\angle BOC = 48^\\circ$,\\\\\n$\\therefore \\angle AOB = \\angle BOC = \\boxed{120^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52796090.png"
  },
  {
    "solution": "\\boxed{33^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$. An arc is drawn with point B as the center and BC as the radius, intersecting segment AB at point D; an arc is drawn with point A as the center and AD as the radius, intersecting segment AC at point E. Connect CD. If $\\angle A=24^\\circ$, what is the degree measure of $\\angle ACD$?\n",
    "completion": "\\textbf{Solution:} Given $\\because \\angle ACD=90^\\circ$ and $\\angle A=24^\\circ$,\\\\\n$\\therefore \\angle B=66^\\circ$.\\\\\nSince $BD=BC$,\\\\\n$\\therefore \\angle BCD=\\angle BDC=\\frac{180^\\circ-66^\\circ}{2}=57^\\circ$.\\\\\nThus, $\\angle ACD=90^\\circ-\\angle BCD=90^\\circ-57^\\circ=\\boxed{33^\\circ};$",
    "image_path": "dataset/mm_math/images/MM_Math/52913907_115.png"
  },
  {
    "solution": "\\boxed{\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, quadrilateral $ABCD$ is inscribed in circle $\\odot O$, with $AC$ being the diameter, and point $C$ as the midpoint of the minor arc $BD$. $AB=AD$, if $\\angle ACD=60^\\circ$, and $AD=\\sqrt{3}$, find the length of $BD$.\n",
    "completion": "\\textbf{Solution:} Given $\\because$ $\\overset{\\frown}{AD}=\\overset{\\frown}{AD}$, $\\angle ACD=60^\\circ$,\\\\\n$\\therefore$ $\\angle ABD=\\angle ACD=60^\\circ$,\\\\\n$\\because$ $AB=AD$,\\\\\n$\\therefore$ $\\triangle ABD$ is an equilateral triangle,\\\\\n$\\because$ $AD=\\sqrt{3}$,\\\\\n$\\therefore$ $BD=AD=\\boxed{\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51433720_99.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $AB$ is the diameter of $\\odot O$, $C$ is the midpoint of $\\overset{\\frown}{BD}$, and $AD$ is perpendicular to the line passing through point $C$ at point $E$. $CE$ is the tangent to $\\odot O$. If $\\angle BAD=60^\\circ$ and $AB=4\\sqrt{3}$, what is the length of $CE$?\n",
    "completion": "\\textbf{Solution:} Draw $BC$.\\\\\nSince $AB$ is the diameter of circle $O$,\\\\\nit follows that $\\angle ACB=90^\\circ$.\\\\\nSince $\\angle BAD=60^\\circ$,\\\\\nit follows that $\\angle CAD=\\angle BAC=30^\\circ$.\\\\\nSince $AB=4\\sqrt{3}$,\\\\\nit follows that $AC=AB\\cos\\angle BAC=4\\sqrt{3}\\times \\frac{\\sqrt{3}}{2}=6$,\\\\\nthus $CE=AC\\sin\\angle CAD=6\\times \\frac{1}{2}=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55502234_65.png"
  },
  {
    "solution": "\\boxed{\\frac{4\\sqrt{13}−8}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\square ABCD$, $A(-10,0)$, $B(-2,0)$, $C(0,4)$. Points $P$ and $Q$ are on the line segments $OA$ and $CD$, respectively, with $OP=2DQ$. Connect $PQ$ and $PC$, and let $DQ=x$. There exists an $x$ such that the distance from point $Q$ to $PC$ is $4$. What is the value of $x$ obtained?\n",
    "completion": "\\textbf{Solution:} There exists $x$ such that the distance from Q to PC is 4, for the following reasons: Draw line QE$\\perp$PC at point E, then QE=4, $\\angle$CEQ=$\\angle$POC=$90^\\circ$, \\\\\n$\\because$OC=4, \\\\\n$\\therefore$QE=OC, \\\\\n$\\because$AB$\\parallel$CD, \\\\\n$\\therefore$$\\angle$QCE=$\\angle$CPO, \\\\\nIn $\\triangle$QCE and $\\triangle$CPO, \\\\\n$\\left\\{\\begin{array}{l}\n\\angle QCE=\\angle CPO \\\\\n\\angle CEQ=\\angle POC \\\\\nQE=OC\n\\end{array}\\right.$, \\\\\n$\\therefore$$\\triangle$QCE$\\cong$$\\triangle$CPO, \\\\\n$\\therefore$CQ=PC, \\\\\n$\\therefore$CQ$^2$=PC$^2$, \\\\\n$\\because$CQ=CD−DQ=8−x, PC$^2$=OC$^2$+OP$^2$=$4^2$+$(2x)^2$, \\\\\n$\\therefore$(8−x)$^2$=$4^2$+$(2x)^2$, \\\\\nSolving this, we find: $x=\\frac{4\\sqrt{13}−8}{3}$ or $x=\\frac{−4\\sqrt{13}−8}{3}<0$ (discard the negative solution), \\\\\n$\\therefore x=\\boxed{\\frac{4\\sqrt{13}−8}{3}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53498608_80.png"
  },
  {
    "solution": "\\boxed{3.4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, point $O$ is the midpoint of diagonal $AC$. Through point $O$, $EF \\perp AC$ intersects $BC$ and $AD$ at points $E$ and $F$, respectively. Connect $AE$ and $CF$. Quadrilateral $AECF$ is a rhombus; if $AB = 3$ and $BC = 5$, find the length of $AE$.\n",
    "completion": "\\textbf{Solution:} Let $AE=CE=x$, so $BE=5-x$, \\\\\nsince quadrilateral ABCD is a rectangle, \\\\\ntherefore, $\\angle B=90^\\circ$.\\\\\nIn $\\triangle ABE$, by the Pythagorean theorem, $AB^2+BE^2=AE^2$,\\\\\nwhich gives $3^2+(5-x)^2=x^2$,\\\\\nsolving this, we get $x=\\boxed{3.4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53706374_110.png"
  },
  {
    "solution": "\\boxed{(-10, -8)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated in Figure 1, it is known that line $l$: $y=-x+6$ intersects the $x$ axis at point $A$ and the $y$ axis at point $B$. Line $m$ intersects the $y$ axis at point $C(0, -2)$ and intersects line $l$ at point $D(t, 1)$. As shown in Figure 2, point $P$ is on line $l$ and to the left of the $y$ axis. A line is drawn from point $P$ parallel to the $y$ axis intersecting line $m$ at point $Q$ and the $x$ axis at point $G$. When ${S}_{\\triangle PCG}=2{S}_{\\triangle QCG}$, determine the coordinates of points $P$ and $Q$.\n",
    "completion": "\\textbf{Solution:} Let $P(a, 6-a)$. Since point $P$ is on the left side of the x-axis, we have $a<0$. Given $PQ\\parallel$ to the y-axis, let $G(a,0)$, and $Q(a, \\frac{3}{5}a-2)$. As illustrated, points $P$ and $Q$ are on opposite sides of the x-axis.\n\nSince the area of $\\triangle PCG$ is $S_{\\triangle PCG}=\\frac{1}{2}PG\\cdot(-a)$, and the area of $\\triangle QCG$ is $S_{\\triangle QCG}=\\frac{1}{2}GQ\\cdot(-a)$, and given that $S_{\\triangle PCG}=2S_{\\triangle QCG}$,\n\nit follows that $PG=2QG$,\n\nleading to $6-a=2\\left(2-\\frac{3}{5}a\\right)$.\n\nSolving for $a$, we find $a=\\boxed{-10}$,\n\nTherefore, $6-a=6-(-10)=\\boxed{16}$, and $\\frac{3}{5}a-2=\\frac{3}{5}\\times (-10)-2=\\boxed{-8}$.\n\nThus, the coordinates of point $P$ are $\\boxed{(-10, 16)}$, and the coordinates of point $Q$ are $\\boxed{(-10, -8)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50496938_57.png"
  },
  {
    "solution": "\\boxed{48}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, given a rhombus OABC, where the coordinates of point $A$ are $(10, 0)$, and the hyperbola $y=\\frac{k}{x} (x>0)$ passes through point $C$, and $OB \\cdot AC = 160$, what is the value of $k$?\n",
    "completion": "\\textbf{Solution:} Construct $CD\\perp OA$ through point $C$.\\\\\nSince quadrilateral OABC is a rhombus,\\\\\nand $OB\\cdot AC=160$,\\\\\ntherefore, the area of the rhombus OABC is 80,\\\\\nwhich means $OA\\cdot CD=80$.\\\\\nSince $OA=OC=10$\\\\\ntherefore, $CD=8$.\\\\\nIn $\\triangle OCD$, $OC=10$, $CD=8$, \\\\\ntherefore, $OD=\\sqrt{10^{2}-8^{2}}=6$, that is $C(6, 8)$,\\\\\nthus the value of $k$ is $\\boxed{48}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51615096.png"
  },
  {
    "solution": "\\boxed{26^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, A, B, and C are three points on the circle $\\odot O$, and $\\angle OAB=64^\\circ$, what is the degree measure of $\\angle ACB$?\n",
    "completion": "\\textbf{Solution:} Since $OA = OB$,\\\\\nit follows that $\\angle OAB = \\angle OBA = 64^\\circ$,\\\\\nthus $\\angle AOB = 180^\\circ - 64^\\circ - 64^\\circ = 52^\\circ$,\\\\\ntherefore, $\\angle ACB = \\frac{1}{2}\\angle AOB = \\boxed{26^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52623268.png"
  },
  {
    "solution": "\\boxed{3+\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in a rhombus $ABCD$ with $AB=2$ and $\\angle A=120^\\circ$, perpendiculars are drawn from the center of symmetry $O$ of the rhombus $ABCD$ to sides $AB$ and $BC$, intersecting the sides at points $E$, $F$, $G$, and $H$. What is the area of the rhombus $ABCD$? What is the perimeter of the quadrilateral $EFGH$?\n",
    "completion": "\\textbf{Solution:} Connect $BD$ and $AC$ as shown in the diagram:\\\\\nSince quadrilateral $ABCD$ is a rhombus with $\\angle BAD=120^\\circ$,\\\\\nit follows that $AB=BC=CD=AD=2$, $\\angle BAO=\\angle DAO=60^\\circ$, and $BD\\perp AC$,\\\\\ntherefore $\\angle ABO=\\angle CBO=30^\\circ$,\\\\\nthus $OA=\\frac{AB}{2}=1$, $OB=\\sqrt{3}\\times OA=\\sqrt{3}$, meaning $AC=2\\times OA=2$, $BD=2\\times OB=2\\sqrt{3}$,\\\\\nhence the area of rhombus $ABCD$ is $\\frac{1}{2}AC\\cdot BD=\\frac{1}{2}\\times 2\\times 2\\sqrt{3}=\\boxed{2\\sqrt{3}}$;\\\\\nSince $OE\\perp AB$ and $OF\\perp BC$,\\\\\nit means $\\angle BEO=\\angle BFO=90^\\circ$,\\\\\nin $\\triangle BEO$ and $\\triangle BFO$,\\\\\n$\\left\\{\\begin{array}{l}\n\\angle BEO=\\angle BFO \\\\\n\\angle EBO=\\angle FBO \\\\\nBO=BO \\\\\n\\end{array}\\right.$,\\\\\nthus $\\triangle BEO\\cong \\triangle BFO(AAS)$,\\\\\nmeaning $OE=OF$ and $BE=BF$,\\\\\nsince $\\angle EBF=60^\\circ$,\\\\\nit follows that $\\triangle BEF$ is an equilateral triangle,\\\\\nthus $EF=BE=\\sqrt{3}\\times \\frac{\\sqrt{3}}{2}=\\frac{3}{2}$,\\\\\nBy the same reasoning, $\\triangle DGH$, $\\triangle OEH$, $\\triangle OFG$ are all equilateral triangles,\\\\\nhence $EF=GH=\\frac{3}{2}$, $EH=FG=\\frac{\\sqrt{3}}{2}$,\\\\\ntherefore, the perimeter of quadrilateral $EFGH$ is $3+\\sqrt{3}$,\\\\\nThus, the answer is: $\\boxed{3+\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52189128.png"
  },
  {
    "solution": "\\boxed{(0,1) \\cup (3, +\\infty)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y=kx+b$ ($k\\ne 0$) intersects with the graph of the inverse proportion function $y=\\frac{6}{x}$ ($x>0$) at points $A(m, 6)$ and $B(3, n)$. Based on the graph, directly write the range of $x$ values for which $kx+b<\\frac{6}{x}$ holds.\n",
    "completion": "\\textbf{Solution:} For the values of $x$ that satisfy $kx+b<\\frac{6}{x}$, the range of $x$ is $0<x<1$ or $x>3$, \\\\\nthus the answer is $x \\in \\boxed{(0,1) \\cup (3, +\\infty)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53043725_111.png"
  },
  {
    "solution": "\\boxed{180^\\circ - \\angle C}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, $\\angle BAD=121^\\circ$, $\\angle B=\\angle D=90^\\circ$, points $M$ and $N$ are respectively on $BC$ and $CD$. When $\\angle MAN=\\angle C$, what is the degree measure of $\\angle AMN+\\angle ANM$?\n",
    "completion": "\\textbf{Solution:} Given that, $\\triangle MAN$ and $\\triangle C$ are similar,\\\\\n$\\because$ in any triangle, the sum of the interior angles equals $180^\\circ$,\\\\\n$\\therefore \\angle AMN + \\angle ANM = 180^\\circ - \\angle MAN = 180^\\circ - \\angle C = \\boxed{180^\\circ - \\angle C}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52653516_88.png"
  },
  {
    "solution": "\\boxed{\\frac{168}{125}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle BAC=90^\\circ$, $AB=3$, $AC=4$. Point $D$ is the midpoint of $BC$. $\\triangle ABD$ is folded along $AD$ to obtain $\\triangle AED$. Connect $CE$. What is the distance from point $E$ to $BC$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, draw line $EB$, and construct $EH\\perp BC$ at point $H$, \\\\\nsince $\\angle BAC=90^\\circ$, $AB=3$, $AC=4$,\\\\\nthen $BC=\\sqrt{AB^2+AC^2}=\\sqrt{9+16}=5$,\\\\\nsince point $D$ is the midpoint of $BC$,\\\\\nthus $AD=BD=CD=2.5$,\\\\\nsince by folding $\\triangle ABD$ along $AD$ we get $\\triangle AED$,\\\\\nthus $AE=AB=3$, $BD=DE=CD$,\\\\\nthus $\\angle CEB=90^\\circ$, $AD$ perpendicularly bisects $BE$,\\\\\nthus $EO=BO$,\\\\\nsince $S_{\\triangle ADB}=\\frac{1}{2}S_{\\triangle ABC}=\\frac{1}{2}\\times \\frac{1}{2}\\times 3\\times 4=3$,\\\\\nthus $\\frac{1}{2}\\times \\frac{5}{2}\\times BO=3$,\\\\\nthus $BO=\\frac{12}{5}$,\\\\\nthus $BE=\\frac{24}{5}$,\\\\\nthus $DO=\\sqrt{BD^2-BO^2}=\\sqrt{\\frac{25}{4}-\\frac{144}{25}}=\\frac{7}{10}$,\\\\\nsince $\\sin\\angle DBO=\\frac{DO}{DB}=\\frac{EH}{BE}=\\frac{\\frac{7}{10}}{\\frac{5}{2}}=\\frac{EH}{\\frac{24}{5}}$,\\\\\nthus $EH=\\boxed{\\frac{168}{125}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53066224.png"
  },
  {
    "solution": "\\boxed{160^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that lines AB and CD intersect at point O, with OE $\\perp$ CD. If $\\angle BOD : \\angle BOC = 2 : 7$ and OF bisects $\\angle AOD$, what is the measure of $\\angle EOF$?\n",
    "completion": "\\textbf{Solution:}\\\\\nSince $\\angle$BOD : $\\angle$BOC = 2 : 7,\\\\\nthen $\\angle$BOC = $180^\\circ \\times \\frac{7}{7+2} = 140^\\circ$,\\\\\nthus $\\angle$AOD = $140^\\circ$,\\\\\nSince OF bisects $\\angle$AOD,\\\\\nthen $\\angle$DOF = $\\frac{1}{2}\\angle$AOD = $70^\\circ$,\\\\\nSince OE$\\perp$CD,\\\\\nthen $\\angle$EOF = $90^\\circ + 70^\\circ = \\boxed{160^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52792885_152.png"
  },
  {
    "solution": "\\boxed{25^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $O$ is on line $AB$, and $OC$, $OD$ bisect $\\angle AOE$, $\\angle BOE$ respectively. It can be deduced that $\\angle COD=90^\\circ$. (No solution required)\n\nGiven that point $O$ is on line $AB$, and $\\angle COD=90^\\circ$, $OE$ bisects $\\angle BOC$. As shown in the figure, if $\\angle AOC=50^\\circ$, what is the degree measure of $\\angle DOE$?\n",
    "completion": "\\textbf{Solution:} Given $\\because$ $\\angle COD=90^\\circ$, $\\therefore$ $\\angle AOC+\\angle BOD=90^\\circ$. $\\because$ $\\angle AOC=50^\\circ$, $\\therefore$ $\\angle BOD=40^\\circ$. $\\therefore$ $\\angle COB=\\angle COD+\\angle BOD=90^\\circ+40^\\circ=130^\\circ$. $\\because$ $OE$ bisects $\\angle BOC$, $\\therefore$ $\\angle COE=\\frac{1}{2}\\angle BOC=\\frac{1}{2}\\times 130^\\circ=65^\\circ$. $\\therefore$ $\\angle DOE=\\angle COD-\\angle COE=90^\\circ-65^\\circ=\\boxed{25^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51323829_158.png"
  },
  {
    "solution": "\\boxed{\\left(\\frac{4}{3}, \\frac{10}{3}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the line $y=\\frac{5}{2}x+5$ intersects the $x$-axis at point $A$ and the $y$-axis at point $B$. Another line passing through point $B$ intersects the positive $x$-axis at point $C$, and the area of $\\triangle ABC$ is $15$. If $M$ is a point on segment $BC$ and the area of $\\triangle ABM$ is equal to the area of $\\triangle AOB$, what are the coordinates of $M$?\n",
    "completion": "\\textbf{Solution:} From the given problem, we have $S_{\\triangle ACM} = S_{\\triangle ABC} - S_{\\triangle ABM} = S_{\\triangle ABC} - S_{\\triangle ABO} = 15 - \\frac{1}{2} \\times 2 \\times 5 = 10$,\\\\\n$\\therefore S_{\\triangle ACM} = \\frac{1}{2} \\times 6 \\times y_m = 10$. Solving this, we find: $y_m = \\frac{10}{3}$,\\\\\n$\\therefore \\frac{10}{3} = -\\frac{5}{4}x + 5$,\\\\\nSolving for $x_m$, we obtain: $x_m = \\frac{4}{3}$,\\\\\n$\\therefore M\\boxed{\\left(\\frac{4}{3}, \\frac{10}{3}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52019535_94.png"
  },
  {
    "solution": "\\boxed{120^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, paper triangle $ABC$ is folded along $DE$ such that point $A$ falls on point $A'$, and $A'B$ bisects $\\angle ABC$, $A'C$ bisects $\\angle ACB$. If $\\angle BA'C=120^\\circ$, what is the degree measure of $\\angle 1+\\angle 2$?\n",
    "completion": "\\textbf{Solution:} Let's connect $AA'$,\\\\\nsince $A'B$ bisects $\\angle ABC$, $A'C$ bisects $\\angle ACB$,\\\\\nthus $\\angle A'BC = \\frac{1}{2}\\angle ABC$, $\\angle A'CB = \\frac{1}{2}\\angle ACB$,\\\\\nsince $\\angle BA'C = 120^\\circ$,\\\\\nthus $\\angle A'BC + \\angle A'CB = 180^\\circ - 120^\\circ = 60^\\circ$,\\\\\nthus $\\angle ABC + \\angle ACB = 120^\\circ$,\\\\\nthus $\\angle BAC = 180^\\circ - 120^\\circ = 60^\\circ$,\\\\\nsince it's folded along DE,\\\\\nthus $\\angle DAA' = \\angle DA'A$, $\\angle EAA' = \\angle EA'A$,\\\\\nsince $\\angle 1 = \\angle DAA' + \\angle DA'A = 2\\angle DAA'$, $\\angle 2 = \\angle EAA' + \\angle EA'A = 2\\angle EAA'$,\\\\\nthus $\\angle 1 + \\angle 2 = 2\\angle DAA' + 2\\angle EAA' = 2\\angle BAC = 2\\times 60^\\circ = \\boxed{120^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51545384.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the inverse proportion function $y=\\frac{2}{x}$ passes through the midpoint $D$ of side AB of the rectangle OABC. What is the area of the rectangle OABC?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $DE \\perp OC$ at point $E$,\\\\\nsince the graph of the inverse proportion function $y=\\frac{2}{x}$ passes through the midpoint $D$ of side $AB$ of rectangle $OABC$,\\\\\ntherefore, $S_{\\text{rectangle }OABC}=2S_{\\text{rectangle }OADE}$,\\\\\nalso, since $k=2$,\\\\\ntherefore, $S_{\\text{rectangle }OADE}=2$,\\\\\ntherefore, $S_{\\text{rectangle }OABC}=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51435972.png"
  },
  {
    "solution": "\\boxed{28}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point B is a point on line segment AC, and $AB=21$, $BC=\\frac{1}{3}AB$. Calculate the length of line segment AC.\n",
    "completion": "\\textbf{Solution:} We have $BC = \\frac{1}{3}AB = 7$, \\\\\n$AC = AB + BC = AB + \\frac{1}{3}AB = 21 + 7 = \\boxed{28}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52251794_178.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points $C$ and $E$ are two points on line segment $AB$, point $D$ is the midpoint of line segment $AB$, and $AB=6$, $CD=1$. If $AE:EC=1:3$, what is the length of $EC$?\n",
    "completion": "\\textbf{Solution:} Let $AE=x$, then $EC=3x$. \\\\\nSince $AC=AD+DC=AE+EC$, \\\\\nit follows that $3+1=x+3x$, solving this gives: $x=1$, \\\\\ntherefore, $EC=3x=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51561259_161.png"
  },
  {
    "solution": "\\boxed{BC \\text{ is the tangent to the circle } O}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, $AB$ is a chord of $\\odot O$. Through point $O$, construct $OC\\perp OA$, with $OC$ intersecting $AB$ at point $P$, and $CP=BC$. Point $Q$ is a point on the arc $\\overset{\\frown}{AmB}$. Prove that: $BC$ is a tangent to $\\odot O$.\n",
    "completion": "\\textbf{Solution} Proof: Connect OB,\\\\\nsince OA = OB,\\\\\nit follows that $\\angle OAB = \\angle OBA$,\\\\\nsince PC = CB,\\\\\nit follows that $\\angle CPB = \\angle PBC$,\\\\\nsince $\\angle APO = \\angle CPB$,\\\\\nit follows that $\\angle APO = \\angle CBP$,\\\\\nsince OC $\\perp$ OA,\\\\\nit follows that $\\angle AOP = 90^\\circ$,\\\\\nthus, $\\angle OAP + \\angle APO = 90^\\circ$,\\\\\nthus, $\\angle CBP + \\angle ABO = 90^\\circ$,\\\\\nthus, $\\angle CBO = 90^\\circ$,\\\\\nthus, \\boxed{BC \\text{ is the tangent to the circle } O}.",
    "image_path": "dataset/mm_math/images/MM_Math/52622752_92.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, where $M$ is a point on $BC$, $EM\\perp AM$ intersects the extension of $AD$ at point $E$. $\\triangle ABM \\sim \\triangle EMA$. If $AB=4$, $BM=3$, what is the value of $\\sin E$?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABM \\sim \\triangle EMA$,\\\\\nit follows that $\\angle E = \\angle BAM$,\\\\\nin $\\triangle ABM$, $AM = \\sqrt{BM^2+AB^2} = \\sqrt{3^2+4^2} = 5$,\\\\\nthus $\\sin \\angle BAM = \\frac{BM}{AM} = \\frac{3}{5}$,\\\\\nhence $\\sin E = \\boxed{\\frac{3}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51445931_110.png"
  },
  {
    "solution": "\\boxed{x^{2}-2x-3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the quadratic function $y=x^2+bx+c$ intersects the x-axis at points $A$ and $B$, and intersects the y-axis at point $C$. Furthermore, it is symmetrical about the line $x=1$, and the coordinates of point $A$ are $(-1,0)$. What is the expression of the quadratic function?\n",
    "completion": "\\textbf{Solution:} From the given, we have:\n\n\\[\n\\left\\{\\begin{array}{l}\n-\\frac{b}{2a}=1\\\\ \n1-b+c=0\n\\end{array}\\right.\n\\]\n\nTherefore, \n\n\\[\n\\left\\{\\begin{array}{l}\nb=-2\\\\ \nc=-3\n\\end{array}\\right.\n\\]\n\nHence, \\(y=x^{2}-2x-3=\\boxed{x^{2}-2x-3}\\)",
    "image_path": "dataset/mm_math/images/MM_Math/51369956_188.png"
  },
  {
    "solution": "\\boxed{135^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, two right-angle triangles are overlapped at point $O$ with their vertices coinciding, where the vertex $C$ of one triangle falls on the edge $OA$ of the other triangle. It is known that $\\angle ABO=\\angle DCO=90^\\circ$, $\\angle AOB=45^\\circ$, $\\angle COD=60^\\circ$. Draw the bisector of $\\angle AOD$ intersecting side $CD$ at point $E$. As shown in Figure 2, if point $C$ does not fall on side $OA$, when $\\angle COE=15^\\circ$, find the degree measure of $\\angle BOD$.\n",
    "completion": "Solution: Since $\\angle COD=60^\\circ$ and $\\angle COE=15^\\circ$, \\\\\ntherefore $\\angle DOE=\\angle COD-\\angle COE=45^\\circ$\\\\\nSince $OE$ bisects $\\angle AOD$,\\\\\ntherefore $\\angle AOD=2\\angle DOE=90^\\circ$\\\\\nSince $\\angle AOB=45^\\circ$\\\\\ntherefore $\\angle BOD=\\angle AOD+\\angle AOB=\\boxed{135^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/51495039_154.png"
  },
  {
    "solution": "\\boxed{27^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 2, in $\\triangle ABC$, $AD$ and $BE$ bisect $\\angle BAC$ and $\\angle ABC$ respectively. If $\\angle C = 60^\\circ$ and $\\angle ADE$ is $6^\\circ$ larger than $\\angle BED$, what is the measure of $\\angle BED$?\n",
    "completion": "\\textbf{Solution:} Given that $AD$ and $BE$ bisect $\\angle BAC$ and $\\angle ABC$ respectively,\n\n$\\therefore \\angle 1=\\angle 2$, $\\angle 3=\\angle 4$,\n\nalso, given that $\\angle C=60^\\circ$,\n\n$\\therefore \\angle BAC+\\angle ABC=180^\\circ - \\angle C=180^\\circ - 60^\\circ=120^\\circ$,\n\n$\\therefore \\angle 1+\\angle 2+\\angle 3+\\angle 4=120^\\circ$,\n\n$\\therefore 2\\angle 1+2\\angle 3=120^\\circ$,\n\n$\\therefore \\angle 1+\\angle 3=60^\\circ$,\n\nFrom the figure, it is known that $\\triangle ABF$ and $\\triangle DEF$ are vertically opposite triangles,\n\n$\\therefore \\angle 1+\\angle 3=\\angle ADE+\\angle BED=60^\\circ$ (1),\n\nalso, given that $\\angle ADE$ is $6^\\circ$ larger than $\\angle BED$,\n\n$\\therefore \\angle ADE-\\angle BED=6^\\circ$ (2),\n\nSolving (1) and (2) together yields $\\begin{cases} \\angle ADE+\\angle BED=60^\\circ \\\\ \\angle ADE-\\angle BED=6^\\circ \\end{cases}$,\n\nSolution obtained is: $\\begin{cases} \\angle ADE=33^\\circ \\\\ \\angle BED=27^\\circ \\end{cases}$,\n\n$\\therefore \\angle BED=\\boxed{27^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53413072_100.png"
  },
  {
    "solution": "\\boxed{0.8\\sin20^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $O$ is the midpoint of the seesaw $AB$. The support column $OC$ is perpendicular to the ground $MN$, with the foot of the perpendicular being point $C$. When one end $B$ of the seesaw touches the ground, the angle between seesaw $AB$ and the ground $MN$ is $20^\\circ$. Given that $AB=1.6m$, what is the length of $OC$?\n",
    "completion": "\\textbf{Solution:} Since O is the midpoint of AB, and AB=1.6,\\\\\nit follows that OB=$\\frac{1}{2}$AB=0.8,\\\\\nIn $\\triangle OCB$, $\\sin\\angle OBC=\\frac{OC}{OB}$,\\\\\nTherefore, OC=OB$\\cdot\\sin\\angle OBC=0.8\\sin20^\\circ=\\boxed{0.8\\sin20^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/53043139.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB=6$. Extend the line segment $AB$ to point $C$, and $D$ is the midpoint of $AC$. If $BC=2AB$, what is the length of $BD$?\n",
    "completion": "\\textbf{Solution:} Since $BC = 2AB$ and $AB = 6$,\\\\\nit follows that $BC = 12$,\\\\\nhence $AC = AB + BC = 18$,\\\\\nGiven that $D$ is the midpoint of $AC$,\\\\\nit means $AD = \\frac{1}{2}AC = 9$,\\\\\ntherefore $BD = AD - AB = \\boxed{3};$",
    "image_path": "dataset/mm_math/images/MM_Math/52632926_190.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated, $ \\triangle ABC$ and $ \\triangle A_1B_1C_1$ are similar with the center of similarity being point $O$. Given that $OA: OA_1 = 1:2$ and the area of $ \\triangle ABC$ is 3, what is the area of $ \\triangle A_1B_1C_1$?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABC \\sim \\triangle A_1B_1C_1$,\\\\\nit follows that $\\triangle ABC \\sim \\triangle A_1B_1C_1$, $AC \\parallel A_1C_1$,\\\\\nthus $\\triangle AOC \\sim \\triangle A_1OC_1$,\\\\\ntherefore $\\frac{AC}{A_1C_1} = \\frac{OA}{OA_1} = \\frac{1}{2}$,\\\\\nthus, the area ratio of $\\triangle ABC$ to $\\triangle A_1B_1C_1$ is $1\\colon 4$,\\\\\nsince the area of $\\triangle ABC$ is $3$,\\\\\nthe area of $\\triangle A_1B_1C_1$ is $\\boxed{12}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53578726.png"
  },
  {
    "solution": "\\boxed{\\frac{25}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AD$ is the angle bisector, and point $E$ is a point on side $AC$, satisfying $\\angle ADE = \\angle B$. $\\triangle ADB \\sim \\triangle AED$; if $AE = 3$ and $AD = 5$, find the length of $AB$.\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ADB \\sim \\triangle AED$, \\\\\nit follows that $\\frac{AD}{AE}=\\frac{AB}{AD}$, \\\\\nsince $AE=3$, and $AD=5$, \\\\\nthus $\\frac{5}{3}=\\frac{AB}{5}$, \\\\\ntherefore $AB=\\boxed{\\frac{25}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52872073_87.png"
  },
  {
    "solution": "\\boxed{(24,0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within the rectangle $ABCO$, point $C$ is on the $x$-axis, and point $A$ is on the $y$-axis, with point $B$ having the coordinates $(-9, 12)$. The rectangle $ABCO$ is folded along line $BD$, such that point $A$ falls at point $E$ on the diagonal $OB$, and line $BD$ intersects lines $OA$ and the $x$-axis at points $D$ and $F$, respectively. On the $x$-axis, there exists a point $M$, such that the quadrilateral formed by vertices $A$, $B$, $F$, and $M$ is a parallelogram. If such a point $M$ exists, find the coordinates of point $M$ that satisfy this condition.\n",
    "completion": "\\textbf{Solution:} From equation (2), we have $OD=\\frac{15}{2}$, thus $D\\left(0,\\frac{15}{2}\\right)$.\\\\\nLet the equation of line BD be $y=kx+b$,\\\\\nSince $B(-9,12)$, $D\\left(0,\\frac{15}{2}\\right)$,\\\\\nThen $\\begin{cases}-9k+b=12,\\\\ b=\\frac{15}{2},\\end{cases}$ yields $\\begin{cases}k=-\\frac{1}{2},\\\\ b=\\frac{15}{2},\\end{cases}$\\\\\nTherefore, the equation of $BD$ is: $y=-\\frac{1}{2}x+\\frac{15}{2}$.\\\\\nWhen $y=0$, $x=15$,\\\\\nTherefore, $OF=15$.\\\\\nAlso, since $AB=9$,\\\\\nTherefore, the coordinates of the point M that satisfy the conditions are $\\boxed{(6,0)}$ or $\\boxed{(24,0)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52056028_209.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that AD is the median of side BC in $\\triangle ABC$. A line is drawn through the centroid G, $GE\\parallel AC$, intersecting BC at point E. If $DE=2$, what is the length of BC?\n",
    "completion": "\\textbf{Solution:} Since $G$ is the centroid,\\\\\nhence $DG : DA = 1 : 3$,\\\\\nsince $GE \\parallel AC$,\\\\\nhence $DE : DC = DG : DA = 1 : 3$,\\\\\nsince $DE = 2$,\\\\\nhence $CD = 6$,\\\\\ntherefore $BC = 2CD = \\boxed{12}$",
    "image_path": "dataset/mm_math/images/MM_Math/52861887.png"
  },
  {
    "solution": "\\boxed{x\\leq 1, {y}_{1}\\ge {y}_{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the linear function $y_1=kx-2$ intersects the x-axis at point $B(-2, 0)$, and $y_2=x+b$ intersects the x-axis at point $C(4, 0)$. The graphs of these two functions intersect at point $A$. Based on the graph, directly write the range of values for $x$ when $y_1\\ge y_2$.\n",
    "completion": "\\textbf{Solution:} From the graph, it is known that at point A or to the left of point A, ${y}_{1}\\ge {y}_{2}$. \\\\\nHence, when $\\boxed{x\\leq 1, {y}_{1}\\ge {y}_{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52381120_150.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the Cartesian coordinate system, the coordinates of point A are $(6,4)$. Lines $AB$ and $AC$ are drawn from point A and are perpendicular to the x-axis at point B and to the y-axis at point C, respectively. The graph of the linear function $y=kx+b$ passes through point $P(2,3)$. When $k=2$, how long is the segment intercepted by the line $y=kx+b$ on the rectangle $OBAC$?\n",
    "completion": "\\textbf{Solution:} For the line $y=kx+b$, the length of the segment intercepted by the rectangle $OBAC$ is $\\boxed{2\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52342947_115.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, AB is the diameter of $\\odot O$, with chord CDAB. The foot of the perpendicular from C to AB is point E. If the radius of $\\odot O$ is 5, and CD equals 8, what is the length of AE?\n",
    "completion": "\\textbf{Solution:} Connect OC, as shown in the diagram\\\\\nSince AB is the diameter of $\\odot$O, and CDAB is perpendicular at point E, with CD=8,\\\\\nit follows that $CE=\\frac{1}{2}CD=\\frac{1}{2}\\times 8=4$,\\\\\nSince $AO=CO=5$,\\\\\nit follows that $OE=\\sqrt{CO^{2}-CE^{2}}=\\sqrt{5^{2}-4^{2}}=3$,\\\\\ntherefore, $AE=5-3=\\boxed{2}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51351118.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{21}-2\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the semicircle \\(O\\) with \\(AB\\) as the diameter, \\(C\\) is the trisection point of the semicircle. Point \\(P\\) is a moving point on arc \\(BC\\), and lines \\(CP\\) and \\(AP\\) are connected. Construct \\(OM\\) perpendicular to \\(CP\\) and intersecting \\(AP\\) at \\(N\\), connect \\(BN\\). If \\(AB=12\\), what is the minimum value of \\(NB\\)?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect AC and OC. \\\\\n$\\because$ C is the trisecting point of the semicircle, \\\\\n$\\therefore$ $\\angle$AOC = $60^\\circ$, \\\\\n$\\because$ OA = OC, \\\\\n$\\therefore$ $\\triangle$AOC is an equilateral triangle, \\\\\nConstruct the circumscribed circle $\\odot$T of $\\triangle$AOC, and connect TA=TC, TN, TB. \\\\\n$\\because$ OM$\\perp$PC, \\\\\n$\\therefore$ CM = PM, \\\\\n$\\therefore$ NC = NP, \\\\\n$\\therefore$ $\\angle$NPC = $\\angle$NCP = $\\frac{1}{2}\\angle$AOC = $30^\\circ$, \\\\\n$\\therefore$ $\\angle$CNM = $60^\\circ$, \\\\\n$\\therefore$ $\\angle$CNO = $120^\\circ$, \\\\\n$\\because$ $\\angle$CNO + $\\angle$OAC = $180^\\circ$, \\\\\n$\\therefore$ point N is on $\\odot$T, the trajectory is $\\overset{\\frown}{OC}$, \\\\\nDraw TH$\\perp$AB at H through point T. \\\\\nIn the right triangle $\\triangle$ATH, AH=OH=3, $\\angle$TAH=$30^\\circ$, \\\\\n$\\therefore$ TH = AH$\\cdot \\tan30^\\circ$ = $\\sqrt{3}$, \\\\\n$\\therefore$ AT = TN = 2HN = 2$\\sqrt{3}$, \\\\\nIn the right triangle $\\triangle$BHT, BT = $\\sqrt{TH^2+BH^2}=\\sqrt{(\\sqrt{3})^2+9^2}=2\\sqrt{21}$, \\\\\n$\\because$ BN $\\geq$ BT−TN, \\\\\n$\\therefore$ BN $\\geq 2\\sqrt{21}-2\\sqrt{3}$, \\\\\n$\\therefore$ The minimum value of BN is $\\boxed{2\\sqrt{21}-2\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52483220.png"
  },
  {
    "solution": "\\boxed{3, 2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure: it is known that the line $y_1=kx+b$ passes through point $A(5,0)$ and $B(1,4)$, and intersects with the line $y_2=2x-4$ at point $C$. What are the coordinates of point $C$?\n",
    "completion": "\\textbf{Solution:} Given the problem, we have:\n\\[\n\\left\\{\n\\begin{array}{l}\ny = -x + 5 \\\\\ny = 2x - 4\n\\end{array}\n\\right.\n\\]\nSolving these equations, we get:\n\\[\n\\left\\{\n\\begin{array}{l}\nx = 3 \\\\\ny = 2\n\\end{array}\n\\right.\n\\]\n$\\therefore C(\\boxed{3, 2})$",
    "image_path": "dataset/mm_math/images/MM_Math/50561230_54.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{7}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, points $D$ and $E$ are the midpoints of sides $AB$ and $AC$, respectively. Line $CF$ is drawn parallel to $AB$ through point $C$ and intersects the extension of $DE$ at point $F$. Line $BE$ is drawn. Quadrilateral $BCFD$ is a parallelogram. When $AB=BC$, if $BD=2$ and $BE=3$, find the length of $AC$.\n",
    "completion": "\\textbf{Solution:} Since $AB=BC$ and $E$ is the midpoint of $AC$,\nthus $BE\\perp AC$.\\\\\nSince $AB=2DB=4$ and $BE=3$,\\\\\nthus $AE=\\sqrt{4^2-3^2}=\\sqrt{7}$,\\\\\ntherefore $AC=2AE=2\\sqrt{7}=\\boxed{2\\sqrt{7}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52284290_110.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, fold the rectangular paper \\(ABCD\\) such that side \\(DC\\) falls onto the diagonal \\(AC\\), with the fold line being \\(CE\\), and point \\(D\\) falls on the diagonal at point \\(D'\\). If \\(AB=3\\) and \\(AD=4\\), what is the length of \\(ED\\)?\n",
    "completion": "\\textbf{Solution:} Given that $AB=3$, $AD=4$, thus $DC=3$, $BC=4$. Therefore, $AC=\\sqrt{AB^{2}+BC^{2}}=5$. By the principle of folding, we have $\\triangle DEC\\cong \\triangle D'EC$, hence $D'C=DC=3$, $DE=D'E$. Let $ED=x$, then $D'E=x$, $AD'=AC-CD'=2$, $AE=4-x$. In the triangle $\\triangle AED'$: $(AD')^{2}+(ED')^{2}=AE^{2}$, therefore $2^{2}+x^{2}=(4-x)^{2}$. Solving this, we find $x=\\boxed{\\frac{3}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52814867.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within circle $O$, $AB$ is the diameter, and chord $CD \\perp AB$ at $E$. Connect $OC$ and $OD$. If the diameter is $10$ and $CD=8$, then what is the length of $BE$?\n",
    "completion": "Solution: Since $CD \\perp AB$ at $E$, and $CD=8$, with $AB$ being the diameter, \\\\\ntherefore $CE=\\frac{1}{2}CD=\\frac{1}{2}\\times 8=4$, and $\\angle OEC=90^\\circ$\\\\\ntherefore $OE=\\sqrt{OC^{2}-CE^{2}}=\\sqrt{5^{2}-4^{2}}=3$\\\\\ntherefore $BE=OB-OE=5-3=\\boxed{2}$",
    "image_path": "dataset/mm_math/images/MM_Math/51367517.png"
  },
  {
    "solution": "\\boxed{160^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $OD$ bisects $\\angle AOB$, ray $OC$ is inside $\\angle AOD$, $\\angle BOC = 2\\angle AOC$, $\\angle AOB = 120^\\circ$, find the complement of $\\angle COD$.\n",
    "completion": "\\textbf{Solution:} Since $\\angle AOB = \\angle BOC + \\angle AOC$ and $\\angle BOC = 2\\angle AOC$,\\\\\nit follows that $\\angle AOB = 2\\angle AOC + \\angle AOC$, that is $3\\angle AOC = 120^\\circ$,\\\\\nhence $\\angle AOC = 40^\\circ$.\\\\\nSince OD bisects $\\angle AOB$,\\\\\nit follows that $\\angle AOD = \\angle BOD = \\frac{1}{2}\\angle AOB = \\frac{1}{2} \\times 120^\\circ = 60^\\circ$.\\\\\nSince $\\angle COD = \\angle AOD - \\angle AOC$,\\\\\nit follows that $\\angle COD = 60^\\circ - 40^\\circ = 20^\\circ$.\\\\\nTherefore, the supplement of $\\angle COD$ is $180^\\circ - \\angle COD = 180^\\circ - 20^\\circ = \\boxed{160^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50166938.png"
  },
  {
    "solution": "\\boxed{\\frac{35}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in right-angled triangle $ABC$, $\\angle C=90^\\circ$, $AC=8\\text{cm}$, and $BC=6\\text{cm}$. Now, a moving point $P$ starts from point $B$ and moves towards the endpoint $A$ along line segment $BA$, and a moving point $Q$ starts from point $A$ and moves towards the endpoint along the polyline $AC-CB$. If the speed of point $P$ is $1\\text{cm/s}$, and the speed of point $Q$ is $1\\text{cm/s}$. They start at the same time, and when one point reaches the endpoint, the other point also stops moving. Let the time of movement be $t$ seconds. As shown in figure 2, $Q$ is on $CB$. Is there a moment that makes the triangle with vertices $B$, $P$, $Q$ similar to $\\triangle ABC$? If it exists, find the value of $t$; if not, please explain the reason.\n",
    "completion": "\\textbf{Solution:} As shown in Figure 3, when $\\triangle BPQ\\sim \\triangle BAC$, \\\\\n$\\therefore \\frac{BP}{AB}=\\frac{BQ}{BC}$. \\\\\n$\\because BQ=14-t$ and $BP=t$,\\\\\n$\\therefore \\frac{t}{10}=\\frac{14-t}{6}$,\\\\\n$\\therefore t= \\boxed{\\frac{35}{4}}$,\\\\\nWhen $\\triangle BQP\\sim \\triangle BAC$,\\\\\n$\\therefore \\frac{BP}{BC}=\\frac{BQ}{BA}$,\\\\\n$\\therefore \\frac{t}{6}=\\frac{14-t}{10}$,\\\\\n$\\therefore t= \\frac{21}{4}$,\\\\\n$\\because Q$ is on $CB$,\\\\\n$\\therefore t>8$,\\\\\n$\\therefore t= \\frac{21}{4}$ is discarded,\\\\\n$\\therefore when $t= \\frac{35}{4}$, $Q$ is on $CB$, and the triangle formed by points $B$, $P$, and $Q$ is similar to $\\triangle ABC$.",
    "image_path": "dataset/mm_math/images/MM_Math/51368352_98.png"
  },
  {
    "solution": "\\boxed{77.5^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, the perpendicular bisectors $DE$ and $FG$ of $AB$ and $AC$, respectively, intersect $BC$ at points $E$ and $G$. If point $G$ is to the left of point $E$ and $\\angle GAE=25^\\circ$, then what is the measure of $\\angle BAC$ in degrees?\n",
    "completion": "\\textbf{Solution:} \nGiven that $DE$ is the perpendicular bisector of $AB$, and $FG$ is the perpendicular bisector of $AC$,\\\\\nthus $EA=EB$, $GA=GC$,\\\\\ntherefore, $\\angle B=\\angle BAE$, $\\angle C=\\angle GAC$,\\\\\nsince $\\angle B+\\angle BAE+\\angle GAC+\\angle C-\\angle GAE=180^\\circ$, $\\angle GAE=25^\\circ$,\\\\\nit follows that $2\\angle B+2\\angle C-25^\\circ=180^\\circ$,\\\\\ntherefore, $\\angle B+\\angle C=102.5^\\circ$,\\\\\nhence, $\\angle BAC=180^\\circ-(\\angle B+\\angle C)=\\boxed{77.5^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52774845.png"
  },
  {
    "solution": "\\boxed{y=-x-3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, it is known that the parabola $y=x^2-2x-3$ has its vertex at point $A$, intersects the y-axis at point $C$, and the line segment $CB$ is parallel to the x-axis, intersecting the parabola at another point $B$. Find the coordinates of point $B$ and the equation of the line $AC$.\n",
    "completion": "Given the problem, we have $y=x^{2}-2x-3=(x-1)^{2}-4$, \\\\\n$\\therefore$ the vertex $A(1, -4)$, \\\\\nLet $x=0$, then $y=-3$, \\\\\n$\\therefore C(0, -3)$, \\\\\n$\\because CB\\parallel x$-axis, \\\\\n$\\therefore B(2, -3)$, \\\\\nLet the equation of line $AC$ be $y=kx+b$, \\\\\n$\\left\\{\\begin{array}{l}\nk+b=-4 \\\\\nb=-3\n\\end{array}\\right.$, \\\\\nSolving yields $\\left\\{\\begin{array}{l}\nk=-1 \\\\\nb=-3\n\\end{array}\\right.$, \\\\\n$\\therefore y=-x-3$.\n\nThe coordinates of point $B$ are $B=\\boxed{(2, -3)}$, the equation of line $AC$ is $\\boxed{y=-x-3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52512802_169.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y_1 = x + 1$ intersects the x-axis and y-axis at points A and B, respectively. The line $y_2 = -2x + 4$ intersects the x-axis and y-axis at points C and D, respectively. The two lines intersect at point E. What is the area of $\\triangle ACE$?\n",
    "completion": "\\textbf{Solution:} Solve when $y_1=x+1=0$, we get: $x=-1$, $\\therefore A(-1,0)$. When $y_2=-2x+4=0$, we get: $x=2$, $\\therefore C(2,0)$, $\\therefore AC=2-(-1)=3$, $S_{\\triangle ACE}=\\frac{1}{2}\\times AC\\times y_E =\\frac{1}{2}\\times 3\\times 2 =\\boxed{3}.$",
    "image_path": "dataset/mm_math/images/MM_Math/52380902_164.png"
  },
  {
    "solution": "\\boxed{\\frac{10\\sqrt{26}}{13}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $E$ is the midpoint of $BC$, and $DF \\perp AE$ with the foot of the perpendicular being $F$. $\\triangle ABE \\sim \\triangle DFA$; If $AB = 10$ and $BC = 4$, find the length of $DF$.\n",
    "completion": "\\textbf{Solution:} Given that $E$ is the midpoint of $BC$, and $BC=4$. Since quadrilateral $ABCD$ is a rectangle, it follows that $AD=BC=4$. Hence, $BE=EC=\\frac{1}{2}BC=2$. Since $\\angle ABE=90^\\circ$, we have $AE=\\sqrt{AB^{2}+BE^{2}}=2\\sqrt{26}$. Given that $\\triangle ABE\\sim \\triangle DFA$, we have $\\frac{AE}{AD}=\\frac{AB}{DF}$. Thus, $DF=\\frac{AB\\cdot AD}{AE}=\\boxed{\\frac{10\\sqrt{26}}{13}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51433236_143.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of $\\odot O$, with the chord $CD\\perp AB$ at $E$, and $F$ is a point on the extension of $AB$. Connect $CF$ and $DF$. Given that $OE = 3$, $BE = 2$, find the length of $CD$.\n",
    "completion": "\\textbf{Solution:} Connect OC, \\\\\n$\\because$ CD$\\perp$AB, \\\\\n$\\therefore$ CE = DE, \\\\\n$\\therefore$ OC = OB = OE + BE = 3 + 2 = 5, \\\\\nIn $\\triangle OCE$, $\\angle OEC = 90^\\circ$, by the Pythagorean theorem, we have: \\\\\nCE$^{2}$ = OC$^{2}$ - OE$^{2}$, \\\\\n$\\therefore$ CE$^{2}$ = 5$^{2}$ - 3$^{2}$, \\\\\n$\\therefore$ CE = 4, \\\\\n$\\therefore$ CD = 2CE = \\boxed{8}.",
    "image_path": "dataset/mm_math/images/MM_Math/51433293_101.png"
  },
  {
    "solution": "\\boxed{\\frac{21}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, point D is the midpoint of the hypotenuse $AB$, $CD=5$, $BC=6$. Connect $CD$, fold $\\triangle BCD$ along $CD$ so that point B falls on point E, point F is a point on the right-angle side $AC$, connect $DF$, fold $\\triangle ADF$ along $DF$ so that point A coincides with point E. What is the area of $\\triangle EFC$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle ACB=90^\\circ$ and point D is the midpoint of hypotenuse $AB$ with $CD=5$ and $BC=6$,\\\\\n$\\therefore$ $AB=10$,\\\\\n$\\therefore$ $AC=\\sqrt{AB^2-BC^2}=8$,\\\\\nFrom folding, we know that $\\angle B=\\angle DEC$, $\\angle A=\\angle DEF$, $CE=BC=6$, and $AF=EF$,\\\\\n$\\because$ $\\angle A+\\angle B=90^\\circ$,\\\\\n$\\therefore$ $\\angle DEF+\\angle DEC=90^\\circ$, which means $\\angle FEC=90^\\circ$,\\\\\n$\\therefore$ $EF^2+CE^2=CF^2$,\\\\\nLet $AF=EF=x$, then $CF=AC-AF=8-x$,\\\\\n$\\therefore$ $x^2+6^2=(8-x)^2$,\\\\\nSolving for $x$ yields $x=\\frac{7}{4}$,\\\\\n$\\therefore$ $EF=\\frac{7}{4}$,\\\\\n$\\therefore$ The area of $\\triangle EFC$ is $\\frac{1}{2}CE\\times EF=\\frac{1}{2}\\times 6\\times \\frac{7}{4}=\\boxed{\\frac{21}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53033463.png"
  },
  {
    "solution": "\\boxed{(\\frac{18}{25}, \\frac{116}{25})}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=-\\frac{1}{2}x+5$ intersects with the coordinate axes at points A and B, and intersects with the graph of the inverse proportion function $y=\\frac{k}{x}\\ (x>0)$ at points C and D, with $CD=3AC$. Point E is a point on the line AB. Connect OE and construct a right triangle OEF to the right of OE, where $\\angle OEF=90^\\circ$, $\\angle OFE=\\angle ABO$. If side OF intersects the graph of the inverse proportion function at point G, and $OG=GF$, what is the value of k? What are the coordinates of point E?\n",
    "completion": "\\textbf{Solution:} Let the line $y=-\\frac{1}{2}x+5$ intersect the coordinate axes at points A and B,\\\\\n$\\therefore A(0, 5), B(10, 0)$. Let the coordinates of point C be $(a, b)$. Draw CM$\\perp y$-axis at point M, and through point D draw DN$\\perp y$-axis at point N. Then $b=-\\frac{1}{2}a+5$, CM=a, AM=5−b,\\\\\n$\\triangle AMC\\sim \\triangle AND$\\\\\n$\\therefore \\frac{AM}{AN}=\\frac{AC}{AD}$. Since $CD=3AC$, AD=AC+CD\\\\\n$\\therefore \\frac{5−b}{AN}=\\frac{1}{4}$\\\\\n$\\therefore AN=4(5−b)$,\\\\\n$\\therefore ON=OA−AN=5−4(5−b)=4b−15$,\\\\\n$\\therefore$ the coordinates of point D are $(4a, 4b−15)$,\\\\\nSince points C, D lie on the graph of the inverse proportion $y=\\frac{k}{x}(x>0)$,\\\\\n$\\therefore k=ab=4a(4b−15)$. Solving for b, we get $b=4$, substituting $b=4$ into,\\\\\n$b=-\\frac{1}{2}a+5$ gives $a=2$. Solving for $k$, we get $k=\\boxed{8}$. Draw BF. Since $\\angle OFE=\\angle ABO$, the points O, B, F, E lie on the same circle. Since $\\angle OEF=90^\\circ$ and OG=GF, point G is the center of the circle, and OF is the diameter, hence $\\angle OBF=90^\\circ$. Since B(10, 0), the x-coordinate of point G is 5. When x=5, \\\\\n$y=\\frac{8}{x}=\\frac{8}{5}$, hence the coordinates of point G are\\\\\n$(5, \\frac{8}{5})$. Since OG=GF, the coordinates of point F are\\\\\n$(10, \\frac{16}{5})$. Let the coordinates of point E be\\\\\n$(x, -\\frac{1}{2}x+5)$. By the Pythagorean theorem, OE$^2$+EF$^2$=OF$^2$, so\\\\\n$x^2+(-\\frac{1}{2}x+5)^2+(10-x)^2+(-\\frac{1}{2}x+5-\\frac{16}{5})^2=10^2+(\\frac{16}{5})^2$ Solving for x,\\\\\n$x=\\frac{18}{25}$ or $10$ (discard), hence the coordinates of point E are\\\\\n$(\\frac{18}{25}, \\frac{116}{25})$, therefore the answer is:\\\\\nThe coordinates of point E are $\\boxed{(\\frac{18}{25}, \\frac{116}{25})}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51407086.png"
  },
  {
    "solution": "\\boxed{-2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the right triangle has its right-angled vertex at the coordinate origin, $\\angle OAB=30^\\circ$. If point A lies on the graph of the inverse proportional function $y=\\frac{6}{x}$ $(x>0)$, what is the value of $k$ in the inverse proportional function $y=\\frac{k}{x}$ passing through point B?\n",
    "completion": "\\textbf{Solution:} Draw $BC\\perp x$ axis at point $C$ through point $B$, and draw $AD\\perp x$ axis at point $D$ through point $A$, as shown in the figure.\\\\\n$\\because \\angle BOA=90^\\circ$,\\\\\n$\\therefore \\angle BOC+\\angle AOD=90^\\circ$,\\\\\n$\\because \\angle AOD+\\angle OAD=90^\\circ$,\\\\\n$\\therefore \\angle BOC=\\angle OAD$,\\\\\nAlso, $\\because \\angle BCO=\\angle ADO=90^\\circ$,\\\\\n$\\therefore \\triangle BCO\\sim \\triangle ODA$,\\\\\n$\\therefore \\frac{OB}{OA}=\\tan30^\\circ=\\frac{\\sqrt{3}}{3}$,\\\\\n$\\therefore \\frac{{S}_{\\triangle BCO}}{{S}_{\\triangle ODA}}=\\frac{1}{3}$,\\\\\n$\\because \\frac{1}{2}\\times AD\\times DO=\\frac{1}{2}xy=3$,\\\\\n$\\therefore {S}_{\\triangle BCO}=\\frac{1}{2}\\times BC\\times CO=\\frac{1}{3}{S}_{\\triangle ODA}=1$,\\\\\n$\\because$ The graph of the inverse proportion function that passes through point $B$ is in the second quadrant,\\\\\ntherefore, the equation of the inverse proportion function is: $y=-\\frac{2}{x}$,\\\\\n$\\therefore k=\\boxed{-2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51379205.png"
  },
  {
    "solution": "\\boxed{(0,1)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=-\\frac{1}{3}x+1$ intersects the x-axis and y-axis at points A and B, respectively. Find the coordinates of points A and B.\n",
    "completion": "\\textbf{Solution:} For $y = -\\frac{1}{3}x + 1$, let $x = 0$, then $y = 1$, let $y = 0$, that is $-\\frac{1}{3}x + 1 = 0$, solving gives $x = 3$,\\\\\nhence the coordinates of points A and B are \\boxed{(3,0)} and \\boxed{(0,1)}, respectively.",
    "image_path": "dataset/mm_math/images/MM_Math/51051573_102.png"
  },
  {
    "solution": "\\boxed{\\frac{20}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $DE\\perp AC$ at $E$, let $\\angle ADE=\\alpha$ and $\\cos\\alpha =\\frac{3}{5}$, with $AB=5$. What is the length of $AD$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rectangle, it follows that $\\angle B=\\angle BAC=90^\\circ$, BC=AD. Therefore, $\\angle BAC+\\angle DAE=90^\\circ$.\\\\\nSince $DE\\perp AC$, it follows that $\\angle ADE+\\angle DAE=90^\\circ$, and therefore, $\\angle BAC= \\angle ADE=\\alpha$.\\\\\nIn right triangle $\\triangle ABC$, since $\\cos\\alpha =\\frac{3}{5}$ and $AB=5$, it follows that $AC=\\frac{AB}{\\cos\\alpha}=\\frac{25}{3}$.\\\\\nTherefore, AD=BC= $\\sqrt{AC^{2}-AB^{2}}=\\sqrt{\\left(\\frac{25}{3}\\right)^{2}-5^{2}}=\\boxed{\\frac{20}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50890999.png"
  },
  {
    "solution": "\\boxed{x<\\frac{3}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given the functions $y_{1}=2x$ and $y_{2}=ax+4$ ($a\\neq 0$) intersect at point $A(m,3)$, what is the solution set for the inequality $2x<ax+4$ in terms of $x$?\n",
    "completion": "Solution: The graph of the function $y_1=2x$ intersects with the graph of $y_2=ax+4$ ($a\\neq 0$) at point A$\\left(\\frac{3}{2}, 3\\right)$, \\\\\n$\\therefore 3=2\\times\\frac{3}{2}$, \\\\\n$\\therefore m=\\frac{3}{2}$, \\\\\n$\\therefore$ point A$\\left(\\frac{3}{2}, 3\\right)$, \\\\\n$\\because 2x<ax+4$, that is $y_1<y_2$, \\\\\n$\\therefore \\boxed{x<\\frac{3}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52095101.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the storage center of a supermarket has a slope $AB$ with a gradient of $1:2$. The height at the top, $AC=4\\text{m}$, and $B$, $C$ are on the same horizontal ground. Rectangle $DEFG$ represents the side view of a rectangular container, which is being transported up the slope. At this moment, $GB=6\\text{m}$, and Xiao Ming, who is $1.5\\text{m}$ tall, stands at point $B$ and sees a chandelier directly above point $D$ at a height of $1.5\\text{m}$. Find the height of point $D$ from the ground and calculate the tangent value of Xiao Ming's elevation angle $\\alpha$.\n",
    "completion": "\\textbf{Solution:} Let line $DS \\perp BC$ at point S and intersect with $AB$ at point H, \\\\\n$\\because \\angle DGH=\\angle BSH=90^\\circ$, $\\angle GHD=\\angle BHS$, \\\\\n$\\therefore \\angle GDH=\\angle SBH$, \\\\\n$\\therefore \\tan\\alpha =\\tan\\angle GDH=\\tan\\angle SBH=\\frac{AC}{BC}=\\frac{GH}{DG}=\\frac{1}{2}$, \\\\\n$\\because DG=EF=2\\text{m}$, \\\\\n$\\therefore GH=1\\text{m}$, \\\\\n$\\therefore DH=\\sqrt{GH^2+DG^2}=\\sqrt{1^2+2^2}=\\sqrt{5}\\text{m}$, $BH=GB-GH=6-1=5\\text{m}$, \\\\\nLet $HS=x\\text{m}$, then $BS=2x\\text{m}$, by the Pythagorean theorem: $HS^2+BS^2=BH^2$, \\\\\nhence, $x^2+(2x)^2=5^2$, \\\\\nSolving gives: $x=\\sqrt{5}$ (negative value discarded), \\\\\n$\\therefore DS=DH+HS=\\sqrt{5}+\\sqrt{5}=2\\sqrt{5}\\text{m}$, thus the height of point D from the ground is $\\boxed{2\\sqrt{5}\\text{m}}$, \\\\\n$\\therefore BS=2x=2\\sqrt{5}\\text{m}$, \\\\\nLet the angle of elevation for Xiaoming be $\\alpha$, \\\\\nthen $\\tan\\alpha =\\frac{DS+1.5-1.5}{BS}=\\frac{2\\sqrt{5}+1.5-1.5}{2\\sqrt{5}}=\\boxed{1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53065573_34.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, we have $AB=3$ and $BC=4$. Point $E$ is a point on side $BC$. Connect $AE$. Fold $\\angle B$ along $AE$, making point $B$ fall on point $B^{\\prime}$. When $\\triangle CEB^{\\prime}$ is a right-angled triangle, what is the length of $BE$?\n",
    "completion": "\\textbf{Solution:} As shown in Figure 1, when A, B', and C are collinear, $\\angle EB'C=90^\\circ$.\n\nQuadrilateral ABCD is a rectangle,\n\n$\\therefore \\angle B=90^\\circ$,\n\n$\\therefore AC=\\sqrt{AB^{2}+CB^{2}}=\\sqrt{3^{2}+4^{2}}=5$,\n\n$\\because AB=AB' =3$,\n\n$\\therefore CB' =5-3=2$, let $BE=EB' =x$, then $EC=4-x$,\n\nIn $\\triangle CEB'$, $CE^{2}=B'E^{2}+B'C^{2}$,\n\n$\\therefore (4-x)^{2}=2^{2}+x^{2}$,\n\n$\\therefore x=\\boxed{\\frac{3}{2}}$,\n\nAs shown in Figure 2, when point B' lies on AD, $\\angle CEB' =90^\\circ$,\n\nAt this time, quadrilateral ABEB' is a square,\n\n$\\therefore BE=AB=3$,\n\nIn conclusion, the values of $BE$ that meet the conditions are $\\frac{3}{2}$ or 3.",
    "image_path": "dataset/mm_math/images/MM_Math/51366537.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the diameter $AB$ of $\\odot O$ is perpendicular to the chord $CD$, with the foot of the perpendicular being point $E$. Also, $AD=5$, and $\\cos\\angle BCD=\\frac{4}{5}$. What is the length of the chord $CD$?\n",
    "completion": "\\textbf{Solution:} Since $AB \\perp CD$,\\\\\nthus $CE = DE = \\frac{1}{2}CD$,\\\\\nSince $\\cos\\angle BCD = \\frac{4}{5}$,\\\\\nthus $\\cos\\angle A= \\cos\\angle BCD = \\frac{4}{5}$,\\\\\nIn $\\triangle ADE$, $AD=5$, and $\\cos\\angle A= \\frac{4}{5}$,\\\\\nthus $AE=4$,\\\\\nthus $DE=3$, then $CD=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51322362_125.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{\\sqrt{3}}{3}x+3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, it is known that the coordinates of point $C$ are $(\\sqrt{3}, -2)$, and the line $AB$ intersects the $x$ axis and $y$ axis at point $A$ and point $B$ respectively, with the coordinates of point $B$ being $(0, 3)$, and $\\angle BAO = 30^\\circ$. What is the analytical equation of line $AB$?\n",
    "completion": "\\textbf{Solution:} Since $B(0, 3)$, it follows that $OB=3$. In the right triangle $\\triangle AOB$, we have $\\angle AOB=90^\\circ$ and $\\angle BAO=30^\\circ$. Therefore, $AB=2OB=6$ and $OA=\\sqrt{AB^{2}-OB^{2}}=3\\sqrt{3}$. Hence, $A(3\\sqrt{3}, 0)$. Let the equation of line $AB$ be $y=kx+b$. Substituting points $A(3\\sqrt{3}, 0)$ and $B(0, 3)$ gives $\\begin{cases}3\\sqrt{3}k+b=0\\\\ b=3\\end{cases}$. Solving, we find $\\begin{cases}k=-\\frac{\\sqrt{3}}{3}\\\\ b=3\\end{cases}$. Thus, the equation of line $AB$ is $y=-\\frac{\\sqrt{3}}{3}x+3$; hence, the answer is $\\boxed{y=-\\frac{\\sqrt{3}}{3}x+3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50482871_121.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, CD is the median to side AB, $\\angle B=45^\\circ$, $\\tan\\angle ACB=3$, $AC= \\sqrt{10}$. Find: the area of $\\triangle ABC$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, construct $AH \\perp BC$ at $H$. In $\\triangle ACH$, since $\\tan \\angle ACB = 3$ and $AC= \\sqrt{10}$, let $CH=x$ and $AH=3x$. According to the Pythagorean theorem, we find $AC= \\sqrt{10}x$, thus $CH=1$ and $AH=3$. In $\\triangle ABH$, $\\angle B=45^\\circ$, so $BH=AH=3$,\\\\\ntherefore $S_{\\triangle ABC}= \\frac{1}{2} \\times 4 \\times 3 = \\boxed{6}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51051567_106.png"
  },
  {
    "solution": "\\boxed{4.8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the given figure, in $\\triangle ABC$, $AH \\perp BC$ at $H$, with $BC=12$ and $AH=8$. Points $D$ and $E$ are located on $AB$ and $AC$ respectively, while points $G$ and $F$ are on $BC$. The quadrilateral $DEFG$ is a square. What is the length of the side $DE$ of the square?\n",
    "completion": "Solution:As shown in the figure,\n\n$\\because$ square DEFG,\n\n$\\therefore$ DE=DG, $\\angle$NDG=$\\angle$DGH=90$^\\circ$, DE$\\parallel$BC,\n\n$\\because$ AH$\\perp$BC,\n\n$\\therefore$ AH$\\perp$DE,\n\n$\\therefore$ $\\angle$DNH=90$^\\circ$,\n\n$\\therefore$ quadrilateral DNHG is a rectangle,\n\n$\\therefore$ DG=NH=DE=x,\n\n$\\therefore$ AN=AH−NH=8−x,\n\n$\\because$ DE$\\parallel$BC,\n\n$\\therefore$ $\\triangle$ADE$\\sim$$\\triangle$ABC,\n\n$\\therefore$ $\\frac{DE}{BC}=\\frac{AN}{AH}$, that is, $\\frac{x}{12}=\\frac{8−x}{8}$\n\nSolve it: x=4.8,\n\nUpon verification, x=4.8 is a solution of the equation,\n\n$\\therefore$ DE=\\boxed{4.8}.",
    "image_path": "dataset/mm_math/images/MM_Math/52694085.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the right-angled triangle $\\triangle ABO$ where $\\angle O = 90^\\circ$, a circle with center $O$ and radius $OB$ intersects $AB$ at point $C$, and intersects $OA$ at point $D$. If $C$ is the midpoint of $AB$, what is the measure of $\\angle A$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, connect OC, where $\\angle AOB=90^\\circ$, and C is the midpoint of AB, \\\\\ntherefore $OC=AC=BC$, \\\\\nsince $OB=OC$, \\\\\nhence $\\triangle BOC$ is an equilateral triangle, \\\\\ntherefore $\\angle B=60^\\circ$, \\\\\nthus $\\angle A=90^\\circ-60^\\circ=\\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51029323_109.png"
  },
  {
    "solution": "\\boxed{M\\left(-\\frac{1}{2}, \\frac{3}{2}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, quadrilateral $ABCD$ is a rhombus, with $A(4, 0)$ and $B(0, 3)$. Draw the center $M$ of the rhombus, and write down the coordinates of point $M$.\n",
    "completion": "\\textbf{Solution:} Connect $AC$ and $BD$ intersecting at $M$, then $M$ is the center of the rhombus $ABCD$.\\\\\nSince quadrilateral $ABCD$ is a rhombus,\\\\\ntherefore, $M$ is the midpoint of $BD$,\\\\\nsince $B(0,3)$, $D(-1,0)$,\\\\\ntherefore, \\boxed{M\\left(-\\frac{1}{2}, \\frac{3}{2}\\right)}.",
    "image_path": "dataset/mm_math/images/MM_Math/52423104_80.png"
  },
  {
    "solution": "\\boxed{20^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $O$ is a point on line $AB$, $\\angle COD$ is a right angle, and $OE$ bisects $\\angle BOC$. If $\\angle AOC=40^\\circ$, what is the measure of $\\angle DOE$ in degrees?\n",
    "completion": "\\textbf{Solution:} Since $\\angle AOC=40^\\circ$,\\\\\nit follows that $\\angle BOC=140^\\circ$,\\\\\nalso, since OE bisects $\\angle BOC$,\\\\\nit implies $\\angle COE= \\frac{1}{2} \\times 140^\\circ=70^\\circ$,\\\\\nsince $\\angle COD=90^\\circ$,\\\\\nit follows that $\\angle DOE=90^\\circ-70^\\circ=\\boxed{20^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50938639_219.png"
  },
  {
    "solution": "\\boxed{\\frac{2}{3}\\pi}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the side length of each of the four small squares is 1. If an arc is drawn with O as the center and OG as the radius, intersecting AB and CD at points E and F, respectively, what is the length of the arc EF?\n",
    "completion": "\\textbf{Solution:} From the given information, we can deduce that $OE=OG=OF=2$\\\\\n$\\therefore$ $OD=\\frac{1}{2}OF$\\\\\n$\\therefore$ In $\\triangle ODF$ with a right angle at $D$, $\\sin\\angle OFD=\\frac{OD}{OF}=\\frac{1}{2}$\\\\\n$\\therefore$ $\\angle OFD=30^\\circ$\\\\\nSimilarly, we can deduce: $\\angle OEA=30^\\circ$\\\\\nSince $AB\\parallel OG\\parallel DC$\\\\\n$\\therefore \\angle EOG=\\angle OEA=30^\\circ$, $\\angle FOG=\\angle OFD=30^\\circ$\\\\\n$\\therefore$ $\\angle EOF=\\angle EOG+\\angle FOG=60^\\circ$\\\\\nTherefore, $\\overset{\\frown}{EF}=\\frac{n\\pi r}{180}=\\frac{60\\pi \\times 2}{180}=\\boxed{\\frac{2}{3}\\pi}$",
    "image_path": "dataset/mm_math/images/MM_Math/52483324.png"
  },
  {
    "solution": "\\boxed{-4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown, it is known that $\\triangle ABC$ and $\\triangle A'B'C'$ are similar figures with point $C$ as the center of similitude, and the ratio of the perimeters of $\\triangle ABC$ and $\\triangle A'B'C'$ is $1\\colon 2$. The coordinate of point $C$ is $(-1,0)$. If the x-coordinate of the corresponding point $B'$ of point $B$ is $5$, then what is the x-coordinate of point $B$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, draw $\\overline{BD} \\perp$ x-axis at point D, and draw $\\overline{B^{\\prime}H} \\perp$ x-axis at point H, then $\\overline{BD} \\parallel \\overline{B^{\\prime}H}$,\\\\\n$\\therefore \\angle DBC = \\angle HB^{\\prime}C$, $\\angle BDC = \\angle B^{\\prime}HC$,\\\\\n$\\therefore \\triangle BCD \\sim \\triangle B^{\\prime}CH$,\\\\\n$\\therefore \\frac{CD}{CH} = \\frac{BC}{B^{\\prime}C}$,\\\\\n$\\because$ the perimeter ratio of $\\triangle ABC$ to $\\triangle A^{\\prime}B^{\\prime}C$ is $1:2$,\\\\\n$\\therefore \\frac{BC}{B^{\\prime}C} = \\frac{1}{2}$,\\\\\n$\\therefore \\frac{CD}{CH} = \\frac{1}{2}$,\\\\\n$\\because$ the coordinates of point C are ($-1$, $0$), and the x-coordinate of the corresponding point $B^{\\prime}$ of point B is 5,\\\\\n$\\therefore$ OC = 1, OH = 5,\\\\\n$\\therefore$ CH = 6,\\\\\n$\\therefore CD = \\frac{1}{2}CH$ = 3,\\\\\n$\\therefore$ OD = OC + CD = 1 + 3 = 4,\\\\\n$\\therefore$ the x-coordinate of point B is $\\boxed{-4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51284428.png"
  },
  {
    "solution": "\\boxed{\\frac{32\\pi}{3} - 8\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\odot O$ is a circle centered at the origin $O$ with a radius of $4\\sqrt{2}$. The coordinates of point $P$ are $(2, 2)$. If the chord $AB$ passes through point $P$, what is the minimum area of the shaded region in the figure?\n",
    "completion": "\\textbf{Solution:} Given that when OP$\\perp$A'B', the area of the shaded region is minimized,\\\\\n$\\because$P(2,2),\\\\\n$\\therefore$OP = 2 $\\sqrt{2}$,\\\\\n$\\because$OA' = OB' = 4 $\\sqrt{2}$,\\\\\n$\\therefore \\cos \\angle$A'OP = $\\cos \\angle$B'OP = $\\frac{1}{2}$,\\\\\n$\\therefore$$\\angle$A'OP = $\\angle$B'OP = 60$^\\circ$,\\\\\n$\\therefore$$\\angle$A'OB' = 120$^\\circ$, A'$\\text{P}$ = 4 $\\sqrt{2}$ $\\times$ $\\frac{\\sqrt{3}}{2}$ = 2 $\\sqrt{6}$,\\\\\n$\\therefore$A'$\\text{B}'$ = 4 $\\sqrt{6}$,\\\\\n$\\therefore$S$_{\\text{shaded}}$ = S$_{\\text{sector}}$$_{OA'B'}$ - S$_{\\triangle}$$_{A'OB'}$ = $\\frac{120\\pi \\times {(4\\sqrt{2})}^2}{360}$ - $\\frac{1}{2} \\times 4\\sqrt{6} \\times 2\\sqrt{2}$ = $\\boxed{\\frac{32\\pi}{3} - 8\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51130891.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, point D is on side $AB$, and $\\angle ACD = \\angle ABC$. If $AC = \\sqrt{3}$ and $AD = 1$, what is the length of $DB$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle ACD = \\angle ABC$ and $\\angle A = \\angle A$,\\\\\nit follows that $\\triangle ABC \\sim \\triangle ACD$,\\\\\nhence, $\\frac{AD}{AC} = \\frac{AC}{AB}$,\\\\\ngiven that AC = $\\sqrt{3}$ and AD = 1,\\\\\nit implies that $\\frac{1}{\\sqrt{3}} = \\frac{\\sqrt{3}}{AB}$,\\\\\ntherefore, AB = 3,\\\\\nthus, BD = AB - AD = 3 - 1 = \\boxed{2}.",
    "image_path": "dataset/mm_math/images/MM_Math/51213318.png"
  },
  {
    "solution": "\\boxed{\\left(4, 5\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: It is known that the parabola $y=\\frac{1}{4}x^{2}+1$ has the following property: the distance from any point on the parabola to the fixed point $F(0, 2)$ is equal to its distance to the x-axis, as illustrated in the figure. The point $M$ has coordinates $(\\sqrt{3}, 3)$. Let $P$ be a moving point on the parabola $y=\\frac{1}{4}x^{2}+1$. When the area of $\\triangle POF$ is 4, find the coordinates of point $P$?\n",
    "completion": "\\textbf{Solution:} Let the coordinates of point $P$ be $\\left(x, \\frac{1}{4}x^{2}+1\\right)$,\n\nsince the coordinates of point $F$ are $\\left(0, 2\\right)$,\n\ntherefore $OF=2$,\n\ntherefore when the area of $\\triangle POF$ is 4, $\\frac{1}{2}\\times 2\\times |x|=4$,\n\nsolving this, we get: $x=\\pm 4$,\n\ntherefore $y=\\frac{1}{4}\\times (\\pm 4)^{2}+1=5$,\n\nhence, the coordinates of point $P$ are $\\boxed{\\left(-4, 5\\right)}$ or $\\boxed{\\left(4, 5\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52978121_181.png"
  },
  {
    "solution": "\\boxed{15}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given that the line segment $AD=80$, and points B and C are on the line segment AD. If $AB=10$, $BC=20$, and point P moves from B towards D at a speed of 3 units per second, taking $t$ seconds, where point Q is the midpoint of CD, find the value of $t$ when $AP=AQ$.\n",
    "completion": "\\textbf{Solution:} Given that $AD=80$, $AB=10$, and $BC=20$, \\\\\nhence, $CD=80-10-20=50$, \\\\\nsince $Q$ is the midpoint of $CD$, \\\\\nthus, $CQ=\\frac{1}{2}CD=25$, and $AQ=10+20+25=55$, \\\\\nafter $t$ seconds, $AP=10+3t$, \\\\\nwhen $AP=AQ$, $10+3t=55$, \\\\\ntherefore, $t=\\boxed{15}$, \\\\\nwhen $AP=AQ$, the value of $t$ is \\boxed{15}.",
    "image_path": "dataset/mm_math/images/MM_Math/51439120_173.png"
  },
  {
    "solution": "\\boxed{40^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, point $D$ is on side $BC$, and $AD=AB$, $AE\\parallel BC$, $\\angle BAD=\\angle CAE$. Connect $DE$ and intersect $AC$ at point $F$. If $\\angle B=70^\\circ$, what is the degree measure of $\\angle C$?\n",
    "completion": "\\textbf{Solution:} Since $ AD=AB$ and $ \\angle B=70^\\circ$,\\\\\nit follows that $ \\angle B=\\angle ADB=70^\\circ$,\\\\\nwhich means $ \\angle BAD=180^\\circ-\\angle B-\\angle ADB=40^\\circ$,\\\\\nthus $ \\angle BAD=\\angle CAE=40^\\circ$,\\\\\nand since $ AE\\parallel BC$,\\\\\nit follows that $ \\angle C=\\boxed{40^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55403716_80.png"
  },
  {
    "solution": "\\boxed{\\frac{48}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $P$ is a point on $AB$, and $PB>PC$. Connect $PC$ and fold rectangle $ABCD$ along $PC$, such that point $B$ falls onto ${B}^{\\prime}$. Extend ${B}^{\\prime}C$ to intersect the extension of $AB$ at point $Q$. Given that $AB=10, BC=4$, and $PB=6$, find the length of $BQ$?\n",
    "completion": "\\textbf{Solution:} Let $BQ=x$, $CQ=y$,\n\nSince $ABCD$ is folded along $PC$, and point $B$ falls onto $B'$, with $BC=4$, $PB=6$,\n\nTherefore, $PB'=PB=6$, $\\angle PB'Q=\\angle PBC=90^\\circ$,\n\nSince $\\angle Q=\\angle Q$,\n\nTherefore, $\\triangle PB'Q \\sim \\triangle CBQ$,\n\nThus, $\\frac{BC}{CQ} = \\frac{PB'}{PQ}$,\n\nTherefore, $\\frac{4}{y}=\\frac{6}{6+x}$,\n\nSolving gives $y=\\frac{2x+12}{3}$,\n\nBy the Pythagorean theorem, we get ${x}^{2}+16={y}^{2}$,\n\nTherefore, ${x}^{2}+16=\\left(\\frac{2x+12}{3}\\right)^{2}$,\n\nSolving gives $x=\\boxed{\\frac{48}{5}}$, $x=0$ (discard),\n\nTherefore, $BQ=\\boxed{\\frac{48}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53102165_74.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, point N(0,6) and point M is on the negative half-axis of the x-axis, with ON = 3OM. A is a point on segment MN, AB$\\perp$x-axis with B as its foot, and AC$\\perp$y-axis with C as its foot. If the x-coordinate of point A is $-1$, what is the area of rectangle ABOC?\n",
    "completion": "\\textbf{Solution:} In the equation $y=3x+6$, when $x=-1$, we have $y=3$,\\\\\n$\\therefore OB=1$, $AB=3$,\\\\\n$\\therefore$ the area of the rectangle ABOC is $1\\times 3=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52225566_111.png"
  },
  {
    "solution": "\\boxed{75^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within the circle $\\odot O$ where the radius is 1, the chord $AB = \\sqrt{2}$. Construct an equilateral triangle $\\triangle ABC$ inside $\\odot O$ with $AB$ as one side. Rotate $\\triangle ABC$ counterclockwise around point A. When side $AC$ first touches $\\odot O$ tangentially, what is the angle of rotation?\n",
    "completion": "\\textbf{Solution:} When $AC$ is tangent to $\\odot O$, connect $CO$ and extend it to meet $AB$ at point $M$, connect $AO$,\\\\\nsince $\\triangle ABC$ is an equilateral triangle,\\\\\nthus, $CM\\perp AB$,\\\\\nsince $AB = \\sqrt{2}$,\\\\\nthus, $AM = \\frac{\\sqrt{2}}{2}$,\\\\\nsince $OA = 1$,\\\\\nthus, $OM = \\frac{\\sqrt{2}}{2}$,\\\\\ntherefore, $\\angle OAM = 45^\\circ$,\\\\\nsince $\\angle OAC'=90^\\circ$,\\\\\nthus, $\\angle BAC'=135^\\circ$,\\\\\nsince $\\angle C'AB'=60^\\circ$,\\\\\nthus, $\\angle BAB'=75^\\circ$,\\\\\ntherefore, when $AC$ is tangent to $\\odot O$ for the first time, the angle of rotation is $\\boxed{75^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52601694.png"
  },
  {
    "solution": "\\boxed{(2,\\sqrt{3})}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the vertices \\(A\\), \\(B\\), and \\(C\\) of the rhombus \\(ABCD\\) are on the coordinate axes. If the coordinates of point \\(B\\) are \\((-1, 0)\\) and \\(\\angle BCD = 120^\\circ\\), what are the coordinates of point \\(D\\)?\n",
    "completion": "\\textbf{Solution:} Given that point B has the coordinates $(-1,0)$,\\\\\nthus OB=1,\\\\\nin diamond $ABCD$, $\\angle BCD=120^\\circ$, $\\angle ABC+\\angle BCD=180^\\circ$,\\\\\nthus $\\angle ABC=60^\\circ$, $\\angle BAD=120^\\circ$,\\\\\nthus $\\angle BAC=30^\\circ$, $\\angle DAO=90^\\circ$,\\\\\nthus AB=2OB=2,\\\\\nthus OA$=\\sqrt{AB^{2}-OB^{2}}=\\sqrt{2^{2}-1^{2}}=\\sqrt{3}$,\\\\\nsince AD=AB=2,\\\\\nthus the coordinates of point D are $\\boxed{(2,\\sqrt{3})}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52424464.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the figure, inside $\\triangle ABC$, the median from $AB$ intersects side $AC$ at point $E$, with $BE=5$ and $CE=3$. What is the length of $AC$?\n",
    "completion": "Solution: Since the perpendicular bisector of AB intersects side AC at point E, and BE = 5, \\\\\nit follows that AE = BE = 5. \\\\\nSince CE = 3 and $AC = AE + CE$, \\\\\nit follows that $AC = \\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52861525.png"
  },
  {
    "solution": "\\boxed{16}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the ratio of the radii of two concentric circles is $3\\colon 5$. $AB$ is the diameter of the larger circle, and the chord $BC$ of the larger circle is tangent to the smaller circle. If $AC=12$, what is the length of $BC$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, let chord BC be tangent to the smaller circle at point D, and connect OD,\\\\\n$\\therefore$OD$\\perp$BC,\\\\\n$\\therefore$BD=CD,\\\\\n$\\because$O is the midpoint of AB,\\\\\n$\\therefore$OD is the median of $\\triangle ABC$,\\\\\n$\\therefore$OD=$\\frac{1}{2}$AC=$\\frac{1}{2}\\times 12=6$,\\\\\n$\\because$the ratio of the radii of the two concentric circles is 3:5,\\\\\n$\\therefore$the radius of the larger circle is 10,\\\\\n$\\therefore$AB=20,\\\\\n$\\because$AB is the diameter of the larger circle,\\\\\n$\\therefore$$\\angle$ACB=$90^\\circ$,\\\\\n$\\therefore$BC=$\\sqrt{AB^{2}-AC^{2}}=\\sqrt{20^{2}-12^{2}}=\\boxed{16}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52551789.png"
  },
  {
    "solution": "\\boxed{\\frac{4}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\triangle ABC$, $ \\angle ACB=90^\\circ$, $AC=3$, and $BC=4$. Fold side $AC$ along $CE$, making point $A$ fall onto point $D$ on $AB$; then fold side $BC$ along $CF$, making point $B$ fall onto point $B'$ on the extension line of $CD$. The two creases intersect hypotenuse $AB$ at points $E$ and $F$, respectively. What is the length of segment $B'F$?\n",
    "completion": "\\textbf{Solution:} Let's consider right-angled triangle $\\triangle ABC$ where $\\angle ACB=90^\\circ$, $AC=3$, and $BC=4$,\\\\\nhence $AB=5$,\\\\\nbased on the properties of folding, we know $AC=CD$, $\\angle A=\\angle CDE$, $CE\\perp AB$,\\\\\nthus $B'D=BC-CD=4-3=1$, $\\angle DCE+\\angle B'CF=\\angle ACE+\\angle BCF$,\\\\\nsince $\\angle ACB=90^\\circ$,\\\\\nthus $\\angle ECF=45^\\circ$,\\\\\nhence $\\triangle ECF$ is an isosceles right triangle,\\\\\nthus $EF=CE$, $\\angle EFC=45^\\circ$,\\\\\nthus $\\angle BFC=\\angle B'FC=135^\\circ$,\\\\\nthus $\\angle B'FD=90^\\circ$,\\\\\nsince $S_{\\triangle ABC}=\\frac{1}{2}AC\\cdot BC=\\frac{1}{2}AB\\cdot CE$,\\\\\nthus $AC\\cdot BC=AB\\cdot CE$,\\\\\nthus $CE=\\frac{12}{5}$,\\\\\nthus $EF=\\frac{12}{5}$, $ED=AE=\\sqrt{AC^2-CE^2}=\\frac{9}{5}$,\\\\\nthus $DF=EF-ED=\\frac{3}{5}$,\\\\\nthus $B'F=\\sqrt{B'^2D^2-DF^2}=\\boxed{\\frac{4}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52827840.png"
  },
  {
    "solution": "\\boxed{\\frac{36\\sqrt{7}}{7}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in rectangle $ABCD$, $AB=12$, and point $E$ is on side $BC$. Connect $DE$ and $AE$, $\\angle DEC=30^\\circ$, and $AE\\perp DE$. Fold $\\triangle CDE$ along $DE$ to obtain $\\triangle DEF$. Connect $AF$. What is the distance from $E$ to $AF$?\n",
    "completion": "\\textbf{Solution}: Since quadrilateral ABCD is a rectangle, we have $\\angle ABC=\\angle BAC=\\angle ADC=\\angle BCD=90^\\circ$, and CD=AB=12. Draw FM$\\perp$BC at M, intersecting AD at N. Draw AH$\\perp$EF at H through point A, and draw EG$\\perp$AF at G through point E. Then, quadrilaterals ABMN and CDNM are rectangles, thus $AB=MN=CD=12$, AN=BM, and by the folding property, CE=EF. Also, $\\angle CED=\\angle DEF=30^\\circ$, thus DE=2CD=24, and $\\angle CEF=60^\\circ$. By the Pythagorean theorem, CE=EF=$\\sqrt{DE^2-CD^2}=\\sqrt{24^2-12^2}=12\\sqrt{3}$, which makes $\\triangle CEF$ an equilateral triangle. Thus, ME=CM=$\\frac{1}{2}$CE=$6\\sqrt{3}$. By the Pythagorean theorem again, FM=$\\sqrt{EF^2-ME^2}=\\sqrt{(12\\sqrt{3})^2-(6\\sqrt{3})^2}=18$. Since AE$\\perp$DE and $\\angle CDE=\\angle DEF=30^\\circ$, it follows that $\\angle AEH=\\angle AEB=60^\\circ$, and $\\angle ABE=\\angle AHE=90^\\circ$, making AH=AB=12 and $\\angle HAE=\\angle BAE=30^\\circ$. Hence, AE=2BE. By applying the Pythagorean theorem: $AB^2+BE^2=AE^2$ gives $12^2+BE^2=4BE^2$, from which we get BE=$4\\sqrt{3}$. Consequently, AN=BE+ME=$4\\sqrt{3}+6\\sqrt{3}=10\\sqrt{3}$, and FN=FM−MN=18−12=6. Therefore, AF=$\\sqrt{AN^2+FN^2}=\\sqrt{(10\\sqrt{3})^2+6^2}=4\\sqrt{21}$. From $S_{\\triangle AEF}=\\frac{1}{2}AF\\cdot EG=\\frac{1}{2}EF\\cdot AH$, we derive $\\frac{1}{2}\\times 4\\sqrt{21}\\cdot EG=\\frac{1}{2}\\times 12\\sqrt{3}\\times 12$, solving for EG yields $\\boxed{\\frac{36\\sqrt{7}}{7}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50497001.png"
  },
  {
    "solution": "\\boxed{\\frac{8\\sqrt{3}}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated in the diagram, two strips with a width of 2 and parallel opposite edges are placed crosswise on top of each other. By rotating one of the strips, we obtain a quadrilateral ABCD, where $\\angle ABC=60^\\circ$. What is the area of the quadrilateral ABCD?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw CF$\\perp$AB at point F, and draw AE$\\perp$BC at point E through point A.\\\\\nAccording to the problem, we know: AE=CF=2,\\\\\n$\\because$ $\\angle$ABC=60$^\\circ$,\\\\\n$\\therefore$ $\\angle$FCB=30$^\\circ$,\\\\\n$\\therefore$ let BF=x, then BC=2x,\\\\\n$\\therefore$ $4x^{2}-x^{2}=4$,\\\\\n$\\therefore$ x=$\\frac{2\\sqrt{3}}{3}$,\\\\\n$\\therefore$ BC=$\\frac{4\\sqrt{3}}{3}$,\\\\\n$\\because$ AB$\\parallel$CD, AD$\\parallel$BC,\\\\\n$\\therefore$ Quadrilateral ABCD is a parallelogram,\\\\\n$\\therefore$ $S_{\\square ABCD}$=BC·AE=$\\frac{4\\sqrt{3}}{3} \\times 2=\\boxed{\\frac{8\\sqrt{3}}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51942337.png"
  },
  {
    "solution": "\\boxed{22^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, lines AB and CD intersect at point O, $\\angle COE$ is a right angle, $OF$ bisects $\\angle AOE$, $\\angle COF=34^\\circ$, what are the measures of $\\angle AOC$ and $\\angle BOD$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle COE=90^\\circ$ and $\\angle COF=34^\\circ$,\\\\\nit follows that $\\angle EOF=\\angle COE-\\angle COF=56^\\circ$,\\\\\nGiven that OF is the bisector of $\\angle AOE$,\\\\\nit implies $\\angle AOE=2\\angle EOF=112^\\circ$,\\\\\nthus, $\\angle AOC=112^\\circ-90^\\circ=\\boxed{22^\\circ}$,\\\\\nSince $\\angle BOD$ and $\\angle AOC$ are vertical angles,\\\\\nwe conclude that $\\angle BOD=\\boxed{22^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/42265661.png"
  },
  {
    "solution": "\\boxed{70^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the quadrilateral $ABCD$ is a rhombus, with points $E$ and $F$ on sides $AD$ and $CD$ respectively, and $AE=CF$, $BA=BE$. If $\\angle EBF=30^\\circ$, what is the measure of $\\angle C$?\n",
    "completion": "\\textbf{Solution}: Let the measure of $\\angle C$ be $x$, thus $\\angle A=\\angle C=x$,\\\\\nin $\\triangle ABE$ and $\\triangle CBF$,\\\\\nsince $\\left\\{\\begin{array}{l}BA=BC\\\\ \\angle A=\\angle C\\\\ AE=CF\\end{array}\\right.$,\\\\\nit follows that $\\triangle ABE\\cong \\triangle CBF$ (SAS),\\\\\ntherefore $\\angle ABE=\\angle CBF=180^\\circ-2x$,\\\\\nthus $\\angle ABC=2\\left(180^\\circ-2x\\right)+30^\\circ$,\\\\\nsince $\\angle ABC=\\angle D$,\\\\\nit follows that $\\angle ABC+\\angle D+\\angle A+\\angle C=360^\\circ$,\\\\\nthat is: $2\\left[2\\left(180^\\circ-2x\\right)+30^\\circ\\right]+2x=360^\\circ$,\\\\\nsolving gives: $x=\\boxed{70^\\circ}$,\\\\\nthus, the measure of $\\angle C$ is: $70^\\circ$.",
    "image_path": "dataset/mm_math/images/MM_Math/53333443.png"
  },
  {
    "solution": "\\boxed{\\frac{130}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: A student performed the following operations on a rectangular piece of paper ABCD: as shown in the figure, first fold along the line AG so that point B falls on point P on the diagonal AC, then fold along the line CH so that point D falls on point Q on the diagonal AC. If $AB=5$ and $BC=12$, what is the area of the quadrilateral $AGCH$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral ABCD is a rectangle, with $AB=5$ and $BC=12$. Thus, $AB=CD=5$ and $\\angle B=90^\\circ$. Therefore, $AC=\\sqrt{AB^2+BC^2}=13$. From the property of folding, we know that AP=AB=5, BG=PG, and $\\angle B=\\angle APG=90^\\circ$. Similarly, CQ=CD=5, thus CP=8 and $\\angle CPG=90^\\circ$. Let CG=x, then BG=PG=12-x. Therefore, by the Pythagorean theorem, we have $(12-x)^2+8^2=x^2$. Solving this, we get $x=\\frac{26}{3}$, i.e., CG$=\\frac{26}{3}$. Similarly, AH$=\\frac{26}{3}$, hence CG=AH. Since quadrilateral ABCD is a rectangle, we have AD$\\parallel$BC. Consequently, quadrilateral AGCH is a parallelogram. Therefore, the area of quadrilateral AGCH is $S_{\\text{quadrilateral} AGCH}=CG\\cdot AB=\\boxed{\\frac{130}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51722351.png"
  },
  {
    "solution": "\\boxed{7\\text{cm}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, \\(C\\) is a point on line segment \\(AB\\), \\(M\\) is the midpoint of line segment \\(AC\\), and \\(N\\) is the midpoint of line segment \\(BC\\). If \\(AM = \\frac{5}{4} BC = 5 \\, \\text{cm}\\), find the length of \\(MN\\).\n",
    "completion": "\\textbf{Solution:} Since $M$ is the midpoint of segment $AC$, $AM=\\frac{5}{4}BC=5\\text{cm}$,\\\\\nthus $AM=CM=5\\text{cm}$, and $BC=4\\text{cm}$.\\\\\nAdditionally, since $N$ is the midpoint of segment $BC$,\\\\\nthus $CN=\\frac{1}{2}BC=2\\text{cm}$,\\\\\ntherefore $MN=MC+CN=\\boxed{7\\text{cm}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/54962545_11.png"
  },
  {
    "solution": "\\boxed{x^{2}+2x-3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=ax^2+bx-3$ $(a \\neq 0)$ intersects the x-axis at points $A(-3,0)$ and $B(1,0)$, and intersects the y-axis at point C. Find the equation of this parabola.\n",
    "completion": "\\textbf{Solution:} Because the parabola $y=ax^{2}+bx-3\\left(a\\ne 0\\right)$ intersects the x-axis at point $A(-3,0)$ and point $B(1,0)$, \\\\\ntherefore $\\begin{cases}9a-3b-3=0\\\\ a+b-3=0\\end{cases}$, \\\\\nsolving the system of equations, we get $\\begin{cases}a=1\\\\ b=2\\end{cases}$, \\\\\ntherefore, the expression of the parabola is $y=x^{2}+2x-3=\\boxed{x^{2}+2x-3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52720059_107.png"
  },
  {
    "solution": "\\boxed{1.5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, $BC\\parallel AD$, $\\angle ADC=90^\\circ$, $BC=8\\,cm$, and $AD=CD=10\\,cm$. Point $E$ moves uniformly along the direction of $BC$ from point $B$ at a speed of $2\\,cm/s$; simultaneously, point $F$ moves uniformly along the direction of $DA$ from point $D$ at a speed of $3\\,cm/s$. A line through point $E$ is drawn such that $EH\\perp AD$ with the foot of the perpendicular at $H$, and $EH$ intersects $AC$ at point $G$, and $FG$ is connected. Let the movement time be $t(s)\\left(0<t<\\frac{10}{3}\\right)$, answer the following question: When $\\triangle CEG\\cong \\triangle AHG$, what is the value of $t$?\n",
    "completion": "\\textbf{Solution:} Given the problem, \\\\\nsince $AD=10$ and $DH=8-2t$, \\\\\nthus $AH=AD-DH=10-(8-2t)=2+2t$, \\\\\nsince $\\triangle CEG\\cong \\triangle AHG$, \\\\\nthus $CE=AH$, \\\\\nthus $8-2t=2+2t$, \\\\\nsolving for $t$ yields $t=\\boxed{1.5}$. \\\\\nTherefore, when $\\triangle CEG\\cong \\triangle AHG$, the value of $t$ is \\boxed{1.5}.",
    "image_path": "dataset/mm_math/images/MM_Math/52211373_85.png"
  },
  {
    "solution": "\\boxed{-\\frac{3}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the function $y=-2x+3$ intersects with the graph of the function $y=-\\frac{1}{2}x+m$ at point $P(n, -2)$. The line $y=-2x+3$ intersects with the y-axis at point A, and the line $y=-\\frac{1}{2}x+m$ intersects with the y-axis at point B. Determine the values of $m$ and $n$.\n",
    "completion": "\\textbf{Solution:} Since the graph of the function $y=-2x+3$ passes through the point $P(n, -2)$,\\\\\nit follows that $-2=-2n+3$. Solving this, we find $n=\\boxed{\\frac{5}{2}}$.\\\\\nThus, the coordinates of point $P$ are $\\left(\\frac{5}{2}, -2\\right)$.\\\\\nSince the graph of the function $y=-\\frac{1}{2}x+m$ passes through point $P\\left(\\frac{5}{2}, -2\\right)$,\\\\\nit follows that $-2=-\\frac{1}{2}\\times \\frac{5}{2}+m$. Solving this, we find $m=\\boxed{-\\frac{3}{4}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52355125_190.png"
  },
  {
    "solution": "\\boxed{40^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, if $\\triangle ABC \\cong \\triangle ADF$ and $\\angle B=60^\\circ$, $\\angle C=20^\\circ$, with point D on side BC, then what is the degree measure of $\\angle DAC$?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABC \\cong \\triangle ADF$,\\\\\nit follows that $AB = AD$,\\\\\ntherefore, $\\angle ADB = \\angle B = 60^\\circ$,\\\\\nthus, $\\angle DAC = \\angle ADB - \\angle C = \\boxed{40^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/52604042.png"
  },
  {
    "solution": "\\boxed{40^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given, as shown in the figure, $AB \\parallel CD$, $\\angle ABE = 80^\\circ$, $EF$ bisects $\\angle BEC$, $EF \\perp EG$, find the degree measure of $\\angle DEG$.\n",
    "completion": "\\textbf{Solution:} Given $AB\\parallel CD$, $\\angle ABE=80^\\circ$,\\\\\nthus, $\\angle BEC=180^\\circ-\\angle ABE=100^\\circ$,\\\\\nsince EF bisects $\\angle BEC$,\\\\\nthus, $\\angle CEF= \\frac{1}{2} \\angle BEC=50^\\circ$,\\\\\nsince $EF\\perp EG$,\\\\\nthus, $\\angle FEG=90^\\circ$,\\\\\nhence, $\\angle DEG=180^\\circ-\\angle CEF-\\angle FEG=\\boxed{40^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/47362705.png"
  },
  {
    "solution": "\\boxed{1<AD<6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AD$ is the median of $\\triangle ABC$, $CE \\parallel AB$ intersects the extension of $AD$ at point $E$, $AB=5$, $AC=7$, then what are the possible values of $AD$?\n",
    "completion": "\\textbf{Solution:} Since $AD$ is the median of $\\triangle ABC$, \\\\\nthus $CD=BD$, \\\\\nsince $CE \\parallel AB$, \\\\\nthus $\\angle DCE=\\angle DBA$, \\\\\nin $\\triangle CDE$ and $\\triangle BDA$, \\\\\n$\\left\\{\\begin{array}{l}\n\\angle CDE=\\angle BDA \\\\\nCD=BD \\\\\n\\angle DCE=\\angle DBA\n\\end{array}\\right.$, \\\\\ntherefore $\\triangle CDE\\cong \\triangle BDA$ (ASA), \\\\\nthus $EC=AB=5$, \\\\\nsince $7-5<AE<7+5$, \\\\\nthus $2<2AD<12$, \\\\\nthus $\\boxed{1<AD<6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52513495.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated, in triangle $ABC$, $\\angle ACB=90^\\circ$, $\\angle A=20^\\circ$, $\\triangle ABC \\cong \\triangle A'B'C'$. If $A'B'$ precisely passes through point B, and $A'C'$ intersects $AB$ at $D$, what is the measure of $\\angle BDC$?\n",
    "completion": "\\textbf{Solution:} Given $\\triangle ABC \\cong \\triangle A'B'C'$,\n\nit follows that $\\angle A = \\angle A' = 20^\\circ$, $\\angle ACB = \\angle A'CB' = 90^\\circ$, CB = CB',\n\nsince $\\angle B' = \\angle CBA = 70^\\circ$,\n\nand because CB = CB',\n\nit follows that $\\angle B' = \\angle B'BC = 70^\\circ$,\n\nthus $\\angle BCB' = 180^\\circ - 70^\\circ - 70^\\circ = 40^\\circ$,\n\nhence $\\angle BCD = 90^\\circ - 40^\\circ = 50^\\circ$,\n\nconsequently $\\angle BDC = 180^\\circ - \\angle CBD - \\angle BCD = 180^\\circ - 50^\\circ - 70^\\circ = \\boxed{60^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52991699.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the circle $\\odot O$, $CD$ is a diameter, $AB$ is a chord, and $AB\\perp CD$ at $M$, with $AB=8$ and $OC=5$, what is the length of $MD$?\n",
    "completion": "\\textbf{Solution:} Draw connection OA, \\\\\nsince $CD$ is the diameter, $AB$ is a chord, and $AB\\perp CD$ at $M$, with $AB=8$, \\\\\nthus $AM=BM=4$, \\\\\nsince $OC=5$, \\\\\nthus $OA=OD=5$, \\\\\nthus $OM=\\sqrt{OA^{2}-AM^{2}}=\\sqrt{5^{2}-4^{2}}=3$. \\\\\nTherefore, $DM=OD-OM=5-3=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53080627.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{2}{3}x+4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given the figure, in the Cartesian coordinate system, the graph of the function $y=kx+2$ passes through the point $A(3,0)$. After moving the graph upwards by 2 units, it intersects the $x$ axis at point $B$ and the $y$ axis at point $C$. What is the linear equation represented by the graph passing through points $B$ and $C$?\n",
    "completion": "\\textbf{Solution:}  Substituting $A(3, 0)$ into $y=kx+2$ gives: $3k+2=0$, \\\\\nSolving for $k$, we find $k=-\\frac{2}{3}$, \\\\\n$\\therefore y=-\\frac{2}{3}x+2$. \\\\\n$\\because$ Shifting the graph of the function $y=-\\frac{2}{3}x+2$ up by 2 units results in $y=-\\frac{2}{3}x+4$, \\\\\n$\\therefore$ The linear equation passing through points B and C is $\\boxed{y=-\\frac{2}{3}x+4}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52250984_102.png"
  },
  {
    "solution": "\\boxed{(3, 2)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system, the squares $ABCD$ and $BEFG$ are similar figures with the origin $O$ as the center of similarity, and the ratio of similarity is $\\frac{1}{3}$. The points $A$, $B$, and $E$ are located on the $x$-axis. If the side length of square $BEFG$ is 6, what are the coordinates of point $C$?\n",
    "completion": "\\textbf{Solution:} Given that square ABCD and square BEFG are similar figures with center at the origin O, and the ratio of similarity is $\\frac{1}{3}$,\\\\\n$\\therefore$ $\\frac{BC}{EF}=\\frac{1}{3}$,\\\\\nGiven that EF = 6,\\\\\n$\\therefore$ BC=2,\\\\\nGiven that $\\triangle OBC\\sim \\triangle OEF$,\\\\\n$\\therefore$ $\\frac{OB}{OE}=\\frac{BC}{EF}$,\\\\\n$\\therefore$ $\\frac{OB}{OB+6}=\\frac{1}{3}$,\\\\\n$\\therefore$ OB=3,\\\\\n$\\therefore$ the coordinates of point C are $\\boxed{(3, 2)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51179561.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in quadrilateral $ABCD$, $AB= BC$, and $\\angle ABC=\\angle CDA=90^\\circ$, $BE\\perp AD$ at point $E$, and the area of quadrilateral $ABCD$ is 8. What is the length of $BE$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, draw line BF $\\perp$ CD, intersecting the extension of DC at point F, \\\\\n$\\because$ BE $\\perp$ AD, \\\\\n$\\therefore$ $\\angle$AEB = $\\angle$BED = $\\angle$F = 90$^\\circ$, \\\\\n$\\because$ $\\angle$CDA = 90$^\\circ$, \\\\\n$\\therefore$ Quadrilateral DEBF is a rectangle, \\\\\n$\\therefore$ $\\angle$EBF = $\\angle$ABC = 90$^\\circ$, \\\\\n$\\therefore$ $\\angle$ABE = $\\angle$CBF, \\\\\n$\\because$ AB = BC, \\\\\n$\\therefore$ $\\triangle$BCF $\\cong$ $\\triangle$BAE (AAS), \\\\\n$\\therefore$ BE = BF, \\\\\n$\\therefore$ Quadrilateral DEBF is a square, \\\\\n$\\therefore$ Area of Quadrilateral ABCD = Area of Square BEDF = 8, \\\\\n$\\therefore$ BE = \\boxed{2\\sqrt{2}}.",
    "image_path": "dataset/mm_math/images/MM_Math/51614529.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the length of line segment $AF$ is $13cm$, and points $B$, $C$, $D$, $E$ are sequentially on $AF$, with $AB=BC=CD$. Point $E$ is the midpoint of $DF$, and $CE=5cm$. Find the length of $BE$.\n",
    "completion": "\\textbf{Solution:} Let $AB=BC=CD=x\\, \\text{cm}$, thus $BD=2x\\, \\text{cm}$,\\\\\nsince $AF$ is $13\\, \\text{cm}$ long,\\\\\ntherefore $DF=13-3x$,\\\\\nsince $E$ is the midpoint of $DF$,\\\\\ntherefore $DE=\\frac{1}{2}(13-3x)$,\\\\\nsince $CE=5$,\\\\\ntherefore $x+\\frac{1}{2}(13-3x)=5$,\\\\\ntherefore $x=3$,\\\\\ntherefore $BC=3$,\\\\\ntherefore $BE=BC+CE=\\boxed{8}\\, (\\text{cm})$.",
    "image_path": "dataset/mm_math/images/MM_Math/51434278.png"
  },
  {
    "solution": "\\boxed{6.5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, we know that $ABCD$ is a rectangle, with $AD = 12$ and $CD = 9$. Points $R$ and $P$ are fixed on $DC$ and $BC$, respectively. Points $E$ and $F$ are the midpoints of $AP$ and $RP$, respectively. If $CR = 4$, what is the length of $EF$?\n",
    "completion": "\\textbf{Solution:} Draw $AR$,\\\\\nIn $\\triangle ADR$, $AD = 12$, $DR=CD-CR=9-4=5$,\\\\\nBy the Pythagorean theorem, we have:\\\\\n$AR= \\sqrt{12^2+5^2}=13$ ,\\\\\nSince points $E$ and $F$ are the midpoints of $AP$ and $RP$ respectively,\\\\\nThen $EF$ is the median of $\\triangle APR$,\\\\\nTherefore, $EF=\\frac{1}{2}AR=\\boxed{6.5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53437332.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{4}{x}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y=-\\frac{1}{2}x+1$ intersects with the graph of the inverse proportion function $y=\\frac{k}{x}$ at points $A$ and $B$. The x-coordinate of point $A$ is $-2$, and the x-coordinate of point $B$ is $4$. The graph of the linear function intersects with the y-axis at point $C$. What is the expression of the inverse proportion function?\n",
    "completion": "\\textbf{Solution:} When $x=-2$, $y=-\\frac{1}{2}\\times (-2)+1=2$,\\\\\n$\\therefore \\boxed{A(-2,2)}$,\\\\\n$\\because$ the graph of the inverse proportion function $y=\\frac{k}{x}$ passes through point A,\\\\\n$\\therefore k=2\\times (-2)=-4$,\\\\\n$\\therefore$ the expression for the inverse proportion function is $\\boxed{y=-\\frac{4}{x}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52602046_155.png"
  },
  {
    "solution": "\\boxed{36}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $B$ is a point on line segment $AC$, and $AB=27\\text{cm}$, $BC=\\frac{1}{3}AB$. Try to find the length of line segment $AC$?\n",
    "completion": "\\textbf{Solution:} Given that $AB=27\\text{cm}$, and $BC=\\frac{1}{3}AB=9\\text{cm}$,\\\\\nthus $AC=AB+BC=27+9=\\boxed{36}\\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51292247_202.png"
  },
  {
    "solution": "\\boxed{40^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $CD$ is the altitude to the hypotenuse $AB$, and $\\angle ACD=40^\\circ$. What is the measure of $\\angle B$?\n",
    "completion": "\\textbf{Solution:} Given that in the right-angled triangle $ABC$, $\\angle ACB=90^\\circ$, $\\angle ACD=40^\\circ$, it follows that $\\angle DCB=50^\\circ$. Since $CD$ is the altitude from the hypotenuse $AB$, we have $\\angle CDB=90^\\circ$. In the right-angled triangle $CDB$: $\\angle B=180^\\circ-\\angle DCB-\\angle CDB=180^\\circ-50^\\circ-90^\\circ=\\boxed{40^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55403711.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure (1), it is known that $AM\\parallel CN$, where point B is a point in the plane and $AB\\perp BC$ at point B. Through point B, construct $BD\\perp AM$ at point D, and let $\\angle BCN=\\alpha$. As shown in Figure (2), if points E and F are on $DM$, and $BE$, $BF$, and $CF$ are connected such that $BE$ bisects $\\angle ABD$ and $BF$ bisects $\\angle DBC$, what is the measure of $\\angle EBF$ in degrees?\n",
    "completion": "\\textbf{Solution:} Extend $DB$, intersect $NC$ at point H, as shown,\\\\\nsince $AM\\parallel CN$, $BD\\perp AM$,\\\\\ntherefore $DH\\perp NC$.\\\\\nTherefore $\\angle BHC=90^\\circ$.\\\\\nSince $\\angle BCN=\\alpha $,\\\\\ntherefore $\\angle HBC=90^\\circ-\\alpha $.\\\\\nSince $AB\\perp BC$,\\\\\ntherefore $\\angle ABC=90^\\circ$.\\\\\nTherefore $\\angle ABD=180^\\circ-\\angle ABC-\\angle HBC=\\alpha $.\\\\\nSince $BE$ bisects $\\angle ABD$,\\\\\ntherefore $\\angle DBE=\\angle ABE=\\frac{1}{2}\\alpha $.\\\\\nSince $\\angle HBC=90^\\circ-\\alpha $,\\\\\ntherefore $\\angle DBC=180^\\circ-\\angle HBC=90^\\circ+\\alpha $.\\\\\nSince $BF$ bisects $\\angle DBC$,\\\\\ntherefore $\\angle DBF=\\angle CBF=\\frac{1}{2}\\angle DBC=45^\\circ+\\frac{1}{2}\\alpha $.\\\\\nTherefore $\\angle EBF=\\angle DBF-\\angle DBE=45^\\circ+\\frac{1}{2}\\alpha -\\frac{1}{2}\\alpha =\\boxed{45^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51321729_57.png"
  },
  {
    "solution": "\\boxed{\\sqrt{15}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In quadrilateral $ABCD$, $AD\\parallel BC$, $DE\\perp BC$ with the foot of the perpendicular being point $E$, connect $AC$ intersecting $DE$ at point $F$, point $G$ is the midpoint of $AF$, and $\\angle ACD=2\\angle ACB$. If $DG=4$ and $EC=1$, what is the length of $DE$?\n",
    "completion": "\\textbf{Solution:} Given that $AD \\parallel BC$ and $DE \\perp BC$,\\\\\nit follows that $AD \\perp DE$, and $\\angle CAD = \\angle ACB$,\\\\\nGiven that $\\angle ACD = 2\\angle ACB$,\\\\\nit follows that $\\angle ACD = 2\\angle CAD$,\\\\\nGiven that $AD \\perp DE$ and $G$ is the midpoint of $AF$,\\\\\nit follows that $GA = GD$,\\\\\ntherefore, $\\angle GAD = \\angle GDA$,\\\\\nthus, $\\angle DGC = \\angle DCG$,\\\\\ntherefore, $DC = DG = 4$,\\\\\nthus, $DE= \\sqrt{CD^{2}-EC^{2}} = \\boxed{\\sqrt{15}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52991704.png"
  },
  {
    "solution": "\\boxed{y = \\frac{1}{2}x - 2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, in the Cartesian coordinate system xOy, point $A(-1, 0)$ and point $B(4, 0)$ are given. Point $C$ is located on the negative semi-axis of the y-axis. The lines $AC$ and $BC$ are connected and it is given that $\\angle ACO = \\angle CBO$. What is the equation of the line $BC$?\n",
    "completion": "\\textbf{Solution:} Given $\\because$ $\\angle AOC = \\angle BOC = 90^\\circ$ and $\\angle ACO = \\angle CBO$,\n\\\\\n$\\therefore$ $\\triangle AFC \\sim \\triangle CFG$,\n\\\\\n$\\therefore$ $\\frac{OA}{OC} = \\frac{OC}{OB}$,\n\\\\\n$\\because$ OA = 1 and OB = 4,\n\\\\\n$\\therefore$ OC = 2, and the coordinates of point C are (0, $-2$),\n\\\\\nLet the equation of line BC be $y = kx - 2$,\n\\\\\nSubstituting point $B(4, 0)$ into the equation yields $0 = 4k - 2$,\n\\\\\nSolving for $k$ gives: $k = \\frac{1}{2}$,\n\\\\\n$\\therefore$ the equation of line BC is $\\boxed{y = \\frac{1}{2}x - 2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51152254_116.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle A=90^\\circ$, $\\angle B=30^\\circ$, $CD$ bisects $\\angle ACB$ and intersects $AB$ at point $D$, $DE\\parallel BC$, and intersects $AC$ at point $E$. If $BC=9$, what is the length of $AE$?\n",
    "completion": "\\textbf{Solution:} In the right-angled triangle $ABC$, $\\angle B=30^\\circ$, \\\\\n$\\therefore AC= \\frac{1}{2}BC= \\frac{9}{2}$, \\\\\n$\\angle ACB=90^\\circ-\\angle B=60^\\circ$, \\\\\n$\\because CD$ bisects $\\angle ACB$, \\\\\n$\\therefore \\angle ACD= \\frac{1}{2}\\angle ACB=30^\\circ$, \\\\\n$\\because DE\\parallel BC$, \\\\\n$\\therefore \\angle EDC=\\angle DCB=\\angle ACD=30^\\circ$, \\\\\n$\\therefore CE=DE$, \\\\\n$\\therefore \\angle ADC=60^\\circ$, \\\\\n$\\therefore \\angle ADE=\\angle ADC-\\angle EDC=30^\\circ$, \\\\\n$\\therefore AE=\\frac{1}{2}DE=\\frac{1}{2}EC$, \\\\\n$\\therefore AE= \\frac{1}{3}AC= \\frac{1}{3}\\times \\frac{9}{2}=\\boxed{\\frac{3}{2}}$",
    "image_path": "dataset/mm_math/images/MM_Math/53033975.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Definition: If a moving point $P$ satisfies the condition $PA=4PB$ with respect to the two endpoints of a line segment $AB$, then $P$ is called the $KZ$ point of the line segment $AB$, but point $P$ is not the $KZ$ point of line segment $BA$. As shown in Figure 1, in $\\triangle ABC$, $\\angle C=90^\\circ$, $AB=17$. If point $C$ is the $KZ$ point of the line segment $AB$, what is the length of $AC$?\n",
    "completion": "\\textbf{Solution:} Given that point $C$ is on the line segment $AB$ as the $KZ$ point, \\\\\n$\\therefore$ $AC=4BC$, \\\\\nLet $BC=m$, then $AC=4m$, \\\\\nGiven that $\\angle C=90^\\circ$, \\\\\n$\\therefore$ $AC^{2}+BC^{2}=AB^{2}$, \\\\\n$\\therefore$ $m^{2}+(4m)^{2}=17^{2}$, \\\\\n$\\therefore$ $m^{2}=1$, \\\\\nSince $m>0$, \\\\\n$\\therefore$ $m=1$, \\\\\n$\\therefore$ $AC=\\boxed{4}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52991464_111.png"
  },
  {
    "solution": "\\boxed{\\frac{25}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, $AC=5$, and $D$ is the midpoint of $AB$. $BE\\perp CD$ intersects the extension of $CD$ at $E$. If $BE=4$, what is the length of $AB$?\n",
    "completion": "\\textbf{Solution:} Construct $CF \\perp AB$ at point $F$,\\\\\nSince $BE \\perp CD$,\\\\\n$\\therefore \\angle E=\\angle DFC=90^\\circ$,\\\\\nSince point $D$ is the midpoint of $AB$,\\\\\n$\\therefore CD$ is the median of right triangle $ABC$,\\\\\n$\\therefore CD=AD=BD$,\\\\\nIn $\\triangle CDF$ and $\\triangle BDE$, $\\left\\{\\begin{array}{l}\\angle CFD=\\angle E\\\\ \\angle CDF=\\angle BDE\\\\ CD=BD\\end{array}\\right.$\\\\\n$\\therefore \\triangle CDF \\cong \\triangle BDE$ (AAS)\\\\\n$\\therefore CF=BE=4$,\\\\\nIn right triangle $ACF$, $AF=\\sqrt{AC^{2}-CF^{2}}=\\sqrt{5^{2}-4^{2}}=3$;\\\\\nLet $AB=x$, then $BF=x-3$,\\\\\nIn right triangle $ABC$, $BC^{2}=x^{2}-9$,\\\\\nIn right triangle $BCF$, $BC^{2}=16+(x-3)^{2}$\\\\\n$\\therefore 16+(x-3)^{2}=x^{2}-9$,\\\\\nSolving for $x$: $x=\\frac{25}{3}$.\\\\\n$\\therefore AB=\\boxed{\\frac{25}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52815640.png"
  },
  {
    "solution": "\\boxed{64}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, it is known that point O is on line AB, with rays OD and OC on opposite sides of line AB, and $ \\angle COD=80^\\circ$. Ray OE is always the bisector of $ \\angle AOD$. As shown in Figure 2, let $ \\angle AOD=n^\\circ$. It is known that $ \\angle DOE=2\\angle AOC$. Find the value of $n$.\n",
    "completion": "\\textbf{Solution:} Given that,\\\\\n$\\because OE$ bisects $\\angle AOD$,\\\\\n$\\therefore \\angle DOE=\\frac{n^\\circ}{2}$,\\\\\n$\\because \\angle COD=80^\\circ$,\\\\\n$\\therefore \\angle AOC=80^\\circ-n^\\circ$,\\\\\n$\\because \\angle DOE=2\\angle AOC$,\\\\\n$\\therefore \\frac{n}{2}=2(80-n)$,\\\\\n$\\therefore n=\\boxed{64}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53414133_152.png"
  },
  {
    "solution": "\\boxed{\\frac{16}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $P$ is a point on side $AB$, and $PB>PC$. Connect $PC$, and fold rectangle $ABCD$ along $PC$, such that point $B$ falls onto $B'$. Extend $B'C$ to intersect the extended line of $AB$ at $Q$, given that $AB=10, BC=4$. If $DP \\perp PC$, find the length of $BQ$.\n",
    "completion": "\\textbf{Solution:} Let $PB = m$, $PA = n$, then $m + n = 10$,\\\\\nsince $DP \\perp PC$, and rectangle $ABCD$,\\\\\ntherefore $\\angle A = \\angle B = \\angle DPC = 90^\\circ$,\\\\\nsince $\\angle BPC = 90^\\circ - \\angle APD$, $\\angle ADP = 90^\\circ - \\angle APD$,\\\\\ntherefore $\\angle BPC = \\angle ADP$,\\\\\ntherefore $\\triangle BPC \\sim \\triangle ADP$,\\\\\ntherefore $\\frac{BP}{DA} = \\frac{BC}{AP}$,\\\\\ntherefore $\\frac{m}{4} = \\frac{4}{n}$, that is, $mn = 16$,\\\\\ntherefore $(m - n)^{2} = (m + n)^{2} - 4mn = 100 - 64 = 36$,\\\\\ntherefore $m - n = 6$,\\\\\nsince $m + n = 10$,\\\\\ntherefore $m = 8 = PB$.\\\\\nLet $BQ = x$, $CQ = y$,\\\\\nsince $ABCD$ is folded along $PC$, point B falls on ${B}^{\\prime}$, $BC = 4$, $PB = 6$.\\\\\ntherefore $P{B}^{\\prime} = PB = 6$, $\\angle P{B}^{\\prime}Q = \\angle PBC = 90^\\circ$,\\\\\nsince $\\angle Q = \\angle Q$,\\\\\ntherefore $\\triangle P{B}^{\\prime}Q \\sim \\triangle CBQ$,\\\\\ntherefore $\\frac{BC}{CQ} = \\frac{P{B}^{\\prime}}{PQ}$,\\\\\ntherefore $\\frac{4}{y} = \\frac{8}{8 + x}$,\\\\\nsolving gives $y = \\frac{x + 8}{2}$,\\\\\nby Pythagorean theorem, ${x}^{2} + 16 = {y}^{2}$,\\\\\ntherefore ${x}^{2} + 16 = \\left(\\frac{x + 8}{2}\\right)^{2}$,\\\\\nsolving gives $x = \\boxed{\\frac{16}{3}}, x = 0$ (discard $x = 0$).",
    "image_path": "dataset/mm_math/images/MM_Math/53102165_75.png"
  },
  {
    "solution": "\\boxed{y=\\frac{4}{3}x-2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given, as shown in the figure, the graph of a linear function passes through points $P(3, 2)$ and $B(0, -2)$, and intersects the x-axis at point $A$. What is the analytical expression of the linear function?\n",
    "completion": "\\textbf{Solution:} Let the linear function be $y=kx+b$. Substituting points $P(3,2)$ and $B(0, -2)$ into $y=kx+b$, we get\n\\[\n\\begin{cases}\n3k+b=2\\\\\nb=-2\n\\end{cases}\n\\]\nSolving these equations, we obtain\n\\[\n\\begin{cases}\nk=\\frac{4}{3}\\\\\nb=-2\n\\end{cases}\n\\]\n$\\therefore$ The equation of the linear function is $\\boxed{y=\\frac{4}{3}x-2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50677300_146.png"
  },
  {
    "solution": "\\boxed{2.4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, $AD$ bisects $\\angle BAC$ and intersects $BC$ at point $D$, $DE\\perp AB$ with the foot at $E$. If $BC=4$ and $DE=1.6$, what is the length of $BD$?\n",
    "completion": "\\textbf{Solution:} Since $AD$ bisects $\\angle BAC$, $DE\\perp AB$, and $\\angle C=90^\\circ$,\\\\\nit follows that $CD=DE$,\\\\\nSince $DE=1.6$,\\\\\nit follows that $CD=1.6$,\\\\\ntherefore $BD=BC-CD=4-1.6=\\boxed{2.4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53033289.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given the parabola: $y=ax^2-6ax-16a$ (where $a>0$) intersects the x-axis at points $A$ and $B$ (with $A$ to the left of $B$), and intersects the y-axis at point $C$. Point $G$ is the midpoint of segment $AC$. Find the coordinates of points $A$ and $B$, and the axis of symmetry of the parabola.\n",
    "completion": "\\textbf{Solution:} Given the problem, we have $y=ax^2-6ax-16a=a(x^2-6x-16)$ \\\\\nSetting $y=0$, then $x^2-6x-16=0$, \\\\\nSolving yields $x=-2$, $x=8$ \\\\\nTherefore, $\\boxed{A(-2, 0)}$, $\\boxed{B(8, 0)}$ \\\\\nAlso, since $y=a(x-3)^2-25a$, \\\\\nTherefore, the axis of symmetry is the line $x=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51545315_150.png"
  },
  {
    "solution": "\\boxed{65^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC \\cong \\triangle ADE$, where D is on side $BC$, $\\angle E = 35^\\circ$, $\\angle DAC = 30^\\circ$. What is the measure of $\\angle BDA$?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABC\\cong \\triangle ADE$\\\\\n$\\therefore \\angle E=\\angle C=35^\\circ$\\\\\nSince in $\\triangle ADC$, $\\angle DAC=30^\\circ$, $\\angle C=35^\\circ$\\\\\n$\\therefore \\angle BDA=30^\\circ+35^\\circ=\\boxed{65^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/52603529.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, $AC=BC=8$. Fold $\\triangle ABC$ so that point A coincides with point D on side BC, with EF as the crease. If $AE=5$, what is the value of $\\sin\\angle BFD$?\n",
    "completion": "\\textbf{Solution:} Given that in $\\triangle ABC$, $\\angle ACB=90^\\circ$ and AC=BC=8, \\\\\nit follows that $\\angle A=\\angle B$, \\\\\nBy the properties of folding: $\\triangle AEF\\cong \\triangle DEF$, \\\\\nHence, $\\angle EDF=\\angle A$, \\\\\nThus, $\\angle EDF=\\angle B$, \\\\\nTherefore, $\\angle CDE+\\angle BDF+\\angle EDF=\\angle BFD+\\angle BDF+\\angle B=180^\\circ$, \\\\\nHence, $\\angle CDE=\\angle BFD$. \\\\\nMoreover, since AE=DE=5, \\\\\nit follows that CE=8−5=3, \\\\\nTherefore, in right-angled $\\triangle ECD$, $\\sin\\angle CDE= \\frac{CE}{ED}=\\frac{3}{5}$, \\\\\nTherefore, $\\sin\\angle BFD=\\boxed{\\frac{3}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51445481.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{37}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $BC=6$, $AB=3$, $R$ is on side $CD$, and $CR=2$. Let $P$ be a moving point on $BC$, and $E$, $F$ be the midpoints of $AP$, $RP$, respectively. As $P$ moves from $B$ to $C$, what is the length of segment $EF$?\n",
    "completion": "\\textbf{Solution}: As shown in the figure, connect AR.\\\\\n$\\because$ Quadrilateral ABCD is a rectangle,\\\\\n$\\therefore \\angle D=90^\\circ$,\\\\\n$\\because BC=6, AB=3, CR=2$,\\\\\n$\\therefore AD=6, DR=1$,\\\\\n$\\therefore AR=\\sqrt{6^2+1^2}=\\sqrt{37}$,\\\\\n$\\because$E and F are the midpoints of AP and RP respectively,\\\\\n$\\therefore EF=\\frac{1}{2}AR=\\frac{1}{2}\\times \\sqrt{37}=\\boxed{\\frac{\\sqrt{37}}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53232779.png"
  },
  {
    "solution": "\\boxed{\\frac{10}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, two lines are intersected by parallel lines $l_{1}$, $l_{2}$, $l_{3}$, with points A, B, C, D, E, F being the points of intersection. Given that $AB=5$, $BC=6$, $EF=4$, what is the length of $DE$?\n",
    "completion": "\\textbf{Solution:} Given that two lines are intersected by the parallel lines $l_{1}$, $l_{2}$, $l_{3}$,\\\\\nit follows that $\\frac{AB}{DE}=\\frac{BC}{CF}$,\\\\\nwhich means $\\frac{5}{DE}=\\frac{6}{4}$,\\\\\nsolving for $DE$ gives $DE=\\boxed{\\frac{10}{3}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52865964.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is a \"Gougu Circle and Square Diagram\" (also known as \"Zhao Shuang's Chord Diagram\") created by ancient Chinese mathematician Zhao Shuang. It consists of four congruent right-angled triangles and a small square EFGH seamlessly put together to form a larger square ABCD. If $\\angle ABE=30^\\circ$ and $EF=\\sqrt{3}-1$, find the length of AB.\n",
    "completion": "\\textbf{Solution:} Given $\\because \\triangle ABE \\cong \\triangle BCF$, $\\therefore AE=BF$, $\\because \\angle ABE=30^\\circ$, let $AE=x$, $\\therefore \\tan \\angle ABE = \\frac{AE}{BE} = \\frac{x}{x+\\sqrt{3}-1} = \\frac{\\sqrt{3}}{3}$, solving this, we get $x=1$, $\\therefore AE=1$, $BE=BF+EF= \\sqrt{3}$, $\\therefore AB=\\sqrt{AE^2+BE^2}=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51385818_97.png"
  },
  {
    "solution": "\\boxed{\\frac{4}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the coordinates of points $A$ and $B$ are (3, 0) and (0, 4), respectively. Point $C$ is a point on the positive $x$-axis. Line $BC$ is drawn. The line passing through point $A$ and perpendicular to $AB$ intersects with the line passing through point $C$ and perpendicular to $BC$ at point $D$. Line $BD$ is drawn. What is the value of $\\sin\\angle BDC$?\n",
    "completion": "\\textbf{Solution:} Since $BA \\perp AD$ and $BC \\perp CD$,\\\\\nit follows that $\\angle BAD = \\angle BCD = 90^\\circ$.\\\\\nHence, points A, B, C, D are concyclic.\\\\\nTherefore, $\\angle BDA = \\angle BCA$.\\\\\nSince $\\angle BDA + \\angle DBA = \\angle BCA + \\angle CBO = 90^\\circ$,\\\\\nit implies that $\\angle DBA = \\angle CBO$.\\\\\nConsequently, $\\angle DBA - \\angle CBA = \\angle CBO - \\angle CBA$,\\\\\nwhich means $\\angle DBC = \\angle ABO$.\\\\\nMoreover, $\\angle DBC + \\angle BDC = \\angle ABO + \\angle BAO = 90^\\circ$,\\\\\nthus $\\angle BDC = \\angle BAO$.\\\\\nGiven that the coordinates of points A and B are (3, 0) and (0, 4) respectively,\\\\\nit follows that $BO = 4$, $OA = 3$, and $AB = \\sqrt{4^2+3^2} = 5$.\\\\\nHence, $\\sin \\angle BAO = \\frac{BO}{AB} = \\frac{4}{5}$.\\\\\nTherefore, $\\sin \\angle BDC = \\boxed{\\frac{4}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51083573.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given: As shown in the figure, segment $AB=6\\text{cm}$, point $P$ is a moving point on segment $AB$, with sides $AP$, $BP$ to construct equilateral $\\triangle APC$ and equilateral $\\triangle BPD$ respectively on $AB$, connect $CD$, point $M$ is the midpoint of $CD$. When point $P$ moves from point $A$ to point $B$, what is the length of the path traced by point $M$ in centimeters?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, extend AC and BD to meet at H. Through point M, draw GN $\\parallel$ AB intersecting AH at G and BH at N,\\\\\n$\\because$ $\\triangle APC$ and $\\triangle BPD$ are both equilateral triangles,\\\\\n$\\therefore$ $\\angle A=\\angle B=\\angle DPB=\\angle CPA=60^\\circ$,\\\\\n$\\therefore$ AH $\\parallel$ PD and BH $\\parallel$ CP,\\\\\n$\\therefore$ quadrilateral CPDH is a parallelogram,\\\\\n$\\therefore$ CD and HP bisect each other,\\\\\n$\\therefore$ M is the midpoint of PH,\\\\\nThus, as P moves, M always remains at the midpoint of HP, so the trajectory of M is indeed the median of $\\triangle HAB$, that is, the segment GN,\\\\\n$\\therefore$ $GN=\\frac{1}{2}AB=\\boxed{3}$ cm.",
    "image_path": "dataset/mm_math/images/MM_Math/52423387.png"
  },
  {
    "solution": "\\boxed{(5, 4)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system $xoy$, if the coordinates of points A and B of the rhombus $ABCD$ are $(-3, 0)$ and $(2, 0)$ respectively, and point D is on the positive direction of the y-axis, then what are the coordinates of point C?\n",
    "completion": "\\textbf{Solution:} Given that in the rhombus ABCD, the coordinates of vertices A and B are respectively (-3, 0) and (2, 0), and point D is on the y-axis,\\\\\n$\\therefore$AB=AO+OB=5,\\\\\n$\\therefore$AD=AB=CD=5,\\\\\n$\\therefore$DO=$\\sqrt{AD^{2}-AO^{2}}=\\sqrt{5^{2}-3^{2}}=4$,\\\\\n$\\therefore$the coordinates of point C are: $\\boxed{(5, 4)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52250626.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB=AC$, $\\angle BAC=120^\\circ$, and the perpendicular bisector of $AB$ intersects $BC$ at point $D$. What is the degree measure of $\\angle ADC$?\n",
    "completion": "\\textbf{Solution:} Given $AB=AC$ and $\\angle BAC=120^\\circ$,\\\\\nit follows that $\\angle B=\\angle C= \\frac{1}{2} (180^\\circ-\\angle BAC)= \\frac{1}{2} (180^\\circ-120^\\circ)=30^\\circ$,\\\\\nsince $DE$ is the perpendicular bisector of $AB$,\\\\\nit implies that $AD=BD$,\\\\\ntherefore $\\angle BAD=\\angle B=30^\\circ$,\\\\\nthus $\\angle ADC=\\angle B+\\angle BAD=30^\\circ+30^\\circ=\\boxed{60^\\circ};$",
    "image_path": "dataset/mm_math/images/MM_Math/52837330_101.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, CD is the altitude from C to the hypotenuse AB, $BC=\\sqrt{7}$, and $AC=3$. What is the value of $\\sin\\angle ACD$?\n",
    "completion": "\\textbf{Solution:} Given $\\triangle ABC$, where $\\angle ACB = 90^\\circ$,\\\\\n$\\therefore AB = \\sqrt{3^2 + (\\sqrt{7})^2} = 4$,\\\\\n$\\angle ACD + \\angle BCD = 90^\\circ$,\\\\\n$\\because CD$ is the height from the hypotenuse $AB$,\\\\\n$\\therefore CD \\perp AB$,\\\\\n$\\therefore \\angle B + \\angle BCD = 90^\\circ$,\\\\\n$\\therefore \\angle ACD = \\angle B$,\\\\\n$\\therefore \\sin \\angle ACD = \\sin \\angle B = \\frac{AC}{AB} = \\boxed{\\frac{3}{4}}$",
    "image_path": "dataset/mm_math/images/MM_Math/53110835.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $C$ is on line segment $AB$, it is known that $AB=8$, $CB=3$. If $D$ is the midpoint of line segment $AB$, what is the length of line segment $CD$?\n",
    "completion": "\\textbf{Solution:} Given that D is the midpoint of line segment AB, and AB=8,\\\\\nit follows that $BD=\\frac{1}{2}AB=4$,\\\\\nsince $CB=3$,\\\\\nthus $CD=BD-CB=\\boxed{1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53364494.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $DE\\parallel BC$ and $AD: DB = 2: 1$. The area of $\\triangle ABC$ is $27$. What is the area of $\\triangle ADE$?\n",
    "completion": "\\textbf{Solution:} Since $AD:DB = 2:1$, \\\\\nit follows that $\\frac{AD}{AB} = \\frac{2}{3}$. \\\\\nSince $DE \\parallel BC$, \\\\\nit implies $\\triangle ADE \\sim \\triangle ABC$. \\\\\nTherefore, $\\frac{S_{\\triangle ADE}}{S_{\\triangle ABC}} = \\left( \\frac{AD}{AB} \\right)^2$, that is $\\frac{S_{\\triangle ADE}}{27} = \\frac{4}{9}$, \\\\\nSolving this we get: $S_{\\triangle ADE} = \\boxed{12}$, \\\\\nThus, the area of $\\triangle ADE$ is $12$.",
    "image_path": "dataset/mm_math/images/MM_Math/52536496.png"
  },
  {
    "solution": "\\boxed{\\frac{60}{13}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated in the figure, in $\\triangle ABC$, $\\angle ACB = 90^\\circ$, $AC = 5$, $BC = 12$. Let $D$ be a moving point on $AB$. Construct $DE \\perp AC$ at point $E$ and $DF \\perp BC$ at point $F$. Connect $EF$. What is the minimum value of the line segment $EF$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, connect CD.\\\\\n$\\because$ $\\angle ACB = 90^\\circ$, AC = 5, BC = 12,\\\\\n$\\therefore$ AB = $\\sqrt{AC^2 + BC^2}$ = $\\sqrt{5^2 + 12^2}$ = 13,\\\\\n$\\because$ DE $\\perp$ AC, DF $\\perp$ BC, $\\angle C = 90^\\circ$,\\\\\n$\\therefore$ Quadrilateral CFDE is a rectangle,\\\\\n$\\therefore$ EF = CD,\\\\\nBy the principle that the perpendicular segment is the shortest, when CD $\\perp$ AB, the value of segment EF is the smallest,\\\\\nAt this moment, $S_{\\triangle ABC} = \\frac{1}{2} \\text{BC} \\cdot \\text{AC} = \\frac{1}{2} \\text{AB} \\cdot \\text{CD}$,\\\\\nwhich means $\\frac{1}{2} \\times 12 \\times 5 = \\frac{1}{2} \\times 13 \\cdot \\text{CD}$,\\\\\nSolving yields: CD = $\\frac{60}{13}$,\\\\\n$\\therefore$ EF = $\\boxed{\\frac{60}{13}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52344186.png"
  },
  {
    "solution": "\\boxed{0}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the right triangle $\\triangle ABC$ the two perpendicular sides satisfy $BC=6$ and $AC=8$. Three squares are constructed on the three sides of right triangle $\\triangle ABC$. If the areas of the four shaded regions are denoted as $S_{1}$, $S_{2}$, $S_{3}$, and $S_{4}$ respectively, what is the value of $S_{4}$? What is the value of $S_{2}+S_{3}-S_{1}$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral ACFE and quadrilateral ABGD are squares, and the right-angle sides of $\\triangle ACB$ are BC and AC,\n\n$\\therefore$AE=AC, $\\angle EAD+\\angle DAC=\\angle CAB+\\angle DAC=90^\\circ$, $\\angle E=\\angle ACB=90^\\circ$, AD=AB, AB=BG, $\\angle ABK=\\angle BGH=90^\\circ$, $\\angle KAB+\\angle ABC=\\angle HBG+\\angle ABC=90^\\circ$,\n\n$\\therefore$$\\angle EAD=\\angle CAB$, $\\angle KAB=\\angle HBG$,\n\nGiven that in $\\triangle EAD$ and $\\triangle CAB$,\n\n$\\left\\{\\begin{array}{l}\n\\angle E=\\angle ACB \\\\\n\\angle EAD=\\angle CAB \\\\\nAE=AC\n\\end{array}\\right.$\n\n$\\therefore$$\\triangle EAD\\cong \\triangle CAB$,\n\nGiven BC=6, AC=8,\n\n$\\therefore$S$_{4}$=S$_{\\triangle ABC}=\\frac{1}{2}AC\\cdot BC=\\frac{1}{2}\\times 6\\times 8=\\boxed{24}$,\n\nGiven that in $\\triangle ABK$ and $\\triangle BGH$,\n\n$\\left\\{\\begin{array}{l}\n\\angle ABK=\\angle BGH \\\\\nAB=BG \\\\\n\\angle KAB=\\angle HBG\n\\end{array}\\right.$\n\n$\\therefore$$\\triangle ABK\\cong \\triangle BGH$,\n\n$\\therefore$S$_{\\triangle ABC}=S_{1}$,\n\nLet the area of quadrilateral ADHC be $m$, and the area of $\\triangle BCK$ be $n$,\n\nGiven quadrilateral ACFE, quadrilateral ABGD, and quadrilateral BJIC are all squares, and the right-angle sides of $\\triangle ACB$ are BC and AC,\n\n$\\therefore$${S}_{3}+{S}_{4}+m=AC^{2}$, ${S}_{2}+n=BC^{2}$, ${S}_{1}+m+n+{S}_{\\triangle ABC}=AB^{2}$, $AC^{2}+BC^{2}=AB^{2}$\n\n$\\therefore$S$_{2}$+S$_{3}$−S$_{1}$\n\n=$\\left({S}_{3}+{S}_{4}+m\\right)+\\left({S}_{2}+n\\right)−\\left({S}_{\\triangle ABC}+m+n+{S}_{1}\\right)$\n\n=$AC^{2}+BC^{2}−AB^{2}$\n\n=$\\boxed{0}$",
    "image_path": "dataset/mm_math/images/MM_Math/52836358.png"
  },
  {
    "solution": "\\boxed{32}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, to measure the distance between points $A$ and $B$ on the side of a pond, Xiao Jun selects a point $P$ on one side of the pond. The midpoints of $PA$ and $PB$ are $D$ and $E$ respectively, and the length of $DE$ is $16$ meters. What is the distance between points $A$ and $B$ in meters?\n",
    "completion": "\\textbf{Solution:} Since D and E are the midpoints of PA and PB, respectively, \\\\\nit follows that DE is the midline of $\\triangle PAB$, \\\\\ntherefore, AB $=$ 2DE, \\\\\nsince DE $=$ 16 meters, \\\\\nit follows that AB $=$ \\boxed{32} meters.",
    "image_path": "dataset/mm_math/images/MM_Math/53728938.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in parallelogram $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$. Let $M$ be the midpoint of $AD$, and connect $CM$ to intersect $BD$ at point $N$, with $ON=1$, and $\\triangle DMN \\sim \\triangle BCN$; what is the length of $BD$?\n",
    "completion": "\\textbf{Solution:} By the given information, we have:\n\\begin{align*}\n\\because \\triangle DMN\\sim \\triangle BCN, \\\\\n\\therefore \\frac{MD}{CB}=\\frac{DN}{BN}, \\\\\n\\because M\\text{ is the midpoint of }AD, \\\\\n\\therefore MD=\\frac{1}{2}AD=\\frac{1}{2}BC, \\\\\n\\text{i.e., } \\frac{MD}{CB}=\\frac{1}{2}, \\\\\n\\therefore \\frac{DN}{BN}=\\frac{1}{2}, \\\\\n\\text{i.e., } BN=2DN, \\\\\n\\text{let } OB=OD=x, \\text{ thus } BD=2x, BN=OB+ON=x+1, DN=x-1, \\\\\n\\therefore x+1=2(x-1), \\\\\n\\text{Solving, we find } x=3, \\\\\n\\therefore BD=2x=\\boxed{6}.\n\\end{align*}",
    "image_path": "dataset/mm_math/images/MM_Math/52866019_122.png"
  },
  {
    "solution": "\\boxed{20}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC$ and $AD\\perp BC$ at point $D$. If $AB=6$ and $CD=4$, what is the perimeter of $\\triangle ABC$?\n",
    "completion": "\\textbf{Solution:} \n\nIn triangle $\\triangle ABC$, we have $AB=AC$, and $AD \\perp BC$ at point D, \n\n$\\therefore BC=2CD=8$,\n\n$\\therefore$ the perimeter of $\\triangle ABC$ is: $AB+AC+BC=6+6+8=\\boxed{20}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52808244.png"
  },
  {
    "solution": "\\boxed{(5,4)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system, if the vertices $A$ and $B$ of the rhombus $ABCD$ are at $(-3, 0)$ and $(2, 0)$ respectively, and point $D$ is on the y-axis, what are the coordinates of point $C$?\n",
    "completion": "Solution: Since the coordinates of points A and B are respectively $(-3,0)$ and $(2,0)$, \\\\\nit follows that $AB=5$ \\\\\nSince quadrilateral $ABCD$ is a rhombus, \\\\\nit follows that $AB=AD=BC=5$, and $DC\\parallel AB$ \\\\\nSince $AO=3$, \\\\\nit follows that $DO=\\sqrt{AB^{2}-AO^{2}}=\\sqrt{5^{2}-3^{2}}=4$ \\\\\nTherefore, $C(5,4)$ \\\\\nHence, the coordinates of point C are $\\boxed{(5,4)}$",
    "image_path": "dataset/mm_math/images/MM_Math/52422315.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, $AD \\parallel BC$, $\\angle A=90^\\circ$, $BD=BC$. Point $E$ is the midpoint of $CD$, the ray $BE$ extends and intersects the extension line of $AD$ at point $F$, and $CF$ is connected. If $AD=1$, $BC=2$, find the length of $BF$.\n",
    "completion": "\\textbf{Solution:} Since quadrilateral BCFD is a rhombus,\\\\\nit follows that BD=DF=BC=2,\\\\\nin $\\triangle BAD$, AB=$\\sqrt{BD^{2}-AD^{2}}=\\sqrt{3}$,\\\\\nSince AF=AD+DF=1+2=3, in $\\triangle BAF$, BF=$\\sqrt{AB^{2}+AF^{2}}=\\boxed{2\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53599308_95.png"
  },
  {
    "solution": "\\boxed{80^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $O$ is a point on $AB$, $OD$ bisects $\\angle BOC$, $\\angle 1=20^\\circ$. What is the degree measure of $\\angle 2$?\n",
    "completion": "\\textbf{Solution:} Given that $\\because \\angle 1=20^\\circ$,\\\\\n$\\therefore \\angle BOC=180^\\circ-\\angle 1=180^\\circ-20^\\circ=160^\\circ$,\\\\\n$\\because$ OD bisects $\\angle BOC$,\\\\\n$\\therefore \\angle 2=\\boxed{80^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52794245.png"
  },
  {
    "solution": "\\boxed{\\sqrt{10}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, with $D$ being a point on side $AC$ and $\\angle ABC = \\angle ADB = 45^\\circ$, if $AD=2$ and $CD=3$, then what is the length of $AB$?\n",
    "completion": "\\textbf{Solution:} Given that, $\\angle ABC = \\angle ADB = 45^\\circ$, $\\angle BAD = \\angle CAB$,\\\\\n$\\therefore \\triangle ABD \\sim \\triangle ACB$,\\\\\n$\\therefore \\frac{AD}{AB} = \\frac{AB}{AC}$,\\\\\n$\\therefore AB^2 = AD \\cdot AC$,\\\\\n$\\because AD = 2$, CD = 3,\\\\\n$\\therefore AC = 5$,\\\\\n$\\therefore AB^2 = 2 \\times 5 = 10$, that is: $AB = \\boxed{\\sqrt{10}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51402990.png"
  },
  {
    "solution": "\\boxed{x<1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y = -x + b$ intersects with the graph of $y = kx - 1$ at point P, and intersects the x-axis at point B. It is known that the ordinate of point P is 3, and the abscissa of point B is 4. What is the solution set of the inequality $-x + b > kx - 1$?\n",
    "completion": "\\textbf{Solution:}\n\nSince the linear function $y=-x+b$ intersects the x-axis at point B and the x-coordinate of point B is 4, \n\nwe have B(4,0).\n\nSubstituting B(4,0) into $y=-x+b$, we get: $0=-4+b$.\n\nSolving this, we find $b=4$.\n\nTherefore, the equation of line BP is $y=-x+4$.\n\nGiven that the y-coordinate of point $P$ is 3,\n\nwe therefore have $3=-x+4$.\n\nThus, $x=1$.\n\nHence, P(1,3).\n\nFrom the graph, it is known that when $x<1$, the graph of the line $y=kx-1$ is below the graph of the function $y=-x+b$,\n\ntherefore, the solution set of the inequality $-x+b>kx-1$ is $x<1$,\n\nso the answer is: $\\boxed{x<1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52189388.png"
  },
  {
    "solution": "\\boxed{30}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rhombus $ABCD$, the diagonals $AC=6$ and $BD=10$. What is the area of rhombus $ABCD$?\n",
    "completion": "\\textbf{Solution:} Given that in the rhombus $ABCD$, the diagonals $AC=6$ and $BD=10$, \\\\\ntherefore, the area of rhombus $ABCD$ is $= \\frac{1}{2}AC \\cdot BD = \\boxed{30}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51968646.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the musical staff is composed of five equidistant and equal-length parallel horizontal lines. There are three points, A, B, and C, all located on the lines. If the length of segment AB equals \\( \\frac{3}{2} \\), what is the length of segment BC?\n",
    "completion": "\\textbf{Solution:} Draw a perpendicular line through point A parallel to the horizontal line, intersecting the horizontal line through point B at D and the horizontal line through point C at E,\\\\\nGiven that:\\\\\nAD＝2DE, $ \\frac{AB}{BC}=\\frac{AD}{DE}$,\\\\\nSince AB＝$\\frac{3}{2}$,\\\\\nTherefore, $\\frac{\\frac{3}{2}}{BC}$ ＝2,\\\\\nThus, BC＝$\\boxed{\\frac{3}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52861885.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that point $C$ is on the line segment $AB$, and $M$, $N$ are the midpoints of $AC$, $BC$ respectively. If the length of segment $AC=12$ and $BC=8$, find the length of segment $MN$.\n",
    "completion": "\\textbf{Solution:} By the given information, we have \\(AC=12\\) and \\(BC=8\\), with \\(M\\), \\(N\\) being the midpoints of \\(AC\\), \\(BC\\) respectively, \\\\\n\\(\\therefore MC= \\frac{1}{2}AC\\), \\(NC= \\frac{1}{2}BC\\), \\\\\n\\(\\therefore MN= \\frac{1}{2}AC+ \\frac{1}{2}BC= \\frac{1}{2} \\times 12+ \\frac{1}{2} \\times 8=\\boxed{10}.\\)",
    "image_path": "dataset/mm_math/images/MM_Math/54927380_129.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, AB is the diameter of circle $O$, with C and D being two points on the circle. $OC \\parallel BD$, and $OC$ intersects $AD$ at point E. $AC = CD$; if $OE = 2$ and $AD = 8$, find the radius of circle $O$.\n",
    "completion": "\\textbf{Solution:} From (1) we know that $OC\\perp AD$,\\\\\n$\\because AD=8$,\\\\\n$\\therefore AE=\\frac{1}{2}AD=4$,\\\\\n$\\because OE=2$,\\\\\n$\\therefore OA=\\sqrt{OE^{2}+AE^{2}}=2\\sqrt{5}$,\\\\\n$\\therefore$ the radius of circle $O$ is $\\boxed{2\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52537026_84.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that in rectangle $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$. Lines parallel to $AC$ and $BD$ are drawn through points $D$ and $C$ respectively, intersecting at point $E$. Quadrilateral $OCED$ is a rhombus. If $AB=3$ and $BC=4$, what is the area of the rhombus $OCED$?\n",
    "completion": "\\textbf{Solution:} Given $AB=3$, $BC=4$,\\\\\nthus, the area of rectangle $ABCD$ equals $3\\times 4=12$,\\\\\nsince $S_{\\triangle ODC}=\\frac{1}{4}S_{\\text{rectangle}ABCD}=3$,\\\\\ntherefore, the area of the quadrilateral $OCED$ equals $2S_{\\triangle ODC}=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52380930_141.png"
  },
  {
    "solution": "\\boxed{y=-x+4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, the line AB intersects the x-axis and the y-axis at points A and B, respectively, and point C is on the negative half-axis of the x-axis. The coordinates of these three points are A(4,0), B(0,4), and C(-1,0), respectively. Find the equation of line AB.\n",
    "completion": "\\textbf{Solution:} Since line AB passes through points A(4,0) and B(0,4), \\\\\nlet the equation of line AB be $y=kx+4$, \\\\\nsubstituting A(4,0) into it gives: $4k+4=0$, \\\\\nsolving for $k$ gives: $k=-1$, \\\\\ntherefore, the equation of line AB is $\\boxed{y=-x+4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51353152_179.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $A$ is a point on the circle $\\odot O$, $OD\\perp$ chord $BC$ at point $D$. If $\\angle BAC=60^\\circ$ and $OD=1$, then what is the length of $BC$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect OC.\\\\\nSince $\\angle BAC = 60^\\circ$,\\\\\nthus $\\angle BOC = 2\\angle BAC = 120^\\circ$,\\\\\nSince OB = OC,\\\\\nthus $\\angle OBC = \\angle OCB = 30^\\circ$,\\\\\nSince OD$\\perp$BC,\\\\\nthus BD = CD,\\\\\nIn $\\triangle BOD$, with $\\angle OBC = 30^\\circ$, and OD = 1,\\\\\nthus BD = $\\sqrt{3}$OD = $\\sqrt{3}$,\\\\\ntherefore BC = 2BD = $\\boxed{2\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52808101.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $DE$ is the perpendicular bisector of side $AC$ of $\\triangle ABC$, intersecting $BC$ at point $D$ and intersecting $AC$ at point $E$. If $BC = 8\\text{cm}$ and $AD = 5\\text{cm}$, what is the length of $BD$?\n",
    "completion": "\\textbf{Solution:} Since $DE$ is the perpendicular bisector of side $AC$ in $\\triangle ABC$, and $AD=5\\text{cm}$, \\\\\nthen $DC=AD=5\\text{cm}$, \\\\\nthus $BD=BC-DC=8-5=\\boxed{3}\\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52828907.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, $AD \\parallel CB$, and $O$ is the midpoint of the diagonal $AC$. A line passing through point $O$ intersects the sides $AD$ and $BC$ of quadrilateral $ABCD$ at points $M$ and $N$, respectively. The segments $CM$ and $AN$ are then connected. Quadrilateral $ANCM$ is a parallelogram; when $MN$ bisects $\\angle AMC$,\n",
    "completion": "\\textbf{Solution:}\\\\\n(1) Since $MN$ bisects $\\angle AMC$,\\\\\nit follows that $\\angle AMN=\\angle CMN$,\\\\\nand since $AD\\parallel BC$,\\\\\nit follows that $\\angle AMN=\\angle CNM$,\\\\\ntherefore, $\\angle CMN=\\angle CNM$,\\\\\nhence $CM=CN$,\\\\\nthus parallelogram $ANCM$ is a rhombus;\\\\\n(2) Since quadrilateral $ABCD$ is a rectangle,\\\\\nit implies that $\\angle ABN=90^\\circ$, $BC=AD=8$, and $AM=CN$,\\\\\nthus $AB=\\sqrt{AC^{2}-BC^{2}}=\\sqrt{(4\\sqrt{5})^{2}-8^{2}}=4$, $AM=AN=NC=AD-DM$, $BN=DM$,\\\\\nin $\\triangle ABN$ right-angled at $B$, by the Pythagorean theorem, we have $AN^{2}=AB^{2}+BN^{2}$,\\\\\nthus, $(8-DM)^{2}=4^{2}+DM^{2}$, solving this yields $DM=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51863620_108.png"
  },
  {
    "solution": "\\boxed{120^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a regular hexagon and a regular pentagon are placed as depicted, with the common vertex $O$, and the side $AB$ of the hexagon and the side $DE$ of the pentagon lying on the same straight line. What is the measure of $\\angle ABO$?\n",
    "completion": "\\textbf{Solution}: From what is given in the problem, we have $\\angle ABO = \\frac{1}{6} \\times (6-2) \\times 180^\\circ = \\boxed{120^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52905816_132.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within circle $O$ with a radius of $5$, and chord $AB=8$, $P$ is a moving point on chord $AB$. What is the minimum value of $OP$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, connect $OA$ and draw $OC\\perp AB$ through point $O$ with the foot at $C$. Hence, $AC=BC=\\frac{1}{2}AB=4$,\\\\\nIn $\\triangle AOC$,\\\\\n$OC=\\sqrt{OA^2-AC^2}=3$,\\\\\nwhich means the minimum value for point $P$ moving on $AB$ is $\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53009484.png"
  },
  {
    "solution": "\\boxed{(0,4)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the parabola $y=-x^2+2x+4$ intersects the y-axis at point $C$, and its vertex is point $D$. Find the coordinates of points $C$ and $D$.\n",
    "completion": "\\textbf{Solution:} The equation of the parabola is $y=-x^2+2x+4=-(x-1)^2+5$,\\\\\n$\\therefore$ the coordinates of point $D$ are \\boxed{(1,5)}.\\\\\nLet $x=0$, then $y=4$,\\\\\n$\\therefore$ the coordinates of point $C$ are \\boxed{(0,4)}.",
    "image_path": "dataset/mm_math/images/MM_Math/52660399_100.png"
  },
  {
    "solution": "\\boxed{70^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, point D is on side AB, and $\\triangle CBD$ is folded along CD such that point B coincides exactly with point E on side AC. If $\\angle A=25^\\circ$, then what is the measure of $\\angle CDE$?\n",
    "completion": "\\textbf{Solution:} Given $\\because \\angle ACB=90^\\circ$ and $\\angle A=25^\\circ$,\\\\\n$\\therefore \\angle B=65^\\circ$,\\\\\nFrom the folding, we have: $\\angle BCD=\\angle ACD= \\frac{1}{2}\\angle ACB=45^\\circ$ and $\\angle CED=\\angle B=65^\\circ$,\\\\\n$\\therefore \\angle CDE=180^\\circ-\\angle CED-\\angle ACD=\\boxed{70^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53059019.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, $AB$ is the diameter of the circle $\\odot O$, with chord $CD\\perp AB$. Point E is located on the extension line of $CA$, and line $DE$ intersects $\\odot O$ at point F. Lines $AF$ and $CF$ are drawn. Given that $AB=5$ and $CD=4$, and that $CF$ passes through the center of the circle, calculate the length of $CE$.\n",
    "completion": "\\textbf{Solution:} As shown in Figure 2, suppose $AB$ intersects $CD$ at $H$. \\\\\nSince $AB$ is a diameter and $AB\\perp CD$, $CD=4$ \\\\\nThus, $CH=DH=2$, \\\\\nSince $OC=\\frac{1}{2}CF=\\frac{1}{2}AB=\\frac{5}{2}$, $\\angle OHC=90^\\circ$, \\\\\nThus, $OH=\\sqrt{OC^{2}-CH^{2}}=\\sqrt{\\left(\\frac{5}{2}\\right)^{2}-2^{2}}=\\frac{3}{2}$, \\\\\nTherefore, $HA=OH+OA=4$, \\\\\nAs a result, $AC=\\sqrt{CH^{2}+AH^{2}}=\\sqrt{2^{2}+4^{2}}=2\\sqrt{5}$, \\\\\nSince $CF$ is a diameter, \\\\\nThus, $\\angle CDF=\\angle AHC=90^\\circ$, \\\\\nTherefore, $AH\\parallel DE$, \\\\\nTherefore, $\\triangle AHC\\sim \\triangle EDC$, \\\\\nThus, $\\frac{CH}{CD}=\\frac{AC}{CE}$ \\\\\nSince $CH=HD$, \\\\\nThus, $CD=2CH$ \\\\\nTherefore, $CE=2AC=4\\sqrt{5}$, \\\\\nTherefore, $CE=\\boxed{4\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52991580_101.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the length of line segment $AB=23$, $BC=15$, and $M$ is the midpoint of $AC$. What is the length of line segment $CM$?\n",
    "completion": "\\textbf{Solution:} Since $AB=23$, $BC=15$,\\\\\nthus $AC=AB-BC=23-15=8$.\\\\\nAlso, since $M$ is the midpoint of $AC$,\\\\\nthus $CM=\\frac{1}{2}AC=\\frac{1}{2}\\times 8=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53363922_83.png"
  },
  {
    "solution": "\\boxed{70^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $a\\parallel b$, $\\angle 1=\\angle 2$, $\\angle 3=40^\\circ$, then what is the degree measure of $\\angle 4$?\n",
    "completion": "\\textbf{Solution:} Given that $a \\parallel b$,\\\\\nit follows that $\\angle 1 + \\angle 2 + \\angle 3 = 180^\\circ$,\\\\\nsince $\\angle 3 = 40^\\circ$,\\\\\nthen $\\angle 1 + \\angle 2 = 180^\\circ - 40^\\circ = 140^\\circ$,\\\\\ngiven that $\\angle 1 = \\angle 2$,\\\\\nthus $\\angle 2 = 70^\\circ$,\\\\\nsince $a \\parallel b$,\\\\\nit concludes that $\\angle 4 = \\angle 2 = \\boxed{70^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52579991.png"
  },
  {
    "solution": "\\boxed{5\\sin A}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$. If $AB=5$, what is the length of $BC$?\n",
    "completion": "\\textbf{Solution:} In right-angled triangle $\\triangle ABC$, where $\\angle C=90^\\circ$ and $AB=5$, \\\\\nit follows that $\\sin A= \\frac{BC}{AB}$, \\\\\nthus $BC=AB\\sin A=5\\sin A$. \\\\\nTherefore, $BC=\\boxed{5\\sin A}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51446529.png"
  },
  {
    "solution": "\\boxed{15^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, place a right-angled triangle plate as indicated, keeping \\(AB \\parallel DE\\). What is the degree measure of \\(\\angle ACD\\)?\n",
    "completion": "\\textbf{Solution:} From the given, we have: $\\angle A= 30^\\circ$, $\\angle D= 45^\\circ$,\\\\\nAs shown in the figure, let AB intersect CD at point F,\\\\\nSince AB$\\parallel$DE,\\\\\nTherefore, $\\angle AFD=\\angle D= 45^\\circ$,\\\\\nSince $\\angle AFD$ is an exterior angle of $\\triangle ACF$,\\\\\nTherefore, $\\angle ACD=\\angle AFD-\\angle A = \\boxed{15^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52603281.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In $\\triangle ABC$, $AB=6$, $\\angle B$ is an acute angle and $\\cos B=\\frac{1}{2}$, $\\tan C=3\\sqrt{3}$. What is the length of $BC$?\n",
    "completion": "\\textbf{Solution:} Draw $AH\\perp BC$ at H through point A,\\\\\nsince $\\cos B=\\frac{1}{2}$,\\\\\ntherefore $\\frac{BH}{AB}=\\frac{1}{2}$,\\\\\nsince $AB=6$,\\\\\ntherefore $BH=3$,\\\\\nin $Rt\\triangle ABH$, $AH=\\sqrt{AB^{2}-BH^{2}}=\\sqrt{6^{2}-3^{2}}=3\\sqrt{3}$,\\\\\nsince $\\tan C=3\\sqrt{3}$,\\\\\ntherefore $\\frac{AH}{CH}=3\\sqrt{3}$,\\\\\nthus $\\frac{3\\sqrt{3}}{CH}=3\\sqrt{3}$,\\\\\nsolving gives $CH=1$,\\\\\ntherefore $BC=BH+CH=3+1=\\boxed{4}$",
    "image_path": "dataset/mm_math/images/MM_Math/53080305_99.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the side length of the square paper $ABCD$ is 4, and point $F$ is on side $AD$. Connect $BF$, fold the paper along the line $BF$, and the corresponding point of point $A$ is point $G$. Extend line $AG$ to intersect $CD$ at point $E$. If $DE=3$, what is the length of $GE$?\n",
    "completion": "\\textbf{Solution:}  \nSince quadrilateral $ABCD$ is a square,  \n$\\therefore \\angle D=\\angle DAB=90^\\circ$, $AD=BA$,  \nSince it is folded,  \n$\\therefore AH\\perp BF$,  \n$\\therefore \\angle AHB=90^\\circ$,  \n$\\therefore \\angle ABF+\\angle BAH=90^\\circ$,  \nSince $\\angle BAH+\\angle DAE=90^\\circ$,  \n$\\therefore \\angle DAE=\\angle ABF$,  \n$\\therefore \\triangle DAE\\cong \\triangle ABF$ (ASA),  \n$\\therefore AE=BF$, $AF=DE$,  \nSince $AB=4$, $DE=4$,  \n$\\therefore$ $AF=3$, $BF=\\sqrt{AB^2+AF^2}=\\sqrt{3^2+4^2}=5$,  \n$\\therefore$ $AH=\\frac{AF\\cdot AB}{BF}=\\frac{3\\times 4}{5}=\\frac{12}{5}$,  \n$\\therefore AG=2AH=\\frac{24}{5}$,  \n$\\therefore$ $GE=AE-AG=5-\\frac{24}{5}=\\boxed{\\frac{1}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50496915.png"
  },
  {
    "solution": "\\boxed{15}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, $AD$ bisects $\\angle CAB$, if $AB=10$, $CD=3$, what is the area of $\\triangle ABD$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $DE \\perp AB$ at point E, \\\\\nsince $\\angle C=90^\\circ$, $AD$ bisects $\\angle BAC$, \\\\\nthus $DE=CD=3$, \\\\\ntherefore, the area of $\\triangle ABD = \\frac{1}{2}AB\\cdot DE=\\frac{1}{2}\\times 10\\times 3=\\boxed{15}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52905905.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle BAC = 90^\\circ$, $AB = 15$, $AC = 20$, and $AD \\perp BC$ with D being the foot of the perpendicular. Find the lengths of AD and BD.\n",
    "completion": "\\textbf{Solution:} In the right-angled triangle $ABC$, where $\\angle BAC = 90^\\circ$, $AB = 15$, and $AC = 20$,\\\\\naccording to the Pythagorean theorem, we have $BC = \\sqrt{AB^2 + AC^2} = \\sqrt{15^2 + 20^2} = 25$,\\\\\nsince the area of $\\triangle ABC$ is calculated as ${S}_{\\triangle ABC} = \\frac{1}{2}AB \\cdot AC$ and ${S}_{\\triangle ABC} = \\frac{1}{2}BC \\cdot AD$,\\\\\nit follows that $\\frac{1}{2}AB \\cdot AC = \\frac{1}{2}BC \\cdot AD$.\\\\\nTherefore, $AD = \\frac{AB \\cdot AC}{BC} = \\frac{15 \\times 20}{25} = \\boxed{12}$;\\\\\nsince $AD \\perp BC$,\\\\\nit follows that $\\angle ADB = 90^\\circ$.\\\\\nIn the right-angled triangle $ADB$, according to the Pythagorean theorem, we have $BD = \\sqrt{AB^2 - AD^2} = \\sqrt{15^2 - 12^2} = 9$,\\\\\ntherefore, the lengths of $AD$ and $BD$ are $12$ and $9$ respectively.",
    "image_path": "dataset/mm_math/images/MM_Math/51366600.png"
  },
  {
    "solution": "\\boxed{\\frac{3\\sqrt{10}}{10}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the vertices of $\\triangle ABC$ are all on the vertices of small squares with equal side lengths. What is the value of $\\cos\\angle BAC$?\n",
    "completion": "\\[ \\textbf{Solution:} \\text{As shown in the figure, construct } CD\\perp AB \\text{ at point } D, \\]\n\\[ \\text{Given that, } S_{\\triangle ABC}=\\frac{1}{2}\\times 2\\times 2=2, \\quad AB=\\sqrt{2^{2}+4^{2}}=2\\sqrt{5}, \\quad AC=2\\sqrt{2}, \\]\n\\[ \\therefore S_{\\triangle ABC}=\\frac{1}{2}AB\\cdot CD=2, \\quad CD=\\frac{2S_{\\triangle ABC}}{AB}=\\frac{4}{2\\sqrt{5}}=\\frac{2\\sqrt{5}}{5}, \\]\n\\[ \\text{In } \\triangle ADC, \\quad AD=\\sqrt{AC^{2}-CD^{2}}=\\frac{6\\sqrt{5}}{5}, \\]\n\\[ \\therefore \\cos\\angle BAC=\\frac{AD}{AC}=\\frac{\\frac{6\\sqrt{5}}{5}}{2\\sqrt{2}}=\\boxed{\\frac{3\\sqrt{10}}{10}}. \\]",
    "image_path": "dataset/mm_math/images/MM_Math/51083566.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points D, B, E are three points on the line segment AC, with D being the midpoint of the line segment AB. If point E is the midpoint of BC and $BE= \\frac{1}{5} AC=2cm$, find the length of the line segment DE.\n",
    "completion": "\\textbf{Solution:}\n\nSince point $E$ is the midpoint of $BC$, \\\\\nthus $BC=2BE=2\\times \\frac{1}{5}=\\frac{2}{5}$cm \\\\\nTherefore $AB=AC−BC=2−\\frac{2}{5}=\\frac{8}{5}$cm \\\\\nSince $D$ is the midpoint of line segment $AB$, \\\\\ntherefore $BD=\\frac{1}{2}AB=\\frac{1}{2}\\times \\frac{8}{5}=\\frac{4}{5}$ \\\\\nTherefore $DE=BD+BE=\\frac{4}{5}+\\frac{1}{5}=\\boxed{1}$cm",
    "image_path": "dataset/mm_math/images/MM_Math/51179538_122.png"
  },
  {
    "solution": "\\boxed{135^\\circ-\\frac{\\alpha}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $ \\triangle ABC$ is inscribed in $ \\odot O$, $ \\angle ABC>90^\\circ$. The bisector of its exterior angle $ \\angle EAC$ intersects $ \\odot O$ at point D. Connect $ DB, DC$, where $ DB$ intersects $ AC$ at point F. Given $ DB=DC$ and $ DA=DF$, when $ \\angle ABC=\\alpha $, find the measure of $ \\angle DFC$ in degrees (expressed in an algebraic formula involving $ \\alpha$)?\n",
    "completion": "\\textbf{Solution:} Given that $DA=DF$,\\\\\nit follows that $\\angle DAF=\\angle DFA$,\\\\\nhence, $\\angle DAF=\\angle DFA=\\angle CBD=\\angle BCD$,\\\\\ntherefore, $\\triangle DAF\\sim \\triangle DBC$,\\\\\nwhich means that $\\angle ADF=\\angle BDC$,\\\\\ngiven that $\\angle ABC=\\alpha$ and knowing that the opposite angles in a cyclic quadrilateral are supplementary,\\\\\nit follows that $\\angle ADC=180^\\circ-\\alpha$,\\\\\nthus, $\\angle ADF=90^\\circ-\\frac{\\alpha}{2}$,\\\\\nresulting in $\\angle DAF=\\angle DFA=\\left(180^\\circ-\\angle ADF\\right)\\div 2=45^\\circ+\\frac{\\alpha}{4}$,\\\\\ntherefore, $\\angle DFC=180^\\circ-\\angle DFA=\\boxed{135^\\circ-\\frac{\\alpha}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55497770_68.png"
  },
  {
    "solution": "\\boxed{\\left(-\\frac{1}{2}, -\\frac{9}{4}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the parabola $y=x^{2}+x-2$ intersects the x-axis at points A and B, and intersects the y-axis at point C. Find the coordinates of the vertex D.\n",
    "completion": "\\textbf{Solution:} Let $y=x^2+x-2=\\left(x+\\frac{1}{2}\\right)^2-\\frac{9}{4}$,\\\\\n$\\therefore$ the coordinates of the vertex D are $\\boxed{\\left(-\\frac{1}{2}, -\\frac{9}{4}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52747348_68.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AD\\parallel BE\\parallel CF$, the lines $l_1, l_2$ intersect these three parallel lines at points A, B, C, and points D, E, F, respectively. If $AB=9, BC=6, EF=4$, what is the length of DE?\n",
    "completion": "\\textbf{Solution:} From the property that parallel lines divide line segments proportionally, we know that $\\frac{AB}{BC}=\\frac{DE}{EF}$\\\\\n$\\therefore$ $\\frac{9}{6}=\\frac{DE}{4}$\\\\\nSolving this, we find $DE=\\boxed{6}$",
    "image_path": "dataset/mm_math/images/MM_Math/51446259.png"
  },
  {
    "solution": "\\boxed{B(1, 0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the parabola $y=-(x+1)^{2}+4$ intersects with the x-axis at points A and B, and intersects with the y-axis at point C. Find the coordinates of points A, B, and C.\n",
    "completion": "\\textbf{Solution:} The parabola $y=-(x+1)^{2}+4$ intersects the x-axis at points A and B, and intersects the y-axis at point C.\\\\\n$\\because y=-(x+1)^{2}+4=-x^{2}-2x+3$,\\\\\nLet $x=0$, then $y=3$,\\\\\n$\\therefore \\boxed{C(0, 3)}$,\\\\\nLet $y=0$,\\\\\nthen $-(x+1)^{2}+4=0$,\\\\\nsolving for $x$ gives $x_{1}=1, x_{2}=-3$,\\\\\n$\\therefore \\boxed{A(-3, 0)}, \\boxed{B(1, 0)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53066218_97.png"
  },
  {
    "solution": "\\boxed{25^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, point $O$ is a point on line $AB$. A ray $OC$ is drawn passing through point $O$, such that $\\angle BOC=65^\\circ$. A right-angled triangle ruler is placed with its right-angle vertex at point $O$. As shown in Figure 1, when one side $ON$ of the triangle ruler $MON$ coincides with ray $OB$, what is the degree measure of $\\angle MOC$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle MON=90^\\circ$ and $\\angle BOC=65^\\circ$, \\\\\nit follows that $\\angle MOC=\\angle MON-\\angle BOC=90^\\circ-65^\\circ=\\boxed{25^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51291351_162.png"
  },
  {
    "solution": "\\boxed{24}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rhombus $ABCD$, $AC$ and $BD$ intersect at point $O$. If $AC=8$ and $BD=6$, then what is the area of the rhombus $ABCD$?\n",
    "completion": "\\textbf{Solution:} Since the quadrilateral $ABCD$ is a rhombus, \\\\\n$\\therefore AC\\perp BD$, \\\\\nSince $AC=8, BD=6$, \\\\\n$\\therefore$ the area of the rhombus $ABCD$ is $\\frac{1}{2}AC\\times BD=\\frac{1}{2}\\times 8\\times 6=\\boxed{24}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52395148.png"
  },
  {
    "solution": "\\boxed{\\frac{11}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, we have $AB=AC=10$ and $BC=12$. Let points $D$ and $E$ be on the sides $BC$ and $AC$, respectively (point $D$ does not coincide with the endpoints $B$ and $C$), and suppose $\\angle ADE = \\angle B$. When $\\triangle ADE$ is an isosceles triangle, find the length of $BD$.\n",
    "completion": "\\textbf{Solution:}\n\n(1) When $DA=DE$,\\\\\n$\\therefore \\angle DAE=\\angle AED$,\\\\\n$\\because AB=AC=10$,\\\\\n$\\therefore \\angle B=\\angle C=\\angle ADE$.\\\\\n$\\because \\angle AED=\\angle EDC+\\angle C=\\angle EDC+\\angle ADE$,\\\\\n$\\therefore \\angle DAE=\\angle EDC+\\angle ADE$,\\\\\n$\\therefore \\angle EAD=\\angle ADC$,\\\\\n$\\therefore CD=AC=10$,\\\\\n$\\therefore x=BD=BC-CD=12-10=2$,\\\\\nTherefore, when $x=2$, $\\triangle ADE$ is an isosceles triangle,\\\\\n(2) When $AE=DE$, $\\triangle ADE$ is an isosceles triangle, i.e., $\\angle DAE=\\angle ADE=\\angle B$\\\\\nAlso, $\\angle ACD=\\angle BCA$,\\\\\n$\\therefore \\triangle ADC\\sim \\triangle BAC$,\\\\\n$\\therefore \\frac{AC}{BC}=\\frac{CD}{AC}$,\\\\\n$\\therefore DC\\cdot BC=AC^{2}$,\\\\\n$\\therefore DC= \\frac{25}{3}$,\\\\\n$\\therefore x=BD=12-DC=12-\\frac{25}{3}=\\frac{11}{3}$;\\\\\nIn summary: $\\triangle ADE$ is an isosceles triangle when $x=2$ or $x=\\frac{11}{3}$.\\\\\n$\\therefore \\boxed{2}$ or $\\boxed{\\frac{11}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52601788_185.png"
  },
  {
    "solution": "\\boxed{\\sqrt{3} - \\frac{1}{3}\\pi}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the sector $AOB$, $\\angle AOB=90^\\circ$, taking point A as the center, and the length of OA as the radius, let $\\overset{\\frown}{OC}$ intersect $\\overset{\\frown}{AB}$ at point C. If $OA=2$, what is the area of the shaded region?\n",
    "completion": "\\textbf{Solution:} Connect OC, AC, and construct CD $\\perp$ OA at D,\\\\\n$\\because$OA = OC = AC,\\\\\n$\\therefore$ $\\triangle$AOC is an equilateral triangle,\\\\\n$\\therefore$ $\\angle$OAC = $60^\\circ$,\\\\\n$\\because$ CD $\\perp$ OA, $\\angle$CDO = $90^\\circ$, OD = AD = $\\frac{1}{2}OA = 1$,\\\\\n$\\therefore$ CD = OD$\\times\\tan60^\\circ = 1 \\times \\sqrt{3} = \\sqrt{3}$,\\\\\n$S\\triangle$OAC = $\\frac{1}{2}OA \\cdot CD = \\frac{1}{2} \\times 2 \\times \\sqrt{3} = \\sqrt{3}$,\\\\\n$\\therefore$ $\\angle$BOC = $30^\\circ$,\\\\\n$S_{\\text{sector } OBC} = \\frac{30\\pi \\times 2^2}{360} = \\frac{1}{3}\\pi$,\\\\\n$S_{\\text{sector } OAC} = \\frac{60\\pi \\times 2^2}{360} = \\frac{2}{3}\\pi$,\\\\\nthus, the area of the shaded part equals $\\frac{1}{3}\\pi - (\\frac{2}{3}\\pi - \\sqrt{3}) = \\boxed{\\sqrt{3} - \\frac{1}{3}\\pi}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51322768.png"
  },
  {
    "solution": "\\boxed{150^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in triangle $AOB$ and triangle $COD$, we have $\\angle AOB = \\angle COD$, with $\\angle AOB = 90^\\circ$. When the two triangles are combined as shown in the figure and $\\angle BOD = 30^\\circ$, what is the measure of $\\angle AOC$?\n",
    "completion": "\\textbf{Solution:} \\\\\nSince $\\angle AOC = \\angle AOB + \\angle BOC = 90^\\circ + \\angle BOC$,\\\\\n$\\angle DOB = \\angle DOC - \\angle BOC = 90^\\circ - \\angle BOC$,\\\\\nTherefore, $\\angle AOC + \\angle DOB = 90^\\circ + \\angle BOC + 90^\\circ - \\angle BOC = 180^\\circ$,\\\\\nSince $\\angle BOD = 30^\\circ$,\\\\\nTherefore, $\\angle AOC = \\boxed{150^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/17008253_57.png"
  },
  {
    "solution": "\\boxed{80^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, OC is the bisector of $\\angle AOB$, OD bisects $\\angle AOC$, and $\\angle COD=20^\\circ$. What is the measure of $\\angle AOB$?\n",
    "completion": "\\textbf{Solution:} Given that $OD$ bisects $\\angle AOC$ and $\\angle COD = 20^\\circ$, we can deduce that $\\angle AOC = 2\\angle COD = 40^\\circ$, \\\\ \nand since $OC$ is the bisector of $\\angle AOB$, it follows that $\\angle AOB = 2\\angle AOC = 80^\\circ$. \\\\\nTherefore, $\\angle AOB = \\boxed{80^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52732849.png"
  },
  {
    "solution": "\\boxed{1+\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that in rectangle $ABCD$, $AB=2$. A point $E$ is chosen on $BC$, and $\\triangle ABE$ is folded upward along $AE$, causing point $B$ to coincide with point $F$ on $AD$. If quadrilateral $EFDC$ is similar to rectangle $ABCD$, what is the length of $AD$?\n",
    "completion": "\\textbf{Solution:} Given that in rectangle ABCD, AF is obtained by folding AB,\\\\\nthus ABEF is a square.\\\\\nAlso, since AB = 2, thus AF = AB = EF = 2.\\\\\nLet AD = x, then FD = x - 2.\\\\\nSince quadrilateral EFDC is similar to rectangle ABCD,\\\\\nthus $\\frac{EF}{FD} = \\frac{AD}{AB}$, that is $\\frac{2}{x-2} = \\frac{x}{2}$\\\\\nSolving this, we get $x_{1}=1+\\sqrt{5}$, $x_{2}=1−\\sqrt{5}$ (negative value is discarded).\\\\\nUpon verification, $x_{1}=1+\\sqrt{5}$ is the solution of the original equation.\\\\\nTherefore, AD $ =\\boxed{1+\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53065988.png"
  },
  {
    "solution": "\\boxed{\\frac{5}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, there is a rectangular sheet of paper ABCD with point E on the side AB. Fold $\\triangle BCE$ along the straight line CE, so that point B falls on the diagonal AC, denoted as point F. If $AB=4$ and $BC=3$, what is the length of $AE$?\n",
    "completion": "\\textbf{Solution:} As shown in Figure 1, in the rectangular paper ABCD, $\\because$AB＝4, BC＝3, thus by the Pythagorean theorem, we get AC＝5. From the fold, it follows: FC＝BC＝3, $\\angle$EFC＝$\\angle$B＝$90^\\circ$, BE＝FE. $\\therefore$ $AF=AC-FC=5-3=2$. Let AE＝x, then $BE=4-x=FE$. In $\\triangle AFE$, $2^2+(4-x)^2=x^2$, solving for x, we get: $x=\\frac{5}{2}$.\\\\\n$\\therefore$ $AE=\\boxed{\\frac{5}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52122243_160.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Could you please provide the text you need to be translated into English and formatted in LaTeX?\n",
    "completion": "Solution: Since $DE$ is the perpendicular bisector of $AB$, \\\\\n$\\therefore DA=DB$, \\\\\n$\\therefore \\angle DBA=\\angle A=30^\\circ$, \\\\\n$\\therefore \\angle CBD=\\angle ABC-\\angle ABD=75^\\circ-30^\\circ=\\boxed{45^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51545371_113.png"
  },
  {
    "solution": "\\boxed{24}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the quadrilateral OABC is a square. An equilateral $\\triangle$ABD is formed with AB as the side, where point D is precisely on the x-axis. The diagonal CA is extended to intersect the x-axis at point E. Given that the coordinates of point E are $\\left(4, 0\\right)$. What is the area of the square OABC?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, \\\\\n$\\because \\angle BDO=45^\\circ$, $\\angle BHD=90^\\circ$, \\\\\n$\\therefore \\angle HBD=45^\\circ$, \\\\\n$\\therefore \\triangle BHD$ is an isosceles right triangle, \\\\\n$\\because BH=2\\sqrt{3}$, \\\\\n$\\therefore BD=2\\sqrt{6}$, \\\\\n$\\therefore AB=BD=2\\sqrt{6}$, \\\\\n$\\therefore$ the area of the square OABC is: ${\\left(2\\sqrt{6}\\right)}^{2}=\\boxed{24}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53359725_139.png"
  },
  {
    "solution": "\\boxed{5+5\\sqrt{3}+\\frac{5\\pi}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the right-angled triangle $\\triangle ABC$ with $\\angle C=90^\\circ$, point D is on AB. The circle $O$ with diameter AD intersects BC at point E, and AC at point F, with AE bisecting $\\angle BAC$. BC is a tangent to circle $O$. Given that $\\angle EAB=30^\\circ$ and $OD=5$, find the perimeter of the shaded area in the diagram.\n",
    "completion": "\\textbf{Solution:} Given that $\\because \\angle EAB=30^\\circ$, \\\\\n$\\therefore \\angle EOD=60^\\circ$, \\\\\n$\\therefore \\angle OEB=90^\\circ$, \\\\\n$\\therefore \\angle B=30^\\circ$, \\\\\n$\\therefore OB=2OE=2OD=10$, \\\\\n$\\therefore BD=5$, \\\\\n$\\therefore BE= \\sqrt{OB^{2}-OE^{2}} = \\sqrt{10^{2}-5^{2}}=5\\sqrt{3}$, \\\\\n$\\therefore$ the length of arc DE is $= \\frac{60^\\circ \\pi \\times 5}{180} = \\frac{5\\pi }{3}$, \\\\\n$\\therefore C_{\\text{shaded}}= BD+BE+l_{\\overset{\\frown}{DE}} = 5+5\\sqrt{3}+\\frac{5\\pi}{3} = \\boxed{5+5\\sqrt{3}+\\frac{5\\pi}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51433589_112.png"
  },
  {
    "solution": "\\boxed{100^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, $\\triangle ABC$ is an equilateral triangle, where D is a point on side $AC$ that does not coincide with point A. Extend $BC$ to point E such that $CE=AD$, and extend $AC$ to F so that $CF=AC$. Connect $EF$ and $BD$. If $\\angle ABD=20^\\circ$, what are the degrees of $\\angle CFE$ and $\\angle CEF$?\n",
    "completion": "Solution: Since $\\triangle ABC$ is an equilateral triangle, we have $AB=AC$, and $\\angle A=\\angle ACB=60^\\circ$. Therefore, $\\angle ECF=60^\\circ$, which means $\\angle A=\\angle ECF$. Given that $CF=AC$, we find $AB=CF$. Since $CE=AD$, we deduce $\\triangle ABD \\cong \\triangle CFE$, thus $\\angle CFE=\\angle ABD=20^\\circ$. Consequently, $\\angle CEF=180^\\circ-\\angle CFE-\\angle ECF=\\boxed{100^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51458337_83.png"
  },
  {
    "solution": "\\boxed{\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in $\\square ABCD$, $\\angle ABC=60^\\circ$, points E and F are located on the extensions of AD and BA, respectively, with $CE \\parallel BD$ and $EF \\perp AB$, and $BC = 1$. What is the length of EF?\n",
    "completion": "\\textbf{Solution:}\n\nSince quadrilateral ABCD is a parallelogram, \\\\\n$\\therefore$AD = BC, and BC $\\parallel$ AD, \\\\\nSince CE $\\parallel$ BD, \\\\\n$\\therefore$ quadrilateral BCED is a parallelogram, \\\\\n$\\therefore$ DE = BC = AD = 1, hence D is the midpoint of AE, \\\\\n$\\therefore$ AE = 2, \\\\\nSince EF $\\perp$ AB, \\\\\n$\\therefore$ $\\angle$EFA = $90^\\circ$, \\\\\nSince AD $\\parallel$ BC, \\\\\n$\\therefore$ $\\angle$EAF = $\\angle$ABC = $60^\\circ$, $\\angle$AEF = $30^\\circ$, \\\\\n$\\therefore$ AF = $\\frac{1}{2}$AE = 1, \\\\\n$\\therefore$ EF = $\\sqrt{AE^{2} - AF^{2}} = \\sqrt{2^{2} - 1^{2}} = \\boxed{\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53359457.png"
  },
  {
    "solution": "\\boxed{4 - 2\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, ABCD is a square with a side length of 4. Point E is on the diagonal BD, and $ \\angle BAE=22.5^\\circ$, $EF\\perp AB$ with the foot of the perpendicular at F. What is the length of EF?\n",
    "completion": "\\textbf{Solution:} In the square $ABCD$, $\\angle ABD = \\angle ADB = 45^\\circ$,\\\\\n$\\because \\angle BAE = 22.5^\\circ$,\\\\\n$\\therefore \\angle DAE = 90^\\circ - \\angle BAE = 67.5^\\circ$.\\\\\nIn $\\triangle ADE$, $\\angle AED = 180^\\circ - 45^\\circ - 67.5^\\circ = 67.5^\\circ$,\\\\\n$\\therefore \\angle DAE = \\angle ADE$.\\\\\n$\\therefore AD = DE = 4$.\\\\\n$\\because$ The side length of the square is 4,\\\\\n$\\therefore BD = 4\\sqrt{2}$.\\\\\n$\\therefore BE = BD - DE = 4\\sqrt{2} - 4$.\\\\\n$\\because EF \\perp AB, \\angle ABD = 45^\\circ$,\\\\\n$\\therefore \\triangle BEF$ is an isosceles right triangle.\\\\\n$\\therefore EF = \\frac{\\sqrt{2}}{2}BE = \\frac{\\sqrt{2}}{2} \\times (4\\sqrt{2} - 4) = \\boxed{4 - 2\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53333373.png"
  },
  {
    "solution": "\\boxed{\\frac{8\\sqrt{3}}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line MN is an axis of symmetry of the rectangle ABCD, with point E located on side AD. When $\\triangle BAE$ is folded along BE, the corresponding point of A, denoted as F, falls on the line MN. If $AB=4$, what is the length of BE?\n",
    "completion": "\\textbf{Solution:} Given that MN is an axis of symmetry of rectangle ABCD, \\\\\nit follows that $AM=BM=\\frac{1}{2}AB=2$, and $MN\\parallel AD$, \\\\\nhence, $\\angle AEF+\\angle MFE=180^\\circ$, $\\angle BMF=\\angle BAE=90^\\circ$, \\\\\nby the properties of folding, we have $BF=AB=4$, $\\angle AEB=\\angle FEB$, $\\angle BFE=\\angle A=90^\\circ$, \\\\\ntherefore, $\\sin\\angle BFM=\\frac{AB}{BF}=\\frac{1}{2}$, \\\\\nhence, $\\angle BFM=30^\\circ$, \\\\\nthus, $\\angle MFE=60^\\circ$, \\\\\nand so, $\\angle AEF=120^\\circ$, \\\\\ntherefore, $\\angle AEB=60^\\circ$, \\\\\nthus, $BE=\\frac{AB}{\\sin\\angle AEB}=\\boxed{\\frac{8\\sqrt{3}}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51910248.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC=6$, and $AD$ is the median to side $BC$. If $\\angle BAC=120^\\circ$, what is the length of $AD$?\n",
    "completion": "\\textbf{Solution:} Given that $AB=AC$ and $AD$ is the median of side $BC$,\\\\\nthus $AD\\perp BC$ and $\\angle BAD=\\angle CAD$,\\\\\nsince $\\angle BAC=120^\\circ$,\\\\\ntherefore, $\\angle BAD=60^\\circ$,\\\\\nIn right triangle $ABD$, $\\angle B=90^\\circ-60^\\circ=30^\\circ$,\\\\\nthus, $AD=\\frac{1}{2}AB=\\boxed{3}$",
    "image_path": "dataset/mm_math/images/MM_Math/52950952.png"
  },
  {
    "solution": "\\boxed{\\frac{2}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB = 90^\\circ$, $CB = CA$, $CE \\perp AB$ at point $E$, point $F$ is a point on $CE$, extend $AF$ to intersect $BC$ at point $D$, $CG \\perp AD$ at point $G$, connect $EG$. As shown in Figure 2, if $GC = 2$ and $GE = 2\\sqrt{2}$, find the length of $GD$.\n",
    "completion": "\\textbf{Solution:} Since $CB=CA$, and $\\angle ACB=90^\\circ$,\\\\\nit follows that $\\angle BAC=45^\\circ$,\\\\\nSince $CE\\perp AB$, and $CG\\perp AD$,\\\\\nit implies that $\\angle AGC=\\angle AEC=90^\\circ$, and $\\angle ACE=45^\\circ$,\\\\\nTherefore, points A, C, G, E are concyclic,\\\\\nhence $\\angle AGE=\\angle ACE=45^\\circ$,\\\\\nAs shown in Figure 2, draw EH$\\perp$AD at point H through point E,\\\\\nthus $\\triangle EGH$ is an isosceles right triangle,\\\\\n$EH=GH=GE\\cdot \\sin 45^\\circ=2\\sqrt{2}\\times \\frac{\\sqrt{2}}{2}=2$,\\\\\nSince $CG=2$,\\\\\nit follows that $CG=EH$,\\\\\nSince $\\angle CGF=\\angle EHF=90^\\circ$, and $\\angle CFG=\\angle EFH$,\\\\\ntherefore $\\triangle CFG\\cong \\triangle EFH$ (by AAS),\\\\\nthus $FG=FH=1$, and $CF=EF$,\\\\\nIn $\\triangle CFG$, $CF=\\sqrt{CG^2+FG^2}=\\sqrt{2^2+1^2}=\\sqrt{5}$,\\\\\nthus $CE=2CF=2\\sqrt{5}$,\\\\\nthus $AC=\\frac{CE}{\\sin\\angle CAE}=\\frac{2\\sqrt{5}}{\\sin 45^\\circ}=2\\sqrt{10}$,\\\\\nthus $AG=\\sqrt{AC^2-CG^2}=\\sqrt{(2\\sqrt{10})^2-2^2}=6$,\\\\\nSince $\\angle CGD=\\angle AGC=90^\\circ$,\\\\\nit follows that $\\angle CAG+\\angle ACG=90^\\circ$,\\\\\nSince $\\angle ACG+\\angle DCG=90^\\circ$,\\\\\nit implies that $\\angle CAG=\\angle DCG$,\\\\\nTherefore, $\\triangle CAG\\sim \\triangle DCG$,\\\\\nthus $\\frac{DG}{CG}=\\frac{CG}{AG}$,\\\\\nthus $DG=\\frac{CG^2}{AG}=\\frac{2^2}{6}=\\boxed{\\frac{2}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51212645_95.png"
  },
  {
    "solution": "\\boxed{\\frac{a}{b}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, $AB=c$, $AC=b$, $BC=a$, what is the value of $\\tan A$?\n",
    "completion": "\\textbf{Solution:} Since in the right triangle $ABC$, $\\angle C=90^\\circ$, $AC=b$, $BC=a$, \\\\\ntherefore $\\tan A = \\frac{a}{b}$, thus $\\tan A = \\boxed{\\frac{a}{b}}$",
    "image_path": "dataset/mm_math/images/MM_Math/55574667.png"
  },
  {
    "solution": "\\boxed{y=\\frac{6}{5}x}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in $\\triangle ABC$, $\\angle C=90^\\circ$, $AC=6$, $BC=8$, $D$ is the midpoint of side $BC$, $E$ is a moving point on side $AB$, and $\\angle DEF=90^\\circ$, line $EF$ intersects the ray $BC$ at point $F$. Let $BE=x$, the area of $\\triangle BED$ is $y$. Determine the functional relationship of $y$ in terms of $x$, and state the domain of the independent variable $x$.\n",
    "completion": "\\textbf{Solution:} As shown in Figure 1, draw $EH\\perp CB$ at $H$ through point $E$.\\\\\n$\\therefore EH\\parallel AC$\\\\\nIn $\\triangle ABC$, $\\angle C=90^\\circ$, $AC=6$, $BC=8$,\\\\\n$\\therefore AB=\\sqrt{AC^2+BC^2}=10$,\\\\\n$\\because E$ can be a moving point on $AB$ that coincides with $A$ or $B$,\\\\\nwhen $x=5$, $E$ is the midpoint of $AB$, $ED\\parallel AC$, $\\angle DEF=90^\\circ$, at this time $EF\\parallel BC$, $EF$ has no intersection with $BC$,\\\\\nLet the distance from $C$ to $AB$ be $h$,\\\\\nthen $h=\\frac{AC\\times CB}{AB}=\\frac{24}{5}$\\\\\nWhen $ED\\perp AB$, $DE=\\frac{1}{2}h=\\frac{12}{5}$,\\\\\nat this time $BE=\\sqrt{DB^2-DE^2}=\\frac{16}{5}$, according to the figure, when $\\frac{16}{5}<BE<5$, $EF$ has no intersection with $BC$,\\\\\n$\\therefore 0\\le x\\le \\frac{16}{5}$ or $5<x\\le 10$\\\\\n$\\because EH\\parallel AC$\\\\\n$\\therefore \\triangle BEH\\sim \\triangle BAC$\\\\\n$\\therefore \\frac{EH}{AC}=\\frac{BE}{AB}$\\\\\n$EH=\\frac{AC\\cdot BE}{AB}=\\frac{3}{5}x$\\\\\n$\\therefore {S}_{\\triangle DEB}=\\frac{1}{2}BD\\cdot EH=\\frac{1}{2}\\times 4\\times \\frac{3}{5}x=\\frac{6}{5}x$,\\\\\n$\\therefore \\boxed{y=\\frac{6}{5}x}(0<x\\le \\frac{16}{5}$ or $5<x\\le 10)$",
    "image_path": "dataset/mm_math/images/MM_Math/53078900_115.png"
  },
  {
    "solution": "\\boxed{y = -x - 1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the line $y=kx+b$ ($k \\neq 0$) intersects with the hyperbola $y=\\frac{m}{x}$ ($m \\neq 0$) at points $A(2, -3)$ and $B(n, 2)$. What are the expressions for the functions of the line and the hyperbola respectively?\n",
    "completion": "\\textbf{Solution:} Since the hyperbola $y = \\frac{m}{x}$ (where $m \\neq 0$) passes through point $A(2, -3)$,\nit follows that $m = -6$.\nTherefore, the equation of the hyperbola is $y = -\\frac{6}{x}$.\n\nGiven that point $B(n, 2)$ lies on the hyperbola $y = -\\frac{6}{x}$,\nthe coordinates of point $B$ are $(-3, 2)$.\n\nSince the line $y = kx + b$ passes through points $A(2, -3)$ and $B(-3, 2)$,\nwe have the system of equations $\\begin{cases} 2k + b = -3 \\\\ -3k + b = 2 \\end{cases}$.\n\nSolving this system yields $\\begin{cases} k = -1 \\\\ b = -1 \\end{cases}$.\n\nTherefore, the equation of the line is $\\boxed{y = -x - 1}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52623645_227.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y_1=x+3$ intersects with the $x$-axis and the $y$-axis at points $A$ and $C$ respectively, and the line $y_2=-x+3$ intersects with the $x$-axis and the $y$-axis at points $B$ and $C$ respectively. If point $P(m, 2)$ is a point inside (including on the edges of) $\\triangle ABC$, then what is the difference between the maximum and minimum values of $m$?\n",
    "completion": "\\textbf{Solution:} Let point P$(m,2)$ be a point inside (including on the edge of) $\\triangle ABC$, \\\\\n$\\therefore$ point P lies on the line $y=2$, as shown in the figure, \\\\\nWhen P is the intersection of line $y=2$ and line $y_{2}$, $m$ takes its maximum value, \\\\\nWhen P is the intersection of line $y=2$ and line $y_{1}$, $m$ takes its minimum value, \\\\\n$\\because$ in $y_{1}=-x+3$, letting $y=2$, we get $x=1$, \\\\\nin $y_{2}=x+3$, letting $y=2$, we get $x=-1$, \\\\\n$\\therefore$ the maximum value of $m$ is $1$, and the minimum value of $m$ is $-1$. \\\\\nThus, the difference between the maximum and minimum values of $m$ is: $1-(-1)=\\boxed{2}$",
    "image_path": "dataset/mm_math/images/MM_Math/54409220.png"
  },
  {
    "solution": "\\boxed{22.5^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the square $ABCD$, $E$ is a point on the diagonal $BD$, and $BE=BC$, then what is the value of $\\angle DCE$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a square, \\\\\n$\\therefore \\angle CBE=45^\\circ$, $\\angle BCD=90^\\circ$, \\\\\nSince BE=BC, \\\\\n$\\therefore \\angle BCE= \\frac{1}{2} \\times (180^\\circ-45^\\circ)=67.5^\\circ$, \\\\\nTherefore, $\\angle DCE=\\angle BCD-\\angle BCE=90^\\circ-67.5^\\circ=\\boxed{22.5^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51863567.png"
  },
  {
    "solution": "\\boxed{120^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ is an arbitrary triangle. Equilateral triangles $ABD$ and $ACE$ are constructed externally on sides $AB$ and $AC$, respectively. Lines $CD$ and $BE$ are drawn and intersect at point $P$. $\\triangle DAC \\cong \\triangle BAE$; what is the measure of $\\angle BPC$?\n",
    "completion": "\\textbf{Solution:} From (1) we have $\\angle ADC=\\angle ABE$, \\\\\n$\\therefore \\angle BPC=\\angle PBD+\\angle PDB=\\angle ABD+\\angle ABE+\\angle PDB=\\angle ABD+\\angle ADC+\\angle PDB=\\angle ABD+\\angle ADB$, \\\\\n$\\because \\angle ABD=\\angle ADB=60^\\circ$, \\\\\n$\\therefore \\angle BPC=\\boxed{120^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52861540_93.png"
  },
  {
    "solution": "\\boxed{\\frac{4\\pi }{9}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that two chords $AB$ and $CD$ of the circle $\\odot O$ intersect at the point $M$ with $AB=CD$. Let the radius of $\\odot O$ be $r$. $DM=BM$; if $\\angle DMB=100^\\circ$, $r=1$, what is the length of the arc $\\overset{\\frown}{BC}$?\n",
    "completion": "Solution: Connect $OB$ and $OC$,\\\\\nsince $DM=BM$, $\\angle DMB=100^\\circ$,\\\\\nthus $\\angle MDB=\\angle MBD=40^\\circ$,\\\\\ntherefore $\\angle BOC=80^\\circ$,\\\\\nsince $r=1$,\\\\\nhence the length of arc $\\overset{\\frown}{BC} = \\frac{n\\pi r}{180} = \\frac{80\\pi }{180} = \\boxed{\\frac{4\\pi }{9}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52871756_106.png"
  },
  {
    "solution": "\\boxed{\\frac{15}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB=10$, point C is the midpoint of segment AB, and point D is the midpoint of segment CB. Find the length of AD.\n",
    "completion": "\\textbf{Solution:} Since point C is the midpoint of the line segment AB, \\\\\nit follows that $AC=BC=\\frac{1}{2}AB$, \\\\\nsince $AB=10$, \\\\\nit follows that $AC=BC=\\frac{1}{2}\\times 10=5$. \\\\\nSince point D is the midpoint of the line segment BC, \\\\\nit follows that $CD=\\frac{1}{2}BC$, \\\\\ntherefore $CD=\\frac{1}{2}\\times 5=\\frac{5}{2}$. \\\\\nThus, $AD=AC+CD=5+\\frac{5}{2}=\\boxed{\\frac{15}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51212476.png"
  },
  {
    "solution": "\\boxed{x<-2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system xOy, the lines $l_{1}$ and $l_{2}$ are respectively the graphs of the functions $y=k_{1}x+b_{1}$ and $y=k_{2}x+b_{2}$. Then, what is the estimated solution set for the inequality $k_{1}x+b_{1}>k_{2}x+b_{2}$ with respect to $x$?\n",
    "completion": "\\textbf{Solution:} When $x<-2$, $k_1x+b_1>k_2x+b_2$,\\\\\nthus, the solution set of the inequality $k_1x+b_1>k_2x+b_2$ is $x<-2$.\\\\\nTherefore, the answer is: $\\boxed{x<-2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53729287.png"
  },
  {
    "solution": "\\boxed{24}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in square $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$. $E$ is the midpoint of $OB$. Connect $AE$ and extend it to intersect $BC$ at point $F$. If the area of $\\triangle BEF$ is $1$, what is the area of square $ABCD$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral ABCD is a square,\\\\\nit follows that OB=OD and AD$\\parallel$BC,\\\\\nthus, $\\triangle BEF\\sim \\triangle DEA$,\\\\\nwhich means $\\frac{BE}{ED}=\\frac{EF}{AE}$,\\\\\nsince E is the midpoint of OB,\\\\\nit implies $\\frac{BE}{ED}=\\frac{EF}{AE}=\\frac{1}{3}$,\\\\\nhence, the area ratio $S_{\\triangle BEF}: S_{\\triangle AEB}=EF:AE=\\frac{1}{3}$,\\\\\ngiven that the area of $\\triangle BEF$ is 1,\\\\\nthus, the area of $\\triangle AEB$ is 3,\\\\\nconsidering $\\frac{BE}{ED}=\\frac{1}{3}$,\\\\\nthe area ratio $S_{\\triangle AEB}: S_{\\triangle AED}=\\frac{1}{3}$,\\\\\nas a result, the area of $\\triangle AED$ is 9,\\\\\nthus, the area of $\\triangle ABD$=9+3=12,\\\\\ntherefore, the area of square ABCD=$12\\times2=\\boxed{24}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51284463.png"
  },
  {
    "solution": "\\boxed{120^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, lines AB and CD intersect at O, OA bisects $\\angle COE$, and $\\angle COE \\colon \\angle BOE = 2 \\colon 5$, then what is the degree measure of $\\angle EOD$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle COE : \\angle BOE = 2 : 5$,\\\\\nlet $\\angle COE = 2x$, then $\\angle BOE = 5x$,\\\\\nSince OA bisects $\\angle COE$,\\\\\nthus $\\angle AOE = \\angle COA = \\angle BOD = x$,\\\\\nthus $\\angle DOE = 4x$,\\\\\nSince $\\angle COE + \\angle DOE = 180^\\circ$,\\\\\nthus $2x + 4x = 180^\\circ$,\\\\\nthus $x = 30^\\circ$,\\\\\nthus $\\angle DOE = 4x = \\boxed{120^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/54792296.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given the graph as shown, the graph of the linear function $y=ax+b$ passes through points $(1,2)$ and $(-1,6)$. Find: the area enclosed by the graph of the linear function and the two coordinate axes.\n",
    "completion": "\\textbf{Solution:} The graph of the linear function intersects the y-axis and x-axis at points A and B, respectively.\n\nGiven $y=-2x+4$, we have\n\nCoordinates of point A $(0,4)$, coordinates of point B $(2,0)$,\n\nwhich means $OA=4$, $OB=2$.\n\nTherefore, the area of $\\triangle AOB$ is $\\frac{1}{2}OA \\cdot OB=\\frac{1}{2} \\times 4 \\times 2=\\boxed{4}$.\n\nThus, the area enclosed by the graph of the linear function and the two coordinate axes is \\boxed{4}.",
    "image_path": "dataset/mm_math/images/MM_Math/52344480_214.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $l_{1} \\parallel l_{2}$, $l_{2} \\parallel l_{3}$, line $a$ and $b$ intersect $l_{1}$, $l_{2}$, $l_{3}$ at points A, B, C and points D, E, F respectively. If $AB = 3$, $DE = 2$, $BC = 6$, what is the value of $EF$?\n",
    "completion": "\\textbf{Solution}: Since $l_{1} \\parallel l_{2} \\parallel l_{3}$,\\\\\nit follows that $\\frac{AB}{BC}=\\frac{DE}{EF}$,\\\\\nthus $\\frac{3}{6}=\\frac{2}{EF}$\\\\\nSolving yields: EF=\\boxed{4}",
    "image_path": "dataset/mm_math/images/MM_Math/52992231.png"
  },
  {
    "solution": "\\boxed{45}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, it is known that point O is on line AB, and rays OD and OC are on the upper and lower sides of line AB, respectively, with $\\angle COD=80^\\circ$. OE is always the bisector of $\\angle AOD$. As illustrated in Figure 3, under the condition of (2), ray OP starts rotating counterclockwise from OB around point O at a speed of $4^\\circ$ per second, and ray OQ starts rotating clockwise from OE around point O at a speed of $2^\\circ$ per second. Rays OP and OQ start rotating simultaneously, and the rotation time is denoted as t seconds $\\left(0\\leq t\\leq 45\\right)$. When $\\angle DOQ$ and $\\angle EOP$ are complementary, find the value of the rotation time t.\n",
    "completion": "\\textbf{Solution:} When $n=64$, we have $\\angle EOD=\\angle AOE=32^\\circ$ and $\\angle EOB=180^\\circ-32^\\circ=148^\\circ$.\n\n(1) When $0\\leq t \\leq 16$\n\nSince $\\angle QOD=32^\\circ-2t^\\circ$ and $\\angle EOP=148^\\circ-4t^\\circ$,\n\nit follows that $32-2t+148-4t=90$.\n\nTherefore, $t=\\boxed{15}$.\n\n(2) For $16 < t < 37$\n\nSince $\\angle QOD=2t^\\circ-32^\\circ$ and $\\angle EOP=148^\\circ-4t^\\circ$,\n\nit follows that $2t-32+148-4t=90$.\n\nTherefore, $t=13$ (discarded).\n\n(3) For $37 \\leq t \\leq 45$\n\n$\\angle QOD=2t^\\circ-32^\\circ$ and $\\angle EOP=4t^\\circ-148^\\circ$,\n\nit follows that $2t-32+4t-148=90$.\n\nTherefore, $t=\\boxed{45}$.\n\nIn summary, $t=\\boxed{15}$ seconds or $\\boxed{45}$ seconds.",
    "image_path": "dataset/mm_math/images/MM_Math/53414133_153.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle A=90^\\circ$, BD bisects $\\angle ABC$ and intersects AC at point D, AB=4, BC=12, and AD=3. If point P moves on BC, when DP is minimized, what is the area of $\\triangle CDP$?\n",
    "completion": "\\textbf{Solution}: Since $\\angle A=90^\\circ$, \\\\\nit follows that BD=$\\sqrt{AB^{2}+AD^{2}}=\\sqrt{4^{2}+3^{2}}=5$, \\\\\nWhen DP is at its minimum, DP=3, and DP$\\perp$BC, \\\\\nthus $\\angle$DPB=$\\angle$DPC=$90^\\circ$, \\\\\nso PB=$\\sqrt{BD^{2}-DP^{2}}=\\sqrt{5^{2}-3^{2}}=4$, \\\\\nthus CP=BC-PB=12-4=8, \\\\\ntherefore, the area of $\\triangle CDP$=$\\frac{1}{2}$CP$\\times$DP=$\\frac{1}{2}\\times8\\times3=\\boxed{12}$, \\\\\ni.e., when DP is at its minimum, the area of $\\triangle CDP$ is \\boxed{12}.",
    "image_path": "dataset/mm_math/images/MM_Math/52095800_114.png"
  },
  {
    "solution": "\\boxed{\\frac{9}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, point $D$ is on line $AB$, $DE\\parallel BC$, $BE\\perp AB$. $\\triangle DEB\\sim \\triangle BAC$; If $BE=2$, $AC=3$, and the area of $\\triangle BDE$ is 1, find the area of $\\triangle ABC$.\n",
    "completion": "\\textbf{Solution:} From (1), we can obtain $ \\triangle DEB\\sim \\triangle BAC$,\\\\\n$\\therefore \\frac{S_{\\triangle BDE}}{S_{\\triangle ABC}}=\\left(\\frac{BE}{AC}\\right)^2=\\left(\\frac{2}{3}\\right)^2=\\frac{4}{9}$,\\\\\n$\\because S_{\\triangle BDE}=1$,\\\\\n$\\therefore \\frac{1}{S_{\\triangle ABC}}=\\frac{4}{9}$, solving this we get: $S_{\\triangle ABC}=\\boxed{\\frac{9}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53102192_69.png"
  },
  {
    "solution": "\\boxed{\\frac{14}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, $AC=4$, $BC=3$. $\\triangle ABC$ is rotated $90^\\circ$ counterclockwise around point $C$ to obtain $\\triangle DEC$. $P$ is a moving point on line segment $DE$. Taking point $P$ as the center and the length of $PD$ as the radius, a circle $\\odot P$ is drawn. When $\\odot P$ is tangent to the sides of $\\triangle ABC$, what is the radius of $\\odot P$?\n",
    "completion": "\\textbf{Solution:}\n\n(1) When $\\odot P$ is tangent to AC at point F, as shown in the figure, connect PF, then PF $\\perp$ AC,\\\\\nLet DP=PF=x,\\\\\n$\\because$ $\\angle ACB=90^\\circ$, AC=4, BC=3,\\\\\n$\\therefore$ AB=$\\sqrt{AC^2+BC^2}=5$,\\\\\n$\\because$ $\\triangle ABC$ is rotated 90$^\\circ$ counterclockwise around point C to get $\\triangle DEC$,\\\\\n$\\therefore$ $\\triangle DCE \\cong \\triangle ACB$,\\\\\n$\\therefore$ DE=AB=5, PE=DE−DP=5−x, CE=CB=3, DC=AC=4,\\\\\n$\\because$ $\\angle ACB=90^\\circ$, PF$\\perp$AC,\\\\\n$\\therefore$ PF$\\parallel$DC,\\\\\n$\\therefore$ $\\triangle EPF \\sim \\triangle EDC$,\\\\\n$\\therefore$ $\\frac{EP}{ED}=\\frac{PF}{DC}$, that is, $\\frac{5−x}{5}=\\frac{x}{4}$, solving for x yields $x=\\boxed{\\frac{20}{9}}$, i.e., the radius is $\\frac{20}{9}$;\\\\\n(2) When $\\odot P$ is tangent to AB at point F, as shown in the figure, BF$\\perp$AB,\\\\\n$\\because$ $\\angle A=\\angle BDF$, $\\angle ABC=\\angle DBF$,\\\\\n$\\therefore$ $\\triangle ABC \\sim \\triangle DBF$,\\\\\n$\\therefore$ $\\frac{AB}{DB}=\\frac{AC}{DF}$, that is, $\\frac{5}{7}=\\frac{4}{DF}$, solving for DF yields DF=$\\frac{28}{5}$,\\\\\n$\\therefore$ PD=PF=$\\frac{1}{2}$DF=$\\boxed{\\frac{14}{5}}$, i.e., the radius is $\\frac{14}{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53136696.png"
  },
  {
    "solution": "\\boxed{y=6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the function $y= \\frac{k}{x}$ ($x>0$) passes through the midpoint of one side of the rectangle $OBCD$ and also passes through the vertex $P$ of the rectangle $OAPE$. If the area of the shaded part is $6$, what is the value of $k$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, let $M$ be the midpoint of $CD$, and draw $EF \\perp x$-axis at $N$, \\\\\n$\\therefore$ Rectangle $DONM$ is equal to half of the area of rectangle $OBCD$, \\\\\n$\\because$ Points $M$ and $P$ both lie on the graph of the inverse proportion function, \\\\\n$\\therefore$ The area of rectangle $DOMN$ is equal to the area of rectangle $EOAP$, \\\\\n$\\therefore$ The area of rectangle $EOAP$ is equal to half the area of rectangle $OBCD$, \\\\\n$\\therefore$ The area of rectangle $EOAP$ = area of the shaded part = 6, \\\\\n$\\therefore k=x\\boxed{y=6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52082449.png"
  },
  {
    "solution": "\\boxed{\\frac{32}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $D$, $E$, and $F$ are points on $AB$, $BC$ respectively, and $DE\\parallel AC$, $AE\\parallel DF$. Given that $BD: AD=3:2$ and $BF=6$, find the lengths of $EF$ and $FC$.\n",
    "completion": "\\textbf{Solution:} Since $AE\\parallel DF$, \\\\\nthen $\\frac{BD}{AD}=\\frac{BF}{EF}$, that is $\\frac{6}{EF}=\\frac{3}{2}$, \\\\\ntherefore $EF=\\boxed{4}$, \\\\\nthus $BE=BF+EF=6+4=10$, \\\\\nsince $DE\\parallel AC$, \\\\\nthen $\\frac{BD}{AD}=\\frac{BE}{CE}$, that is $\\frac{10}{CE}=\\frac{3}{2}$, \\\\\ntherefore $CE=\\frac{20}{3}$, \\\\\nthus $CF=CE+EF=\\boxed{\\frac{32}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52602147.png"
  },
  {
    "solution": "\\boxed{y = 2x + 10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $l_{1} \\colon y_{1} = -2x + 6$ intersects the x-axis and the y-axis at points A and B, respectively. The line $l_{2}$ passes through the point C $(-5, 0)$ and intersects the line $l_{1}$ at the point D $(a, 8)$, and intersects the y-axis at the point E. Find the analytical expression of the line $l_{2}$?\n",
    "completion": "\\textbf{Solution:} Given that line $l_{1}$ passes through point $D(-1,8)$,\n\n\\[\n\\therefore 8 = -2a + 6,\n\\]\n\\[\n\\therefore a = -1,\n\\]\n\\[\n\\therefore D(-1,8),\n\\]\nGiven that line $l_{2}$ passes through points $C(-5,0)$ and $D(-1,8)$,\n\n\\[\n\\therefore \\begin{cases}\n-5k + b = 0 \\\\\n-k + b = 8\n\\end{cases},\n\\]\nsolving this, we get \n\\[\n\\begin{cases}\nk = 2 \\\\\nb = 10\n\\end{cases},\n\\]\n\\[\n\\therefore \\text{the equation of line}\\ l_{2}\\ \\text{is}\\ \\boxed{y = 2x + 10};\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/52381542_160.png"
  },
  {
    "solution": "\\boxed{7}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $P_{1}\\colon y=x^{2}+2x+m$ is translated to obtain parabola $P_{2}\\colon y=x^{2}-5x+n$. The two parabolas intersect the $y$-axis at points $C$ and $D$, respectively. The intersection point $E$ of parabolas $P_{1}$ and $P_{2}$ has an abscissa of $1$. A line parallel to the $x$-axis passing through point $E$ intersects parabolas $P_{1}$ and $P_{2}$ at points $A$ and $B$, respectively. What is the length of segments $AB$ and $CD$?\n",
    "completion": "\\textbf{Solution:} Given the information, the parabola $P_2\\colon y=x^2-5x+n$ has an axis of symmetry represented by the line $x=\\frac{5}{2}$,\\\\\n$\\because AB\\parallel x$ axis,\\\\\n$\\therefore$ point B is symmetric to point E with respect to the axis of symmetry $x=\\frac{5}{2}$,\\\\\n$\\therefore$ the x-coordinate of point B is 4,\\\\\n$\\therefore AB=4-(-3)=\\boxed{7}$;\\\\\n$\\because$ point E is the intersection point of parabola $P_1$ and parabola $P_2$,\\\\\n$1^2+2\\times 1+m=1^2-5\\times 1+n$,\\\\\n$\\therefore m=n-7$,\\\\\n$\\therefore n-m=7$,\\\\\nLet $x=0$,\\\\\nthen $C(0, m), D(0, n)$,\\\\\n$\\therefore CD=n-m=\\boxed{7}$;\\\\\nThus, the lengths of segments AB and CD are both 7.",
    "image_path": "dataset/mm_math/images/MM_Math/52731763_80.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in rhombus ABCD, DE $\\perp$ AB, $\\cos A = \\frac{3}{5}$, and BE = 2, then what is the value of $\\tan \\angle DBE$?\n",
    "completion": "\\textbf{Solution:} Let the side length of rhombus $ABCD$ be $t$.\\\\\n$\\because BE = 2$,\\\\\n$\\therefore AE = t - 2$.\\\\\n$\\therefore \\cos A = \\frac{3}{5} = \\frac{AE}{AD} = \\frac{AB - BE}{AD}$,\\\\\n$\\therefore \\frac{3}{5} = \\frac{t - 2}{t}$,\\\\\n$\\therefore t = 5$.\\\\\n$\\therefore AE = 5 - 2 = 3$.\\\\\n$\\therefore DE = \\sqrt{AD^2 - AE^2} = \\sqrt{5^2 - 3^2} = 4$.\\\\\n$\\therefore \\tan \\angle DBE = \\frac{DE}{BE} = \\frac{4}{2} = \\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50890998.png"
  },
  {
    "solution": "\\boxed{\\frac{5}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the right-angled triangle $ABC$ with $\\angle A=90^\\circ$, $AB=3$, and $AC=4$, triangle $ABC$ is folded along $BD$ such that point $A$ precisely falls on $BC$. What is the length of $CD$?\n",
    "completion": "\\textbf{Solution:} Let $CD=x$, then $AD=A^{\\prime}D=4-x$. In the right-angled triangle $ABC$, $BC= \\sqrt{AB^{2}+AC^{2}} =5$. Hence $A^{\\prime}C=BC-AB=BC-A^{\\prime}B=5-3=2$. In the right-angled triangle $A^{\\prime}DC$: $AD^{2}+AC^{2}=CD^{2}$. That is, $(4-x)^{2}+2^{2}=x^{2}$. Solving this, we find: $x=\\boxed{\\frac{5}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51419668.png"
  },
  {
    "solution": "\\boxed{y=\\frac{12}{x}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y=kx+2$ intersects with the graph of the hyperbolic function $y=\\frac{m}{x}$ ($x>0$) at point P, which lies in the first quadrant. $PA$ is perpendicular to the $x$-axis at point A. The graph of the linear function intersects the $x$-axis and $y$-axis at points C and D, respectively, and the area of $\\triangle COD$ is 1, and $\\frac{CO}{OA}=\\frac{1}{2}$. Find the analytical expressions of the linear function and the hyperbolic function?\n",
    "completion": "\\textbf{Solution:} Given $PA\\perp CA$, $DO\\perp CO$,\\\\\n$\\therefore \\angle PAC=\\angle COD=90^\\circ$\\\\\nSince $\\angle DCO=\\angle DCO$,\\\\\n$\\therefore Rt\\triangle PAC\\sim Rt\\triangle DOC$\\\\\nGiven $\\frac{CO}{OA}=\\frac{1}{2}$, and $OD=2$,\\\\\n$\\therefore \\frac{OD}{PA}=\\frac{CO}{CA}=\\frac{1}{3}$\\\\\nWe find $PA=6$\\\\\nGiven ${S}_{\\triangle COD}=1$, we have: $\\frac{1}{2}OC\\cdot OD=1$,\\\\\nWe find $OC=1$\\\\\n$\\therefore OA=2$, $P(2, 6)$\\\\\nSubstitute $P(2, 6)$ into $y=kx+2$ and $y=\\frac{m}{x}$\\\\\nThus, the linear function is $y=2x+2$, and the inverse proportion function is $\\boxed{y=\\frac{12}{x}}\\left(x>0\\right)$",
    "image_path": "dataset/mm_math/images/MM_Math/53043682_122.png"
  },
  {
    "solution": "\\boxed{P\\left(\\frac{21}{2}, 6\\right), Q\\left(\\frac{21}{2}, \\frac{5}{3}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, in the Cartesian coordinate system, point O is the origin, and quadrilateral $ABCO$ is a rhombus. The coordinate of point A is $(-5, 12)$, and point C is on the positive half-axis of the x-axis. The line AC intersects the y-axis at point M, and the side AB intersects the y-axis at point H. As shown in Figure 2, a moving point P starts from point A, moving along the polyline $A-B-C$ towards the endpoint C. A line $PQ\\parallel y$-axis is drawn through point P intersecting AC at point Q, let the x-coordinate of point P be $a$, and the length of the segment PQ be $l$.\n",
    "completion": "\\textbf{Solution}: Since $PQ\\parallel y$-axis and the x-coordinate of point P is a,\\\\\nit follows that $Q(a, -\\frac{2}{3}a+\\frac{26}{3})$,\\\\\n(1) When $-5\\le a\\le 8$, point P is on AB, P$(a, 12)$,\\\\\nthus $l=12-\\left(-\\frac{2}{3}a+\\frac{26}{3}\\right)=\\frac{2}{3}a+\\frac{10}{3}$;\\\\\nWhen $8<a\\le 13$, point P is on BC,\\\\\nsince the coordinates of point B are $(8, 12)$ and the coordinates of point C are $(13, 0)$,\\\\\nlet the equation of line BC be: $y=mx+n$ (where $m\\neq 0$),\\\\\nthen $\\begin{cases}8m+n=12\\\\ 13m+n=0\\end{cases}$, solving this yields: $\\begin{cases}m=-\\frac{12}{5}\\\\ n=\\frac{156}{5}\\end{cases}$,\\\\\nthus the equation of line BC is: $y=-\\frac{12}{5}x+\\frac{156}{5}$,\\\\\nthus at this time P$(a, -\\frac{12}{5}a+\\frac{156}{5})$,\\\\\nthus $l=-\\frac{12}{5}a+\\frac{156}{5}-\\left(-\\frac{2}{3}a+\\frac{26}{3}\\right)=-\\frac{26}{15}a+\\frac{338}{15}$,\\\\\nsummarizing, $l=\\begin{cases}\\frac{2}{3}a+\\frac{10}{3} & (-5\\le a\\le 8)\\\\ -\\frac{26}{15}a+\\frac{338}{15} & (8<a\\le 13)\\end{cases}$;\\\\\n(2) Possible;\\\\\nIn the equation $y=-\\frac{2}{3}x+\\frac{26}{3}$, let $x=0$, yielding $y=\\frac{26}{3}$,\\\\\nthus $OM=\\frac{26}{3}$,\\\\\nthus $MN=\\frac{1}{2}OM=\\frac{13}{3}$,\\\\\nsince $PQ\\parallel MN$,\\\\\nthus when $PQ=MN$, quadrilateral PQNM is a parallelogram,\\\\\nwhen $-5\\le a\\le 8$, we have $\\frac{2}{3}a+\\frac{10}{3}=\\frac{13}{3}$,\\\\\nsolving yields: $a=\\frac{3}{2}$,\\\\\nthus $P\\left(\\frac{3}{2}, 12\\right), Q\\left(\\frac{3}{2}, \\frac{23}{3}\\right)$;\\\\\nwhen $8<a\\le 13$, we have $-\\frac{26}{15}a+\\frac{338}{15}=\\frac{13}{3}$,\\\\\nsolving yields: $a=\\frac{21}{2}$,\\\\\nthus $P\\left(\\frac{21}{2}, 6\\right), Q\\left(\\frac{21}{2}, \\frac{5}{3}\\right)$,\\\\\nsummarizing, \\boxed{P\\left(\\frac{3}{2}, 12\\right), Q\\left(\\frac{3}{2}, \\frac{23}{3}\\right)} or \\boxed{P\\left(\\frac{21}{2}, 6\\right), Q\\left(\\frac{21}{2}, \\frac{5}{3}\\right)}.",
    "image_path": "dataset/mm_math/images/MM_Math/52395385_100.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $C$ is the midpoint of $\\overset{\\frown}{ACB}$, $\\angle AOC=4\\angle B$, and $OC=4$. What is the degree measure of $\\angle A$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect $AC, BC$\\\\\n$\\because \\overset{\\frown}{AB}=\\overset{\\frown}{AB}$,\\\\\n$\\therefore \\angle AOB=2\\angle ACB$,\\\\\n$\\because OA=OB$,\\\\\n$\\therefore \\angle OAB=\\angle OBA$,\\\\\n$\\therefore \\angle AOB=180^\\circ-2\\angle B$,\\\\\n$\\because C$ is the midpoint of $\\overset{\\frown}{ACB}$,\\\\\n$\\therefore \\overset{\\frown}{AC}=\\overset{\\frown}{BC}$,\\\\\n$\\therefore \\angle AOC=\\angle BOC$,\\\\\n$\\because \\angle AOC=4\\angle B$,\\\\\n$\\therefore \\angle AOC+\\angle BOC+\\angle AOB=360^\\circ$,\\\\\nthat is, $4\\angle B+4\\angle B+180^\\circ-2\\angle B=360^\\circ$,\\\\\n$\\therefore 6\\angle B=180^\\circ$,\\\\\n$\\therefore \\angle A=\\angle B=30^\\circ$;\\\\\nThus, the measure of $\\angle A$ is $\\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52906015_79.png"
  },
  {
    "solution": "\\boxed{\\frac{9}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=ax^2+bx+3$ ($a\\neq 0$) intersects the x-axis at points $A(1,0)$ and $B(-3,0)$, and intersects the y-axis at point $C$. Line $BC$ is drawn, intersecting the axis of symmetry of the parabola at point $E$, with the vertex being point $D$. What is the area of $\\triangle BOC$?\n",
    "completion": "\\textbf{Solution:} Given equation (1), $y=-x^{2}-2x+3$,\\\\\n$\\therefore$ the coordinates of point C are $(0,3)$,\\\\\n$\\therefore$ OC$=3$,\\\\\n$\\because$ the coordinates of point B are $(-3,0)$,\\\\\n$\\therefore$ OB$=3$,\\\\\n$\\because$ $\\angle$BOC$=90^\\circ$,\\\\\n$\\therefore$ the area of $\\triangle BOC$ is $\\frac{OB \\cdot OC}{2}=\\frac{3\\times 3}{2}=\\boxed{\\frac{9}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53059187_105.png"
  },
  {
    "solution": "\\boxed{75^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, a set of triangular plates are pieced together, with sides $OA$ and $OC$ coinciding with line $EF$, where $\\angle AOB=45^\\circ$ and $\\angle COD=60^\\circ$. What is the degree measure of $\\angle BOD$ in Figure 1?\n",
    "completion": "\\textbf{Solution:} Given that $\\because$ $\\angle AOB=45^\\circ$ and $\\angle COD=60^\\circ$,\\\\\n$\\therefore$ $\\angle BOD=180^\\circ-\\angle AOB-\\angle COD=180^\\circ-45^\\circ-60^\\circ=\\boxed{75^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51407218_196.png"
  },
  {
    "solution": "\\boxed{x>3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y_1=-x+3$ intersects the x-axis and the y-axis at points $B$ and $C$ respectively, and the parabola $y_2=ax^2+bx+c$ that passes through points $B$ and $C$ has another intersection point with the x-axis at point $A$, with its axis of symmetry being the line $x=2$. When $y_1<y_2$, write down directly the range of values for $x$.\n",
    "completion": "\\textbf{Solution:} Solve: $x<0$ or $x>3$.\n\nDirectly write the range of $x$ as \\boxed{x<0} or \\boxed{x>3}.",
    "image_path": "dataset/mm_math/images/MM_Math/52871717_294.png"
  },
  {
    "solution": "\\boxed{(0,1)\\cup(2,\\infty)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y=kx+b$ intersects with the graph of the inverse proportion function $y=\\frac{4}{x}$ ($x>0$) at points $A(m, 4)$ and $B(2, n)$. It also intersects with the coordinate axes at points $M$ and $N$. Based on the graph, write directly the solution set of the inequality $kx+b-\\frac{4}{x}<0$.\n",
    "completion": "\\textbf{Solution:} According to the graph, we directly obtain the solution set of the inequality $kx+b-\\frac{4}{x}<0$ as $0<x<1$ or $x>2$.\\\\\nHence, the answer is $x\\in(0,1)\\cup(x>2)$.\\\\\nTherefore, the solution set is $x\\in\\boxed{(0,1)\\cup(2,\\infty)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52601975_177.png"
  },
  {
    "solution": "\\boxed{2 \\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral \\(ABCD\\), \\(AD \\parallel BC\\), diagonal \\(BD\\) bisects diagonal \\(AC\\) perpendicularly at point \\(O\\). Quadrilateral \\(ABCD\\) is a rhombus; if \\(\\angle DBC = 30^\\circ\\) and \\(BC = 2\\), find the area of quadrilateral \\(ABCD\\).\n",
    "completion": "Solution: Since $AC \\perp BD$, \\\\\nit follows that $\\angle BOC = 30^\\circ$, \\\\\nGiven that $\\angle DBC = 30^\\circ$, and $BC = 2$, \\\\\nwe find that $OC = 1$, and $OB = \\sqrt{3}$, \\\\\nTherefore, the area of rhombus ABCD $= \\frac{1}{2} AC \\cdot BD = \\frac{1}{2} \\times 2OC \\times 2 \\times OB = \\frac{1}{2} \\times 2 \\times 2 \\sqrt{3} = \\boxed{2 \\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52019457_154.png"
  },
  {
    "solution": "\\boxed{34^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points A, B, C are on circle $O$, and $\\angle ABC=28^\\circ$. A tangent line to circle $O$ at point C intersects the extension of OA at point D. What is the measure of $\\angle D$?\n",
    "completion": "\\textbf{Solution:} Given that line CD is tangent to circle O,\\\\\nthus OC $\\perp$ CD,\\\\\ntherefore $\\angle$OCD = $90^\\circ$,\\\\\nBy the Inscribed Angle Theorem, we know that $\\angle$COD = $2\\angle$CBA = $56^\\circ$,\\\\\ntherefore $\\angle$D = $90^\\circ$ - $\\angle$OCD = $90^\\circ$ - $56^\\circ$ = $\\boxed{34^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/52601659.png"
  },
  {
    "solution": "\\boxed{140^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $O$ is the circumcenter of $\\triangle ABC$, and $\\angle ABC=40^\\circ$, $\\angle ACB=70^\\circ$, what is the measure of $\\angle BOC$?\n",
    "completion": "\\textbf{Solution:} Given: $\\because$ $\\angle ABC=40^\\circ$ and $\\angle ACB=70^\\circ$,\\\\\n$\\therefore$ $\\angle BAC=180^\\circ-40^\\circ-70^\\circ=70^\\circ$,\\\\\n$\\because$ O is the circumcenter of $\\triangle ABC$,\\\\\n$\\therefore$ the circle with center O and radius OB is the circumcircle of $\\triangle ABC$,\\\\\n$\\therefore$ $\\angle BOC=2\\angle BAC=\\boxed{140^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51261695.png"
  },
  {
    "solution": "\\boxed{100}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, on a flat ground, to measure the distance between two points A and B located beside a pond, a point O is determined first. The midpoints of OA and OB are marked as C and D respectively. It is measured that CD = 50 m. What is the distance between A and B?\n",
    "completion": "Solution: Since points C and D are the midpoints of OA and OB, respectively, and CD equals 50m, \\\\\nit follows that CD is the midsegment of $\\triangle$OAB, \\\\\ntherefore, AB equals $2 \\times CD = \\boxed{100}$ (m).",
    "image_path": "dataset/mm_math/images/MM_Math/51817942.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $DE \\parallel BC$, $\\frac{AD}{AB} = \\frac{1}{3}$. If the area of $\\triangle ABC$ is $9$, then what is the area of quadrilateral $BCED$?\n",
    "completion": "\\textbf{Solution:} Given that DE$\\parallel$ BC, it follows that $\\triangle ADE \\sim \\triangle ABC$. Therefore, $\\frac{S_{\\triangle ADE}}{S_{\\triangle ABC}} = \\left(\\frac{AD}{AB}\\right)^2 = \\left(\\frac{1}{3}\\right)^2 = \\frac{1}{9}$.\\\\\nSince $S_{\\triangle ABC} = 9$, it implies that $S_{\\triangle ADE} = 1$. Consequently, the area of quadrilateral BCDE equals $9 - 1 = \\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55497749.png"
  },
  {
    "solution": "\\boxed{7}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=5$, $AC=8$, and $\\angle A=60^\\circ$. Find the length of $BC$.\n",
    "completion": "\\textbf{Solution:} Construct $CD \\perp AB$ at point C, with D being the foot of the perpendicular.\\\\\nIn $\\triangle ACD$,\\\\\nsince $\\angle A=60^\\circ$ and $AC=8$,\\\\\nthen $\\angle ACD=30^\\circ$,\\\\\nthus $AD= \\frac{1}{2}AC=4$,\\\\\nand $CD=\\sqrt{8^2-4^2}=4\\sqrt{3}$, $BD=AB-AD=1$.\\\\\nTherefore, in $\\triangle BCD$, $BC=\\sqrt{CD^2+BD^2}=\\sqrt{48+1}=\\boxed{7}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52602114_147.png"
  },
  {
    "solution": "\\boxed{3cm}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the straight line $CD$ is the perpendicular bisector of the line segment $AB$, and $P$ is a point on the line $CD$. Given that the length of segment $PA$ is $3\\,\\text{cm}$, what is the length of segment $PB$?\n",
    "completion": "\\textbf{Solution}: Since line CD is the perpendicular bisector of segment AB, \\\\\nit follows that PA=PB, \\\\\nSince PA=3cm, \\\\\nit follows that PB=\\boxed{3cm}.",
    "image_path": "dataset/mm_math/images/MM_Math/52905795.png"
  },
  {
    "solution": "\\boxed{\\frac{4}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, $AC=3$, and $BC=4$, what is the value of $\\sin A$?\n",
    "completion": "\\textbf{Solution:} In $Rt\\triangle ABC$, where $\\angle C=90^\\circ$, $AC=3$, and $BC=4$,\\\\\nby the Pythagorean theorem, we have $AB=\\sqrt{AC^2+BC^2}=5$,\\\\\ntherefore $\\sin A=\\frac{BC}{AB}=\\boxed{\\frac{4}{5}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51262418.png"
  },
  {
    "solution": "\\boxed{}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, put together a set of triangles as demonstrated below. Write down the degrees of $\\angle A$, $\\angle B$, $\\angle BCD$, $\\angle D$, and $\\angle AED$.\n",
    "completion": "\\textbf{Solution:} From the problem, we get \\\\\n$\\angle A=30^\\circ$, \\\\\n$\\angle B=90^\\circ$, \\\\\n$\\angle BCD=150^\\circ$, \\\\\n$\\angle D=45^\\circ$, \\\\\n$\\angle AED=135^\\circ$.",
    "image_path": "dataset/mm_math/images/MM_Math/54927701_8.png"
  },
  {
    "solution": "\\boxed{20}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated in the diagram, in $\\triangle ABC$, where $AB=AC$, the circle $\\odot O$ with diameter $AB$ intersects $BC$ at point $D$. A line $EF$ passing through point $D$ intersects $AC$ at point $F$ and the extension of $AB$ at point $E$, and $\\angle BAC=2\\angle BDE$. $DF$ is a tangent to $\\odot O$; when $CF=4$ and $BE=6$, find the length of $AF$.\n",
    "completion": "\\textbf{Solution:} Since $AB=AC$ and $AD\\perp BC$, we have $BD=CD$. Since $BO=AO$, it follows that $OD$ is the median line of triangle $ABC$, thus $OD\\parallel AC$ and $AC=2OD$. Therefore, $\\triangle EOD\\sim \\triangle EAF$, which implies $\\frac{OD}{AF}=\\frac{EO}{EA}$. Let $OD=x$, then $AC=2x$. Since $CF=4$ and $BE=6$, we have $OA=OB=x$, $AF=AC-CF=2x-4$, $EO=x+6$, and $EA=2x+6$. So, $\\frac{x}{2x-4}=\\frac{x+6}{2x+6}$. Solving this, we find $x=12$. Checking, we find $x=12$ is the solution of the given fractional equation. Therefore, $AF=2x-4=\\boxed{20}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51368305_101.png"
  },
  {
    "solution": "\\boxed{\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, with point O as the center, draw an arc with any length as the radius, which intersects the ray $OA$ at point B. Then, with point B as the center and the length of $BO$ as the radius, draw another arc. If the two arcs intersect at point C, draw ray $OC$. What is the value of $\\tan\\angle AOC$?\n",
    "completion": "\\textbf{Solution:} \nConnect $BC$, \\\\\ngiven the problem, we have: $OB=OC=BC$, \\\\\nthus, $\\triangle OBC$ is an equilateral triangle, \\\\\nhence, $\\tan\\angle AOC=\\tan60^\\circ=\\boxed{\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53019659.png"
  },
  {
    "solution": "\\boxed{4-2\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle CDE$, $CD=1$ and $\\angle CDE=45^\\circ$, respectively construct squares $ABCD$ and $CEFG$ externally on sides $CD$ and $CE$. If $AE=BD$, what is the value of $EF^2$?\n",
    "completion": "\\textbf{Solution:} Draw $EN \\perp AD$ and $EM \\perp CD$ at $E$,\\\\\nsince $\\angle CDN=45^\\circ$,\\\\\nthus, quadrilateral $DMEN$ is a square,\\\\\nlet $DN=x$, thus $EN=DM=EM=x$,\\\\\nsince $AE=BD=\\sqrt{2}CD=\\sqrt{2}$,\\\\\nin $\\triangle AEN$, since $AN^{2}+EN^{2}=AE^{2}$,\\\\\ni.e., $(1+x)^{2}+x^{2}=2$,\\\\\nwe get $x=\\frac{\\sqrt{3}-1}{2}$ or $x=\\frac{-\\sqrt{3}-1}{2}$ (discard),\\\\\nsince in $\\triangle CME$, since $CM^{2}+EM^{2}=CE^{2}$,\\\\\ni.e., $CE^{2}=\\left(\\frac{\\sqrt{3}-1}{2}+1\\right)^{2}+\\left(\\frac{\\sqrt{3}-1}{2}\\right)^{2}=4-2\\sqrt{3}$,\\\\\nsince quadrilateral $CEFG$ is a square,\\\\\nthus $EF^{2}=\\boxed{4-2\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50475276.png"
  },
  {
    "solution": "\\boxed{24}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC=2\\sqrt{5}$, $\\angle BAC=\\alpha^\\circ$, $\\tan\\angle ABC=\\frac{1}{2}$, $G$ is the midpoint of $BC$, and $D$ is a moving point in the plane with $DG=\\frac{\\sqrt{5}}{5}$. If line segment $BD$ rotates counterclockwise by $\\alpha^\\circ$ around point $D$ to obtain $DB'$, what is the maximum area of quadrilateral $BACB'$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect AD, AG, and draw GH perpendicular to AB at point H through point G,\\\\\nsince $AB=AC=2\\sqrt{5}$ and $BG=GC$,\\\\\nthus, $AG\\perp BC$,\\\\\nsince $\\tan\\angle ABC=\\frac{AG}{BG}=\\frac{1}{2}$,\\\\\nthus, $AG=2$ and $BG=4$,\\\\\nsince $\\sin\\angle ABG=\\sin\\angle GBH$,\\\\\nthus, $\\frac{GH}{BC}=\\frac{AG}{AB}$,\\\\\nthus, $\\frac{GH}{4}=\\frac{2}{2\\sqrt{5}}$,\\\\\nthus, $GH=\\frac{4\\sqrt{5}}{5}$,\\\\\nsince $AB=AC$, $DB=DB'$, $\\angle BAC=\\angle BDB'$,\\\\\nthus, $\\angle ABC=\\angle DBB'$, $\\frac{AB}{BC}=\\frac{BD}{BB'}$,\\\\\nthus, $\\angle ABD=\\angle CBB'$,\\\\\nthus, $\\triangle ABD\\sim \\triangle CBB'$,\\\\\nthus, $\\frac{S_{\\triangle ABD}}{S_{\\triangle CBB'}}=\\left(\\frac{AB}{BC}\\right)^2=\\left(\\frac{2\\sqrt{5}}{4}\\right)^2=\\frac{5}{16}$,\\\\\nsince $DG=\\frac{\\sqrt{5}}{5}$,\\\\\nthus, the motion trajectory of point G is a circle with G as the center and $\\frac{\\sqrt{5}}{5}$ as the radius. When point D is on the extension line of HG, the area of $\\triangle ABD$ is maximized, with the maximum value $=\\frac{1}{2}\\times 2\\sqrt{5}\\times \\left(\\frac{4\\sqrt{5}}{5}+\\frac{\\sqrt{5}}{5}\\right)=5$,\\\\\nthus, the maximum area of $\\triangle BCB'$ is $16$,\\\\\nthus, the maximum area of quadrilateral $BACB'$ is $\\frac{1}{2}\\times 8\\times 2+16=\\boxed{24}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51029388.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, $E$ is a point on the side $BC$ of the equilateral triangle $\\triangle ABC$ (not coinciding with point $B$ or $C$). Connect $AE$ and construct an equilateral triangle $\\triangle AEF$ to the right using $AE$ as a side, and connect $CF$. It is known that the function relationship between the area ($S$) of $\\triangle ECF$ and the length ($x$) of $BE$ is shown in Figure 2 ($P$ is the vertex of the parabola). When the area of $\\triangle ECF$ is maximized, what is the degree measure of $\\angle FEC$?\n",
    "completion": "\\textbf{Solution:} Let the side length of the equilateral $\\triangle ABC$ be $a$. Draw $FD\\perp BC$ at $D$, \\\\\n$\\because$ $\\triangle ABC$ is equilateral, $\\triangle AEF$ is equilateral, \\\\\n$\\therefore AB=AC$, $AE=AF$, $\\angle BAC=\\angle ABC=\\angle ACB=\\angle EAF=\\angle AEF=60^\\circ$, \\\\\n$\\therefore \\angle BAE=\\angle CAF$, \\\\\n$\\therefore \\triangle ABE\\cong \\triangle ACF$ (SAS), \\\\\n$\\therefore CF=BE=x$, $\\angle ACF=\\angle ABE=60^\\circ$, $\\angle FCD=180^\\circ-\\angle ACB-\\angle ACF=60^\\circ$, \\\\\n$FD=CF\\cdot\\sin60^\\circ=\\frac{\\sqrt{3}}{2}x$, \\\\\n$S_{\\triangle ECF}=\\frac{1}{2}CE\\times FD = -\\frac{\\sqrt{3}}{4}x^2+\\frac{\\sqrt{3}}{4}ax$ \\\\\n$\\therefore$ When $S_{\\triangle ECF}$ is maximized, $x= \\frac{1}{2}a$, i.e., E is the midpoint of BC, the maximum value of $S_{\\triangle ECF}$ is $\\frac{\\sqrt{3}}{16}a^2$ \\\\\n$\\because$ E is the midpoint of BC \\\\\n$\\therefore AE\\perp BC$, $\\angle AEB=90^\\circ$ \\\\\n$\\therefore \\angle FEC=180^\\circ-\\angle AEB-\\angle AEF=30^\\circ$ \\\\\nHence, the measure of $\\angle FEC$ is $\\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51212719_87.png"
  },
  {
    "solution": "\\boxed{125^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, $\\angle AOB$ is a straight angle, $\\angle AOC=80^\\circ$, $\\angle BOD=30^\\circ$, $OM$ and $ON$ are the angle bisectors of $\\angle AOC$ and $\\angle BOD$ respectively. What is the degree measure of $\\angle MON$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle AOB$ is a straight angle,\\\\\n$\\therefore \\angle AOB=180^\\circ$,\\\\\nsince OM and ON are the bisectors of $\\angle AOC$ and $\\angle BOD$, respectively, and $\\angle AOC=80^\\circ$, $\\angle BOD=30^\\circ$,\\\\\n$\\therefore \\angle AOM=\\frac{1}{2}\\angle AOC=40^\\circ$, $\\angle BON=\\frac{1}{2}\\angle BOD=15^\\circ$,\\\\\n$\\therefore \\angle MON=\\angle AOB-\\angle AOM-\\angle BON=180^\\circ-40^\\circ-15^\\circ=\\boxed{125^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51495371.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the pentagon $ABCDE$, $AB \\perp BC$, $AE \\perp DE$, $AB=1$, $BC=2$, $CD=2\\sqrt{5}$, $DE=3$, $AE=4$. Find the area of the pentagon $ABCDE$.\n",
    "completion": "\\textbf{Solution:} Draw $AC$ and $AD$,\\\\\nsince $AB\\perp BC$ and $AE\\perp DE$,\\\\\nthus $\\angle B=90^\\circ$ and $\\angle E=90^\\circ$,\\\\\nhence $AB^2+BC^2=AC^2$ and $AE^2+DE^2=AD^2$,\\\\\nsince $AB=1$, $BC=2$, $DE=3$, $AE=4$,\\\\\nthus $1^2+2^2=AC^2$ and $4^2+3^2=AD^2$,\\\\\ntherefore $AC=\\sqrt{5}$ and $AD=5$,\\\\\nsince $CD=2\\sqrt{5}$,\\\\\nthus $AC^2+CD^2=AD^2$,\\\\\nthus $\\angle ACD=90^\\circ$,\\\\\nthus the area of pentagon $ABCDE$ $=S_{\\triangle ABC}+S_{\\triangle ADE}+S_{\\triangle ACD}=\\frac{1}{2}\\cdot AB\\cdot BC+\\frac{1}{2}\\cdot AE\\cdot DE+\\frac{1}{2}\\cdot AC\\cdot CD$\\\\\n$=\\frac{1}{2}\\times 1\\times 2+\\frac{1}{2}\\times 4\\times 3+\\frac{1}{2}\\times \\sqrt{5}\\times 2\\sqrt{5}=1+6+5=\\boxed{12}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52731268.png"
  },
  {
    "solution": "\\boxed{x \\leq 2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated in the Cartesian coordinate system, the line $l_1\\colon y_1=kx+b$, passing through point $A(-6,0)$, intersects with the line $l_2\\colon y_2=2x$ at point $B(m,4)$. Utilize the graphical representation of functions to directly determine the range of $x$ values for which $y_2\\leq y_1$.\n",
    "completion": "\\textbf{Solution:} By using the function graph, we can directly write the range of $x$ when $y_2 \\leq y_1$ as $\\boxed{x \\leq 2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52188279_131.png"
  },
  {
    "solution": "\\boxed{3\\text{cm}^2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, it is known that $S_{\\triangle ABD} : S_{\\triangle ACD} = 2 : 1$, point E is the midpoint of AB, and the area of $\\triangle ABC$ is $9\\,\\text{cm}^2$. What is the area of $\\triangle AED$?\n",
    "completion": "\\textbf{Solution:} Given that $S_{\\triangle ABD} : S_{\\triangle ACD} = 2 : 1$,\\\\\nit follows that $S_{\\triangle ACD} = \\frac{1}{2}S_{\\triangle ABD}$,\\\\\nsince point E is the midpoint of AB,\\\\\nthus $S_{\\triangle ADE} = S_{\\triangle BDE} = \\frac{1}{2}S_{\\triangle ABD}$,\\\\\nhence $S_{\\triangle ADE} = \\frac{1}{3}S_{\\triangle ABC} = \\frac{1}{3} \\times 9 = \\boxed{3\\text{cm}^2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52733477.png"
  },
  {
    "solution": "\\boxed{(-2,6)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $l_{1}$ intersects the x-axis and y-axis at points A and B, respectively, with coordinates $(2, 0)$ and $(0, 3)$. The line $l_{2}: y=\\frac{1}{2}x+3$ passing through point B intersects the x-axis at point C. Point $D(n, 6)$ is a point on the line $l_{1}$. Connect $CD$. What are the coordinates of C and D?\n",
    "completion": "\\textbf{Solution}: When $y=0$, $\\frac{1}{2}x+3=0$, solving this gives $x=-6$,\\\\\nhence the coordinates of point C are $\\boxed{(-6,0)}$,\\\\\nsubstituting point D$(n,6)$ into $y=-\\frac{3}{2}x+3$ we get $-\\frac{3}{2}n+3=6$, solving for $n$ gives $n=-2$,\\\\\ntherefore, the coordinates of point D are $\\boxed{(-2,6)}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51699679_176.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points A, B, and C are three points on a circle $\\odot$O with a radius of 4. If $\\angle BAC=45^\\circ$, what is the length of chord BC?\n",
    "completion": "\\textbf{Solution:} Connect OB and OC.\\\\\nSince $\\angle$BOC = 2$\\angle$BAC and $\\angle$BAC = 45$^\\circ$,\\\\\nit follows that $\\angle$BOC = 90$^\\circ$,\\\\\nSince OB = OC = 4,\\\\\nit follows that BC = $\\sqrt{4^2 + 4^2} = 4\\sqrt{2}$,\\\\\nTherefore, the answer is: $BC = \\boxed{4\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51318401.png"
  },
  {
    "solution": "\\boxed{\\frac{193}{7}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given in the diagram, $AB$ is the diameter of $\\odot O$, and points $C$, $D$ are on $\\odot O$, with $BC\\perp AC$ and $OD\\parallel BC$, $AC$ intersects with $BD$, $OD$ at point $E$, $F$ respectively. Point $D$ is the midpoint of $\\overset{\\frown}{AC}$; if $DF=7$, $AC=24$, what is the diameter of $\\odot O$?\n",
    "completion": "Solution: From the given information, we have\n\n\\[\n\\because OF\\perp AC,\n\\]\n\\[\n\\therefore AF=\\frac{1}{2}AC=12,\n\\]\n\\[\n\\because DF=7,\n\\]\n\\[\n\\therefore OF=OD-DF=OA-7,\n\\]\n\\[\n\\because OA^2=AF^2+OF^2,\n\\]\n\\[\n\\therefore OA^2=12^2+(OA-7)^2,\n\\]\n\\[\n\\therefore OA=\\frac{193}{14},\n\\]\n\\[\n\\therefore \\text{The diameter of the circle } O \\text{ is } \\boxed{\\frac{193}{7}}.\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/52861807_112.png"
  },
  {
    "solution": "\\boxed{3\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, the length of side $AB$ is 3. Points $E$ and $F$ are located on $AD$ and $BC$ respectively. Connect $BE$, $DF$, $EF$, and $BD$. If quadrilateral $BEDF$ is a rhombus, and $EF=AE+FC$, then what is the length of side $BC$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral ABCD is a rectangle,\\\\\n$\\therefore$ $\\angle A=90^\\circ$, AD＝BC,\\\\\n$\\therefore$ BA$\\perp$BF,\\\\\nGiven that quadrilateral BEDF is a rhombus,\\\\\n$\\therefore$ EF$\\perp$BD, BE＝BF＝DE＝DF,\\\\\n$\\therefore$ $\\triangle$ BEF is an isosceles triangle, AE＝CF,\\\\\n$\\therefore$ $\\angle$ EBO＝$\\angle$ DBF,\\\\\nGiven that EF＝AE+FC, AE＝CF, EO＝FO,\\\\\n$\\therefore$ EF＝2AE＝2EO,\\\\\n$\\therefore$ AE＝EO＝CF＝FO,\\\\\n$\\therefore$ From knowledge that angles form an axisymmetric figure, AB＝BO＝3, $\\angle$ ABE＝$\\angle$ EBO,\\\\\n$\\therefore$ $\\angle$ ABE＝$\\angle$ EBD＝$\\angle$ DBC＝$30^\\circ$,\\\\\nLet OE＝x, then BE＝2x,\\\\\nAccording to the Pythagorean theorem, $BE^2-OE^2=OB^2$,\\\\\n$4x^2-x^2=3^2$,\\\\\nSolving, we get x＝$\\sqrt{3}$,\\\\\n$\\therefore$ BE＝2$\\sqrt{3}$,\\\\\n$\\therefore$ BF＝BE＝2$\\sqrt{3}$,\\\\\n$\\therefore$ CF＝AE＝$\\sqrt{3}$,\\\\\n$\\therefore$ BC＝BF+CF＝$3\\sqrt{3}$,\\\\\nTherefore, the answer is: BC＝\\boxed{3\\sqrt{3}}.",
    "image_path": "dataset/mm_math/images/MM_Math/52210824.png"
  },
  {
    "solution": "\\boxed{4cm}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, line segment $AB=8\\,\\text{cm}$, ray $AN \\perp AB$ with the foot of the perpendicular at point A. Point C is a moving point on the ray, and isosceles right triangles are formed with $AC$ and $BC$ as the legs, resulting in $\\triangle ACD$ and $\\triangle BCE$ respectively. $DE$ is connected and intersects ray $AN$ at point M. What is the length of $CM$?\n",
    "completion": "\\textbf{Solution:} Construct $EF \\perp AN$ at $F$, as shown in the figure.\\\\\nSince $AN \\perp AB$, and $\\triangle BCE$ and $\\triangle ACD$ are isosceles right triangles,\\\\\nthus $\\angle BAC = \\angle BCE = \\angle ACD =\\angle CFE = 90^\\circ$, $BC = CE$, $AC = CD$.\\\\\nTherefore, $\\angle ABC + \\angle ACB = 90^\\circ$, $\\angle FCE + \\angle ACB = 90^\\circ$,\\\\\nthus $\\angle ABC = \\angle FCE$,\\\\\nIn $\\triangle ABC$ and $\\triangle FCE$\\\\\n$\\left\\{\\begin{array}{l} \\angle BAC = \\angle CFE \\\\ \\angle ABC = \\angle FCE \\\\ BC = CE \\end{array}\\right.$\\\\\nTherefore, $\\triangle ABC \\cong \\triangle FCE$\\\\\nTherefore, $AB = FC = 8\\,cm$, $AC = FE$\\\\\nTherefore, $CD = FE$\\\\\nIn $\\triangle DCM$ and $\\triangle EFM$\\\\\n$\\left\\{\\begin{array}{l} \\angle DMC = \\angle EMF \\\\ \\angle DCM = \\angle EFM = 90^\\circ \\\\ CD =\nFE \\end{array}\\right.$\\\\\nTherefore, $\\triangle DCM \\cong \\triangle EFM$\\\\\nTherefore, $CM = FM = \\frac{1}{2}FC = \\boxed{4cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52978752.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y=ax+b$ intersects with the graph of the reciprocal function $y=\\frac{k}{x}$ ($k\\ne 0$) at points $A(1, -3)$ and $B(m, -1)$. Connect $OA$ and $OB$. What is the area of $\\triangle OAB$?\n",
    "completion": "\\textbf{Solution:} Let the intersection point of line AB with the y-axis be C. Then, the coordinates of point C are $(0, -4)$. \\\\\nThe area of triangle $OAB$, $S_{\\triangle OAB}=S_{\\triangle BOC}-S_{\\triangle AOC}=\\frac{1}{2}\\times 4\\times 3- \\frac{1}{2}\\times 4\\times 1=\\boxed{4}.$",
    "image_path": "dataset/mm_math/images/MM_Math/52602121_134.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the quadrilateral $ABCD$ is inscribed in circle $\\odot O$, $BD$ bisects $\\angle ABC$, through point $D$ draw $DE\\parallel AB$, intersecting $BC$ at point $E$, connect $AE$ and intersect $BD$ at point $F$. Given that $\\angle AFD = \\angle ADB + \\angle CDE$, if $AB^2 = BF \\cdot BD$ and $AD = 2$, find the length of $CB$.\n",
    "completion": "\\textbf{Solution}: Given from equation (1), we have  $\\angle AFD=180^\\circ-3\\alpha$,\\\\\n$\\therefore$ $\\angle FAB=180^\\circ-4\\alpha$,\\\\\n$\\because$ it is about an inscribed quadrilateral,\\\\\n$\\therefore$ $\\angle CDE+2\\alpha +\\alpha +2\\alpha =180^\\circ$, that is $\\angle CDE=180^\\circ-5\\alpha$,\\\\\n$\\therefore$ $\\angle CDF=\\angle CDE+\\angle EDB=\\alpha +180^\\circ-5\\alpha =180^\\circ-4\\alpha$,\\\\\n$\\therefore$ $\\angle BAF=\\angle BDC=180^\\circ-4\\alpha$, and $\\angle ABF=\\angle FBC=\\alpha$,\\\\\n$\\therefore$ $\\triangle ABF\\sim \\triangle DBC$,\\\\\n$\\therefore$ $\\frac{AB}{DC}=\\frac{FB}{BC}$, thus $BC=\\frac{FB\\cdot DC}{AB}$,\\\\\n$\\therefore$ $\\angle AFB=\\angle DCB=180^\\circ-(180^\\circ-3\\alpha )=3\\alpha$,\\\\\n$\\because$ $\\angle DAF+\\angle FAB+\\angle DCB=180^\\circ$, that is $\\angle DAF+(180^\\circ-4\\alpha )+3\\alpha =180^\\circ$,\\\\\n$\\therefore$ $\\angle DAF=\\angle FDE=\\angle DBE=\\alpha$,\\\\\n$\\therefore$ $\\triangle ADE\\sim \\triangle BCD$,\\\\\n$\\therefore$ $\\frac{AD}{BC}=\\frac{AE}{BD}$, thus $BC=\\frac{AD\\cdot BD}{AE}$,\\\\\n$\\therefore$ $BC=\\frac{FB\\cdot DC}{AB}=\\frac{AD\\cdot BD}{AE}$,\\\\\nand $\\triangle ABE\\cong \\triangle BAD$ (by SAS),\\\\\n$\\because$ $AB^{2}=BF\\cdot BD$, $AD=2$, and $AB=AE$, $AD=EC=DC=2$\\\\\n$\\therefore$ $CB=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52991542_119.png"
  },
  {
    "solution": "\\boxed{118^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of the semicircle $O$, and $\\angle BAC=28^\\circ$. What is the measure of $\\angle D$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect CB, \\\\\n$\\because$ AB is the diameter of the semicircle O, and $\\angle BAC=28^\\circ$, \\\\\n$\\therefore \\angle B=\\angle ACB - \\angle BAC=90^\\circ - 28^\\circ = 62^\\circ$, \\\\\nAlso, $\\because$ quadrilateral ABCD is inscribed in the circle, \\\\\n$\\therefore \\angle D + \\angle ABC= 180^\\circ$, \\\\\n$\\therefore \\angle D=180^\\circ - 62^\\circ = \\boxed{118^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53059655.png"
  },
  {
    "solution": "\\boxed{\\frac{9}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the line $y=-2x+3$ intersects the $y$-axis at point $A$ and the $x$-axis at point $B$. Point $P$ is a moving point on segment $AB$. Perpendicular lines are drawn from $P$ to the $x$-axis and the $y$-axis, and the feet of the perpendiculars are denoted as $C$ and $D$, respectively. Connect $CD$. When $CD$ is minimized, what is the value of $PA\\cdot PB$?\n",
    "completion": "\\textbf{Solution:} Since $PC\\perp OB$, $PD\\perp OA$, and $\\angle DOB=90^\\circ$, \\\\\nit follows that quadrilateral $OCPD$ is a rectangle. \\\\\nBy connecting $OP$, we have $CD=OP$. \\\\\nTherefore, when $OP\\perp AB$, $OP$ is at its shortest, meaning $CD$ is also at its shortest. \\\\\nSince $y=-2x+3$, \\\\\nwhen $x=0$, $y=3$, and when $y=0$, $x=\\frac{3}{2}$. \\\\\nThus, $A(0,3)$ and $B\\left(\\frac{3}{2},0\\right)$, \\\\\nwhich means $OA=3$ and $OB=\\frac{3}{2}$, \\\\\nand so $AB=\\sqrt{OA^2+OB^2}=\\sqrt{3^2+\\left(\\frac{3}{2}\\right)^2}=\\frac{3}{2}\\sqrt{5}$. \\\\\n$\\tan\\angle OBA=\\frac{OP}{PB}=\\frac{OA}{OB}=2$, \\\\\nLetting $PB=x$, then $PO=2x$. \\\\\nTherefore, $x^2+4x^2=\\left(\\frac{3}{2}\\right)^2$, \\\\\nSolving this, we find $x=\\frac{3}{10}\\sqrt{5}$, \\\\\nHence, $PB=\\frac{3}{10}\\sqrt{5}$ and $PA=AB-PB=\\frac{3\\sqrt{5}}{2}-\\frac{3\\sqrt{5}}{10}=\\frac{6}{5}\\sqrt{5}$, \\\\\nTherefore, $PA\\cdot PB=\\frac{3\\sqrt{5}}{10}\\times \\frac{6\\sqrt{5}}{5}=\\boxed{\\frac{9}{5}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53058713.png"
  },
  {
    "solution": "\\boxed{7.5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $C$ is a point on line segment $AB$, with $AC = 10\\, \\text{cm}$, $CB = 5\\, \\text{cm}$. $M$, $N$ are the midpoints of $AC$ and $BC$, respectively. What is the length of line segment $MN$ in centimeters?\n",
    "completion": "\\textbf{Solution:} Given that $AC=10\\text{cm}$, $CB=5\\text{cm}$, and $M$, $N$ are the midpoints of $AC$, $BC$ respectively,\\\\\nthus, $CM=\\frac{1}{2}AC=5\\text{cm}$, $CN=\\frac{1}{2}CB=2.5\\text{cm}$,\\\\\nhence $MN=CM+CN=\\boxed{7.5}\\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51439013.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the parallelogram $ABCD$, point $E$ is on the side $AD$, and $CE$ and $BD$ intersect at point $F$, with $BF = 3DF$. What is the value of $AE\\colon ED$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a parallelogram, \\\\\nit follows that AD$\\parallel$BC and AD=BC. \\\\\nTherefore, $\\frac{BC}{ED}=\\frac{BF}{DF}$. \\\\\nSince BF=3DF, \\\\\nit follows that $\\frac{BF}{DF}=3$. \\\\\nTherefore, $\\frac{BC}{ED}=3$. \\\\\nTherefore, $\\frac{AD}{ED}=3$, which means $AE\\colon ED=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51379484_90.png"
  },
  {
    "solution": "\\boxed{x<-3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y=kx+b$ intersects the x-axis at point $A(-3,0)$. What is the solution set of the inequality $kx+b>0$ with respect to $x$?\n",
    "completion": "\\textbf{Solution:} Since the solution set of the inequality $kx+b>0$ represents the range of the independent variable $x$ when $y>0$, \\\\\ntherefore, from the graph in the question, it indicates the range of the coordinate $x$ where the graph of the linear function is above the x-axis, \\\\\ntherefore, according to the graph in the question, when $x<-3$, the graph of the linear function is above the x-axis, \\\\\ntherefore, when $x<-3$, we have $kx+b>0$, \\\\\nHence, the answer is: $\\boxed{x<-3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52386687.png"
  },
  {
    "solution": "\\boxed{y=3x-1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the inverse proportion function $y=\\frac{2m}{x}$ and the graph of the linear function $y=kx-1$ intersect at points $A(m, 2m)$ and $B$. Find the expression for the linear function.\n",
    "completion": "\\textbf{Solution:} Since point $A(m,2m)$ lies on the graph of the inverse proportion function, it follows that $2m = \\frac{2m}{m}$, so $m=1$, hence $A(1,2)$.\\\\\nFurthermore, since $A(1,2)$ lies on the graph of the linear function $y=kx-1$, it follows that $2=k-1$, which gives $k=3$,\\\\\nthus, the expression for the linear function is: $\\boxed{y=3x-1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52622749_215.png"
  },
  {
    "solution": "\\boxed{\\frac{4\\pi }{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, quadrilateral $ABCD$ is inscribed in circle $\\odot O$, with $AE$ being the diameter of $\\odot O$. If the radius of $\\odot O$ is $6$, and $\\angle ADC - \\angle ABC = 40^\\circ$, what is the length of $\\overset{\\frown}{CE}$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect $OC$.\n\nSince quadrilateral $ABCD$ is inscribed in $\\odot O$,\n\nit follows that $\\angle ADC+\\angle ABC=180^\\circ$,\n\nand since $\\angle ADC-\\angle ABC=40^\\circ$,\n\nwe have $\\angle ADC=110^\\circ$, $\\angle ABC=70^\\circ$.\n\nTherefore, $\\angle AOC=2\\angle ABC=140^\\circ$,\n\nwhich leads to $\\angle COE=180^\\circ-\\angle AOC=40^\\circ$,\n\nSince the radius of $\\odot O$ is 6,\n\nthe length of the arc $\\overset{\\frown}{CE}$ is $\\frac{40\\pi \\times 6}{180}=\\boxed{\\frac{4\\pi }{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52872102.png"
  },
  {
    "solution": "\\boxed{105^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a set of triangles is stacked together. What is the value of $\\angle \\alpha$ in the figure?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, $\\angle C=90^\\circ$, $\\angle DAE=45^\\circ$, $\\angle BAC=60^\\circ$,\\\\\n$\\therefore \\angle CAO=\\angle BAC−\\angle DAE=60^\\circ−45^\\circ=15^\\circ$,\\\\\n$\\therefore \\angle \\alpha =\\angle C+\\angle CAO=90^\\circ+15^\\circ=\\boxed{105^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51378976.png"
  },
  {
    "solution": "\\boxed{78^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $OB$ is the bisector of $\\angle AOC$, $OD$ is the bisector of $\\angle COE$. If $\\angle AOB = 42^\\circ$, $\\angle DOE = 36^\\circ$, what is the degree measure of $\\angle BOD$?\n",
    "completion": "\\textbf{Solution:} Since $OB$ is the bisector of $\\angle AOC$ and $\\angle AOB=42^\\circ$,\\\\\nand $OD$ is the bisector of $\\angle COE$ and $\\angle DOE=36^\\circ$,\\\\\n$\\therefore$ $\\angle AOB=\\angle BOC=42^\\circ$,\\\\\n$\\angle COD=\\angle DOE=36^\\circ$,\\\\\n$\\therefore$ $\\angle BOD=\\angle BOC+\\angle DOC=42^\\circ+36^\\circ=\\boxed{78^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51380002_130.png"
  },
  {
    "solution": "\\boxed{\\frac{540^\\circ}{7}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the isosceles $\\triangle ABC$, $AB=AC$, $\\angle A<90^\\circ$, $CD$ is the altitude of $\\triangle ABC$, $BE$ is the angle bisector of $\\triangle ABC$, and $CD$ intersects $BE$ at point $P$. As the size of $\\angle A$ changes, the shape of $\\triangle EPC$ also changes. When $\\triangle EPC$ is an isosceles triangle, what is the degree measure of $\\angle ACB$?\n",
    "completion": "\\textbf{Solution:} Let $ \\angle A=\\alpha $ and $ \\angle EPC=\\beta $. \\\\\n(1) If $ EP=EC$, \\\\\nthen $ \\angle ECP=\\angle EPC=\\beta $, \\\\\nand $ \\angle ABC=\\angle ACB=90^\\circ-\\frac{\\alpha }{2}$, $ \\angle ABC+\\angle BCD=90^\\circ$, \\\\\nthus, $ 90^\\circ-\\frac{\\alpha }{2}+\\left(90^\\circ-\\frac{\\alpha }{2}-\\beta \\right)=90^\\circ$, \\\\\nfrom (2), we know $ \\beta =45^\\circ+\\frac{\\alpha }{4}$, \\\\\n$ \\therefore 90^\\circ-\\frac{\\alpha }{2}+90^\\circ-\\frac{\\alpha }{2}-\\left(45^\\circ+\\frac{\\alpha }{4}\\right)=90^\\circ$, \\\\\nsolving gives: $ \\alpha =36^\\circ$, \\\\\n$ \\therefore \\angle ACB=90^\\circ-\\frac{\\alpha }{2}=\\boxed{72^\\circ}$; \\\\\n(2) If $ PC=PE$, \\\\\nthen $ \\angle PCE=\\angle PEC=\\left(180^\\circ-\\beta \\right)\\div 2=90^\\circ-\\frac{\\beta }{2}$, \\\\\nfrom (1), we have: $ \\angle ABC+\\angle BCD=90^\\circ$, \\\\\n$ \\therefore 90^\\circ-\\frac{\\alpha }{2}+90^\\circ-\\frac{\\alpha }{2}-\\left(90^\\circ-\\frac{\\beta }{2}\\right)=90^\\circ$, \\\\\nsince $ \\beta =45^\\circ+\\frac{\\alpha }{4}$, \\\\\n$ \\therefore 90^\\circ-\\frac{\\alpha }{2}+90^\\circ-\\frac{\\alpha }{2}-\\left(90^\\circ-\\frac{45^\\circ+\\frac{\\alpha }{4}}{2}\\right)=90^\\circ$, \\\\\nsolving gives: $ \\alpha =\\frac{180^\\circ}{7}$, \\\\\n$ \\therefore \\angle ACB=90^\\circ-\\frac{\\alpha }{2}=\\boxed{\\frac{540^\\circ}{7}}$; \\\\\n(3) If $ CP=CE$, \\\\\nthen $ \\angle EPC=\\angle PEC=\\beta$, $ \\angle PCE=180^\\circ-2\\beta$, \\\\\nfrom (1), we have: $ \\angle ABC+\\angle BCD=90^\\circ$, \\\\\n$ \\therefore 90^\\circ-\\frac{\\alpha }{2}+90^\\circ-\\frac{\\alpha }{2}-\\left(180^\\circ-2\\beta \\right)=90^\\circ$, \\\\\nsince $ \\beta =45^\\circ+\\frac{\\alpha }{4}$, \\\\\n$ \\therefore 90^\\circ-\\frac{\\alpha }{2}+90^\\circ-\\frac{\\alpha }{2}-\\left[180^\\circ-2\\left(45^\\circ+\\frac{\\alpha }{4}\\right)\\right]=90^\\circ$, \\\\\nsolving gives: $ x=0$, which does not make sense, \\\\\nIn summary: when $ \\triangle EPC$ is an isosceles triangle, the degree of $ \\angle ACB$ is either $ \\boxed{72^\\circ}$ or $ \\boxed{\\frac{540^\\circ}{7}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52861546_103.png"
  },
  {
    "solution": "\\boxed{\\frac{4}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: The ancient Chinese mathematician Zhao Shuang, while annotating the \"Zhou Bi Suan Jing\", assembled 4 congruent right-angled triangles into a square (as illustrated), and utilized it to prove the Pythagorean theorem. This illustration is referred to as the \"Gougu Diagram\". If the area of the small square within the \"Gougu Diagram\" is 1, and the area of each right-angled triangle is $6$, and $\\alpha$ is one of the acute angles in the right-angled triangle, then what is the value of $\\tan \\alpha$?\n",
    "completion": "\\textbf{Solution:} Let the length of the shorter leg in the right-angled triangle be $x$, then the longer leg is $x+1$. $\\because$ The area of each right-angled triangle is 6. $\\therefore$ $\\frac{1}{2}x(x+1)=6$\\\\\nSolving gives $x_{1}=3$, $x_{2}=-4$ (discard). $\\therefore$ The shorter leg is 3, the longer leg is 4. $\\therefore$ $\\tan\\alpha = \\frac{4}{3}$\\\\\nThus, the answer is: $\\tan\\alpha = \\boxed{\\frac{4}{3}}$",
    "image_path": "dataset/mm_math/images/MM_Math/55497824.png"
  },
  {
    "solution": "\\boxed{3\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the figure, the quadrilateral $ABCD$ is inscribed in circle $\\odot O$, $\\angle ADC=60^\\circ$, and the radius of $\\odot O$ is $3\\,cm$. What is the length of the chord $AC$?\n",
    "completion": "\\textbf{Solution:} Draw $OA$ and $OC$, and construct $OE \\perp AC$ at $E$,\\\\\nsince $\\angle ADC=60^\\circ$,\\\\\nthus $\\angle AOC=2\\angle ADC=120^\\circ$,\\\\\nsince $OA=OC$,\\\\\nthus $\\angle AOE=\\angle COE=60^\\circ$, $AC=2AE$,\\\\\nthus $\\sin\\angle AOE=\\frac{AE}{AO}$,\\\\\nthus $AE=AO\\cdot \\sin\\angle AOE$,\\\\\nthus $AE=3\\sin60^\\circ=\\frac{3\\sqrt{3}}{2}$,\\\\\nthus $AC=2AE=\\boxed{3\\sqrt{3}}$,",
    "image_path": "dataset/mm_math/images/MM_Math/53149424.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point B is the midpoint of segment AD, C is on segment BD and satisfies BD = 3CD. If the sum of the lengths of all segments in the figure is 30, what is the length of segment BC?\n",
    "completion": "\\textbf{Solution:} Let $CD=x$, then $BC=2x$, $AB=BD=3x$,\\\\\nthe sum of all line segments is $AB+AC+AD+BC+BD+CD$\\\\\n$=AB+(AB+BC)+(AB+BD)+BC+BD+CD$\\\\\n$=3x+5x+6x+2x+3x+x$\\\\\n$=20x$,\\\\\nAccording to the problem, $20x=30$,\\\\\nsolving this gives $x=1.5$,\\\\\nthus, $BC=2 \\times 1.5=\\boxed{3}$,\\\\\nmeaning the length of segment BC is 3.",
    "image_path": "dataset/mm_math/images/MM_Math/51367686.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $C$ is a point on line segment $AB$, and points $M$ and $N$ are the midpoints of line segments $AC$ and $BC$, respectively. If point $P$ is the midpoint of line segment $AB$, $AC=6\\,\\text{cm}$, and $CP=2\\,\\text{cm}$, what is the length of line segment $PN$ in cm?\n",
    "completion": "\\textbf{Solution:} Since $AC=6\\text{cm}$, $CP=2\\text{cm}$,\\\\\nit follows that $AP=AC+CP=8\\text{cm}$,\\\\\nSince $P$ is the midpoint of segment $AB$,\\\\\nit follows that $AB=2AP=16\\text{cm}$,\\\\\nTherefore, $CB=AB-AC=16-6=10\\text{cm}$,\\\\\nSince $N$ is the midpoint of segment $CB$,\\\\\nit follows that $CN= \\frac{1}{2} CB=5\\text{cm}$,\\\\\nTherefore, $PN=CN-CP=5-2=\\boxed{3}\\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50810423_197.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ is an equilateral triangle, with $P$ and $Q$ located on sides $AC$ and $BC$ respectively, such that $AP=CQ$. Lines $AQ$ and $BP$ intersect at point $O$. Since $\\triangle ACQ \\cong \\triangle BAP$, we are asked to find the measure of $\\angle BOQ$.\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ACQ\\cong \\triangle BAP$,\\\\\nit follows that $\\angle CAQ=\\angle ABP$,\\\\\nthus $\\angle BOQ=\\angle BAO+\\angle ABP=\\angle BAO+\\angle CAQ=BAC=60^\\circ$;\\\\\ntherefore, the measure of $\\angle BOQ$ is $\\boxed{60^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52867084_154.png"
  },
  {
    "solution": "\\boxed{2+\\sqrt{6}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, $\\angle A=90^\\circ$, $AB=AD$, and $E$, $F$ are the midpoints of $AB$, $AD$, respectively. If $EF=\\sqrt{2}$, $BC=\\sqrt{11}$, and $CD=\\sqrt{3}$, then what is the area of quadrilateral $ABCD$?\n",
    "completion": "\\textbf{Solution:} Connect BD, \\\\\n$\\because$ E and F are the midpoints of AB and AD respectively, \\\\\n$\\therefore$ BD=2EF=2$\\sqrt{2}$, \\\\\n$\\because$ $\\angle$A=90$^\\circ$ and AB=AD, \\\\\n$\\therefore$ 2AB$^{2}$=BD$^{2}$=8, \\\\\n$\\therefore$ AB=AD=2, \\\\\n$\\therefore$ Area of $\\triangle ABD$=$\\frac{1}{2}$×2×2=2, \\\\\n$\\because$ BC=$\\sqrt{11}$, CD=$\\sqrt{3}$, and BD=2$\\sqrt{2}$, \\\\\n$\\therefore$ BC$^{2}$=CD$^{2}$+BD$^{2}$, \\\\\n$\\therefore$ $\\angle$BDC=90$^\\circ$, \\\\\n$\\therefore$ Area of $\\triangle ABD$=$\\frac{1}{2}$×$\\sqrt{3}$×2$\\sqrt{2}$=$\\sqrt{6}$, \\\\\n$\\therefore$ Area of the quadrilateral ABCD = Area of $\\triangle ABD$ + Area of $\\triangle ABD$ = 2 + $\\sqrt{6}$. \\\\\nTherefore, the answer is: $S_{\\text{quadrilateral} ABCD} = \\boxed{2+\\sqrt{6}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51998807.png"
  },
  {
    "solution": "\\boxed{\\frac{4}{9}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ and $\\triangle A'B'C'$ are similar figures with the point $O$ as the center of similitude. If $OA:OA'=2:3$, what is the ratio of the areas of $\\triangle ABC$ and $\\triangle A'B'C'$?\n",
    "completion": "\\textbf{Solution:} Given that $\\triangle ABC$ and $\\triangle A'B'C'$ are similar figures with point O as the center of similitude,\\\\\nit follows that $\\triangle ABC \\sim \\triangle A'B'C'$, $AC \\parallel A'C'$,\\\\\ntherefore $\\angle ACO = \\angle A'C'O$, $\\angle CAO = \\angle C'A'O$,\\\\\nhence $\\triangle AOC \\sim \\triangle A'OC'$,\\\\\nthus $AC : A'C' = OA : OA' = 2 : 3$,\\\\\ntherefore $\\frac{S_{\\triangle ABC}}{S_{\\triangle A'B'C'}} = \\left(\\frac{AC}{A'C'}\\right)^2 = \\boxed{\\frac{4}{9}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51444993.png"
  },
  {
    "solution": "\\boxed{\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $E$ is a point on side $CD$. The triangle $ADE$ is folded along $AE$ so that point $D$ precisely falls on point $F$ on side $BC$. $\\triangle ABF \\sim \\triangle FCE$; if $AB = 3\\sqrt{3}$ and $AD = 6$, what is the length of $EC$?\n",
    "completion": "\\textbf{Solution:} By folding, we have $AF=AD=6$,\\\\\n$\\therefore BF=\\sqrt{AF^{2}-AB^{2}}=3$,\\\\\n$\\therefore FC=6-3=3$,\\\\\n$\\because \\triangle ABF\\sim \\triangle FCE$,\\\\\n$\\therefore \\frac{AB}{FC}=\\frac{BF}{EC}$,\\\\\n$\\frac{3\\sqrt{3}}{3}=\\frac{3}{EC}$,\\\\\n$\\therefore EC=\\boxed{\\sqrt{3}}$",
    "image_path": "dataset/mm_math/images/MM_Math/52866379_113.png"
  },
  {
    "solution": "\\boxed{y=-x^{2}+2x+3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, in the Cartesian coordinate system, the parabola $y=ax^2-2ax-3a$ ($a<0$) intersects the x-axis at points A and B (A is to the left of B), and intersects the y-axis at point C. Point D is the vertex of the parabola, and the axis of symmetry intersects the x-axis at point E, with $OB=OC$. Find the analytical expression of the parabola?\n",
    "completion": "\\textbf{Solution:} To solve for the parabola $y=ax^{2}-2ax-3a$, let $y=0$ to get $ax^{2}-2ax-3a=0$. Solving this yields $x=-1$ or $3$, \\\\\n\\[\\therefore \\boxed{A(-1,0)}, \\boxed{B(3,0)},\\] \\\\\n\\[\\therefore OA=1, OB=OC=3,\\] \\\\\n\\[\\therefore \\boxed{C(0,3)},\\] \\\\\n\\[\\therefore -3a=3,\\] \\\\\n\\[\\therefore a=-1,\\] \\\\\n\\[\\therefore\\] the equation of the parabola is \\boxed{y=-x^{2}+2x+3}.",
    "image_path": "dataset/mm_math/images/MM_Math/51367443_123.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $OABC$, we have $OA=3$ and $OC=4$. The lines containing $OA$ and $OC$ are taken as the $x$-axis and $y$-axis, respectively, establishing a Cartesian coordinate system. Let $D$ be a variable point on side $CB$ (not coinciding with $C$ or $B$), and the graph of the inverse proportion function $y=\\frac{k}{x}$ $(k>0)$ passes through point $D$ and intersects side $BA$ at point $E$. Construct line $DE$. Find the value of $\\frac{BD}{BE}$?\n",
    "completion": "Solution: Since points D and E lie on the graph of the inverse proportion function $y=\\frac{k}{x}$ ($k>0$),\\\\\nit follows that the coordinates of D are $\\left(\\frac{k}{4}, 4\\right)$, and the coordinates of E are $\\left(3, \\frac{k}{3}\\right)$.\\\\\nGiven also that the coordinates of point B are (3, 4),\\\\\nit follows that $BD=3-\\frac{k}{4}$ and $BE=4-\\frac{k}{3}$,\\\\\ntherefore, $\\frac{BD}{BE}=\\frac{3-\\frac{k}{4}}{4-\\frac{k}{3}}=\\boxed{\\frac{3}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52934017_239.png"
  },
  {
    "solution": "\\boxed{(2,6)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=ax^2-a$ ($a>0$) intersects the x-axis at points $A$ and $B$ (point $A$ is to the left of point $B$), and intersects the y-axis at point $C$, with $OC=2OA$. Is there a point $P$ on the parabola such that the incenter of $\\triangle PAC$ lies on the x-axis? If it exists, find the coordinates of point $P$; if not, please explain why.\n",
    "completion": "\\textbf{Solution:} Existence is proven,\\\\\nAssume there exists a point $P$ such that the intersection of the angle bisectors of the three interior angles of $\\triangle PAC$ lies on the $x$-axis, then the $x$-axis is the angle bisector of $\\angle PAC$.\\\\\n$\\therefore$ The reflection point of $C$ about the $x$-axis must lie on line $PA$. Denote it as $C'$,\\\\\n$\\because$ $C(0, -2)$,\\\\\n$\\therefore$ $C'(0, 2)$,\\\\\n$\\therefore$ Line $AP$ passes through $A(-1, 0), C'(0, 2)$,\\\\\nLet the equation of line $AP$ be $y=kx+2$,\\\\\n$\\therefore$ $-k+2=0$, solving yields $k=2$,\\\\\n$\\therefore$ The equation of line $AP$ is $y=2x+2$,\\\\\n$\\because$ Line $AP$ intersects the quadratic function $y=2x^2-2$ at point $P$,\\\\\n$\\therefore$ $2x+2=2x^2-2$,\\\\\nSolving gives: $x=2$ or $x=-1$,\\\\\nWhen $x=2$, $y=6$,\\\\\nWhen $x=-1$, $y=0$, which corresponds to point $A$,\\\\\n$\\therefore$ There exists a point $P$ such that the intersection of the angle bisectors of the three interior angles of $\\triangle PAC$ lies on the $x$-axis, and the coordinates of point $P$ are $\\boxed{(2,6)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52917460_255.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, AB is the diameter of $\\odot O$, point C is on the extension line of AB, AD bisects $\\angle CAE$ and intersects $\\odot O$ at point D, and AE $\\perp$ CD at point E. Line CE is a tangent to $\\odot O$. If BC = 3 and CD = 3$\\sqrt{2}$, find the length of radius OB and line segment AE.\n",
    "completion": "\\textbf{Solution:} Let $OD=OB=x$. In $\\triangle COD$, $OD^2+CD^2=OC^2$, \\\\\n$\\therefore x^2+(3\\sqrt{2})^2=(x+3)^2$, \\\\\n$\\therefore x=\\boxed{1.5}$, \\\\\n$\\therefore OB=1.5$, \\\\\n$\\because OD\\parallel AE$, \\\\\n$\\therefore \\frac{OD}{AE}=\\frac{OC}{AC}$, \\\\\n$\\therefore \\frac{1.5}{AE}=\\frac{4.5}{6}$, \\\\\n$\\therefore AE=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52936361_162.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given points $C$ and $D$ are on the line segment $AB$, where $AB=22$ and $CD=8$. If point $C$ is the midpoint of line segment $AB$, find the length of $BD$.\n",
    "completion": "\\textbf{Solution:} Since $AB=22$ and $C$ is the midpoint of $AB$,\\\\\nit follows that $BC=11$,\\\\\nSince $CD=8$,\\\\\nit follows that $BD=BC-CD=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52837995_154.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in rectangle $ABCD$, $AC$ intersects $BD$ at point $O$. If $\\angle COB=120^\\circ$ and $AB=6$, what is the length of the diagonal $BD$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rectangle,\\\\\n$ \\therefore BD = 2OB$, $AC = 2OA$, $AC = BD$,\\\\\n$ \\therefore OA = OB$,\\\\\nSince $\\angle BOC = 120^\\circ$,\\\\\n$ \\therefore \\angle AOB = 60^\\circ$,\\\\\n$ \\therefore \\triangle AOB$ is an equilateral triangle,\\\\\n$ \\therefore OB = AB = 6$,\\\\\n$ \\therefore BD = 2OB = \\boxed{12}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52284176.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y_{1}=-x-1$ intersects the x-axis at point A and the y-axis at point B. It has a point of intersection, M($-2$, m), with the graph of the inverse proportional function $y_{2}= \\frac{k}{x}$. What is the area of $\\triangle MOB$?\n",
    "completion": "\\[ \\textbf{Solution:} \\text{Given the problem, when } x=0, \\text{ we have } y=-1, \\text{ meaning the coordinates of B are }(0, -1), \\text{ thus } OB=1, \\]\n\\[ \\text{Since the coordinates of M are } (-2, 1), \\]\n\\[ \\text{Therefore, the distance from point M to line OB is } 2, \\]\n\\[ \\text{Therefore, the area of }\\triangle MOB \\text{ is } \\frac{1}{2} \\times 1 \\times 2 = \\boxed{1}. \\]",
    "image_path": "dataset/mm_math/images/MM_Math/51457952_273.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in parallelogram $ABCD$, $AB=8$, $DE\\perp AB$ with the foot of the perpendicular at E, $BE=6$. Let $CE$ be connected. If $\\angle CEB=\\angle ADE$, then what is the length of $DE$?\n",
    "completion": "\\textbf{Solution:}\n\nGiven: In quadrilateral $ABCD$, $AB=8$, $BE=6$,\\\\\nthus $AB\\parallel CD$, $CD=AB=8$, $AE=2$,\\\\\nthus $\\angle CDE=\\angle AED, \\angle DCE=\\angle CEB$,\\\\\ngiven that $\\angle CEB=\\angle ADE$,\\\\\nthus $\\angle DCE=\\angle ADE$,\\\\\ntherefore, $\\triangle ADE \\sim \\triangle ECD$,\\\\\nthus $\\frac{AE}{DE}=\\frac{DE}{CD}$,\\\\\nthus $\\frac{2}{DE}=\\frac{DE}{8}$,\\\\\nthus $DE^2=16$,\\\\\nthus $DE=\\pm 4$,\\\\\nUpon verification: $DE=4$ is consistent with the problem, $DE=-4$ is not consistent and is discarded,\\\\\ntherefore, $DE=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51379211.png"
  },
  {
    "solution": "\\boxed{\\frac{27}{10}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in quadrilateral $ABCD$, $AB \\parallel DC$ and $AB=DC$. The diagonals $AC$ and $BD$ intersect at point $O$. $AC$ bisects $\\angle BAD$, and a line is drawn from point $C$ perpendicular to $AB$, intersecting the extension of $AB$ at point $E$. Line $OE$ is then drawn. Quadrilateral $ABCD$ is a rhombus. If $AB = \\sqrt{10}$ and $BD = 2$, what is the area of $\\triangle AOE$?\n",
    "completion": "\\textbf{Solution:} Thus, the area of $\\triangle AOE$ is $\\boxed{\\frac{27}{10}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52866241_125.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $ \\angle AOD=60^\\circ$, $ \\angle BOC$ is a right angle, and OC bisects $ \\angle AOD$. What is the measure of $ \\angle BOD$ in degrees?\n",
    "completion": "\\textbf{Solution:} \\\\\n$\\because$ OC bisects $\\angle AOD$,\\\\\n$\\therefore \\angle COD=\\frac{1}{2}\\angle AOD$,\\\\\n$\\because \\angle AOD=60^\\circ$,\\\\\n$\\therefore \\angle COD=30^\\circ$,\\\\\n$\\because \\angle BOC$ is a right angle,\\\\\n$\\therefore \\angle BOC=90^\\circ$,\\\\\n$\\therefore \\angle BOD=\\angle BOC-\\angle COD=90^\\circ-30^\\circ=\\boxed{60^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/46945329.png"
  },
  {
    "solution": "\\boxed{-3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given the partial graph of the parabola $y = -x^2 + bx - c$ as shown in the figure. Find the values of $b$ and $c$.\n",
    "completion": "Solution: Substituting $(1, 0)$ into $y = -x^2 + bx - c$ yields\n\\[\n\\left\\{\n\\begin{array}{l}\n-1 + b + c = 0 \\\\\n-c = 3\n\\end{array}\n\\right.\n\\]\nSolving yields $b = \\boxed{-2}$, $c = \\boxed{-3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52693909_97.png"
  },
  {
    "solution": "\\boxed{\\left(-7,-\\frac{16}{3}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=\\frac{1}{3}x^{2}+\\frac{8}{3}x-3$ intersects the x-axis at points $A$ and $B$, and intersects the y-axis at point $C$. Point $D$ is a moving point on the parabola within the third quadrant. When $\\angle ACD+2\\angle ABC=180^\\circ$, what are the coordinates of point $D$?\n",
    "completion": "\\textbf{Solution:} See figure below. \\\\\nWhen $y=0$, $\\frac{1}{3}x^{2}+\\frac{8}{3}x-3=0$, \\\\\nSolving this, we get: $x=-9$ or $1$, \\\\\n$\\therefore$ A$(-9,0)$, B$(1,0)$, \\\\\n$\\therefore$ AB$=10$, \\\\\nWhen $x=0$, $y=3$, \\\\\n$\\therefore$ C$(0,-3)$, \\\\\n$\\because$ AC$^{2}=9^{2}+3^{2}=90$, $BC^{2}=1^{2}+3^{2}=10$, $AB^{2}=100$, \\\\\n$\\therefore$ AC$^{2}+BC^{2}=AB^{2}$, \\\\\n$\\therefore$ $\\angle$ACB$=90^\\circ$, \\\\\n$\\therefore$ $\\angle$BAC$+\\angle$ABC$=90^\\circ$, \\\\\n$\\therefore$ $2\\angle$BAC$+2\\angle$ABC$=180^\\circ$, \\\\\n$\\because$ $\\angle$ACD$+2\\angle$ABC$=180^\\circ$ \\\\\n$\\therefore$ $2\\angle$BAC$=\\angle$ACD, \\\\\nConstruct AE$\\perp$x-axis, intersecting the extension of CD at E, and construct the bisector of $\\angle$ACD, intersecting AE at F, then $\\angle$ACF$=\\angle$BAC, \\\\\n$\\therefore$ CF$\\parallel$AB, \\\\\n$\\therefore$ CF$\\perp$AE, \\\\\n$\\therefore$ AF$=EF=BC=3$, \\\\\n$\\therefore$ E$(-9,-6)$, \\\\\nAssume the equation of the line CD is $y=kx-3$, \\\\\nSubstituting the coordinates of E, we get $-6=-9k-3$, \\\\\n$\\therefore$ $k=\\frac{1}{3}$, \\\\\n$\\therefore$ The equation of line CD is $y=\\frac{1}{3}x-3$, \\\\\n$\\therefore$ $\\left\\{\\begin{array}{l}y=\\frac{1}{3}x-3\\\\ y=\\frac{1}{3}x^{2}+\\frac{8}{3}x-3\\end{array}\\right.$ \\\\\nSolving these, we get: $\\left\\{\\begin{array}{l}x_{1}=0\\\\ y_{1}=-3\\end{array}\\right.$, $\\left\\{\\begin{array}{l}x_{2}=-7\\\\ y_{2}=-\\frac{16}{3}\\end{array}\\right.$ \\\\\n$\\therefore$ The point D is $\\boxed{\\left(-7,-\\frac{16}{3}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51369891.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, point $C$ is a moving point on $AB$. On the same side of $AB$, isosceles right triangles $ACD$ and $CBE$ are constructed with $AC$ and $BC$ as hypotenuses, respectively. Line segment $DE$ is drawn, and point $F$ is located on $DE$ with $CF$ connected and $FD=FC$. Point $F$ is the midpoint of $DE$. A perpendicular line to $AB$ is drawn from point $F$, and point $G$ is the foot of the perpendicular. What is the value of $\\frac{FG}{AB}$?\n",
    "completion": "\\textbf{Solution:} Construct $DM \\perp AB$ at $M$ through point $D$, and construct $EN \\perp AB$ at $N$ through point $E$,\\\\\nsince $\\triangle ADC$ and $\\triangle BCE$ are isosceles right triangles,\\\\\ntherefore $AC=2DM$, $BC=2EN$,\\\\\nthus $AB=AC+BC=2(DM+EN)$,\\\\\nsince $FG\\perp AB$,\\\\\nthus $DM\\parallel FG\\parallel EN$,\\\\\nsince $DF=EF$,\\\\\nthus $MG=NG$,\\\\\ntherefore $FG=\\frac{1}{2}(DM+EN)$,\\\\\ntherefore $\\frac{FG}{AB}=\\frac{\\frac{1}{2}(DM+EN)}{2(DM+EN)}=\\boxed{\\frac{1}{4}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/55588163_85.png"
  },
  {
    "solution": "\\boxed{\\left(0, 0\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $x=t$ intersects the line $y=-x$ and the line $y=\\frac{1}{2}x+2$ at points E and D (D is above E), respectively. Point N is a moving point on the y-axis, and $\\triangle NDE$ is an isosceles right triangle. Directly write the value of $t$ and the coordinates of point N.\n",
    "completion": "\\textbf{Solution:} Write directly the value of $t$ and the coordinates of point $N$.\\\\\nWhen $t=-\\frac{4}{5}$, the coordinates of point N are $\\left(0, \\frac{8}{5}\\right)$ or $\\left(0, \\frac{4}{5}\\right)$;\\\\\nWhen $t=-\\frac{4}{7}$, the coordinates of point N are $\\left(0, \\frac{8}{7}\\right)$;\\\\\nWhen $t=4$, $\\triangle NDE$ is an isosceles right triangle, at this point the coordinates of N are \\boxed{\\left(0, 0\\right)}.",
    "image_path": "dataset/mm_math/images/MM_Math/52395484_114.png"
  },
  {
    "solution": "\\boxed{\\frac{15}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, a right-angled triangle board is placed on line $AD$ (right-angled triangle board $OBC$ and right-angled triangle board $MON$, with $\\angle OBC=90^\\circ$, $\\angle BOC=45^\\circ$, $\\angle MON=90^\\circ$, $\\angle MNO=30^\\circ$). Keeping the triangle board $OBC$ stationary, the triangle board $MON$ is rotated clockwise around point $O$ at a speed of $6^\\circ$ per second until the side $OM$ coincides with the line $AD$ for the first time. The rotation time is recorded as $t$ seconds. At what value of $t$ does $OM$ bisect $\\angle AOC$?\n",
    "completion": "\\textbf{Solution}: When $t = \\boxed{\\frac{15}{4}}$ seconds, $OM$ bisects $\\angle AOC$.",
    "image_path": "dataset/mm_math/images/MM_Math/53833477_95.png"
  },
  {
    "solution": "\\boxed{22.5^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the two diagonals $AC, BD$ of the square $ABCD$ intersect at point $O$, and point $E$ is on $BD$, with $BE=BC$. What is the degree measure of $\\angle ACE$?\n",
    "completion": "\\textbf{Solution:} In square $ABCD$, $\\angle CBD=\\angle ACB=45^\\circ$,\\\\\n$\\because BE=BC$,\\\\\n$\\therefore \\angle BCE=\\frac{1}{2}\\left(180^\\circ-\\angle CBD\\right)=67.5^\\circ$,\\\\\n$\\therefore \\angle ACE=\\angle BCE-\\angle ACB=\\boxed{22.5^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52189656.png"
  },
  {
    "solution": "\\boxed{70^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in $\\triangle ABC$, point D is on the extension of line $BC$. A line through point D, $DF\\perp AC$ at point F, is extended to intersect line $AB$ at point E and the bisector of $\\angle ACB$ at point N. Point M is the intersection of $CN$ and $AB$, with $\\angle BMC=80^\\circ$ and $\\angle B=40^\\circ$. What is the measure of $\\angle AEF$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle BMC=80^\\circ$ and $\\angle B=40^\\circ$, \\\\\nit follows that $\\angle BCM=180^\\circ-\\angle B-\\angle BMC=60^\\circ$ and $\\angle AMC=180^\\circ-\\angle BMC=100^\\circ$. \\\\\nSince $DF\\perp AC$, \\\\\nit implies that $\\angle EFC=90^\\circ$. \\\\\nSince $CN$ bisects $\\angle ACB$, \\\\\nit implies that $\\angle ACN=\\angle BCN=60^\\circ$. \\\\\nHence, $\\angle MEF=360^\\circ-\\angle AMC-\\angle ACM-\\angle EFC=110^\\circ$, \\\\\nthus, $\\angle AEF=180^\\circ-\\angle MEF=\\boxed{70^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53066397_124.png"
  },
  {
    "solution": "\\boxed{\\frac{180}{11}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Since you have not provided a specific text to translate, I'm unable to fulfill this request accurately. Please provide the text you want translated to English in standard LaTeX format, and I would be happy to assist.\n",
    "completion": "\\textbf{Solution:} Let point P be on AB, as shown in Figure 3. When $PO=PQ$, draw $PE\\perp x$-axis at E, draw $PF\\perp OA$ at F\\\\\n$\\therefore \\angle OPE= \\frac{1}{2}\\angle OPQ = \\frac{1}{2}\\angle OAB$,\\\\\n$\\because PE\\parallel OA$,\\\\\n$\\therefore \\angle AOP=\\angle OPE= \\frac{1}{2} \\angle OAB$,\\\\\nConstruct $\\angle OPG=\\angle AOP$,\\\\\n$\\therefore \\angle APG=2\\angle AOP=\\angle OAB$,\\\\\n$\\because \\tan\\angle OAB= \\frac{PF}{AF} = \\frac{4}{3}$,\\\\\n$\\therefore$ let $PF=4a$, $AF=3a$, $OG=PG=AP=5a$,\\\\\n$\\therefore OF=OG+FG=8a$,\\\\\n$\\therefore P(4a,8a)$, $B(8,4)$,\\\\\n$\\therefore$ the equation of line AB is: $y= -\\frac{3}{4}x+10$,\\\\\n$\\therefore -\\frac{3}{4} \\times 4a+10=8a$,\\\\\n$\\therefore a= \\frac{10}{11}$,\\\\\n$\\therefore PF=4a= \\frac{40}{11}$,\\\\\n$\\therefore OQ=2PF= \\boxed{\\frac{80}{11}}$,\\\\\nAs shown in Figure 4, when $OQ=PQ$, draw $PE\\perp OP$ at E,\\\\\n$\\therefore \\angle POQ=\\angle OPQ=\\angle OAB$, $OE= \\frac{1}{2} OP$,\\\\\n$\\because \\angle AOP+\\angle POQ=90^\\circ$,\\\\\n$\\therefore \\angle OAB+\\angle AOP=90^\\circ$,\\\\\n$\\therefore \\angle APO=90^\\circ$,\\\\\n$\\because \\tan\\angle OAB= \\frac{4}{3}$,\\\\\n$\\therefore$ let $AP=3a$, $OP=4a$,\\\\\n$\\therefore OA=5a=10$,\\\\\n$\\therefore a=2$,\\\\\n$\\therefore OP=4a=8$,\\\\\n$\\therefore OE=4$,\\\\\n$\\because \\frac{OQ}{OE} = \\frac{5}{3}$,\\\\\n$\\therefore OQ= \\boxed{\\frac{20}{3}}$,\\\\\nWhen P is on CD, as in Figure 5, when $OQ=PQ$, draw $QE\\perp OP$ at E,\\\\\nFrom above, we know: $OP\\perp CD$, $OP=8+10=18$,\\\\\n$\\therefore OE=9$,\\\\\n$\\therefore OQ=9\\times \\frac{5}{3} =\\boxed{15}$,\\\\\nAs in Figure 6, when $PQ=PO$, draw $PE\\perp OQ$ at E,\\\\\n$\\because$ line AB: $y= -\\frac{3}{4}x+10$, $CD\\parallel AB$, $C(14,12)$,\\\\\n$\\therefore$ the equation of line CD is: $y= -\\frac{3}{4}x+ \\frac{45}{2}$,\\\\\nLet $P(a,2a)$,\\\\\n$\\therefore -\\frac{3}{4} \\times a+ \\frac{45}{2} =2a$,\\\\\n$\\therefore 2a= \\frac{180}{11}$,\\\\\n$\\therefore OQ=2a= \\boxed{\\frac{180}{11}}$,\\\\\nOverall, during the entire movement of point P, when $\\triangle OPQ$ is an isosceles triangle with $PQ$ as the base, $OQ= \\boxed{\\frac{80}{11}}$ or $\\boxed{\\frac{20}{3}} or \\boxed{15} or \\boxed{\\frac{180}{11}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51317370_136.png"
  },
  {
    "solution": "\\boxed{2+\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the radius of $\\odot O$ is $\\sqrt{5}$. AB and CD are two parallel chords of $\\odot O$, with $\\angle CDE=30^\\circ$ and $AD=2$. What is the length of chord BE?\n",
    "completion": "\\textbf{Solution:} Connect OC, OE, BC, CE, and draw CH$\\perp$BE intersecting BE at point H,\\\\\n$\\because$AB$\\parallel$CD,\\\\\n$\\therefore$BC = AD = 2,\\\\\n$\\because$$\\angle$CDE = $30^\\circ$,\\\\\n$\\therefore$$\\angle$COE= $60^\\circ$, $\\angle$CBE = $\\angle$CDE = $30^\\circ$,\\\\\n$\\therefore$$\\triangle$OCE is an equilateral triangle,\\\\\n$\\therefore$CE= $\\sqrt{5}$, in $Rt\\triangle BCH$,\\\\\n$CH=\\frac{1}{2}BC=1$\\\\\n$\\therefore$BH= $\\sqrt{2^2-1^2}=\\sqrt{3}$, in $Rt\\triangle$CEH\\\\\nHE= $\\sqrt{(\\sqrt{5})^2-1^2}=2$\\\\\n$\\therefore$BE=2+ $\\sqrt{3}$.\\\\\nTherefore, the answer is: BE=\\boxed{2+\\sqrt{3}}.",
    "image_path": "dataset/mm_math/images/MM_Math/52950180.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within the Cartesian coordinate system, the vertex C of the parabola passing through points $A(0, 4)$ and $B(5, 9)$ is located on the positive half of the x-axis. Let $P(x, y)$ be a point on segment AB, where $1\\le x\\le 4$. Construct $PM\\parallel y$-axis to intersect the parabola at point M. Find the maximum and minimum values of PM.\n",
    "completion": "\\textbf{Solution:} Let the equation of line $AB$ be: $y=kx+b$, \\\\\nSubstituting points $A\\left(0, 4\\right)$ and $B\\left(5, 9\\right)$ into $y=kx+b$ yields: $\\left\\{\\begin{array}{l}b=4 \\\\ 5k+b=9\\end{array}\\right.$, \\\\\nSolving this, we get: $\\left\\{\\begin{array}{l}k=1 \\\\ b=4\\end{array}\\right.$, \\\\\nTherefore, the equation of line $AB$ is: $y=x+4$, \\\\\nSince point $P$ is on the line segment $AB$ and point $M$ is on the parabola $y=\\left(x−2\\right)^{2}$, with $1\\le x\\le 4$, and $PM\\parallel y$ axis, \\\\\nTherefore, when $x=1$, $P\\left(1, 5\\right)$, $M\\left(1, 5\\right)$, \\\\\nThus, $PM=5−1=4$, \\\\\nWhen $x=4$, $P\\left(4, 8\\right)$, $M\\left(4, 4\\right)$, \\\\\nThus, $PM=8−4=4$, \\\\\nWhen $x=2$, $P\\left(2, 6\\right)$, $M\\left(2, 0\\right)$, \\\\\nThus, $PM=6−0=6$, \\\\\nLet $P\\left(n, n+4\\right)$, $M\\left(n, n^{2}−4n+4\\right)$, \\\\\nThus, $PM=n+4−\\left(n^{2}−4n+4\\right)=−n^{2}+5n=−\\left(n−\\frac{5}{2}\\right)^{2}+\\frac{25}{4}$, \\\\\nTherefore, when $n=\\frac{5}{2}$, the maximum value of $PM$ is: $\\boxed{\\frac{25}{4}}$, \\\\\nTherefore, the maximum value of $PM$ is $\\frac{25}{4}$, and the minimum value is $\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52514156_167.png"
  },
  {
    "solution": "\\boxed{22.5^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, let $ABCD$ be a square, and let points $E$ and $F$ be on the diagonal $AC$. Given that $\\angle EBF = 45^\\circ$, when lines $BE$ and $BF$ are connected, triangles $ABE$ and $GBE$ are symmetric about line $BE$. Point $G$ is on $BD$, and line $FG$ is connected. What is the measure of $\\angle FBC$?\n",
    "completion": "\\textbf{Solution:} Let the intersection point of the diagonals be $O$.\\\\\n$\\because \\angle ABO=\\angle EBF=45^\\circ$\\\\\n$\\therefore \\angle ABE+\\angle EBO=\\angle EBO+\\angle DBF$\\\\\n$\\therefore \\angle ABE=\\angle DBF$\\\\\n$\\because \\triangle ABE$ and $\\triangle GBE$ are symmetric with respect to the line $BE$\\\\\n$\\therefore \\angle ABE=\\angle EBO$\\\\\n$\\therefore \\angle EBO=\\angle OBF=\\frac{1}{2}\\angle EBF=22.5^\\circ$\\\\\n$\\therefore \\angle FBC=45^\\circ-\\angle OBF=\\boxed{22.5^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/52065370_99.png"
  },
  {
    "solution": "\\boxed{75^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a set of triangular boards are stacked together as depicted. What is the degree measure of $\\angle \\alpha$ in the figure?\n",
    "completion": "\\textbf{Solution:} As shown in the figure,\\\\\n$\\because$ $\\angle DBC = 90^\\circ$ and $\\angle C = 45^\\circ$ and $\\angle A = 30^\\circ$,\\\\\n$\\therefore$ $\\angle BDC = 90^\\circ - \\angle C = 45^\\circ$,\\\\\n$\\therefore$ $\\angle ADE = \\angle BDC = 45^\\circ$,\\\\\n$\\therefore$ $\\angle \\alpha = \\angle A + \\angle ADE = 30^\\circ + 45^\\circ = \\boxed{75^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53033966.png"
  },
  {
    "solution": "\\boxed{\\frac{4}{7}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the centroid of $\\triangle ABC$ is $G$, and point $D$ is on the extension line of $CB$, with $BD = \\frac{1}{2}BC$. The line passing through points $D$ and $G$ intersects $AC$ at point $E$. What is the value of $\\frac{AE}{AC}$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, extend CG and intersect AB at F, extend AG and intersect BC at H, connect FH which intersects DE at N,\\\\\n$\\because$ G is the centroid of $\\triangle ABC$,\\\\\n$\\therefore$ $FH$ is the midline of $\\triangle ABC$,\\\\\n$\\therefore$ $FH\\parallel AC$,\\\\\n$\\because$ $BD=\\frac{1}{2}BC$,\\\\\n$\\therefore$ $BD=BH=CH$,\\\\\n$\\because$ $NH\\parallel EC$,\\\\\n$\\therefore$ $\\frac{DH}{DC}=\\frac{NH}{EC}=\\frac{2}{3}$,\\\\\n$\\therefore$ $EC=\\frac{3}{2}NH$,\\\\\n$\\because$ $NH\\parallel AE$,\\\\\n$\\therefore$ $\\frac{NH}{EA}=\\frac{HG}{AG}=\\frac{1}{2}$, thus $AE=2NH$,\\\\\n$\\therefore$ $\\frac{AE}{CE}=\\frac{2NH}{\\frac{3}{2}NH}=\\frac{4}{3}$,\\\\\n$\\therefore$ $\\frac{AE}{AC}=\\boxed{\\frac{4}{7}}$",
    "image_path": "dataset/mm_math/images/MM_Math/52871901.png"
  },
  {
    "solution": "\\boxed{80^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $\\triangle ABC \\cong \\triangle ADE$. If $\\angle E = 70^\\circ$ and $\\angle D = 30^\\circ$, what is the degree measure of $\\angle BAC$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle E=70^\\circ$ and $\\angle D=30^\\circ$,\\\\\nit follows that $\\angle EAD=180^\\circ-\\angle E-\\angle D=180^\\circ-70^\\circ-30^\\circ=80^\\circ$,\\\\\nSince $\\triangle ABC \\cong \\triangle ADE$,\\\\\nit follows that $\\angle BAC=\\angle EAD=\\boxed{80^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/52720421.png"
  },
  {
    "solution": "\\boxed{20}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, O is the intersection point of the diagonals of rectangle ABCD, DE $\\parallel$ AC, and CE $\\parallel$ BD. Quadrilateral OCED is a rhombus. If AB = 6 and BC = 8, find the perimeter of the quadrilateral OCED.\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rectangle,\\\\\nit follows that $\\angle ABC=90^\\circ$,\\\\\nSince AB=6, and BC=8,\\\\\nthen in $\\triangle ABC$, by the Pythagorean theorem, we have: AC=10,\\\\\nwhich means OC$=\\frac{1}{2}$AC=5,\\\\\nSince quadrilateral OCED is a rhombus,\\\\\nit follows that OC=OD=DE=CE=5,\\\\\ntherefore, the perimeter of quadrilateral OCED is $5+5+5+5=\\boxed{20}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51968684_120.png"
  },
  {
    "solution": "\\boxed{110}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in $\\triangle ABC$, $\\angle BAC=90^\\circ$, $AB=3$, and $AC=4$. Squares $ABED$, $ACHI$, and $BCGF$ are constructed externally on sides $AB$, $AC$, and $BC$ of $\\triangle ABC$, respectively. Lines $ED$ and $HI$ intersect at point $J$. Line $KF$ is drawn through point $F$ parallel to $HI$ and intersects $DE$ at point $K$. Line $GM$ is drawn through point $G$ parallel to $DE$ and intersects $HI$ and $KF$ at points $M$ and $L$, respectively. What is the area of quadrilateral $KLMJ$?\n",
    "completion": "\\textbf{Solution:} In the right triangle $ABC$, $\\angle BAC=90^\\circ$, $AB=3$, $AC=4$. By the Pythagorean theorem, we have $BC=5$.\\\\\nSince quadrilaterals $ABED$, $ACHI$, $BCGF$ are all squares,\\\\\nit follows that each quadrilateral has four $90^\\circ$ angles, with four sides being parallel and equal.\\\\\nSince $KF\\parallel HI$ and $GM\\parallel DE$,\\\\\nit follows that $\\angle JKL=\\angle KJM=90^\\circ$.\\\\\nTherefore, quadrilaterals $KLMJ$ and $DGIA$ are rectangles.\\\\\nHence, $ED=DA=AB=3$, $AI=AC=IH=4$, $BC=BF=CG=5$.\\\\\nIt follows that $DJ=AI=4$, $JI=DA=3$.\\\\\nExtending $AC$ to $O$ gives $CO\\perp ML$.\\\\\nSince $\\angle ACB+\\angle GCO=90^\\circ=\\angle ACB+\\angle ABC$,\\\\\nit follows that $\\angle ABC=\\angle GCO$.\\\\\nSince $\\angle BAC=\\angle COG$ and $BC=CG$,\\\\\nit follows that $\\triangle ABC\\cong \\triangle OCG$ (AAS).\\\\\nHence, $AB=OC=3$.\\\\\nThus, $\\angle COG=90^\\circ$.\\\\\nSince $\\angle JML=90^\\circ$,\\\\\nit follows that quadrilateral $COMH$ is a rectangle.\\\\\nTherefore, $CO=HM=3$. Similarly, it can be shown that quadrilateral $EKQB$ is a rectangle.\\\\\nIt follows that $KE=QB=4$.\\\\\nHence, the area of quadrilateral $KLMJ$ is $HJ\\cdot JM=(KE+ED+DJ)\\cdot (JI+IH+HM)=(4+3+4)\\times (3+4+3)=\\boxed{110}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52344789.png"
  },
  {
    "solution": "\\boxed{7}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the area of $\\triangle ABF$ is equal to that of the quadrilateral $CDFE$, with point $E$ on side $BC$, $AE\\parallel CD$ intersecting $BD$ at point $F$, $CD=12$, $EF=9$. What is the length of $AF$?\n",
    "completion": "\\textbf{Solution:}\\\\\n$\\because$ The area of $\\triangle ABF$ is equal to the area of the quadrilateral $CDFE$,\\\\\n$\\therefore$ The area of $\\triangle BEA$ is equal to the area of $\\triangle BCD$,\\\\\n$\\because$ $AE\\parallel CD$,\\\\\n$\\therefore$ $\\triangle BEF\\sim \\triangle BCD$,\\\\\n$\\therefore$ $\\frac{S_{\\triangle BEF}}{S_{\\triangle BCD}}=\\left(\\frac{EF}{CD}\\right)^2$\\\\\n$\\because$ $CD=12$ and $EF=9$,\\\\\n$\\therefore$ $\\frac{S_{\\triangle BEF}}{S_{\\triangle BCD}}=\\left(\\frac{9}{12}\\right)^2=\\frac{9}{16}$,\\\\\n$\\therefore$ $\\frac{S_{\\triangle BEF}}{S_{\\triangle BEA}}=\\frac{9}{16}$\\\\\n$\\therefore$ $\\frac{EF}{EA}=\\frac{9}{16}$\\\\\n$\\therefore$ $\\frac{AF}{EF}=\\frac{7}{9}$\\\\\n$\\therefore$ $AF=\\frac{7}{9}EF=\\frac{7}{9}\\times 9 = \\boxed{7}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53080399.png"
  },
  {
    "solution": "\\boxed{20^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, extend the side BC of the rectangle ABCD to point E, such that CE = BD, and connect AE. If $\\angle ADB = 40^\\circ$, what is the degree measure of $\\angle E$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect AC and let the intersection of AC and BD be O,\\\\\n$\\because$ Quadrilateral ABCD is a rectangle,\\\\\n$\\therefore$ AC=BD, AD$\\parallel$BC, OA=OB=OC=OD,\\\\\nAlso, $\\because$ CE=BD,\\\\\n$\\therefore$ AC=CE,\\\\\n$\\therefore$ $\\angle$CAE=$\\angle$E,\\\\\n$\\because$ AD$\\parallel$BC,\\\\\n$\\therefore$ $\\angle$ADB=$\\angle$CBD=40$^\\circ$,\\\\\n$\\because$ OB=OC,\\\\\n$\\therefore$ $\\angle$ACB=$\\angle$CBD=40$^\\circ$,\\\\\n$\\therefore$ $\\angle$ACB=$\\angle$CAE+$\\angle$E=2$\\angle$E=40$^\\circ$,\\\\\n$\\therefore$ $\\angle$E=\\boxed{20^\\circ}.",
    "image_path": "dataset/mm_math/images/MM_Math/51770540_139.png"
  },
  {
    "solution": "\\boxed{127.5^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Assemble a set of triangles as shown in the figure, where points B, C, and D lie on the same straight line, $\\angle ACB=45^\\circ$, $\\angle DCE=60^\\circ$. If CM and CN bisect $\\angle ACB$ and $\\angle DCE$ respectively, what is the measure of $\\angle MCN$?\n",
    "completion": "\\textbf{Solution:} Hence, $\\angle MCN = \\boxed{127.5^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51438029_47.png"
  },
  {
    "solution": "\\boxed{CF = AF}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $E$ is a point on semicircle $O$, $C$ is the midpoint of arc $\\overset{\\frown}{BE}$, and diameter $AB \\parallel$ chord $DC$, intersecting $AE$ at point $F$. Prove that: $CF=AF$.\n",
    "completion": "\\textbf{Solution:} Proof: \\\\\n$\\because C$ is the midpoint of $\\overset{\\frown}{BE}$, \\\\\n$\\therefore \\angle EAC = \\angle CAB$, \\\\\n$\\because$ diameter AB$\\parallel$chord DC, \\\\\n$\\therefore \\angle DCA = \\angle CAB$, \\\\\n$\\therefore \\angle DCA = \\angle EAC$, \\\\\n$\\therefore \\boxed{CF = AF}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52733131_139.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{2}{3}x-2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the points $A(-6, 2)$ and $B(n, -4)$ are two intersection points of the linear function $y=kx+b$ and the inverse proportion function $y=\\frac{m}{x}$. Find the expressions for the inverse proportion function and the linear function.\n",
    "completion": "\\textbf{Solution:} Substitute $A(-6, 2)$ into $y=\\frac{m}{x}$, thus we get $m=2\\times(-6)=-12$, \\\\\n$\\therefore$ the inverse proportion function equation is $y=-\\frac{12}{x}$, \\\\\nSubstituting $B(n, -4)$ into $y=-\\frac{12}{x}$, it yields $-4=-\\frac{12}{n}$, \\\\\nSolving that gives $n=3$, \\\\\nSubstituting $A(-6, 2)$ and $B(3, -4)$ into $y=kx+b$, we get \\\\\n$\\begin{cases}\n-6k+b=2\\\\ \n3k+b=-4\n\\end{cases}$, resulting in: $\\begin{cases}\nk=-\\frac{2}{3}\\\\ \nb=-2\n\\end{cases}$, \\\\\n$\\therefore$ the linear function equation is $\\boxed{y=-\\frac{2}{3}x-2}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52602496_139.png"
  },
  {
    "solution": "\\boxed{y=-2x+2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the line $l_{1}$ passes through points $A(1,0)$ and $B(0,2)$. What is the analytical expression of line $l_{1}$?\n",
    "completion": "\\textbf{Solution:} Let the equation of line $l_{1}$ be $y=kx+b$. According to the problem, we have\n\\[\n\\left\\{\n\\begin{array}{l}\nk+b=0, \\\\\nb=2.\n\\end{array}\n\\right.\n\\]\nSolving these equations, we find\n\\[\n\\left\\{\n\\begin{array}{l}\nk=-2, \\\\\nb=2.\n\\end{array}\n\\right.\n\\]\n$\\therefore$ The equation of line $l_{1}$ is \\boxed{y=-2x+2}.",
    "image_path": "dataset/mm_math/images/MM_Math/50561277_49.png"
  },
  {
    "solution": "\\boxed{\\frac{21}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point C is on the line segment AB, with $AC=12\\,cm$ and $CB=9\\,cm$. M and N are the midpoints of AC and BC, respectively. What is the length of the line segment MN in cm?\n",
    "completion": "\\textbf{Solution:} Given that M and N are the midpoints of AC and BC, respectively,\n\n\\[\n\\therefore MC=\\frac{1}{2}AC, \\quad CN=\\frac{1}{2}CB,\n\\]\n\n\\[\n\\therefore MC+CN=\\frac{1}{2}AC+\\frac{1}{2}CB=\\frac{1}{2}(AC+CB),\n\\]\n\nwhich means $MN=\\frac{1}{2}(AC+CB)=\\frac{1}{2}\\times(12+9)=\\boxed{\\frac{21}{2}}\\text{cm}.$",
    "image_path": "dataset/mm_math/images/MM_Math/55429566_73.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $l_{1}\\colon y=-x-b$ intersects the $x$ and $y$ axes at points $A(6, 0)$ and $B$, respectively. The line $l_{2}$ passing through point $B$ intersects the negative semi-axis of the $x$ axis at point $C$, with $OB:OC=3:1$. Find the value of $S_{\\triangle AOB}-S_{\\triangle BOC}$.\n",
    "completion": "\\textbf{Solution:} Given by the problem, we have $S_{\\triangle AOB}-S_{\\triangle BOC}=\\frac{1}{2}\\cdot OA\\cdot OB-\\frac{1}{2}\\cdot OC\\cdot OB=\\frac{1}{2}\\times 6\\times 6-\\frac{1}{2}\\times 6\\times 2=\\boxed{12}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52354930_195.png"
  },
  {
    "solution": "\\boxed{40^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, line segment $AC$ intersects line segment $BD$ at point $O$. If $\\angle A = 70^\\circ$, $\\angle B = 30^\\circ$, $\\angle C = 60^\\circ$, what is the degree measure of $\\angle D$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle AOB = \\angle COD$,\\\\\nit follows that $\\angle A + \\angle B = \\angle C + \\angle D$,\\\\\ngiven $\\angle A = 70^\\circ$, $\\angle B = 30^\\circ$, and $\\angle C = 60^\\circ$,\\\\\nthus $70^\\circ + 30^\\circ = 60^\\circ + \\angle D$,\\\\\ntherefore, $\\angle D = \\boxed{40^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52950416.png"
  },
  {
    "solution": "\\boxed{\\frac{5}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the line \\(l\\) passes through points \\(A(0, 1)\\) and \\(B(-2, 0)\\). If point \\(M(3, m)\\) is on line \\(l\\), find the value of \\(m\\).\n",
    "completion": "\\textbf{Solution:} Since $M(3,m)$ lies on line $l$,\\\\\ntherefore $m=\\frac{1}{2}\\times3+1=\\boxed{\\frac{5}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51662963_203.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=\\frac{1}{2}x-2$ intersects the x-axis at point A. An isosceles right triangle $OAB$ is constructed above the x-axis with $OA$ as its hypotenuse. When the line is translated to the left along the x-axis until point B lies on the translated line, what is the distance by which the line has been translated?\n",
    "completion": "\\textbf{Solution:} As shown in the figure below,\n\nDraw a perpendicular line from B to the x-axis, with the foot at D, let the intersection of the translated line with the x-axis be C,\n\nFor the line $y=\\frac{1}{2}x-2$, let y=0, solving for x gives x=4, $\\therefore$ the coordinates of point A are $(4,0)$\n\n$\\therefore$ OA=4\n\n$\\because$ $\\triangle$OAB is an isosceles right triangle, BD$\\perp$x-axis\n\n$\\therefore$ it is easy to find that OD=2, BD=2\n\n$\\therefore$ B$(2,2)$;\n\nLet the equation of the translated line be: $y=\\frac{1}{2}x+b$, substituting B$(2,2)$ into it gives $2=1+b$, solving for b gives $b=1$,\n\nTherefore, the equation of the translated line is $y=\\frac{1}{2}x+1$, setting its y=0 gives\n\n$0=\\frac{1}{2}x+1$\n\nSolving it gives x=$-2$\n\n$\\therefore$ C$(0,-2)$,\n\n$\\therefore$ OC=2\n\n$\\therefore$ The distance of translation is OA+OC=4+2=\\boxed{6}.",
    "image_path": "dataset/mm_math/images/MM_Math/51662552.png"
  },
  {
    "solution": "\\boxed{36^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, regular pentagon ABCDE is inscribed in circle $O$, and point P is on the minor arc $\\overset{\\frown}{BC}$ (point P does not coincide with point C). What is the measure of $\\angle CPD$?\n",
    "completion": "\\textbf{Solution:} Draw lines OC and OD,\\\\\n$ \\because $ the regular pentagon ABCDE is inscribed in $\\odot$O,\\\\\n$ \\therefore \\angle COD=\\frac{360^\\circ}{5}=72^\\circ$,\\\\\n$ \\therefore \\angle CPD=\\frac{1}{2}\\angle COD=\\boxed{36^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52765389.png"
  },
  {
    "solution": "\\boxed{72^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, $E$ and $F$ are points on $CB$ and $CD$ respectively, such that $\\angle AEB = \\angle AFD$ and $BE = DF$. Quadrilateral $ABCD$ is a rhombus. If $\\triangle ABE \\cong \\triangle ADF$, what is the measure of $\\angle B$?\n",
    "completion": "\\textbf{Solution:} From (1), we know that $\\triangle ABE \\cong \\triangle ADF$, $\\therefore \\angle BAE = \\angle DAF$, $\\because \\triangle ABE \\cong \\triangle AEF$, $\\therefore \\angle BAE = \\angle EAF$, $AB = AE$, $\\therefore \\angle B = \\angle AEB$, let $\\angle B = \\angle AEB = x$, then $\\angle BAE = \\angle EAF = \\angle DAF = 180^\\circ - 2x$, $\\because$ quadrilateral $ABCD$ is a parallelogram, $\\therefore AD \\parallel BC$, $\\therefore \\angle B + \\angle BAD = 180^\\circ$, that is $x + 3(180^\\circ - 2x) = 180^\\circ$, solving yields: $x = \\boxed{72^\\circ}$, that is, the measure of $\\angle B$ is $72^\\circ$.",
    "image_path": "dataset/mm_math/images/MM_Math/50474822_148.png"
  },
  {
    "solution": "\\boxed{\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ is inscribed in $\\odot O$. If $\\angle A=50^\\circ$ and $\\angle B=70^\\circ$, what is the length of $AB$?\n",
    "completion": "\\textbf{Solution:}\n\nAs shown in the figure, connect OA and OB, and draw $OD\\perp AB$,\n\n$\\because$ $\\angle A=50^\\circ$, $\\angle B=70^\\circ$,\n\n$\\therefore$ $\\angle C=180^\\circ-50^\\circ-70^\\circ=60^\\circ$,\n\n$\\therefore$ $\\angle AOB=2\\angle C=120^\\circ$,\n\n$\\because$ $OA=OB$,\n\n$\\therefore$ $\\triangle AOB$ is an isosceles triangle,\n\n$\\therefore$ $\\angle AOD=\\frac{1}{2}\\angle AOB=60^\\circ$, $AD=BD=\\frac{1}{2}AB$,\n\n$\\therefore$ $\\angle DAO=30^\\circ$,\n\n$\\therefore$ $OD=\\frac{1}{2}$, $AD=\\sqrt{OA^2-OD^2}=\\sqrt{1^2-\\left(\\frac{1}{2}\\right)^2}=\\frac{\\sqrt{3}}{2}$,\n\n$\\therefore$ $AB=2AD=\\boxed{\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51323908.png"
  },
  {
    "solution": "\\boxed{55^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in square $ABCD$, $AB=\\sqrt{5}$. Let $E$ be a point inside square $ABCD$, with $DE=AB$ and $\\angle EDC=\\alpha$ $\\left(0^\\circ<\\alpha<90^\\circ\\right)$. Connect $CE$ and $AE$, draw $DF\\perp AE$ with the foot being point $F$, and extend $CE$ to meet $DF$ at point $G$. Connect $AG$. When $\\alpha=20^\\circ$, find the measure of $\\angle DAE$ in degrees.\n",
    "completion": "\\textbf{Solution}: Since quadrilateral $ABCD$ is a square,\\\\\nit follows that $\\angle ADC=90^\\circ$, and $AB=AD$,\\\\\nsince $\\angle CDE=20^\\circ$,\\\\\nit follows that $\\angle ADE=70^\\circ$,\\\\\nsince $DE=AB$,\\\\\nit follows that $DA=DE$,\\\\\ntherefore $\\angle DAE=\\angle DEA=\\frac{1}{2}(180^\\circ-70^\\circ)=\\boxed{55^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50481469_137.png"
  },
  {
    "solution": "\\boxed{x\\leq 1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=x+2$ intersects with the line $y=ax+c$ at point $P(m,3)$. What is the solution to the inequality $x+2\\le ax+c$ with respect to $x$?\n",
    "completion": "\\textbf{Solution:} It is known from the problem that point P also lies on $y=x+2$. Hence, when $y=3$, it follows that $x=1$.\n\n$\\therefore$ When $x\\leq 1$, $x+2\\le ax+c$.\n\nTherefore, the answer is $\\boxed{x\\leq 1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52381301.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, it is known that $\\triangle ABC$, $\\angle CAB = 45^\\circ$, $AB = 7$, $AC = 3\\sqrt{2}$, and $CD \\perp AB$ at point D. Point E is a moving point on side BC, a circle $\\odot O$ is drawn with DE as the diameter, intersecting BC at F, and intersecting AB at G. Connect DF and FG. $\\angle BCD = \\angle FDB$. When point E is on segment BF, and $\\triangle DFG$ is an isosceles triangle, find the length of DG.\n",
    "completion": "\\textbf{Solution:}\n\n(i) When $DF = DG$,\n\n$\\because \\angle CAB = 45^\\circ$, $CD \\perp AB$, $AC = 3\\sqrt{2}$,\n\n$\\therefore AD = CD = 3$,\n\n$\\because AB = 7$,\n\n$\\therefore BD = 7 - 3 = 4$,\n\n$\\therefore BC = \\sqrt{3^2 + 4^2} = 5$,\n\n$\\therefore DF = \\frac{3 \\times 4}{5} = \\frac{12}{5}$,\n\n$\\therefore DG = \\frac{12}{5}$,\n\n(ii) When $DF = FG$, draw $FH \\perp BD$ intersecting $BD$ at point $H$,\n\n$\\therefore \\triangle DFH \\sim \\triangle CBD$,\n\n$\\therefore \\frac{DH}{CD} = \\frac{DF}{CB}$,\n\n$\\therefore DH = DF \\times \\frac{3}{5} = \\frac{12}{5} \\times \\frac{3}{5} = \\frac{36}{25}$,\n\n$\\therefore DG = 2DH = \\frac{72}{25}$,\n\n(iii) When $FG = DG$, $\\angle 1 = \\angle 2$,\n\n$\\because \\angle 1 + \\angle 3 = \\angle 2 + \\angle 4 = 90^\\circ$,\n\n$\\therefore \\angle 3 = \\angle 4$,\n\n$\\therefore FG = GB = DG$,\n\n$\\therefore DG = \\frac{1}{2}BD = 2$,\n\nHence, the lengths of $DG$ are $\\boxed{\\frac{12}{5}}$, $\\boxed{\\frac{72}{25}}$, $\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51209242_90.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is a chord of the circle $\\odot O$, C is the midpoint of the arc $AB$, and $OC$ intersects $AB$ at point D. If $AB=8$ cm and $CD=2$ cm, what is the radius of $\\odot O$ in cm?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram below, connect OA,\n\n$\\because$ C is the midpoint of arc $AB$,\n\n$\\therefore$ $CD\\perp AB$, and $CD$ bisects $AB$,\n\nLet the radius of $\\odot O$ be $r$,\n\n$\\because$ $AB=8$cm, $CD=2$cm,\n\n$\\therefore$ $OD=r-2$, $AD=4$,\n\n$\\because$ $AO^{2}=AD^{2}+OD^{2}$,\n\n$\\therefore$ $r^{2}=4^{2}+(r-2)^{2}$,\n\n$\\therefore$ $r^{2}=16+r^{2}-4r+4$,\n\n$\\therefore$ $r=\\boxed{5}$cm.",
    "image_path": "dataset/mm_math/images/MM_Math/53102217.png"
  },
  {
    "solution": "\\boxed{40^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $OB$ bisects $\\angle AOC$, and the complement of $\\angle AOC$ is $30^\\circ$ less than $\\angle BOC$. What is the measure of $\\angle AOB$?\n",
    "completion": "\\textbf{Solution:} Let $\\angle BOC = \\angle AOB = x$, then $\\angle AOC = 2x$\\\\\nFrom the problem, we get: $90^\\circ - 2x = x - 30^\\circ$\\\\\n$3x = 120^\\circ$\\\\\n$x = 40^\\circ$\\\\\nThus, the measure of $\\angle AOB$ is $\\boxed{40^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53206051_132.png"
  },
  {
    "solution": "\\boxed{-x^{2}+2x+3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, the parabola \\(y = -x^2 + bx + c\\) intersects the x-axis at points A and B, and intersects the y-axis at point C \\((0, 3)\\). The axis of symmetry is the line \\(x=1\\), intersecting the x-axis at point D, and the vertex of the parabola is point E. What is the analytical expression of this parabola?\n",
    "completion": "\\textbf{Solution:} Since the parabola $y=-x^{2}+bx+c$ intersects the y-axis at point C(0, 3),\\\\\nlet $x=0$, then we have $c=3$,\\\\\nSince the axis of symmetry is the line $x=1$,\\\\\nit follows that $-\\frac{b}{2\\times (-1)}=1$,\\\\\nthus, $b=2$,\\\\\ntherefore, the equation of the parabola is $y=-x^{2}+2x+3=\\boxed{-x^{2}+2x+3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53080571_41.png"
  },
  {
    "solution": "\\boxed{B(0, -1)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the lines $y=2x+3$ and $y=-2x-1$ intersect. Find the coordinates of points $A$ and $B$, which are the respective points of intersection of each line with the y-axis.\n",
    "completion": "\\textbf{Solution:} In $y=2x+3$, when $x=0$, we have $y=3$, that is point \\boxed{A(0, 3)}; \\\\\nIn $y=-2x-1$, when $x=0$, we have $y=-1$, that is point \\boxed{B(0, -1)};",
    "image_path": "dataset/mm_math/images/MM_Math/51699291_161.png"
  },
  {
    "solution": "\\boxed{(10, -4)$ or $(-10, 4)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the Cartesian coordinate system, given point $A(2, 3)$ and $B(5, -2)$, with the origin $O$ as the center of dilation, if $\\triangle ABO$ is enlarged to twice its original size, what are the coordinates of point $B'$?\n",
    "completion": "\\textbf{Solution:} Given that the origin O is the center of similitude, and $\\triangle ABO$ is magnified by a factor of 2 with $B(5, -2)$, \\\\\n\\textbf{therefore,} the coordinates of the corresponding point $B'$ of point $B$ are\\\\\n$(5\\times 2, -2\\times 2)$ or $(5\\times (-2), -2\\times (-2))$, that is, $(10, -4)$ or $(-10, 4)$,\\\\\nHence, the answer is: $\\boxed{(10, -4)$ or $(-10, 4)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51445173.png"
  },
  {
    "solution": "\\boxed{\\frac{AE}{BE}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that in rectangle $ABCD$, the diagonals $AC$ and $BD$ intersect at point $O$, and point $E$ is a moving point on side $AB$. The line $CE$ intersects $BD$ at point $F$. Connect $OE$. If point $F$ is the midpoint of $OB$, then what is the value of $\\frac{AE}{BE}$?\n",
    "completion": "\\textbf{Solution:} Given the information, it is insufficient to provide a detailed mathematical solution process. Therefore, it is impossible to directly determine the specific value of \\boxed{\\frac{AE}{BE}}.",
    "image_path": "dataset/mm_math/images/MM_Math/52977191_46.png"
  },
  {
    "solution": "\\boxed{72^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC$, and $\\odot O$ is the circumcircle of $\\triangle ABC$. The extension of line $BO$ intersects side $AC$ at point $D$. When $\\triangle BCD$ is an isosceles triangle, find the measure of $\\angle BCD$.\n",
    "completion": "\\textbf{Solution}: Let $\\angle ABD=\\alpha$, \\\\\nsince in $\\triangle ABC$, $BO=AO$, \\\\\ntherefore $\\angle ABO=\\angle BAO$, \\\\\ntherefore $\\angle BAC=2\\angle ABD=2\\alpha$, \\\\\ntherefore $\\angle BDC=3\\alpha$. \\\\\nFurthermore, since $\\triangle BCD$ is an isosceles triangle, \\\\\n(1) If $BD=BC$, then $\\angle C = \\angle BDC = 3\\alpha$. \\\\\nSince $AB=AC$, \\\\\ntherefore $\\angle ABC = \\angle C = 3\\alpha$. \\\\\nThus, in $\\triangle ABC$, $3\\alpha + 3\\alpha + 2\\alpha = 180^\\circ$. \\\\\nSolving gives $\\alpha = 22.5^\\circ$. \\\\\ntherefore $\\angle BCD = 67.5^\\circ$. \\\\\n(2) If $BC=CD$, then $\\angle BDC = \\angle CBD = 3\\alpha$. \\\\\nSince in $\\triangle ABC$, $\\angle ABC + \\angle BCA + \\angle BAD = 180^\\circ$, \\\\\ntherefore $4\\alpha + 4\\alpha + 2\\alpha = 180^\\circ$. \\\\\nSolving gives $\\alpha = 18^\\circ$. \\\\\ntherefore $\\angle BCD = 4\\alpha = 72^\\circ$. \\\\\n(3) If $DB=DC$, then point D coincides with A, and this case does not exist. \\\\\nTo summarize, the value of $\\angle BCD$ is either $67.5^\\circ$ or $\\boxed{72^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52745971_83.png"
  },
  {
    "solution": "\\boxed{7\\,\\text{cm}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the right-angled triangle $ABC$, $\\angle BAC=90^\\circ$, $AB=AC$, perpendicular lines $BD$ and $CE$ are drawn from points $B$ and $C$ respectively, through point $A$, with the feet of the perpendiculars being $D$ and $E$. If $BD=4\\,\\text{cm}$ and $CE=3\\,\\text{cm}$, what is the length of $DE$ in centimeters?\n",
    "completion": "\\textbf{Solution:} Since $BD \\perp AD$ and $CE \\perp AD$,\\\\\nit follows that $\\angle D = \\angle E = 90^\\circ$,\\\\\nhence $\\angle DAB + \\angle ABD = 90^\\circ$,\\\\\nSince $\\angle BAC = 90^\\circ$,\\\\\nit follows that $\\angle DAB + \\angle CAE = 90^\\circ$,\\\\\nhence $\\angle ABD = \\angle CAE$,\\\\\nSince $AB = AC$,\\\\\nit follows that $\\triangle DAB \\cong \\triangle ECA$ (AAS),\\\\\nhence $BD = AE = 4\\,\\text{cm}, AD = CE = 3\\,\\text{cm}$,\\\\\ntherefore $DE = AE + AD = \\boxed{7\\,\\text{cm}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52733493.png"
  },
  {
    "solution": "\\boxed{2+2\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, point O is a point inside $\\triangle ABC$. Connect OB, OC, and join the midpoints D, E, F, G of AB, OB, OC, AC respectively, to get quadrilateral DEFG. Quadrilateral DEFG is a parallelogram. If $\\angle OBC=45^\\circ$, $\\angle OCB=30^\\circ$, and OC$=8$, what is the length of EF?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw OM$\\perp$BC at M, \\\\\nIn Rt$\\triangle$OCM, $\\angle$OCM=$30^\\circ$, OC=8, \\\\\n$\\therefore$OM=$\\frac{1}{2}$OC=4, \\\\\n$\\therefore$CM=$\\sqrt{OC^{2}-OM^{2}}=4\\sqrt{3}$, \\\\\nIn Rt$\\triangle$OBM, \\\\\n$\\because$$\\angle$OBC=$45^\\circ$, \\\\\n$\\therefore$$\\angle$OBM=$\\angle$BOM=$45^\\circ$ \\\\\n$\\therefore$BM=OM=4, \\\\\n$\\therefore$BC=4+$4\\sqrt{3}$, \\\\\n$\\because$EF=$\\frac{1}{2}$BC, \\\\\n$\\therefore$EF=$2+2\\sqrt{3}$. \\\\\nHence, the length of EF is $\\boxed{2+2\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52395259_108.png"
  },
  {
    "solution": "\\boxed{8 + 8\\sqrt{13}\\text{cm}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the quadrilateral $ABCD$, $AB \\parallel CD$, $\\angle BCD = 90^\\circ$, $AB = AD = 10 \\text{cm}$, $BC = 8 \\text{cm}$. Point $P$ starts from point $A$ and moves at a constant speed of $3\\text{cm/s}$ along the polyline $ABCD$ direction; point $Q$ starts from point $D$ and moves at a constant speed of $2\\text{cm/s}$ along the segment $DC$ direction. It is known that both points start simultaneously, and when one point reaches the end, the other point also stops moving, let the motion time be $t (s)$. When the quadrilateral $PBQD$ is a parallelogram, what is the perimeter of the quadrilateral $PBQD$?\n",
    "completion": "\\textbf{Solution:} Draw lines $PD$ and $BP$, \\\\\nwhen the quadrilateral $PBQD$ forms a parallelogram, we have $PB = DQ$, \\\\\ngiven that point P starts from point A and moves in the direction of the polyline $ABCD$ at a constant speed of $3\\text{cm/s}$; point Q starts from point D and moves along the line segment $DC$ at a constant speed of $2\\text{cm/s}$, \\\\\nlet the time of movement for point P be t seconds, then $AP = 3t$, $DQ = 2t$, \\\\\n$\\therefore PB = 10 - 3t$, \\\\\n$\\therefore 10 - 3t = 2t$, solving this equation, we find $t = 2$, \\\\\n$\\therefore PB = DQ = 10 - 3 \\times 2 = 4$, hence $CQ = 16 - 4 = 12$, \\\\\nin $\\triangle BCQ$, $BQ = \\sqrt{BC^2 + CQ^2} = \\sqrt{8^2 + 12^2} = 4\\sqrt{13}$, \\\\\n$\\therefore$ the perimeter of the parallelogram $PBQD$ is: $2(PB + BQ) = 2 \\times (4 + 4\\sqrt{13}) = 8 + 8\\sqrt{13}\\text{cm}$, \\\\\n$\\therefore$ the perimeter of the quadrilateral $PBQD$ is $\\boxed{8 + 8\\sqrt{13}\\text{cm}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53490651_87.png"
  },
  {
    "solution": "\\boxed{(3, 4)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the rhombus $ABCD$, point $B$ is on the $x$-axis, and the coordinates of point $C$ are $(6, 2)$, and the coordinates of point $A$ are $(0, 2)$. What are the coordinates of point $D$?\n",
    "completion": "\\textbf{Solution:} Connect AC, BD, with the intersection point as E, \\\\\n$\\because$ Quadrilateral ABCD is a rhombus, \\\\\n$\\therefore$ AC$\\perp$BD, 2AE=AC, 2BE=BD, quadrilateral AEBO is a rectangle, \\\\\n$\\therefore$ OA=BE, AE=OB, \\\\\n$\\because$ the coordinates of point C are (6, 2), and the coordinates of point A are (0, 2), \\\\\n$\\therefore$ AC$\\parallel$x-axis, \\\\\n$\\therefore$ OA=2, AC=6, \\\\\n$\\therefore$ BE=DE=OA=2, OB=3, \\\\\n$\\therefore$ BD=2DE=4, \\\\\n$\\therefore$ the coordinates of point D are: $\\boxed{(3, 4)}$",
    "image_path": "dataset/mm_math/images/MM_Math/52251407.png"
  },
  {
    "solution": "\\boxed{20}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $BC=120$, the height $AD=60$, and the square $EFGH$ with one side on $BC$, points $E$, $F$ respectively on $AB$, $AC$, $AD$ intersects $EF$ at point $N$, what is the length of $AN$?\n",
    "completion": "\\textbf{Solution:} Let the side lengths of square EFGH be EF = EH = x,\\\\\nsince EFGH is a square,\\\\\nthus $\\angle$HEF = $\\angle$EHG = $90^\\circ$, and EF$\\parallel$BC,\\\\\nthus $\\triangle$AEF $\\sim$ $\\triangle$ABC,\\\\\nsince AD is the altitude of $\\triangle$ABC,\\\\\nthus $\\angle$HDN = $90^\\circ$,\\\\\nthus quadrilateral EHDN is a rectangle,\\\\\nthus DN = EH = x,\\\\\nsince $\\triangle$AEF $\\sim$ $\\triangle$ABC,\\\\\nthus  $\\frac{AN}{AD}=\\frac{EF}{BC}$ (the ratio of altitudes is equal to the scale factor),\\\\\nsince BC = 120, AD = 60,\\\\\nthus AN = 60−x,\\\\\nthus $\\frac{60−x}{60}$ = $\\frac{x}{120}$,\\\\\nsolving for x gives: x = 40,\\\\\nthus AN=60−x=60−40=\\boxed{20}.",
    "image_path": "dataset/mm_math/images/MM_Math/51317388.png"
  },
  {
    "solution": "\\boxed{2.25}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: The diagram shows a schematic of an arched bridge, with the points where the arch intersects the deck being O and B, respectively. Taking point O as the origin and the horizontal line OB as the x-axis, we establish a Cartesian coordinate system. The arch of the bridge can be approximated by the parabola $y=-0.01(x-20)^2+4$. The point C, where the arch intersects the pier AC, is precisely at the water's surface, and $AC \\perp x$-axis. If OA is 5 meters, what is the height of the deck above the water surface AC?\n",
    "completion": "\\textbf{Solution:} Since $OA=5$ and $CA\\perp x$-axis, \\\\\nit follows that the coordinates of point A are $(-5, 0)$, \\\\\nwhen $x=-5$, $y=-0.01(-5-20)^{2}+4=-2.25$, \\\\\nthus point C has coordinates $(-5, -2.25)$, \\\\\nhence $AC=|-2.25|=\\boxed{2.25}$ meters.",
    "image_path": "dataset/mm_math/images/MM_Math/52620595.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $O$ is a point on $AC$. Taking point O as the center and $OC$ as the radius, a circle is drawn tangent to $BC$ at point C. A line is drawn through point O such that $OE \\perp AB$ and intersects $AB$ at point E. Through point A, $AD \\perp BO$ is drawn, and it intersects the extension of $BO$ at point D, and $\\angle AOD = \\angle BAD$. Given $OE=OC$, if $BC=6$ and $\\tan \\angle ABC = \\frac{4}{3}$, find the length of $AD$.\n",
    "completion": "\\textbf{Solution:} Given that $BC=6$, $\\tan\\angle ABC =\\frac{4}{3}$, and $\\angle ACB=90^\\circ$,\\\\\nit follows that $AC=BC\\cdot\\tan\\angle ABC=8$\\\\\nThus, $AB =\\sqrt{6^2+8^2}=10$\\\\\nSince $OE=OC$,\\\\\nit implies $BE=BC=6$\\\\\nHence, $AE=AB-BE=4$\\\\\nLet $OC=OE=x$, then in $\\triangle AEO$, $(8-x)^2=4^2+x^2$ yields $x=3$,\\\\\nTherefore, $OB= \\sqrt{OC^2+BC^2}=3\\sqrt{5}$\\\\\nGiven that $S_{\\triangle BOA}= \\frac{1}{2} AB \\cdot OE= \\frac{1}{2} BO \\cdot AD$,\\\\\nit follows that $AB \\cdot OE= BO \\cdot AD$\\\\\nTherefore, $AD =\\boxed{2\\sqrt{5}}$",
    "image_path": "dataset/mm_math/images/MM_Math/50951684_131.png"
  },
  {
    "solution": "\\boxed{65^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within circle $\\odot O$, the diameter $AB$ intersects with the chord $CD$ at point $P$, $\\angle CAB=40^\\circ$, $\\angle APD=65^\\circ$. What is the measure of $\\angle DAB$?\n",
    "completion": "\\textbf{Solution:} Given $\\because \\angle APD = \\angle C + \\angle CAB$,\\\\\n$\\therefore \\angle C = \\angle APD - \\angle CAB = 65^\\circ - 40^\\circ = 25^\\circ$,\\\\\n$\\therefore \\angle B = \\angle C = 25^\\circ$.\\\\\nBecause $AB$ is the diameter of $\\odot O$, $\\therefore \\angle ADB = 90^\\circ$,\\\\\n$\\therefore \\angle DAB = 90^\\circ - \\angle B = 90^\\circ - 25^\\circ = \\boxed{65^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52814915_121.png"
  },
  {
    "solution": "\\boxed{\\frac{98}{25}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rhombus ABCD, $AC=48$, $BD=14$, and $DH \\perp AB$ at point H, what is the length of line segment BH?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rhombus, and AC=48, BD=14, \\\\\ntherefore, AO=24, OB=7, \\\\\nsince AC $\\perp$ BD, \\\\\ntherefore, AB=$\\sqrt{AO^{2}+OB^{2}}=\\sqrt{24^{2}+7^{2}}=25$, \\\\\nsince DH $\\perp$ AB, \\\\\nsince the area of rhombus ABCD = $\\frac{1}{2}AC\\cdot BD=AB\\cdot DH$, \\\\\ntherefore $\\frac{1}{2} \\times 48 \\times 14 = 25DH$, \\\\\ntherefore $DH=\\frac{336}{25}$, \\\\\ntherefore $BH=\\sqrt{BD^{2}-DH^{2}}=\\sqrt{14^{2}-\\left(\\frac{336}{25}\\right)^{2}}=\\boxed{\\frac{98}{25}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52188948.png"
  },
  {
    "solution": "\\boxed{x \\in (-4, 0) \\cup (1, +\\infty)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the graph of the inverse proportion function \\(y_{1}=\\frac{k}{x}\\) and the graph of the linear function \\(y_{2}=x+b\\) intersect at point \\(A(1, 4)\\) and point \\(B(-4, n)\\). Directly write out the range of values of the independent variable \\(x\\) when \\(y_{2}>y_{1}\\).\n",
    "completion": "\\textbf{Solution:} When $-4 < x < 0$ or $x > 1$, we have $y_{2} > y_{1}$.\n\nWe directly express the range of the independent variable $x$ as \\boxed{x \\in (-4, 0) \\cup (1, +\\infty)}.",
    "image_path": "dataset/mm_math/images/MM_Math/52977772_192.png"
  },
  {
    "solution": "\\boxed{16}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points A, C, E, B, D are on the same straight line, and $AB=CD$, with point E being the midpoint of line segment AD. If $AB=11$ and $CE=3$, find the length of the line segment AD.\n",
    "completion": "\\textbf{Solution:} Given that $CE=3, CE=BE$,\\\\\nhence $BC=CE+BE=3+3=6$,\\\\\nsince $AB=11$,\\\\\nthus $AC=AB-BC=11-6=5$,\\\\\nsince $CD=AB=11$,\\\\\ntherefore $AD=AC+CD=5+11=\\boxed{16}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53223520_102.png"
  },
  {
    "solution": "\\boxed{\\sqrt{10}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, for square $ABCD$ with side length $3$, point $E$ is on side $AB$, and $BE=1$. If point $P$ moves along diagonal $BD$, what is the minimum value of $PA+PE$?\n",
    "completion": "\\textbf{Solution:} Given the figure, since ABCD is a square, thus points A and C are symmetrical about BD. Draw line CE intersecting BD at point P, and connect AP. Therefore, $AP=CP$, which implies $AP+PE=CP+PE=CE$. Hence, the minimum value of $AP+PE$ is the length of CE; in $\\triangle BCE$, we have $CE=\\sqrt{BE^2+BC^2}=\\sqrt{3^2+1^2}=\\sqrt{10}$. Consequently, the answer is $\\boxed{\\sqrt{10}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51614754.png"
  },
  {
    "solution": "\\boxed{\\frac{4}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle ABCD, E is a moving point on BC, and $\\triangle ABE$ is folded along AE to obtain $\\triangle AFE$, where point F is inside rectangle ABCD. Extend AF to intersect CD at point G, with $AB=3$ and $AD=4$. As shown in Figure 2, when point E is the midpoint of BC, find the length of line segment GC.\n",
    "completion": "\\textbf{Solution:} As shown in Diagram 4, connect GE,\\\\\n$\\because$ E is the midpoint of BC,\\\\\n$\\therefore$ BE=EC,\\\\\n$\\because$ $\\triangle ABE$ is folded along AE to obtain $\\triangle AFE$,\\\\\n$\\therefore$ BE=EF,\\\\\n$\\therefore$ EF=EC,\\\\\n$\\because$ in rectangle ABCD,\\\\\n$\\therefore$ $\\angle C=90^\\circ$,\\\\\n$\\therefore$ $\\angle EFG=90^\\circ$,\\\\\n$\\because$ in $\\triangle GFE$ and $\\triangle GCE$,\\\\\n$\\left\\{\\begin{array}{l}EG=EG \\\\ EF=EC\\end{array}\\right.$ \\\\\n$\\therefore$ $\\triangle GFE \\cong \\triangle GCE$ (HL),\\\\\n$\\therefore$ GF=GC;\\\\\nLet GC=x, then AG=3+x, DG=3-x,\\\\\nin $\\triangle ADG$, $4^2+(3-x)^2=(3+x)^2$,\\\\\nsolving gives $x=\\boxed{\\frac{4}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51684393_168.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, the side $DE$ of rectangle $DEFG$ is on side $AB$, and the vertices $F$ and $G$ are on sides $BC$ and $AC$, respectively. If the areas of $\\triangle BEF$, $\\triangle ADG$, and $\\triangle CFG$ are respectively 1, 2, and 3, then what is the area of rectangle $DEFG$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, draw $CH\\perp AB$, intersecting AB and GF at points H and M, respectively.\\\\\nSince $CH\\perp AB$,\\\\\nit follows that $CM\\perp GF$,\\\\\nSince quadrilateral DEFG is a rectangle,\\\\\nit follows that $GF\\parallel AB$,\\\\\nhence $\\triangle CGF\\sim \\triangle CAB$,\\\\\ntherefore $\\frac{GF}{AB}=\\frac{CM}{CH}$,\\\\\nGiven that the areas of $\\triangle BEF$ and $\\triangle ADG$ are 1 and 2, respectively, with equal heights,\\\\\nit follows that $\\frac{BE}{AD}=\\frac{1}{2}$,\\\\ \nLet $DE=x$, $EF=y$, $BE=a$, then $AD=2a$, the height of $\\triangle CGF$ is $CM=\\frac{6}{x}$, $CH=y+\\frac{6}{x}$,\\\\\n$\\frac{x}{x+3a}=\\frac{\\frac{6}{x}}{y+\\frac{6}{x}}$,\\\\\nwhich simplifies to: $18a=x^2y$ (1),\\\\\nSince $\\frac{1}{2}ay=1$,\\\\\nit follows that $ay=2$ (2),\\\\\nSubstituting (2) into (1) yields: $x^2y^2=36$,\\\\\ntherefore $xy=6$ (disregarding $xy=-6$),\\\\\nthus, the area of the rectangle is $\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51353485.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, point $C$ is any point on segment $AB$ (point $C$ is not coincident with points $A$ and $B$). Equilateral triangles $\\triangle ACD$ and $\\triangle BCE$ are constructed on the same side of line $AB$ with $AC$ and $BC$ as sides, respectively. $AE$ and $CD$ intersect at point $M$, $BD$ and $CE$ intersect at point $N$, and $AE$ and $BD$ intersect at point $F$. $\\triangle ACE \\cong \\triangle DCB$; what is the measure of $\\angle AFC$?\n",
    "completion": "\\textbf{Solution:} Given $\\triangle ACE \\cong \\triangle DCB$, we have $\\angle MAC=\\angle NDC$,\\\\\nalso, $\\angle ACM=60^\\circ$, and $\\angle AMC=\\angle DMF$,\\\\\ntherefore, $\\angle DFM=\\angle ACM=60^\\circ$,\\\\\nthus, $\\angle AFB=180^\\circ-\\angle DFM=120^\\circ$,\\\\\ndraw $CG \\perp AE$ at point G, and $CH \\perp BD$ at point H from point C.\\\\\nSince $\\triangle ACE \\cong \\triangle DCB$,\\\\\nthus, $DB=AE$, and $S_{\\triangle ACE}=S_{\\triangle DCB}$,\\\\\ntherefore, $\\frac{1}{2}DB \\cdot CH=\\frac{1}{2}AE \\cdot CG$,\\\\\nthus, $CG=CH$,\\\\\ntherefore, $CF$ bisects $\\angle AFB$.\\\\\nThus, $\\angle AFC=\\frac{1}{2}\\angle AFB=\\frac{1}{2}\\times 120^\\circ=\\boxed{60^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53106988_93.png"
  },
  {
    "solution": "\\boxed{80^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the perpendicular bisectors of line segments $AB$ and $BC$, denoted as $ {l}_{1}$ and $ {l}_{2}$, intersect at point $O$. If $\\angle 1=40^\\circ$, what is the measure of $\\angle AOC$?\n",
    "completion": "\\textbf{Solution:} Draw line BO and extend BO to point P, \\\\\n$\\because$ the perpendicular bisectors $l_{1}$ and $l_{2}$ of line segments AB and BC intersect at point O, and $l_{1}$, $l_{2}$ intersect AB, BC at points D, E respectively, \\\\\n$\\therefore$ AO=OB=OC, $\\angle$BDO=$\\angle$BEO=$90^\\circ$, \\\\\n$\\therefore$ $\\angle$DOE + $\\angle$ABC=$180^\\circ$, \\\\\n$\\because$ $\\angle$DOE + $\\angle$1=$180^\\circ$, \\\\\n$\\therefore$ $\\angle$ABC=$\\angle$1=$40^\\circ$, \\\\\n$\\because$ OA=OB=OC, \\\\\n$\\therefore$ $\\angle$A=$\\angle$ABO, $\\angle$OBC=$\\angle$C, \\\\\n$\\because$ $\\angle$AOP=$\\angle$A+$\\angle$ABO, $\\angle$COP=$\\angle$C+$\\angle$OBC, \\\\\n$\\therefore$ $\\angle$AOC=$\\angle$AOP+$\\angle$COP=$\\angle$A+$\\angle$ABC+$\\angle$C=2$\\angle$ABC=2$\\times$40$^\\circ$=$\\boxed{80^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51545480.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, $AB \\parallel CD$, $AD=4$, and $AB=BC=BD=6$, what is the sine of $\\angle ACD$?\n",
    "completion": "\\textbf{Solution:} Given that $AB=BC=BD=6$,\\\\\nthus points A, D, C are on the circle with center B and radius equal to the length of AB, as shown in the diagram, extend AB to meet the circle B at point E,\\\\\nsince $AB\\parallel CD$,\\\\\ntherefore $\\angle CDB=\\angle ABD$, $\\angle DCB=\\angle CBE$,\\\\\nsince BD=BC,\\\\\ntherefore $\\angle CDB=\\angle BCD$,\\\\\nthus $\\angle ABD=\\angle CBE$,\\\\\ntherefore $AD=CE=4$,\\\\\nsince AE is the diameter of circle B,\\\\\ntherefore $\\angle ACE=90^\\circ$,\\\\\nin the right triangle $\\triangle ACE$, AE=$2AB=12$,\\\\\ntherefore $\\sin\\angle CAE=\\frac{CE}{AE}=\\frac{4}{12}=\\frac{1}{3}$,\\\\\nsince $\\angle ACD=\\angle CAE$,\\\\\ntherefore $\\sin\\angle ACD=\\boxed{\\frac{1}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51446059.png"
  },
  {
    "solution": "\\boxed{45}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points B and C divide line segment AD into three parts in the ratio 2:3:4, and $CD = 20$ cm. What is the length of line segment AD in centimeters?\n",
    "completion": "\\textbf{Solution:} \nLet $AB=2x$, $BC=3x$, and $CD=4x$\\\\\nSince $CD=20$\\\\\n$\\therefore 4x=20$\\\\\n$\\therefore x=5$\\\\\nThen $AD=AB+BC+CD=2x+3x+4x=9x=9\\times 5=\\boxed{45}$ cm.",
    "image_path": "dataset/mm_math/images/MM_Math/53206048_101.png"
  },
  {
    "solution": "\\boxed{P\\left(2, -6\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system \\(xOy\\), the parabola \\(y=ax^2+bx-4\\) intersects the \\(x\\)-axis at points \\(A\\) and \\(B\\) (with \\(B\\) to the right of \\(A\\)) and intersects the \\(y\\)-axis at point \\(C\\). It is known that \\(OA=1\\), \\(OB=4OA\\), and segment \\(BC\\) is drawn. As shown in Figure 1, point \\(P\\) is a moving point on the parabola below \\(BC\\), and segments \\(BP\\) and \\(CP\\) are drawn. When the area of \\(\\triangle BCP\\) equals the area of \\(\\triangle BOC\\), what are the coordinates of point \\(P\\)?\n",
    "completion": "\\textbf{Solution:} Given $y=ax^2+bx-4$ intersects at point $C$,\\\\\nhence $C(0, -4)$,\\\\\nLet $y=kx+b$, substituting $B(4, 0)$ and $C(0, -4)$, we get\\\\\n\\[\\left\\{\\begin{array}{l}\n0=4k+b \\\\\n-4=b\n\\end{array}\\right.,\\]\\\\\nfrom which we find\\\\\n\\[\\left\\{\\begin{array}{l}\nk=1 \\\\\nb=-4\n\\end{array}\\right.,\\]\\\\\ntherefore, $y=x-4$,\\\\\nDraw $PH\\perp x$-axis from point $P$, intersecting the $x$-axis at $H$, and denote the intersection with $BC$ as point $D$,\\\\\nLet $P(m, m^2-3m-4)$,\\\\\nsubstituting $x=m$ into $y=x-4$, we obtain $y=m-4$,\\\\\nthus $D(m, m-4)$,\\\\\ntherefore, $DP=m-4-(m^2-3m-4)=-m^2+4m$,\\\\\nsince $S_{\\triangle BOC}=\\frac{1}{2}OB\\cdot OC=\\frac{1}{2}\\times 4\\times 4=8$,\\\\\ntherefore, $S_{\\triangle BCP}=8$,\\\\\nthus, $S_{\\triangle PDC}+S_{\\triangle PDB}=8$,\\\\\nwhich leads to $\\frac{1}{2}PD\\cdot OH+\\frac{1}{2}PD\\cdot BH=8$, that is $\\frac{1}{2}PD\\cdot OB=8$,\\\\\nhence $(-m^2+4m)\\times 4\\times \\frac{1}{2}=8$,\\\\\nsolving this gives $m=2$,\\\\\ntherefore, \\boxed{P\\left(2, -6\\right)}.",
    "image_path": "dataset/mm_math/images/MM_Math/51446444_220.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within circle $O$, $AB$ is a diameter, and point $D$ lies on circle $O$. Point $C$ is the midpoint of arc $\\stackrel{\\frown}{AD}$. Chord $CM$ is perpendicular to $AB$ at point $F$. Connect $AD$, which intersects $CF$ at point $P$, and connect $BC$. Given that $\\angle DAB=30^\\circ$, what is the measure of $\\angle ABC$?\n",
    "completion": "\\textbf{Solution}: Draw line BD, since AB is the diameter of $\\odot$O,\\\\\ntherefore $\\angle ADB=90^\\circ$,\\\\\nsince $\\angle DAB=30^\\circ$,\\\\\ntherefore $\\angle ABD=90^\\circ-30^\\circ=60^\\circ$,\\\\\nsince C is the midpoint of $\\overset{\\frown}{AD}$,\\\\\ntherefore $\\angle ABC=\\angle DBC=\\frac{1}{2}\\angle ABD=30^\\circ$;\\\\\nthus, the degree of $\\angle ABC$ is $\\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53102087_123.png"
  },
  {
    "solution": "\\boxed{\\frac{9}{25}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in parallelogram $ABCD$, point $E$ is on $DC$, with $DE:EC = 3:2$. Connecting $AE$ and intersecting $BD$ at point $F$, what is the ratio of $S_{\\triangle DEF}$ to $S_{\\triangle BAF}$?\n",
    "completion": "\\textbf{Solution:} Given, DE: EC = 3: 2,\\\\\nTherefore, DE: DC = 3: 5,\\\\\nGiven parallelogram ABCD,\\\\\nTherefore, AB $\\parallel$ CD and AB = CD,\\\\\nTherefore, $\\frac{DE}{AB} = \\frac{DE}{CD} = \\frac{3}{5}$, $\\triangle$ DEF $\\sim$ $\\triangle$ BAF,\\\\\nTherefore, $\\frac{S_{\\triangle DEF}}{S_{\\triangle BAF}} = \\left(\\frac{DE}{AB}\\right)^2 = \\left(\\frac{3}{5}\\right)^2 = \\boxed{\\frac{9}{25}}$,",
    "image_path": "dataset/mm_math/images/MM_Math/53066022.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in $ \\triangle ABC$, $ BA=BC$, point E lies on $ BC$, and $ AE\\perp BC$, $\\cos\\angle B=\\frac{4}{5}$, $ EC=3$, find the length of $ BE$ and the value of $\\tan\\angle CAE$.\n",
    "completion": "\\textbf{Solution:} Given $AE\\perp BC$ and $\\cos\\angle B=\\frac{4}{5}$,\n\\[\\therefore \\frac{BE}{AB}=\\frac{4}{5},\\] suppose $BE=4x$, then $AB=5x$,\n\\[\\because BA=BC,\\] thus $BC=5x$,\n\\[\\because EC=3\\text{ and }EC=BC-BE,\\]\n\\[\\therefore 5x-4x=3,\\] that is $x=3$,\n\\[\\therefore BE=4x=\\boxed{12},\\] $BA=BC=5x=15$,\n\\[\\therefore AE=\\sqrt{AB^{2}-BE^{2}}=\\sqrt{15^{2}-12^{2}}=9,\\]\n\\[\\therefore \\tan\\angle CAE=\\frac{CE}{AE}=\\frac{3}{9}=\\boxed{\\frac{1}{3}}.\\]",
    "image_path": "dataset/mm_math/images/MM_Math/52977771.png"
  },
  {
    "solution": "\\boxed{90^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$ and $\\triangle ECF$, we have $\\angle ACB = \\angle ECF = 90^\\circ$, and also $AC = BC$, $EC = FC$. Connecting $AE$ and $BF$, they intersect at point $O$. $\\triangle ACE \\cong \\triangle BCF$. What is the measure of $\\angle AOB$?\n",
    "completion": "\\textbf{Solution:} In $\\triangle BOM$ and $\\triangle ACM$, \\\\\nsince $\\angle AMC = \\angle BMO$, $\\angle CAE = \\angle BEF$, \\\\\nit follows that $\\angle BOM = \\angle ACM$, \\\\\nsince $\\angle ACB = 90^\\circ$, \\\\\nit implies $\\angle AOB = \\boxed{90^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53059820_134.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points A, B, and C are on the same straight line. Point $D$ is the midpoint of line segment $AC$, and point $E$ is the midpoint of line segment $BC$. As shown in Figure 2, point $C$ is on the extension of line segment $AB$. If $DE=5$, find the length of line segment $AB$.\n",
    "completion": "\\textbf{Solution:} Since $D$ is the midpoint of the line segment $AC$ and $E$ is the midpoint of the line segment $BC$, \\\\\nthus $DC = \\frac{1}{2} AC$, $CE = \\frac{1}{2} BC$, \\\\\nthus $DE = DC - CE = \\frac{1}{2} (AC - BC) = \\frac{1}{2} AB = 5$, \\\\\nthus $AB = \\boxed{10}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55393350_78.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the diagonals $AC$ and $BD$ of $ABCD$ intersect at point $O$. $E$ is the midpoint of $AD$. Connect $OE$. Given $AC=8$, $BC=10$, if $AC \\perp CD$, then what is the length of $OE$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral ABCD is a parallelogram,\\\\\nit follows that $AO=OC$, and $AD=BC=10$,\\\\\nsince $AC\\perp CD$,\\\\\nit follows that $\\angle ACD=90^\\circ$,\\\\\ntherefore $CD=\\sqrt{AD^{2}-AC^{2}}=\\sqrt{10^{2}-8^{2}}=6$,\\\\\nsince E is the midpoint of AD,\\\\\nit follows that OE is the median of $\\triangle ACD$,\\\\\nthus $OE=\\frac{1}{2}CD=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52160354.png"
  },
  {
    "solution": "\\boxed{3:5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, D and E are points on AB and AC, respectively. AF bisects $\\angle BAC$ and intersects DE at point G and intersects BC at point F. If $\\angle AED=\\angle B$, and $AG: GF=3:2$, what is the ratio of $DE: BC$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle DAE = \\angle CAB$ and $\\angle AED = \\angle B$,\\\\\nit follows that $\\triangle ADE \\sim \\triangle ACB$,\\\\\nsince $GA$ and $FA$ are the angle bisectors of $\\triangle ADE$ and $\\triangle ABC$ respectively,\\\\\ntherefore $\\frac{DE}{BC} = \\frac{AG}{AF}$ (the ratio of the corresponding angle bisectors is equal to the similarity ratio),\\\\\n$AG:FG = 3:2$,\\\\\nthus, $AG:AF = 3:5$,\\\\\nhence, $DE:BC = \\boxed{3:5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52620528.png"
  },
  {
    "solution": "\\boxed{\\frac{3\\sqrt{10}}{10}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, each small square in the grid has a side length of \\(1\\), and the vertices of \\(\\triangle ABC\\) are all located at grid points. What is the sine value of \\(\\angle ABC\\)?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, \\\\\n$\\because AC=\\sqrt{2^2+4^2}=2\\sqrt{5}$, $AB=\\sqrt{2^2+4^2}=2\\sqrt{5}$, \\\\\n$\\therefore AC=AB$, \\\\\n$\\because CD=BD=\\sqrt{1^2+1^2}=\\sqrt{2}$, \\\\\n$\\therefore AD\\perp BC$, \\\\\nIn $\\triangle ABD$ with a right angle, \\\\\n$AD=\\sqrt{AB^2−BD^2}=3\\sqrt{2}$, \\\\\n$\\therefore \\sin\\angle ABC=\\sin\\angle ABD=\\frac{AD}{AB}=\\frac{3\\sqrt{2}}{2\\sqrt{5}}=\\boxed{\\frac{3\\sqrt{10}}{10}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51106899.png"
  },
  {
    "solution": "\\boxed{80^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that points A, B, and C are sequentially on the circle $\\odot o$. Given that $\\angle B - \\angle A = 40^\\circ$, what is the degree measure of $\\angle AOB$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, $AC$ intersects $OB$ at point $D$,\\\\\n$\\because$ $\\angle A + \\angle O + \\angle ADO = 180^\\circ$, $\\angle B + \\angle C + \\angle CDB = 180^\\circ$, $\\angle ADO = \\angle CDB$,\\\\\n$\\therefore$ $\\angle A + \\angle O = \\angle B + \\angle C$,\\\\\n$\\therefore$ $\\angle B - \\angle A = \\angle O - \\angle C = 40^\\circ$,\\\\\n$\\because$ $\\overset{\\frown}{AB} = \\overset{\\frown}{AB}$,\\\\\n$\\therefore$ $\\angle O = 2\\angle C$,\\\\\n$\\therefore$ $2\\angle C - \\angle C = 40^\\circ$,\\\\\nSolving this: $\\angle C = 40^\\circ$,\\\\\n$\\therefore$ $\\angle AOB = 2 \\times 40^\\circ = \\boxed{80^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52815675.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the line $l\\colon y=kx-1$ ($k\\neq 0$) intersects the graph of the function $y=\\frac{m}{x}$ ($x>0$) at point $A(3,2)$. After the line $l$ is translated upwards along the y-axis by $t$ ($t>0$) units, the resulting line intersects the $x$ axis and the $y$ axis at points $P$ and $Q$, respectively, and intersects the graph of the function $y=\\frac{m}{x}$ ($x>0$) at point $C$.\n",
    "completion": "\\textbf{Solution:}\\\\\n(1) Given the equation of line $l$ as $y=x-1$ and the equation of the hyperbola as $y=\\frac{6}{x}$,\\\\\nwhen $t=2$, the shifted equation of line $l$ is $y=x+1$,\\\\\nwhen $x=0$, $y=1$, and when $y=0$, $x=-1$,\\\\\n$\\therefore$ points $P(-1, 0), Q(0, 1)$,\\\\\ncombining $y=x+1$ and $y=\\frac{6}{x}$ yields:\\\\\n$\\begin{cases}\ny=x+1\\\\ \ny=\\frac{6}{x}\n\\end{cases}$, solving gives:\\\\\n$\\begin{cases}\nx_1=2\\\\ \ny_1=3\n\\end{cases}, \\begin{cases}\nx_2=-3\\\\ \ny_2=-2\n\\end{cases}$ (discarded),\\\\\n$\\therefore$ point $C(2, 3)$,\\\\\n$\\therefore QC=\\sqrt{(2-0)^2+(3-1)^2}=\\boxed{2\\sqrt{2}}$;\\\\\n(2) $1+\\frac{\\sqrt{2}}{2}<t<2$",
    "image_path": "dataset/mm_math/images/MM_Math/53078816_46.png"
  },
  {
    "solution": "\\boxed{18cm^{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a rectangle $ABCD$ with a length of $5\\,cm$ and a width of $3\\,cm$ is first translated $2\\,cm$ to the right and then $1\\,cm$ downwards to obtain rectangle $A'B'C'D'$. What is the area of the shaded part in $cm^{2}$?\n",
    "completion": "\\textbf{Solution:} It can be inferred from the problem that the blank part is a rectangle, with a length of $ \\left(5-2\\right)cm=3cm$ and a width of $ \\left(3-1\\right)cm=2cm$, \\\\\n$\\therefore$ the area of the shaded part $ =5\\times 3\\times 2-2\\times 3\\times 2=\\boxed{18cm^{2}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51212141.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C = 90^\\circ$, $AB = 5$, and $BC = 3$. Point P starts from point A and moves along $AC-CB-BA$ with a velocity of 2 units per second, and stops at point A. When point P does not coincide with the vertices of $\\triangle ABC$, a perpendicular line is drawn from point P to the side it is on, intersecting another side of $\\triangle ABC$ at point Q. Let the motion time of point P be $t$ seconds. When one vertex of $\\triangle APQ$ is equidistant from the hypotenuse and one leg of $\\triangle ABC$, write down the value of $t$ directly.\n",
    "completion": "\\textbf{Solution:} The value of $t$ can be directly written as $t=\\boxed{\\frac{5}{4}}$, $t=\\boxed{\\frac{8}{3}}$, $t=\\boxed{4}$, and $t=\\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51318739_102.png"
  },
  {
    "solution": "\\boxed{}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graphs of the linear functions $y=k_1x+b_1$ and $y=kx+b$ intersect the $x$-axis at points $A(-1, 0)$ and $B(2, 0)$, respectively. What is the solution set of the system of inequalities in $x$ given by\n\\[\n\\begin{cases} \nk_1x+b_1>0 \\\\\nkx+b>0 \n\\end{cases}\n\\]\n?\n",
    "completion": "\\textbf{Solution:} Let the equations of the linear functions be $y=k_1x+b_1$ and $y=kx+b$, with their graphs intersecting the $x$-axis at points $A(-1,0)$ and $B(2,0)$, respectively.\\\\\nBased on the graphs, we know that the solution set for $y=k_1x+b_1>0$ is: $x>-1$,\\\\\nand the solution set for $y=kx+b>0$ is: $x<2$,\\\\\nTherefore, the solution set for the system of inequalities $\\begin{cases}k_1x+b_1>0\\\\ kx+b>0\\end{cases}$ is $-1<x<2$,\\\\\nThus, the answer is: $-1<x<2$.",
    "image_path": "dataset/mm_math/images/MM_Math/52250423.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point C is a point on the line segment $AB$, point M is the midpoint of line segment $AC$, and point N is the midpoint of line segment $BC$. Given that $AB=20$, what is the length of $MN$?\n",
    "completion": "\\textbf{Solution:} Let $M$ be the midpoint of segment $AC$, and $N$ be the midpoint of segment $BC$,\\\\\n$\\therefore$ $MC=\\frac{1}{2}AC$, $NC=\\frac{1}{2}BC$,\\\\\n$\\therefore$ $MN=MC+NC=\\frac{1}{2}\\left(AC+BC\\right)=\\frac{1}{2}AB$,\\\\\nGiven that $AB=20$,\\\\\n$\\therefore$ $MN=\\frac{1}{2}\\times 20=\\boxed{10}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55377885.png"
  },
  {
    "solution": "\\boxed{\\frac{-\\sqrt{2}+\\sqrt{10}}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the square $ABCD$, point $E$ is a point on the diagonal $AC$. A line through point $E$ is drawn perpendicular to $AC$, intersecting sides $AD$ and $AB$ at points $F$ and $H$, respectively. Lines $CF$ and $CH$ are then drawn. It is given that $CF=CH$; if the side length of the square $ABCD$ is $1$, when the areas of triangles $\\triangle AFH$ and $\\triangle CDF$ are equal, what is the length of $AE$?\n",
    "completion": "\\textbf{Solution:} Let $AE=x$, then $AF=\\sqrt{2}x$, $DF=1-\\sqrt{2}x$, $FH=2AE=2x$,\\\\\nsince the area of $\\triangle AFH$ is equal to the area of $\\triangle CDF$,\\\\\ntherefore $\\frac{1}{2}\\cdot 2x\\cdot x=\\frac{1}{2}\\times 1\\cdot (1-\\sqrt{2}x)$,\\\\\nwhich simplifies to $2x^2+\\sqrt{2}x-1=0$, solving this we get $x_1=\\frac{-\\sqrt{2}+\\sqrt{10}}{4}$, $x_2=\\frac{-\\sqrt{2}-\\sqrt{10}}{4}$ (discard this solution),\\\\\ntherefore $AE=\\boxed{\\frac{-\\sqrt{2}+\\sqrt{10}}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50496676_98.png"
  },
  {
    "solution": "\\boxed{18^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, $\\angle B=36^\\circ$. With $C$ as the center and $CA$ as the radius, the circle intersects $AB$ at point $D$ and $BC$ at point $E$. What is the degree measure of the central angle opposite to the arc $DE$?\n",
    "completion": "\\textbf{Solution:} Connect CD, as shown in the diagram: \\\\\n$\\because$ $\\angle ACB=90^\\circ$, $\\angle B=36^\\circ$,\\\\\n$\\therefore$ $\\angle A=90^\\circ-\\angle B=54^\\circ$,\\\\\n$\\because$ CA=CD,\\\\\n$\\therefore$ $\\angle CDA=\\angle A=54^\\circ$,\\\\\n$\\therefore$ $\\angle ACD=180^\\circ-54^\\circ-54^\\circ=72^\\circ$,\\\\\n$\\therefore$ $\\angle DCE=90^\\circ-\\angle ACD=18^\\circ$,\\\\\nHence, the answer is: $\\boxed{18^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52537049.png"
  },
  {
    "solution": "\\boxed{\\frac{125}{16}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, rectangle $OABC$ has vertices at $A\\left(0, 3\\right)$ and $C\\left(-1, 0\\right)$. Now, rotate rectangle $OABC$ clockwise by $90^\\circ$ around the origin $O$ to obtain rectangle $O{A}^{\\prime}{B}^{\\prime}{C}^{\\prime}$. The line $B{B}^{\\prime}$ intersects the $x$-axis at point $M$ and the $y$-axis at point $N$. The parabola $y=ax^{2}+bx+c$ passes through points $C$, $M$, and $N$. Let point $P$ be a moving point on the parabola and above the line $B{B}^{\\prime}$. Determine the coordinates of point $P$ and the maximum value of the area of $\\triangle PMN$ when the area of $\\triangle PMN$ is maximized.\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $PQ \\perp x$-axis through point P, intersecting $BB'$ at point Q. Let $P\\left(m, -\\frac{1}{2}m^2+2m+\\frac{5}{2}\\right)$, then $Q\\left(m, -\\frac{1}{2}m+\\frac{5}{2}\\right)$,\\\\\n$\\therefore PQ=-\\frac{1}{2}m^2+2m+\\frac{5}{2}-\\left(-\\frac{1}{2}m+\\frac{5}{2}\\right)=-\\frac{1}{2}m^2+\\frac{5}{2}m=-\\frac{1}{2}\\left(m^2-5m+\\frac{25}{4}\\right)+\\frac{25}{8}$;\\\\\nWhen $m=\\frac{5}{2}$, $PQ$ attains its maximum value of $\\frac{25}{8}$, at this moment $-\\frac{1}{2}\\left(\\frac{5}{2}-5\\right)\\left(\\frac{5}{2}+1\\right)=\\frac{35}{8}$,\\\\\n$\\therefore P\\left(\\frac{5}{2}, \\frac{35}{8}\\right)$\\\\\n$\\therefore$ The maximum area of $\\triangle PMN$ is: $S_{\\triangle PMN}=\\frac{1}{2}\\times OM\\times PQ=\\frac{1}{2}\\times 5\\times \\frac{25}{8}=\\boxed{\\frac{125}{16}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52602539_164.png"
  },
  {
    "solution": "\\boxed{220^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the hexagon $ABCDEF$, if $\\angle 1 + \\angle 2 + \\angle 3 = 140^\\circ$, then $\\angle 4 + \\angle 5 + \\angle 6 = ?$ degrees.\n",
    "completion": "\\textbf{Solution:} Given that the sum of the exterior angles of any polygon equals $360^\\circ$, we have $\\angle 1+\\angle 2+\\angle 3+\\angle 4+\\angle 5+\\angle 6=360^\\circ$,\\\\\nsince $\\angle 1+\\angle 2+\\angle 3=140^\\circ$,\\\\\ntherefore $\\angle 4+\\angle 5+\\angle 6=360^\\circ-140^\\circ=\\boxed{220^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53728747.png"
  },
  {
    "solution": "\\boxed{\\frac{6\\sqrt{13}}{65}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, point $F$ is on $CD$, and line $BF$ intersects $AC$ at point $E$. If $\\frac{BC}{AB}=\\frac{2}{3}$ and point $F$ is the midpoint of $CD$, connecting $AF$, find the value of $\\sin\\angle CAF$.\n",
    "completion": "\\textbf{Solution:}\nLet a perpendicular be drawn from point F to AC at H, \\\\\nLet $BC = 2a$, then $AB = CD = 3a$, \\\\\nBy the Pythagorean theorem, $AC = \\sqrt{BC^{2} + CD^{2}} = \\sqrt{13}a$, \\\\\nSince point F is the midpoint of $CD$, \\\\\nThus, $DF = \\frac{3}{2}a$, \\\\\nThen $AF = \\sqrt{AD^{2} + DF^{2}} = \\frac{5}{2}a$, \\\\\nSince the area of $\\triangle AFC = \\frac{1}{2} AC \\cdot FH = \\frac{1}{2} CF \\cdot AD$, \\\\\nTherefore, $\\frac{1}{2} \\times \\sqrt{13}a \\cdot FH = \\frac{1}{2} \\times \\frac{3}{2} \\times a \\times 2a$, \\\\\nSolving this, we find $FH = \\frac{3\\sqrt{13}}{13}a$, \\\\\nThus, $\\sin\\angle CAF = \\frac{FH}{AF} = \\frac{\\frac{3\\sqrt{13}}{13}a}{\\frac{5}{2}a} = \\boxed{\\frac{6\\sqrt{13}}{65}}$",
    "image_path": "dataset/mm_math/images/MM_Math/53065875_82.png"
  },
  {
    "solution": "\\boxed{\\frac{24}{25}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\sin\\angle BAC=\\frac{5}{13}$, $AB=13$, $AC=7.2$, $BD\\perp AC$ with the foot of the perpendicular at point D, point E is the midpoint of BD, and AE intersects BC at point F. Find the tangent of $\\angle CBD$?\n",
    "completion": "\\textbf{Solution:} Since $BD\\perp AC$,\\\\\nit follows that $\\angle D=90^\\circ$,\\\\\ntherefore $\\sin\\angle BAD=\\frac{BD}{AB}=\\frac{5}{13}$,\\\\\nhence $BD=\\frac{5}{13}AB=5$,\\\\\nthus $AD=\\sqrt{AB^{2}-BD^{2}}=12$,\\\\\ntherefore $CD=AD-AC=4.8$,\\\\\nhence $\\tan\\angle CBD=\\frac{CD}{BD}=\\frac{4.8}{5}=\\boxed{\\frac{24}{25}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52746343_93.png"
  },
  {
    "solution": "\\boxed{y=(-2+\\sqrt{3})x+\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system, it is known that point $A(1,0)$. The $x$-axis is rotated clockwise around point $A$ by $60^\\circ$ to intersect the $y$-axis at point $B$, and then point $B$ is rotated clockwise around point $A$ by $90^\\circ$ to obtain point $C$. What is the equation of line $BC$?\n",
    "completion": "\\textbf{Solution}: Given the problem statement, since the $x$-axis rotates $60^\\circ$ clockwise about point $A$ and intersects the $y$-axis at point $B$,\n\ntherefore $\\angle OAB = 60^\\circ$,\n\nsince point $A(1, 0)$,\n\ntherefore $OA=1$,\n\ntherefore $AB=2$, $OB=\\sqrt{3}$,\n\ntherefore $B(0, \\sqrt{3})$,\n\nsince point $B$ rotates $90^\\circ$ clockwise about point $A$ to get to point $C$,\n\ntherefore $AC=AB=2$, $\\angle BAC=90^\\circ$,\n\ndraw $CG \\perp x$-axis at point $G$,\n\ntherefore $\\angle CAG=30^\\circ$,\n\ntherefore $CG=1$, $AG=\\sqrt{3}$,\n\ntherefore $C(1+\\sqrt{3}, 1)$,\n\nlet the equation of line $BC$ be $y=kx+b$,\n\nthen $\\left\\{\\begin{array}{l}b=\\sqrt{3}\\\\ (1+\\sqrt{3})k+b=1\\end{array}\\right.$,\n\nsolving gets $\\left\\{\\begin{array}{l}k=-2+\\sqrt{3}\\\\ b=\\sqrt{3}\\end{array}\\right.$,\n\ntherefore $y=(-2+\\sqrt{3})x+\\sqrt{3}$;\n\nthus, the equation of line $BC$ is $\\boxed{y=(-2+\\sqrt{3})x+\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50682648_121.png"
  },
  {
    "solution": "\\boxed{22.5^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB\\parallel CD$, $\\angle A=45^\\circ$, $\\angle C=\\angle E$, what is the measure of $\\angle C$?\n",
    "completion": "\\textbf{Solution:} Since $AB\\parallel CD$ and $\\angle A=45^\\circ$,\\\\\nit follows that $\\angle DOE=\\angle A=45^\\circ$\\\\\nGiven that $\\angle C=\\angle E$ and $\\angle C+\\angle E=\\angle DOE$,\\\\\nwe have $\\angle C=\\frac{1}{2}\\angle DOE=\\frac{1}{2}\\times 45^\\circ=\\boxed{22.5^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/52603089.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{3}}{6}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, $\\angle ABC=30^\\circ$, and D is the midpoint of AC, then what is the value of $\\tan \\angle DBC$?\n",
    "completion": "\\textbf{Solution:} In $\\triangle ABC$, $\\angle C=90^\\circ$, and $\\angle ABC=30^\\circ$,\\\\\nhence $\\tan\\angle ABC=\\frac{AC}{BC}=\\tan30^\\circ=\\frac{\\sqrt{3}}{3}$. If we let $AC=\\sqrt{3}a$, then $BC=3a$,\\\\\nsince D is the midpoint of AC,\\\\\nthus $DC=\\frac{1}{2}AC=\\frac{\\sqrt{3}}{2}a$,\\\\\ntherefore $\\tan\\angle DBC=\\frac{DC}{BC}=\\frac{\\frac{\\sqrt{3}}{2}a}{3a}=\\boxed{\\frac{\\sqrt{3}}{6}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51322352.png"
  },
  {
    "solution": "\\boxed{25^\\circ or 155^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $AB\\parallel CD$, $AD\\parallel BC$, $\\angle DCE=90^\\circ$, point E is on the line segment AB, $\\angle FCG=90^\\circ$, point F is on the line AD, $\\angle AHG=90^\\circ$. Given the conditions in (2), point C (point C does not coincide with points B and H) starts from point B and moves in the direction of ray BG, with other conditions unchanged, find the degree measure of $\\angle BAF$?\n",
    "completion": "\\textbf{Solution:} Discuss in two cases:\\\\\n(1) As shown in figure a, when point C is on segment BH, point F is on the extension of DA, then $\\angle ECF=\\angle DCG=\\angle B=25^\\circ$.\\\\\nSince AD$\\parallel$BC,\\\\\ntherefore $\\angle BAF=\\angle B=25^\\circ$;\\\\\n(2) As shown in figure b, when point C is on the extension of BH, point F is on segment AD.\\\\\nSince $\\angle B=25^\\circ$, AD$\\parallel$BC,\\\\\ntherefore $\\angle BAF=180^\\circ-25^\\circ=155^\\circ$.\\\\\nIn summary, the degree of $\\angle BAF$ is $\\boxed{25^\\circ or 155^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51434356_69.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: \"The cross section of a dam in a reservoir is a trapezoid, and the slope of a ramp inside the dam is $i=1\\colon \\sqrt{3}$. What is the angle of this slope?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, \\\\\nsince the slope is $i=1\\colon \\sqrt{3}$, \\\\\ntherefore $\\tan A= \\frac{1}{\\sqrt{3}}=\\frac{\\sqrt{3}}{3}$, \\\\\nsince $\\tan 30^\\circ= \\frac{\\sqrt{3}}{3}$, \\\\\ntherefore, the angle of the slope is $30^\\circ$, \\\\\nhence, the answer is: $A=\\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51458068.png"
  },
  {
    "solution": "\\boxed{155^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Lines AB and CD intersect at point O, and OE bisects $\\angle BOD$. As shown in Figure 1, if $\\angle BOC=130^\\circ$, what is the degree measure of $\\angle AOE$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle BOC= 130^\\circ$,\\\\\nit follows that $\\angle AOD=\\angle BOC=130^\\circ$,\\\\\n$\\angle BOD=180^\\circ-\\angle BOC=50^\\circ$.\\\\\nSince OE bisects $\\angle BOD$,\\\\\nit follows that $\\angle DOE=25^\\circ$,\\\\\ntherefore $\\angle AOE=\\angle AOD+\\angle DOE=\\boxed{155^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/54962530_119.png"
  },
  {
    "solution": "\\boxed{65^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC$, $D$ is the midpoint of side $BC$, $\\angle B=25^\\circ$. Find the measure of $\\angle DAC$?\n",
    "completion": "\\textbf{Solution:} Since $AB=AC$,\\\\\nit follows that $\\angle C=\\angle B=25^\\circ$,\\\\\nSince $AD\\perp BC$,\\\\\nit implies $\\angle ADB=90^\\circ$,\\\\\nthus $\\angle DAC=90^\\circ-25^\\circ=\\boxed{65^\\circ};$",
    "image_path": "dataset/mm_math/images/MM_Math/52604307_109.png"
  },
  {
    "solution": "\\boxed{-4x+32}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point \\(P(x, y)\\) is a moving point in the first quadrant and is on the line \\(y = -2x + 8\\), where the line intersects the x-axis at point \\(A\\). Let the area of \\(\\triangle APO\\) be \\(S\\), express \\(S\\) with an algebraic formula involving \\(x\\), and write down the range of values for \\(x\\).\n",
    "completion": "\\textbf{Solution:} Given that point P lies on the line $y=-2x+8$,\\\\\nthus P$(x, -2x+8)$.\\\\\nTherefore, $S_{\\triangle APO}=OA \\times (-2x+8) = 4 \\times (-2x+8) = -4x+32$,\\\\\nwhere the range of $x$ is $0<x<4$,\\\\\nhence $S_{\\triangle APO}=\\boxed{-4x+32}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52422771_84.png"
  },
  {
    "solution": "\\boxed{\\frac{10}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $E$ is on the side $CD$ of rectangle $ABCD$. The rectangle $ABCD$ is folded along $AE$ so that point $D$ falls on point $F$ on side $BC$. If $AB=6$ and $AD=10$, what is the length of $DE$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral $ABCD$ is a rectangle, \\\\\n$\\therefore AD=BC=10$, $AB=CD=6$,\\\\\nFolding rectangle $ABCD$ along line $AE$, vertex $D$ precisely falls on point $F$ on side $BC$,\\\\\n$\\therefore AF=AD=10$, $EF=DE$,\\\\\nIn $\\triangle ABF$, $BF=\\sqrt{AF^{2}-AB^{2}}=\\sqrt{10^{2}-6^{2}}=8$,\\\\\n$\\therefore CF=BC-BF=10-8=2$,\\\\\nLet $CE=x$, then $DE=EF=6-x$,\\\\\nIn $\\triangle ECF$, $CE^{2}+FC^{2}=EF^{2}$,\\\\\n$\\therefore x^{2}+2^{2}=(6-x)^{2}$,\\\\\nSolving for $x$ gives $x=\\frac{8}{3}$,\\\\\n$\\therefore DE=6-x=\\boxed{\\frac{10}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52204763.png"
  },
  {
    "solution": "\\boxed{\\frac{53}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, the lengths of sides $AB$ and $BC$ ($AB<BC$) are the two roots of the equation ${x}^{2}-7x+12=0$. Point $P$ starts from point $A$ and moves along the edge $A\\to B\\to C\\to A$ of $\\triangle ABC$ at a speed of $1$ unit per second. The movement lasts for $t$ seconds. When point $P$ moves to side $AC$, there exists a point $P$ such that $\\triangle CDP$ is an isosceles triangle. What is the value of the movement time $t$?\n",
    "completion": "\\textbf{Solution:} There exists a point $P$ such that $\\triangle CDP$ is an isosceles triangle.\\\\\n(1) When $PC=CD=3$, $t=\\frac{3+4+3}{1}=10$ seconds;\\\\\n(2) When $PD=PC$ (i.e., $P$ is the midpoint of the diagonal $AC$), $AB=3$, $BC=4$.\\\\\n$\\therefore AC=\\sqrt{AB^2+BC^2}=5$, $CP=\\frac{1}{2}AC=2.5$,\\\\\n$\\therefore t=\\frac{3+4+2.5}{1}=9.5$ seconds;\\\\\n(3) When $PD=CD=3$, construct $DQ\\perp AC$ at $Q$,\\\\\n$DQ=\\frac{\\frac{1}{2}\\times 3\\times 4}{\\frac{1}{2}\\times 5}=\\frac{12}{5}$, $PQ=\\sqrt{3^2-\\left(\\frac{12}{5}\\right)^2}=\\frac{9}{5}$,\\\\\n$\\therefore PC=2PQ=\\frac{18}{5}$,\\\\\n$\\therefore t=\\frac{3+4+\\frac{18}{5}}{1}=\\frac{53}{5}$ seconds,\\\\\nIt is known that when $t$ is $10$ seconds, $9.5$ seconds, or $\\frac{53}{5}$ seconds, $\\triangle CDP$ is an isosceles triangle. Therefore, the possible values for the duration $t$ are $10$ seconds, $9.5$ seconds, and $\\boxed{\\frac{53}{5}}$ seconds.",
    "image_path": "dataset/mm_math/images/MM_Math/51722482_165.png"
  },
  {
    "solution": "\\boxed{\\frac{108}{25}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, point D is on side $AB$, and points F, E are on side $AC$, with $DF \\parallel BE$, $\\frac{AF}{AE}=\\frac{AE}{AC}$, and $DE\\parallel BC$. Given $\\frac{AF}{FE}=\\frac{3}{2}$ and the area of $\\triangle ABC$ is 12, find the value of the area of $\\triangle ADE$.\n",
    "completion": "\\textbf{Solution:} Given $\\because \\frac{AF}{FE}=\\frac{3}{2}$,\\\\\n$\\therefore \\frac{AD}{BD}=\\frac{AF}{EF}=\\frac{3}{2}$,\\\\\n$\\because DE\\parallel BC$,\\\\\n$\\therefore \\frac{AE}{CE}=\\frac{AD}{BD}=\\frac{3}{2}$,\\\\\n$\\therefore \\frac{AE}{AC}=\\frac{3}{5}$.\\\\\n$\\because DE\\parallel BC$,\\\\\n$\\therefore \\triangle ADE\\sim \\triangle ABC$.\\\\\n$\\therefore \\frac{S_{\\triangle ADE}}{S_{\\triangle ABC}}=\\left(\\frac{AE}{AC}\\right)^2=\\left(\\frac{3}{5}\\right)^2$, hence $\\frac{S_{\\triangle ADE}}{S_{\\triangle ABC}}=\\frac{9}{25}$,\\\\\n$\\because S_{\\triangle ABC}=12$,\\\\\n$\\therefore \\frac{S_{\\triangle ADE}}{12}=\\frac{9}{25}$, hence $S_{\\triangle ADE}=\\boxed{\\frac{108}{25}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52917533_58.png"
  },
  {
    "solution": "\\boxed{\\frac{34}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Definition: If a moving point P satisfies the condition that the distances from P to the two endpoints of a line segment $AB$ are in the ratio $PA = 4PB$, then P is called the $KZ$ point of the line segment $AB$. However, point P is not considered a $KZ$ point of the line segment $BA$. As illustrated in the diagram, within a rhombus $ABCD$ where $AB=8$ and $\\angle B=120^\\circ$, points E and F are located on $BC$ and $CD$ respectively, satisfying $\\angle AEF=120^\\circ$. By connecting $AF$, assuming point E is the $KZ$ point of line segment $AF$, what is the length of $DF$?\n",
    "completion": "\\textbf{Solution:} As shown in Figure 3, take $CJ$ on $CB$ such that $CJ=CF$.\\\\\nSince quadrilateral $ABCD$ is a rhombus,\\\\\nit follows that $AB=BC=CD=AD=8$, $AB\\parallel CD$,\\\\\nhence $\\angle B+\\angle C=180^\\circ$,\\\\\nsince $\\angle B=120^\\circ$,\\\\\nit follows that $\\angle C=60^\\circ$,\\\\\nsince $CJ=CF$,\\\\\nit follows that $\\triangle CFJ$ is an equilateral triangle,\\\\\nthus $FJ=CF=CJ$, $\\angle CJF=60^\\circ$,\\\\\ntherefore $\\angle FJE=120^\\circ=\\angle B$,\\\\\nsince point E is the $KZ$ point on the line segment $AF$,\\\\\nit follows that $AE=4EF$,\\\\\nsince $\\angle AEC=\\angle AEF+\\angle FEJ=\\angle B+\\angle BAE$, $\\angle AEF=\\angle B=120^\\circ$,\\\\\nit follows that $\\angle FEJ=\\angle BAE$,\\\\\ntherefore $\\triangle ABE\\sim \\triangle EJF$,\\\\\nthus $\\frac{AB}{EJ}=\\frac{BE}{FJ}=\\frac{AE}{EF}=4$,\\\\\nthus $EJ=2$, $BE=4FJ=4CJ$,\\\\\nthus $BC=BE+EJ+CJ=4CJ+2+CJ=8$,\\\\\ni.e., $CJ=\\frac{6}{5}$,\\\\\ntherefore $CF=CJ=\\frac{6}{5}$,\\\\\nthus $DF=CD-CF=8-\\frac{6}{5}=\\boxed{\\frac{34}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52991464_113.png"
  },
  {
    "solution": "\\boxed{75^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the equilateral $\\triangle ABC$, $D$ is the midpoint of side $BC$. An arc is drawn with $A$ as the center and $AD$ as the radius, intersecting side $AC$ at point $E$. What is the degree measure of $\\angle ADE$?\n",
    "completion": "\\textbf{Solution:} Let's consider in equilateral $\\triangle ABC$, with D being the midpoint of side BC,\\\\\n$\\therefore$ $\\angle DAC=\\frac{1}{2}\\angle BAC=\\frac{1}{2}\\times 60^\\circ=30^\\circ$,\\\\\nAccording to the problem statement, we know $AD=AE$,\\\\\n$\\therefore$ $\\angle ADE=\\angle AED$,\\\\\n$\\therefore$ $\\angle ADE=\\frac{1}{2}\\times (180^\\circ-\\angle DAC)=\\frac{1}{2}\\times (180^\\circ-30^\\circ)=\\boxed{75^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53033454.png"
  },
  {
    "solution": "\\boxed{\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, fold a rectangular paper ABCD in the method illustrated twice, to form an isosceles right-angled triangle BEF. If \\(BC = 1\\), what is the length of \\(AB\\)?\n",
    "completion": "\\textbf{Solution:} As illustrated,\\\\\nsince quadrilateral ABCD is a rectangle,\\\\\nthus $\\angle A = \\angle ADC = \\angle C = \\angle B = 90^\\circ$,\\\\\nthus $BC = AD = 1$, $CD = AB$\\\\\nAfter the first fold,\\\\\n$\\angle A = \\angle DA^{\\prime}E = 90^\\circ$,\\\\\nquadrilateral DAEA$^{\\prime}$ is a square,\\\\\nthus $\\angle ADE = 45^\\circ = \\angle AED$\\\\\nthus $AE = AD = 1$,\\\\\nthus $DE = \\sqrt{{1}^{2}+{1}^{2}} = \\sqrt{2}$;\\\\\nAfter the second fold,\\\\\n$CD = DE = \\sqrt{2}$\\\\\nthus $AB = \\boxed{\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51614709.png"
  },
  {
    "solution": "\\boxed{\\frac{24}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, within the rhombus $ABCD$, the diagonals $AC=6$ and $BD=8$. A line is drawn from point A such that $AE\\perp BC$. What is the length of $AE$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral ABCD is a rhombus,\\\\\n$\\therefore$ $CO=\\frac{1}{2}AC=3$, $BO=\\frac{1}{2}BD=4$, and $AO\\perp BO$,\\\\\n$\\therefore$ $BC=\\sqrt{CO^{2}+BO^{2}}=\\sqrt{9+16}=5$,\\\\\nSince the area of rhombus $ABCD$, $S_{\\text{rhombus} ABCD}=\\frac{1}{2}AC\\cdot BD=BC\\cdot AE$,\\\\\n$\\therefore$ $AE=\\frac{\\frac{1}{2}AC\\cdot BD}{BC}=\\boxed{\\frac{24}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52386838.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point E is a point on side CD of rhombus ABCD. Triangle ADE is folded along AE such that the corresponding point of D, point F, exactly lies on side BC. Let $ \\frac{DE}{CE}=k$. If point F is the midpoint of side BC, what is the value of $k$?\n",
    "completion": "Solution: Given that point F is the midpoint of side BC in rhombus ABCD, we need to find the value of \\( k \\) such that \\( \\frac{DE}{CE} = k \\).\n\nSince F is the midpoint of BC, we know that \\( BF = FC \\).\n\nWhen triangle ADE is folded along AE, point D maps to point F. Since F is the midpoint of BC, and D maps to F, it implies that E must be the midpoint of CD. This is because the fold symmetry ensures that DE and CE are equal when D maps to the midpoint of BC.\n\nTherefore, \\( DE = CE \\), which means \\( \\frac{DE}{CE} = 1 \\).\n\nThus, the value of \\( k \\) is \\(\\boxed{1}\\).",
    "image_path": "dataset/mm_math/images/MM_Math/53565373_57.png"
  },
  {
    "solution": "\\boxed{62.5^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $O$ is on line $AB$, $CO \\perp AB$, and $OE$ bisects $\\angle BOD$, with $OF \\perp OE$. If $\\angle 2 - \\angle 1 = 20^\\circ$, what is the degree measure of $\\angle BOE$?\n",
    "completion": "\\textbf{Solution:} Given $CO \\perp AB$,\n\nit follows that $\\angle COA = 90^\\circ$, which means $\\angle 2 = 90^\\circ - \\angle 1$.\n\nAlso, given $\\angle 2 - \\angle 1 = 20^\\circ$,\n\nit results in $\\angle 2 = 20^\\circ + \\angle 1$,\n\nhence $90^\\circ - \\angle 1 = 20^\\circ + \\angle 1$.\n\nSolving this, we find $\\angle 1 = 35^\\circ$,\n\ntherefore $\\angle 2 = 55^\\circ$,\n\nand thus $\\angle BOD = 180^\\circ - \\angle 2 = 125^\\circ$.\n\nSince $OE$ bisects $\\angle BOD$,\n\nit means $\\angle BOE = \\angle DOE = \\frac{1}{2} \\angle BOD = \\boxed{62.5^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51457845_185.png"
  },
  {
    "solution": "\\boxed{80^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that ship A is in the direction of 30 degrees north of east from lighthouse P, ship B is in the direction of 70 degrees south of east from lighthouse P, and ship C is on the bisector of $\\angle APB$. What is the degree measure of $\\angle APB$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle APN=30^\\circ$ and $\\angle BPS=70^\\circ$, \\\\\n$\\therefore \\angle APB=180^\\circ-\\angle APN-\\angle BPS=180^\\circ-30^\\circ-70^\\circ=80^\\circ$.\\\\\n$\\therefore$ The measure of $\\angle APB$ is $\\boxed{80^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55431959_70.png"
  },
  {
    "solution": "\\boxed{\\frac{50}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ is inscribed in $\\odot O$, with $AB$ being the diameter of $\\odot O$, $AD=16$, and $CE=6$. Connect $OC$, and the chord $AD$ intersects $OC$ and $BC$ at points $E$ and $F$, respectively, where point $E$ is the midpoint of $AD$. $\\angle CAD=\\angle CBA$. What is the length of $AB$?\n",
    "completion": "\\textbf{Solution:} Given that $AB$ is the diameter,\\\\\nthen $\\angle ACB=90^\\circ$,\\\\\nsince $AE=DE=\\frac{1}{2}AD=8$,\\\\\nit follows that $OC\\perp AD$,\\\\\nthus $\\angle AEC=90^\\circ$,\\\\\ngiven $CE=6$,\\\\\nit follows that $AC=\\sqrt{AE^2+CE^2}=10$,\\\\\nsince $\\angle AEC=\\angle ACB$ and $\\angle CAD=\\angle CBA$,\\\\\ntherefore $\\triangle AEC\\sim \\triangle BCA$,\\\\\nthus $\\frac{CE}{AC}=\\frac{AC}{AB}$,\\\\\nwhich gives $\\frac{6}{10}=\\frac{10}{AB}$,\\\\\ntherefore $AB=\\boxed{\\frac{50}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52660728_81.png"
  },
  {
    "solution": "\\boxed{56^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given that $O$ is a point on line $AB$, and $OC \\perp OE$. As shown in Figure (1), if $\\angle COA = 34^\\circ$, what is the measure of $\\angle BOE$?\n",
    "completion": "\\textbf{Solution:} Since $OC\\perp OE$ and $\\angle COA = 34^\\circ$, \\\\\nit follows that $\\angle BOE = 180^\\circ-90^\\circ-34^\\circ = \\boxed{56^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51438726_164.png"
  },
  {
    "solution": "\\boxed{4.5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that point $D$ is a point on line segment $AB$, point $C$ is the midpoint of line segment $AB$, if $AB=8\\,\\text{cm}, BD=3\\,\\text{cm}$. If point $E$ is a point on line $AB$, and $BE=\\frac{1}{3}BD$, point $F$ is the midpoint of $BE$, what is the length of line segment $CF$ in $\\,\\text{cm}$?\n",
    "completion": "\\textbf{Solution}:\n\n(1) When point E is on the left side of point B\\\\\n$\\because BE=\\frac{1}{3}BD$\\\\\n$\\therefore BE=\\frac{1}{3}\\times 3=1\\,\\text{cm}$\\\\\n$\\because F$ is the midpoint of $BE$\\\\\n$\\therefore BF=\\frac{1}{2}BE=0.5\\,\\text{cm}$\\\\\n$\\therefore CF=CB−BF=4−0.5=\\boxed{3.5}\\,\\text{cm}$\\\\\n(2) When point E is on the right side of point B\\\\\n$\\because BE=\\frac{1}{3}BD$\\\\\n$\\therefore BE=\\frac{1}{3}\\times 3=1\\,\\text{cm}$\\\\\n$\\because F$ is the midpoint of $BE$\\\\\n$\\therefore BF=\\frac{1}{2}BE=0.5\\,\\text{cm}$\\\\\n$\\therefore CF=CB+BF=4+0.5=\\boxed{4.5}\\,\\text{cm}$\\\\\nIn summary, the length of segment $CF$ can be \\boxed{3.5}\\,\\text{cm} or \\boxed{4.5}\\,\\text{cm}.",
    "image_path": "dataset/mm_math/images/MM_Math/47110741_185.png"
  },
  {
    "solution": "\\boxed{70^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC \\cong \\triangle EDC$, $BC \\perp CD$, and points A, D, E lie on the same straight line. Given that $\\angle ACB = 25^\\circ$, what is the measure of $\\angle ADC$?\n",
    "completion": "\\textbf{Solution:} Since BC$\\perp$CD and $\\angle$ACB$=25^\\circ$,\\\\\nit follows that $\\angle$ACD$=65^\\circ$,\\\\\nSince $\\triangle ABC \\cong \\triangle EDC$,\\\\\nit implies AC=CE and $\\angle$DCE= $\\angle$ACB$=25^\\circ$,\\\\\nthus $\\angle$ACE$=90^\\circ$,\\\\\ntherefore $\\angle$E=$\\angle$CAE$= \\frac{1}{2}\\times(180^\\circ-90^\\circ)=45^\\circ$,\\\\\nconsequently, $\\angle$ADC=$\\angle$DCE+$\\angle$E$=\\boxed{70^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51419792.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the side AB of rectangle ABCD is parallel to the $y$-axis, and the coordinates of vertex $A$ are $(1, 2)$. Points $B$ and $D$ lie on the graph of the inverse proportion function $y=\\frac{6}{x} \\, (x>0)$. What is the area of rectangle ABCD?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rectangle, with vertex A's coordinates being (1, 2),\\\\\nwe assume the coordinates of points B and D to be (1, y) and (x, 2) respectively,\\\\\nsince points B and D lie on the graph of the inverse proportion function $y=\\frac{6}{x}$ (x>0),\\\\\nwe get y=6, x=3,\\\\\nthus, point B is (1, 6), and point D is (3, 2),\\\\\nhence, AB=6-2=4, AD=3-1=2,\\\\\ntherefore, the area of rectangle ABCD is AB$\\cdot$AD=4$\\times$2=$\\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51615105.png"
  },
  {
    "solution": "\\boxed{\\sqrt{13}-\\frac{3}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $AB=3$ and $AD=4$. Let $E$ be a moving point on line segment $BC$, and construct point $F$ which is the reflection of point B across line $AE$. Connect $EF$ and $DF$. Let $G$ be the midpoint of $DF$. When points $D$, $F$, and $E$ are collinear, what is the length of $CE$? During the entire motion of $E$, what is the minimum distance between points $C$ and $G$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, connect AF, take the midpoint H of AD, and connect GH, GC, HC,\\\\\nQuadrilateral ABCD is a rectangle, thus AB=CD=3, AD=BC=4, $\\angle$B=90$^\\circ$,\\\\\nBy the property of folding, it can be obtained that AF=AB=3, BE=EF, $\\angle$AFE=$\\angle$B=90$^\\circ$,\\\\\nLet CE=x, then BE=EF=4−x,\\\\\nWhen points E, F, and D are collinear, AF$\\perp$DE,\\\\\nIn $\\triangle AFD$, DF=$\\sqrt{AD^{2}-AF^{2}}=\\sqrt{7}$,\\\\\nIn $\\triangle DCE$, DE$^{2}$=DC$^{2}$+CE$^{2}$,\\\\\n$\\therefore$ ($\\sqrt{7}$+4−x)$^{2}$=x$^{2}$+9,\\\\\nSolving yields: x=$\\boxed{\\sqrt{7}}$;\\\\\n$\\because$H is the midpoint of AD, G is the midpoint of DF,\\\\\n$\\therefore$HG is the midline in $\\triangle DAF$,\\\\\n$\\therefore$HG=$\\frac{1}{2}$AF=$\\frac{3}{2}$,\\\\\nIn $\\triangle DHC$, HC=$\\sqrt{DH^{2}+DC^{2}}=\\sqrt{13}$,\\\\\nWhen point G is not on line HC,\\\\\nIn $\\triangle GHC$, GC>HC−HG,\\\\\nWhen point G is on line HC, GC=HC−HG,\\\\\n$\\therefore$The minimum value of CG is $\\boxed{\\sqrt{13}-\\frac{3}{2}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53538278.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the length of line segment $AB=10\\, \\text{cm}$. Extend $AB$ to point $C$ such that $BC=6\\, \\text{cm}$. Let points $M$ and $N$ be the midpoints of $AC$ and $BC$, respectively. Find the lengths of the line segments $BM$ and $MN$.\n",
    "completion": "\\textbf{Solution:} Given $\\because$ AB=10, BC=6\\\\\n$\\therefore$ AC=16\\\\\nAlso, $\\because$ M is the midpoint of AC, $\\therefore$ MC=AM=8\\\\\n$\\because$ N is the midpoint of BC, $\\therefore$ BN=NC=3\\\\\nBM=AB−AM=10−8=\\boxed{2}\\\\\nMN=BM+BN=2+3=\\boxed{5}",
    "image_path": "dataset/mm_math/images/MM_Math/51212541.png"
  },
  {
    "solution": "\\boxed{48}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points $A$ and $B$ are two points in the Cartesian coordinate system. Connect $OA$ and $OB$, with $OA=5$, $OB=10$, and $OA \\perp OB$. If the x-coordinate of point $A$ is $-4$, and the graph of the inverse proportional function $y=\\frac{k_2}{x}$ passes through point $B$, while the graph of the inverse proportional function $y=\\frac{k_1}{x}$ passes through point $A$. What are the values of $k_1$ and $k_2$?\n",
    "completion": "\\textbf{Solution:} Construct perpendicular lines from points A and B to the x-axis, with feet at M and N, respectively, as shown in the figure. \\\\\nAccording to the problem, we know: OM = 4, OA = 5 \\\\\nBy the Pythagorean theorem, we have: $AM = \\sqrt{OA^{2} - OM^{2}} = \\sqrt{5^{2} - 4^{2}} = 3$ \\\\\n$\\therefore$ The coordinates of point A are (-4, 3) \\\\\nSubstituting the coordinates of point A into $y = \\frac{k_1}{x}$ gives: $\\frac{k_1}{-4} = 3$ \\\\\n$\\therefore k_1 = \\boxed{-12}$ \\\\\n$\\because$ OA$\\perp$OB, AM$\\perp$OM, BN$\\perp$ON \\\\\n$\\therefore \\angle AOB = \\angle AMO = \\angle BNO = 90^{\\circ}$ \\\\\n$\\therefore \\angle AOM + \\angle BON = 90^{\\circ}$, $\\angle AOM + \\angle MAO = 90^{\\circ}$ \\\\\n$\\therefore \\angle MAO = \\angle BON$ \\\\\n$\\because \\angle AMO = \\angle BNO = 90^{\\circ}$ \\\\\n$\\therefore \\triangle AMO \\sim \\triangle ONB$ \\\\\n$\\therefore \\frac{AM}{ON} = \\frac{OA}{OB} = \\frac{5}{10} = \\frac{1}{2}$ \\\\\n$\\therefore ON = 2AM = 6$ \\\\\nIn $\\triangle BON$, by the Pythagorean theorem, we get: $BN = \\sqrt{OB^{2} - ON^{2}} = \\sqrt{100 - 36} = 8$ \\\\\n$\\therefore$ The coordinates of point B are (6, 8) \\\\\nSubstituting the coordinates of point B into $y = \\frac{k_2}{x}$ gives: $\\frac{k_2}{6} = 8$ \\\\\n$\\therefore k_2 = \\boxed{48}$",
    "image_path": "dataset/mm_math/images/MM_Math/52601676_118.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $AB$ is the diameter of $\\odot O$, the bisector of the acute angle $\\angle DAB$ intersects $\\odot O$ at point $C$. Draw $CD \\perp AD$ at $D$, and the extension of line $CD$ intersects the extension of $AB$ at point $E$. The line $CD$ is the tangent of $\\odot O$; when $AB=2BE$, and $CE=\\sqrt{3}$, find the length of $AD$.\n",
    "completion": "\\textbf{Solution}: Since diameter AB $= 2$BE,\\\\\nit follows that OE $= 2$OC,\\\\\nIn $\\triangle EOC$, let CO $= x$, thus OE $= 2x$,\\\\\nBy the Pythagorean theorem: CE $= \\sqrt{3}x$,\\\\\nAlso, since CE $= \\sqrt{3}$, it follows that $x = 1$, hence OC $= 1$,\\\\\nSince OC $\\parallel$ AD, it follows that $\\triangle EOC \\sim \\triangle EAD$,\\\\\nHence $\\frac{OC}{AD} = \\frac{OE}{AE}$, that is, $\\frac{1}{AD} = \\frac{2}{3}$,\\\\\nSolving gives $AD = \\boxed{\\frac{3}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51212783_84.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{3}}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AC>AB$, $AD$ is the angle bisector, $AE$ is the median, $BF\\perp AD$ at point $G$, intersects $AE$ at point $F$, and intersects $AC$ at point $M$, the extension line of $EG$ intersects $AB$ at point $H$. If $\\angle BAC=60^\\circ$, what is the value of $\\frac{FG}{DG}$?\n",
    "completion": "\\textbf{Solution:} Extend EH to P such that GH=HP, and connect AP and BP,\\\\\nsince BF$\\perp$AD,\\\\\nit follows that $\\angle$ABG+$\\angle$BAG=$90^\\circ$,\\\\\n$\\angle$AMG+$\\angle$MAG=$90^\\circ$,\\\\\nsince AD is the angle bisector,\\\\\nit follows that $\\angle$BAG=$\\angle$MAG,\\\\\nthus, $\\angle$ABG=$\\angle$AMG,\\\\\nthus, AB=AM,\\\\\nthus, BG=MG,\\\\\nsince BE=EC,\\\\\nit follows that GE$\\parallel$AC,\\\\\nthus, AH=BH,\\\\\nthus, Quadrilateral APBG is a parallelogram,\\\\\nthus, AP=BG, and AP$\\parallel$BG,\\\\\nthus, $\\frac{GF}{PA}=\\frac{GE}{PE}$,\\\\\nthus, $\\frac{GF}{BG}=\\frac{GE}{PE}$,\\\\\nSimilarly, $\\frac{GD}{AG}=\\frac{GE}{PE}$,\\\\\nthus, $\\frac{GF}{BG}=\\frac{GD}{AG}$,\\\\\nthus, $\\frac{GF}{GD}=\\frac{BG}{AG}$,\\\\\nsince $\\angle$BAC=$60^\\circ$ and AD is the angle bisector,\\\\\nit follows that $\\angle$BAG=$30^\\circ$,\\\\\nIn $\\triangle$ABG, $\\frac{BG}{AG}=\\tan 30^\\circ=\\frac{\\sqrt{3}}{3}$,\\\\\nthus, $\\frac{FG}{DG}=\\boxed{\\frac{\\sqrt{3}}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51099866.png"
  },
  {
    "solution": "\\boxed{100}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral ABCD, AC and BD intersect at point O, where O is the midpoint of AC, AB$\\parallel$DC, AC=20, and BD=10. Quadrilateral ABCD is a parallelogram; if AC$\\perp$BD, find the area of parallelogram ABCD.\n",
    "completion": "\\textbf{Solution:} It follows from equation (1) that quadrilateral ABCD is a parallelogram. Furthermore, since $AC \\perp BD$, the parallelogram ABCD is thereby a rhombus. Given that $AC=20$, and $BD=10$, \\\\\nthe area of quadrilateral ABCD is $=\\frac{1}{2}AC \\times BD=\\frac{1}{2} \\times 20 \\times 10=\\boxed{100}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52065947_97.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the circle $\\odot O$, $PA$ is a diameter, $PC$ is a chord, $PH$ bisects $\\angle APB$ and intersects $\\odot O$ at point $H$. A line is drawn through $H$ such that $HB\\perp PC$ and it intersects the extension of $PC$ at point $B$. $HB$ is a tangent to the circle $\\odot O$; if $HB=4$ and $BC=2$, what is the diameter of $\\odot O$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw OE$\\perp$PC with E being the foot of the perpendicular from O, \\\\\nsince OE$\\perp$PC, OH$\\perp$BH, BP$\\perp$BH, \\\\\nthus, quadrilateral EOHB is a rectangle, \\\\\ntherefore, OE=BH=4, OH=BE, \\\\\ntherefore, CE=OH-2, \\\\\nsince OE$\\perp$PC, \\\\\ntherefore, PE=EC=OH-2=OP-2, \\\\\nin $\\triangle POE$, $OP^{2}=PE^{2}+OE^{2}$, \\\\\ntherefore, $OP^{2}=(OP-2)^{2}+16$, \\\\\ntherefore, OP=5, \\\\\ntherefore, AP=2OP=\\boxed{10}, \\\\\ntherefore, the diameter of $\\odot O$ is \\boxed{10}.",
    "image_path": "dataset/mm_math/images/MM_Math/53080618_61.png"
  },
  {
    "solution": "\\boxed{y_2=x-3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the graphs of the functions $y_1=2x+b$ and $y_2=ax-3$ intersect at the point $P(-2, -5)$. The graphs of these two functions intersect the x-axis at points A and B, respectively. Determine the analytic expressions of these two functions.\n",
    "completion": "\\textbf{Solution:} Since substituting point $P(-2, -5)$ into $y_1=2x+b$ gives $-5=2\\times (-2)+b$, solving this equation yields $b=-1$.\\\\\nSubstituting point $P(-2, -5)$ into $y_2=ax-3$ gives $-5=a\\times (-2)-3$, solving this equation yields $a=1$.\\\\\nTherefore, the equations of the two functions are $\\boxed{y_1=2x-1}$ and $\\boxed{y_2=x-3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53706598_202.png"
  },
  {
    "solution": "\\boxed{100^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, AB is a chord of $\\odot O$ (AB is not a diameter). Taking point A as the center and the length of AB as the radius, an arc is drawn to intersect $\\odot O$ at point C. Connect AC, BC, OB, and OC. If $\\angle ABC=65^\\circ$, what is the degree measure of $\\angle BOC$?\n",
    "completion": "\\textbf{Solution:} From the given, we can derive: AB=AC, \\\\\n$\\because$ $\\angle$ABC=65°, \\\\\n$\\therefore$ $\\angle$ACB=65°, \\\\\n$\\therefore$ $\\angle$A=50°, \\\\\n$\\therefore$ $\\angle$BOC=\\boxed{100^\\circ}",
    "image_path": "dataset/mm_math/images/MM_Math/52623584.png"
  },
  {
    "solution": "\\boxed{3\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in rhombus $ABCD$ with a perimeter of 24, $\\angle BAD=120^\\circ$, points E and F are moving points on sides $AB, AD$ respectively, and point P is a moving point on diagonal $BD$. What is the minimum value of the segment $PE+PF$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect $AC$, intersecting $BD$ at point $O$. Draw a perpendicular line from $AE'$ to $BC$ at foot $E'$, intersecting $BD$ at point $P$.\\\\\nWhen point $F$ coincides with point $A$, draw a perpendicular line from $E'$ to $BD$, extending it to intersect $AB$ at point $E$. The sum of $PE+PF$ achieves its minimum value, which is $AE'$.\\\\\nSince the perimeter of the rhombus $ABCD$ is 24 and $\\angle BAD=120^\\circ$\\\\\nTherefore, $AB=BC=6$ and $\\angle BAC=\\frac{1}{2}\\angle BAD=60^\\circ$,\\\\\nThus, $\\triangle ABC$ is an equilateral triangle,\\\\\nTherefore, $AB=AC$,\\\\\nSince $AE'\\perp BC$ at foot $E'$,\\\\\nTherefore, $BE'=CE'=\\frac{1}{2}BC=3$,\\\\\nIn $\\triangle ABC$, by Pythagoras' theorem,\\\\\n$AE'=\\sqrt{AB^2-BE'^2}=\\sqrt{6^2-3^2}=3\\sqrt{3}$.\\\\\nHence, the minimum value of $PE+PF$ is $\\boxed{3\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52416277.png"
  },
  {
    "solution": "\\boxed{(6,6)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, the graph of the linear function $y=kx-3$ ($k\\neq 0$) intersects the y-axis at point B and intersects the graph of the inverse proportion function $y=\\frac{m}{x}$ ($x>0$) at point A (8, 1). Assuming condition (2), by translating $\\triangle OCD$ a certain distance in the direction of ray BA, we obtain $\\triangle O'CD'$. If the corresponding point of point O, O', falls exactly on the graph of the inverse proportion function (as shown in Figure 2), determine the coordinates of points O' and D'.\n",
    "completion": "\\textbf{Solution:} Connect $OO'$, due to the translation we have: $OO' \\parallel AC$, \\\\\ntherefore, the equation of line $OO'$ is $y= \\frac{1}{2}x$, \\\\\nsolving simultaneously, we get: $\\left\\{\\begin{array}{l}y=\\frac{8}{x}\\\\ y=\\frac{1}{2}x\\end{array}\\right.$, \\\\\nthis yields: $\\left\\{\\begin{array}{l}x=4\\\\ y=2\\end{array}\\right.$ or $\\left\\{\\begin{array}{l}x=-4\\\\ y=-2\\end{array}\\right.$ (discard as it is irrelevant to the question), \\\\\ntherefore, the coordinates of $O'$ are $\\boxed{(4,2)}$, \\\\\nwhich means point $O$ (0,0) is translated 4 units to the right and 2 units up to obtain $O'$ (4,2), \\\\\nfurthermore, from (2) we know the coordinates of D are (2,4), \\\\\ntherefore, point D (2,4) translated 4 units to the right and 2 units up gives the coordinates of $D'$ as $\\boxed{(6,6)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51445655_276.png"
  },
  {
    "solution": "\\boxed{\\frac{10}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, it is known that $AB$ is the diameter of circle $O$, and chord $CD$ intersects diameter $AB$ at point $E$, with $AC=CD$. Extend line $CO$ to intersect chord $AD$ at point $F$, and connect $AC$, $BC$, $BD$. Given $AB=10$ and $BE=2$, what is the length of $BD$?\n",
    "completion": "\\textbf{Solution:} Since the diameter AB $= 10$, it follows that OA $=$ OB $=$ OC $= 5$ and $\\angle ADB = \\angle ACB = 90^\\circ$. Since BE $= 2$, it follows that OE $=$ OB - EB $= 3$. Since AC $=$ CD, it implies $\\overset{\\frown}{AC} = \\overset{\\frown}{CD}$, thus CF $\\perp$ AD, leading to AF $=$ DF. Since $\\angle AFC = \\angle ADB = 90^\\circ$, it follows that CF $\\parallel$ BD,\n\\[\n\\therefore \\frac{OC}{BD}=\\frac{OE}{EB},\n\\]\n\\[\n\\therefore \\frac{5}{BD}=\\frac{3}{2}, \\therefore BD=\\boxed{\\frac{10}{3}}.",
    "image_path": "dataset/mm_math/images/MM_Math/51317366_154.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, $AB \\parallel CD$, $\\angle B = 90^\\circ$, $AB = 1$, $CD = 2$, $BC = m$. Point $P$ is a moving point on side $BC$. If $\\triangle PAB$ is similar to $\\triangle PCD$, and there are exactly two points $P$ that satisfy this condition, what is the value of $m$?\n",
    "completion": "\\textbf{Solution}: When $\\angle BAP = \\angle CPD$, then $\\triangle PAB \\sim \\triangle PDC$,\\\\\n$\\therefore \\frac{AB}{DC} = \\frac{BP}{PC}$, that is, $\\frac{1}{2} = \\frac{x}{m-x}$,\\\\\n$\\therefore x = \\frac{m}{3}$;\\\\\nWhen $\\angle BAP = \\angle CPD$, then $\\triangle PAB \\sim \\triangle DPC$,\\\\\n$\\therefore \\frac{AB}{CP} = \\frac{BP}{CD}$, that is, $\\frac{1}{m-x} = \\frac{x}{2}$,\\\\\n$\\therefore x^2 - mx + 2 = 0$,\\\\\n$\\because$ there are exactly two points P that satisfy the condition,\\\\\n$\\therefore$ the value of $x$ from (1) is a solution in (2), $\\left(\\frac{m}{3}\\right)^2 - m \\times \\frac{m}{3} + 2 = 0$, yielding $m = 3$ or $-3$ (discard);\\\\\nEquation in (2) has two equal real roots,\\\\\n$\\therefore \\Delta = m^2 - 4 \\times 2 = 0$, yielding $m = 2\\sqrt{2}$ or $-2\\sqrt{2}$ (discard).\\\\\nHence, the answer is: $m = \\boxed{3}$ or $\\boxed{2\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51368294.png"
  },
  {
    "solution": "\\boxed{40^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, a rectangular sheet of paper $ABCD$ is folded along $EF$, such that vertices $C$ and $D$ respectively fall on points $C'$ and $D'$. The line $CE$ intersects $AF$ at point $G$, and $\\angle CEF = 70^\\circ$. What is the value of $\\angle GFD'$?\n",
    "completion": "\\textbf{Solution:} Since $AD \\parallel BC$,\\\\\nit follows that $\\angle DFE + \\angle CEF = 180^\\circ$, $\\angle AFE = \\angle CEF$,\\\\\ngiven that $\\angle CEF = 70^\\circ$,\\\\\nthus $\\angle DFE = 180^\\circ - \\angle CEF = 180^\\circ - 70^\\circ = 110^\\circ$, $\\angle AFE = 70^\\circ$,\\\\\nFrom the fold, it is known that: $\\angle D'FE = \\angle DFE = 110^\\circ$, therefore $\\angle GFD' = \\angle D'FE - \\angle AFE = 110^\\circ - 70^\\circ = \\boxed{40^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51212158.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, segment $AC=6\\,\\text{cm}$, segment $BC=15\\,\\text{cm}$, point $M$ is the midpoint of $AC$, and a point $N$ is taken on $CB$ such that $CN\\colon NB=1\\colon 2$, find the length of $MN$.\n",
    "completion": "\\textbf{Solution:} Since $M$ is the midpoint of $AC$, and $AC=6$cm,\\\\\nit follows that $MC=\\frac{1}{2}AC=6\\times\\frac{1}{2}=3$cm.\\\\\nAlso, since $CN:NB=1:2$, and $BC=15$cm,\\\\\nit follows that $CN=15\\times\\frac{1}{3}=5$cm,\\\\\ntherefore $MN=MC+CN=3+5=\\boxed{8}$cm.",
    "image_path": "dataset/mm_math/images/MM_Math/53206178_98.png"
  },
  {
    "solution": "\\boxed{3\\colon 5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given points $A$, $B$, $C$, and $D$ lie on the same straight line, with point $C$ being the midpoint of segment $AB$, and point $D$ lies on segment $AB$. If point $E$ is a point on segment $AB$, and $AE=2BE$, when $AD:BD=2:3$, what is the value of $CD:CE$?\n",
    "completion": "\\textbf{Solution:} Let $AD=2x$, $BD=3x$, then $AB=5x$.\\\\\nSince point C is the midpoint of line segment AB,\\\\\nit follows that $AC=\\frac{1}{2}AB=\\frac{5}{2}x$,\\\\\nthus $CD=AC-AD=\\frac{1}{2}x$,\\\\\nsince $AE=2BE$,\\\\\nit follows that $AE=\\frac{2}{3}AB=\\frac{10}{3}x$,\\\\\nthus $CE=AE-AC=\\frac{10}{3}x-\\frac{5}{2}x=\\frac{5}{6}x$,\\\\\nthus $CD\\colon CE=\\frac{1}{2}x\\colon \\frac{5}{6}x=\\boxed{3\\colon 5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50810294_200.png"
  },
  {
    "solution": "\\boxed{120^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, let P be any point on side BC of the equilateral $\\triangle ABC$. Point D lies on the extension of side BA. Rotate line segment PD $60^\\circ$ counterclockwise around point P to obtain line segment PE. Connect BE. What is the measure of $\\angle CBE$ in degrees?\n",
    "completion": "\\textbf{Solution:} Draw line $DF \\parallel AC$ passing through point $D$ and intersecting the extension of $BC$ at point $F$, as shown in the figure: \\\\\n$\\because$ $\\triangle ABC$ is an equilateral triangle,\\\\\n$\\therefore$ $\\angle ABC = \\angle BAC = \\angle BCA = 60^\\circ$,\\\\\n$\\therefore$ $\\angle BAC = \\angle BDF = \\angle BCA = \\angle F = 60^\\circ$,\\\\\n$\\therefore$ $\\triangle DBF$ is an equilateral triangle,\\\\\n$\\therefore$ $DB = DF$,\\\\\n$\\because$ rotating segment $PD$ $60^\\circ$ counterclockwise around point $P$ results in segment $PE$,\\\\\n$\\therefore$ $\\triangle PDE$ is an equilateral triangle,\\\\\n$\\therefore$ $DE = DP$, $\\angle EDP = 60^\\circ$,\\\\\n$\\therefore$ $\\angle EDB + \\angle BDP = \\angle PDF + \\angle BDP = 60^\\circ$,\\\\\n$\\therefore$ $\\angle EDB = \\angle PDF$,\\\\\n$\\therefore$ $\\triangle EDB \\cong \\triangle PDF$ (SAS),\\\\\n$\\therefore$ $\\angle EBD = \\angle F = 60^\\circ$,\\\\\n$\\therefore$ $\\angle CBE = \\angle ABC + \\angle EBD = \\boxed{120^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/51368145.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the sides of the rhombus OABC are such that OC is on the x-axis, and B is at (18, 6). The graph of the inverse proportion function $y=\\frac{k}{x} (k \\neq 0)$ passes through point A and intersects OB at point E. What is the length of the sides of the rhombus OABC?\n",
    "completion": "\\textbf{Solution:} Draw line $BF \\perp x$-axis at point F, according to the given information, we have $BF=6$ and $OF=18$.\\\\\nSince quadrilateral OABC is a rhombus,\\\\\nit follows that $OC=BC$.\\\\\nIn $\\triangle BCF$, $6^{2}+(18-BC)^{2}=BC^{2}$.\\\\\nSolving this, we find $BC=\\boxed{10}$,\\\\\nthus, the side length of rhombus OABC is $\\boxed{10}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51433106_155.png"
  },
  {
    "solution": "\\boxed{35^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, AB is the diameter of $\\odot O$, CD is a chord of $\\odot O$, $\\angle CAB=55^\\circ$, then what is the degree measure of $\\angle D$?\n",
    "completion": "\\textbf{Solution:} Given that $AB$ is the diameter of $\\odot O$,\\\\\nit follows that $\\angle ACB=90^\\circ$,\\\\\ngiven $\\angle CAB=55^\\circ$,\\\\\nthus $\\angle B=90^\\circ-\\angle CAB=35^\\circ$,\\\\\ntherefore, $\\angle D=\\angle B=\\boxed{35^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52765424.png"
  },
  {
    "solution": "\\boxed{14}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the hyperbola $y=\\frac{k}{x}$ passes through the midpoint A of the hypotenuse of the right triangle $\\triangle BOC$, and intersects with BC at point D. Given that the area of $\\triangle BOD$ is 21, what is the value of $k$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, construct $AE\\perp$ to the $x$-axis, then the area of $\\triangle AOE$ and $\\triangle DOC$ is $\\frac{1}{2}|k|$, \\\\\n$\\therefore$ the area of the quadrilateral $BAEC$ and $\\triangle BOD$ is $21$, \\\\\n$\\because$ $AE\\perp x$ axis, \\\\\n$\\angle BCO=90^\\circ$, and point A is the midpoint of OB, \\\\\n$\\therefore$ $\\angle AOE = \\angle BOC$, \\\\\n$\\angle AEO = \\angle BCO = 90^\\circ$ \\\\\n$\\therefore$ $\\triangle AOE \\sim \\triangle BOC$, \\\\\n$\\therefore$ $\\frac{{S}_{\\triangle AOE}}{{S}_{\\triangle BOC}} = \\left(\\frac{OA}{OB}\\right)^{2} = \\frac{1}{4}$, \\\\\n$\\because$ the area of quadrilateral $BAEC$ times the area of $\\triangle AOE$ equals the area of $\\triangle BOC$, \\\\\n$\\therefore$ $\\frac{{S}_{\\triangle AOE}}{{S}_{\\text{quadrilateral }BAEC}} = \\frac{{S}_{\\triangle AOE}}{21} = \\frac{1}{3}$, \\\\\n$\\therefore$ ${S}_{\\triangle AOE} = 7$, \\\\\n$\\therefore$ $\\frac{1}{2}|k| = 7$, \\\\\nSolving for $k$, we get: $k = \\pm 14$, \\\\\n$\\because$ the inverse proportion function passes through the first quadrant, \\\\\n$\\therefore$ $k = \\boxed{14}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51434121.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=8$, $\\angle B=90^\\circ$. Point $P$ is a moving point on side $AB$, and through point $P$, $PD \\parallel AC$ intersects $BC$ at $D$, and $PE \\parallel BC$ intersects $AC$ at $E$. When point $P$ is the midpoint of $AB$, $PE=3$. $\\triangle APE \\sim \\triangle PBD$; find the value of $AP$ for which the quadrilateral $PDCE$ is a rhombus.\n",
    "completion": "\\textbf{Solution:} Let $AP=x$, then $BP=8-x$\\\\\nSince $PE\\parallel BC$, $PD\\parallel AC$\\\\\nTherefore, quadrilateral PDCE is a parallelogram\\\\\nTherefore, $CD=PE$\\\\\nSince $\\triangle APE\\sim \\triangle PBD$\\\\\nTherefore, $\\frac{AP}{PB}=\\frac{PE}{BD}$\\\\\ni.e., $\\frac{x}{8-x}=\\frac{PE}{6-CD}$\\\\\n$\\frac{x}{8-x}=\\frac{PE}{6-PE}$\\\\\nTherefore, $PE=\\frac{3}{4}x$\\\\\nTherefore, $CD=PE=\\frac{3}{4}x$\\\\\nTherefore, $BD=BC-CD=6-\\frac{3}{4}x$\\\\\nWhen $PE=PD$, parallelogram PDCE becomes a rhombus.\\\\\nBy the Pythagorean theorem, $PB^{2}+BD^{2}=PD^{2}$\\\\\nTherefore, $(8-x)^{2}+(6-\\frac{3}{4}x)^{2}=(\\frac{3}{4}x)^{2}$\\\\\nSimplify to get $x^{2}-25x+100=0$\\\\\nSolving this equation gives $x_{1}=5$, $x_{2}=20$\\\\\n$x=20>8$ is not feasible, disregard it.\\\\\nTherefore, when $AP=5$, quadrilateral PDCE is a rhombus.\\\\\nSolutions 2, 3, and 4 are omitted, as they are identical to Solution 1.\\\\\nTherefore, when $AP=\\boxed{5}$, quadrilateral PDCE is a rhombus.",
    "image_path": "dataset/mm_math/images/MM_Math/51212675_81.png"
  },
  {
    "solution": "\\boxed{49^\\circ54'}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the straight lines $AB$ and $CD$ intersect at the point $O$, $OE \\perp CD$ with the foot of the perpendicular at point $O$. If $\\angle BOE = 40^\\circ 6'$, what is the degree measure of $\\angle AOC$?\n",
    "completion": "Solution: Since $OE \\perp CD$, \\\\\ntherefore $\\angle EOC = 90^\\circ$.\\\\\nSince $\\angle BOE = 40^\\circ6'$, \\\\\ntherefore $\\angle AOC = 180^\\circ - 90^\\circ - 40^\\circ6' = \\boxed{49^\\circ54'}$.",
    "image_path": "dataset/mm_math/images/MM_Math/54962529.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AD$ is the altitude of $\\triangle ABC$, $E$ is a point on $AC$, $BE$ intersects $AD$ at point $F$, and it is given that $BF=AC$, $FD=CD$. $\\triangle BFD \\cong \\triangle ACD$; what is the measure of $\\angle ABD$?\n",
    "completion": "\\textbf{Solution:} Given that $Rt\\triangle BFD\\cong Rt\\triangle ACD$,\\\\\nit follows that $BD=AD$,\\\\\nsince $\\angle ADB=90^\\circ$,\\\\\nit results in $\\angle ABD=\\angle BAD=\\frac{1}{2}(180^\\circ-\\angle ADB)=\\frac{1}{2}\\times (180^\\circ-90^\\circ)=45^\\circ$,\\\\\nhence, the measure of $\\angle ABD$ is $\\boxed{45^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52950966_87.png"
  },
  {
    "solution": "\\boxed{-\\frac{8}{5}t+\\frac{28}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, $AB=5$, and $BC=3$. Point $P$ starts from point $A$ and moves along $AC-CB-BA$ at a speed of 2 units per second until it stops at point $A$. When point $P$ does not coincide with the vertices of $\\triangle ABC$, a perpendicular line is drawn from point $P$ to the side it is on, intersecting another side of $\\triangle ABC$ at point $Q$. Let the movement time of point $P$ be $t$ seconds. When point $P$ is moving along the right-angle sides of $\\triangle ABC$, what is the distance from point $P$ to side $AB$? (Expressed in an algebraic equation containing $t$)\n",
    "completion": "\\textbf{Solution:} Let the distance from point P to side $AB$ be $h$.\n\\begin{enumerate}\n    \\item When point P moves on side $AC$, draw $PH\\perp AB$ at H, as shown in figure 1:\\\\\n    $\\because \\sin A=\\frac{PH}{AP}=\\frac{BC}{AB}=\\frac{3}{5}$, $AP=2t$,\\\\\n    $\\therefore PH=h=AP\\cdot \\sin A=2t\\times \\frac{3}{5}=\\boxed{\\frac{6}{5}t}$;\n    \\item When point P moves on side $BC$, draw $PH\\perp AB$ at H, as shown in figure 2:\\\\\n    $\\because \\sin B=\\frac{PH}{PB}=\\frac{AC}{AB}=\\frac{4}{5}$, $PB=7-2t$,\\\\\n    $\\therefore PH=h=PB\\cdot \\sin B=(7-2t)\\times \\frac{4}{5}=\\boxed{-\\frac{8}{5}t+\\frac{28}{5}}$;\n\\end{enumerate}\nIn conclusion, the distance from point P to side $AB$ is $\\boxed{\\frac{6}{5}t}$ or $\\boxed{-\\frac{8}{5}t+\\frac{28}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51318739_100.png"
  },
  {
    "solution": "\\boxed{2.8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated, fold the equilateral triangle $ABC$ such that point $A$ falls on point $D$ on side $BC$ (not coinciding with $B$ or $C$), with the fold line being $EF$. Under the condition mentioned in (2), find the length of $BE$.\n",
    "completion": "\\textbf{Answer:} Given $\\because \\triangle BDE\\sim \\triangle CFD$, and the perimeter of $\\triangle BDE$ is 14, and the perimeter of $\\triangle CFD$ is 10,\\\\\n$\\therefore \\frac{BE}{CD}=\\frac{14}{10}$,\\\\\n$\\because DC=2$,\\\\\n$\\therefore \\frac{BE}{2}=\\frac{14}{10}$,\\\\\n$\\therefore BE=\\boxed{2.8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53110772_170.png"
  },
  {
    "solution": "\\boxed{y=\\frac{1}{2}x}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, the point $A(4, n)$ lies on the graph of the inverse proportion function $y=\\frac{8}{x}$ ($x>0$). Find the value of $n$ and the equation of the line $OA$.\n",
    "completion": "\\textbf{Solution:} Let's substitute point A$(4,n)$ into $y=\\frac{8}{x}$, we get $n=2$;\\\\\nLet the equation of line OA be $y=kx$. Substitute $\\left\\{\\begin{array}{l}x=4\\\\ y=2\\end{array}\\right.$ into it, we have\\\\\n$4k=2$, solving this yields: $k=\\frac{1}{2}$,\\\\\n$\\therefore$ the equation of line OA is $y=\\frac{1}{2}x$;\\\\\n$\\therefore$ the value of $n$ is $\\boxed{2}$, and the equation of line $OA$ is $\\boxed{y=\\frac{1}{2}x}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52250480_91.png"
  },
  {
    "solution": "\\boxed{(1,4)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle AOB$ is an isosceles right triangle. If $A(-4, 1)$, what are the coordinates of point $B$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, draw AE$\\perp$x-axis at E through point A, and draw BF$\\perp$x-axis at F through point B,\\\\\n$\\therefore \\angle$AEO=$\\angle$OFB=$90^\\circ$,\\\\\n$\\therefore \\angle$AOE+$\\angle$OAE=$90^\\circ$,\\\\\nAlso, $\\because \\angle$AOB=$90^\\circ$,\\\\\n$\\therefore \\angle$AOE+$\\angle$BOF=$90^\\circ$,\\\\\n$\\therefore \\angle$OAE=$\\angle$BOF,\\\\\n$\\because$AO=OB,\\\\\n$\\therefore \\triangle$OAE$\\cong \\triangle$BOF (AAS),\\\\\n$\\therefore$OF=AE, BF=OE,\\\\\n$\\because$the coordinates of point A are (-4, 1),\\\\\n$\\therefore$OF=AE=1, BF=OE=4,\\\\\n$\\therefore$the coordinates of point B are $\\boxed{(1,4)}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51368165_109.png"
  },
  {
    "solution": "\\boxed{70^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $AB=AD$, $AE=AC=BC$, $\\angle 1=\\angle 2$, $\\angle C=40^\\circ$. What is the measure of $\\angle ADE$ in degrees?\n",
    "completion": "\\textbf{Solution:} Since $\\angle 1=\\angle 2$,\\\\\nit follows that $\\angle 1+\\angle DAC=\\angle 2+\\angle DAC$,\\\\\nwhich means $\\angle BAC=\\angle DAE$,\\\\\nIn $\\triangle ABC$ and $\\triangle ADE$,\\\\\nsince $\\left\\{\\begin{array}{l}AB=AD\\\\ \\angle BAC=\\angle DAE\\\\ AC=AE\\end{array}\\right.$,\\\\\nit follows that $\\triangle ABC\\cong \\triangle ADE$ (SAS),\\\\\nhence $\\angle C=\\angle E=40^\\circ$, $AE=BC=DE$,\\\\\ntherefore $\\angle ADE=\\angle EAD=\\frac{1}{2}\\left(180^\\circ-\\angle E\\right)=\\frac{1}{2}\\left(180^\\circ-40^\\circ\\right)=\\boxed{70^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52838009.png"
  },
  {
    "solution": "\\boxed{P_4\\left(2, 0\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the equation of line $l_{1}$ is $y=kx+b \\left(k \\ne 0\\right)$, and the equation of line $l_{2}$ is $y=x$. Line $l_{1}$ passes through point $A(6, 0)$ and $B(0, 3)$, and line $l_{1}$ intersects with line $l_{2}$ at point C. Is there a point P on the x-axis such that $\\triangle POC$ is an isosceles triangle? Find the coordinates of point P.\n",
    "completion": "\\textbf{Solution:} There exists a point $P$ on the $x$-axis such that $\\triangle POC$ is an isosceles triangle. The reasoning is as follows:\\\\\nBased on the problem statement, there are three scenarios:\\\\\n(1) When point $O$ is the vertex angle, $OP=OC$. The two intersection points $P_1$ and $P_2$ of the circle centered at $O$ with radius $OC$ and the $x$-axis are the desired points $P$. It follows that $OC=2\\sqrt{2}$.\\\\\n$\\therefore$ The coordinates of point $P_1$ are: $\\left(-2\\sqrt{2}, 0\\right)$, and the coordinates of point $P_2$ are: $\\left(2\\sqrt{2}, 0\\right)$.\\\\\n(2) When point $C$ is the vertex angle, $CP=CO$. The intersection point $P_3$ of the circle centered at $C$ with radius $CO$ and the $x$-axis is the desired point $P$. It follows that the coordinates of point $P_3$ are: $\\left(4, 0\\right)$.\\\\\n(3) When point $P$ is the vertex angle, the intersection point $P_4$ of the perpendicular bisector of $OC$ and the $x$-axis is the desired point $P$. It follows that the coordinates of point $P_4$ are: $\\left(2, 0\\right)$.\\\\\nCombining (1), (2), and (3), it can be concluded that there are four points $P$ that form an isosceles triangle $\\triangle POC$, which are: $\\boxed{P_1\\left(-2\\sqrt{2}, 0\\right)}$, $\\boxed{P_2\\left(2\\sqrt{2}, 0\\right)}$, $\\boxed{P_3\\left(4, 0\\right)}$, $\\boxed{P_4\\left(2, 0\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52386525_136.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within $\\triangle ABC$, AD and AE are the median and altitude from vertex A to side BC, respectively, with AE$=6$ and the area of $\\triangle ABD=S_{\\triangle ABD}=15$. What is the length of CD?\n",
    "completion": "\\textbf{Solution:} Since $S_{\\triangle ABC}=15$ and AE is the altitude on side BC, we have\n\\[\n\\frac{1}{2} \\times BD \\times AE=15,\n\\]\nwhich gives\n\\[\n\\frac{1}{2} \\times 6 \\times BD=15,\n\\]\nsolving this yields $BD=5$. Since AD is the median to side BC, it follows that $CD=BD=\\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52095985.png"
  },
  {
    "solution": "\\boxed{120^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points D and E are the trisection points of $BC$, and $\\triangle ADE$ is an equilateral triangle. What is the measure of $\\angle BAC$?\n",
    "completion": "Solution: Since $E$ is the trisecting point of $BC$, and $\\triangle ADE$ is an equilateral triangle,\\\\\n$\\therefore BD=DE=EC=AD=AE$, $\\angle ADE=\\angle AED=60^\\circ$,\\\\\n$\\therefore \\angle B=\\angle BAD=\\angle C=\\angle EAC=30^\\circ$,\\\\\n$\\therefore \\angle BAC=180^\\circ-\\angle B-\\angle C=\\boxed{120^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53033290.png"
  },
  {
    "solution": "\\boxed{1 + \\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, AC is the diagonal of square ABCD, E is a point on BC, with $BE=1$. By folding $\\triangle ABE$ along AE so that point B falls on point F on AC, what is the length of AB?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a square, \\\\\n$\\therefore$ AB = BC, $\\angle$BCD = $\\angle$B = $90^\\circ$, $\\angle$ECF = $\\frac{1}{2}$ $\\angle$BCD = $45^\\circ$,\\\\\nBy the property of folding: $\\angle$AFE = $\\angle$B = $90^\\circ$, FE = BE = 1,\\\\\n$\\therefore$ $\\angle$CFE = $90^\\circ$,\\\\\n$\\therefore$ $\\triangle$ CEF is an isosceles right-angled triangle,\\\\\n$\\therefore$ CE = $\\sqrt{2}$ FE = $\\sqrt{2}$,\\\\\n$\\therefore$ BC = BE + CE = 1 + $\\sqrt{2}$,\\\\\n$\\therefore$ AB = BC = \\boxed{1 + \\sqrt{2}};",
    "image_path": "dataset/mm_math/images/MM_Math/51863562.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the line $y=kx+5$ intersects the $x$-axis at point $A$ and the $y$-axis at point $B$, and the coordinates of point $A$ are $(5,0)$. The line $y=2x-4$ intersects the $x$-axis at point $D$ and intersects the line $AB$ at point $C$. What is the area of $\\triangle ADC$?\n",
    "completion": "\\textbf{Solution:} Substituting $y=0$ into $y=2x-4$, we get $2x-4=0$. Solving this yields $x=2$. Thus, the coordinates of point D are $(2,0)$. Since A$(5,0)$, it follows that $AD=3$. Given the coordinates of point C as $C(3,2)$, it follows that the area of $\\triangle ADC$ is $\\frac{1}{2} \\cdot 3 \\cdot 2 = \\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52416375_91.png"
  },
  {
    "solution": "\\boxed{\\frac{13}{12}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: During the comprehensive practice class, Xiaohui used two right-angled triangle paper pieces as shown in the figure: $\\angle A=90^\\circ$, $AD=2\\text{cm}$, $AB=3\\text{cm}$, and overlapped the hypotenuses to form a quadrilateral. Then, points $E$ and $F$ were chosen on $CB$ and $CD$, respectively, and lines $AE$ and $BF$ were drawn such that $AE\\perp BF$. If the quadrilateral formed is as shown in the figure, then what is the value of $\\frac{BF}{AE}$?\n",
    "completion": "\\textbf{Solution:} Therefore, $\\therefore \\frac{BF}{AE} = \\boxed{\\frac{13}{12}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51545297_62.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, it is known that the radius of circle $O$ is 5, and the chord $AB=4\\sqrt{5}$. Points C and D are on the major arc $\\overset{\\frown}{BCA}$ (points B, C, and D are arranged clockwise), $\\overset{\\frown}{CD}=\\overset{\\frown}{AB}$. Point F is a moving point on $\\overset{\\frown}{CD}$, connect CF and extend it to intersect the extension of chord AD at point E. $BC\\parallel AD$. As shown in Figure 2, connect $AC$, $BF$, $AF$. When $\\angle BFA$ is equal to one of the interior angles of $\\triangle ACE$, and the length of $BF$ can be determined, find all possible lengths of $BF$ that satisfy the condition.\n",
    "completion": "\\textbf{Solution}: In figure (2), \\\\\n$\\because \\overset{\\frown}{CD}=\\overset{\\frown}{AB}$, \\\\\n$\\therefore \\angle BFA=\\angle CAE$, $AB=CD$, \\\\\nThus, regardless of the value of $BF$, we always have $\\angle BFA=\\angle CAE$, which is inconsistent with the problem statement; \\\\\nWhen $\\angle BFA=\\angle ACE$, $\\because \\angle ACE=\\angle ABF$ \\\\\n$\\therefore \\angle BFA=\\angle ABF$, \\\\\n$\\therefore AB=AF$, thus $\\triangle BAF$ is an isosceles triangle, \\\\\nAs shown in the figure, draw $AH\\perp BF$ at H through A, then $BH=HF=\\frac{1}{2}BF$, \\\\\nExtend $AH$ to meet $\\odot O$ at P, connect $BP$, then $AP$ is the diameter, \\\\\n$\\therefore \\angle ABP=90^\\circ$, and since $AB=4\\sqrt{5}$, $AP=2AO=10$, \\\\\n$\\therefore BP=\\sqrt{AP^{2}-AB^{2}}=2\\sqrt{5}$, \\\\\nFrom $S_{\\triangle ABP}=\\frac{1}{2}AP\\cdot BH=\\frac{1}{2}AB\\cdot BP$, we get: $BH=\\frac{AB\\cdot BP}{AP}=\\frac{4\\sqrt{5}\\times 2\\sqrt{5}}{10}=4$, \\\\\n$\\therefore BF=2BH=\\boxed{8}$; \\\\\nWhen $\\angle BFA=\\angle E$, $\\because \\angle AFC=\\angle BFA+\\angle BFC=\\angle E+\\angle FAE$, \\\\\n$\\therefore \\angle BFC=\\angle FAE$ \\\\\n$\\therefore \\overset{\\frown}{BC}=\\overset{\\frown}{DF}$, thus $\\overset{\\frown}{BF}=\\overset{\\frown}{CD}$, \\\\\n$\\therefore BF=CD=AB=4\\sqrt{5}$, \\\\\nTo sum up, the length of $BF$ that satisfies the condition is either $8$ or $\\boxed{4\\sqrt{5}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52933588_182.png"
  },
  {
    "solution": "\\boxed{y=\\frac{x^2+6x+13}{7+x}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In quadrilateral $ABCD$, $AD \\parallel BC$, $AB= \\sqrt{5}$, $AD=2$, $DC= 2\\sqrt{5}$, and $\\tan \\angle ABC=2$ (as shown in the figure). Point $E$ is on the ray $AD$, and point $F$ is on side $BC$. $BE$ and $EF$ are connected, and $\\angle BEF=\\angle DCB$. When point $E$ is on the extension line of segment $AD$, let $DE=x$, $BF=y$. Find the analytical expression of $y$ as a function of $x$.\n",
    "completion": "\\textbf{Solution:} Construct line $EN\\parallel DC$ through point E, intersecting the extension of BC at point N.\\\\\nSince $DE\\parallel CN$,\\\\\ntherefore quadrilateral DCNE is a parallelogram,\\\\\nthus $DE=CN$, $\\angle DCB=\\angle ENB$,\\\\\nSince $\\angle FEB=\\angle DCB$,\\\\\nthus $\\angle FEB=\\angle ENB$,\\\\\nAlso, since $\\angle EBF=\\angle NBE$,\\\\\nthus $\\triangle BEF\\sim \\triangle BNE$,\\\\\nthus $\\frac{BF}{BE}=\\frac{BE}{BN}$,\\\\\nthus $BE^2=BF\\cdot BN$,\\\\\nConstruct $EM\\perp BC$ through point E, with M being the foot of the perpendicular. We find $EM=2$, $BM=x+3$.\\\\\nthus $BE^2=ME^2+BM^2=(x+3)^2+2^2=x^2+6x+13$,\\\\\nthus $y(7+x)=x^2+6x+13$,\\\\\nthus $y=\\frac{x^2+6x+13}{7+x}$\\\\\nWhen $y=7$, $x^2-x-36=0$,\\\\\nSolving yields: $x=\\frac{1+\\sqrt{145}}{2}$, (neglect the negative root)\\\\\nthus \\boxed{y=\\frac{x^2+6x+13}{7+x}} \\left(0<x\\le \\frac{1+\\sqrt{145}}{2}\\right).",
    "image_path": "dataset/mm_math/images/MM_Math/51379498_105.png"
  },
  {
    "solution": "\\boxed{\\frac{2-\\sqrt{2}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In real life, we often see many \"standard\" rectangles, such as the covers of our textbooks, A4 printing paper, etc. In fact, the ratio of the length to the width of these rectangles is $\\sqrt{2}: 1$. We might as well call such rectangles \"standard rectangles\". In the \"standard rectangle\" $ABCD$, as shown in the figure, point $Q$ is on $DC$, and $DQ=AD$. If $G$ is a moving point on side $BC$, when the perimeter of $\\triangle AGQ$ is minimized, what is the value of $\\frac{CG}{GB}$?\n",
    "completion": "\\textbf{Solution:} Let DC=$\\sqrt{2}x$, DQ=AD=$x$,\\\\\n$\\therefore$ $CQ=(\\sqrt{2}-1)x$\\\\\n$\\because$ In rectangle ABCD,\\\\\n$\\therefore$ $\\angle D=\\angle DCB=\\angle B=90^\\circ$, $AB=DC=\\sqrt{2}x$, $BC=AD=x$,\\\\\n$\\therefore$ $AQ=\\sqrt{AD^2+DQ^2}=\\sqrt{2}x$,\\\\\nAs shown in the figure, construct the symmetric point E of point Q with respect to BC, and connect AE to intersect BC at point M,\\\\\n$\\therefore$ GQ=GE, CQ=CE=$(\\sqrt{2}-1)x$\\\\\n$\\therefore$ AQ+QG+AG=$\\sqrt{2}x+AG+EG\\ge \\sqrt{2}x+AE$,\\\\\n$\\therefore$ When points A, G, and E are collinear, the perimeter of $\\triangle AGQ$ is minimized,\\\\\nAt this time, point G should be located at point M in the figure;\\\\\n$\\because$ In rectangle ABCD, $\\angle QCG=90^\\circ$,\\\\\n$\\therefore$ Point E lies on the extension line of QC,\\\\\n$\\therefore$ CE$\\parallel$AB,\\\\\n$\\therefore$ $\\frac{CM}{MB}=\\frac{CE}{AB}=\\frac{(\\sqrt{2}-1)x}{\\sqrt{2}x}=\\frac{2-\\sqrt{2}}{2}$,\\\\\nwhich means $\\frac{CG}{GB}=\\boxed{\\frac{2-\\sqrt{2}}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53308020.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\triangle ABC$, $CD$ is the altitude of $ \\triangle ABC$, $AE$ is the angle bisector of $ \\triangle ABC$, and $CD$ intersects $AE$ at point $G$. Given $\\angle BCD = 50^\\circ$ and $\\angle BEA = 110^\\circ$, find the measure of $\\angle ACD$.\n",
    "completion": "\\textbf{Solution:} By the given information, $CD$ is the height of $\\triangle ABC$,\\\\\nhence $\\angle BDC=\\angle ADC=90^\\circ$.\\\\\nTherefore, $\\angle B=90^\\circ-\\angle BCD=90^\\circ-50^\\circ=40^\\circ$.\\\\\nIn $\\triangle ABE$, $\\angle BAE=180^\\circ-\\angle B-\\angle BEA=180^\\circ-40^\\circ-110^\\circ=30^\\circ$.\\\\\nSince $AE$ is the angle bisector of $\\triangle ABC$,\\\\\nit follows that $\\angle BAC=2\\angle BAE=60^\\circ$.\\\\\nTherefore, $\\angle ACD=90^\\circ-\\angle BAC=\\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51369206.png"
  },
  {
    "solution": "\\boxed{\\frac{12}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=ax^{2}+bx+c$ passes through three points $A(-1,0)$, $B(4,0)$, and $C(0,3)$. $D$ is a moving point on the parabola above the line $BC$, and $DE\\perp BC$ at point $E$. What is the maximum length of the line segment $DE$?\n",
    "completion": "\\textbf{Solution:} Let the equation of line $BC$ be $y=kx+b$, with the system $\\left\\{\\begin{array}{l}4k+b=0\\\\ b=3\\end{array}\\right.$. Solving, we find $\\left\\{\\begin{array}{l}k=-\\frac{3}{4}\\\\ b=3\\end{array}\\right.$, $\\therefore y=-\\frac{3}{4}x+3$.\n\nLet $D(a,-\\frac{3}{4}a^2+\\frac{9}{4}a+3)$ (where $0<a<4$). Construct $DM\\perp x$-axis, intersecting $BC$ at point $M$, to have $M(a,-\\frac{3}{4}a+3)$.\n\nHence, $DM= (-\\frac{3}{4}a^2+\\frac{9}{4}a+3)-(-\\frac{3}{4}a+3)=-\\frac{3}{4}a^2+3a$.\n\nGiven $\\angle DME=\\angle OCB$ and $\\angle DEM=\\angle BOC$,\n\nit follows that $\\triangle DEM \\sim \\triangle BOC$,\n\nhence $\\frac{DE}{DM}=\\frac{OB}{BC}$.\n\nGiven that $OB=4$, $OC=3$,\n\nit follows that $BC=5$,\n\nhence $DE=\\frac{4}{5}DM$,\n\nthus $DE=-\\frac{3}{5}a^2+\\frac{12}{5}a=-\\frac{3}{5}(a-2)^2+\\frac{12}{5}$.\n\nWhen $a=2$, $DE$ reaches its maximum value, which is $\\boxed{\\frac{12}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52551774_96.png"
  },
  {
    "solution": "\\boxed{70^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $AB$ intersects with the line $MN$ at point $O$, $OC \\perp AB$, $OA$ bisects $\\angle MOD$, if $\\angle BON = 20^\\circ$, what is the measure of $\\angle COD$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle BON=20^\\circ$\\\\\nThus, $\\angle AOM=\\angle BON=20^\\circ$\\\\\nSince $OA$ bisects $\\angle MOD$,\\\\\nThus, $\\angle AOD=\\angle AOM=20^\\circ$\\\\\nSince $OC\\perp AB$\\\\\nThus, $\\angle COD=90^\\circ-\\angle AOD=90^\\circ-20^\\circ=\\boxed{70^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/53253509.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that points $C$, $D$, and $E$ lie on line $AB$, and point $D$ is the midpoint of segment $AC$, point $E$ is the midpoint of segment $DB$. If point $C$ is on segment $EB$ and $DB=6$, $CE=1$, what is the length of segment $AB$?\n",
    "completion": "\\textbf{Solution:} Since point $E$ is the midpoint of line segment $DB$ and $DB=6$,\\\\\nit follows that $DE=\\frac{1}{2}DB=\\frac{1}{2}\\times 6=3$,\\\\\nsince $EC=1$,\\\\\nthen $DC=DE+EC=3+1=4$,\\\\\nsince point $D$ is the midpoint of line segment $AC$,\\\\\nit follows that $AD=DC=4$,\\\\\ntherefore $AB=AD+DB=4+6=\\boxed{10}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51429265.png"
  },
  {
    "solution": "\\boxed{110^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of $\\odot O$, with points $C, D$ on $\\odot O$, and $\\angle ACD=35^\\circ$, then what is the degree measure of $\\angle BOD$?\n",
    "completion": "\\textbf{Solution:} Since $\\overset{\\frown}{AD}=\\overset{\\frown}{AD}$,\\\\\n$\\therefore \\angle AOD=2\\angle ACD=2\\times 35^\\circ=70^\\circ$,\\\\\n$\\therefore \\angle BOD=180^\\circ-\\angle AOD=180^\\circ-70^\\circ=\\boxed{110^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52774651.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, if $DE\\parallel BC$, $\\frac{AD}{BD}=\\frac{1}{2}$, and $DE=4\\,\\text{cm}$, then what is the length of $BC$ in $\\,\\text{cm}$?\n",
    "completion": "\\textbf{Solution:} Given $DE\\parallel BC$,\\\\\ntherefore $\\triangle ADE \\sim \\triangle ABC$,\\\\\nthus $\\frac{DE}{BC} = \\frac{AD}{AB}$,\\\\\nalso, given $\\frac{AD}{DB} = \\frac{1}{2}$,\\\\\nthus $\\frac{AD}{AB}=\\frac{1}{3}$,\\\\\nhence $\\frac{4}{BC}=\\frac{1}{3}$,\\\\\ntherefore $BC=\\boxed{12}\\,\\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52623488.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC \\cong \\triangle DEF$, where point A corresponds to point D, and point B corresponds to point E. If $\\angle A=70^\\circ$ and $\\angle B=50^\\circ$, what is the measure of $\\angle 1$?\n",
    "completion": "\\textbf{Solution:} In $\\triangle ABC$, $\\angle A=70^\\circ$, $\\angle B=50^\\circ$,\\\\\nthus $\\angle C=180^\\circ-\\angle A-\\angle B=180^\\circ-70^\\circ-50^\\circ=60^\\circ$,\\\\\nsince $\\triangle ABC \\cong \\triangle DEF$,\\\\\nit follows that $\\angle 1=\\angle C=\\boxed{60^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/52836818.png"
  },
  {
    "solution": "\\boxed{\\frac{10}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of $\\odot O$, point $F$ is on $\\odot O$, $AE$ bisects $\\angle CAD$ and intersects $\\odot O$ at point $E$. Through point $E$, construct $ED\\perp AF$, and extend it to meet the extension of $AF$ at point $D$. Extend $DE$ and $AB$ to meet at point $C$. $CD$ is a tangent to $\\odot O$; given that $AB=10, \\frac{ED}{AD}=\\frac{1}{2}$, find the length of $BC$.\n",
    "completion": "\\textbf{Solution:} Draw BE. Since AB is the diameter,\\\\\n$\\angle AEB=90^\\circ=\\angle D$, and $\\angle DAE=\\angle BAE$,\\\\\nthus $\\triangle ADE\\sim \\triangle AEB$,\\\\\nthus $\\frac{AD}{AE}=\\frac{AE}{AB}=\\frac{DE}{BE}$,\\\\\nand $\\frac{ED}{AD}=\\frac{1}{2}$,\\\\\nthus $\\frac{DE}{AD}=\\frac{BE}{AE}=\\frac{1}{2}$, hence AE=2BE, and AB=10,\\\\\nin $\\triangle ABE$, $AE^2+BE^2=AB^2$, that is $(2BE)^2+BE^2=10^2$,\\\\\nsolving gives: BE=$2\\sqrt{5}$, then AE=$4\\sqrt{5}$,\\\\\nthus $\\frac{AD}{4\\sqrt{5}}=\\frac{4\\sqrt{5}}{10}=\\frac{DE}{2\\sqrt{5}}$,\\\\\nsolving gives: AD=8, DE=4,\\\\\nsince OE$\\parallel$AD,\\\\\n$\\triangle COE\\sim \\triangle CAD$,\\\\\nthus $\\frac{CO}{CA}=\\frac{OE}{AD}$, let BC=x,\\\\\nthus $\\frac{x+5}{x+10}=\\frac{5}{8}$, solving gives: x=$\\boxed{\\frac{10}{3}}$,\\\\\nupon verification: x=$\\frac{10}{3}$ is the solution to the original equation,\\\\\ntherefore, the length of BC is $\\boxed{\\frac{10}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51446284_106.png"
  },
  {
    "solution": "\\boxed{\\frac{18}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in $\\triangle ABC$, where $AB=AC=10$ and $BC=12$, points $D$ and $E$ are located on sides $BC$ and $AC$ respectively (with point $D$ being distinct from endpoints $B$ and $C$), and it is given that $\\angle ADE = \\angle B$. $\\triangle ABD \\sim \\triangle DCE$; Let $BD=x$ and $CE=y$, determine the value of $x$ for which $y$ attains its maximum value. What is the maximum value of $y$?\n",
    "completion": "Solution: Let $BD=x$, $CE=y$, \\\\\nTherefore, $DC=BC-BD=12-x$, \\\\\nFrom (1), we know that $\\triangle ABD \\sim \\triangle DCE$, \\\\\nTherefore, $\\frac{AB}{CD}=\\frac{BD}{CE}$, \\\\\nTherefore, $\\frac{10}{12-x}=\\frac{x}{y}$, \\\\\nTherefore, $y=-\\frac{1}{10}x^2+\\frac{6}{5}x=-\\frac{1}{10}(x-6)^2+\\frac{18}{5}$, \\\\\nSince $0<x<12$ and $-\\frac{1}{10}<0$, \\\\\nTherefore, when $x=\\boxed{6}$, $y$ reaches its maximum value, which is $\\boxed{\\frac{18}{5}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52601788_184.png"
  },
  {
    "solution": "\\boxed{67.5^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ is an isosceles right triangle with $\\angle ABC=90^\\circ$. $\\triangle ABC$ is folded along a straight line such that the corresponding point of vertex $A$, denoted as ${A}^{\\prime}$, lands exactly on side $BC$. This crease intersects $AB$ and $AC$ at points $D$ and $E$, respectively. The bisector of $\\angle ABC$ intersects ${A}^{\\prime}D$ at point $F$. Line $EF$ is drawn. If $CE={A}^{\\prime}B$, what is the degree measure of $\\angle {A}^{\\prime}EF$?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABC$ is an isosceles right-angled triangle, we fold $\\triangle ABC$ along a straight line such that vertex $A'$ falls exactly on side $BC$, where this fold line crosses $AB$ and $AC$ at points D and E, respectively,\\\\\n$\\therefore$ $\\angle A = \\angle C = \\angle DA'E = 45^\\circ$,\\\\\n$\\angle FA'B + \\angle EA'C = 180^\\circ - 45^\\circ = 135^\\circ$,\\\\\nSince BF bisects $\\angle ABC$,\\\\\n$\\therefore$ $\\angle A'BF = \\frac{1}{2} \\angle ABC = 45^\\circ$,\\\\\n$\\therefore$ $\\angle BFA' + \\angle BA'F = 180^\\circ - 45^\\circ = 135^\\circ$,\\\\\n$\\therefore$ $\\angle BFA' = \\angle EA'C$,\\\\\nIn $\\triangle A'FB$ and $\\triangle EA'C$,\\\\\n$\\left\\{\\begin{array}{l}\n\\angle BFA' = \\angle EA'C\\\\\n\\angle A'BF = \\angle C\\\\\nA'B = CE\n\\end{array}\\right.$\\\\\n$\\therefore$ $\\triangle A'FB \\cong \\triangle EA'C$ (AAS),\\\\\n$\\therefore$ $A'F = A'E$,\\\\\n$\\therefore$ $\\angle A'EF = \\angle A'FE = \\frac{1}{2}(180^\\circ - \\angle DA'E) = \\frac{1}{2}(180^\\circ - 45^\\circ) = \\boxed{67.5^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52838018.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $P$ is on the positive $y$-axis, $\\odot P$ intersects the $x$-axis at points $B$ and $C$. With $AC$ as one of its right-angled sides, an isosceles $\\triangle ACD$ is constructed. $BD$ intersects the $y$-axis and $\\odot P$ at points $E$ and $F$, respectively. $AC$ and $FC$ are connected, and $AC$ intersects $BD$ at point $G$. $\\angle ACF=\\angle ADB$; $CF=DF$; what is the degree measure of $\\angle DBC$?\n",
    "completion": "\\textbf{Solution:} We have $\\angle DBC = \\boxed{45^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53010347_74.png"
  },
  {
    "solution": "\\boxed{\\frac{2}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the rectangle $ABCD$, where $AB=8$ and $BC=6$, a circular arc $BH$ is drawn with vertex $C$ as the center and the length of $BC$ as the radius. A tangent $PE$ to the arc $BH$ is drawn passing through the midpoint $P$ of $AB$, with $E$ as the point of tangency. Connect $AE$ and extend it to intersect $CD$ at point $F$. What is the numerical value of $\\tan\\angle DAF$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, connect CE, CP, and BE. \\\\\nSince PE and BC are the tangents to circle C, \\\\\nand PE = PB, \\\\\nand BC = BE, \\\\\ntherefore, CP is the perpendicular bisector of BE, \\\\\ntherefore, CP $\\perp$ BE, \\\\\nSince PA = PB = PE, \\\\\ntherefore, $\\angle$AEB$=90^\\circ$, that is, BE $\\perp$ AF, \\\\\ntherefore, AF $\\parallel$ BC, \\\\\nsince AP $\\parallel$ FC, \\\\\ntherefore, the quadrilateral APCF is a parallelogram, \\\\\nthus, FC = AP = 2, \\\\\ntherefore, FD = CD - FC = 8 - 4 = 4, \\\\\ntherefore, $\\tan\\angle$DAF$= \\frac{FD}{AD}=\\frac{4}{6}=\\boxed{\\frac{2}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51109760.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{10}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the side length of square $ABCD$ is $4$, and $E$, $F$, $G$ are moving points on sides $AB$, $BC$, $AD$, respectively, with $AE=BF$. Fold $\\triangle BEF$ inward along $EF$ to $\\triangle B^{\\prime}EF$, and connect $BB^{\\prime}$, $B^{\\prime}G$, and $GC$. What is the minimum value of $B^{\\prime}G+GC$ when $BB^{\\prime}$ is at its maximum?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, when the quadrilateral B$^{\\prime}$EBF is a square, BB$^{\\prime}$ is maximized. \\\\\n$\\therefore$ BE = BF, \\\\\n$\\because$ AE = BF, \\\\\n$\\therefore$ AE = BE, \\\\\n$\\therefore$ E, F are respectively the midpoints of sides AB, BC, \\\\\nDraw a line through B$^{\\prime}$ perpendicular to CD at point H, \\\\\nThen AE = BE = BF = B$^{\\prime}$H = CH = $\\frac{1}{2}$ BC = 2, \\\\\nConstruct the symmetric point C$^{\\prime}$ of C with respect to AD, and connect B$^{\\prime}$C$^{\\prime}$, GC$^{\\prime}$, \\\\\n$\\therefore$ CG = GC$^{\\prime}$, \\\\\n$\\therefore$ B$^{\\prime}$G + GC = B$^{\\prime}$G + GC$^{\\prime}$ $\\ge$ B$^{\\prime}$C$^{\\prime}$, that is, the minimum value of B$^{\\prime}$G + GC is B$^{\\prime}$C$^{\\prime}$, \\\\\nIn the right triangle $\\triangle$B$^{\\prime}$C$^{\\prime}$H, B$^{\\prime}$H = 2, HC$^{\\prime}$ = CC$^{\\prime}$ − CH = 8 − 2 = 6, \\\\\nBy the Pythagorean theorem, B$^{\\prime}$C$^{\\prime}$ = $\\sqrt{2^{2}+6^{2}} = 2\\sqrt{10}$, \\\\\n$\\therefore$ The minimum value of B$^{\\prime}$G + GC is $\\boxed{2\\sqrt{10}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50496823.png"
  },
  {
    "solution": "\\boxed{-4,0}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: The location of rectangle $ABCD$ in the Cartesian coordinate system is shown in the diagram, where $F$ is a point on $AB$. Fold $\\triangle BCF$ along $CF$ so that point $B$ precisely falls on the intersection point $E$ of $AD$ and the $y$-axis. Connect $CE$, if the lengths of $AE, AB$ satisfy $\\sqrt{AE-4}+(AB-8)^{2}=0$. What are the coordinates of points $A$ and $B$?\n",
    "completion": "\\textbf{Solution:} Solving from $\\sqrt{AE-4}+(AB-8)^{2}=0$, we have:\\\\\n$AE-4=0$ and $AB-8=0$\\\\\n$\\therefore AE=4$\\\\\n$AB=8$\\\\\n$\\therefore A(\\boxed{-4,8})$\\\\\n$B(\\boxed{-4,0})$",
    "image_path": "dataset/mm_math/images/MM_Math/53359586_200.png"
  },
  {
    "solution": "\\boxed{\\frac{126}{25}-\\frac{27}{20}\\pi}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, circle $\\odot O$ with diameter $BC$ intersects $AC$ at point $E$. A line through point $E$ is drawn such that $EF\\perp AB$ at point $F$, and extending $EF$ to intersect the extension of $CB$ at point $G$, with $\\angle ABG=2\\angle C$. If $\\sin\\angle EGC=\\frac{3}{5}$, and the radius of $\\odot O$ is $3$, find the area of the shaded region in the figure. ($\\sin 36^\\circ=\\frac{3}{5}$)\n",
    "completion": "\\textbf{Solution:} Let's consider Right triangles $\\triangle EOG$ and $\\triangle BFG$, where $\\sin\\angle EGC=\\frac{OE}{OG}=\\frac{BF}{BG}=\\frac{3}{5}$. This means $\\frac{3}{OG}=\\frac{3}{5}$. Solving this, we get: $OG=5$, hence $BG=OG-OB=5-3=2$, implying $\\frac{BF}{2}=\\frac{3}{5}$, hence $BF=\\frac{6}{5}$, and thus $EG=\\sqrt{OG^2-OE^2}=\\sqrt{5^2-3^2}=4$. Let $EF=x$, then $GF=4-x$. Since $BF\\parallel OE$, we have $\\frac{GB}{OB}=\\frac{GF}{EF}$, that is $\\frac{2}{3}=\\frac{4-x}{x}$. Solving this, we get: $x=EF=\\frac{12}{5}$. Given $\\sin\\angle EGC=\\frac{3}{5}$, we find $\\angle G=36^\\circ$, hence $\\angle EOB=90^\\circ-36^\\circ=54^\\circ$. The area of the shaded region $S_{\\text{shaded}}=S_{\\text{trapezoid} BFEO}-S_{\\text{sector} EOB}=\\frac{1}{2}\\left(\\frac{6}{5}+3\\right)\\times \\frac{12}{5}-\\frac{54^\\circ\\pi \\times 3^2}{360}=\\boxed{\\frac{126}{25}-\\frac{27}{20}\\pi}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53059948_27.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, quadrilateral ABCD is a parallelogram. The circle $O$ with diameter AD intersects AB at point E. Connect DE. It is given that DA $= 2\\sqrt{2}$, DE $= \\sqrt{7}$, and DC $= 5$. A line $l$ is drawn through point E. Through point C, draw CH $\\perp$ $l$, with H being the foot of the perpendicular. If $l \\parallel$ AD, and $l$ intersects the circle $O$ at another point F, connect DF. What is the length of DF?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect DF, \\\\\n$\\because$AD$\\parallel$l, \\\\\n$\\therefore$ $\\angle$ADE=$\\angle$DEF, \\\\\n$\\therefore$ AE=DF, \\\\\n$\\because$AD is the diameter of circle O, \\\\\n$\\therefore$ $\\angle$AED=$90^\\circ$, \\\\\n$\\therefore$ DF=AE=$\\sqrt{AD^{2}-DE^{2}}=\\boxed{1}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51351656_169.png"
  },
  {
    "solution": "\\boxed{9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point B is a point on line segment AC, and $AB=27\\text{cm}$, $BC=\\frac{1}{3}AB$. If point O is the midpoint of line segment AC, what is the length of line segment $OB$ in cm?\n",
    "completion": "\\textbf{Solution:} From (1) we know: $AC=24\\,\\text{cm}$,\\\\\nsince point O is the midpoint of line segment AC,\\\\\ntherefore $OC=\\frac{1}{2}AC=\\frac{1}{2}\\times 24=12\\,\\text{cm}$.\\\\\nTherefore $OB=OC-BC=12-9=\\boxed{9}\\,\\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51292247_203.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, if $\\triangle ABC \\cong \\triangle DEF$ with points $B$, $E$, $C$, and $F$ lying on the same straight line, and $BC=7$, $EC=4$, then what is the length of $CF$?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABC \\cong \\triangle DEF$ and $BC = 7$, \\\\\n$\\therefore EF = BC = 7$, \\\\\n$\\therefore CF = EF - EC = \\boxed{3}$",
    "image_path": "dataset/mm_math/images/MM_Math/52603474.png"
  },
  {
    "solution": "\\boxed{\\left(-\\frac{4}{5}, -\\frac{3}{5}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the line $l_{1}: y=2x-1$ intersects the $x$-axis and $y$-axis at points $A$ and $B$, respectively, and the line $l_{2}: y=-\\frac{1}{2}x+1$ intersects the $x$-axis and $y$-axis at points $P$ and $C$, respectively. Connect $AC$, and lines $l_{1}$ and $l_{2}$ intersect at point $D$. If point $E$ is on line $l_{1}$ and $F$ is any point in the coordinate plane, investigate whether there exists a quadrilateral with $B$, $C$, $E$, $F$ as vertices which is a rectangle? Write down the coordinates of point $F$.\n",
    "completion": "\\textbf{Solution:} Existence of coordinates for $F$ are either $\\left(1, -1\\right)$ or $\\left(-\\frac{4}{5}, -\\frac{3}{5}\\right)$,\n\n(1) When $BC$ is the side of the rectangle, the quadrilateral $BCEF$ is a rectangle, \\\\\nthen $CE \\perp BC$ \\\\\n$\\therefore CE \\parallel x$-axis, $BF \\parallel x$-axis \\\\\n$\\because C\\left(0, 1\\right)$, $B\\left(0, -1\\right)$ \\\\\nLet $E\\left(m, 1\\right)$, $\\because E$ is on line $l_1\\colon y = 2x - 1$, \\\\\n$\\therefore 2m - 1 = 1$ \\\\\nSolving, we get $m = 1$ \\\\\n$\\therefore x_F = 1$ \\\\\n$\\therefore F \\left(1, -1\\right)$; \\\\\n(2) When $BC$ is the diagonal of the rectangle, \\\\\n$\\because OB = OC = 1$, $O$ is the intersection point of the diagonal, $C\\left(0, 1\\right), D\\left(\\frac{4}{5}, \\frac{3}{5}\\right), B\\left(0, -1\\right)$, \\\\\n$\\therefore BC=2, CD=\\sqrt{\\left(\\frac{4}{5}\\right)^2+\\left(1-\\frac{3}{5}\\right)^2}=\\frac{2\\sqrt{5}}{5}, BD=\\sqrt{\\left(\\frac{4}{5}\\right)^2+\\left(1+\\frac{3}{5}\\right)^2}=\\frac{4\\sqrt{5}}{5}$, \\\\\n$\\therefore CD^2+BD^2=\\frac{4}{5}+\\frac{16}{5}=4=BC^2$, \\\\\n$\\therefore \\triangle BCD$ is a right triangle, with $\\angle ADC = 90^\\circ$ \\\\\n$\\therefore$ When $BC$ is the diagonal of rectangle $BECF$, points $D, E$ coincide, \\\\\n$\\therefore O$ is also the midpoint of $DF$, \\\\\nThrough central symmetry, we get $F\\left(-\\frac{4}{5}, -\\frac{3}{5}\\right)$ \\\\\nIn conclusion, the coordinates of $F$ are either $\\boxed{\\left(1, -1\\right)}$ or $\\boxed{\\left(-\\frac{4}{5}, -\\frac{3}{5}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52386314_132.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $O$ is the midpoint of the diagonal $AC$ of rectangle $ABCD$, and $E$ is a point on $AB$. After folding $\\triangle BCE$ along $CE$, point $B$ coincides exactly with point $O$. If $BC = 3$, what is the length of the crease $CE$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, we connect $OB$, \\\\\nsince quadrilateral $ABCD$ is a rectangle, and O is the midpoint of the diagonal AC of rectangle ABCD, \\\\\nit follows that $BO=CO$, and $\\angle ABC=90^\\circ$, \\\\\nsince folding $\\triangle BCE$ along CE exactly overlaps point B with point O, \\\\\nit follows that $BC=OC$, and $\\angle BCE=\\angle OCE$, \\\\\nthus, $OB=OC=BC$, \\\\\nwhich means $\\triangle OBC$ is an equilateral triangle, \\\\\ntherefore, $\\angle OCB=60^\\circ$, \\\\\ntherefore, $\\angle BCE=\\angle OCE=30^\\circ$, \\\\\nIn the right triangle $\\triangle BCE$, $CE=2EB$, and $CB=3$, \\\\\n$CE^2=BC^2+EB^2$, \\\\\nwhich implies $CE^2=\\left(\\frac{1}{2}CE\\right)^2+3^2$, \\\\\nsolving this gives us $CE=\\boxed{2\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52386508.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=kx+b$ ($k\\neq 0$) intersects the x-axis at point $A(6,0)$ and the y-axis at point $B(0,3)$, and intersects the line $y=x$ at point $C$. What is the area of $\\triangle BOC$?\n",
    "completion": "\\textbf{Solution:} Given the conditions, we have $B(0, 3)$, $O(0, 0)$, $C(2, 2)$,\\\\\n$\\therefore S_{\\triangle BOC}=\\frac{1}{2}OB\\cdot|x_{C}|=\\frac{1}{2}\\times 3\\times 2=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50677506_133.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the line $l_1: y=x+n-2$ and the line $l_2: y=mx+n$ intersect at the point $P(1, 2)$. Find the values of $m$ and $n$.\n",
    "completion": "\\textbf{Solution:} Since point $P$ is the intersection of two lines, \\\\\n$\\therefore$ by substituting point $P\\left(1, 2\\right)$ into $y=x+n-2$ and $y=mx+n$ respectively, we get $2=1+n-2$, $2=m+n$, \\\\\nsolving these, we find $m=\\boxed{-1}$, $n=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50589403_158.png"
  },
  {
    "solution": "\\boxed{48}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $D$, $E$, and $F$ are the midpoints of $BC$, $AD$, and $BE$ respectively. If the area of $\\triangle BFD$ is 6, what is the area of $\\triangle ABC$?\n",
    "completion": "\\textbf{Solution:} Since D, E, and F are the midpoints of BC, AD, and BE, respectively,\n\n$\\therefore {S}_{\\triangle ABC}=2{S}_{\\triangle ABD}$, ${S}_{\\triangle ABD}=2{S}_{\\triangle BDE}$, and ${S}_{\\triangle BDE}=2{S}_{\\triangle BFD}$,\n\n$\\therefore {S}_{\\triangle ABC}=8{S}_{\\triangle BFD}=\\boxed{48}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52655234.png"
  },
  {
    "solution": "\\boxed{\\frac{16}{5}\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\odot O$ is the circumcircle of $\\triangle ABC$, where the diameter $BD$ intersects $AC$ at point E. A tangent to $\\odot O$ through point D intersects the extension of $BC$ at point F, with $\\angle F=\\angle BAC$. Given $DF\\parallel AC$, $AB = 8$, and $CF = 2$, what is the length of $AC$?\n",
    "completion": "\\textbf{Solution:} Draw $CD$,\\\\\n$\\because DF\\parallel AC$ and $\\angle ODF=90^\\circ$,\\\\\n$\\therefore \\angle BEC=\\angle ODF=90^\\circ$,\\\\\n$\\therefore$ diameter $BD\\perp AC$ at E,\\\\\n$\\therefore AE=CE=\\frac{1}{2}AC$,\\\\\n$\\therefore AB=BC=8$,\\\\\n$\\because BD$ is the diameter of $\\odot O$,\\\\\n$\\therefore \\angle BCD=90^\\circ$,\\\\\n$\\therefore \\angle DBC+\\angle BDC=90^\\circ$,\\\\\n$\\because \\angle DBC+\\angle F=90^\\circ$,\\\\\n$\\therefore \\angle BDC=\\angle F$,\\\\\n$\\because \\angle BCD=\\angle FCD=90^\\circ$,\\\\\n$\\therefore \\triangle BCD\\sim \\triangle DCF$,\\\\\n$\\therefore \\frac{BC}{DC}=\\frac{DC}{CF}$, i.e., $\\frac{8}{DC}=\\frac{DC}{2}$,\\\\\n$\\therefore DC=4$,\\\\\n$\\therefore BD=\\sqrt{BC^2+CD^2}=\\sqrt{8^2+4^2}=4\\sqrt{5}$,\\\\\n$\\because$ in $\\triangle BCD$, $S_{\\triangle BCD}=\\frac{1}{2}BC\\cdot CD=\\frac{1}{2}BD\\cdot CE$,\\\\\n$\\therefore \\frac{1}{2}\\times 8\\times 4=\\frac{1}{2}\\times 4\\sqrt{5}\\cdot CE$,\\\\\n$\\therefore CE=\\frac{8}{5}\\sqrt{5}$,\\\\\n$\\therefore AC=2CE=\\boxed{\\frac{16}{5}\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53078823_122.png"
  },
  {
    "solution": "\\boxed{100^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $OB$ is a ray inside $\\angle AOC$, $OM$ is a ray inside $\\angle AOB$, and $ON$ is a ray inside $\\angle BOC$. In Figure 2, if $\\angle AOC=140^\\circ$, $\\angle AOM=\\angle NOC=\\frac{1}{4}\\angle AOB$, and $\\angle BOM:\\angle BON=3:2$, what is the degree measure of $\\angle MON$?\n",
    "completion": "\\textbf{Solution:} Let $\\angle AOM=\\angle NOC=x$, thus $\\angle AOB=4x$,\\\\\n$\\therefore \\angle BOM=\\angle AOB-\\angle AOM=3x$,\\\\\n$\\because \\angle BOM\\colon \\angle BON=3\\colon 2$,\\\\\n$\\therefore \\angle BON=2x$,\\\\\n$\\therefore \\angle AOC=\\angle AOB+\\angle BON+\\angle NOC=7x=140^\\circ$,\\\\\n$\\therefore x=20^\\circ$,\\\\\n$\\therefore \\angle MON=\\angle BOM+\\angle BON=5x=\\boxed{100^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53403960_73.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of a semicircle, and $BC$ is a chord of the semicircle. The arc $\\overset{\\frown}{BC}$ is folded along the chord $BC$ to intersect the diameter $AB$ at point D. When $AD=4$ and $BD=6$, what is the length of $BC$?\n",
    "completion": "\\textbf{Solution:} We have $BC=\\boxed{4\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51458277_55.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that in the circle $\\odot O$ with diameter $AB=8\\sqrt{3}$, radius $OC\\perp AB$, point D is the trisection point of the semicircle $\\overset{\\frown}{AB}$, and point P is a moving point on radius $OC$. When the value of $PB+PD$ is minimized, what is the length of $PO$?\n",
    "completion": "\\textbf{Solution:} Draw $DO$, $DA$ intersects $OC$ at point P,\\\\\nsince $OC\\perp AB$ and point O is the midpoint of $AB$,\\\\\ntherefore, the point symmetrical to B with respect to $OC$ is point A,\\\\\ntherefore, the intersection point P of $DA$ and $OC$ makes the value of $PB+PD$ the smallest,\\\\\nsince point D is the trisection point of the semicircle $\\overset{\\frown}{AB}$,\\\\\ntherefore, $\\angle DOB=60^\\circ$,\\\\\ntherefore, $\\angle DAB=30^\\circ$,\\\\\nsince $\\angle AOP=90^\\circ$, $OA=\\frac{1}{2}AB$, $AB=8\\sqrt{3}$, $\\angle PAO=30^\\circ$,\\\\\ntherefore, $OA=4\\sqrt{3}$,\\\\\ntherefore, $OP=OA\\cdot\\tan30^\\circ=4\\sqrt{3}\\times \\frac{\\sqrt{3}}{3}=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53080529.png"
  },
  {
    "solution": "\\boxed{9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $B$ is on line segment $AC$, and $AB=27\\text{cm}$, $BC=\\frac{1}{3}AB$. If point $O$ is the midpoint of line segment $AC$, find the length of line segment $OB$.\n",
    "completion": "\\textbf{Solution:} Since point O is the midpoint of AC, \\\\\n$\\therefore$ OC=$\\frac{1}{2}$AC=$\\frac{1}{2} \\times 36=18$, \\\\\n$\\therefore$ OB=OC−BC=18−9=\\boxed{9}.",
    "image_path": "dataset/mm_math/images/MM_Math/53206769_126.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: The straight line $y=-\\frac{1}{3}x+1$ intersects the x-axis and y-axis at points A and B, respectively. Given that the coordinates of point G are $(2,7)$, the straight line BG is drawn through points G and B, and AG is connected. What is the value of $\\tan(\\angle AGB)$?\n",
    "completion": "\\textbf{Solution:} Given the coordinates of A, B, and G, $BG^{2}=2^{2}+(7-1)^{2}=40$, similarly $AB^{2}=10$, $AG^{2}=50$, thus $AG^{2}=AB^{2}+BG^{2}$,\\\\\n$\\therefore \\triangle ABG$ is a right triangle,\\\\\n$\\therefore \\tan\\angle AGB=\\frac{AB}{BG}=\\frac{\\sqrt{10}}{\\sqrt{40}}=\\boxed{\\frac{1}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51051573_103.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, $\\angle BAD=\\angle ABD=\\angle BCD=45^\\circ$, and $BC=\\sqrt{2}$. What is the area of $\\triangle ABC$?\n",
    "completion": "\\textbf{Solution:} Draw $BE\\perp DC$ at point $E$, and draw $CF\\perp AB$, intersecting the extension line of $AB$ at point $F$,\\\\\n$\\therefore$ $\\angle$DFB=$\\angle$BFC=90$^\\circ$, $\\because$ $\\angle$BAD=$\\angle$ABD=$\\angle$BCD=45$^\\circ$, $\\therefore$ $\\angle$ADB=90$^\\circ$, $AD=BD$, let $BD=x$, then $AB=\\sqrt{2}x$, in $\\triangle BEC$, $BE=CE$, $\\therefore$ 2$BE^{2}$=$BC^{2}$=$(\\sqrt{2})^{2}$, solving gives: $BE=1$, $\\because$ $\\angle$CDB+$\\angle$CBD=180$^\\circ$-45$^\\circ$=135$^\\circ$, $\\angle$CBF+$\\angle$DBC=180$^\\circ$-45$^\\circ$=135$^\\circ$, $\\therefore$ $\\angle$CBF=$\\angle$CDB, $\\therefore$ $\\triangle BDE\\sim \\triangle CBF$, $\\therefore$ $\\frac{BE}{CF}=\\frac{BD}{CB}$, $\\therefore$ $\\frac{1}{CF}=\\frac{x}{\\sqrt{2}}$, solving gives: $CF=\\frac{\\sqrt{2}}{x}$, $\\therefore$ the area of $\\triangle ABC$ is $\\frac{1}{2}AB\\cdot CF=\\frac{1}{2}\\times \\sqrt{2}x\\times \\frac{\\sqrt{2}}{x}=\\boxed{1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52634686.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=kx+b$ intersects the $x$-axis at point $B$ and the $y$-axis at point $C(0,3)$, and intersects $y=2x$ at point $A(a,2)$. Point $M$ is on the line $OA$. What is the area of $\\triangle OAB$?\n",
    "completion": "\\textbf{Solution:} Let $y=0$, we have $-x+3=0$. Solving this, we obtain: $x=3$, $\\therefore B(3,0)$, $\\therefore OB=3$, $\\therefore$ the area of $\\triangle OAB$ is $=\\frac{1}{2}\\times 3\\times 2=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52354898_183.png"
  },
  {
    "solution": "\\boxed{3\\sqrt{6}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in square $ABCD$, points $E$ and $F$ are respectively on side $AD$ and diagonal $AC$. Connect $EF$, and $\\angle AEF = \\angle CAB$, $\\triangle AEF \\sim \\triangle ACD$; if $AF = 2CF$, $AE = 4$, $DE = 5$, find the length of $AC$?\n",
    "completion": "\\textbf{Solution:} Given the problem statement,\\\\\n$\\because \\triangle AEF\\sim \\triangle ACD$,\\\\\n$\\therefore \\frac{AE}{AC}=\\frac{AF}{AD}$.\\\\\n$\\because AF=2CF$,\\\\\n$\\therefore AF=\\frac{2}{3}AC$.\\\\\n$\\because AE=4, DE=5$,\\\\\n$\\therefore \\frac{4}{AC}=\\frac{\\frac{2}{3}AC}{4+5}$.\\\\\n$\\therefore \\frac{2}{3}AC^2=36$.\\\\\nSolving gives $AC=\\boxed{3\\sqrt{6}}$ (Negative values discarded).",
    "image_path": "dataset/mm_math/images/MM_Math/51212753_74.png"
  },
  {
    "solution": "\\boxed{\\frac{12}{7}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, quadrilateral $OABC$ is a rectangle, with points $A$ and $C$ located on the $x$-axis and $y$-axis, respectively. Triangle $ODE$ is obtained by rotating triangle $OCB$ $90^\\circ$ clockwise around point $O$. Point $D$ is located on the $x$-axis. The line $BD$ intersects the $y$-axis at point $F$, and intersects $OE$ at point $H$. The coordinates of point $B$ are $(-2, 4)$. Find the area of $\\triangle DEH$.\n",
    "completion": "\\textbf{Solution:} From the given, we know the coordinates of point E are (4, 2)\\\\\nLet the equation of line OE be: $y=mx$, then we have $4m=2$, solving this gives: $m= \\frac{1}{2}$\\\\\n$\\therefore y= \\frac{1}{2} x$\\\\\n$\\because \\left\\{\\begin{array}{l}y=\\frac{1}{2}x\\\\ y=-\\frac{2}{3}x+\\frac{8}{3}\\end{array}\\right.$\\\\\n$\\therefore \\left\\{\\begin{array}{l}x=\\frac{16}{7}\\\\ y=\\frac{8}{7}\\end{array}\\right.$\\\\\n$\\therefore H\\left( \\frac{16}{7}, \\frac{8}{7}\\right)$\\\\\n$S_{\\triangle DEH} = \\frac{1}{2}\\times DE\\times \\left(4-\\frac{16}{7}\\right) = \\frac{1}{2}\\times 2\\times \\left(4-\\frac{16}{7}\\right) = \\boxed{\\frac{12}{7}}$",
    "image_path": "dataset/mm_math/images/MM_Math/52251770_75.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that point C is the midpoint of line segment AB, and point D is the midpoint of line segment BC. Given AB = 16 cm, find the length of line segment AD.\n",
    "completion": "Solution: Since $AB=16\\,cm$, and $C$ is the midpoint of $AB$, \\\\\ntherefore $AC=BC=\\frac{1}{2}AB=8\\,cm$, \\\\\nSince $D$ is the midpoint of $BC$, \\\\\ntherefore $CD=\\frac{1}{2}BC=4\\,cm$ \\\\\nthus $AD=AC+CD=\\boxed{12}\\,cm$.",
    "image_path": "dataset/mm_math/images/MM_Math/51251856.png"
  },
  {
    "solution": "\\boxed{\\frac{38}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the equilateral triangle $ABC$, point $E$ is a point on side $CB$ (not coinciding with point $C$), and point $F$ is a point on side $AC$. Given $BC=10$, $EC=6$, and $\\angle AEF=60^\\circ$, what is the length of $AF$?\n",
    "completion": "\\textbf{Solution:} Given from the problem, $\\triangle ABC$ is an equilateral triangle,\n\\[\n\\therefore \\angle B = \\angle C = 60^\\circ, \\quad AC = BC = AB = 10,\n\\]\n\\[\n\\because EC = 6,\n\\]\n\\[\n\\therefore BE = 4,\n\\]\n\\[\n\\because \\angle AEC = \\angle BAE + \\angle B,\n\\]\nthat is, $\\angle AEF + \\angle CEF = \\angle BAE + \\angle B$,\nand considering $\\angle AEF = 60^\\circ$, $\\angle B = 60^\\circ$,\n\\[\n\\therefore \\angle BAE = \\angle CEF,\n\\]\n\\[\n\\because \\angle B = \\angle C,\n\\]\n\\[\n\\therefore \\triangle ABE \\sim \\triangle ECF,\n\\]\n\\[\n\\therefore \\frac{BE}{CF} = \\frac{AB}{EC}, \\text{ that is, } \\frac{4}{CF} = \\frac{10}{6},\n\\]\n\\[\n\\therefore CF = \\frac{12}{5},\n\\]\n\\[\n\\therefore AF = AC - CF = 10 - \\frac{12}{5} = \\boxed{\\frac{38}{5}}\\].",
    "image_path": "dataset/mm_math/images/MM_Math/51446274_125.png"
  },
  {
    "solution": "\\boxed{24}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $C$ is a point on line segment $AB$, and $M$ and $N$ are the midpoints of $AC$ and $BC$, respectively. If the ratio of the lengths of line segments $CM$ and $CN$ is $2:1$, and the length of line segment $CN$ is $4\\,\\text{cm}$, find the length of line segment $AB$.\n",
    "completion": "Solution: Since the ratio of the lengths of segments $CM$ and $CN$ is $2\\colon 1$ and $CN=4\\,\\text{cm}$,\\\\\n$\\therefore$ segment $CM=8\\,\\text{cm}$.\\\\\nSince $M$ and $N$ are the midpoints of $AC$ and $BC$ respectively,\\\\\n$\\therefore$ $AC=2CM=16\\,\\text{cm}$, $BC=2CN=8\\,\\text{cm}$,\\\\\n$\\therefore$ $AB=AC+BC=16+8=\\boxed{24}\\,\\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53319958_81.png"
  },
  {
    "solution": "\\boxed{-4\\leq b\\leq 2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the Cartesian coordinate system $xOy$, it is known that the line $y=-\\frac{1}{2}x+2$ intersects the $x$-axis and $y$-axis at points $A$ and $B$ respectively. The graph of the line $y=kx-1$ intersects the $y$-axis at point $C$ and intersects the given line at point $D$, where the $x$-coordinate of point $D$ is $2$. If the graph of the line $y=kx-1$ is translated upwards or downwards, intersecting the given line at point $E$, and assuming the intercept of the resulting function graph after translation is $b$, if the intersection point $E$ always falls on the line segment $AB$, what is the range of values for $b$?\n",
    "completion": "\\textbf{Solution:} Let $x=0$, then $y=-\\frac{1}{2}x+2=2$,\\\\\nLet $y=0$, then $-\\frac{1}{2}x+2=0$,\\\\\n$\\therefore x=4$,\\\\\n$\\because$ the line $y=-\\frac{1}{2}x+2$ intersects the $x$-axis and $y$-axis at points $A$ and $B$ respectively,\\\\\n$\\therefore$ the coordinates of $A$ are $(4, 0)$, and the coordinates of $B$ are $(0, 2)$,\\\\\nAssume the equation of the line $y=kx-1$ after translation is $y=x+b$,\\\\\nWhen the line $y=x+b$ passes through point $A$, $b=-4$,\\\\\nWhen the line $y=x+b$ passes through point $B$, $b=2$,\\\\\nFrom the graph, when the intersection point $E$ always falls on the segment $AB$, the range of $b$ is $-4\\leq b\\leq 2$,\\\\\nTherefore, the range of $b$ is $\\boxed{-4\\leq b\\leq 2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50446789_234.png"
  },
  {
    "solution": "\\boxed{70^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rhombus $ABCD$, if $ \\angle A = \\angle B + 40^\\circ$, then what is the measure of $\\angle B$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rhombus,\\\\\ntherefore $\\angle A+\\angle B=180^\\circ$.\\\\\nSince $\\angle A=\\angle B+40^\\circ$,\\\\\ntherefore, we find: $\\angle B=\\boxed{70^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53437330.png"
  },
  {
    "solution": "\\boxed{-4<x<-2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=-x+m$ intersects with $y=nx+4n\\;\\left(n\\neq 0\\right)$ at a point with an abscissa of $-2$. What is the solution set of the inequality $-x+m>nx+4n>0$ with respect to $x$?\n",
    "completion": "\\textbf{Solution:} Given that the line $y=-x+m$ intersects with $y=nx+4n (n\\ne 0)$ at a point where the x-coordinate is $-2$,\\\\\nit follows that the solution set of the inequality $-x+m>nx+4n$ with respect to $x$ is $x<-2$,\\\\\nthus, when $y=nx+4n=0$, we have $x=-4$,\\\\\ntherefore, the solution set for the inequality $-x+m>nx+4n>0$ is $-4<x<-2$.\\\\\nHence, the answer is: $\\boxed{-4<x<-2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52381492.png"
  },
  {
    "solution": "\\boxed{\\sqrt{7}+1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ is an equilateral triangle with a side length of $1$. By folding $\\triangle ABC$ along the line $AC$, we obtain $\\triangle AB^{\\prime}C$. Then, by translating $\\triangle AB^{\\prime}C$ along the line $AC$, we obtain $\\triangle A^{\\prime}B^{\\prime\\prime}C^{\\prime}$. What is the minimum perimeter of $\\triangle BB^{\\prime\\prime}C^{\\prime}$?\n",
    "completion": "\\textbf{Solution:} Let $E$ be the symmetric point of $B'$ with respect to point $B$, and let the extension of $B'B$ intersect $AC$ at point $O$. Connect $EC$,\\\\\n$\\because$ translating $\\triangle ABC'$ along line $AC$, we obtain $\\triangle AB''C'$,\\\\\n$\\therefore$ we can keep $\\triangle ABC'$ stationary and translate point $B$ parallel to line $AC$,\\\\\n$\\therefore$ the problem of finding the minimum perimeter of $\\triangle BB''C'$ is transformed into finding the minimum perimeter of $\\triangle BB'C$,\\\\\n$\\therefore$ when points $E$, $B$, and $C$ are collinear, the sum of $BB'+BC$ is minimized, being equal to the length of $CE$,\\\\\n$\\because$ both $\\triangle ABC$ and $\\triangle AB'C$ are equilateral triangles,\\\\\n$\\therefore$ $AB=BC=CB'=AB'$,\\\\\n$\\therefore$ the quadrilateral $ABCB'$ is a rhombus,\\\\\n$\\therefore$ $BB'\\perp AC$, and $OC=\\frac{1}{2}AC=\\frac{1}{2}$,\\\\\n$\\therefore$ $BO=B'O=\\frac{\\sqrt{3}}{2}$,\\\\\n$\\therefore$ $OE=BE+OB=\\sqrt{3} +\\frac{\\sqrt{3}}{2}=\\frac{3\\sqrt{3}}{2}$,\\\\\nIn $\\triangle CEO$, by the Pythagorean theorem:\\\\\n$CE=\\sqrt{OE^2+OC^2}=\\sqrt{\\left(\\frac{3\\sqrt{3}}{2}\\right)^2+\\left(\\frac{1}{2}\\right)^2}=\\sqrt{7}$,\\\\\n$\\therefore$ the minimum perimeter of $\\triangle BB'C$ is: $\\sqrt{7}+1$,\\\\\nwhich means the minimum perimeter of $\\triangle BB''C'$ is: $\\boxed{\\sqrt{7}+1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52019294.png"
  },
  {
    "solution": "\\boxed{-4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, parallelogram $OABC$ has its side $OA$ on the $x$-axis, and vertex $C$ is on the graph of the inverse proportion function $y=\\frac{k}{x}$. $BC$ intersects the $y$-axis at point $D$, and $D$ is the midpoint of $BC$. If the area of the parallelogram $OABC$ is $8$, what is the value of $k$?\n",
    "completion": "\\textbf{Solution:} Since $D$ is the midpoint of $BC$, and the area of parallelogram $OABC$ is 8,\\\\\nit follows that $S_{\\triangle OCD}=\\frac{1}{4}\\times 8=2$,\\\\\nwhich means $\\frac{1}{2}\\left|k\\right|=2$,\\\\\nthus $k=\\pm 4$,\\\\\nsince the graph of the inverse proportion function is in the second quadrant,\\\\\nit follows that $k<0$,\\\\\nhence, $k=\\boxed{-4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53043145.png"
  },
  {
    "solution": "\\boxed{y=-\\left(x-5\\right)\\left(x+1\\right)=-x^{2}+4x+5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola intersects the x-axis at $A(5, 0)$ and $B(-1, 0)$, and intersects the positive semi-axis of the y-axis at point $C$. Lines $BC$ and $AC$ are drawn, and it is known that $\\sin\\angle BAC=\\frac{\\sqrt{2}}{2}$. What is the analytical equation of the parabola?\n",
    "completion": "\\textbf{Solution:} Given $A\\left(5, 0\\right)$, it is known that $OA=5$. In $\\triangle AOC$, $\\sin\\angle BAC=\\frac{\\sqrt{2}}{2}$, \\\\\n$\\therefore \\angle BAC=45^\\circ$, \\\\\n$\\therefore OA=OC=5$, i.e., point C is (0, 5), \\\\\nAccording to the problem, let $y=a\\left(x-5\\right)\\left(x+1\\right)$, substituting point C into it results in: \\\\\n$-5a=5$, \\\\\nSolving this gives: $a=-1$, \\\\\n$\\therefore$ The equation of the parabola is $\\boxed{y=-\\left(x-5\\right)\\left(x+1\\right)=-x^{2}+4x+5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51368197_112.png"
  },
  {
    "solution": "\\boxed{\\frac{5\\sqrt{2}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the cartesian coordinate system $xOy$, the parabola $y=ax^2+bx-4$ intersects the $x$-axis at points $A$ and $B$ (with $B$ to the right of $A$) and intersects the $y$-axis at point $C$. It is given that $OA=1$ and $OB=4OA$. Connect $BC$. As shown in figure 2, let point $N$ be a point on segment $OC$. Find the minimum value of $AN+\\frac{\\sqrt{2}}{2}CN$?\n",
    "completion": "\\textbf{Solution:} Let point $N$ be constructed such that $NM\\perp BC$ at point $M$, and draw $AH\\perp BC$ at $H$,\\\\\n$\\because OB=OC$, $\\angle BOC=90^\\circ$\\\\\n$\\therefore \\triangle BOC$ is an isosceles right triangle, $\\angle BCO=\\angle OBC=45^\\circ$\\\\\n$\\because NM\\perp BC$\\\\\n$\\therefore \\angle NMC=90^\\circ$,\\\\\n$\\therefore \\angle CNM=90^\\circ-\\angle NCM=90^\\circ-\\angle OCB=45^\\circ$,\\\\\n$\\therefore \\triangle NCM$ is an isosceles right triangle,\\\\\n$\\therefore NM=NC\\cdot \\sin45^\\circ=\\frac{\\sqrt{2}}{2}NC$,\\\\\n$\\therefore AN+\\frac{\\sqrt{2}}{2}CN=AN+NE \\geq AH$,\\\\\nWhen points A, $N$, and $M$ are collinear, the minimum value of $AN+NM$ $=AH$,\\\\\n$\\because AB=OA+OB=5$,\\\\\n$\\because \\angle ABH=45^\\circ$, $AH\\perp BC$,\\\\\n$\\therefore \\angle BAH=90^\\circ-\\angle ABH=45^\\circ=\\angle ABH$,\\\\\n$\\therefore \\triangle AEB$ is an isosceles right triangle, $\\angle ABE=45^\\circ$,\\\\\n$\\therefore AH=AB\\cdot \\sin45^\\circ=\\boxed{\\frac{5\\sqrt{2}}{2}}$\\\\\n$\\therefore$ The minimum value of $AN+\\frac{\\sqrt{2}}{2}CN$ $\\boxed{\\frac{5\\sqrt{2}}{2}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51446444_221.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, rhombus $ABCD$ has a side length of $2\\sqrt{3}$, and $\\angle ABC=60^\\circ$. The diagonals $AC$ and $BD$ intersect at point $O$. Point $E$ is a moving point on line $AD$, and $CE$ is connected. After rotating the line segment $EC$ clockwise around point $C$ by the angle of $\\angle BCD$, we get the corresponding line segment $CF$ (i.e., $\\angle ECF=\\angle BCD$). What is the minimum length of $DF$?\n",
    "completion": "\\textbf{Solution:} Let's connect BE and construct BH $\\perp$ AD, intersecting the extension line of DA at H. In diamond ABCD, $\\angle ABC=60^\\circ$, thus $\\angle BCD=120^\\circ$. Since $\\angle ECF=120^\\circ$, it follows that $\\angle BCD=\\angle ECF$, and thus $\\angle BCE=\\angle DCF$. By rotation, we get: EC=FC. In $\\triangle BEC$ and $\\triangle DFC$, we have:\n\\[\n\\left\\{\n\\begin{array}{l}\nBC=DC \\\\\n\\angle BCE=\\angle DCF \\\\\nEC=FC\n\\end{array}\n\\right.\n\\]\nTherefore, $\\triangle DCF \\cong \\triangle BCE$ (SAS), hence $DF=BE$. Thus, finding the minimum value of DF is equivalent to finding the minimum value of BE. Since in $\\triangle AHB$, $\\angle BAH=60^\\circ$ and AB= $2\\sqrt{3}$, it follows that BH= $3$. When E coincides with H, the minimum value of BE is 3. Therefore, the minimum value of DF is $\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50496425.png"
  },
  {
    "solution": "\\boxed{3.90km}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, there is a schematic diagram of a simulated rocket launch setup. A carrier rocket is launched from the ground at point L. When the rocket reaches point A, the distance $AR$ measured from the radar station located on the ground at point R is $5km$, with an elevation angle of $39^\\circ$; about $1.5s$ later, when the rocket reaches point B, the measured elevation angle is $45^\\circ$ (Reference data: $\\sin 39^\\circ \\approx 0.63$, $\\cos 39^\\circ \\approx 0.78$, $\\tan 39^\\circ \\approx 0.8$). Determine the horizontal distance from the ground radar station R to the launch site L.\n",
    "completion": "\\textbf{Solution:} In $\\triangle ARL$, we have $RL=AR\\cdot \\cos 39^\\circ \\approx 5\\times 0.78=3.90$ ($km$),\\\\\nThus, the horizontal distance from the radar station to the launch site is $RL=\\boxed{3.90km}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53043157_36.png"
  },
  {
    "solution": "\\boxed{\\frac{11}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, point \\(P\\) is the centroid of \\(\\triangle ABC\\). Through point \\(P\\), draw \\(DE\\parallel AC\\) intersecting \\(BC, AB\\) at \\(D, E\\) respectively, and \\(EF\\parallel BC\\) intersecting \\(AC\\) at point \\(F\\). If \\(BC=11\\), what is the length of \\(EF\\)?\n",
    "completion": "Solution: Connect BP extending it to meet AC at point G,\\\\\nsince DE$\\parallel$AC, and EF$\\parallel$BC,\\\\\nthus quadrilateral EFCD is a parallelogram,\\\\\ntherefore EF=CD;\\\\\nsince point P is the centroid,\\\\\ntherefore $\\frac{BP}{PG}=2$,\\\\\nsince ED$\\parallel$AC,\\\\\nthus $\\frac{BP}{PG}=\\frac{BD}{CD}=2$,\\\\\ntherefore $\\frac{BD}{EF}=2$\\\\\nsince BD+CD=BC=11 which implies 2EF+EF=11\\\\\nSolving gives: $EF=\\boxed{\\frac{11}{3}}$",
    "image_path": "dataset/mm_math/images/MM_Math/52807995.png"
  },
  {
    "solution": "\\boxed{18}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the line $y=-\\frac{3}{4}x+m$ intersects with the x-axis and y-axis at points B and A, respectively, where the coordinate of point B is $(12, 0)$. The line $y=\\frac{3}{8}x$ intersects with line AB at point C. What is the area of $\\triangle BOC$?\n",
    "completion": "\\textbf{Solution:} According to the problem, we have:\n\\[\n\\left\\{\n\\begin{array}{l}\ny=\\frac{3}{8}x \\\\\ny=-\\frac{3}{4}x+9\n\\end{array}\n\\right.\n\\]\nSolving this, we get:\n\\[\n\\left\\{\n\\begin{array}{l}\nx=8 \\\\\ny=3\n\\end{array}\n\\right.\n\\]\nTherefore, point C is at (8, 3), and hence the area of $\\triangle BOC$ is $\\frac{1}{2} \\times 12 \\times 3 = \\boxed{18}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52381062_171.png"
  },
  {
    "solution": "\\boxed{y=\\sqrt{3}x-\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the vertices of rectangle $OABC$ with vertex $A$ on the positive y-axis and point $C$ on the positive x-axis. Given $AO=\\sqrt{3}$ and $\\angle BAC=30^\\circ$, point $D$ is located on line segment $OC$. When $\\triangle AOD$ is folded along line $AD$, point $O$ falls on point $E$ on the diagonal $AC$ of the rectangle. What is the equation of line $DE$?\n",
    "completion": "\\textbf{Solution:} Let $EH\\perp y$-axis at $H$, and $EG\\perp x$-axis at $G$. Then, quadrilateral EHOG is a rectangle, \\\\\n$\\therefore EG=HO$. By folding, $AE=AO=\\sqrt{3}$.\\\\\n$\\because AC=2\\sqrt{3}$,\\\\\n$\\therefore E$ is the midpoint of $AC$,\\\\\n$\\therefore G$ is the midpoint of $OC$, and $H$ is the midpoint of $AO$,\\\\\n$\\therefore HE, EG$ are midlines,\\\\\n$HE=OG=GC=\\frac{1}{2}OC=\\frac{3}{2}$,\\\\\n$EG=HO=AH=\\frac{1}{2}AO=\\frac{\\sqrt{3}}{2}$,\\\\\n$\\therefore$ Point $E\\left(\\frac{3}{2}, \\frac{\\sqrt{3}}{2}\\right)$.\\\\\nIn $\\triangle ADO$, $\\angle OAD=\\frac{1}{2}\\angle OAC=30^\\circ$, $AO=\\sqrt{3}$,\\\\\n$\\therefore OD=1$,\\\\\n$\\therefore$ Point $D\\left(1, 0\\right)$.\\\\\nLet the equation of line $ED$ be $y=kx+b$.\\\\\nSubstituting the coordinates of $D$, $E$ gives $\\left\\{\\begin{array}{l}\\frac{3}{2}k+b=\\frac{\\sqrt{3}}{2} \\\\ k+b=0\\end{array}\\right.$,\\\\\nSolving gives, $\\left\\{\\begin{array}{l}k=\\sqrt{3} \\\\ b=-\\sqrt{3}\\end{array}\\right.$.\\\\\nThus, the equation of line $ED$ is: $y=\\sqrt{3}x-\\sqrt{3}$.\\\\\nHence, the answer is $\\boxed{y=\\sqrt{3}x-\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50519550_188.png"
  },
  {
    "solution": "\\boxed{1\\colon 2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given $ \\angle AOB$ and $ \\angle COD$, the ray $ OE$ bisects $ \\angle AOD$; the positions of $ \\angle AOB$ and $ \\angle COD$ are as shown in Figure 2. Given that $ \\angle COD = \\frac{1}{2}\\angle AOB$, find the value of $ \\angle COE : \\angle DOB$.\n",
    "completion": "\\textbf{Solution:} From the given information, we have\\\\\n$\\because OE$ bisects $\\angle AOD$,\\\\\n$\\therefore \\angle EOD=\\frac{1}{2}\\left(\\angle AOB+\\angle BOD\\right)$\\\\\n$\\because \\angle COD=\\frac{1}{2}\\angle AOB$,\\\\\n$\\therefore \\angle EOC=\\angle EOD-\\angle COD$\\\\\n$=\\frac{1}{2}\\angle AOB+\\frac{1}{2}\\angle BOD-\\frac{1}{2}\\angle AOB$\\\\\n$=\\frac{1}{2}\\angle BOD$\\\\\n$\\therefore \\angle COE\\colon \\angle DOB=\\boxed{1\\colon 2}$",
    "image_path": "dataset/mm_math/images/MM_Math/52916518_82.png"
  },
  {
    "solution": "\\boxed{\\frac{3\\sqrt{10}}{10}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the vertices of $\\triangle ABC$ are all on the grid points of the square grid paper. What is the value of $\\sin C$?\n",
    "completion": "\\textbf{Solution:} As the figure shows, connect AD.\nAccording to the problem:\n\\[AC^{2}=1^{2}+7^{2}=50,\\]\n\\[CD^{2}=1^{2}+2^{2}=5,\\]\n\\[AD^{2}=6^{2}+3^{2}=45,\\]\n\\[\\therefore AD^{2}+CD^{2}=AC^{2},\\]\n\\[\\therefore \\triangle ACD \\text{ is a right-angled triangle},\\]\n\\[\\therefore \\angle ADC=90^\\circ,\\]\nIn $Rt\\triangle ACD$, $AD=3\\sqrt{5}$, $AC=5\\sqrt{2}$,\n\\[\\therefore \\sin C=\\frac{AD}{AC}=\\frac{3\\sqrt{5}}{5\\sqrt{2}}=\\boxed{\\frac{3\\sqrt{10}}{10}}.\\]",
    "image_path": "dataset/mm_math/images/MM_Math/53043061.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, two right-angled triangles $\\triangle ABC$ and $\\triangle BCD$ are placed as shown in the figure, with $\\angle BCA=45^\\circ$, $\\angle D=30^\\circ$, and the two hypotenuses intersecting at point O. What is the ratio of the areas of $\\triangle AOB$ to $\\triangle COD$?\n",
    "completion": "\\textbf{Solution:} Let $BC=x$,\\\\\nIn right triangle $\\triangle ABC$,\\\\\n$\\because$ $\\angle BCA=45^\\circ$,\\\\\n$\\therefore \\tan \\angle BCA = \\frac{AB}{BC}$,\\\\\n$\\therefore AB=BC=x$,\\\\\nIn right triangle $\\triangle BCD$, $\\angle D=30^\\circ$,\\\\\n$\\therefore \\tan D = \\frac{BC}{DC}=\\frac{\\sqrt{3}}{3}$,\\\\\n$\\therefore DC= \\sqrt{3}BC=\\sqrt{3}x$,\\\\\nFurthermore, $\\because AB\\parallel CD$,\\\\\n$\\therefore \\triangle AOB \\sim \\triangle COD$,\\\\\n$\\therefore \\frac{S_{\\triangle AOB}}{S_{\\triangle COD}}=\\left(\\frac{AB}{CD}\\right)^2=\\left(\\frac{x}{\\sqrt{3}x}\\right)^2=\\boxed{\\frac{1}{3}}$,\\\\",
    "image_path": "dataset/mm_math/images/MM_Math/52601761.png"
  },
  {
    "solution": "\\boxed{7}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, we have $AB=AC$, and the points $D$, $A$, and $E$ all lie on the straight line $m$. Moreover, it is given that $\\angle BDA=\\angle AEC=\\angle BAC=\\alpha$. If $DE=10$ and $BD=3$, find the length of $CE$.\n",
    "completion": "\\textbf{Solution:} Since $\\angle AEC = \\angle BAC = \\alpha$,\\\\\nit follows that $\\angle ECA + \\angle CAE = 180^\\circ - \\alpha$,\\\\\nand $\\angle BAD + \\angle CAE = 180^\\circ - \\alpha$,\\\\\nhence $\\angle ECA = \\angle BAD$,\\\\\nIn $\\triangle BAD$ and $\\triangle ACE$,\\\\\nwe have $\\left\\{ \\begin{array}{l} \\angle BDA = \\angle AEC \\\\ \\angle BAD = \\angle ACE \\\\ AB = AC \\end{array} \\right.$\\\\\nTherefore, $\\triangle BAD \\cong \\triangle ACE$ (AAS),\\\\\nwhich implies CE = AD, and AE = BD = 3,\\\\\nSince DE = AD + AE = 10,\\\\\nit follows that AD = DE - AE = DE - BD = 10 - 3 = 7.\\\\\nTherefore, CE = \\boxed{7}.",
    "image_path": "dataset/mm_math/images/MM_Math/52653340.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ is an isosceles right triangle with $\\angle ACB=90^\\circ$ and $AC=BC=6$. Point $D$ is on the line segment $BC$, and point $E$ is on the line segment $AD$. Now, taking $CE$ as the right angle side and $C$ as the right angle vertex, an isosceles right triangle $\\triangle ECF$ is constructed below $CE$, and $BF$ is connected. As shown in figure 3, if $\\angle BAD=15^\\circ$ and $DF$ is connected, when $E$ moves to make $\\angle ACE=30^\\circ$, what is the area of $\\triangle CDF$?\n",
    "completion": "\\textbf{Solution:} Let $FH \\perp BC$ at $H$.\\\\\nSince $\\angle ACE = \\angle CAE = 30^\\circ$,\\\\\nit follows that $AE = EC$,\\\\\nsince $\\triangle ACE \\cong \\triangle BCF$,\\\\\nit follows that $BF = AE$ and $CF = CE$,\\\\\nthus $CF = BF$ and $\\angle FCB = \\angle CBF = 30^\\circ$,\\\\\nsince $FC = FB$ and $FH \\perp BC$,\\\\\nit follows that $CH = BH = 3$ and $FH = \\sqrt{3}$, $CF = BF = 2\\sqrt{3}$,\\\\\nsince $\\angle CED = \\angle CAE + \\angle ACE = 60^\\circ$ and $\\angle ECD = 90^\\circ - 30^\\circ = 60^\\circ$,\\\\\nthus $\\triangle ECD$ is an equilateral triangle,\\\\\nthus $EC = CF = CD = 2\\sqrt{3}$,\\\\\ntherefore, the area of $\\triangle CDF$ is $\\frac{1}{2} \\times CD \\times FH = \\frac{1}{2} \\times 2\\sqrt{3} \\times \\sqrt{3} = \\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55403719_78.png"
  },
  {
    "solution": "\\boxed{15}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $P$ is located on the positive half of the y-axis, and $\\odot P$ intersects the x-axis at points $B$ and $C$. An isosceles triangle $\\triangle ACD$ is formed with $AC$ as the right angle side. $BD$ intersects the y-axis and $\\odot P$ at points $E$ and $F$ respectively. Connect $AC$ and $FC$, with $AC$ and $BD$ intersecting at point $G$. If $OB=3$ and $OA=6$, then what is the area of $\\triangle GDC$?\n",
    "completion": "\\textbf{Solution:} Hence, the area of $\\triangle GDC$ is \\boxed{15}.",
    "image_path": "dataset/mm_math/images/MM_Math/53010347_75.png"
  },
  {
    "solution": "\\boxed{y=-x^{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, the nozzle B of a certain irrigation equipment is $1.25m$ above the ground. The parabolic water flow reaches its maximum height of $2.25m$ from the ground at a distance of $1m$ from the bottom of the nozzle A, as shown in Figure 2, where a Cartesian coordinate system is established. What is the quadratic function equation corresponding to this parabolic water flow?\n",
    "completion": "\\textbf{Solution:}\n\nSince the parabolic water flow corresponds to a quadratic function graph passing through the origin,\n\nwe assume the quadratic function relationship of the parabolic water flow to be $y=ax^{2}$;\n\nAccording to the problem statement, the distance from point B to the x-axis is $1m$, and the distance from point B to the y-axis is $2.25-1.25=1m$,\n\ntherefore, the coordinates of point B are $(-1,-1)$,\n\nsubstituting $B(-1,-1)$ into $y=ax^{2}$, we get $-1=a(-1)^{2}$,\n\nsolving this gives $a=-1$,\n\ntherefore, the quadratic function relationship of the parabolic water flow is $\\boxed{y=-x^{2}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52977560_184.png"
  },
  {
    "solution": "\\boxed{\\frac{2}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In square $ABCD$, where $AB = 2$, let $E$ be the midpoint of $AB$. If $\\triangle ADE$ is folded along $DE$ to form $\\triangle FDE$, and $FH \\perp BC$ with the foot of the perpendicular being $H$, then what is the length of $FH$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw MN$\\parallel$BC through point F, intersecting AB at M and CD at N,\\\\\n$\\because$ Quadrilateral ABCD is a square\\\\\n$\\therefore \\angle B = \\angle C = 90^\\circ$\\\\\n$\\because$MN$\\parallel$BC\\\\\n$\\therefore \\angle AMF = \\angle B = 90^\\circ$, $\\angle DNF=\\angle C= 90^\\circ$\\\\\n$\\therefore \\angle EMF = \\angle DNF = 90^\\circ$\\\\\nBy folding: AD= DF = 2, AE = EF, $\\angle A = \\angle EFD= 90^\\circ$,\\\\\n$\\therefore \\angle EFM + \\angle DFN= \\angle DFN+ \\angle NDF= 90^\\circ$\\\\\n$\\therefore \\angle EFM=\\angle NDF$\\\\\n$\\therefore \\triangle EMF\\sim \\triangle FND$\\\\\n$\\therefore \\frac{EF}{DF}=\\frac{FM}{DN}=\\frac{EM}{FN}$\\\\\n$\\because$In square ABCD, AB = 2, E is the midpoint of AB\\\\\n$\\therefore$AE= BE = EF = 1\\\\\nLet FH = BM=x, then EM= 1 – x, FN = 2EM = 2 (1 − x)= 2 – 2x\\\\\n$\\therefore$FM = 2 − FN = 2 −(2 −2 x)= 2 x\\\\\nIn $\\triangle EMF$, by Pythagorean theorem:\\\\\nEF$^2$= EM$^2$+ FM$^2$\\\\\n$\\therefore 1^2=(1−x)^2 +(2x)^2$\\\\\nSolving yields: $x_1=0$, $x_2 = \\frac{2}{5}$\\\\\n$\\therefore FH= \\boxed{\\frac{2}{5}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51635786.png"
  },
  {
    "solution": "\\boxed{\\frac{15\\sqrt{34}}{17}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $AB=3$, $AD=5$, point $E$ is on $BC$, and point $P$ is a moving point inside the rectangle, connecting $PD$, $PB$; when point $P$ moves along segment $BD$, what is the minimum value of $PE+DE$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, construct the symmetric point $D'$ of point D with respect to BC, and connect $ED'$. The minimum value is achieved when points $P'$, $E'$, and $D'$ are collinear and perpendicular to BD. In the rectangle ABCD, $\\angle BCD=90^\\circ$. In $\\triangle D'P'D$, $\\angle D'P'D=90^\\circ$.\\\\\nTherefore, $\\triangle BCD \\sim \\triangle D'P'D$\\\\\nTherefore, $\\frac{DD'}{DB} = \\frac{D'P'}{BC}$\\\\\nTherefore, $D'P' = \\frac{DD' \\cdot BC}{BD}$\\\\\nAlso, since $DD' = DC + CD' = 3 + 3 = 6$, and $BD = \\sqrt{BC^2 + CD^2} = \\sqrt{34}$\\\\\nTherefore, $D'P' = \\frac{6 \\cdot BC}{\\sqrt{34}}$\\\\\nBy symmetry, $D'E' = DE'$\\\\\nTherefore, $P'E'+DE' = D'E + P'E' = D'P'$\\\\\nTherefore, the minimum value of $PE+DE$ is $\\boxed{\\frac{15\\sqrt{34}}{17}}$",
    "image_path": "dataset/mm_math/images/MM_Math/50496942_82.png"
  },
  {
    "solution": "\\boxed{\\frac{24}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in quadrilateral $ABCD$, $AB\\parallel CD$, $AD\\parallel BC$, $AC$ bisects $\\angle DAB$. Line $BD$ intersects $AC$ at point $O$, and a line through point $C$ and perpendicular to $AB$ intersects the extension of line $AB$ at point $E$. Quadrilateral $ABCD$ is a rhombus; if $OA=4$ and $OB=3$, find the length of $CE$.\n",
    "completion": "\\textbf{Solution:} From equation (1), we have $AO\\perp OB$ and $AC=2OA=8$,\\\\\nthus $AB=\\sqrt{OA^2+OB^2}=\\sqrt{4^2+3^2}=5$,\\\\\nsince $CE\\perp AB$ and $\\angle OAB=\\angle EAC$,\\\\\nit follows that $\\triangle OAB\\sim \\triangle EAC$,\\\\\nhence $\\frac{OB}{AB}=\\frac{CE}{AC}$, that is $\\frac{3}{5}=\\frac{CE}{8}$,\\\\\nthus $CE=\\boxed{\\frac{24}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53110876_173.png"
  },
  {
    "solution": "\\boxed{y=-x+3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system, line $l_1$ intersects the x-axis at point $A$ and the y-axis at point $B$. The coordinates of point $B$ are $(0, 3)$. Line $l_2: y=2x$ intersects line $l_1$ at point $C$, and the x-coordinate of point $C$ is 1. What is the equation of line $l_1$?\n",
    "completion": "\\textbf{Solution:} Since line $l_{2}\\colon y=2x$ intersects with line $l_{1}$ at point $C$, and the x-coordinate of point $C$ is 1, therefore $y=2$, thus point $C(1, 2)$. Let the equation of line $l_{1}$ be $y=kx+b$. Substituting the coordinates of point $C$ yields: $\\begin{cases}k+b=2\\\\ b=3\\end{cases}$, solving this gives: $\\begin{cases}k=-1\\\\ b=3\\end{cases}$, therefore the equation of line $l_{1}$ is $y=-x+3$; hence, the equation of line $l_{1}$ is $\\boxed{y=-x+3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50481348_86.png"
  },
  {
    "solution": "\\boxed{230^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle B=50^\\circ$, $\\angle DAC$ and $\\angle ACF$ are two exterior angles of this triangle, then what is the sum of $\\angle DAC + \\angle ACF$ in degrees?\n",
    "completion": "\\textbf{Solution:} Given: $ \\angle DAC=\\angle B+\\angle ACB$, $ \\angle FCA=\\angle B+\\angle BAC$,\\\\\n$\\therefore \\angle DAC+\\angle ACF=\\angle B+\\angle B+\\angle BAC+\\angle BCA$,\\\\\n$\\because \\angle B=50^\\circ$, $\\angle B+\\angle BAC+\\angle BCA=180^\\circ$,\\\\\n$\\therefore \\angle DAC+\\angle ACF=180^\\circ+50^\\circ=\\boxed{230^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51212039.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AD$ is the altitude from $A$ to side $BC$, and $BE$ is the altitude from $B$ to side $AC$, with $AD$ and $BE$ intersecting at point $F$. If $BF=AC$, $BD=8$, and $CD=3$, what is the length of segment $AF$?\n",
    "completion": "\\textbf{Solution:}\n\nSince AD is the height from vertex A to side BC, and BE is the height from vertex B to side AC,\n\n$\\therefore \\angle ADC=\\angle FDB=90^\\circ$, $\\angle AEB=90^\\circ$,\n\n$\\therefore \\angle 1+\\angle C=90^\\circ$, $\\angle 1+\\angle 2=90^\\circ$,\n\n$\\therefore \\angle 2=\\angle C$,\n\nSince $\\angle 2=\\angle 3$,\n\n$\\therefore \\angle 3=\\angle C$,\n\nIn $\\triangle ADC$ and $\\triangle BDF$,\n\n$\\left\\{\\begin{array}{l}\n\\angle 3=\\angle C \\\\\n\\angle FDB=\\angle CDA \\\\\nBF=AC\n\\end{array}\\right.$,\n\n$\\therefore \\triangle BDF \\cong \\triangle ADC$ (AAS),\n\n$\\therefore FD=CD$, $AD=BD$,\n\nSince CD=3, BD=8,\n\n$\\therefore AD=8$, $DF=3$,\n\n$\\therefore AF=8-3=\\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52836935.png"
  },
  {
    "solution": "\\boxed{\\left( \\frac{2}{5}, \\frac{7}{5} \\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y=k_1x+b$ intersects the graph of the inverse proportion function $y=\\frac{k_2}{x}$ at points $A(1, 2)$ and $B(-2, n)$. If point $P$ is on segment $AB$, and the area ratio $S_{\\triangle AOP} : S_{\\triangle BOP} = 1 : 4$, find the coordinates of point $P$.\n",
    "completion": "\\textbf{Solution:} Let $P(x, x+1)$,\n\n$\\because$ the ratio of areas $S_{\\triangle AOP} : S_{\\triangle BOP} = 1 : 4$,\n\n$\\therefore$ the ratio of segments $AP : PB = 1 : 4$,\n\nwhich means $PB = 4PA$,\n\n$\\therefore (x+2)^2+(x+1+1)^2=16[(x-1)^2+(x+1-2)^2]$,\n\nsolving yields $x_1 = \\frac{2}{5}$, $x_2 = 2$ (discard),\n\n$\\therefore$ the coordinates of point $P$ are $\\boxed{\\left( \\frac{2}{5}, \\frac{7}{5} \\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51402407_32.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{2}a}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in square $ABCD$, $E$ is any point on diagonal $BD$. A line is drawn through point $E$ so that $EF\\perp BC$ at point $F$, and $EG\\perp CD$ at point $G$. If the perimeter of the square $ABCD$ is $a$, what is the perimeter of quadrilateral $EFCG$?\n",
    "completion": "\\textbf{Solution:} Given square $ABCD$, \\\\\n$\\therefore \\angle BDC=45^\\circ$, $\\angle C=90^\\circ$, \\\\\n$\\because$ $EF \\perp BC$, $EG \\perp CD$, \\\\\n$\\therefore \\angle C = \\angle CGE = \\angle CFE = \\angle DGE = 90^\\circ$, \\\\\n$\\therefore \\angle EDG = \\angle DEG = 45^\\circ$, quadrilateral $CGEF$ is a rectangle, \\\\\n$\\therefore EG = DG$, \\\\\n$\\therefore CG+EG=CG+DG=DC$ \\\\\n$\\because$ The perimeter of square $ABCD$ is $a$, \\\\\n$\\therefore$ The perimeter of quadrilateral $EFCG$ is $2DC = \\boxed{\\frac{1}{2}a}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51614857.png"
  },
  {
    "solution": "\\boxed{(4,2)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a rectangular piece of paper $OABC$ is placed in the Cartesian coordinate system, with $OA$ and $OC$ lying on the positive halves of the x-axis and y-axis, respectively. The line $AC$ is drawn, and it is given that $AC=4\\sqrt{5}$ and $OA=2CO$. If any line passing through a fixed point $M$ always divides the area of rectangle $OABC$ into two equal parts, what are the coordinates of point $M$?\n",
    "completion": "\\textbf{Solution:} The coordinates of point M are $\\boxed{(4,2)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52342628_124.png"
  },
  {
    "solution": "\\boxed{y=x^{2}+4x-1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, it is known that the parabola $y=x^{2}+bx+c$ intersects the line AB at points A and B, where A ($-3$, $-4$) and B ($0$, $-1$), and the axis of symmetry of the parabola intersects the x-axis at point Q. What is the function expression of this parabola?\n",
    "completion": "\\textbf{Solution:} Since the parabola passes through point A ($-3$, $-4$) and point B ($0$, $-1$), \\\\\nit follows that $\\begin{cases} 9-3b+c=-4 \\\\ c=-1 \\end{cases}$, hence $b=4$ and $c=-1$, \\\\\nthus, the equation of the parabola is $\\boxed{y=x^{2}+4x-1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52694121_90.png"
  },
  {
    "solution": "\\boxed{x>-1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the lines $y=-x+b$ and $y=kx$ are represented in the same Cartesian coordinate system. What is the solution set of the inequality $-x+b<kx$ with respect to $x$?\n",
    "completion": "\\textbf{Solution:} From the graph, it is known that the graphs of the two lines intersect at ($-1$, $-3$), \\\\\n$\\because$ $-x+b<kx$, \\\\\n$\\therefore$ select the part to the right of the intersection point, \\\\\nthus, for $x>-1$, $-x+b<kx$, \\\\\nso the answer is: $\\boxed{x>-1}$",
    "image_path": "dataset/mm_math/images/MM_Math/52212222.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y_{1} = 2x - 2$ intersects the y-axis at point $A$, and the line $y_{2} = -2x + 6$ intersects the y-axis at point $B$, with the two lines intersecting at point $C$. Find the area of $\\triangle ABC$.\n",
    "completion": "Solution: Since let $x=0$, then ${y}_{1}=-2$, ${y}_{2}=6$\\\\\nTherefore, $A(0, 2)$, $B(0, 6)$.\\\\\nTherefore, $AB=8$\\\\\nTherefore, the area of $\\triangle ABC$ is $\\frac{1}{2}\\times 8\\times 2=\\boxed{8}.",
    "image_path": "dataset/mm_math/images/MM_Math/52422271_67.png"
  },
  {
    "solution": "\\boxed{\\frac{5}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ABC=90^\\circ$, $AB=3$, and $AC=4$. Now, fold $\\triangle ABC$ along $BD$ such that point A lies exactly on $BC$. What is the length of $CD$?\n",
    "completion": "\\textbf{Solution:} Let $CD=x$, then $AD=A'D=4-x$,\\\\\nIn $\\triangle ABC$ right-angled at $B$,\\\\\n$BC=\\sqrt{AB^{2}+AC^{2}}=\\sqrt{3^{2}+4^{2}}=5$,\\\\\n$\\therefore A'C=BC-A'B=BC-AB=5-3=2$,\\\\\nIn $\\triangle A'DC$ right-angled at $D$,\\\\\n$A'D^{2}+A'C^{2}=CD^{2}$  \\\\\ni.e., ${(4-x)}^{2}+2^{2}=x^{2}$\\\\\nSolving gives: $x=\\boxed{\\frac{5}{2}}$",
    "image_path": "dataset/mm_math/images/MM_Math/52867062.png"
  },
  {
    "solution": "\\boxed{P\\left(\\frac{n-m}{4},\\frac{n+3m}{4}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the plane Cartesian coordinate system $xOy$, it is known that the graph of the linear function $y=x+m$ intersects the x-axis and y-axis at points A and B, respectively. It intersects the graph of the linear function $y=-3x+n\\ (n>m)$ at point P. Also, the graph of the linear function $y=-3x+n$ intersects the x-axis and y-axis at points C and D, respectively. Express the coordinates of point P in an algebraic expression involving $m$ and $n$.\n",
    "completion": "\\textbf{Solution:} Given that the graph of the linear function $y=x+m$ intersects the $x$-axis and $y$-axis at points $A$ and $B$, respectively. Therefore, $A(-m,0)$ and $B(0,m)$, which implies $OA=OB$. Hence, $\\angle PAC=\\angle ABO=45^\\circ$. Since the graph of the linear function $y=x+m$ intersects with the graph of the linear function $y=-3x+n$ at point $P$, by solving the system of equations \n\\[\\left\\{\\begin{array}{l}\ny=x+m\\\\ \ny=-3x+n \n\\end{array}\\right.,\\]\nwe get \n\\[\\left\\{\\begin{array}{l}\nx=\\frac{n-m}{4}\\\\ \ny=\\frac{n+3m}{4}\n\\end{array}\\right..\\]\nTherefore, $\\boxed{P\\left(\\frac{n-m}{4},\\frac{n+3m}{4}\\right)}$, which leads to $\\angle PAC=45^\\circ$, $\\boxed{P\\left(\\frac{n-m}{4},\\frac{n+3m}{4}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52205271_96.png"
  },
  {
    "solution": "\\boxed{9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, it is known that the length of segment $AB=20$. Point $C$ is on segment $AB$, point $D$ is on segment $CB$, and point $E$ is the midpoint of segment $DB$, with $DE=3$. If $CE=8$, what is the length of $AC$?\n",
    "completion": "\\textbf{Solution:} We have $E$ as the midpoint of $DB$, with $DE=3$, and $CE=8$, \\\\ \nthus $DB=2DE=6$, and $CD=CE-DE=5$, \\\\ \ntherefore, $AC=AB-CD-DB=20-6-5=\\boxed{9}$.",
    "image_path": "dataset/mm_math/images/MM_Math/42461901_214.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{1}{12}(x-6)^2+4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a goalkeeper at point $O$ kicks a high ball, which starts its flight from point $A$, 1 meter above the ground ($A$ is on the y-axis). Player B notices the ball directly above his head at its highest point $M$, 6 meters away from $O$ and approximately 4 meters above the ground. After the ball lands, it bounces back. Experimental calculations show that the parabolic trajectory of the ball after bouncing on the grass has the same shape as its original trajectory, but its maximum height is reduced to half of the original maximum height. Determine the expression of the parabola for the ball's trajectory from its launch to its first landing.\n",
    "completion": "\\textbf{Solution:} As shown in the figure, assume the expression of the parabola when it first hits the ground is $y=a(x-6)^2+4$.\\\\\nGiven: when $x=0$, $y=1$.\\\\\nThus, $1=36a+4$, $\\therefore a=-\\frac{1}{12}$.\\\\\n$\\therefore$ The expression is $\\boxed{y=-\\frac{1}{12}(x-6)^2+4}$ (or $\\boxed{y=-\\frac{1}{12}x^2+x+1}$).",
    "image_path": "dataset/mm_math/images/MM_Math/52977497_174.png"
  },
  {
    "solution": "\\boxed{\\left(-\\frac{3}{7}, 0\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the Cartesian coordinate system, the parabola $y=-x^2-2x+3$ intersects the x-axis at points $A$ and $B$ (where $A$ is to the left of $B$) and intersects the y-axis at point $C$, with the vertex denoted as $D$. As shown in the diagram, find a point $E$ on the x-axis such that the perimeter of $\\triangle CDE$ is minimized. What are the coordinates of point $E$?\n",
    "completion": "\\textbf{Solution:} Let's construct point $C'$ symmetric to point C with respect to the x-axis, and connect $C'D$ intersecting the x-axis at point E. At this moment, the perimeter of $\\triangle CDE$ is minimized, as shown in figure 1,\\\\\n$\\because C(0, 3),$\\\\\n$\\therefore C'(0, -3),$\\\\\nSuppose the equation of line $C'D$ is $y=kx+b$,\\\\\nthen $\\begin{cases}4=-k+b\\\\-3=b\\end{cases}$, solving this we get: $\\begin{cases}k=-7\\\\b=-3\\end{cases}$,\\\\\n$\\therefore$ the equation of line $C'D$ is $y=-7x-3$,\\\\\nwhen in $y=-7x-3$, $y=0$, then $x=-\\frac{3}{7}$,\\\\\n$\\therefore$ when the perimeter of $\\triangle CDE$ is minimized, the coordinates of point E are $\\boxed{\\left(-\\frac{3}{7}, 0\\right)}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52720132_80.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, there is a square $ABCD$ with side length $6\\sqrt{2}$. $E$ is the midpoint of side $CD$. There is a moving point $F$ on the diagonal $BD$. When $\\triangle ABF$ is similar to $\\triangle DEF$, what is the value of $BF$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral ABCD is a square, \\\\\nit follows that BC=CD=$6\\sqrt{2}$, and $\\angle C=90^\\circ$.\\\\\nSince E is the midpoint of side CD, \\\\\nit follows that DE=CE=$\\frac{1}{2}$CD=$3\\sqrt{2}$.\\\\\nIn $\\triangle BCD$, by the Pythagorean theorem, we have \\\\\nBD=$\\sqrt{BC^{2}+CD^{2}}=\\sqrt{(6\\sqrt{2})^{2}+(6\\sqrt{2})^{2}}=12$.\\\\\nLet BF=x, then DF=12-x.\\\\\n(1) When $\\triangle ABF\\sim \\triangle FDE$,\\\\\nfrom $\\frac{DF}{BA}=\\frac{DE}{BF}$, we get $\\frac{12-x}{6\\sqrt{2}}=\\frac{3\\sqrt{2}}{x}$,\\\\\nwhich gives $x=\\boxed{6}$.\\\\\n(2) When $\\triangle ABF\\sim \\triangle EDF$,\\\\\nfrom $\\frac{DF}{BF}=\\frac{DE}{BA}$, we get $\\frac{12-x}{x}=\\frac{3\\sqrt{2}}{6\\sqrt{2}}$,\\\\\nwhich gives $x=8$.\\\\\nIn conclusion, the value of BF can be either 6 or 8.",
    "image_path": "dataset/mm_math/images/MM_Math/52602484.png"
  },
  {
    "solution": "\\boxed{65^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the cyclic pentagon $ABCDE$, if $\\angle C + \\angle CDE + \\angle E + \\angle EAB = 425^\\circ$, what is the measure of $\\angle CDA$?\n",
    "completion": "\\textbf{Solution:} Since the sum of the interior angles of pentagon $ABCDE$ is $(5-2)\\times 180^\\circ=540^\\circ$, \\\\\nthus $\\angle EAB+\\angle B+\\angle C+\\angle CDE+\\angle E=540^\\circ$, \\\\\nbecause $\\angle EAB+\\angle C+\\angle CDE+\\angle E=425^\\circ$, \\\\\ntherefore $\\angle B=540^\\circ-425^\\circ=115^\\circ$, \\\\\nsince quadrilateral $ABCD$ is inscribed in circle $O$, \\\\\ntherefore $\\angle B+\\angle CDA=180^\\circ$, \\\\\nthus $\\angle CDA=180^\\circ-115^\\circ=\\boxed{65^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51318709.png"
  },
  {
    "solution": "\\boxed{75^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in $\\triangle ABC$, $AB=AC$, $\\angle B=25^\\circ$, and point D is on side $BC$. Fold $\\triangle ACD$ along $AD$ to get $\\triangle ADE$. If $AD=DE$, then what is the measure of $\\angle AFB$?\n",
    "completion": "\\textbf{Solution:} Given that in $\\triangle ABC$, $AB=AC$, and $\\angle B=25^\\circ$, \\\\\nthus, $\\angle C=\\angle B=25^\\circ$, \\\\\nby folding, we know $\\angle E=\\angle C=25^\\circ$, \\\\\nsince $AD=DE$, \\\\\nhence, $\\angle DAE=\\angle E=25^\\circ$, \\\\\nsince $\\angle DAE=\\angle DAC=25^\\circ$, \\\\\nthus, $\\angle CAF=50^\\circ$, \\\\\ntherefore, $\\angle AFB=\\angle C+\\angle CAF=\\boxed{75^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52827130.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\triangle ABC$, with $ \\odot O$ passing through vertices A and C of $ \\triangle ABC$ and intersecting side AB at point D. A line is drawn through point D such that $ DF\\parallel BC$, intersecting $ AC$ at point E and $ \\odot O$ at point F. Line $ DC$ is drawn, with point C being the midpoint of arc $ DF$, and $ BC$ being the tangent of $ \\odot O$. If the radius of $ \\odot O$ is 3, and $ DF=4\\sqrt{2}$, what is the value of $ CE\\cdot CA$?\n",
    "completion": "\\textbf{Solution:} Draw line $OC$ intersecting $DF$ at $M$, and connect $OD$, $AF$, as shown in the figure:\\\\\nSince $C$ is the midpoint of arc $DF$,\\\\\n$\\therefore$ $OC\\perp DF$,\\\\\n$\\therefore$ $DM=MF=\\frac{1}{2}DF=2\\sqrt{2}$,\\\\\nSince the radius of circle $O$ is 3,\\\\\n$\\therefore$ $OM=\\sqrt{OD^{2}-DM^{2}}=\\sqrt{3^{2}-\\left(2\\sqrt{2}\\right)^{2}}=1$,\\\\\n$\\therefore$ $CM=OC-OM=3-1=2$,\\\\\n$\\therefore$ $DC^{2}=DM^{2}+CM^{2}=\\left(2\\sqrt{2}\\right)^{2}+2^{2}=12$,\\\\\nSince $\\overset{\\frown}{DC}=\\overset{\\frown}{FC}$,\\\\\n$\\therefore$ $\\angle DAC=\\angle CAF$,\\\\\nSince $\\angle CDF=\\angle CAF$,\\\\\n$\\therefore$ $\\angle CDF=\\angle DAC$,\\\\\nSince $\\angle DCE=\\angle ACD$,\\\\\n$\\therefore$ $\\triangle DCE\\sim \\triangle ACD$,\\\\\n$\\therefore$ $\\frac{CD}{AC}=\\frac{CE}{CD}$,\\\\\n$\\therefore$ $CD^{2}=CE\\cdot CA$,\\\\\n$\\therefore$ $CE\\cdot CA=\\boxed{12}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53148281_125.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Point $O$ is the midpoint of line segment $AB$, where $OB=14\\,\\text{cm}$. Point $P$ divides the line segment $AB$ into two parts such that the ratio of $AP$ to $PB$ is $5:2$. Find the length of line segment $OP$.\n",
    "completion": "\\textbf{Solution:} Since point O is the midpoint of line segment AB, $OB=14\\,\\text{cm}$,\\\\\ntherefore $AB=2OB=28\\,\\text{cm}$,\\\\\nsince $AP\\colon PB=5\\colon 2$,\\\\\ntherefore $BP= \\frac{2}{7}AB=8\\,\\text{cm}$,\\\\\ntherefore $OP=OB-BP=14-8=\\boxed{6}\\,\\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/42460301_217.png"
  },
  {
    "solution": "\\boxed{24}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the rhombus $ABCD$, $AB=5$ and $AC+BD=14$. What is the area of the rhombus $ABCD$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral $ABCD$ is a rhombus,\\\\\n$\\therefore OA=\\frac{1}{2}AC$, $OB=\\frac{1}{2}BD$, and $AC\\perp BD$,\\\\\n$\\therefore \\angle AOB=90^\\circ$,\\\\\n$\\therefore OA^{2}+OB^{2}=AB^{2}=25$,\\\\\n$\\because AC+BD=14$,\\\\\n$\\therefore OA+OB=7$,\\\\\n$\\therefore (OA+OB)^{2}=7^{2}=49$,\\\\\nwhich means $OA^{2}+2AO\\cdot OB+OB^{2}=49$,\\\\\n$\\therefore 2OA\\cdot OB=49-25=24$,\\\\\n$\\therefore S_{\\text{rhombus} ABCD}=\\frac{1}{2}AC\\cdot BD=2OA\\cdot OB=\\boxed{24}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53538341.png"
  },
  {
    "solution": "\\boxed{\\frac{2\\sqrt{5}}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $M$ is a point on side $BC$. Connect $AM$ and draw $DE \\perp AM$ with $E$ as the foot of the perpendicular. If $DE=DC=1$ and $AE=2EM$, what is the length of $BM$?\n",
    "completion": "\\textbf{Solution}: Since rectangle ABCD, with DE$\\perp$AM, \\\\\nit follows that $\\angle$C=$\\angle$AED=$\\angle$B=$90^\\circ$, with AB=DE=CD=1, and AD$\\parallel$BC,\\\\\nhence, $\\angle$DAE=$\\angle$AEM. In $\\triangle$ADE and $\\triangle$MAB\n\\[ \\left\\{\\begin{array}{l}\n\\angle B=\\angle AED\\\\\n\\angle DAE=\\angle AMB\\\\\nDE=AB\n\\end{array}\\right. \\]\nit follows that $\\triangle$ADE$\\cong$$\\triangle$MAB (by AAS),\\\\\nthus, AE=BM=2EM;\\\\\nIn the right triangle ABM, AB$^{2}$+BM$^{2}$=AM$^{2}$, that is, 1$^{2}$+($2EM$)$^{2}$=($3EM$)$^{2}$,\\\\\nSolving for this yields: $EM=\\frac{\\sqrt{5}}{5}$,\\\\\ntherefore, BM=\\boxed{\\frac{2\\sqrt{5}}{5}}.",
    "image_path": "dataset/mm_math/images/MM_Math/52251668.png"
  },
  {
    "solution": "\\boxed{20°}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the isosceles triangle $ABC$, where $AB=AC$ and $BD\\perp AC$, with $\\angle ABC=70^\\circ$, what is the value of $\\angle DBC$?\n",
    "completion": "\\textbf{Solution:} \\\\\nSince AB = AC, \\\\\nthus $\\angle$ACB = $\\angle$ABC = 70°, \\\\\nSince BD $\\perp$ AC, \\\\\nthus $\\angle$BDC = 90°, \\\\\nhence $\\angle$DBC = 90° - $\\angle$ACB = \\boxed{20°}.",
    "image_path": "dataset/mm_math/images/MM_Math/52845720.png"
  },
  {
    "solution": "\\boxed{21^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points B and D are on the ray AM, points C and E are on the ray AN, and AB=BC=CD=DE. Given that $\\angle EDM=84^\\circ$, what is the measure of $\\angle A$ in degrees?\n",
    "completion": "Solution: Since $AB=BC=CD=DE$, \\\\\nit follows that $\\angle A=\\angle BCA$, $\\angle CBD=\\angle BDC$, $\\angle ECD=\\angle CED$,\\\\\naccording to the exterior angle theorem, $\\angle A+\\angle BCA=\\angle CBD$, $\\angle A+\\angle CDB=\\angle ECD$, $\\angle A+\\angle CED=\\angle EDM$,\\\\\nadditionally, since $\\angle EDM=84^\\circ$,\\\\\nthus, $\\angle A+3\\angle A=84^\\circ$,\\\\\nsolving this gives $\\angle A=\\boxed{21^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52793522.png"
  },
  {
    "solution": "\\boxed{120^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, OC is the bisector of $\\angle AOB$, and $\\angle COD = 20^\\circ$. If $\\angle BOD = 2\\angle AOD$, what is the degree measure of $\\angle AOB$?\n",
    "completion": "\\textbf{Solution:} Let $\\angle AOD=x$, then $\\angle BOD=2x$,\\\\\nthus $\\angle AOB=\\angle AOD+\\angle BOD=3x$,\\\\\nsince $OC$ is the bisector of $\\angle AOB$,\\\\\nthus $\\angle AOC=\\frac{1}{2}\\angle AOB=\\frac{3}{2}x$,\\\\\nthus $\\frac{3}{2}x-x=20^\\circ$,\\\\\nsolving for $x=40^\\circ$,\\\\\nthus $\\angle AOB=3x=\\boxed{120^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52837816_95.png"
  },
  {
    "solution": "\\boxed{120^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, line segment AB passes through the center O and intersects the circle $\\odot O$ at points A and C, $\\angle B=30^\\circ$. The line BD is tangent to the circle $\\odot O$ at point D. What is the degree measure of $\\angle ADB$?\n",
    "completion": "\\textbf{Solution:} Connect OD, as shown in the diagram,\\\\\nBD is tangent to $\\odot$O at D,\\\\\n$\\therefore$ $\\angle$ODB$=90^\\circ$,\\\\\n$\\because$ $\\angle$B$=30^\\circ$\\\\\n$\\therefore$ $\\angle$DOB$=60^\\circ$\\\\\n$\\therefore$ $\\angle$A$=30^\\circ$\\\\\n$\\therefore$ $\\angle$ADB$=180^\\circ-\\angle$B$-\\angle$A=$180^\\circ-30^\\circ-30^\\circ$=\\boxed{120^\\circ}",
    "image_path": "dataset/mm_math/images/MM_Math/52978026.png"
  },
  {
    "solution": "\\boxed{(6，2\\sqrt{3})}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate plane, the quadrilateral OABC is a square. An equilateral $\\triangle ABD$ is constructed with AB as the side, and point D is exactly on the x-axis. The diagonal CA intersects the x-axis at point E. If the coordinates of point E are $\\left(4, 0\\right)$, what are the coordinates of point B?\n",
    "completion": "\\textbf{Solution:} In square $OABC$, $OC=BC$, $\\angle OCA = \\angle BCA = \\angle OAC = 45^\\circ$,\n\\[\n\\because EC = EC,\n\\]\n\\[\n\\therefore \\triangle OEC \\cong \\triangle BEC \\text{ (SAS)},\n\\]\n\\[\n\\therefore \\angle OEC = \\angle BEC, \\text{ BE=OE},\n\\]\n\\[\n\\because \\angle EOA = \\angle EDA = 15^\\circ,\n\\]\n\\[\n\\therefore \\angle OEC = 60^\\circ,\n\\]\n\\[\n\\therefore \\angle BED = 180^\\circ - 60^\\circ - 60^\\circ = 60^\\circ,\n\\]\nAs shown in the figure, draw BH $\\perp$ ED at point H,\n\\[\n\\therefore \\angle EBH = 30^\\circ,\n\\]\n\\[\n\\because E(4,0),\n\\]\n\\[\n\\therefore OE = 4 = EB,\n\\]\n\\[\n\\therefore EH = \\frac{1}{2} EB = 2,\n\\]\nAccording to the Pythagorean theorem, we get BH=$2\\sqrt{3}$,\n\\[\n\\therefore \\text{The coordinates of point B are} \\boxed{(6，2\\sqrt{3})};\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/53359725_138.png"
  },
  {
    "solution": "\\boxed{\\frac{30}{13}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the right-angled $\\triangle ABC$, $\\angle C=90^\\circ$, $AC=8$, $BC=6$. The point $M$ starts from point $C$ and moves along the segment $CA$ towards point $A$, connecting $BM$, and $MN \\perp BM$ intersects side $AB$ at point $N$. If $CM=2$, what is the length of segment $AN$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $NH \\perp AC$ at point $H$ through point $N$, \\\\\n$\\because \\angle C=90^\\circ$, $AC=8$, $BC=6$, \\\\\n$\\therefore AB=\\sqrt{AC^2+BC^2}=10$, \\\\\n$\\because \\sin A=\\frac{NH}{AN}=\\frac{BC}{AB}=\\frac{3}{5}$, $\\tan A=\\frac{NH}{AH}=\\frac{BC}{AC}=\\frac{3}{4}$, \\\\\n$\\therefore AN=\\frac{5}{3}NH$, $AH=\\frac{4}{3}NH$, \\\\\n$\\because \\angle C=\\angle BMN=\\angle MNH=90^\\circ$, \\\\\n$\\therefore \\angle BMC+\\angle NMH=90^\\circ$, \\\\\n$\\therefore \\angle CBM=90^\\circ-\\angle BMC=\\angle NMH$, \\\\\n$\\therefore \\tan\\angle CBM=\\frac{CM}{BC}=\\tan\\angle NHM=\\frac{NH}{MH}$, \\\\\nthus, $\\frac{2}{6}=\\frac{NH}{8-2-\\frac{4}{3}NH}$, \\\\\nSolving gives: $NH=\\frac{18}{13}$, \\\\\n$\\therefore AN=\\frac{5}{3}NH=\\frac{5}{3} \\times \\frac{18}{13}=\\boxed{\\frac{30}{13}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53065536.png"
  },
  {
    "solution": "\\boxed{16}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that points $A$, $B$, and $C$ are on the same line segment, $M$ is the midpoint of segment $AC$, $N$ is the midpoint of segment $BC$, and $AM=5\\, \\text{cm}$, $CN=3\\, \\text{cm}$. What is the length of line segment $AB$ in centimeters?\n",
    "completion": "\\textbf{Solution:} Since $M$ is the midpoint of line segment $AC$ and $N$ is the midpoint of line segment $BC$,\\\\\nit follows that $MC = AM = 5\\,\\text{cm}$, $BN = CN = 3\\,\\text{cm}$,\\\\\ntherefore $AB = AM + MC + CN + NB = \\boxed{16}\\,\\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53206252.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, fold side AD of the rectangle so that point D falls on point F on side BC. If $ AB=8\\,cm$ and $BC=10\\,cm$, what is the length of EC in cm?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rectangle,\\\\\nhence AB=CD=8cm, AD=BC=10cm, $\\angle$B=$\\angle$C=$90^\\circ$,\\\\\nBy the property of folding: AF=AD=10cm, DE=EF,\\\\\nIn $\\triangle$ABF, BF=$\\sqrt{AF^{2}-AB^{2}}=6cm$,\\\\\nthus CF=BC-BF=4cm,\\\\\nLet EC=x, then DE=EF=8-x,\\\\\nIn $\\triangle$EFC, since EF$^{2}$=EC$^{2}$+CF$^{2}$,\\\\\nthus (8-x)$^{2}$=x$^{2}$+4$^{2}$,\\\\\nthus x=\\boxed{3}cm,",
    "image_path": "dataset/mm_math/images/MM_Math/53333928.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the trapezoid $ABCD$, $\\angle ABC = 90^\\circ$, $AD \\parallel BC$, $BC = 2AD$, and the diagonals $AC$ and $BD$ intersect at point $E$. Point $F$ is a point on the line segment $EC$, and $\\angle BDF = \\angle BAC$. If $BC = 6$ and $\\sin\\angle BAC = \\frac{2}{3}$, what is the length of $FC$?\n",
    "completion": "\\textbf{Solution:}\n\nSince $\\angle ABC=90^\\circ$, $BC=6$, and $\\sin\\angle BAC=\\frac{2}{3}$,\n\nit follows that $\\frac{BC}{AC}=\\frac{2}{3}$, hence $AC=9$,\n\nthus $AB=\\sqrt{AC^2-BC^2}=3\\sqrt{5}$,\n\nsince $BC=2AD$,\n\ntherefore $AD=3$,\n\ngiven that $AD\\parallel BC$,\n\nit follows that $\\angle BAD=90^\\circ$,\n\nhence $BD=\\sqrt{AB^2+AD^2}=3\\sqrt{6}$,\n\nbecause $\\triangle EAD\\sim \\triangle ECB$,\n\nit follows that $\\frac{EA}{EC}=\\frac{ED}{EB}=\\frac{AD}{BC}=\\frac{3}{6}=\\frac{1}{2}$,\n\nthus $\\frac{EA}{AC}=\\frac{1}{3}$, $\\frac{ED}{BD}=\\frac{1}{3}$,\n\ntherefore $EA=\\frac{1}{3}AC=3$, $ED=\\frac{1}{3}BD=\\sqrt{6}$,\n\nhence $EC=6$, $BE=2\\sqrt{6}$,\n\nsince $EB^2=EF\\cdot EC$,\n\nit follows that $(2\\sqrt{6})^2=6EF$,\n\ntherefore $EF=4$,\n\ntherefore $FC=EC-EF=6-4=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51379378_81.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points $C$, $E$ are two points on line segment $AB$, point $D$ is the midpoint of line segment $AB$, and $AB=6$, $CD=1$. What is the length of $BC$?\n",
    "completion": "\\textbf{Solution:} Since point D is the midpoint of line segment AB, and AB$=6$, \\\\\nit follows that BD$=\\frac{1}{2}$AB$=3$, \\\\\nSince CD$=1$, \\\\\nit follows that BC$=BD-CD=3-1=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51653655_166.png"
  },
  {
    "solution": "\\boxed{0.09}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given that in $\\triangle ABC$, we have $AB=AC=5$ and $BC=8$. Point $E$ is a moving point on ray $CA$, and point $O$ is a moving point on side $BC$, with $OC=OE$. Ray $OE$ intersects ray $BA$ at point $D$. As shown in Figure 1, if $OC=2$, find the value of $\\frac{{S}_{\\triangle ADE}}{{S}_{\\triangle ODB}}$.\n",
    "completion": "\\textbf{Solution:} Given $AB=AC$,\\\\\nit follows $\\angle B=\\angle C$,\\\\\nsince $OC=OE$,\\\\\nit follows $\\angle OEC=\\angle C$,\\\\\nhence, $\\angle B=\\angle OEC$,\\\\\nthus, $\\triangle ABC\\sim \\triangle OEC$,\\\\\nimplying $\\frac{AC}{OC}=\\frac{BC}{CE}$,\\\\\nwhich gives $\\frac{5}{2}=\\frac{8}{CE}$,\\\\\ntherefore $CE=3.2$,\\\\\nthus, $AE=1.8$;\\\\\nSince $\\angle AED=\\angle OEC=\\angle B, \\angle D=\\angle D$,\\\\\nit implies $\\triangle OBD\\sim \\triangle AED$,\\\\\ngiving $\\frac{AE}{OB}=\\frac{1.8}{6}=0.3$,\\\\\ntherefore $\\frac{S_{\\triangle ADE}}{S_{\\triangle ODB}}=\\boxed{0.09}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51319009_112.png"
  },
  {
    "solution": "\\boxed{x<-1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the graphs of the functions $y_1=kx$ and $y_2=ax+3$ intersect at point $A(-1, 2)$. What is the solution set of the inequality $kx>ax+3$ with respect to $x$?\n",
    "completion": "\\textbf{Solution:} By observing the graph, we can see that when $x<-1$, the line $y_1=kx$ is above the line $y_2=ax+3$. \\\\\nTherefore, the solution set for the inequality $kx>ax+3$ with respect to $x$ is $x<-1$.\\\\\nHence, the answer is: $\\boxed{x<-1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51661989.png"
  },
  {
    "solution": "\\boxed{9\\pi-12\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the sector $AOB$, $\\angle AOB=90^\\circ$, with radius $OA=6$. When the sector $AOB$ is folded along the line passing through point $B$, point $O$ falls exactly on point $D$ on the arc $AB$. What is the area of the entire shaded region? (Express in terms of $\\pi$)\n",
    "completion": "\\textbf{Solution:} Draw $OD$. According to the properties of folding, $CD=CO$, $BD=BO$, $\\angle DBC=\\angle OBC$,\\\\\n$\\therefore OB=OD=BD$, i.e., $\\triangle OBD$ is an equilateral triangle,\\\\\n$\\therefore \\angle DBO=60^\\circ$, $\\therefore \\angle CBO= \\frac{1}{2} \\angle DBO=30^\\circ$,\\\\\n$\\because \\angle AOB=90^\\circ$, $\\therefore OC=OB \\cdot \\tan \\angle CBO=6 \\times \\frac{\\sqrt{3}}{3} =2\\sqrt{3}$,\\\\\n$\\therefore S_{\\triangle BDC}=S_{\\triangle OBC}= \\frac{1}{2} \\times OB \\times OC= \\frac{1}{2} \\times 6 \\times 2\\sqrt{3} =6\\sqrt{3}$,\\\\\n$S_{\\text{sector} AOB}= \\frac{90}{360} \\pi \\times 6^2=9\\pi$,\\\\\n$\\therefore$ the area of the entire shaded region is: $S_{\\text{sector} AOB}-S_{\\triangle BDC}-S_{\\triangle OBC}=9\\pi-6\\sqrt{3}-6\\sqrt{3} =9\\pi-12\\sqrt{3}$.\\\\\nTherefore, the answer is: $9\\pi-12\\sqrt{3}=\\boxed{9\\pi-12\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52949253.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, the graph of the direct proportion function $y=2x$ intersects with the graph of the linear function $y=kx+b$ at point $A(m, 2)$. The graph of the linear function passes through point $B(-2, -1)$ and intersects with the $x$-axis at point $C$. What is the area of $\\triangle AOC$?\n",
    "completion": "\\textbf{Solution:} Let $y=0$, then $x+1=0$, thus $x=-1$.\\\\\nHence, the coordinates of point C are $(-1,0)$,\\\\\ntherefore, the area of $\\triangle AOC$ is $=\\frac{1}{2} \\times 1 \\times 2 = \\boxed{1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53183208_208.png"
  },
  {
    "solution": "\\boxed{\\left(\\frac{8}{3},0\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the Cartesian coordinate system, the vertices of rectangle OABC are as follows: vertex O is the origin of coordinates, vertices A and C are on the x-axis and y-axis respectively, with $OA=4$ and $OC=3$. D is the midpoint of side AB, and E is a moving point on side OA. When the perimeter of $\\triangle CDE$ is minimized, what are the coordinates of point E?\n",
    "completion": "\\textbf{Solution:} Construct point F symmetrical to point D with respect to the x-axis, as shown in the figure,\\\\\n$\\because$ Quadrilateral OABC is a rectangle,\\\\\n$\\therefore$OC＝BD＝3, the coordinates of point C are $\\left(0,3\\right)$,\\\\\n$\\because$D is the midpoint of side AB,\\\\\n$\\therefore$AD＝$\\frac{3}{2}$,\\\\\n$\\because$OA＝4,\\\\\n$\\therefore$The coordinates of point D are $\\left(4,\\frac{3}{2}\\right)$, then the coordinates of point F are $\\left(4,-\\frac{3}{2}\\right)$,\\\\\nAccording to the property of axial symmetry: EF＝ED,\\\\\n$\\therefore$P$_{\\triangle CDE}$＝CD+CE+DE＝CD+CE+EF, where CD is a constant,\\\\\nWhen the sum of CE+EF is minimized, the perimeter of $\\triangle CDE$ is minimal, at this time points C, E, F are collinear,\\\\\nLet the equation of line CF be: $y=kx+b\\left(k\\ne 0\\right)$,\\\\\nSubstituting $\\left(0,3\\right)$ and $\\left(4,-\\frac{3}{2}\\right)$ into the equation yields:\\\\\n$\\begin{cases}b=3\\\\ 4k+b=-\\frac{3}{2}\\end{cases}$, solving gives: $\\begin{cases}k=-\\frac{9}{8}\\\\ b=3\\end{cases}$,\\\\\n$\\therefore$The equation of line CF is: $y=-\\frac{9}{8}x+3$,\\\\\nSetting $y=0$, we get: $-\\frac{9}{8}x+3=0$,\\\\\nSolving for $x$ gives: $x=\\frac{8}{3}$,\\\\\n$\\therefore$The coordinates of point E are $\\boxed{\\left(\\frac{8}{3},0\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51663250.png"
  },
  {
    "solution": "\\boxed{24\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, in $ \\triangle DEF$, $\\angle DEF=90^\\circ$, $\\angle D=30^\\circ$. Let B be a moving point on the hypotenuse DF. Construct AB $\\perp$ DF at B, intersecting side DE (or side EF) at point A. Let BD $= x$, and the area of $\\triangle ABD = y$. Figure 2 is the graph of the function of y with respect to x. What is the maximum area of $\\triangle ABD$?\n",
    "completion": "\\textbf{Solution:} It can be observed from the diagram that point A reaches point E, the area of $\\triangle ABD$ is maximized. At this juncture, $DB=12$, \\\\\n$AB=DB\\cdot \\tan 30^\\circ=12\\times \\frac{\\sqrt{3}}{3}=4\\sqrt{3}$, \\\\\n$\\therefore S_{\\triangle ABD}=\\frac{1}{2}\\times 12\\times 4\\sqrt{3}=\\boxed{24\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51321576.png"
  },
  {
    "solution": "\\boxed{13^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle B=42^\\circ$, $\\angle C=68^\\circ$, $AD$ is the altitude of $\\triangle ABC$, $AE$ is the angle bisector of $\\triangle ABC$. Find the degree measure of $\\angle EAD$.\n",
    "completion": "\\textbf{Solution:} Given that $\\angle B=42^\\circ$ and $\\angle C=68^\\circ$, \\\\\nit follows that $\\angle BAC=180^\\circ-68^\\circ-42^\\circ=70^\\circ$\\\\\nSince AE is the angle bisector of $\\triangle ABC$,\\\\\nit follows that $\\angle CAE=70^\\circ \\div 2=35^\\circ$\\\\\nSince AD is the altitude of $\\triangle ABC$,\\\\\nit follows that $\\angle ADB=90^\\circ$\\\\\nTherefore, $\\angle DAC=90^\\circ-\\angle C=90^\\circ-68^\\circ=22^\\circ$\\\\\nTherefore, $\\angle EAD=\\angle EAC-\\angle DAC=35^\\circ-22^\\circ=\\boxed{13^\\circ}$,\\\\\nthat is, the measure of $\\angle EAD$ is $13^\\circ$",
    "image_path": "dataset/mm_math/images/MM_Math/52603636.png"
  },
  {
    "solution": "\\boxed{x<-3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the line $l_1\\colon y = 3x+1$ and the line $l_2\\colon y = mx+n$ intersect at point $P(a, -8)$. What is the solution set of the inequality $3x+1 < mx+n$ with respect to $x$?\n",
    "completion": "\\[\n\\textbf{Solution:} \\text{Let} \\because \\text{line } l_1\\colon y = 3x+1 \\text{ and line } l_2\\colon y = mx+n \\text{ intersect at point } P(a, -8)\\\\\n\\therefore 3a+1=-8 \\\\\n\\therefore a=-3 \\\\\n\\therefore P(-3, -8)\\\\\n\\text{According to the problem, when } x<-3, 3x+1<mx+n\\\\\n\\therefore \\text{the solution set of the inequality } 3x+1<mx+n \\text{ with respect to } x \\text{ is } \\boxed{x<-3}\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/51699535.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $AB=8$, $BC=4\\sqrt{2}$, $DE\\perp AC$ at point $F$, and intersects $AB$ at point $E$. Find the length of $DE$.\n",
    "completion": "\\textbf{Solution:} In rectangle $ABCD$,\\\\\n$\\therefore$ $\\angle DAB = \\angle ADC = 90^\\circ$, $CD = AB = 8$, $AD = BC = 4\\sqrt{2}$,\\\\\n$\\therefore$ $\\angle DAF + \\angle ACD = 90^\\circ$,\\\\\n$\\because$ $DE \\perp AC$ at point F, $\\therefore$ $\\angle ADE + \\angle DAF = 90^\\circ$,\\\\\n$\\therefore$ $\\angle ACD = \\angle EDA$,\\\\\n$\\therefore$ $\\triangle ACD \\sim \\triangle EDA$,\\\\\n$\\therefore$ $\\frac{AD}{EA} = \\frac{CD}{AD}$,\\\\\n$\\therefore$ $EA = \\frac{{\\left(4\\sqrt{2}\\right)}^2}{8} = 4$,\\\\\n$\\therefore$ $DE = \\sqrt{AD^2 + EA^2} = \\sqrt{{\\left(4\\sqrt{2}\\right)}^2 + 4^2} = \\boxed{4\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52977745.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within circle $\\odot O$, point C is on the extension line of diameter AB, point D is a point on the circle above diameter AB, and CD is connected such that $\\angle A = \\angle BDC$. CD is a tangent of $\\odot O$. If CE bisects $\\angle ACD$ and intersects AD, BD at points E, F, respectively, when $DE = 2$, what is the length of $EF$?\n",
    "completion": "\\textbf{Solution:} Since $CE$ bisects $\\angle ACD$,\\\\\nit follows that $\\angle DCE = \\angle ACE$.\\\\\nAdditionally, since $\\angle A=\\angle BDC$,\\\\\nit implies that $\\angle A+\\angle ACE=\\angle BDC+\\angle DCE$,\\\\\nwhich means that $\\angle DEF=\\angle DFE$.\\\\\nGiven that $\\angle ADB=90^\\circ$ and $DE=2$,\\\\\nit follows that $DF=DE=2$,\\\\\nhence $EF=\\sqrt{DE^2+DF^2}=2\\sqrt{2}=\\boxed{2\\sqrt{2}}$",
    "image_path": "dataset/mm_math/images/MM_Math/52602184_151.png"
  },
  {
    "solution": "\\boxed{80^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $\\angle AOB=120^\\circ$, and OC is a ray inside $\\angle AOB$, and $\\angle AOC : \\angle BOC=1 : 2$. Find the measures of $\\angle AOC$ and $\\angle BOC$.\n",
    "completion": "\\textbf{Solution:} Since $\\angle AOC : \\angle BOC = 1 : 2$ and $\\angle AOB = 120^\\circ$,\n\nthus $\\angle AOC = \\frac{1}{3} \\angle AOB = \\frac{1}{3} \\times 120^\\circ = \\boxed{40^\\circ}$,\n\nand $\\angle BOC = \\frac{2}{3} \\angle AOB = \\frac{2}{3} \\times 120^\\circ = \\boxed{80^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51438377_168.png"
  },
  {
    "solution": "\\boxed{\\frac{7}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, there is a right-angled triangle paper piece, where $\\angle C=90^\\circ$, $BC=6$, and $AC=8$. Now, fold $\\triangle ABC$ as shown in the diagram, making points A and B coincide. The folding line is DE. What is the length of $CE$?\n",
    "completion": "\\textbf{Solution:} In the right-angled triangle $ACB$, we have $AC=8$ and $BC=6$,\n\\[\n\\therefore AB= \\sqrt{AC^{2}+BC^{2}}=\\sqrt{6^{2}+8^{2}}=10.\n\\]\nAccording to the principle of folding invariance, $\\triangle EDA\\cong \\triangle EDB$,\n\\[\n\\therefore EA=EB.\n\\]\nTherefore, in the right-angled triangle $BCE$, let $CE=x$, then $BE=AE=8-x$,\n\\[\n\\therefore BE^{2}=BC^{2}+CE^{2},\n\\]\n\\[\n\\therefore (8-x)^{2}=6^{2}+x^{2},\n\\]\nsolving this, we get $x=\\boxed{\\frac{7}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52991476.png"
  },
  {
    "solution": "\\boxed{6\\sqrt{11}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the circle $\\odot O$ with a radius of $6\\, \\text{cm}$, $C$ and $D$ are the trisection points of diameter $AB$, with points $E$ and $F$ located on the semicircles on either side of $AB$, respectively, where $\\angle BCE = \\angle BDF = 60^\\circ$. By connecting $AE$ and $BF$, what is the area of the two shaded regions in the figure in cm$^{2}$?\n",
    "completion": "\\textbf{Solution:} Construct the axial symmetric figure of $\\triangle DBF$ as $\\triangle CAG$, and draw $AM \\perp CG$, $ON \\perp CE$, \\\\\nSince $\\triangle DBF$ and its axial symmetric figure $\\triangle CAG$, \\\\\nBecause points C and D are trisection points of the diameter AB, then point H coincides with point C \\\\\n$\\therefore \\triangle ACG \\cong \\triangle BDF$, \\\\\n$\\therefore \\angle ACG = \\angle BDF = 60^\\circ$, \\\\\nSince $\\angle ECB = 60^\\circ$, \\\\\n$\\therefore$ Points G, C, and E are collinear, \\\\\nSince $AM \\perp CG$ and $ON \\perp CE$, \\\\\n$\\therefore AM \\parallel ON$, \\\\\n$\\therefore \\frac{AM}{ON} = \\frac{AC}{OC}$, \\\\\nIn $\\triangle ONC$, $\\angle OCN = 60^\\circ$, \\\\\n$\\therefore ON = \\sin \\angle OCN \\cdot OC = \\frac{\\sqrt{3}}{2} \\cdot OC$, \\\\\nSince $OC = OA - AC = 6 - 12 \\div 3 = 2$, \\\\\n$\\therefore ON = \\frac{\\sqrt{3}}{2} \\times 2 = \\sqrt{3}$, \\\\\nSimilarly: $AM = 2\\sqrt{3}$, \\\\\nSince $ON \\perp GE$, \\\\\n$\\therefore NE = GN = \\frac{1}{2}GE$, \\\\\nConnect OE, \\\\\nIn $\\triangle ONE$, $NE = \\sqrt{OE^2 - ON^2} = \\sqrt{6^2 - (\\sqrt{3})^2} = \\sqrt{33}$, \\\\\n$\\therefore GE = 2NE = 2\\sqrt{33}$, \\\\\n$\\therefore S_{\\triangle AGE} = \\frac{1}{2}GE \\cdot AM = \\frac{1}{2} \\times 2\\sqrt{3} \\times 2\\sqrt{33} = 6\\sqrt{11}$, \\\\\n$\\therefore$ The total area of the two shaded parts in the figure is $\\boxed{6\\sqrt{11}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52991371.png"
  },
  {
    "solution": "\\boxed{\\frac{25}{8}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In $\\triangle ABC$, $AC = BC = 10\\text{cm}$, $CD \\perp AB$, $CD = 8\\text{cm}$. Point $P$ moves from $B$ towards $A$ at a speed of $1\\text{cm/s}$, simultaneously, points $Q$ and $R$ move towards $B$ and $C$ from $C$ and $A$ respectively, at a speed of $2\\text{cm/s}$. The motion of all points stops when any one of them stops. The motion time is $t$ ($t$ is in seconds, $0 < t < 5$). At what value of $t$ will point $P$ be on the perpendicular bisector of $BQ$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, point $P$ is on the perpendicular bisector of $BQ$, \\\\\n$\\therefore PQ=PB$, $PE\\perp BQ$, $BE=QE=\\frac{1}{2}BQ$. \\\\\n$\\because CQ=2t\\,\\text{cm}$, \\\\\n$\\therefore BQ=BC-CQ=(10-2t)\\,\\text{cm}$, \\\\\n$\\therefore BE=QE=(5-t)\\,\\text{cm}$. \\\\\n$\\because \\angle PBE=\\angle CBD$, $\\angle PEB=\\angle CDB=90^\\circ$, \\\\\n$\\therefore \\triangle PBE\\sim \\triangle CBD$, \\\\\n$\\therefore \\frac{BE}{BD}=\\frac{PB}{BC}$, namely $\\frac{5-t}{6}=\\frac{t}{10}$, \\\\\nSolving, we get: $t=\\boxed{\\frac{25}{8}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53043392_176.png"
  },
  {
    "solution": "\\boxed{\\frac{75}{16}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the function $y=-2x+3$ intersects with the graph of the function $y=-\\frac{1}{2}x+m$ at point $P(n, -2)$. The line $y=-2x+3$ intersects the y-axis at point A, and the line $y=-\\frac{1}{2}x+m$ intersects the y-axis at point B. Calculate the area of $\\triangle ABP$.\n",
    "completion": "\\textbf{Solution:} Given that in the equation $y=-2x+3$, when $x=0$, $y=3$, thus point $A(0, 3)$ is found. Also, in the equation $y=-\\frac{1}{2}x-\\frac{3}{4}$, when $x=0$, $y=-\\frac{3}{4}$, hence point $B\\left(0, -\\frac{3}{4}\\right)$ is determined. Consequently, the length of $AB=\\frac{15}{4}$. Therefore, the area of the triangle $\\triangle ABP$ is calculated as $\\frac{1}{2}\\times \\frac{15}{4}\\times \\frac{5}{2}=\\boxed{\\frac{75}{16}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52355125_192.png"
  },
  {
    "solution": "\\boxed{8\\,\\text{cm}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the segment \\(AC=6\\,cm\\), the segment \\(BC=15\\,cm\\), point M is the midpoint of \\(AC\\), and a point N is taken on \\(CB\\) such that \\(CN=\\frac{1}{2}NB\\). What is the length of \\(MN\\)?\n",
    "completion": "\\textbf{Solution:} Given that $M$ is the midpoint of $AC$, and $AC=6\\,\\text{cm}$,\\\\\nit follows that $MC=\\frac{1}{2}AC=\\frac{1}{2}\\times 6=3\\,\\text{cm}$,\\\\\nsince $CN=\\frac{1}{2}NB$ and $BC=15\\,\\text{cm}$,\\\\\nthen $CN=\\frac{1}{3}BC=\\frac{1}{3}\\times 15=5\\,\\text{cm}$,\\\\\ntherefore, $MN=MC+NC=3\\,\\text{cm}+5\\,\\text{cm}=\\boxed{8\\,\\text{cm}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53042311.png"
  },
  {
    "solution": "\\boxed{C\\left(2, 2\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=kx+b$ ($k\\neq 0$) intersects the x-axis at point $A(6,0)$ and the y-axis at point $B(0,3)$, and intersects the line $y=x$ at point $C$. Find the values of $k$ and $b$, and the coordinates of point $C$?\n",
    "completion": "\\textbf{Solution:} Given the line $y=kx+b$ ($k\\neq 0$) intersects the x-axis at point A(6, 0) and the y-axis at point B(0, 3),\\\\\n$\\therefore \\left\\{\\begin{array}{l}\n6k+b=0\\\\ \nb=3\n\\end{array}\\right.$\\\\\nSolving this system, we get: $\\left\\{\\begin{array}{l}\nk=-\\frac{1}{2}\\\\ \nb=3\n\\end{array}\\right.$,\\\\\nHence, the equation of line $AB$ is: $y=-\\frac{1}{2}x+3$,\\\\\nBy solving the system: $\\left\\{\\begin{array}{l}\ny=-\\frac{1}{2}x+3\\\\ \ny=x\n\\end{array}\\right.$,\\\\\nWe find: $\\left\\{\\begin{array}{l}\nx=2\\\\ \ny=2\n\\end{array}\\right.$,\\\\\nTherefore, $C(2, 2).$\\\\\n$\\therefore k=\\boxed{-\\frac{1}{2}}, b=\\boxed{3}, \\boxed{C\\left(2, 2\\right)}.$",
    "image_path": "dataset/mm_math/images/MM_Math/50677506_132.png"
  },
  {
    "solution": "\\boxed{\\frac{4}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $E$ and $F$ are the midpoints of sides $AB$ and $AD$, respectively, and $BF$ intersects $EC$ and $ED$ at points $M$ and $N$, respectively. Given that $AB=4$ and $BC=6$, what is the length of $MN$?\n",
    "completion": "\\textbf{Solution:} As shown in Figure 1, extend CE and DA to intersect at point Q, \\\\\n$\\because$ Quadrilateral ABCD is a rectangle, and BC=6, \\\\\n$\\therefore$ $\\angle$BAD=$90^\\circ$, AD=BC=6, and AD$\\parallel$ BC, \\\\\n$\\because$ F is the midpoint of AD, \\\\\n$\\therefore$ AF=DF=3, \\\\\nIn $\\triangle$BAF, by the Pythagorean theorem, we have: $BF=\\sqrt{AB^{2}+AF^{2}}=\\sqrt{4^{2}+3^{2}}=5$, \\\\\n$\\because$ AD $\\parallel$ BC, \\\\\n$\\therefore$ $\\angle$Q=$\\angle$ECB, \\\\\n$\\because$ E is the midpoint of AB, and AB=4, \\\\\n$\\therefore$ AE=BE=2, \\\\\nIn $\\triangle$QAE and $\\triangle$CBE, \\\\\n$\\begin{array}{c}\n\\angle QEA=\\angle BEC\\\\\n\\angle Q=\\angle ECB\\\\\nAE=BE\n\\end{array}$\\\\\n$\\therefore$ $\\triangle$QAE$\\cong$$\\triangle$CBE (AAS), \\\\\n$\\therefore$ AQ=BC=6, thus QF=6+3=9, \\\\\n$\\because$ AD $\\parallel$ BC, \\\\\n$\\therefore$ $\\triangle$QMF$\\sim$$\\triangle$CMB, \\\\\n$\\therefore$ $\\frac{FM}{BM}=\\frac{QF}{BC}=\\frac{9}{6}$, \\\\\n$\\because$ BF=5, \\\\\n$\\therefore$ BM=2, FM=3, \\\\\nAs shown in Figure 2, extend BF and CD to intersect at W, \\\\\nSimilarly, AB=DW=4, CW=8, and BF=FW=5, \\\\\n$\\because$ AB $\\parallel$ CD, \\\\\n$\\therefore$ $\\triangle$BNE$\\sim$$\\triangle$WND, \\\\\n$\\therefore$ $\\frac{BN}{NW}=\\frac{BE}{DW}$, \\\\\n$\\therefore$ $\\frac{BN}{5-BN+5}=\\frac{2}{4}$, \\\\\nSolving this, we get: BN= $\\frac{10}{3}$, \\\\\n$\\therefore$ MN=BN−BM= $\\frac{10}{3}-2= \\boxed{\\frac{4}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51433235.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: The photo wall made up of four squares is shown in Figure 1, and its planar geometry is shown in Figure 2. In it, the areas of the squares $ABCD$, $DEFG$, and $BIJK$ are $4\\,\\text{dm}^2$, $4\\,\\text{dm}^2$, and $16\\,\\text{dm}^2$ respectively. What is the area of square $AGHI$?\n",
    "completion": "\\textbf{Solution:} Draw $DM \\perp AG$ at point $M$, and draw $IN \\perp BA$ intersecting the extension of $BA$ at point $N$,\\\\\nsince the areas of squares $ABCD$, $DEFG$, and $BIJK$ are $4 \\, \\text{square meters}$, $4 \\, \\text{square meters}$, and $16 \\, \\text{square meters}$ respectively,\\\\\ntherefore, $AD = 2$, $DG = 2$, $BI = 4$, $\\angle IAG = \\angle BAD = 90^\\circ$,\\\\\ntherefore, $\\angle IAB + \\angle MAD = 180^\\circ$,\\\\\nalso, since $\\angle IAB + \\angle IAN = 180^\\circ$,\\\\\ntherefore, $\\angle IAN = \\angle MAD$,\\\\\nlet $AG = 2x$,\\\\\nsince $DA = DG = 2$, $DM \\perp AG$,\\\\\ntherefore, $\\cos \\angle MAD = \\frac{AM}{AD} = \\frac{x}{2}$,\\\\\nsince $\\cos \\angle IAN = \\frac{AN}{AI} = \\frac{AN}{2x}$,\\\\\ntherefore, $\\frac{x}{2} = \\frac{AN}{2x}$,\\\\\ntherefore, $AN = x^2$,\\\\\ntherefore, $NI^2 = AI^2 - AN^2 = (2x)^2 - (x^2)^2 = 4x^2 - x^4$,\\\\\nsince $BI = 4$, $BN = BA + AN = 2 + x^2$, $\\angle BNI = 90^\\circ$,\\\\\ntherefore, $4^2 = (2+x^2)^2 + 4x^2 - x^4$,\\\\\nsolving this, we get $x^2 = \\frac{3}{2}$,\\\\\ntherefore, the area of square $AGHI$ is: $(2x)^2 = 4x^2 = 4 \\times \\frac{3}{2} = \\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52514048.png"
  },
  {
    "solution": "\\boxed{11}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AC = BC = 13$. Place $\\triangle ABC$ in the Cartesian coordinate system, with the coordinates of points A and B being $(2, 0)$ and $(12, 0)$, respectively. Translate $\\triangle ABC$ along the x-axis to the left such that point C lies on the line $y = -x + 8$. What is the distance of this translation of $\\triangle ABC$?\n",
    "completion": "\\textbf{Solution:} Construct CD $\\perp$ x-axis at point D, as shown in the figure.\\\\\nSince the coordinates of points A and B are respectively (2, 0) and (12, 0), and AC=BC=13,\\\\\nit follows that AD=BD=$\\frac{1}{2}$AB=5,\\\\\nhence CD=$\\sqrt{AC^{2}-AD^{2}}=12$.\\\\\nTherefore, the coordinates of point C are (7, 12).\\\\\nWhen y=12, we have 12=$-x+8$,\\\\\nsolving this gives: x=$-4$,\\\\\nthus, the coordinates of point C after translation are ($-4$, 12).\\\\\nTherefore, $\\triangle ABC$ is translated 7-(-4)=\\boxed{11} units length to the left along the x-axis.",
    "image_path": "dataset/mm_math/images/MM_Math/52381306.png"
  },
  {
    "solution": "\\boxed{3\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the sector $ABC$ with a radius of 6, $\\angle AOB=120^\\circ$, C is a moving point on the arc $\\overset{\\frown}{AB}$ (not coinciding with A or B), $OD\\perp AC$, $OE\\perp BC$, with the feet of the perpendiculars being points D and E, respectively. What is the length of $DE$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, connect $AB$, and draw $OF\\perp AB$ at $F$,\\\\\n$\\because \\angle AOB=120^\\circ, OA=OB$,\\\\\n$\\therefore \\angle OAB=\\angle OBA=\\frac{180^\\circ-\\angle AOB}{2}=30^\\circ$,\\\\\n$\\therefore OF=\\frac{1}{2}OA=3$,\\\\\nIn $\\triangle AOF$, by the Pythagorean theorem, $AF=\\sqrt{OA^{2}-OF^{2}}=3\\sqrt{3}$,\\\\\n$\\therefore$, by the perpendicular diameter theorem, $AB=2AF=6\\sqrt{3}$,\\\\\n$\\because OD\\perp AC, OE\\perp BC$,\\\\\n$\\therefore$ D and E are respectively the midpoints of $AC$ and $BC$,\\\\\n$\\therefore DE$ is the midline of $\\triangle ABC$,\\\\\n$\\therefore DE=\\frac{1}{2}AB=3\\sqrt{3}$;\\\\\nThus, the length of $DE$ is $\\boxed{3\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52871720_104.png"
  },
  {
    "solution": "\\boxed{15}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCO$, point $C$ is on the x-axis, point $A$ is on the y-axis, and the coordinates of point $B$ are $(-9, 12)$. Rectangle $ABCO$ is folded along line $BD$ so that point $A$ falls at point $E$ on the diagonal $OB$, and line $BD$ intersects $OA$ and the x-axis at points $D$ and $F$, respectively. What is the length of segment $BO$?\n",
    "completion": "\\textbf{Solution:} As shown in Figure 3, since quadrilateral $ABCO$ is a rectangle, \\\\\nthus $\\angle BCO=90^\\circ$. \\\\\nIn $\\triangle BCO$ right-angled at $C$, \\\\\n$BO^{2}=BC^{2}+OC^{2}$, $BO^{2}=12^{2}+9^{2}$, \\\\\nhence, $BO=\\boxed{15}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52056028_207.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, \\(C\\) is the midpoint of segment \\(AB\\), \\(D\\) is a point on \\(CB\\), \\(E\\) is the midpoint of segment \\(DB\\), \\(EB = 1\\,cm\\), and \\(EB = \\frac{1}{5}AB\\), then what is the length of segment \\(CE\\) in centimeters?\n",
    "completion": "Solution: Since $EB=1\\text{cm}$, and $EB=\\frac{1}{5}AB$,\\\\\nit follows that $AB=5EB=5\\text{cm}$, \\\\\nSince C is the midpoint of the segment $AB$, \\\\\nwe have $CB=\\frac{1}{2}AB=\\frac{5}{2}\\text{cm}$, \\\\\nthus $CE=CB-EB=\\frac{5}{2}-1=\\boxed{\\frac{3}{2}}\\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53027060.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, point $E$ is the midpoint of side $AB$. Lines $AC$ and $DE$ intersect at point $F$. If the area of $\\triangle CDF$ is 4, what is the area of $\\triangle AED$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral ABCD is a rectangle,\\\\\nit follows that AB $\\parallel$ CD and AB = CD,\\\\\nwhich means that $\\triangle AEF \\sim \\triangle CDF$,\\\\\ngiven that point E is the midpoint of AB,\\\\\nhence CD = AB = 2AE,\\\\\nthus $\\frac{EF}{DF} = \\frac{AE}{CD} = \\frac{1}{2}$,\\\\\ngiven that the area of $\\triangle CDF$ is 4,\\\\\nit follows that the area of $\\triangle AEF$ is 1,\\\\\nhence the area of $\\triangle ADF$ is 2,\\\\\ntherefore, the area of $\\triangle AED$ is $\\boxed{3}$,",
    "image_path": "dataset/mm_math/images/MM_Math/52623482.png"
  },
  {
    "solution": "\\boxed{26}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points $A$ and $B$ are located at the ends of a pond. Xiaocong wants to measure the distance between $A$ and $B$ with a rope, but the rope is not long enough. A classmate helped him with an idea: firstly, select a point $C$ on the ground that can be directly reached from both $A$ and $B$. Find the midpoints $D$ and $E$ of $AC$ and $BC$, respectively, and measure the length of $DE$ to be $13\\,m$. Then, what is the distance between $A$ and $B$ in meters?\n",
    "completion": "\\textbf{Solution:} Since $D$ and $E$ are the midpoints of $AC$ and $BC$, respectively, \\\\\ntherefore, $AB=2DE=\\boxed{26}m$.",
    "image_path": "dataset/mm_math/images/MM_Math/52303724.png"
  },
  {
    "solution": "\\boxed{4.5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the length of line segment $AB=20$, $M$ is the midpoint of line segment $AB$, $P$ is any point on line segment $AB$, and $N$ is the midpoint of line segment $PB$. When the length of line segment $MP=1$, what is the length of line segment $NB$?\n",
    "completion": "\\textbf{Solution:} Since $MP=1$,\\\\\nTherefore, when P is on the left side of M, $BP=MB+MP=11$,\\\\\nSince N is the midpoint of the line segment $PB$,\\\\\nTherefore, $NB=\\frac{1}{2}PB=5.5$,\\\\\nWhen P is on the right side of M, $BP=MB-MP=9$,\\\\\nSince N is the midpoint of the line segment $PB$,\\\\\nTherefore, $NB=\\frac{1}{2}PB=\\boxed{4.5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50056753_142.png"
  },
  {
    "solution": "\\boxed{y=-x+8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system, the vertices of the rectangle $OABC$ are such that vertex $O$ coincides with the origin, and vertices $A$ and $C$ are respectively on the coordinate axes, with the coordinates of vertex $B$ being $\\left(6, 4\\right)$. $E$ is the midpoint of $AB$, and the line passing through points $D\\left(8, 0\\right)$ and $E$ intersects $BC$ and the y-axis at points $F$ and $G$, respectively. What is the equation of the line $DE$?\n",
    "completion": "\\textbf{Solution:} Let the equation of $DE$ be: $y=kx+b\\ (k\\ne 0)$,\\\\\nsince $B(6, 4)$ and $E$ is the midpoint of $AB$,\\\\\nthus $E(6, 2)$,\\\\\nsubstituting $E(6, 2)$ and $D(8, 0)$ into the equation gives $\\left\\{\\begin{array}{l}6k+b=2 \\\\ 8k+b=0 \\end{array}\\right.$,\\\\\nsolving this yields $\\left\\{\\begin{array}{l}k=-1 \\\\ b=8 \\end{array}\\right.$,\\\\\ntherefore $\\boxed{y=-x+8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51663234_192.png"
  },
  {
    "solution": "\\boxed{\\frac{7}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the diagonals AC and BD of the rhombus ABCD intersect at point O. Point E is a point on side AD, and $OD = OE$. If $AC = 8$ and $BD = 6$, then what is the length of AE?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw ON $\\perp$ AD at point N,\\\\\n$\\because$ Quadrilateral ABCD is a rhombus,\\\\\n$\\therefore$ OD=$\\frac{1}{2}$BD=3, OA=$\\frac{1}{2}$AC=4, AC$\\perp$BD,\\\\\n$\\therefore$ AD=$\\sqrt{OA^{2}+OD^{2}}$=5,\\\\\n$\\because$ OD=OE, ON$\\perp$AD at point N,\\\\\n$\\therefore$ $\\angle$OND=90$^\\circ$, DN=EN,\\\\\n$\\therefore$ S$_{\\triangle AOD}$=$\\frac{1}{2}$AD·ON=$\\frac{1}{2}$OD·OA,\\\\\n$\\therefore$ 5ON=12,\\\\\n$\\therefore$ ON=$\\frac{12}{5}$,\\\\\n$\\therefore$ DN=$\\sqrt{OD^{2}-ON^{2}}=\\sqrt{3^{2}-\\left(\\frac{12}{5}\\right)^{2}}=\\frac{9}{5}$,\\\\\n$\\therefore$ DE=2DN=$\\frac{18}{5}$,\\\\\n$\\therefore$ AE=AD-DE=5-$\\frac{18}{5}=\\boxed{\\frac{7}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52205277.png"
  },
  {
    "solution": "\\boxed{3\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the side length of rhombus ABCD is 6, and $\\angle D = 120^\\circ$. Point E is the midpoint of AB, and point P is a moving point on diagonal AC. What is the minimum value of $BP+EP$?\n",
    "completion": "\\textbf{Solution:} Connect $DP$, $BD$, \\\\\nsince quadrilateral $ABCD$ is a rhombus, \\\\\nthus point $B$ and point $D$ are symmetric with respect to $AC$, \\\\\nthus $DP=PB$, \\\\\nthus $PB+PE=DP+PE \\geq DE$, \\\\\nwhen $DE \\perp AB$, the value of $PB+PE$ is minimal, \\\\\nsince $\\angle D=120^\\circ$, \\\\\nthus $\\angle DAB=60^\\circ$, \\\\\nsince $AD=AB$, \\\\\nthus $\\triangle ABD$ is an equilateral triangle, \\\\\nsince $E$ is the midpoint of $AB$, \\\\\nthus $DE \\perp AB$, \\\\\nsince $AB=6$, \\\\\nthus $DE=3\\sqrt{3}$, \\\\\nthus the minimum value of $PB+PE$ is $\\boxed{3\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52387091.png"
  },
  {
    "solution": "\\boxed{80^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the figure, in $\\triangle ABC$, $AB=AC$, $AD\\perp BC$ at point D, and point E is on side $AB$, with $EF\\parallel AC$ intersecting the extension of $AD$ at point F. If $\\angle C=40^\\circ$, what is the degree measure of $\\angle AEF$?\n",
    "completion": "\\textbf{Solution:} Given the problem,\\\\\n$\\because AB=AC$,\\\\\n$\\angle C=40^\\circ$,\\\\\n$\\therefore \\angle B=\\angle C=40^\\circ$,\\\\\n$\\therefore \\angle BAC=180^\\circ-\\angle B-\\angle C=100^\\circ$,\\\\\n$\\because EF\\parallel AC$,\\\\\n$\\therefore \\angle BAC+\\angle AEF=180^\\circ$,\\\\\n$\\therefore \\angle AEF=\\boxed{80^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51369004_101.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the line segment $AB=3\\text{cm}$. Extend $AB$ to point $C$ such that $BC=2AB$, then extend $BA$ to point $D$ such that $AD=AC$. Find the length of the line segment $BD$.\n",
    "completion": "\\textbf{Solution}: Since $AB = 3\\,cm$, and $BC = 2AB$, \\\\\ntherefore $BC = 2 \\times 3 = 6\\,cm$, and $AC = AB + BC = 3 + 6 = 9\\,cm$, \\\\\nsince $AD = AC = 9\\,cm$, \\\\\ntherefore $BD = AD + AB = 9 + 3 = \\boxed{12}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53010651.png"
  },
  {
    "solution": "\\boxed{(1,2)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y_{1}=x+1$ intersects the x-axis and y-axis at points A and B, respectively, and the line $y_{2}=-2x+4$ intersects the x-axis and y-axis at points C and D, respectively. The two lines intersect at point E. What are the coordinates of point E?\n",
    "completion": "\\textbf{Solution:} Since $\\begin{cases}y_1=x+1\\\\ y_2=-2x+4\\end{cases}$,\\\\\nwe have $\\begin{cases}x=1\\\\ y=2\\end{cases}$,\\\\\nthus, the coordinates of point E are $\\boxed{(1,2)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52380902_163.png"
  },
  {
    "solution": "\\boxed{-2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated, the lines $l_{1}$ and $l_{2}$ are the graphs of the functions $y=k_{1}x+b_{1}$ and $y=k_{2}x+b_{2}$, respectively. Based on the graph, what is the solution set of the inequality $k_{1}x+b_{1}>k_{2}x+b_{2}$ with respect to $x$?\n",
    "completion": "\\textbf{Solution:} When $x<-2$, $k_{1}x+b_{1}>k_{2}x+b_{2}$,\\\\\nthus the solution set of the inequality $k_{1}x+b_{1}>k_{2}x+b_{2}$ is $x<-2$.\\\\\nTherefore, the answer is: $x=\\boxed{-2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52055750.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In a Cartesian coordinate system, suppose the inverse proportional function $y_{1}=\\frac{k_{1}}{x}$ ($k_{1}\\ne 0$) and the linear function $y_{2}=k_{2}x+b$ ($k_{2}\\neq 0$) both pass through points $A$ and $B$, with coordinates of point $A$ being $(1, m)$ and coordinates of point $B$ being $(-2, -2)$. If the graph of the function $y_{2}$ is translated downwards by $n$ ($n>0$) units, and intersects with the graph of the function $y_{1}$ at points $(p_{1}, q_{1})$ and $(p_{2}, q_{2})$, given that $p_{1}=-1$, find the values of $n$ and $p_{2}\\times q_{2}$ at this time.\n",
    "completion": "\\textbf{Solution:} The equation of the function after translating the graph of $y_{2}$ downwards by $n$ ($n>0$) units is $y=2x+2-n$; When $p_{1}=-1$ (at point $(-1, q_{1})$), since translating the graph of $y_{2}$ downwards by $n$ ($n>0$) units intersects with the graph of $y_{1}$ at points $(-1, q_{1})$ and $(p_{2}, q_{2})$, hence $-1q_{1}=4$, $p_{2}\\times q_{2}=4$, solving these we find $q_{1}=-4$, thus point $(-1, -4)$, hence $-2+2-n=-4$; solving for this gives $n=\\boxed{4}$; therefore, at this instance $n=4$, $p_{2}\\times q_{2}=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52457688_183.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, points D and E are the midpoints of AB and BC, respectively. If AC=6, what is the length of DE?\n",
    "completion": "\\textbf{Solution:} Since $D$ and $E$ are the midpoints of $AB$ and $BC$, respectively,\\\\\ntherefore $DE$ is the midline of $\\triangle ABC$,\\\\\nand since $AC=6$,\\\\\ntherefore $DE=\\frac{1}{2}BC=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52690697.png"
  },
  {
    "solution": "\\boxed{80^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the figure, in $\\triangle ABC$, $\\angle B=40^\\circ$. If $\\triangle ABC$ is rotated counterclockwise around point A to $\\triangle ADE$, such that point B falls on point D, which is on the extension line of BC, then what is the measure of $\\angle BDE$?\n",
    "completion": "\\textbf{Solution:} Since it is known from the property of rotation that $AB=AD$ and $\\angle ADE=\\angle B=40^\\circ$, in $\\triangle ABD$, as $AB=AD$, according to the property of isosceles triangles, we have $\\angle ADB=\\angle B=40^\\circ$. Therefore, $\\angle BDE=\\angle ADE+\\angle ADB=80^\\circ$.\\\\\nHence, the answer is: $\\boxed{80^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52344564.png"
  },
  {
    "solution": "\\boxed{9.6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ABC = \\angle ACB$, D is the midpoint of BC, AD is connected, E is a point on AB, P is a point on AD, EP and BP are connected, $AC=10$, $BC=12$, what is the minimum value of $EP+BP$?\n",
    "completion": "\\textbf{Solution:} Connect PC,\n\n$\\because$ $\\angle ABC = \\angle ACB$,\n\n$\\therefore$ AB = AC,\n\n$\\because$ D is the midpoint of BC,\n\n$\\therefore$ AD is the perpendicular bisector of BC, BD = $\\frac{1}{2}$BC = 6\n\n$\\therefore$ BP = CP,\n\n$AD = \\sqrt{AB^{2} - BD^{2}} = \\sqrt{10^{2} - 6^{2}} = 8$\n\n$\\therefore$ EP + BP = EP + CP\n\nTo minimize the value of EP + BP, utilizing the fact that the shortest distance between two points is a straight line and the shortest segment is the perpendicular, it is known that when points E, P, C are collinear and CE$\\perp$AB, the value of EP + BP is minimized, with the minimum value being the length of EC;\n\n$\\because$ $S_{\\triangle ABC} = \\frac{1}{2}AB \\cdot CE = \\frac{1}{2}CB \\cdot AD$,\n\n$\\therefore$ 10CE = 12 $\\times$ 8\n\nSolving for it: CE = \\boxed{9.6}.",
    "image_path": "dataset/mm_math/images/MM_Math/51419899.png"
  },
  {
    "solution": "\\boxed{3cm}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $AB \\parallel CF$, and E is the midpoint of $DF$. If $AB = 8\\,cm$ and $CF = 5\\,cm$, then how many centimeters is $BD$?\n",
    "completion": "\\textbf{Solution:} Given that $AB\\parallel CF$,\\\\\nthus $\\angle A=\\angle ACF$, $\\angle F=\\angle ADE$,\\\\\nsince E is the midpoint of $DF$,\\\\\nthus $DE=EF$,\\\\\nin $\\triangle ADE$ and $\\triangle CFE$, $\\left\\{\\begin{array}{l}\n\\angle A=\\angle ECF\\\\\n\\angle ADE=\\angle F\\\\\nDE=EF\n\\end{array}\\right.$,\\\\\ntherefore $\\triangle ADE\\cong \\triangle CFE$ (by AAS),\\\\\nthus $CF=AD=5cm$,\\\\\nthus $BD=AB-AD=8-5=\\boxed{3cm}$,",
    "image_path": "dataset/mm_math/images/MM_Math/52991481.png"
  },
  {
    "solution": "\\boxed{26}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in parallelogram $ABCD$, $E$ and $F$ are points on sides $AD$ and $BC$ respectively, and $\\angle ABE = \\angle CDF$. Quadrilateral $BEDF$ is a parallelogram; connect $CE$. If $CE$ bisects $\\angle DCB$, $CF = 3$, and $DE = 5$, what is the perimeter of parallelogram $ABCD$?\n",
    "completion": "\\textbf{Solution:} Given that CE bisects $\\angle BCD$, it follows that $\\angle 1 = \\angle 2$. Furthermore, since $AD \\parallel BC$, it implies $\\angle 1 = \\angle 3$, which leads to $\\angle 2 = \\angle 3$, and consequently, $ED = DC = 5$. In quadrilateral $EBFD$, it can be deduced that $BF = ED = 5$. Hence, the perimeter of the quadrilateral $ABCD$ equals $2 \\times (5 + 3) + 2 \\times 5 = \\boxed{26}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52387533_138.png"
  },
  {
    "solution": "\\boxed{-9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the side length of the square $ABCD$ is $3$, with the side $AD$ on the negative x-axis. The graph of the inverse proportional function $y = \\frac{k}{x}$ (where $x < 0$) passes through point $B$ and the midpoint $E$ of side $CD$. What is the value of $k$?\n",
    "completion": "Let $B(m, 3)$, then $C(m-3, 3)$, \\\\\n$\\because$ Point E is the midpoint of CD, \\\\\n$\\therefore$ The coordinates of E are $\\left(m-3, \\frac{3}{2}\\right)$. \\\\\n$\\because$ B and E both lie on the graph of $y=\\frac{k}{x}$, \\\\\n$\\therefore$ $(m-3)\\cdot\\frac{3}{2}=m\\cdot3$, \\\\\nSolving for m gives $m=\\boxed{-3}$, \\\\ \n$\\therefore$ $B(-3, 3)$, \\\\\n$\\therefore$ $k=-3\\times3=\\boxed{-9}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52623347.png"
  },
  {
    "solution": "\\boxed{\\frac{52}{9}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given the graph of the quadratic function $y=ax^2+bx+c$ with its vertex at $M(2, 1)$, and it passes through point $N(3, 2)$. If the graph of the linear function $y=-\\frac{4}{3}x-4$ intersects the x-axis at point $A$ and the y-axis at point $B$. Let $P$ be a moving point on the parabola. Draw line $PQ\\parallel y$ axis intersecting line $AB$ at point $Q$. A circle with diameter $PQ$ intersects line $AB$ at point $D$. Let the x-coordinate of point $P$ be $n$. What is the value of $n$ when the length of segment $DQ$ is minimized, and what is the minimum value?\n",
    "completion": "\\textbf{Solution:} Given $P\\left(n, n^{2}-4n+5\\right)$, $Q\\left(n, -\\frac{4}{3}n-4\\right)$. \\\\\nTherefore, $PQ=n^{2}-4n+5-\\left(-\\frac{4}{3}n-4\\right)=n^{2}-\\frac{8}{3}n+9=\\left(n-\\frac{4}{3}\\right)^{2}+\\frac{65}{9}$. \\\\\nHence, when $n=\\frac{4}{3}$, $PQ$ reaches its minimum value of $\\frac{65}{9}$. \\\\\nIt is easy to prove that $\\triangle DPQ \\sim \\triangle OAB$, \\\\\ntherefore $\\frac{DQ}{PQ}=\\frac{OB}{AB}$, \\\\\nresulting in $DQ=\\frac{4}{5}PQ$. \\\\\nThus, when $n=\\frac{4}{3}$, $DQ$ reaches its minimum value, which is $\\boxed{\\frac{52}{9}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52457074_127.png"
  },
  {
    "solution": "\\boxed{y=-x-1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within the square $OABC$, point $O$ is the origin, point $C(-3, 0)$, point $A$ is on the positive y-axis, points $E$ and $F$ are on $BC$ and $CO$ respectively, with $CE=CF=2$. The graph of the linear function $y=kx+b(k\\ne 0)$ passes through points $E$ and $F$, intersecting the y-axis at point $G$. The graph of the reciprocal function $y=\\frac{m}{x}(m\\ne 0)$ passing through point $E$ intersects $AB$ at point $D$. What are the equations of the reciprocal function and the linear function?\n",
    "completion": "\\textbf{Solution:} Given the problem statement, point $C(-3, 0)$, hence the side length of the square $OABC$ is 3, implying $OA=AB=BC=CO=3$. Since $CE=CF=2$, it follows that $OF=1$. Thus, $E(-3, 2)$ and $F(-1, 0)$. Substituting $E(-3, 2)$ and $F(-1, 0)$ into $y=kx+b$ yields $\\begin{cases}-3k+b=2\\\\ -k+b=0\\end{cases}$. Solving this system of equations, we get $\\begin{cases}k=-1\\\\ b=-1\\end{cases}$. Therefore, the linear function is $y=-x-1$. Substituting $E(-3, 2)$ into $y=\\frac{m}{x}$ gives $2=\\frac{m}{-3}$. Solving for $m$ yields $m=-6$. Therefore, the inverse proportionality function is $y=-\\frac{6}{x}$. Answer: The inverse proportionality function is $\\boxed{y=-\\frac{6}{x}}$, and the linear function is $\\boxed{y=-x-1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52949741_252.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $BD$ is the median. If $AB=2\\sqrt{10}$, $BC=10$, and $BD=3\\sqrt{5}$, what is the value of $\\tan\\angle ABC$?\n",
    "completion": "\\textbf{Solution:} Construct a line from point A such that $AE \\perp BC$ at E, and $DF \\perp BC$ at F,\\\\\ntherefore $AE \\parallel DF$, $\\angle AEB = \\angle DFB = 90^\\circ$\\\\\ntherefore $\\triangle AEC \\sim \\triangle DFC$,\\\\\n$AB^2 - BE^2 = AE^2$,\\\\\n$BD^2 - BF^2 = DF^2$,\\\\\ntherefore $\\frac{DF}{AE} = \\frac{CF}{CE} = \\frac{CD}{AC}$,\\\\\nsince BD is the median,\\\\\ntherefore $CD = AD = \\frac{1}{2}AC$,\\\\\ntherefore $\\frac{DF}{AE} = \\frac{CF}{CE} = \\frac{CD}{AC} = \\frac{1}{2}$,\\\\\ntherefore $AE = 2DF$, $CE = 2CF$,\\\\\ntherefore $EF = CF$, let $EF = CF = x$, then $BE = BC - CE = 10 - 2x$, $BF = BC - CF = 10 - x$,\\\\\nsince $AB^2 - BE^2 = AE^2$, $BD^2 - BF^2 = DF^2$,\\\\\ntherefore $AE^2 = (2\\sqrt{10})^2 - (10 - 2x)^2 = 40x - 60 - 4x^2$,\\\\\n$DF^2 = (3\\sqrt{5})^2 - (10 - x)^2 = 20x - 55 - x^2$,\\\\\nsince $AE = 2DF$,\\\\\ntherefore $AE^2 = 4DF^2$,\\\\\ntherefore $40x - 60 - 4x^2 = 80x - 220 - 4x^2$, solving for $x = 4$,\\\\\ntherefore $BE = 10 - 2x = 2$, $AE = \\sqrt{AB^2 - BE^2} = 6$,\\\\\ntherefore $\\tan \\angle ABC = \\frac{AE}{BE} = \\boxed{3}$",
    "image_path": "dataset/mm_math/images/MM_Math/50933575.png"
  },
  {
    "solution": "\\boxed{65^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the square $ABCD$, point $F$ is on the side $CD$, and $BF$ intersects $AC$ at point $E$. $\\triangle ABE \\cong \\triangle ADE$, If $\\angle CBF = 20^\\circ$, then what is the measure of $\\angle AED$?\n",
    "completion": "By the given information, we have $\\angle AED = \\boxed{65^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53313466_136.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the equilateral triangle $\\triangle ABC$, the point $E$ is the midpoint of side $AC$, and point $P$ is a moving point on the median $AD$ of $\\triangle ABC$. If $AD = 6$, what is the minimum value of $EP+CP$?\n",
    "completion": "\\textbf{Solution:} Construct point F as the symmetric point of E with respect to AD, and connect CF, which intersects AD at point P,\\\\\nsince $\\triangle ABC$ is an equilateral triangle, and AD is the median to side BC,\\\\\ntherefore $AD\\perp BC$,\\\\\nthus AD is the perpendicular bisector of BC,\\\\\nsince the corresponding point of E with respect to AD is point F,\\\\\ntherefore CF is the minimum value of EP+CP.\\\\\nSince $\\triangle ABC$ is an equilateral triangle, and E is the midpoint of side AC,\\\\\nthus F is the midpoint of AB,\\\\\ntherefore CF is the median of $\\triangle ABC$,\\\\\nthus $CF=AD=\\boxed{6}$,\\\\\nwhich means the minimum value of EP+CP is 6.",
    "image_path": "dataset/mm_math/images/MM_Math/51367153.png"
  },
  {
    "solution": "\\boxed{x > -1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y_1=x+b$ intersects with $y_2=kx-1$ at point P. What is the solution set of the inequality $x+b>kx-1$ with respect to $x$?\n",
    "completion": "\\textbf{Solution:} From the graph, we know that the coordinates of the intersection point of the two lines are $\\left(-1, \\frac{1}{2}\\right)$. Additionally, when $x > -1$, the graph of the function $y = x + b$ is above the graph of the function $y = kx - 1$,\\\\\n$\\therefore$ the solution set for the inequality in terms of $x$, $x + b > kx - 1$, is $x > -1$.\\\\\nTherefore, the answer is: $\\boxed{x > -1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51699660.png"
  },
  {
    "solution": "\\boxed{9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, segment \\(AB=20\\), point \\(C\\) is the midpoint of segment \\(AB\\), point \\(D\\) is the midpoint of segment \\(BC\\), and point \\(E\\) is on segment \\(AC\\) such that \\(CE=\\frac{2}{5}AC\\). Find the length of segment \\(ED\\).\n",
    "completion": "\\textbf{Solution:} Since $CE=\\frac{2}{5}AC$,\\\\\nthen $CE=\\frac{2}{5}\\times 10=4$,\\\\\nthus $ED=CE+CD=4+5=\\boxed{9}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51251486_75.png"
  },
  {
    "solution": "\\boxed{72^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the straight lines AB and CD intersect at point O, with $\\angle 1: \\angle 2 = 2:3$. Find the measure of $\\angle BOD$ in degrees?\n",
    "completion": "\\textbf{Solution:} Let us assume $\\because$ $\\angle 1\\colon \\angle 2=2\\colon 3$,\\\\\n$\\therefore$ let $\\angle 1=2x$, $\\angle 2=3x$,\\\\\n$\\because$ $\\angle 1+\\angle 2=180^\\circ$,\\\\\n$\\therefore$ $2x+3x=180^\\circ$,\\\\\n$\\therefore$ $x=36^\\circ$,\\\\\n$\\therefore$ $\\angle 1=72^\\circ$,\\\\\n$\\therefore$ $\\angle BOD=\\angle 1=\\boxed{72^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55190300.png"
  },
  {
    "solution": "\\boxed{\\frac{2\\sqrt{5}}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the three vertices of $\\triangle ABC$ are all on the grid points of a square. What is the value of $\\cos A$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure: Draw $BD\\perp AC$ at D,\\\\\nBy the Pythagorean theorem,\\\\\n$AB=\\sqrt{3^2+3^2}=3\\sqrt{2}$, $AC=\\sqrt{3^2+1^2}=\\sqrt{10}$,\\\\\n$\\because$ ${S}_{\\triangle ABC}=\\frac{1}{2}AB\\cdot CD=\\frac{1}{2}BC\\times 3$, thus $CD=\\frac{3\\times 2}{3\\sqrt{2}}=\\sqrt{2}$,\\\\\nIn triangle $\\triangle ABD$, $\\angle ADC=90^\\circ$, $AC=\\sqrt{10}$, $CD=\\sqrt{2}$,\\\\\n$AD=\\sqrt{AC^2-CD^2}=\\sqrt{(\\sqrt{10})^2-(\\sqrt{2})^2}=2\\sqrt{2}$,\\\\\n$\\therefore$ $\\cos A=\\frac{AD}{AC}=\\frac{2\\sqrt{2}}{\\sqrt{10}}=\\boxed{\\frac{2\\sqrt{5}}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53043749.png"
  },
  {
    "solution": "\\boxed{22.5^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $OC \\perp AB$ at point O, $OD$ bisects $\\angle BOC$, $OE$ bisects $\\angle AOD$. Find the degree measure of $\\angle COE$.\n",
    "completion": "\\textbf{Solution:} From (1), we have\n\\[\n\\angle AOD = \\angle AOC + \\angle COD = 90^\\circ + 45^\\circ = 135^\\circ,\n\\]\n\\[\n\\because OE \\text{ bisects } \\angle AOD,\n\\]\n\\[\n\\therefore \\angle AOE = \\frac{1}{2}\\angle AOD = 67.5^\\circ\n\\]\n\\[\n\\therefore \\angle COE = 90^\\circ - 67.5^\\circ = \\boxed{22.5^\\circ}\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/53364075_83.png"
  },
  {
    "solution": "\\boxed{x>1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the linear function $y_{1}=x+b$ and the inverse proportion function $y_{2}=\\frac{k}{x}$ intersect at points $A(1, a)$ and $D(-4, -1)$, and intersect the $y$-axis and $x$-axis at points $B$ and $C$, respectively. Based on the graph, please directly write down the range of $x$ values when $y_{1}>y_{2}$.\n",
    "completion": "\\textbf{Solution:} When $y_{1}>y_{2}$, the graph of the linear function $y_{1}=x+3$ is above the graph of the hyperbolic function $y_{2}=\\frac{4}{x}$,\\\\\nsince the intersection points of the two functions are $A(1, 4)$ and $D(-4, 1)$,\\\\\nto the right of point D, to the left of the y-axis, and to the right of point A meet the condition,\\\\\ntherefore, \\boxed{-4<x<0} or \\boxed{x>1}.",
    "image_path": "dataset/mm_math/images/MM_Math/51379335_198.png"
  },
  {
    "solution": "\\boxed{(2,4)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated in the figure, the graph of the linear function $y=-2x+8$ intersects with the graph of the function $y=\\frac{k}{x}$ ($x>0$) at points $A(m,6)$ and $B(n,2)$. $AC$ is perpendicular to the $y$-axis at point $C$, and $BD$ is perpendicular to the $x$-axis at point $D$. Let $P$ be a point on the line segment $AB$. Lines $PC$ and $PD$ are drawn. If the areas of triangles $\\triangle PCA$ and $\\triangle PDB$ are equal, find the coordinates of point $P$.\n",
    "completion": "\\textbf{Solution:} Let the coordinates of point $P$ on the line $y=-2x+8$ be $(x, -2x+8)$, \\\\\nby the equality of areas of $\\triangle PCA$ and $\\triangle PDB$, we get: $\\frac{1}{2}AC\\times |y_A-y_P|=\\frac{1}{2}BD\\times |y_B-y_P|$, \\\\\n$\\therefore \\frac{1}{2}\\times 1\\times[6-(-2x+8)]=\\frac{1}{2}\\times 2\\times(3-x)$, \\\\\nSolving this yields: $x=2$, \\\\\nthus $y=-2x+8=4$, \\\\\n$\\therefore$ the coordinates of point $P$ are $\\boxed{(2,4)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53078921_55.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $AC$ and $BD$ intersect at point $O$, and $E$ and $F$ are the midpoints of $AO$ and $AD$ respectively. If $EF=4$ and $AB=8$, what is the measure of $\\angle ACB$ in degrees?\n",
    "completion": "\\textbf{Solution:} Given that E and F are the midpoints of AO and AD respectively,\\\\\nit follows that EF is the midline of $\\triangle AOD$,\\\\\nhence OD = 2EF = $2\\times 4$ = 8,\\\\\nsince quadrilateral ABCD is a rectangle,\\\\\nit implies OB = OD = OA = OC = 8,\\\\\nwhich means AC = 16,\\\\\ngiven AB = 8,\\\\\nit follows AC = 2AB,\\\\\nsince $\\angle$ABC = 90$^\\circ$,\\\\\nit results in $\\angle$ACB = \\boxed{30^\\circ}.",
    "image_path": "dataset/mm_math/images/MM_Math/52425085.png"
  },
  {
    "solution": "\\boxed{\\left(4,0\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $A$ is located at $(1, \\sqrt{3})$ in the Cartesian coordinate system. $AB$ is a line passing through point $A$, and $B$ is the intersection of this line with the $x$-axis. A parallelogram $AOBC$ is formed using $OA$ and $OB$ as adjacent sides. If $OD$ is the bisector of $\\angle AOB$ and $D$ is the midpoint of $AC$. If $P$ is a moving point on the line $AB$, and $S_{\\triangle POD} = \\frac{1}{2} S_{\\text{parallelogram} AOBC}$, please directly write down the coordinates of the point $P$ that satisfies the condition.\n",
    "completion": "\\textbf{Solution:}\n\n(1) Let the intersection of AB and the y-axis be M, and the intersection with OD be E, connect MD and BD. Set $x=0$, then $y=-\\frac{\\sqrt{3}}{3}x+\\frac{4\\sqrt{3}}{3}=\\frac{4\\sqrt{3}}{3}$.\\\\\nTherefore, the intersection M of line AB and the y-axis is $\\left(0,\\frac{4\\sqrt{3}}{3}\\right)$, therefore MO$=\\frac{4\\sqrt{3}}{3}$. Let the equation of OD be $y=mx$, substitute the coordinates of point D $\\left(3,\\sqrt{3}\\right)$ to get $m=\\frac{3}{\\sqrt{3}}=\\sqrt{3}$. Therefore, the equation of OD is $y=\\sqrt{3}x$, by solving the system\n\\[\n\\begin{cases}\ny=\\sqrt{3}x\\\\\ny=-\\frac{\\sqrt{3}}{3}x+\\frac{4\\sqrt{3}}{3}\n\\end{cases}\n\\]\nwe get\n\\[\n\\begin{cases}\nx=2\\\\\ny=\\frac{2\\sqrt{3}}{3}\n\\end{cases}\n\\]\nTherefore, the coordinates of point E are $\\left(2,\\frac{2\\sqrt{3}}{3}\\right)$, therefore OE$=\\sqrt{2^2+\\left(\\frac{2\\sqrt{3}}{3}\\right)^2}=\\frac{4\\sqrt{3}}{3}$, ME$=\\sqrt{(2-0)^2+\\left(\\frac{2\\sqrt{3}}{3}-\\frac{4\\sqrt{3}}{3}\\right)^2}=\\frac{4\\sqrt{3}}{3}$, AM$=\\sqrt{(1-0)^2+\\left(\\sqrt{3}-\\frac{4\\sqrt{3}}{3}\\right)^2}=\\frac{2\\sqrt{3}}{3}$, AE$=\\sqrt{(2-1)^2+\\left(\\frac{2\\sqrt{3}}{3}-\\sqrt{3}\\right)^2}=\\frac{2\\sqrt{3}}{3}$.\\\\\nTherefore, OM=OE=EM, AE=AM$=\\frac{2\\sqrt{3}}{3}$, therefore $S_{\\triangle OAM}$=$S_{\\triangle OAE}$, $S_{\\triangle AMD}$=$S_{\\triangle AED}$, therefore $S_{\\triangle ODM}$=$S_{\\triangle OAM}$+$S_{\\triangle OAE}$+$S_{\\triangle AMD}$+$S_{\\triangle AED}$=2$S_{\\triangle OAD}$=$\\frac{1}{2}S_{\\text{parallelogram }AOBC}$. Therefore P and M coincide, so the coordinates of point P are $\\boxed{\\left(0,\\frac{4\\sqrt{3}}{3}\\right)}$;\\\\\n(2) Because point D is the midpoint of AC, therefore $S_{\\triangle OBD}$=$\\frac{1}{2}S_{\\text{parallelogram }AOBC}$, therefore P coincides with B, so the coordinates of point P are $\\boxed{\\left(4,0\\right)}$.\\\\\nTherefore, the coordinates of point P are $\\boxed{\\left(0,\\frac{4\\sqrt{3}}{3}\\right)}$ or $\\boxed{\\left(4,0\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50613254_112.png"
  },
  {
    "solution": "\\boxed{24}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the diagonals AC and BD of the rhombus ABCD intersect at point O, with BE $\\parallel$ AC and AE $\\parallel$ BD. OE intersects AB at point F. If OE $= 5$ and AC $= 8$, what is the area of the rhombus ABCD.\n",
    "completion": "\\textbf{Solution:}\n\nSince $BE\\parallel AC$ and $AE\\parallel BD$, \\\\\nit follows that quadrilateral AOBE is a parallelogram, \\\\\nSince quadrilateral ABCD is a rhombus, \\\\\nit follows that $AC\\perp BD$, \\\\\ntherefore, parallelogram AOBE is a rectangle, \\\\\ntherefore, $AB=OE=5$, \\\\\nSince $AC=8$, \\\\\nit follows that $AO=\\frac{1}{2}AC=4$, \\\\\nthus, $OB=\\sqrt{AB^{2}-AO^{2}}=\\sqrt{5^{2}-4^{2}}=3$, \\\\\ntherefore, the area of ABCD, $S_{ABCD}=OB\\cdot AC=3\\times 8=\\boxed{24}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52189463.png"
  },
  {
    "solution": "\\boxed{1.5\\pi}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, circle $\\odot O$ is inscribed in square $ABCD$ with $O$ as its center. Let $\\angle MON=90^\\circ$ with its two sides intersecting $BC$ and $CD$ at points $M$ and $N$ respectively, and intersecting $\\odot O$ at points $E$ and $F$. If $CM+CN=6$, what is the length of arc $EF$?\n",
    "completion": "\\textbf{Solution:} Let circle $O$ be tangent to side $CD$ of square $ABCD$ at point $G$ and to $BC$ at point $H$, as shown in the solution diagram. Connect $OG$ and $OH$. Then, quadrilateral $OHCG$ is a square. \\\\\nSince $\\angle GON+\\angle NOH=90^\\circ$ \\\\\nand $\\angle HOM+\\angle NOH=90^\\circ$, \\\\\nthus $\\angle GON=\\angle HOM$, \\\\\nAlso, since $\\angle OGN=\\angle OHM=90^\\circ$, \\\\\nand $OG=OH$, \\\\\nthus $\\triangle OGN \\cong \\triangle OHM$ (ASA), \\\\\ntherefore $GN=HM$ \\\\\nHence, the radius of circle $O$ is $= \\frac{1}{2}(CM+CN)=3$ \\\\\nTherefore, $\\overset{\\frown}{EF}=\\frac{90}{180}\\times 3\\pi =\\boxed{1.5\\pi}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53147973.png"
  },
  {
    "solution": "\\boxed{120^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within quadrilateral $ABCD$, points $E$ and $F$ are the trisection points of the diagonal $BD$. Connect $AE$, $CF$, $AF$, and $CE$. The quadrilateral $AECF$ forms a parallelogram. If quadrilateral $AECF$ is a rhombus, and $AE=BE$, find the measure of $\\angle BAD$.\n",
    "completion": "\\textbf{Solution:} Since quadrilateral AECF is a rhombus, it follows that AE=AF=CF=CE. Also, since AE=BE, it follows that AE=BE=EF=AF=DF, which means that $\\angle EAB=\\angle EBA$, $\\angle EAF=\\angle EFA$, and $\\angle FAD=\\angle FDA$. Therefore, $\\triangle AEF$ is an equilateral triangle, which implies that $\\angle EAF=\\angle AFE=\\angle AEF=60^\\circ$. Hence, $\\angle BAE=30^\\circ$ and $\\angle FAD=30^\\circ$, leading to $\\angle BAD=\\boxed{120^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50496781_95.png"
  },
  {
    "solution": "\\boxed{19}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $DE$ is the perpendicular bisector of $AC$. If $AE=3$ and the perimeter of $\\triangle ABD$ is $13$, what is the perimeter of $\\triangle ABC$?\n",
    "completion": "\\textbf{Solution:} Given that $DE$ is the perpendicular bisector of $AC$ and $AE=3$,\n\n$\\therefore DA=DC$, $AC=2AE=6$,\n\nSince the perimeter of $\\triangle ABD$ is $13$,\n\n$\\therefore AB+BD+AD=AB+BD+DC=AB+BC=13$,\n\n$\\therefore$ the perimeter of $\\triangle ABC$ $=AB+BC+AC=13+6=\\boxed{19}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52653333.png"
  },
  {
    "solution": "\\boxed{(-6,6)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the graph of the linear function $y=-\\frac{1}{2}x+b$ passes through point $A(0,3)$. Point $P$ is a moving point on this line. Through point $P$, $PM$ is drawn perpendicular to the $x$-axis at point $M$, and $PN$ is drawn perpendicular to the $y$-axis at point $N$. On quadrilateral $PMON$, the following segments are intercepted: $PC=\\frac{1}{3}MP$, $MB=\\frac{1}{3}OM$, $OE=\\frac{1}{3}ON$, and $ND=\\frac{1}{3}NP$. Quadrilateral $BCDE$ is a parallelogram. Is there such a point $P$ on the line $y=-\\frac{1}{2}x+b$ that makes quadrilateral $BCDE$ a square? If such a point exists, please find the coordinates of all such points $P$; if it does not exist, please explain why.\n",
    "completion": "\\textbf{Solution:} Let the coordinates of point $P$ be $(x,y)$,\n\nwhen $\\triangle OBE \\cong \\triangle MCB$, quadrilateral $BCDE$ becomes a square,\n\n$OE=BM$,\n\nwhen point $P$ is in the first quadrant, i.e., $\\frac{1}{3}y=\\frac{1}{3}x$, $x=y$.\n\nPoint $P$ lies on the line,\n\n$\\left\\{\\begin{array}{l}\ny=\\frac{1}{2}x+3 \\\\\ny=x\n\\end{array}\\right.$,\n\nsolving we get $\\left\\{\\begin{array}{l}\nx=\\boxed{2} \\\\\ny=2\n\\end{array}\\right.$,\n\nwhen point $P$ is in the second quadrant, $-x=y$\n\n$\\left\\{\\begin{array}{l}\ny=\\frac{1}{2}x+3 \\\\\ny=-x\n\\end{array}\\right.$,\n\nsolving we get $\\left\\{\\begin{array}{l}\nx=-6 \\\\\ny=6\n\\end{array}\\right.$\n\nThere exists such a point $P$ on the line $y=-\\frac{1}{2}x+b$ that makes quadrilateral $BCDE$ a square, with $P$'s coordinates being $\\boxed{(2,2)}$ or $\\boxed{(-6,6)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52395032_112.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in parallelogram $ABCD$, point $E$ lies on side $BC$, and $AE$ intersects $BD$ at point $F$. If $BE=2$, $EC=3$, and $DF=10$, then what is the value of $BF$?\n",
    "completion": "\\textbf{Solution:} Given that $BE=2, EC=3$, \\\\\nit follows that $BC=BE+CE=2+3=5$, \\\\\nSince quadrilateral $ABCD$ is a parallelogram, \\\\\nit follows that $AD\\parallel BC$ and $AD=BC=5$, \\\\\nTherefore, $\\triangle BEF\\sim \\triangle DAF$, \\\\\nwhich implies $\\frac{BF}{DF}=\\frac{BE}{AD}$, i.e., $\\frac{BF}{10}=\\frac{2}{5}$, \\\\\nThus, $BF=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55604334.png"
  },
  {
    "solution": "\\boxed{\\left(4 +\\sqrt{7},0\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=ax^2+bx-3$ intersects the x-axis at points $A(-1,0)$ and $B(3,0)$. The line $l$ passing through point $A$ intersects the parabola at point $C(2,n)$. Given a point $F$ moving on the parabola, is there a point $D$ on the x-axis such that the quadrilateral with vertices $A$, $C$, $D$, $F$ is a parallelogram? If such a point $D$ exists, please write down the coordinates of all points $D$ that satisfy the condition; if not, please explain why.\n",
    "completion": "\\textbf{Solution:} Exist; The coordinates of point $D$ meeting the conditions are $\\boxed{(-3,0)}$ or $\\boxed{(1,0)}$ or $\\boxed{\\left(4 -\\sqrt{7},0\\right)}$ or $\\boxed{\\left(4 +\\sqrt{7},0\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51446075_228.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle BAC=90^\\circ$, squares $AGFB$, $BCDE$, and $ACMN$ are constructed upwards on sides $AB$, $BC$, and $AC$ respectively, with point $E$ on $FG$. If $AC=2$ and $BC=\\sqrt{13}$, what is the area of the shaded region in the diagram?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, we connect $DM$, and draw $DQ\\perp CG$ through point $D$,\n\\[\\therefore \\angle DQC=90^\\circ,\\]\n\\[\\because\\] Quadrilateral $BCDE$ is a square,\n\\[\\therefore BC=CD,\\; \\angle ACB+\\angle QCD=90^\\circ,\\]\nAlso \\[\\because \\angle BAC=90^\\circ,\\]\n\\[\\therefore \\angle ACB+\\angle ABC=90^\\circ\\]\n\\[\\therefore \\angle ABC=\\angle QCD\\]\nIn $\\triangle ABC$ and $\\triangle QCD$: \\[\\left\\{\\begin{array}{l}\\angle BAC=\\angle DQC \\\\ \\angle ABC=\\angle QCD \\\\ BC=CD \\end{array}\\right.\\]\n\\[\\therefore \\triangle ABC \\cong \\triangle QCD \\text{ (AAS)}\\]\n\\[\\therefore DQ=AC\\]\nAlso, \\[\\because ACMN\\] is a square,\n\\[\\therefore CM=AC,\\; \\angle QCM=90^\\circ,\\; MN\\parallel AC\\]\n\\[\\therefore QD\\parallel CM,\\; DQ=CM\\]\n\\[\\therefore CMDQ\\] is a parallelogram,\n\\[\\therefore MD\\parallel AC\\]\n\\[\\therefore\\] Points $D$, $N$, and $M$ lie on a straight line,\nTherefore, $\\triangle DCM$ is also a right triangle and $\\triangle ABC \\cong \\triangle DCM$, given that the quadrilateral $BCDE$ is a square, $AGFB$ is a Square, $ACMN$ is a square, $\\triangle DCM$, $\\triangle ABC$ are congruent triangles, analogously to Zhao Shuang's chord diagram it is known $\\triangle ABC \\cong \\triangle DCM \\cong \\triangle FBE$, thus proving $\\triangle DNK \\cong \\triangle EGP$ (the proof is omitted here)\\\\\nThen: $S_{\\text{shaded}}=S_{\\triangle FBE}+S_{\\triangle EGP}+S_{CMNK}=S_{\\triangle FBE}+S_{\\triangle DNK}+S_{CMNK}=S_{\\triangle FBE}+S_{\\triangle DCM}=2S_{\\triangle ABC}$\\\\\n\\[\\because AC=2,\\; BC=\\sqrt{13}\\]\n\\[\\therefore AB=\\sqrt{BC^2-AC^2}=\\sqrt{(\\sqrt{13})^2-2^2}=3\\]\n\\[\\therefore S_{\\text{shaded}}=2S_{\\triangle ABC}=2\\times \\frac{1}{2}AB\\cdot AC=2\\times \\frac{1}{2}\\times 2\\times 3=\\boxed{6}\\]",
    "image_path": "dataset/mm_math/images/MM_Math/53078524.png"
  },
  {
    "solution": "\\boxed{8.4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a billboard CD stands on top of a building. Xiao Ming and his classmates measured the angle of elevation to the bottom of the billboard, D, as $53^\\circ$ from point A at the foot of the hill. Walking up the slope AB to point B, they measured the angle of elevation to the top of the billboard, C, as $45^\\circ$. It is known that the slope of the hill, AB, has a gradient of $i=1\\colon \\sqrt{3}$ and $AB=12$ meters, $AE=24$ meters. (The height of the protractor is negligible, and the result should be accurate to 0.1 meter, reference data: $\\sqrt{2}\\approx 1.41$, $\\sqrt{3}\\approx 1.73$, $\\sin 53^\\circ\\approx \\frac{4}{5}$, $\\cos 53^\\circ\\approx \\frac{3}{5}$, $\\tan 53^\\circ\\approx \\frac{4}{3}$). How tall is the billboard CD, in meters?\n",
    "completion": "\\textbf{Solution:} In right triangle $ABM$,\\\\\n$\\therefore NE=BM=\\frac{1}{2}AB=6$ (meters),\\\\\n$AM=\\frac{\\sqrt{3}}{2}AB=6\\sqrt{3}$ (meters),\\\\\n$\\therefore ME=AM+AE=\\left(6\\sqrt{3}+24\\right)$ meters,\\\\\n$\\because \\angle CBN=45^\\circ$,\\\\\n$\\therefore CN=BN=ME=\\left(6\\sqrt{3}+24\\right)$ meters,\\\\\n$\\therefore CE=CN+NE=\\left(6\\sqrt{3}+30\\right)$ meters,\\\\\nIn right triangle $ADE$, $\\angle DAE=53^\\circ$, $AE=24$ meters,\\\\\n$\\therefore DE=AE\\cdot\\tan53^\\circ\\approx 24\\times \\frac{4}{3}=32$ (meters),\\\\\n$\\therefore CD=CE-DE$\\\\\n$=6\\sqrt{3}+30-32$\\\\\n$=6\\sqrt{3}-2$\\\\\n$\\approx \\boxed{8.4}$ (meters)",
    "image_path": "dataset/mm_math/images/MM_Math/51445934_98.png"
  },
  {
    "solution": "\\boxed{\\frac{9}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the vertices $B$ and $C$ of rectangle $ABCD$ are located on the positive half of the $x$-axis, with $AB=8$ and $BC=6$. The diagonals $AC$ and $BD$ intersect at point $E$. The graph of the inverse proportion function $y=\\frac{k}{x}$ (for $x>0$) passes through point $E$ and intersects $AB$ and $CD$ at points $F$ and $G$, respectively. Connect $EG$. If $BF+BE=11$, what is the area of $\\triangle CEG$?\n",
    "completion": "\\textbf{Solution:} \nGiven that $AC=\\sqrt{AB^2+BC^2}=\\sqrt{6^2+8^2}=10$,\\\\\nhence $BE=EC=5$,\\\\\nGiven that $BF+BE=11$,\\\\\nhence $BF=6$,\\\\\nLet $OB=t$, then $F(t,6)$, and $E(t+3,4)$,\\\\\nSince the graph of the inverse proportion function $y=\\frac{36}{x}$ passes through points E and F,\\\\\nthus $6t=4(t+3)$,\\\\\nSolving this, we find $t=6$,\\\\\nthus $k=6t=36$,\\\\\nHence, the equation of the inverse proportion function is $y=\\frac{36}{x}$,\\\\\nwhen $x=12$, $y=\\frac{36}{12}=3$,\\\\\nthus $G(12,3)$,\\\\\ntherefore, the area of $\\triangle CEG$ equals $\\frac{1}{2}\\times 3\\times 3=\\boxed{\\frac{9}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53078897_58.png"
  },
  {
    "solution": "\\boxed{\\sqrt{13}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the rectangle $ABCD$, points $E$ and $F$ are located on sides $AD$ and $DC$ respectively. Given that $\\triangle ABE \\sim \\triangle DEF$, $AB=6$, $AE=9$, and $DE=2$, find the length of $EF$.\n",
    "completion": "\\textbf{Solution}: Given $\\triangle ABE \\sim \\triangle DEF$, with AB=6, AE=9, and DE=2,\\\\\nthen $\\frac{AB}{DE}=\\frac{AE}{DF}$, which means $\\frac{6}{2}=\\frac{9}{DF}$, thus solving for DF gives DF=3,\\\\\nsince quadrilateral ABCD is a rectangle,\\\\\nthen $\\angle D=90^\\circ$,\\\\\nby the Pythagorean theorem,\\\\\n$EF=\\sqrt{DE^{2}+DF^{2}}=\\sqrt{2^{2}+3^{2}}=\\boxed{\\sqrt{13}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51317428.png"
  },
  {
    "solution": "\\boxed{C}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: How many of the following plane figures, when rotated around the dashed line for one complete turn, can form a geometric solid in the shape of a vase as shown in the figure?\n",
    "completion": "\\textbf{Solution:}\\\\\nA. The shape after rotation does not match the required 3D figure, thus it is not applicable;\\\\\nB. The shape after rotation is a cylinder, which is not the required 3D figure, thus it is not applicable;\\\\\nC. The shape after rotation matches the required 3D figure, thus it is applicable;\\\\\nD. The shape after rotation does not match the required 3D figure, thus it is not applicable;\\\\\nHence, the answer is: \\boxed{C}",
    "image_path": "dataset/mm_math/images/MM_Math/55470821.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given that $AB$ is the diameter of $\\odot O$, and $BD$ is a tangent to $\\odot O$ at point $B$. Through point $C$ on $\\odot O$, construct $CD\\parallel AB$, intersecting $BD$ at point $D$. Connect $AC$ and $BC$. As shown in Figure (2), when $CD$ intersects $\\odot O$ at point $E$, connect $BE$. If $\\angle EBD=30^\\circ$, what are the measures of $\\angle BCD$ and $\\angle DBC$?\n",
    "completion": "\\textbf{Solution:} Given that,\\\\\n$\\because$ AB is the diameter of $\\odot$O, and DB is the tangent of $\\odot$O touching at B,\\\\\n$\\therefore$ DB$\\perp$AB,\\\\\n$\\therefore$ $\\angle$DBA$=90^\\circ$,\\\\\n$\\angle$DEB$=\\angle$EBA,\\\\\n$\\therefore$ $\\angle$BDC$=90^\\circ$,\\\\\n$\\because$ $\\angle$EBD$=30^\\circ$,\\\\\n$\\therefore$ $\\angle$DEB$=60^\\circ$,\\\\\n$\\therefore$ $\\angle$EBA$=60^\\circ$,\\\\\n$\\therefore$ $\\angle$ACE$=120^\\circ$,\\\\\n$\\because$ AB is the diameter of $\\odot$O,\\\\\n$\\therefore$ $\\angle$BCA$=90^\\circ$,\\\\\n$\\therefore$ $\\angle$BCD$=\\boxed{30^\\circ}$,\\\\\n$\\therefore$ $\\angle$DBC$=\\boxed{60^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52720307_181.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, $AC=3$, and $BC=4$. Squares $ABMN$, $ACDE$, and $BCFG$ are constructed on the sides of $\\triangle ABC$ on the same side as $AB$. Point $M$ is on side $FG$, $MN$ intersects $CF$ at point $P$, and $AN$ intersects $CD$ at point $Q$. $\\triangle ABQ \\cong \\triangle NAP$; what is the area of quadrilateral $CPNQ$?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABQ \\cong \\triangle NAP$,\\\\\nit follows that $S_{\\triangle ABQ} = S_{\\triangle NAP}$,\\\\\nhence $S_{\\triangle ABC} = S_{\\text{quadrilateral} CPNQ}$,\\\\\nSince $\\angle ACB = 90^\\circ$, $AC = 3$, and $BC = 4$,\\\\\nit follows that $S_{\\triangle ABC} = \\frac{1}{2}AC \\cdot BC = 6$,\\\\\nthus $S_{\\text{quadrilateral} CPNQ} = \\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53182655_155.png"
  },
  {
    "solution": "\\boxed{x<3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graphs of the functions $y=2x$ and $y=ax+4$ intersect at point $A(3,m)$. What is the solution set of the inequality $2x<ax+4$?\n",
    "completion": "\\textbf{Solution:} Observing the graph, we have: when $x<3$, then $2x<ax+4$,\\\\\nwhich means the solution set of the inequality $2x<ax+4$ is $\\boxed{x<3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52211165.png"
  },
  {
    "solution": "\\boxed{20^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, observe the compass and straightedge constructions in the diagram. If $\\angle FMO = 50^\\circ$, what is the measure of $\\angle FOE$?\n",
    "completion": "\\textbf{Solution:} From the drawing, we know that PQ perpendicularly bisects FM, \\\\\n$\\therefore$ point E is the midpoint of the arc $\\overset{\\frown}{FM}$, \\\\\n$\\therefore$ $\\overset{\\frown}{FE} = \\overset{\\frown}{EM}$, \\\\\n$\\therefore$ $\\angle MOE = \\angle BOE = \\frac{1}{2}\\angle AOB$, \\\\\nAlso, $\\because \\angle FMO = 50^\\circ$, $\\angle OFM = 90^\\circ$, \\\\\n$\\therefore$ $\\angle AOB = 40^\\circ$, \\\\\n$\\therefore$ $\\angle FOE = \\boxed{20^\\circ}$,",
    "image_path": "dataset/mm_math/images/MM_Math/52917444.png"
  },
  {
    "solution": "\\boxed{75^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AD \\perp BC$ at point $D$, and $AE$ bisects $\\angle CAD$ and intersects $BC$ at $E$. If $\\angle C= 60^\\circ$, then what is the measure of $\\angle DEA$ in degrees?\n",
    "completion": "\\textbf{Solution:} Given that $AD$ is the height from $D$ to side $BC$, and $\\angle C=60^\\circ$,\\\\\n$\\therefore \\angle CAD=90^\\circ-\\angle C=90^\\circ-60^\\circ=30^\\circ$;\\\\\nGiven that $AE$ is the bisector of $\\angle BAC$,\\\\\n$\\therefore \\angle CAE=\\angle DAE= \\frac{1}{2}\\angle DAC=15^\\circ$,\\\\\n$\\therefore \\angle DEA=\\angle CAE+\\angle C=15^\\circ+60^\\circ=\\boxed{75^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52511705.png"
  },
  {
    "solution": "\\boxed{40^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, fold the corner where vertex D of rectangle ABCD is located, so that point D falls on point F on side BC, with the crease being AE. If $\\angle DAE=25^\\circ$, what is the measure of $\\angle EFC$?\n",
    "completion": "\\textbf{Solution}: As shown in the figure, since quadrilateral $ABCD$ is a rectangle, thus $AD \\parallel BC$ and $\\angle D=90^\\circ$. By folding, it is known that $\\triangle ADE \\cong \\triangle AFE$. Therefore, $\\angle FAE=\\angle DAE=25^\\circ$, and $\\angle AFE=\\angle D=90^\\circ$. Therefore, $\\angle AFB=\\angle DAF=2\\angle DAE=50^\\circ$. Therefore, $\\angle EFC=180^\\circ-\\angle AFB-\\angle AFE=180^\\circ-50^\\circ-90^\\circ=\\boxed{40^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50481650_131.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, AB is the diameter of $\\odot O$, CD is a chord, and CD $\\perp$ AB at point E. Connect DO and extend it to intersect $\\odot O$ at point F. Connect AF, which intersects CD at point G, with CG = AG, and connect AC. AC $\\parallel$ DF; if AB = 12, what are the lengths of AC and GD?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect CO,\\\\\nsince AB$\\perp$CD,\\\\\nthus $\\overset{\\frown}{AC} $= $\\overset{\\frown}{AD}$, CE=DE,\\\\\nsince $\\angle$DCA=$\\angle$CAF,\\\\\nthus $\\overset{\\frown}{AD} $= $\\overset{\\frown}{CF}$,\\\\\nthus $\\overset{\\frown}{AC} $= $\\overset{\\frown}{AD} $= $\\overset{\\frown}{CF}$,\\\\\nthus $\\angle$AOD=$\\angle$AOC=$\\angle$COF,\\\\\nsince DF is the diameter,\\\\\nthus $\\angle$AOD=$\\angle$AOC=$\\angle$COF=$60^\\circ$,\\\\\nsince OA=OC,\\\\\nthus $\\triangle$AOC is an equilateral triangle,\\\\\nthus AC=AO=6, $\\angle$CAO=$60^\\circ$,\\\\\nsince CE$\\perp$AO,\\\\\nthus AE=EO=3, $\\angle$ACD=$30^\\circ$,\\\\\nthus CE=3$\\sqrt{3}$=DE,\\\\\nsince AG$^{2}$=GE$^{2}$+AE$^{2}$,\\\\\nthus AG$^{2}$=（3$\\sqrt{3}$−AG）$^{2}$+9,\\\\\nthus AG=2$\\sqrt{3}$,\\\\\nthus GE=$\\sqrt{3}$,\\\\\nthus DG=4$\\sqrt{3}$.\\\\\nTherefore, the lengths of AC and GD are respectively AC=\\boxed{6} and DG=\\boxed{4\\sqrt{3}}.",
    "image_path": "dataset/mm_math/images/MM_Math/52807144_146.png"
  },
  {
    "solution": "\\boxed{23^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, AB is the diameter of the semicircle O, and point D is the midpoint of the arc AC. If $\\angle BAC = 44^\\circ$, then what is the value of $\\angle DAC$?\n",
    "completion": "\\textbf{Solution:} Given that $AB$ is the diameter of the semicircle $O$,\\\\\nit follows that $\\angle ACB = 90^\\circ$,\\\\\nsince $\\angle BAC = 44^\\circ$,\\\\\nthen $\\angle B = 90^\\circ - \\angle BAC = 46^\\circ$,\\\\\nas quadrilateral $ABCD$ is inscribed in semicircle $O$,\\\\\nthen $\\angle D = 180^\\circ - \\angle B = 134^\\circ$,\\\\\ngiven that point $D$ is the midpoint of arc $AC$,\\\\\nit follows that $ \\overset{\\frown}{AD} =  \\overset{\\frown}{DC}$,\\\\\nthus $AD = DC$,\\\\\ntherefore $\\angle DAC = \\angle DCA = \\frac{1}{2} (180^\\circ - \\angle D) = \\boxed{23^\\circ}$,",
    "image_path": "dataset/mm_math/images/MM_Math/53124717.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the circle $\\odot O$, $AB$ is a diameter, and point C is a point on the circle. The minor arc is folded along the chord $AC$ intersecting $AB$ at point D, and $CD$ is connected. As shown in Figure 1, if point D coincides with the center O and $AC = \\sqrt{3}$, what is the radius $r$ of $\\odot O$?\n",
    "completion": "\\textbf{Solution:} Let the point $D$ be symmetric to the chord $AC$, and call this symmetric point $F$. Connect $DF$ and let it intersect $AC$ at point $E$,\n\nthus $DE=EF, DF\\perp AC, AE=EC$,\n\nsince $AC=\\sqrt{3}$,\n\ntherefore $AE=EC=\\frac{\\sqrt{3}}{2}$,\n\nlet $DE=EF=\\frac{r}{2}$,\n\nthen $AD=DF=r$,\n\naccording to the Pythagorean theorem, we get ${r}^{2}-\\left(\\frac{r}{2}\\right)^{2}=\\left(\\frac{\\sqrt{3}}{2}\\right)^{2}$,\n\nsolving this yields $r=\\boxed{1}$,\n\ntherefore, the radius $r$ of the circle is \\boxed{1}.",
    "image_path": "dataset/mm_math/images/MM_Math/52871917_107.png"
  },
  {
    "solution": "\\boxed{8\\sqrt{3}-12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rhombus $ABCD$, $\\angle ABC=60^\\circ$. Quadrilateral $EFGH$ is a rectangle. If $FA=FB=2\\sqrt{2}$, what is the area of rectangle $EFGH$?\n",
    "completion": "\\textbf{Solution:} Draw $AM\\perp BC$ at M and $GN\\perp BC$ at N through point A and G respectively, and connect GM,\\\\\n$\\because$ Quadrilateral EFGH is a rectangle,\\\\\n$\\therefore \\angle AFB=\\angle AED=\\angle BGC=\\angle CHD=90^\\circ$,\\\\\n$\\because FA=FB=2\\sqrt{2}$,\\\\\n$\\therefore AB=\\sqrt{AF^{2}+BF^{2}}=4$, $\\angle ABF=\\angle BAF=45^\\circ$,\\\\\n$\\because$ Quadrilateral ABCD is a rhombus, $\\angle ABC=60^\\circ$,\\\\\n$\\therefore AB=BC=CD=AD$, $\\angle ABC=\\angle ADC=60^\\circ$, $\\angle BAD=\\angle BCD=120^\\circ$,\\\\\n$\\therefore \\angle CBG=15^\\circ$, $\\angle DAF=75^\\circ$,\\\\\n$\\therefore \\angle CDH=\\angle DCH=45^\\circ$, $\\angle ADE=15^\\circ$, $\\angle BCG=75^\\circ$,\\\\\n$\\therefore \\angle BAF=\\angle DCH=\\angle ABF=\\angle CDH$, $\\angle ADE=\\angle CBG$, $\\angle DAE=\\angle BCG$,\\\\\nIn $\\triangle ABF$ and $\\triangle CDH$,\\\\\n$\\left\\{\\begin{array}{l}\n\\angle BAF=\\angle DCH\\\\\nAB=CD\\\\\n\\angle ABF=\\angle CDH\n\\end{array}\\right.$,\\\\\n$\\therefore \\triangle ABF \\cong \\triangle CDH(ASA)$,\\\\\nSimilarly, $\\triangle BCG \\cong \\triangle DAE(ASA)$,\\\\\n$\\because AM\\perp BC$, $\\angle ABC=60^\\circ$,\\\\\n$\\therefore \\angle BAM=30^\\circ$,\\\\\n$\\therefore BM=\\frac{1}{2}AB=2$,\\\\\n$\\therefore AM=\\sqrt{3}BM=2\\sqrt{3}$, $BC=2BM$,\\\\\n$\\because \\angle BGC=90^\\circ$,\\\\\n$\\therefore BM=CM=GM=2$,\\\\\n$\\therefore \\angle CMG=2\\angle CBG=30^\\circ$,\\\\\n$\\because GN\\perp BC$,\\\\\n$\\therefore GN=\\frac{1}{2}GM=1$,\\\\\n$\\therefore S_{\\text{rhombus}ABCD}=BC\\cdot AM=4\\times 2\\sqrt{3}=8\\sqrt{3}$,\\\\\n$S_{\\triangle ABF}=\\frac{1}{2}AF\\cdot BF=\\frac{1}{2}\\times 2\\sqrt{2}\\times 2\\sqrt{2}=4$,\\\\\n$S_{\\triangle BCG}=\\frac{1}{2}BC\\cdot GN=\\frac{1}{2}\\times 4\\times 1=2$,\\\\\n$\\therefore S_{\\text{rectangle}EFGH}=S_{\\text{rhombus}ABCD}-2S_{\\triangle ABF}-2S_{\\triangle BCG}=8\\sqrt{3}-12$.\\\\\nTherefore, the answer is: $S_{\\text{rectangle}EFGH}=\\boxed{8\\sqrt{3}-12}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52284136.png"
  },
  {
    "solution": "\\boxed{(3,2)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the line $y=kx+5$ intersects the $x$-axis at point $A$ and the $y$-axis at point $B$, and the coordinates of point $A$ are $(5,0)$. The line $y=2x-4$ intersects the $x$-axis at point $D$ and intersects line $AB$ at point $C$. What are the coordinates of point $C$?\n",
    "completion": "\\textbf{Solution:} Given the line $y=kx+5$ passes through the point $A(5,0)$, we have $5k+5=0$. Solving this equation gives $k=-1$. Therefore, the equation of line $AB$ is $y=-x+5$. By solving the system of equations of lines $AB$ and $CD$:\n\\[\n\\begin{cases}\ny=-x+5 \\\\\ny=2x-4\n\\end{cases}\n\\]\nwe find:\n\\[\n\\begin{cases}\nx=3 \\\\\ny=2\n\\end{cases}\n\\]\nTherefore, the coordinates of point $C$ are $\\boxed{(3,2)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52416375_89.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $\\triangle ABD \\cong \\triangle CFD$, $AD \\perp BC$ at point $D$, and $CE \\perp AB$; given that $BC=7$ and $AD=5$, find the length of $AF$.\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABD \\cong \\triangle CFD$, \\\\\n$\\therefore BD=DF$,\\\\\nSince $BC=7$, $AD=DC=5$,\\\\\n$\\therefore BD=BC-CD=2$,\\\\\n$\\therefore AF=AD-DF=5-2=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52991713_112.png"
  },
  {
    "solution": "\\boxed{\\frac{25}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, $AD\\parallel BC\\parallel EF$. If $AE=3$, $AB=8$, $CD=10$, then what is the length of $CF$?\n",
    "completion": "\\[\n\\textbf{Solution:} \\quad \\text{Since } AD\\parallel BC\\parallel EF, \n\\]\n\\[\n\\therefore \\frac{AE}{AB}=\\frac{DF}{CD},\n\\]\n\\[\n\\text{Since } AE=3, AB=8, CD=10,\n\\]\n\\[\n\\therefore \\frac{3}{8}=\\frac{DF}{10},\n\\]\n\\[\n\\text{Solving for } DF, \\text{ we get } DF=\\frac{15}{4},\n\\]\n\\[\n\\therefore CF=CD-DF=10-\\frac{15}{4}=\\boxed{\\frac{25}{4}}.\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/52917561.png"
  },
  {
    "solution": "\\boxed{\\frac{12}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the straight lines $l_1 \\parallel l_2 \\parallel l_3$. If $AB = 3, BC = 5, EF = 4$, then what is the length of DE?\n",
    "completion": "\\textbf{Solution:} Since $l_{1}\\parallel l_{2}\\parallel l_{3}$,\\\\\nit follows that $\\frac{AB}{BC}=\\frac{DE}{EF}$,\\\\\nsince $AB=3$, $BC=5$, $EF=4$,\\\\\nit follows that $\\frac{3}{5}=\\frac{DE}{4}$, hence $DE=\\boxed{\\frac{12}{5}}$",
    "image_path": "dataset/mm_math/images/MM_Math/52827667.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $D$ and $E$ are the midpoints of sides $AB$ and $AC$, respectively, and $F$ is a point on the extension of side $CA$, with $\\angle F = \\angle C$. If $BC = 8$, what is the length of $FD$?\n",
    "completion": "\\textbf{Solution:} Since $D$ and $E$ are the midpoints of the sides $AB$ and $AC$ respectively,\n\\newline\nit follows that $DE=\\frac{1}{2}BC$ and $DE \\parallel BC$,\n\\newline\nGiven that $BC=8$,\n\\newline\nit follows that $DE=4$,\n\\newline\nSince $DE \\parallel BC$,\n\\newline\nit follows that $\\angle AED=\\angle C$,\n\\newline\nSince $\\angle F=\\angle C$,\n\\newline\nit follows that $\\angle AED=\\angle F$,\n\\newline\nTherefore, $DF=DE=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52949550_135.png"
  },
  {
    "solution": "\\boxed{\\left(\\frac{3}{2}, \\frac{75}{16}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=-\\frac{3}{4}x^{2}+bx+c$ intersects the x-axis at point $A(4,0)$ and the y-axis at point $B(0,3)$. Point $M(m,0)$ is a moving point on segment $OA$. The line passing through point $M$ and perpendicular to the x-axis intersects line $AB$ and the parabola at points $P$ and $N$, respectively. Find the equation of the parabola and write down the coordinates of its axis of symmetry and vertex.\n",
    "completion": "\\textbf{Solution:} Since the parabola $y=-\\frac{3}{4}x^{2}+bx+c$ intersects the $x$ axis at point $A(4,0)$ and the $y$ axis at point $B(0,3)$,\\\\\nwe have $\\begin{cases}-\\frac{3}{4}\\times 4^{2}+4b+c=0\\\\ c=3\\end{cases}$,\\\\\nsolving this, we get $\\begin{cases}b=\\frac{9}{4}\\\\ c=3\\end{cases}$,\\\\\ntherefore, the equation of the parabola is $y=-\\frac{3}{4}x^{2}+\\frac{9}{4}x+3$,\\\\\nsince $y=-\\frac{3}{4}x^{2}+\\frac{9}{4}x+3=-\\frac{3}{4}(x-\\frac{3}{2})^{2}+\\frac{75}{16}$,\\\\\ntherefore, the axis of symmetry of this parabola is $x=\\frac{3}{2}$,\\\\\nand the vertex coordinates are $\\boxed{\\left(\\frac{3}{2}, \\frac{75}{16}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51403288_129.png"
  },
  {
    "solution": "\\boxed{\\frac{73}{16}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of $\\odot O$, $CD$ is a chord of $\\odot O$, and $CD \\perp AB$ at point $E$. $\\angle BCO = \\angle D$; Given $BE = 8\\,\\text{cm}, CD = 6\\,\\text{cm}$, what is the radius of $\\odot O$?\n",
    "completion": "\\textbf{Solution:}\n\nBecause $AB$ is the diameter of the circle $O$, and $CD \\perp AB$,\n\ntherefore $CE=DE=\\frac{1}{2}CD=\\frac{1}{2}\\times 6=3$,\n\nbecause $\\angle B=\\angle D$ and $\\angle BEC=\\angle DEA$,\n\ntherefore $\\triangle BCE \\sim \\triangle DAE$,\n\ntherefore $AE: CE=DE: BE$,\n\ntherefore $AE: 3=3: 8$,\n\nsolving gives $AE=\\frac{9}{8}$,\n\ntherefore $AB=AE+BE=\\frac{9}{8}+8=\\frac{73}{8}$,\n\ntherefore the radius of circle $O$ is $\\boxed{\\frac{73}{16}}$ (cm).",
    "image_path": "dataset/mm_math/images/MM_Math/53042861_121.png"
  },
  {
    "solution": "\\boxed{\\frac{40000}{49}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given: As shown in the figure, in $\\triangle ABC$, $AD$ is the altitude from $A$ to side $BC$. There is an inscribed rectangle $PQMN$ within this triangle, one side of the rectangle lies on $BC$, and the other two vertices are on $AB$ and $AC$, respectively. If $BC=100$, $AD=40$, and $PQ=PN$ with $\\angle BAC=90^\\circ$, directly provide the value of $BN\\cdot CM$.\n",
    "completion": "\\textbf{Solution:} We have $BN \\cdot CM = \\boxed{\\frac{40000}{49}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52602086_188.png"
  },
  {
    "solution": "\\boxed{x\\geq 3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=kx+b$ $(k<0)$ passes through $A(3, 1)$. What is the range of $x$ when $kx+b\\le \\frac{1}{3}x$?\n",
    "completion": "\\textbf{Solution:} Since the direct proportion function $y= \\frac{1}{3}x$ also passes through point A, \\\\\nit follows that the solution set for $kx+b\\leq \\frac{1}{3}x$ is $\\boxed{x\\geq 3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50560362.png"
  },
  {
    "solution": "\\boxed{C(2t+3,t)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, in the Cartesian coordinate system, point $A(0,6)$, and point $B$ is a moving point on the positive x-axis. Connect $AB$, take the midpoint $M$ of $AB$, and rotate the segment $MB$ $90^\\circ$ clockwise around point $B$ to obtain the segment $BC$. Draw a perpendicular line from point $B$ to the x-axis, intersecting line $AC$ at point $D$. Let the coordinates of point $B$ be $(2t,0)$. Find the coordinates of point $C$ (expressed in terms of $t$).\n",
    "completion": "\\textbf{Solution:} As shown in Figure 1, let $ME\\perp OB$ at $E$, and $CF\\perp$ the x-axis at $F$.\\\\\nSince $B(2t,0)$,\\\\\nit follows that $OB=2t$,\\\\\nSince $ME\\parallel OA$ and $AM=BM$,\\\\\nit follows that $OE=EB=t$ and $ME=\\frac{1}{2}OA=3$,\\\\\nSince $\\angle MEB=\\angle CFB=\\angle CBM=90^\\circ$,\\\\\nit follows that $\\angle MBE+\\angle CBF=90^\\circ$ and $\\angle MBE+\\angle BME=90^\\circ$,\\\\\nTherefore, $\\angle BME=\\angle CBF$,\\\\\nSince $BM=BC$,\\\\\nit follows that $\\triangle MEB\\cong \\triangle BFC$ (AAS),\\\\\nTherefore, $BF=ME=3$ and $CF=BE=t$,\\\\\nTherefore, $OF=OB+BF=2t+3$,\\\\\nTherefore, \\boxed{C(2t+3,t)}.",
    "image_path": "dataset/mm_math/images/MM_Math/53490616_91.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the regular hexagon $ABCDEF$ is inscribed in the circle $\\odot O$. Draw $OM \\perp BC$ at point $M$. If the radius of $\\odot O$ is 4, what is the length of the apothem $OM$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect OB and OC.\\\\\nSince hexagon ABCDEF is a regular hexagon,\\\\\nit follows that $\\angle BOC=60^\\circ$, and OB=OC=4,\\\\\ntherefore $\\triangle OBC$ is an equilateral triangle,\\\\\nhence BC=OB=OC=4,\\\\\nsince OM$\\perp$BC,\\\\\nit follows that BM=CM=2,\\\\\nin $\\triangle OBM$, OM=$\\sqrt{OB^{2}-BM^{2}}=\\sqrt{4^{2}-2^{2}}=\\boxed{2\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52755173.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the rhombus $ABCD$, the diagonals $AC$ and $BD$ intersect at point $O$, and $E$ is the midpoint of $CD$. Connect $OE$. If $OE=4$, what is the length of $AB$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rhombus, \\\\\nit follows that OD = OB, and AB = BC, \\\\\nsince DE = EC, \\\\\nit follows that OE is the median of $\\triangle BDC$, \\\\\nhence, AB = BC = 2OE = \\boxed{8}.",
    "image_path": "dataset/mm_math/images/MM_Math/52160201.png"
  },
  {
    "solution": "\\boxed{\\left(0,\\frac{15}{4}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given a linear function $y=kx+b$ whose graph intersects with the graph of the inverse proportion function $y=\\frac{m}{x}$ $(x>0)$ at point A, and intersects with the x-axis at point B$(5,0)$. If $OB=AB$, and the area of $\\triangle OAB$ is $\\frac{15}{2}$. If point P is on the y-axis, and point Q is on the graph of the inverse proportion function $y=\\frac{m}{x}$ $(x>0)$, and the quadrilateral ABPQ is precisely a parallelogram, directly write down the coordinates of point P.\n",
    "completion": "\\textbf{Solution:} According to the given problem, PQ is located to the top left of AB. Let P be $(0,t)$, then PQ can be seen as moving AB 5 units to the left, and then t units up.\\\\\n$\\therefore$ The coordinates of point Q are $(4,3+t)$\\\\\n$\\because$ Point Q is on the graph of $y= \\frac{27}{x}$,\\\\\n$\\therefore$ $3+t=\\frac{27}{4}$\\\\\n$\\therefore$ $t=\\frac{15}{4}$\\\\\n$\\therefore$ $P\\boxed{\\left(0,\\frac{15}{4}\\right)}$",
    "image_path": "dataset/mm_math/images/MM_Math/50496638_62.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, within $ \\triangle ABC$, $ AD$ and $ AF$ are the median and the altitude of $ \\triangle ABC$, respectively, and $ BE$ is the angle bisector of $ \\triangle ABD$. Given that the area of $ \\triangle ABC$ is 40 and $ BD=5$, find the length of $ AF$.\n",
    "completion": "\\textbf{Solution:} Since $AD$ is the median of $\\triangle ABC$ and $BD=5$,\\\\\nit follows that $BC=2BD=2\\times 5=10$.\\\\\nGiven $AF\\perp BC$ and the area of $\\triangle ABC$ is $40$,\\\\\nwe have $\\frac{1}{2}BC\\cdot AF=\\frac{1}{2}\\times 10\\cdot AF=40$,\\\\\nwhich leads to $AF=\\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52655159_115.png"
  },
  {
    "solution": "\\boxed{\\frac{2\\sqrt{5}}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $AB=6$, $BC=2$. Point $B$ is on line $l$, and the distance from point $A$ to line $l$, denoted as $AE=4$. What is the distance from point $C$ to line $l$, denoted as $CF$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral $ABCD$ is a rectangle,\\\\\nit follows that $\\angle ABC=90^\\circ$,\\\\\nhence $\\angle ABE+\\angle FBC=90^\\circ$,\\\\\nsince $\\angle EAB+\\angle ABE=90^\\circ$,\\\\\nit follows that $\\angle EAB=\\angle FBC$,\\\\\nalso, since $\\angle AEB=\\angle BFC$,\\\\\nit follows that $\\triangle AEB\\sim \\triangle BFC$,\\\\\ntherefore $\\frac{AB}{BC}=\\frac{AE}{BF}$, that is, $\\frac{6}{2}=\\frac{4}{BF}$,\\\\\nhence $BF=\\frac{4}{3}$,\\\\\nthus $CF=\\sqrt{BC^{2}-BF^{2}}=\\boxed{\\frac{2\\sqrt{5}}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52977738.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rhombus $ABCD$, the diagonals $AC$ and $BD$ intersect at point $O$. If $AB=3$ and $BD=4$, what is the area of rhombus $ABCD$?\n",
    "completion": "\\textbf{Solution:}\\\\\nSince quadrilateral $ABCD$ is a rhombus and $BD=4$,\\\\\nit follows that $OB=2$,\\\\\nSince $AB=3$,\\\\\nwe have $OA=\\sqrt{AB^{2}-OB^{2}}=\\sqrt{3^{2}-2^{2}}=\\sqrt{5}$,\\\\\nTherefore, $AC=2OA=2\\sqrt{5}$,\\\\\nSo, the area of rhombus $ABCD$ is $S_{\\text{rhombus}ABCD}=\\frac{1}{2}AC\\cdot BD=\\frac{1}{2}\\times 2\\sqrt{5}\\times 4=4\\sqrt{5}$,\\\\\nHence, the answer is: $S_{\\text{rhombus}ABCD}=\\boxed{4\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50481644.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $A$ is on the graph of the inverse proportion function $y=\\frac{a}{x}$ in the first quadrant, and point $B$ is on the positive half-axis of the $x$-axis. Given $OA=AB$ and the area of $\\triangle AOB$ is 2, what is the value of $a$?\n",
    "completion": "\\textbf{Solution:} Refer to the diagram, draw $AC\\perp OB$ at point C through point A, \\\\\nLet the coordinates of point A be $A(m,n)$, hence $OC=m$, and $AC=n$, \\\\\nSince $OA=AB$, \\\\\nIt follows that $OB=2OC=2m$, \\\\\nGiven that the area of $\\triangle AOB$ is 2, \\\\\nIt follows that $\\frac{1}{2}OB\\cdot AC=\\frac{1}{2}\\times 2m\\cdot n=2$, \\\\\nUpon rearranging: $mn=2$, \\\\\nSubstituting point $A(m,n)$ into the inverse proportion function $y=\\frac{a}{x}$ yields：$a=mn=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53043280.png"
  },
  {
    "solution": "\\boxed{(2, 3)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=-2x+7$ intersects the x-axis at point C and the y-axis at point B, and intersects the line $y=\\frac{3}{2}x$ at point A. What are the coordinates of point A?\n",
    "completion": "\\textbf{Solution:} Solve the system of equations: $\\begin{cases}y=-2x+7\\\\ y=\\frac{3}{2}x\\end{cases}$ to get: $\\begin{cases}x=2\\\\ y=3\\end{cases}$,\\\\\n$\\therefore$ The coordinates of point $A$ are $\\boxed{(2, 3)}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52342217_126.png"
  },
  {
    "solution": "\\boxed{1.5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $D$ is inside $\\triangle ABC$, with $CD$ bisecting $\\angle ACB$, $BE \\perp CD$ at point $D$, and intersecting $AC$ at point $E$, with $\\angle A = \\angle ABE$. If $AC = 7$ and $BC = 4$, what is the length of $BD$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, since $CD$ bisects $\\angle ACB$ and $BE\\perp CD$,\\\\\nit follows that $\\angle BCD=\\angle ECD$ and $\\angle BDC=\\angle EDC=90^\\circ$,\\\\\nsince $CD=CD$,\\\\\nit follows that $\\triangle BCD\\cong \\triangle ECD\\ (ASA)$,\\\\\nhence $BC=CE=4$, $BD=DE=\\frac{1}{2}BE$,\\\\\nsince $\\angle A=\\angle ABE$,\\\\\nit follows that $\\triangle ABE$ is an isosceles triangle,\\\\\nthus $AE=BE=AC−EC=AC−BC=3$,\\\\\ntherefore $BD=DE=\\frac{1}{2}BE=\\boxed{1.5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53044180.png"
  },
  {
    "solution": "\\boxed{(-1,\\frac{-3-\\sqrt{17}}{2})}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=ax^2+bx-3$ passes through points $A$, $B$, and $C$, where $A(-3,0)$, $C(1,0)$, and point $B$ is on the $y$-axis. Point $P$ is a moving point on the parabola below the line $AB$ (not coinciding with $A$ or $B$). Point $Q$ is a moving point on the axis of symmetry of the parabola. Is there a point $Q$ for which the triangle with vertices $A$, $B$, and $Q$ is a right-angled triangle? If such a point exists, find the coordinates of point $Q$; if it does not exist, please explain why.\n",
    "completion": "\\textbf{Solution:} The point $Q$ exists, and the reasons are as follows,\\\\\n$\\because x=-\\frac{2}{2\\times 1}=-1$,\\\\\n$\\therefore$ the axis of symmetry of the parabola is the line $x=-1$,\\\\\nLet $Q(-1,a)$,\\\\\n$\\because B(0,-3), A(-3,0)$,\\\\\n(1) When $\\angle QAB=90^\\circ$, it holds that $AQ^2+AB^2=BQ^2$,\\\\\n$\\therefore 2^2+a^2+3^2+3^2=1^2+(3+a)^2$,\\\\\nSolving this, we find: $a=2$,\\\\\n$\\therefore Q_1(-1,2)$,\\\\\n(2) When $\\angle QBA=90^\\circ$, it holds that $BQ^2+AB^2=AQ^2$,\\\\\n$\\therefore 1^2+(3+a)^2+3^2+3^2=2^2+a^2$,\\\\\nSolving this, we find: $a=-4$,\\\\\n$\\therefore Q_2(-1,-4)$,\\\\\n(3) When $\\angle AQB=90^\\circ$, it holds that $BQ^2+AQ^2=AB^2$,\\\\\n$\\therefore 1^2+(3+a)^2+2^2+a^2=3^2+3^2$,\\\\\nSolving this, we find: $a_1=\\frac{-3+\\sqrt{17}}{2}$ or $a_1=\\frac{-3-\\sqrt{17}}{2}$,\\\\\n$\\therefore Q_3(-1,\\frac{-3+\\sqrt{17}}{2})$, $Q_4(-1,\\frac{-3-\\sqrt{17}}{2})$,\\\\\nTo conclude: the coordinates of point $Q$ are $\\boxed{(-1,2)}$, $\\boxed{(-1,-4)}$, $\\boxed{(-1,\\frac{-3+\\sqrt{17}}{2})}$, or $\\boxed{(-1,\\frac{-3-\\sqrt{17}}{2})}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51402562_52.png"
  },
  {
    "solution": "\\boxed{\\frac{9}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, rectangle $ABCD \\sim $ rectangle $BCEF$. If $AB=8$ and $BC=6$, what is the value of $CE$?\n",
    "completion": "\\textbf{Solution:} Since rectangle $ABCD \\sim$ rectangle $BCEF$, \\\\\ntherefore $\\frac{AB}{BC} = \\frac{BC}{CE}$, \\\\\nthus $CE = \\frac{BC^2}{AB} = \\frac{6^2}{8} = \\boxed{\\frac{9}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52872176.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the side length of the square cardboard ABCD is 4, with E and F being the midpoints of AB and BC, respectively. If it is cut along the dashed line as shown in the lower left figure and assembled into a \"small villa\" as shown below, what is the area of the shaded part?\n",
    "completion": "\\textbf{Solution:} According to the diagram, the top of the \"small villa\" is an isosceles triangle, whose area is half of the square ABCD's area, and the area of the \"small villa's\" bottom is half of the square ABCD's area as well, and the bottom consists of two equal rectangles.\\\\\nTherefore, the area of the shaded part in the diagram is $\\frac{1}{4}$ of the square ABCD's area,\\\\\ni.e., the area of the shaded part = $\\frac{1}{4}S_{\\text{square ABCD}}=\\frac{1}{4}\\times 4^{2}=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51614617.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\angle AOB=90^\\circ$, $OM$ bisects $\\angle AOC$, and $ON$ bisects $\\angle BOC$. Let $\\angle BOC=(2x)^\\circ$, find the measure of $\\angle MON$?\n",
    "completion": "\\textbf{Solution:} Given that, $\\angle AOB=90^\\circ$, \\\\\n$\\therefore \\angle AOC=\\angle AOB+\\angle BOC=90^\\circ+(2x)^\\circ$, \\\\\n$\\because OM$ bisects $\\angle AOC$ and $ON$ bisects $\\angle BOC$,\\\\\n$\\therefore \\angle MOC=\\frac{1}{2}\\angle AOC=45^\\circ+x^\\circ$, $\\angle NOC=\\frac{1}{2}\\angle BOC=\\frac{1}{2}\\times (2x)^\\circ=x^\\circ$, \\\\\n$\\therefore \\angle MON=\\angle MOC-\\angle NOC=\\boxed{45^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53365775_79.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{7}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, within quadrilateral $ABCD$, $AC \\perp AD$. Construct $\\angle ECA = \\angle ACD$, where $CE$ intersects $AB$ at point $O$ and extends to intersect the extended line of $DA$ at point $E$, and connect $BE$. Quadrilateral $ACBE$ is a rectangle; connect $OD$. If $AB=4$ and $\\angle ACD=60^\\circ$, what is the length of $OD$?\n",
    "completion": "\\textbf{Solution:} Draw $OF \\perp AE$ at point F. In $\\triangle ACD$, $\\angle ADC = 90^\\circ - \\angle ACD = 90^\\circ - 60^\\circ = 30^\\circ$,\\\\\nsince parallelogram ABCD,\\\\\ntherefore $\\angle ADC = \\angle ABC = 30^\\circ$,\\\\\nsince rectangle ABCD,\\\\\ntherefore $AO = CO = 2$, $\\angle ACB = 90^\\circ$,\\\\\ntherefore $\\angle BAC = 90^\\circ - \\angle ABC = 90^\\circ - 30^\\circ = 60^\\circ$,\\\\\ntherefore $\\triangle AOC$ is an equilateral triangle,\\\\\ntherefore $AC = 2$;\\\\\ntherefore $BC = AE = AD = \\sqrt{AB^2 - AC^2} = \\sqrt{4^2 - 2^2} = 2\\sqrt{3}$,\\\\\nsince $AC \\perp AE$,\\\\\ntherefore $OF \\parallel AC$,\\\\\ntherefore $OF$ is the midline of $\\triangle ACE$,\\\\\ntherefore $OF = \\frac{1}{2}AC = 1$, AF$=\\frac{1}{2}AE=\\sqrt{3}$;\\\\\ntherefore $DF = AF + AD = 2\\sqrt{3} + \\sqrt{3} = 3\\sqrt{3}$,\\\\\nin $\\triangle OFD$,\\\\\n$OD=\\sqrt{OF^2+DF^2}=\\sqrt{1^2+(3\\sqrt{3})^2}=\\boxed{2\\sqrt{7}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52188291_113.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the diameter $AB=6$ of circle $\\odot O$, with chord $CD$ of $\\odot O$ perpendicular to $AB$ at point $P$, and $BP:AP=1:5$, what is the length of $CD$?\n",
    "completion": "\\textbf{Solution:} Given that the diameter $AB$ of the circle $\\odot O$ equals 6,\\\\\nit follows that $OB=\\frac{1}{2}AB=3$,\\\\\nsince the ratio $BP:AP=1:5$,\\\\\nwe have $BP=\\frac{1}{6}AB=\\frac{1}{6} \\times 6=1$,\\\\\nthus $OP=OB-BP=3-1=2$,\\\\\nby joining $OC$,\\\\\nand since $CD\\perp AB$,\\\\\nit implies $CD=2PC$, and $\\angle OPC=90^\\circ$,\\\\\ntherefore, $PC=\\sqrt{OC^{2}-OP^{2}}=\\sqrt{3^{2}-2^{2}}=\\sqrt{5}$,\\\\\nresulting in $CD=2PC=2\\sqrt{5}$.\\\\\nHence, the length of $CD$ is $\\boxed{2\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51318298.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle AOB$ is an isosceles right triangle. $AN \\perp$ the y-axis, with the foot of the perpendicular being N, and $BM \\perp$ the y-axis, with the foot of the perpendicular being point M. Point P is the midpoint of AB. Connect PM and find the degree measure of $\\angle PMO$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, we extend $MP$ to intersect $AN$ at $H$,\\\\\nsince $AH \\perp y$-axis, and $BM \\perp y$-axis,\\\\\nit follows that $BM \\parallel AN$,\\\\\nhence, $\\angle MBP = \\angle HAP$, and $\\angle AHP = \\angle BMP$,\\\\\nsince point $P$ is the midpoint of $AB$,\\\\\nit implies that $AP = BP$,\\\\\ntherefore, $\\triangle APH \\cong \\triangle BPM$ (AAS),\\\\\nthus, $AH = BM$,\\\\\nsince the coordinates of point $A$ are $(-4,1)$, and the coordinates of point $B$ are $(1,4)$,\\\\\nit follows that $AN = 4$, $OM = 4$, $BM = 1$, $ON = 1$,\\\\\nthus, $HN = AN - AH = AN - BM = 3$, $MN = OM - ON = 3$,\\\\\nhence, $HN = MN$,\\\\\ntherefore, $\\angle NHM = \\angle NMH = 45^\\circ$, that is $\\angle PMO = \\boxed{45^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51368165_110.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $BC=\\sqrt{6}+\\sqrt{2}$, $\\angle C=45^\\circ$, $AB=\\sqrt{2}AC$, what is the length of $AC$?\n",
    "completion": "\\textbf{Solution:} Draw line AD perpendicular to BC through point A, with D being the foot of the perpendicular, as shown in the diagram.\\\\\nLet AC = x, then AB = $\\sqrt{2}x$.\\\\\nIn $\\triangle ACD$, AD = AC$\\cdot\\sin C = \\frac{\\sqrt{2}}{2}x$,\\\\\nCD = AC$\\cdot\\cos C = \\frac{\\sqrt{2}}{2}x$;\\\\\nIn $\\triangle ABD$, AB = $\\sqrt{2}x$, AD = $\\frac{\\sqrt{2}}{2}x$,\\\\\n$\\therefore$BD = $\\sqrt{AB^{2}-AD^{2}} = \\frac{\\sqrt{6}}{2}x$,\\\\\n$\\therefore$BC = BD + CD = $\\frac{\\sqrt{6}}{2}x + \\frac{\\sqrt{2}}{2}x = \\sqrt{6} + \\sqrt{2}$,\\\\\n$\\therefore x = \\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51446652.png"
  },
  {
    "solution": "\\boxed{1\\colon 3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, there is an isosceles right-angled triangle board, its hypotenuse $BC=8\\, \\text{cm}$. Inside it, $\\triangle DEF$ has its sides parallel to the sides of $OABC$ respectively, and its hypotenuse $EF=4\\, \\text{cm}$. What is the ratio of the area of $\\triangle DEF$ to the area of the shaded part?\n",
    "completion": "\\textbf{Solution.} Given by the problem, we have: $\\frac{S_{\\triangle DEF}}{S_{\\triangle ABC}}=\\left(\\frac{EF}{BC}\\right)^2=\\left(\\frac{4}{8}\\right)^2=\\frac{1}{4}$,\\\\\n$\\therefore$ $\\frac{S_{\\triangle DEF}}{S_{\\text{shaded area}}}=\\frac{1}{3}$,\\\\\nthat is, the ratio of the area of $\\triangle DEF$ to the area of the shaded portion is $\\boxed{1\\colon 3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52976990.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, point $D$ is the midpoint of the hypotenuse $AB$, and $DE$ bisects $\\angle ADC$, $BC=4$. What is the length of $DE$?\n",
    "completion": "Solution: Since $\\triangle ABC$ is a right-angled triangle and point D is the midpoint of the hypotenuse AB, \\\\\n$\\therefore CD=AD$, meaning $\\triangle ACD$ is an isosceles triangle, \\\\\nSince $DE$ bisects $\\angle ADC$ and $\\triangle ACD$ is an isosceles triangle, \\\\\n$\\therefore$ point E is the midpoint of AC, \\\\\nSince $BC=4$, \\\\\n$\\therefore DE=\\frac{1}{2}BC=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52387151.png"
  },
  {
    "solution": "\\boxed{50^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, place the right-angle vertex of the triangular cardboard on one side of the ruler. If $\\angle 1=20^\\circ$ and $\\angle 3=30^\\circ$, what is the value of $\\angle 2$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, \\\\\n\\(\\because\\) \\(\\angle 1=20^\\circ\\), \\(\\angle 3=30^\\circ\\), \\\\\n\\(\\therefore\\) \\(\\angle 4=\\angle 1+\\angle 3=20^\\circ+30^\\circ=50^\\circ\\), \\\\\n\\(\\because\\) The two sides of the straightedge are parallel to each other, \\\\\n\\(\\therefore\\) \\(\\angle 2=\\angle 4= \\boxed{50^\\circ}\\).",
    "image_path": "dataset/mm_math/images/MM_Math/52095901.png"
  },
  {
    "solution": "\\boxed{(6,0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a rectangular paper ABCO is placed in the Cartesian coordinate system, with BC = 10. After folding the paper, point B exactly falls on the x-axis, denoted as $B'$, and the fold line is CE. Given that the ratio $OC:OB'=4:3$, what are the coordinates of point $B'$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral OABC is a rectangle, $\\angle$AOC$=90^\\circ$. Given OC$\\colon$ OB'$=4\\colon 3$, it follows that B$^{\\prime }$C$\\colon$ OB$^{\\prime }$=$5\\colon 3$. Given B$^{\\prime }$C=BC=10, it implies OB$^{\\prime }$=6. Therefore, the coordinates of point B$^{\\prime}$ are: $\\boxed{(6,0)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50677242_153.png"
  },
  {
    "solution": "\\boxed{1-\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, it is known that the parabola $y=x^2-2x-3$ has its vertex at point $A$, intersects the $y$-axis at point $C$, and the line segment $CB\\parallel x$-axis intersects the parabola at another point $B$. When the independent variable $x$ of the quadratic function $y=x^2-2x-3$ satisfies $m\\le x\\le m+2$, the maximum value of this function is $p$, the minimum value is $q$, and $p-q=2$. What is the value of $m$?\n",
    "completion": "\\textbf{Solution:} Given the problem, the axis of symmetry of the parabola $y=x^2-2x-3$ is the line $x=1$,\n\n(1) When $m>1$, \\\\\nat $x=m$, we have $q=m^2-2m-3$, \\\\\nat $x=m+2$, we have $p=(m+2)^2-2(m+2)-3$, \\\\\n$\\therefore p-q=(m+2)^2-2(m+2)-3-m^2+2m+3=2$, \\\\\nsolving this gives $m=\\frac{1}{2}$ (discard);\n\n(2) When $m+2<1$, i.e., $m<-1$, \\\\\nat $x=m$, we have $p=m^2-2m-3$, \\\\\nat $x=m+2$, we have $q=(m+2)^2-2(m+2)-3$, \\\\\n$\\therefore p-q=m^2-2m-3-(m+2)^2+2(m+2)+3=2$, \\\\\nsolving this gives $m=-\\frac{1}{2}$ (discard);\n\n(3) When $m\\le 1\\le m+1$, i.e., $0\\le m\\le 1$, \\\\\nat $x=1$, we have $q=-4$, \\\\\nat $x=m+2$, we have $p=(m+2)^2-2(m+2)-3$, \\\\\n$\\therefore p-q=(m+2)^2-2(m+2)-3+4=2$, \\\\\nsolving this gives $m=\\sqrt{2}-1$ or $m=-\\sqrt{2}-1$ (discard);\n\n(4) When $m+1<1\\le m+2$, i.e., $-1\\le m<0$, \\\\\nat $x=1$, we have $q=-4$, \\\\\nat $x=m$, we have $p=m^2-2m-3$, \\\\\n$\\therefore p-q=m^2-2m-3+4=2$, \\\\\nsolving this gives $m=\\sqrt{2}+1$ (discard) or $m=-\\sqrt{2}+1$,\n\nTo conclude: The values for $m$ are $m=\\boxed{\\sqrt{2}-1}$ or $m=\\boxed{1-\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52512802_170.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $BC= \\frac{1}{2} AB$, and $D$ is the midpoint of $AC$. If $DB=1$, what is the length of $AB$?\n",
    "completion": "\\textbf{Solution:} Given $\\because BC= \\frac{1}{2} AB$,\\\\\n$\\therefore AC=AB+BC= \\frac{3}{2}AB$,\\\\\n$\\because D$ is the midpoint of $AC$,\\\\\n$\\therefore CD= \\frac{1}{2}AC=\\frac{1}{2}\\times \\frac{3}{2}AB=\\frac{3}{4}AB$,\\\\\n$\\therefore DB=CD-BC= \\frac{3}{4}AB-\\frac{1}{2}AB=\\frac{1}{4}AB$,\\\\\nwhich means $\\frac{1}{4}AB=1$,\\\\\n$\\therefore AB=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/40552727.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AD \\perp BC$, $CE \\perp AB$, with the feet of the perpendiculars being $D$ and $E$ respectively, $AD$ and $CE$ intersect at point $H$. Given that $AE=CE=5$ and $CH=2$, what is the length of $BE$?\n",
    "completion": "\\textbf{Solution:} Since $AD \\perp BC$, $CE \\perp AB$,\\\\\ntherefore $\\angle AEH = \\angle BEC = \\angle HDC = 90^\\circ$,\\\\\nthus $\\angle EAH + \\angle EHA = \\angle ECB + \\angle DHC = 90^\\circ$\\\\\nSince $\\angle EHA = \\angle DHC$,\\\\\nthus $\\angle EAH = \\angle ECB$,\\\\\nIn $\\triangle AEH$ and $\\triangle CEB$,\\\\\n$\\left\\{ \\begin{array}{l} \\angle EAH = \\angle ECB \\\\ CE = AE \\\\ \\angle AEH = \\angle BEC \\end{array} \\right.$,\\\\\ntherefore, $\\triangle AEH \\cong \\triangle CEB$ (ASA),\\\\\nthus $BE = EH = CE - CH = 5 - 2 = \\boxed{3}$,\\\\",
    "image_path": "dataset/mm_math/images/MM_Math/52653621.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $C$ and $D$ are points on the circle $\\odot O$ and are located on either side of the diameter $AB$, with $BC=2\\sqrt{3}$ and $\\angle ABC=30^\\circ$. When $CD\\perp AB$ at point $E$, what is the length of $CD$?\n",
    "completion": "\\textbf{Solution:} Given that $CD \\perp AB$ and $AB$ is the diameter of the circle,\\\\\ntherefore $CD = 2CE$ and $\\angle CEB = 90^\\circ$. Also, given that $\\angle ABC = 30^\\circ$,\\\\\nthus $BC = 2CE$,\\\\\ntherefore $CD = BC = 2\\sqrt{3}$;\\\\\nHence, the length of $CD$ is $\\boxed{2\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52991389_144.png"
  },
  {
    "solution": "\\boxed{(6,4) \\text{ or } (3,3)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, the line $y=-\\frac{1}{2}x+2$ intersects the $x$ axis at point $A$ and the $y$ axis at point $B$. Point $P$ lies on the line $y=\\frac{1}{3}x+b$ that passes through point $B$. What are the coordinates of point $P$ when $\\triangle PAB$ is an isosceles right triangle?\n",
    "completion": "\\textbf{Solution:} Solve: For $y=-\\frac{1}{2}x+2$, let $x=0$, then $y=2$,\\\\\nlet $y=0$, then $-\\frac{1}{2}x+2=0$, solve to get: $x=4$,\\\\\n$\\therefore$ point A $(4,0)$, B $(0,2)$,\\\\\n$\\therefore$ $OB=2$, $OA=4$,\\\\\nSubstituting point B $(0,2)$ into $y=\\frac{1}{3}x+b$, we get: $b=2$,\\\\\n$\\therefore$ the equation of line PB is $y=\\frac{1}{3}x+2$,\\\\\nAccording to the question, $\\angle PBA \\neq 90^\\circ$,\\\\\n(1) When $\\angle BAP'=90^\\circ$ and $AB=AP'$, draw $AP'\\perp AB$ at A, draw $P'H'\\perp$ axis through $P'$,\\\\\n$\\therefore$ $\\angle AOB=\\angle P'H'A=90^\\circ$, $\\angle OAB+\\angle P'AH'=90^\\circ$,\\\\\n$\\because$ $\\angle OAB+\\angle OBA=90^\\circ$,\\\\\n$\\therefore$ $\\angle OBA=\\angle P'AH'$,\\\\\nAnd $AB=AP'$,\\\\\n$\\therefore$ $\\triangle AOB \\cong \\triangle P'AH'$ (AAS),\\\\\n$\\therefore$ $AH'=OB=2$, $P'H'=OA=4$,\\\\\n$\\therefore$ $OH'=OA+AH'=6$,\\\\\n$\\therefore$ $P'(6,4)$,\\\\\nSubstitute $x=6$ into $y=\\frac{1}{3}x+2$, get $y=4$,\\\\\n$\\therefore$ point $P'$ is on the line $y=\\frac{1}{3}x+2$, which is consistent with the question.\\\\\n(2) When $\\angle BPA=90^\\circ$ and $BP=AP$, draw $AP\\perp BP$ at point $P$, draw $PH\\perp y$ axis through $P$, and draw $PQ\\perp x$ axis through $P$,\\\\\n$\\therefore$ $\\angle PHO=\\angle PQO=\\angle HOQ=90^\\circ$,\\\\\n$\\therefore$ quadrilateral $OHPQ$ is a rectangle,\\\\\n$\\therefore$ $PH=OQ$, $PQ=OH$, $\\angle HPB+\\angle BPQ=90^\\circ$,\\\\\n$\\because$ $\\angle APQ+\\angle BPQ=90^\\circ$,\\\\\n$\\therefore$ $\\angle HPB=\\angle APQ$,\\\\\nAnd $\\because$ $BP=AP$,\\\\\n$\\therefore$ $\\triangle HBP \\cong \\triangle QAP$ (AAS),\\\\\n$\\therefore$ $HP=PQ$, $HB=QA$,\\\\\n$\\therefore$ quadrilateral $OHPQ$ is a square,\\\\\n$\\because$ $OH+OQ=(OB+HB)+OQ=OB+AQ+OQ=OB+(AQ+OQ)=OB+OA=4+2=6$,\\\\\n$\\therefore$ $PH=PQ=3$,\\\\\n$\\therefore$ $P(3,3)$,\\\\\nSubstitute $x=3$ into $y=\\frac{1}{3}x+2$ to get: $y=3$,\\\\\n$\\therefore$ point $P$ is on the line $y=\\frac{1}{3}x+2$, which is consistent with the question.\\\\\nIn summary, the coordinates of point $P$ are $(6,4)$ or $(3,3)$.\\\\\nTherefore, the answer is: $\\boxed{(6,4) \\text{ or } (3,3)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52303856.png"
  },
  {
    "solution": "\\boxed{x>0}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=kx+b$ intersects the coordinate axes at points $A(2, 0)$ and $B(0, -3)$, respectively. What is the solution set of the inequality $kx+b+3>0$?\n",
    "completion": "Solution: Since the line $y=kx+b$ intersects the axes at two points, $A(2,0)$ and $B(0,-3)$, we have:\n\\[\n\\begin{cases}\n2k + b = 0 \\\\\nb = -3\n\\end{cases}\n\\]\nThus,\n\\[\n\\begin{cases}\nk=\\frac{3}{2} \\\\\nb=-3\n\\end{cases}\n\\]\nTherefore, $\\frac{3}{2}x+(-3)+3=\\frac{3}{2}x>0$,\\\\\nImplying $x>0$.\\\\\nHence, the answer is $\\boxed{x>0}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51699530.png"
  },
  {
    "solution": "\\boxed{150^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the quadrilateral $ABCD$ is inscribed in circle $\\odot O$, with $\\angle BCD = 105^\\circ$. What is the measure of $\\angle BOD$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral ABCD is inscribed in circle O,\\\\\nit follows that $\\angle A + \\angle BCD = 180^\\circ$,\\\\\ntherefore $\\angle A = 180^\\circ - \\angle BCD = 180^\\circ - 105^\\circ = 75^\\circ$,\\\\\nhence $\\angle BOD = 2\\angle A = 2 \\times 75^\\circ = \\boxed{150^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52949310.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ is an equilateral triangle with $AB=3$. Point $D$ is a moving point on side $BC$, $\\angle ADE=60^\\circ$, and $DE$ intersects $AC$ at point $E$. What is the maximum value of the length of segment $CE$?\n",
    "completion": "\\textbf{Solution:}\n\nSince $\\triangle ABC$ is an equilateral triangle, \\\\\ntherefore, AB=BC=3, $\\angle B=\\angle C=60^\\circ$, \\\\\nsince $\\angle ADC=\\angle ADE+\\angle EDC=\\angle B+\\angle BAD$, $\\angle ADE=60^\\circ$, \\\\\ntherefore, $\\angle BAD=\\angle CDE$, \\\\\nthus, $\\triangle ABD\\sim \\triangle DCE$, \\\\\nso, AB$\\colon$CD=BD$\\colon$CE, \\\\\nlet BD=x, thus CD=3−x, \\\\\nso, 3$\\colon$(3−x)=x$\\colon$CE, \\\\\nthus, CE$=-\\frac{1}{3}x^2+x=-\\frac{1}{3}\\left(x-\\frac{3}{2}\\right)^2+\\frac{3}{4}$ \\\\\nSince $a=-\\frac{1}{3}<0$, the graph opens downward, \\\\\nthus, the maximum value of CE is $\\boxed{\\frac{3}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52874571.png"
  },
  {
    "solution": "\\boxed{\\left(0, \\frac{5}{2}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given the graph of the function $y=-\\frac{1}{2}x+b$ intersects the x-axis and y-axis at points A and B, respectively, and intersects the graph of the function $y=2x$ at point M(2, 4); there is a moving point P on the x-axis, through which a perpendicular line to the x-axis is drawn, intersecting the graphs of the functions $y=-\\frac{1}{2}x+b$ and $y=2x$ at points C and D, respectively. Is there a point E on the y-axis such that $\\triangle OEM$ is an isosceles triangle? If it exists, find the coordinates of point E; if not, explain why.\n",
    "completion": "\\textbf{Solution:} Let $E(0, m)$, since $M(2, 4)$. Therefore, $OM^2 = 2^2 + 4^2 = 20$, that is: $OE^2 = m^2$, $EM^2 = 2^2 + (m-4)^2 = m^2 - 8m + 20$,\n\\begin{enumerate}\n    \\item When $OE=OM$, $OE^2 = OM^2$, therefore $m^2 = 20$. Solving this gives $m = \\pm\\sqrt{20}$, therefore, the coordinates of point E are $(0, \\sqrt{20})$ or $(0, -\\sqrt{20})$;\n    \\item When $OE=EM$, $OE^2 = EM^2$, therefore $m^2 = m^2 - 8m + 20$. Solving this gives $m = \\frac{5}{2}$, therefore, the coordinates of point E are $\\left(0, \\frac{5}{2}\\right)$;\n    \\item When $OM=EM$, draw $MF \\perp OE$, where point F is the midpoint of OE, \\\\\n    $OF = \\sqrt{20 - 4} = 4$, $OE = 8$. \\\\\n    When E coincides with O (which is not according to the question, discard) \\\\\n    (Or: When $OM=EM$, $OM^2 = EM^2$, therefore $20 = m^2 - 8m + 20$, solving this gives $m = 8$ or $0$ (discard), \\\\\n    therefore, the coordinates of point E are $(0, 8)$);\n\\end{enumerate}\nHence, the coordinates of point E are $\\boxed{(0, \\sqrt{20})}$, $\\boxed{(0, -\\sqrt{20})}$, $\\boxed{\\left(0, \\frac{5}{2}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52251693_83.png"
  },
  {
    "solution": "\\boxed{35^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\angle B = \\angle C = 90^\\circ$, $M$ is the midpoint of $BC$, $DM$ bisects $\\angle ADC$, and $\\angle ADC = 110^\\circ$, find the degree measure of $\\angle MAB$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $MN \\perp AD$ at N,\\\\\n$\\because$ $\\angle B = \\angle C = 90^\\circ$,\\\\\n$\\therefore$ $AB \\parallel CD$,\\\\\n$\\therefore$ $\\angle DAB = 180^\\circ - \\angle ADC = 70^\\circ$,\\\\\n$\\because$ $DM$ bisects $\\angle ADC$, $MN \\perp AD$, $MC \\perp CD$,\\\\\n$\\therefore$ $\\angle DMN = \\angle DMC$, $\\angle DNM = \\angle DCM = 90^\\circ$,\\\\\nIn $\\triangle DMN$ and $\\triangle DMC$,\\\\\n$\\left\\{\\begin{array}{l}\n\\angle DMN = \\angle DMC \\\\\n\\angle DNM = \\angle DCM \\\\\nDM = DM\n\\end{array}\\right.$\\\\\n$\\therefore$ $\\triangle DMN \\cong \\triangle DMC$ (AAS),\\\\\n$\\therefore$ $MN = MC$.\\\\\n$\\because$ M is the midpoint of $BC$,\\\\\n$\\therefore$ $MC = MB$,\\\\\n$\\therefore$ $MN = MB$,\\\\\nAnd since $MN \\perp AD$, $MB \\perp AB$,\\\\\n$\\therefore$ $\\angle MAB = \\frac{1}{2}\\angle DAB = \\boxed{35^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52978477.png"
  },
  {
    "solution": "\\boxed{3\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\triangle ABC$, $ \\angle ACB=90^\\circ$, $AC=BC$, points D and E are on AB and AC respectively. CD and ED are connected, $ED=CD$, $\\tan\\angle ADE=\\frac{1}{3}$, $BD=\\sqrt{2}$. What is the length of $AB$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $EH \\perp AB$ at H, and draw $DM \\perp AC$ at M, and draw $DG \\perp BC$ at G, \\\\\n$\\because \\angle ACB=90^\\circ$ and $AC=BC$, \\\\\n$\\therefore$ quadrilateral $CGDM$ is a rectangle, $\\angle A=\\angle B=45^\\circ$, \\\\\n$\\therefore DG=CM$, \\\\\n$\\because BD=\\sqrt{2}$, \\\\\n$\\therefore DG=BG=BD\\cdot \\sin 45^\\circ=\\sqrt{2}\\times \\frac{\\sqrt{2}}{2}=1$, \\\\\n$\\therefore CM=1$, \\\\\n$\\because DE=DC$ and $DM\\perp AC$, \\\\\n$\\therefore EM=CM=1$, \\\\\n$\\because$ $\\tan \\angle ADE=\\frac{1}{3}=\\frac{EH}{DH}$, \\\\\nLet $EH=x$, then $DH=3x$, \\\\\n$\\because \\angle A=45^\\circ$ and $EH\\perp AH$, \\\\\n$\\therefore AH=EH=x$, \\\\\nFrom $\\sin A=\\sin 45^\\circ=\\frac{AH}{AE}=\\frac{\\sqrt{2}}{2}$, \\\\\n$\\therefore AE=\\sqrt{2}x$, \\\\\nSimilarly: $\\cos A=\\cos 45^\\circ=\\frac{AC}{AB}=\\frac{\\sqrt{2}}{2}$, \\\\\n$\\therefore AB=\\sqrt{2}AC$, \\\\\n$\\therefore x+3x+\\sqrt{2}=\\sqrt{2}\\left(\\sqrt{2}x+2\\right)$, \\\\\nSolving gives: $x=\\frac{\\sqrt{2}}{2}$, \\\\\n$\\therefore AB=4x+\\sqrt{2}=4\\times \\frac{\\sqrt{2}}{2}+\\sqrt{2}=\\boxed{3\\sqrt{2}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51318308.png"
  },
  {
    "solution": "\\boxed{\\sqrt{13}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $AB=3$, $AC=1$, and $\\angle D=90^\\circ$. If $\\triangle DEC$ is centrally symmetric to $\\triangle ABC$ with respect to point $C$, what is the length of $AE$?\n",
    "completion": "\\textbf{Solution:} Given that $\\triangle DEC$ and $\\triangle ABC$ are centrally symmetric with respect to point C,\\\\\ntherefore $\\triangle DEC \\cong \\triangle ABC$,\\\\\nsince $AB=3$ and $AC=1$,\\\\\nthen $DE=3$ and $CD=1$,\\\\\nthus $AD=2$,\\\\\nalso, since $\\angle D=90^\\circ$,\\\\\nit follows that $AE= \\sqrt{DE^2+AD^2}=\\sqrt{3^2+2^2}=\\boxed{\\sqrt{13}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51827844.png"
  },
  {
    "solution": "\\boxed{\\frac{5}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the diagonals \\(AC\\) and \\(BD\\) of the rhombus \\(ABCD\\) intersect at point \\(O\\), where \\(AC=8\\) and \\(BD=6\\). If point \\(E\\) is the midpoint of \\(BC\\), what is the length of \\(OE\\)?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rhombus,\\\\\nthen OC=OA=$\\frac{1}{2}AC=4$ and OB=OD=$\\frac{1}{2}BD=3$, $\\angle$BOC=$90^\\circ$.\\\\\nThus, AB=$\\sqrt{OA^{2}+OB^{2}}=5$.\\\\\nSince point E is the midpoint of AB, and O is the midpoint of AC,\\\\\nthen OE is the median of $\\triangle ABC$,\\\\\nthus OE=$\\frac{1}{2}AB=\\boxed{\\frac{5}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52344345.png"
  },
  {
    "solution": "\\boxed{\\frac{8\\pi }{3}-4\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of $\\odot O$, point $C$ is a point on $\\odot O$, connect $BC$ and $AC$. Point $E$ is the midpoint of $BC$, connect and extend $OE$ to intersect the circle at point $D$. $OD\\parallel AC$. If $DE=2$ and $BE=2\\sqrt{3}$, find the area of the shaded region.\n",
    "completion": "\\textbf{Solution:} Connect OC, and set the radius of the circle as $r$, then $OE=r-2$,\\\\\n$\\because O{E}^{2}+B{E}^{2}=O{B}^{2}$,\\\\\n$\\therefore {\\left(r-2\\right)}^{2}+{\\left(2\\sqrt{3}\\right)}^{2}={r}^{2}$,\\\\\nSolving this, we get $r=4$,\\\\\n$\\therefore OE=2=\\frac{1}{2}OB$,\\\\\n$\\therefore \\angle ABC=30^\\circ$,\\\\\n$\\therefore \\angle COA=60^\\circ$,\\\\\nFrom (1), we have $BE=CE=2\\sqrt{3}$,\\\\\n$\\therefore BC=4\\sqrt{3}$,\\\\\n$\\therefore {S}_{\\triangle BOC}=\\frac{1}{2}OE\\cdot BC=4\\sqrt{3}$,\\\\\n$\\because \\triangle BOC$ and $\\triangle AOC$ have the same base and height,\\\\\n$\\therefore {S}_{\\triangle AOC}={S}_{\\triangle BOC}=4\\sqrt{3}$,\\\\\n$\\therefore {S}_{\\text{shaded}}={S}_{\\text{sector} AOC}-{S}_{\\triangle AOC}=\\frac{60^\\circ \\cdot \\pi \\times {4}^{2}}{360}-4\\sqrt{3}=\\boxed{\\frac{8\\pi }{3}-4\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53065830_109.png"
  },
  {
    "solution": "\\boxed{24}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $C$ is a point on line segment $AB$, and $P$, $Q$ are the midpoints of line segments $AC$ and $BC$ respectively. If $PQ = 12$, find the length of $AB$.\n",
    "completion": "\\textbf{Solution:} Since $P$ and $Q$ are the midpoints of line segments $AC$ and $BC$ respectively,\\\\\nit follows that $AP=PC=\\frac{1}{2}AC$ and $CQ=QB=\\frac{1}{2}CB$,\\\\\nsince $PQ=PC+CQ=\\frac{1}{2}AC+\\frac{1}{2}CB=\\frac{1}{2}AB=12$,\\\\\nthus $AB=\\boxed{24}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53205929.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of circle $O$, and points $C$ and $D$ are on circle $O$, satisfying: point $C$ is the midpoint of arc $AD$, and $AD$ intersects $OC$ at point $E$. Given $AD=8$ and $EC=2$, what is the radius of circle $O$?\n",
    "completion": "\\textbf{Solution:} Since point C is the midpoint of arc AD,\\\\\nit follows that $OC\\perp AD$,\\\\\nthus $AE=\\frac{1}{2}AD=4$,\\\\\nLet the radius of circle O be $r$, then $OA=OC=r, OE=r-2$,\\\\\nSince in $\\triangle AOE$, $OE^2+AE^2=OA^2$,\\\\\nit follows that $(r-2)^2+4^2=r^2$,\\\\\nSolving for $r$ gives $r=\\boxed{5}$,\\\\\ntherefore, the radius of circle O is \\boxed{5}.",
    "image_path": "dataset/mm_math/images/MM_Math/53148601_133.png"
  },
  {
    "solution": "\\boxed{18^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $\\angle AOC : \\angle BOC = 1 : 5$. OD is the angle bisector of $\\angle AOB$, and $\\angle COD = 36^\\circ$. What is the degree measure of $\\angle AOC$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle AOC : \\angle BOC = 1 : 5$,\\\\\nlet $\\angle AOC = x$, then $\\angle BOC = 5x$,\\\\\nthus $\\angle AOB = \\angle AOC + \\angle BOC = x + 5x = 6x$,\\\\\nsince OD bisects $\\angle AOB$,\\\\\nthus $\\angle AOD = \\frac{1}{2} \\angle AOB = 3x$,\\\\\nthus $\\angle COD = \\angle AOD - \\angle AOC$,\\\\\nthus $3x - x = 36^\\circ$,\\\\\nsolving this gives $x = 18^\\circ$,\\\\\nthus $\\angle AOC = \\boxed{18^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53124841.png"
  },
  {
    "solution": "\\boxed{3\\sqrt{10}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the quadrilateral $ABCD$ is inscribed in circle $\\odot O$, with the diagonal $AC$ being the diameter of $\\odot O$. Through point $C$, draw $CE \\perp AC$ to intersect the extension of line $AD$ at point $E$. Let $F$ be the midpoint of $CE$, and connect $DB$, $DF$. $DF$ is the tangent to $\\odot O$. If $\\tan\\angle ABD=3$, find the value of $\\frac{AC}{DE}$.\n",
    "completion": "\\textbf{Solution:} \\\\\nSince $\\angle ACD = \\angle ABD$, \\\\\nit follows that $\\tan \\angle ACD = \\frac{AD}{CD} = 3$, \\\\\nLet $AD=3$ and $CD=1$, \\\\\nthus $AC = \\sqrt{AD^2 + CD^2} = \\sqrt{10}$, \\\\\nSince $\\angle ACD + \\angle DCE = \\angle DEC + \\angle DCE$, \\\\\nit follows that $\\angle ACD = \\angle DEC$, \\\\\nthus $\\tan \\angle DEC = \\frac{DC}{DE} = 3$, \\\\\ntherefore $DE = \\frac{1}{3}$, \\\\\nhence $\\frac{AC}{DE} = \\frac{\\sqrt{10}}{\\frac{1}{3}} = \\boxed{3\\sqrt{10}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51010356_125.png"
  },
  {
    "solution": "\\boxed{26\\text{cm}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, assume $\\odot O$ is the cross-section of a water pipe, with the width of the water surface $AB=48\\,cm$, and the maximum depth of the water being $36\\,cm$. What is the radius of $\\odot O$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $OD\\perp AB$ at D, extend DO to intersect circle O at C, and connect AO,\\\\\nAccording to the problem, $AB=48\\text{cm}$, $CD=36\\text{cm}$,\\\\\n$\\therefore$ by the Perpendicular Diameter Theorem, $AD=\\frac{1}{2}AB=24$,\\\\\nLet $OA=OC=x\\text{cm}$, then $OD=CD-OD=(36-x)$,\\\\\n$\\because$ $AD^2+OD^2=AO^2$,\\\\\n$\\therefore$ $24^2+(36-x)^2=x^2$,\\\\\nSolving for $x$, we get $x=26$,\\\\\n$\\therefore$ the radius of circle O is $\\boxed{26\\text{cm}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51446274_124.png"
  },
  {
    "solution": "\\boxed{75^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, two right-angled triangles are stacked on point $O$ with two of their vertices overlapping, where the vertex $C$ of one triangle is on the side $OA$ of the other triangle. It is known that $\\angle ABO=\\angle DCO=90^\\circ$, $\\angle AOB=45^\\circ$, $\\angle COD=60^\\circ$. The bisector of $\\angle AOD$ intersects side $CD$ at point $E$. What is the degree measure of $\\angle BOE$?\n",
    "completion": "\\textbf{Solution:} Given that $\\because \\angle AOD=60^\\circ$, and $OE$ bisects $\\angle AOD$,\\\\\n$\\therefore \\angle AOE=\\frac{1}{2}\\angle AOD=30^\\circ$\\\\\n$\\because \\angle AOB=45^\\circ$\\\\\n$\\therefore \\angle BOE=\\angle AOE+\\angle AOB=\\boxed{75^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/51495039_153.png"
  },
  {
    "solution": "\\boxed{4\\text{cm}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the right triangle $ABC$, where $\\angle C=90^\\circ$, the bisector of $\\angle ABC$, $BD$, intersects $AC$ at $D$. If $CD = 4\\, \\text{cm}$, what is the distance from point $D$ to $AB$, denoted as $DE$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle C=90^\\circ$ and BD bisects $\\angle ABC$, with $DE \\perp AB$,\\\\\nit follows that $DE=CD$,\\\\\nSince $CD=4\\text{cm}$,\\\\\nthe distance from point D to AB, DE, is $\\boxed{4\\text{cm}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52754262.png"
  },
  {
    "solution": "\\boxed{71^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $\\angle AOD$ and $\\angle BOC$ are both right angles, $\\angle AOC = 38^\\circ$, and $OE$ bisects $\\angle BOD$. What is the measure of $\\angle COE$ in degrees?\n",
    "completion": "\\textbf{Solution}: Since $\\angle AOD$ is a right angle and $\\angle AOC=38^\\circ$,\\\\\nit follows that $\\angle COD=\\angle AOD-\\angle AOC=90^\\circ-38^\\circ=52^\\circ$.\\\\\nSince $\\angle BOC$ is also a right angle,\\\\\nwe have $\\angle BOD=\\angle BOC-\\angle COD=90^\\circ-52^\\circ=38^\\circ$.\\\\\nGiven that OE bisects $\\angle BOD$,\\\\\nit follows that $\\angle DOE= \\frac{1}{2} \\angle BOD= \\frac{1}{2} \\times 38^\\circ=19^\\circ$.\\\\\nTherefore, $\\angle COE=\\angle COD+\\angle DOE=52^\\circ+19^\\circ=\\boxed{71^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/10472128.png"
  },
  {
    "solution": "\\boxed{\\frac{7}{9}\\pi}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, $AB$ is the diameter of $\\odot O$, and $C$ is a point on $\\odot O$. If $\\angle BCO=35^\\circ$ and $AO=2$, then what is the length of $\\overset{\\frown}{AC}$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle BCO=35^\\circ$,\\\\\nit follows that $\\angle AOC=2\\angle BCO=70^\\circ$,\\\\\nand given that $AO=2$,\\\\\nthus, the arc $\\overset{\\frown}{AC}=\\frac{n\\pi r}{180}=\\frac{70\\pi \\times 2}{180}=\\frac{140\\pi }{180}=\\boxed{\\frac{7}{9}\\pi}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53019906.png"
  },
  {
    "solution": "\\boxed{\\frac{7}{8}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, quadrilateral $ABCD$ is a rectangle. Fold $\\triangle ACD$ along $AC$ to $\\triangle ACD^{\\prime}$. $AD^{\\prime}$ intersects $BC$ at point $E$. Given $AD=4$ and $DC=3$, find the length of $BE$.\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rectangle, \\\\\nit follows that AB=DC=3, BC=AD=4, AD$\\parallel$BC, and $\\angle$B=90$^\\circ$, \\\\\nSince $\\triangle$ACD is folded along AC to $\\triangle$ACD$^{\\prime}$, and AD$^{\\prime}$ intersects BC at point E, \\\\\nit follows that $\\angle$DAC=$\\angle$D$^{\\prime}$AC, \\\\\nSince AD$\\parallel$BC, it follows that $\\angle$DAC=$\\angle$ACB, \\\\\nhence $\\angle$D$^{\\prime}$AC=$\\angle$ACB, therefore AE=EC, \\\\\nLet BE=x, then EC=4−x, and AE=4−x, \\\\\nIn $\\triangle$ABE, since AB$^{2}$+BE$^{2}$=AE$^{2}$, \\\\\nit follows that 3$^{2}$+x$^{2}$=$(4−x)^{2}$, solving this gives x= $\\boxed{\\frac{7}{8}}$, \\\\\nthus, the length of BE is $\\frac{7}{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50482069.png"
  },
  {
    "solution": "\\boxed{36m}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, locations A and B are separated by a pond. Xiao Ming used the following method to measure the distance between A and B: He selected a point C outside line $AB$, then on lines $AC$ and $BC$ he found points M and N, respectively, such that $AM=2CM$ and $BN=2CN$. He measured the length of $MN$ to be $12m$. Based on this, what is the distance between A and B?\n",
    "completion": "\\textbf{Solution:} \\\\\nSince $AM=2CM$ and $BN=2CN$, \\\\\nit follows that $\\frac{AM}{CM}=2$ and $\\frac{BN}{CN}=2$, thus $\\frac{AC}{CM}=3$ \\\\\nBecause $\\angle ACB=\\angle MCN$, \\\\\nit implies that $\\triangle ACB\\sim \\triangle MCN$, \\\\\nhence $\\frac{AB}{MN}=3$, \\\\\nGiven that $MN=12m$, \\\\\nit follows that $AB=\\boxed{36m}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52796915.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC$, the perpendicular bisector $MN$ of $AB$ intersects $AC$ at point $D$ and intersects $AB$ at point $E$, $\\angle A=36^\\circ$, $AD=5$, find the degree measure of $\\angle DBC$ and the length of side $BC$.\n",
    "completion": "\\textbf{Solution:} Since in $\\triangle ABC$, $AB=AC$ and $\\angle A=36^\\circ$,\\\\\nit follows that $\\angle ABC=\\angle C=\\frac{180^\\circ-\\angle A}{2}=72^\\circ$,\\\\\nSince the perpendicular bisector of $AB$, $MN$, intersects $AC$ at point $D$,\\\\\nit follows that $AD=BD$,\\\\\nTherefore, $\\angle ABD=\\angle A=36^\\circ$,\\\\\nTherefore, $\\angle DBC=\\angle ABC-\\angle ABD=36^\\circ$,\\\\\nSince $\\angle BDC=\\angle ABD+\\angle A=72^\\circ$,\\\\\nit follows that $\\angle BDC=\\angle C=72^\\circ$,\\\\\nTherefore, $BC=BD$,\\\\\nAlso, since $AD=BD$,\\\\\nit follows that $BC=AD=\\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52827578.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{1}{2}(x-4)^2+8=-\\frac{1}{2}x^2+4x}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a small ball M is thrown from point O on slope $OA$. The trajectory of the ball is a part of a parabola. A Cartesian coordinate system is established as shown in the figure. The slope can be described by the linear function $y=\\frac{1}{2}x$. If the highest point reached by the ball has coordinates $(4, 8)$, find the expression for the parabola.\n",
    "completion": "\\textbf{Solution:} Given that the highest point the ball reaches is at the coordinates $(4, 8)$,\\\\\nwe can assume the expression of the parabola to be $y=a(x-4)^2+8$. Since the parabola passes through the origin,\\\\\nwe have $0=a(0-4)^2+8$, solving for $a$ yields: $a=-\\frac{1}{2}$,\\\\\nthus, the expression of the parabola is $\\boxed{y=-\\frac{1}{2}(x-4)^2+8=-\\frac{1}{2}x^2+4x}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52934216_234.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{10}}{10}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, if point A has the coordinates $(3,1)$, what is the value of $\\sin\\alpha$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram: draw AB $\\perp$ x-axis at point B,\\\\\n$\\because$ the coordinates of point A are (3, 1),\\\\\n$\\therefore$ OB = 3, and AB = 1,\\\\\nIn right triangle ABO, according to the Pythagorean theorem, AO = $\\sqrt{OB^{2}+AB^{2}}=\\sqrt{3^{2}+1^{2}}=\\sqrt{10}$,\\\\\n$\\therefore \\sin\\alpha = \\frac{AB}{OA}=\\frac{1}{\\sqrt{10}}=\\frac{\\sqrt{10}}{10}$,\\\\\nTherefore, $\\sin\\alpha = \\boxed{\\frac{\\sqrt{10}}{10}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51103840.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $E$, $F$, $G$, $H$ are points on the sides of the square $ABCD$, and $AG=BE=CH=DF$, $EF$ and $GH$ cut the square into four pieces which are then reassembled into the quadrilateral $MQPN$. If the ratio of the area of the square $ABCD$ to that of the quadrilateral $MQPN$ is $9\\colon 10$, then what is the ratio $AG\\colon GD$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, connect $EH, FG, GE, HF$,\n\\[\n\\because \\text{quadrilateral ABCD is a square},\n\\]\n\\[\n\\therefore \\angle A=\\angle B=\\angle C=\\angle D=90^\\circ, AB=BC=CD=DA.,\n\\]\n\\[\n\\because AG=BE=CH=DF,\n\\]\n\\[\n\\therefore DG=AE=BH=CF,\n\\]\n\\[\n\\therefore \\triangle AEG\\cong \\triangle BHE\\cong \\triangle CFH\\cong \\triangle DGF,\n\\]\n\\[\n\\therefore EG=FG=EH=HF,\n\\]\n\\[\n\\therefore \\text{quadrilateral EHFG is a rhombus},\n\\]\n\\[\n\\because \\triangle DGF\\cong \\triangle AEG,\n\\]\n\\[\n\\therefore \\angle DGF=\\angle AEG,\n\\]\n\\[\n\\because \\angle AEG+\\angle AGE=90^\\circ,\n\\]\n\\[\n\\therefore \\angle DGF+\\angle AGE=90^\\circ,\n\\]\n\\[\n\\therefore \\angle EGF=90^\\circ,\n\\]\n\\[\n\\therefore \\text{quadrilateral EHFG is a square},\n\\]\n\\[\n\\therefore GH\\perp EF,\n\\]\nCombining the above, it is known that both the quadrilateral MQPN and the quadrilateral A'B'C'D' are squares, $AG={A}^{\\prime}{G}^{\\prime}$, $DG={D}^{\\prime}{G}^{\\prime}$,\n\\[\n\\therefore {A}^{\\prime}{D}^{\\prime}={A}^{\\prime}{G}^{\\prime}-{G}^{\\prime}{D}^{\\prime}=AG-GD.\n\\]\n\\[\n\\because \\text{The ratio of the area of square ABCD to the area of quadrilateral MQPN is } 9\\colon 10,\n\\]\n\\[\n\\therefore \\text{The ratio of the area of square ABCD to the area of quadrilateral A'B'C'D' is } 9\\colon 1,\n\\]\n\\[\n\\therefore {\\left(AG+GD\\right)}^{2}\\colon {\\left({A}^{\\prime}{G}^{\\prime}-{G}^{\\prime}{D}^{\\prime}\\right)}^{2}=9\\colon 1,\n\\]\n\\[\n\\therefore \\left(AG+GD\\right)\\colon \\left(AG-GD\\right)=3\\colon 1,\n\\]\n\\[\n\\therefore AG=2GD,\n\\]\n\\[\n\\therefore AG\\colon GD=\\boxed{2}.\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/53080813.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $O$ is a point on line $AB$, $\\angle AOE = \\angle COD$, ray $OC$ bisects $\\angle BOE$, and $\\angle EOC = 50^\\circ$. What is the measure of $\\angle DOE$ in degrees?\n",
    "completion": "\\textbf{Solution:} Since OC bisects $\\angle BOE$ and $\\angle EOC=50^\\circ$, \\\\\n$\\therefore \\angle BOE=2\\angle EOC=100^\\circ$,\\\\\n$\\therefore \\angle AOE=180^\\circ-\\angle BOE=80^\\circ$,\\\\\nSince $\\angle AOE=\\angle COD$, \\\\\n$\\therefore \\angle COD=80^\\circ$,\\\\\n$\\therefore \\angle DOE=\\angle COD-\\angle COE=80^\\circ-50^\\circ=\\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52796148_176.png"
  },
  {
    "solution": "\\boxed{n > 3 \\text{ or } 0 < n < \\frac{3}{5} \\text{ or } n < 0}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given that in the Cartesian coordinate system, $A(-3,0)$, $B(1,0)$, point $A$ is rotated counter-clockwise around the origin to fall on point $C$ on the negative half of the $y$-axis. The parabola $y=ax^2+bx+c$ passes through points $A$, $B$, and $C$, connecting $AC$. Let point $N$ be a moving point on line $BC$, with the $x$-coordinate of $N$ denoted as $n$. If the triangle formed by points $A$, $C$, and $N$ is an obtuse triangle, determine the range of values for $n$.\n",
    "completion": "\\textbf{Solution:}\n\nDraw $AQ\\perp AC$ through point A, intersecting $BC$ at point Q. Draw $AP\\perp BC$ through point A at point P. \n\nSince $B(1, 0)$, $C(0, -3)$, \n\nlet the equation of $BC$ be: $y=kx+b(k\\ne 0)$, \n\nsolving the system\n\\[\\begin{cases}\nk + b = 0 \\\\\nb = -3\n\\end{cases},\\] \nwe get \n\\[\\begin{cases}\nk = 3 \\\\\nb = -3\n\\end{cases}.\\] \n\nThus, the equation of $BC$ is $y = 3x - 3$. \n\nSince the equation of $AC$ is $y = -x - 3$, \n\nthus $\\angle CAB=45^\\circ$. \n\nSince $AQ \\perp AC$, \n\nthus $\\angle QAB=45^\\circ$. \n\nLet the equation of $AQ$ be $y = x + b$. \n\nSubstituting $A(-3, 0)$ into it, we get $b = 3$, \n\nthus the equation of $AQ$ is $y = x + 3$. \n\nSolving the system of equations of lines $BC$ and $AQ$, \n\\[\\begin{cases}\ny = 3x - 3 \\\\\ny = x + 3\n\\end{cases},\\] we find $x = 3$. \n\nThus, the x-coordinate of point Q is 3. \n\nLet $AP$ intersect the y-axis at point E. \n\nSince $\\angle BAP + \\angle ABP = \\angle OBC + \\angle ABP = 90^\\circ$, \n\nthus $\\angle BAP = \\angle OBC$. \n\nSince $\\angle APB = \\angle BOC = 90^\\circ$ and $OA = OC$, \n\nthus $\\triangle AOE \\cong \\triangle COB$, \n\ntherefore $OE = OB = 1$, \n\nthus E(-1, 0). \n\nTherefore, the equation of $AP$ is $y = \\frac{1}{3}x - 1$. \n\nSolving the system of equations of lines $BC$ and $AP$, \n\\[\\begin{cases}\ny = 3x - 3 \\\\\ny = \\frac{1}{3}x - 1\n\\end{cases},\\] we find $x = \\frac{3}{5}$. \n\nThus, the x-coordinate of point P is $\\frac{3}{5}$. \n\n(1) When point N is above point Q, $\\angle NAC$ is obtuse, $\\triangle CAN$ is an obtuse triangle, in this case, $n > 3$. \n\n(2) When point N coincides with point Q or point P, $\\triangle CAN$ is a right triangle, which does not meet the criteria. \n\n(3) When point N is on the line segment $PQ$ (excluding the endpoints), $\\triangle CAN$ is an acute triangle, which does not meet the criteria. \n\n(4) When point N is on the line segment $PC$ (excluding the endpoints), $\\angle ANC$ is obtuse, $\\triangle CAN$ is an obtuse triangle, in this case, $0 < n < \\frac{3}{5}$. \n\n(5) When point N coincides with point C, it cannot form a triangle, which does not meet the criteria. \n\n(6) When point N is below point C, $\\angle ACN$ is obtuse, $\\triangle CAN$ is an obtuse triangle, in this case, $n < 0$. \n\nOverall, the range of $n$ is $n > 3$ or $0 < n < \\frac{3}{5}$ or $n < 0$ (also can be written as $n > 3$ or $n < \\frac{3}{5}$ and $n \\neq 0$). \n\nTherefore, the range of $n$ is $\\boxed{n > 3 \\text{ or } 0 < n < \\frac{3}{5} \\text{ or } n < 0}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52602789_89.png"
  },
  {
    "solution": "\\boxed{x<-3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=kx+b(k\\ne 0)$ passes through point $A(-3, 2)$. What is the solution set of the inequality $kx+b<2$ with respect to $x$?\n",
    "completion": "\\textbf{Solution:} From the diagram, it can be observed that when $x<-3$, $kx+b<2$, \\\\\nthus, the solution set is $\\boxed{x<-3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52387150.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given that $C$ is the midpoint of the segment $AB$, and $D$ is on the segment $CB$, with $DA=6$, $DB=4$, find the length of $CD$.\n",
    "completion": "\\textbf{Solution:} Since $DA=6$ and $DB=4$, \\\\\n$\\therefore$ $AB=10$, \\\\\nSince $C$ is the midpoint of line segment $AB$, \\\\\n$\\therefore$ $AC=5$, \\\\\nSince $DA=6$, \\\\\n$\\therefore$ $CD=\\boxed{1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52795772.png"
  },
  {
    "solution": "\\boxed{(\\frac{7}{4},6)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, quadrilateral $OABC$ is a rectangle, where point $A$ is on the $x$-axis, and point $C$ is on the $y$-axis, with the coordinates of point $B$ being $(8, 6)$. If $\\triangle OAB$ is folded along $OB$, with the corresponding point of $A$ being $E$, and $OE$ intersects $BC$ at point $D$, what are the coordinates of point $D$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral OABC is a rectangle, with point A on the x-axis and point C on the y-axis, and the coordinates of point B are $(8,6)$\n\n$\\therefore$ OC=AB=6, and BC=OA=8, with $\\angle OCB=90^\\circ$, and BC$\\parallel$OA\n\n$\\therefore$$\\angle AOB = \\angle OBC$\n\nSince triangle $\\triangle OAB$ is folded along OB, the corresponding point for A is E\n\n$\\therefore$$\\angle EOB=\\angle AOB$\n\n$\\therefore$$\\angle OBC=\\angle EOB$\n\n$\\therefore$OD=BD\n\nLet CD=x, then\n\n$OD=DB=BC-CD=8-x$\n\nIn $\\triangle OCD$, a right triangle,\n\n$OC^{2}+CD^{2}=OD^{2}$\n\n$\\therefore x^{2}+6^{2}=(8-x)^{2}$\n\nSolving gives: $x=\\frac{7}{4}$\n\n$\\therefore$ The coordinates of point D are $\\boxed{(\\frac{7}{4},6)}$",
    "image_path": "dataset/mm_math/images/MM_Math/52720458.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the length of segment $AD=30\\text{cm}$, points $C$ and $B$ are both on segment $AD$, and point $E$ is the midpoint of segment $AB$. If $BD=6\\text{cm}$, what is the length of segment $AE$ in $\\text{cm}$?\n",
    "completion": "\\textbf{Solution:} Given that $AD=30cm$, $BD=6cm$,\\\\\nhence $AB=AD-BD=30-6=24cm$,\\\\\nsince point E is the midpoint of AB,\\\\\nthus $AE= \\frac{1}{2}AB= \\frac{1}{2} \\times 24=\\boxed{12}cm$.",
    "image_path": "dataset/mm_math/images/MM_Math/51445267_107.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given: As shown in the trapezoid $ABCD$, $AD\\parallel BC$, and diagonals $AC$ and $BD$ intersect at point $E$. Point $F$ is on side $AB$, and $EF\\parallel BC$. It is known that $AD=3$, $BC=6$. Question: What is the length of $EF$?\n",
    "completion": "\\textbf{Solution:} Given that $EF\\parallel BC$,\\\\\nit follows that $\\triangle AFE\\sim \\triangle ABC$,\\\\\nthus $\\frac{EF}{BC}=\\frac{AF}{AB}$,\\\\\nGiven that $AD\\parallel BC$ and $EF\\parallel BC$,\\\\\nit implies $EF\\parallel AD$,\\\\\ntherefore $\\triangle BFE\\sim \\triangle BAD$,\\\\\nhence $\\frac{EF}{AD}=\\frac{BF}{AB}$,\\\\\ntherefore $\\frac{EF}{BC}+\\frac{EF}{AD}=\\frac{AF}{AB}+\\frac{BF}{AB}=\\frac{AF+BF}{AB}=1$,\\\\\nwhich means $\\frac{EF}{BC}+\\frac{EF}{AD}=1$,\\\\\nGiven that $AD=3$ and $BC=6$,\\\\\ntherefore $\\frac{EF}{6}+\\frac{EF}{3}=1$,\\\\\nSolving this equation yields $EF=\\boxed{2}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52917645_63.png"
  },
  {
    "solution": "\\boxed{5.8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within the rectangular paper ABCD, where E is a point on AD, fold $ \\triangle CDE$ along CE to overlap with $ \\triangle CFE$. If point F exactly falls on AB, and $ AF=4$, $ BC=10$, then what is the length of $ DE$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral ABCD is a rectangle,\\\\\nit follows that AD = BC = 10,\\\\\nlet AE = x, then DE = AD - AE = 10 - x,\\\\\nsince $\\triangle CDE$ is folded along CE to $\\triangle CFE$,\\\\\nit follows that EF = DE = 10 - x,\\\\\nin $\\triangle AEF$, $AF^{2}$ + $AE^{2}$ = $EF^{2}$,\\\\\nthus $4^{2}$ + $x^{2}$ = $(10 - x)^{2}$,\\\\\nsolving for x, we get x = 4.2,\\\\\nthus AE = 4.2,\\\\\nhence DE = AD - AE = \\boxed{5.8},",
    "image_path": "dataset/mm_math/images/MM_Math/52386627.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{281}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In rectangle $ABCD$, where $AB=3$ and $AD=5$, point $E$ is on line $BC$, and point $P$ is a moving point inside the rectangle. Lines $PD$ and $PB$ are connected as shown in Figure 3. If point $Q$ is the midpoint of $AD$, and $P$ is a moving point inside the rectangle, with lines $PQ$, $PB$, and $PC$ connected, the question is whether $PQ+\\sqrt{2}PB+PC$ has a minimum value. If it does, please write the answer directly; if not, please explain why.\n",
    "completion": "\\textbf{Solution:} As shown in the figure, rotate triangle $BPC$ clockwise by $90^\\circ$ to get the new triangle $B{P}^{\\prime }{C}^{\\prime }$, and then draw ${C}^{\\prime }N$ $\\perp$ AD at N through point ${C}^{\\prime }$, and connect ${C}^{\\prime }Q$ intersecting BC at M.\\\\\nAccording to the properties of rotation, we know that $BP=B{P}^{\\prime }$, $BC=B{C}^{\\prime }$, $\\angle PB{P}^{\\prime }=\\angle CB{C}^{\\prime }=90^\\circ$.\\\\\n$\\therefore P{P}^{\\prime }=\\sqrt{BP^{2}+{P}^{\\prime }{B}^{2}}=\\sqrt{2}BP$.\\\\\n$\\therefore PQ+\\sqrt{2}PB+PC=PQ+P{P}^{\\prime }+{P}^{\\prime }{C}^{\\prime }$.\\\\\nHence, when points Q, P, ${P}^{\\prime }$, ${C}^{\\prime }$ are collinear, $PQ+\\sqrt{2}PB+PC=PQ+P{P}^{\\prime }+{P}^{\\prime }{C}^{\\prime }$ reaches its minimum value.\\\\\n$\\because \\angle CB{C}^{\\prime }=90^\\circ$,\\\\\n$\\therefore$ points A, B, ${C}^{\\prime }$ are collinear,\\\\\nIn right triangle ${C}^{\\prime }AQ$, ${C}^{\\prime }Q=\\sqrt{A{{C}^{\\prime }}^{2}+A{Q}^{2}}$.\\\\\n$\\therefore {C}^{\\prime }A=AB+B{C}^{\\prime }=AB+BC=8$.\\\\\n$\\because$ point $Q$ is the midpoint of $AD$,\\\\\n$\\therefore AQ=\\frac{5}{2}$,\\\\\n$\\therefore {C}^{\\prime }Q=\\sqrt{A{{C}^{\\prime }}^{2}+A{Q}^{2}}=\\frac{\\sqrt{281}}{2}$.\\\\\n$\\therefore$ The minimum value of $PQ+\\sqrt{2}PB+PC$ is $\\boxed{\\frac{\\sqrt{281}}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50496942_83.png"
  },
  {
    "solution": "\\boxed{\\frac{108}{25}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In parallelogram $ABCD$, the diagonal $AC$ is perpendicular to side $CD$, and $\\frac{AB}{AC} = \\frac{3}{4}$. The perimeter of quadrilateral $ABCD$ is $16$. Point $E$ is on the extension of side $AD$, and point $F$ is on the ray $AB$, with $\\angle CED = \\angle CDF$. If $BF:FA = 1:2$, what is the area of $\\triangle CDE$?\n",
    "completion": "\\textbf{Solution:} Given point $F$ on the ray $AB$, we can deduce: $\\triangle CDE\\sim \\triangle DAF$, \\\\\n$\\therefore \\frac{S_{\\triangle CDE}}{S_{\\triangle DAF}}=\\left(\\frac{DC}{AD}\\right)^{2}$, \\\\\n(1) When point $F$ is on side $AB$, \\\\\n$\\because BF:FA=1:2$, $AB=3$, $\\therefore AF=2$, \\\\\nAccording to the problem, we have $S_{\\triangle DAF}=\\frac{1}{2}AF\\cdot AC$, \\\\\n$\\because AC=4$, \\\\\n$\\therefore S_{\\triangle DAF}=\\frac{1}{2}AF\\cdot AC=4$, \\\\\n$\\therefore \\frac{S_{\\triangle CDE}}{4}=\\left(\\frac{3}{5}\\right)^{2}$, \\\\\n$\\therefore S_{\\triangle CDE}=\\boxed{\\frac{36}{25}}$, \\\\\n(2) When point $F$ is on the extension of $AB$, \\\\\n$\\because BF:FA=1:2$, $AB=3$, \\\\\n$\\therefore AF=6$, \\\\\nAccording to the problem, we have $S_{\\triangle DAF}=\\frac{1}{2}AF\\cdot AC$, \\\\\n$\\therefore S_{\\triangle DAF}=\\frac{1}{2}AF\\cdot AC=12$, \\\\\n$\\therefore \\frac{S_{\\triangle CDE}}{12}=\\left(\\frac{3}{5}\\right)^{2}$, \\\\\n$\\therefore S_{\\triangle CDE}=\\boxed{\\frac{108}{25}}$, \\\\\nTo sum up, the area of $\\triangle CDE$ is either \\boxed{\\frac{36}{25}} or \\boxed{\\frac{108}{25}}.",
    "image_path": "dataset/mm_math/images/MM_Math/51379461_95.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AD$ is the altitude on side $BC$, $CE$ is the median on side $AB$, $G$ is the midpoint of $EC$, connecting $DG$, $CD = AE$, $AD = 6$, and $BD = 8$. Find the length of $CD$?\n",
    "completion": "\\textbf{Solution:} Since $AD$ is the altitude from $A$ to side $BC$, and $AD=6$, \\\\\n$BD=8$, \\\\\ntherefore $\\angle ADB=90^\\circ$, and $AB=\\sqrt{AD^{2}+BD^{2}}=\\sqrt{6^{2}+8^{2}}=10$, \\\\\nsince $CE$ is the median to side $AB$, \\\\\ntherefore $BE=AE=\\frac{1}{2}AB=5$, \\\\\nsince $CD=AE$, \\\\\ntherefore $CD=\\boxed{5}$",
    "image_path": "dataset/mm_math/images/MM_Math/53080378_112.png"
  },
  {
    "solution": "\\boxed{\\angle C<55^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown, it is known that points O and C are on the same side of line $m$. Draw $\\odot O$ intersecting $m$ at points A and B. Connect AC, BC, OA, and OB. If point C is outside of $\\odot O$ and $\\angle AOB=110^\\circ$, then what could be the possible degree of $\\angle C$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, let $AC$ intersect $\\odot O$ at point $D$, and connect $BD$,\n\n$\\because$ $\\angle AOB=110^\\circ$,\n\n$\\therefore$ $\\angle ADB=55^\\circ$,\n\n$\\because$ $\\angle ADB=\\angle C+\\angle CBD$,\n\n$\\therefore$ $\\angle C<55^\\circ$,\n\n$\\therefore$ $A$ meets the requirements, $BCD$ does not.\n\nThus, the answer is: $\\boxed{\\angle C<55^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51209211.png"
  },
  {
    "solution": "\\boxed{\\frac{25}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\angle CAB=\\angle CBD$, $AB=4$, $AC=6$, $BD=7.5$, $BC=5$, find the length of $CD$.\n",
    "completion": "\\textbf{Solution:} Given $AB=4$, $AC=6$, $BD=7.5$, $BC=5$,\\\\\n$\\therefore \\frac{AC}{BD}=\\frac{AB}{BC}=\\frac{4}{5}$,\\\\\n$\\because \\angle CAB=\\angle CBD$,\\\\\n$\\therefore \\triangle ABC \\sim \\triangle BCD$,\\\\\n$\\therefore \\frac{BC}{CD}=\\frac{4}{5}$,\\\\\n$\\therefore CD=\\frac{5}{4}BC=\\frac{5}{4}\\times 5=\\boxed{\\frac{25}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53081448.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $D$ is on side $BC$ of $\\triangle ABC$, with $AC \\parallel BE$, $BC = BE$, and $\\angle ABC = \\angle E$. $\\triangle ABC \\cong \\triangle DEB$; if $BE = 9$ and $AC = 4$, find the length of $CD$.\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABC\\cong \\triangle DEB$,\\\\\ntherefore $AC=DB=4$,\\\\\ntherefore $CD=BC-BD=9-4=\\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51435258_114.png"
  },
  {
    "solution": "\\boxed{120^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $O$ is a point inside $\\triangle ABC$, with $\\angle A=60^\\circ$. $BO$ and $CO$ are the angle bisectors of $\\angle ABC$ and $\\angle ACB$, respectively. What is the measure of $\\angle BOC$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle A=60^\\circ$, \\\\\ntherefore $\\angle ABC+\\angle ACB=180^\\circ-60^\\circ=120^\\circ$, \\\\\nsince $BO$ and $CO$ are the angle bisectors of $\\angle ABC$ and $\\angle ACB$ respectively, \\\\\ntherefore $\\angle OBC+\\angle OCB=\\frac{1}{2}\\left(\\angle ABC+\\angle ACB\\right)=60^\\circ$, \\\\\nhence $\\angle BOC=180^\\circ-\\left(\\angle OBC+\\angle OCB\\right)=180^\\circ-60^\\circ=\\boxed{120^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52916222.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in trapezoid $ABCD$, where $AD \\parallel BC$, and the diagonals $AC$ and $BD$ intersect at point $O$, the area of $\\triangle AOD$ is 9, and the area of $\\triangle BOC$ is 16. What is the area of $\\triangle AOB$?\n",
    "completion": "\\textbf{Solution:} Since $AD \\parallel BC$,\n\nit follows that $\\triangle AOD \\sim \\triangle COB$,\n\nand given $S_{\\triangle AOD} = 9$, $S_{\\triangle BOC} = 16$,\n\nthen $\\frac{AO}{OC} = \\sqrt{\\frac{S_{\\triangle AOD}}{S_{\\triangle BOC}}} = \\frac{3}{4}$.\n\nSince $\\triangle AOB$ and $\\triangle BOC$ have the same height,\n\nit follows that $S_{\\triangle AOB} = \\boxed{12}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52917597.png"
  },
  {
    "solution": "\\boxed{32^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, $\\angle ABC = \\angle ADC = 90^\\circ$, $E$ is the midpoint of diagonal $AC$, connecting $BE$, $ED$, $BD$. If $\\angle BAD = 58^\\circ$, what is the degree measure of $\\angle EBD$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle ABC = \\angle ADC = 90^\\circ$,\\\\\npoints A, B, C, D lie on the same circle with E as the center and AC as the diameter.\\\\\nSince $\\angle BAD = 58^\\circ$,\\\\\n$\\angle DEB = 116^\\circ$.\\\\\nSince DE = BE = $\\frac{1}{2}$AC,\\\\\n$\\angle EBD = \\angle EDB = \\boxed{32^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52913823.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $AB=8$. The rectangle is folded along the line $AC$, with point $B$ falling onto point $E$, and $AE$ intersects $CD$ at point $F$. If $AF=\\frac{25}{4}$, what is the length of $AD$?\n",
    "completion": "\\textbf{Solution:} Given $DC\\parallel AB$,\\\\\n$\\therefore \\angle FCA=\\angle CAB$, also $\\angle FAC=\\angle CAB$,\\\\\n$\\therefore \\angle FAC=\\angle FCA$,\\\\\n$\\therefore FA=FC=\\frac{25}{4}$,\\\\\n$\\therefore FD=FE$,\\\\\nGiven $DC=AB=8$, $AF=\\frac{25}{4}$,\\\\\n$\\therefore FD=FE=8-\\frac{25}{4}=\\frac{7}{4}$,\\\\\n$\\therefore AD=BC=EC=\\sqrt{FC^{2}-FE^{2}}=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52423378.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, isosceles right triangles are constructed externally on each side of $\\triangle ABC$ using the sides of $\\triangle ABC$ as the right angles. If $AB = \\sqrt{3}$, what is the area of the shaded region in the figure?\n",
    "completion": "\\textbf{Solution:} Since $ABC$ is a right-angled triangle,\\\\\nwe have $AC^{2}+BC^{2}=AB^{2}=3$,\\\\\nthus, the area of the shaded region $S_{\\text{shaded}}= \\frac{1}{2} AC^{2}+ \\frac{1}{2} BC^{2}+ \\frac{1}{2} AB^{2}= \\frac{1}{2} (AC^{2}+BC^{2})+ \\frac{1}{2} AB^{2}= \\frac{1}{2} AB^{2}+ \\frac{1}{2} AB^{2}=AB^{2}=\\boxed{3}.$",
    "image_path": "dataset/mm_math/images/MM_Math/51369020.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $BE$ bisects $\\angle ABC$, and $AE\\perp BE$ at point $E$. The area of $\\triangle BCE$ is $2$. What is the area of $\\triangle ABC$?\n",
    "completion": "\\textbf{Solution}: Extend AE to intersect BC at D, \\\\\n$\\because$ BE bisects $\\angle ABC$, \\\\\n$\\therefore$ $\\angle ABE=\\angle DBE$, \\\\\n$\\because$ $AE\\perp BE$, \\\\\n$\\therefore$ $\\angle AEB=\\angle DEB=90^\\circ$, \\\\\nIn $\\triangle ABE$ and $\\triangle DBE$, \\\\\n$\\begin{array}{c}\n\\angle ABE=\\angle DBE\\\\\nEB=EB\\\\\n\\angle AEB=\\angle DEB\n\\end{array}$, \\\\\n$\\therefore$ $\\triangle ABE\\cong \\triangle DBE$, \\\\\n$\\therefore$ $AE=ED$, \\\\\n$\\therefore$ ${S}_{\\triangle ABE}={S}_{\\triangle DBE}$, \\\\\n${S}_{\\triangle ACE}={S}_{\\triangle DCE}$, \\\\\n$\\therefore$ ${S}_{\\triangle ABC}=2+{S}_{\\triangle CBE}=\\boxed{4}$",
    "image_path": "dataset/mm_math/images/MM_Math/51434967.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\angle 2 : \\angle 3 : \\angle 4 = 3 : 9 : 7$, then what is the degree measure of $\\angle 1$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle 2 : \\angle 3 : \\angle 4 = 3 : 9 : 7$,\\\\\nlet $\\angle 2 = 3\\beta$, $\\angle 3 = 9\\beta$, and $\\angle 4 = 7\\beta$,\\\\\nsince $\\angle 2 + \\angle 3 = 180^\\circ$,\\\\\nthus $3\\beta + 9\\beta = 180^\\circ$,\\\\\ntherefore $\\beta = 15^\\circ$,\\\\\ntherefore $\\angle 2 = 45^\\circ$ and $\\angle 4 = 105^\\circ$,\\\\\ntherefore $\\angle 1 = 180^\\circ - \\angle 2 - \\angle 4 = \\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52837433.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $O$ is the intersection point of the diagonals of rhombus $ABCD$, with $CE\\parallel BD$ and $EB\\parallel AC$. Line $OE$ is drawn and intersects $BC$ at point $F$. Quadrilateral $OBEC$ is a rectangle. Line $DG$ is drawn perpendicular to the extension of line $BA$ at point $G$, and line $OG$ is connected. If $OC:OB=1:2$ and $OE=4\\sqrt{5}$, find the length of $OG$.\n",
    "completion": "\\textbf{Solution:} \\text{ In rhombus } ABCD, DO = OB,\n\\[\n\\text{and since } DG \\perp BA,\n\\]\n\\[\n\\therefore \\angle DGB = 90^\\circ,\n\\]\n\\[\n\\therefore OG = \\frac{1}{2}BD = OB,\n\\]\n\\[\n\\text{In rectangle } OBEC, OE = BC = 4\\sqrt{5},\n\\]\n\\[\n\\text{and since } OC:OB = 1:2,\n\\]\n\\[\n\\therefore \\text{let } OC = x, OB = 2x,\n\\]\n\\[\n\\text{By the Pythagorean theorem: } OC^2 + OB^2 = BC^2, \\text{ namely } x^2 + (2x)^2 = (4\\sqrt{5})^2,\n\\]\n\\[\n\\text{Solving, we get } x = 4, \\text{ thus } OB = 8,\n\\]\n\\[\n\\therefore OG = \\boxed{8}\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/51737166_155.png"
  },
  {
    "solution": "\\boxed{19}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, DE is the perpendicular bisector of AC, and AE $= 3\\,cm$. If the perimeter of $\\triangle ABD$ is $13\\,cm$, what is the perimeter of $\\triangle ABC$ in centimeters?\n",
    "completion": "\\textbf{Solution:} Since DE is the perpendicular bisector of AC,\\\\\nit follows that AD = CD, and AC = 2AE = 6cm,\\\\\nFurthermore, since the perimeter of $\\triangle ABD$ equals AB + BD + AD = 13cm,\\\\\nit implies that AB + BD + CD = 13cm,\\\\\nwhich means AB + BC = 13cm,\\\\\nTherefore, the perimeter of $\\triangle ABC$ equals AB + BC + AC = 13 + 6 = \\boxed{19}cm.",
    "image_path": "dataset/mm_math/images/MM_Math/52693948.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, Xiaoying and her classmate are swinging. When the swing AB is at rest, the lower end B' is 0.6m above the ground. When the swing moves to the position AB, the horizontal distance from the lower end B to the rest position EB is 2.4m, and it is 1.4m above the ground. Find the length of the swing AB.\n",
    "completion": "\\textbf{Solution:} Let the length of AB=AB$^{\\prime }$ be $x$ meters. According to the problem, we can deduce that B$^{\\prime }$E = 1.4 - 0.6 = 0.8 (m),\\\\\nhence, AE = AB - 0.8,\\\\\nin $\\triangle AEB$, since AE$^{2}$ + BE$^{2}$ = AB$^{2}$,\\\\\ntherefore, (x-0.8)$^{2}$ + 2.4$^{2}$ = x$^{2}$\\\\\nSolving this, we find x = \\boxed{4},\\\\\nhence, the length of the swing AB is 4m.",
    "image_path": "dataset/mm_math/images/MM_Math/52913906.png"
  },
  {
    "solution": "\\boxed{18}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the isosceles $\\triangle ABC$ has a base $BC=20$ and an area of $120$. Point $F$ is on side $BC$, and $BF=3FC$. $EG$ is the perpendicular bisector of the leg $AC$. If point $D$ moves on $EG$, what is the minimum perimeter of $\\triangle CDF$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure,\\\\\n$\\because$ EG is the perpendicular bisector of the line segment AC,\\\\\n$\\therefore$ DA=DC,\\\\\n$\\therefore$ DF+DC=AD+DF,\\\\\n$\\therefore$ when A, D, F are collinear, DF+DC is minimized, and the minimum value is the length of the line segment AF.\\\\\n$\\because$ $ \\frac{1}{2}\\cdot BC\\cdot AH=120$,\\\\\n$\\therefore$ AH=12,\\\\\n$\\because$ AB=AC, AH$\\perp$BC,\\\\\n$\\therefore$ BH=CH=10,\\\\\n$\\because$ BF=3FC,\\\\\n$\\therefore$ CF=FH=5,\\\\\n$\\therefore$ $AF=\\sqrt{AH^{2}+HF^{2}}=\\sqrt{12^{2}+5^{2}}=13$,\\\\\n$\\therefore$ the minimum value of DF+DC is 13,\\\\\n$\\therefore$ the shortest perimeter of $\\triangle CDF$=13+5=\\boxed{18}.",
    "image_path": "dataset/mm_math/images/MM_Math/52731259.png"
  },
  {
    "solution": "\\boxed{y=\\frac{1}{2}x+2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the graph of the quadratic function $y=-x^2+bx+c$ intersects the coordinate axes at points $A$, $B$, and $C$. The coordinates of point $B$ are $(1,0)$, the coordinates of point $C$ are $(0,4)$, and the coordinates of point $D$ are $(0,2)$. Point $P$ is a moving point on the graph of the quadratic function. Find the analytical expressions for the quadratic function and the line $AD$.\n",
    "completion": "Solution: Substitute B(1, 0) and C(0, 4) into $y=-x^{2}+bx+c$, we get\n\\[\n\\left\\{\n\\begin{array}{l}\n0=-1+b+c\\\\\n4=c\n\\end{array}\n\\right.,\n\\]\nSolving this, we have\n\\[\n\\left\\{\n\\begin{array}{l}\nb=-3\\\\\nc=4\n\\end{array}\n\\right.,\n\\]\n$\\therefore$ The quadratic function can be written as $y=-x^{2}-3x+4$.\n\nSubstitute $y=0$ into $y=-x^{2}-3x+4$, we have $-x^{2}-3x+4=0$, solving this we find $x_{1}=-4$, $x_{2}=1$,\n\n$\\therefore$ \\boxed{A(-4, 0)}.\n\nLet's assume the equation of line AD as $y=kx+b'$.\n\n$\\because$ D(0, 2),\n\n$\\therefore$\n\\[\n\\left\\{\n\\begin{array}{l}\n0=-4k+b' \\\\\n2=b'\n\\end{array}\n\\right.,\n\\]\nSolving this, we find:\n\\[\n\\left\\{\n\\begin{array}{l}\nk=\\frac{1}{2}\\\\\nb'=2\n\\end{array}\n\\right.,\n\\]\n$\\therefore$ The equation of line AD is \\boxed{y=\\frac{1}{2}x+2}.",
    "image_path": "dataset/mm_math/images/MM_Math/51321592_200.png"
  },
  {
    "solution": "\\boxed{120^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C = 90^\\circ$, $\\angle A = 30^\\circ$, and $BD = 2CD$, what is the measure of $\\angle ADB$ in degrees?\n",
    "completion": "\\textbf{Solution:} Since $\\angle C=90^\\circ$ and BD=2CD,\\\\\nit follows that $\\angle DBC=30^\\circ$,\\\\\nhence $\\angle ADB=\\angle C+\\angle DBC=90^\\circ+30^\\circ=\\boxed{120^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52874609.png"
  },
  {
    "solution": "\\boxed{40}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, square $ABEF$ and square $ACGD$ are constructed externally on sides $AB$ and $AC$ of $\\triangle ABC$, respectively. Lines $BD$, $CF$, and $DF$ are drawn, with $CF$ intersecting $BD$ at $O$ and $AD$ at $H$. If $AB=2$ and $AC=4$, what is the value of $BC^2+DF^2$?\n",
    "completion": "\\textbf{Solution:} We have $BC^2 + DF^2 = \\boxed{40}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53359756_129.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{21}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In Figure 1, within rectangle $ABCD$, $AC$ and $BD$ intersect at point $O$, and $E$ is a point on $AD$, with $CE$ and $BD$ intersecting at point $F$. Given that $AE = CD$, and $BD \\perp CE$, (1) find the degree measure of $\\angle DEC$. (2) As shown in Figure 2, connect $AF$. When $BC = 3$, find the length of $AF$.\n",
    "completion": "\\textbf{Solution:}\n\n(1)\\\\\nSince AE = CE,\\\\\nit follows that $\\angle$EAC = $\\angle$ECA,\\\\\nSince AD $\\parallel$ BC,\\\\\nit follows that $\\angle$EAC = $\\angle$ACB,\\\\\nSince OB = OC,\\\\\nit follows that $\\angle$ACB = $\\angle$DBC,\\\\\nHence, $\\angle$ACB = $\\angle$DBC = $\\angle$ECA,\\\\\nSince BD $\\perp$ CE,\\\\\nit follows that $\\angle$BFC = 90$^\\circ$,\\\\\nTherefore, $\\angle$ECA + $\\angle$ACB + $\\angle$DBC = 90$^\\circ$,\\\\\nTherefore, $\\angle$ECA = $\\angle$ACB = $\\angle$DBC = 30$^\\circ$,\\\\\nTherefore, $\\angle$DEC = $\\angle$ECB = \\boxed{60^\\circ},\\\\\n\n(2) Draw FH $\\perp$ AD at point H,\\\\\nIn $\\triangle$BCD, $\\angle$DBC = 30$^\\circ$, BC = 3,\\\\\nit follows that BD = 2$\\sqrt{3}$,\\\\\nIn $\\triangle$BFC, $\\angle$FBC = 30$^\\circ$, BC = 3,\\\\\nit follows that BF = $\\frac{3}{2}\\sqrt{3}$,\\\\\nTherefore, DF = BD - BF = $\\frac{\\sqrt{3}}{2}$,\\\\\nIn $\\triangle$DHF, $\\angle$FDH = $\\angle$DBC = 30$^\\circ$,\\\\\nit follows that FH = $\\frac{1}{2}$DF = $\\frac{\\sqrt{3}}{4}$,\\\\\nTherefore, HD = $\\sqrt{3}$HF = $\\frac{3}{4}$,\\\\\nTherefore, AH = AD - DH = 3 - $\\frac{3}{4}$ = $\\frac{9}{4}$,\\\\\nTherefore, AF = $\\sqrt{AH^2 + HF^2} = \\sqrt{\\left(\\frac{9}{4}\\right)^2 + \\left(\\frac{\\sqrt{3}}{4}\\right)^2} = \\boxed{\\frac{\\sqrt{21}}{2}}$",
    "image_path": "dataset/mm_math/images/MM_Math/53009704_164.png"
  },
  {
    "solution": "\\boxed{6-\\pi}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AC$ is the diameter of the circle $\\odot O$, $P$ is the midpoint of the semicircle $AC$. Line $AP$ is extended to point $B$ such that $PB=AP$. Lines $CP$, $CB$, and $OP$ are drawn. $BC$ is a tangent of circle $\\odot O$. If $AC=4$, what is the area of the shaded section in the figure?\n",
    "completion": "\\textbf{Solution:} Given $AC=4$,\\\\\nthus $BC=AC=4$, $OA=OP=2$,\\\\\nsince $P$ is the midpoint of the semicircle $AC$,\\\\\nhence $AC\\perp OP$,\\\\\ntherefore $S_{\\text{shaded}}=S_{\\triangle ABC}-S_{\\triangle AOP}-S_{\\text{sector} POC}=\\frac{1}{2}\\times 4\\times 4-\\frac{1}{2}\\times 2\\times 2-\\frac{90^\\circ\\pi \\times 2^2}{360}=\\boxed{6-\\pi}$",
    "image_path": "dataset/mm_math/images/MM_Math/53019920_79.png"
  },
  {
    "solution": "\\boxed{30\\cdot\\tan 65^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, to measure the height of a building, at point A, which is 30 meters away from the base of the building at point O, the angle of elevation to the top of the building at point B is measured to be $\\angle OAB = 65^\\circ$. What is the height of this building?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, in the right-angled triangle $\\triangle$ABO,\\\\\nsince $\\angle$AOB$=90^\\circ$ and $\\angle$A$=65^\\circ$ with AO$=30$m,\\\\\nthen $\\tan 65^\\circ = \\frac{OB}{AO}$,\\\\\nthus, OB$=30\\cdot\\tan 65^\\circ$ meters.\\\\\nTherefore, the answer is: $BO=\\boxed{30\\cdot\\tan 65^\\circ}$ meters.",
    "image_path": "dataset/mm_math/images/MM_Math/51320365.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, with side $AB$ of $\\triangle ABC$ as the diameter, draw $\\odot O$, which intersects side $AC$ at point $D$, and $BC$ is a tangent of $\\odot O$. A chord $DE \\perp AB$ at point $F$, and connect $BE$. It is given that $\\angle ABE = \\angle C$. If point $F$ is the midpoint of $OB$, and $OF=1$, what is the length of the line segment $ED$?\n",
    "completion": "\\textbf{Solution:} Connect $OE$,\\\\\nsince point $F$ is the midpoint of $OB$,\\\\\nit follows that $OF=\\frac{1}{2}OB=\\frac{1}{2}OE$,\\\\\nand since $OF=1$,\\\\\nit follows that $OE=2$. By the Pythagorean theorem, $EF=\\sqrt{3}$,\\\\\nand since $DE\\perp AB$,\\\\\nit follows that $DE=2EF=2\\sqrt{3}$,\\\\\nthus, the length of segment $ED$ is $DE=\\boxed{2\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53042797_115.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, within circle $\\odot O$ of radius 5, chord $AB=6$, $OC\\perp AB$ at point $D$, and intersects $\\odot O$ at point $C$, what is the length of $CD$?\n",
    "completion": "\\textbf{Solution:} Draw a line from point O to point A,\\\\\nsince AB=6, and OC $\\perp$ AB at point D,\\\\\ntherefore, AD=$\\frac{1}{2}$AB=$\\frac{1}{2}\\times 6=3$,\\\\\nsince the radius of circle O is 5,\\\\\ntherefore, OD=$\\sqrt{OA^{2}-AD^{2}}=\\sqrt{5^{2}-3^{2}}=4$,\\\\\ntherefore, CD=OC-OD=5-4=\\boxed{1}.",
    "image_path": "dataset/mm_math/images/MM_Math/51351765.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the length of line segment $AB=10$. Extend $AB$ to point $C$ such that $BC=2AB$. Let points $M$ and $N$ be the midpoints of line segments $AC$ and $AB$, respectively. Find the length of $MN$.\n",
    "completion": "\\textbf{Solution:} Given that $BC=2AB$ and $AB=10$,\\\\\nthen $BC=2\\times 10=20$.\\\\\nTherefore, $AC=AB+BC=10+20=30$.\\\\\nSince points $M$ and $N$ are the midpoints of line segments $AC$ and $AB$ respectively,\\\\\nthen $AM=\\frac{1}{2}AC=\\frac{1}{2}\\times 30=15$.\\\\\nAnd $AN=\\frac{1}{2}AB=\\frac{1}{2}\\times 10=5$.\\\\\nTherefore, $MN=AM-AN=15-5=\\boxed{10}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52732940_85.png"
  },
  {
    "solution": "\\boxed{y=\\frac{25x-5x^2}{16+x^2-5x}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In rectangle $ABCD$, $AB=4$, $BC=5$, $P$ is a moving point on ray $BC$, draw $PE\\perp AP$, $PE$ intersects ray $DC$ at point $E$, ray $AE$ intersects ray $BC$ at point $F$. Let $BP=x$, $CF=y$. As shown in the figure, when point $P$ is on side $BC$ (point $P$ does not coincide with points $B$ or $C$), what is the function relationship between $y$ and $x$?\n",
    "completion": "\\textbf{Solution:} Given the construction, \\\\\n$\\because EC\\parallel AB$, \\\\\n$\\therefore \\triangle ECF\\sim \\triangle ABF$, \\\\\n$\\therefore \\frac{FC}{FB}=\\frac{EC}{AB}$, \\\\\nthen $CE=\\frac{FC\\cdot AB}{FB}=\\frac{4y}{5+y}$, \\\\\n$\\because \\triangle PCE\\sim \\triangle ABP$, \\\\\n$\\therefore \\frac{CE}{BP}=\\frac{PC}{AB}$, \\\\\nthen $CE=\\frac{BP\\cdot PC}{AB}=\\frac{x(5-x)}{4}$, \\\\\n$\\therefore \\frac{4y}{5+y}=\\frac{x(5-x)}{4}$, \\\\\nUpon rearrangement, we get: $y=\\frac{25x-5x^2}{16+x^2-5x} \\left(0<x<5\\right)$, \\\\\nHence, the function of $y$ in terms of $x$ is $\\boxed{y=\\frac{25x-5x^2}{16+x^2-5x}}\\left(0<x<5\\right)$.",
    "image_path": "dataset/mm_math/images/MM_Math/52977987_135.png"
  },
  {
    "solution": "\\boxed{6\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, quadrilateral $ABCD$ is a parallelogram, with $AD=BD$. Draw $CE\\parallel BD$, intersecting the extension of $AD$ at point $E$. Quadrilateral $BDEC$ is a rhombus; draw $BE$. If $AB=4$ and $AD=7$, what is the length of $BE$?\n",
    "completion": "\\textbf{Solution:} Extend BE to intersect CD at O,\\\\\nsince quadrilateral ABCD is a parallelogram,\\\\\nthus CD = AB = 4.\\\\\nFrom (1), we know quadrilateral BDEC is a rhombus,\\\\\nthus DO = CO = $\\frac{1}{2}$CD = 2, and BO = $\\frac{1}{2}$BE, with CD$\\perp$BE.\\\\\nIn $\\triangle BDO$, BD = AD = 7, and DO = 2,\\\\\nthus BO = $\\sqrt{BD^{2}-DO^{2}}$ = $\\sqrt{7^{2}-2^{2}}$ = $3\\sqrt{5}$,\\\\\ntherefore, BE = 2BO = $\\boxed{6\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52159940_155.png"
  },
  {
    "solution": "\\boxed{80^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ is an equilateral triangle. Points $P$ and $Q$ are taken on sides $AC$ and $BC$, respectively, such that $AP=CQ$, and $\\angle ABP=20^\\circ$. Lines $AQ$ and $BP$ intersect at point $O$. What is the measure of $\\angle AQB$?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABC$ is an equilateral triangle,\\\\\nit follows that AB=AC, $\\angle BAP=\\angle ACQ=60^\\circ$,\\\\\nIn $\\triangle BAP$ and $\\triangle ACQ$,\\\\\n$\\left\\{\\begin{array}{l}\nAB=AC\\\\\n\\angle BAP=\\angle ACQ\\\\\nAP=CQ\n\\end{array}\\right.$,\\\\\nthus $\\triangle BAP \\cong \\triangle ACQ$ (SAS),\\\\\ntherefore $\\angle CAQ=\\angle ABP=20^\\circ$,\\\\\nthus $\\angle AQB=\\angle C+\\angle CAQ=60^\\circ+20^\\circ=\\boxed{80^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51385938.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the rhombus $ABCD$, the diagonals $AC$ and $BD$ intersect at point $O$, and $BE \\parallel AC$, $AE \\parallel BD$. Connect $EO$. If $CD=6$, find the length of $OE$.\n",
    "completion": "\\textbf{Solution:} \\\\\nSince quadrilateral AEBO is a rectangle, \\\\\nit follows that $EO = AB$, \\\\\nSince quadrilateral ABCD is a rhombus, \\\\\nit follows that $AB = CD$. \\\\\nTherefore, $EO = CD = \\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50482317_129.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in parallelogram $ABCD$, point $E$ is the midpoint of $AB$. $CE$ and $BD$ intersect at point $O$. If $S_{\\triangle EOB} = 1$, find the area of quadrilateral $AEOD$.\n",
    "completion": "\\textbf{Solution:} Let's consider parallelogram ABCD, where point E is the midpoint of AB,\\\\\nhence CD$\\parallel$AB, and CD=AB=2BE,\\\\\ntherefore, $\\triangle DOC\\sim \\triangle BOE$,\\\\\nthus, $\\frac{OC}{OE}=\\frac{CD}{BE}=2$,\\\\\ntherefore, $\\frac{S_{\\triangle BOC}}{S_{\\triangle EOB}}=\\frac{OC}{OE}=2$, $ \\frac{S_{\\triangle DOC}}{S_{\\triangle EOB}}=\\left(\\frac{OC}{OE}\\right)^2=4$\\\\\nSince $S_{\\triangle EOB}=1$,\\\\\nthus, $S_{\\triangle BOC}=2$, $S_{\\triangle DOC}=4$,\\\\\ntherefore, $S_{\\triangle BCD}=6$,\\\\\ntherefore, $S_{\\triangle DAB}=6$,\\\\\nthus, the area of quadrilateral AEOD is: $S_{\\triangle DAB}-S_{\\triangle EOB}=6-1=\\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51351875.png"
  },
  {
    "solution": "\\boxed{56^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the lines $AB$ and $CD$ intersect at point $O$, $\\angle COE$ is a right angle, $OF$ bisects $\\angle AOE$, $\\angle COF = 28^\\circ$. Find the measure of $\\angle BOE$.\n",
    "completion": "\\textbf{Solution:} Since $\\angle COE$ is a right angle and $\\angle COF=28^\\circ$,\\\\\nit follows that $\\angle EOF=90^\\circ-28^\\circ=62^\\circ$,\\\\\nsince $OF$ bisects $\\angle AOE$,\\\\\nit follows that $\\angle AOF=\\angle EOF=62^\\circ$,\\\\\ntherefore, $\\angle EOB=180^\\circ-62^\\circ-62^\\circ=\\boxed{56^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53206394.png"
  },
  {
    "solution": "\\boxed{\\frac{9}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, a rectangular paper ABCD is folded along MN such that point A lands on point A' on side BC, and the corresponding point for D is D'. Connect A'D' and intersect side CD at point E. Connect CD'. If AB = 9, AD = 6, and point A' is the midpoint of BC, then what is the length of segment ED'?\n",
    "completion": "\\textbf{Solution:} Fold the rectangle $ABCD$ along $MN$, so that point $A$ falls on the side $BC$ at point $A'$,\\\\\n$\\therefore AM=A'M$, $\\angle MA'D'=\\angle A=90^\\circ$,\\\\\nLet $AM=A'M=x$, then $BM=9-x$,\\\\\n$\\because A'$ is the midpoint of $BC$,\\\\\n$\\therefore A'B=A'C=\\frac{1}{2}BC=\\frac{1}{2}AD=3$,\\\\\nIn $\\triangle A'BM$,\\\\\n$A'B^2+BM^2=A'M^2$,\\\\\nthat is $3^2+(9-x)^2=x^2$,\\\\\nSolving for $x$ gives: $x=5$,\\\\\n$\\therefore A'M=5$, $MB=4$,\\\\\n$\\because \\angle MA'B+\\angle EA'C=90^\\circ$, $\\angle A'EC+\\angle EA'C=90^\\circ$,\\\\\n$\\therefore \\angle MA'B=\\angle A'EC$,\\\\\n$\\because \\angle B=\\angle A'CE=90^\\circ$,\\\\\n$\\therefore \\triangle A'BM\\sim \\triangle ECA'$,\\\\\n$\\therefore \\frac{A'E}{A'M}=\\frac{A'C}{BM}$, that is $\\frac{A'E}{5}=\\frac{3}{4}$,\\\\\n$\\therefore A'E=\\frac{15}{4}$,\\\\\n$\\therefore ED'=A'D'-A'E=AD-A'E=6-\\frac{15}{4}=\\boxed{\\frac{9}{4}}$",
    "image_path": "dataset/mm_math/images/MM_Math/52096361.png"
  },
  {
    "solution": "\\boxed{\\begin{cases}x=3\\\\ y=5\\end{cases}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=x+2$ intersects with the line $y=kx+6$ at point $P(3, n)$. Then, the system of equations is\n",
    "completion": "\\textbf{Solution:} Let $\\because$ the line $y=x+2$ passes through the point $P(3, n)$,\\\\\n$\\therefore n=3+2=5$,\\\\\n$\\therefore P(3, 5)$,\\\\\n$\\because$ the line $y=x+2$ intersects with the line $y=kx+6$ at point $P$,\\\\\n$\\therefore$ the solution to the system of equations $\\begin{cases}y=x+2\\\\ y=kx+6\\end{cases}$ is $\\boxed{\\begin{cases}x=3\\\\ y=5\\end{cases}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50677500.png"
  },
  {
    "solution": "\\boxed{\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the Cartesian coordinate system, $Q$ is a moving point on the line $y=-\\frac{1}{2}x+2$. When $Q$ is rotated clockwise by $90^\\circ$ around point $P(1, 0)$ to get point $Q'$, and $OQ'$ is drawn, what is the minimum value of $OQ'$?\n",
    "completion": "\\textbf{Solution:} Draw $QM \\perp x$-axis at point M through point Q and draw $Q^{\\prime}N \\perp x$-axis at point N through point $Q^{\\prime}$, \\\\\n$\\therefore$ $\\angle PMQ = \\angle Q^{\\prime}NP = 90^\\circ$, \\\\\n$\\because$ $\\angle QPM + \\angle MPQ^{\\prime} = 90^\\circ$, $\\angle QPM + \\angle PQM = 90^\\circ$, \\\\\n$\\therefore$ $\\angle MPQ^{\\prime} = \\angle PQM$, \\\\\nIn $\\triangle PQM$ and $\\triangle Q^{\\prime}NP$, \\\\\n$\\left\\{ \\begin{array}{l}\n\\angle PMQ = \\angle Q^{\\prime}NP, \\\\\n\\angle MPQ^{\\prime} = \\angle PQM, \\\\\nPQ = PQ^{\\prime}\n\\end{array} \\right.$ \\\\\n$\\therefore$ $\\triangle PQM \\cong \\triangle Q^{\\prime}NP$ (AAS), \\\\\n$\\therefore$ PN = QM, $Q^{\\prime}N$ = PM, \\\\\nLet point Q be $\\left(m, -\\frac{1}{2}m + 2\\right)$, \\\\\n$\\therefore$ PM = $|m - 1|$, QM = $\\left|-\\frac{1}{2}m + 2\\right|$, \\\\\n$\\therefore$ ON = $\\left|3 - \\frac{1}{2}m\\right|$, \\\\\n$\\therefore$ point $Q^{\\prime} \\left(3 - \\frac{1}{2}m, 1 - m\\right)$, \\\\\n$\\therefore$ $OQ^{\\prime}{}^{2} = \\left(3 - \\frac{1}{2}m\\right)^{2} + \\left(1 - m\\right)^{2} = \\frac{5}{4}(m - 2)^{2} + 5$, \\\\\n$\\because$ $\\frac{5}{4} > 0$, \\\\\n$\\therefore$ the parabola opens upwards, \\\\\n$\\therefore$ when $m = 2$, $OQ^{\\prime}{}^{2}$ has its minimum value, which is 5, \\\\\n$\\therefore$ the minimum value of $OQ^{\\prime}$ is $\\boxed{\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51369705.png"
  },
  {
    "solution": "\\boxed{13}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given the figure, $AB=16\\text{cm}$, where $C$ is a point on $AB$ and $AC=10\\text{cm}$. Point $D$ is the midpoint of $AC$, and point $E$ is the midpoint of $BC$. Find the lengths of the line segments $DE$ and $AE$.\n",
    "completion": "\\textbf{Solution:} Since $D$ is the midpoint of $AC$, and $AC = 10$ cm,\n\n$\\therefore DC = \\frac{1}{2}AC = 5$ cm.\n\nAlso, since $AB = 16$ cm,\n\n$\\therefore BC = AB - AC = 6$ cm.\n\nSince $E$ is the midpoint of $BC$,\n\n$\\therefore CE = \\frac{1}{2}BC = 3$ cm.\n\nTherefore, $DE = DC + CE = \\boxed{8}$ cm, $AE = AC + CE = \\boxed{13}$ cm.",
    "image_path": "dataset/mm_math/images/MM_Math/51284639.png"
  },
  {
    "solution": "\\boxed{(1,0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the square ABCD with a side length of 2 has its side AB on the x-axis, and the midpoint of AB is the origin O. Fix points A and B, and push the square in the direction of the arrow until point D falls on the positive half-axis of the y-axis at point $D'$. Point E is a moving point on the x-axis. When $ED'+EC'$ takes the minimum value, what are the coordinates of point E?\n",
    "completion": "\\textbf{Solution:} Construct point $F$ as the reflection of point $D'$ across the $x$ axis, and connect $C'F$. The intersection of $C'F$ and the $x$ axis is the desired point $E$.\\\\\nSince the $x$ axis bisects $D'F$ perpendicularly,\\\\\nit follows that $ED'=EF$\\\\\nTherefore, $ED'+EC'=EF+EC'=FC'$\\\\\nThis implies that at point $E$, the sum $ED'+EC'$ is minimized.\\\\\nSince $AD'=2$, $AO=1$, and $\\angle AOD'=90^\\circ$,\\\\\nit follows that $OD'=\\sqrt{AD{'}^{2}-AO^{2}}=\\sqrt{2^{2}-1^{2}}=\\sqrt{3}$,\\\\\nTherefore, $F(0,-\\sqrt{3})$, $C'(2,\\sqrt{3})$,\\\\\nThus, the equation of line $FC'$ is: $y=\\sqrt{3}x-\\sqrt{3}$,\\\\\nHence, the coordinates of point $E$ are $\\boxed{(1,0)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50496574.png"
  },
  {
    "solution": "\\boxed{4\\colon 3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, point E is on BC, and $BE=3EC$. D is the midpoint of AC, AE and BD intersect at point F, what is the value of $AF: EF$?\n",
    "completion": "\\textbf{Solution:} Construct $EH\\parallel AC$ through point E and intersect BD at point H, as shown in the figure:\\\\\nSince $EH\\parallel AC$,\\\\\nthus $\\triangle BEH\\sim \\triangle BCD$\\\\\ntherefore $\\frac{EH}{CD}=\\frac{BE}{BC}$,\\\\\nsince $BE=3EC$,\\\\\ntherefore $\\frac{EH}{CD}=\\frac{3CE}{4CE}=\\frac{3}{4}$,\\\\\nsince $D$ is the midpoint of $AC$,\\\\\ntherefore $AD=CD$,\\\\\ntherefore $\\frac{EH}{CD}=\\frac{EH}{AD}=\\frac{3}{4}$,\\\\\nsince $EH\\parallel AC$,\\\\\ntherefore $\\frac{AF}{EF}=\\frac{AD}{EH}=\\frac{4}{3}$,\\\\\ntherefore $AF\\colon EF=\\boxed{4\\colon 3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53032947.png"
  },
  {
    "solution": "\\boxed{18}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the diagram, within $\\triangle ABC$, where $AB=10$, $AC=8$, the angle bisectors of $\\angle ABC$ and $\\angle ACB$ intersect at point D. A line through D, $EF\\parallel BC$, intersects $AB$ at point E, and intersects $AC$ at point F. What is the perimeter of $\\triangle AEF$?\n",
    "completion": "\\textbf{Solution:} Since $EF\\parallel BC$\\\\\n$\\therefore \\angle BDE=\\angle DBC$\\\\\nSince $BD$ bisects $\\angle ABC$\\\\\n$\\therefore \\angle DBC=\\angle DBE$\\\\\n$\\therefore \\angle BDE=\\angle DBE$\\\\\n$\\therefore BE=DE$\\\\\nSimilarly, it can be proved that $CF=DF$\\\\\nSince $AB=10$, $AC=8$\\\\\n$\\therefore$ the perimeter of $\\triangle AEF = AE+AF+DE+DF=AE+AF+BE+CF=AB+AC=\\boxed{18}$",
    "image_path": "dataset/mm_math/images/MM_Math/52655307.png"
  },
  {
    "solution": "\\boxed{C(-3, 4)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=x+7$ intersects with the line $y=-2x-2$ at point C, and they intersect the y-axis at points A and B, respectively. What are the coordinates of points A, B, and C?\n",
    "completion": "\\textbf{Solution:} For the line $y=x+7$, when $x=0$, we have $y=0+7=7$, therefore $A(0, 7)$. On the line $y=-2x-2$, when $x=0$, we have $y=0-2=-2$, therefore $B(0, -2)$. Solving the system of equations $\\begin{cases}y=x+7\\\\ y=-2x-2\\end{cases}$ yields $\\begin{cases}x=-3\\\\ y=4\\end{cases}$, therefore $C(-3, 4)$. Consequently, the coordinates of points $A$, $B$, and $C$ are $\\boxed{A(0, 7)}$, $\\boxed{B(0, -2)}$, and $\\boxed{C(-3, 4)}$, respectively.",
    "image_path": "dataset/mm_math/images/MM_Math/50561111_61.png"
  },
  {
    "solution": "\\boxed{25}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$ and $\\triangle DCE$, we have $AC = DE$, $\\angle B = \\angle DCE = 90^\\circ$, and the points A, C, D are collinear in that order, with $AB \\parallel DE$. Therefore, $\\triangle ABC \\cong \\triangle DCE$. Connecting $AE$, when $BC = 5$ and $AB = 12$, find the length of $AD$.\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABC \\cong \\triangle DCE$,\\\\\nit follows that $BC=EC$, $AC=DE$, and $AB=CD$.\\\\\nIn the right triangle $\\triangle ABC$, with $AB=12$ and $BC=5$,\\\\\nby the Pythagorean theorem, we have $AC=\\sqrt{AB^2+BC^2}=\\sqrt{12^2+5^2}=13$,\\\\\nthus, $AD=AC+CD=13+12=\\boxed{25}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52720370_100.png"
  },
  {
    "solution": "\\boxed{55^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, $OC$ is outside $\\angle AOB$, and $OM$, $ON$ are the angle bisectors of $\\angle AOC$ and $\\angle BOC$, respectively. If $\\angle AOB=110^\\circ$ and $\\angle BOC=60^\\circ$, what is the measure of $\\angle MON$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle AOB=110^\\circ$ and $\\angle BOC=60^\\circ$,\\\\\nthus $\\angle AOC=\\angle AOB+\\angle BOC=170^\\circ$,\\\\\nsince OM bisects $\\angle AOC$,\\\\\nthus $\\angle MOC=\\angle MOA=\\frac{1}{2}\\angle AOC=85^\\circ$,\\\\\nsince ON bisects $\\angle BOC$,\\\\\nthus $\\angle BON=\\angle CON=30^\\circ$,\\\\\ntherefore $\\angle MON=\\angle COM-\\angle CON=\\boxed{55^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53010670_127.png"
  },
  {
    "solution": "\\boxed{\\frac{120}{13}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in an isosceles $\\triangle ABC$ with $AB=AC=13$ and $BC=10$, $D$ is the midpoint of side $BC$, and $M$, $N$ are moving points on $AD$ and $AB$, respectively. What is the minimum value of $BM+MN$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $BH\\perp AC$ at H, intersecting AD at point $M'$, and draw $M'N'\\perp AB$ at $N'$, then $BM'+M'N'$ is the minimum value sought.\\\\\nSince $AB=AC$, and D is the midpoint on side BC,\\\\\nthus AD is the bisector of $\\angle BAC$,\\\\\ntherefore $M'H=M'N'$,\\\\\nthus BH is the shortest distance from point B to line AC (perpendicular segment is the shortest),\\\\\nsince $AB=AC=13$, $BC=10$, and D is the midpoint on side BC,\\\\\nthus $AD\\perp BC$,\\\\\ntherefore $AD=12$,\\\\\nsince the area of $\\triangle ABC=\\frac{1}{2}AC\\times BH=\\frac{1}{2}BC\\times AD$,\\\\\nthus $13\\times BH=10\\times 12$,\\\\\nsolving yields: $BH=\\boxed{\\frac{120}{13}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52603482.png"
  },
  {
    "solution": "\\boxed{\\frac{10}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, we have $AB=10$ and $AD=6$. Let E be a point on side $BC$. Fold $\\triangle CDE$ along $DE$ so that point C falls on point F on side $AB$. What is the length of $CE$?\n",
    "completion": "\\textbf{Solution:} Given the property of folding, we can determine that $CE=EF$,\\\\\nfrom the properties of a rectangle, we have $\\angle B=90^\\circ$, $BC=AD=6$,\\\\\nlet $CE=EF=x$, then $BE=6-x$, $BF=AB-AF=2$,\\\\\nin $\\triangle BEF$, by the Pythagorean theorem, we have $EF^2=BE^2+BF^2$,\\\\\n$\\therefore$ $2^2+(6-x)^2=x^2$,\\\\\nsolving for $x$, we get $x=\\frac{10}{3}$,\\\\\n$\\therefore$ $CE=\\boxed{\\frac{10}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52978448_135.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y=kx+b$ intersects with the graph of the inverse proportion function $y=\\frac{m}{x}$ at points $A(1,n)$ and $B(-3,-2)$. What is the area of $\\triangle AOB$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, let the intersection of $AB$ and the $x$-axis be $C$, \\\\\nfor the linear function $y=2x+4$, \\\\\nwhen $y=0$, $2x+4=0$, solving this we find $x=-2$, i.e., $C(-2,0)$, \\\\\nthus $S_{\\triangle AOB}=S_{\\triangle AOC}+S_{\\triangle BOC}=\\frac{1}{2}\\times 2\\times 6+\\frac{1}{2}\\times 2\\times 2=\\boxed{8}$, \\\\\nmeaning the area of $\\triangle AOB$ is $\\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53066101_84.png"
  },
  {
    "solution": "\\boxed{y=\\frac{6}{x}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system, the vertices $A$ and $C$ of rectangle $ABCD$ are located on the negative half of the $x$-axis and the positive half of the $y$-axis, respectively. The $y$-axis bisects side $AB$, and the coordinate of point $A$ is $(-2,0)$, with $AB=5$. What is the expression of the inverse ratio function that passes through point $B$?\n",
    "completion": "\\textbf{Solution:} Construct BF $\\perp$ x-axis through point B, with F being the foot of the perpendicular. Assume AB intersects with the y-axis at point E,\\\\\n$\\because$ The coordinates of point A are $(-2, 0)$,\\\\\n$\\therefore$ OA = 2,\\\\\n$\\because$ The y-axis bisects side AB, and AB = 5,\\\\\n$\\therefore$ AE = BE = $\\frac{1}{2}$AB = $\\frac{5}{2}$,\\\\\n$\\because$ BF $\\parallel$ y-axis,\\\\\n$\\therefore$ $\\angle$AOE = $\\angle$AFB, $\\angle$AEO = $\\angle$ABF,\\\\\n$\\therefore$ $\\triangle$AOE $\\sim$ $\\triangle$AFB,\\\\\n$\\therefore$ $\\frac{AO}{AF} = \\frac{AE}{AB} = \\frac{1}{2}$,\\\\\n$\\therefore$ AF = 2AO = 4,\\\\\n$\\therefore$ OF = AF - OA = 4 - 2 = 2,\\\\\n$\\therefore$ BF = $\\sqrt{AB^{2} - AF^{2}} = \\sqrt{5^{2} - 4^{2}} = 3$,\\\\\n$\\therefore$ B(2, 3),\\\\\nAssume the expression for the inverse proportion function passing through point B is $y = \\frac{k}{x}$,\\\\\nSubstituting B(2, 3) into $y = \\frac{k}{x}$ yields:\\\\\n3 = $\\frac{k}{2}$,\\\\\n$\\therefore$ k = 6,\\\\\n$\\therefore$ The expression for the inverse proportion function passing through point B is: $\\boxed{y=\\frac{6}{x}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51403201.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $P$ is a point inside $\\triangle ABC$, with $AP \\perp BP$, $BP=12$, and $CP=15$. Points $D$, $E$, $F$, and $G$ are the midpoints of $AP$, $BP$, $BC$, and $AC$, respectively. If the perimeter of quadrilateral $DEFG$ is 28, what is the length of $AP$?\n",
    "completion": "\\textbf{Solution:} By the problem, $DE$, $EF$, $FG$, and $GD$ are the medians of $\\triangle PAB$, $\\triangle BPC$, $\\triangle CAB$, and $\\triangle ACP$, respectively, \\\\\n$\\therefore$DE=GF=$\\frac{1}{2}$AB, EF=GD=$\\frac{1}{2}$PC, \\\\\n$\\because$ The perimeter of the quadrilateral $DEFG$ is 28, \\\\\n$\\therefore$AB+PC=28, \\\\\n$\\because$PC=15, \\\\\n$\\therefore$AB=13, \\\\\nIn $\\triangle PAB$, $\\angle APB=90^\\circ$, BP=12, \\\\\n$\\therefore$AP=$\\sqrt{AB^{2}-BP^{2}}=\\sqrt{169-144}=\\boxed{5}$",
    "image_path": "dataset/mm_math/images/MM_Math/53538182.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{13} - 5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, within rectangle $ABCD$, $AB=8$, $BC=12$. Point $E$ is on $AB$, with $AE=5$. $P$ is a point on $AD$, and the rectangle is folded along $PE$ such that point $A$ falls on point ${A}^{\\prime}$. Connect $AC$ and let it intersect $PE$ at point $F$, with $AP=x$. If point ${A}^{\\prime}$ lies on the bisector of $\\angle BAC$, what is the length of $FC$?\n",
    "completion": "\\textbf{Solution:} As shown in Figure 1,\\\\\n$\\because AA'$ bisects $\\angle BAC$,\\\\\n$\\therefore \\angle EAA' = \\angle FAA'$,\\\\\nFrom the folding, it is known that $AA' \\perp EF$,\\\\\n$\\therefore \\angle EAA' + \\angle AEF = 90^\\circ$, $\\angle AFE + \\angle FAA' = 90^\\circ$,\\\\\n$\\therefore \\angle AEF = \\angle AFE$,\\\\\n$\\therefore AE = AF = 5$,\\\\\n$\\therefore CF = AC - AF = 4\\sqrt{13} - 5 = \\boxed{4\\sqrt{13} - 5}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52793389_153.png"
  },
  {
    "solution": "\\boxed{\\frac{54}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the side \\(OB\\) of the quadrilateral \\(AOBD\\) coincides with the positive half of the x-axis, \\(AD \\parallel OB\\), \\(DB \\perp x\\)-axis, and the diagonals \\(AB\\), \\(OD\\) intersect at point \\(M\\). It is given that \\(AD:OB = 2:3\\), and the area of \\(\\triangle AMD\\) is 4. If the graph of the inverse proportion function \\(y=\\frac{k}{x}\\) exactly passes through point \\(M\\), what is the value of \\(k\\)?\n",
    "completion": "\\textbf{Solution:} Construct MH $\\perp$ OB at H.\\\\\nSince AD $\\parallel$ OB,\\\\\n$\\therefore$ $\\triangle ADM \\sim \\triangle BOM$,\\\\\n$\\therefore$ $\\frac{S_{\\triangle ADM}}{S_{\\triangle BOM}} = \\left(\\frac{AD}{OB}\\right)^{2} = \\frac{4}{9}$,\\\\\nSince $S_{\\triangle ADM} = 4$,\\\\\n$\\therefore$ $S_{\\triangle BOM} = 9$,\\\\\nSince DB $\\perp$ OB, MH $\\perp$ OB,\\\\\n$\\therefore$ MH $\\parallel$ DB,\\\\\n$\\therefore$ $\\frac{OH}{HB} = \\frac{OM}{DM} = \\frac{OB}{AD} = \\frac{3}{2}$,\\\\\n$\\therefore$ OH = $\\frac{3}{5}$OB,\\\\\n$\\therefore$ $S_{\\triangle MOH} = \\frac{3}{5} \\times S_{\\triangle OBM} = \\frac{27}{5}$,\\\\\nSince $\\frac{k}{2} = \\frac{27}{5}$,\\\\\n$\\therefore$ k = $\\boxed{\\frac{54}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51446501.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, with $BC=4$, D and E are respectively the midpoints of $AB$ and $AC$, G and H are respectively the midpoints of $AD$ and $AE$, then what is the length of $GH$?\n",
    "completion": "\\textbf{Solution:} Since $D$ and $E$ are the midpoints of $AB$ and $AC$ respectively,\\\\\nhence $DE$ is the median of $\\triangle ABC$,\\\\\nthus $DE=\\frac{1}{2}BC=\\frac{1}{2}\\times 4=2$,\\\\\nsince $G$ and $H$ are the midpoints of $AD$ and $AE$ respectively,\\\\\nhence $GH$ is the median of $\\triangle ADE$,\\\\\nthus $GH=\\frac{1}{2}DE=\\frac{1}{2}\\times 2=\\boxed{1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53572646.png"
  },
  {
    "solution": "\\boxed{(0,3)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given the quadratic function $y=x^{2}-4x+3$, let its intersection points with the $x$-axis be A and B (with point A to the left of point B), and its intersection point with the $y$-axis be C. Find the coordinates of points A, B, and C.\n",
    "completion": "\\textbf{Solution:} Since $y=x^{2}-4x+3=(x-1)(x-3)=0$, \\\\\nwe find $x_{1}=1, x_{2}=3$, \\\\\nthus, the intersections of the quadratic function $y=x^{2}-4x+3$ and the x-axis are $\\boxed{A(1,0)}$ and $\\boxed{B(3,0)}$.\\\\\nSetting $x=0$, we have $y=3$, which means the coordinates of point $C$ are $\\boxed{(0,3)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52992235_171.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABCD$, point E is the midpoint of side BC, and the diagonals AC and BD intersect at point O. The perimeter of $\\triangle ABC$ is 16. If we connect OE, what is the perimeter of $\\triangle OEC$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral $ABCD$ is a parallelogram, \\\\\nthus $OA=OC=\\frac{1}{2}AC$, \\\\\nsince point $E$ is the midpoint of side $BC$, \\\\\nthus $BE=CE=\\frac{1}{2}BC$, and $OE$ is the median of $\\triangle ABC$, \\\\\nthus $OE=\\frac{1}{2}AB$, \\\\\nsince the perimeter of $\\triangle ABC$ is 16, \\\\\nthus $AB+BC+AC=16$, \\\\\nthus the perimeter of $\\triangle OEC$ is $OE+CE+OC=\\frac{1}{2}AB+\\frac{1}{2}BC+\\frac{1}{2}AC=\\frac{1}{2}\\times 16=\\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52387288.png"
  },
  {
    "solution": "\\boxed{8^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AD$ is the altitude from $A$ to side $BC$, $AE$ and $BF$ are the angle bisectors of $\\angle BAC$ and $\\angle ABC$ respectively, $\\angle BAC=56^\\circ$, $\\angle C=70^\\circ$. What is the measure of $\\angle DAE$?\n",
    "completion": "\\textbf{Solution:} Given that $AE$ bisects $\\angle BAC$, \\\\\nwe have $\\angle EAC = \\frac{1}{2}\\angle BAC = \\frac{1}{2}\\times 56^\\circ = 28^\\circ$. \\\\\nSince $AD$ is the height from $A$ to $BC$, \\\\\nit follows that $\\angle ADC = 90^\\circ$. \\\\\nHence, in $\\triangle ADC$, $\\angle DAC = 90^\\circ - \\angle C = 70^\\circ = 20^\\circ$. \\\\\nTherefore, $\\angle DAE = \\angle EAC - \\angle DAC = 28^\\circ - 20^\\circ = \\boxed{8^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52845737_114.png"
  },
  {
    "solution": "\\boxed{20}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C = 90^\\circ$, $AD$ bisects $\\angle BAC$ and intersects $BC$ at point $D$. If $BD : CD = 3 : 2$ and the distance from point $D$ to $AB$ is 8, what is the length of $BC$?\n",
    "completion": "\\textbf{Solution}: Draw $DE\\perp AB$ at point $E$ through point $D$,\\\\\nsince $AD$ bisects $\\angle BAC$, $\\angle C=90^\\circ$, and $DE\\perp AB$,\\\\\nit follows that $CD=DE=8$,\\\\\nalso, since $BD:CD=3:2$,\\\\\nit follows that $BD=12$,\\\\\nhence, $BC=BD+DC=\\boxed{20}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52837436.png"
  },
  {
    "solution": "\\boxed{\\sqrt{10}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In rectangle $ABCD$, $AB=9$ and $AD=3$, $M$ and $N$ are points on $AB$ and $CD$, respectively. When quadrilateral $MBCN$ is folded along $MN$, point $B$ coincides exactly with point $D$, and point $C$ coincides with point $E$. Connect $BN$. What is the length of the fold $MN$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect BD, \\\\\n$\\because$ Quadrilateral ABCD is a rectangle, \\\\\n$\\therefore$ $\\angle A=90^\\circ$, \\\\\nIn $\\triangle ABD$, AB=9, AD=3, \\\\\n$\\therefore$ $BD=\\sqrt{AD^2+AB^2}=3\\sqrt{10}$, \\\\\nFrom (2), we get: AM=4, \\\\\n$\\therefore$BM=5, \\\\\n$\\because$ ${S}_{\\text{rhombus }BMDN}=\\frac{1}{2}MN\\cdot BD=BM\\cdot AD$, \\\\\n$\\therefore$ $\\frac{1}{2}MN \\times 3\\sqrt{10}=5\\times 3$, \\\\\nSolving this gives: $MN=\\boxed{\\sqrt{10}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52302802_140.png"
  },
  {
    "solution": "\\boxed{54^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a ruler is placed together with a set square that includes a $30^\\circ$ angle. If $\\angle 1 = 24^\\circ$, what is the measure of $\\angle 2$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, \\\\\nsince $\\angle$BEF is an exterior angle of $\\triangle AEF$, $\\angle 1=24^\\circ$, $\\angle F=30^\\circ$, \\\\\ntherefore $\\angle$BEF=$\\angle 1+\\angle F=54^\\circ$, \\\\\nsince AB$\\parallel$CD, \\\\\ntherefore $\\angle 2=\\angle$BEF=$\\boxed{54^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52653511.png"
  },
  {
    "solution": "\\boxed{x^{2}+2x-3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=ax^2+bx+c$ ($a \\neq 0$) has its axis of symmetry as the line $x=-1$, intersects the x-axis at points A and B, and intersects the y-axis at point C, where OA = OC. The coordinates of point A are ($-3$, $0$). What is the equation of the parabola?\n",
    "completion": "\\textbf{Solution:} Let $x=0$, then $y=c$, \\\\\n$\\therefore OC=-c$, \\\\\n$\\because OA=OC$, \\\\\n$\\therefore 3=-c$, i.e., $c=-3$. \\\\\n$\\because$ The axis of symmetry is the line $x=-1$, and the coordinates of point A are $(-3,0)$, \\\\\naccording to the problem, we get: $\\begin{cases}-\\frac{b}{2a}=-1\\\\ 9a-3b+c=0\\end{cases}$, \\\\\nSolving this, we get: $\\begin{cases}a=1\\\\ b=2\\end{cases}$. \\\\\n$\\therefore$ The equation of the parabola is $y=x^{2}+2x-3$. Therefore, the answer is $y=x^{2}+2x-3=\\boxed{x^{2}+2x-3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52765835_125.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AE$ and $BD$ are two altitudes of $\\triangle ABM$, where $AE$ and $BD$ intersect at point $C$, and $AE=BE$. $\\triangle AME \\cong \\triangle BCE$; what is the measure of $\\angle MDE$ in degrees?\n",
    "completion": "\\textbf{Solution:} Draw $EF\\perp ED$ through point E and intersect $BC$ at point F,\\\\\nsince $\\angle DEF=\\angle AEB$,\\\\\ntherefore $\\angle DEA=\\angle BEF$,\\\\\nin $\\triangle AED$ and $\\triangle BEF$,\\\\\n$\\left\\{\\begin{array}{l}\n\\angle DEA=\\angle BEF \\\\\nAE=BE \\\\\n\\angle DAE=\\angle FBE\n\\end{array}\\right.$,\\\\\ntherefore $\\triangle AED\\cong \\triangle BEF$ (ASA),\\\\\nthus $ED=EF$,\\\\\ntherefore $\\angle EDF=\\angle EFD=45^\\circ$,\\\\\nsince $\\angle BDE=90^\\circ$,\\\\\nthus $\\angle MDE=\\boxed{45^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52918922_100.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the Cartesian coordinate system, the graph of the linear function $y=-\\frac{3}{4}x+6$ intersects with the coordinate axes at points $A$ and $B$. Segment $AE$ bisects $\\angle BAO$ and intersects the x-axis at point $E$. A line is drawn from point $B$ perpendicular to $AE$, denoted as $BF$, with $F$ as the foot of the perpendicular. The segment $OF$ is then connected. Determine the shape of $\\triangle OFB$ and calculate the area of $\\triangle OFB$.\n",
    "completion": "\\textbf{Solution:} Draw a line $FM$ perpendicular to the x-axis from point F, intersecting the x-axis at point M,\\\\\nsince $A(0,6), E(3,0)$,\\\\\nin $\\triangle AOE$, $AE= \\sqrt{OA^2+OE^2}=\\sqrt{6^2+3^2}=3\\sqrt{5}$,\\\\\nsince $BF\\perp AE$,\\\\\ntherefore $\\angle AOE=\\angle BFE$,\\\\\nalso, since $\\angle AEO=\\angle BEF$,\\\\\ntherefore $\\triangle AOE \\sim \\triangle BEF$,\\\\\ntherefore $\\frac{BF}{OA}=\\frac{BE}{AE}$,\\\\\ntherefore $\\frac{BF}{6}=\\frac{5}{3\\sqrt{5}}$,\\\\\nsolving, we find $BF= 2\\sqrt{5}$,\\\\\nsince $\\angle FMB=\\angle BFE, \\angle MBF=\\angle FBE$,\\\\\ntherefore $\\triangle BMF \\sim \\triangle BFE$,\\\\\ntherefore $\\frac{BF}{BE}=\\frac{BM}{BF}$,\\\\\ntherefore $\\frac{2\\sqrt{5}}{5}=\\frac{BM}{2\\sqrt{5}}$,\\\\\nsolving, we find $BM=4$,\\\\\nin $\\triangle BMF$, $FM= \\sqrt{BF^2-BM^2}=2$,\\\\\nalso, since the coordinates of point B are $(8,0)$,\\\\\ntherefore $OM=BM=4$ and $FM\\perp x$-axis,\\\\\ntherefore $OF=BF$,\\\\\ntherefore $\\triangle OFB$ is an isosceles triangle,\\\\\ntherefore the area of $\\triangle OFB$ is $\\frac{1}{2}OB\\cdot FM=\\frac{1}{2}\\times 8\\times 2=\\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50675755_177.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the side length of square paper $ABCD$ is 12, where $E$ is a point on side $CD$. Connect $AE$ and fold the paper so that point $A$ falls exactly on point $G$ on $AE$, creating a crease $BF$ that intersects with $AD$ at point $F$. When $E$ is the midpoint of $CD$, what is the length of $AF$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a square, \\\\\nit follows that $AB=AD$, and $\\angle BAD=\\angle D=90^\\circ$, \\\\\nSince $\\angle ABF+\\angle BAG=90^\\circ$, and $\\angle DAE+\\angle BAG=90^\\circ$, \\\\\nit follows that $\\angle ABF=\\angle DAE$, \\\\\nTherefore, $\\triangle ABF\\cong \\triangle DAE$ (by ASA), \\\\\nHence, $AP=DE$, $BF=AE$ \\\\\nSince E is the midpoint of CD, it follows that $DC=12$, therefore $AF=DE=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52211582_74.png"
  },
  {
    "solution": "\\boxed{\\frac{5}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, with O as the center and AB as the diameter, a circle is drawn. A point on $\\odot O$ is chosen, and AB is extended to point D. Line DC is drawn, and a tangent to $\\odot O$ at point A intersects the extension of DC at point E, with $\\angle DCB = \\angle DAC$. CD is a tangent to $\\odot O$; if AD = 6 and $\\tan\\angle DCB = \\frac{2}{3}$, find the length of AE.\n",
    "completion": "\\textbf{Solution:} Since $EA$ is the tangent to circle $\\odot O$,\\\\\n$\\therefore EC=EA$, and $OE\\perp AC$,\\\\\n$\\therefore \\angle BAC=\\angle OEA$ (the complementary angles of equal angles are equal),\\\\\n$\\therefore \\angle CDB=\\angle OEA$.\\\\\nSince $\\tan\\angle DCB= \\frac{2}{3}$,\\\\\n$\\therefore \\tan\\angle OEA= \\frac{OA}{AE}=\\frac{2}{3}$,\\\\\nSince right $\\triangle DCO\\sim \\triangle DAE$,\\\\\n$\\therefore \\frac{DC}{DA}=\\frac{OC}{AE}=\\frac{OA}{AE}=\\frac{2}{3}$,\\\\\n$\\therefore CD= \\frac{2}{3}\\times6=4$,\\\\\nIn the right $\\triangle DAE$, let $AE=x$,\\\\\n$\\therefore (x+4)^{2}=x^{2}+6^{2}$,\\\\\nSolving this, we get $x=\\boxed{\\frac{5}{2}}$.\\\\\nThat is, the length of $AE$ is $\\boxed{\\frac{5}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52623507_122.png"
  },
  {
    "solution": "\\boxed{B(8, 0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the Cartesian coordinate system, the parabola $y = ax^2 - 6ax - 16a \\, (a \\neq 0)$ intersects the $x$-axis at points $A$ and $B$, and intersects the $y$-axis at point $C$. The line $BC$ is drawn, and it is known that point C is at $(0,4)$. Find the coordinates of points $A$ and $B$, and the equation of the parabola?\n",
    "completion": "\\textbf{Solution:}\n\nGiven $y=-\\frac{1}{4}x^2+\\frac{3}{2}x+4$ passes through point $C(0, 4)$,\\\\\n$\\therefore -16a=4$,\\\\\nsolving for $a$ gives $a=-\\frac{1}{4}$,\\\\\n$\\therefore y=-\\frac{1}{4}x^2+\\frac{3}{2}x+4$;\\\\\nLet $y=0$, i.e., $-\\frac{1}{4}x^2+\\frac{3}{2}x+4=0$,\\\\\nsolving yields: $x_1=-2, x_2=8$\\\\\n$\\therefore \\boxed{A(-2, 0)}, \\boxed{B(8, 0)}$",
    "image_path": "dataset/mm_math/images/MM_Math/53148288_71.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AD$ and $AF$ are respectively the median and height of $\\triangle ABC$, $BE$ is the angle bisector of $\\triangle ABD$. If $\\angle BED=40^\\circ$ and $\\angle BAD=25^\\circ$, find the measure of $\\angle BAF$.\n",
    "completion": "\\textbf{Solution:} In $\\triangle ABE$, $\\angle BED$ is an exterior angle, and $\\angle BED=40^\\circ$, $\\angle BAD=25^\\circ$,\\\\\n$\\therefore \\angle ABE=\\angle BED-\\angle BAD=40^\\circ-25^\\circ=15^\\circ$.\\\\\nSince $BE$ is the angle bisector of $\\triangle ABD$,\\\\\n$\\therefore \\angle ABC=2\\angle ABE=2\\times 15^\\circ=30^\\circ$.\\\\\nSince $AF\\perp BC$,\\\\\n$\\therefore \\angle AFB=90^\\circ$,\\\\\nIn $Rt\\triangle ABF$, $\\angle BAF=90^\\circ-\\angle ABC=90^\\circ-30^\\circ=\\boxed{60^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52655159_116.png"
  },
  {
    "solution": "\\boxed{(2, 3)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, the line $y=-\\frac{1}{2}x+2$ intersects the x-axis at point B and the y-axis at point C. A parabola with the axis of symmetry $x=\\frac{3}{2}$ passes through points B and C and intersects the negative x-axis at point A. Let P be a point on the parabola. Connect AP, and rotate line segment AP clockwise by $90^\\circ$ around point A to obtain line segment AQ. When the distance from point Q to the axis of symmetry is $\\frac{1}{2}$, what are the coordinates of point P?\n",
    "completion": "\\textbf{Solution:}\n\nLet $PH \\perp AB$, $QG \\perp AB$, and $QM \\perp l$, where H, G, M are the feet of the perpendiculars. Due to the symmetry of the parabola, it can be derived that $A(-1, 0)$,\n\nsince $AP = AQ$, $\\angle APH = \\angle QAG = 90^\\circ - \\angle PAH$, $\\angle AHP = \\angle QGA = 90^\\circ$,\n\nthus $\\triangle AHP \\cong \\triangle QGA \\ (AAS)$,\n\ntherefore, $PH = AG$,\n\nsince the distance from A to the axis of symmetry is $\\frac{5}{2}$ and the distance from Q to the axis of symmetry is $\\frac{1}{2}$,\n\nwhen Q is on the left side of the axis of symmetry, $PH = AG = 2$,\n\nwhen Q is on the right side of the axis of symmetry, $PH = AG = 3$,\n\ntherefore, $-\\frac{1}{2}x^2 + \\frac{3}{2}x + 2 = 2$ or $-\\frac{1}{2}x^2 + \\frac{3}{2}x + 2 = 3$,\n\nsolving yields $x_1 = 0$, $x_2 = 3$ or $x_1 = 1$, $x_1 = 2$\n\nthus, the coordinates of point P are $\\boxed{(0, 2)}$ or $\\boxed{(3, 2)}$ or $\\boxed{(1, 3)}$ or $\\boxed{(2, 3)}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52977381_195.png"
  },
  {
    "solution": "\\boxed{x^2-4x}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the parabola passes through points $O(0, 0)$ and $A(5, 5)$, and its axis of symmetry is $x=2$. Find the analytic equation of the parabola.\n",
    "completion": "Solution: Suppose the equation of the parabola is $y=ax^2+bx+c$, then:\n\\[\n\\begin{cases}\nc=0 \\\\\n5=25a+5b+c \\\\\n-\\frac{b}{2a}=2\n\\end{cases}\n\\]\nSolving for:\n\\[\n\\begin{cases}\na=1 \\\\\nb=-4 \\\\\nc=0\n\\end{cases}\n\\]\n$\\therefore y=x^2-4x=\\boxed{x^2-4x}$",
    "image_path": "dataset/mm_math/images/MM_Math/52747459_65.png"
  },
  {
    "solution": "\\boxed{(4, 1)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, rotating $\\triangle ABC$ about point $C(0, -1)$ by $180^\\circ$ results in $\\triangle A'B'C'$. If the coordinates of point $A$ are $(-4, -3)$, what are the coordinates of point $A'$?\n",
    "completion": "\\textbf{Solution:} Let $A'$E$\\perp$y-axis at point E, and AD$\\perp$y-axis at point D, then $\\angle A'$EC=$\\angle$ADC, \\\\\n$\\because$ $\\angle A'$CE=$\\angle$ACD, AC=$A'$C, \\\\\n$\\therefore$ $\\triangle A'$EC$\\cong \\triangle$ADC (AAS), \\\\\n$\\therefore$ AD=$A'$E, CE=CD, \\\\\n$\\because$ the coordinates of point A are (-4, -3), OC=1, \\\\\n$\\therefore$ AD=4, OD=3, \\\\\n$\\therefore$ AD=$A'$E=4, CD=2, \\\\\n$\\therefore$ CE=2, \\\\\n$\\therefore$ OE=1, \\\\\n$\\therefore$ the coordinates of point $A'$ are $\\boxed{(4, 1)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51212132.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $C$ is the midpoint of the line segment $AB$, and $D$ is on the line segment $CB$, and $DA=6$, $DB=4$. Find the length of $AB$.\n",
    "completion": "\\textbf{Solution:} Since $DA=6, DB=4$,\\\\\nit follows that $AB=6+4=\\boxed{10}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55000668_18.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the equation of line ${l}_{1}$ is $y=kx+b(k \\ne 0)$, and the equation of line ${l}_{2}$ is $y=x$. Line ${l}_{1}$ passes through points $A(6, 0)$ and $B(0, 3)$, and line ${l}_{1}$ intersects line ${l}_{2}$ at point C. What is the area of $\\triangle COB$?\n",
    "completion": "\\textbf{Solution:} As shown in Figure 1, draw $CD\\perp OB$ at point D through point C,\\\\\nSolve $y=-\\frac{1}{2}x+3$ and $y=x$ simultaneously,\\\\\n$\\begin{cases}y=-\\frac{1}{2}x+3\\\\ y=x\\end{cases}$,\\\\\nto get $\\begin{cases}x=2\\\\ y=2\\end{cases}$.\\\\\nThus, the coordinates of the intersection point C of line $l_1$ and line $l_2$ are: (2, 2)\\\\\n$\\therefore CD=x_C=2$\\\\\nObviously, $OB=y_B=3$\\\\\n$\\therefore S_{\\triangle COB}=\\frac{1}{2}OB\\cdot CD=\\frac{1}{2}\\times 3\\times 2=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52386525_135.png"
  },
  {
    "solution": "\\boxed{9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the length of segment $AB=4$. The segment $AB$ is extended to point C such that $BC=\\frac{1}{2}AB$. The segment $AB$ is also extended in the opposite direction to point $D$ such that $AC=2AD$. Find the length of segment $CD$.\n",
    "completion": "\\textbf{Solution:} Given that $AB=4\\text{cm}$, $BC=\\frac{1}{2}AB$,\\\\\nit follows that $BC=2\\text{cm}$, $AC=AB+BC=4+2=6\\text{cm}$,\\\\\nsince $AC=2AD$,\\\\\nit implies that $AD=3\\text{cm}$,\\\\\ntherefore $CD=AD+AC=3+6=\\boxed{9}\\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53119305_116.png"
  },
  {
    "solution": "\\boxed{54^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $C$ is on line $AB$, and $CD$ bisects $\\angle ACE$. If $\\angle 1 = 63^\\circ$, then what is the measure of $\\angle BCE$ in degrees?\n",
    "completion": "\\textbf{Solution:} Since $CD$ bisects $\\angle ACE$, and $\\angle 1=63^\\circ$,\n\nit follows that $\\angle ACE=2\\angle 1=126^\\circ$,\n\nSince $\\angle ACE+\\angle BCE=180^\\circ$,\n\nit follows that $\\angle BCE=180^\\circ−126^\\circ=\\boxed{54^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52375758.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{2}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown, it is known that $\\triangle ABC$ and $\\triangle ADE$ are both isosceles right-angled triangles, with $\\angle BAC=\\angle ADE=90^\\circ$, and $AB=AC=1$, $AD=DE=\\sqrt{5}$. Point D lies on line BC, and the extension of line segment $EA$ intersects line BC at point F. What is the length of $FB$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw AH $\\perp$ BC at point H,\\\\\n$\\because$ $\\angle$BAC=$90^\\circ$, AB=AC=1,\\\\\n$\\therefore$ BC=$\\sqrt{2}$,\\\\\n$\\because$ AH$\\perp$BC,\\\\\n$\\therefore$ BH=CH=$\\frac{\\sqrt{2}}{2}$,\\\\\n$\\therefore$ AH=$\\frac{\\sqrt{2}}{2}$,\\\\\n$\\because$ AD=DE=$\\sqrt{5}$,\\\\\n$\\therefore$ DH=$\\sqrt{AD^{2}-AH^{2}}=\\sqrt{5-\\frac{1}{2}}=\\frac{3\\sqrt{2}}{2}$,\\\\\n$\\therefore$ CD=DH−CH=$\\sqrt{2}$,\\\\\n$\\because$ $\\angle$ABC=$\\angle$ACB=$45^\\circ$,\\\\\n$\\therefore$ $\\angle$ABF=$\\angle$ACD=$135^\\circ$,\\\\\n$\\because$ $\\angle$DAE=$45^\\circ$,\\\\\n$\\therefore$ $\\angle$DAF=$135^\\circ$,\\\\\n$\\because$ $\\angle$BAC=$90^\\circ$,\\\\\n$\\therefore$ $\\angle$BAF+$\\angle$DAC=$45^\\circ$,\\\\\n$\\because$ $\\angle$BAF+$\\angle$F=$45^\\circ$,\\\\\n$\\therefore$ $\\angle$F=$\\angle$DAC,\\\\\n$\\therefore$ $\\triangle$ABF$\\sim$ $\\triangle$DCA,\\\\\n$\\therefore$ $\\frac{AB}{CD}=\\frac{BF}{AC}$,\\\\\n$\\therefore$ $\\frac{1}{\\sqrt{2}}=\\frac{BF}{1}$,\\\\\n$\\therefore$ BF=$\\boxed{\\frac{\\sqrt{2}}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51402480.png"
  },
  {
    "solution": "\\boxed{(\\sqrt{5}, 8-2\\sqrt{5})}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=-2x+8$ intersects the $x$-axis and the $y$-axis at points $A$ and $B$, respectively, and the line $y=\\frac{1}{2}x+3$ intersects the $y$-axis at point $C$. The two lines intersect at point $D$. As shown in Figure 3, under the condition of (2), point $F$ is on the line segment $BC$, and point $G$ is on the line segment $AB$. Connect $CG$ and $FG$. When $CG=FG$ and $\\angle CGF=\\angle ABC$, what are the coordinates of point $G$?\n",
    "completion": "\\textbf{Solution:} Given $BC=AC$,\\\\\nit follows that $\\angle ABC=\\angle CAB$,\\\\\nsince $\\angle CGF=\\angle ABC$ and $\\angle AGF=\\angle ABC+\\angle BFG=\\angle AGC+\\angle CGF$,\\\\\nit implies $\\angle AGC=\\angle BFG$ and $FG=CG$, $\\angle ABC=\\angle CAB$,\\\\\nthus $\\triangle ACG\\cong \\triangle BGF$ (By the AAS criterion),\\\\\nimplying $BG=AC=5$,\\\\\nlet point $G(a, -2a+8)$,\\\\\nhence $(-2a+8-8)^2+(a-0)^2=5^2$,\\\\\nleading to $a=\\pm \\sqrt{5}$,\\\\\nsince point $G$ lies on segment $AB$,\\\\\nit must be $a=\\sqrt{5}$,\\\\\ntherefore, the coordinates of point $G$ are $\\boxed{(\\sqrt{5}, 8-2\\sqrt{5})}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52250730_100.png"
  },
  {
    "solution": "\\boxed{-\\frac{1}{18}x^2+\\frac{1}{4}x+3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the parabola $y=ax^2+bx+3$ (where $a\\neq 0$) intersects the x-axis at points $A(-4,0)$ and $B(6,0)$, and intersects the y-axis at point $C$. Suppose $G$ is a moving point between $A$ and $C$ on the parabola, and a line $GD$ is drawn through $G$ parallel to the x-axis, intersecting the parabola at point $D$. Through points $D$ and $G$, perpendiculars to the x-axis are drawn, with the feet of the perpendiculars being $E$ and $F$, respectively, forming rectangle $DEFG$. What is the equation of this parabola?\n",
    "completion": "Solution: Substituting $A(-4,0)$, $B(6,0)$ into $y=ax^2+bx+3$, we get\n\\[\n\\begin{cases}\n16a-4b+3=0\\\\\n36a+6b+3=0\n\\end{cases}\n\\]\nSolving for $a$ and $b$, we find\n\\[\n\\begin{cases}\na=-\\frac{1}{18}\\\\\nb=\\frac{1}{4}\n\\end{cases}\n\\]\n$\\therefore$ The equation of the parabola is $y=-\\frac{1}{18}x^2+\\frac{1}{4}x+3=\\boxed{-\\frac{1}{18}x^2+\\frac{1}{4}x+3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52550670_87.png"
  },
  {
    "solution": "\\boxed{15^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: A pair of set squares OAB and OCD are placed as shown in the figure, where ($\\angle ABO=30^\\circ$, $\\angle ODC=45^\\circ$). If OM and ON bisect $\\angle AOD$ and $\\angle BOC$ respectively. Rotate set square OCD clockwise around point O in the figure, if $\\angle AOC=90^\\circ$, find the degree of the rotation angle $\\alpha$.\n",
    "completion": "\\textbf{Solution:} Given $\\because \\angle AOC=90^\\circ$ and $\\angle COD=45^\\circ$,\\\\\n$\\therefore \\angle AOD=\\angle AOC-\\angle COD=45^\\circ$,\\\\\n$\\therefore$ the angle of rotation $\\angle BOD=\\angle AOB-\\angle AOD=60^\\circ-45^\\circ=\\alpha =\\boxed{15^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/8928654_73.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, point E is on the extension line of BC. The angle bisectors of $\\angle ABC$ and $\\angle ACE$ intersect at point D, $\\angle D=15^\\circ$. What is the degree measure of $\\angle A$?\n",
    "completion": "\\textbf{Solution:} \\\\\nSince the bisector of $\\angle ABC$ intersects the bisector of $\\angle ACE$ at point D, \\\\\nTherefore, $\\angle CBD=\\frac{1}{2}\\angle ABC$, $\\angle ECD=\\frac{1}{2}\\angle ACE$, \\\\\nSince $\\angle ACE=\\angle A +\\angle ABC$, $\\angle ECD=\\angle CBD+\\angle D$, \\\\\nTherefore, $\\angle CBD + \\angle D=\\frac{1}{2}(\\angle A + \\angle ABC)=\\frac{1}{2}(\\angle A+2\\angle CBD)$, \\\\\nTherefore, $\\angle D=\\frac{1}{2}\\angle A$, \\\\\nSince $\\angle D=15^\\circ$, \\\\\nTherefore, $\\angle A=\\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51435316.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ is an inscribed triangle of circle $\\odot O$ with $AB$ as the diameter. $BD$ is tangent to $\\odot O$ at point $B$, and intersects the extension line of $AC$ at point $D$. $E$ is the midpoint of $BD$, and $CE$ intersects the extension line of $BA$ at point $F$. Given that $BD=8$, and $BE = \\frac{1}{3}EF$. $FC$ is a tangent line of $\\odot O$; find the length of $AF$?\n",
    "completion": "\\textbf{Solution:} Given $BD=8$ and point $E$ is the midpoint of $BD$,\\\\\nhence $BE= \\frac{1}{2}BD=4$,\\\\\nsince $BE= \\frac{1}{3}EF$,\\\\\nthus $EF=3BE=12$,\\\\\nin $\\triangle FBE$, $BF= \\sqrt{EF^{2}-BE^{2}}=\\sqrt{12^{2}-4^{2}}=8\\sqrt{2}$,\\\\\nfrom (1) we have $\\angle OCF=\\angle ABD=90^\\circ$,\\\\\nsince $\\angle F=\\angle F$,\\\\\nthus $\\triangle FOC\\sim \\triangle FEB$,\\\\\nhence $\\frac{OC}{BE}=\\frac{OF}{EF}$,\\\\\nlet $OC=x$, then $OF=BF-OB=8\\sqrt{2}-x$,\\\\\nhence $\\frac{x}{4}=\\frac{8\\sqrt{2}-x}{12}$,\\\\\nthus $x=2\\sqrt{2}$,\\\\\nhence $AF=8\\sqrt{2}-2x=4\\sqrt{2}$;\\\\\nTherefore, the length of $AF$ is $\\boxed{4\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51446035_118.png"
  },
  {
    "solution": "\\boxed{3.6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In rectangle $ABCD$, $AB = 12$, and $BC = 8$. The rectangle is folded along $MN$ so that point $C$ precisely falls on the midpoint $F$ of side $AD$. If the circle with center $O$, which is the center of symmetry of the rectangle, is tangent to $FN$ at point $G$, what is the radius of circle $\\odot O$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect OF, OG, ON, and draw OH$\\perp$DC at point H through O,\\\\\n$\\because$ point O is the center of symmetry for the rectangle, and AB=12, BC=8, F is the midpoint of AD,\\\\\n$\\therefore$ OF=DH=HC=$\\frac{1}{2}$DC=6, OH=FD=$\\frac{1}{2}$AD=4,\\\\\n$\\because$ circle O is tangent to FN at point G,\\\\\n$\\therefore$ OG$\\perp$FN,\\\\\nBy the property of folding, we get: FN=NC, let FN=NC=a, then DN=12−a\\\\\nIn the right-angled triangle FDN, FN$^{2}$=FD$^{2}$+DN$^{2}$, which gives a$^{2}$=4$^{2}$+(12−a)$^{2}$, solving for a, we get a=$\\frac{20}{3}$,\\\\\n$\\therefore$ DN=$\\frac{16}{3}$,\\\\\n$\\therefore$ NH=DH−DN=6−$\\frac{16}{3}$=$\\frac{2}{3}$,\\\\\nIn the right-angled triangle OHN, by Pythagoras' theorem, ON$^{2}$=NH$^{2}$+OH$^{2}$=$\\frac{4}{9}$+16,\\\\\nLet FG=b, then GN=$\\frac{20}{3}$−b,\\\\\nIn the right-angled triangles OGF and OGN, by Pythagoras' theorem, OF$^{2}$−FG$^{2}$=OG$^{2}$=ON$^{2}$−GN$^{2}$,\\\\\n$\\therefore$ 6$^{2}$−b$^{2}$=$\\frac{4}{9}$+16−($\\frac{20}{3}$−b)$^{2}$,\\\\\nSolving for b, we get b=$\\frac{24}{5}$,\\\\\n$\\therefore$ OG$^{2}$=36−($\\frac{24}{5}$)$^{2}$,\\\\\nSolving for OG, we obtain OG=\\boxed{3.6}, which means the radius is 3.6.",
    "image_path": "dataset/mm_math/images/MM_Math/51445130.png"
  },
  {
    "solution": "\\boxed{y = -2x + 6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the graph of the linear function $y = -\\frac{3}{4}x + 6$ intersects the coordinate axes at points $A$ and $B$. The line $AE$ bisects the angle $\\angle BAO$ and intersects the x-axis at point $E$. What is the equation of the line $AE$?\n",
    "completion": "**Solution:**\n\nLet line $EG \\perp AB$ pass through point E and intersect AB at point G,\n\n$\\because$ AE bisects $\\angle BAO$,\n\n$\\therefore \\angle OAE = \\angle BAE$,\n\nAlso, $\\because \\angle AOE = \\angle AGE = 90^\\circ$, AE = AE,\n\n$\\therefore \\triangle AOE \\cong \\triangle AGE$ (AAS),\n\n$\\therefore OE = GE$, AO = AG = 6,\n\nAlso, $\\because$ The coordinates of point A are (0, 6), and the coordinates of point B are (8, 0),\n\nIn $\\triangle AOB$, $AB= \\sqrt{OA^2 + OB^2} = \\sqrt{6^2 + 8^2} = 10$,\n\n$\\therefore BG = AB - AG = 4$,\n\nLet OE = GE = x, then BE = 8 - x,\n\nIn $\\triangle BEG$, $4^2 + x^2 = (8 - x)^2$,\n\nSolving for $x$, we get: $x=3$,\n\n$\\therefore$ The coordinates of point E are (3, 0),\n\nLet the equation of line AE be $y = kx + b$ ($k \\neq 0$),\n\nSubstituting E(3, 0) and A(0, 6) into the equation, we get:\n\n$\\begin{cases} 3k + b = 0 \\\\ b = 6 \\end{cases}$,\n\nSolving for $k$ and $b$, we get:\n\n$\\begin{cases} k = -2 \\\\ b = 6 \\end{cases}$,\n\n$\\therefore$ The equation of line AE is $\\boxed{y = -2x + 6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50675755_176.png"
  },
  {
    "solution": "\\boxed{y_{2}=-x+4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, the graph of the linear function $y_{1}=x+2$ is the straight line $l_{1}$, and the graph of the linear function $y_{2}=kx+b$ is the straight line $l_{2}$. The two lines intersect at point $A(1, a)$. It is known that the intersections of the lines $l_{1}$ and $l_{2}$ with the $x$-axis are points $B$ and $C$, respectively, and the line $l_{2}$ intersects the $y$-axis at point $E(0, 4)$. What is the equation of the line $l_{2}$?\n",
    "completion": "\\textbf{Solution:} Since the linear function $y_{2}=kx+b$ passes through the points $E(0,4)$ and $A(1, 3)$,\\\\\nwe have $\\left\\{\\begin{array}{l}\n3=k+b \\\\\n4=b\n\\end{array}\\right.$,\\\\\nsolving this, we get $\\left\\{\\begin{array}{l}\nk=-1 \\\\\nb=4\n\\end{array}\\right.$,\\\\\n$\\therefore$ the equation of line $l_{2}$ is $\\boxed{y_{2}=-x+4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52395370_109.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, there are two congruent rectangles AEFG and ABCD placed as shown. The line containing CD intersects AE and GF respectively at points H and M. If AB = 3, BC = $ \\sqrt{3} $, and CH = MH, what is the length of the segment MH?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $HQ \\perp GF$, \\\\\n$\\therefore \\angle HQM = 90^\\circ$, \\\\\n$\\because$ Rectangle $AEFG$ and rectangle $ABCD$ are congruent, \\\\\n$\\therefore AD = EF = HQ$, $\\angle ADH = \\angle MQH = \\angle QHA = 90^\\circ$, \\\\\n$\\therefore \\angle QHM + \\angle AHD = \\angle QHM + \\angle QMH$, \\\\\n$\\therefore \\angle AHD = \\angle QHM$, \\\\\n$\\therefore \\triangle ADH \\cong \\triangle MQH$ (AAS), \\\\\n$\\therefore AH = MH$, \\\\\n$\\because AB = 3$, \\\\\n$\\therefore$ Let $DH = x$, then $CH = MH = AH = 3 - x$, \\\\\n$\\because BC = \\sqrt{3}$, \\\\\n$\\therefore AD = \\sqrt{3}$, \\\\\nIn $\\triangle ADH$, $AD^2 + DH^2 = AH^2$, \\\\\n$\\therefore 3 + x^2 = (3 - x)^2$, \\\\\nUpon simplification, we find: $x = 1$, \\\\\n$\\therefore MH = 3 - 1 = \\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51942332.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, all quadrilaterals in the figure are squares, and the triangle is a right-angled triangle. The areas of the two smaller squares are respectively 1 and 2. What is the area of the largest square?\n",
    "completion": "\\textbf{Solution}: Since the areas of the two small squares are respectively 1 and 2,\\\\\nit follows that they are the squares of the two legs of a right triangle,\\\\\nthus, according to the Pythagorean theorem, we have: $S_{\\text{large square}}=S_{\\text{area of the small square is 1}}+S_{\\text{area of the small square is 2}}=\\boxed{3}$",
    "image_path": "dataset/mm_math/images/MM_Math/51968610.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle A=90^\\circ$, $AB=AC$, $BD$ bisects $\\angle ABC$, $CE \\perp BD$ at $E$. If $BD=8$, what is the length of $CE$?\n",
    "completion": "\\textbf{Solution:} Extend BA and CE to intersect at point F, \\\\\nsince $\\angle BAC = 90^\\circ$ and $CE \\perp BD$, \\\\\ntherefore, $\\angle BAC = \\angle BEC = \\angle FAC$, \\\\\nsince $\\angle ABD + \\angle ADB = 90^\\circ$ and $\\angle CDE + \\angle ACF = 90^\\circ$, \\\\\nsince $\\angle ADB = \\angle CDE$, \\\\\ntherefore, $\\angle ABD = \\angle ACF$, \\\\\nin $\\triangle ABD$ and $\\triangle ACF$, \\\\\n$\\left\\{\\begin{array}{l}\n\\angle BAD = \\angle CAF \\\\\nAB = AC \\\\\n\\angle ABD = \\angle ACF\n\\end{array}\\right.$ \\\\\ntherefore, $\\triangle ABD \\cong \\triangle ACF$, \\\\\ntherefore, BD = CF = 8, \\\\\nsince BD bisects $\\angle ABC$, \\\\\ntherefore, $\\angle ABE = \\angle CBE$, \\\\\nsince CE$\\perp$BD, \\\\\ntherefore, $\\angle BEF = \\angle BEC = 90^\\circ$ \\\\\nin $\\triangle BEF$ and $\\triangle BEC$, \\\\\n$\\left\\{\\begin{array}{l}\n\\angle FBE = \\angle CBE \\\\\nBE = BE \\\\\n\\angle BEF = \\angle BEC\n\\end{array}\\right.$ \\\\\ntherefore, $\\triangle BEF \\cong \\triangle BEC$, \\\\\ntherefore, EF = EC, \\\\\ntherefore, EC = $\\frac{1}{2}$CF = \\boxed{4}.",
    "image_path": "dataset/mm_math/images/MM_Math/52603850.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $CD$ is the altitude from the right angle of $\\triangle ABC$ to the hypotenuse $AB$. By folding $\\triangle BCD$ along $CD$ such that point $B$ coincides precisely with the midpoint $E$ of $AB$, what is the measure of $\\angle A$?\n",
    "completion": "\\textbf{Solution:} Given that $\\triangle BCD$ is folded along CD, \\\\\nit follows that $BC = CE$, \\\\\nsince E is the midpoint of AB and $\\angle ACB=90^\\circ$, \\\\\nit follows that $CE = BE = AE$, \\\\\nthus, $\\triangle BEC$ is an equilateral triangle, \\\\\nhence, $\\angle B = 60^\\circ$, \\\\\ntherefore, $\\angle A = 90^\\circ - 60^\\circ = \\boxed{30^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/52861952.png"
  },
  {
    "solution": "\\boxed{(0, 5)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, quadrilateral $OABC$ is a rectangular piece of paper placed in a Cartesian coordinate system, where $O$ is the origin, point $A$ is on the negative half-axis of the $x$-axis, and point $C$ is on the positive half-axis of the $y$-axis, with $OA=10$ and $OC=8$. Point $D$ is taken on side $OC$, and the paper is folded along $AD$, with point $O$ landing on point $E$ on side $BC$. What are the coordinates of point $D$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral ABCD is a rectangle,\\\\\nthus AB=OC=8 and BC=OA=10,\\\\\nsince the paper is folded along AD, bringing point O to point E on side BC,\\\\\nthus AE=AO=10 and DE=DO,\\\\\nin right triangle $\\triangle ABE$, where AB=8 and AE=10,\\\\\nthus BE=$\\sqrt{AE^{2}-AB^{2}}=6$,\\\\\nthus CE=BC-BE=4,\\\\\nlet OD=x, then DE=x, and DC=8-x,\\\\\nin right triangle $\\triangle CDE$, since $DE^{2}=CD^{2}+CE^{2}$,\\\\\nthus $x^{2}=(8-x)^{2}+4^{2}$,\\\\\nthus x=5,\\\\\nthus the coordinates of point D are \\boxed{(0, 5)}.",
    "image_path": "dataset/mm_math/images/MM_Math/53359599.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within $\\triangle ABC$, $D$ and $E$ are the midpoints of sides $AB$ and $AC$, respectively. If the area of $\\triangle ADE$ is $\\frac{1}{2}$, what is the area of quadrilateral $DBCE$?\n",
    "completion": "\\textbf{Solution:} Since D and E are the midpoints of AB and AC, respectively,\\\\\nit follows that DE$\\parallel$BC, and AE = CE = $\\frac{1}{2}$AB,\\\\\ntherefore, $\\frac{AE}{AC}$ = $\\frac{1}{2}$, and $\\triangle$ADE$\\sim$$\\triangle$ABC,\\\\\nthus, $\\frac{S_{\\triangle ADE}}{S_{\\triangle ABC}}$ = $\\left(\\frac{AE}{AC}\\right)^2$ = $\\left(\\frac{1}{2}\\right)^2$ = $\\frac{1}{4}$,\\\\\nhence, S$_{\\triangle ABC}$ = 4S$_{\\triangle ADE}$ = 4$\\times$$\\frac{1}{2}$ = 2,\\\\\ntherefore, S$_{\\text{quadrilateral DBCE}}$ = S$_{\\triangle ABC}$ - S$_{\\triangle ADE}$ = 2 - $\\frac{1}{2}$ = $\\boxed{\\frac{3}{2}}$,",
    "image_path": "dataset/mm_math/images/MM_Math/52550292.png"
  },
  {
    "solution": "\\boxed{\\frac{27}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the straight lines $l_{1} \\parallel l_{2} \\parallel l_{3}$. If $AB = 6$, $BC = 10$, and $EF = 9$, then what is the length of $DE$?\n",
    "completion": "\\textbf{Solution:} Since $l_{1} \\parallel l_{2} \\parallel l_{3}$, and $AB = 6$, $BC = 10$, \\\\\nit follows that $\\frac{DE}{EF} = \\frac{AB}{BC} = \\frac{6}{10} = \\frac{3}{5}$, \\\\\nSince $EF = 9$, \\\\\nthen $DE = \\boxed{\\frac{27}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52601696.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In $\\triangle ABC$, the perpendicular bisector of $AB$ intersects $AB$ and $BC$ at points $D$ and $E$, respectively, and the perpendicular bisector of $AC$ intersects $AC$ and $BC$ at points $F$ and $G$, respectively. Given that $BG=9$, $CE=11$, and the perimeter of $\\triangle AEG$ is 16, what is the length of $EG$?\n",
    "completion": "Solution: Since $DE$ bisects $AB$ perpendicularly, and $GF$ bisects $AC$ perpendicularly, \\\\\ntherefore $AE=BE$, and $AG=CG$, \\\\\nsince the perimeter of $\\triangle AEG$ is 16, \\\\\ntherefore $AE+EG+AG=16$, \\\\\nhence $BE+EG+CG=16$, \\\\\nthus $9-EG+EG+11-EG=16$, \\\\\nhence $EG=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52693978.png"
  },
  {
    "solution": "\\boxed{15}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that point C is on line segment AB, and points M and N are on line segments AC and BC, respectively, with $AM=2MC$ and $BN=2NC$. If $MN=5$, what is the length of line segment AB?\n",
    "completion": "\\textbf{Solution:} Since $AM=2MC$ and $BN=2NC$,\\\\\nit follows that $MC= \\frac{1}{3} AC$ and $NC= \\frac{1}{3} BC$,\\\\\nThus $MN=MC+NC= \\frac{1}{3} AC+ \\frac{1}{3} BC= \\frac{1}{3} AB$,\\\\\nIf $MN=5$, then $AB=3MN=15$,\\\\\nAnswer: The length of $AB$ is $\\boxed{15}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53206849_48.png"
  },
  {
    "solution": "\\boxed{110^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle BAC$ is an obtuse angle, and $AF$, $CE$ are the altitudes of this triangle, with $P$ being the midpoint of $AC$. If $\\angle B=35^\\circ$, what is the degree measure of $\\angle EPF$?\n",
    "completion": "\\textbf{Solution:} Given that $AF \\perp BC$,\\\\\nit follows that $\\angle AFB = \\angle AFC = 90^\\circ$,\\\\\nsince $\\angle B = 35^\\circ$,\\\\\nthus $\\angle EAF = \\angle B + \\angle AFB = 125^\\circ$,\\\\\ngiven $CE \\perp BE$,\\\\\nit follows that $\\angle AEC = \\angle AFC = 90^\\circ$,\\\\\nhence $PE = AP = PC = PF$,\\\\\nthus $\\angle AEP = \\angle EAP$, $\\angle AFP = \\angle FAP$,\\\\\ntherefore $\\angle AEP + \\angle EAF + \\angle AFP = 250^\\circ$,\\\\\nhence $\\angle EPF = 360^\\circ - 250^\\circ = \\boxed{110^\\circ}$,\\\\",
    "image_path": "dataset/mm_math/images/MM_Math/52303782.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, point $P$ is a point on the diagonal $BD$ of the rhombus $ABCD$. Line $CP$ is extended to meet $AD$ at point $E$ and the extended line of $BA$ at point $F$. It is given that $\\triangle APD \\cong \\triangle CPD$; $\\triangle APE \\sim \\triangle FPA$; If $PE = 4$ and $PF = 12$, what is the length of $PC$?\n",
    "completion": "\\textbf{Solution:} Given that $\\triangle APE \\sim \\triangle FPA$,\\\\\n$\\therefore \\frac{AP}{PF} = \\frac{PE}{PA}$,\\\\\n$\\therefore PA^{2} = PE \\cdot PF$,\\\\\nSince $\\triangle APD \\cong \\triangle CPD$,\\\\\n$\\therefore PA = PC$,\\\\\n$\\therefore PC^{2} = PE \\cdot PF$,\\\\\nGiven that $PE = 4$, $PF = 12$,\\\\\n$\\therefore PC = 4\\sqrt{3}$.\\\\\nThus, the length of $PC$ is $\\boxed{4\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51458264_157.png"
  },
  {
    "solution": "\\boxed{(-\\sqrt{2}, 0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated in the Cartesian coordinate system, given that the vertices O and B of the rhombus OABC are (0, 0) and (2, 2), respectively, what are the coordinates of the intersection point D of the diagonals of the rhombus after it has been rotated \\(135^\\circ\\) counterclockwise about point O?\n",
    "completion": "\\textbf{Solution:} Given the vertices of the rhombus OABC are O(0,0) and B(2,2), the coordinate of point D is found to be (1,1).\\\\\n$\\therefore OD=\\sqrt{1^2+1^2}=\\sqrt{2}$, the angle between OB and the positive y-axis is $45^\\circ$,\\\\\nand the angle between the positive y-axis and the negative x-axis is $90^\\circ$,\\\\\n$\\therefore$ when the rhombus is rotated $135^\\circ$ counterclockwise around point O, i.e., when line segment OD is rotated $135^\\circ$ counterclockwise around point O, OD is then on the negative x-axis,\\\\\n$\\therefore$ the coordinates of the intersection of the diagonals of the rhombus D are ($-\\sqrt{2}$, 0).\\\\\nThus, the answer is: $(−\\sqrt{2}, 0) = \\boxed{(-\\sqrt{2}, 0)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52344910.png"
  },
  {
    "solution": "\\boxed{30}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, to facilitate pedestrians crossing a certain footbridge, the municipal government constructed ramps at both ends of the 10-meter-high footbridge. The design of the slope satisfies $\\sin A = \\frac{1}{3}$. What is the length of the ramp AC?\n",
    "completion": "\\textbf{Solution:} In $\\triangle ABC$, where $\\angle ABC = 90^\\circ$ and $BC = 10$ meters, with $\\sin A = \\frac{1}{3}$,\\\\\nthen $\\frac{BC}{AC} = \\frac{1}{3}$, that is, $\\frac{10}{AC} = \\frac{1}{3}$,\\\\\nSolving this, we get: $AC = \\boxed{30}$ (meters).",
    "image_path": "dataset/mm_math/images/MM_Math/51174755.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As the figure shows, $AD$ is the median of $\\triangle ABC$, and $E$ is a point on $AD$ with $DE=2AE$. The extension of $CE$ intersects $AB$ at point $F$. If $AF=1.6\\text{cm}$, then what is the length of $AB$ in $\\text{cm}$?\n",
    "completion": "\\textbf{Solution:} Let DM be drawn parallel to CF and intersect AB at M,\\\\\nsince $DM\\parallel CF$,\\\\\nthen $\\frac{BD}{DC}=\\frac{BM}{MF}$,\\\\\nsince AD is the median of $\\triangle ABC$,\\\\\nthen D is the midpoint of BC,\\\\\nthus BD=DC,\\\\\nthus $\\frac{BD}{DC}=\\frac{BM}{MF}=1$,\\\\\nthus BM=MF,\\\\\nsince EF$\\parallel$MD,\\\\\nthen $\\triangle AFE\\sim \\triangle AMD$,\\\\\nalso since DE=2AE,\\\\\nthen $\\frac{AF}{AM}=\\frac{AE}{AD}=\\frac{AE}{DE+AE}=\\frac{AE}{2AE+AE}=\\frac{1}{3}$,\\\\\nthus AM=3AF,\\\\\nsince AM=AF+MF=3AF,\\\\\nthen MF=2AF,\\\\\nsince BM=MF,\\\\\nthen BM=MF=2AF,\\\\\nthus AB=AF+MF+BF=5AF,\\\\\nsince AF=1.6,\\\\\nthus AB=$5\\times 1.6$=\\boxed{8},",
    "image_path": "dataset/mm_math/images/MM_Math/51367392.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the length of the fishing rod $AC$ is $10\\mathrm{m}$, and the length of the fishing line $BC$ exposed above the water surface is $6\\mathrm{m}$. A fisher wants to check the situation on the hook, so he rotates the fishing rod $AC$ to the position $A{C}^{\\prime}$, at this time, the length of the fishing line ${B}^{\\prime}{C}^{\\prime}$ exposed above the water surface is $8\\mathrm{m}$. What is the length of $B{B}^{\\prime}$?\n",
    "completion": "\\textbf{Solution:} Given $\\because$AC=10m, BC=6m, $\\angle$ABC$=90^\\circ$,\\\\\n$\\therefore$AB= $\\sqrt{AC^{2}-BC^{2}}=\\sqrt{10^{2}-6^{2}}=8$ m,\\\\\n$\\because$AC$^{\\prime }$=10m, B$^{\\prime }$C$^{\\prime }$=8m, $\\angle$AB$^{\\prime }$C$^{\\prime }$=$90^\\circ$,\\\\\n$\\therefore$AB$^{\\prime }$= $\\sqrt{AC{'}^{2}-B'C{'}^{2}}=\\sqrt{10^{2}-8^{2}}=6$ m,\\\\\n$\\therefore$BB$^{\\prime }$=AB−AB$^{\\prime }$=$\\boxed{2}$m;",
    "image_path": "dataset/mm_math/images/MM_Math/51367758.png"
  },
  {
    "solution": "\\boxed{\\frac{6}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the right-angled $\\triangle ABC$, $\\angle BAC=90^\\circ$, AB=3, AC=4, P is a moving point on side BC, PE$\\perp$AB at E, PF$\\perp$AC at F, M is the midpoint of EF, what is the minimum value of AM?\n",
    "completion": "\\textbf{Solution:} Connect AP, \\\\\n$\\because$ PE $\\perp$ AB, PF $\\perp$ AC, \\\\\n$\\therefore \\angle$AEP = $\\angle$AFP = $90^\\circ$, \\\\\nAlso, $\\because \\angle$BAC = $90^\\circ$, \\\\\n$\\therefore$ Quadrilateral AEPF is a rectangle, \\\\\n$\\therefore$ AP = EF, \\\\\n$\\because \\angle$BAC = $90^\\circ$, M is the midpoint of EF, \\\\\n$\\therefore$ AM = $\\frac{1}{2}$EF = $\\frac{1}{2}$AP, \\\\\n$\\because$ in $\\triangle$ABC, $\\angle$BAC = $90^\\circ$, AB = 3, AC = 4, \\\\\n$\\therefore$ BC = $\\sqrt{AB^2 + AC^2}$ = 5, \\\\\nWhen AP $\\perp$ BC, the value of AP is minimal, \\\\\nAt this moment, $S_{\\triangle ABC}$ = $\\frac{1}{2} \\times 3 \\times 4$ = $\\frac{1}{2} \\times 5 \\times$ AP, \\\\\nSolving this, we get AP = $\\frac{12}{5}$, \\\\\n$\\therefore$ The minimum value of AP is $\\frac{12}{5}$, \\\\\n$\\therefore$ The minimum value of AM is $\\boxed{\\frac{6}{5}}$,",
    "image_path": "dataset/mm_math/images/MM_Math/51737179.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within square $ABCD$, point $E$ is on diagonal $AC$, $EF \\perp AB$ at point $F$, $EG \\perp BC$ at point $G$. Connect $FG$. If $AB = 8$, what is the minimum value of $FG$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, connect BE, \\\\\n$\\because$ Quadrilateral ABCD is a square, \\\\\n$\\therefore \\angle B=90^\\circ$, AB=BC=8, \\\\\n$\\therefore$AC=8$\\sqrt{2}$, \\\\\n$\\because$EF$\\perp$AB, EG$\\perp$BC, \\\\\n$\\therefore \\angle$BFE=$\\angle$FEG=$\\angle$B=$90^\\circ$, \\\\\n$\\therefore$ Quadrilateral BFEG is a rectangle, \\\\\n$\\therefore$FG=BE, \\\\\n$\\therefore$ When BE is the shortest, FG is also the shortest, \\\\\n$\\therefore$ When BE$\\perp$AC, BE is the shortest, \\\\\n$\\therefore$AE=CE, \\\\\n$\\therefore$BE=$\\frac{1}{2}$AC=4$\\sqrt{2}$, \\\\\n$\\therefore$ The minimum value of FG is $\\boxed{4\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51998806.png"
  },
  {
    "solution": "\\boxed{6+6\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in an equilateral triangle $\\triangle ABC$, points $D$ and $E$ are the midpoints of sides $AB$ and $AC$ respectively. Extend $BC$ to point $F$ such that $CF = \\frac{1}{2}BC$, and connect $DE$, $CD$, and $EF$. Quadrilateral $DCFE$ is a parallelogram. If $AB=6$, what is the perimeter of parallelogram $DCFE$?\n",
    "completion": "\\textbf{Solution:} Given that $\\triangle ABC$ is an equilateral triangle and $AB=6$, with $D$ being the midpoint of $AB$, it follows that $BC=6$ and $CD\\perp AB$ with $BD=\\frac{1}{2}AB=3$. Therefore, $DC=\\sqrt{BC^2-BD^2}=\\sqrt{6^2-3^2}=3\\sqrt{3}$. From statement (1), we know that $DE$ is the median of $\\triangle ABC$, and quadrilateral $DCFE$ is a parallelogram, thus $DE=CF=\\frac{1}{2}BC=3$ and $EF=DC=3\\sqrt{3}$. Therefore, the perimeter of the parallelogram $DCFE$ is $2(DE+DC)=2(3+3\\sqrt{3})=\\boxed{6+6\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51770657_137.png"
  },
  {
    "solution": "\\boxed{y=3x+8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the plane Cartesian coordinate system, line $L_1\\colon y=-\\frac{1}{2}x+8$ intersects the $x$-axis and the $y$-axis at points $B$ and $C$, respectively, and intersects line $L_2\\colon y=x$ at point $A$. If $D$ is a point on line $L_2$, and the area of $\\triangle BOD$ is $32$, find the function expression of line $CD$.\n",
    "completion": "\\textbf{Solution:} Given the line $L_1: y=-\\frac{1}{2}x+8$ intersects with the x-axis at point $B$ and with the y-axis at point $C$, \\\\\n$\\therefore B(16, 0)$, $C(0, 8)$, \\\\\n$\\therefore OB=16$, \\\\\nthen the height of $\\triangle BOD$ is: $32\\times 2\\div16=4$, \\\\\n$\\because D$ is a point on the line $L_2$, \\\\\n$\\therefore$ Coordinates of point D are: $(4, 4)$ or $(-4, -4)$, \\\\\nLet the equation of line $CD$ be: $y=kx+b$, \\\\\nThen $\\left\\{\\begin{array}{l} b=8 \\\\ 4=4k+b \\end{array}\\right.$ or $\\left\\{\\begin{array}{l} b=8 \\\\ -4=-4k+b \\end{array}\\right.$, \\\\\nSolving, we get: $\\left\\{\\begin{array}{l} k=-1 \\\\ b=8 \\end{array}\\right.$ or $\\left\\{\\begin{array}{l} k=3 \\\\ b=8 \\end{array}\\right.$, \\\\\n$\\therefore$ The function expression of line $CD$ is: \\boxed{y=-x+8} or \\boxed{y=3x+8}.",
    "image_path": "dataset/mm_math/images/MM_Math/52019348_98.png"
  },
  {
    "solution": "\\boxed{117}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the four vertices of the square paper $ABCD$ lie on four parallel lines $a$, $b$, $c$, $d$, respectively. The distances between two adjacent lines among these four lines are ${h}_{1}$, ${h}_{2}$, ${h}_{3}$ (${h}_{1}>0$, ${h}_{2}>0$, ${h}_{3}>0$) respectively. If ${h}_{1}=6$ and ${h}_{2}=3$, what is the area $S$ of the square $ABCD$?\n",
    "completion": "\\textbf{Solution}: Draw $BM \\perp a$ at point M through point B, and draw $CN \\perp a$ at point N through point C, as shown in the figure:\n\nThus, we have $\\angle BMA=\\angle ANC=90^\\circ$,\n\n$\\therefore \\angle MBA+\\angle MAB=90^\\circ$,\n\nIn square $ABCD$, $AB=AC$, $\\angle BAC=90^\\circ$,\n\n$\\therefore \\angle MAB+\\angle CAN=90^\\circ$,\n\n$\\therefore \\angle CAN=\\angle MBA$,\n\n$\\therefore \\triangle MAB \\cong \\triangle NCA\\left(AAS\\right)$,\n\n$\\therefore AN=MB$,\n\nSince $h_{1}=6$, $h_{2}=3$,\n\n$\\therefore AN=MB=6$, $NC=6+3=9$,\n\nBy the Pythagorean theorem, we get $AC=\\sqrt{6^2+9^2}=\\sqrt{117}$,\n\n$\\therefore$ the area $S$ of square $ABCD$ is $\\boxed{117}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53538336.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral \\(ABCD\\), \\(AD \\parallel BC\\), \\(\\angle BAD=90^\\circ\\), and diagonals \\(BD \\perp DC\\), \\(AD=4\\), \\(BC=9\\), what is the length of \\(BD\\)?\n",
    "completion": "\\textbf{Solution:} Given that $AD \\parallel BC$ and $BD \\perp DC$, \\\\\n$\\therefore \\angle ADB = \\angle DBC$, and $\\angle BDC = 90^\\circ$, \\\\\n$\\therefore \\angle BAD = \\angle BDC = 90^\\circ$, \\\\\n$\\therefore \\triangle ADB \\sim \\triangle DBC$, \\\\\n$\\therefore \\frac{AD}{BD} = \\frac{BD}{BC}$, \\\\\n$\\therefore BD^{2} = AD \\cdot BC = 4 \\times 9 = 36$, \\\\\n$\\therefore BD = \\boxed{6}$ (negative values are discarded).",
    "image_path": "dataset/mm_math/images/MM_Math/51445485.png"
  },
  {
    "solution": "\\boxed{3\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in diamond ABCD, AB=4, $\\angle B=60^\\circ$, AE$\\perp$BC, and AF$\\perp$CD, with the perpendicular feet being points E and F respectively. If we connect EF, what is the area of $\\triangle AEF$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw line AH $\\perp$ EF at point H,\\\\\n$\\because$ rhombus ABCD,\\\\\n$\\therefore$ BC=CD, $\\angle$B=$\\angle$D=$60^\\circ$,\\\\\n$\\therefore$ $\\angle$BAD=$180^\\circ-60^\\circ=120^\\circ$,\\\\\n$\\because$ AE$\\perp$BC, AF$\\perp$CD,\\\\\n$\\therefore$ $\\angle$AEB=$\\angle$AFD=$90^\\circ$,\\\\\n$\\therefore$ $\\angle$BAE=$\\angle$DAF=$30^\\circ$,\\\\\n$\\therefore$ $\\angle$EAF=$\\angle$BAD-$\\angle$BAE-$\\angle$DAF=$120^\\circ-30^\\circ-30^\\circ=60^\\circ$,\\\\\nAlso, $\\because$ BC$\\cdot$AE=CD$\\cdot$AF,\\\\\n$\\therefore$ AE=AF,\\\\\n$\\therefore$ $\\triangle$AEF is an equilateral triangle,\\\\\n$\\therefore$ AE=EF=AF, EH=HF=$\\frac{1}{2}$EF, AH=$\\sqrt{3}$EH, $\\because$ AB=4,\\\\\n$\\therefore$ in $\\triangle$RtAEB, BE=$\\frac{1}{2}$AB=2,\\\\\n$\\therefore$ EF=AF=AE=$\\sqrt{3}$BE=2$\\sqrt{3}$,\\\\\n$\\therefore$ AH=$\\sqrt{3}$EH=3,\\\\\n$\\therefore$ Area of $\\triangle$AEF=$\\frac{1}{2}$EF$\\cdot$AH=$\\frac{1}{2} \\times 2\\sqrt{3} \\times 3 = \\boxed{3\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51435948.png"
  },
  {
    "solution": "\\boxed{(4,2)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the line $l_1$ intersects the x-axis at point A and the y-axis at point B, with the coordinates of point B being $\\left(0, 3\\right)$. The line $l_2$: $y=2x$ intersects the line $l_1$ at point C, where the x-coordinate of point C is 1. Is there a point E in the plane such that the quadrilateral with vertices O, A, C, and E is a parallelogram? If such a point exists, directly write down the coordinates of the point E that meet the conditions.\n",
    "completion": "\\textbf{Solution:} There exists a point E with coordinates $\\boxed{(4,2)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51662414_212.png"
  },
  {
    "solution": "\\boxed{3 - \\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the square $ABCD$ and the equilateral triangle $\\triangle AEF$ are both inscribed in the circle $O$, with $EF$ intersecting $BC$ and $CD$ at points $G$ and $H$ respectively. If $AE=6$, what is the radius of circle $O$? And what is the length of $EG$?\n",
    "completion": "\\textbf{Solution:} As shown in Figure 1, draw lines OA, OE, and through point O draw line OP $\\perp$ AE at P, \\\\\nthus AP = PE = $\\frac{1}{2}$AE = 3, \\\\\n$\\because$ $\\triangle AEF$ is an equilateral triangle, \\\\\n$\\therefore$ $\\angle AOE = 120^\\circ$, \\\\\n$\\because$ OA = OE, \\\\\n$\\therefore$ $\\angle OAP = 30^\\circ$, \\\\\n$\\therefore$ OA = $\\frac{AP}{\\cos 30^\\circ}$ = $2\\sqrt{3}$; \\\\\nConnect BD and AC, with AC intersecting EF at Q, and connect OF, \\\\\nthen AC $\\perp$ EF, \\\\\n$\\therefore$ EQ = $\\frac{1}{2}$EF = 3, \\\\\nIn $\\triangle OQF$, $\\angle OFQ = 30^\\circ$, \\\\\n$\\therefore$ OQ = $\\frac{1}{2}$OF = $\\sqrt{3}$, \\\\\n$\\therefore$ CQ = OC - OQ = $\\sqrt{3}$, \\\\\n$\\because$ Quadrilateral ABCD is a square, \\\\\n$\\therefore$ $\\angle GCQ = 45^\\circ$, \\\\\n$\\therefore$ GQ = CQ = $\\sqrt{3}$, \\\\\n$\\therefore$ EG = EQ - QG = $3 - \\sqrt{3}$, \\\\\nHence, the answer is: The radius of circle O is $\\boxed{2\\sqrt{3}}$; The length of $EG$ is $\\boxed{3 - \\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52874743.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the side length of the square $ABCD$ is $2\\, \\text{cm}$. What is the area of the shaded region?\n",
    "completion": "\\textbf{Solution:} It is known from symmetry that $S_{\\text{shaded area}} = \\frac{1}{2}S_{\\text{square} ABCD} = \\frac{1}{2} \\times 2 \\times 2 = \\boxed{2}\\text{cm}^2$.",
    "image_path": "dataset/mm_math/images/MM_Math/51614615.png"
  },
  {
    "solution": "\\boxed{\\frac{16}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y_1=x+2$ intersects the x-axis and y-axis at points A and B, respectively, and the line $y_2=kx+4$ intersects the x-axis and y-axis at points C and D, respectively. The two lines intersect at point $E\\left(\\frac{2}{3}, n\\right)$. What is the area of $\\triangle ACE$?\n",
    "completion": "\\textbf{Solution:} Given the line $y_1=x+2$, when $y=0$, we have $0=x+2$. Solving this, we get $x=-2$, hence the coordinates of point A are $(-2,0)$. For the line $y_2=-2x+4$, when $y=0$, we have $0=-2x+4$. Solving this, we get $x=2$, hence the coordinates of point C are $(2,0)$. Therefore, the length of $AC=2-(-2)=4$. As shown in the figure, draw $EH\\perp AC$ at point H. Since the coordinates of point E are $E\\left(\\frac{2}{3}, \\frac{8}{3}\\right)$, it follows that $EH=\\frac{8}{3}$. Therefore, the area of $\\triangle ACE$ is $\\frac{1}{2} AC\\times EH=\\frac{1}{2}\\times 4\\times \\frac{8}{3}=\\boxed{\\frac{16}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52381906_142.png"
  },
  {
    "solution": "\\boxed{\\sqrt{3}\\leq m\\leq 3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated in the diagram, the sides of the rhombus $ABCD$ are of length $\\sqrt{3}$, with $\\angle ABC=60^\\circ$. Let $M$ be any point on line segment $CD$ (which can coincide with points $C$ or $D$). Perpendicular lines to ray $BM$ are drawn from points $A$, $C$, and $D$, meeting $BM$ at points $E$, $F$, and $G$, respectively. Let $AE+CF+DG=m$. What is the range of possible values for $m$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, connect AC and BD, which intersect at point O, and also connect AM,\\\\\nsince quadrilateral ABCD is a rhombus and $\\angle ABC=60^\\circ$,\\\\\ntherefore, AC$\\perp$BD, AO＝CO, BO＝DO, AB＝BC＝$\\sqrt{3}$, $\\angle ABO=30^\\circ$,\\\\\ntherefore, AO＝$\\frac{1}{2}$AB＝$\\frac{\\sqrt{3}}{2}$,\\\\\nthus, AC＝$\\sqrt{3}$,\\\\\nsince BO＝$\\sqrt{AB^2-AO^2}=\\sqrt{3-\\frac{3}{4}}=\\frac{3}{2}$,\\\\\nthus, BD＝3,\\\\\nthus, the area of rhombus ABCD ＝$\\frac{AC\\times BD}{2}=\\frac{3\\sqrt{3}}{2}$,\\\\\nsince $S_{\\triangle ABM}=\\frac{1}{2}\\times$BM$\\times$AE, $S_{\\triangle BCM}=\\frac{1}{2}\\times$BM$\\times$CF, $S_{\\triangle BMD}=\\frac{1}{2}\\times$BM$\\times$DG,\\\\\nthus, $S_{\\triangle ABM}$+$S_{\\triangle BCM}$+$S_{\\triangle BMD}=\\frac{1}{2}S_{\\text{rhombus ABCD}}+\\frac{1}{2}S_{\\text{rhombus ABCD}}=\\frac{1}{2}\\times$BM$\\times$(AE+CF+DG),\\\\\nthus, $\\frac{3\\sqrt{3}}{2}＝\\frac{1}{2}\\times$BM$\\times$m,\\\\\nsince $\\sqrt{3}\\leq$BM$\\leq$3,\\\\\nthus, $\\sqrt{3}\\leq$m$\\leq$3,\\\\\ntherefore, the answer is: $\\boxed{\\sqrt{3}\\leq m\\leq 3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50496469.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{5}-1}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Quadrilateral $ABCD$ is a rectangular piece of paper, with point $E$ located on side $AD$. When $\\triangle ABE$ is folded along $BE$ such that point $A$ lies on the diagonal $BD$ of the rectangle, and $CF$ is drawn. Given that $DE=1$, explore the following question: As shown in Figure 2, when points $C$, $E$, and $F$ lie on the same straight line, what is the length of $AE$?\n",
    "completion": "\\textbf{Solution:} When points $C$, $E$, and $F$ are on the same line, we have $AE=\\boxed{\\frac{\\sqrt{5}-1}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52601962_65.png"
  },
  {
    "solution": "\\boxed{100}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, AB is the diameter of $\\odot O$, point C is a point on the semicircle above AB, point D is the midpoint of the semicircle below AB. The segments AC, BC, and AD are connected, and through point D, DE is drawn parallel to AB and intersects the extension of CB at point E. If AD equals $5 \\sqrt{2}$, what is the maximum value of the product of AC and CE?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect CD and BD, \\\\\n$\\because$ DE$\\parallel$AB, \\\\\n$\\therefore$ $\\angle$ABC=$\\angle$E, \\\\\n$\\because$ $\\angle$ABC=$\\angle$CDA, \\\\\n$\\therefore$ $\\angle$E =$\\angle$CDA, \\\\\n$\\because$ point D is the midpoint of the semicircle below AB, \\\\\n$\\therefore$ $\\angle$ACD=$\\angle$DCE, \\\\\n$\\therefore$ $\\triangle ACD\\sim \\triangle DCE$, \\\\\n$\\therefore$ $\\frac{CD}{CE}=\\frac{AC}{CD}$, \\\\\n$\\therefore$ $CD^{2}=AC\\cdot CE$, \\\\\n$\\because$ point D is the midpoint of the semicircle below AB, \\\\\n$\\therefore$ $\\angle$ADB=$90^\\circ$, BD=AD=$5\\sqrt{2}$, \\\\\n$\\therefore$ $AB=\\sqrt{AD^{2}+BD^{2}}=10$, \\\\\n$\\because$ $CD\\le AB=10$, \\\\\n$\\therefore$ $AC\\cdot CE\\le 10^{2}=\\boxed{100}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51458349.png"
  },
  {
    "solution": "\\boxed{\\frac{5}{12}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $P(12, a)$ lies on the graph of the inverse proportion function $y=\\frac{60}{x}$. Line $PH$ is perpendicular to the $x$-axis at point $H$. What is the value of $\\tan \\angle POH$?\n",
    "completion": "\\textbf{Solution:} Since $P(12,a)$ lies on the graph of the inverse proportion function $y=\\frac{60}{x}$,\\\\\nthus $a=\\frac{60}{12}=5$,\\\\\ntherefore, $P(12,5)$,\\\\\ntherefore, $OH=12$, $PH=5$,\\\\\nthus, $\\tan \\angle POH=\\frac{PH}{OH}=\\boxed{\\frac{5}{12}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52551786.png"
  },
  {
    "solution": "\\boxed{(8,0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the line AC intersects the x-axis at point C, and intersects the y-axis at point A. The line AB intersects the x-axis at point B, and intersects the y-axis at point A. Given that A(0, 4) and B(2, 0). If $S_{\\triangle ABC} = 12$, find the coordinates of point C.\n",
    "completion": "\\textbf{Solution:} Let C$(x,0)$, since A$(0,4)$ and B$(2,0)$, therefore OA$=4$, OB$=2$. Given that the area $S_{\\triangle ABC}=12$, thus $\\frac{1}{2}$BC$\\cdot$OA$=12$, therefore BC$=6$, hence $|x-2|=6$. Solving this gives $x=8$ or $x=-4$. Therefore, the coordinates of C are $\\boxed{(-4,0)}$ or $\\boxed{(8,0)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50555230_99.png"
  },
  {
    "solution": "\\boxed{\\frac{10}{3}\\pi}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, AD is the bisector of the exterior angle $\\angle EAC$ of $\\triangle ABC$, and AD intersects the circumcircle $O$ of $\\triangle ABC$ at point D. $DB = DC$; if $\\angle CAB=30^\\circ$, and $BC=4$, what is the length of the minor arc $\\overset{\\frown}{DC}$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, connect OB, OC, and OD.\\\\\nBy the Circumference Angle Theorem, $\\angle COB=2\\angle CAB=60^\\circ$, $\\angle CDB=\\angle CAB=30^\\circ$,\\\\\n$\\therefore \\triangle COB$ is an equilateral triangle,\\\\\n$\\therefore OC=BC=4$,\\\\\n$\\because DC=DB$, $\\angle CDB=30^\\circ$,\\\\\n$\\therefore \\angle DCB=75^\\circ$,\\\\\n$\\therefore \\angle DCO=15^\\circ$,\\\\\n$\\therefore \\angle COD=150^\\circ$,\\\\\nTherefore, the length of the minor arc $\\overset{\\frown}{DC}$ is $= \\frac{150\\pi \\times 4}{180}=\\boxed{\\frac{10}{3}\\pi}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52536503_87.png"
  },
  {
    "solution": "\\boxed{y=\\frac{1}{2}x^2-\\frac{1}{2}x-3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola passes through points A$(-2, 0)$ and C$(0, -3)$, and its axis of symmetry is the line $x=\\frac{1}{2}$. Find the equation of the parabola.\n",
    "completion": "\\textbf{Solution:} Let the equation of the parabola be $y=ax^2+bx+c$,\\\\\nsince the axis of symmetry is the line $x=\\frac{1}{2}$,\\\\\nthus $-\\frac{b}{2a}=\\frac{1}{2}$,\\\\\nthus $b=-a$,\\\\\nthus $y=ax^2-ax+c$,\\\\\nSubstituting points A ($-2$, $0$) and C ($0$, $-3$),\\\\\nthus $\\begin{cases}c=-3 \\\\ 4a+2a+c=0\\end{cases}$,\\\\\nSolving yields $\\begin{cases}c=-3 \\\\ a=\\frac{1}{2}\\end{cases}$,\\\\\nthus $y=\\frac{1}{2}x^2-\\frac{1}{2}x-3$;\\\\\nHence, the equation of the parabola is $\\boxed{y=\\frac{1}{2}x^2-\\frac{1}{2}x-3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53080876_149.png"
  },
  {
    "solution": "\\boxed{130^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $BD$ bisects $\\angle ABC$, and $CD$ bisects $\\angle ACD$. If $\\angle A = 80^\\circ$, what is the degree measure of $\\angle D$?\n",
    "completion": "\\textbf{Solution:}\n\nGiven that $\\angle A=80^\\circ$,\\\\\nit follows that $\\angle ABC + \\angle ACB = 180^\\circ - \\angle A = 100^\\circ$,\\\\\nSince BD bisects $\\angle ABC$ and CD bisects $\\angle ACB$,\\\\\nwe have $\\angle CBD = \\frac{1}{2} \\angle ABC$ and $\\angle BCD = \\frac{1}{2} \\angle ACB$,\\\\\nIn $\\triangle BCD$,\\\\\n$\\angle D = 180^\\circ - (\\angle CBD + \\angle BCD)$\\\\\n$= 180^\\circ - (\\frac{1}{2} \\angle ABC + \\frac{1}{2} \\angle ACB)$\\\\\n$= 180^\\circ - \\frac{1}{2} (\\angle ABC + \\angle ACB)$\\\\\n$= 180^\\circ - \\frac{1}{2} \\times 100^\\circ$\\\\\n$= 180^\\circ - 50^\\circ$\\\\\n$= \\boxed{130^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52855863.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points $C$, $D$, and $E$ are on the line segment $AB$, with $AB=12$ and $AC=4$. $D$ and $E$ are the midpoints of $AB$ and $CB$ respectively. Find the length of $DE$.\n",
    "completion": "\\textbf{Solution:} Given $\\because AB=12$, $AC=4$, \\\\\n$\\therefore BC=AB-AC=12-4=8$, \\\\\n$\\because D$ and $E$ are the midpoints of $AB$ and $CB$, respectively, \\\\\n$\\therefore BD= \\frac{1}{2}AB= \\frac{1}{2}\\times 12=6$, $BE=CE= \\frac{1}{2}BC= \\frac{1}{2}\\times 8=4$, \\\\\n$\\therefore DE=BD-BE=6-4=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52773591_161.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $OB$, $OC$, $OD$ are three rays within $\\angle AOE$, with $OB$ bisecting $\\angle AOE$, and $OD$ bisecting $\\angle COE$. If $\\angle DOE = 20^\\circ$ and $\\angle AOE + \\angle BOD = 220^\\circ$, what is the degree measure of $\\angle BOD$?\n",
    "completion": "\\textbf{Solution:} From the given information, we have\\\\\n$\\because$ OB bisects $\\angle AOE$,\\\\\n$\\therefore \\angle AOE=2\\angle BOE$.\\\\\n$\\because \\angle AOE+\\angle BOD=220^\\circ$,\\\\\n$\\therefore 2\\angle BOE+\\angle BOD=220^\\circ$.\\\\\n$\\because \\angle BOE-\\angle BOD=\\angle DOE=20^\\circ$,\\\\\n$\\therefore 2\\angle BOE-2\\angle BOD=40^\\circ$, i.e., $2\\angle BOE=40^\\circ+2\\angle BOD$,\\\\\n$\\therefore 2\\angle BOE+\\angle BOD=40^\\circ+3\\angle BOD=220^\\circ$,\\\\\n$\\therefore 3\\angle BOD=180^\\circ$,\\\\\n$\\therefore \\angle BOD=\\boxed{60^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/54927250_124.png"
  },
  {
    "solution": "\\boxed{\\frac{4}{3}\\pi}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of $\\odot O$, with $C$ and $D$ being two points on $\\odot O$, $\\angle BAC = \\angle DAC$. A line $EF \\perp AD$ is drawn through point $C$, intersecting the extension of $AD$ at point $E$, and $BC$ is connected. $EF$ is the tangent of $\\odot O$; if $DE=2$ and $BC=4$, what is the length $l$ of the minor arc $\\overset{\\frown}{BC}$?\n",
    "completion": "\\textbf{Solution:} Draw $OD$, $DC$,\\\\\nsince $\\angle DAC = \\frac{1}{2}\\angle DOC$, $\\angle OAC = \\frac{1}{2}\\angle BOC$, $\\angle DAC = \\angle OCA = \\angle OAC$,\\\\\nthus $\\angle DOC = \\angle BOC$, $DC=BC=4$,\\\\\ntherefore, in $\\triangle CED$, $\\sin\\angle ECD = \\frac{DE}{DC} = \\frac{1}{2}$,\\\\\nthus $\\angle ECD = 30^\\circ$,\\\\\nalso, since $EF$ is a tangent to $\\odot O$,\\\\\nthus $\\angle OCD = 60^\\circ$,\\\\\ntherefore, $\\triangle COD$ is an equilateral triangle,\\\\\nthus $\\angle BOC = \\angle COD = 60^\\circ$,\\\\\ntherefore, $\\triangle BOC$ is an equilateral triangle,\\\\\nthus $OC = BC = 4$,\\\\\ntherefore, $l = \\frac{60\\pi \\times 4}{180} = \\boxed{\\frac{4}{3}\\pi}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52536729_94.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{4}{x}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, for the rectangle $ABCD$, the lengths of sides $AD$ and $AB$ are $3$ and $8$, respectively. Point $E$ is the midpoint of $DC$. The graph of the inverse proportional function $y=\\frac{m}{x}$ ($x<0$) passes through point $E$ and intersects $AB$ at point $F$. If $AF-AE=2$, find the expression of the inverse proportional function.\n",
    "completion": "\\textbf{Solution:} Given $\\because AD=3, DE=4$, \\\\\n$\\therefore AE=\\sqrt{AD^2+DE^2}=5$ \\\\\n$\\because AF-AE=2$ \\\\\n$\\therefore AF=7, BF=1$ \\\\\nLet the coordinates of point $E$ be $(a, 4)$, then the coordinates of point $F$ are $(a-3, 1)$,\\\\\n$\\because E, F$ are on the graph of the function $y=\\frac{m}{x} (x<0)$,\\\\\n$\\therefore 4a=a-3$, solving this gives $a=-1$,\\\\\n$\\therefore E(-1, 4)$ \\\\\n$\\therefore m=-1\\times 4=-4$ \\\\\n$\\therefore \\boxed{y=-\\frac{4}{x}} (x<0)$.",
    "image_path": "dataset/mm_math/images/MM_Math/51615141_76.png"
  },
  {
    "solution": "\\boxed{1\\colon 3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a rectangular field $ABCD$ has the ratio of its length $AB$ to its width $BC$ as $\\sqrt{2} : 1$. $DE \\perp AC$, and $BF \\perp AC$, with the perpendicular feet being points $E$ and $F$, respectively. Now, there is a plan to plant flowers and grass in the quadrilateral region $DEBF$. What is the ratio of the area of quadrilateral $DEBF$ to the area of the rectangle $ABCD$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral $ABCD$ is a rectangle,\\\\\n$ \\therefore AD\\parallel BC$, $AD=BC$, $\\angle ABC=90^\\circ$,\\\\\n$ \\therefore \\angle DAE=\\angle BCF$.\\\\\nSince $BF\\perp AC$, $DE\\perp AC$,\\\\\n$ \\therefore \\angle AED=\\angle CFB=90^\\circ$, $BF\\parallel DE$.\\\\\nIn $\\triangle ADE$ and $\\triangle CBF$,\\\\\n$\\left\\{\\begin{array}{l}\n\\angle DAE=\\angle BCF\\\\\n\\angle AED=\\angle CFB\\\\\nAD=CB\n\\end{array}\\right.$,\\\\\n$ \\therefore \\triangle ADE \\cong \\triangle CBF$ (AAS),\\\\\n$ \\therefore DE=BF$, $AE=CF$,\\\\\nSince $BF\\parallel DE$,\\\\\n$ \\therefore $ Quadrilateral $DEBF$ is a parallelogram,\\\\\nLet $AD=BC=x$, then $CD=AB=\\sqrt{2}x$,\\\\\n$ \\therefore AC=\\sqrt{AB^2+BC^2}=\\sqrt{(\\sqrt{2}x)^2+x^2}=\\sqrt{3}x$,\\\\\nSince $DE\\perp AC$ at point $E$,\\\\\n$ \\therefore S_{\\triangle ADC}=\\frac{1}{2}AD\\cdot CD=\\frac{1}{2}AC\\cdot DE$,\\\\\n$ \\therefore x\\cdot \\sqrt{2}x=\\sqrt{3}x\\cdot DE$,\\\\\n$ \\therefore DE=\\frac{\\sqrt{6}}{3}x$,\\\\\nIn $\\triangle ADE$, $AE=\\sqrt{x^2-\\left(\\frac{\\sqrt{6}}{3}x\\right)^2}=\\frac{\\sqrt{3}}{3}x$,\\\\\n$CF=\\frac{\\sqrt{3}}{3}x$,\\\\\n$ \\therefore EF=AC-AE-CF=\\frac{\\sqrt{3}}{3}x$,\\\\\n$ \\therefore S_{\\text{quadrilateral }DEBF}=EF\\cdot DE=\\frac{\\sqrt{3}}{3}x\\cdot \\frac{\\sqrt{6}}{3}x=\\frac{\\sqrt{2}}{3}x^2$,\\\\\nSince $S_{\\text{rectangle }ABCD}=x\\cdot \\sqrt{2}x=\\sqrt{2}x^2$,\\\\\n$ \\therefore $ The ratio of the area of quadrilateral $DEBF$ to rectangle $ABCD$ is $\\boxed{1\\colon 3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51722492.png"
  },
  {
    "solution": "\\boxed{C}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the solid shape is composed of a rectangular prism and a cylinder. What shape is obtained when viewing this figure from above?\n",
    "completion": "\\textbf{Solution:} From the given problem, we can deduce that the top view is a rectangle with a circle in the middle. Thus, the answer is \\boxed{C}.",
    "image_path": "dataset/mm_math/images/MM_Math/55483153.png"
  },
  {
    "solution": "\\boxed{36^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, D is a point on side $BC$, and $AB=AD=DC$. If $AC=BC$, find the measure of $\\angle C$.\n",
    "completion": "\\textbf{Solution:} Let $\\angle C=x$,\\\\\nsince $AD=DC$,\\\\\nthus $\\angle DAC=\\angle C=x$,\\\\\ntherefore $\\angle ADB=2x$,\\\\\nsince $AB=AD$,\\\\\nthus $\\angle B=\\angle ADB=2x$,\\\\\nsince $AC=BC$,\\\\\nthus $\\angle BAC=\\angle B=2x$,\\\\\nin $\\triangle ABC$, $2x+2x+x=180^\\circ$,\\\\\ntherefore $x=\\boxed{36^\\circ}$,\\\\\ntherefore the degree of $\\angle C$ is $\\boxed{36^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53044190_144.png"
  },
  {
    "solution": "\\boxed{\\sqrt{3} \\times 2^{2020}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, there is a right-angled triangle $Rt\\triangle AOB$, with $\\angle ABO=90^\\circ$ and $\\angle AOB=30^\\circ$. The right-angle side $OB$ is on the positive half of the $y$-axis, point $A$ is in the first quadrant, and $OA=1$. Rotate the $Rt\\triangle AOB$ counter-clockwise around the origin by $30^\\circ$, and at the same time, double the lengths of its sides (i.e., $OA_{1}=2OA$). This results in $Rt\\triangle OA_{1}B_{1}$. Similarly, rotate $Rt\\triangle OA_{1}B_{1}$ around the origin $O$ counter-clockwise by $30^\\circ$, and at the same time, double the lengths of its sides, resulting in $Rt\\triangle OA_{2}B_{2}$, ..., Following this rule, we obtain $Rt\\triangle OA_{2021}B_{2021}$. What is the length of $OB_{2021}$?\n",
    "completion": "\\textbf{Solution:} In right triangle $\\triangle AOB$, we have $\\angle AOB=30^\\circ$ and $OA = 1$, \\\\\ntherefore, $OB = OA \\cdot \\cos{\\angle AOB} = \\frac{\\sqrt{3}}{2}$, \\\\\naccording to the problem, $OB_1 = 2OB = \\frac{\\sqrt{3}}{2} \\times 2$, \\\\\n$OB_2 = 2OB_1 = \\frac{\\sqrt{3}}{2} \\times 2^2$, \\ldots \\\\\n$OB_n = \\frac{\\sqrt{3}}{2} \\times 2^n = \\sqrt{3} \\times 2^{n-1}$, \\\\\ntherefore, the length of $OB_{2021}$ is: $\\sqrt{3} \\times 2^{2020} = \\boxed{\\sqrt{3} \\times 2^{2020}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51367519.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{13}}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, rhombus $ABCD$ has a side length of $4$, and points $E$ and $F$ are located on $AB$ and $AD$, respectively. $AC$ intersects $EF$ at point $G$. If $BE = AF = 1$ and $\\angle BAD = 120^\\circ$, what is the length of $FG$?\n",
    "completion": "\\textbf{Solution:} Construct $EM \\parallel BC$ intersecting $AC$ at $M$, and $EN \\perp BC$ at $N$, as shown in the diagram:\\\\\nSince the side length of rhombus $ABCD$ is 4, and $\\angle BAD = 120^\\circ$,\\\\\ntherefore, $AB = BC = 4$, $\\angle BAC = \\angle FAC = \\frac{1}{2}\\angle BAD = 60^\\circ$, $AD \\parallel BC$,\\\\\ntherefore, $\\triangle ABC$ is an equilateral triangle,\\\\\ntherefore, $\\angle B = \\angle ACB = 60^\\circ$, $BC = AC$,\\\\\nsince $EM \\parallel BC$,\\\\\ntherefore, $EM \\parallel AD$, $\\angle AEM = \\angle B = 60^\\circ = \\angle BAC$,\\\\\ntherefore, $\\triangle AEM$ is an equilateral triangle,\\\\\ntherefore, $AM = AE = AB - BE = 4 - 1 = 3$,\\\\\nsince $EM \\parallel AD$,\\\\\ntherefore, $\\triangle AGF \\sim \\triangle MGE$,\\\\\ntherefore, $\\frac{FG}{EG} = \\frac{AF}{EM} = \\frac{1}{3}$,\\\\\ntherefore, $FG = \\frac{1}{4}EF$,\\\\\nIn $\\triangle BCE$ and $\\triangle ACF$,\\\\\n$\\begin{cases} BC = AC \\\\ \\angle B = \\angle FAC \\\\ BE = AF \\end{cases}$,\\\\\ntherefore, $\\triangle BCE \\cong \\triangle ACF$ (SAS),\\\\\ntherefore, $CE = CF$, $\\angle BCE = \\angle ACF$,\\\\\ntherefore, $\\angle ACF + \\angle ACE = \\angle ACF + \\angle ACE = \\angle ACB = 60^\\circ$,\\\\\ntherefore, $\\triangle CEF$ is an equilateral triangle,\\\\\ntherefore, $EF = CE$,\\\\\nsince $EN \\perp BC$, $\\angle B = 60^\\circ$,\\\\\ntherefore, $\\angle BEN = 30^\\circ$,\\\\\ntherefore, $BN = \\frac{1}{2}BE = \\frac{1}{2}$,\\\\\ntherefore, $EN = \\sqrt{3}BN = \\frac{\\sqrt{3}}{2}$, $CN = BC - BN = 4 - \\frac{1}{2} = \\frac{7}{2}$,\\\\\ntherefore, $EF = CE = \\sqrt{EN^2 + CN^2} = \\sqrt{\\left(\\frac{\\sqrt{3}}{2}\\right)^2 + \\left(\\frac{7}{2}\\right)^2} = \\sqrt{13}$,\\\\\ntherefore, $FG = \\frac{1}{4}EF = \\boxed{\\frac{\\sqrt{13}}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51433787.png"
  },
  {
    "solution": "\\boxed{3\\text{cm}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the length of segment $AB=4$. Extend $AB$ to point C such that $BC=\\frac{1}{2}AB$, and extend $AB$ in the opposite direction to point $D$ so that $AC=2AD$. If $Q$ is the midpoint of $AB$, and $P$ is a point on the segment $CD$; and $BP=\\frac{1}{2}BC$, calculate the length of the segment $PQ$.\n",
    "completion": "\\textbf{Solution:} As shown in the figure, when $P$ is on segment $AB$, we have $BP_1=\\frac{1}{2}BC$\\\\\nSince $Q$ is the midpoint of $AB$,\\\\\nit follows that $BQ= \\frac{1}{2}AB=2\\text{cm}$,\\\\\nSince $BP_1=\\frac{1}{2}BC$,\\\\\nit follows that $BP_1=1\\text{cm}$,\\\\\nTherefore, $P_1Q=BQ-BP_1=2-1=1\\text{cm}$.\\\\\nWhen $P$ is on segment $BC$, $BP_2=\\frac{1}{2}BC=1$,\\\\\n$P_2Q=BQ+BP_2=2+1=3\\text{cm}$.\\\\\nTherefore, the length of segment $PQ$ is $\\boxed{1}$ or $\\boxed{3\\text{cm}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53119305_117.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in a grid of small squares with equal side lengths, points A, B, C, and D are all on the vertices of these small squares. Lines AB and CD intersect at point P. What is the value of $\\frac{AP}{PB}$? What is the value of $\\tan\\angle APD$?\n",
    "completion": "\\textbf{Solution}:\n\nSince quadrilateral BCED is a square, \\\\\nit follows that DB$\\parallel$AC, \\\\\nwhich means $\\triangle DBP\\sim \\triangle CAP$, \\\\\nthus, $\\frac{AP}{PB} = \\frac{AC}{DB} = \\boxed{3}$, \\\\\nConnect BE, intersecting CD at point F, \\\\\nSince quadrilateral BCED is a square, \\\\\nit follows that DF=CF= $ \\frac{1}{2}$ CD, BF= $ \\frac{1}{2}$ BE, and CD=BE, BE$\\perp$CD, \\\\\nthus, BF=CF. According to the problem, AC$\\parallel$BD, \\\\\nit follows that $\\triangle ACP\\sim \\triangle BDP$, \\\\\nthus, DP$\\colon$ CP=BD$\\colon$ AC=1$\\colon$ 3, \\\\\nwhich means DP$\\colon$ DF=1$\\colon$ 2, therefore, DP=PF= $ \\frac{1}{2}$ CF= $ \\frac{1}{2}$ BF, \\\\\nIn $\\triangle PBF$ right-angled at B, $\\tan\\angle BPF= \\frac{BF}{PF} = \\boxed{2}$, \\\\\nSince $\\angle APD=\\angle BPF$, \\\\\nit follows that $\\tan\\angle APD=2$.",
    "image_path": "dataset/mm_math/images/MM_Math/51103675.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, within the rectangle $ABCD$, point $E$ is the midpoint of $AB$, $EF\\perp AB$ intersects $CD$ at point $F$, and point $M$ is located on $AD$. Connect $BM$, and fold $\\triangle ABM$ along $BM$. When the corresponding point $A'$ of point $A$ exactly falls on $EF$, what is the degree measure of $\\angle CBA'$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect $AA'$.\\\\\nSince point E is the midpoint of $AB$, $EF\\perp AB$ and intersects $CD$ at point F,\\\\\nit follows that $EF$ bisects $AB$ perpendicularly,\\\\\nthus $A'A=A'B$,\\\\\ndue to the properties of folding, it is known that $A'B=AB$,\\\\\ntherefore $A'A=A'B=AB$,\\\\\nhence $\\triangle ABA'$ is an equilateral triangle,\\\\\nthus $\\angle ABA'=60^\\circ$,\\\\\nin rectangle $ABCD$, $\\angle ABC=90^\\circ$,\\\\\nthus $\\angle CBA'=\\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52423810.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points A, B, and C are on the circle $\\odot O$, with $OA\\perp OB$. What is the degree measure of $\\angle ACB$?\n",
    "completion": "\\textbf{Solution:} Since $OA \\perp OB$, \\\\\n$\\therefore$ $\\angle AOB=90^\\circ$,\\\\\n$\\therefore$ $\\angle ACB=\\frac{1}{2}\\angle AOB=45^\\circ$,\\\\\nThus, the answer is: $\\boxed{45^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52905989.png"
  },
  {
    "solution": "\\boxed{(x+1)^2+4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: The partial graph of a quadratic function is shown in the figure, with the axis of symmetry being the line \\(x=-1\\). Find the analytic expression of this quadratic function.\n",
    "completion": "\\textbf{Solution:} Let the equation of the parabola be $y=a(x+1)^2+k$, \\\\\nSubstituting $(-3,0)$ and $(0,3)$ into $y=a(x+1)^2+k$, we get $\\begin{cases}0=4a+k\\\\ 3=a+k\\end{cases}$, \\\\\nSolving this, we obtain $\\begin{cases}a=-1\\\\ k=4\\end{cases}$, \\\\\n$\\therefore y=-\\boxed{(x+1)^2+4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52605434_112.png"
  },
  {
    "solution": "\\boxed{20^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AD$ and $BC$ intersect at point $O$, with $AD=BC$, and $\\angle C=\\angle D=90^\\circ$. We have $AC=BD$; if $\\angle ABC=35^\\circ$, what is the degree measure of $\\angle CAO$?\n",
    "completion": "\\textbf{Solution:} In $\\triangle ACB$, $\\angle ABC=35^\\circ$, \\\\\n$\\therefore \\angle CAB=90^\\circ-35^\\circ=55^\\circ$, \\\\\nFrom (1) it is known $\\triangle ACB \\cong \\triangle BDA$, \\\\\n$\\therefore \\angle BAD=\\angle ABC=35^\\circ$, \\\\\n$\\therefore \\angle CAO=\\angle CAB-\\angle BAD=55^\\circ-35^\\circ=\\boxed{20^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52913903_110.png"
  },
  {
    "solution": "\\boxed{9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, $E$, $F$, $G$, and $H$ are respectively the midpoints of $AB$, $BD$, $CD$, and $AC$. Given that $AD=4$ and $BC=5$, what is the perimeter of quadrilateral $EFGH$?\n",
    "completion": "\\textbf{Solution:} Given that E, F, G, and H are the midpoints of AB, BD, CD, and AC, respectively,\n\n$\\therefore EF = \\frac{1}{2}AD = 2, FG = \\frac{1}{2}BC = 2.5$, $GH = \\frac{1}{2}AD = 2, EH = \\frac{1}{2}BC = 2.5$,\n\n$\\therefore$ the perimeter of the quadrilateral EFGH = EF + FG + GH + HE = \\boxed{9}",
    "image_path": "dataset/mm_math/images/MM_Math/53490596.png"
  },
  {
    "solution": "\\boxed{7}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, point $D$ is the midpoint of segment $AC$, and point $E$ is the midpoint of segment $BC$, with $BC=\\frac{4}{7}AC$. If $BC=8$, what is the length of $DC$?\n",
    "completion": "\\textbf{Solution:} Since $BC=8$ and $BC=\\frac{4}{7}AC$,\\\\\nit follows that $AC=8\\div\\frac{4}{7}=14$,\\\\\nGiven that point $D$ is the midpoint of line segment $AC$,\\\\\nit follows that $DC=\\frac{1}{2}AC=\\frac{1}{2}\\times 14=\\boxed{7}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52531561_131.png"
  },
  {
    "solution": "\\boxed{\\angle AOE}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the lines AB, CD, and EF intersect at point O. Write the vertical angles of $\\angle AOC$ and $\\angle BOF$.\n",
    "completion": "\\textbf{Solution:} From the given, we understand that the vertical angle of $\\angle AOC$ is \\boxed{\\angle BOD}, and the vertical angle of $\\angle BOF$ is \\boxed{\\angle AOE}.",
    "image_path": "dataset/mm_math/images/MM_Math/54927579_136.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{5}}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $DE$ parallel to $BC$ divides $\\triangle ABC$ into two parts with an area ratio of 1$\\colon$4. What is the value of $\\frac{AD}{AB}$?\n",
    "completion": "\\textbf{Solution:} Given that line DE divides $\\triangle ABC$ into two parts with an area ratio of 1:4,\\\\\nit follows that $S_{\\triangle ADE} = \\frac{1}{5}S_{\\triangle ABC}$,\\\\\nSince DE$\\parallel$BC,\\\\\nit implies that $\\triangle ADE \\sim \\triangle ABC$,\\\\\nthus $\\frac{S_{\\triangle ADE}}{S_{\\triangle ABC}}=\\left(\\frac{AD}{AB}\\right)^2=\\frac{1}{5}$,\\\\\ntherefore $\\frac{AD}{AB}=\\boxed{\\frac{\\sqrt{5}}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51367391.png"
  },
  {
    "solution": "\\boxed{\\left(0,\\frac{\\sqrt{15}}{15}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB=AC$, $A\\left(0, \\sqrt{15}\\right)$, $C(1, 0)$, D is a point on the ray AO, and a moving point P starts from A, moving along the path $A-D-C$. The speed on AD is $4$ units per second, and the speed on CD is $1$ unit per second. What are the coordinates of D when the total moving time is minimal?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $DH\\perp AB$ at H, and $CM\\perp AB$ at $M$, intersecting AO at ${D}^{\\prime}$.\\\\\nSince the travel time $t=\\frac{AD}{4}+CD$,\\\\\nand since $AB=AC$, $AO\\perp BC$,\\\\\nit follows that $BO=OC=1$,\\\\\ngiven $A(0,\\sqrt{15})$, $C(1,0)$, $AB=AC$, $AO\\perp BC$,\\\\\nit follows that $AB=AC=\\sqrt{OA^2+OB^2}=\\sqrt{15+1}=4$,\\\\\nSince $\\angle DAH=\\angle BAO$, $\\angle DHA=\\angle AOB=90^\\circ$,\\\\\nit follows that $\\triangle AHD\\sim \\triangle AOB$,\\\\\ntherefore $\\frac{AD}{AB}=\\frac{DH}{OB}$,\\\\\nhence $DH=\\frac{1}{4}AD$,\\\\\ntherefore $\\frac{1}{4}AD+CD=CD+DH$,\\\\\nhence, when C, D, H are collinear and coincide with CM, the travel time is minimized,\\\\\nthus $\\frac{1}{2}BC\\cdot AO=\\frac{1}{2}AB\\cdot CM$,\\\\\nhence $CM=\\frac{\\sqrt{15}}{2}$,\\\\\ntherefore $AM=\\sqrt{AC^2-CM^2}=\\sqrt{4^2-\\left(\\frac{\\sqrt{15}}{2}\\right)^2}=\\frac{7}{2}$,\\\\\nGiven $AD^{\\prime}=4MD^{\\prime}$, let $MD^{\\prime}=m$, then $AD^{\\prime}=4m$,\\\\\nwe get: $16m^2-m^2=\\frac{49}{4}$\\\\\nHence $m=\\frac{7\\sqrt{15}}{30}$ or $-\\frac{7\\sqrt{15}}{30}$ (discard latter),\\\\\nTherefore $AD^{\\prime}=\\frac{14\\sqrt{15}}{15}$\\\\\nThus $D\\left(0,\\frac{\\sqrt{15}}{15}\\right)$,\\\\\nTherefore, the answer is $\\boxed{\\left(0,\\frac{\\sqrt{15}}{15}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53110839.png"
  },
  {
    "solution": "\\boxed{50^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, with point O as the endpoint, make rays OA, OB, OC, OD, and OE in clockwise order. Moreover, OB is the bisector of $\\angle AOC$ and OD is the bisector of $\\angle COE$. When $\\angle AOD = n^\\circ$, then $\\angle BOE = (150 - n)^\\circ$. What is the degree measure of $\\angle BOD$?\n",
    "completion": "\\textbf{Solution:} Let $\\angle EOD = \\angle DOC = x^\\circ$ and $\\angle AOB = \\angle BOC = y^\\circ$. According to the problem statement, we have $\\angle AOD = \\angle AOB + \\angle BOC + \\angle COD = y^\\circ + y^\\circ + x^\\circ = n^\\circ$ and $\\angle BOE = \\angle BOC + \\angle COD + \\angle DOE = y^\\circ + x^\\circ + x^\\circ = (150 - n)^\\circ$. Hence, $\\angle AOD + \\angle BOE = 3x^\\circ + 3y^\\circ = 150^\\circ$, therefore $x^\\circ + y^\\circ = 50^\\circ$. Since $\\angle BOD = \\angle BOC + \\angle COD = x^\\circ + y^\\circ$, it follows that $\\angle BOD = \\boxed{50^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52579788_201.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{5}+4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $F$ lies on diagonal $AC$ of the parallelogram $ABCD$. Lines parallel to $AB$ and $AC$ are drawn passing through points $F$ and $B$ respectively, intersecting at point $E$. Line $BF$ is drawn, with given $\\angle ABF = \\angle FBC + \\angle DAC$. Quadrilateral $ABEF$ is a rhombus; if $BE=6$, $AD=10$, and $\\tan \\angle CBE = \\frac{1}{2}$, find the length of $AC$.\n",
    "completion": "\\textbf{Solution:} Construct DH$\\perp$AC at point H, as shown in the diagram:\\\\\nSince BE$\\parallel$AC,\\\\\nthen $\\angle 1=\\angle CBE$,\\\\\nSince AD$\\parallel$BC,\\\\\nthen $\\angle 2=\\angle 1$,\\\\\nthus $\\angle 2=\\angle CBE$,\\\\\nthus $\\tan \\angle 2=\\tan \\angle CBE=\\frac{1}{2}$,\\\\\nIn $\\triangle ADH$, $\\tan \\angle 2=\\frac{DH}{AH}=\\frac{1}{2}$,\\\\\nLet DH$=x$ (where $x>0$), then AH$=2x$,\\\\\nBy Pythagoras' theorem, $x^{2}+(2x)^{2}=10^{2}$,\\\\\nSolving this gives: $x=2\\sqrt{5}$,\\\\\ntherefore AH$=2x=4\\sqrt{5}$, DH$=2\\sqrt{5}$,\\\\\nSince quadrilateral ABEF is a rhombus,\\\\\nthen CD$=$AB$=$BE$=6$,\\\\\nIn $\\triangle CDH$, CH$=\\sqrt{CD^{2}-DH^{2}}=\\sqrt{6^{2}-(2\\sqrt{5})^{2}}=4$,\\\\\ntherefore AC$=$AH$+$CH$=4\\sqrt{5}+4=\\boxed{4\\sqrt{5}+4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53080940_128.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the figure, within $\\triangle ABC$ and $\\triangle ADE$, we have $\\angle BAC=\\angle DAE=90^\\circ$, $AC=AD=3$, and $AB=AE=5$. Connect $BD$ and $CE$, and rotate $\\triangle ADE$ around point $A$ for one complete circle. During the rotation, when $\\angle DBA$ is at its maximum, what is the area of $\\triangle ACE$?\n",
    "completion": "\\textbf{Solution:} From the given information, we know that the trajectory of point D is a circle with center at point A and radius equal to AD. \\\\\nWhen the line BD is tangent to the trajectory circle of point D, $\\angle DBA$ reaches its maximum value, at which point $\\angle BDA=90^\\circ$, as shown in the figure. \\\\\nDraw line CF $\\perp$ AE at F. \\\\\nSince $\\angle DAE=90^\\circ$ and $\\angle BAC=90^\\circ$, \\\\\nit follows that $\\angle CAF=\\angle BAD$. \\\\\nIn $\\triangle ABD$, by the Pythagorean theorem, BD=$\\sqrt{5^2-3^2}=4$. \\\\\nTherefore, from $\\sin \\angle CAF=\\sin \\angle BAD$, we get: \\\\\n$\\frac{CF}{AC}=\\frac{BD}{AB}$, \\\\\nthat is, $\\frac{CF}{3}=\\frac{4}{5}$, \\\\\nsolving this yields: CF=$\\frac{12}{5}$, \\\\\nthus, the area of triangle ACE = $\\frac{1}{2}\\times \\frac{12}{5}\\times 5=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51910249.png"
  },
  {
    "solution": "\\boxed{32^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that circle $O$ is the circumscribed circle of $\\triangle ABD$, where $AB$ is the diameter of circle $O$, and $CD$ is a chord of circle $O$. Given that $\\angle ABD=58^\\circ$, what is the measure of $\\angle BCD$?\n",
    "completion": "\\textbf{Solution:} Given that AB is the diameter of circle O, \\\\\nthus, $\\angle ADB=90^\\circ$, \\\\\nwhich means $\\angle A=90^\\circ-\\angle ABD=90^\\circ-58^\\circ=32^\\circ$, \\\\\nsince $\\overset{\\frown}{BD}=\\overset{\\frown}{BD}$, \\\\\ntherefore, $\\angle BCD=\\angle A=\\boxed{32^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53059861.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, quadrilateral $ABCD$ is inscribed in circle $\\odot O$, with $AC$ being the diameter of $\\odot O$, and $\\angle ADB = \\angle CDB$. If $AB = 2\\sqrt{2}$ and $AD = 2$, find the length of $CD$.\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABC$ is an isosceles right-angled triangle,\\\\\nit follows that $BC=AB=2\\sqrt{2}$,\\\\\nhence $AC=\\sqrt{AB^{2}+BC^{2}}=4$,\\\\\nIn $\\triangle ADC$, where $\\angle ADC=90^\\circ$ and $AD=2$,\\\\\nit therefore follows that $CD=\\sqrt{AC^{2}-AD^{2}}=\\boxed{2\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53080605_94.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within square $ABCD$, point $P$ is on side $AB$, $AE\\perp DP$ at point $E$, and $CF\\perp DP$ at point $F$. If $AE=4$ and $CF=7$, what is the length of $EF$?\n",
    "completion": "\\textbf{Solution:} Since $ABCD$ is a square,\\\\\ntherefore, $\\angle ADC=90^\\circ$, and $CD=DA$,\\\\\ntherefore, $\\angle CDF+\\angle ADE=90^\\circ$,\\\\\nsince $AE\\perp DP$, and $CF\\perp DP$,\\\\\ntherefore, $\\angle AED=\\angle DFC=90^\\circ$,\\\\\ntherefore, $\\angle DAE+\\angle ADE=90^\\circ$,\\\\\ntherefore, $\\angle CDF=\\angle DAE$,\\\\\nin $\\triangle CDF$ and $\\triangle DAE$,\\\\\nsince $\\angle DFC=\\angle AED=90^\\circ$, $\\angle CDF=\\angle DAE$, $CD=DA$,\\\\\ntherefore, $\\triangle CDF \\cong \\triangle DAE$ (AAS),\\\\\ntherefore, $CF=DE=7$, $DF=AE=4$,\\\\\ntherefore, $EF=DE-DF=7-4=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53706814.png"
  },
  {
    "solution": "\\boxed{N(-3, 0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the vertices of rectangle $OABC$ are such that vertex $A$ is on the positive half of the $y$-axis, and point $C$ is on the positive half of the $x$-axis, with $AO=\\sqrt{3}$ and $\\angle BAC=30^\\circ$. Point $D$ is a point on segment $OC$, and triangle $AOD$ is folded along line $AD$, placing point $O$ at point $E$ on the diagonal $AC$ of the rectangle. There exists a point $M$ on line $DE$ and a point $N$ on one of the coordinate axes, such that the quadrilateral formed by vertices $N$, $A$, $M$, and $C$ is a rhombus. Write down the coordinates of point $N$.\n",
    "completion": "\\textbf{Solution:} Consider two situations:\n1. When the diagonal of the rhombus is $AC$, the line $ED$ is the perpendicular bisector of $AC$. The vertex $N$ of the rhombus must be on the line $ED$, and point $N$ only has two opportunities on the coordinate axis:\n   (1) When line $ED$ intersects with the $x$-axis, point $N$ coincides with point $D$, forming the rhombus $AN_1CM_1$, as shown in the figure $N_1(1, 0)$;\n   (2) When line $ED$ intersects with the $y$-axis, $y=\\sqrt{3}x-\\sqrt{3}$, let $x=0$, then $y=-\\sqrt{3}$, forming the rhombus $AN_2CM_2$, as shown in the figure $N_2(0, -\\sqrt{3})$;\n2. When the side of the rhombus is $AC$:\n   (1) When $N$, $M$ are above side $AC$, point $N$ is on the $y$-axis, $AN=AC=2\\sqrt{3}$, $N(0, 3\\sqrt{3})$.\n   $y=\\sqrt{3}x-\\sqrt{3}$, let $x=3$, then $y=2\\sqrt{3}$.\n   $CM_3=2\\sqrt{3}$. $M_3(0, 3\\sqrt{3})$. $AN$ is parallel and equal to $CM_3$ forming the rhombus $ACM_3N_3$. As shown in the figure $N_3(0, 3\\sqrt{3})$.\n   (2) When $N$, $M$ are below side $AC$, point $N$ is on the $x$-axis, $ON_4=OC=3$, $N_4(-3, 0)$.\n   $y=\\sqrt{3}x-\\sqrt{3}$, let $x=0$, then $y=-\\sqrt{3}$.\n   $AO=OM_4=\\sqrt{3}$, forming the rhombus $AN_4M_4C$. As shown in the figure $N_4(-3, 0)$.\n\n$\\therefore$ The coordinates of $N$ are: \\boxed{N(1, 0)}, \\boxed{N(0, -\\sqrt{3})}, \\boxed{N(0, 3\\sqrt{3})}, or \\boxed{N(-3, 0)}.",
    "image_path": "dataset/mm_math/images/MM_Math/50519550_189.png"
  },
  {
    "solution": "\\boxed{15^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, right-angled triangle $ABC$ has its right-angled vertex $C$ on line $DE$, and $CF$ bisects $\\angle BCD$. As shown in Figure (1), if $\\angle BCE=30^\\circ$, find the degree measure of $\\angle ACF$.\n",
    "completion": "\\textbf{Solution:} Given the problem statement, since $\\angle ACB=90^\\circ,\\angle BCE=30^\\circ$,\\\\\n$\\therefore \\angle ACD=180^\\circ-90^\\circ-30^\\circ=60^\\circ,\\angle BCD=180^\\circ-30^\\circ=150^\\circ$.\\\\\nAlso, since $CF$ bisects $\\angle BCD$,\\\\\n$\\therefore \\angle DCF=\\angle BCF=\\frac{1}{2}\\angle BCD=75^\\circ$,\\\\\n$\\therefore \\angle ACF=\\angle DCF-\\angle ACD=75^\\circ-60^\\circ=\\boxed{15^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/46955179_185.png"
  },
  {
    "solution": "\\boxed{x \\geq -1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: The graph of the linear function $y=ax+b$ and the direct proportion function $y=kx$ are shown in the same Cartesian coordinate system as in the figure. What is the solution set of the inequality $ax+b\\geq kx$ with respect to $x$?\n",
    "completion": "\\textbf{Solution:} The intersection point of the two lines is $(-1, 2)$. Given that for $x \\geq -1$, the line $y=kx$ lies below the line $y=ax+b$, therefore, the solution set of the inequality $ax+b \\geq kx$ is $x \\geq -1$.\\\\\nHence, the solution set is $\\boxed{x \\geq -1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50555372.png"
  },
  {
    "solution": "\\boxed{160^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that line $AB$ intersects with line $CD$ at point $O$, $OE \\perp CD$, $\\angle AOC = 40^\\circ$, and $OF$ is the bisector of $\\angle AOD$. What is the degree measure of $\\angle EOF$?\n",
    "completion": "\\textbf{Solution:} Since the straight lines $AB$ and $CD$ intersect at the point $O$,\\\\\nit follows that $\\angle AOC=\\angle BOD=40^\\circ$,\\\\\nthus $\\angle AOD=180^\\circ-\\angle BOD=140^\\circ$,\\\\\nsince $OF$ is the bisector of $\\angle AOD$,\\\\\nit follows that $\\angle AOF=\\angle FOD=70^\\circ$,\\\\\ntherefore $\\angle EOF=\\angle EOD+\\angle FOD=\\boxed{160^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/46869571_217.png"
  },
  {
    "solution": "\\boxed{\\frac{9}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of $\\odot O$, and $OC$ is the radius. Connect $AC$ and $BC$. Extend $OC$ to point $D$ such that $\\angle CAD = \\angle B$. Draw $DM\\perp AD$ meeting the extension of $AC$ at point $M$. $AD$ is the tangent line of $\\odot O$. If $AD=6$ and $\\tan\\angle CAD=\\frac{1}{2}$, find the length of the radius of $\\odot O$.\n",
    "completion": "\\textbf{Solution:}\\\\\nBecause $\\tan\\angle CAD=\\frac{1}{2}=\\frac{DM}{AD}$, and $AD=6$,\\\\\nthen $DM=3$,\\\\\nSince $OA=OC$,\\\\\nit follows that $\\angle OAC=\\angle OCA$,\\\\\nGiven $AD\\perp OA$ and $DM\\perp AD$,\\\\\nit implies $OA\\parallel DM$,\\\\\nthus $\\angle M=\\angle OAC$,\\\\\nSince $\\angle OCA=\\angle DCM$,\\\\\nit results in $\\angle DCM=\\angle M$,\\\\\ntherefore $DC=DM=3$,\\\\\nIn $\\triangle OAD$, using $OA^2+AD^2=OD^2$, which means $OA^2+6^2=(OC+3)^2=(OA+3)^2$,\\\\\nthus $OA=\\frac{9}{2}$,\\\\\nHence, the radius of $\\odot O$ is $\\boxed{\\frac{9}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51446366_122.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of $\\odot O$, $C$ and $D$ are points on $\\odot O$, and $\\angle CDB=30^\\circ$. A tangent to $\\odot O$ passing through point $C$ intersects the extension of $AB$ at point $E$. What is the value of $\\sin E$?\n",
    "completion": "\\textbf{Solution:} Draw the line OC, \\\\\nsince $CE$ is a tangent to the circle $O$, \\\\\nit follows that $OC \\perp CE$, \\\\\nthat is, $\\angle OCE = 90^\\circ$, \\\\\nsince $\\angle CDB = 30^\\circ$, and angles $\\angle COB$ and $\\angle CDB$ are the central and inscribed angles that intercept the same arc $BC$, respectively, \\\\\nit follows that $\\angle COB = 2\\angle CDB = 60^\\circ$, \\\\\ntherefore $\\angle E = 90^\\circ - \\angle COB = 30^\\circ$, \\\\\ntherefore $\\sin\\angle E = \\boxed{\\frac{1}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51403118.png"
  },
  {
    "solution": "\\boxed{\\begin{cases}x=-2\\\\ y=-1\\end{cases}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, given the functions $y=ax+b$ and $y=kx$ intersect at point $P$, according to the graph, what is the solution to the system of linear equations $\\begin{cases}y=ax+b,\\\\ y=kx?\\end{cases}$\n",
    "completion": "\\textbf{Solution:} Since the intersection point P of the graphs of the functions $y=ax+b$ and $y=kx$ is at ($-2$, $-1$), \\\\\ntherefore, the solution to the system of linear equations $\\begin{cases}y=ax+b\\\\ y=kx\\end{cases}$ is $\\boxed{\\begin{cases}x=-2\\\\ y=-1\\end{cases}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53182976.png"
  },
  {
    "solution": "\\boxed{(-3,-4- \\sqrt{6})}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, it is known that the parabola $y=x^{2}+bx+c$ intersects with line AB at points A and B, where A($-3$, $-4$) and B($0$, $-1$), and the axis of symmetry of the parabola intersects with the x-axis at point Q. The parabola is translated 2 units to the right to obtain the parabola $y=a_{1}x^{2}+b_{1}x+c_{1}$ ($a_{1}\\neq 0$). The translated parabola intersects with the original parabola at point C. Point D is a point on the axis of symmetry of the original parabola. In the Cartesian coordinate system, is there a point E such that the quadrilateral with vertices B, C, D, E is a rhombus? If it exists, please find the coordinates of point E; if it does not exist, please explain the reason.\n",
    "completion": "\\textbf{Solution:} The vertex of the parabola $y=x^{2}+4x-1$ is $(-2,-5)$, and its axis of symmetry is the line $x=-2$,\\\\\nThe parabola after translation is $y=x^{2}-5$, by\n\\[\n\\begin{cases}\ny=x^{2}+4x-1\\\\\ny=x^{2}-5\n\\end{cases}\n\\]\nwe get\n\\[\n\\begin{cases}\nx=-1\\\\\ny=-4\n\\end{cases}\n\\]\n$\\therefore$ The coordinates of point C are $(-1,-4)$, let the coordinates of point D be $(-2,t)$,\\\\\nThe quadrilateral formed by vertices B, C, D, E is a rhombus, that is, $\\triangle BDC$ is an isosceles triangle,\\\\\n(1) When BC is the diagonal, by the midpoint formula we get $E(1,-5-t)$, $\\because BD=CD$,\\\\\n$\\therefore 1^{2}+(-4-t)^{2}=2^{2}+(-1-t)^{2}$, solving for $t=-2$, $\\therefore E(1,-3)$;\\\\\n(2) When BD is the diagonal, by the midpoint formula we get $E(-1,t+3)$, $\\because BC=CD$,\\\\\n$\\therefore (-1)^{2}+(-3)^{2}=(-1)^{2}+(t+4)^{2}$,\\\\\nsolving for $t_{1}=-1$, $t_{2}=-7$ (discard), $\\therefore E(-1,2)$;\\\\\n(3) When BE is the diagonal, by the midpoint formula we get $E(-3,-3+t)$,\\\\\n$\\because BC=BD$, $\\therefore (-1)^{2}+(-3)^{2}=(-2)^{2}+(t+1)^{2}$,\\\\\nsolving for $t_{1}=-1+ \\sqrt{6}$, $t_{2}=-1- \\sqrt{6}$,\\\\\n$\\therefore E(-3,-4+ \\sqrt{6})$ or $E(-3,-4- \\sqrt{6})$,\\\\\nIn summary, the coordinates of point E are $\\boxed{(1,-3)}$ or $\\boxed{(-1,2)}$ or $\\boxed{(-3,-4+ \\sqrt{6})}$ or $\\boxed{(-3,-4- \\sqrt{6})}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52694121_92.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated in the figure, $AB \\parallel CD \\parallel EP$, given that $\\frac{AC}{CE} = \\frac{3}{2}$ and $BD = 12$, what is the length of $DF$?\n",
    "completion": "\\textbf{Solution:} Given $\\because AB\\parallel CD\\parallel EP$, $\\frac{AC}{CE}=\\frac{3}{2}$, and $BD=12$, \\\\\n$\\therefore \\frac{AC}{CE}=\\frac{BD}{DF}=\\frac{3}{2}$, \\\\\n$\\therefore \\frac{12}{DF}=\\frac{3}{2}$, \\\\\n$\\therefore DF=\\boxed{8}$",
    "image_path": "dataset/mm_math/images/MM_Math/52966957.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the angle bisector of $\\angle BAC$ intersects the circumcircle of $\\triangle ABC$ at point $D$ and intersects $BC$ at point $F$. The angle bisector of $\\angle ABC$ meets $AD$ at point $E$. Given $DE=DB$, if $\\angle BAC=90^\\circ$ and $BD=4$, what is the radius of the circumcircle of $\\triangle ABC$?\n",
    "completion": "\\textbf{Solution:} Draw line CD, as shown,\\\\\n$\\because$ $\\angle BAC=90^\\circ$,\\\\\n$\\therefore$ BC is the diameter,\\\\\n$\\therefore$ $\\angle BDC=90^\\circ$,\\\\\n$\\because$ $\\angle 1=\\angle 2$,\\\\\n$\\therefore$ DB=BC,\\\\\n$\\therefore$ $\\triangle DBC$ is an isosceles right triangle,\\\\\n$\\therefore$ BC=$\\sqrt{2}$BD=$4\\sqrt{2}$,\\\\\n$\\therefore$ the radius of the circumscribed circle of $\\triangle ABC$ is $\\boxed{2\\sqrt{2}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52745974_84.png"
  },
  {
    "solution": "\\boxed{40}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $BC=20$, points $B_1$, $B_2$, $B_3$, $B_4$ and points $C_1$, $C_2$, $C_3$, $C_4$ are the quintisection points of $AB$, $AC$ respectively. What is the value of $B_1C_1+B_2C_2+B_3C_3+B_4C_4$?\n",
    "completion": "\\[ \\text{Solution:} \\quad \\because \\frac{AB_4}{AB} = \\frac{AC_4}{AC} = \\frac{4}{5}, \\]\n\\[ \\angle A = \\angle A, \\]\n\\[ \\therefore \\triangle AB_4C_4 \\sim \\triangle ABC, \\]\n\\[ \\therefore \\frac{B_4C_4}{BC} = \\frac{AB_4}{AB} = \\frac{4}{5}, \\]\n\\[ \\therefore B_4C_4 = \\frac{4}{5}BC = \\frac{4}{5} \\times 20 = 16, \\]\nSimilarly, $B_1C_1 = 4$,\n\\[ B_2C_2 = 8, \\]\n\\[ B_3C_3 = 12, \\]\n\\[ \\therefore B_1C_1 + B_2C_2 + B_3C_3 + B_4C_4 = \\boxed{40}. \\]",
    "image_path": "dataset/mm_math/images/MM_Math/51369938.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that in $\\triangle ABC$, $AB=AC$, $AD$ bisects $\\angle BAC$, and point $E$ is the midpoint of $AB$. If $AC=6$, what is the length of $DE$?\n",
    "completion": "\\textbf{Solution:} \\\\\nSince $AB=AC$ and $AD$ bisects $\\angle BAC$, \\\\\nit follows that $D$ is the midpoint of $BC$, \\\\\nSince $E$ is the midpoint of $AB$, \\\\\nit follows that $DE$ is the median of the triangle, \\\\\nGiven $AC=6$, \\\\\nit follows that $DE=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53009913.png"
  },
  {
    "solution": "\\boxed{30}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in a Cartesian coordinate system, the line $y=x-5$ intersects the x-axis at point A and the y-axis at point B. The parabola $y=ax^2+bx+c$ passes through points A and B, with another intersection point C on the x-axis, and $\\tan\\angle CBO=\\frac{1}{5}$. If point P is the vertex of the parabola, what is the area of the quadrilateral $APBC$?\n",
    "completion": "\\textbf{Solution:} From the given problem, we have \n\n\\[\\because y=x^2-4x-5=(x-2)^2-9,\\]\n\n\\[\\therefore\\] the vertex \\(P(2, -9)\\),\n\nDraw \\(AP\\) and \\(BP\\), and through point B draw \\(BD\\parallel x\\) axis intersecting \\(AP\\) at \\(D\\),\n\nLet the equation of line \\(AP\\) be \\(y=kx+b(k\\ne 0)\\),\n\nSubstituting \\(A(5, 0)\\) and \\(P(2, -9)\\) yields: \\[\\left\\{\\begin{array}{l} 5k+b=0 \\\\ 2k+b=-9\\end{array}\\right.,\\]\n\nSolving this, we get: \\[\\left\\{\\begin{array}{l} k=3 \\\\ b=-15\\end{array}\\right.,\\]\n\n\\[\\therefore\\] the equation of line \\(AP\\) is \\(y=3x-15\\),\n\nSubstituting \\(y=-5\\) into \\(y=3x-15\\) yields: \\(-5=3x-15\\),\n\nSolving this, we get: \\(x=\\frac{10}{3}\\),\n\n\\[\\therefore D\\left(\\frac{10}{3}, -5\\right),\\]\n\n\\[\\because AC=5-(-1)=6\\], \\(BD=\\frac{10}{3}\\),\n\n\\[\\therefore {S}_{\\text{quadrilateral}APBC}={S}_{\\text{trapezoid}ADBC}+{S}_{\\triangle BDP}\\]\n\n\\[=\\frac{1}{2}\\times \\left(6+\\frac{10}{3}\\right)\\times 5+\\frac{1}{2}\\times \\frac{10}{3}\\times \\left[-5-\\left(-9\\right)\\right]\\]\n\n\\[=\\frac{1}{2}\\times \\frac{28}{3}\\times 5+\\frac{1}{2}\\times \\frac{10}{3}\\times 4\\]\n\n\\[=\\frac{70}{3}+\\frac{20}{3}\\]\n\n\\[=\\boxed{30}.\\]",
    "image_path": "dataset/mm_math/images/MM_Math/52797118_50.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $P$ is a point inside the equilateral triangle $ABC$. When $\\triangle ABP$ is rotated $60^\\circ$ clockwise around point $B$, it becomes $\\triangle CBP'$. If $PB = 3$, what is the length of $PP'$?\n",
    "completion": "\\textbf{Solution:} We have, since rotating $\\triangle ABP$ $60^\\circ$ clockwise about point B yields $\\triangle CBP'$,\\\\\nthus $\\angle PBP'=60^\\circ$ and $BP=BP'$,\\\\\ntherefore, $\\triangle BPP'$ is an equilateral triangle,\\\\\nhence $PP'=BP=\\boxed{3}$",
    "image_path": "dataset/mm_math/images/MM_Math/52891125.png"
  },
  {
    "solution": "\\boxed{\\frac{15}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $DE \\parallel BC$, $G$ is a point on $BC$, and line $AG$ intersects $DE$ at point $F$. It is known that $AF = 2$, $AG = 6$, and $EC = 5$. What is the length of $AC$?\n",
    "completion": "\\textbf{Solution:} Since $DE \\parallel BC$,\\\\\nit follows that $\\frac{AF}{FG} = \\frac{AE}{EC}$, that is, $\\frac{2}{4} = \\frac{AE}{5}$,\\\\\ntherefore, $AE = \\frac{5}{2}$,\\\\\nhence, $AC = AE + EC = \\frac{5}{2} + 5 = \\boxed{\\frac{15}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52550914.png"
  },
  {
    "solution": "\\boxed{1<x<3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the line $y_{1}=2x-2$ intersects the y-axis at point A, and the graph of the line $y_{2}=-2x+6$ intersects the y-axis at point B. The two lines intersect at point C. What is the range of $x$ when both $y_{1}>0$ and $y_{2}>0$ are satisfied?\n",
    "completion": "\\textbf{Solution:} When both $y_{1}>0$ and $y_{2}>0$ hold, the range of $x$ is $x=\\boxed{1<x<3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52423206_70.png"
  },
  {
    "solution": "\\boxed{70^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point D is on the extension line of BC, DE $\\perp$ AB at point E, and intersects AC at point F. If $\\angle A = 35^\\circ$ and $\\angle D = 15^\\circ$, what is the degree measure of $\\angle ACB$?\n",
    "completion": "\\textbf{Solution:} Since $DE \\perp AB$ and $\\angle A=35^\\circ$,\\\\\nit follows that $\\angle CFD = \\angle AFE = 90^\\circ - \\angle A = 90^\\circ - 35^\\circ = 55^\\circ$,\\\\\nAlso, since $\\angle D = 15^\\circ$,\\\\\nit follows that $\\angle ACB = \\angle D + \\angle CFD = 15^\\circ + 55^\\circ = \\boxed{70^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52807545.png"
  },
  {
    "solution": "\\boxed{63^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, O is a point on line AB, and $\\angle COD$ is a right angle. OE bisects $\\angle BOD$, and OF bisects $\\angle AOE$. If $\\angle BOC=54^\\circ$, what are the degrees of $\\angle COE$ and $\\angle DOF$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle COD=90^\\circ$ and $\\angle BOC=54^\\circ$,\\\\\nit follows that $\\angle BOD=90^\\circ-54^\\circ=36^\\circ$,\\\\\nSince OE bisects $\\angle BOD$,\\\\\nit follows that $\\angle DOE=\\angle BOE=18^\\circ$,\\\\\nthus $\\angle COE=\\angle BOC+\\angle BOE=54^\\circ+18^\\circ=\\boxed{72^\\circ}$, $\\angle AOE=180^\\circ-\\angle BOE=180^\\circ-18^\\circ=162^\\circ$.\\\\\nSince OF bisects $\\angle AOE$,\\\\\nit follows that $\\angle EOF=\\frac{1}{2}\\angle AOE=81^\\circ$\\\\\nthus $\\angle DOF=\\angle EOF-\\angle DOE=81^\\circ-18^\\circ=\\boxed{63^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/51208154.png"
  },
  {
    "solution": "\\boxed{\\left(2, 1\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the Cartesian coordinate system, let us consider a parabola $y=\\frac{1}{2}x^{2}+bx+c$ (where $b$ and $c$ are constants). The vertex of the parabola is given by the coordinates $\\left(\\frac{3}{2}, -\\frac{25}{8}\\right)$. This parabola intersects the $x$-axis at points $A$ and $B$ (with $A$ located to the left of $B$) and intersects the $y$-axis at point $C$. Point $C$ and point $D$ are symmetric with respect to the $x$-axis. The line segment $AD$ is drawn, and so is the straight line $BD$. When point $P$ is on the parabola and point $Q$ is on the line $BD$, and the quadrilateral formed by points $C$, $D$, $P$, and $Q$ is a parallelogram, write down the coordinates of point $Q$ directly.\n",
    "completion": "\\textbf{Solution:} The coordinates of point $Q$ are $\\boxed{\\left(2, 1\\right)}$, $\\left(1-\\sqrt{17}, \\frac{3+\\sqrt{17}}{2}\\right)$, $\\left(1+\\sqrt{17}, \\frac{3-\\sqrt{17}}{2}\\right)$, $\\left(-2, 3\\right)$.",
    "image_path": "dataset/mm_math/images/MM_Math/53079018_51.png"
  },
  {
    "solution": "\\boxed{-6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the area of the rhombus OABC is 12. Point $B$ is on the $y$-axis, and point $C$ is on the graph of the inverse proportion function $y=\\frac{k}{x}$. What is the value of $k$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, we connect AC to intersect the y-axis at point D,\\\\\n$\\because$ rhombus OABC,\\\\\n$\\therefore$ AC$\\perp$OB, and CD=AD=$\\frac{1}{2}$AC, BD=OD=$\\frac{1}{2}$BO,\\\\\n$\\because$ the area of rhombus OABC is 12,\\\\\n$\\therefore$ AC$\\cdot$BO=24,\\\\\n$\\therefore$ CD$\\cdot$OD=6,\\\\\n$\\therefore$ $\\left|k\\right|=6$,\\\\\nFurthermore, $\\because$ the graph of the inverse proportional function is located in the second quadrant,\\\\\n$\\therefore$ k$<$0, that is, $k=\\boxed{-6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51435977.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that point $D$ is the midpoint of side $AC$ of $\\triangle ABC$, $AE \\parallel BC$, line $ED$ intersects $AB$ at point $G$ and extends to intersect $BC$ at point $F$. If $\\frac{BG}{GA} = 3$ and $BC = 8$, what is the length of $AE$?\n",
    "completion": "\\textbf{Solution:} Since $AE \\parallel BC$, \\\\\nit follows that $\\triangle AEG \\sim \\triangle BFG$, and $\\triangle AED \\sim \\triangle CFD$, \\\\\nthus, $\\frac{AE}{BF}=\\frac{AG}{BG}=\\frac{1}{3}$, and $\\frac{AE}{CF}=\\frac{AD}{CD}=1$, \\\\\nwhich means $AE=CF$, \\\\\nAlso, given $BC=8$, \\\\\ntherefore $\\frac{AE}{8+AE}=\\frac{1}{3}$\\\\\n$AE=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52917522.png"
  },
  {
    "solution": "\\boxed{(-\\sqrt{3}, 1)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the radius $OA=2$ of the regular hexagon ABCDEF. What are the coordinates of point B?\n",
    "completion": "\\textbf{Solution:} Draw OB, let BC intersect the x-axis at M, \\\\\n$\\because$ the radius of the regular hexagon ABCDEF, $OA=2$, \\\\\n$\\therefore$ OB=OA=AB=2, $\\angle$AOB=$\\angle$ABO=$\\angle$OBC=$60^\\circ$, \\\\\n$\\therefore$ $\\angle$BOM=$30^\\circ$, $\\angle$BMO=$90^\\circ$, \\\\\n$\\therefore$ BM=$\\frac{1}{2}$OB=1, \\\\\n$\\therefore$ OM=$\\sqrt{OB^{2}-BM^{2}}=\\sqrt{2^{2}-1^{2}}=\\sqrt{3}$, \\\\\n$\\therefore$ the coordinates of point B are ($-\\sqrt{3}$, 1), \\\\\nHence the answer is: $\\boxed{(-\\sqrt{3}, 1)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51402793.png"
  },
  {
    "solution": "\\boxed{24}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in rhombus $ABCD$, the midpoint of side $AB$ is $M$, and the diagonals intersect at point $O$. If $OM = 3\\,\\text{cm}$, what is the perimeter of the rhombus $ABCD$ in centimeters?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral $ABCD$ is a rhombus, \\\\\nit follows that AO=CO, and AB=BC=CD=AD, \\\\\nsince point M is the midpoint of side $AB$ and $OM=3$, \\\\\nit follows that AD=2OM=6, \\\\\ntherefore, the perimeter of rhombus $ABCD$ is $4AD=\\boxed{24}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52354541.png"
  },
  {
    "solution": "\\boxed{\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in rhombus $ABCD$, $\\angle A=120^\\circ$, and the length of diagonal $BD$ is 3. What is the length of the sides of the rhombus?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect AC, intersecting BD at point O.\\\\\n$\\because$ Quadrilateral ABCD is a rhombus, hence AC$\\perp$BD, OB=OD=$\\frac{1}{2}$BD=$\\frac{1}{2}\\times 3=\\frac{3}{2}$,\\\\\n$\\angle$BAC=$\\frac{1}{2}\\angle BAD=\\frac{1}{2}\\times 120^\\circ=60^\\circ$,\\\\\n$\\therefore \\angle$ABO=$90^\\circ-\\angle$BAO=$30^\\circ$\\\\\n$\\therefore$AB=2AO\\\\\nIn $\\triangle AOB$, let AO=x, then AB=2x\\\\\nBy the Pythagorean theorem, we have: $AO^{2}+BO^{2}=AB^{2}$\\\\\n$\\therefore x^{2}+\\left(\\frac{3}{2}\\right)^{2}=(2x)^{2}$\\\\\nSolving for x, we get $x=\\frac{\\sqrt{3}}{2}$\\\\\nHence, AB=2x=$\\boxed{\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52189283.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, two congruent circles $\\odot O_1$ and $\\odot O_2$ intersect at points A and B, and $\\odot O_1$ passes through the center of $\\odot O_2$. What is the measure of $\\angle O_1AB$?\n",
    "completion": "\\textbf{Solution:} Connect $O_1O_2$, $AO_2$, and $O_1B$, \\\\\nbecause $O_1B= O_1A$ \\\\\nthus, $\\angle O_1AB=\\angle O_1BA=\\frac{1}{2}\\angle AO_2O_1$ \\\\\nbecause $\\odot O_1$ and $\\odot O_2$ are congruent circles, \\\\\nthus, $AO_1=O_1O_2=AO_2$, \\\\\ntherefore, $\\triangle AO_2O_1$ is an equilateral triangle, \\\\\nthus, $\\angle AO_2O_1=60^\\circ$, \\\\\ntherefore, $\\angle O_1AB= \\frac{1}{2}\\angle AO_2O_1 =\\frac{1}{2}\\times 60^\\circ =\\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51319015.png"
  },
  {
    "solution": "\\boxed{74^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=37^\\circ$, the perpendicular bisector of side $BC$ intersects $AC$ and $BC$ at points $D$ and $E$, respectively, and $AB=CD$, then what is the degree measure of $\\angle A$?\n",
    "completion": "\\textbf{Solution:} Draw BD as shown in the diagram: \\\\\n$\\because$ DE is the perpendicular bisector of BC, and AB=CD, \\\\\n$\\therefore$ BD=CD=AB, \\\\\n$\\because$ $\\angle$C=37$^\\circ$, \\\\\n$\\therefore$ $\\angle$DBC=$\\angle$C=37$^\\circ$, \\\\\n$\\therefore$ $\\angle$ADB=2$\\angle$C=74$^\\circ$, \\\\\n$\\because$ AB=BD, \\\\\n$\\therefore$ $\\angle$A=$\\angle$ADB=\\boxed{74^\\circ},",
    "image_path": "dataset/mm_math/images/MM_Math/51379083.png"
  },
  {
    "solution": "\\boxed{16}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=ax^2+bx+c\\ (a\\ne 0)$ intersects the y-axis at point $C(0, 4)$ and intersects the $x$-axis at points A and B, where the coordinates of point A are $(-2, 0)$. The axis of symmetry of the parabola $x=1$ intersects the parabola at point D and intersects the line $BC$ at point E. Suppose point F is a moving point on the parabola above the line $BC$. Determine whether there exists a point F that maximizes the area of the quadrilateral $ABFC$. If such a point exists, find the coordinates of point F and the maximum area; if not, please explain why.\n",
    "completion": "\\textbf{Solution:} Existence is proven as follows: As shown in Figure 1, draw $FH \\perp x$-axis at point $H$ intersecting $BC$ at point $G$, and let $F(x, -\\frac{1}{2}x^2+x+4)$,\\\\\n$\\because$ Point $B$ is symmetric to point $A(-2, 0)$ with respect to the line $x=1$,\\\\\n$\\therefore B(4, 0)$, and $AB=6$,\\\\\nLet the equation of line $BC$ be $y=kx+4$, then $4k+4=0$,\\\\\nSolving gives $k=-1$,\\\\\n$\\therefore y=-x+4$,\\\\\n$\\therefore G(x, -x+4)$,\\\\\n$\\therefore FG=-\\frac{1}{2}x^2+2x$,\\\\\n$\\therefore S_{\\triangle FBC}=\\frac{1}{2} \\times 4(-\\frac{1}{2}x^2+2x)=-x^2+4x$,\\\\\n$\\therefore S_{\\text{quadrilateral} ABFC}=-x^2+4x+\\frac{1}{2} \\times 6 \\times 4=-(x-2)^2+16$,\\\\\n$\\therefore$ When $x=2$, the maximum area of $\\text{quadrilateral } ABFC=\\boxed{16}$, and $F(2, 4)$,\\\\\n$\\therefore$ The coordinates of point $F$ are $(2, 4)$, and the maximum area of quadrilateral $ABFC$ is $\\boxed{16}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53080245_26.png"
  },
  {
    "solution": "\\boxed{15^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $\\angle AOB$ is a right angle, $OC$ is the bisector of $\\angle AOB$, and $\\angle AOD = 2\\angle BOD$. What is the measure of $\\angle COD$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle AOB$ is a right angle, and OC bisects $\\angle AOB$,\n\n$\\therefore$ $\\angle AOB=90^\\circ$, $\\angle AOC=\\angle BOC=45^\\circ$,\n\nSince $\\angle AOD=2\\angle BOD$, and $\\angle AOD+\\angle BOD=90^\\circ$,\n\n$\\therefore$ $\\angle DOB=30^\\circ$,\n\n$\\therefore$ $\\angle COD=\\angle BOC−\\angle DOB=45^\\circ−30^\\circ=\\boxed{15^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51508815.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, $AB\\parallel CD$, $DE$ bisects $\\angle ADC$ and intersects $BC$ at point $E$, $AE\\perp DE$, $AB=4$, $AE=3$, and $\\sin\\angle CDE=\\frac{3}{5}$. Then, what is the length of $CD$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, extend $AE,DC$ to intersect at point $F$,\\\\\nsince $DE$ bisects $\\angle ADC$ and $DE\\perp AE$,\\\\\ntherefore $\\angle ADE=\\angle FDE$, $AE=EF$,\\\\\nlet $\\angle ADE=\\angle FDE=\\alpha$\\\\\nsince $DE\\perp AE$, $\\sin\\angle CDE=\\frac{3}{5}$, $AB=4$, $AE=3$\\\\\ntherefore $\\frac{AE}{AD}=\\sin\\alpha =\\frac{3}{5}$\\\\\ntherefore $AD=5$\\\\\ntherefore $DE=\\sqrt{AD^{2}-AE^{2}}=4$\\\\\nsince $AB\\parallel CD$\\\\\ntherefore $\\angle F=\\angle BAE=90^\\circ-\\alpha$\\\\\nsince $DE\\perp AE$\\\\\ntherefore $\\angle DAE=90^\\circ-\\alpha$,\\\\\ntherefore $\\angle F=\\angle DAE=\\angle EAB$\\\\\ntherefore $DF=AD=5$\\\\\nIn $\\triangle ABE$ and $\\triangle FCE$,\\\\\n$\\left\\{\\begin{array}{l}\n\\angle BAE=\\angle F\\\\\n\\angle AEB=\\angle FEC\\\\\nAE=FE\n\\end{array}\\right.$,\\\\\ntherefore $\\triangle ABE\\cong \\triangle FCE$,\\\\\ntherefore $CF=AB=4$,\\\\\ntherefore $DC=DF-CF=5-4=\\boxed{1}$,",
    "image_path": "dataset/mm_math/images/MM_Math/53079125.png"
  },
  {
    "solution": "\\boxed{\\pi-2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, given that the circumference of $\\odot O$ is $4\\pi$ and the length of $\\overset{\\frown}{AB}$ is $\\pi$, what is the area of the shaded region in the figure?\n",
    "completion": "\\textbf{Solution:} Since the circumference of circle O is $4\\pi$,\\\\\ntherefore, the diameter of circle O is 4,\\\\\nthus, the radius of circle O is 2,\\\\\nsince the length of arc AB is $\\pi$,\\\\\ntherefore, the length of arc AB is $\\frac{1}{4}$ of the circumference of circle O,\\\\\nhence, $\\angle AOB=90^\\circ$,\\\\\ntherefore, the area of the shaded region $= \\frac{90^\\circ \\cdot \\pi \\times 2^2}{360} - \\frac{1}{2} \\times 2 \\times 2 = \\boxed{\\pi-2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53080746.png"
  },
  {
    "solution": "\\boxed{\\frac{8}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $AE$ is the bisector of $\\angle BAC$, and $ED\\parallel CA$. Given that $BE=2$, $EC=3$, and $AC=4$, what is the length of $AD$?\n",
    "completion": "\\textbf{Solution:} Given that \\\\\n$\\because ED\\parallel CA$, \\\\\n$\\therefore \\triangle BED\\sim \\triangle BCA$, $\\angle DEA=\\angle EAC$, \\\\\n$\\therefore \\frac{BE}{BC}=\\frac{ED}{CA}$, \\\\\n$\\because BE=2$, $EC=3$, $AC=4$, \\\\\n$\\therefore BC=5$, \\\\\n$\\therefore ED=\\frac{BE\\cdot CA}{BC}=\\frac{8}{5}$, \\\\\n$\\because AE$ bisects $\\angle BAC$, \\\\\n$\\therefore \\angle EAC=\\angle DAE$, \\\\\n$\\therefore \\angle DAE=\\angle DEA$, \\\\\n$\\therefore \\triangle ADE$ is an isosceles triangle, \\\\\n$\\therefore ED=AD=\\boxed{\\frac{8}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51212743.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, given the line segment AB, arcs are drawn with A and B as centers, respectively, and with radii greater than half the length of segment AB. The arcs intersect at points C and D. Connect AC, BC, BD, CD, and AD. Given that $AB=3$ and $CD=4$, what is the perimeter of the quadrilateral $ACBD$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, AC and CD intersect at point O.\\\\\nFrom the construction steps, it is known that CD is the perpendicular bisector of line segment AB,\\\\\n$\\because$AC=AD=BC=BD,\\\\\n$\\therefore$The quadrilateral ACBD is a rhombus,\\\\\n$\\because$AB=3, CD=4,\\\\\n$\\therefore$AO=$\\frac{1}{2}$AB=$\\frac{3}{2}$, OD=$\\frac{1}{2}$CD=2,\\\\\n$\\therefore$AD=$\\sqrt{AO^{2}+OB^{2}}$=$\\sqrt{\\left(\\frac{3}{2}\\right)^{2}+2^{2}}$=$\\frac{5}{2}$,\\\\\n$\\therefore$The perimeter of rhombus ACBD=4AD=4$\\times\\frac{5}{2}$=$\\boxed{10}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51510277.png"
  },
  {
    "solution": "\\boxed{72^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the pentagon $ABCDE$ is a regular pentagon. If $l_{1} \\parallel l_{2}$, then what is the value of $\\angle \\alpha - \\angle \\beta$?\n",
    "completion": "\\textbf{Solution} Let line $BF$ pass through point $B$ and be parallel to $l_{1}$, as shown in the figure:\\\\\nSince $l_{1}\\parallel l_{2}$,\\\\\nit follows that $BF\\parallel l_{1}\\parallel l_{2}$,\\\\\nthus, $\\angle \\beta =\\angle CBF$ and $\\angle ABF+\\angle \\alpha =180^\\circ$,\\\\\ntherefore, $\\angle \\alpha =180^\\circ-\\angle ABF$,\\\\\nalso, since pentagon $ABCDE$ is a regular pentagon,\\\\\nit follows that $\\angle ABC=\\frac{180^\\circ\\times (5-2)}{5}=108^\\circ$,\\\\\nalso, since $\\angle ABC=\\angle ABF+\\angle CBF$,\\\\\nit follows that $108^\\circ=\\angle \\beta +180^\\circ-\\angle \\alpha$\\\\\ntherefore, $\\angle \\alpha -\\angle \\beta =\\boxed{72^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53102103.png"
  },
  {
    "solution": "\\boxed{y=-x^2-2x+3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the parabola $y=-x^2+bx+c$ passes through points A, B ($-3$, $0$), and C ($0$, $3$). Find the analytic equation of the parabola.\n",
    "completion": "\\textbf{Solution:} Substitute B ($-3$, $0$) and C ($0$, $3$) into $y=-x^2+bx+c$, yielding\n\\[\n\\left\\{\n\\begin{array}{l}\n-9-3b+c=0 \\\\\nc=3\n\\end{array}\n\\right.\n\\]\nTherefore,\n\\[\n\\left\\{\n\\begin{array}{l}\nb=-2 \\\\\nc=3\n\\end{array}\n\\right.\n\\]\nTherefore, the equation of the parabola is: \\boxed{y=-x^2-2x+3}.",
    "image_path": "dataset/mm_math/images/MM_Math/52602225_172.png"
  },
  {
    "solution": "\\boxed{\\frac{14}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=ax^2+bx+6$ intersects the x-axis at points $A(2,0)$ and $B(8,0)$, and intersects the y-axis at point $C$. Point $P$ is a moving point on the parabola, when $\\angle PCB =\\frac{1}{2}\\angle BCO$, find the x-coordinate of point $P$.\n",
    "completion": "\\textbf{Solution:} When point $P$ is on the parabola below $BC$, then $\\angle PCB =\\frac{1}{2}\\angle BCO$ i.e., $CP$ bisects $\\angle BCO$. As shown in the figure, draw $CP$ to bisect $\\angle BCO$, intersecting the x-axis at point $D$, and draw $DE\\perp BC$, with $E$ being the foot of the perpendicular,\\\\\n$\\because CP$ bisects $\\angle BCO$, $DE\\perp BC$,\\\\\n$\\therefore OD=DE$, $\\angle DCO=\\angle DCE$,\\\\\n$\\because OD=DE$, $\\angle DCO=\\angle DCE$, $\\angle COD=\\angle CED=90^\\circ$,\\\\\n$\\therefore \\triangle DOC\\cong \\triangle DEC, CO=CE=6$,\\\\\n$\\therefore BC=\\sqrt{CO^2+BO^2}=\\sqrt{6^2+8^2}=10, BE=BC−CE=4$,\\\\\nLet $OD=DE=m$, $BD=8−m$,\\\\\nBy the Pythagorean theorem: $BE^2+DE^2=BD^2$, i.e., $4^2+m^2=(8−m)^2$,\\\\\nSolving this yields: $m=3$, i.e., $OD=DE=3$, the coordinates of $D$ are $(3,0)$,\\\\\nLet the equation of $CD$ be: $y=kx+b(k\\ne 0)$, substituting points $C, D$ we get:\\\\\n$\\begin{cases} 6=b \\\\ 0=3k+b \\end{cases}$, thus solving we get: $\\begin{cases} k=-2 \\\\ b=6 \\end{cases}$, so, the equation of $CD$ is: $y=-2x+6$,\\\\\n$\\because P$ is the intersection point of line $CD$ and the parabola,\\\\\n$\\therefore$ we solve together to get: $-2x+6=\\frac{3}{8}x^2-\\frac{15}{4}x+6$,\\\\\nSolving for $x=0$ (discard) or $x=\\boxed{\\frac{14}{3}}$, hence the x-coordinate of $P$ is $x=\\frac{14}{3}$,\\\\\nWhen $P$ is on the parabola above $BC$, then $\\angle PCB =\\frac{1}{2}\\angle BCO$. As shown in the figure, draw $\\angle PCB =\\frac{1}{2}\\angle BCO$ intersecting the parabola at point $P$, extend $DE$ to intersect $CP$ at point $F$, draw $EH\\perp$x-axis from $E$ intersecting at point $H$,\\\\\n$\\because \\angle PCB =\\frac{1}{2}\\angle BCO$, $\\angle DCB=\\angle DCO$,\\\\\n$\\therefore \\angle PCB=\\angle DCB$,\\\\\n$\\because \\angle PCB=\\angle DCB, CE=CE, \\angle DEC=\\angle FEC$,\\\\\n$\\therefore \\triangle DEC\\cong \\triangle FEC, DE=DF$,\\\\\n$\\because \\angle CBO=\\angle EBH, \\angle COB=\\angle EHB=90^\\circ$,\\\\\n$\\therefore \\triangle EHB\\sim \\triangle COB$,\\\\\n$\\therefore \\frac{BE}{BC}=\\frac{EH}{CO}=\\frac{BH}{BO}, \\frac{4}{10}=\\frac{EH}{6}=\\frac{BH}{8}$,\\\\\nHence $EH=\\frac{12}{5}, BH=\\frac{16}{5}, OH=BO−BH=\\frac{24}{5}$,\\\\\n$\\therefore E(\\frac{24}{5}, \\frac{12}{5})$,\\\\\nLet $F$ be $(m, n)$, by $DE=DF$ we get $\\frac{m+3}{2}=\\frac{24}{5}, \\frac{n+0}{2}=\\frac{12}{5}$, solving for: $m=\\frac{33}{5}, n=\\frac{24}{5}$,\\\\\nThat is, $F$ is $(\\frac{33}{5}, \\frac{24}{5})$,\\\\\nLet the equation of $CF$ be: $y=kx+b(k\\ne 0)$, substituting $C, F$ we find:\\\\\n$\\begin{cases} 6=b \\\\ \\frac{24}{5}=\\frac{33}{5}k+b \\end{cases}$, solving, we get: $\\begin{cases} k=-2 \\\\ b=6 \\end{cases}$, therefore, the equation of $CF$ is: $y=-2x+6$,\\\\",
    "image_path": "dataset/mm_math/images/MM_Math/51284444_184.png"
  },
  {
    "solution": "\\boxed{\\sqrt{6} - \\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that there is an equilateral triangle $\\triangle ABC$, point $D$ is on the outside of $AC$, $BD$ is rotated clockwise by $60^\\circ$ around point $B$ to $BF$, point $F$ corresponds to point $D$, connect $AF$, $AD$, with $AD=2$, $\\angle CBD=15^\\circ$, and $\\angle AFB=30^\\circ$, then what is the length of $AF$?\n",
    "completion": "\\[ \\textbf{Solution:} \\quad \\text{As shown in the figure, connect } DF, \\text{ extend } FA \\text{ to intersect } BD \\text{ at } E, \\]\n\\[ \\because BD \\text{ rotates clockwise around point } B \\text{ by } 60^\\circ \\text{ to } BF, \\]\n\\[ \\therefore BD = BF, \\angle DBF = 60^\\circ, \\]\n\\[ \\therefore \\triangle BDF \\text{ is an equilateral triangle, } \\angle BFD = 60^\\circ. \\]\n\\[ \\because \\angle BFA = 30^\\circ, \\]\n\\[ \\therefore \\angle DFA = 30^\\circ, \\]\n\\[ \\therefore FE \\text{ perpendicularly bisects } BD, \\]\n\\[ \\therefore AD = AB, \\]\n\\[ \\because \\angle ABC = 60^\\circ, \\angle CBD = 15^\\circ, \\]\n\\[ \\therefore \\angle ABD = 45^\\circ \\]\n\\[ \\therefore \\triangle DAB \\text{ is an isosceles right triangle, } \\]\n\\[ \\therefore DE = AE = \\frac{\\sqrt{2}}{2}AD = \\sqrt{2}, \\]\n\\[ \\therefore BD = DF = 2\\sqrt{2} \\]\n\\[ \\therefore EF = \\sqrt{DF^{2} - DE^{2}} = \\sqrt{6} \\]\n\\[ AF = EF - AE = \\boxed{\\sqrt{6} - \\sqrt{2}}. \\]",
    "image_path": "dataset/mm_math/images/MM_Math/52867037.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the radius of the semicircle is 5. The vertex of the $30^\\circ$ angle of the triangle plate is placed on the semicircle, and the two sides of this angle intersect with the semicircle at points A and B, respectively. What is the length of $AB$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure,\\\\\nlet point O be the center of the circle, connect $AO$, $BO$,\\\\\n$\\therefore$ $AO=BO=5$,\\\\\n$\\because$ $\\angle ACB=30^\\circ$,\\\\\n$\\therefore$ $\\angle AOB=60^\\circ$,\\\\\n$\\therefore$ $\\triangle AOB$ is an equilateral triangle,\\\\\n$\\therefore$ $AB=AO=BO=\\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53042887.png"
  },
  {
    "solution": "\\boxed{(-1,4)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: The partial graph of a quadratic function is shown in the figure, with its axis of symmetry being the line $x = -1$. What are the coordinates of the vertex of this graph?\n",
    "completion": "\\textbf{Solution:} Since $y=-\\left(x+1\\right)^{2}+4$,\\\\\nthe vertex of the parabola is at $\\boxed{(-1,4)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52605434_113.png"
  },
  {
    "solution": "\\boxed{\\frac{2\\sqrt{13}}{13}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=x+b$ intersects the $x$ axis at point $C(4,0)$ and the $y$ axis at point $B$, and it intersects the hyperbola $y=\\frac{m}{x}$ (where $x<0$) at point $A(-1,n)$. Connect $OA$. What is the sine value of $\\angle OAB$?\n",
    "completion": "\\textbf{Solution:} Construct $OM \\perp AC$ at point $M$ through point $O$,\\\\\nwhen $x=0$, $y=-4$, that is, $B(0, -4)$.\\\\\nSince $OC=OB=4$,\\\\\nTherefore, $\\triangle OCB$ is an isosceles right triangle,\\\\\nThus, $\\angle OBC=\\angle OCB=45^\\circ$\\\\\nTherefore, in $\\triangle OMB$ $\\sin 45^\\circ=\\frac{OM}{OB}$,\\\\\nThus, $OM=4\\times \\frac{\\sqrt{2}}{2}=2\\sqrt{2}$.\\\\\nTherefore, in the right triangle $AOM$,\\\\\n$AO=\\sqrt{(-1-0)^{2}+(-5-0)^{2}}=\\sqrt{26}$,\\\\\n$\\sin\\angle OAB=\\frac{OM}{OA}=\\boxed{\\frac{2\\sqrt{2}}{\\sqrt{26}}}=\\boxed{\\frac{2\\sqrt{13}}{13}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51379219_205.png"
  },
  {
    "solution": "\\boxed{\\frac{8\\pi }{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, inside the circle $\\odot O$, $AB$ is a diameter, point $D$ is a point on $\\odot O$, point $C$ is the midpoint of the arc $\\overset{\\frown}{AD}$, the chord $CM\\perp AB$ at point $F$, connect $AD$, intersecting $CF$ at point $P$, and connect $BC$, with $\\angle DAB=30^\\circ$. If $CM=8\\sqrt{3}$, find the length of the arc $\\overset{\\frown}{AC}$ (the result should retain $\\pi$).\n",
    "completion": "\\textbf{Solution:} Connect OC, then $\\angle AOC=2\\angle ABC=60^\\circ$,\\\\\n$\\because$ CM $\\perp$ diameter AB at point F,\\\\\n$\\therefore$ CF=$\\frac{1}{2}$CM=$4\\sqrt{3}$, $\\angle CFO=90^\\circ$,\\\\\n$\\therefore$ in $\\triangle COF$, $\\angle OCF=30^\\circ$,\\\\\n$\\therefore$ OC=2OF, $OF^2+CF^2=OC^2$, that is $OF^2+(4\\sqrt{3})^2=(2OF)^2$,\\\\\nSolving gives: OF=4, $\\therefore$ OC=8,\\\\\n$\\therefore$ the length of $\\overset{\\frown}{AC}$ is $\\frac{60^\\circ\\pi \\times 8}{180}=\\boxed{\\frac{8\\pi }{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53102087_124.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, there is a slope behind a teaching building in the mountainous area, where the slope surface BC is parallel to the ground AD. The slope ratio of the slope AB is $i=1\\colon \\frac{5}{12}$, and AB=26 meters. To prevent landslides and ensure safety, the school decides to modify this slope. According to geological surveys, when the slope angle does not exceed $53^\\circ$, the stability of the slope against landslides can be ensured. To eliminate safety hazards, the school plans to modify the slope AB to AF (as shown in the diagram). What is the minimum length of BF? (The result should be accurate to 1 meter, reference data: $\\sin{53^\\circ} \\approx 0.8$, $\\cos{53^\\circ} \\approx 0.6$, $\\tan{53^\\circ} \\approx 1.33$)\n",
    "completion": "\\textbf{Solution:} Construct FH $\\perp$ AD at H,\\\\\nthen quadrilateral FHEB is a rectangle,\\\\\n$\\therefore$ FH = BE, BF = HE,\\\\\n$\\because$ the slope ratio of slope AB is $i=1\\colon \\frac{5}{12}$,\\\\\n$\\therefore$ BE : AE = 12 : 5,\\\\\nLet BE = 12x meters, then AE = 5x meters,\\\\\nIn $\\triangle ABE$, $AB^{2}$ = $AE^{2}$ + $BE^{2}$, that is $26^{2}$ = $(12x)^{2}$ + $(5x)^{2}$,\\\\\nSolving gives: $x_{1}$ = 2, $x_{2}$ = -2 (discard),\\\\\nThen AE = 10 meters, BE = FH = 24 meters,\\\\\nIn $\\triangle FAH$, $\\tan\\angle FAH$ = $\\frac{FH}{AH}$,\\\\\n$\\therefore$ AH = $\\frac{FH}{\\tan\\angle FAH}$ $\\approx$ $\\frac{24}{1.33} \\approx 18$ (meters),\\\\\n$\\therefore$ BF = HE = AH - AE = 18 - 10 = \\boxed{8} (meters),\\\\\nAnswer: BF is at least 8 meters.",
    "image_path": "dataset/mm_math/images/MM_Math/50934100.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in rectangle $ABCD$, $AD=4$, $AB=8$, diagonals $AC$ and $BD$ intersect at point $O$. Points $E$ and $F$ are located on the extensions of $CD$ and $DA$ respectively, and $DE=4$, $AF=2$. The segments $EF$ and $OE$ are drawn, with $G$ being the midpoint of $EF$. The segment $OE$ intersects $AD$ at point $H$, and segment $GH$ is drawn. Given that $H$ is the midpoint of $OE$, find the length of $GH$.\n",
    "completion": "\\textbf{Solution:} Connect OF. \\\\\nSince quadrilateral ABCD is a rectangle, \\\\\nit follows that $ \\angle ADC=90^\\circ$. \\\\\nSince $OM\\parallel DC$, \\\\\nit follows that $ \\angle FMO=\\angle ADC$, \\\\\ntherefore $ \\angle FMO=90^\\circ$. \\\\\nSince $AD=4$ and M is the midpoint of AD, \\\\\nit follows that $AM=\\frac{1}{2}AD=2$. \\\\\nSince $AF=2$, \\\\\nit follows that $FM=4$. \\\\\nTherefore, in $ \\triangle FOM$, $ \\angle FMO=90^\\circ$, $ OM=4$, and $ FM=4$. \\\\\nBy the Pythagorean theorem, we get $ OF=\\sqrt{OM^{2}+FM^{2}}=4\\sqrt{2}$. \\\\\nSince G is the midpoint of EF and H is the midpoint of OE, \\\\\nit follows that GH is the midline of $ \\triangle FEO$, \\\\\ntherefore $GH=\\frac{1}{2}OF=2\\sqrt{2}$. \\\\\nWhat is the length of $GH$? $GH=\\boxed{2\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52211035_141.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, rectangle ABCD is folded along the straight line DE, with vertex A landing on side BC at point F. Given that $BE=3$ and $CD=8$, what is the length of BF?\n",
    "completion": "\\textbf{Solution:} From the property of the fold, we know that $AE=EF$, $AE=AB-BE=8-3=5$, \\\\\nin $\\triangle BEF$, $EF=AE=5$, $BE=3$, \\\\\nby the Pythagorean theorem, we have $BF=\\sqrt{EF^2-BE^2}=\\sqrt{5^2-3^2}=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52386538.png"
  },
  {
    "solution": "\\boxed{40^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, lines AB, CD, and EF intersect at point O. If $\\angle COF = 120^\\circ$ and $\\angle AOD = 100^\\circ$, what is the measure of $\\angle AOF$ in degrees?\n",
    "completion": "\\textbf{Solution:} Since $\\angle COF=120^\\circ$,\\\\\nit follows that $\\angle COE=180^\\circ-120^\\circ=60^\\circ$,\\\\\nthus $\\angle DOF=\\angle COE=60^\\circ$,\\\\\nand since $\\angle AOD=100^\\circ$,\\\\\nit follows that $\\angle AOF=\\angle AOD-\\angle DOF=100^\\circ-60^\\circ=\\boxed{40^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55191221.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, $BC=3\\sqrt{2}$, and $AC=4\\sqrt{2}$. The bisectors of $\\angle CAB$ and $\\angle CBA$ intersect at point $P$. Points $D$ and $E$ are located on sides $AC$ and $BC$, respectively (neither coincides with point $C$), and it is given that $\\angle DPE=45^\\circ$. What is the distance from point $P$ to side $AB$? What is the perimeter of $\\triangle CDE$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $PG \\perp AB$ at $G$, $PH \\perp AC$ at $H$, and $PM \\perp BC$ at $M$. Take a point $M'$ on $AH$ such that $M'H=EM$, and connect $PM'$,\\\\\n$\\because$ the angle bisectors of $\\angle CAB$ and $\\angle CBA$ intersect at point $P$,\\\\\n$\\therefore$ $PG=PM=PH$,\\\\\n$\\because$ $\\angle PMC=\\angle C=\\angle PHC=90^\\circ$,\\\\\n$\\therefore$ quadrilateral $PMCH$ is a square,\\\\\n$\\therefore$ $CM=CH=PH=PM$, $\\angle MPH=90^\\circ$,\\\\\n$\\because$ $\\angle DPE=45^\\circ$,\\\\\n$\\therefore$ $\\angle MPE+\\angle DPH=45^\\circ$,\\\\\nIn $\\triangle PME$ and $\\triangle PHM'$,\\\\\n$\\begin{cases} PM=PH \\\\ \\angle PME=\\angle PHM'=90^\\circ \\\\ ME=HM' \\end{cases}$,\\\\\n$\\therefore$ $\\triangle PME \\cong \\triangle PHM'$ (SAS),\\\\\n$\\therefore$ $PE=PM'$, $\\angle MPE=\\angle HPM'$,\\\\\n$\\therefore$ $\\angle HPM'+\\angle DPH=\\angle DPH+\\angle EPM=45^\\circ=\\angle DPE$,\\\\\nIn $\\triangle DPM'$ and $\\triangle DPE$,\\\\\n$\\begin{cases} PD=PD \\\\ \\angle DPM'=\\angle DPE \\\\ PE=PM' \\end{cases}$,\\\\\n$\\therefore$ $\\triangle DPM' \\cong \\triangle DPE$ (SAS),\\\\\n$\\therefore$ $DE=DM'=DH+HM'=DH+ME$,\\\\\n$\\therefore$ the perimeter of $\\triangle CDE=CD+CE+DE=CD+CE+DH+EM=CH+CM$,\\\\\n$\\because$ $\\angle PGB=\\angle PMB=90^\\circ$, $\\angle PBG=\\angle PBM$,\\\\\n$\\therefore$ $\\angle BPG=\\angle BPM$,\\\\\n$\\therefore$ $BG=BM$,\\\\\nSimilarly, we get: $AG=AH$,\\\\\nIn $\\triangle ACB$, $\\angle C=90^\\circ$, $AC=4\\sqrt{2}$, $BC=3\\sqrt{2}$,\\\\\nBy Pythagorean theorem: $AB=\\sqrt{(4\\sqrt{2})^2+(3\\sqrt{2})^2}=5\\sqrt{2}$,\\\\\n$\\therefore$ $AG+BG=AH+BM=5\\sqrt{2}$,\\\\\n$\\because$ $AC+BC=3\\sqrt{2}+4\\sqrt{2}=7\\sqrt{2}$,\\\\\n$\\therefore$ $CH+CM=AC+BC−AH−BM=7\\sqrt{2}−5\\sqrt{2}=2\\sqrt{2}$,\\\\\n$\\therefore$ the distance from point $P$ to side $AB$ is $PG=\\boxed{\\sqrt{2}}$, the perimeter of $\\triangle CDE$ is $\\boxed{2\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52019253.png"
  },
  {
    "solution": "\\boxed{(0,-4)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system, the graph of the linear function $y=kx+b$ passes through the point $A(-2,6)$ and intersects the $x$-axis at point $B$, and intersects the graph of the direct proportion function $y=3x$ at point $C$, with the $x$-coordinate of point $C$ being $1$. If point $D$ is on the $y$-axis, and the area of $\\triangle BCD$ is twice the area of $\\triangle BOC$, find the coordinates of point $D$.\n",
    "completion": "\\textbf{Solution:} From (1), we know the linear function $y=kx+b$ is $y=-x+4$. When $y=0$, we have $0=-x+4$, solving this gives $x=4$, $\\therefore B(4, 0)$, $\\therefore S_{\\triangle BCO}=\\frac{1}{2}\\times 4\\times 3=6$, $\\because S_{\\triangle BCD}=2S_{\\triangle BOC}$, $\\therefore S_{\\triangle BCD}=12$. Assume the coordinates of point D are (0, m), when point D is on the positive y-axis,\\\\\n$S_{\\triangle DBC}=S_{\\triangle DOB}-S_{\\triangle DCO}-S_{\\triangle BOC}=\\frac{1}{2}\\times 4m-\\frac{1}{2}\\times 1\\times m-6=\\frac{3}{2}m-6$, $\\because S_{\\triangle BCD}=12$,\\\\\n$\\therefore \\frac{3}{2}m-6=6\\times 2$, $\\therefore m=12$, $\\therefore$ the coordinates of point D are $\\boxed{(0,12)}$; when point D is on the negative y-axis,\\\\\n$S_{\\triangle DBC}=S_{\\triangle DOB}-S_{\\triangle DCO}+S_{\\triangle BOC}=\\frac{1}{2}\\times 4\\times (-m)+6-\\frac{1}{2}\\times 1\\times (-m)=-\\frac{3}{2}m+6$, $\\because S_{\\triangle BCD}=12$,\\\\\n$\\therefore -\\frac{3}{2}m+6=6\\times 2$, $\\therefore m=-4$, $\\therefore$ the coordinates of point D are $\\boxed{(0,-4)}$,\\\\\nIn conclusion, the coordinates of point D are $\\boxed{(0,12)}$ or $\\boxed{(0,-4)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52381468_176.png"
  },
  {
    "solution": "\\boxed{11}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $D$ is a point inside $\\triangle ABC$ with $BD \\perp CD$, $AD=6$, $BD=4$, $CD=3$, and $E$, $F$, $G$, $H$ are the midpoints of $AB$, $AC$, $CD$, and $BD$, respectively. What is the perimeter of the quadrilateral $EFGH$?\n",
    "completion": "\\textbf{Solution:} Given $BD\\perp DC$, with $BD=4$ and $CD=3$, by the Pythagorean theorem, we have $BC=\\sqrt{BD^2+CD^2}=5$.\\\\\nSince $E$, $F$, $G$, $H$ are the midpoints of $AB$, $AC$, $CD$, and $BD$, respectively,\\\\\nit follows that $HG=\\frac{1}{2}BC=EF$ and $EH=FG=\\frac{1}{2}AD$.\\\\\nGiven $AD=6$,\\\\\ntherefore $EF=HG=2.5$ and $EH=GF=3$,\\\\\nhence the perimeter of quadrilateral $EFGH$ is $EF+FG+HG+EH=2\\times(2.5+3)=\\boxed{11}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53232634.png"
  },
  {
    "solution": "\\boxed{2.5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Sorry, it seems like you didn't provide any specific text to translate. Could you please share the text you'd like to be translated?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral $ABCD$ is a rectangle,\\\\\nit follows that $\\angle BAD=90^\\circ$, $OD= \\frac{1}{2}BD$, and $AD=BC=8$,\\\\\nthus $BD=\\sqrt{AB^{2}+AD^{2}}=\\sqrt{6^{2}+8^{2}}=10\\text{cm}$,\\\\\nwhich means $OD=5\\text{cm}$,\\\\\nsince points $E$ and $F$ are the midpoints of $AO$ and $AD$ respectively,\\\\\nit follows that $EF$ is the midline of $\\triangle AOD$,\\\\\ntherefore $EF= \\frac{1}{2}OD=\\boxed{2.5}\\text{cm}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52354879.png"
  },
  {
    "solution": "\\boxed{36}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the quadrilateral $ABCD$ is given with $AB=3$, $BC=4$, $CD=12$, $AD=13$, and $\\angle B=90^\\circ$. Find: The area of the quadrilateral $ABCD$.\n",
    "completion": "\\textbf{Solution:} For $\\triangle ABC$, we have $S_{\\triangle ABC}=\\frac{1}{2}AB\\cdot BC=\\frac{1}{2}\\times 3\\times 4=6$, \\\\\nSince, in $\\triangle ACD$, $CD=12$, $AD=13$, and $AC=5$, \\\\\nThus, $CD^2+AC^2=12^2+5^2=13^2=AD^2$, \\\\\nTherefore, $\\triangle ACD$ is a right-angled triangle, with $\\angle ACD=90^\\circ$, \\\\\nHence, $S_{\\triangle ACD}=\\frac{1}{2}AC\\cdot CD=\\frac{1}{2}\\times 5\\times 12=30$. \\\\\nTherefore, $S_{\\text{Quadrilateral} ABCD}=S_{\\triangle ABC}+S_{\\triangle ACD}=6+30=\\boxed{36}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53033473_99.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that point $P$ is a point on diagonal $BD$ of square $ABCD$, and $AP=3$, $PF \\perp CD$ at point $F$, $PE \\perp BC$ at point $E$. When connecting $EF$, what is the length of $EF$?\n",
    "completion": "\\textbf{Solution}: Let's connect CP.\\\\\nSince square ABCD,\\\\\nthus $\\angle C=90^\\circ$, AB=BC, and $\\angle ABP=\\angle CBP=45^\\circ$. Let's connect CP again,\\\\\nIn $\\triangle ABP$ and $\\triangle CBP$\\\\\n\\[\\left\\{\\begin{array}{l}\nAB=BC\\\\\n\\angle ABP=\\angle CBP\\\\\nBP=BP\n\\end{array}\\right.\\]\\\\\nthus $\\triangle ABP\\cong \\triangle CBP$ (SAS)\\\\\nthus AP=CP=3\\\\\nSince PF$\\perp$CD, and PE$\\perp$BC,\\\\\nthus $\\angle PFC=\\angle PEC=90^\\circ$,\\\\\nthus quadrilateral CEPF is a rectangle,\\\\\nthus EF=PC=\\boxed{3}.",
    "image_path": "dataset/mm_math/images/MM_Math/50475019.png"
  },
  {
    "solution": "\\boxed{x\\le -1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of a linear function intersects the y-axis at the point $\\left(0, 2\\right)$ and the x-axis at the point $\\left(-1, 0\\right)$. What is the range of the independent variable $x$ when $y \\le 0$?\n",
    "completion": "\\textbf{Solution:} From the graph, it is observed that when $y\\le 0$, the graph of the linear function that meets the condition is below the x-axis,\\\\\nas shown by the bold part in the figure,\\\\\n$\\because$ this part of the graph is to the left of the line $x=-1$,\\\\\n$\\therefore$ when $y\\le 0$, the range of the independent variable x is $x\\le -1$.\\\\\nHence, the answer is: \\boxed{x\\le -1}.",
    "image_path": "dataset/mm_math/images/MM_Math/52690851.png"
  },
  {
    "solution": "\\boxed{50^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ABC=80^\\circ$, $\\angle ACB=50^\\circ$. What is the measure of $\\angle A$ in degrees?\n",
    "completion": "\\textbf{Solution}: Since $\\angle ABC=80^\\circ$ and $\\angle ACB=50^\\circ$, \\\\\nit follows that $\\angle A=180^\\circ - \\angle ABC - \\angle ACB=\\boxed{50^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/54368654_148.png"
  },
  {
    "solution": "\\boxed{(3,0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the Cartesian coordinate system $xOy$, A and B are two points on the x-axis, while C and D are two points on the y-axis. The parabolic segment $C_1$ passing through points A, C, B and the parabolic segment $C_2$ passing through points A, D, B combine to form a closed curve, which we call the “egg curve”. It is known that the coordinates of point C are $(0,)$, and point M is the vertex of the parabola $C_2\\colon y=mx^2-2mx-3m$ ($m<0$). What are the coordinates of points A and B?\n",
    "completion": "\\textbf{Solution:} Let $y=0$, then $mx^{2}-2mx-3m=0$,\\\\\nsince $m<0$, it follows that $x^{2}-2x-3=0$, solving this we get: $x_{1}=-1$, $x_{2}=3$.\\\\\nTherefore, the coordinates of A are $\\boxed{(-1,0)}$, and the coordinates of B are $\\boxed{(3,0)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52977501_201.png"
  },
  {
    "solution": "\\boxed{41^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, DE is the median line of $\\triangle ABC$, the bisector of $\\angle ABC$ intersects DE at point F. If $\\angle DFB=32^\\circ$ and $\\angle A=75^\\circ$, then what is the measure of $\\angle AED$ in degrees?\n",
    "completion": "\\textbf{Solution}: Since $DE$ is the median line of $\\triangle ABC$,\\\\\nthus $DE\\parallel BC$,\\\\\ntherefore $\\angle ADE=\\angle ABC$, $\\angle DFB=\\angle FBC$.\\\\\nSince $\\angle DFB=32^\\circ$,\\\\\nthus $\\angle DFB=\\angle FBC=32^\\circ$.\\\\\nSince the bisector of $\\angle ABC$ intersects $DE$ at point F,\\\\\nthus $\\angle DBF=\\angle FBC=\\frac{1}{2}\\angle ABC=32^\\circ$,\\\\\ntherefore $\\angle ADE=\\angle ABC=2\\times 32^\\circ=64^\\circ$.\\\\\nSince $\\angle A=75^\\circ$,\\\\\nthus $\\angle AED=180^\\circ−\\angle A−\\angle ADE=180^\\circ−75^\\circ−64^\\circ=\\boxed{41^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52251027.png"
  },
  {
    "solution": "\\boxed{\\sqrt{3}-1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, $\\angle A=30^\\circ$, and $BC=2$. Let $D$ be any point on $AC$, and let $F$ be the midpoint of $AB$. Draw line $BD$. Let $E$ be a point on $BD$ such that $CD^{2}=DE\\cdot BD$. Draw line $EF$. What is the minimum value of $EF$?\n",
    "completion": "\\textbf{Solution:} In $\\triangle CED$ and $\\triangle BDC$,\\\\\n$\\because CD^2=DE\\cdot BD$,\\\\\n$\\therefore \\frac{CD}{DB}=\\frac{DE}{DC}$,\\\\\n$\\because \\angle EDC=\\angle CDB$,\\\\\n$\\therefore \\triangle CDE\\sim \\triangle BDC$,\\\\\n$\\therefore \\angle DEC=\\angle DCB=90^\\circ$,\\\\\n$\\therefore \\angle BEC=180^\\circ-\\angle DEC=90^\\circ$,\\\\\nAs shown in the diagram, let Q be the midpoint of BC, then EQ= $\\frac{1}{2}$ BC=1,\\\\\n$\\because \\angle ACB=90^\\circ$ and $\\angle A=30^\\circ$, BC=2,\\\\\n$\\therefore AB=2BC=4$,\\\\\n$\\therefore AC= \\sqrt{AB^2-BC^2}=\\sqrt{4^2-2^2}=2\\sqrt{3}$,\\\\\n$\\because F$ is the midpoint of AB, and Q is the midpoint of BC,\\\\\n$\\therefore FQ= \\frac{1}{2} AC= \\sqrt{3}$,\\\\\n$\\because EF\\geq FQ-EQ$, when and only when points E, F, and Q are collinear, EF can reach $\\sqrt{3}-1$,\\\\\n$\\therefore The minimum value of EF is \\boxed{\\sqrt{3}-1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51155310.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ABC = 90^\\circ$, $BC = 3$. D is a point on the extension line of $AC$ with $AC = 3CD$. A line through point D is drawn, $DH \\parallel AB$, intersecting the extension line of $BC$ at point H. If $\\angle CBD = \\angle A$, what is the length of $AB$?\n",
    "completion": "\\textbf{Solution:} From method one, we have\\\\\n$\\because \\angle A=\\angle CBD, \\angle ABC=\\angle BHD$,\\\\\n$\\therefore \\triangle ABC\\sim \\triangle BHD$,\\\\\n$\\therefore \\frac{BC}{HD}=\\frac{AB}{BH}$,\\\\\n$\\because \\triangle ABC\\sim \\triangle DHC$,\\\\\n$\\therefore \\frac{DH}{AB}=\\frac{DC}{AC}=\\frac{1}{3}$,\\\\\n$\\therefore AB=3DH$,\\\\\n$\\therefore \\frac{3}{DH}=\\frac{3DH}{4}$,\\\\\n$\\therefore DH=2$,\\\\\n$\\therefore AB=\\boxed{6}$.\\\\\nFrom method two, we have,\\\\\n$\\because \\angle CBD =\\angle A, \\angle ADB=\\angle ADB$,\\\\\n$\\therefore \\triangle CDB\\sim \\triangle BDA$,\\\\\n$\\therefore \\frac{CD}{BD}=\\frac{BD}{AD}$, hence $BD^2=CD\\cdot AD$,\\\\\n$\\therefore BD^2=CD\\cdot 4CD=4CD^2$,\\\\\n$\\therefore BD=2CD$,\\\\\n$\\because \\triangle CDB\\sim \\triangle BDA$,\\\\\n$\\therefore \\frac{CD}{BD}=\\frac{BC}{AB}$,\\\\\n$\\therefore \\frac{CD}{2CD}=\\frac{3}{AB}$,\\\\\n$\\therefore AB=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52623501_108.png"
  },
  {
    "solution": "\\boxed{\\frac{15}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the line $AB$ intersects the parabola $C\\colon y=ax^2+2x+c$ at points $A(-1, 0)$ and $B(2, 3)$. Is there a fixed point $F$ on the axis of symmetry of the parabola $C$ such that the distance from any point $P$ on the parabola $C$ to point $F$ is equal to the distance to the line $y=\\frac{17}{4}$? If such a point exists, determine the coordinates of the fixed point $F$; if not, please explain why.\n",
    "completion": "\\textbf{Solution:}\n\nThere exists a point $F$, \\\\\n$\\because y = -x^2 + 2x + 3 = - (x - 1)^2 + 4$, \\\\\n$\\therefore$ the axis of symmetry is the line $x = 1$, \\\\\nwhen $y = 0$, $x_1 = -1$, $x_2 = 3$, \\\\\n$\\therefore$ the parabola intersects the positive $x$-axis at point $C(3, 0)$, \\\\\nPerpendiculars to the line $y = \\frac{17}{4}$ through points $B$, $C$ are drawn, with foot of perpendiculars being $N$, $H$, respectively, \\\\\nThere exists a point $F$ on the axis of symmetry of the parabola such that for any point $P$ on the parabola $C$, the distance from $P$ to point $F$ is equal to its distance to the line $y = \\frac{17}{4}$. Let $F(1,t)$, connect $BF$, $CF$, \\\\\nthen $BF = BN = \\frac{17}{4} - 3 = \\frac{5}{4}$, $CF = CH = \\frac{17}{4}$, \\\\\nBased on the problem statement, we can form the equations: $\\begin{cases}\n(2 - 1)^2 + (t - 3)^2 = \\left(\\frac{5}{4}\\right)^2 \\\\\n(3 - 1)^2 + t^2 = \\left(\\frac{17}{4}\\right)^2\n\\end{cases}$, \\\\\nSolving, we get $t = \\frac{15}{4}$, \\\\\n$\\therefore F\\left(1, \\boxed{\\frac{15}{4}}\\right)$.",
    "image_path": "dataset/mm_math/images/MM_Math/53101984_150.png"
  },
  {
    "solution": "\\boxed{\\left(0, \\frac{12}{7}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $l\\colon y=\\frac{4}{3}x+b$ passes through point $A(-3,0)$ and intersects the $y$-axis at point $B$. The bisector of $\\angle OAB$ intersects the $y$-axis at point $C$. A perpendicular line to line $AB$ passing through point $C$ intersects the $x$-axis at point $E$, and the foot of the perpendicular is point $D$. Let point $P$ be a moving point on the $y$-axis. When the value of $PA+PD$ is minimized, please write down the coordinates of point $P$ directly.\n",
    "completion": "\\textbf{Solution:} The coordinate of point $P$ is $\\boxed{\\left(0, \\frac{12}{7}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52189067_135.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a right-angled triangle paper $ABO$ is placed in the Cartesian coordinate system. It is known that point $A\\left(\\sqrt{3}, 0\\right)$, point $B\\left(0, 1\\right)$, and point $O\\left(0, 0\\right)$. $P$ is a moving point on side $AB$ (point $P$ does not coincide with points $A$ or $B$). By folding the paper along $OP$, we obtain the corresponding point $A'$ of point $A$. As shown in Figure 2, when $P$ is the midpoint of $AB$, what is the length of $A'B$?\n",
    "completion": "\\textbf{Solution:} In right triangle $Rt\\triangle AOB$, we have $OA=\\sqrt{3}$ and $OB=1$, therefore $AB=\\sqrt{OA^{2}+OB^{2}}=2$. Since $P$ is the midpoint of $AB$, we have $AP=BP=1$ and $OP=\\frac{1}{2}AB=1$. Thus, $OP=OB=BP$, which means $\\triangle BOP$ is an equilateral triangle, therefore $\\angle OBP=\\angle BOP=\\angle BPO=60^\\circ$. Consequently, $\\angle OPA=180^\\circ-\\angle BPO=120^\\circ$. According to the property of reflection, $\\angle OPA'=\\angle OPA=120^\\circ$ and $PA'=PA=1$, which gives $\\angle BPA'=\\angle OBP=60^\\circ$. Therefore, $OB\\parallel PA'$. Moreover, since $OB=PA'=1$, the quadrilateral $OPA'B$ is a parallelogram,\\\\\nhence, $A'B=OP=\\boxed{1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53572668_145.png"
  },
  {
    "solution": "\\boxed{y = 2x + 2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the graph of the inverse proportion function $y_{1}=\\frac{k}{x}$ intersects with the graph of the linear function $y_{2}=ax+b$ at points $A(1, 4)$ and $B(m, -2)$. Find the value of $m$ and the equation of the linear function.\n",
    "completion": "\\textbf{Solution:} Since the graph of the inverse proportion function $y_1 = \\frac{k}{x}$ intersects with the graph of the linear function $y_2 = ax + b$ at points $A(1, 4)$ and $B(m, -2)$,\\\\\nwe have $k = 1 \\times 4 = 4$,\\\\\nthus, the equation of the inverse proportion function is: $y = \\frac{4}{x}$,\\\\\nhence, $-2 = \\frac{4}{m}$, solving this gives $m = \\boxed{-2}$, that is point $B(-2, -2)$,\\\\\ntherefore $\\left\\{\\begin{array}{l} 4 = a + b \\\\ -2 = -2a + b \\end{array}\\right.$, solving this set gives $\\left\\{\\begin{array}{l} a = 2 \\\\ b = 2 \\end{array}\\right.$,\\\\\nthus, the equation of the linear function is: $\\boxed{y = 2x + 2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51445145_139.png"
  },
  {
    "solution": "\\boxed{23^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given $\\angle AOB$ and $\\angle COD$, the ray $OE$ bisects $\\angle AOD$; as shown in Figure 1, given $\\angle AOB = 180^\\circ$ and $\\angle COD = 90^\\circ$, if $\\angle DOB = 46^\\circ$, what is the measure of $\\angle COE$ in degrees?\n",
    "completion": "\\textbf{Solution:} Given $\\angle AOB=180^\\circ$ and $\\angle DOB=46^\\circ$, \\\\\nit follows that $\\angle AOD=134^\\circ$. \\\\\nSince $OE$ bisects $\\angle AOD$, \\\\\nwe have $\\angle EOD=\\frac{1}{2}\\angle AOD=\\frac{1}{2}\\times 134^\\circ=67^\\circ$. \\\\\nGiven $\\angle COD=90^\\circ$, \\\\\nit follows that $\\angle COE=\\angle COD-\\angle EOD=90^\\circ-67^\\circ=\\boxed{23^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52916518_81.png"
  },
  {
    "solution": "\\boxed{24}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB \\perp BC$ at point $B$, $AB \\perp AD$ at point $A$, and point $E$ is the midpoint of $CD$. Given that $BC=5$, $AD=12$, and $BE=12.5$, what is the length of $AB$?\n",
    "completion": "\\textbf{Solution:} Extend BE to meet AD at point G, as shown in the diagram: \\\\\n$\\because$ AB$\\perp$BC at point B, AB$\\perp$AD at point A,\\\\\n$\\therefore$ BC$\\parallel$AD, $\\angle$A=90$^\\circ$,\\\\\n$\\therefore$ $\\angle$C=$\\angle$D, $\\angle$CBE=$\\angle$DGE\\\\\n$\\because$ point E is the midpoint of CD,\\\\\n$\\therefore$ CE=DE,\\\\\nIn $\\triangle$BCE and $\\triangle$GDE,\\\\\n$ \\left\\{\\begin{array}{l}\\begin{array}{c}\\angle C=\\angle D\\\\ \\angle CBE=\\angle DGE\\\\ CE=DE\\end{array}\\\\ \\end{array}\\right.$\\\\\n$\\therefore$ $\\triangle$BCD$\\cong$ $\\triangle$GDE (AAS),\\\\\n$\\therefore$ BE=GE=12.5, BC=DG=5,\\\\\n$\\therefore$ BG=BE+GE=25,\\\\\n$\\because$ AD=12,\\\\\n$\\therefore$ AG=AD−DG=7,\\\\\n$\\therefore$ $AB=\\sqrt{BG^{2}−AG^{2}}=\\sqrt{25^{2}−7^{2}}=\\boxed{24}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52814874.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, CE is the angle bisector of $\\triangle ABC$, and $\\angle AEC=105^\\circ$. What is the measure of $\\angle B$ in degrees?\n",
    "completion": "\\textbf{Solution:} Since CE bisects $\\angle ACB$ and $\\angle ACB=90^\\circ$, \\\\\nit follows that $\\angle BCE=\\frac{1}{2}\\angle ACB=45^\\circ$, \\\\\nGiven that $\\angle AEC=105^\\circ$, \\\\\nit implies that $\\angle B=\\angle AEC-\\angle BCE=\\boxed{60^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/52837520.png"
  },
  {
    "solution": "\\boxed{\\frac{8}{3}\\pi}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of $\\odot O$, the chord $CD\\perp AB$ at point $E$, and chord $AF$ intersects chord $CD$ at point $G$, with $AG=CG$. A perpendicular line to $BF$ through point $C$ intersects the extension of $BF$ at point $H$. $CH$ is the tangent of $\\odot O$; if $FH=2$ and $BF=4$, what is the length of arc $CD$?\n",
    "completion": "\\textbf{Solution}: From (1) we have: $BH \\perp CH$, $OC \\perp CH$,\\\\\n$\\therefore OC \\parallel BH$,\\\\\n$\\because CH \\parallel AF$,\\\\\n$\\therefore$ Quadrilateral $CMFH$ is a parallelogram,\\\\\n$\\because OC \\perp CH$,\\\\\n$\\therefore \\angle OCH = 90^\\circ$,\\\\\n$\\therefore$ Quadrilateral $CMFH$ is a rectangle,\\\\\n$\\therefore OC \\perp AF$, $CM = HF = 2$,\\\\\n$\\therefore AM = FM$,\\\\\n$\\because$ Point $O$ is the midpoint of $AB$,\\\\\n$\\therefore OM = \\frac{1}{2} BF = 2$,\\\\\n$\\therefore CM = OM$,\\\\\n$\\therefore OC = 4$, $AM$ perpendicularly bisects $OC$,\\\\\n$\\therefore AC = AO$,\\\\\nand $AO = OC$,\\\\\n$\\therefore AC = OC = OA$,\\\\\n$\\therefore \\triangle AOC$ is an equilateral triangle,\\\\\n$\\therefore \\angle AOC = 60^\\circ$,\\\\\n$\\because \\overset{\\frown}{AC} = \\overset{\\frown}{AD}$,\\\\\n$\\therefore \\angle AOD = \\angle AOC = 60^\\circ$,\\\\\n$\\therefore \\angle COD = 120^\\circ$,\\\\\n$\\therefore$ The length of arc $CD$ is $\\frac{120 \\times \\pi \\times 4}{180} = \\boxed{\\frac{8}{3}\\pi}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51403003_142.png"
  },
  {
    "solution": "\\boxed{\\frac{10}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, vertices $A$ and $C$ of rectangle $OABC$ lie on the positive semiaxis of the $x$-axis and $y$-axis, respectively, with point $B(4,3)$. The graph of the inverse proportion function $y=\\frac{k}{x}$ ($x>0$) intersects lines $AB$ and $BC$ at points $D$ and $E$, respectively, with $BD=1$. Point $P$ is a moving point on line segment $OA$. When $\\angle PDO=45^\\circ$, find the length of line segment $OP$.\n",
    "completion": "\\textbf{Solution}: In the figure, draw $PF\\perp OD$ at point F, then $\\triangle PDF$ is an isosceles right triangle.\\\\\n$\\because$OA=4, AD=2,\\\\\n$\\therefore$OD=$\\sqrt{OA^2+AD^2}=2\\sqrt{5}$.\\\\\nLet AP=m, then OP=4−m,\\\\\n$\\therefore$PD=$\\sqrt{AD^2+AP^2}=\\sqrt{4+m^2}$.\\\\\n$\\because$$\\triangle PDF$ is an isosceles right triangle,\\\\\n$\\therefore$DF=PF=$\\frac{\\sqrt{2}}{2}PD=\\frac{\\sqrt{8+2m^2}}{2}$,\\\\\n$\\therefore$OF=OD−DF=$2\\sqrt{5}-\\frac{\\sqrt{8+2m^2}}{2}$.\\\\\n$\\because$OF$^2$+PF$^2$=OP$^2$, that is, $(2\\sqrt{5}-\\frac{\\sqrt{8+2m^2}}{2})^2+(\\frac{\\sqrt{8+2m^2}}{2})^2=(4−m)^2$,\\\\\nAfter simplification, we get: $3m^2+16m-12=0$,\\\\\nSolving, we find: $m_1=\\frac{2}{3}$, $m_2=-6$ (discard as it is not meaningful for this problem),\\\\\n$\\therefore$OP=4−m=$\\boxed{\\frac{10}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51446330_269.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, $\\angle B=30^\\circ$, and $AD$ bisects $\\angle BAC$. If $BC=9$, what is the distance from point D to line $AB$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $DE \\perp AB$ at point E\\\\\n$\\therefore \\angle DEB=90^\\circ$\\\\\n$\\because \\angle C=90^\\circ$, $\\angle B=30^\\circ$\\\\\n$\\therefore \\angle BAC=60^\\circ$, $BD=2ED$, $CD\\perp AC$\\\\\n$\\because AD$ bisects $\\angle BAC$\\\\\n$\\therefore CD=DE$, $\\angle DAE=\\frac{1}{2}\\angle BAC=30^\\circ$\\\\\n$\\therefore \\angle DAB=\\angle B$\\\\\n$\\therefore BD=AD=2CD$\\\\\n$\\because BC=9=CD+BD=3ED$\\\\\n$\\therefore DE=\\boxed{3}$",
    "image_path": "dataset/mm_math/images/MM_Math/52836682.png"
  },
  {
    "solution": "\\boxed{\\sqrt{34}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $DE\\parallel BC$, intersecting $AB$ and $AC$ at $D$ and $E$ respectively, and $CD\\perp BE$, with $CD=3$ and $BE=5$. Calculate the value of $BC+DE$.\n",
    "completion": "\\textbf{Solution:} Construct EF$\\parallel$DC intersecting the extension of BC at F,\\\\\nsince DE$\\parallel$BC, EF$\\parallel$DC,\\\\\nthus quadrilateral DCFE is a parallelogram,\\\\\nthus EF=CD=3, CF=DE,\\\\\nsince CD$\\perp$BE,\\\\\nthus EF$\\perp$BE,\\\\\nthus BC+DE=BC+CF=BF=$\\sqrt{BE^{2}+EF^{2}}$=$\\sqrt{5^{2}+3^{2}}$=$\\boxed{\\sqrt{34}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52210865.png"
  },
  {
    "solution": "\\boxed{y=-x+2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given the figure in the Cartesian coordinate system $xOy$, it is known that the parabola $y=x^{2}+bx$ passes through point A $(2, 0)$ and point $B\\left(-1, m\\right)$, with vertex at point D. What is the equation of line AB?\n",
    "completion": "\\textbf{Solution:}\n\nGiven the parabola $y=x^{2}+bx$ passes through point A(2, 0),\n\n\\[\n\\therefore 2^{2}+2b=0, \\text{ solving for } b, \\text{ we get } b=-2,\n\\]\n\n\\[\n\\therefore \\text{the equation of the parabola is } y=x^{2}-2x,\n\\]\n\nWhen $x=-1$, $y=3$,\n\n\\[\n\\therefore \\text{the coordinates of point B are } B(-1, 3),\n\\]\n\nLet the equation of line AB be $y=kx+m\\ (k\\neq 0)$,\n\nSubstituting A(2, 0) and $B(-1, 3)$ into it, we get:\n\n\\[\n\\begin{cases}\n2k+m=0, \\\\\n-k+m=3,\n\\end{cases}\n\\]\nsolving these yields:\n\\[\n\\begin{cases}\nk=-1, \\\\\nm=2,\n\\end{cases}\n\\]\n\n\\[\n\\therefore \\text{the equation of line AB is } \\boxed{y=-x+2};\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/51403345_145.png"
  },
  {
    "solution": "\\boxed{(3,0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the graph of the parabola $y=x^2+bx+c$ intersects the x-axis at point $A(-1,0)$ and intersects the y-axis at point $C(0,3)$. Find the analytical expression of the parabola and the coordinates of the other intersection point $B$ with the x-axis.\n",
    "completion": "Solution: Substituting points $A(-1,0)$ and $C(0,-3)$ into the parabola $y=x^{2}+bx+c$ yields the system of equations\n\\[\n\\left\\{\n\\begin{array}{l}\n1-b+c=0\\\\\nc=-3\n\\end{array}\n\\right.\n\\]\nSolving the system of equations, we get:\n\\[\n\\left\\{\n\\begin{array}{l}\nb=-2\\\\\nc=-3\n\\end{array}\n\\right.\n\\]\nHence, the function equation is $y=x^{2}-2x-3$. When $y=0$, $x^{2}-2x-3=0$, solving this gives $x_{1}=-1$, $x_{2}=3$;\\\\\nThe coordinate of another intersection point $B$ is $\\boxed{(3,0)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52874723_282.png"
  },
  {
    "solution": "\\boxed{6\\pi +9\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of circle $O$, $AC$ is a chord of $\\odot O$, $D$ is the midpoint of arc $BC$, draw $DE\\perp AC$ at point $E$, and intersect the extension line of $AB$ at point $F$, connect $DA$. If $DA=DF=6\\sqrt{3}$, calculate the area of the shaded region. (Keep the result in terms of $\\pi$)\n",
    "completion": "\\textbf{Solution:} Given $DA=DF$, \\\\\nit follows that $\\angle F=\\angle BAD$, \\\\\nthus, we have $\\angle CAD=\\angle BAD$, \\\\\nhence, $\\angle F=\\angle BAD=\\angle CAD$, \\\\\nsince $\\angle F+\\angle BAD+\\angle CAD=90^\\circ$, \\\\\nit implies $\\angle F=\\angle BAD=\\angle CAD=30^\\circ$, \\\\\ntherefore, $\\angle BOD=2\\angle BAD=60^\\circ$, and $OF=2OD$, \\\\\ngiven $DF=6\\sqrt{3}$, \\\\\nit leads to $(2OD)^{2}-OD^{2}=(6\\sqrt{3})^{2}$, \\\\\nsolving yields: $OD=6$, \\\\\nthus, $S_{\\text{shaded}}=S_{\\text{sector} BOD}+S_{\\triangle AOD}=\\frac{60\\pi \\times 6^{2}}{360}+\\frac{1}{2}\\times 6\\times \\frac{\\sqrt{3}}{2}\\times 6=6\\pi +9\\sqrt{3}=\\boxed{6\\pi +9\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52457436_96.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in $\\triangle ABC$, $\\angle B=80^\\circ$, $\\angle C=70^\\circ$. If $\\triangle ABC$ is folded along $EF$, with point $A$ falling on $A'$ inside the shape, what is the measure of $\\angle 1+\\angle 2$?\n",
    "completion": "\\textbf{Solution:} Given that in $\\triangle ABC$, $\\angle B=80^\\circ$ and $\\angle C=70^\\circ$, \\\\\ntherefore $\\angle A=180^\\circ−\\angle B−\\angle C=180^\\circ−80^\\circ−70^\\circ=30^\\circ$, \\\\\nhence $\\angle A'=30^\\circ$, \\\\\nthus $\\angle A'EF+\\angle A'FE=180^\\circ−\\angle A'=180^\\circ−30^\\circ=150^\\circ$, \\\\\nsince $\\triangle AFE$ is created by folding $\\triangle A'FE$, \\\\\nthus $\\angle AEF+\\angle AFE=\\angle A'EF+\\angle A'FE=180^\\circ−\\angle A'=150^\\circ$, \\\\\ntherefore, $\\angle 1+\\angle 2=360^\\circ−\\angle B−\\angle C−\\left(\\angle AEF+\\angle AFE\\right)=360^\\circ−80^\\circ−70^\\circ−150^\\circ=\\boxed{60^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52634681.png"
  },
  {
    "solution": "\\boxed{y=x-4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, the line $y=-\\frac{1}{2}x+2$ intersects with the $x$ axis and the $y$ axis at points $A$ and $B$, respectively. Point $E$ is a point on the negative half of the $y$ axis, and the area of $\\triangle ABE$ is 12. What is the equation of the line AE?\n",
    "completion": "\\textbf{Solution:} Since for $y=-\\frac{1}{2}x+2$, we set $y=-\\frac{1}{2}x+2=0$ and find $x=4$. Setting $x=0$ gives $y=2$. Therefore, the coordinates of points $A$ and $B$ are $(4, 0)$ and $(0, 2)$, respectively. Hence, $OB=2$. The area of triangle $\\triangle ABE$ is given by $12=\\frac{1}{2}\\times EB\\times AO=\\frac{1}{2}\\times (2+OE)\\times 4$. Solving this yields $OE=4$. Thus, point $E$ is $(0, -4)$. Let the equation of line $AE$ be $y=kx-4$. Substituting the coordinates of point $A$ into the equation yields $0=4k-4$. Solving this, we find $k=1$. Therefore, the equation of line $AE$ is given by $\\boxed{y=x-4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50560383_73.png"
  },
  {
    "solution": "\\boxed{25^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, quadrilateral $ABCD$ is a kite, with diagonals $AC$ and $BD$ intersecting at point $O$, and $DH \\perp AB$ at point $H$. Connect $OH$, and $\\angle CAD=25^\\circ$. What is the measure of $\\angle DHO$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rhombus, \\\\\nit follows that AD=AB, BO=OD, $\\angle$DAO=$\\angle$BAO=25$^\\circ$, AC$\\perp$BD, \\\\\ntherefore $\\angle$ABD=65$^\\circ$, \\\\\nsince DH$\\perp$AB, BO=DO, \\\\\nit follows that HO=DO, \\\\\nthus $\\angle$DHO=$\\angle$BDH=90$^\\circ$-$\\angle$ABD=\\boxed{25^\\circ}.",
    "image_path": "dataset/mm_math/images/MM_Math/52425264.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{5}}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, there are two islands C and D in the sea. A fishing boat at point A in the sea measures that island D is in the northeast direction, and the distance is $30\\sqrt{2}$ nautical miles. The fishing boat then sails from the west to the east for a while and reaches point B. At this point, it's measured that island C is exactly to the north of point B and the distance is $75$ nautical miles. It's also measured that the distance between point B and island D is $30\\sqrt{5}$ nautical miles. Find the value of $\\sin\\angle ABD$.\n",
    "completion": "\\textbf{Solution:} Construct $DE \\perp AB$ at $E$,\\\\\nIn $\\triangle AED$, $AD=30\\sqrt{2}$, $\\angle DAE=45^\\circ$,\\\\\n$\\therefore DE=30\\sqrt{2}\\times\\sin45^\\circ=30$,\\\\\nIn $\\triangle BED$, $BD=30\\sqrt{5}$,\\\\\n$\\therefore \\sin\\angle ABD=\\frac{ED}{BD}=\\frac{30}{30\\sqrt{5}}=\\boxed{\\frac{\\sqrt{5}}{5}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53080937_133.png"
  },
  {
    "solution": "\\boxed{24}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure of a newly launched shoe rack from a Taobao store, it can be abstracted into a diagram where $l_1\\parallel l_2\\parallel l_3$. Lines $AC$ and $DF$ are intersected by $l_1$, $l_2$, $l_3$. If $AB=30\\,\\text{cm}$, $BC=50\\,\\text{cm}$, and $EF=40\\,\\text{cm}$, what is the length of $DE$?\n",
    "completion": "Solution: Since $l_1 \\parallel l_2 \\parallel l_3$,\\\\\ntherefore $\\frac{AB}{BC}=\\frac{DE}{EF}$\\\\\nTherefore $\\frac{30}{50}=\\frac{DE}{40}$\\\\\nSolving, we find that $DE=\\boxed{24}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52304567.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\triangle ABC$, D and F are the trisection points on side $AB$, and E and G are the trisection points on side $AC$. If $DE=2$, then how much is $BC$?\n",
    "completion": "\\textbf{Solution:} Since $D,F$ are the trisection points on side $AB$, and $E,G$ are the trisection points on side $AC$,\\\\\nit follows that $DE\\parallel FG\\parallel BC$,\\\\\nwhich means $\\triangle ADE\\sim \\triangle ABC$,\\\\\nthus, $\\frac{AD}{AB}=\\frac{DE}{BC}$,\\\\\nleading to $\\frac{1}{3}=\\frac{DE}{BC}$,\\\\\nsince $DE=2$,\\\\\nit follows that $BC=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52871905.png"
  },
  {
    "solution": "\\boxed{\\frac{7}{17}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, $\\sin B= \\frac{5}{13}$, D is on side $BC$, and $\\angle ADC=45^\\circ$, $AC=5$. Find the tangent value of $\\angle BAD$.\n",
    "completion": "\\textbf{Solution:} Since $\\angle C=90^\\circ$ and $\\angle ADC=45^\\circ$ with $AC=5$,\\\\\nit follows that $AC=CD=5$ and $AD=5\\sqrt{2}$,\\\\\nSince $\\sin B= \\frac{5}{13}$,\\\\\nthen $AB=\\frac{AC}{\\sin B}=13$,\\\\\nSince $\\angle C=90^\\circ$ and $CD=5$,\\\\\nthen $BC=12$ and $BD=7$,\\\\\nDraw $BE\\perp AD$ to extend $AD$ at $E$,\\\\\nSince $\\angle BDE=\\angle ADC=45^\\circ$,\\\\\nit follows that $BE=DE=\\frac{7}{\\sqrt{2}}=\\frac{7\\sqrt{2}}{2}$,\\\\\nTherefore, $AE=AD+DE=5\\sqrt{2}+\\frac{7\\sqrt{2}}{2}=\\frac{17\\sqrt{2}}{2}$,\\\\\nThus, $\\tan \\angle BAD = \\boxed{\\frac{7}{17}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51176403.png"
  },
  {
    "solution": "\\boxed{80^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $\\angle BDC=142^\\circ$, $\\angle B=34^\\circ$, and $\\angle C=28^\\circ$. What is the degree measure of $\\angle A$?\n",
    "completion": "\\textbf{Solution:} Construct ray AE passing through points A and D, \\\\\n$\\because$ $\\angle CDE = \\angle CAD + \\angle C$, $\\angle BDE = \\angle BAD + \\angle B$, \\\\\n$\\therefore$ $\\angle BDC = \\angle CDE + \\angle BDE = \\angle CAD + \\angle C + \\angle BAD + \\angle B = 142^\\circ$, \\\\\n$\\therefore$ $\\angle CAB + 34^\\circ + 28^\\circ = 142^\\circ$, \\\\\n$\\therefore$ $\\angle CAB = \\boxed{80^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52733094.png"
  },
  {
    "solution": "\\boxed{13}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, C is a point on the semicircle O with AB as the diameter. Lines AC and BC are connected, and squares ACDE and BCFG are constructed externally on sides AC and BC, respectively. The midpoints of DE, FG, $\\overset{\\frown}{AC}$, and $\\overset{\\frown}{BC}$ are M, N, P, Q, respectively. If $MP+NQ=14$ and $AC+BC=18$, what is the length of AB?\n",
    "completion": "\\textbf{Solution:} \\\\\nLet's connect OP and OQ, \\\\\nsince the midpoints of DE, FG, $\\overset{\\frown}{AC}$, and $\\overset{\\frown}{BC}$ are respectively M, N, P, and Q, \\\\\ntherefore OP $\\perp$ AC, and OQ $\\perp$ BC, \\\\\nthus H and I are the midpoints of AC and BC, respectively, \\\\\nhence OH + OI = $\\frac{1}{2}$(AC + BC) = 9, \\\\\nsince MH + NI = AC + BC = 18, and MP + NQ = 14, \\\\\ntherefore PH + QI = 18 - 14 = 4, \\\\\nthus AB = OP + OQ = OH + OI + PH + QI = 9 + 4 = \\boxed{13},",
    "image_path": "dataset/mm_math/images/MM_Math/52814833.png"
  },
  {
    "solution": "\\boxed{84^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, D is a point on $BC$, $\\angle 1=\\angle 2$, $\\angle 3=\\angle 4$, $\\angle BAC=108^\\circ$. What is the measure of $\\angle DAC$?\n",
    "completion": "\\textbf{Solution:} Let $ \\angle 1=\\angle 2=x$,\n\n$ \\therefore \\angle 3=\\angle 4=\\angle 1+\\angle 2=2x$,\n\n$ \\because \\angle DAC+\\angle 3+\\angle 4=180^\\circ$,\n\n$ \\therefore \\angle DAC=180^\\circ−4x$,\n\n$ \\because \\angle BAC=108^\\circ$,\n\n$ \\therefore x+180^\\circ−4x=108^\\circ$,\n\n$ \\therefore x=24^\\circ$,\n\n$ \\therefore \\angle DAC=180^\\circ−4\\times 24^\\circ=\\boxed{84^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/52950309.png"
  },
  {
    "solution": "\\boxed{100^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, fold $\\triangle ABC$ along its median $DE$, with point $A$ landing on point $F$. If $\\angle C = 120^\\circ$ and $\\angle A = 20^\\circ$, what is the degree measure of $\\angle FEB$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle C=120^\\circ$ and $\\angle A=20^\\circ$,\\\\\nit follows that $\\angle B=40^\\circ$,\\\\\nsince DE is the median line of $\\triangle ABC$,\\\\\nit implies that DE$\\parallel$BC,\\\\\nthus, $\\angle B+\\angle DEB=180^\\circ$, and $\\angle B=\\angle AED=\\angle DEF=40^\\circ$\\\\\nwhich leads to $\\angle DEB=140^\\circ$,\\\\\nhence, $\\angle FEB=\\angle DEB-\\angle DEF=\\boxed{100^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52211191.png"
  },
  {
    "solution": "\\boxed{\\left(13, -2\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=-\\frac{1}{2}x+3$ intersects the $x$-axis and $y$-axis at points $A$ and $B$ respectively, and intersects the line $y=x$ at point $C$. The line passing through point $C$ that bisects the area of $\\triangle AOC$ intersects the $x$-axis at point $D$. Let $P$ be a moving point on line $AB$. Is there a point $Q$ in the plane such that the quadrilateral with vertices $A$, $D$, $P$, $Q$ is a parallelogram, and its area is equal to the area of $\\triangle AOC$? Write down the coordinates of point $Q$.\n",
    "completion": "\\textbf{Solution:} There exists a point $Q$, with coordinates $\\left(-1, 2\\right)$ or $\\left(5, 2\\right)$ or $\\left(7, -2\\right)$ or $\\left(13, -2\\right)$, which makes the quadrilateral formed by vertices $A$, $D$, $P$, $Q$ a parallelogram, and its area equals to the area of $\\triangle AOC$.\\\\\nSince $S_{\\triangle AOC}=2S_{\\triangle ADC}$ and $S_{\\text{parallelogram}}=S_{\\triangle AOC}$,\\\\\nit follows that $S_{\\text{parallelogram}}=2S_{\\triangle PAD}$.\\\\\nWhen $P$ coincides with point $C$ above the $x$-axis,\\\\\nsince $A\\left(6, 0\\right)$, $D\\left(3, 0\\right)$,\\\\\nthen $AD=3$,\\\\\nsince $P_1(2, 2)$,\\\\\nby the property of translation, it's known that $Q_1\\left(-1, 2\\right)$ or $Q_2\\left(5, 2\\right)$;\\\\\nWhen point $P$ lies on line $AB$ below the $x$-axis, and the distance to the $x$-axis is 2,\\\\\nlet $P_2(x, -\\frac{1}{2}x+3)$, then $-\\frac{1}{2}x+3=-2$,\\\\\nsolve to get $x=10$,\\\\\nthus $P(10, -2)$,\\\\\nby the property of translation, it's known that $Q_3\\left(7, -2\\right)$ or $Q_4\\left(13, -2\\right)$;\\\\\nIn conclusion, the coordinates of point $Q$ are $\\boxed{\\left(-1, 2\\right)}$ or $\\boxed{\\left(5, 2\\right)}$ or $\\boxed{\\left(7, -2\\right)}$ or $\\boxed{\\left(13, -2\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50595478_132.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the diagonals AC and BD of the quadrilateral $ABCD$ intersect at point O. P is the midpoint of side AB, and $OP=2$. What is the length of BC?\n",
    "completion": "\\textbf{Solution:} \\\\\nSince $ABCD$ is a square, \\\\\nit follows that $OA = OC$, \\\\\nhence, $O$ is the midpoint of $AC$, \\\\\nSince $P$ is the midpoint on side $AB$, \\\\\nit follows that $OP$ is the median of $\\triangle ABC$, \\\\\nhence, $BC = 2OP = 2\\times2 = \\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52386544.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB=9$, and $C$ is the midpoint of $AB$. Point $D$ is on the line segment $AC$, and $AD\\colon CB=1\\colon 3$. What is the length of $DC$?\n",
    "completion": "\\textbf{Solution:} Since $C$ is the midpoint of $AB$ and $AB = 9$, \\\\\nthen $AC = BC = \\frac{1}{2} AB = 4.5$, \\\\\nand since the ratio $AD : CB = 1 : 3$, \\\\\nit follows that $AD : AC = 1 : 3$, \\\\\nhence $AD = \\frac{1}{3} AC = 1.5$, \\\\\nthus $DC = AC - AD = 4.5 - 1.5 = \\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55301095.png"
  },
  {
    "solution": "\\boxed{\\angle DAB=60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the diagonals of the rhombus $ABCD$ intersect at point $O$. Extend $AB$ to point $E$ such that $BE=AB$, and connect $CE$. Given $BD=EC$; at what angle $\\angle DAB$ will the quadrilateral $BECD$ form a rhombus? Please explain the reason.\n",
    "completion": "\\textbf{Solution:} When $\\angle DAB=60^\\circ$, the quadrilateral $BECD$ is a rhombus.\\\\\nThe reason is as follows:\\\\\n$\\because$ Quadrilateral $ABCD$ is a rhombus,\\\\\n$\\therefore$ $AD=AB$,\\\\\n$\\because$ $\\angle DAB=60^\\circ$,\\\\\n$\\therefore$ $\\triangle ADB$, $\\triangle DCB$ are equilateral triangles,\\\\\n$\\therefore$ $DC=DB$,\\\\\n$\\because$ Quadrilateral $BECD$ is a parallelogram,\\\\\n$\\therefore$ Quadrilateral $BECD$ is a rhombus.\\\\\n$\\therefore$ \\boxed{\\angle DAB=60^\\circ}",
    "image_path": "dataset/mm_math/images/MM_Math/52344234_120.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the figure, within $\\triangle ABC$, $\\angle C=90^\\circ$. If $BC=20$, $AD$ bisects $\\angle BAC$ and intersects $BC$ at point D, and $BD:CD=3:2$, then what is the length of distance $DE$ from point D to line segment $AB$?\n",
    "completion": "\\textbf{Solution:} Given that $BC=20$ and $BD:CD=3:2$, \\\\\nthus $CD=\\frac{2}{5}CB=8$, \\\\\nsince $AD$ bisects $\\angle BAC$ and $\\angle C=90^\\circ$, with $DE$ being the distance from point D to line segment $AB$, \\\\\nhence $DE=CD=\\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53044021.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As depicted, the rectangle $DEFG$ has its side $DE$ on side $BC$ of $\\triangle ABC$, with points $G$ and $E$ respectively on sides $AB$ and $AC$. Given that $BC=6\\,cm$, $DE=3\\,cm$, $EF=2\\,cm$, what is the area of $\\triangle ABC$ in $\\,cm^{2}$?\n",
    "completion": "\\textbf{Solution:} Draw $AH\\perp BC$ at $H$, intersecting $GF$ at $M$,\\\\\nsince $AH\\perp BC$ and quadrilateral $DEFG$ is a rectangle,\\\\\nthus, quadrilateral $HEFM$ is a rectangle, so $MH=EF=2\\,cm$,\\\\\nsince quadrilateral $DEFG$ is a rectangle,\\\\\nthus, $GF\\parallel BC$, $DG=EF=2\\,cm$, $GF=DE=3\\,cm$,\\\\\ntherefore, $\\triangle AGF\\sim \\triangle ABC$,\\\\\nthus, $\\frac{AM}{AH}=\\frac{GF}{BC}$, that is $\\frac{AM}{AM+2}=\\frac{3}{6}$, solving gives: $AM=2\\,cm$,\\\\\nthus, $AH=AM+MH=4\\,cm$,\\\\\ntherefore, ${S}_{\\triangle ABC}=\\frac{1}{2}BC\\cdot AH=\\frac{1}{2}\\times 6\\times 4=\\boxed{12}\\,cm^{2}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53148264.png"
  },
  {
    "solution": "\\boxed{\\frac{15}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In $\\triangle ABC$, $AC=8$, $BC=6$, and $\\angle C=90^\\circ$. Now, $\\triangle ABC$ is folded as shown in the figure, making point A coincident with point B. The fold line is $DE$. What is the length of $DE$?\n",
    "completion": "\\textbf{Solution:} Given that $AC=8$, $BC=6$, and $\\angle C=90^\\circ$,\\\\\nthus $AB=\\sqrt{AC^2+BC^2}=10$,\\\\\nGiven folding $\\triangle ABC$ such that point A coincides with point B, with the fold line being $DE$,\\\\\nthus $BE=AE$, $AD=BD=5$, and $DE\\perp AB$,\\\\\nIn right triangle $BEC$, $BE^2=BC^2+CE^2$,\\\\\nthus $BE^2=36+(8-BE)^2$,\\\\\nthus $BE=\\frac{25}{4}$,\\\\\nIn right triangle $BDE$, $DE=\\sqrt{BE^2-BD^2}=\\boxed{\\frac{15}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52890844.png"
  },
  {
    "solution": "\\boxed{D\\left(0,\\frac{1}{2}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given in the Cartesian coordinate system $xOy$ (as shown in the figure), the line $y=2x+2$, intersects the x-axis and y-axis at points A and B, respectively, and the coordinates of point C are $\\left(3, 2\\right)$. Connect AC and intersect the y-axis at point D. Find the coordinates of point D.\n",
    "completion": "\\textbf{Solution:} Let the equation of line $AC$ be $y=kx+b$,\\\\\n$\\therefore \\left\\{\\begin{array}{l}\n-k+b=0 \\\\\n3k+b=2\n\\end{array}\\right.$,\\\\\nSolving, we get $\\left\\{\\begin{array}{l}\nk=\\frac{1}{2} \\\\\nb=\\frac{1}{2}\n\\end{array}\\right.$,\\\\\n$\\therefore y=\\frac{1}{2}x+\\frac{1}{2}$,\\\\\nLet $x=0$, then $y=\\frac{1}{2}$,\\\\\n$\\therefore \\boxed{D\\left(0,\\frac{1}{2}\\right)}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52977945_47.png"
  },
  {
    "solution": "\\boxed{20^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given point O on line AB, construct ray OC such that $\\angle BOC=40^\\circ$. Place the right-angled vertex of a right triangle at O, that is, $\\angle DOE=90^\\circ$. As shown in figure 2, rotate the right triangle DOE clockwise around point O to a certain position. If OE exactly bisects $\\angle AOC$, then what is the measure of $\\angle COD$?\n",
    "completion": "\\textbf{Solution:} We have $\\angle COD = \\boxed{20^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52916141_150.png"
  },
  {
    "solution": "\\boxed{\\frac{7}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that trapezoid $ABCD$ satisfies $AD \\parallel BC$. The diagonal $AC$ intersects the midline $EF$ at point $G$. If $AD = 3$ and $EF = 5$, what is the length of $EG$?\n",
    "completion": "\\textbf{Solution:} Since $EF$ is the midline of trapezoid $ABCD$,\\\\\nit follows that $CF=DF$,\\\\\nand since $AD\\parallel BC$,\\\\\nit also follows that $\\triangle CFG\\sim \\triangle CDA$,\\\\\nthus $\\frac{GF}{AD}=\\frac{CF}{CD}=\\frac{1}{2}$,\\\\\ntherefore $GF=\\frac{1}{2}AD=\\frac{3}{2}$,\\\\\nhence $EG=EF-GF=5-\\frac{3}{2}=\\boxed{\\frac{7}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52942169.png"
  },
  {
    "solution": "\\boxed{2 \\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the radius $OA$ of $\\odot O$ is 2, and $B$ is a moving point on $\\odot O$ (not coinciding with point $A$). A tangent line $BC$ to $\\odot O$ passing through point $B$ is drawn, where $BC=OA$. The lines $OC$ and $AC$ are connected. When $\\triangle OAC$ is a right triangle, what is the length of its hypotenuse?\n",
    "completion": "\\textbf{Solution:} Since BC is the tangent of $\\odot$O, \\\\\n$\\therefore \\angle$OBC$=90^\\circ$, \\\\\nSince BC=OA, \\\\\n$\\therefore$ OB=BC=2, \\\\\n$\\therefore \\triangle$OBC is an isosceles right triangle, \\\\\n$\\therefore \\angle$BCO=$45^\\circ$, \\\\\n$\\therefore \\angle$ACO$\\leq 45^\\circ$, \\\\\nSince when $\\triangle$OAC is a right triangle, (1) $\\angle$AOC=$90^\\circ$, connect OB, \\\\\n$\\therefore$ OC=$ \\sqrt{2}$ OB=$2 \\sqrt{2}$, \\\\\n$\\therefore$ AC=$ \\sqrt{OA^{2}+OC^{2}}=\\sqrt{2^{2}+(2\\sqrt{2})^{2}}$=$2 \\sqrt{3}$; \\\\\n(2) When $\\triangle$OAC is a right triangle, $\\angle$OAC=$90^\\circ$, connect OB, \\\\\nSince BC is the tangent of $\\odot$O, \\\\\n$\\therefore \\angle$CBO=$\\angle$OAC=$90^\\circ$, \\\\\nSince BC=OA=OB, \\\\\n$\\therefore \\triangle$OBC is an isosceles right triangle, \\\\\n$\\therefore OC=2\\sqrt{2}$, \\\\\nHence, the answer is: $AC=\\boxed{2 \\sqrt{3}}$ or $2 \\sqrt{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52807541.png"
  },
  {
    "solution": "\\boxed{20^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, AB is the diameter of the circle $O$, and point C is on the circle. If $\\angle ABC = 70^\\circ$, what is the degree measure of $\\angle BAC$?\n",
    "completion": "\\textbf{Solution:} Given that $AB$ is the diameter of circle $O$,\\\\\nit follows that $\\angle C=90^\\circ$,\\\\\nsince $\\angle ABC=70^\\circ$,\\\\\nit follows that $\\angle BAC=90^\\circ-70^\\circ=\\boxed{20^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52872165.png"
  },
  {
    "solution": "\\boxed{0.462}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: A rectangular piece of iron, with a length of 25 cm and a width of 15 cm, has small squares with a side length of 2 cm cut from each of its four corners. Then, by folding up the four sides, a lidless iron box is made. What is the volume of the iron box in liters?\n",
    "completion": "\\textbf{Solution:} By the problem statement, we have: the length of the iron box $= 25 - 2 \\times 2 = 21\\,\\text{cm}$, the width of the iron box $= 15 - 2 \\times 2 = 11\\,\\text{cm}$, the height of the iron box $= 2\\,\\text{cm}$,\\\\\n$\\therefore$ the volume of the iron box $= 21 \\times 11 \\times 2 = 462\\,\\text{cm}^3 = 0.462$ litres.\\\\\nThus, the volume of the iron box is $\\boxed{0.462}$ litres.",
    "image_path": "dataset/mm_math/images/MM_Math/51081262.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\angle BAC=30^\\circ$, $AM$ is the bisector of $\\angle BAC$, through $M$ draw $ME\\parallel BA$ intersecting $AC$ at $E$, draw $MD\\perp BA$ with $D$ as the foot of the perpendicular, and $ME=10\\,\\text{cm}$, then how long is $MD$ in $cm$?\n",
    "completion": "\\textbf{Solution:} Draw $MF \\perp AC$ at $F$,\\\\\nsince $AM$ is the angle bisector of $\\angle BAC$,\\\\\nthus $MD=MF$, $\\angle BAM = \\angle CAM$,\\\\\nsince $ME \\parallel BA$,\\\\\nthus $\\angle AME = \\angle BAM$,\\\\\nthus $\\angle CAM = \\angle AME = \\frac{1}{2} \\angle BAC = \\frac{1}{2} \\times 30^\\circ = 15^\\circ$,\\\\\nsince $\\angle CEM$ is the exterior angle of $\\triangle AME$,\\\\\nthus $\\angle CEM = \\angle CAM + \\angle AME = 15^\\circ + 15^\\circ = 30^\\circ$,\\\\\nin right-angled $\\triangle MEF$, $\\angle FEM = 30^\\circ$,\\\\\nthus $MF = \\frac{1}{2} ME = \\frac{1}{2} \\times 10 = \\boxed{5} \\, cm$,\\\\\nthus $MD = MF = 5cm$.",
    "image_path": "dataset/mm_math/images/MM_Math/52457333.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in square \\(ABCD\\), \\(E\\) and \\(F\\) are located on \\(BD\\) and \\(AD\\) respectively. Connect \\(EF\\) and construct an isosceles right triangle \\(EFG\\) on the right side of \\(EF\\) such that \\(\\angle EFG=90^\\circ\\), and connect \\(DG\\). \\(DG\\perp BD\\); If \\(AB=8\\), \\(FD=2\\), \\(BE=2\\sqrt{2}\\), write down the length of segment \\(DG\\) directly.\n",
    "completion": "\\textbf{Solution:} The length of the line segment $DG$ is $\\boxed{4\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53359763_132.png"
  },
  {
    "solution": "\\boxed{P_2\\left(\\frac{2}{3},\\frac{11}{9}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system, the parabola $y=-x^2+bx+c$ passes through points $A(-3,0)$ and $B(1,0)$, and intersects the $y$-axis at point $C$. Line segment $BC$ is drawn, and point $P$ is a moving point on the parabola above the $x$-axis. A perpendicular line from $P$ to the $x$-axis is drawn, with the foot of the perpendicular being point $E$. Is there a point $P$ such that the triangle formed by $A$, $P$, and $E$ is similar to $\\triangle BOC$? If such a point exists, find the coordinates of point $P$ and explain the reasoning.\n",
    "completion": "\\textbf{Solution:} Let $A(-3,0)$, $B(1,0)$, $C(0,3)$,\\\\\ntherefore $OC=3$, $OB=1$,\\\\\nlet $P(m,-m^2-2m+3)$,\\\\\nthus $PE=-m^2-2m+3$, $AE=m+3$,\\\\\nBased on the given conditions, we have $\\frac{m+3}{-m^2-2m+3}=\\frac{1}{3}$,\\\\\nSolving this, we find $m_1=-2$, $m_2=-3$ (discard),\\\\\nthus $-m^2-2m+3=-(-2)^2-2\\times(-2)+3=3$,\\\\\nthus $P_1(-2,3)$,\\\\\nor $\\frac{m+3}{-m^2-2m+3}=\\frac{3}{1}$,\\\\\nSolving this, we find $m_1=\\frac{2}{3}$, $m_2=-3$ (discard),\\\\\nthus $-m^2-2m+3=-\\left(\\frac{2}{3}\\right)^2-2\\times\\frac{2}{3}+3=\\frac{11}{9}$,\\\\\nthus $P_2\\left(\\frac{2}{3},\\frac{11}{9}\\right)$,\\\\\nIn conclusion, the coordinates of point $P$ are $\\boxed{P_1(-2,3)}$ or $\\boxed{P_2\\left(\\frac{2}{3},\\frac{11}{9}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51403139_39.png"
  },
  {
    "solution": "\\boxed{15}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given: As shown in the figure, in trapezoid $ABCD$, $AD\\parallel BC$, diagonals $AC$ and $BD$ intersect at point $E$, point $F$ is on side $AB$, and $EF\\parallel BC$. It is known that $AD=3$, $BC=6$. Fill in the blank: If ${S}_{\\triangle AEF}=5$, then ${S}_{\\triangle ECD}$ equals what?\n",
    "completion": "\\textbf{Solution:} From the given, it is obtained that the area of $\\triangle ECD$ is $\\boxed{15}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52917645_64.png"
  },
  {
    "solution": "\\boxed{7.5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in rectangle $ABCD$, $BC>CD$, and $BC$ and $CD$ are the two roots of the quadratic equation ${x}^{2}-14x+48=0$. Connect $BD$. A moving point $P$ starts from $B$ and moves towards $D$ along $BD$ at a speed of $1$ unit per second. At the same time, another moving point $Q$ starts from $D$ and moves along the ray $DA$ at the same speed. When point $P$ reaches point $D$, point $Q$ stops moving. Let the movement time be $t$ seconds. Construct $\\triangle PQM$ with $PQ$ as the hypotenuse, such that point $M$ lies on the segment $BD$. What is the maximum area of $\\triangle PDQ$?\n",
    "completion": "\\textbf{Solution:} Draw $PE \\perp AD$ at point E, \\\\\n$\\therefore \\angle DEP=\\angle DAB=90^\\circ$,\\\\\n$\\therefore PE \\parallel AB$,\\\\\n$\\therefore \\triangle DEP \\sim \\triangle DAB$,\\\\\n$\\therefore \\frac{EP}{AB}=\\frac{PD}{BD}$, given $AB=CD=6$, $DQ=BP=t$, then $PD=10-t$,\\\\\n$\\therefore \\frac{EP}{6}=\\frac{10-t}{10}$, solving this: $PE=6-\\frac{3}{5}t$,\\\\\n$\\therefore S_{\\triangle PQD}=\\frac{1}{2}DQ \\cdot PE=\\frac{1}{2}t\\left(6-\\frac{3}{5}t\\right)=-\\frac{3}{10}(t-5)^2+\\frac{15}{2}$,\\\\\n$\\because$ the parabola opens downwards,\\\\\nwhen $t=5$, the maximum area is $\\boxed{7.5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52774682_58.png"
  },
  {
    "solution": "\\boxed{\\frac{48}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\triangle ABC$, $ \\angle ABC=\\angle ACB$, D is the midpoint of BC, AD is connected, E is a point on AB, P is a point on AD, EP and BP are connected, $ AC=10$, $ BC=12$, what is the minimum value of $ EP+BP$?\n",
    "completion": "\\textbf{Solution:} Given $\\because$ $\\triangle ABC$ is an isosceles triangle, and AD is the median to side BC,\\\\\n$\\therefore$ AD bisects BC at a right angle,\\\\\n$\\therefore$ points B and C are symmetric about AD,\\\\\n$\\therefore$ BP=CP,\\\\\nA perpendicular from point C to AB is drawn, and the perpendicular foot is point E, with CE intersecting AD at point P, (the perpendicular distance is the shortest distance between a point and a line),\\\\\nAs shown in the figure, at this time, the sum of $BP+EP$ is minimal, and it is equal to the length of CE,\\\\\n$\\because$ D is the midpoint of BC, and BC=12,\\\\\n$\\therefore$ CD=$\\frac{1}{2} \\times 12=6$,\\\\\n$\\therefore$ AD=$\\sqrt{AC^{2}-CD^{2}}=8$,\\\\\n$\\because$ $\\angle ABC=\\angle ACB$,\\\\\n$\\therefore$ AB=AC=10,\\\\\n$\\because$ $S_{\\triangle ABC}=\\frac{1}{2}BC \\cdot AD=\\frac{1}{2}AB \\cdot CE$,\\\\\n$\\therefore$ CE=$\\frac{BC \\cdot AD}{AB}=\\frac{12 \\times 8}{10}=\\frac{48}{5}$,\\\\\n$\\therefore$ the minimum value of $BP+EP$ is $\\boxed{\\frac{48}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52836214.png"
  },
  {
    "solution": "\\boxed{\\left(0, \\frac{6}{5}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=-\\frac{1}{2}x+3$ intersects the $x$-axis and $y$-axis at points $A$ and $B$, respectively, and intersects the line $y=x$ at point $C$. The line through point $C$ that bisects the area of $\\triangle AOC$ intersects the $x$-axis at point $D$. Point $E$ is on the $y$-axis. When the perimeter of $\\triangle DCE$ is minimized, find the coordinates of point $E$ (proof of minimum perimeter is not required).\n",
    "completion": "\\textbf{Solution:} Let point $D$ be symmetrically reflected across the $y$-axis to point $F$, and draw line $FC$ intersecting the $y$-axis at point $E$. Connecting $ED$, point $E$ is the required point. Due to symmetry, we have $F(-3, 0)$. Assume the equation of line $FC$ is $y=kx+b$. Substituting points $F(-3, 0)$ and $C(2, 2)$ into this equation, we get $\\therefore \\begin{cases}2k+b=2\\\\ -3k+b=0\\end{cases}$, and solving these gives $\\begin{cases}k=\\frac{2}{5}\\\\ b=\\frac{6}{5}\\end{cases}$. $\\therefore$ $y=\\frac{2}{5}x+\\frac{6}{5}$. Letting $x=0$, we find $y=\\frac{6}{5}$, $\\therefore$ the coordinates of point $E$ are $\\boxed{\\left(0, \\frac{6}{5}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50595478_131.png"
  },
  {
    "solution": "\\boxed{x<0}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given: As shown in the figure, the graph of the linear function $y=-2x+10$ intersects with the graph of the inverse proportion function $y=\\frac{k}{x}$ at points $A$ and $B$ (with $A$ located to the right of $B$), where the abscissa of point $A$ is $4$. Observing the graph, directly write out the solution set of the inequality $-2x+10-\\frac{k}{x}>0$ with respect to $x$.\n",
    "completion": "\\textbf{Solution:} Observing the graph, the solution set of the inequality $-2x+10-\\frac{k}{x}>0$ with respect to $x$ is: \\boxed{1<x<4} or \\boxed{x<0}.",
    "image_path": "dataset/mm_math/images/MM_Math/53080947_109.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, $\\odot O$ is the inscribed circle of $\\triangle ABC$, and D, E, F are the points of tangency. Quadrilateral ODCE is a square; if $AB=5$ and $AC=3$, what is the radius of the inscribed circle $\\odot O$?\n",
    "completion": "\\textbf{Solution:} Let the radius of circle $ \\odot O$ be $r$,\\\\\nsince quadrilateral $ODCE$ is a square,\\\\\ntherefore $OD=DC=CE=r$,\\\\\nin $\\triangle ABC$, $BC=\\sqrt{AB^{2}-AC^{2}}=4$,\\\\\ntherefore $BD=4-r$, $AE=3-r$,\\\\\nsince circle $ \\odot O$ is tangent to the sides of $\\triangle ABC$ at points $D$, $E$, and $F$,\\\\\ntherefore $AE=AF=3-r$, $BF=BD=4-r$,\\\\\nand since $AB=AF+BF=5$,\\\\\ntherefore $3-r+4-r=5$, solving this gives $r=\\boxed{1}$\\\\\ntherefore, the radius of the inscribed circle is \\boxed{1}.",
    "image_path": "dataset/mm_math/images/MM_Math/52719793_160.png"
  },
  {
    "solution": "\\boxed{90^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, quadrilaterals ABCD, DCFE, and EFGH are three squares with side length of 1. Connect AC, AF, and AG. It follows that $\\triangle ACF \\sim \\triangle GCA$. Find the degree measure of $\\angle 1 + \\angle 2 + \\angle 3$?\n",
    "completion": "\\textbf{Solution:} From (1) we have $\\angle 1 = \\angle \\text{FAC}$.\\\\\nSince quadrilateral ABCD is a square,\\\\\ntherefore $\\angle 3 = 45^\\circ$,\\\\\ntherefore $\\angle 2 + \\angle \\text{FAC} = \\angle 3 = 45^\\circ$,\\\\\ntherefore $\\angle 1 + \\angle 2 = 45^\\circ$,\\\\\ntherefore $\\angle 1 + \\angle 2 + \\angle 3 = \\boxed{90^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53081381_96.png"
  },
  {
    "solution": "\\boxed{\\frac{8}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $E$ and $F$ are the midpoints of the sides $CD$ and $AD$ respectively, $CF$ intersects $EA$ and $EB$ at points $M$ and $N$ respectively. Given $AB=8$ and $BC=12$, what is the length of $MN$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, extend BE and AD to meet at point Q. Since quadrilateral ABCD is a rectangle, $BC=12$, thus $\\angle BAD=90^\\circ$, $AD=BC=12$, $AD\\parallel BC$. Since F is the midpoint of AD, it follows that $AF=DF=6$. In $\\triangle DCF$, by the Pythagorean theorem, we get: $CF=\\sqrt{DF^2+DC^2}=\\sqrt{6^2+8^2}=10$. Since $AD\\parallel BC$, it follows that $\\angle Q=\\angle EBC$. Since E is the midpoint of CD and $CD=8$, it follows that $CE=DE=4$. In $\\triangle DQE$ and $\\triangle CBE$, we have $\\begin{cases}\n\\angle DEQ=\\angle BEC\\\\ \n\\angle Q=\\angle EBC\\\\ \nDE=CE\n\\end{cases}$, thus $\\triangle DQE\\cong \\triangle CBE$ (AAS), hence $DQ=BC=12$, that is $QF=12+6=18$. Since $AD\\parallel BC$, it follows that $\\triangle QNF\\sim \\triangle BNC$, hence $\\frac{FN}{CN}=\\frac{QF}{BC}=\\frac{18}{12}=\\frac{3}{2}$. Since $CF=10$, we get $CN=4$, $FN=6$. As illustrated, extend CF and BA to meet at W. Since $WB\\parallel CD$, it follows that $\\angle W=\\angle DCF$. In $\\triangle WAF$ and $\\triangle CDF$, we have $\\begin{cases}\n\\angle W=\\angle DCF\\\\ \nAF=DF\\\\ \n\\angle AFW=\\angle CFD\n\\end{cases}$, thus $\\triangle WAF\\cong \\triangle CDF$ (AAS), hence $CD=AW=8$, $BW=16$, $CF=FW=10$. Because $AB\\parallel CD$, it follows that $\\triangle CME\\sim \\triangle WMA$, hence $\\frac{CM}{MW}=\\frac{CE}{AW}$, solving this we get: $CM=\\frac{20}{3}$, hence $MN=CM-CN=\\frac{20}{3}-4=\\boxed{\\frac{8}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51368292.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the figure, within $\\triangle ADC$, $\\angle C=90^\\circ$. Point B is on the extension line of $CD$. Connect $AB$. Point E is on $AC$. Connect $DE$. $AD$ bisects $\\angle BAC$. $CE = 2AE$, $DB = DE$, and $CD = 3$. What is the length of $AC$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, draw $DF\\perp AB$ at $F$ through point $D$, \\\\\nsince $\\angle C=90^\\circ$ and $AD$ bisects $\\angle BAC$, with $CD=3$, \\\\\ntherefore $DF=CD=3$, \\\\\nsince $DB=DE$, \\\\\ntherefore by the HL theorem, $Rt\\triangle EDC\\cong Rt\\triangle BDF$, \\\\\nhence $BF=CE$, \\\\\nlet $AE=x$, then $CE=2x=BF$, \\\\\nsince $CD=DF$ and $AD=AD$, \\\\\ntherefore, by the Pythagorean theorem, we get: $AC=AF=3x$, \\\\\nhence $AB=5x$, \\\\\nsince $\\angle B=\\angle B$ and $\\angle BFD=\\angle ACB=90^\\circ$, \\\\\ntherefore $\\triangle BFD\\sim \\triangle BCA$, \\\\\ntherefore $\\frac{DF}{AC}=\\frac{BD}{AB}=\\frac{BF}{BC}$, \\\\\ntherefore $\\frac{BD}{5x}=\\frac{3}{3x}$, \\\\\nthus $BD=5$, \\\\\ntherefore from $BF^2+DF^2=BD^2$, we obtain: $(2x)^2=5^2-3^2=16$, \\\\\nsolving gives: $x=2$, \\\\\nthus $AC=3x=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53066099.png"
  },
  {
    "solution": "\\boxed{\\frac{-1+\\sqrt{17}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in square $ABCD$, $E$ is a point on the diagonal $AC$, $FH \\perp AC$ at point $E$, intersecting $AD$, $AB$ at points $F$, $H$ respectively. $CF=CH$; if $AH= \\frac{1}{3} CH$ and $AB=4$, what is the length of $AH$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral ABCD is a square, it follows that $BC=AB=4$ and $\\angle B=90^\\circ$. Let $AH=x$, then $BH=AB-AH=4-x$. Since $AH=\\frac{1}{3}CH$, it follows that $CH=3x$. In $\\triangle CBH$, by the Pythagorean theorem, we have $CH^{2}-BH^{2}=BC^{2}$, which simplifies to $(3x)^{2}-(4-x)^{2}=16$. This reduces to $x^{2}+x-4=0$. Solving for $x$52251693, we find $x_{1}=\\frac{-1+\\sqrt{17}}{2}$, $x_{2}=\\frac{-1-\\sqrt{17}}{2}$ (discard this solution). Therefore, $x=\\boxed{\\frac{-1+\\sqrt{17}}{2}}$, meaning the length of AH is $\\boxed{\\frac{-1+\\sqrt{17}}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50496809_92.png"
  },
  {
    "solution": "\\boxed{x>4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given the line $y=kx+b$ passes through the points $A(6, 0)$, $B(2, 2)$. According to the graph, write the solution set for the inequality $x-3>kx+b$ with respect to $x$.\n",
    "completion": "\\textbf{Solution:} From the graph, we can conclude that $\\boxed{x>4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52056013_217.png"
  },
  {
    "solution": "\\boxed{9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $BD = \\frac{1}{5}BC$, $AE = \\frac{1}{4}AD$, and $CF = \\frac{1}{2}CE$. Given that the area of $\\triangle ABC = 30$, what is the area of $\\triangle DEF$?\n",
    "completion": "\\textbf{Solution:}\\\\\n$\\because BD= \\frac{1}{5} BC$,\\\\\n$\\therefore CD= \\frac{4}{5} BC$,\\\\\n$\\therefore S_{\\triangle ACD}= \\frac{4}{5} S_{\\triangle ABC}= \\frac{4}{5} \\times 30 = 24$,\\\\\n$\\because AE= \\frac{1}{4} AD$,\\\\\n$\\therefore DE= \\frac{3}{4} AD$,\\\\\n$\\therefore S_{\\triangle CDE}= \\frac{3}{4} S_{\\triangle ACD}= \\frac{3}{4} \\times 24 = 18$,\\\\\n$\\because CF= \\frac{1}{2} CE$,\\\\\n$\\therefore EF= \\frac{1}{2} CE$,\\\\\n$\\therefore S_{\\triangle DEF}= \\frac{1}{2} S_{\\triangle CDE}= \\frac{1}{2} \\times 18 = \\boxed{9}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52815571.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, where $AB=5$, $AC=7$, and $AB\\perp AC$, $EF$ is the perpendicular bisector of $BC$, and point $P$ is a moving point on line $EF$. What is the minimum perimeter of $\\triangle ABP$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, connect PC,\n\n$\\because$ AB=5,\n\n$\\therefore$ the perimeter of $\\triangle ABP$ is AB+PA+PB=5+PA+PB,\n\nTo minimize the perimeter of $\\triangle ABP$, it is required to minimize the value of PA+PB,\n\n$\\because$ EF is a perpendicular bisector of BC,\n\n$\\therefore$ PC=PB,\n\n$\\therefore$ PA+PB=PA+PC,\n\nAccording to the principle that the shortest distance between two points is a straight line, when points A, P, and C are collinear,\n\nthat is, when point P is on the side AC, PA+PC reaches its minimum value, which is AC,\n\n$\\therefore$ the minimum value of PA+PB is AC=7,\n\n$\\therefore$ the minimum perimeter of $\\triangle ABP$ is 5+7=\\boxed{12},",
    "image_path": "dataset/mm_math/images/MM_Math/52693980.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle B=90^\\circ$, $AC=13$, and $\\tan A=\\frac{5}{12}$. $CD$ is the median to side $AB$. $\\triangle ABC$ is rotated about point D, and points A, B, C correspond to points $A'$, $B'$, $C'$, respectively. $C'D$ intersects side $AC$ at point E. During the rotation, if $\\frac{AD}{AB}=\\frac{DE}{BC}$, then what is the value of $\\frac{AE}{AC}$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle B=90^\\circ$,\\\\\nit follows that $\\tan A=\\frac{BC}{AB}=\\frac{5}{12}$.\\\\\nLet $BC=5x$, then $AB=12x$,\\\\\nsince $AC^2=AB^2+BC^2$,\\\\\nwe find that $13^2=(12x)^2+(5x)^2$,\\\\\nresulting in: $x=1$ (negative values discarded),\\\\\nthus $BC=5$, and therefore $AB=12$.\\\\\nSince $CD$ is the midline on side $AB$,\\\\\nit follows that $AD=BD=\\frac{1}{2}AB=6$,\\\\\nthus $\\frac{AD}{AB}=\\frac{6}{12}=\\frac{1}{2}=\\frac{DE}{BC}$,\\\\\nonly when $\\triangle ADE\\sim \\triangle ABC$ does it satisfy $\\frac{AD}{AB}=\\frac{DE}{BC}=\\frac{1}{2}$, as shown in the figure,\\\\\nthus $\\frac{AE}{AC}=\\boxed{\\frac{1}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52917564.png"
  },
  {
    "solution": "\\boxed{y=-x+3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the line $l_{1}\\colon y=kx+b$ and the line $l_{2}\\colon y=mx+n$ intersect at point $A(1,2)$, the line $l_{2}$ intersects with the $y$-axis at point $B(0,3)$, and the line $l_{1}$ intersects with the $x$-axis at point $C(-1,0)$. What are the equations of the lines $l_{1}$ and $l_{2}$?\n",
    "completion": "\\textbf{Solution:} Let's consider the line $ l_{1}: y=kx+b$. It intersects the x-axis at point C ($-1$, 0) and intersects at point A ($1, 2$), then\n\\[\n\\begin{cases}\n0=-k+b \\\\\n2=k+b\n\\end{cases}\n\\]\nSolving these, we get\n\\[\n\\begin{cases}\nk=1 \\\\\nb=1\n\\end{cases}\n\\]\nTherefore, the equation of the line $ l_{1}$ is: $\\boxed{y=x+1}$.\n\nFor line $ l_{2}: y=mx+n$ which intersects at point A ($1, 2$) and intersects the y-axis at point B ($0, 3$), we have\n\\[\n\\begin{cases}\n2=m+n \\\\\n3=n\n\\end{cases}\n\\]\nSolving this gives\n\\[\n\\begin{cases}\nm=-1 \\\\\nn=3\n\\end{cases}\n\\]\nTherefore, the equation of the line $ l_{2}$ is $\\boxed{y=-x+3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52188903_157.png"
  },
  {
    "solution": "\\boxed{35^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=85^\\circ$, $\\angle B=30^\\circ$. Arcs are drawn with radii greater than half of the length of $AB$, centered at points A and B, respectively. The arcs intersect at points M and N, intersect $BC$ at point D. When $AD$ is connected, what is the measure of $\\angle CAD$?\n",
    "completion": "\\textbf{Solution:} Given that in $\\triangle ABC$, $\\angle B=30^\\circ$, $\\angle C=85^\\circ$,\\\\\nhence $\\angle BAC=180^\\circ-\\angle B-\\angle C=65^\\circ$,\\\\\nFrom the construction, it is known that $MN$ is the perpendicular bisector of $AB$,\\\\\nhence $DA=DB$,\\\\\nthus $\\angle DAB=\\angle B=30^\\circ$,\\\\\ntherefore, $\\angle CAD=\\angle BAC-\\angle DAB=\\boxed{35^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53080052.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: \\textit{As shown in the figure, in an equilateral $\\triangle ABC$, $BD$ is the bisector of $\\angle ABC$, point $E$ is the midpoint of $BC$, and point $P$ is a movable point on $BD$. Lines $PE$ and $PC$ are drawn. When the value of $PE + PC$ is at its minimum, what is the degree measure of $\\angle EPC$?}\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABC$ is an equilateral triangle, and BD is the bisector of $\\angle ABC$, \\\\\ntherefore, point D is the midpoint of AC, and BD$\\perp$AC, \\\\\nthus, points A and C are symmetric with respect to BD, \\\\\nas shown in the figure, draw AE, intersecting BD at P, the length of line segment AE is the minimum value of PE+PC, \\\\\nsince point E is the midpoint of side BC, \\\\\nthus AE$\\perp$BC, \\\\\nsince $\\angle ABC=60^\\circ$, and BD is the bisector of $\\angle ABC$, \\\\\nthus $\\angle PBE=30^\\circ$, \\\\\ntherefore, $\\angle BPE=60^\\circ$, \\\\\nsince in $\\triangle BPE$ and $\\triangle CPE$, \\\\\n$ \\left\\{\\begin{array}{l}PE=PE \\\\ \\angle PEB=\\angle PEC=90^\\circ \\\\ BE=CE \\end{array}\\right.$, \\\\\ntherefore, $\\triangle BPE\\cong \\triangle CPE$ (SAS), \\\\\ntherefore, $\\angle EPC=\\angle BPE=60^\\circ$. \\\\\nHence, the answer is: $\\boxed{60^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52836690.png"
  },
  {
    "solution": "\\boxed{\\frac{S}{6}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC$, $AD$ is the height from $A$ to side $BC$, point $E$ is on $AD$, and $AE=\\frac{1}{3}AD$. If the area of $\\triangle ABC$ is $s$, what is the area of $\\triangle ABE$?\n",
    "completion": "Given that in $\\triangle ABC$, $\\angle ACB=90^\\circ$ and AB=AC, with AD being the altitude on the side BC,\\\\\nit follows that AD is also the median on side BC, that is BD=DC,\\\\\nFurthermore, given that the area of $\\triangle ABC$ equals S,\\\\\nthus, the area of $\\triangle ABD$ equals $\\frac{S}{2}$,\\\\\nSince AE=$\\frac{1}{3}AD$,\\\\\nconsequently, the area of $\\triangle ABE$ equals $\\boxed{\\frac{S}{6}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51385685.png"
  },
  {
    "solution": "\\boxed{\\frac{15}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, $AC=8$, and $BC=6$. Points D and E are movable points on sides $BC$ and $AB$, respectively, with $BE=BD$. Connect $DE$ and fold $\\triangle BDE$ along $DE$ such that point B falls on point F. Connect $AF$. As shown in Figure 2, when point F is on side $AC$, what is the length of $BE$?\n",
    "completion": "\\textbf{Solution:} Solving: By the property of folding, we get $BE=BD=FE=FD$, \\\\\n$\\therefore$ quadrilateral $BEFD$ is a rhombus, \\\\\n$\\therefore$ $EF\\parallel BC$. \\\\\n$\\because$ point F is on side $AC$, \\\\\n$\\therefore$ $\\triangle AEF\\sim \\triangle ABC$, \\\\\n$\\therefore$ $\\frac{EF}{BC}=\\frac{AE}{AB}=\\frac{AF}{AC}$. \\\\\n$\\because$ $AC=8, BC=6, \\angle ACB=90^\\circ$, \\\\\n$\\therefore$ in $Rt\\triangle ABC$, $AB=\\sqrt{6^2+8^2}=10$. \\\\\nLet $BE=EF=x$, \\\\\nthen $\\frac{x}{6}=\\frac{10-x}{10}$, \\\\\n$\\therefore$ $x=\\frac{15}{4}$, \\\\\n$\\therefore$ $BE=\\boxed{\\frac{15}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52991872_104.png"
  },
  {
    "solution": "\\boxed{-x^2+2x+3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system, the vertices A and B of the square ABCD are on the x-axis, and the parabola $y=-x^2+bx+c$ passes through point A and $C(4, -5)$, and intersects with line DC at another point E. Determine the equation of the parabola.\n",
    "completion": "\\textbf{Solution}: Since quadrilateral ABCD is a square, and $C(4, -5)$,\n\nit follows that $AD=AB=5$, and $B(4, 0)$,\n\nthus $OA=1$,\n\ntherefore $A(-1, 0)$,\n\nSubstituting the coordinates of points A and C into $y=-{x}^{2}+bx+c$ yields: $\\begin{cases}-16+4b+c=-5 \\\\ -1-b+c=0\\end{cases}$,\n\nSolving this, we find: $\\begin{cases}b=2 \\\\ c=3\\end{cases}$,\n\nTherefore, the equation of the parabola is $y=-{x}^{2}+2x+3=\\boxed{-x^2+2x+3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52132998_101.png"
  },
  {
    "solution": "\\boxed{2.4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the figure, within $\\triangle ABC$, $\\angle C=90^\\circ$, $AC=4$, $BC=3$. Let point P be a moving point on the hypotenuse $AB$. Draw $PE\\perp AC$ at point E and $PF\\perp BC$ at point F. Connect $EF$. Then, the minimum value of the line segment $EF$ is\n",
    "completion": "\\textbf{Solution:} Draw $PC$,\\\\\nsince $PE\\perp AC$ and $PF\\perp BC$,\\\\\ntherefore $\\angle PEC = \\angle PFC = 90^\\circ$,\\\\\nthus quadrilateral $ECFP$ is a rectangle,\\\\\nhence $EF=PC$,\\\\\nthus, when $PC$ is at its minimum, $EF$ is also at its minimum,\\\\\nwhich is when $CP\\perp AB$, $PC$ is at its minimum,\\\\\nsince $AC=4$ and $BC=3$,\\\\\ntherefore $AB=5$,\\\\\nthus, the minimum value of $PC$ is: $ \\frac{AC\\cdot BC}{AB}=\\frac{4\\times 3}{5}=\\boxed{2.4}$\\\\\ntherefore, the minimum length of segment $EF$ is 2.4.",
    "image_path": "dataset/mm_math/images/MM_Math/53706851.png"
  },
  {
    "solution": "\\boxed{\\frac{6\\sqrt{10}}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=ax^2+2ax+c$ intersects the x-axis at points $A$ and $B$ (with point $A$ on the left side of point $B$), and intersects the y-axis at point $C(0, 3)$. The y-coordinate of its vertex $D$ is $4$. Point $F$ is located on the extension line of segment $CB$, and $\\angle AFC = \\angle DAB$. What is the length of $CF$?\n",
    "completion": "\\textbf{Solution:} Draw $DM\\perp x$-axis through point D, and draw $AN\\perp BC$ through point A,\\\\\n$\\because D(-1, 4)$, $\\because M(-1, 0)$,\\\\\n$\\because A\\left(-3, 0\\right)$,\\\\\n$\\therefore AM=2$, $DM=4$,\\\\\n$\\therefore \\tan\\angle DAB=\\frac{DM}{AM}=\\frac{4}{2}=2$,\\\\\n$\\because AN\\perp BC$,\\\\\n$\\therefore \\angle ANB=90^\\circ$,\\\\\n$\\therefore \\angle ANB=\\angle BOC$,\\\\\n$\\because \\angle ABN=\\angle OBC$,\\\\\n$\\therefore \\triangle OBC\\sim \\triangle NBA$,\\\\\n$\\therefore \\frac{OB}{NB}=\\frac{OC}{NC}=\\frac{BC}{BA}$,\\\\\n$\\therefore \\frac{1}{NB}=\\frac{3}{NC}=\\frac{\\sqrt{10}}{4}$,\\\\\n$NB=\\frac{2\\sqrt{10}}{5}$, $NA=\\frac{6\\sqrt{10}}{5}$,\\\\\n$\\because \\angle DAB=\\angle AFC$,\\\\\n$\\therefore \\tan\\angle DAB=\\tan\\angle AFC=2$,\\\\\n$\\therefore \\tan\\angle AFC=\\frac{AN}{NF}=2$,\\\\\n$\\therefore \\frac{\\frac{6\\sqrt{10}}{5}}{NF}=2$,\\\\\n$NF=\\frac{3\\sqrt{10}}{5}$,\\\\\nWhen point F is on the extension of CB, F can only be in the fourth quadrant, hence $CF=NF+CN$,\\\\\n$\\because BC=\\sqrt{10}NB=\\frac{2\\sqrt{10}}{5}$,\\\\\n$\\because CN=BC-NB=\\sqrt{10}-\\frac{2\\sqrt{10}}{5}=\\frac{3\\sqrt{10}}{5}$,\\\\\n$\\therefore CF=\\frac{3\\sqrt{10}}{5}+\\frac{3\\sqrt{10}}{5}=\\boxed{\\frac{6\\sqrt{10}}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51319005_106.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the diameter $CD$ of the circle $\\odot O$ is perpendicular to the chord $AB$ at point $E$, and $BE=4$, $DE=8$, then what is the length of $CE$?\n",
    "completion": "\\textbf{Solution:} Let the radius of the circle $\\odot O$ be $r$. Thus, $OB=r$ and $OE = DE - OD = 8 - r$,\\\\\nsince $AB \\perp CD$,\\\\\nin $\\triangle OEB$, we have $OB^{2} = OE^{2} + BE^{2}$,\\\\\nwhich leads to $r^{2} = (8 - r)^{2} + 4^{2}$,\\\\\nsolving this gives $r = 5$,\\\\\nthus, $CD = 2r = 10$,\\\\\ntherefore, $CE = CD - DE = 10 - 8 = \\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52991362.png"
  },
  {
    "solution": "\\boxed{37.5^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given in the figure, two acute vertices of a set squares are coincident, with $\\angle AOB=45^\\circ$, $\\angle COD=30^\\circ$. $OM$ and $ON$ are the bisectors of $\\angle AOC$ and $\\angle BOD$ respectively. When $\\angle COD$ is rotated around point $O$ to the position shown in the figure, and $\\angle BOC=n^\\circ$, what is the degree measure of $\\angle MON$?\n",
    "completion": "\\textbf{Solution:} Given $\\because \\angle BOC = n^\\circ$, then $\\angle AOC = 45^\\circ + n^\\circ$ and $\\angle BOD = 30^\\circ + n^\\circ$. Furthermore, OM and ON are the angle bisectors of $\\angle AOC$ and $\\angle BOD$ respectively, \\\\\n$\\therefore \\angle BON = \\frac{1}{2} \\angle BOD = \\frac{1}{2} (30^\\circ + n^\\circ) = 15^\\circ + \\frac{1}{2} n^\\circ$, \\\\\n$\\angle MOB = \\frac{1}{2} \\angle AOC - \\angle BOC = \\frac{1}{2} (45^\\circ + n^\\circ) - n^\\circ = 22.5^\\circ - \\frac{1}{2} n^\\circ$, \\\\\n$\\therefore \\angle MON = \\angle MOB + \\angle BON = 15^\\circ + \\frac{1}{2} n^\\circ + 22.5^\\circ - \\frac{1}{2} n^\\circ = \\boxed{37.5^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53195491_130.png"
  },
  {
    "solution": "\\boxed{\\left(-\\frac{4}{3}, \\frac{4}{3}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $x=t$ intersects with the line $y=-x$ and the line $y=\\frac{1}{2}x+2$ at points E and D (where D is above E), respectively. The line $y=-x$ and the line $y=\\frac{1}{2}x+2$ intersect at point M. What are the coordinates of point M?\n",
    "completion": "\\textbf{Solution:} The coordinates of point M are $\\boxed{\\left(-\\frac{4}{3}, \\frac{4}{3}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52395484_112.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that both quadrilaterals $ABCD$ and $CEFG$ are squares. Point $K$ is on $BC$, and $CD$ is extended to point $H$ such that $DH=BK=CE$. Connect $AK, KF, HF, AH$. $AK=AH$; Quadrilateral $AKFH$ is a square; If the area of quadrilateral $AKFH$ is $10$, and $CE=1$, what is the distance between points $A$ and $E$?\n",
    "completion": "\\textbf{Solution:} Since the area of quadrilateral $AKFH$ is 10, \\\\\nand from (2) we know that quadrilateral $AKFH$ is a square,\\\\\ntherefore $KF=\\sqrt{10}$,\\\\\nsince $EF=CE=1$,\\\\\ntherefore $KE=\\sqrt{KF^{2}-EF^{2}}=\\sqrt{10-1}=3$,\\\\\nthus $AB=KE=3$,\\\\\nsince $BK=EF=1$,\\\\\ntherefore $BE=BK+KE=4$,\\\\\nthus $AE=\\sqrt{AB^{2}+BE^{2}}=\\sqrt{3^{2}+4^{2}}=\\boxed{5}$,\\\\\nhence the distance between points $A$ and $E$ is \\boxed{5}.",
    "image_path": "dataset/mm_math/images/MM_Math/52189680_123.png"
  },
  {
    "solution": "\\boxed{\\frac{2}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\angle ABD = \\angle CBE = 90^\\circ$, $AB = BD$, $\\angle CAB = \\angle E$. If $BE = 10$, $AD = 4\\sqrt{2}$ , what is the value of $\\frac{AC}{DE}$?\n",
    "completion": "\\textbf{Solution:} \\\\\nSince $\\angle ABD=90^\\circ$ and $AB=BD$, $AD=4\\sqrt{2}$, \\\\\ntherefore $2AB^{2}=2BD^{2}=AD^{2}$, \\\\\nhence $2AB^{2}=32$, solving this yields: $AB=BD=4$ (taking the positive value); \\\\\nSince $\\angle ABD=\\angle CBE=90^\\circ$, \\\\\nit follows that $\\angle ABD-\\angle CBD=\\angle CBE-\\angle CBD$, \\\\\ntherefore $\\angle ABC=\\angle DBE$, \\\\\nand since $\\angle CAB=\\angle E$, \\\\\nit can be concluded that $\\triangle ABC \\sim \\triangle EBD$, \\\\\ntherefore $\\frac{AC}{DE}=\\frac{AB}{BE}=\\frac{4}{10}=\\boxed{\\frac{2}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52304570.png"
  },
  {
    "solution": "\\boxed{x \\in (-\\infty, -1) \\cup (3, +\\infty)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the functional relationship between $y$ and $x$ is as illustrated. Specifically, when $x \\geq 0$, $y = x$; when $x < 0$, $y = -2x + 1$. Then, what is the range of $x$ when the function value $y > 3$?\n",
    "completion": "\\textbf{Solution:} Substitute $y=3$ into $y=x$, then $x=3$.\\\\\nSubstitute $y=3$ into $y=-2x+1$ to get, $3=-2x+1$, solving this gives $x=-1$.\\\\\nTherefore, the points of intersection between the line $y=3$ and the graph of the function are $(3,3)$ and $(-1,3)$.\\\\\nObserving the graph, when the function value $y>3$, the range of $x$ is $x<-1$ or $x>3$,\\\\\nthus, the answer is: $\\boxed{x \\in (-\\infty, -1) \\cup (3, +\\infty)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50682699.png"
  },
  {
    "solution": "\\boxed{\\sqrt{6}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the rhombus $ABCD$, $P$ is a point on the diagonal $AC$. If $AB = 4$, $PA = 5$, and $PC = 2$, what is the length of $PB$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, connect BD, intersecting AC at point O,\\\\\nit is easy to obtain that BD perpendicularly bisects AC\\\\\n$\\because AC=7$\\\\\n$\\therefore AO=\\frac{7}{2}$, $PO=\\frac{3}{2}$,\\\\\nin $\\triangle ABO$ with a right angle at B, $BO=\\sqrt{4^2-\\left(\\frac{7}{2}\\right)^2}=\\frac{\\sqrt{15}}{2}$\\\\\n$\\therefore$ in $\\triangle PBO$ with a right angle at B, $PB=\\sqrt{BO^2+PO^2}=\\boxed{\\sqrt{6}}$",
    "image_path": "dataset/mm_math/images/MM_Math/52211277.png"
  },
  {
    "solution": "\\boxed{(-2, 6)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=ax^2+bx+4$ intersects the x-axis at points $A(-4, 0)$ and $B(1, 0)$, and intersects the y-axis at point C. Point P is a moving point on the parabola above line $AC$. Draw $PE\\perp AC$ at point E. Find the maximum value of $\\frac{\\sqrt{2}}{2}PE$ and the coordinates of point P at this time.\n",
    "completion": "\\textbf{Solution:} Given that the quadratic function $y=-x^2-3x+4$ intersects with the y-axis at point C,\\\\\nhence the coordinates of point C are $(0, 4)$,\\\\\nlet the equation of line $AC$ be $y=kx+4$,\\\\\nsince line $AC$ passes through point $A(-4, 0)$,\\\\\ntherefore $0=-4k+4$,\\\\\nsolving for $k$ gives $k=1$,\\\\\nthus the equation of line $AC$ is $y=-x+4$,\\\\\ndraw $PF\\perp x$-axis at point F, intersecting line $AC$ at point D,\\\\\nlet point $P(x, -x^2-3x+4)$, then $D(x, x+4)$,\\\\\nhence $PD=-x^2-4x=-(x+2)^2+4$,\\\\\nthus when $x=-2$, $PD$ is maximized, with the maximum value being $4$.\\\\\nSince $A(-4, 0)$, $C(0, 4)$,\\\\\ntherefore $OA=OC$,\\\\\nhence $\\angle OAC=45^\\circ$,\\\\\nsince $PF\\perp x$-axis,\\\\\nthus $\\angle ADF=\\angle PDE=45^\\circ$,\\\\\nsince $PE\\perp AC$,\\\\\ntherefore $\\triangle PDE$ is an isosceles right triangle,\\\\\nthus $\\sqrt{2}PE=PD$,\\\\\nhence $\\frac{\\sqrt{2}}{2}PE=\\frac{1}{2}PD$,\\\\\ntherefore the maximum value of $\\frac{\\sqrt{2}}{2}PE$ is $\\frac{1}{2}\\times 4=2$,\\\\\nat this moment, the coordinates of point P are $(-2, 6)$,\\\\\ntherefore, the maximum value of $\\frac{\\sqrt{2}}{2}PE$ is $\\boxed{2}$, with the coordinates of point P being $\\boxed{(-2, 6)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53107240_276.png"
  },
  {
    "solution": "\\boxed{2.5m}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, the nozzle B of an irrigation equipment is $1.25m$ above the ground. The parabolic water flow emitted reaches its maximum height of $2.25m$ above the ground at a distance of $1m$ from the bottom of the nozzle A, as illustrated in Figure 2 by establishing a Cartesian coordinate system. Find the horizontal distance between the nozzle and the point where the water flow touches the ground.\n",
    "completion": "\\textbf{Solution:} Let $y=-2.25$, i.e., $-{x}^{2}=-2.25$,\\\\\nSolving gives: $x_{1}=1.5$, $x_{2}=-1.5$ (discard),\\\\\n$1+1.5=2.5m$,\\\\\nThe horizontal distance between the nozzle and the point where the stream of water hits the ground is $\\boxed{2.5m}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52977560_185.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, quadrilateral $ABCD$ is inscribed in circle $O$, $\\angle ABC=135^\\circ$, $OE \\perp AC$, $\\angle AOE = \\angle D$. If $AC=4$, what is the radius of circle $O$?\n",
    "completion": "\\textbf{Solution:} From (1), we know that $\\triangle AOC$ is an isosceles right triangle, \\\\\nbecause $AC=4$, \\\\\ntherefore $OA=OC=2\\sqrt{2}$, \\\\\nwhich means the radius of the circle $O$ is $\\boxed{2\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52906012_83.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{1}{2}x-2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, in the Cartesian coordinate system $xOy$, the line $l_1$ intersects the $x$-axis at point $A\\left(m, 0\\right)$ and intersects the $y$-axis at point $B\\left(0, mn\\right)$ ($m<0$, $n>0$). If $m=-4$ and $n=\\frac{1}{2}$, what is the equation of the line $l_1$?\n",
    "completion": "\\textbf{Solution:} Given the problem, we know:\\\\\nPoint A ($-4$, $0$), Point B ($0$, $-2$),\\\\\nLet the equation of line $l_{1}$ be: $y=kx+b$,\\\\\n$\\begin{cases} b=-2 \\\\ -4k+b=0 \\end{cases}$, solving this gives: $\\begin{cases} b=-2 \\\\ k=-\\frac{1}{2} \\end{cases}$,\\\\\n$\\therefore y=-\\frac{1}{2}x-2$;\\\\\n$\\therefore$ The equation of line $l_{1}$ is $\\boxed{y=-\\frac{1}{2}x-2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51983386_179.png"
  },
  {
    "solution": "\\boxed{48}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the diagonals \\(AC\\) and \\(BD\\) of rhombus \\(ABCD\\) intersect at point \\(O\\). \\(E\\) is the midpoint of \\(CD\\). If \\(OE = 6\\), what is the perimeter of the rhombus?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rhombus, \\\\\n$\\therefore$ AC $\\perp$ BD, AB = BC = CD = DA, \\\\\n$\\therefore$ $\\triangle$COD is a right-angled triangle, \\\\\nSince OE = 6, and point E is the midpoint of the segment CD, \\\\\n$\\therefore$ CD = 2OE = 12, \\\\\n$\\therefore$ the perimeter of the rhombus ABCD = 4CD = $4\\times12 = \\boxed{48}$",
    "image_path": "dataset/mm_math/images/MM_Math/52395347.png"
  },
  {
    "solution": "\\boxed{80^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, quadrilateral $ABCD$ is inscribed in circle $\\odot O$, where $\\angle A=100^\\circ$. What is the measure of $\\angle C$?\n",
    "completion": "\\textbf{Solution}: Since quadrilateral ABCD is inscribed in a circle, \\\\\n$\\therefore$ $\\angle A + \\angle C = 180^\\circ$, \\\\\nSince $\\angle A = 100^\\circ$, \\\\\n$\\therefore$ $\\angle C = 180^\\circ - \\angle A = 180^\\circ - 100^\\circ = \\boxed{80^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/51545248.png"
  },
  {
    "solution": "\\boxed{\\frac{12}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, $AC=4$, $AB=5$. Point $D$ is on $AC$ and $AD=3$, $DE\\perp AB$ at point $E$. Find the length of $AE$.\n",
    "completion": "\\textbf{Solution:} Since DE$\\perp$AB at point E, and $\\angle C=90^\\circ$,\\\\\nit follows that $\\angle$AED＝$\\angle$C,\\\\\nsince $\\angle$A＝$\\angle$A,\\\\\nit follows that $\\triangle$ ADE$\\sim$ $\\triangle$ ABC,\\\\\nthus, $\\frac{AD}{AB}=\\frac{AE}{AC}$,\\\\\n$\\frac{3}{5}=\\frac{AE}{4}$,\\\\\nAE＝$\\boxed{\\frac{12}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51350897.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated, points A, B, C are located at the intersections of the grid lines of a square. If $\\triangle ABC$ is rotated counterclockwise around point A to obtain $\\triangle AC'B'$, then what is the value of $\\tan B''$?\n",
    "completion": "\\textbf{Solution:} To solve, rotate $\\triangle ABC$ counter-clockwise about point A to get $\\triangle AC'B'$. Based on the characteristic of rotation, $\\angle B=\\angle B'$. In $\\triangle ABC$, points A, B, and C are at the intersections of the square grid lines, thus $\\tan B' = \\tan B= \\boxed{\\frac{1}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51213313.png"
  },
  {
    "solution": "\\boxed{y=x^{2}-2x-3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola intersects the x-axis at points $A$ and $B(3, 0)$, and intersects the y-axis at point $C(0, -3)$. The vertex of the parabola is $D$. What is the analytical equation of the parabola?\n",
    "completion": "Solution: (1) Substituting point B$(3,0)$ and C$(0,-3)$ into $y=ax^{2}-2x+c$, we get\n\\[\n\\left\\{\n\\begin{array}{l}\n9a-6+c=0\\\\\nc=-3\n\\end{array}\n\\right.\n\\]\n$\\therefore$\n\\[\n\\left\\{\n\\begin{array}{l}\na=1\\\\\nc=-3\n\\end{array}\n\\right.\n\\]\n$\\therefore$ The equation of the parabola is $\\boxed{y=x^{2}-2x-3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52733221_71.png"
  },
  {
    "solution": "\\boxed{\\frac{36}{11}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle B=90^\\circ$, $AB=12\\,cm$, and $BC=16\\,cm$. Point $P$ starts from point $A$ and moves towards point $B$ along side $AB$ at a speed of $1\\,cm/s$. Point $Q$ starts from point $B$ and moves towards point $C$ along side $BC$ at a speed of $2\\,cm/s$. If $P$ and $Q$ start moving from points $A$ and $B$ simultaneously, and when one point reaches the end, the other point also stops moving. Let the moving time be $t$ seconds. Explore after how many seconds, the triangle with vertices $B$, $P$, $Q$ is similar to $\\triangle CBA$?\n",
    "completion": "\\textbf{Solution:} Let's assume after $x$ seconds, the two triangles become similar.\\\\\n(1) When $\\frac{BP}{BA}=\\frac{BQ}{BC}$, it implies $\\triangle BPQ\\sim \\triangle BAC$, which gives $\\frac{12-x}{12}=\\frac{2x}{16}$,\\\\\nSolving this, we get: $x=\\boxed{\\frac{24}{5}}$,\\\\\n(2) When $\\frac{BP}{BC}=\\frac{BQ}{BA}$, it implies $\\triangle BQP\\sim \\triangle BAC$, which gives $\\frac{12-x}{16}=\\frac{2x}{12}$,\\\\\nSolving this, we get: $x=\\boxed{\\frac{36}{11}}$,\\\\\nTherefore, after $\\boxed{\\frac{24}{5}}$ seconds or $\\boxed{\\frac{36}{11}}$ seconds, the triangle formed by the points $B$, $P$, $Q$ becomes similar to $\\triangle CBA$.",
    "image_path": "dataset/mm_math/images/MM_Math/52622502_116.png"
  },
  {
    "solution": "\\boxed{20}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the parallelogram \\(ABCD\\), the diagonals \\(AC\\) and \\(BD\\) intersect at point \\(O\\). Point \\(E\\) is the midpoint of \\(AD\\). If \\(OE = 10\\), what is the length of \\(AB\\)?\n",
    "completion": "\\textbf{Solution:} Since the quadrilateral $ABCD$ is a parallelogram, \\\\\n$\\therefore OB=OD$, which means point $O$ is the midpoint of $BD$, \\\\\nGiven that point $E$ is the midpoint of $AD$, \\\\\n$\\therefore OE=\\frac{1}{2}AB$, \\\\\nGiven $OE=10$, \\\\\n$\\therefore AB=2OE=\\boxed{20}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53359990.png"
  },
  {
    "solution": "\\boxed{30}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AE$ is the median to side $BC$. A line is drawn through point $C$ such that $CD \\perp AE$, and it intersects the extension line of $AE$ at point $D$. Line $BD$ is drawn. If $AB = BD$ and the area of $\\triangle BCD$ is 10, what is the area of $\\triangle ABC$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, construct $BF\\perp AE$ at point $F$,\n\nsince $CD\\perp AE$,\n\n$\\therefore \\angle BFE=\\angle CDE=90^\\circ$,\n\nsince $AE$ is the median to side $BC$,\n\n$\\therefore BE=CE$,\n\nIn $\\triangle BEF$ and $\\triangle CED$,\n\n$\\left\\{\\begin{array}{l}\n\\angle BFE=\\angle CDE \\\\\n\\angle BEF=\\angle CED \\\\\nBE=CE\n\\end{array}\\right.$\n\n$\\therefore \\triangle BEF \\cong \\triangle CED$ (AAS),\n\n$\\therefore FE=DE$, $BF=CD$,\n\nsince $AB = BD$,\n\n$\\therefore BF= DF$,\n\nLet $FE=DE=m$, $BF=CD=n$, then $AF= DF= 2FE= 2m$,\n\n$\\therefore AE=AF+FE= 3m$,\n\nsince $S_{\\triangle BCD}=S_{\\triangle BDE}+S_{\\triangle CDE}=\\frac{1}{2}DE\\cdot BF+\\frac{1}{2}DE\\cdot CD=\\frac{1}{2}mn+\\frac{1}{2}mn=mn$\n\nand $S_{\\triangle BCD}=10$,\n\n$\\therefore mn=10$,\n\n$\\therefore S_{\\triangle ABC}=S_{\\triangle ABE}+S_{\\triangle ACE}=\\frac{1}{2}AE\\cdot BF+\\frac{1}{2}AE\\cdot CD=\\frac{1}{2}\\times 3mn+\\frac{1}{2}\\times 3mn=3mn=\\boxed{30}$",
    "image_path": "dataset/mm_math/images/MM_Math/52836829.png"
  },
  {
    "solution": "\\boxed{y= -\\frac{1}{2}x+8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, a rectangular paper ABCO is placed in a Cartesian coordinate system, where BC = 10. The paper is folded so that point B precisely lands on the x-axis, denoted as $B'$, with CE being the crease. Given that the ratio $OC:OB'=4:3$, what is the equation of the line on which the crease CE lies?\n",
    "completion": "\\textbf{Solution:} After folding the paper, point B lands precisely on point B' on the x-axis, with CE being the fold line. Therefore, $\\triangle CBE \\cong \\triangle CB^{\\prime}E$, which implies that BE=B$^{\\prime}E$, and CB$^{\\prime}=CB=OA$. Given OB$^{\\prime}=6$ and the ratio OC:OB'=4:3, it follows that OC=8. Let AE=a, then EB$^{\\prime}=EB=8-a$, and AB$^{\\prime}=AO-OB^{\\prime}=10-6=4$. By the Pythagorean theorem, we obtain $a^{2}+4^{2}=(8-a)^{2}$. Solving for a yields $a=3$. Therefore, the coordinates of point E are $(10,3)$, and the coordinates of point C are $(0,8)$. Let the equation of line CE be $y=kx+b$. According to the given conditions, we have\n\\[\n\\begin{cases}\n3=10k+b\\\\ \n8=b\n\\end{cases}\n\\]\nSolving this, we find:\n\\[\n\\begin{cases}\nk=-\\frac{1}{2}\\\\ \nb=8\n\\end{cases}\n\\]\nTherefore, the equation of line CE is $y= -\\frac{1}{2}x+8$, so the answer is $\\boxed{y= -\\frac{1}{2}x+8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50677242_154.png"
  },
  {
    "solution": "\\boxed{a\\tan\\beta + a\\tan\\alpha}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the horizontal distance between teaching buildings $AB$ and $CD$ is $BD=a\\,m$. At point $C$ on top of teaching building $CD$, the angle of elevation to point $A$ on top of teaching building $AB$ is measured to be $\\alpha$, and the angle of depression to point $B$ at the bottom of teaching building $AB$ is measured to be $\\beta$. What is the height of teaching building $AB$?\n",
    "completion": "\\textbf{Solution:} Construct $CE\\perp AB$,\\\\\nsince $BD=a\\,m$,\\\\\nthus $CE=BD=a\\,m$,\\\\\nin $\\triangle BCE$, $BE=CE\\tan \\beta = a\\tan\\beta$,\\\\\nin $\\triangle ACE$, $AE=CE\\tan \\alpha = a\\tan\\alpha$,\\\\\ntherefore $AB=a\\tan\\beta + a\\tan\\alpha$,\\\\\nhence the height of $AB$ is $\\boxed{a\\tan\\beta + a\\tan\\alpha}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52797152.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, $M$ is on side $AD$, $E$ and $F$ are the midpoints of $BM$ and $CM$ respectively. If $AM = 2DM$, then what is the value of $\\frac{EF}{DM}$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a parallelogram, \\\\\nwe have BC=AD, \\\\\nand since AM=2DM, \\\\\nit follows that AD=3DM, that is BC=3DM, \\\\\nsince E and F are the midpoints of BM and CM respectively, \\\\\nthen EF is the midsegment of $\\triangle$BCM, \\\\\ntherefore EF=$\\frac{1}{2}$BC=$\\frac{3}{2}$DM, \\\\\nthus $\\frac{EF}{DM}=\\boxed{\\frac{3}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52303855.png"
  },
  {
    "solution": "\\boxed{9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ and $\\triangle A'B'C'$ are similar, with the center of similarity at point O, $OA=2AA'$, and the area of $\\triangle ABC$ is 4. What is the area of $\\triangle A'B'C'$?\n",
    "completion": "\\textbf{Solution:} Since $OA = 2AA'$,\\\\\nit follows that $\\frac{OA}{OA'}=\\frac{2}{3}$,\\\\\nSince $\\triangle ABC$ and $\\triangle A'B'C'$ are similar,\\\\\nit implies that $\\triangle ABC\\sim \\triangle A'B'C'$, and $AB\\parallel A'B'$,\\\\\ntherefore $\\triangle OAB\\sim \\triangle OA'B'$,\\\\\nwhich leads to $\\frac{AB}{A'B'}=\\frac{OA}{OA'}=\\frac{2}{3}$,\\\\\nthus $\\frac{S_{\\triangle OAB}}{S_{\\triangle OA'B'}}=\\left(\\frac{2}{3}\\right)^2=\\frac{4}{9}$,\\\\\nGiven that the area of $\\triangle ABC$ is 4,\\\\\nit follows that the area of $\\triangle A'B'C'$ is $\\boxed{9}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53080091.png"
  },
  {
    "solution": "\\boxed{1.5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the graph of the linear function $y=-x+m$ intersects with the graph of the inverse proportion function $y=\\frac{k}{x}$ ($x>0$) at points $A$ and $B$. It is known that $A(1,2)$. Connect $AO$ and $BO$, what is the area of $\\triangle AOB$?\n",
    "completion": "\\textbf{Solution:} Solve the system of equations\n\\[\n\\left\\{\\begin{array}{l}\ny=-x+3\\\\ \ny=\\frac{2}{x}\n\\end{array}\\right.\n\\]\nto get\n\\[\n\\left\\{\\begin{array}{l}\nx=1\\\\ \ny=2\n\\end{array}\\right.\n\\]\nor\n\\[\n\\left\\{\\begin{array}{l}\nx=2\\\\ \ny=1\n\\end{array}\\right.\n\\]\n$\\therefore$ For point B $(2, 1)$, let the intersection of the line $y=-x+3$ with the $y$-axis be point C, $\\therefore$ C $(0, 3)$. Therefore, the area of $\\triangle AOB$ equals the area of $\\triangle COB$ minus the area of $\\triangle COA$, which is $\\frac{1}{2}\\times3\\times2-\\frac{1}{2}\\times3\\times1=\\boxed{1.5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52694023_95.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the right-angled triangle $ABC$, where $\\angle C=90^\\circ$, the bisector of $\\angle CAB$ intersects $BC$ at point $D$, and $DE$ is the perpendicular bisector of $AB$ with the foot of the perpendicular being $E$. What is the measure of $\\angle CAD$?\n",
    "completion": "\\textbf{Solution:} Since DE is the perpendicular bisector of AB,\\\\\nit follows that AD=BD,\\\\\nhence, $\\angle B=\\angle EAD$,\\\\\nmoreover, since AD is the bisector of $\\angle CAB$,\\\\\nit follows that $\\angle CAD=\\angle EAD$,\\\\\nlet $\\angle CAD=x$, then $3x=90^\\circ$,\\\\\ntherefore, $x=30^\\circ$,\\\\\nthus, $\\angle CAD=\\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52095924_110.png"
  },
  {
    "solution": "\\boxed{6\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, $\\angle A=60^\\circ$, and $AC=6$. The triangle $\\triangle ABC$ is rotated counterclockwise around point C to obtain $\\triangle A'B'C'$, where point A' happens to be on side AB. What is the distance between point B' and point B?\n",
    "completion": "\\textbf{Solution:} Draw line $B'B$.\\\\\n$\\because$ Triangle $ABC$ is rotated counterclockwise around point $C$ to obtain $\\triangle A'B'C$,\\\\\n$\\therefore$ $AC=A'C$, $AB=A'B$, $\\angle A=\\angle CA'B'=60^\\circ$,\\\\\n$\\therefore$ $\\triangle AA'C$ is an equilateral triangle,\\\\\n$\\therefore$ $\\angle AA'C=60^\\circ$,\\\\\n$\\therefore$ $\\angle B'A'B=180^\\circ-60^\\circ-60^\\circ=60^\\circ$,\\\\\n$\\because$ Triangle $ABC$ is rotated counterclockwise around point $C$ to obtain $\\triangle A'B'C$,\\\\\n$\\therefore$ $\\angle ACA'=\\angle BAB'=60^\\circ$, $BC=B'C$, $\\angle CB'A'=\\angle CBA=90^\\circ-60^\\circ=30^\\circ$,\\\\\n$\\therefore$ $\\triangle BCB'$ is an equilateral triangle,\\\\\n$\\therefore$ $\\angle CB'B=60^\\circ$,\\\\\n$\\because$ $\\angle CB'A'=30^\\circ$,\\\\\n$\\therefore$ $\\angle A'B'B=30^\\circ$,\\\\\n$\\therefore$ $\\angle B'BA'=180^\\circ-60^\\circ-30^\\circ=90^\\circ$,\\\\\n$\\because$ $\\angle ACB=90^\\circ$, $\\angle A=60^\\circ$, $AC=6$,\\\\\n$\\therefore$ $AB=12$,\\\\\n$\\therefore$ $A'B=AB−AA'=AB−AC=6$,\\\\\n$\\therefore$ $B'B=6\\sqrt{3}$,\\\\\nTherefore, the answer is: \\boxed{6\\sqrt{3}}.",
    "image_path": "dataset/mm_math/images/MM_Math/51212179.png"
  },
  {
    "solution": "\\boxed{33}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the perimeter of $\\triangle ABC$ is 22, $PB$ and $PC$ bisect $\\angle ABC$ and $\\angle ACB$ respectively, $PD \\perp BC$ at D, and $PD = 3$, what is the area of $\\triangle ABC$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, connect AP, and through point P, draw PE$\\perp$AB at point E, and PF$\\perp$AC at point F,\\\\\n$\\because$ PB and PC bisect $\\angle ABC$ and $\\angle ACB$ respectively,\\\\\n$PD\\perp BC$ at D,\\\\\n$\\therefore$ PD=PE, PD=PF,\\\\\n$\\therefore$ PD=PE=PF=3,\\\\\n$\\because$ the perimeter of $\\triangle ABC$ is 22,\\\\\n$\\therefore$ the area of $\\triangle ABC$ is\\\\\n$\\frac{1}{2}AB\\times PE+\\frac{1}{2}BC\\times PD+\\frac{1}{2}AC\\times PF=\\frac{1}{2}\\times PD\\times (AB+BC+AC)=\\frac{1}{2}\\times 3\\times 22=\\boxed{33}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51435002.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, point $D$ is on side $AB$, $E$ is the midpoint of side $AC$, $CF \\parallel AB$, and $CF$ intersects the extension line of $DE$ at point $F$. If $AB=4$ and $CF=3$, what is the length of $BD$?\n",
    "completion": "\\textbf{Solution:} Given that $CF\\parallel AB$, \\\\\nthus $\\angle A=\\angle ACF$, $\\angle F=\\angle ADE$, \\\\\nsince E is the midpoint of side $AC$, \\\\\nthen $AE=CE$, \\\\\nin $\\triangle ADE$ and $\\triangle CFE$, \\\\\n$\\left\\{\\begin{array}{l}\n\\angle A=\\angle ACF \\\\\n\\angle ADE=\\angle F \\\\\nAE=CE\n\\end{array}\\right.$, \\\\\ntherefore $\\triangle ADE\\cong \\triangle CFE$ (AAS), \\\\\nthus $AD=CF=3$, \\\\\nhence $BD=AB-AD=4-3=\\boxed{1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53106541.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in $\\triangle ABC$, $\\angle C=90^\\circ$, $AC=4$, and $BC=8$, then what is the value of $\\tan B$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle C=90^\\circ$, $AC=4$, and $BC=8$, \\\\\ntherefore $\\tan B=\\frac{AC}{BC}=\\boxed{\\frac{1}{2}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51082675.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that point C is a point on line segment AB, with $AC=12\\, \\text{cm}$ and $CB=\\frac{2}{3}AC$. D is the midpoint of AB. Find the length of DC.\n",
    "completion": "Solution:\\\\\nSince $AC=12$\\\\\nThen $CB=\\frac{2}{3}AC=\\frac{2}{3}\\times 12=8$\\\\\nThen $AB=AC+CB=12+8=20$\\\\\nAlso, since $D$ is the midpoint of $AB$\\\\\nThen $DB=\\frac{1}{2}AB=\\frac{1}{2}\\times 20=10$\\\\\nTherefore $DC=DB-CB=10-8=\\boxed{2}$",
    "image_path": "dataset/mm_math/images/MM_Math/52795906.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that in the parallelogram $ABCD$, $G$ is a point on the extension line of $AB$. Connect $DG$, which intersects $AC$ and $BC$ at points $E$ and $F$ respectively, and $AE: EC = 3: 2$. If $AB = 10$, what is the length of $BG$?\n",
    "completion": "Solution: Since quadrilateral ABCD is a parallelogram, \\\\\n$\\therefore$AB$\\parallel$CD, \\\\\n$\\therefore$$\\angle GAE=\\angle CDE$, $\\angle AGE=\\angle CDE$, \\\\\n$\\therefore$$\\triangle AGE\\sim \\triangle CDE$, \\\\\n$\\therefore$$\\frac{AG}{CD}=\\frac{AE}{CE}=\\frac{3}{2}$, \\\\\nFurthermore, since AB=CD=10, \\\\\n$\\therefore$AG=$\\frac{3}{2}$CD=$\\frac{3}{2}\\times 10=15$, \\\\\n$\\therefore$BG=AG−AB=15−10=\\boxed{5};",
    "image_path": "dataset/mm_math/images/MM_Math/51379520_100.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that in $\\triangle ABC$, $\\angle C = 90^\\circ$, $AB = 4$, and $AC = \\sqrt{7}$. Find the length of $BC$.\n",
    "completion": "\\textbf{Solution:} In $\\triangle ABC$, $\\angle C=90^\\circ$, $AB=4$, and $AC=\\sqrt{7}$.\\\\\n$BC=\\sqrt{AB^2-AC^2}=\\sqrt{4^2-(\\sqrt{7})^2}=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51262135_108.png"
  },
  {
    "solution": "\\boxed{-10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point A is on the curve $y_1=\\frac{2}{x}$ ($x>0$), and point B is on the hyperbola $y_2=\\frac{k}{x}$ ($x<0$), with AB parallel to the x-axis. Point C is a point on the x-axis, and lines AC and BC are connected. If the area of $\\triangle ABC$ is 6, what is the value of $k$?\n",
    "completion": "\\textbf{Solution:} Let the coordinates of point $A$ be $\\left(a,\\frac{2}{a}\\right)$, then the coordinates of point $B$ are $\\left(\\frac{ak}{2},\\frac{2}{a}\\right)$\\\\\nThus, ${S}_{\\triangle ABC}=\\frac{1}{2}\\times \\left|a-\\frac{ak}{2}\\right|\\times \\frac{2}{a}=6$\\\\\nWe find $k=-10$, $k=14$ (discard)\\\\\nTherefore, the answer is: $k=\\boxed{-10}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51379286.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle B=90^\\circ$, $\\angle C=30^\\circ$, $DE$ is perpendicular to and bisects $AC$, intersecting $BC$ at point $E$, with $CE=2$, then how much is $BC$?\n",
    "completion": "\\textbf{Solution:} Since DE is the perpendicular bisector of AC, and CE=2,\\\\\nit follows that AE=CE,\\\\\nwhich means $\\angle C=\\angle$EAC$=30^\\circ$,\\\\\ntherefore $\\angle$BAE$=30^\\circ$,\\\\\nhence $EB=\\frac{1}{2}EA=1$,\\\\\nthus BC=BE+EC$=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51368725.png"
  },
  {
    "solution": "\\boxed{x>2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the straight line $y=kx+b$ ($k\\neq 0$) passes through points $(2,0)$ and $(0,-3)$. What is the range of values of $x$ when $y>0$?\n",
    "completion": "\\textbf{Solution:} Let the equation of the line be $y=kx+b$. This line intersects the x-axis at the point $(2,0)$, and it passes through the first, third, and fourth quadrants. From the graph, it is evident that\\\\\nwhen $x>2$, the value of $y$ is greater than $0$,\\\\\ntherefore, the range of $x$ is $\\boxed{x>2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51663244.png"
  },
  {
    "solution": "\\boxed{16}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the quadrilateral $ABCD$, where $AB\\parallel CD$ and $\\angle BCD=90^\\circ$, with $AB=AD=10\\text{cm}$ and $BC=8\\text{cm}$. Point $P$ starts moving from point $A$, moving at a constant speed of $3\\text{cm/s}$ along the polyline in the direction of $ABCD$; point $Q$ starts moving from point $D$, moving at a constant speed of $2\\text{cm/s}$ along the line segment in the direction of $DC$. It is known that both points start moving at the same time, and when one point reaches its endpoint, the other point also stops moving. Let the movement time be $t\\text{(s)}$. Find the length of $CD$, in cm.\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $AE\\perp DC$ at E,\\\\\nsince $AB\\parallel CD$ and $\\angle BCD=90^\\circ$, with $AB=AD=10\\text{cm}$ and $BC=8\\text{cm}$, and $AE\\perp DC$,\\\\\nthus, quadrilateral $ABCE$ is a rectangle, that is, $AE=BC=8$ and $AB=AD=CE=10$,\\\\\ntherefore, in $\\triangle ADE$, $DE=\\sqrt{AD^{2}-AE^{2}}=\\sqrt{10^{2}-8^{2}}=6$,\\\\\nthus, $CD=DE+EC=6+10=\\boxed{16}$,\\\\\nhence, the length of $CD$ is $16\\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53490651_86.png"
  },
  {
    "solution": "\\boxed{17}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, the goalkeeper kicks a high ball from point $O$, and the ball starts its flight from point $A$, 1 meter above the ground ($A$ is on the $y$-axis). The player Yi discovers that the ball reaches its highest point, $M$, right above his head at a height of approximately 4 meters from the ground, and this point is 6 meters away from $O$ at point $B$. After landing, the ball bounces again. According to experimental measurements, the parabolic trajectory of the ball after bouncing on the lawn retains the same shape as the original trajectory, but its maximum height is reduced to half of the original height. To reach the second landing point $D$, how much further does the player Yi need to run?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, the distance after the soccer ball bounces for the second time is $CD$.\\\\\nAccording to the problem: $CD=EF$ (which means the parabola $AEMFC$ is translated downward by $2$ units).\\\\\n$\\therefore 2=-\\frac{1}{12}(x-6)^{2}+4$ solving this gives $x_{1}=6-2\\sqrt{6}$, $x_{2}=6+2\\sqrt{6}$.\\\\\n$\\therefore CD=|x_{1}-x_{2}|=4\\sqrt{6}\\approx 10$.\\\\\n$\\therefore BD=13-6+10=17$ (meters).\\\\\nAnswer: He should run another $\\boxed{17}$ meters forward.",
    "image_path": "dataset/mm_math/images/MM_Math/52977497_176.png"
  },
  {
    "solution": "\\boxed{100^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the inscribed angle $\\angle ACB = 130^\\circ$, then what is the degree measure of the central angle $\\angle AOB$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, take point D on the major arc AB, and connect AD and BD,\\\\\n$\\because$ Quadrilateral ACBD is a cyclic quadrilateral,\\\\\n$\\therefore$ $\\angle$D=180$^\\circ$−$\\angle$ACB=180$^\\circ$−130$^\\circ$=50$^\\circ$,\\\\\n$\\because$ $ \\overset{\\frown}{AB}=\\overset{\\frown}{AB}$  , \\\\\n$\\therefore$ $\\angle$AOB=2$\\angle$D=2×50$^\\circ$=\\boxed{100^\\circ}.",
    "image_path": "dataset/mm_math/images/MM_Math/52861153.png"
  },
  {
    "solution": "\\boxed{55^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\angle ACD$ is the exterior angle of $\\triangle ABC$, and $CE$ bisects $\\angle ACD$. If $\\angle A=70^\\circ$ and $\\angle B=40^\\circ$, what is the measure of $\\angle ECD$?\n",
    "completion": "\\textbf{Solution:}\\\\\nSince $\\angle A=70^\\circ$ and $\\angle B=40^\\circ$,\\\\\nit follows that $\\angle ACD=\\angle A+\\angle B=110^\\circ$,\\\\\nand since CE bisects $\\angle ACD$,\\\\\nwe have $\\angle ECD= \\frac{1}{2}\\angle ACD=\\boxed{55^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52655062.png"
  },
  {
    "solution": "\\boxed{\\frac{16}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $\\triangle ABC \\sim \\triangle AMN$, where point M is the midpoint of AC, $AB=6$, and $AC=8$. What is the length of $AN$?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABC \\sim \\triangle AMN$, \\\\\nthen $\\frac{AB}{AM} = \\frac{AC}{AN}$, \\\\\nSince M is the midpoint of AC, and AB$=6$, AC$=8$, \\\\\nthen AM$=$MC$=4$, \\\\\nthus $\\frac{6}{4} = \\frac{8}{AN}$, \\\\\nsolving gives AN$=\\boxed{\\frac{16}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51351735.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points A and B are on the graph of the linear function $y=kx+1$, where point A is in the first quadrant and point B is on the x-axis. Point D is on the positive half of the x-axis, and the coordinates of point C are $(1, -2)$. Quadrilateral OADC is a rhombus. What is the value of $k$?\n",
    "completion": "\\textbf{Solution:} Given the coordinates of point C as $\\left(1, -2\\right)$, and using the properties of a rhombus,\\\\\n$\\therefore$ the coordinates of point A are $\\left(1, 2\\right)$, and of D $\\left(2, 0\\right)$,\\\\\nsubstituting point A $\\left(1, 2\\right)$ into $y=kx+1$ gives $k+1=2$,\\\\\nsolving for $k$ yields $k=\\boxed{1}$;\\\\\n$\\therefore$ the equation of line AB is $y=x+1$.",
    "image_path": "dataset/mm_math/images/MM_Math/52211814_80.png"
  },
  {
    "solution": "\\boxed{15}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the figure, in $ \\triangle ABC$, $ \\angle BAC=90^\\circ$, AD is the altitude to side BC, if $\\sin\\angle CAD=\\frac{3}{5}$ and $ BC=25$, find the length of AC.\n",
    "completion": "\\textbf{Solution:} According to the problem, $\\because$ $\\angle BAC=90^\\circ$ and AD is the height from A to side BC, $\\therefore$ $\\angle BAD+\\angle CAD=90^\\circ$ and $\\angle BAD+\\angle B=90^\\circ$, $\\therefore$ $\\angle CAD=\\angle B$, $\\therefore$ $\\sin B=\\sin CAD=\\frac{3}{5}$. In $Rt\\triangle ABC$ with $\\angle BAC=90^\\circ$, $\\therefore$ $\\sin B=\\frac{AC}{BC}=\\frac{AC}{25}=\\frac{3}{5}$,\\\\\n$\\therefore$ $AC=\\boxed{15}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51032247.png"
  },
  {
    "solution": "\\boxed{y=2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y_{1}=2x-2$ intersects the $y$ axis at point $A$, and the line $y_{2}=-2x+6$ intersects the $y$ axis at point $B$. The two lines intersect at point $C$. Solve the system of equations $\\left\\{\\begin{array}{l}2x-y=2 \\\\ 2x+y=6 \\end{array}\\right.$.\n",
    "completion": "\\textbf{Solution:} The solution to the system of equations $\\left\\{\\begin{array}{l}2x-y=2 \\\\ 2x+y=6 \\end{array}\\right.$ is: $\\left\\{\\begin{array}{l}x=\\boxed{2} \\\\ \\boxed{y=2} \\end{array}\\right.$.",
    "image_path": "dataset/mm_math/images/MM_Math/52422271_65.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points C and D are two points on line segment AB, with AC:BC = 3:2, and point D being the midpoint of AB. If AB = 40, find the length of line segment CD.\n",
    "completion": "\\textbf{Solution}: Since $AC:BC=3:2$ and $AB=40$, \\\\\nit follows that $AC=40\\times \\frac{3}{3+2}=24$, \\\\\nsince point D is the midpoint of AB, \\\\\nit follows that $AD=\\frac{1}{2}AB=20$, \\\\\ntherefore $CD=AC-AD=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51354427_193.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, within the square $ABCD$, $BD$ is the diagonal. Point $E$ is the midpoint of $CD$, and through point $E$ a line $EF\\perp BD$ is drawn, where the foot of the perpendicular is point $F$. If $BC= 2\\sqrt{2}$, what is the length of $EF$?\n",
    "completion": "\\textbf{Solution:} In the square $ABCD$, we have $CD=BC=2\\sqrt{2}$ and $\\angle C=90^\\circ$,\\\\\n$\\therefore$ $BD=\\sqrt{BC^2+CD^2}=4$,\\\\\n$\\because$ point $E$ is the midpoint of $CD$,\\\\\n$\\therefore$ $CE=DE=\\sqrt{2}$,\\\\\n$\\therefore$ $BE=\\sqrt{BC^2+CE^2}=\\sqrt{10}$,\\\\\nLet $DF=x$, then $BF=4-x$,\\\\\n$\\because$ $EF\\perp BD$,\\\\\n$\\therefore$ $\\angle DFE=\\angle BFE=90^\\circ$,\\\\\n$\\therefore$ $DE^2-DF^2=EF^2=BE^2-BF^2$,\\\\\n$\\therefore$ $\\left(\\sqrt{10}\\right)^2-\\left(4-x\\right)^2=\\left(\\sqrt{2}\\right)^2-x^2$,\\\\\nSolving gives: $x=1$,\\\\\ni.e., $DF=1$,\\\\\n$\\therefore$ $EF=\\sqrt{DE^2-DF^2}=\\boxed{1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52160170.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in $\\triangle ABC$, $D$ and $E$ are points on sides $AB$ and $AC$ respectively, with $DE \\parallel BC$. If $\\frac{S_{\\triangle ADE}}{S_{\\text{quadrilateral } DBCE}} = 1:8$, what is the value of $\\frac{DE}{BC}$?\n",
    "completion": "\\textbf{Solution:} Since the ratio of the area of $\\triangle ADE$ to the area of quadrilateral DBCE is $1:8$, \\\\\nit follows that the area of $\\triangle ADE$ is $\\frac{1}{9}$ the area of $\\triangle ABC$, \\\\\nthus $\\frac{{S}_{\\triangle ADE}}{{S}_{\\triangle ABC}}=\\frac{1}{9}$. \\\\\nSince $DE \\parallel BC$, \\\\\nthen $\\triangle ADE \\sim \\triangle ABC$, \\\\\nthus $\\frac{{S}_{\\triangle ADE}}{{S}_{\\triangle ABC}}=\\left(\\frac{DE}{BC}\\right)^{2}$, \\\\\nhence $\\left(\\frac{DE}{BC}\\right)^{2}=\\frac{1}{9}$, \\\\\ntherefore $\\frac{DE}{BC}=\\boxed{\\frac{1}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52872167.png"
  },
  {
    "solution": "\\boxed{2+2\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=ax^2+bx-3a$ intersects the negative half of the x-axis at point $A(-1,0)$, and another intersection point with the x-axis is $B$, and it intersects the positive half of the y-axis at point $C(0,3)$. The axis of symmetry of the parabola intersects line $BC$ at point $M$ and intersects the x-axis at point $G$. There exists a point $P$ on the axis of symmetry of the parabola, and when point $P$ is above the x-axis, it satisfies $\\angle APB = \\angle ABC$. What is the length of $PG$?\n",
    "completion": "\\textbf{Solution:} Let $y=0$, we get: $-x^2+2x+3=0$, solving this gives $x_1=-1$, $x_2=3$,\\\\\n$\\therefore$OB=OC=3,\\\\\n$\\therefore\\angle ABC=45^\\circ$,\\\\\n$\\because\\angle APB=\\angle ABC=45^\\circ$, and PA=PB,\\\\\n$\\therefore\\angle PBA=\\frac{1}{2}(180^\\circ-45^\\circ)=67.5^\\circ$,\\\\\n$\\therefore\\angle MPB=\\frac{1}{2}\\angle APB=22.5^\\circ$,\\\\\n$\\because\\angle MBP=67.5^\\circ-45^\\circ=22.5^\\circ$,\\\\\n$\\therefore\\angle MPB=\\angle MBP$,\\\\\n$\\therefore$MP=MB, in Rt$\\triangle BMG$, we have BG=MG=2, by the Pythagorean theorem, we find: BM=$2\\sqrt{2}$,\\\\\n$\\therefore$MP=$2\\sqrt{2}$,\\\\\n$\\therefore$PG=MG+MP=2+$2\\sqrt{2}$.\\\\\nHence, the length of $PG$ is $\\boxed{2+2\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52622585_239.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, points D, E, and F are located on sides AB, AC, and BC respectively, with $DE \\parallel BC$ and $EF \\parallel AB$, and $AD = 2BD$. What is the value of $\\frac{CF}{CB}$?\n",
    "completion": "\\textbf{Solution:} Since $AD=2BD$, \\\\\nit follows that $BD: AB=1: 3$, \\\\\nSince $DE\\parallel BC$, \\\\\nit follows that $CE: AC=BD: AB=1: 3$, \\\\\nSince $EF\\parallel AB$, \\\\\nit follows that $CF: CB=CE: AC=1: 3$. \\\\\nTherefore, $\\frac{CF}{CB} = \\boxed{\\frac{1}{3}}$",
    "image_path": "dataset/mm_math/images/MM_Math/52623765.png"
  },
  {
    "solution": "\\boxed{\\left(-\\frac{\\sqrt{3}}{2},\\frac{3}{2}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Rectangle $OABC$ is in the Cartesian coordinate system, with $OA = \\sqrt{3}$ and $AB = 1$. Folding it along $OB$, point $A$ falls at point ${A}^{\\prime}$. What are the coordinates of ${A}^{\\prime}$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, construct ${A}^{\\prime}D \\perp OA$ at D,\\\\\nsince in rectangle $OABC$, $OA=\\sqrt{3}$, $AB=1$,\\\\\nthen $OB=\\sqrt{OA^{2}+AB^{2}}=2$, that is $AB=\\frac{1}{2}OB$,\\\\\ntherefore $\\angle AOB=30^\\circ$,\\\\\nfrom the folding, $\\angle {A}^{\\prime}OB=\\angle AOB=30^\\circ$, $O{A}^{\\prime}=OA=\\sqrt{3}$,\\\\\ntherefore $\\angle {A}^{\\prime}OD=60^\\circ$,\\\\\ntherefore $\\angle O{A}^{\\prime}D=30^\\circ$,\\\\\ntherefore $OD=\\frac{1}{2}O{A}^{\\prime}=\\frac{\\sqrt{3}}{2}$,\\\\\ntherefore ${A}^{\\prime}D=\\sqrt{O{{A}^{\\prime}}^{2}−OD^{2}}=\\sqrt{{\\left(\\sqrt{3}\\right)}^{2}−{\\left(\\frac{\\sqrt{3}}{2}\\right)}^{2}}=\\frac{3}{2}$,\\\\\ntherefore, the coordinates of ${A}^{\\prime}$ are $\\boxed{\\left(-\\frac{\\sqrt{3}}{2},\\frac{3}{2}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53080470.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the running trajectory of the \"Tube Cart\" water barrel is a circle centered at $O$. It is known that the center $O$ is always above the water surface, and when the chord $AB$ cut by the water surface is $6\\, \\text{m}$, the maximum depth of the water barrel below the water surface is $1\\, \\text{m}$ (that is, the maximum distance from a point on the part of the circle below the water surface to the water surface). Find the radius of the circle in meters.\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $OD\\perp AB$ at point $E$, intersecting the circle at point $D$, \\\\\nthen $AE=\\frac{1}{2}AB=3$ meters, $DE=1$ meter, \\\\\nlet the radius of the circle be $r$ meters, \\\\\nin the right triangle $\\triangle AOE$, we have $AE^{2}+OE^{2}=OA^{2}$, \\\\\n$\\therefore 3^{2}+(r-1)^{2}=r^{2}$, \\\\\nsolving this equation, we find $r=\\boxed{5}$, \\\\\n$\\therefore$ the radius of the circle is $\\boxed{5}$ meters.",
    "image_path": "dataset/mm_math/images/MM_Math/53009890_132.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given: As shown in the figure, within $\\triangle ABC$, $AD$ is the altitude from $A$ to side $BC$, $E$ is the midpoint of side $AC$, $BC=14$, $AD=12$, and $\\sin B=\\frac{4}{5}$. Find: The length of segment $DC$.\n",
    "completion": "\\textbf{Solution:} In $\\triangle ABC$, since AD is the altitude on side BC,\\\\\ntherefore AD $\\perp$ BC.\\\\\ntherefore $\\sin B= \\frac{AD}{AB}=\\frac{4}{5}$.\\\\\nsince AD＝12,\\\\\ntherefore $AB=\\frac{5}{4}AD=15$.\\\\\nIn $\\triangle ABD$, since $BD=\\sqrt{AB^{2}−AD^{2}}=\\sqrt{15^{2}−12^{2}}=9$,\\\\\ntherefore CD＝BC−BD＝14−9＝\\boxed{5}.",
    "image_path": "dataset/mm_math/images/MM_Math/53043832_88.png"
  },
  {
    "solution": "\\boxed{(3,0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=-\\frac{3}{4}x^{2}+bx+c$ intersects the x-axis at point $A(4,0)$ and the y-axis at point $B(0,3)$. Point $M(m,0)$ is a moving point on line segment $OA$, and the line passing through point $M$ and perpendicular to the x-axis intersects line $AB$ and the parabola at points $P$ and $N$, respectively. If the triangle with vertices $B$, $P$, $N$ is similar to $\\triangle ABO$, what are the coordinates of point $M$?\n",
    "completion": "\\textbf{Solution:} Given that $A(4,0)$, $B(0,3)$, and $P(m, -\\frac{3}{4}m+3)$,\\\\\nhence $AB= \\sqrt{4^2+3^2}=5$, and $BP= \\sqrt{m^2+\\left(-\\frac{3}{4}m+3-3\\right)^2}=\\frac{5}{4}m$,\\\\\nmeanwhile $NP=-\\frac{3}{4}m^2+3m$,\\\\\nsince $PN\\parallel OB$,\\\\\nit follows that $\\angle BPN=\\angle ABO$,\\\\\nwhen $\\frac{PB}{OB}=\\frac{PN}{AB}$, $\\triangle BPN\\sim \\triangle OBA$,\\\\\nthat is $\\frac{\\frac{5}{4}m}{3}=\\frac{-\\frac{3}{4}m^2+3m}{5}$,\\\\\nafter simplification, we get $9m^2-11m=0$, solving this yields $m_1=0$ (discard), $m_2=\\frac{11}{9}$,\\\\\nat this point, the coordinates of point M are $\\boxed{\\left(\\frac{11}{9},0\\right)}$;\\\\\nwhen $\\frac{PB}{AB}=\\frac{PN}{OB}$, $\\triangle BPN\\sim \\triangle ABO$,\\\\\nthat is $\\frac{\\frac{5}{4}m}{5}=\\frac{-\\frac{3}{4}m^2+3m}{3}$,\\\\\nafter simplification, we get $2m^2-5m=0$, solving this yields $m_1=0$ (discard), $m_2=3$,\\\\\nat this point, the coordinates of point M are $\\boxed{(3,0)}$;\\\\\nIn conclusion, the coordinates of point M are either $\\boxed{\\left(\\frac{11}{9},0\\right)}$ or $\\boxed{(3,0)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51403288_131.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of $\\odot O$, and point C is the midpoint of the arc $BE$. Draw $CD\\perp AB$ at point G, intersecting $\\odot O$ at point D. If $BE=8, BG=2$, what is the radius of $\\odot O$?\n",
    "completion": "\\textbf{Solution:} Let us connect $OD$, as shown in the diagram. Assume the radius of $\\odot O$ to be $r$,\\\\\nsince $CD\\perp AB$,\\\\\nit follows that $\\overset{\\frown}{BC}=\\overset{\\frown}{BD}$, and $CG=DG$,\\\\\nsince point C is the midpoint of arc BE,\\\\\nit follows that $\\overset{\\frown}{CE}=\\overset{\\frown}{CB}$,\\\\\ntherefore $\\overset{\\frown}{BE}=\\overset{\\frown}{CD}$,\\\\\ntherefore $CD=BE=8$,\\\\\ntherefore $DG=\\frac{1}{2}CD=4$,\\\\\nIn $\\triangle ODG$ with right angle at $D$,\\\\\nsince $OG=r-2$ and $OD=r$,\\\\\nit follows that $4^2+(r-2)^2=r^2$, solving this gives $r=\\boxed{5}$,\\\\\nwhich means the radius of $\\odot O$ is $5$.",
    "image_path": "dataset/mm_math/images/MM_Math/53065818.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{5}}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the Cartesian coordinate system $xOy$, a line $y = \\frac{1}{2}x + b$ (where $b$ is a constant and $b < 2$) is drawn through point $P(0, 2)$ perpendicular to it, with the foot of the perpendicular at point $Q$. What is the value of $\\sin\\angle OPQ$?\n",
    "completion": "\\textbf{Solution}: Given $\\because \\angle PBQ = \\angle ABO$,\\\\\n$PQ \\perp AQ$, $BO \\perp AO$\\\\\n$\\therefore \\angle PBQ + \\angle OPQ = \\angle ABO + \\angle OAB$\\\\\n$= 90^\\circ$\\\\\n$\\therefore \\angle OPQ = \\angle OAB$\\\\\nFrom $y=\\frac{1}{2}x+b$, setting $x=0$, $y=b$,\\\\\nSetting $y=0$,\\\\\n$x=-2b$\\\\\n$\\therefore AO=2b$, $BO=b$\\\\\nIn $\\triangle AOB$ right,\\\\\n$AB=\\sqrt{AO^2 + BO^2} = \\sqrt{5}b$\\\\\n$\\therefore \\sin\\angle OPQ =\\sin\\angle BAO =\\frac{BO}{AB} =\\frac{b}{\\sqrt{5}b} = \\boxed{\\frac{\\sqrt{5}}{5}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51406936.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it's known that $AB$ is the diameter of $\\odot O$, $PD$ is tangent to $\\odot O$ at point C, and intersects the extension line of $AB$ at point D, and $\\angle D=2\\angle CAD$. Find the size of $\\angle D$.\n",
    "completion": "\\textbf{Solution:} Connect $OC$.\\\\\nSince $\\overset{\\frown}{BC}=\\overset{\\frown}{BC}$,\\\\\nit follows that $ \\angle CAD=\\frac{1}{2}\\angle COB$, i.e., $ \\angle COB=2\\angle CAD$.\\\\\nSince $ \\angle D=2\\angle CAD$,\\\\\nit follows that $ \\angle COB=\\angle D$.\\\\\nSince $ PD$ is tangent to $\\odot O$,\\\\\nit follows that $ OC\\perp PD$, i.e., $ \\angle OCD=90^\\circ$.\\\\\nTherefore, $ \\angle COB+\\angle D=90^\\circ$.\\\\\nTherefore, $ 2\\angle D=90^\\circ$.\\\\\nTherefore, $ \\angle D=\\angle COB=\\boxed{45^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51403374_130.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{6}{x}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the graph of the direct proportion function $y_1=-\\frac{3}{2}x$ and the graph of the inverse proportion function $y_2=\\frac{k}{x}$ intersect at point A in the second quadrant, and the x-coordinate of point A is $-2$. What is the analytic expression of the inverse proportion function?\n",
    "completion": "\\textbf{Solution:} Given that the graph of the direct proportion function $y=-\\frac{3}{2}x$ passes through point A, and the x-coordinate of point A is $-2$,\\\\\n\\[\\therefore\\] the y-coordinate of point A is 3, thus the coordinates of point A are $\\left(-2, 3\\right)$.\\\\\n\\[\\because\\] The graph of the inverse proportion function $y=\\frac{k}{x}$ passes through point $A\\left(-2, 3\\right)$,\\\\\n\\[\\therefore\\] $3=\\frac{k}{-2}$.\\\\\n\\[\\therefore\\] $k=-6$.\\\\\n\\[\\therefore\\] The equation of the inverse proportion function is $\\boxed{y=-\\frac{6}{x}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52602886_83.png"
  },
  {
    "solution": "\\boxed{4\\pi-2-2\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, quadrilateral $ABCD$ is a square with a side length of 4. A semicircle $\\overset{\\frown}{AB}$ is drawn with $B$ as the center and $BC$ as the radius. Point $E$ is located inside the quadrilateral, and $BE\\perp CE$, $\\angle BCE=30^\\circ$. Connect $AE$. Calculate the area of the shaded region.\n",
    "completion": "\\textbf{Solution:} Construct $EM\\perp BC$ at point $M$, and $EN\\perp AB$ at point $N$, as shown in the diagram.\n\nSince $BE\\perp CE$,\n\n$\\therefore \\angle BEC=90^\\circ$,\n\nSince $\\angle BCE=30^\\circ$,\n\n$\\therefore \\angle EBC=60^\\circ$,\n\nSince $EM\\perp BC$,\n\n$\\therefore$ in $\\triangle EMC$,\n\n$\\tan \\angle ECM=\\frac{EM}{MC}=\\tan30^\\circ=\\frac{\\sqrt{3}}{3}$,\n\n$\\therefore MC=\\sqrt{3}EM$,\n\n$\\therefore$ in $\\triangle EBM$,\n\n$\\tan \\angle EBM=\\frac{EM}{BM}=\\tan60^\\circ=\\sqrt{3}$,\n\n$\\therefore BM=\\frac{\\sqrt{3}}{3}EM$,\n\nSince $BM+MC=BC=4$,\n\n$\\therefore \\frac{\\sqrt{3}}{3}EM+\\sqrt{3}EM=4$,\n\n$\\therefore EM=\\sqrt{3}$,\n\n$\\therefore BM=\\frac{\\sqrt{3}}{3}EM=\\frac{\\sqrt{3}}{3}\\times\\sqrt{3}=1$,\n\nSince $NE\\perp AB$, $EM\\perp BC$, and $\\angle ABC=90^\\circ$,\n\n$\\therefore$ quadrilateral $BMEN$ is a rectangle,\n\n$\\therefore NE=BM=1$,\n\nSince $AB=BC=4$, $\\angle ABC=90^\\circ$,\n\n$\\therefore Area_{\\triangle ABE}=\\frac{1}{2}\\times AB\\times NE=\\frac{1}{2}\\times 4\\times 1=2$, $Area_{\\triangle BEC}=\\frac{1}{2}\\times BC\\times EM=\\frac{1}{2}\\times 4\\times\\sqrt{3}=2\\sqrt{3}$,\n\n$Area_{\\text{sector} ABC}=\\pi\\times AB^{2}\\times\\frac{90^\\circ}{360^\\circ}=\\pi\\times4^{2}\\times\\frac{1}{4}=4\\pi$\n\n$\\therefore Area_{\\text{shaded}}=4\\pi-2-2\\sqrt{3}=\\boxed{4\\pi-2-2\\sqrt{3}}$",
    "image_path": "dataset/mm_math/images/MM_Math/52715402.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In $\\triangle ABC$, $AB=6$, $BC=4$, and $\\angle B$ is an acute angle with $\\cos B=\\frac{1}{2}$. What is the measure of $\\angle B$ in degrees?\n",
    "completion": "\\textbf{Solution:} Since $\\angle B$ is an acute angle and $\\cos B = \\frac{1}{2}$,\\\\\nit follows that $\\angle B = \\boxed{60^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51322742_128.png"
  },
  {
    "solution": "\\boxed{24}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, rectangle $ABCD$, where $AB=8$, $AD=6$, and $F$ is a moving point on side $AB$. Connect $CF$, and draw $DG \\perp CF$ at $D$ with the foot of the perpendicular being point $G$, intersecting $BC$ at point $E$. Draw $AH \\perp DE$ through $A$ with the foot of the perpendicular being point $H$, and connect $CH$. What is the maximum area of quadrilateral $AGCH$?\n",
    "completion": "\\textbf{Solution:}\n\nSince quadrilateral $ABCD$ is a rectangle,\n\n$\\therefore$ $\\angle ADC=90^\\circ$,\n\nand since $DG\\perp CF$, $AH\\perp DE$,\n\n$\\therefore$ $\\angle AHD=\\angle DGC=90^\\circ$,\n\n$\\therefore$ $\\angle HAD+\\angle ADH=\\angle GDC+\\angle ADH=90^\\circ$,\n\n$\\therefore$ $\\angle HAD=\\angle GDC$,\n\nand $\\angle AHD=\\angle DGC=90^\\circ$,\n\n$\\therefore$ $\\triangle AHD\\sim \\triangle DGC$,\n\n$\\therefore$ $\\frac{DH}{CG}=\\frac{AH}{DG}=\\frac{AD}{DC}=\\frac{3}{4}$,\n\nLet $AH=3x$, $DG=4x$, $DH=3a$, $CG=4a$,\n\n$\\therefore$ $GH=DG-DH=4x-3a$,\n\n$\\therefore$ $S_{\\text{quadrilateral }AGCH}=S_{\\triangle AHG}+S_{\\triangle HGC}$,\n\n$=\\frac{1}{2}AH\\cdot GH+\\frac{1}{2}CG\\cdot GH$\n\n$=\\frac{1}{2}GH(AH+CG)$\n\n$=\\frac{1}{2}(4x-3a)(3x+4a)$\n\n$=\\frac{1}{2}(12x^2+7ax-12a^2)$\n\n$=-6a^2+\\frac{7}{2}ax+6x^2$\n\n$=-6(a-\\frac{7}{24}x)^2+\\frac{625}{96}x^2$,\n\n$\\therefore$ when $a=\\frac{7}{24}x$, $S_{\\text{quadrilateral }AGCH}$ reaches its maximum value of $\\frac{625}{96}x^2$,\n\nSince in $Rt\\triangle ADH$,\n\n$AH^2+DH^2=AD^2$, that is $(3x)^2+(3a)^2=36$,\n\n$\\therefore$ $x^2+a^2=4$,\n\nSubstituting $a=\\frac{7}{24}x$ into $x^2+a^2=4$,\n\nwe get $x^2=\\frac{2304}{625}$,\n\n$\\therefore$ $\\frac{625}{96}x^2=\\frac{625}{96}\\times \\frac{2304}{625}=\\boxed{24}$,\n\nThe maximum area of quadrilateral $AGCH$ is: $\\boxed{24}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53080160.png"
  },
  {
    "solution": "\\boxed{\\sqrt{13}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $D$ is a point on $AB$ such that $AD=DC$. $DE$ bisects $\\angle ADC$ and intersects $AC$ at point $E$. $DF$ bisects $\\angle BDC$ and intersects $BC$ at point $F$, where $\\angle DFC=90^\\circ$. Quadrilateral $CEDF$ is a rectangle; if $\\angle B=30^\\circ$ and $AD=2$, connect $BE$ and find the length of $BE$.\n",
    "completion": "\\textbf{Solution:} From (1), it is known that quadrilateral CEDF is a rectangle,\\\\\n$\\therefore \\angle CED = \\angle ECF = 90^\\circ$,\\\\\n$\\therefore \\angle A = 90^\\circ - \\angle B = 90^\\circ - 30^\\circ = 60^\\circ$, DE$\\perp$AC,\\\\\n$\\because AD = DC$,\\\\\n$\\therefore CE = AE$, $\\triangle ACD$ is an equilateral triangle,\\\\\n$\\therefore \\angle ACD = 60^\\circ$, AC = AD = 2,\\\\\n$\\therefore AE = CE = 1$,\\\\\n$\\therefore DE = \\sqrt{AD^2 - AE^2} = \\sqrt{3}$,\\\\\n$\\angle DCB = \\angle ECF - \\angle ACD = 90^\\circ - 60^\\circ = 30^\\circ$,\\\\\n$\\therefore \\angle DCB = \\angle B$,\\\\\n$\\therefore DB = DC = AD$,\\\\\n$\\therefore DE$ is the median of $\\triangle ABC$,\\\\\n$\\therefore BC = 2DE = 2\\sqrt{3}$,\\\\\nIn $\\triangle BCE$, by the Pythagorean theorem: $BE = \\sqrt{CE^2 + BC^2} = \\sqrt{13}$,\\\\\nHence, the length of BE is $\\boxed{\\sqrt{13}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52211079_78.png"
  },
  {
    "solution": "\\boxed{129^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the pentagon ABCDE is a regular pentagon, and F, G are points on the sides CD, DE, respectively, with BF$\\parallel$AG. If $\\angle$CFB$=57^\\circ$, what is the value of $\\angle$AGD?\n",
    "completion": "\\textbf{Solution:} Since pentagon ABCDE is a regular pentagon, \\\\\n$\\therefore$ $\\angle$D=$180^\\circ \\times 3 \\div 5=108^\\circ$, \\\\\nSince $\\angle$CFB=$57^\\circ$, \\\\\n$\\therefore$ $\\angle$BFD=$180^\\circ - \\angle$CFB=$180^\\circ - 57^\\circ =123^\\circ$, \\\\\nSince BF$\\parallel$AG, \\\\\n$\\therefore$ $\\angle$BAG+$\\angle$FBA=$180^\\circ$, \\\\\nIn pentagon ABFDG, \\\\\n$\\therefore$ $540^\\circ=\\angle$BAG+$\\angle$FBA+$\\angle$BFD+$\\angle$D+$\\angle$AGD=$180^\\circ+123^\\circ+108^\\circ+\\angle$AGD, \\\\\n$\\therefore$ $\\angle$AGD=\\boxed{129^\\circ}.",
    "image_path": "dataset/mm_math/images/MM_Math/51747881.png"
  },
  {
    "solution": "\\boxed{6-2\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within the square $ABCD$, points $M$ and $N$ are the midpoints of $AD$ and $BC$ respectively, point $P$ is on $CD$. After folding $\\triangle BCP$ along $BP$, the corresponding point of $C$, point $Q$, falls on $MN$. If $AB=6$, what is the length of $PD$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect CQ, \\\\\n$\\because$ Quadrilateral ABCD is a square, \\\\\n$\\therefore$ AD=BC=AB, $AD\\parallel BC$, $\\angle D=90^\\circ$, \\\\\n$\\because$ M and N are the midpoints of AD and BC respectively, \\\\\n$\\therefore$ DM=$\\frac{1}{2}AD=CN=\\frac{1}{2}BC=BN$, \\\\\n$\\therefore$ Quadrilateral CNMD is a rectangle, \\\\\n$\\therefore$ $\\angle QNB=\\angle QNC=90^\\circ$, \\\\\nIn $\\triangle QNB$ and $\\triangle QNC$, \\\\\n$\\left\\{\\begin{array}{l}BN=CN\\\\ \\angle QNB=\\angle QNC\\\\ QN=QN\\end{array}\\right.$, \\\\\n$\\therefore$ $\\triangle QNB\\cong \\triangle QNC$ (SAS), \\\\\n$\\therefore$ QB=QC, \\\\\nBy the property of folding, it is known that BC=BQ, $\\angle QBP=\\angle CBP$, \\\\\n$\\therefore$ BC=BQ=CQ, \\\\\n$\\therefore$ $\\triangle QBC$ is an equilateral triangle, \\\\\n$\\therefore$ $\\angle QBC=60^\\circ$, \\\\\n$\\therefore$ $\\angle PBC=30^\\circ$, \\\\\n$\\therefore$ PB=2PC, \\\\\n$\\because$ $BC^2+PC^2=BP^2$, \\\\\n$\\therefore$ $4PC^2=PC^2+6^2$, \\\\\n$\\therefore$ $PC=2\\sqrt{3}$, \\\\\n$\\therefore$ $PD=CD-PC=6-2\\sqrt{3}$, \\\\\nHence, the answer is: $PD=\\boxed{6-2\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51968614.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, fold the rectangular paper $ABCD$ along $EF$ so that vertex $C$ coincides with vertex $A$, and point $D$ falls at point $G$. If $AB=4$ and $BC=8$, find the area of the overlapping part of the paper after folding.\n",
    "completion": "Solution: From the folding, we have: $AE=EC$, $\\angle AEF=\\angle CEF$. Let $BE=x$, then $AE=EC=8-x$. In $\\triangle ABE$, we have $AB^2+BE^2=AE^2$,\n\\[\\therefore x^2+4^2=(8-x)^2\\], solving this gives $x=3$,\n\\[\\therefore AE=EC=8-x=5\\],\n\\[\\because rectangle ABCD\\],\n\\[\\therefore AD\\parallel BC\\],\n\\[\\therefore \\angle AFE=\\angle CEF\\],\n\\[\\therefore \\angle AEF=\\angle AFE\\],\n\\[\\therefore AE=AF=5\\],\nTherefore, the area of the overlapped part of the paper after folding, $\\triangle AEF$, is $\\frac{1}{2}AB\\cdot AF=\\frac{1}{2}\\times 4\\times 5=\\boxed{10}$",
    "image_path": "dataset/mm_math/images/MM_Math/52303866_132.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{2}\\alpha}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given that $O$ is a point on the line $AB$, $\\angle COD$ is a right angle, and $OE$ bisects $\\angle BOC$. As shown in Figure 1, if $\\angle AOC = \\alpha$, what is the degree measure of $\\angle DOE$ (expressed as an algebraic expression containing $\\alpha$)?\n",
    "completion": "\\textbf{Solution:} Given $\\because$ $\\angle AOC = \\alpha$, $\\angle COD$ is a right angle,\\\\\n$\\therefore$ $\\angle BOC = 180^\\circ - \\angle AOC = 180^\\circ - \\alpha$, $\\angle COD = 90^\\circ$,\\\\\n$\\therefore$ $\\angle BOD = 180^\\circ - \\angle AOC - \\angle COD = 90^\\circ - \\alpha$.\\\\\nSince OE bisects $\\angle BOC$,\\\\\n$\\therefore$ $\\angle BOE = \\frac{1}{2} \\angle BOC = 90^\\circ - \\frac{1}{2}\\alpha$,\\\\\n$\\therefore$ $\\angle DOE = \\angle BOE - \\angle BOD = 90^\\circ - \\frac{1}{2}\\alpha - (90^\\circ - \\alpha) = \\boxed{\\frac{1}{2}\\alpha}$.",
    "image_path": "dataset/mm_math/images/MM_Math/54927325_12.png"
  },
  {
    "solution": "\\boxed{y=\\frac{3}{8}x^2-\\frac{15}{4}x+6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=ax^2+bx+6$ intersects the x-axis at points $A(2,0)$ and $B(8,0)$, and intersects the y-axis at point $C$. Find the analytical expression of the parabola.\n",
    "completion": "\\textbf{Solution:} By substituting points A(2,0) and B(8,0) according to the problem, we have:\n\\[\n\\begin{cases}\n0=4a+2b+6\\\\ \n0=64a+8b+6\n\\end{cases},\n\\]\nfrom which we derive:\n\\[\n\\begin{cases}\na=\\frac{3}{8}\\\\ \nb=-\\frac{15}{4}\n\\end{cases},\n\\]\nTherefore, the equation of the parabola is: $\\boxed{y=\\frac{3}{8}x^2-\\frac{15}{4}x+6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51284444_183.png"
  },
  {
    "solution": "\\boxed{x<-1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y_1=kx+b$ passes through points A and B, and the line $y_2=2x$ passes through point A. What is the solution set of the inequality $2x<kx+b$?\n",
    "completion": "\\textbf{Solution:} Since $A(-1, -2)$,\\\\\nby observing the graph, the solution set of the inequality $2x<kx+b$ is $x<-1$.\\\\\nTherefore, the answer is: \\boxed{x<-1}.",
    "image_path": "dataset/mm_math/images/MM_Math/52386221.png"
  },
  {
    "solution": "\\boxed{(-3, 2)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system, the graph of the linear function \\(y=\\frac{1}{2}x+1\\) intersects the x-axis and y-axis at points A and B, respectively. A square ABCD is constructed in the second quadrant with AB as one of its sides. What are the coordinates of points C and D?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, let $CE\\perp y$ axis, $DF\\perp x$ axis,\\\\\n$\\therefore \\angle CEB=\\angle AFD=\\angle AOB=90^\\circ$.\\\\\n$\\because$ Quadrilateral $ABCD$ is a square,\\\\\n$\\therefore BC=AB=AD$, $\\angle DAB=\\angle ABC=90^\\circ$,\\\\\n$\\therefore \\angle DAF+\\angle BAO=90^\\circ$, $\\angle ABO+\\angle CBE=90^\\circ$,\\\\\n$\\because \\angle DAF+\\angle ADF=90^\\circ$, $\\angle BAO+\\angle ABO=90^\\circ$,\\\\\n$\\therefore \\angle BAO=\\angle ADF=\\angle CBE$,\\\\\n$\\therefore \\triangle BCE\\cong \\triangle DAF\\cong \\triangle ABO(AAS)$,\\\\\n$\\therefore BE=DF=OA=2$, $CE=AF=OB=1$,\\\\\n$\\therefore OE=OB+BE=3$, $OF=OA+AF=3$,\\\\\n$\\therefore$ The coordinates of point C are $\\boxed{(-1, 3)}$, the coordinates of point D are $\\boxed{(-3, 2)}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53706609_106.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the diagonals of rhombus $ABCD$ intersect at point $O$. $DE \\parallel AC$ and $CE \\parallel BD$. Connect $OE$. Quadrilateral $OCED$ is a rectangle. If $AC=6$ and $BD=8$, find the length of $OE$.\n",
    "completion": "\\textbf{Solution}: Since the quadrilateral $ABCD$ is a rhombus, $AC=6, BD=8$, it follows that $OC=\\frac{1}{2}AC=3, OD=\\frac{1}{2}BD=4, AC\\perp BD$, hence $CD=\\sqrt{OC^{2}+OD^{2}}=5$. Since the quadrilateral $OCED$ is a rectangle, it follows that $OE = \\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52354552_96.png"
  },
  {
    "solution": "\\boxed{\\left(2, \\frac{3+\\sqrt{17}}{2}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=x^2+bx+c$ intersects the x-axis at points $A$ and $B(3,0)$, and intersects the y-axis at point $C(0,3)$. Given these conditions, when $MN$ reaches its maximum value, is there a point $P$ on the axis of symmetry $l$ of the parabola such that $\\triangle PBN$ is an isosceles triangle? If such a point exists, please directly write down the coordinates of all such points $P$; if it does not exist, please explain why.\n",
    "completion": "\\textbf{Solution:} The coordinates of point $P$ are $\\boxed{\\left(2, \\frac{1}{2}\\right)}$, $\\boxed{\\left(2, -\\frac{\\sqrt{14}}{2}\\right)}$, $\\boxed{\\left(2, \\frac{\\sqrt{14}}{2}\\right)}$, $\\boxed{\\left(2, \\frac{3-\\sqrt{17}}{2}\\right)}$, or $\\boxed{\\left(2, \\frac{3+\\sqrt{17}}{2}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53124516_63.png"
  },
  {
    "solution": "\\boxed{15^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $E$ is on the side $BC$ of the rhombus $ABCD$, and $\\angle DAE=\\angle B=70^\\circ$, then what is the degree measure of $\\angle CDE$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral $ABCD$ is a rhombus,\\\\\nit follows that $AD\\parallel BC$,\\\\\nthus $\\angle AEB=\\angle DAE=\\angle B=70^\\circ$,\\\\\nwhich means $AE=AB=AD$,\\\\\nIn $\\triangle AED$, $AE=AD$, and $\\angle DAE=70^\\circ$,\\\\\nthus $\\angle ADE=55^\\circ$,\\\\\nAlso, since $\\angle B=70^\\circ$,\\\\\nit follows that $\\angle ADC=70^\\circ$,\\\\\nthus $\\angle CDE=\\angle ADC-\\angle ADE=15^\\circ$.\\\\\nTherefore, the answer is: $\\boxed{15^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/52395010.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle OABC, point A is on the x-axis, and point C is on the y-axis. The coordinates of point B are $\\left(6, 8\\right)$. Rectangle OABC is folded along line BD so that point C lands exactly on point E on the diagonal OB. The line of the fold BD intersects the y-axis and the x-axis at points D and F, respectively. What is the length of line segment OE?\n",
    "completion": "\\textbf{Solution:} Since in rectangle OABC, point A is on the x-axis and point C is on the y-axis, with point B having the coordinates (6, 8),\\\\\nit follows that OA = 6, AB = 8, and $\\angle$OAB$=90^\\circ$,\\\\\nthus OB $= \\sqrt{OA^{2} + AB^{2}} = \\sqrt{6^{2} + 8^{2}} = 10$,\\\\\nfrom folding, we know BE = BC = 6,\\\\\ntherefore, OE = OB - BE = 10 - 6 = \\boxed{4};",
    "image_path": "dataset/mm_math/images/MM_Math/52094893_170.png"
  },
  {
    "solution": "\\boxed{2\\text{cm}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the length of line segment $AC=12\\, \\text{cm}$, and point $B$ is on line segment $AC$, satisfying $BC= \\frac{1}{2} AB$. If $D$ is the midpoint of $AB$, and $E$ is the midpoint of $AC$, find the length of $DE$.\n",
    "completion": "\\textbf{Solution:}\\\\\nSince $D$ is the midpoint of $AB$,\\\\\nthus $AD=\\frac{1}{2}AB=4\\text{cm}$,\\\\\nSince $E$ is the midpoint of $AC$,\\\\\nthus $AE=\\frac{1}{2}AC=6\\text{cm}$,\\\\\nTherefore, $DE=AE-AD=\\boxed{2\\text{cm}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51457698_191.png"
  },
  {
    "solution": "\\boxed{14}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line segment AB is divided into three parts in the ratio 2:4:7 by points C and D. Points M and N are the midpoints of AC and DB, respectively. If MN equals 17 cm, what is the length of BD in cm?\n",
    "completion": "\\textbf{Solution:} Since segment AB is divided into three parts with ratios 2:4:7 by points C and D,\\\\\nlet AC = 2x, CD = 4x, and BD = 7x.\\\\\nSince M and N are the midpoints of segments AC and DB, respectively,\\\\\nthen CM = $\\frac{1}{2}$AC = x and DN = $\\frac{1}{2}$BD = $\\frac{7}{2}$x.\\\\\nGiven MN = 17 cm, then x + 4x + $\\frac{7}{2}$x = 17,\\\\\nthus x = 2,\\\\\ntherefore BD = \\boxed{14} cm.",
    "image_path": "dataset/mm_math/images/MM_Math/55192320.png"
  },
  {
    "solution": "\\boxed{120^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, diagonals $AC$ and $BD$ of rectangle $ABCD$ intersect at point $O$. If $\\angle OAD = 30^\\circ$, what is the measure of $\\angle AOD$?\n",
    "completion": "\\textbf{Solution:} Given a rectangle ABCD, where diagonals AC and BD intersect at point O,\\\\\nit follows that $OD=OB=OA=OC$,\\\\\nsince $\\angle OAD=30^\\circ$,\\\\\nit follows that $\\angle ODA=\\angle OAD=30^\\circ$,\\\\\nthus, $\\angle AOD=180^\\circ-\\left(\\angle ODA+\\angle OAD\\right)=180^\\circ-\\left(30^\\circ+30^\\circ\\right)=\\boxed{120^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53333516.png"
  },
  {
    "solution": "\\boxed{2\\text{cm}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, it is known that $BA=BC$, $\\angle B=120^\\circ$, and the perpendicular bisector $DE$ of $AB$ intersects $AC$ at point D. If $AC=6\\,cm$, what is the length of $AD$?\n",
    "completion": "\\textbf{Solution:} Connect $BD$,\\\\\nsince the perpendicular bisector $DE$ of $AB$ intersects $AC$ at point D,\\\\\nit follows that $AD=BD$,\\\\\nhence $\\angle A=\\angle ABD$,\\\\\nsince $BA=BC$, $\\angle B=120^\\circ$,\\\\\nit follows $\\angle A=\\angle C=\\frac{1}{2}\\left(180^\\circ-120^\\circ\\right)=30^\\circ$,\\\\\nthus $\\angle ABD=30^\\circ$,\\\\\nhence $\\angle CBD=90^\\circ$,\\\\\nthus $CD=2BD$,\\\\\nhence $CD=2AD$,\\\\\nthus $AC=AD+CD=AD+2AD=3AD$,\\\\\nfurther, since $AC=6\\text{cm}$,\\\\\nit follows $AD=\\boxed{2\\text{cm}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52905589.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, points $D, E, F$ are on the sides $AB$, $AC$, and $BC$ respectively, with $DE \\parallel BC$, $EF \\parallel AB$, and $AD: DB = 3:5$, what is the ratio of $BF: CF$?\n",
    "completion": "\\textbf{Solution:} Given that DE$\\parallel$BC,\\\\\nit follows that $\\frac{AD}{BD}=\\frac{AE}{CE}=\\frac{3}{5}$;\\\\\nsince EF$\\parallel$AB,\\\\\nit also follows that $\\frac{AE}{CE}=\\frac{BF}{CF}=\\boxed{\\frac{3}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52774652.png"
  },
  {
    "solution": "\\boxed{(2, 0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=x+7$ intersects with the line $y=-2x-2$ at point $C$. They intersect the y-axis at points $A$ and $B$, respectively. Point $F$ is on the positive x-axis, such that the area of $\\triangle ABC$ equals the area of $\\triangle AFC$. Find the coordinates of point $F$.\n",
    "completion": "\\textbf{Solution:} Since $S_{\\triangle AFC} = S_{\\triangle ABC}$, \\\\\nit follows that $BF \\parallel AC$, and thus the equation of line $BF$ is $y=x-2$. When $y=0$, $x-2=0$, which implies $x=2$, \\\\\nhence, the coordinates of $F$ are $\\boxed{(2, 0)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50561111_62.png"
  },
  {
    "solution": "\\boxed{9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $O$ is a point on line $AB$, $\\angle AOC=40^\\circ$, $OD$ bisects $\\angle AOC$, $\\angle DOE=90^\\circ$. Please answer the following question: How many angles in the figure are less than $180^\\circ$?\n",
    "completion": "\\textbf{Solution:} The angles in the figure that are less than $180^\\circ$ are: $\\angle AOD$, $\\angle AOC$, $\\angle AOE$, $\\angle DOC$, $\\angle DOE$, $\\angle DOB$, $\\angle COE$, $\\angle COB$, $\\angle EOB$, for a total of $\\boxed{9}$ angles.",
    "image_path": "dataset/mm_math/images/MM_Math/51495163_61.png"
  },
  {
    "solution": "\\boxed{(10, 3)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the rectangle $AOBC$ has sides $OB$ and $OA$ on the x-axis and y-axis, respectively, with point $D$ on side $BC$. The rectangle is folded along $AD$ such that point $C$ precisely lands on point $E$ on side $OB$. If point $A(0, 8)$ and point $B(10, 0)$, what are the coordinates of point $D$?\n",
    "completion": "\\textbf{Solution:}\n\nSince rectangle AOBC,\\\\\nthus AC=OB, $\\angle$AOB=$\\angle$DBE=$90^\\circ$,\\\\\nSince point A (0, 8), point B (10, 0),\\\\\nthus OA=8, AC=OB=10,\\\\\nSince the rectangle is folded along AD, point C perfectly falls on side OB,\\\\\nthus AC=AE=10, DC=DE,\\\\\nIn Rt$\\triangle$AOE,\\\\\n$OE=\\sqrt{AE^{2}-OA^{2}}=\\sqrt{10^{2}-8^{2}}=6$,\\\\\nthus BE=OB−OE=10−6=4,\\\\\nLet DC=DE=x, then DB=8−x,\\\\\nIn Rt$\\triangle$DEB, $DE^{2}=BE^{2}+DB^{2}$ implies $x^{2}=4^{2}+(8−x)^{2}$,\\\\\nSolving this: $x=5$,\\\\\nthus DB=8−5=3,\\\\\nthus the coordinates of point D are $\\boxed{(10, 3)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53224455.png"
  },
  {
    "solution": "\\boxed{20m}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points A and B are located at the two ends of a pond. Xiao Cong wants to measure the distance between A and B with a rope, but the rope is not long enough. A classmate helps him come up with an idea: first, find a point C on the ground that can be directly reached from both A and B, locate the midpoints D and E of AC and BC, respectively, and measure the length of DE as $10m$. What is the distance between A and B?\n",
    "completion": "\\textbf{Solution:} Since D and E are respectively the midpoints of AC and BC, \\\\\ntherefore DE is the midline of $\\triangle ABC$, \\\\\ntherefore AB=2DE, \\\\\nsince DE=10m, \\\\\ntherefore AB=\\boxed{20m}",
    "image_path": "dataset/mm_math/images/MM_Math/52304469.png"
  },
  {
    "solution": "\\boxed{\\frac{4}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, points $D$ and $E$ are located on sides $AB$ and $AC$ respectively. If $\\frac{AD}{AC}=\\frac{AE}{AB}=\\frac{2}{3}$ and $BC=2$, what is the length of $DE$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle DAE = \\angle CAB$,\\\\\nit follows that $\\triangle DAE \\sim \\triangle CAB$,\\\\\nhence $\\frac{DE}{BC} = \\frac{AE}{AB} = \\frac{2}{3}$,\\\\\ntherefore $DE = \\frac{2}{3} \\times BC = \\frac{2}{3} \\times 2 = \\boxed{\\frac{4}{3}}$",
    "image_path": "dataset/mm_math/images/MM_Math/55497782.png"
  },
  {
    "solution": "\\boxed{\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure (1), both quadrilateral ABCD and quadrilateral AEFG are squares. Rotate the square AEFG around point A and connect BE and CF. What is the value of $FC\\colon BE$?\n",
    "completion": "\\textbf{Solution:} We have $FC: BE = \\boxed{\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51445446_39.png"
  },
  {
    "solution": "\\boxed{y=\\frac{6}{x}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, the graph of the linear function $y=kx+b$ intersects with the graph of the inverse proportion function $y=\\frac{m}{x}$ at points A and B, and intersects with the x-axis at point C. The coordinates of point A are (2, 3), and the coordinates of point B are (-6, n). What are the analytic expressions for the linear function and the inverse proportion function?\n",
    "completion": "\\textbf{Solution:} Given that the graph of the inverse proportion function $y=\\frac{k}{x}$ passes through point A(2, 3),\\\\\nwe have $k=6$,\\\\\nthus, the expression for the inverse proportion function is $y=\\frac{6}{x}$.\\\\\nSince point B(-6, n) lies on the graph of the inverse proportion function $y=\\frac{6}{x}$,\\\\\nit follows that $n=-1$.\\\\\nSubstituting A(2, 3) and B(-6, -1) into $y=kx+b$, we get\n\\[\n\\left\\{\n\\begin{array}{l}\n2k+b=3, \\\\\n-6k+b=-1,\n\\end{array}\n\\right.\n\\]\nFrom which we derive\n\\[\n\\left\\{\n\\begin{array}{l}\nk=\\frac{1}{2}, \\\\\nb=2,\n\\end{array}\n\\right.\n\\]\nTherefore, the expression for the linear function is $\\boxed{y=\\frac{1}{2}x+2}$ and the expression for the inverse proportion function is $\\boxed{y=\\frac{6}{x}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50496315_68.png"
  },
  {
    "solution": "\\boxed{\\left(\\frac{-6 - \\sqrt{2}}{2}, \\frac{6 + \\sqrt{2}}{4}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, in the Cartesian coordinate system $xOy$, the parabola $y=ax^2+bx$ passes through the point $\\left(-2, 5\\right)$, and intersects with the line $y=-\\frac{1}{2}x$ at point $A$ in the second quadrant. A perpendicular line $AB$ is drawn from point $A$ to the $x$-axis, with the foot of the perpendicular being point $B\\left(-4, 0\\right)$. Let $P$ be a moving point on the parabola above line $OA$, and draw a line $PC\\perp x$-axis at point $C$, intersecting $OA$ at point $D$. Connect $OP$ and $PA$. Connect $PB$ intersecting $OA$ at point $E$, as shown in Figure 2. Can the line segments $PB$ and $AD$ bisect each other? If so, please find the coordinates of point $E$; if not, please explain why.\n",
    "completion": "\\textbf{Solution:} Solution: The line segment $PB$ and $AD$ bisect each other. The reason is as follows: Connect $BD$, $\\because$ the line segments $PB$ and $AD$ bisect each other, $\\therefore$ the quadrilateral $ABDP$ is a parallelogram, $\\therefore$ $PD=AB$, $\\because A(-4, 2)$, $PD = -t^2 - 4t$, $\\therefore$ $-t^2 - 4t = 2$, $\\therefore$ $t = -2 + \\sqrt{2}$ or $t = -2 - \\sqrt{2}$. When $t = -2 + \\sqrt{2}$, then $D(-2 + \\sqrt{2}, 1 - \\frac{\\sqrt{2}}{2})$, $\\because E$ is the midpoint of $AD$, $\\therefore$ the coordinates of point E are $\\boxed{\\left(\\frac{-6 + \\sqrt{2}}{2}, \\frac{6 - \\sqrt{2}}{4}\\right)}$. When $t = -2 - \\sqrt{2}$, then $D(-2 - \\sqrt{2}, 1 + \\frac{\\sqrt{2}}{2})$, $\\because E$ is the midpoint of $AD$, $\\therefore$ the coordinates of point E are $\\boxed{\\left(\\frac{-6 - \\sqrt{2}}{2}, \\frac{6 + \\sqrt{2}}{4}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51323941_198.png"
  },
  {
    "solution": "\\boxed{16}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram of rhombus ABCD, where $\\angle BAD=120^\\circ$ and $AC=4$, what is the perimeter of the rhombus?\n",
    "completion": "\\textbf{Solution:} In rhombus ABCD, we have AB=AC\\\\\n$\\because$ $\\angle$BAD = 120$^\\circ$\\\\\n$\\therefore$ $\\angle$ABC = 60$^\\circ$\\\\\n$\\therefore$ $\\triangle ABC$ is an equilateral triangle\\\\\nThat is, AB=AC=BC=4\\\\\nThus, the perimeter of the rhombus is \\boxed{16}",
    "image_path": "dataset/mm_math/images/MM_Math/51863557.png"
  },
  {
    "solution": "\\boxed{9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in parallelogram $ABCD$, $AC$ and $BD$ intersect at point $O$. Point $E$ is the midpoint of $OA$. Connect $BE$ and extend it to intersect $AD$ at point $F$. Given that the area of $\\triangle AEF$ is $3$, what is the area of $\\triangle ABE$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral $ABCD$ is a parallelogram,\\\\\nit follows that $AD\\parallel BC$ and $OA=OC$,\\\\\ntherefore $\\triangle AFE\\sim \\triangle CBE$,\\\\\nwhich implies $\\frac{AE}{EC}=\\frac{EF}{BE}$,\\\\\nsince point E is the midpoint of $OA$,\\\\\nit follows that $AE=\\frac{1}{3}EC$,\\\\\nhence, $EF=\\frac{1}{3}BE$,\\\\\ngiven that $S_{\\triangle AEF}=3$,\\\\\nthus, $S_{\\triangle AEB}=\\boxed{9}$",
    "image_path": "dataset/mm_math/images/MM_Math/53080328.png"
  },
  {
    "solution": "\\boxed{2+\\sqrt{7}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the quadratic function $y=x^2+bx+c$ intersects the x-axis at points $A$ and $B$, and intersects the y-axis at point $C$. It is also symmetric about the line $x=1$, and the coordinate of point $A$ is $(-1,0)$. When $a \\leq x \\leq a+1$, the minimum value of the quadratic function $y=x^2+bx+c$ is $2a$. Find the value of $a$.\n",
    "completion": "Solution:\n(1) For $a\\le 0$, when $x=a+1$, y reaches its minimum value, which is\n\\[\\left(a+1\\right)^{2}-2\\times \\left(a+1\\right)-3=a^{2}-4=2a\\]\n\\[\\therefore a=1\\pm \\sqrt{5}\\] (discard the positive value)\n\n(2) For $0<a\\le 1$, when $x=1$, y reaches its minimum value, which is\n\\[1^{2}-2\\times 1-3=-4=2a\\]\n\\[\\therefore a=-1\\] (discard)\n\n(3) For $a>1$, when $x=a$, y reaches its minimum value, which is\n\\[a^{2}-2a-3=2a\\]\n\\[\\therefore a=2\\pm \\sqrt{7}\\] (discard the negative value)\n\nHence, $a= \\boxed{1-\\sqrt{5}}$ or $a= \\boxed{2+\\sqrt{7}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51369956_190.png"
  },
  {
    "solution": "\\boxed{\\frac{2}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within $\\triangle ABC$, point D and point E are located on sides $AC$ and $AB$ respectively, $EC$ and $BD$ intersect at point O, and $\\angle ABD = \\angle ACE$, connect $DE$. $\\triangle EOD \\sim \\triangle BOC$; If $\\frac{DE}{BC} = \\frac{2}{3}$, find the value of $\\frac{AE}{AC}$.\n",
    "completion": "\\textbf{Solution:} From (1), we have $ \\triangle ADE\\sim \\triangle ABC$, \\\\\n$\\because \\frac{DE}{BC}=\\frac{2}{3}$, \\\\\n$\\therefore \\frac{AE}{AC}=\\frac{DE}{BC}=\\boxed{\\frac{2}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52977977_152.png"
  },
  {
    "solution": "\\boxed{\\frac{-1+\\sqrt{17}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, within square $ABCD$, $E$ is a point on diagonal $AC$, $FH \\perp AC$ at point $E$, intersecting $AD$, $AB$ at points $F$, $H$ respectively. Given that $CF = CH$ and $AH = \\frac{1}{3}CH$ with $AB = 4$, determine the length of $AH$.\n",
    "completion": "\\textbf{Solution:} Let $AH=x$, then $CH=3x$ and $HB=4-x$.\\\\\nSince quadrilateral $ABCD$ is a square,\\\\\nit follows that $BC=AB=4$ and $\\angle B=90^\\circ$.\\\\\nIn $\\triangle CBH$, by the Pythagorean theorem, we have\\\\\n$CH^2=HB^2+BC^2$,\\\\\nwhich implies $(3x)^2=(4-x)^2+4^2$,\\\\\nsolving for $x$ gives $x=\\frac{-1+\\sqrt{17}}{2}$ or $x=\\frac{-1-\\sqrt{17}}{2}$ (discard),\\\\\nthus, $AH=\\boxed{\\frac{-1+\\sqrt{17}}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50474816_151.png"
  },
  {
    "solution": "\\boxed{9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ is similar to $\\triangle DEF$, with point O being the center of similitude. Given that $OA=1$ and $OD=3$, if the perimeter of $\\triangle ABC$ is 3, what is the perimeter of $\\triangle DEF$?\n",
    "completion": "\\textbf{Solution}: Since $\\triangle ABC$ is similar to $\\triangle DEF$ and $OA=1$, $OD=3$, \\\\\nthus $OA: OD=1: 3$ and $AC\\parallel DF$, \\\\\ntherefore $\\triangle OAC\\sim \\triangle ODF$, \\\\\nhence $OA: OD=AC: DF=1: 3$, \\\\\nsince $\\triangle ABC$ is similar to $\\triangle DEF$, \\\\\ntherefore $\\triangle ABC\\sim \\triangle DEF$, \\\\\nthus the perimeter ratio of $\\triangle ABC$ to $\\triangle DEF$ is $1: 3$, \\\\\ntherefore the perimeter of $\\triangle DEF$ is $3\\times 3=\\boxed{9}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51809615.png"
  },
  {
    "solution": "\\boxed{7}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that in $\\triangle ABC$, $AD=BD$, $F$ is the intersection of the height $AD$ and $BE$, $CD=2$, $AF=3$. What is the length of segment $BC$?\n",
    "completion": "\\textbf{Solution:}\\\\\nSince $AD$ and $BE$ are the altitudes of the triangle,\\\\\nit follows that $\\angle C+\\angle DBF=90^\\circ$, $\\angle C+\\angle CAD=90^\\circ$, $AD\\perp BC$,\\\\\nwhich means that $\\angle DBF=\\angle DAC$, $\\angle ADB=\\angle ADC=90^\\circ$,\\\\\nConsidering triangles $\\triangle ACD$ and $\\triangle BFD$,\\\\\nwe have $\\left\\{\\begin{array}{l}\n\\angle DBF=\\angle DAC\\\\\nAD=BD\\\\\n\\angle BDF=\\angle ADC=90^\\circ\n\\end{array}\\right.$,\\\\\ntherefore, $\\triangle ACD\\cong \\triangle BFD$ (by ASA),\\\\\nwhich implies that $CD=FD$,\\\\\nGiven that $CD=2$, $AF=3$,\\\\\nit follows that $BD=AD=FD+AF=2+3=5$,\\\\\nthus $BC=5+2=\\boxed{7}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53106604.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the side $OC$ of the rhombus $OABC$ is on the positive half of the $x$-axis, and the coordinates of point $B$ are $\\left(8, 4\\right)$. What is the length of the side of this rhombus?\n",
    "completion": "\\textbf{Solution}: As shown in the figure, point B is drawn such that $BE\\perp x$ axis at point E. Let the side length of the diamond be $x$, \\\\\n$\\because B(8,4)$, \\\\\n$\\therefore CE=8-x$, $BE=4$, \\\\\nIn $\\triangle CBE$, $CB^{2}=CE^{2}+BE^{2}$, \\\\\nthus, $x^{2}=(8-x)^{2}+4^{2}$, solving for $x$ gives $x=\\boxed{5}$, \\\\\n$\\therefore$ the side length of the diamond is \\boxed{5}.",
    "image_path": "dataset/mm_math/images/MM_Math/52601775_189.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{97}}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the right triangle $\\triangle ABC$ has $\\angle ACB=90^\\circ$, with $AC=BC=3$. A square $CDEF$ with $C$ as the vertex (the four vertices $C$, $D$, $E$, and $F$ are arranged counterclockwise) can freely rotate around point $C$, and $CD=2$. Connect $AF$ and $BD$. During the rotation of square $CDEF$, what is the minimum value of $BD+\\frac{2}{3}AD$?\n",
    "completion": "\\textbf{Solution}: As shown in the figure, take a point \\(M\\) on \\(AC\\) such that \\(CM=\\frac{4}{3}\\). Connect \\(DM\\) and \\(AD\\).\\\\\n\\(\\because\\) \\(CD=2\\), \\(CM=\\frac{4}{3}\\), \\(CA=3\\),\\\\\n\\(\\therefore\\) \\(CD^{2}=CM\\cdot CA\\),\\\\\n\\(\\therefore\\) \\(\\frac{CD}{CA}=\\frac{CM}{CD}\\),\\\\\n\\(\\because\\) \\(\\angle DCM=\\angle ACD\\),\\\\\n\\(\\therefore\\) \\(\\triangle DCM \\sim \\triangle ACD\\),\\\\\n\\(\\therefore\\) \\(\\frac{DM}{AD}=\\frac{CD}{AC}=\\frac{2}{3}\\),\\\\\n\\(\\therefore\\) \\(DM=\\frac{2}{3}AD\\),\\\\\n\\(\\therefore\\) \\(BD+\\frac{2}{3}AD=BD+DM\\),\\\\\n\\(\\therefore\\) when \\(B\\), \\(D\\), \\(M\\) are collinear, the value of \\(BD+\\frac{2}{3}AD\\) is minimized,\\\\\n\\(\\therefore\\) The minimum value = \\(\\sqrt{CB^{2}+CM^{2}}=\\sqrt{3^{2}+(\\frac{4}{3})^{2}}=\\boxed{\\frac{\\sqrt{97}}{3}}\\).",
    "image_path": "dataset/mm_math/images/MM_Math/51322658.png"
  },
  {
    "solution": "\\boxed{y=\\frac{6}{x}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the graph of the linear function $y=2x+4$ intersects the $x$-axis at point $A$ and the $y$-axis at point $B$. It intersects the graph of the hyperbolic function $y=\\frac{k}{x}$ ($k\\neq 0$) at points $C$ and $D$. The coordinates of point $C$ are ($n$, $6$). What is the expression of this hyperbolic function?\n",
    "completion": "\\textbf{Solution:} Since point C $(n, 6)$ lies on the graph of the linear function $y=2x+4$,\\\\\nit follows that $6=2n+4$, solving this, we get $n=1$,\\\\\ntherefore, the coordinates of point C are $(1, 6)$.\\\\\nSubstituting the coordinates of point C $(1,6)$ into $y=\\frac{k}{x}$, we obtain $k=6$,\\\\\ntherefore, the expression for the inverse proportion function is $\\boxed{y=\\frac{6}{x}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52720084_75.png"
  },
  {
    "solution": "\\boxed{(2, 4)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=-2x+8$ intersects the $x$-axis and the $y$-axis at points $A$ and $B$, respectively, and the line $y=\\frac{1}{2}x+3$ intersects the $y$-axis at point $C$. The two lines intersect at point $D$. What are the coordinates of point $D$?\n",
    "completion": "\\textbf{Solution:} According to the problem, we can establish the following system of equations:\n\\[\n\\begin{cases}\ny = -2x + 8\\\\\ny = \\frac{1}{2}x + 3\n\\end{cases}\n\\]\nSolving this, we obtain:\n\\[\n\\begin{cases}\nx = 2\\\\\ny = 4\n\\end{cases}\n\\]\nTherefore, the coordinates of point $D$ are $\\boxed{(2, 4)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52250730_98.png"
  },
  {
    "solution": "\\boxed{\\left(\\frac{460}{3}\\right)^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\angle AOD = 130^\\circ$, $\\angle BOC : \\angle COD = 1 : 2$, and $\\angle AOB$ is $\\frac{1}{3}$ of the supplement of $\\angle COD$. In the plane, the ray OM satisfies $\\angle AOM = 2\\angle DOM$. What is the measure of $\\angle AOM$?\n",
    "completion": "\\textbf{Solution:} When ray OM is inside $\\angle AOD$,\\\\\n$\\because \\angle AOD = 130^\\circ$, and $\\angle AOM = 2\\angle DOM$,\\\\\n$\\therefore \\angle AOM = \\frac{2}{3}\\angle AOD = \\boxed{\\left(\\frac{260}{3}\\right)^\\circ}$.\\\\\nWhen ray OM is outside $\\angle AOD$,\\\\\n$\\because \\angle AOD = 130^\\circ$, and $\\angle AOM = 2\\angle DOM$, $\\angle AOD + \\angle AOM + \\angle DOM = 360^\\circ$,\\\\\n$\\therefore \\angle AOM = \\boxed{\\left(\\frac{460}{3}\\right)^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51324353_155.png"
  },
  {
    "solution": "\\boxed{3-\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: We call a rectangle whose width to length ratio is the golden ratio $\\left(\\frac{\\sqrt{5}-1}{2}\\right)$ a golden rectangle. As shown in the figure, in the golden rectangle $ABCD$, where $AB<BC$, $BC=2$, and the bisector of $\\angle ABC$ intersects side $AD$ at point $E$, what is the length of $DE$?\n",
    "completion": "\\textbf{Solution:} Given a golden rectangle $ABCD$, \\\\\nthe ratio of width to length is the golden ratio $\\left(\\frac{\\sqrt{5}-1}{2}\\right)$, and since AD$\\parallel$BC, AD=BC=2, \\\\\nthus, AB=$\\left(\\frac{\\sqrt{5}-1}{2}\\right)$BC=$\\frac{\\sqrt{5}-1}{2}\\times 2=\\sqrt{5}-1$, \\\\\nbecause BE is the bisector of $\\angle ABC$, \\\\\nit follows that $\\angle$ABE=$\\angle$CBE, \\\\\nand since AD$\\parallel$BC, \\\\\nit also follows that $\\angle$AEB=$\\angle$CBE=$\\angle$ABE, \\\\\nhence, AE=AB=$\\sqrt{5}-1$, \\\\\ntherefore, DE=AD−AE=2−($\\sqrt{5}-1$)=$\\boxed{3-\\sqrt{5}}$．",
    "image_path": "dataset/mm_math/images/MM_Math/51379323.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, within $\\triangle ABC$, $\\angle C = 90^\\circ$, $BE$ bisects $\\angle ABC$, $ED$ perpendicularly bisects $AB$ at foot $D$. If $AC=12$, what is the length of $AE$?\n",
    "completion": "\\textbf{Solution:} Given that $BE$ bisects $\\angle ABC$,\\\\\nhence, $\\angle CBE = \\angle ABE$,\\\\\nand since $ED$ perpendicularly bisects $AB$ at $D$,\\\\\nit follows that $EA=EB$,\\\\\nthus $\\angle A = \\angle ABE$,\\\\\ntherefore $\\angle CBE = \\angle A = \\angle ABE = \\frac{1}{3}\\left(180^\\circ-\\angle C\\right)= \\frac{1}{3}\\left(180^\\circ-90^\\circ\\right)=30^\\circ$,\\\\\nhence, $BE=2EC$, that is, $AE=2EC$,\\\\\nand given $AE+EC=AC=12$,\\\\\nit follows that $AE=\\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52653260.png"
  },
  {
    "solution": "\\boxed{(-6, 6)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, in the Cartesian coordinate system, $\\odot P$ is the inscribed circle of $\\triangle AOC$, with points D, E, and F as the tangency points. The line CD intersects $\\odot P$ at point G. The radius of $\\odot P$ is 2, $EG$ is parallel to the x-axis, $AB=AC$, and the parabola passes through points A, B, and C. As shown in Figure 2, point M is on the parabola and above the line AC, $MN$ is parallel to the y-axis and intersects AC at point N. A perpendicular line from point N to BC is drawn, forming $NK\\perp BC$ at the foot K. Let $t=MN+\\frac{2}{5}\\sqrt{10}NK$, find the maximum value of $t$ and the coordinates of point M at that time.\n",
    "completion": "\\textbf{Solution:} In the right-angled triangle $Rt\\triangle BCO$, we have $BC=\\sqrt{OB^{2}+OC^{2}}=\\sqrt{2^{2}+6^{2}}=2\\sqrt{10}$. Let the equation of the line $AC$ be $y=kx+b$. Then, we have the system of equations $\\left\\{\\begin{array}{l}-8k+b=0 \\\\ b=6\\end{array}\\right.$. Solving this, we get $\\left\\{\\begin{array}{l}k=\\frac{3}{4} \\\\ b=6\\end{array}\\right.$, therefore, the equation of the line $AC$ is $y=\\frac{3}{4}x+6$. Let $M(m, -\\frac{3}{8}m^{2}-\\frac{9}{4}m+6)$, since $MN$ is parallel to the $y$ axis and intersects $AC$ at $N$, we have $N(m, \\frac{3}{4}m+6)$, therefore $MN=-\\frac{3}{8}m^{2}-\\frac{9}{4}m+6-(\\frac{3}{4}m+6)=-\\frac{3}{8}m^{2}-3m$, $CN=-\\frac{5}{4}m$. Since $AB=AC$, we have $\\angle CBO=\\angle NCK$, and since $NK\\perp BC$ with the foot at $K$, we have $\\angle NKC=\\angle COB=90^\\circ$, therefore, $\\triangle NCK\\sim \\triangle CBO$, which implies $\\frac{NK}{CN}=\\frac{OC}{BC}$, hence $\\frac{NK}{-\\frac{5}{4}m}=\\frac{6}{2\\sqrt{10}}$, leading to $NK=-\\frac{3\\sqrt{10}}{8}m$. Therefore, $t=MN+\\frac{2}{5}\\sqrt{10}NK=-\\frac{3}{8}m^{2}-3m+\\frac{2\\sqrt{10}}{5}\\times (-\\frac{3\\sqrt{10}}{8}m)=-\\frac{3}{8}m^{2}-\\frac{9}{2}m=-\\frac{3}{8}(m+6)^{2}+\\frac{27}{2}$. Therefore, when $m=-6$, $t$ achieves its maximum value $\\boxed{\\frac{27}{2}}$, and at this point the coordinates of $M$ are $\\boxed{(-6, 6)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52949747_133.png"
  },
  {
    "solution": "\\boxed{55^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, the diagonals $AC$ and $BD$ intersect at point $O$. If $\\angle DOC = 110^\\circ$, what is the measure of $\\angle DAC$ in degrees?\n",
    "completion": "\\textbf{Solution}: Given rectangle ABCD,\\\\\nhence DO=AO,\\\\\nthus $\\angle$ADO=$\\angle$DAO,\\\\\nsince $\\angle$DOC=110$^\\circ$,\\\\\nthus 2$\\angle$DAO=110$^\\circ$, i.e., $\\angle$DAO=55$^\\circ$,\\\\\ntherefore, $\\angle$DAC=\\boxed{55^\\circ}.",
    "image_path": "dataset/mm_math/images/MM_Math/51614396.png"
  },
  {
    "solution": "\\boxed{y=\\frac{4}{x}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y=kx+b$ intersects with the graph of the inverse proportion function $y=\\frac{m}{x}$ at points $P(2,a)$ and $Q(-1, -4)$; determine the expressions for the linear function and the inverse proportion function.\n",
    "completion": "\\textbf{Solution:} Substitute $Q(-1, -4)$ into $y=\\frac{4}{x}$, we get $m=4$\\\\\nTherefore, $y=\\frac{4}{x}$\\\\\nSubstitute $P(2, a)$ into $y=\\frac{4}{x}$, we obtain $a=2$\\\\\nTherefore, $P(2, 2)$\\\\\nSubstituting $Q(-1, -4)$ and $P(2, 2)$ into $y=kx+b$ yields:\\\\\n$\\begin{cases}\n-k+b=-4\\\\\n2k+b=2\n\\end{cases}$\\\\\nSolving these, we get: $\\begin{cases}\nk=2\\\\\nb=-2\n\\end{cases}$\\\\\nHence, the equation of the linear function is $\\boxed{y=2x-2}$, and the expression for the inverse proportion function is $\\boxed{y=\\frac{4}{x}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52797108_42.png"
  },
  {
    "solution": "\\boxed{110^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in $\\triangle ABC$, $AB=AC$ and $\\angle B=40^\\circ$. Point D moves along the line segment $BC$ (D does not coincide with B or C). Connect $AD$ and construct $\\angle ADE=40^\\circ$, with $DE$ intersecting line segment $AC$ at $E$. As point D moves, the shape of $\\triangle ADE$ also changes. When $\\triangle ADE$ is an isosceles triangle, what is the degree measure of $\\angle BDA$?\n",
    "completion": "\\textbf{Solution:} Since AB = AC,\\\\\nit follows that $\\angle B = \\angle C = 40^\\circ$,\\\\\n(1) When AD = AE, $\\angle ADE = \\angle AED = 40^\\circ$,\\\\\nsince $\\angle AED > \\angle C$,\\\\\nthis case does not hold;\\\\\n(2) When DA = DE, that is $\\angle DAE = \\angle DEA = \\frac{1}{2}(180^\\circ - 40^\\circ) = 70^\\circ$,\\\\\nsince $\\angle BAC = 180^\\circ - 40^\\circ - 40^\\circ = 100^\\circ$,\\\\\nit follows that $\\angle BAD = 100^\\circ - 70^\\circ = 30^\\circ$;\\\\\ntherefore $\\angle BDA = 180^\\circ - 30^\\circ - 40^\\circ = \\boxed{110^\\circ}$;\\\\\n(3) When EA = ED, $\\angle ADE = \\angle DAE = 40^\\circ$,\\\\\nit follows that $\\angle BAD = 100^\\circ - 40^\\circ = 60^\\circ$,\\\\\ntherefore $\\angle BDA = 180^\\circ - 60^\\circ - 40^\\circ = 80^\\circ$;\\\\\ntherefore, when $\\triangle ADE$ is an isosceles triangle, the degree of $\\angle BDA$ is $110^\\circ$ or $80^\\circ$.",
    "image_path": "dataset/mm_math/images/MM_Math/53010115.png"
  },
  {
    "solution": "\\boxed{6\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $OA$ is the radius of the circle $\\odot O$, and $BC$ is a chord of $\\odot O$, with $OA=6$. If $OA$ is perpendicular bisector of $BC$, what is the length of $BC$?\n",
    "completion": "\\textbf{Solution:} Let's connect $OB$ and suppose $OA$ intersects $BC$ at point $D$.\\\\\nSince $OA$ is perpendicular bisector to $BC$,\\\\\nit follows that $BC=2BD$ and $OD=\\frac{1}{2}OA=3$,\\\\\nAlso, since $OB=OA=6$,\\\\\nthen $BD=\\sqrt{OB^{2}-OD^{2}}=\\sqrt{6^{2}-3^{2}}=3\\sqrt{3}$,\\\\\nthus, $BC=2BD=6\\sqrt{3}$.\\\\\nTherefore, the length of $BC$ is $\\boxed{6\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52602642.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $AB \\parallel CD \\parallel EF$. If $AC = 6$, $CE = 3$, and $DF = 2$, then what is the length of $BD$?\n",
    "completion": "\\textbf{Solution:} Given that $AB\\parallel CD\\parallel EF$,\\\\\nit follows that $\\frac{AC}{EC}=\\frac{BD}{FD}$,\\\\\nsince $AC=6$, $CE=3$, $DF=2$,\\\\\nwe have $\\frac{6}{3}=\\frac{BD}{2}$,\\\\\nthus, $BD=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52456882.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\angle AOB$ is a right angle, $\\angle BOC$ is an acute angle, and $OM$ bisects $\\angle AOC$, $ON$ bisects $\\angle BOC$. If $\\angle BOC=36^\\circ$, what is the measure of $\\angle MON$?\n",
    "completion": "\\textbf{Solution}: Since $\\angle AOB$ is a right angle and $\\angle BOC=36^\\circ$,\\\\\nit follows that $\\angle AOC=\\angle AOB+\\angle BOC=90^\\circ+36^\\circ=126^\\circ$,\\\\\nSince OM bisects $\\angle AOC$ and ON bisects $\\angle BOC$.\\\\\nThus, $\\angle MOC= \\frac{1}{2}\\angle AOC=63^\\circ$, $\\angle NOC= \\frac{1}{2}\\angle BOC=18^\\circ$,\\\\\nTherefore, $\\angle MON=\\angle MOC-\\angle NOC=63^\\circ-18^\\circ=\\boxed{45^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52794787_107.png"
  },
  {
    "solution": "\\boxed{(-6,0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y = \\frac{2}{3}x + 4$ intersects the $x$-axis at point $B$ and the $y$-axis at point $A$. What are the coordinates of point $A$ and point $B$?\n",
    "completion": "\\textbf{Solution:} Given $y= \\frac{2}{3}x+4$, let $x=0$, then $y=4$,\\\\\n$\\therefore A(0,4)$;\\\\\nLet $y=0$, then $\\frac{2}{3}x+4=0$, solving for $x$ gives $x=-6$,\\\\\n$\\therefore B(-6,0)$;\\\\\nHence, the coordinates of point $A$ are $A=\\boxed{(0,4)}$, and the coordinates of point $B$ are $B=\\boxed{(-6,0)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50682487_125.png"
  },
  {
    "solution": "\\boxed{16}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, point D is on the right angle side $BC$, $AD$ bisects $\\angle BAC$, $DE$ is the perpendicular bisector of $AB$, $CD=8\\,cm$, then what is the length of $BD$ in $cm$?\n",
    "completion": "Solution: Since $DE$ is the perpendicular bisector of side $AB$ of $\\triangle ABC$,\\\\\nit follows that $AD=BD$,\\\\\ntherefore, $\\angle B=\\angle DAB$,\\\\\nsince $AD$ bisects $\\angle BAC$ and $\\angle C=90^\\circ$,\\\\\nit follows that $\\angle DAB=\\angle DAC$ and $DC=DE=8$,\\\\\ntherefore, $\\angle B=\\angle DAB=\\angle DAC$,\\\\\nhence, $\\angle B=30^\\circ$,\\\\\nin right triangle $Rt\\triangle BDE$ with $\\angle B=30^\\circ$ and $DE=8$,\\\\\nit follows that $DB=\\boxed{16}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52827542.png"
  },
  {
    "solution": "\\boxed{\\left(-\\frac{3}{2},-\\frac{25}{4}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=x^{2}+bx+c$ passes through points $A(-4,0)$ and $C(0,-4)$, with another intersection point with the x-axis being $B$. If the vertex is $D$, what are the coordinates of point $D$?\n",
    "completion": "\\textbf{Solution:} Since the equation of the parabola is $y=x^{2}+3x-4=\\left(x+\\frac{3}{2}\\right)^{2}-\\frac{25}{4}$,\\\\\nthe coordinates of the vertex D of the parabola are $\\boxed{\\left(-\\frac{3}{2},-\\frac{25}{4}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52871856_290.png"
  },
  {
    "solution": "\\boxed{\\sqrt{6}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rhombus $ABCD$, $DE \\perp AB$ with the foot of the perpendicular being point $E$, and $CE$ is drawn. If $AE=2$ and $\\angle DCE=30^\\circ$, what is the side length of the rhombus?\n",
    "completion": "\\textbf{Solution}: Let $DE = x$. Since quadrilateral $ABCD$ is a rhombus, it follows that $AB \\parallel CD$, and therefore $\\angle CDE = \\angle AED = 90^\\circ$. Given that $\\angle DCE = 30^\\circ$, it follows that $CD = \\sqrt{3}DE = \\sqrt{3}x$. Consequently, $AD = CD = \\sqrt{3}x$. In right triangle $\\triangle AED$, $AD^{2} = DE^{2} + AE^{2}$, which implies $(\\sqrt{3}x)^{2} = x^{2} + 2^{2}$. Solving for $x$ gives $x = \\sqrt{2}$, therefore $CD = \\sqrt{3}x = \\sqrt{6}$. Hence, the answer is: $\\boxed{\\sqrt{6}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51510392.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points $A$ and $B$ are on the hyperbola $y=\\frac{3}{x}$. Perpendiculars are drawn from points $A$ and $B$ to the $x$-axis and the $y$-axis, respectively. If the area of the shaded region is $S_{\\text{shaded}}=2$, what is the value of $S_{1}+S_{2}$?\n",
    "completion": "\\textbf{Solution:} Given $y= \\frac{3}{x}$,\\\\\nthe area of the shaded region, $S_{\\text{shaded}}=2$,\\\\\ntherefore $S_{1}+S_{\\text{shaded}}=3=S_{2}+S_{\\text{shaded}}$,\\\\\nthus $S_{1}=1$, $S_{2}=1$,\\\\\nhence $S_{1}+S_{2}=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52551848.png"
  },
  {
    "solution": "\\boxed{(9,15)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, place the rectangle OABC within the Cartesian coordinate system with the x-axis and y-axis, positioning OA and OC along the positive halves of the x-axis and y-axis, respectively. Here, AB = 15, and the equation of the line containing the diagonal AC is given by $y=-\\frac{5}{3}x+b$. If the rectangle OABC is folded along BE such that point A lands on point D on side OC, what are the coordinates of point B?\n",
    "completion": "\\textbf{Solution:} Given that $AB=15$ and the quadrilateral OABC is a rectangle, $\\therefore OC=AB=15$. Thus, $C(0,15)$. By substituting into $y=-\\frac{5}{3}x+b$, we find $b=15$. Therefore, the equation of line AC is $y=-\\frac{5}{3}x+15$. Setting $y=0$ gives $x=9$. Therefore, $A(9,0)$ and $B=\\boxed{(9,15)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50675997_171.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB=24$, point $C$ is the midpoint of $AB$, and point $D$ is on the line segment $AC$ with $AD = \\frac{1}{3}CB$. Find the lengths of line segments $CD$ and $BD$.\n",
    "completion": "\\textbf{Solution:} Since point C is the midpoint of AB,\\\\\nit follows that $AC=BC=\\frac{1}{2}AB=12$,\\\\\nsince $AD=\\frac{1}{3}CB$,\\\\\nit follows that $AD=\\frac{1}{3}\\times 12=4$,\\\\\ntherefore $CD=AC-AD=\\boxed{8}$,\\\\\nthus, $BD=BC+CD=20$",
    "image_path": "dataset/mm_math/images/MM_Math/53364074.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{2}+2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, quadrilateral $ABCD$ is a rectangular piece of paper, with $AB=4$. Point $G$ is the midpoint of $CD$. First, fold the rectangular paper $ABCD$ along a straight line passing through point $B$ such that point $A$ lands on point $F$ on $BC$, with the fold line being $BE$. Then, unfold the paper. Next, fold the rectangular paper $ABCD$ along $BG$ so that point $C$ precisely lands on point $H$ on $BE$, with the fold line being $BG$. After unfolding the paper, connect $EF$, $HG$, and $BG$. What is the length of $BC$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect $EG$. Due to the properties of folding, it is known that quadrilateral $ABFE$ is a square and quadrilateral $CDEF$ is a rectangle.\\\\\n$\\therefore$AB=AE=4, $BH=BC$, $\\angle BHG=\\angle BCD=90^\\circ$, $CG=HG$.\\\\\n$\\because$ $AB=4$,\\\\\n$\\therefore$ $BE=\\sqrt{AB^2+AE^2}=\\sqrt{4^2+4^2}=4\\sqrt{2}$.\\\\\nAlso, $\\because$ point $G$ is the midpoint of $CD$,\\\\\n$\\therefore$ $CG=HG=DG=2$.\\\\\nIn $Rt\\ \\triangle EHG$ and $Rt\\ \\triangle EDG$,\\\\\n$\\left\\{\\begin{array}{l}HG=DG,\\\\ EG=EG,\\end{array}\\right.$\\\\\n$\\therefore$ $Rt\\ \\triangle EHG\\cong Rt\\ \\triangle EDG$ (HL).\\\\\n$\\therefore$ $EH=ED=FC$.\\\\\nLet $FC=x$, then $BH=BC=4x$.\\\\\n$\\therefore$ $BE=4+2x$. By $4+2x=4\\sqrt{2}$,\\\\\nwe solve to get $x=2\\sqrt{2}-2$.\\\\\n$\\therefore$ $BC=\\boxed{2\\sqrt{2}+2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51531726.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the rhombus $ABCD$, $AC$ and $BD$ intersect at point $O$, $E$ is the midpoint of $AB$, and $DE \\perp AB$. If $BD = 2\\sqrt{3}$, what is the length of $DE$?\n",
    "completion": "\\textbf{Solution:} Let $E$ be the midpoint of $AB$ and $DE\\perp AB$,\\\\\nhence $AD=DB$,\\\\\nsince quadrilateral $ABCD$ is a rhombus,\\\\\nthus $AB=AD$ and $AC\\perp BD$,\\\\\nhence $AD=DB=AB$ and $BO=\\frac{1}{2}BD=\\sqrt{3}$,\\\\\nthus $\\triangle ABD$ is an equilateral triangle,\\\\\nhence $\\angle DAB=60^\\circ$,\\\\\nthus $\\angle OAB=30^\\circ$,\\\\\nhence $AB=2OB=2\\sqrt{3}$,\\\\\nthus $AO=\\sqrt{AB^{2}-OB^{2}}=3$,\\\\\nsince $AO$ and $DE$ are both heights of the equilateral $\\triangle ABD$,\\\\\nhence $DE=AO=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51968607.png"
  },
  {
    "solution": "\\boxed{(2,0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the equation of line $MN$ is $y=-x+3$. Point $A$ is located on line segment $MN$ and satisfies $AN=2AM$. Point $B$ is on the $x$-axis. When $\\triangle AOB$ is an isosceles triangle with $OA$ as one of its equal sides, what are the coordinates of point $B$?\n",
    "completion": "\\textbf{Solution:} As illustrated in the diagram, draw AC $\\perp$ OM and AD $\\perp$ ON,\\\\\nlet x=0, then y=3,\\\\\nlet y=0, then x=3,\\\\\n$\\therefore$ M(0, 3), N(3, 0),\\\\\n$\\therefore$ OM=ON=3,\\\\\n$\\therefore$ MN=$3\\sqrt{2}$, $\\angle M=45^\\circ$,\\\\\n$\\because$ AN = 2AM,\\\\\n$\\therefore$ AM=$\\sqrt{2}$,\\\\\n$\\therefore$ AC=CM=1,\\\\\n$\\therefore$ OC=2,\\\\\n$\\therefore$ OA=$\\sqrt{1^2+2^2}=\\sqrt{5}$,\\\\\nWhen point B is on the positive x-axis, OB=OA=$\\sqrt{5}$, hence point B$_1$ ($\\sqrt{5}$, 0),\\\\\nWhen point B is on the negative x-axis, OB=OA=$\\sqrt{5}$, hence point B$_2$ ($-\\sqrt{5}$, 0),\\\\\nWhen AB = OA, OD=$\\sqrt{(\\sqrt{5})^2-2^2}=1$,\\\\\n$\\therefore$ OB=2OD=2,\\\\\n$\\therefore$ point B$_3$ (2, 0),\\\\\n$\\therefore$ the coordinates of point B are $\\boxed{(2,0)}$ or ($\\sqrt{5}$, 0) or ($-\\sqrt{5}$, 0).",
    "image_path": "dataset/mm_math/images/MM_Math/51419802.png"
  },
  {
    "solution": "\\boxed{\\left(2, 4\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=\\frac{1}{2}x$ intersects the graph of the inverse proportion function $y=\\frac{k}{x}\\ (x>0)$ at point $A(4, m)$, and point $B(n, 2n)$ is another point on the graph of the inverse proportion function. The line AB intersects the x-axis at point C. Find the analytical expression of the inverse proportion function and the coordinates of point B?\n",
    "completion": "\\textbf{Solution:} Given that the line $y=\\frac{1}{2}x$ intersects the graph of the inverse proportionality function $y=\\frac{k}{x}\\left(x>0\\right)$ at point $A\\left(4, m\\right)$,\\\\\nthen $m=\\frac{1}{2}\\times 4=2$,\\\\\nthus the coordinates of point A are $\\left(4, 2\\right)$,\\\\\nthereby $k=2\\times 4=8$,\\\\\nhence the equation of the inverse proportionality function is: $y=\\frac{8}{x}$,\\\\\nGiven that point $B\\left(n, 2n\\right)$ lies on the graph of the inverse proportionality function,\\\\\nthus $2n=\\frac{8}{n}$,\\\\\nsolving this gives: $n_{1}=2$, $n_{2}=-2$ (discard),\\\\\nupon verification, these are the roots of the equation\\\\\ntherefore, the coordinates of point B are $\\boxed{\\left(2, 4\\right)}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52602854_166.png"
  },
  {
    "solution": "\\boxed{0<n\\le 2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the parabola $y=-\\frac{1}{2}x^2+bx+c$ intersects the x-axis at points $A(-1, 0)$ and $B(4, 0)$, and intersects the y-axis at point C. Let the expression of line $BC$ be $y=mx+c$. If the first-degree equation in $x$, $x^2-2(b-m)x+2n=0$, has two positive real roots, directly write the range of values for $n$.\n",
    "completion": "\\textit{Solution:} We have $\\boxed{0<n\\le 2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53065871_93.png"
  },
  {
    "solution": "\\boxed{(3,0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the axis of symmetry of the parabola is the line $x=1$, and it intersects the $x$-axis at the point $A(-1,0)$. What are the coordinates of the other intersection point?\n",
    "completion": "\\textbf{Solution:} The axis of symmetry of the parabola is the line $x=1$. Given that the coordinates of point $A$ are $(-1,0)$,\\\\\nthe symmetry of the parabola allows us to find the coordinates of the other intersection point with the $x$-axis as $\\boxed{(3,0)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51445765.png"
  },
  {
    "solution": "\\boxed{(1, 2)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system $xOy$, for rectangle $ABCD$, edge $AB$ lies on the $x$-axis, $AB = 3$, and $AD = 2$. The line passing through point $C$ described by $y = x - 2$ intersects the $x$-axis and the $y$-axis at points $E$ and $F$, respectively. What are the coordinates of point $D$?\n",
    "completion": "\\textbf{Solution:} Let the coordinates of point $C$ be $(4, 2)$,\\\\\n$\\because$ point $C$ is on the line $y=x-2$,\\\\\n$\\therefore 2=4-2$,\\\\\nwhich means the coordinates of point $C$ are $(4, 2)$,\\\\\n$\\because$ quadrilateral $ABCD$ is a rectangle,\\\\\n$\\therefore AB=CD=3$, $AD=BC=2$,\\\\\n$\\therefore$ the coordinates of point $D$ are $\\boxed{(1, 2)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50561201_57.png"
  },
  {
    "solution": "\\boxed{70^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the rhombus $ABCD$, $E$ and $F$ are respectively on $BC$ and $DC$, with $BE=DF$ and $AE=AB$. If $\\angle EAF=30^\\circ$, what is the degree measure of $\\angle D$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral $ABCD$ is a rhombus,\\\\\nit follows that $AB=AD$, $\\angle B=\\angle D$,\\\\\nGiven that $BE=DF$,\\\\\nit follows that $\\triangle ABE\\cong \\triangle ADF$ (SAS),\\\\\ntherefore, $\\angle BAE=\\angle DAF$,\\\\\nLet $\\angle BAE=\\angle DAF=x$,\\\\\nSince $\\angle EAF=30^\\circ$,\\\\\nit follows that $\\angle BAD=2x+30^\\circ$,\\\\\nGiven that $AE=AB$,\\\\\nit follows that $\\angle B=\\angle AEB=\\frac{180^\\circ-x}{2}$,\\\\\nSince $\\angle B+\\angle BAD=180^\\circ$,\\\\\nit follows that $\\frac{180^\\circ-x}{2}+30^\\circ+2x=180^\\circ$,\\\\\nSolving gives: $x=40^\\circ$,\\\\\ntherefore, $\\angle D=\\angle B=\\frac{180^\\circ-40^\\circ}{2}=\\boxed{70^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/50481483.png"
  },
  {
    "solution": "\\boxed{3\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, draw $BE\\perp CD$ at point $E$, connect $AE$, and let $F$ be a point on $AE$ with $\\angle AFB=\\angle D$. $\\triangle ABF\\sim \\triangle EAD$. If $AB=4\\sqrt{3}$, $AD=6$, and $\\angle BAE=30^\\circ$, find the length of $BF$.\n",
    "completion": "\\textbf{Solution:} \\\\\nSince $BE\\perp CD$, \\\\\ntherefore, $\\angle BEC=90^\\circ$, \\\\\nsince $AB\\parallel CD$, \\\\\ntherefore, $\\angle ABE=\\angle BEC=90^\\circ$, \\\\\nsince $\\angle BAE=30^\\circ$, \\\\\ntherefore, $\\cos30^\\circ=\\frac{AB}{AE}$, \\\\\nthus, $AE=8$, \\\\\nsince $\\triangle ABF\\sim \\triangle EAD$, \\\\\ntherefore, $\\frac{BF}{AD}=\\frac{AB}{AE}$, \\\\\nthus, $\\frac{BF}{6}=\\frac{4\\sqrt{3}}{8}$, \\\\\nwe find: $BF=\\boxed{3\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51322983_136.png"
  },
  {
    "solution": "\\boxed{9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points $E$ and $F$ are located on sides $AD$ and $CD$ of the square $ABCD$, respectively, and $OE \\perp OF$. Given that $AD = 6$, what is the area of the shaded region in the figure?\n",
    "completion": "\\textbf{Solution:} \n\nSince quadrilateral ABCD is a square, \\\\\n$\\therefore \\angle EDO = \\angle FCO$, AC $\\perp$ BD, OD = $ \\frac{1}{2}$ BD, OC = $ \\frac{1}{2}$ AC, and AC = BD, \\\\\n$\\therefore \\angle DOC = 90^\\circ$, and OD = OC, \\\\\nSince OE $\\perp$ OF, \\\\\n$\\therefore \\angle EOF = 90^\\circ$, \\\\\n$\\therefore \\angle DOE = \\angle COF$, \\\\\nThus, $\\triangle ODE \\cong \\triangle OCF$ (ASA), \\\\\nTherefore, the area of the shaded part in the figure = Area of $\\triangle AOD$ = $ \\frac{1}{4}$ Area of square ABCD, \\\\\nSince AD = 6, \\\\\nTherefore, the area of the shaded part in the figure = $ \\frac{1}{4} \\times 6^{2}$ = $\\boxed{9}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52204223.png"
  },
  {
    "solution": "\\boxed{(6, -7)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the parabola $y=-\\frac{1}{2}x^2+bx+c$ intersects the x-axis at points $A(-1, 0)$ and $B(4, 0)$, and intersects the y-axis at point $C$. If point $P$ is a point on the parabola and $\\tan\\angle BAP=1$, what are the coordinates of point $P$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram:\n\n(1) When point P is above the x-axis, draw $PE\\perp x$ axis at point E,\n\n$\\because$ $\\tan\\angle BAP=\\frac{PE}{AE}=1$,\n\n$\\therefore$ $PE=AE$,\n\n$\\because$ $A(-1, 0)$,\n\n$\\therefore$ $OA=1$,\n\n$\\because$ $AE=PE=OE+OA=OE+1$,\n\nLet $OE=m$, $PE=m+1$,\n\n$\\therefore$ the coordinates of point P are $(m, m+1)$,\n\n$\\because$ point P is on the parabola,\n\n$\\therefore$ $-\\frac{1}{2}m^{2}+\\frac{3}{2}m+2=m+1$,\n\nSolving this, we get: $m_{1}=2$, $m_{2}=-1$ (irrelevant, discard),\n\n$\\therefore$ the coordinates of point P are $\\boxed{(2, 3)}$;\n\n(2) When point $P'$ is below the x-axis, draw $P'F\\perp x$ axis at point F,\n\n$\\because$ $\\tan\\angle BAP=\\frac{P'F}{AF}=1$,\n\n$\\therefore$ $P'F=AF$,\n\n$\\because$ $OA=1$,\n\n$\\therefore$ $AF=P'F=OF+OA=OF+1$,\n\nLet $OF=n$, then $P'F=n+1$,\n\n$\\therefore$ the coordinates of point $P'$ are $(n, -n-1)$,\n\n$\\because$ point $P'$ is on the parabola,\n\n$\\therefore$ $-\\frac{1}{2}n^{2}+\\frac{3}{2}n+2=-n-1$,\n\nSolving this, we get: $n_{1}=6$, $n_{2}=-1$ (irrelevant, discard),\n\n$\\therefore$ the coordinates of point $P'$ are $\\boxed{(6, -7)}$,\n\nIn conclusion: the coordinates of point P are $\\boxed{(2, 3)}$ or $\\boxed{(6, -7)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53065871_92.png"
  },
  {
    "solution": "\\boxed{(0,\\sqrt{13}-2)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, the parabola $y=ax^2+bx+3$ intersects the $x$-axis at points $A(3,0)$ and $B(-1,0)$, and intersects the $y$-axis at point $C$. Point $P$ is a moving point on the parabola in the first quadrant, and point $F$ is a moving point on the $y$-axis. Connect $PA$, $PF$, and $AF$. As shown in Figure 2, is there a point $F$ such that $\\triangle AFP$ is an isosceles right triangle with $AP$ as one of its legs? If such a point F exists, find the coordinates of all such points $F$; if not, please explain why.\n",
    "completion": "\\textbf{Solution:} Let $P(m, -m^{2}+2m+3)$ ($-1<m<3$), $F(0, n)$, \\\\\nsince $A(3,0)$, \\\\\ntherefore $OA=3$, $OF=|n|$, \\\\\n(1) When $AP=AF$, $\\angle PAF=90^\\circ$, draw $PD\\perp x$-axis at point $D$, \\\\\nthen $\\angle ADP=90^\\circ=\\angle AOF$, \\\\\nthus $\\angle PAD+\\angle APD=90^\\circ$, \\\\\nsince $\\angle PAD+\\angle FAO=90^\\circ$, \\\\\ntherefore $\\angle APD=\\angle FAO$, \\\\\nin $\\triangle APD$ and $\\triangle FAO$, \\\\\n$\\left\\{\\begin{array}{l}\n\\angle ADP=\\angle AOF \\\\\n\\angle APD=\\angle FAO \\\\\nAP=AF\n\\end{array}\\right.$ \\\\\nthus $\\triangle APD \\cong \\triangle FAO$ (AAS), \\\\\nthus $PD=OA$, $AD=OF$, \\\\\nsince $PD=-m^{2}+2m+3$, $AD=3-m$, \\\\\ntherefore $-m^{2}+2m+3=3$, \\\\\nsolving for $m$: $m=0$ or $2$, \\\\\nwhen $m=0$, $P(0,3)$, $AD=3$, \\\\\nthus $OF=3$, i.e., $|n|=3$, \\\\\nsince point $F$ is on the negative $y$-axis, \\\\\nthus $n=-3$, \\\\\ntherefore $F(0,-3)$; \\\\\nwhen $m=2$, $P(2,3)$, $AD=1$, \\\\\nthus $OF=1$, i.e., $|n|=1$, \\\\\nsince point $F$ is on the negative $y$-axis, \\\\\nthus $n=-1$, \\\\\ntherefore $F(0,-1)$; \\\\\n(2) When $AP=PF$, $\\angle APF=90^\\circ$, draw $PD\\perp x$-axis at point $D$, $PG\\perp y$-axis at point $G$, \\\\\nthen $\\angle PDA=\\angle PDO=\\angle PGF=90^\\circ$, \\\\\nsince $\\angle PDO=\\angle PGF=\\angle DOG=90^\\circ$, \\\\\nthus quadrilateral $PDOG$ is a rectangle, \\\\\nthus $\\angle FPG+\\angle FPD=90^\\circ$, \\\\\nsince $\\angle APD+\\angle FPD=\\angle APF=90^\\circ$, \\\\\ntherefore $\\angle FPG=\\angle APD$, \\\\\nin $\\triangle FPG$ and $\\triangle APD$, \\\\\n$\\left\\{\\begin{array}{l}\n\\angle PGF=\\angle PDA \\\\\n\\angle FPG=\\angle APD \\\\\nPF=AP\n\\end{array}\\right.$, \\\\\nthus $\\triangle FPG \\cong \\triangle APD$ (AAS), \\\\\nthus $PG=PD$, $FG=AD$, \\\\\nsince $PD=-m^{2}+2m+3$, $AD=3-m$, $PG=m$, \\\\\ntherefore $-m^{2}+2m+3=m$, \\\\\nsolving for $m$: $m=\\frac{1-\\sqrt{13}}{2}$ (discard) or $m=\\frac{1+\\sqrt{13}}{2}$, \\\\\nwhen $m=\\frac{1+\\sqrt{13}}{2}$, $P\\left(\\frac{1+\\sqrt{13}}{2}, \\frac{1+\\sqrt{13}}{2}\\right)$, \\\\\nthus $FG=AD=3-m=3-\\frac{1+\\sqrt{13}}{2}=\\frac{5-\\sqrt{13}}{2}$, \\\\\nthus $F\\left(0, \\sqrt{13}-2\\right)$; \\\\\nIn summary, the coordinates of point $F$ are $(0,-3)$ or $(0,-1)$ or $\\boxed{(0,\\sqrt{13}-2)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52660930_106.png"
  },
  {
    "solution": "\\boxed{25^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AD \\perp BC$ at $D$, $AE$ bisects $\\angle DAC$, $\\angle BAC = 80^\\circ$, $\\angle B = 60^\\circ$, what is the measure of $\\angle DAE$?\n",
    "completion": "Solution: In $\\triangle ABC$, since $\\angle BAC + \\angle B + \\angle C = 180^\\circ$,\\\\\nthen $\\angle C = 180^\\circ - 80^\\circ - 60^\\circ = 40^\\circ$,\\\\\nsince $AD \\perp BC$ at D,\\\\\nthen $\\angle ADC = 90^\\circ$,\\\\\nin $\\triangle ADC$, $\\angle DAC = 90^\\circ - \\angle C = 90^\\circ - 40^\\circ = 50^\\circ$,\\\\\nsince $AE$ bisects $\\angle DAC$,\\\\\nthen $\\angle DAE = \\frac{1}{2} \\angle DAC = \\boxed{25^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/52951344.png"
  },
  {
    "solution": "\\boxed{(6,0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y = -x + m$ intersects the $x$-axis at point $B(4, 0)$ and the $y$-axis at point $A$. Point $C$ is on the $x$-axis, and the area of $\\triangle ABC$ is known to be 4. Find the coordinates of point $C$.\n",
    "completion": "\\textbf{Solution:} Given $x=0$, then $y=-x+4=4$, thus point $A(0,4)$,\\\\\nsince $S_{\\triangle ABC}=4$,\\\\\nthus $\\frac{1}{2} BC \\cdot 4=4$, solving gives $BC=2$,\\\\\nsince point $B(4,0)$,\\\\\nthus $OB=4$,\\\\\nthus $OC=OB-BC=4-2=2$ or $OC=OB+BC=4+2=6$,\\\\\nthus point $\\boxed{C(2,0)}$ or $\\boxed{(6,0)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50482083_80.png"
  },
  {
    "solution": "\\boxed{(9, 2)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the Cartesian coordinate system, point $P$ is in the first quadrant. Circle $\\odot P$ is tangent to both the x-axis and y-axis, and it passes through the vertex $C$ of rectangle $AOBC$, intersecting with $BC$ at point $D$. Given that the radius of $\\odot P$ is 5 and the coordinates of point $A$ are $(0, 8)$, what are the coordinates of point $D$?\n",
    "completion": "\\textbf{Solution:}\\\\\nLet the points of tangency be $G$, $E$, respectively. Connect $PG$, $PE$, $PC$, $PD$, and extend $EP$ to intersect $BC$ at $F$. Then $PG=PE=PC=5$, and quadrilateral $OBFE$ is a rectangle.\\\\\n$\\because$OA=8,\\\\\n$\\therefore$CF=8-5=3,\\\\\n$\\therefore$PF=4,\\\\\n$\\therefore$OB=EF=5+4=9.\\\\\n$\\because$PF passes through the circle's center,\\\\\n$\\therefore$DF=CF=3,\\\\\n$\\therefore$BD=8-3-3=2,\\\\\n$\\therefore$The coordinates of $D$ are $\\boxed{(9, 2)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51213349.png"
  },
  {
    "solution": "\\boxed{x<-1$ and $1<x<2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given that $y$ is a function of $x$, and the graph of the function is as shown, what is the range of the independent variable $x$ when $y>0$?\n",
    "completion": "\\textbf{Solution:} When $y>0$, which means the part above the x-axis,\\\\\n$\\therefore$ the range of the independent variable $x$ is divided into two parts: $\\boxed{x<-1$ and $1<x<2}$.\\\\\nTherefore, the answer is: $x<-1$ and $1<x<2$.",
    "image_path": "dataset/mm_math/images/MM_Math/53333599.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=-\\frac{1}{2}x+b$ intersects the $x$-axis at point $A$ and the $y$-axis at point $B$. It intersects the graph of the function $y=kx$ at point $M(1, 2)$. What are the values of $k$ and $b$?\n",
    "completion": "\\textbf{Solution:} Given point $M(1, 2)$ lies on the line $y=-\\frac{1}{2}x+b$, \\\\\nthus $2=-\\frac{1}{2}\\cdot 1+b$, \\\\\nsolving for $b$ yields $b=\\boxed{\\frac{5}{2}}$, \\\\\nsince the graph of the function $y=kx$ passes through point $M(1, 2)$, \\\\\nit follows that $2=k$ or $k=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50560215_77.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the vertices $A$ and $C$ of the rectangle $ABCO$ are located on the $x$-axis and $y$-axis, respectively. The coordinates of point $B$ are $\\left(4, 3\\right)$. Circle $M$ is the inscribed circle of $\\triangle AOC$. Points $N$ and $P$ are moving points on circle $M$ and the $x$-axis, respectively. What is the minimum value of $BP+PN$?\n",
    "completion": "\\textbf{Solution:} Let point $B^{\\prime}$ be the symmetrical point of B with respect to the x-axis. Draw $MB^{\\prime}$, which intersects $\\odot M$ at point N and the x-axis at point P. Draw $MQ \\perp$ x-axis through point M, intersecting the x-axis at point E. Draw $B^{\\prime}Q \\perp MQ$ through point $B^{\\prime}$, \n\n$\\because$ Point B and point $B^{\\prime}$ are symmetrical with respect to the x-axis,\n\n$\\therefore$ $PB + PN = PB^{\\prime} + PN$,\n\nThe minimum value is obtained when N, P, and $B^{\\prime}$ are on the same line and pass through point M.\n\nIn the right triangle $\\triangle ABC$, $AC= \\sqrt{OA^{2}+OC^{2}}=5$,\n\nSince $\\odot M$ is the inscribed circle of $\\triangle AOC$, let the radius of $\\odot M$ be r,\n\n$\\therefore$ $S_{\\triangle AOC} = \\frac{1}{2}(3r + 4r + 5r) = \\frac{1}{2} \\times 3 \\times 4$,\n\nSolving for r gives $r=1$,\n\n$\\therefore$ $ME=MN=1$,\n\n$\\therefore$ $QB^{\\prime}=4-1=3$, $QM=3+1=4$,\n\n$\\therefore$ $MB^{\\prime}=5$,\n\n$\\therefore$ $PB^{\\prime}+PN=5-1=4$,\n\nHence, the minimum value of $PB + PN$ is $\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52949728.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{2}x+1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the line $l_1$ passes through points $A(0, 1)$ and $B(2, 2)$. The line $l_1$ is translated downward by $m$ units to obtain the line $l_2$, which is known to pass through point $(-1, -2)$ and intersects the x-axis at point C. What is the equation of the line $l_1$?\n",
    "completion": "\\textbf{Solution:} Let the equation of line $l_1$ be $y=kx+b$,\\\\\nsince line $l_1$ passes through points A(0, 1) and B(2, 2),\\\\\nthus $\\begin{cases}b=1\\\\ 2k+b=2\\end{cases}$, solving yields: $\\begin{cases}k=\\frac{1}{2}\\\\ b=1\\end{cases}$,\\\\\ntherefore, the equation of line $l_1$ is $y= \\frac{1}{2}x+1=\\boxed{\\frac{1}{2}x+1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50446864_229.png"
  },
  {
    "solution": "\\boxed{70^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Let $O$ be a point on line $AB$. A ray $OC$ is drawn from point $O$ such that $\\angle BOC=65^\\circ$. Place the right-angled vertex of a right-angled triangle plate at point $O$. When the triangle plate $MON$ is rotated counterclockwise around point $O$ to position 3, $\\angle NOC=\\frac{1}{4}\\angle AOM$, what is the degree measure of $\\angle NOB$?\n",
    "completion": "\\textbf{Solution:} Given that, \\\\\n$\\because \\angle NOC = \\frac{1}{4}\\angle AOM$,\\\\\n$\\therefore \\angle AOM = 4\\angle NOC$.\\\\\n$\\because \\angle BOC = 65^\\circ$,\\\\\n$\\therefore \\angle AOC = \\angle AOB - \\angle BOC = 180^\\circ - 65^\\circ = 115^\\circ$,\\\\\n$\\because \\angle MON = 90^\\circ$,\\\\\n$\\therefore \\angle AOM + \\angle NOC = \\angle AOC - \\angle MON = 115^\\circ - 90^\\circ = 25^\\circ$,\\\\\n$\\therefore 4\\angle NOC + \\angle NOC = 25^\\circ$,\\\\\n$\\therefore \\angle NOC = 5^\\circ$,\\\\\n$\\therefore \\angle NOB = \\angle NOC + \\angle BOC = \\boxed{70^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51291351_164.png"
  },
  {
    "solution": "\\boxed{\\left(\\frac{5+\\sqrt{61}}{2}, \\frac{2\\sqrt{61}-10}{3}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the sides of rhombus $OABC$ are such that side $OC$ is on the positive semi-axis of the $x$-axis, and the coordinates of point $B$ are $ \\left(8, 4\\right)$. If the graph of the inverse proportion function $y=\\frac{k}{x}\\left(x>0\\right)$ passes through point $A$ and intersects side $BC$ at point $D$, find the coordinates of point $D$.\n",
    "completion": "\\textbf{Solution:} Given that the side length of the diamond is 5, \\\\\nit follows that A(3, 4), \\\\\ntherefore, $k=3\\times4=12$, the inverse proportional function can be expressed as $y=\\frac{12}{x}$.\\\\\nGiven points C(5, 0) and B(8, 4), \\\\\nlet the equation of line CB be $y=kx+b$, \\\\\nthen $\\begin{cases}5k+b=0\\\\ 8k+b=4\\end{cases}$, \\\\\nsolving this, we get $\\begin{cases}k=\\frac{4}{3}\\\\ b=-\\frac{20}{3}\\end{cases}$, \\\\\ntherefore, the equation of line CB is: $y=\\frac{4}{3}x-\\frac{20}{3}$,\\\\\nFrom $\\begin{cases}y=\\frac{12}{x}\\\\ y=\\frac{4}{3}x-\\frac{20}{3}\\end{cases}$\\\\\nwe solve to get $\\begin{cases}x=\\frac{5+\\sqrt{61}}{2}\\\\ y=\\frac{2\\sqrt{61}-10}{3}\\end{cases}$ or $\\begin{cases}x=\\frac{5-\\sqrt{61}}{2}\\\\ y=-\\frac{2\\sqrt{61}+10}{3}\\end{cases}$ (discarded for being irrelevant to the problem),\\\\\ntherefore, the coordinates of point D are $\\boxed{\\left(\\frac{5+\\sqrt{61}}{2}, \\frac{2\\sqrt{61}-10}{3}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52601775_190.png"
  },
  {
    "solution": "\\boxed{2m^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the right-angled vertex $O$ of the right-angled triangle $MON$ is placed on the line $AB$, and the ray $OC$ bisects $\\angle AON$. If $\\angle MOC = m^\\circ$, what is the measure of $\\angle BON$ in degrees?\n",
    "completion": "\\textbf{Solution:} We have $\\angle BON = \\boxed{2m^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51481976_181.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, the circle $\\odot O$ with diameter $AB$ intersects $BC$ at point $D$ and the extension line of $CA$ at point $E$. The tangent $DF$ of $\\odot O$ is perpendicular to $AC$ at $F$. $AB=AC$. If $CF=2AF$ and $AE=4$, find the radius of $\\odot O$.\n",
    "completion": "\\textbf{Solution:} Connect BE and AD as shown in the figure, \\\\\n$\\because$ AB is the diameter of $\\odot O$, \\\\\n$\\therefore$ AD$\\perp$BC, BE$\\perp$EC, \\\\\n$\\because$ AB=AC, \\\\\n$\\therefore$ BD=DC, \\\\\n$\\because$ DF$\\perp$AC, BE$\\perp$EC, \\\\\n$\\therefore$ DF$\\parallel$BE, \\\\\n$\\therefore$ $\\frac{CD}{BD}=\\frac{CF}{EF}$, \\\\\n$\\therefore$ CF=EF, \\\\\n$\\because$ $CF=2AF$, AE=4, \\\\\nLet AF=x, then CF=2x, EF=4+x, AC=3x, \\\\\n$\\therefore$ 2x=4+x, solving gives: x=4, \\\\\n$\\therefore$ AC=12, \\\\\n$\\therefore$ AB=AC=12, \\\\\n$\\therefore$ The radius of $\\odot O$ is $\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51366880_121.png"
  },
  {
    "solution": "\\boxed{7.5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the graph of the inverse proportion function $y_{1}=\\frac{k}{x}$ intersects with the graph of the linear function $y_{2}=x+b$ at point $A(1, 4)$ and point $B(-4, n)$. What is the value of $k$ and the area of $\\triangle ABO$?\n",
    "completion": "\\textbf{Solution:} \nGiven point $A(1,4)$ lies on the graph of the inverse proportion function $y_1=\\frac{k}{x}$,\\\\\n$\\therefore k=1\\times 4=4$,\\\\\n$\\therefore$ the expression for the inverse proportion function is $y_1=\\frac{4}{x}$,\\\\\n$\\because$ point $B(-4,-1)$ also lies on the graph of the inverse proportion function $y_1=\\frac{4}{x}$,\\\\\n$\\therefore n=4-4=-1$, that is, $B(-4,-1)$,\\\\\nSubstituting point $A(1,4)$ into the linear function $y_2=x+b$,\\\\\nwe get $1+b=4$, solving for $b$ gives $b=3$,\\\\\n$\\therefore$ the expression for the linear function is $y_2=x+3$;\\\\\nThus, the analytical expression for the inverse proportion function is $y_1=\\frac{4}{x}$, and the linear function is found to be $y_2=x+3$;\\\\\nLet the intersection of the line with the x-axis be C,\\\\\nIn $y_2=x+3$, when $y=0$, we get $x=-3$,\\\\\n$\\therefore$ the intersection of line $y_2=x+3$ and the x-axis is $C(-3,0)$,\\\\\n$\\because$ the segment $OC$ divides $\\triangle AOB$ into $\\triangle AOC$ and $\\triangle BOC$,\\\\\n$\\therefore S_{\\triangle AOB}=S_{\\triangle AOC}+S_{\\triangle BOC}=\\frac{1}{2}\\times 3\\times 4+\\frac{1}{2}\\times 3\\times 1=\\boxed{7.5}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52977772_191.png"
  },
  {
    "solution": "\\boxed{(0, \\frac{5}{3})}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the graph of the linear function $y=kx+b$ passes through two points $A$ and $B$, where $A(-2,-1)$ and $B(1,3)$, and it intersects the $y$-axis at point $D$. What are the equation of the linear function and the coordinates of point $D$?\n",
    "completion": "\\textbf{Solution:} Substitute the coordinates $(-2, -1)$ and $(1, 3)$ into $y=kx+b$, \\\\\nwe have $\\begin{cases}\n-2k+b=-1 \\\\\nk+b=3\n\\end{cases}$\\\\\nSolving this, we find $\\begin{cases}\nk=\\frac{4}{3} \\\\\nb=\\frac{5}{3}\n\\end{cases}$\\\\\nTherefore, the equation of the line is $\\boxed{y=\\frac{4}{3}x+\\frac{5}{3}}$.\\\\\nFurthermore, since the line intersects the y-axis at point $D$\\\\\nThus, by substituting $x=0$ into $y=\\frac{4}{3}x+\\frac{5}{3}$\\\\\nwe get $y=\\frac{5}{3}$,\\\\\nTherefore, the coordinates of point $D$ are $\\boxed{(0, \\frac{5}{3})}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51699361_172.png"
  },
  {
    "solution": "\\boxed{(-1, 1)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the Cartesian coordinate system, the parabola $y=-x^2-2x+3$ intersects the x-axis at points $A$ and $B$ (with $A$ on the left side of $B$) and intersects the y-axis at point $C$, with $D$ being the vertex. As shown in the figure, point $P$ is a moving point on the axis of symmetry of the parabola such that $\\triangle ACP$ is an isosceles triangle. If it exists, find the coordinates of point $P$; if not, please state the reason.\n",
    "completion": "\\textbf{Solution:}\n\nThere exists a point P(-1, t).\n\nSince $A(-3,0)$ and $C(0,3)$,\n\nTherefore, $AC^2=3^2+3^2=18$, $AP^2=[-1-(-3)]^2+t^2=t^2+4$,\n\n$CP^2=[0-(-1)]^2+(t-3)^2=(t-3)^2+1$,\n\n(1) When $AC=AP$, then $18=t^2+4$,\n\nSolving this gives $t=\\sqrt{14}$ or $t=-\\sqrt{14}$,\n\nHence, P(-1, $\\sqrt{14}$) or (-1, $-\\sqrt{14}$);\n\n(2) When $AC=CP$, then $18=(t-3)^2+1$,\n\nSolving this gives $t=3+\\sqrt{17}$ or $t=3-\\sqrt{17}$,\n\nP(-1, $3+\\sqrt{17}$) or (-1, $3-\\sqrt{17}$);\n\n(3) When $AP=CP$, then $t^2+4=(t-3)^2+1$,\n\nSolving this gives $t=1$,\n\nTherefore, P(-1, 1);\n\nIn conclusion, the coordinates of point P are: \\boxed{(-1, \\sqrt{14})} or \\boxed{(-1, -\\sqrt{14})} or \\boxed{(-1, 3+\\sqrt{17})} or \\boxed{(-1, 3-\\sqrt{17})} or \\boxed{(-1, 1)}.",
    "image_path": "dataset/mm_math/images/MM_Math/52720132_81.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $l_1$ passes through points A(0,4) and D(4,0), the line $l_2: y=\\frac{1}{2}x+1$ intersects the x-axis at point C, and the two lines $l_1$ and $l_2$ intersect at point B. Find the area of $\\triangle BCD$.\n",
    "completion": "\\textbf{Solution:} The line $y=\\frac{1}{2}x+1$ intersects the x-axis at point $C$,\\\\\n$\\therefore$ the coordinates of point $C$ are $(-2,0)$,\\\\\n$\\therefore$ $CD=6$,\\\\\n$\\therefore$ the area of $\\triangle ABC=S_{\\triangle ACD}-S_{\\triangle BCD}=\\frac{1}{2}\\times 6\\times 4-\\frac{1}{2}\\times 6\\times 2=6$.\\\\\n$\\therefore$ the area of $\\triangle BCD=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52056352_195.png"
  },
  {
    "solution": "\\boxed{3\\sqrt{10}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $AD=3$, $AB=5$, point $E$ is a moving point on the ray $CD$ (not coincident with $D$). $\\triangle ADE$ is folded along $AE$ to obtain $\\triangle D'AE$. Connect $D'B$. If $\\triangle ABD'$ is a right triangle, then what is the length of $AE$?\n",
    "completion": "\\textbf{Solution:} According to the problem, quadrilateral ABCD is a rectangle, with $AD=3$ and $AB=5$. By folding $\\triangle ADE$ along AE, we obtain $\\triangle D^{\\prime}AE$, then $\\angle D=\\angle ED^{\\prime}A=90^\\circ$, $AD=BC=AD^{\\prime}=3$, $AB=CD=5$,\\\\\n(1) As shown in Figure 1, when point E is on segment CD,\\\\\nsince $\\angle ED^{\\prime}A=\\angle D=\\angle AD^{\\prime}B=90^\\circ$,\\\\\nthen points $B$, $D^{\\prime}$, and $E$ are collinear,\\\\\nsince $S_{\\triangle ABE}=\\frac{1}{2}AB\\cdot AD=\\frac{1}{2}BE\\cdot AD^{\\prime}$,\\\\\nthus $BE=AB=5$,\\\\\nsince $BD^{\\prime}=\\sqrt{AB^{2}-AD^{\\prime 2}}=\\sqrt{5^{2}-3^{2}}=4$,\\\\\nthus $DE=D^{\\prime}E=BE-BD^{\\prime}=5-4=1$;\\\\\ntherefore, in $Rt\\triangle ADE$, $AE=\\sqrt{AD^{2}+DE^{2}}=\\sqrt{3^{2}+1^{2}}=\\sqrt{10}$;\\\\\n(2) As shown in Figure 2, when point E is on the ray extending from CD,\\\\\nsince $\\angle AD^{\\prime}B=\\angle BCE=90^\\circ$, $AD=BC=AD^{\\prime}=3$, $AB=CD=5$,\\\\\nthus $BD^{\\prime}=\\sqrt{AB^{2}-AD^{\\prime 2}}=4$,\\\\\nlet $CE=x$, then $D^{\\prime}E=DE=x+5$,\\\\\nthus $BE=D^{\\prime}E-BD^{\\prime}=x+1$,\\\\\nsince $CE^{2}+BC^{2}=BE^{2}$, that is, $x^{2}+3^{2}=(x+1)^{2}$,\\\\\nsolving for $x$ yields $x=4$,\\\\\nthus $DE=CD+CE=5+4=9$,\\\\\ntherefore, in $Rt\\triangle ADE$, $AE=\\sqrt{AD^{2}+DE^{2}}=\\sqrt{3^{2}+9^{2}}=3\\sqrt{10}$.\\\\\nIn summary, the value of AE is either $\\boxed{\\sqrt{10}}$ or $\\boxed{3\\sqrt{10}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52826908.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the quadrilateral OABC is a square. An equilateral triangle $\\triangle ABD$ is constructed with AB as the side, and point D is precisely on the x-axis. The diagonal CA intersects the x-axis at point E. If the coordinate of point E is $\\left(4, 0\\right)$, write down directly the degree measure of $\\angle BDO$.\n",
    "completion": "\\textbf{Solution:} We have $\\angle BDO = \\boxed{45^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53359725_137.png"
  },
  {
    "solution": "\\boxed{80^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $\\angle BAE=\\angle CAF=110^\\circ$, $\\angle CAE=60^\\circ$, and $AD$ is the angle bisector of $\\angle BAF$. What is the measure of $\\angle BAD$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle$BAE$=110^\\circ$, $\\angle$CAE$=60^\\circ$,\n\ntherefore $\\angle$BAC$=110^\\circ-60^\\circ=50^\\circ$,\n\nand since $\\angle$CAF$=110^\\circ$,\n\ntherefore $\\angle$BAF$=110^\\circ+50^\\circ=160^\\circ$,\n\nand since AD is the angle bisector of $\\angle$BAF,\n\ntherefore $\\angle$BAD= $\\frac{1}{2}$ $\\angle$BAF= $\\frac{1}{2}$ $\\times160^\\circ=\\boxed{80^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/10378022.png"
  },
  {
    "solution": "\\boxed{\\frac{5}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, given $ \\triangle ABC$ with $\\angle CAB=45^\\circ$, $AB=7$, and $AC=3\\sqrt{2}$. $CD\\perp AB$ at point D. E is a moving point on side BC. A circle $\\odot O$ is drawn with DE as its diameter, intersecting BC at F and AB at point G. Connect DF and FG. As depicted in Figure 2, the other intersection point of $\\odot O$ and CD is P. If ray AP passes through point F, find the value of $\\frac{AP}{DE}$.\n",
    "completion": "\\textbf{Solution:} As shown in the figure:\\\\\n$\\because$ Quadrilateral PDEF is a cyclic quadrilateral inside circle $O$\\\\\n$\\therefore \\angle APD=\\angle DEF$\\\\\n$\\because \\angle APD+\\angle PAD=\\angle DEF+\\angle EDF=90^\\circ$\\\\\n$\\therefore \\angle PAD=\\angle EDF$\\\\\nConnect PG\\\\\n$\\because \\angle PAD=\\angle EDF$\\\\\n$\\angle ADP=\\angle DFE=90^\\circ$\\\\\n$\\therefore \\triangle ADP\\sim \\triangle DFE$\\\\\n$\\therefore \\frac{AP}{DE}=\\frac{AD}{DF}=3\\times \\frac{5}{12}=\\boxed{\\frac{5}{4}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51322957_133.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the diagonals AC and BD of rectangle ABCD intersect at point O, with BE $\\parallel$ AC, and CE $\\parallel$ DB. Quadrilateral OBEC is a rhombus; if AD = 4 and AB = 2, find the area of the rhombus OBEC.\n",
    "completion": "\\textbf{Solution:} Given $AD=4$ and $AB=2$,\\\\\ntherefore, the area of rectangle $ABCD=4\\times 2=8$,\\\\\nthus, the area of $\\triangle OBC= \\frac{1}{4}$ area of rectangle $ABCD=2$,\\\\\nhence, the area of the rhombus $OBEC=2\\times$ area of $\\triangle OBC=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50496436_110.png"
  },
  {
    "solution": "\\boxed{7}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle BCA=90^\\circ$, $BC=8$, and $AC=6$. Point $D$ is the midpoint on side $AB$, and point $E$ is a movable point on side $BC$. Connect $DE$, and fold $\\triangle BDE$ along $DE$ to get $\\triangle FDE$. As shown in figure (3), connect $CD$, and the line segment $EF$ intersects with the line segment $CD$ at point $M$. When $\\triangle DFM$ is a right-angled triangle, find the length of $BE$.\n",
    "completion": "\\textbf{Solution:}\\\\\nWhen $\\angle FMD=90^\\circ$,\\\\\n$\\because \\angle F=\\angle B, \\angle FMD=\\angle ACB=90^\\circ$,\\\\\n$\\therefore \\triangle FDM\\sim \\triangle BAC$,\\\\\n$\\therefore \\frac{DF}{AB}=\\frac{DM}{AC}$,\\\\\n$\\therefore \\frac{5}{10}=\\frac{DM}{6}$,\\\\\n$\\therefore DM=3$,\\\\\n$\\therefore CM=CD-DM=2$,\\\\\n$\\because \\angle ECM=\\angle B$,\\\\\n$\\therefore \\angle CME=\\angle ACB=90^\\circ$,\\\\\n$\\therefore \\triangle CEM\\sim \\triangle BAC$,\\\\\n$\\therefore \\frac{CE}{AB}=\\frac{CM}{BC}$,\\\\\n$\\therefore \\frac{CE}{10}=\\frac{2}{8}$,\\\\\n$\\therefore CE=\\frac{5}{2}$,\\\\\n$\\therefore BE=\\frac{11}{2}$;\\\\\nWhen $\\angle FDM=90^\\circ$,\\\\\n$\\because \\angle F=\\angle BCD, \\angle FMD=\\angle CME$,\\\\\n$\\therefore \\angle CEM=\\angle FDM=90^\\circ$,\\\\\n$\\therefore \\angle FED=\\angle BED=45^\\circ$,\\\\\nConstruct $DH\\perp BC$ at $H$,\\\\\nThen $\\triangle BDH\\sim \\triangle BAC$,\\\\\n$\\therefore \\frac{DB}{BA}=\\frac{DH}{AC}$,\\\\\n$\\therefore \\frac{5}{10}=\\frac{DH}{6}$,\\\\\n$\\therefore DH=3, BH=4$,\\\\\n$\\therefore EH=DH=3$,\\\\\n$\\therefore BE=3+4=7$.\\\\\nIn conclusion, $BE=\\boxed{\\frac{11}{2}}$ or \\boxed{7}.",
    "image_path": "dataset/mm_math/images/MM_Math/53151227_95.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $ABC$ is an equilateral triangle with side length $4$, $AD$ is the median of side $BC$, and $F$ is a moving point on side $AD$, with $E$ being the midpoint of side $AC$. What is the measure of $\\angle EFC$ when the perimeter of $\\triangle ECF$ is minimized?\n",
    "completion": "\\textbf{Solution:} Draw $EM\\parallel BC$ through $E$, intersecting $AD$ at $N$,\\\\\n$\\therefore \\angle AMN=\\angle ABC$,\\\\\n$\\because$ the sides of equilateral triangle $ABC$ are of length 4, and $E$ is the midpoint of side $AC$.\\\\\n$\\therefore AC=AB=BC=4$, AE=2, $\\angle ABC=60^\\circ=\\angle AMN$, $\\angle AEM=\\angle ACB=60^\\circ$,\\\\\n$\\therefore EC=2=AE$, $\\triangle AME$ is an equilateral triangle,\\\\\n$\\therefore AM=BM=2$,\\\\\n$\\therefore AM=AE$,\\\\\n$\\because AD$ is the median to side $BC$, and $\\triangle ABC$ is an equilateral triangle,\\\\\n$\\therefore AD\\perp BC$,\\\\\n$\\because EM\\parallel BC$,\\\\\n$\\therefore AD\\perp EM$,\\\\\n$\\because AM=AE$,\\\\\n$\\therefore $E and M are symmetrical about $AD$,\\\\\nConnect $CM$ intersecting $AD$ at $F$, then connect $EF$,\\\\\nunder these conditions, the sum of $EF+CF$ is minimized, and the perimeter of $\\triangle CEF$ is minimized,\\\\\n$\\because \\triangle ABC$ is an equilateral triangle,\\\\\n$\\therefore \\angle ACB=60^\\circ$, AC=BC=AB,\\\\\n$\\because AM=BM$, $CM\\perp AB$,\\\\\n$\\therefore \\angle ECF= \\frac{1}{2}\\angle ACB=30^\\circ$, $\\angle AMC=90^\\circ$,\\\\\n$\\therefore \\angle CME=90^\\circ-60^\\circ=30^\\circ$,\\\\\nBy the property of axial symmetry, we have: $\\angle MEF=\\angle FME=30^\\circ$,\\\\\n$\\therefore \\angle CFE=30^\\circ+30^\\circ=60^\\circ$.\\\\\nThus, the answer is: $\\boxed{60^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/52604077.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated in the rectangle $ABCD$, where $DC=6$, there exists a point $E$ on $DC$. When $\\triangle ADE$ is folded along the line $AE$, such that point $D$ precisely falls on side $BC$, let this point be $F$. If the area of $\\triangle ABF$ is 24, what is the length of $BF$?\n",
    "completion": "\\textbf{Solution:} From the given information, the length of $BF$ is $\\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53065890_51.png"
  },
  {
    "solution": "\\boxed{3\\sqrt{3}+3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\triangle ABC$, $ \\angle BAC=60^\\circ$, $ \\angle ABC=45^\\circ$. AD bisects $ \\angle BAC$ and intersects BC at point D. P is a moving point on line AB. Connect DP and construct the parallelogram DPQB with DP and DB as adjacent sides. Connect CQ. If $ AC=6$, what is the minimum value of CQ?\n",
    "completion": "Solution:As shown in Figure 1, draw $CO \\perp AB$ at O and $DH \\perp AB$ at H.\n\nIn $\\triangle ACO$, $\\angle CAB=60^\\circ$,\n\n$\\therefore \\angle ACO=30^\\circ$,\n\n$\\therefore AO=\\frac{1}{2}AC=3$,\n\n$\\therefore OC=\\sqrt{AC^2-AO^2}=3\\sqrt{3}$,\n\nIn $\\triangle BCO$, $\\angle CBA=45^\\circ$,\n\n$\\therefore OB=CO=3\\sqrt{3}$,\n\n$\\therefore AB=AO+OB=3\\sqrt{3}+3$,\n\nSince $AD$ bisects $\\angle CAB$,\n\n$\\therefore \\angle DAB=\\frac{1}{2}\\angle CAB=30^\\circ$,\n\nIn $\\triangle DHB$, $\\angle CBA=45^\\circ$,\n\nLet $DH=HB=a$,\n\n$\\therefore AD=2DH=2a$,\n\n$\\therefore AH=\\sqrt{AD^2-DH^2}=\\sqrt{3}a$,\n\n$\\therefore AB=AH+BH=\\sqrt{3}a+a$,\n\n$\\therefore \\sqrt{3}a+a=3\\sqrt{3}+3$,\n\n$\\therefore a=3$,\n\n$\\therefore DH=3$,\n\nAs shown in Figure 2, draw $QG \\perp AB$ at G and connect DQ intersecting AB at M,\n\nSince quadrilateral DPQB is a parallelogram,\n\n$\\therefore DM=QM$,\n\nIn $\\triangle QGM$ and $\\triangle DHM$,\n\n$\\left\\{\\begin{array}{l}\n\\angle QGM=\\angle DHM=90^\\circ \\\\\n\\angle QMG=\\angle DMH \\\\\nQM=DM\n\\end{array}\\right.$,\n\n$\\therefore \\triangle QGM \\cong \\triangle DHM$ (AAS),\n\n$\\therefore QG=DH=3$,\n\nHence, the distance from Q to line AB is always 3,\n\nTherefore, point Q moves on a line parallel to AB, and the distance between the two lines is 3,\n\nBased on the shortest distance along the perpendicular,\n\nWhen points C, O, and Q are on the same line, at that moment, CQ is minimal, as shown in Figure 3,\n\nThe minimum value is: $CO+3=3\\sqrt{3}+3$,\n\nTherefore, the answer is: $\\boxed{3\\sqrt{3}+3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52387296.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, $AC=BC$, point $C(2, 0)$, point $B(0, 4)$, and the graph of the inverse proportion function $y=\\frac{k}{x} (k\\ne 0)$ passes through point A. The line $OA$ is translated upwards by $m$ units and then passes through the point $(3, n)$ on the graph of the inverse proportion function $y=\\frac{k}{x} (k\\ne 0)$. Find the value of $m$?\n",
    "completion": "\\textbf{Solution:} Given the problem, \\\\\n$\\because$ $(3,n)$ lies on the curve of $y=\\frac{12}{x}$, \\\\\n$\\therefore$ $n=4$, \\\\\nLet the equation of line $OA$ be $y=k_1x$, \\\\\n$\\therefore$ $2=6k_1$, \\\\\n$\\therefore$ $k_1=\\frac{1}{3}$, \\\\\n$\\therefore$ the equation of line $OA$ is $y=\\frac{1}{3}x$, \\\\\n$\\therefore$ the equation of line $OA$ after being translated upwards by $m$ units is: $y=\\frac{1}{3}x+m$, \\\\\n$\\because$ the line $y=\\frac{1}{3}x+m$ goes through $(3,4)$, \\\\\n$\\therefore$ $4=\\frac{1}{3}\\cdot 3+m$, \\\\\nSolving this, we find $m=\\boxed{1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52977778_228.png"
  },
  {
    "solution": "\\boxed{48cm^{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, what is the area of the shaded part?\n",
    "completion": "\\[ \\textbf{Solution:} \\text{ Given:} \\text{ The length of the shaded part is: } \\sqrt{13^{2}-5^{2}}=12, \\]\n\\[ \\because \\text{ the width is } 4 \\text{ cm}, \\]\n\\[ \\therefore \\text{ the area is: } 4\\times 12=\\boxed{48cm^{2}} \\]",
    "image_path": "dataset/mm_math/images/MM_Math/53436907.png"
  },
  {
    "solution": "\\boxed{40^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $CD$ is the median to the hypotenuse $AB$. If $\\angle A=20^\\circ$, what is the measure of $\\angle BDC$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle ACB=90^\\circ$ and $CD$ is the median to the hypotenuse $AB$,\\\\\nit follows that $BD=CD=AD$,\\\\\ntherefore, $\\angle A=\\angle DCA=20^\\circ$,\\\\\nthus, $\\angle BDC=\\angle A+\\angle DCA=20^\\circ+20^\\circ=\\boxed{40^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53033451.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{32}{3x}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the vertices $A$ and $C$ of rectangle $ABCD$ are located on the negative half-axis of the $x$-axis and the positive half-axis of the $y$-axis, respectively. The $y$-axis bisects side $AB$. The coordinates of point $A$ are $(-2,0)$, and $AB=5$. What is the expression of the inverse proportion function passing through point $D$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, draw $DE\\perp x$-axis at point $E$, and let the intersection point of $AB$ and the $y$-axis be point $F$,\n\nsince quadrilateral $ABCD$ is a rectangle,\n\nthen $AD=BC$, $\\angle BAD=\\angle B=90^\\circ$,\n\nsince the $y$-axis bisects side $AB$, and $AB=5$,\n\nthen $AF=BF=\\frac{5}{2}$,\n\nsince $A(-2,0)$,\n\nthen $OA=2$,\n\ntherefore, in $Rt\\triangle OAF$, $OF=\\sqrt{AF^{2}-OA^{2}}=\\frac{3}{2}$,\n\nin $\\triangle BCF$ and $\\triangle OAF$, $\\left\\{\\begin{array}{l}\\angle BFC=\\angle OFA \\\\ \\angle B=\\angle AOF=90^\\circ \\end{array}\\right.$,\n\nthus $\\triangle BCF\\sim \\triangle OAF$,\n\nthus $\\frac{BC}{OA}=\\frac{BF}{OF}$, that is, $\\frac{BC}{2}=\\frac{\\frac{5}{2}}{\\frac{3}{2}}$,\n\nsolving gives $BC=\\frac{10}{3}$,\n\nthus $AD=\\frac{10}{3}$,\n\nsince $\\angle BAD=90^\\circ$, $DE\\perp x$-axis,\n\nthus $\\angle ADE+\\angle DAE=\\angle FAO+\\angle DAE=90^\\circ$,\n\nthus $\\angle ADE=\\angle FAO$,\n\nin $\\triangle ADE$ and $\\triangle FAO$, $\\left\\{\\begin{array}{l}\\angle ADE=\\angle FAO \\\\ \\angle AED=\\angle FOA=90^\\circ \\end{array}\\right.$,\n\nthus $\\triangle ADE\\sim \\triangle FAO$,\n\nthus $\\frac{DE}{OA}=\\frac{AE}{OF}=\\frac{AD}{AF}$, that is, $\\frac{DE}{2}=\\frac{AE}{\\frac{3}{2}}=\\frac{\\frac{10}{3}}{\\frac{5}{2}}$,\n\nsolving gives $DE=\\frac{8}{3}$, $AE=2$,\n\nthus $OE=OA+AE=4$,\n\nthus $D(-4,\\frac{8}{3})$,\n\nlet the expression for the inverse proportion function passing through point $D$ be $y=\\frac{k}{x}$,\n\nsubstituting point $D(-4,\\frac{8}{3})$ gives: $k=-4\\times \\frac{8}{3}=-\\frac{32}{3}$,\n\nthen the expression for the inverse proportion function passing through point $D$ is $y=-\\frac{32}{3x}$,\n\nhence, the answer is: $\\boxed{y=-\\frac{32}{3x}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51403202.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, within the square $ABCD$, where $AB = 8$, the diagonals $AC$ and $BD$ intersect at point $O$, and $N$ is the midpoint of $AO$. The point $M$ is located on side $BC$, with $BM = 6$. Let $P$ be a point on the diagonal $BD$. What is the maximum value of $PM - PN$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, take point N' as the symmetric point of N with BD as the axis of symmetry. Extend and connect MN' to intersect BD at P, and connect NP.\\\\\nBased on the property of axis symmetry, it is known that $PN = PN'$,\\\\\n$\\therefore PM - PN = PM - PN' \\leq MN'$,\\\\\nWhen points P, M, and N' are collinear, take “=” in the equation,\\\\\n$\\because$ The side length of the square is 8,\\\\\n$\\therefore AC = \\sqrt{2}AB = 8\\sqrt{2}$,\\\\\n$\\because$ O is the midpoint of AC,\\\\\n$\\therefore AO = OC = 4\\sqrt{2}$,\\\\\n$\\because$ N is the midpoint of OA,\\\\\n$\\therefore ON = 2\\sqrt{2}$,\\\\\n$\\therefore O{N'} = C{N'} = 2\\sqrt{2}$,\\\\\n$\\therefore A{N'} = 6\\sqrt{2}$,\\\\\n$\\because BM = 6$,\\\\\n$\\therefore CM = AB - BM = 8 - 6 = 2$,\\\\\n$\\therefore \\frac{CM}{BM} = \\frac{C{N}^{'}}{A{N}^{'}} = \\frac{1}{3}$,\\\\\n$\\therefore PM\\parallel AB\\parallel CD,\\ \\angle CMN' = 90^\\circ$,\\\\\n$\\because \\angle N'CM = 45^\\circ$,\\\\\n$\\therefore \\triangle N'CM$ is an isosceles right triangle,\\\\\n$\\therefore CM = MN' = 2$,\\\\\nthus, the maximum value of $PM - PN$ is $\\boxed{2}$,\\\\",
    "image_path": "dataset/mm_math/images/MM_Math/52251038.png"
  },
  {
    "solution": "\\boxed{-\\frac{1}{2}x+3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $AB=2$, $BC=4$, point $D$ is the midpoint of side $AB$, and the graph of the inverse proportion function $y_{1}=\\frac{k}{x}$ $(x>0)$ passes through point $D$ and intersects $BC$ at point $E$. Find the value of $k$ and the equation of line $DE$?\n",
    "completion": "\\textbf{Solution:} Given that $AB=2$ and $BC=4$,\\\\\nthus $B(4, 2)$.\\\\\nSince point $D$ is the midpoint of side $AB$,\\\\\nthus $D(4, 1)$, hence $k=4$.\\\\\nThe inverse proportion function equation is $y_{1}=\\frac{4}{x}(x>0)$,\\\\\nsince $E$ is a point on $BC$, we have $y_{E}=2$.\\\\\nThus $x_{E}=\\frac{4}{2}=2$, hence $E(2, 2)$.\\\\\nAssuming the equation of line $DE$ is $y_{2}=kx+b$, we get\\\\\n$\\begin{cases}2=2k+b\\\\ 1=4k+b\\end{cases}$, which gives $\\begin{cases}k=-\\frac{1}{2}\\\\ b=3\\end{cases}$,\\\\\nhence the equation of line $DE$ is $y_{2}=-\\frac{1}{2}x+3$,\\\\\ntherefore $k=\\boxed{-\\frac{1}{2}}$ and the equation of line $DE$ is $y_{2}=\\boxed{-\\frac{1}{2}x+3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52251436_127.png"
  },
  {
    "solution": "\\boxed{x<1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the graph, in the Cartesian coordinate system \\(xOy\\), the line \\(l_1: y_1 = 2x\\) intersects with the line \\(l_2: y_2 = -x + 3\\) at point A. When \\(y_1 < y_2\\), directly write the range of values for \\(x\\).\n",
    "completion": "\\textbf{Solution:} From (1), it is known that point A is at $(1, 2)$,\\\\\nAccording to the graph, when ${y}_{1}<{y}_{2}$, the range of x can be directly written as $\\boxed{x<1}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52211015_117.png"
  },
  {
    "solution": "\\boxed{(0,4)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure (1), it is known that the line $y=-2x+4$ intersects the $x$-axis and the $y$-axis at points $A$ and $C$ respectively. A rectangle $OABC$ is constructed in the first quadrant with $OA$ and $OC$ as sides. What are the coordinates of points $A$ and $C$?\n",
    "completion": "\\textbf{Solution:} The coordinates of point $A$ are \\boxed{(2,0)}; the coordinates of point $C$ are \\boxed{(0,4)}.",
    "image_path": "dataset/mm_math/images/MM_Math/51419687_119.png"
  },
  {
    "solution": "\\boxed{3\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the side length of square $ABCD$ is $6$. Point $P$ is a movable point on the side $BC$. With $P$ as the right angle vertex and $AP$ as the right angle side, an isosceles right $\\triangle APE$ is constructed. $M$ is the midpoint of side $AE$. When point $P$ moves from point $B$ to point $C$, what is the length of the path that point $M$ moves?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram below, connect AC and BD which intersect at point O, connect EC, and draw ET$\\perp$BC intersecting the extension line of BC at T.\\\\\n$\\because$ $\\triangle APE$ is an isosceles right triangle,\\\\\n$\\therefore \\angle APE=90^\\circ$, $AP=PE$\\\\\n$\\therefore \\angle APB+\\angle TPE=90^\\circ$.\\\\\n$\\because$ Quadrilateral ABCD is a square, ET$\\perp$BC,\\\\\n$\\therefore \\angle ABP=90^\\circ$, $\\angle PTE=90^\\circ$.\\\\\n$\\therefore \\angle ABP=\\angle PTE$, $\\angle BAP+\\angle APB=90^\\circ$.\\\\\n$\\therefore \\angle BAP=\\angle TPE$.\\\\\n$\\therefore \\triangle ABP\\cong \\triangle PTE$ (AAS).\\\\\n$\\therefore AB=PT$, $PB=ET$.\\\\\n$\\because$ Quadrilateral ABCD is a square,\\\\\n$\\therefore AB=BC$.\\\\\n$\\therefore BC=PT$.\\\\\n$\\therefore BC-PC=PT-BC$, that is, PB=CT.\\\\\n$\\therefore PB=CT=ET$.\\\\\n$\\therefore \\angle TEC=\\angle TCE=45^\\circ$.\\\\\n$\\because$ In square ABCD, AC and BD intersect at point O,\\\\\n$\\therefore O is the midpoint of AC$, $\\angle DBC=45^\\circ$.\\\\\n$\\therefore \\angle DBC=\\angle TCE$.\\\\\n$\\therefore OD\\parallel CE$.\\\\\n$\\because$ M is the midpoint of AE,\\\\\n$\\therefore$ OM is the median line of $\\triangle ACE$.\\\\\n$\\therefore OM\\parallel CE$.\\\\\n$\\therefore$ Point M lies on the line OD.\\\\\n$\\because$ Point P moves along side BC,\\\\\n$\\therefore$ The trajectory of point M is OD.\\\\\n$\\because$ The side length of square ABCD is 6, and AC, BD intersect at point O,\\\\\n$\\therefore AB=6$, $AD=6$, O is the midpoint of BD.\\\\\n$\\therefore BD=\\sqrt{AB^2+AD^2}=6\\sqrt{2}$.\\\\\n$\\therefore OD=\\frac{1}{2}BD=3\\sqrt{2}$.\\\\\nHence, the answer is: $\\boxed{3\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52344793.png"
  },
  {
    "solution": "\\boxed{9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, line $L$ passes through vertex $A$ of the square $ABCD$. Lines are drawn from points $B$ and $D$ such that $DE \\perp l$ at point $E$ and $BF \\perp l$ at point $F$. If $DE=4$ and $BF=5$, what is the length of $EF$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral $ABCD$ is a square,\\\\\nit follows that $AD=AB$ and $\\angle DAB=90^\\circ$,\\\\\ntherefore $\\angle BAF+\\angle DAE=90^\\circ$\\\\\nGiven that $DE\\perp l$,\\\\\nit follows that $\\angle DAE+\\angle ADE=90^\\circ$\\\\\ntherefore, $\\angle BAF=\\angle ADE$\\\\\nGiven that $DE\\perp l$ and $BF\\perp l$,\\\\\nit follows that $\\angle AFB=\\angle AED=90^\\circ$\\\\\nIn triangles $DAE$ and $ABF$,\\\\\ngiven that $\\left\\{\\begin{array}{l}\\angle AED=\\angle AFB \\\\ \\angle BAF=\\angle ADE \\\\ AD=AB \\end{array}\\right.$,\\\\\nit follows that $\\triangle DAE\\cong \\triangle ABF$\\\\\ntherefore, $AF=DE=4$, $AE=BF=5$\\\\\ntherefore, $EF=AF+AE=4+5=\\boxed{9}$",
    "image_path": "dataset/mm_math/images/MM_Math/52386889.png"
  },
  {
    "solution": "\\boxed{F_3(-2, -1)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the Cartesian coordinate system, the graph of the quadratic function $y=ax^2+bx+c\\ (a \\neq 0)$ intersects the x-axis at points $A(-3,0)$ and $B(1,0)$, and intersects the y-axis at point $C(0, -3)$. The vertex of the parabola is $D$, and its axis of symmetry intersects the x-axis at point $E$. On the segment $AC$, is there a point $F$ such that triangle $\\triangle AEF$ is an isosceles triangle? If such a point exists, provide the coordinates of point $F$ directly; if it does not exist, please explain the reason.\n",
    "completion": "\\textbf{Solution}: Let $F_1(-1, -2)$, $F_2(\\sqrt{2}-3, -\\sqrt{2})$, and $F_3(-2, -1)$. $\\boxed{F_1(-1, -2)}$, $\\boxed{F_2(\\sqrt{2}-3, -\\sqrt{2})}$, and $\\boxed{F_3(-2, -1)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53147831_66.png"
  },
  {
    "solution": "\\boxed{\\frac{12}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ is an isosceles triangle, with $AB=BC=5$ and $AC=6$. Point $O$ is the midpoint of $AC$, and a line is drawn through $O$ such that $OD \\perp AB$ at point $D$. Point $P$ lies on ray $BO$, and point $Q$ lies on line segment $BC$ (not coinciding with points $B$ or $C$). Connect $PQ$. Find the lengths of $BO$ and $OD$.\n",
    "completion": "\\textbf{Solution:} Given that $AB=BC=5$, $AC=6$, and $OD\\perp AB$,\\\\\nit follows that $AO=CO=3$,\\\\\nin $\\triangle ABO$, we have $BO=\\boxed{4}$,\\\\\nsince the area $S_{\\triangle ABO}=\\frac{1}{2}\\times 5\\times OD=\\frac{1}{2}\\times 3\\times 4$,\\\\\nthus, $OD=\\boxed{\\frac{12}{5}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52914212_128.png"
  },
  {
    "solution": "\\boxed{\\frac{2}{7}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $AB \\parallel CD \\parallel EF$ and $BD: DF = 2: 5$. What is the value of $\\frac{AC}{AE}$?\n",
    "completion": "\\textbf{Solution:} Given $\\because$ AB$\\parallel$CD$\\parallel$EF,\\\\\n$\\therefore$ AC:CE=BD:DF,\\\\\n$\\because$ BD:DF$=2:5$,\\\\\n$\\therefore$ AC:CE= BD:DF$=2:5$, i.e., CE$=\\frac{5}{2}$AC,\\\\\n$\\therefore$ AE$=\\frac{7}{2}$AC,\\\\\n$\\therefore$ AC:AE$=2:7=\\boxed{\\frac{2}{7}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51284381.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the diameter $CD=8$ of the circle $\\odot O$, $AB$ is a chord of $\\odot O$, and $AB \\perp CD$ at the foot point $M$, $OM=2$. What is the length of $AB$?\n",
    "completion": "\\textbf{Solution:} Connect OB, \\\\\n$\\because$ the diameter $CD=8$, and $AB\\perp CD$, $OM=2$\\\\\n$\\therefore$ BM=$\\sqrt{OB^{2}-OM^{2}}$\\\\\n=$\\sqrt{4^{2}-2^{2}}$\\\\\n=$2\\sqrt{3}$,\\\\\nAccording to the Perpendicular Diameter Theorem, we obtain\\\\\nAB=2BM=$\\boxed{4\\sqrt{3}}$",
    "image_path": "dataset/mm_math/images/MM_Math/52623419.png"
  },
  {
    "solution": "\\boxed{\\sqrt{2}+1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is a chord of $\\odot O$, diameter $CD \\perp AB$, intersecting $AB$ at point $H$, and $OA$ is connected. If $\\angle A=45^\\circ$ and $AB=2$, what is the length of $DH$?\n",
    "completion": "\\textbf{Solution:}\n\nSince $AB$ is a chord of the circle $O$, and the diameter $CD \\perp AB$,\n\nit follows that $H$ is the midpoint of $AB$, meaning $AH=\\frac{1}{2}AB=1$,\n\nMoreover, since $\\angle A=45^\\circ$ and $CD \\perp AB$,\n\nit follows that $\\triangle AHO$ is an isosceles right triangle,\n\nTherefore, $OH=AH=1$, $OA=\\sqrt{OH^2+AH^2}=\\sqrt{2}$,\n\nTherefore, $DH=OD+OH=OA+OH=\\sqrt{2}+1=\\boxed{\\sqrt{2}+1}$",
    "image_path": "dataset/mm_math/images/MM_Math/53042917.png"
  },
  {
    "solution": "\\boxed{37.5^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: When two acute angles of a set of set squares coincide, $\\angle AOB=45^\\circ$, $\\angle COD=30^\\circ$, and $OM$, $ON$ are the angle bisectors of $\\angle AOC$, $\\angle BOD$ respectively. When $\\angle COD$ is rotated around point $O$ to the position shown in Figure 2, and $\\angle BOC=10^\\circ$, what is the degree measure of $\\angle MON$?\n",
    "completion": "\\textbf{Solution:} Given $\\because$ when $\\angle BOC=10^\\circ$, $\\angle AOC=35^\\circ$, $\\angle BOD=20^\\circ$, and OM, ON are the bisectors of $\\angle AOC$, $\\angle BOD$, respectively,\\\\\n$\\therefore \\angle BON= \\frac{1}{2} \\angle BOD=10^\\circ$,\\\\\n$\\therefore \\angle MOC= \\frac{1}{2} \\angle AOC=17.5^\\circ$,\\\\\n$\\therefore \\angle MON=\\angle MOC+\\angle BON+\\angle BOC=17.5^\\circ+10^\\circ+10^\\circ=\\boxed{37.5^\\circ}.",
    "image_path": "dataset/mm_math/images/MM_Math/53195491_129.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AE$ is the altitude on side $BC$. If $AD$ is the median on side $BC$, $AE=3\\,\\text{cm}$, and $S_{\\triangle ABC}=12\\,\\text{cm}^2$. What is the length of $DC$ in $\\text{cm}$?\n",
    "completion": "\\textbf{Solution:} Given that $AE=3\\,\\text{cm}$ and $S_{\\triangle ABC}=12\\,\\text{cm}^2$,\\\\\nit follows that $BC=12\\times 2\\div 3=8\\,\\text{cm}$,\\\\\nsince $AD$ is the median to side $BC$,\\\\\nthus $DC =\\frac{1}{2} BC=4\\,\\text{cm}$;\\\\\nhence, the length of $DC$ is $\\boxed{4}\\,\\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52693985_124.png"
  },
  {
    "solution": "\\boxed{48^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point C is on the circle $\\odot O$. If $\\angle ACB=24^\\circ$, what is the degree measure of $\\angle AOB$?\n",
    "completion": "\\textbf{Solution:} From the Inscribed Angle Theorem, it can be derived that $ \\angle AOB=2\\angle ACB=48^\\circ$\\\\\nTherefore, the answer is: $ \\boxed{48^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52872173.png"
  },
  {
    "solution": "\\boxed{\\frac{8}{7}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in quadrilateral $ABCD$, $BC\\parallel AD$, $\\angle ADC=90^\\circ$, $BC=8\\text{cm}$, $AD=CD=10\\text{cm}$. Point $E$ starts from point $B$ and moves uniformly along the direction of $BC$ at a speed of $2\\text{cm/s}$; at the same time, point $F$ starts from point $D$ and moves uniformly along the direction of $DA$ at a speed of $3\\text{cm/s}$. Draw $EH\\perp AD$ through point $E$ with $H$ as the foot of the perpendicular. $EH$ intersects $AC$ at point $G$, and $FG$ is connected. Let the movement time be $t(s)\\left(0<t<\\frac{10}{3}\\right)$. Solve the following question: During the movement, is there a moment when the quadrilateral formed by points $B$, $E$, $F$, $H$ becomes a parallelogram? If such a moment exists, please write down the value of $t$ directly.\n",
    "completion": "\\textbf{Solution:} Exists; \\\\\n$\\because CB\\parallel AD$, \\\\\n$\\therefore$ when $BE=FH$, the quadrilateral with vertices $B$, $E$, $F$, $H$ is a parallelogram, \\\\\nwhen point $F$ is on segment $DH$, $FH=DH-DF=8-2t-3t=8-5t$, \\\\\nwe get $2t=8-5t$, solving we find $t=\\boxed{\\frac{8}{7}}$, \\\\\nwhen point $F$ is on segment $AH$, $FH=DF-DH=3t-(8-2t)=5t-8$, \\\\\nwe get $2t=5t-8$, solving we find $t=\\boxed{\\frac{8}{3}}$, \\\\\nIn conclusion, when $t=\\boxed{\\frac{8}{3}}$ or $t=\\boxed{\\frac{8}{7}}$, the quadrilateral with vertices $B$, $E$, $F$, $H$ is a parallelogram.",
    "image_path": "dataset/mm_math/images/MM_Math/52211373_87.png"
  },
  {
    "solution": "\\boxed{9\\sqrt{10}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\triangle ABC$, $\\angle ACB=90^\\circ$. $BD=AB$ is taken on the extension line of $BC$, and $AD$ is connected. A perpendicular $BE\\perp AD$ is drawn from point $B$ to point $E$, intersecting $AC$ at point $F$. $DF$ is connected. Point $P$ is a moving point on the ray $BE$. If $AC=9$ and $BC=12$, what is the length of $BP$ when $\\triangle APB$ is similar to $\\triangle AFD$?\n",
    "completion": "\\textbf{Solution:} Connect AP. \\\\\nIn $\\triangle ACB$, $\\angle ACB=90^\\circ$, $AC=9$, $BC=12$, \\\\\n$\\therefore AB=\\sqrt{AC^{2}+BC^{2}}=\\sqrt{9^{2}+12^{2}}=15$, \\\\\n$\\therefore BD=BA=15$, $CD=BD-BC=15-12=3$, \\\\\n$\\therefore AD=\\sqrt{CD^{2}+AC^{2}}=\\sqrt{3^{2}+9^{2}}=3\\sqrt{10}$, \\\\\n$\\therefore BE\\perp AD$, \\\\\n$\\therefore AE=ED$, \\\\\n$\\therefore FA=FD$, \\\\\nLet $FA=FD=x$, then we have $x^{2}=(9-x)^{2}+3^{2}$, \\\\\n$\\therefore x=5$, \\\\\n$\\therefore FA=FD=5$, \\\\\nBecause $\\angle AEF=\\angle FCB=90^\\circ$, $\\angle AFE=\\angle CFB$, \\\\\n$\\therefore \\angle EAF=\\angle ABP$, \\\\\nWhen $\\triangle APB \\sim \\triangle AFD$, $\\frac{BP}{DF}=\\frac{AB}{AD}$, \\\\\n$\\therefore \\frac{BP}{5}=\\frac{15}{3\\sqrt{10}}$, \\\\\n$\\therefore BP=\\frac{5\\sqrt{10}}{2}$. \\\\\nWhen $\\triangle AFD \\sim \\triangle PAB$, $\\frac{AD}{BP}=\\frac{DF}{AB}$, \\\\\n$\\therefore \\frac{3\\sqrt{10}}{BP}=\\frac{5}{15}$, \\\\\n$\\therefore BP=9\\sqrt{10}$. \\\\\nIn conclusion, the length of $BP$ is $\\boxed{\\frac{5\\sqrt{10}}{2}}$ or $\\boxed{9\\sqrt{10}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52457428.png"
  },
  {
    "solution": "\\boxed{\\frac{4}{9}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In parallelogram ABCD shown in the figure, F is the midpoint of BC. Extend AD to E such that $AD: AE = 3: 4$. Connect EF and intersect DC at point G. What is the ratio ${S}_{\\triangle DEG} : {S}_{\\triangle CFG}$?\n",
    "completion": "\\textbf{Solution:} Given that $AD: AE = 3:4$,\\\\\nlet $AD=3x$ and $AE=4x$,\\\\\nhence $DE=AE-AD=4x-3x=x$,\\\\\nsince quadrilateral ABCD is a parallelogram,\\\\\nit follows that $AD\\parallel BC$ and $BC=AD=3x$,\\\\\ngiven that point F is the midpoint of BC,\\\\\nthus $CF=\\frac{1}{2}BC=\\frac{3}{2}x$,\\\\\nsince $AD\\parallel BC$,\\\\\nit results in $\\angle DEG=\\angle CFG$ and $\\angle EDG=\\angle FCG$,\\\\\nhence $\\triangle DEG\\sim \\triangle CFG$,\\\\\nthus $\\frac{{S}_{\\triangle DEG}}{{S}_{\\triangle CFG}}=\\left(\\frac{DE}{CF}\\right)^{2}=\\left(\\frac{x}{\\frac{3}{2}x}\\right)^{2}=\\boxed{\\frac{4}{9}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51446262.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{4}{x}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y=k_1x+b$ intersects with the graph of the inverse proportion function $y=\\frac{k_2}{x}$ at points A and B. The coordinates of point A are $(-1, 4)$, and the coordinates of point B are $(4, n)$. What are the expressions of these two functions?\n",
    "completion": "\\textbf{Solution}: According to the problem, the graph of the function $y=\\frac{k_2}{x}$ passes through points $A(-1, 4)$ and $B(4, n)$,\\\\\n$\\therefore k_2=-1\\times 4=4n$,\\\\\n$\\therefore k_2=-4$, $n=-1$,\\\\\n$\\therefore$ point $B(4, -1)$,\\\\\nFurthermore, since the linear function $y=k_1x+b$ passes through points $A$ and $B$,\\\\\n$\\therefore \\left\\{\\begin{array}{l}\n-k_1+b=4 \\\\\n4k_1+b=-1\n\\end{array}\\right.$, yielding: $\\left\\{\\begin{array}{l}\nk_1=-1 \\\\\nb=3\n\\end{array}\\right.$;\\\\\n$\\therefore$ the expression for the linear function is $\\boxed{y=-x+3}$, and for the inverse proportion function, $\\boxed{y=-\\frac{4}{x}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51446676_223.png"
  },
  {
    "solution": "\\boxed{y_{2}=\\frac{8}{x}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the line $y_1=k_1x+b$ intersects with the hyperbola $y_2=\\frac{k_2}{x}$ at points $A(4, 2)$ and $B(m, -4)$. Find the expressions for functions $y_1$ and $y_2$.\n",
    "completion": "\\textbf{Solution:} Since the line $y_{1}=k_{1}x+b$ intersects the hyperbola $y_{2}=\\frac{k_{2}}{x}$ at points $A(4, 2)$ and $B(m, -4)$,\\\\\nwe have $2=\\frac{k_{2}}{4}$, solving which gives $k_{2}=8$,\\\\\ntherefore, the equation of the hyperbola $y_{2}$ is $y_{2}=\\frac{8}{x}$,\\\\\nsubsequently, substituting $B(m, -4)$ into $y_{2}=\\frac{8}{x}$, we get $-4=\\frac{8}{m}$, solving which gives $m=-2$,\\\\\nthus $B(-2, -4)$, substituting $A(4, 2)$ and $B(-2, -4)$ into $y_{1}=k_{1}x+b$ yields the system $\\left\\{\\begin{array}{l}4k_{1}+b=2 \\\\ -2k_{1}+b=-4\\end{array}\\right.$, from which we solve $\\left\\{\\begin{array}{l}k_{1}=1 \\\\ b=-2\\end{array}\\right.$,\\\\\ntherefore, the equation of the line $y_{1}$ is: \\boxed{y_{1}=x-2}; and the equation of the hyperbola $y_{2}$ is: \\boxed{y_{2}=\\frac{8}{x}}.",
    "image_path": "dataset/mm_math/images/MM_Math/52381008_156.png"
  },
  {
    "solution": "\\boxed{90^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $O$ is a point on the straight line $AB$. An arbitrary ray $OM$ passes through $O$, $OC$ bisects $\\angle AOM$, and $OD$ bisects $\\angle BOM$. What is the measure of $\\angle COD$?\n",
    "completion": "\\textbf{Solution:} Since $OC$ bisects $\\angle AOM$ and $OD$ bisects $\\angle BOM$, \\\\\nthus $\\angle MOC= \\frac{1}{2}\\angle AOM$ and $\\angle MOD= \\frac{1}{2}\\angle BOM$, \\\\\nhence $\\angle COD=\\angle MOC+\\angle MOD= \\frac{1}{2}\\angle AOM+\\frac{1}{2}\\angle BOM=\\frac{1}{2}(\\angle AOM+\\angle BOM)=\\frac{1}{2}\\times 180^\\circ=\\boxed{90^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55426225.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, D and E are the midpoints of sides AB and AC of $\\triangle ABC$, respectively. CD and BE intersect at point O. What is the value of $\\frac{S_{\\triangle COE}}{S_{\\triangle BOC}}$?\n",
    "completion": "\\textbf{Solution:} Since $D$ and $E$ are the midpoints of sides $AB$ and $AC$ of $\\triangle ABC$ respectively,\\\\\nit follows that $DE \\parallel BC$ and $DE = \\frac{1}{2}BC$,\\\\\nthereby $\\triangle DOE \\sim \\triangle COB$,\\\\\nwhich implies $\\frac{DE}{BC}=\\frac{OE}{OB}=\\frac{1}{2}$,\\\\\nresulting in ${S}_{\\triangle COE}:{S}_{\\triangle BOC}=\\frac{OE}{OB}=\\boxed{\\frac{1}{2}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51351895.png"
  },
  {
    "solution": "\\boxed{88^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in pentagon $ABCDE$, $AB \\parallel CD$, and $\\angle 1$, $\\angle 2$, $\\angle 3$ are the exterior angles of $\\angle BAE$, $\\angle AED$, and $\\angle EDC$ respectively. If $\\angle 1=32^\\circ$ and $\\angle 3=60^\\circ$, what is the value of $\\angle 2$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure,\\\\\nsince AB $\\parallel$ CD,\\\\\ntherefore, $\\angle$4 + $\\angle$5 = 180$^\\circ$,\\\\\nalso, since $\\angle$1 + $\\angle$3 = 92$^\\circ$,\\\\\naccording to the exterior angle sum theorem of polygons, $\\angle$1 + $\\angle$2 + $\\angle$3 + $\\angle$4 + $\\angle$5 = 360$^\\circ$,\\\\\nthus, $\\angle$2 = 360$^\\circ$ - 180$^\\circ$ - 92$^\\circ$ = \\boxed{88^\\circ}.",
    "image_path": "dataset/mm_math/images/MM_Math/51636053.png"
  },
  {
    "solution": "\\boxed{}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ABC=90^\\circ$, $\\angle A=60^\\circ$, and $D$ is a point on side $AC$, with $\\angle DBC=30^\\circ$, $AC=12\\, \\text{cm}$. What is the perimeter of $\\triangle ABD$?\n",
    "completion": "\\textbf{Solution:} The information provided by the \\LaTeX\\ code of this problem is insufficient to address the core issue, thus preventing the provision of specific solution steps and results.",
    "image_path": "dataset/mm_math/images/MM_Math/55468485.png"
  },
  {
    "solution": "\\boxed{-2x + 6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, the line \\(l_{1}\\colon y=x+3\\) intersects with the line \\(l_{2}\\) passing through point \\(A(3,0)\\) at point \\(C(1,m)\\), and intersects with the \\(x\\)-axis at point \\(B\\). What is the equation of line \\(l_{2}\\)?\n",
    "completion": "\\textbf{Solution:} Substitute $(1, m)$ into $y = x + 3$ to get $m = 4$, therefore $C(1, 4)$. Let the equation of line $l_2$ be $y = kx + b$. Substituting $(1, 4)$ and $(3, 0)$ into the equation, we get \n\\[\n\\left\\{\n\\begin{array}{l}\nk + b = 4 \\\\\n3k + b = 0\n\\end{array}\n\\right.\n\\]\nSolving the system, we find \n\\[\n\\left\\{\n\\begin{array}{l}\nk = -2 \\\\\nb = 6\n\\end{array}\n\\right.\n\\]\nTherefore, the equation of line $l_2$ is $y = -2x + 6$. Hence, the answer is $y = \\boxed{-2x + 6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50979784_102.png"
  },
  {
    "solution": "\\boxed{72^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in circle $\\odot O$, $AB$ is the diameter, and point $C$ is a point on the circle. The minor arc $AC$ is folded along chord $AC$ to intersect $AB$ at point $D$. Connect $CD$. If $\\angle BAC=18^\\circ$, then what is the measure of $\\angle BDC$?\n",
    "completion": "\\textbf{Solution:} Since $AB$ is the diameter, \\\\\n$\\therefore \\angle ACB=90^\\circ$, \\\\\nSince $\\angle BAC=18^\\circ$, \\\\\n$\\therefore \\angle B=90^\\circ-\\angle BAC=90^\\circ-18^\\circ=72^\\circ$, \\\\\nAccording to the property of reflection, \\\\\nthe inscribed angle corresponding to arc $\\overset{\\frown}{AC}$ is $\\angle B$, \\\\\nthe inscribed angle corresponding to arc $\\overset{\\frown}{ABC}$ is $\\angle ADC$, \\\\\n$\\therefore \\angle ADC+\\angle B=180^\\circ$, \\\\\nSince $\\angle ADC+\\angle BDC=180^\\circ$, \\\\\n$\\therefore \\angle B=\\angle BDC=\\boxed{72^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53149258.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, AC is the diagonal of rectangle ABCD. Fold side AB along AE so that point B falls on point M on AC, and fold side CD along CF so that point D falls on point N on AC. It is easy to prove that quadrilateral AECF is a parallelogram. When $\\angle BAE$ is how many degrees, the quadrilateral AECF becomes a rhombus.\n",
    "completion": "\\textbf{Solution}: When $\\angle BAE = 30^\\circ$, the quadrilateral AECF is a rhombus,\\\\\nReason: It is known from folding that $\\angle BAE = \\angle CAE = 30^\\circ$,\\\\\n$\\because \\angle B = 90^\\circ$,\\\\\n$\\therefore \\angle ACE = 90^\\circ - 30^\\circ - 30^\\circ = 30^\\circ$,\\\\\nwhich means $\\angle CAE = \\angle ACE$,\\\\\n$\\therefore EA = EC$,\\\\\n$\\because$ the quadrilateral AECF is a parallelogram,\\\\\n$\\therefore$ the quadrilateral AECF is a rhombus,\\\\\nTherefore, the answer is: $\\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52395201.png"
  },
  {
    "solution": "\\boxed{\\sqrt{13}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within quadrilateral $ABCD$, point $F$ is the midpoint of side $AD$, and side $BC$ is extended to point $E$ such that $CE = \\frac{1}{2}BC$. When connecting $DE$ and $CF$, the quadrilateral $CEDF$ forms a parallelogram. Given that $AB = 4$, $AD = 6$, and $\\angle B = 60^\\circ$, what is the length of $DE$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, draw $DH\\perp BE$ at point H. In quadrilateral $ABCD$, since $\\angle B=60^\\circ$ and $AB\\parallel DC$, it follows that $\\angle B=\\angle DCE$, hence $\\angle DCE=60^\\circ$. Since $AB=4$, it follows that $CD=AB=4$, thus $CH=\\frac{1}{2}CD=2$, and $DH=2\\sqrt{3}$. In quadrilateral $CEDF$, $CE=DF=\\frac{1}{2}AD=3$, thus $EH=1$. Therefore, in $\\triangle DHE$, according to the Pythagorean theorem, $DE=\\sqrt{(2\\sqrt{3})^2+1^2}=\\sqrt{13}$. Hence, the length of $DE$ is $\\boxed{\\sqrt{13}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51510410_131.png"
  },
  {
    "solution": "\\boxed{-\\frac{3}{x}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the line $l\\colon y=x+4$ intersects the graph of the inverse proportion function $y=\\frac{k}{x}$ ($x<0$) at point $A(-1, n)$, the line $l'$ passes through point $A$ and is symmetric to $l$ with respect to the line $x=-1$. What is the expression of the inverse proportion function?\n",
    "completion": "\\textbf{Solution:} Since point A$(-1, 3)$ is on the line $l\\colon y=x+4$,\\\\\nthe coordinates of point A are $(-1, 3)$.\\\\\nSince point A is also on the graph of the inverse proportion function $y= \\frac{k}{x}$ $(x<0)$,\\\\\nit follows that $k=-3$,\\\\\ntherefore, the expression for the inverse proportion function is $y= \\boxed{-\\frac{3}{x}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55608665_57.png"
  },
  {
    "solution": "\\boxed{-\\frac{1}{2}x^2+3x+8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=ax^2+3x+c\\ (a \\ne 0)$ intersects the $x$-axis at points $A(-2, 0)$ and $B$, and intersects the $y$-axis at point $C(0, 8)$. Point $P$ is a moving point on the parabola above line $BC$. Connect $CP$ and $PB$. The line $BC$ intersects the axis of symmetry $l$ of the parabola at point $E$. What is the analytic expression of the parabola?\n",
    "completion": "\\textbf{Solution:} Let's substitute $A(-2, 0)$, and $C(0, 8)$ into $y=ax^2+3x+c$,\\\\\nhence $\\begin{cases} 4a-6+c=0 \\\\ c=8 \\end{cases}$,\\\\\nsolving this, we get $\\begin{cases} a=-\\frac{1}{2} \\\\ c=8 \\end{cases}$,\\\\\ntherefore $y=-\\frac{1}{2}x^2+3x+8=\\boxed{-\\frac{1}{2}x^2+3x+8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53058938_101.png"
  },
  {
    "solution": "\\boxed{(-2，-2)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the graph, within the Cartesian coordinate system, the parabola $y = \\frac{1}{2}x^{2} + bx + c$ passes through point $A(-4,0)$. Point $M$ is the vertex of the parabola, and point $B$ is on the $y$-axis, with $OA=OB$. The line $AB$ intersects with the parabola at point $C(2,6)$ in the first quadrant. Find the equation of the parabola and the coordinates of the vertex $M$?\n",
    "completion": "\\textbf{Solution:} Let's substitute $A(-4,0)$, $C(2,6)$ into $y=\\frac{1}{2}x^{2}+bx+c$, we get: $\\left\\{\\begin{array}{l}0=\\frac{1}{2}\\times 16-4b+c \\\\ 6=\\frac{1}{2}\\times 4+2b+c\\end{array}\\right.$. Solving these, we find $\\left\\{\\begin{array}{l}b=2 \\\\ c=0\\end{array}\\right.$, \\\\\n$\\therefore$ the equation of the parabola is $y=\\frac{1}{2}x^{2}+2x=\\frac{1}{2}(x+2)^{2}-2$, \\\\\n$\\therefore$ the coordinates of the vertex $M$ are $\\boxed{(-2，-2)}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53043807_114.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In $\\triangle ABC$, $AB=AC$, and $\\angle B=30^\\circ$. The perpendicular bisector of side $AC$ intersects $AC$ and $BC$ at points $E$ and $D$ respectively. If $DE=2$, what is the length of $BC$?\n",
    "completion": "\\textbf{Solution:} Draw AD, \\\\\n$\\because$AB = AC, $\\angle B = 30^\\circ$, \\\\\n$\\therefore$ $\\angle C = 30^\\circ$, $\\angle BAC = 120^\\circ$, \\\\\n$\\because$ DE is the perpendicular bisector of AC, \\\\\n$\\therefore$ DC = DA, $\\angle DEC = 90^\\circ$, \\\\\n$\\therefore$ $\\angle DAC = \\angle DCA = 30^\\circ$, \\\\\n$\\because$ in $\\triangle DEC$, $\\angle C = 30^\\circ$, DE = 2, \\\\\n$\\therefore$ CD = AD = 2DE = 4, \\\\\n$\\because$ $\\angle BAC = 120^\\circ$, $\\angle DAC = 30^\\circ$, \\\\\n$\\therefore$ $\\angle BAD = 90^\\circ$, \\\\\n$\\therefore$ BD = 2AD = 8, \\\\\n$\\therefore$ BC = BD + CD = 8 + 4 = \\boxed{12}, \\\\",
    "image_path": "dataset/mm_math/images/MM_Math/51367325.png"
  },
  {
    "solution": "\\boxed{18}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, points $D$, $E$, and $F$ are the midpoints of sides $AB$, $BC$, and $CA$, respectively, and $AB=8\\,cm$, $AC=10\\,cm$. What is the perimeter of the quadrilateral $ADEF$ in $cm$?\n",
    "completion": "\\textbf{Solution:} Let's consider triangle $\\triangle ABC$, where points D, E, and F are the midpoints of sides AB, BC, and CA respectively, and we have AB = 8cm, AC = 10cm,\\\\\nHence, AD = 4cm, AF = 5cm,\\\\\nSince EF and DE are the midlines of $\\triangle ABC$,\\\\\nit follows that $DE \\parallel \\frac{1}{2}AC$ and $EF \\parallel \\frac{1}{2}AB$\\\\\nAlso, since AB = 8cm and AC = 10cm,\\\\\nit results in EF = 4cm and DE = 5cm,\\\\\nTherefore, the perimeter of the quadrilateral $ADEF$ equals $AD + DE + EF + AF = \\boxed{18}$cm.",
    "image_path": "dataset/mm_math/images/MM_Math/50482450.png"
  },
  {
    "solution": "\\boxed{1:3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that in $\\triangle ABC$, $\\cos A = \\frac{1}{3}$, where $BE$ and $CF$ are the altitudes on sides $AC$ and $AB$ respectively. Connecting $EF$, what is the ratio of the perimeters of $\\triangle AEF$ and $\\triangle ABC$?\n",
    "completion": "\\textbf{Solution:} Let $BE$ and $CF$ be the altitudes from $AC$ and $AB$, respectively,\\\\\nHence, $\\angle AEB=\\angle AFC=90^\\circ$,\\\\\nSince $\\angle A=\\angle A$,\\\\\nTherefore, $\\triangle AEB\\sim \\triangle AFC$,\\\\\nTherefore, $\\frac{AE}{AF}=\\frac{AB}{AC}$,\\\\\nTherefore, $\\frac{AE}{AB}=\\frac{AF}{AC}$,\\\\\nSince $\\angle A=\\angle A$,\\\\\nTherefore, $\\triangle AEF\\sim \\triangle ABC$,\\\\\nTherefore, the ratio of the perimeters of $\\triangle AEF$ and $\\triangle ABC$ is $AE:AB$,\\\\\nSince $\\cos A=\\frac{AE}{AB}=\\frac{1}{3}$,\\\\\nTherefore, the ratio of the perimeters of $\\triangle AEF$ and $\\triangle ABC$ is $AE:AB=\\boxed{1:3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53079123.png"
  },
  {
    "solution": "\\boxed{15^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the quadrilateral ABCD is a square. Perform the following steps:\n",
    "completion": "\\textbf{Solution:} Based on the drawing process, it is known that $AD=AP=PD$,\\\\\n$\\therefore$ $\\triangle ADP$ is an equilateral triangle,\\\\\n$\\therefore$ $\\angle DAP=\\angle ADP=\\angle APD=60^\\circ$,\\\\\n$\\because$ Quadrilateral $ABCD$ is a square,\\\\\n$\\therefore$ $AB=AD$, $\\angle BAD=\\angle ABC=90^\\circ$,\\\\\n$\\therefore$ $AB=AP$, $\\angle BAP=30^\\circ$,\\\\\n$\\therefore$ $\\angle ABP=\\angle APB=\\frac{180^\\circ-30^\\circ}{2}=75^\\circ$,\\\\\n$\\therefore$ $\\angle PBC=\\angle ABC-\\angle ABP=90^\\circ-75^\\circ=\\boxed{15^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52344909.png"
  },
  {
    "solution": "\\boxed{x < -2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the intersection's x-coordinate of the lines $y = -x + m$ and $y = nx + 4n$ ($n \\neq 0$) is $-2$. What is the solution set of the inequality $-x + m > nx + 4n$ with respect to $x$?\n",
    "completion": "\\textbf{Solution:} From the graph, it can be seen that when $x < -2$, the graph of the line $y = -x + m$ is above the graph of the line $y = nx + 4n$ (where $n \\neq 0$).\\\\\nTherefore, the solution set for the inequality $-x + m > nx + 4n$ with respect to $x$ is $x < -2$.\\\\\nThus, the answer is: $\\boxed{x < -2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50590144.png"
  },
  {
    "solution": "\\boxed{6\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is a chord of $\\odot O$, $C$ is a point on $\\odot O$, and $\\angle ACB=60^\\circ$, $OD\\perp AB$ at point $E$, intersecting $\\odot O$ at point $D$. If the radius of $\\odot O$ is $6$, how long is the chord $AB$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, connect OB,\\\\\nthen $\\angle AOB = 2\\angle ACB = 120^\\circ$,\\\\\nsince OD$\\perp$AB,\\\\\nthus $\\angle AOE = \\frac{1}{2}\\angle AOB = 60^\\circ$,\\\\\nsince AO=6,\\\\\nthus in $\\triangle AOE$, $OE = \\frac{1}{2} OA = 3$,\\\\\n$AE = \\sqrt{OA^2 - OE^2} = 3\\sqrt{3}$,\\\\\nthus AB=2AE$=\\boxed{6\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51351489.png"
  },
  {
    "solution": "\\boxed{1+\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $ \\angle AOB=30^\\circ$, point P lies on the bisector of $ \\angle AOB$, the perpendicular bisector of $ OP$ intersects $ OA$ and $ OB$ at points M and N, respectively. Point E is a point on $ OA$ different from point M, and $ PE=ON=2$. What is the area of $ \\triangle POE$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect PM, draw PC $\\perp$ OA at point C through point P, suppose MN intersects OP at point D,\\\\\n$\\because \\angle AOB=30^\\circ$, OP is the angle bisector of $\\angle AOB$,\\\\\n$\\therefore \\angle MOP=15^\\circ$\\\\\n$\\because$ MN is the perpendicular bisector of OP\\\\\n$\\therefore MN\\perp OD$, $MP=MO$\\\\\n$\\therefore \\angle ODM=\\angle ODN=90^\\circ$, $\\angle MOP=\\angle MPO=15^\\circ$\\\\\nAlso, $\\because OD=OD$\\\\\n$\\therefore \\triangle OMD\\cong \\triangle OND$\\\\\n$\\therefore OM=ON$\\\\\n$\\because PE=ON=2$\\\\\n$\\therefore PE=OM$\\\\\n$\\because MP=MO$\\\\\n$\\therefore MP=PE=2$\\\\\n$\\because \\angle PMA=\\angle MOP+\\angle MPO=30^\\circ$\\\\\n$\\therefore \\angle PEC=\\angle PME=30^\\circ$\\\\\n$\\because PC\\perp ME$\\\\\n$\\therefore CE=CM$, $PC=\\frac{1}{2}PE=1$\\\\\nIn $\\triangle CPE$, $CE=\\sqrt{3}$\\\\\n$\\therefore OE=OM+ME=OM+2CE=2+2\\sqrt{3}$\\\\\n$\\therefore {S}_{\\triangle POE}=\\frac{1}{2}\\cdot OE\\cdot PC = \\boxed{1+\\sqrt{3}}$",
    "image_path": "dataset/mm_math/images/MM_Math/52836530.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ is an equilateral triangle, with points $D$ and $E$ located on sides $BC$ and $AC$ respectively, and $\\angle ADE = 60^\\circ$. $\\triangle ABD \\sim \\triangle DCE$; if $AB=3$ and $EC=\\frac{2}{3}$, what is the length of $DC$?\n",
    "completion": "\\textbf{Solution:} From equation (1), we have proved that $\\triangle ABD \\sim \\triangle DCE$,\\\\\n$\\therefore \\frac{BD}{AB}=\\frac{CE}{DC}$\\\\\nLet $CD = x$, then $BD = 3 - x$,\\\\\n$\\therefore \\frac{3-x}{3}=\\frac{\\frac{2}{3}}{x}$\\\\\n$\\therefore x = 1$ or $x = 2$,\\\\\n$\\therefore DC = \\boxed{1}$ or $DC = \\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53009960_97.png"
  },
  {
    "solution": "\\boxed{2-\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in square $ABCD$, $\\triangle BPC$ is an equilateral triangle. The extensions of $BP$ and $CP$ intersect $AD$ at points $E$ and $F$, respectively. Connect $BD$, $DP$, and $CF$. $BD$ and $CF$ intersect at point $H$. What is the value of $\\tan\\angle DBE$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, draw $PM\\perp CD$ at $M$, $PN\\perp BC$ at $N$,\\\\\nAssume the side length of square $ABCD$ is $4a$, $\\triangle BPC$ is an equilateral triangle,\\\\\n$\\therefore \\angle PBC=\\angle PCB=60^\\circ$, $PB=PC=BC=CD=4a$,\\\\\n$\\therefore \\angle PCD=30^\\circ$,\\\\\n$\\therefore CM=PN=PB\\cdot \\sin 60^\\circ=4a\\times \\frac{\\sqrt{3}}{2}=2\\sqrt{3}a$,\\\\\n$PM=PC\\cdot \\sin 30^\\circ=2a$,\\\\\n$\\because DE\\parallel PM$,\\\\\n$\\therefore \\angle EDP=\\angle DPM$,\\\\\n$\\therefore \\angle DBE=\\angle DPM$,\\\\\n$\\therefore \\tan\\angle DBE=\\tan\\angle DPM=\\frac{DM}{PM}=\\frac{4a-2\\sqrt{3}a}{2a}=2-\\sqrt{3}$.\\\\\nHence, the value of $\\tan\\angle DBE$ is $\\boxed{2-\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53019686_125.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $D$ is a point on side $BC$ of $\\triangle ABC$, $\\angle B = \\angle 1$, $\\angle 3 = 80^\\circ$, $\\angle BAC = 70^\\circ$. What is the measure of $\\angle 2$?\n",
    "completion": "\\textbf{Solution:} Given $\\angle 3$ is an exterior angle of $\\triangle ABD$,\\\\\nthus $\\angle 3 = \\angle B + \\angle 1$,\\\\\nsince $\\angle 3 = 80^\\circ$ and $\\angle B = \\angle 1$,\\\\\nit follows that $\\angle 1 = \\frac{1}{2} \\angle 3 = \\frac{1}{2} \\times 80^\\circ = 40^\\circ$,\\\\\ngiven $\\angle BAC = 70^\\circ$,\\\\\nthus $\\angle 2 = \\angle BAC - \\angle 1 = 70^\\circ - 40^\\circ = \\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52891113.png"
  },
  {
    "solution": "\\boxed{20^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in circle $\\odot O$, CD is a tangent with the point of tangency at D, and line CO intersects $\\odot O$ at B and A, where $\\angle A = 35^\\circ$. What is the degree measure of $\\angle C$?\n",
    "completion": "Solution: Connect OD, \\\\\n$\\because$ CD is the tangent line of $\\odot$O, \\\\\n$\\therefore$ $\\angle$ODC = $90^\\circ$, \\\\\nAlso $\\because$ $\\angle$COD = 2$\\angle$A = $70^\\circ$, \\\\\n$\\therefore$ $\\angle$C = $90^\\circ - 70^\\circ = \\boxed{20^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/51368176.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the slope of a hillside is $i=1\\colon \\sqrt{3}$. If Xiaoming climbs from point A to point B and the straight-line distance AB is 10 meters, then what is the vertical ascent height CB in meters?\n",
    "completion": "\\textbf{Solution:} Given that the slope of the hill, $i = 1\\colon\\sqrt{3}$,\\\\\nit follows that $\\tan\\angle BAC = \\frac{1}{\\sqrt{3}} = \\frac{\\sqrt{3}}{3}$,\\\\\ntherefore, $\\angle BAC = 30^\\circ$,\\\\\nand given $AB = 10$,\\\\\nit follows that $BC = \\frac{1}{2} AB = \\frac{1}{2} \\times 10 = \\boxed{5}$ (meters),",
    "image_path": "dataset/mm_math/images/MM_Math/52536535.png"
  },
  {
    "solution": "\\boxed{3:7}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, point $E$ is on side $BC$. Connect $DE$ and extend it to intersect the extension of side $AB$ at point $F$. If $\\frac{CE}{BE}=\\frac{4}{3}$, what is the ratio of the perimeter of $\\triangle BEF$ to the perimeter of $\\triangle ADF$?\n",
    "completion": "\\textbf{Solution:} Given that parallelogram ABCD,\\\\\nthus AD=DC, BC$\\parallel$AD,\\\\\ngiven $\\frac{CE}{BE}=\\frac{4}{3}$,\\\\\nthus $\\frac{BE}{BC}=\\frac{3}{7}$,\\\\\ntherefore, BE:AD=3:7,\\\\\ngiven BE$\\parallel$AD,\\\\\nthus $\\triangle ADF \\sim \\triangle BEF$,\\\\\ntherefore, the perimeter ratio of $\\triangle BEF$ to $\\triangle ADF$ is equal to the similarity ratio $3:7$.\\\\\nHence, the answer is: $\\boxed{3:7}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51385609.png"
  },
  {
    "solution": "\\boxed{70^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of $\\odot O$, and chord $CD\\perp AB$. Point $E$ is on the extension of $CA$. Connect $DE$ to intersect $\\odot O$ at point $F$, and connect $AF, CF$. If the degree of $\\overset{\\frown}{BD}$ is $40^\\circ$, what is the degree of $\\angle AFC$?\n",
    "completion": "\\textbf{Solution:} As shown in Figure 1, connect $OD, AD$, and let $AB$ intersect $CD$ at $H$. \\\\\nSince the degree of $\\overset{\\frown}{BD}$ is $40^\\circ$, \\\\\ntherefore $\\angle BOD=40^\\circ$, \\\\\ntherefore $\\angle DAB=\\frac{1}{2}\\angle DOB=20^\\circ$, \\\\\nsince $AB\\perp CD$, \\\\\ntherefore $\\angle AHD=90^\\circ$, \\\\\ntherefore $\\angle ADH=90^\\circ-\\angle DAB=70^\\circ$, \\\\\ntherefore $\\angle AFC=\\angle ADH=\\boxed{70^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52991580_99.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, $AC \\perp BC$, $AB=13$, $BC=12$, $CD=3$, $AD=4$. What is the length of $AC$?\n",
    "completion": "\\textbf{Solution:} In $\\triangle ABC$,\\\\\nsince $AC \\perp BC$,\\\\\nit follows that $AC = \\sqrt{AB^{2} - BC^{2}} = \\sqrt{169 - 144} = \\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52890848_107.png"
  },
  {
    "solution": "\\boxed{4cm}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $C$ is a point on the line segment $AB$, $M$ is the midpoint of $AC$, and $N$ is the midpoint of $BC$. If $MC$ is $2\\,\\text{cm}$ longer than $NC$, how much longer is $AC$ than $BC$?\n",
    "completion": "\\textbf{Solution:} Given that $M$ is the midpoint of $AC$, and $N$ is the midpoint of $BC$,\n\\\\\nhence $MC=\\frac{1}{2}AC$, $CN=\\frac{1}{2}BC$,\n\\\\\nsince $MC-NC=2cm$,\n\\\\\nit follows that $\\frac{1}{2}AC-\\frac{1}{2}BC=2cm$,\n\\\\\ntherefore $AC-BC=\\boxed{4cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55192328.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{2}a}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that point $C$ is on the line segment $AB$, and $M$, $N$ are the midpoints of $AC$, $BC$ respectively. Let $AB=a$, find the length of the line segment $MN$.\n",
    "completion": "\\textbf{Solution:} From (1) it is known that $MN= \\frac{1}{2}AC+ \\frac{1}{2}BC= \\frac{1}{2}(AC+BC)= \\frac{1}{2}AB$,\\\\\nsince $AB=a$,\\\\\ntherefore $MN= \\frac{1}{2}a=\\boxed{\\frac{1}{2}a}$.",
    "image_path": "dataset/mm_math/images/MM_Math/54927380_130.png"
  },
  {
    "solution": "\\boxed{80^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the pentagon $ABCDE$, $\\angle D=120^\\circ$, the exterior angle adjacent to $\\angle EAB$ is $80^\\circ$, and the exterior angles adjacent to $\\angle DEA$ and $\\angle ABC$ are both $60^\\circ$. What is the measure of $\\angle C$ in degrees?\n",
    "completion": "\\textbf{Solution:} Since the external angle adjacent to $\\angle$EAB is $80^\\circ$, and the external angles adjacent to both $\\angle$DEA and $\\angle$ABC are $60^\\circ$,\\\\\nit follows that $\\angle$DEA=$180^\\circ-60^\\circ=120^\\circ$, $\\angle$ABC=$180^\\circ-60^\\circ=120^\\circ$, $\\angle$EAB=$180^\\circ-80^\\circ=100^\\circ$;\\\\\nThe sum of the interior angles of a pentagon is $(5-2)\\times180^\\circ=540^\\circ$;\\\\\nTherefore, $\\angle$C=$540^\\circ-120^\\circ-120^\\circ-120^\\circ-100^\\circ=\\boxed{80^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50475017.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, in right triangle $ABC$, where $\\angle ACB=90^\\circ, AC=8, BC=6$, $D$ and $E$ are moving points located on sides $BC$ and $AB$ respectively, and $BE=BD$. Connect $DE$. Triangle $BDE$ is folded along $DE$ such that point $B$ coincides with point $F$. Connect $AF$. As shown in Figure 3, during the movement of points $D$ and $E$, when $AF \\parallel DE$, find the length of $AF$.\n",
    "completion": "\\textbf{Solution:} As shown in the figure, join $BF$ intersecting $ED$ at point O, and let $EF$ intersect $AC$ at point G.\\\\\nSince quadrilateral $BEFD$ is a rhombus,\\\\\nit follows that $BF\\perp ED, OB=OF, EF\\parallel BC$.\\\\\nSince $AF\\parallel ED$,\\\\\nit implies $FA\\perp BF, \\frac{BE}{BA}=\\frac{BO}{BF}=\\frac{1}{2}$,\\\\\nthus, $\\angle AFB=90^\\circ, BE=\\frac{1}{2}AB=5$,\\\\\nSince E is the midpoint of $AB$,\\\\\nit follows that $EF=AE=BE=5$.\\\\\nSince $EF\\parallel BC$,\\\\\nit implies $\\frac{EG}{BC}=\\frac{AE}{AB}=\\frac{AG}{AC}=\\frac{1}{2}$,\\\\\nthus, $EG=3, GF=2, AG=4$.\\\\\nIn $\\triangle AGF$ with a right angle at $F$, $AF=\\sqrt{AG^{2}+GF^{2}}=\\sqrt{4^{2}+2^{2}}=\\boxed{2\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52991872_105.png"
  },
  {
    "solution": "\\boxed{40^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the quadrilateral $ABCD$ is inscribed in a circle, with extensions of two pairs of opposite sides intersecting at points $E$ and $F$, respectively, and $\\angle E=40^\\circ$, $\\angle F=60^\\circ$. What is the measure of $\\angle A$ in degrees?\n",
    "completion": "\\textbf{Solution:} Connect EF, as shown in the figure, \\\\\n$\\because$ Quadrilateral ABCD is inscribed in a circle, $\\therefore \\angle$ECD$=\\angle A$, \\\\\n$\\because \\angle$ECD$=\\angle 1+\\angle 2$, $\\therefore \\angle A=\\angle 1+\\angle 2$, \\\\\n$\\because \\angle A+\\angle 1+\\angle 2+\\angle E+\\angle F=180^\\circ$, \\\\\n$\\therefore 2\\angle A+40^\\circ+60^\\circ=180^\\circ$, \\\\\n$\\therefore \\angle A=\\boxed{40^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51403023.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{8}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in $\\triangle ABC$, $\\angle ACB = 90^\\circ$, point O is the midpoint of AB, CD bisects $\\angle BCO$ and intersects AB at point D. Draw AE $\\perp$ CD, intersecting CO and BC at points G and E, respectively. Let the area of $\\triangle AGO$ be $S_{1}$, and the area of $\\triangle AEB$ be $S_{2}$. When $\\frac{S_{1}}{S_{2}} = \\frac{2}{5}$, what is the value of $\\frac{OG}{BC}$?\n",
    "completion": "\\textbf{Solution:} Construct line OF through point O parallel to AE and intersecting CB at point F,\\\\\nsince CD bisects $\\angle$BCO and AE$\\perp$CD,\\\\\nit is easy to prove that $\\triangle$CEG and $\\triangle$COF are isosceles triangles,\\\\\ntherefore CG=CE, CO=CF,\\\\\ntherefore OG=EF,\\\\\nsince $\\frac{{S}_{1}}{{S}_{2}} = \\frac{2}{5}$,\\\\\nlet the area of $\\triangle$ABE be 20a, then S$_{1}$=8a, S$_{2}$=20a,\\\\\nsince OF$\\parallel$AE,\\\\\ntherefore $\\triangle$OBF $\\sim$ $\\triangle$AEB,\\\\\ntherefore $\\frac{{S}_{\\triangle OBF}}{{S}_{\\triangle AEB}}={\\left(\\frac{OB}{AB}\\right)}^{2}=\\frac{1}{4}=\\frac{{S}_{\\triangle OBF}}{20a}$,\\\\\ntherefore S$_{\\triangle OBF}$=5a,\\\\\n$\\frac{{S}_{\\triangle OBF}}{{S}_{\\triangle AGO}}=\\frac{5}{8}=\\frac{OF}{AG}$,\\\\\nsince OF:AE=1:2,\\\\\ntherefore AE=2OF,\\\\\ntherefore $\\frac{5}{8}=\\frac{\\frac{1}{2}AE}{AG}$,\\\\\ntherefore $\\frac{AE}{AG}=\\frac{5}{2}$,\\\\\ntherefore $\\frac{GE}{OF}=\\frac{CE}{CF}=\\frac{2}{5}$,\\\\\nlet CE=2x, then CF=5x, EF=OG=3x,\\\\\ntherefore BF=3x,\\\\\ntherefore BC=CE+EF+BF=2x+3x+3x=8x,\\\\\ntherefore $\\frac{OG}{BC}=\\frac{3x}{8x}=\\boxed{\\frac{3}{8}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51299791.png"
  },
  {
    "solution": "\\boxed{16}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, diagonals AC and BD of square $ABCD$ intersect at point O, and point E is the midpoint of AD. Line OE is connected. If $OA = 1$ and the perimeter of $\\triangle AOE$ is 5, then what is the perimeter of square $ABCD$?\n",
    "completion": "\\textbf{Solution:} Given that $OA=1$ and the perimeter of $\\triangle AOE$ is 5,\\\\\nit follows that $OE+AE=5-1=4$,\\\\\nSince point E is the midpoint of AD,\\\\\nthen $AD=2AE$, and OE is the median of $\\triangle ACD$,\\\\\nhence $CD=2OE$,\\\\\ntherefore, $AD+CD=2AE+2OE=2(AE+OE)=8$,\\\\\nthus, the perimeter of $\\square ABCD$ is: $2(AD+CD)=2 \\times 8=\\boxed{16}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52424166.png"
  },
  {
    "solution": "\\boxed{x>\\frac{5}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the function $y=-2x+3$ intersects with the graph of $y=-\\frac{1}{2}x+m$ at point $P(n, -2)$. The line $y=-2x+3$ intersects with the y-axis at point A, and the line $y=-\\frac{1}{2}x+m$ intersects with the y-axis at point B. Directly write out the solution set for the inequality $-\\frac{1}{2}x+m > -2x+3$.\n",
    "completion": "\\textbf{Solution:} The solution set of the inequality $-\\frac{1}{2}x+m>-2x+3$ is $\\boxed{x>\\frac{5}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52355125_191.png"
  },
  {
    "solution": "\\boxed{36^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of the semicircle $O$, and $C$ and $D$ are two points on the semicircle $O$, with $OD \\parallel BC$, and $OD$ intersects $AC$ at point $E$. If $\\angle B=72^\\circ$, find the degree measure of $\\angle CAD$.\n",
    "completion": "\\textbf{Solution:} Given that $AB$ is the diameter of the semicircle $O$,\\\\\nit follows that $\\angle C=90^\\circ$,\\\\\nsince $\\angle B=72^\\circ$,\\\\\nit follows that $\\angle CAB=90^\\circ-\\angle B=18^\\circ$,\\\\\nsince $OD\\parallel BC$,\\\\\nit follows that $\\angle AOD=\\angle B=72^\\circ$,\\\\\nsince $OA=OD$,\\\\\nit follows that $\\angle OAD=\\angle D=54^\\circ$,\\\\\ntherefore $\\angle CAD=\\angle OAD-\\angle CAB=\\boxed{36^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53009507_138.png"
  },
  {
    "solution": "\\boxed{8\\sqrt{2}-8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, two square pieces of paper with areas of 8 and 16 are placed without overlapping inside the rectangle $ABCD$. What is the area of the blank part in the diagram?\n",
    "completion": "Solution: Given the information, we find that:\nthe length of rectangle ABCD is $\\sqrt{16}+\\sqrt{8}=4+2\\sqrt{2}$, and its width is 4, \n$\\therefore$ the area of rectangle ABCD is $4\\left(4+2\\sqrt{2}\\right)=16+8\\sqrt{2}$,\n$\\therefore$ the area of the blank part is: $16+8\\sqrt{2}-16-8=\\boxed{8\\sqrt{2}-8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53182548.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As the figure shows, $\\angle AOB$ is a right angle, $\\angle AOC=50^\\circ$, $ON$ is the bisector of $\\angle AOC$, and $OM$ is the bisector of $\\angle BOC$. What is the measure of $\\angle MON$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle AOB$ is a right angle,\\\\\nit follows that $\\angle AOB = 90^\\circ$.\\\\\nTherefore, $\\angle BOC = \\angle AOB + \\angle AOC = 90^\\circ + 50^\\circ = 140^\\circ$.\\\\\nSince OM bisects $\\angle BOC$,\\\\\nit follows that $\\angle COM = \\frac{1}{2} \\angle BOC = 70^\\circ$.\\\\\nSince ON bisects $\\angle AOC$,\\\\\nit follows that $\\angle CON = \\frac{1}{2} \\angle AOC = 25^\\circ$.\\\\\nTherefore, $\\angle MON = \\angle COM - \\angle CON = 70^\\circ - 25^\\circ = \\boxed{45^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52794957.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, AB is the diameter of circle $O$, and chord CD $\\perp$ AB at point H. If AB $= 10$ and CD $= 8$, what is the length of BH?\n",
    "completion": "\\textbf{Solution:} Draw line OC.\n\nFrom the given: In right triangle $OCH$, we have $OC=\\frac{1}{2}AB=5$, and $CH=\\frac{1}{2}CD=4$;\n\nBy the Pythagorean theorem, we find: $OH=\\sqrt{OC^{2}-CH^{2}}=\\sqrt{5^{2}-4^{2}}=3$;\n\nTherefore, $BH=OB-OH=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51212775.png"
  },
  {
    "solution": "\\boxed{150^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, point O is on line segment $AB$. Ray $OC$ is drawn from point O such that $\\angle BOC = 120^\\circ$. A right-angle triangle ruler is placed with its right-angle vertex at point O, one side $OM$ on ray $OB$, and the other side $ON$ below line $AB$. The triangle ruler in Figure 1 is rotated clockwise around point O to the position in Figure 3, so that side $OM$ is inside $\\angle BOC$ and exactly bisects $\\angle BOC$. What is the measure of $\\angle CON$?\n",
    "completion": "\\textbf{Solution:} Given $\\because \\angle BOC = 120^\\circ$, and $OM$ bisects $\\angle BOC$,\\\\\n$\\therefore \\angle COM = \\angle BOM = 60^\\circ$,\\\\\n$\\because \\angle MON = 90^\\circ$,\\\\\n$\\therefore \\angle CON = \\angle MON + \\angle COM = 90^\\circ + 60^\\circ = \\boxed{150^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53119273_125.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{2}a}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, let there be an equilateral triangle $ABC$ with side length $a$. Let $P$ be a point on side $AB$, and draw $PE\\perp AC$ at point $E$. Let $Q$ be a point on the extension of $BC$. When $AP=CQ$, line $PQ$ intersects $AC$ at point $D$. What is the length of $DE$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, extend line $AD$ through point $Q$ and draw a perpendicular line to it at point $F$, \\\\\nsince $\\triangle ABC$ is an equilateral triangle, \\\\\ntherefore $\\angle A=\\angle ACB=60^\\circ$, \\\\\nsince $\\angle ACB=\\angle QCF$, \\\\\ntherefore $\\angle QCF=\\angle A=60^\\circ$, \\\\\nalso, since $PE\\perp AC$, $QF\\perp AC$, \\\\\ntherefore $\\angle AEP=\\angle CFQ=90^\\circ$, \\\\\nand $AP=CQ$, \\\\\ntherefore $\\triangle AEP\\cong \\triangle CFQ$ (AAS), \\\\\ntherefore $AE=CF$, $PE=QF$, \\\\\nSimilarly, it can be proved that $\\triangle DEP\\cong \\triangle DFQ$, \\\\\ntherefore $DE=DF$, \\\\\ntherefore $AC=AE+DE+CD=DE+CD+CF=DE+DF=2DE$, \\\\\ntherefore $DE=\\frac{1}{2}AC=\\frac{1}{2}a$. \\\\\nHence, $DE=\\boxed{\\frac{1}{2}a}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53058546.png"
  },
  {
    "solution": "\\boxed{36\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, the two diagonals $AC$ and $BD$ intersect at point $O$. If $AB = OB = 6$, what is the area of the rectangle?\n",
    "completion": "\\textbf{Solution:} Because quadrilateral ABCD is a rectangle, \\\\\n$\\therefore$ $\\angle ABC = 90^\\circ$, OA = $\\frac{1}{2}$AC, OB = $\\frac{1}{2}$BD, AC = BD, \\\\\n$\\therefore$ OA = OB, \\\\\nAlso because AB = OB = 6, \\\\\n$\\therefore$ OA = OB = AB = 6, \\\\\n$\\therefore$ AC = 2OA = 12, \\\\\n$\\therefore$ BC = $\\sqrt{AC^2 - AB^2}$ = $6\\sqrt{3}$, \\\\\n$\\therefore$ the area of rectangle ABCD = BC $\\cdot$ AB = $6\\sqrt{3} \\times 6 = \\boxed{36\\sqrt{3}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52425015.png"
  },
  {
    "solution": "\\boxed{x>1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the Cartesian coordinate system $xOy$, the graphs of the linear functions $y=kx$ and $y=mx+n$ are shown as in the figure. What is the solution set of the linear inequality in one variable with respect to $x$, $(k-m)x-n>0$?\n",
    "completion": "\\textbf{Solution:} From $(k-m)x-n>0$, we obtain: $kx>mx+n$,\\\\\nAccording to the graph, the intersection point of the two functions is $(1, 2)$,\\\\\nTherefore, the solution set of the linear inequality in $x$, $kx>mx+n$, is $x>1$, that is, the solution set of the linear inequality in $x$, $(k-m)x-n>0$, is $\\boxed{x>1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50560884.png"
  },
  {
    "solution": "\\boxed{C(0, 2)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Synthesis and Exploration\\\\\nAs shown in Figure 1, the parabola $y=-\\frac{1}{2}x^{2}+\\frac{3}{2}x+2$ intersects the x-axis at points A and B (where point A is to the left of point B), and intersects the y-axis at point C. Point D is on the parabola in the first quadrant. Please write down the coordinates of points A, B, and C directly.\n",
    "completion": "\\textbf{Solution:} Let \\boxed{A(-1, 0)}, \\boxed{B(4, 0)}, \\boxed{C(0, 2)}",
    "image_path": "dataset/mm_math/images/MM_Math/52866560_296.png"
  },
  {
    "solution": "\\boxed{y=\\frac{1}{6}x-\\frac{1}{6}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=2x$ passing through the origin intersects the inverse proportion function $y_{1}=\\frac{2}{x}$ at point $B$, and intersects another inverse proportion function $y_{2}=\\frac{k}{x}$ at point $C$, with $OB=BC$. The abscissa of point $A$ is $4$ and is a point on $y_{1}=\\frac{2}{x}$. A line is drawn from point $B$ such that $BD\\perp x$ axis, with the foot of the perpendicular at point $D$. Find the analytical expression of the inverse proportion function $y_{2}=\\frac{k}{x}$ and the line $AD$.\n",
    "completion": "\\textbf{Solution:} Solving the system of equations, we get\n\\begin{align*}\n\\begin{cases}\ny=2x \\\\\ny=\\frac{2}{x}\n\\end{cases}\n\\end{align*}\nwhich yields\n\\begin{align*}\n\\begin{cases} x_{1}=1 \\\\ y_{1}=2 \\end{cases}, \\quad\n\\begin{cases} x_{2}=-1 \\\\ y_{2}=-2 \\end{cases}\n\\end{align*}\n\\(\\therefore B(1, 2)\\)\n\n\\(\\because\\) BD \\(\\perp\\) x-axis,\n\n\\(\\therefore D(1, 0)\\)\n\n\\(\\therefore BD=2, OD=1.\\)\n\nWhen \\(x=4\\), \\(y_{1}=\\frac{2}{4}=\\frac{1}{2}\\)\n\n\\(\\therefore A(4, \\frac{1}{2})\\)\n\nThrough point C, construct CE \\(\\perp\\) x-axis at point E,\n\n\\(\\therefore\\) BD \\(\\parallel\\) CE\n\n\\(\\therefore \\triangle BOD \\sim \\triangle COE\\)\n\n\\(\\therefore \\frac{OB}{OC}=\\frac{BD}{CE}=\\frac{OD}{OE}\\)\n\n\\(\\because\\) OB = BC, BD = 2\n\n\\(\\therefore \\frac{2}{CE}=\\frac{1}{OE}=\\frac{1}{2}\\)\n\n\\(\\therefore\\) CE = 4, OE = 2\n\n\\(\\therefore C(2, 4)\\)\n\nSubstituting into \\(y_{2}=\\frac{k}{x}\\), we get \\(\\frac{k}{2}=4\\)\n\n\\(\\therefore k=8\\)\n\n\\(\\therefore y_{2}=\\frac{8}{x}\\)\n\nAssume the equation of line AD is \\(y=kx+b\\)\n\nSubstituting \\((1, 0)\\) and \\((4, \\frac{1}{2})\\), we get\n\\begin{align*}\n\\begin{cases}\nk+b=0 \\\\\n4k+b=\\frac{1}{4}\n\\end{cases}\n\\end{align*}\nSolving, we find\n\\begin{align*}\n\\begin{cases}\nk=\\frac{1}{6} \\\\\nb=-\\frac{1}{6}\n\\end{cases}\n\\end{align*}\n\\(\\therefore\\) The equation of line AD is \\(y=\\frac{1}{6}x-\\frac{1}{6}\\), thus the answer is \\(\\boxed{y=\\frac{1}{6}x-\\frac{1}{6}}\\).",
    "image_path": "dataset/mm_math/images/MM_Math/51322681_142.png"
  },
  {
    "solution": "\\boxed{-\\frac{3}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system, point O is the origin, point A $(-3,0)$ and point B $(0,2)$ are both on the coordinate axes. If the graph of the inverse proportion function $y= \\frac{k}{x}$ passes through the center of symmetry of rectangle AOBC, what is the value of $k$?\n",
    "completion": "\\[\n\\text{Solution:} \\quad \\because \\text{Point A} (-3, 0) \\text{ and Point B} (0, 2) \\text{ are both on the coordinate axes}, \\\\\n\\therefore \\text{the center of rectangle AOBC is} \\left( -\\frac{3}{2} ,1\\right), \\\\\n\\because \\text{the graph of the inverse proportion function } y= \\frac{k}{x} \\text{ passes through the center of symmetry of rectangle AOBC}, \\\\\n\\therefore k= -\\frac{3}{2} \\times 1 = \\boxed{-\\frac{3}{2}}\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/50496768.png"
  },
  {
    "solution": "\\boxed{50^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, AB is the diameter of $\\odot O$. Point D is on the extension line of AB, and DC is tangent to $\\odot O$ at point C. If $\\angle A = 20^\\circ$, what is the measure of $\\angle D$?\n",
    "completion": "\\textbf{Solution:} Draw line OC, \\\\\n$\\because$ DC is tangent to $\\odot$O at point C, \\\\\n$\\therefore$ $\\angle$OCD=$90^\\circ$, \\\\\n$\\because$ $\\angle$A=$20^\\circ$, \\\\\n$\\therefore$ $\\angle$OCA=$20^\\circ$, \\\\\n$\\therefore$ $\\angle$DOC=$40^\\circ$, \\\\\n$\\therefore$ $\\angle$D=$90^\\circ-40^\\circ=\\boxed{50^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51433564.png"
  },
  {
    "solution": "\\boxed{16}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point A is any point on the graph of the inverse proportion function $y=\\frac{7}{x}\\left(x>0\\right)$. Line segment $AB$ is parallel to the x-axis and intersects the graph of the inverse proportion function $y=-\\frac{9}{x}$ at point B. Taking $AB$ as a side, a quadrilateral $\\square ABCD$ is formed with points C and D on the x-axis. What is the area of $\\square ABCD$?\n",
    "completion": "\\textbf{Solution:} Draw lines OA, OB, and extend AB to meet the y-axis at E, as shown in the diagram,\\\\\n$\\because$ AB$\\parallel$x-axis,\\\\\n$\\therefore$ AB$\\perp$y-axis,\\\\\n$\\therefore$ Area$_{\\triangle OEA}=\\frac{1}{2}\\times7=\\frac{7}{2}$, Area$_{\\triangle OBE}=\\frac{1}{2}\\times9=\\frac{9}{2}$,\\\\\n$\\therefore$ Area$_{\\triangle OAB}=\\frac{7}{2}+\\frac{9}{2}=8$,\\\\\n$\\because$ The quadrilateral ABCD is a parallelogram,\\\\\n$\\therefore$ Area$_{\\text{parallelogram ABCD}}=2\\times$Area$_{\\triangle OAB}=\\boxed{16}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52766018.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC=\\sqrt{5}$, and $\\sin B=\\frac{2\\sqrt{5}}{5}$. What is the length of side $BC$?\n",
    "completion": "\\textbf{Solution}: As shown in the figure, draw the height AD of $\\triangle ABC$, \\\\\nsince $AB=AC$, \\\\\ntherefore, point D is the midpoint of BC, which means $BD=CD$. \\\\\nSince in $\\triangle ABD$, $\\sin B = \\frac{AD}{AB} = \\frac{2\\sqrt{5}}{5}$, we get $\\frac{AD}{\\sqrt{5}} = \\frac{2\\sqrt{5}}{5}$, \\\\\nthus $AD=2$, \\\\\ntherefore $BD = \\sqrt{AB^{2} - AD^{2}} = \\sqrt{(\\sqrt{5})^{2} - {2}^{2}} = 1$, \\\\\ntherefore, $BC = 2BD = \\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51403279_129.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y_1=x+1$ intersects with the graph of the inverse proportion function $y_2= \\frac{2}{x}$ at points $A$ and $B$. A line is drawn from point $A$ perpendicular to the $x$-axis, and the foot of the perpendicular is point $H$. Then, line $BH$ is drawn. What are the coordinates of points $A$ and $B$ and the area of $\\triangle ABH$?\n",
    "completion": "\\textbf{Solution:} From the given problem, we have $x+1= \\frac{2}{x}$, therefore $x_{1}=1$, $x_{2}=-2$, then $A(1,2)$, $B(-2,-1)$, $H(1,0)$. Let the intersection point of the line $y=x+1$ and the $x$-axis be $C$, thus $C(-1, 0)$. Therefore, $CH=2$, which leads to the area of $\\triangle ABH = S_{\\triangle ACH} + S_{\\triangle BCH} = \\frac{1}{2} \\times 2 \\times 2 + \\frac{1}{2} \\times 2 \\times 1 = \\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50475225_90.png"
  },
  {
    "solution": "\\boxed{\\frac{5}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the vertices $A$ and $C$ of the rectangle $OABC$ are on the positive semi-axes of the $x$ and $y$ axes, respectively. Point $F$ is a moving point on side $BC$ (not coinciding with points $B$, $C$). The inverse proportion function $y=\\frac{k}{x}$ ($x>0$) passing through point $F$ intersects side $AB$ at point $E(8, m)$, with $AB=4$. If $BE=3AE$,\n- (1) Find the expression of the inverse proportion function;\n- (2) Fold rectangle $OABC$ such that point $O$ coincides with point $F$. The fold intersects the $x$ and $y$ axes at points $H$ and $G$, respectively. Find the length of the line segment $OG$.\n",
    "completion": "\\textbf{Solution:}\n\n(1) Since $BE=3AE$ and $AB=4$,\\\\\nthen $AE=1$ and $BE=3$,\\\\\nthus $E(8,1)$,\\\\\nthus $k=8\\times 1=8$,\\\\\ntherefore, the expression for the inverse proportion function is $y=\\frac{8}{x}$;\\\\\n(2) When $y=4$, then $x=2$,\\\\\nthus $F(2,4)$,\\\\\nthus $CF=2$,\\\\\nlet $OG=x$, then $CG=4-x$ and $FG=x$,\\\\\nby the Pythagorean theorem, we get,\\\\\n$(4-x)^2+2^2=x^2$,\\\\\nsolving for $x$, we find $x=\\frac{5}{2}$,\\\\\nthus $OG=\\boxed{\\frac{5}{2}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52457729_185.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, fold the corner where vertex D of rectangle ABCD is located, so that point D falls on point F on side BC, with the crease being AE. If $AB=8$ and $BC=10$, what is the length of $EF$?\n",
    "completion": "\\textbf{Solution}: Since quadrilateral $ABCD$ is a rectangle, we have $AB=CD=8$, $BC=AD=AF=10$, and $\\angle B=\\angle C=90^\\circ$. Therefore, $BF=\\sqrt{AF^2-AB^2}=\\sqrt{10^2-8^2}=6$. Consequently, $FC=BC-BF=10-6=4$. Let $EF=DE=x$, then $EC=CD-DE=8-x$. In $\\triangle EFC$, by the Pythagorean theorem, we have $FC^2+EC^2=EF^2$. Therefore, $4^2+(8-x)^2=x^2$. Solving this equation yields $x=\\boxed{5}$. Therefore, $EF=\\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50481650_132.png"
  },
  {
    "solution": "\\boxed{x^2+2x-3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the quadratic function $y=ax^2+bx+c$ intersects the x-axis at points B and C, and intersects the y-axis at point A. If the coordinates of point A are $(0, -3)$, $\\angle ABC=45^\\circ$, and $3OC=OA$, find the expression for this quadratic function.\n",
    "completion": "\\textbf{Solution:} Given the coordinates of point A, we know that $AO=3$, \\\\\n$\\because$ $\\angle ABC=45^\\circ$ and $OC=OA$, \\\\\n$\\therefore$ $BO=3$, $CO=1$, \\\\\nTherefore, the coordinates of points B and C are $(-3,0)$ and $(1,0)$, respectively, \\\\\nLet the equation of the parabola be $y=a(x-x_1)(x-x_2)=a(x-1)(x+3)$, \\\\\n$\\because$ the coordinates of point A are $(0,-3)$, thus $-3a=-3$, solving this gives $a=1$, \\\\\nTherefore, the equation of the parabola is $y=x^2+2x-3=\\boxed{x^2+2x-3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52977484_198.png"
  },
  {
    "solution": "\\boxed{x<1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y_1 = x+1$ intersects the x-axis and the y-axis at points A and B, respectively, and the line $y_2 = -2x+4$ intersects the x-axis and the y-axis at points C and D, respectively. The two lines intersect at point E. Based on the figure, answer directly: For what values of $x$ is $y_1 < y_2$?\n",
    "completion": "\\textbf{Solution:} Given $E(1, 2),$ when $y_1<y_2$, the graph of $y_1=x+1$ is below the graph of $y_2=-2x+4$. By analyzing the graphs, we can conclude: $\\boxed{x<1}.$",
    "image_path": "dataset/mm_math/images/MM_Math/52380902_165.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\triangle ABC$, $ AD$ bisects $ \\angle BAC$ and intersects $ BC$ at point $ D$, $ DE \\perp AB$ with the foot of the perpendicular being $ E$. If $ DE=1$ and $ AC=4$, then what is the area of $ \\triangle ADC$?\n",
    "completion": "\\textbf{Solution:} Construct $DF \\perp AC$ at $F$, as shown in the figure,\\\\\nsince $AD$ bisects $\\angle BAC$, $DE\\perp AB$, $DF\\perp AC$,\\\\\ntherefore $DF=DE=1$,\\\\\ntherefore the area of $\\triangle ADC = \\frac{1}{2} \\times 4 \\times 1 = \\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52914196.png"
  },
  {
    "solution": "\\boxed{\\left(\\frac{13}{3}, \\frac{2}{3}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system $xOy$, rectangle $OABC$ has its sides $OA$ and $OC$ along the $x$-axis and $y$-axis, respectively, and point $B$ has coordinates $(3, 2)$. Let $D$ be a moving point on side $OA$, and construct rectangle $BDEF$ to the right side of $BD$ with $DE:BD=2:3$. Connect $CD$ and $AF$. Given $3AF=2CD$, when the coordinates of point $D$ are $(1, 0)$, find the coordinates of point $F$.\n",
    "completion": "\\textbf{Solution:} Draw a perpendicular line from $F$ to the $x$-axis, with $H$ being the foot of the perpendicular.\\\\\n$\\because$ $AB\\perp x$-axis,\\\\\n$\\therefore$ $FH\\parallel AB$.\\\\\n$\\therefore$ $\\angle AFH=\\angle BAF$.\\\\\n$\\because$ $CB\\parallel OA$,\\\\\n$\\therefore$ $\\angle CDO=\\angle BCD$.\\\\\nBy $(1)$, $\\angle BAF=\\angle BCD$,\\\\\n$\\therefore$ $\\angle AFH=\\angle CDO$.\\\\\nMoreover, $\\because$ $\\angle AHF=\\angle COD=90^\\circ$,\\\\\n$\\therefore$ $\\triangle AHF\\sim \\triangle COD$.\\\\\n$\\therefore$ $\\frac{AH}{CO}=\\frac{HF}{OD}=\\frac{AF}{CD}=\\frac{2}{3}$.\\\\\n$\\therefore$ $AH=\\frac{2}{3}CO=\\frac{4}{3}$, $HF=\\frac{2}{3}OD=\\frac{2}{3}$.\\\\\n$\\because$ $OH=OA+AH=\\frac{13}{3}$.\\\\\n$\\therefore$ The coordinates of point $F$ are $\\boxed{\\left(\\frac{13}{3}, \\frac{2}{3}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53058741_159.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: During a project learning session, the teacher asked the students to collaborate on creating questions. Inspired by the Chord Diagram of Zhao Shuang, a study group came up with the following problem for you to solve:\n",
    "completion": "\\textbf{Solution:} Let $AE=x$, then $BE=DG=x+1$. In $\\triangle BEF$, $\\angle FEB=45^\\circ$,\\\\\n$\\therefore$ BE=FB,\\\\\n$\\because$ BF=DH=x+1,\\\\\n$\\therefore$ AH=x+1+1=x+2,\\\\\nAlso, $\\because$ $\\tan\\angle AEH=2$,\\\\\n$\\therefore$ $\\frac{AH}{AE}=2$,\\\\\n$\\therefore$ AH=2AE=2x,\\\\\n$\\therefore$ 2x=x+2,\\\\\nSolving, we get $x=2$,\\\\\n$\\therefore$ AE=\\boxed{2}.",
    "image_path": "dataset/mm_math/images/MM_Math/51284477_139.png"
  },
  {
    "solution": "\\boxed{\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a square is placed on a desk. What is the side length of this square?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, \\\\\n$\\because$ $\\angle B = \\angle D = \\angle ACE = 90^\\circ$, \\\\\n$\\angle BAC + \\angle ACB = 90^\\circ$, $\\angle ECD + \\angle ACB = 90^\\circ$, \\\\\n$\\therefore$ $\\angle BAC = \\angle ECD$, \\\\\n$\\because$ AC = CE, \\\\\n$\\therefore$ $\\triangle ABC \\cong \\triangle CBE$, \\\\\n$\\therefore$ BC = DE = 2, \\\\\n$\\therefore$ AC = $\\sqrt{AB^2 + BC^2} = \\sqrt{1^2 + 2^2} = \\sqrt{5}$, \\\\\n$\\therefore$ the length of the side of the square is $\\boxed{\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51420049.png"
  },
  {
    "solution": "\\boxed{\\frac{4}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, within a $5\\times 4$ grid of small squares, points A, B, and C all lie on the lattice points (lattice points refer to the vertices of the small squares). What is the value of $\\tan\\angle ABC$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, take lattice point D, connect AD and BD, then $\\angle BDA=90^\\circ$, BD=5, AD=4,\\\\\n$\\therefore$ $\\tan\\angle ABC=\\tan\\angle ABD=\\frac{AD}{BD}=\\frac{4}{5}=\\boxed{\\frac{4}{5}}$",
    "image_path": "dataset/mm_math/images/MM_Math/52719806.png"
  },
  {
    "solution": "\\boxed{25^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $O$ is a point on the line $MN$, $\\angle MOC=130^\\circ$, $OA$ bisects $\\angle MOC$, $\\angle AOB=90^\\circ$. What is the measure of $\\angle BOC$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle AOB=90^\\circ$,\\\\\n$\\angle AOC=65^\\circ$,\\\\\n$\\angle AOB=\\angle AOC+\\angle BOC$,\\\\\ntherefore, $\\angle BOC=\\angle AOB-\\angle AOC=25^\\circ$.\\\\\nHence, $\\angle BOC=\\boxed{25^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53119430_124.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $l_1\\parallel l_2\\parallel l_3$, the lines $a, b$ intersect $l_1, l_2, l_3$ at points $A, B, C$ and $D, E, F$ respectively. If $AB:BC=2:3$ and $EF=6$, what is the length of $DE$?\n",
    "completion": "\\textbf{Solution:} From the theorem that parallel lines divide segments proportionally, we have: $\\frac{DE}{EF} = \\frac{AB}{BC}$, that is, $\\frac{DE}{6} = \\frac{2}{3}$, \\\\\nThus, we find $DE = \\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51323834.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{41}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, given that the area of the square $ABCD$ is 64 square centimeters and $DE = 10$ cm, what is the length of $CE$?\n",
    "completion": "\\textbf{Solution:} Since the area of square ABCD is 64 square centimeters, $\\therefore$ $\\angle ADC=90^\\circ$ and $DC=8$ cm. Since $DE=10$ cm, $\\therefore$ $CE=\\sqrt{DE^{2}+DC^{2}}=\\sqrt{8^{2}+10^{2}}=\\boxed{2\\sqrt{41}}$ (cm).",
    "image_path": "dataset/mm_math/images/MM_Math/52977098.png"
  },
  {
    "solution": "\\boxed{110^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, one corner of the rectangular notebook loose-leaf sheet is folded over so that the vertex $A$ falls at $A'$, with $BC$ being the crease. If $\\angle ACB = 35^\\circ$, find the degree measure of $\\angle A'CD$.\n",
    "completion": "\\textbf{Solution:} Given that $\\because \\angle ABC=35^\\circ$, \\\\\n$\\therefore$ by folding we get $\\angle A^{\\prime}BC=\\angle ABC=35^\\circ$, \\\\\n$\\therefore \\angle A^{\\prime}CD=180^\\circ-\\angle ACB-\\angle A^{\\prime}CB \\\\\n=180^\\circ-35^\\circ-35^\\circ \\\\\n=\\boxed{110^\\circ};$",
    "image_path": "dataset/mm_math/images/MM_Math/53206857_138.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given: As shown in the figure, $OA$ and $OB$ are two radii of $\\odot O$, and $OA \\perp OB$. Point $C$ is on $\\odot O$. What is the measure of $\\angle ACB$?\n",
    "completion": "\\textbf{Solution:} Since $OA \\perp OB$,\\\\\nit follows that $\\angle AOB = 90^\\circ$,\\\\\nwhich means $\\angle ACB = \\frac{1}{2} \\angle AOB = \\boxed{45^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52513737.png"
  },
  {
    "solution": "\\boxed{y=x^{2}+2x-3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=x^2+bx+c$ (where \\(b\\) and \\(c\\) are constants) has its vertex at \\(C\\), and intersects the \\(x\\)-axis at points \\(A\\) and \\(B\\), with \\(A(1, 0)\\) and \\(AB = 4\\). Point \\(P\\) is a moving point on segment \\(AB\\), and a line through \\(P\\) parallel to \\(BC\\) intersects \\(AC\\) at point \\(Q\\). What is the equation of this parabola?\n",
    "completion": "\\textbf{Solution:} Since point A is (1,0) and AB = 4, \nit follows that the coordinates of point B are (-3,0). \nSubstituting points A(1,0) and B(-3,0) into the function's equation yields:\n\\[\n\\begin{cases}\n0 = 1 + b + c\\\\\n0 = 9 - 3b + c\n\\end{cases},\n\\]\nfrom which we find that $b=2$ and $c=-3$,\nthus, the equation of the parabola is \\boxed{y=x^{2}+2x-3}.",
    "image_path": "dataset/mm_math/images/MM_Math/53043661_107.png"
  },
  {
    "solution": "\\boxed{\\sqrt{2}+1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point C is a point on the semicircle O with AB as its diameter, and $AB=2$. AD bisects $\\angle BAC$ and intersects BC at point D. CP bisects $\\angle BCA$ and intersects AD at point P. $PF\\perp AC$, $PE\\perp BC$. Find the minimum value of $\\frac{1}{AC}+\\frac{1}{DC}$.\n",
    "completion": "\\textbf{Solution:} Let $PE=PF=CE=CF=x$,\\\\\n$\\because$ $PE\\perp BC$, $AC\\perp BC$,\\\\\n$\\therefore PE\\parallel AC$,\\\\\n$\\therefore \\triangle PED\\sim \\triangle ACD$,\\\\\n$\\therefore \\frac{PE}{AC}=\\frac{PD}{AD}$ (1);\\\\\nSimilarly, $PF\\parallel BC$, $\\triangle PAF\\sim \\triangle DAC$,\\\\\n$\\therefore \\frac{PF}{CD}=\\frac{AP}{AD}$ (2),\\\\\nFrom (1) + (2), we get $\\frac{PE}{AC}+\\frac{PF}{CD}=\\frac{PD}{AD}+\\frac{AP}{AD}=\\frac{AD}{AD}=1$,\\\\\n$\\therefore \\frac{PE}{AC}+\\frac{PF}{CD}=1$,\\\\\nthat is $\\frac{x}{AC}+\\frac{x}{CD}=1$,\\\\\n$\\therefore \\frac{1}{AC}+\\frac{1}{DC}=\\frac{1}{x}$;\\\\\nWhen x is at its maximum, $\\frac{1}{AC}+\\frac{1}{DC}$ is at its minimum;\\\\\n$\\because AD$ bisects $\\angle BAC$,\\\\\n$\\therefore$ Point P is the incenter of $\\triangle ACB$,\\\\\n$\\therefore PE, PF$ are the radii of the incircle;\\\\\nDraw $PH\\perp AB$, with H as the foot,\\\\\nthen it is easy to see $AF=AH$, $BE=BH$,\\\\\n$\\therefore AF+BE=AH+BH=AB$,\\\\\n$\\therefore CE=CF=\\frac{AC+BC-AB}{2}$,\\\\\nLet $AC=b$, $BC=a$, $AB=c=2$,\\\\\n$\\therefore x=\\frac{a+b-c}{2}=\\frac{a+b-2}{2}=\\frac{a+b}{2}-1$,\\\\\n$\\because AC^{2}+BC^{2}=AB^{2}$,\\\\\n$\\therefore a^{2}+b^{2}=4$,\\\\\n$\\because (a-b)^{2}=a^{2}+b^{2}-2ab\\ge 0$,\\\\\n$\\therefore 2ab\\le a^{2}+b^{2}=4$,\\\\\n$\\therefore ab\\le 2$,\\\\\n$\\because (a+b)^{2}=a^{2}+b^{2}+2ab=4+2ab\\le 4+4=8$,\\\\\n$\\therefore a+b\\le 2\\sqrt{2}$,\\\\\n$\\therefore$ the maximum value of $a+b$ is $2\\sqrt{2}$;\\\\\n$\\therefore x=\\frac{a+b}{2}-1=\\frac{2\\sqrt{2}}{2}-1=\\sqrt{2}-1$,\\\\\n$\\therefore$ the maximum value of $x$ is $\\sqrt{2}-1$,\\\\\n$\\therefore \\frac{1}{x}=\\frac{1}{\\sqrt{2}-1}=\\sqrt{2}+1$,\\\\\n$\\therefore$ the minimum value of $\\frac{1}{AC}+\\frac{1}{DC}$ is $\\boxed{\\sqrt{2}+1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51433177_153.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in square ABCD, the side BC is rotated counterclockwise around point B to $BC'$, and $CC'$, $DC'$ are connected. If $\\angle CC'D=90^\\circ$ and $C'D=2$, what is the length of the line segment BC?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, take point O as the midpoint of segment CC$^{\\prime}$ and connect BO. \\\\\nAccording to the condition, BC = BC$^{\\prime}$ \\\\\n$\\therefore$BO$\\perp$CC$^{\\prime}$ \\\\\n$\\therefore$$\\angle$BOC = $90^\\circ$ \\\\\nIn square ABCD, \\\\\nBC = CD, $\\angle$BCD = $90^\\circ$ \\\\\n$\\therefore$$\\angle$OCB + $\\angle$C$^{\\prime}$CD = $90^\\circ$ \\\\\nAlso, since $\\angle$CC$^{\\prime}$D = $90^\\circ$ \\\\\n$\\therefore$$\\angle$C$^{\\prime}$DC + $\\angle$C$^{\\prime}$CD = $90^\\circ$ \\\\\n$\\therefore$$\\angle$OCB = $\\angle$C$^{\\prime}$DC \\\\\nIn Rt$\\triangle$OBC and Rt$\\triangle$C$^{\\prime}$CD, \\\\\n$\\left\\{\\begin{array}{l}\\angle OCB = \\angle C^{\\prime}DC \\\\ \\angle BOC = \\angle CC^{\\prime}D \\\\ BC = CD \\end{array}\\right.$ \\\\\n$\\therefore$Rt$\\triangle$OBC$\\cong$ Rt$\\triangle$C$^{\\prime}$CD (AAS) \\\\\n$\\therefore$OC = C$^{\\prime}$D = 2 \\\\\n$\\therefore$CC$^{\\prime}$ = 2 OC = $2 \\times 2$ = 4 \\\\\n$\\therefore$BO = CC$^{\\prime}$ = 4 \\\\\nIn Rt$\\triangle$BOC, \\\\\nBC = $\\sqrt{BO^{2} + OC^{2}}$ = $\\sqrt{4^{2} + 2^{2}}$ = $\\boxed{2\\sqrt{5}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51807170.png"
  },
  {
    "solution": "\\boxed{x<4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the graph of the function $y=2x+b$ intersects with the graph of the function $y=kx-3$ at point $P$. What is the solution set of the inequality $kx-3>2x+b$?\n",
    "completion": "\\textbf{Solution}: Observing the graph, we find: when $x<4$, the graph of the function $y=2x+b$ is below the graph of the function $y=kx-3$, \\\\\n$\\therefore$ the solution set of the inequality $kx-3>2x+b$ is $\\boxed{x<4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53182760.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{5}}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, it is known that $\\angle C = 90^\\circ$, $AC = 1$, and $BC = 2$, then what is the value of $\\sin B$?\n",
    "completion": "\\textbf{Solution:} Since $AB=\\sqrt{AC^{2}+BC^{2}}=\\sqrt{5}$,\\\\\ntherefore $\\sin B=\\frac{AC}{AB}=\\frac{1}{\\sqrt{5}}=\\frac{\\sqrt{5}}{5}=\\boxed{\\frac{\\sqrt{5}}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53110857.png"
  },
  {
    "solution": "\\boxed{x>1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=kx+b$ ($k\\neq 0$) passes through point $A(1, 2)$. What is the solution set of the inequality $kx+b<2x$ with respect to $x$?\n",
    "completion": "\\textbf{Solution:} Given that $\\because x=1$ leads to $y=2$,\\\\\n$\\therefore$ the graph of the function $y=2x$ also passes through point A $(1, 2)$,\\\\\n$\\therefore$ the graphs of the function $y=kx+b$ (where $k<0$) and $y=2x$ both pass through point A $(1, 2)$.\\\\\nUpon observing the graphs, we find that for $x>1$, $kx+b<2x$.\\\\\nHence, the answer is: $\\boxed{x>1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52050161.png"
  },
  {
    "solution": "\\boxed{\\frac{25}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it's known that the point $A(-1,-3)$ lies on the line $l: y=kx-2$, and point $M(m,y_{1})$ is a moving point on the parabola $y=ax^{2}-4ax+2\\ (a\\neq 0)$. If the parabola intersects the line $l$ at point $A$.\n",
    "completion": "\\textbf{Solution:}\n\n(1) Substituting the point $A(-1,-3)$ into the line $l: y=kx-2$ yields\\\\\n$-k-2=-3$, solving this equation gives $k=1$,\\\\\nsubstituting the point $A(-1,-3)$ into the parabola $y=ax^2-4ax+2$ yields\\\\\n$a+4a+2=-3$, solving this equation gives $a=-1$;\\\\\n(2) Since $a=-1$ and $k=1$,\\\\\nthe parabola is $y=-x^2+4x+2$ and the line $l: y=x-2$,\\\\\nconsider $M(m,y_1)$ as a moving point on the parabola $y=-x^2+4x+2$,\\\\\nhence $y_1=-m^2+4m+2$,\\\\\nsince line segment $MN$ is parallel to the y-axis and intersects line $l$ at point $N$,\\\\\nwe have $N(m,m-2)$,\\\\\n$MN=-m^2+3m+4=-(m-\\frac{3}{2})^2+\\frac{25}{4}$,\\\\\nwhen $m=\\frac{3}{2}$, the maximal length of the line segment $MN$ is $\\boxed{\\frac{25}{4}}$, at this point the coordinates of $M$ are $\\left(\\frac{3}{2},\\frac{23}{4}\\right)$.",
    "image_path": "dataset/mm_math/images/MM_Math/53147494_60.png"
  },
  {
    "solution": "\\boxed{\\frac{5}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given the quadrilateral $ABCD$, where $AB \\parallel CD$, $BC=AD=4$, $AB=CD=10$, and $\\angle DCB=90^\\circ$. Let $E$ be a point on the side $CD$, with $DE=7$. A moving point $P$ starts from point $A$, moving towards point $B$ along side $AB$ at a speed of 1 unit per second. Connect $PE$. Let the time taken for point $P$ to move be $t$ seconds. If $\\triangle BPE$ is a right-angled triangle, find the value of $t$.\n",
    "completion": "\\textbf{Solution:}\n\nSince $BC=AD=4$ and $AB=CD=10$,\n\nit follows that quadrilateral ABCD is a parallelogram.\n\nSince $\\angle DCB=90^\\circ$,\n\nABCD is a rectangle.\n\nTherefore $\\angle A=\\angle D=90^\\circ$.\n\nWhen $\\angle BPE=90^\\circ$, that is $\\angle APE=90^\\circ$,\n\nquadrilateral APED is a rectangle,\n\nthus $AP=DE=7$,\n\nso $t=7\\div 1=\\boxed{7}$ seconds,\n\nWhen $\\angle BEP=90^\\circ$, draw $PF\\perp CD$ at F through point P,\n\nsince $\\angle A=\\angle D=\\angle DFP=90^\\circ$,\n\nquadrilateral ADFP is a rectangle,\n\nhence $DF=AP=t$, $PF=AD=4$,\n\nthus $PE^2=PF^2+EF^2=4^2+(7-t)^2$,\n\nsince in $\\triangle BEP$, $BP^2=BE^2+PE^2$,\n\nit follows that $(10-t)^2=5^2+4^2+(7-t)^2$\n\nSolving yields: $t=\\frac{5}{3}$,\n\nTherefore, when $t=7$ or $t=\\boxed{\\frac{5}{3}}$, $\\triangle BPE$ is a right-angled triangle.",
    "image_path": "dataset/mm_math/images/MM_Math/53339149_122.png"
  },
  {
    "solution": "\\boxed{46^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, with $\\angle ACB=90^\\circ$, fold $\\triangle CBD$ along CD, so that point B precisely lands on point E on side AC. If $\\angle A=22^\\circ$, then what is the measure of $\\angle EDA$?\n",
    "completion": "\\textbf{Solution:} In $\\triangle ABC$, $\\angle ACB=90^\\circ$, $\\angle A=22^\\circ$,\\\\\nthus $\\angle B=90^\\circ-\\angle A=68^\\circ$,\\\\\nBy the property of folding, we have: $\\angle CED=\\angle B=68^\\circ$,\\\\\ntherefore $\\angle EDA=\\angle CED-\\angle A=\\boxed{46^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52603365.png"
  },
  {
    "solution": "\\boxed{x>2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=kx+b$ ($k<0$) passes through the point $(2, 1)$. When $kx+b<\\frac{1}{2}x$, what is the range of values for $x$?\n",
    "completion": "\\textbf{Solution:} Let's solve it: when $x=2$, $y=1$, \\\\\n$\\therefore$ the line $y=\\frac{1}{2}x$ goes through the point $(2,1)$,\\\\\nAs can be seen from the graph, \\\\\nwhen $kx+b<\\frac{1}{2}x$, $x>2$\\\\\nTherefore, the answer is: $\\boxed{x>2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51983511.png"
  },
  {
    "solution": "\\boxed{x<-\\frac{1}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y = kx + b$ passes through the point $P\\left(-\\frac{1}{2}, 1\\right)$. What is the solution set of the inequality $kx + b < 1$?\n",
    "completion": "\\textbf{Solution:} The graph of the linear function $y=kx+b$ passes through the point $P\\left( -\\frac{1}{2}, 1\\right)$.\\\\\nBy analyzing the graph, we can find that the solution set of $kx+b<1$ is: $x<-\\frac{1}{2}$,\\\\\nTherefore, the answer is: $\\boxed{x<-\\frac{1}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52189208.png"
  },
  {
    "solution": "\\boxed{65^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the square $ABCD$, point $F$ is on the side $CD$, and $BF$ intersects $AC$ at point $E$. If $\\angle CBF=20^\\circ$, then what is the measure of $\\angle AED$ in degrees?\n",
    "completion": "\\textbf{Solution:} Given $\\because$ quadrilateral ABCD is a square\\\\\n$\\therefore \\angle ACB = \\angle ACD = \\angle BAC = \\angle CAD = 45^\\circ$, $\\angle ABC = 90^\\circ$, AB = AD\\\\\n$\\because \\angle FBC = 20^\\circ$\\\\\n$\\therefore \\angle ABF = 70^\\circ$\\\\\n$\\therefore$ in $\\triangle ABE$, $\\angle AEB = 65^\\circ$\\\\\nIn $\\triangle ABE$ and $\\triangle ADE$\\\\\n$ \\left\\{\\begin{array}{l}\nAB = AD \\\\\n\\angle BAE = \\angle EAD = 45^\\circ \\\\\nAE = AE\n\\end{array}\\right.$ \\\\\n$\\therefore \\triangle ABE \\cong \\triangle ADE$\\\\\n$\\therefore \\angle AED = \\angle AEB = \\boxed{65^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/51968648.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the line $y=kx+5$ intersects the $x$-axis at point $A$ and the $y$-axis at point $B$, with the coordinates of point $A$ being $(5,0)$. The line $y=2x-4$ intersects the $x$-axis at point $D$ and intersects line $AB$ at point $C$. What is the area of $\\triangle ADC$?\n",
    "completion": "\\textbf{Solution:} Substitute $y=0$ into $y=2x-4$ to obtain $2x-4=0$, solving this yields $x=2$, $\\therefore$ the coordinates of point D are $(2,0)$, $\\because$ A$(5,0)$, $\\therefore$ $AD=3$, $\\because$ the coordinates of point C are $C(3,2)$, $\\therefore$ $S_{\\triangle ADC} = \\frac{1}{2} \\cdot 3 \\cdot 2 = \\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52354168_188.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the line segment $AB=4$. Extend $AB$ to $C$ so that $BC=\\frac{1}{2}AB$. Extend $AB$ in the opposite direction to $D$ so that $AD=\\frac{1}{2}AC$. If $Q$ is the midpoint of $AB$, and $P$ is a point on the line segment $CD$, and $BP=\\frac{1}{2}BC$, what is the length of the line segment $PQ$?\n",
    "completion": "\\textbf{Solution}: Since $AB=4$, and $Q$ is the midpoint of $AB$, \\\\\n$\\therefore QB=\\frac{1}{2}AB=\\frac{1}{2}\\times 4=2$.\\\\\nSince $BC=2$.\\\\\n$\\therefore BP=\\frac{1}{2}BC=\\frac{1}{2}\\times 2=1$.\\\\\nWhen point $P$ is to the right of point $B$, $PQ=QB+BP=2+1=3$;\\\\\nWhen point $P$ is to the left of point $B$, $PQ=QB-BP=2-1=1$.\\\\\nThus, the length of $PQ$ is either $1$ or \\boxed{3}.",
    "image_path": "dataset/mm_math/images/MM_Math/55000649_114.png"
  },
  {
    "solution": "\\boxed{40^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle A = 50^\\circ$, what is the degree measure of $\\angle OCB$?\n",
    "completion": "\\textbf{Solution:} Since $\\overset{\\frown}{CB}=\\overset{\\frown}{CB}$ and $\\angle A=50^\\circ$,\\\\\nit follows that $\\angle BOC=2\\angle A=100^\\circ$\\\\\nBecause $OB=OC$,\\\\\nit is concluded that $\\angle OCB=\\angle OBC=\\frac{1}{2}(180^\\circ-\\angle BOC)=\\boxed{40^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/51261921.png"
  },
  {
    "solution": "\\boxed{-2x+3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=-\\frac{3}{4}x+3$ intersects the x-axis and y-axis at points A and B, respectively. Point C is on the line segment OA, and the line segment OB is folded along BC, with point O falling on point D on the side AB. What is the equation of line BC?\n",
    "completion": "\\textbf{Solution:} Since the line $y=-\\frac{3}{4}x+3$ intersects the $x$ and $y$ axes at points $A$ and $B$ respectively,\n\nlet $x=0$, then $y=3$, let $y=0$, then $x=4$,\n\ntherefore, point $A(4,0)$, point $B(0,3)$,\n\ntherefore, OA=4, OB=3,\n\ntherefore, AB= $\\sqrt{OB^{2}+OA^{2}}=5$,\n\nsince segment $OB$ is folded along $BC$, point $O$ falls on the point $D$ on side $AB$,\n\ntherefore, OB=BD=3, OC=CD, $\\angle BOC=\\angle BDC=90^\\circ$,\n\ntherefore, AD=AB−BD=2,\n\nsince $AC^{2}=AD^{2}+CD^{2}$,\n\ntherefore, $(4−OC)^{2}=2^{2}+OC^{2}$,\n\ntherefore, OC=1.5,\n\ntherefore, point $C(1.5,0)$,\n\nAssume the equation of line $BC$ is: $y=kx+3$,\n\ntherefore, $0=1.5k+3$,\n\ntherefore, $k=−2$,\n\ntherefore, the equation of line $BC$ is: $y=-2x+3$.\n\nHence, the answer is: $y=-2x+3=\\boxed{-2x+3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52252064.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, quadrilateral $ABCD$ is inscribed in circle $\\odot O$, with $DE$ being the diameter of $\\odot O$. Connect $BD$. If $\\angle BCD = 2\\angle BAD$, then what is the measure of $\\angle BDE$?\n",
    "completion": "\\textbf{Solution:} Connect BE, \\\\\n$\\because$ Quadrilateral ABCD is inscribed in circle $O$, \\\\\n$\\therefore \\angle BCD + \\angle BAD = 180^\\circ$, \\\\\n$\\because \\angle BCD = 2\\angle BAD$, \\\\\n$\\therefore \\angle BAD = 60^\\circ$, \\\\\nBy the inscribed angle theorem, we have: $\\angle BED = \\angle BAD = 60^\\circ$, \\\\\n$\\because DE$ is the diameter of circle $O$, \\\\\n$\\therefore \\angle EBD = 90^\\circ$, \\\\\n$\\therefore \\angle BDE = 90^\\circ - 60^\\circ = \\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52512549.png"
  },
  {
    "solution": "\\boxed{32}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the line $AB$ intersects the $x$-axis and $y$-axis at point $A$ and point $B$, respectively, and the line $CD$ intersects the $x$-axis and $y$-axis at point $C$ and point $D$, respectively. The equation of the line $AB$ is $y=-\\frac{5}{8}x+5$, and the equation of the line $CD$ is $y=kx+b$, where $k\\ne 0$. The two lines intersect at point $E\\left(m, \\frac{10}{3}\\right)$, and $OB:OC=5:4$. The line $CD$ is translated downwards by a certain distance so that the translated line passes through point $A$ and intersects the $y$-axis at point $F$. What is the area of the quadrilateral $AEDF$?\n",
    "completion": "\\textbf{Solution:} Let $y=-\\frac{5}{8}x+5$ with $y=0$, we obtain $-\\frac{5}{8}x+5=0$. Solving for $x$ yields $x=8$, $\\therefore$ the coordinates of point A are $(8,0)$. Assume the equation of line CD after moving downwards is $y=\\frac{1}{2}x+m$. Substituting the coordinates of point A, we find $m=-4$, $\\therefore$ the equation of line AF is $y=\\frac{1}{2}x-4$, $\\therefore$ the coordinates of point F are $(0,-4)$. $\\because$ the equation of line CD is $y=\\frac{1}{2}x+2$, $\\therefore$ the intersection with the y-axis gives the coordinates of point D as $(0,2)$. Connecting OE,\\\\\n$\\therefore$ the area of quadrilateral $AEDF = S_{\\triangle ODE}+S_{\\triangle OAE}+S_{\\triangle OAF} = \\frac{1}{2}\\times 2\\times \\frac{8}{3}+\\frac{1}{2}\\times 8\\times \\frac{10}{3}+\\frac{1}{2}\\times 8\\times 4 = \\boxed{32}$",
    "image_path": "dataset/mm_math/images/MM_Math/52304489_230.png"
  },
  {
    "solution": "\\boxed{70^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the quadrilateral $ABCD$ is inscribed in circle $O$. If $\\angle AOB = 40^\\circ$ and $BC \\parallel OA$, then what is the measure of $\\angle ADC$?\n",
    "completion": "\\textbf{Solution:} Given $\\because \\angle AOB=40^\\circ$, and OA=OB,\\\\\n$\\therefore \\angle ABO=70^\\circ$,\\\\\n$\\because BC\\parallel OA$,\\\\\n$\\therefore \\angle OBC=\\angle AOB=40^\\circ$,\\\\\n$\\therefore \\angle ABC=110^\\circ$,\\\\\n$\\therefore \\angle ADC=180^\\circ-110^\\circ=\\boxed{70^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52601840.png"
  },
  {
    "solution": "\\boxed{\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the rectangle $ABCD$, $AB=3$, $BC=3\\sqrt{3}$, and $\\angle CBD=30^\\circ$. Point $M$ is a point on the ray $BD$ (not coinciding with points $B$, $D$). Line $AM$ is drawn, and through point $M$, $MN \\perp AM$ intersects line $BC$ at point $N$. If $\\triangle BMN$ is an isosceles triangle, what is the length of $BN$?\n",
    "completion": "\\textbf{Solution:} Connect AN,\\\\\nsince quadrilateral ABCD is a rectangle, it follows that $\\angle ABN=90^\\circ$,\\\\\nsince $MN\\perp AM$, it follows that $\\angle AMN=90^\\circ$,\\\\\nhence $\\angle BNM+\\angle BAM=180^\\circ$,\\\\\nsince $\\angle BAM<90^\\circ$, it follows that $\\angle BNM>90^\\circ$,\\\\\nsince $\\triangle BMN$ is an isosceles triangle, there only exists the case where $BN=MN$,\\\\\nin $Rt\\triangle ABN$ and $Rt\\triangle AMN$, $\\left\\{\\begin{array}{l}AN=AN\\\\ BN=MN\\end{array}\\right.$,\\\\\nhence $Rt\\triangle ABN\\cong Rt\\triangle AMN\\left(HL\\right)$,\\\\\nthus $AB=AM$,\\\\\nsince $\\angle ABN=90^\\circ$, $\\angle CBD=30^\\circ$, hence $\\angle ABM=60^\\circ$,\\\\\nthus $\\triangle ABM$ is an equilateral triangle, hence $\\angle BAM=60^\\circ$,\\\\\nthus $\\angle BAN=\\angle MAN=\\frac{1}{2}\\angle BAM=30^\\circ$, hence $AN=2BN$,\\\\\nsince $AB^2+BN^2=AN^2$, and $AB=3$,\\\\\nhence $3^2+BN^2=(2BN)^2$, hence $BN=\\boxed{\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52977108.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\triangle ABC$, $ \\angle ACB = 90^\\circ$, D is the midpoint of $ AB$, $ ED \\perp AB$ intersects $ AC$ at point E, $ \\tan \\angle EBC = \\frac{3}{4}$. What is the value of $ \\tan \\angle ABE$?\n",
    "completion": "\\textbf{Solution:} In $\\triangle EBC$, we have $\\angle ECB=90^\\circ$,\\\\\n$\\therefore \\tan\\angle EBC=\\frac{CE}{BC}=\\frac{3}{4}$.\\\\\nLet $CE=3k$, $BC=4k$, so $BE=5k$.\\\\\nSince $D$ is the midpoint of $AB$ and $ED\\perp AB$,\\\\\n$\\therefore AE=BE=5k$,\\\\\n$\\therefore \\angle ABE=\\angle BAE$, $AC=8k$,\\\\\nIn $\\triangle ABC$, $\\angle ACB=90^\\circ$,\\\\\n$\\therefore \\tan\\angle CAB=\\frac{BC}{AC}=\\frac{4k}{8k}=\\frac{1}{2}$.\\\\\n$\\therefore$ The tangent of $\\angle ABE$ is $\\boxed{\\frac{1}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51353489.png"
  },
  {
    "solution": "\\boxed{80^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $OB$ is the bisector of $\\angle AOC$, and $OD$ is the bisector of $\\angle COE$. If $\\angle AOD$ and $\\angle BOD$ are complementary, and $\\angle DOE=30^\\circ$, find the degree measure of $\\angle AOC$.\n",
    "completion": "\\textbf{Solution:} Since $\\angle AOD$ and $\\angle BOD$ are complementary,\\\\\nwe have $\\angle AOD+\\angle BOD=180^\\circ$,\\\\\nFrom the figure, we know: $\\angle AOD=\\angle AOC+\\angle COD$,\\\\\nand $\\angle BOD=\\angle BOC+\\angle COD$,\\\\\nBy the definition of angle bisector, we know: $\\angle BOC=\\frac{1}{2}\\angle AOC$,\\\\\nThus, $\\angle AOC+\\angle DOE+\\frac{1}{2}\\angle AOC+\\angle DOE=180^\\circ$,\\\\\nwhich implies $\\frac{3}{2}\\angle AOC+2\\angle DOE=180^\\circ$,\\\\\nSince $\\angle DOE=30^\\circ$,\\\\\nit follows that $\\frac{3}{2}\\angle AOC+2\\times 30^\\circ=180^\\circ$,\\\\\nhence, $\\angle AOC=\\boxed{80^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51380002_131.png"
  },
  {
    "solution": "\\boxed{25^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, quadrilateral $ABCD$ is inscribed in circle $\\odot O$, with $BE$ being the diameter of circle $\\odot O$. Connect $AE$ and $BD$. If $\\angle BCD=115^\\circ$, what is the measure of $\\angle EBD$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is inscribed in circle $O$, and $\\angle BCD=115^\\circ$, \\\\\nit follows that $\\angle BAD=180^\\circ-115^\\circ=65^\\circ$\\\\\nDraw line DE, \\\\\nSince arc $\\overset{\\frown}{BD}=\\overset{\\frown}{BD}$, \\\\\nit follows that $\\angle BED=65^\\circ$\\\\\nSince BE is the diameter of circle $O$, \\\\\nit follows that $\\angle BDE=90^\\circ$\\\\\nTherefore, $\\angle EBD=90^\\circ-65^\\circ=\\boxed{25^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/53043258.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that in rectangle $ABCD$, point $E$ is on the extension line of side $BC$, and $CE=BD$. Connect $AE$ and intersect $BD$ at $F$. If $\\angle E=20^\\circ$, what is the degree measure of $\\angle AFB$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, connect AC, intersecting BD at point O,\\\\\nsince ABCD is a rectangle, then BD=AC,\\\\\nCE=BD, then CE=AC,\\\\\n$\\angle E=20^\\circ$, thus $\\angle CAE=20^\\circ$,\\\\\n$\\therefore \\angle ACB=\\angle CAE+\\angle E=40^\\circ$,\\\\\n$\\because$ OB=$\\frac{1}{2}$BD, OC=$\\frac{1}{2}$AC,\\\\\n$\\therefore$ OB=OC, $\\angle OBC=\\angle OCB=40^\\circ$,\\\\\n$\\therefore \\angle BOC=180^\\circ-\\angle OBC-\\angle OCB=100^\\circ$,\\\\\n$\\therefore \\angle AOF=\\angle BOC=100^\\circ$,\\\\\n$\\because \\angle OAF=20^\\circ$,\\\\\n$\\therefore \\angle AFO=180^\\circ-\\angle OAF-\\angle AOF=60^\\circ$,\\\\\n$\\therefore \\angle AFB=\\boxed{60^\\circ}$,",
    "image_path": "dataset/mm_math/images/MM_Math/53538185.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{6}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $E$ and $F$ are the midpoints of sides $CD$ and $BC$ of rectangle $ABCD$ respectively, with segments $AC$, $AE$, $EF$ connected. Given that $AE \\perp EF$ and $AC = 6$, what is the length of $AB$?\n",
    "completion": "\\textbf{Solution:} Join BD and AF as shown in the diagram: \\\\\nSince quadrilateral ABCD is a rectangle, and $AC=6$, \\\\\ntherefore $BD=AC=6$, $\\angle ADC=\\angle ECF=90^\\circ$, $CD=AB$, \\\\\nBecause E and F are respectively the midpoints of sides CD and BC of rectangle ABCD, \\\\\nthus $EF=\\frac{1}{2}BD=3$, \\\\\nLet $DE=CE=a$, and $CF=BF=b$, then $AB=CD=2a$, $AD=BC=2b$, \\\\\nBecause $AE\\perp EF$, \\\\\ntherefore, in $\\triangle ADE$, $\\triangle CEF$, $\\triangle AEF$, $\\triangle ABF$, by applying the Pythagorean theorem, we get: \\\\\n$a^2+4b^2+9=4a^2+b^2$, $a^2+b^2=9$, \\\\\nthus, from the above equations, we get: $a^2=6$, \\\\\nSolving gives: $a=\\sqrt{6}$ (negative root is discarded), \\\\\ntherefore, $AB=2\\sqrt{6}$; \\\\\nHence, the answer is: $AB=\\boxed{2\\sqrt{6}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52076179.png"
  },
  {
    "solution": "\\boxed{20}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $AB=24$, $C$ is the midpoint of $AB$, point $D$ is on line segment $AC$, and $AD:CB=1:3$, then what is the length of $DB$?\n",
    "completion": "\\textbf{Solution:} Since $C$ is the midpoint of $AB$ and $AB=24$,\\\\\n$\\therefore$AC=CB=12,\\\\\nSince $AD\\colon CB=1\\colon 3$,\\\\\n$\\therefore$AD=$\\frac{1}{3}$BC=4,\\\\\n$\\therefore$DB=AB−AD=24−4=\\boxed{20}.",
    "image_path": "dataset/mm_math/images/MM_Math/55192327.png"
  },
  {
    "solution": "\\boxed{(6, 0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given: As shown in the diagram, the graph of the direct proportion function $y=2x$ intersects with the graph of the linear function $y=-\\frac{2}{3}x+4$ at the point $A\\left(\\frac{3}{2}, 3\\right)$, and the graph of the linear function $y=-\\frac{2}{3}x+4$ intersects with the $x$-axis at point $B$. Find the coordinates of point $B$.\n",
    "completion": "\\textbf{Solution}: Given that the equation of the linear function is $y=-\\frac{2}{3}x+4$,\\\\\nsetting $y=0$ yields: $0=-\\frac{2}{3}x+4$,\\\\\nsolving for $x$, we get: $x=6$,\\\\\n$\\therefore$ the coordinates of point $B$ are $\\boxed{(6, 0)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51545374_102.png"
  },
  {
    "solution": "\\boxed{}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=-2x+8$ intersects the $x$-axis and the $y$-axis at points $A$ and $B$, respectively, and the line $y=\\frac{1}{2}x+3$ intersects the $y$-axis at point $C$. The two lines intersect at point $D$. As shown in figure 2, draw $AE\\parallel y$-axis from point $A$ to intersect the line $y=\\frac{1}{2}x+3$ at point $E$, and connect $AC$, $BE$. Prove that quadrilateral $ACBE$ is a rhombus.\n",
    "completion": "\\textbf{Solution:} Proof: \\\\\n$ \\because $ the line $y=-2x+8$ intersects the $x$-axis and the $y$-axis at points $A$ and $B$, respectively, \\\\\n$ \\therefore $ point $B(0,8)$, and point $A(4, 0)$, \\\\\n$ \\because $ the line $y=\\frac{1}{2}x+3$ intersects the $y$-axis at point $C$, \\\\\n$ \\therefore $ point $C(0, 3)$, \\\\\n$ \\because $AE$\\parallel$y-axis and intersects the line $y=\\frac{1}{2}x+3$ at point $E$, \\\\\n$ \\therefore $ point $E(4, 5)$ \\\\\n$ \\because $ point $B(0,8)$, point $A(4, 0)$, point $C(0, 3)$, and point $E(4, 5)$, \\\\\n$ \\therefore $BC=5, $AE=5$, $AC=5$, $BE=5$, \\\\\n$ \\therefore $BC=AE=AC=BE, \\\\\n$ \\therefore $ the quadrilateral $ACBE$ is a rhombus $\\boxed{}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52250730_99.png"
  },
  {
    "solution": "\\boxed{3+2\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the line $y=-x+m+1$ intersects with the reciprocal function $y=\\frac{m}{x}$ ($x>0$, $m>0$) at points A and B, intersects the $x$-axis at point C, and intersects the $y$-axis at point D. The coordinates of point A are given as $(1, 3)$. If the line CD is translated 2 units to the right to obtain line EF, and the part of the hyperbola below line CD is folded along line CD, and if the folded image (the dashed part in the figure) has only one common point with line EF, then find the value of $m$.\n",
    "completion": "\\textbf{Solution:} Consider the line $y=-x+m+1$. After translating it 2 units to the right, its equation becomes\n\\[y=-(x-2)+m+1=-x+m+3,\\]\nFor the lines $y=-x+m+1, y=-x+m+3$,\nand the inverse proportional function $y=\\frac{m}{x}$ symmetric about the line $y=x$,\nas illustrated, draw the line $y=x$, which intersects the hyperbola at point $M$, intersects the line $y=-x+m+1$ at point $N$,\nand intersects the line $y=-x+m+3$ at point $P$,\nSet $x=-x+m+1$,\nwe get $x=\\frac{m+1}{2}$,\nthus, the coordinates of point $N$ are $\\left(\\frac{m+1}{2}, \\frac{m+1}{2}\\right)$,\nSet $x=-x+m+3$, we obtain $x=\\frac{m+3}{2}$,\nthus, the coordinates of point $P$ are $\\left(\\frac{m+3}{2}, \\frac{m+3}{2}\\right)$,\nSet $x=\\frac{m}{x}$, solving gives $x=-\\sqrt{m}$ (discard) or $x=\\sqrt{m}$,\nthus, the coordinates of point $M$ are $\\left(\\sqrt{m}, \\sqrt{m}\\right)$,\nBased on the problem statement, points $M$, $P$ are symmetric about point $N$, meaning $N$ is the midpoint of $M$, $P$,\ntherefore $\\sqrt{m}+\\frac{m+3}{2}=2\\times \\frac{m+1}{2}$,\nsolving gives $m_1=3-2\\sqrt{2}$ (discard), $m_2=\\boxed{3+2\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53080174_35.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given that $AB$ is the diameter of $\\odot O$ and $\\angle ACD$ is the inscribed angle corresponding to the arc $\\overset{\\frown}{AD}$, where $\\angle ACD=30^\\circ$. Through point $D$, draw $DE\\perp AB$ with $E$ as the foot of the perpendicular. The extension of $DE$ intersects $\\odot O$ at point $F$. If $AB=4$, find the length of $DF$.\n",
    "completion": "\\textbf{Solution:} Since $\\angle ADB=90^\\circ$ and $\\angle B=30^\\circ$, with $AB=4$,\\\\\nit follows that $AD=\\frac{1}{2} AB=2$.\\\\\nSince $\\angle DAB=60^\\circ$, $DE\\perp AB$, and $AB$ is the diameter,\\\\\nit results in $EF=DE=AD\\sin60^\\circ=\\sqrt{3}$,\\\\\nhence $DF=2DE=2\\sqrt{3}$.\\\\\nTherefore, the length of $DF$ is $\\boxed{2\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51073825_134.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB = 90^\\circ$, $AC = BC$, $BE \\perp CE$ at point $E$, and $AD \\perp CE$ at point $D$. Given $DE = 5$ and $AD = 9$, what is the length of $BE$?\n",
    "completion": "\\textbf{Solution}: We have $\\angle ACB = 90^\\circ$ and $BE \\perp CE$,\\\\\ntherefore, $\\angle BCE + \\angle ACD = 90^\\circ$ and $\\angle BCE + \\angle CBE = 90^\\circ$,\\\\\nthus, $\\angle ACD = \\angle CBE$,\\\\\nin $\\triangle BEC$ and $\\triangle CDA$,\\\\\n\\[\n\\left\\{ \\begin{array}{l}\n\\angle BEC = \\angle CDA \\\\\n\\angle CBE = \\angle ACD \\\\\nBC = AC\n\\end{array} \\right.,\n\\]\ntherefore, $\\triangle ACD \\cong \\triangle CBE$ (by AAS),\\\\\nthus, EC = AD = 9, and BE = DC,\\\\\nsince DE = 5,\\\\\nthus, CD = EC - DE = 4,\\\\\ntherefore, BE = \\boxed{4}.",
    "image_path": "dataset/mm_math/images/MM_Math/52991669.png"
  },
  {
    "solution": "\\boxed{-3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the intersection of the line $y=-x+m$ and $y=nx+4n\\ (n\\neq 0)$ has an abscissa of $-2$. What are the integer solutions of the inequality $-x+m>nx+4n>0$ with respect to $x$?\n",
    "completion": "\\textbf{Solution:} From the given condition, for $y=nx+4n\\ (n\\neq 0)$, then $x=-4$. Thus, the solution set for $nx+4n>0$ is $x>-4$.\\\\\nSince the $x$-coordinate of the intersection point between $y=-x+m$ and $y=nx+4n\\ (n\\neq 0)$ is $-2$, by observing the graph, the solution set for $-x+m>nx+4n$ is $x<-2$.\\\\\nTherefore, the solution set for $-x+m>nx+4n>0$ is $-4<x<-2$.\\\\\nSince $x$ is an integer, thus $x=\\boxed{-3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50481319.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure (1), it is known that $AM \\parallel CN$, and point B is a point in the plane, with $AB \\perp BC$ at point B. A line is drawn through point B such that $BD \\perp AM$ at point D, where $\\angle BCN = \\alpha$. If $\\alpha = 30^\\circ$, what is the measure of $\\angle ABD$ in degrees?\n",
    "completion": "\\textbf{Solution:} Extend $DB$ to intersect $NC$ at point H, as shown in the diagram,\\\\\n$\\because AM\\parallel CN$ and $BD\\perp AM$,\\\\\n$\\therefore DH\\perp NC$.\\\\\n$\\therefore \\angle BHC=90^\\circ$.\\\\\n$\\because \\angle BCN=\\alpha =30^\\circ$,\\\\\n$\\therefore \\angle HBC=90^\\circ-\\angle BCN=60^\\circ$.\\\\\n$\\because AB\\perp BC$,\\\\\n$\\therefore \\angle ABC=90^\\circ$.\\\\\n$\\therefore \\angle ABD=180^\\circ-\\angle ABC-\\angle HBC=\\boxed{30^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51321729_56.png"
  },
  {
    "solution": "\\boxed{\\left(\\frac{\\sqrt{2}}{3},\\frac{2\\sqrt{2}}{3}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the line $l\\colon y=-x+\\sqrt{2}$ intersects the $x$-axis at point $B$ and the $y$-axis at point $A$. With $AB$ as the right angle side, an isosceles right triangle $\\triangle ABC$ is constructed to the right, and with $AO$ as the hypotenuse, an isosceles right triangle $\\triangle ADO$ is constructed to the left. Connect $DC$ to intersect line $l$ at point $E$. What are the coordinates of point $E$?\n",
    "completion": "\\textbf{Solution:} For the line $l\\colon y=-x+\\sqrt{2}$,\n\nwhen $x=0$, $y=\\sqrt{2}$; when $y=0$, $x=\\sqrt{2}$.\n\n$\\therefore A\\left(0,\\sqrt{2}\\right), B\\left(\\sqrt{2},0\\right)$.\n\n$\\therefore OA=OB=\\sqrt{2}$.\n\n$\\therefore \\triangle ABO$ is an isosceles right triangle,\n\n$\\therefore AB=\\sqrt{2}OA=2$, $\\angle ABO=\\angle BAO=45^\\circ$.\n\nSince $\\triangle ABC$ is an isosceles right triangle, $\\angle ABC=90^\\circ$,\n\n$\\therefore BC=AB=2$, $\\angle BAC=\\angle BCA=45^\\circ$.\n\n$\\therefore AC=\\sqrt{2}AB=2\\sqrt{2}$.\n\nSince $\\angle CAB=\\angle ABO=45^\\circ$,\n\n$\\therefore AC\\parallel x$ axis.\n\n$\\therefore C\\left(2\\sqrt{2},\\sqrt{2}\\right)$.\n\nSince $\\triangle ADO$ is an isosceles right triangle, $\\angle ADO=90^\\circ$,\n\n$\\therefore AD=OD$.\n\nDraw $DF\\perp x$ axis at point F, thus $DF=OF=\\frac{1}{2}AO=\\frac{\\sqrt{2}}{2}$.\n\n$\\therefore D\\left(-\\frac{\\sqrt{2}}{2},\\frac{\\sqrt{2}}{2}\\right)$.\n\nLet the function of line CD be $y=kx+b$,\n\naccording to the question, we get\n\n$\\left\\{\\begin{array}{l}\n-\\frac{\\sqrt{2}}{2}k+b=\\frac{\\sqrt{2}}{2} \\\\\n2\\sqrt{2}k+b=\\sqrt{2}\n\\end{array}\\right.$.\n\nSolving it, we find $\\left\\{\\begin{array}{l}\nk=\\frac{1}{5} \\\\\nb=\\frac{3\\sqrt{2}}{5}\n\\end{array}\\right.$.\n\n$\\therefore$ The function of line CD is $y=\\frac{1}{5}x+\\frac{3\\sqrt{2}}{5}$.\n\nCombining the functions of line CD and AB, we get the system of equations\n\n$\\left\\{\\begin{array}{l}\ny=\\frac{1}{5}x+\\frac{3\\sqrt{2}}{5} \\\\\ny=-x+\\sqrt{2}\n\\end{array}\\right.$,\n\nsolving it, we find $\\left\\{\\begin{array}{l}\nx=\\frac{\\sqrt{2}}{3} \\\\\ny=\\frac{2\\sqrt{2}}{3}\n\\end{array}\\right.$.\n\n$\\therefore E\\left(\\frac{\\sqrt{2}}{3},\\frac{2\\sqrt{2}}{3}\\right)$.\n\nThus, the coordinates of point $E$ are $\\boxed{\\left(\\frac{\\sqrt{2}}{3},\\frac{2\\sqrt{2}}{3}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50496997.png"
  },
  {
    "solution": "\\boxed{-\\frac{1}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the equation of line $AB$ is $y=kx+6$, the coordinates of point $D$ are $(8, 0)$, and the symmetric point $C$ of point $O$ with respect to line $AB$ lies on line $AD$. Consider a line $l: y=mx+b$ passing through point $D$. When the angle between line $l$ and line $AB$ is equal to $45^\\circ$, find the corresponding value of $m$.\n",
    "completion": "\\textbf{Solution:} Set if the angle between line $DE$, $DF$ and line $AB$ is $45^\\circ$, i.e., $\\triangle DEF$ is an isosceles right triangle. Construct $EM\\perp DM$ at $M$, $FN\\perp DN$ at $N$, then $\\triangle DEM\\cong \\triangle FDN$,\n\n$\\therefore EM=DN$, $DM=FN$. Line $l$ passes through $D(8, 0)$, hence $0=8m+b$, solving this gives: $b=-8m$,\n\n$\\therefore$ the equation of line $l$ is: $y=mx-8m$. Let the coordinates of $E$ be $(t, -2t+6)$, then $EM=DN=8-t$, $DM=FN=-2t+6$,\n\n$\\therefore$ the coordinates of $F$ are $(2+2t, t-8)$. Point $F$ lies on line $AB$,\n\n$\\therefore t-8=-2(2+2t)+6$, solving this gives: $t=2$,\n\n$\\therefore E(2, 2)$, $F(6, -6)$. When line $l$ passes through point $E$, $2m-8m=2$, solving this gives: $m=\\boxed{-\\frac{1}{3}}$, when line $l$ passes through point $F$, $6m-8m=-6$, solving this gives: $m=\\boxed{3}$.\n\nTherefore, $m=\\boxed{3}$ or $\\boxed{-\\frac{1}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50560467_67.png"
  },
  {
    "solution": "\\boxed{75^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: When a set of triangles is arranged as shown in the figure, what is the value of $ \\angle \\alpha $?\n",
    "completion": "\\textbf{Solution:} By the exterior angle property of triangles, we have $ \\angle \\alpha =45^\\circ+30^\\circ=75^\\circ$\\\\\nTherefore, the answer is: $\\boxed{75^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52604137.png"
  },
  {
    "solution": "\\boxed{\\frac{20}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within rectangle $ABCD$, point $E$ is on side $CD$. Fold $\\triangle BCE$ along $BE$ so that point $C$ falls on point $F$ on side $AD$. Draw $FG \\parallel CD$ intersecting $BE$ at point $G$, and then connect $CG$. Quadrilateral $CEFG$ is a rhombus. If $AB = 6$ and $AD = 10$, find the area of quadrilateral $CEFG$.\n",
    "completion": "\\textbf{Solution:} Given that the figure is folded, \\\\\nwe have $BF=BC=AD=10$, thus $AF= \\sqrt{BF^2-AB^2}=8$, and $DF=AD-AF=2$. Let $DE=x$, then $EC=CD-DE=6-x=EF$. In $\\triangle EDF$, since $DE^2+DF^2=EF^2$, we have $x^2+2^2=(6-x)^2$. Solving this equation gives $x= \\frac{8}{3}$, hence $EC=6-DE=6- \\frac{8}{3}=\\frac{10}{3}$. Therefore, the area of the rhombus CEFG is $EC \\times FD= \\frac{10}{3} \\times 2= \\boxed{\\frac{20}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51614498_107.png"
  },
  {
    "solution": "\\boxed{73^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In $\\triangle ABC$, $\\angle BAC=50^\\circ$, $\\angle B=48^\\circ$, $AD$ is the angle bisector of $\\triangle ABC$. What is the measure of $\\angle ADC$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle BAC=50^\\circ$ and $\\angle B=48^\\circ$, \\\\\nit follows that $\\angle C=180^\\circ-\\angle BAC-\\angle B=82^\\circ$, \\\\\nsince $AD$ is the bisector of $\\triangle ABC$, \\\\\nit implies that $\\angle CAD=\\frac{1}{2}\\angle BAC=25^\\circ$, \\\\\nhence, $\\angle ADC=180^\\circ-\\angle CAD-\\angle C=\\boxed{73^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53066780.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of $\\odot O$, the chord $CD\\perp AB$ at point $H$, $CD=8$, $OA=5$, what is the length of $AH$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect OC. \\\\\n$\\because$ AB is the diameter of $\\odot$O, and chord CD $\\perp$ AB at point H with CD=8, \\\\\n$\\therefore$ $CH=DH=\\frac{1}{2}CD=4$, $\\angle$OHC$=90^\\circ$, \\\\\n$\\because$ OC=OA=5, \\\\\n$\\therefore$ $OH=\\sqrt{OC^{2}-CH^{2}}=3$, \\\\\n$\\therefore$ AH=OA+OH=$\\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51320376.png"
  },
  {
    "solution": "\\boxed{x<\\frac{3}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graphs of the function $y=2x$ and $y=ax+4$ intersect at point $A(m, 3)$. What is the solution set of the inequality $2x<ax+4$?\n",
    "completion": "\\textbf{Solution:} Because the graphs of functions $y=2x$ and $y=ax+4$ intersect at point $A\\left(\\frac{3}{2}, 3\\right)$,\\\\\ntherefore $3=2m$,\\\\\ntherefore $m=\\frac{3}{2}$,\\\\\ntherefore the coordinates of point A are $\\left(\\frac{3}{2}, 3\\right)$,\\\\\nfrom the graph, it is known that the solution set of $2x<ax+4$ is $\\boxed{x<\\frac{3}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52304470.png"
  },
  {
    "solution": "\\boxed{(0,9)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y = \\frac{4}{3}x + 4$ intersects the x-axis and y-axis at points A and B, respectively. Point C is on the y-axis, and line AC is drawn. If $\\triangle ABC$ is an isosceles triangle with AB as its base, please write down the coordinates of point C directly.\n",
    "completion": "\\textbf{Solution:} The coordinates of point C are $\\boxed{(0,9)}$ or $(0,-1)$ or $(0,-4)$.",
    "image_path": "dataset/mm_math/images/MM_Math/52978850_132.png"
  },
  {
    "solution": "\\boxed{6\\sqrt{2}-2\\sqrt{6}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AC=12$, $\\angle C=45^\\circ$, $\\angle B=120^\\circ$, find the length of $BC$.\n",
    "completion": "\\textbf{Solution:} Draw $AD\\perp BC$ through point A, where it intersects the extension of $BC$ at point D, then $\\angle ADC=90^\\circ$,\\\\\nIn $\\triangle ADC$, $\\angle C=45^\\circ$,\\\\\n$\\therefore AD=DC$,\\\\\nAccording to the Pythagorean theorem, we have: $AD^2+DC^2=AC^2$,\\\\\nthat is $2AD^2=AC^2=12^2$,\\\\\nSolving this gives $AD=DC=6\\sqrt{2}$,\\\\\n$\\because \\angle ABC=120^\\circ$,\\\\\n$\\therefore \\angle ABD=60^\\circ$,\\\\\nIn $\\triangle ABD$,\\\\\nlet $BD=x$, then $AB=2x$,\\\\\nAccording to the Pythagorean theorem, we have: $AD^2+DB^2=AB^2$,\\\\\nthat is $(6\\sqrt{2})^2+x^2=(2x)^2$,\\\\\nSolving this gives $x=2\\sqrt{6}$,\\\\\nthat is $BD=2\\sqrt{6}$,\\\\\n$\\therefore BC=DC-DB=6\\sqrt{2}-2\\sqrt{6}=\\boxed{6\\sqrt{2}-2\\sqrt{6}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53019891.png"
  },
  {
    "solution": "\\boxed{-15\\leq b\\leq 15}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the quadrilateral $ABCD$ is a diamond. It is known that $A(4, 0)$, $B(0, 3)$. Move the line from (2) up and down along the $y$-axis by $b$. If the line always intersects with the diamond, directly write out the range of values of $b$.\n",
    "completion": "\\textbf{Solution:} Given from equation (2) the equation of the line passing through points B and D is $y=3x+3$. Translating this line along the y-axis by b units, we get the equation $y=3x+3+b$ (where $b>0$ indicates an upward shift, and $b<0$ indicates a downward shift). When the translated line passes through point A (4,0), we have $0=12+3+b$, hence $b=\\boxed{-15}$. When the translated line passes through point C (-5,3), we have $3=-15+3+b$, thus $b=\\boxed{15}$. Therefore, the line always intersects with the rhombus, so the range of $b$ is $\\boxed{-15\\leq b\\leq 15}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52423104_81.png"
  },
  {
    "solution": "\\boxed{6\\pi -9\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of $\\odot O$, $BC$ is a chord of $\\odot O$, and the radius $OD\\perp BC$ with the foot of the perpendicular being $E$. If $BC=6\\sqrt{3}$ and $DE=3$, find the area of the shaded region (the result should retain $\\pi$).\n",
    "completion": "\\textbf{Solution:} Given that $\\angle CEO=90^\\circ$ and $CO=2OE$,\\\\\n$\\therefore \\angle ECO=30^\\circ$.\\\\\nSince $AB$ is the diameter of $\\odot O$,\\\\\n$\\therefore \\angle ACB=90^\\circ$,\\\\\n$\\therefore \\angle ACO=60^\\circ$.\\\\\nSince $OA=OC$,\\\\\n$\\therefore \\triangle ACO$ is an equilateral triangle,\\\\\n$\\therefore \\angle AOC=60^\\circ$,\\\\\n$\\therefore S_{\\text{shaded}}=S_{\\text{sector} ACO}-S_{\\triangle AOC}=\\frac{60\\pi \\times 6^2}{360}-\\frac{1}{2}\\times 6\\times 6\\times \\frac{\\sqrt{3}}{2}=6\\pi -9\\sqrt{3}$.\\\\\n$\\therefore$ The area of the shaded part is $S_{\\text{shaded}}=\\boxed{6\\pi -9\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53009657_142.png"
  },
  {
    "solution": "\\boxed{70^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the ratio of the measures of $\\angle AOC$ to $\\angle BOC$ is $5\\colon 2$. $OD$ bisects $\\angle AOB$. If $\\angle COD=15^\\circ$, what is the measure of $\\angle AOB$?\n",
    "completion": "\\textbf{Solution:} Given that the ratio of the measures of $\\angle AOC$ to $\\angle BOC$ is $5\\colon 2$,\\\\\nlet $\\angle AOC=5x$ and $\\angle BOC=2x$, then $\\angle AOB=7x$,\\\\\nsince OD bisects $\\angle AOB$,\\\\\nit follows that $\\angle BOD= \\frac{1}{2} \\angle AOB= \\frac{7}{2} x$,\\\\\nsince $\\angle COD=\\angle BOD-\\angle BOC$ and $\\angle COD=15^\\circ$,\\\\\nthen $\\frac{7}{2} x-2x=15^\\circ$,\\\\\nsolving for $x$ gives $x=10^\\circ$,\\\\\ntherefore, $\\angle AOB=7\\times 10^\\circ=\\boxed{70^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51495497.png"
  },
  {
    "solution": "\\boxed{\\frac{9}{7}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the figure, in the isosceles $\\triangle ABC$, where $AB=AC$, the circle $\\odot O$ with diameter $AB$ intersects $BC$ at point $D$, $DE\\perp AC$ with the foot of the perpendicular at $E$, and the extension of $ED$ intersects the extension of $AB$ at point $F$. $EF$ is a tangent to $\\odot O$; If the radius of $\\odot O$ is $\\frac{7}{2}$ and $BD=3$, find the length of $CE$.\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, connect $AD$,\\\\\nsince $AB$ is the diameter of $\\odot O$,\\\\\nthus $AD\\perp BC$.\\\\\nSince $DE\\perp AC$,\\\\\nthus $\\angle ADC=\\angle DEC=90^\\circ$.\\\\\nSince $\\angle C=\\angle C$,\\\\\nthus $\\triangle CDE\\sim \\triangle CAD$,\\\\\nthus $\\frac{CD}{CA}=\\frac{CE}{CD}$.\\\\\nSince $AB=AC=7$,\\\\\nthus $DC=DB=3$.\\\\\nTherefore, $\\frac{3}{7}=\\frac{CE}{3}$\\\\\nTherefore, $CE=\\boxed{\\frac{9}{7}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52797177_145.png"
  },
  {
    "solution": "\\boxed{\\frac{5}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the tangent to the circle $\\odot O$, point $D$ is on $\\odot O$, and line $AD$ intersects $\\odot O$ at $C$. $CE$ is the diameter of $\\odot O$, and line $BC$ is drawn. Given $\\angle A=90^\\circ$ and $CB$ bisects $\\angle ACE$; when $AB=2$ and $AC=1$, what is the radius of $\\odot O$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, connect $BE$,\\\\\nin the right triangle $ABC$, $AB=2$, $AC=1$,\\\\\nby the Pythagorean theorem, we have: $BC=\\sqrt{AB^2+AC^2}=\\sqrt{5}$,\\\\\nsince $CE$ is the diameter of $\\odot O$,\\\\\nthus $\\angle CBE=90^\\circ$,\\\\\ntherefore $\\angle BAC=\\angle EBC$,\\\\\nsince $\\angle ACB=\\angle OCB$,\\\\\nthus $\\triangle BAC\\sim \\triangle EBC$,\\\\\ntherefore $\\frac{AC}{BC}=\\frac{BC}{CE}$, that is, $\\frac{1}{\\sqrt{5}}=\\frac{\\sqrt{5}}{CE}$,\\\\\nsolving this gives: $CE=5$,\\\\\nthus, the radius of $\\odot O$ is $\\boxed{\\frac{5}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51402171_143.png"
  },
  {
    "solution": "\\boxed{18}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in \\( \\triangle ABC\\), \\(DE\\) is the perpendicular bisector of side \\(AC\\). If \\(BC=8\\, \\text{cm}\\) and \\(AB=10\\, \\text{cm}\\), what is the perimeter of \\( \\triangle EBC\\)?\n",
    "completion": "\\textbf{Solution:} Given that $DE$ is the perpendicular bisector of side $AC$ in $\\triangle ABC$,\\\\\nthus $EA=EC$,\\\\\ntherefore $BE+EC=BE+EA=AB$,\\\\\ngiven $BC=8$ cm and $AB=10$ cm,\\\\\nthus the perimeter of $\\triangle EBC$ is $BC+BE+EC=8+10=\\boxed{18}$ cm.",
    "image_path": "dataset/mm_math/images/MM_Math/52874602.png"
  },
  {
    "solution": "\\boxed{\\frac{45}{37}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, $AC=3$, $BC=4$. On segment $AB$, a moving point $M$ starts from point $A$ moving towards point $B$ at a uniform speed, while another moving point $N$ starts from point $B$ moving towards point $A$ at a uniform speed. When one of the points $M$ or $N$ stops moving, the other point stops as well. Perpendiculars are drawn from points $M$ and $N$ to $AB$, intersecting the lines containing the right angle sides $AC$ and $BC$ at points $D$ and $E$ respectively. $DE$ is then connected. If the movement time is $t$ seconds, during the movement, the quadrilateral $DENM$ is always a rectangle (except when points $M$ and $N$ coincide). During the movement, if the speed of point $M$ is 1 unit length per second, find the speed of point $N$. For what value of $t$ seconds does the rectangle $DENM$ become a square?\n",
    "completion": "\\textbf{Solution:} Given that before points M and N meet, since the quadrilateral DENM is a rectangle, it follows that NE = MD. Since $\\triangle AMD \\sim \\triangle ABC$, it follows that $\\frac{AM}{AC} = \\frac{MD}{BC}$. According to the problem, $AM = t$, thus $\\frac{t}{3} = \\frac{MD}{4}$, which implies $MD = \\frac{4}{3}t$; Since $\\triangle BEN \\sim \\triangle BAC$, it follows that $\\frac{BN}{BC} = \\frac{EN}{AC}$, which implies $\\frac{BN}{4} = \\frac{\\frac{4}{3}t}{3}$, so $BN = \\frac{16}{9}t$. Thus, the speed of point N is $\\frac{16}{9}$ units per second. Since at the point where N and M meet, $AM + BM = AB$, it follows that $t + \\frac{16}{9}t = 5$. Solving this equation yields $t = \\frac{9}{5}$, meaning the meeting time for M and N is $\\frac{9}{5}$. Continuing movement after N and M meet, when point N reaches point A, solving $\\frac{16}{9}t = 5$ gives $t = \\frac{45}{16}$, which means the time for point N to reach the bottom point A is $\\frac{45}{16}$. Since the rectangle DENM is a square, therefore DM = MN = EN. Before N and M meet, i.e., when $0 < t < \\frac{9}{5}$, we have $MD = \\frac{4}{3}t$, $BN = \\frac{16}{9}t$, $AM = t$. Thus, $MN = AB - AM - BN = 5 - t - \\frac{16}{9}t = 5 - \\frac{25}{9}t$, resulting in $5 - \\frac{25}{9}t = \\frac{4}{3}t$. Solving this equation yields $t = \\boxed{\\frac{45}{37}}$. After N and M meet, i.e., when $\\frac{9}{5} < t < \\frac{45}{16}$, with $BN = \\frac{16}{9}t$, $AM = t$, $MD = \\frac{4}{3}t$, we get $AN = AB - BN = 5 - \\frac{16}{9}t$, $BM = AB - AM = 5 - t$. Thus, $MN = AB - AN - BM = 5 - \\left( 5 - \\frac{16}{9}t \\right) - \\left( 5 - t \\right) = \\frac{25}{9}t - 5$, leading to $\\frac{25}{9}t - 5 = \\frac{4}{3}t$. Solving this equation yields $t = \\frac{45}{13} > \\frac{45}{16}$, which is not reasonable. Therefore, summarizing the above discussion, the speed of point N is $\\frac{16}{9}$ units per second and when $t = \\boxed{\\frac{45}{37}}$, the rectangle DENM forms a square.",
    "image_path": "dataset/mm_math/images/MM_Math/51433258_213.png"
  },
  {
    "solution": "\\boxed{20}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the parallelogram $ABCD$, $BD$ is the diagonal, and point $O$ is the midpoint of $BD$. It is also given that $AD\\parallel EO$ and $OF\\parallel AB$. The perimeter of quadrilateral $BEOF$ is 10. What is the perimeter of the parallelogram $ABCD$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral $ABCD$ is a parallelogram,\\\\\nit follows that $AD \\parallel BC$,\\\\\nand since point $O$ is the midpoint of $BD$, and $AD \\parallel EO$, $OF \\parallel AB$,\\\\\nit follows that $OE$, $OF$ are respectively the midsegments of triangle $ABD$, triangle $BCD$, with $BC \\parallel EO$,\\\\\nthus, quadrilateral $OEBF$ is a parallelogram, with $AD=2OE$, $CD=2OF$, $OE=BF$, $OF=BE$,\\\\\nand since the perimeter of the quadrilateral $OEBF$ is $10$,\\\\\nwe have $OE+BE+BF+OF=10$,\\\\\ntherefore, $OE+OF=5$,\\\\\nand since quadrilateral $ABCD$ is a parallelogram,\\\\\nit follows that $AB=CD$, $AD=BC$,\\\\\ntherefore, the perimeter of the quadrilateral $ABCD$ is $AB+BC+AD+CD=2(AD+CD)=4(OE+OF)=\\boxed{20}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52019328.png"
  },
  {
    "solution": "\\boxed{8\\text{cm}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in $ \\triangle ABC$, $ \\angle A=40^\\circ, AB=AC$, and the perpendicular bisector of $AB$, $MN$, intersects $AC$ at point D and $AB$ at point E. The perimeter of $ \\triangle ABC$ is $13cm$, and $AB=5cm$. What is the perimeter of $ \\triangle DBC$?\n",
    "completion": "\\textbf{Solution:} Since the perimeter of $\\triangle ABC$ is $13\\text{cm}, AB=5\\text{cm}, AB=AC$,\\\\\nit follows that $BC=13-2\\times 5=3\\text{cm}$,\\\\\nSince $MN$ bisects $AB$ perpendicularly,\\\\\nit implies $AD=BD$,\\\\\nTherefore, the perimeter of $\\triangle DBC=BC+CD+BD=BC+CD+AD=BC+AC=5+3=\\boxed{8\\text{cm}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52793526_109.png"
  },
  {
    "solution": "\\boxed{135^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given: $\\angle AOB=90^\\circ$, construct rays $OC$, $OD$ is the angle bisector of $\\angle AOC$, $OE$ is the angle bisector of $\\angle BOC$. If ray $OC$ rotates around point $O$ outside $\\angle AOB$ and $\\angle AOC$ is an obtuse angle, find the degree measure of $\\angle DOE$.\n",
    "completion": "\\textbf{Solution:} Consider two cases:\n\nAs shown in Figure 3, since $OD$ and $OE$ bisect $\\angle AOC$ and $\\angle BOC$ respectively,\\\\\nit follows that $\\angle COD=\\frac{1}{2}\\angle AOC$, $\\angle COE=\\frac{1}{2}\\angle BOC$,\\\\\nthus $\\angle DOE=\\angle COD-\\angle COE=\\frac{1}{2}\\left(\\angle AOC-\\angle BOC\\right)=45^\\circ$;\n\nAs shown in Figure 4, since $OE$ and $OD$ are the bisectors of $\\angle BOC$ and $\\angle AOC$ respectively,\\\\\nit follows that $\\angle EOC=\\angle BOE$, $\\angle COD=\\angle AOD$,\\\\\nand since $\\angle AOB=90^\\circ$,\\\\\nit follows that $\\angle AOD+\\angle DOC+\\angle COE+\\angle EOB=270^\\circ$,\\\\\nthus $2\\angle DOC+2\\angle COE=270^\\circ$,\\\\\ntherefore $\\angle DOC+\\angle COE=135^\\circ$,\\\\\nleading to $\\angle DOE=\\boxed{135^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/47364521_204.png"
  },
  {
    "solution": "\\boxed{\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in $\\triangle ABC$, it is known that $\\angle C=90^\\circ$, $AC=BC=1$, $AB=\\sqrt{2}$, the bisector of $\\angle BAC$ intersects side $BC$ at point $D$, and $DE\\perp AB$ at point $E$. What is the perimeter of $\\triangle DBE$?\n",
    "completion": "\\textbf{Solution:} Because $DE\\perp AB$ and $\\angle C=90^\\circ$, and $AD$ bisects $\\angle BAC$,\\\\\nhence $DC=DE$,\\\\\nthus $DE+BD=DC+BD=BC=1$,\\\\\nin right triangles $\\triangle ADC$ and $\\triangle ADE$,\\\\\n$\\left\\{\\begin{array}{l}\nAD=AD\\\\\nDC=DE\n\\end{array}\\right.$,\\\\\ntherefore, by HL (Hypotenuse-Leg), $\\triangle ADC \\cong \\triangle ADE$,\\\\\nso, $AE=AC=1$,\\\\\ntherefore $BE=AB-AE=\\sqrt{2}-1$,\\\\\nhence the perimeter of $\\triangle DBE$ is $BE+DE+BD=\\sqrt{2}-1+1=\\boxed{\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52978975.png"
  },
  {
    "solution": "\\boxed{\\frac{14\\sqrt{3}}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in square $ABCD$, $\\angle B=60^\\circ$, $AB=6$, $BC=4$, what is the area of square $ABCD$? Point $E$ is a moving point on $AB$, extend line $ED$ to point $F$ such that $DE=3DF$, construct square $EFGC$ with $EC$ and $EF$ as adjacent sides, connect $EG$, what is the minimum value of $EG$?\n",
    "completion": "\\textbf{Solution:} Construct $CH\\perp AB$ at point H, and let EG intersect CD at point O,\\\\\nsince in $\\square ABCD$, $\\angle B=60^\\circ$ and BC=4,\\\\\ntherefore, CH=$2\\sqrt{3}$,\\\\\nthus, the area of $\\square ABCD$: AB×CH=$12\\sqrt{3}$;\\\\\nsince quadrilateral ECGF is a parallelogram,\\\\\nthus, $EF\\parallel CG$,\\\\\ntherefore, $\\triangle EOD\\sim \\triangle GOC$,\\\\\nthus, $\\frac{EO}{GO}=\\frac{DO}{CO}=\\frac{DE}{CG}$,\\\\\ntherefore, DE＝3DF,\\\\\ntherefore, $\\frac{DE}{EF}=\\frac{3}{4}$,\\\\\nthus, $\\frac{ED}{CG}=\\frac{3}{4}$,\\\\\ntherefore, $\\frac{EO}{GO}=\\frac{3}{4}$,\\\\\nthus, when EO takes its minimum value, EG can also take its minimum value, when EO$\\perp$CD, EO takes the minimum value,\\\\\ntherefore,CH=EO,\\\\\nthus, EO=$2\\sqrt{3}$,\\\\\ntherefore, GO=$\\frac{8\\sqrt{3}}{3}$,\\\\\nthus, the minimum value of EG is $2\\sqrt{3}+\\frac{8\\sqrt{3}}{3}=\\boxed{\\frac{14\\sqrt{3}}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51029309.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, within rectangle $ABCD$, where $AB=8$ and $BC=12$, point E is on $AB$ with $AE=5$. Point P is located on $AD$. The rectangle is folded along $PE$ so that point A lands on point ${A}^{\\prime}$. Line $AC$ is drawn and intersects $PE$ at point F, with $AP=x$. Find the minimum distance between points ${A}^{\\prime}$ and D, and calculate the value of $\\frac{AE}{AP}$ under these conditions;\n",
    "completion": "\\textbf{Solution:} Connect $DE$ and $D{A}^{\\prime}$.\n\nIn $\\triangle ADE$, $\\angle EAD=90^\\circ$, $AE=5$, $AD=BC=12$,\n\n$\\therefore DE=13$,\n\n$\\because$ According to the property of folding: $EA=E{A}^{\\prime}=5$,\n\n$\\therefore$ In $\\triangle D{A}^{\\prime}E$, $D{A}^{\\prime}>DE-E{A}^{\\prime}=8$,\n\nWhen ${A}^{\\prime}$ is on $DE$, $D{A}^{\\prime}=DE-E{A}^{\\prime}=8$,\n\nThat is: $D{A}^{\\prime}\\ge 8$,\n\n$\\therefore$ The minimum value of $D{A}^{\\prime}$ is $\\boxed{8}$,\n\nAt this time, ${A}^{\\prime}$ is on $DE$, that is, E, ${A}^{\\prime}$, D are collinear,\n\nAccording to the property of folding: $\\angle P{A}^{\\prime}E=\\angle PAE=90^\\circ$,\n\nThat is: $\\angle P{A}^{\\prime}E=\\angle P{A}^{\\prime}D=90^\\circ$,\n\nTherefore, $\\triangle P{A}^{\\prime}D$ is a right triangle, thus $PD^2={A}^{\\prime}P^2+{A}^{\\prime}D^2$,\n\nAccording to $PA=P{A}^{\\prime}=x$, we have $PD=AD-PA=12-x$,\n\nThus, $(12-x)^2=x^2+8^2$,\n\nSolving gives $x=\\frac{10}{3}$,\n\n$\\therefore \\frac{AE}{AP}=\\frac{5}{\\frac{10}{3}}=\\boxed{\\frac{3}{2}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52793389_154.png"
  },
  {
    "solution": "\\boxed{9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, $AB \\parallel CD$, $AB = 5$, $DC = 11$, and the sum of $AD$ and $BC$ is $12$. Points $E$, $F$, and $G$ are the midpoints of $BD$, $AC$, and $DC$ respectively. What is the perimeter of $\\triangle EFG$?\n",
    "completion": "\\textbf{Solution:}\n\nAs shown in the figure, extend AE and continue it to intersect CD at point K,\\\\\nsince AB$\\parallel$CD,\\\\\nit follows that $\\angle$BAE$=\\angle$DKE and $\\angle$ABD$=\\angle$EDK,\\\\\nsince points E, F, G are the midpoints of BD, AC, DC respectively.\\\\\nIt follows that BE$=$DE,\\\\\nso, triangle AEB is congruent to triangle KED (AAS),\\\\\nthus, DK$=$AB, AE$=$EK,\\\\\ntherefore, EF is the midline of triangle ACK,\\\\\ntherefore, EF$=\\frac{1}{2}$CK$=\\frac{1}{2}$(DC$-$AB),\\\\\nsince EG is the midline of triangle BCD,\\\\\nit follows that EG$=$BC,\\\\\nalso, since FG is the midline of triangle ACD,\\\\\nit follows that FG$=$AD,\\\\\nhence, EG+GF$=\\frac{1}{2}$(AD+BC),\\\\\nsince AD+BC$=12$, AB$=5$, DC$=11$,\\\\\nthus, EG+GF$=6$, FE$=3$,\\\\\ntherefore, the perimeter of triangle EFG is $6+3=\\boxed{9}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51770639.png"
  },
  {
    "solution": "\\boxed{23^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $ \\triangle ABC$ is inscribed in $ \\odot O$, $ \\angle BAC=60^\\circ$, D is the midpoint of $ BC$, and $ \\angle AOD=166^\\circ$, $ AE$ and $ CF$ are the altitudes on sides $ BC$ and $ AB$, respectively. What is the measure of $ \\angle BCF$ in degrees?\n",
    "completion": "\\textbf{Solution:} Connect $OB, OC$,\\\\\nsince $OB=OC$ and $D$ is the midpoint of $BC$,\\\\\nthus $\\angle BOD=\\frac{1}{2}\\angle BOC=\\angle BAC=60^\\circ$\\\\\n$\\angle AOB=\\angle AOD-\\angle BOD=166^\\circ-60^\\circ=106^\\circ$\\\\\nsince $OA=OB$\\\\\nthus $\\angle ABO=\\frac{180^\\circ-106^\\circ}{2}=37^\\circ$\\\\\n$\\angle OBC=90^\\circ-60^\\circ=30^\\circ$\\\\\nthus $\\angle ABC=\\angle ABO+\\angle OBC=37^\\circ+30^\\circ=67^\\circ$\\\\\nsince $CF \\perp AB$\\\\\nthus $\\angle BCF=90^\\circ-\\angle ABC=90^\\circ-67^\\circ=\\boxed{23^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/52991560.png"
  },
  {
    "solution": "\\boxed{\\frac{32}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, $AB$ is the diameter of $\\odot O$, with points $E$ and $F$ on $\\odot O$, and $\\overset{\\frown}{BF} = 2\\overset{\\frown}{BE}$. Connect $OE$ and $AF$, and draw the tangent to $\\odot O$ at point $B$, which intersects the extensions of $OE$ and $AF$ at points $C$ and $D$, respectively. $\\angle COB = \\angle A$; if $AB = 6$ and $CB = 4$, calculate the length of the line segment $FD$.\n",
    "completion": "\\textbf{Solution:} Connect $BF$, as shown in the figure, \\\\\nsince $CD$ is the tangent to $\\odot O$, \\\\\ntherefore $AB\\perp CD$, \\\\\nhence $\\angle OBC=\\angle ABD=90^\\circ$, \\\\\nsince $\\angle COB=\\angle A$, \\\\\nthus $\\triangle OBC\\sim \\triangle ABD$, \\\\\nthus $\\frac{OB}{AB}=\\frac{BC}{BD}$, that is $\\frac{3}{6}=\\frac{4}{BD}$, solving for $BD=8$, \\\\\nin $\\triangle ABD$ right-angled at $B$, $AD=\\sqrt{AB^{2}+BD^{2}}=\\sqrt{6^{2}+8^{2}}=10$, \\\\\nsince $AB$ is the diameter of $\\odot O$, \\\\\nthus $\\angle AFB=90^\\circ$, \\\\\nsince $\\angle BDF=\\angle ADB$, \\\\\nthus $\\triangle BDF$ right-angled at $B \\sim \\triangle ADB$ right-angled at $B$, \\\\\nthus $\\frac{DF}{DB}=\\frac{DB}{DA}$, that is $\\frac{DF}{8}=\\frac{8}{10}$, \\\\\nsolving for: $DF=\\boxed{\\frac{32}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51403109_123.png"
  },
  {
    "solution": "\\boxed{80^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of $\\odot O$, $AC$ is the tangent of $\\odot O$ with $A$ being the point of tangency, and $BC$ intersects $\\odot O$ at point $D$. Connect $OD$. If $\\angle C=50^\\circ$, what is the degree measure of $\\angle AOD$?\n",
    "completion": "\\textbf{Solution:} Since $AC$ is the tangent to the circle $\\odot O$,\\\\\n$\\therefore \\angle CAB=90^\\circ$,\\\\\nAnd since $\\angle C=50^\\circ$,\\\\\n$\\therefore \\angle ABC=90^\\circ-50^\\circ=40^\\circ$,\\\\\nAnd since $OD=OB$,\\\\\n$\\therefore \\angle BDO=\\angle ABC=40^\\circ$,\\\\\nAnd since $\\angle AOD=\\angle OBD+\\angle OBD$,\\\\\n$\\therefore \\angle AOD=40^\\circ+40^\\circ=\\boxed{80^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51445849.png"
  },
  {
    "solution": "\\boxed{y=\\frac{32}{x}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, point O is the origin, and the coordinates of vertex A of the rhombus OABC are (3, 4). Find the equation of the inverse proportion function $y = \\frac{k}{x}$ that passes through point B.\n",
    "completion": "\\textbf{Solution:} Construct AE$\\perp$x-axis through point A, and BF$\\perp$x-axis through point B, where E and F are the feet of the perpendiculars, as shown in the diagram,\\\\\nsince A(3,4),\\\\\nthus OE=3, AE=4,\\\\\ntherefore AO$= \\sqrt{OE^{2}+AE^{2}}$=5,\\\\\nsince quadrilateral OABC is a rhombus,\\\\\ntherefore AO=AB=OC=5, AB$\\parallel$x-axis,\\\\\ntherefore EF=AB=5,\\\\\ntherefore OF=OE+EF=3+5=8,\\\\\ntherefore B(8,4),\\\\\nsince the inverse proportion function passing through point B is $y=\\frac{k}{x}$,\\\\\nsubstituting the coordinates of point B gives $k=32$,\\\\\ntherefore, the expression for the inverse proportion function is $\\boxed{y=\\frac{32}{x}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51402702_64.png"
  },
  {
    "solution": "\\boxed{16}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, quadrilateral $ABCD$ is a rectangle, and diagonals $AC$ and $BD$ intersect at point $O$. Extend $BC$ to point $E$ so that $CE = BC$, and connect $DE$. It is given that $AC = DE$; If $F$ is the midpoint of $BC$, and connecting $OF$, with $AC = 5$ and $OF = 2$, what is the perimeter of $\\triangle BDE$?\n",
    "completion": "\\textbf{Solution:} \\\\\nSince quadrilateral ABCD is a rectangle,\\\\\n$AC=BD=5$,\\\\\n$DE=5$,\\\\\n$OB=OD=OC=OA=\\frac{5}{2}$,\\\\\nSince F is the midpoint of BC,\\\\\n$\\angle OFB=90^\\circ$,\\\\\nBy the Pythagorean theorem: $BF=\\sqrt{OB^{2}-OF^{2}}=\\sqrt{\\left(\\frac{5}{2}\\right)^{2}-2^{2}}=\\frac{3}{2}$,\\\\\n$BC=CE=2BF=3$,\\\\\n$BE=6$,\\\\\nThe perimeter of $\\triangle BDE$ = BE+BD+DE=6+5+5=\\boxed{16}.",
    "image_path": "dataset/mm_math/images/MM_Math/53359711_142.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $OP$ bisects $\\angle AOB$, $E$ is a point on $OA$, $OE=4$, and the distance from $P$ to $OB$ is 2. What is the area of $\\triangle OPE$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw PD$\\perp$OB at point D and PC$\\perp$OA at C, \\\\\nsince $OP$ is the bisector of $\\angle AOB$, and the distance from P to OB is 2, \\\\\ntherefore $PC=PD=2$, \\\\\nsince $OE=4$, \\\\\ntherefore the area of $\\triangle OPE$ is $\\frac{1}{2}OE\\cdot PC=\\frac{1}{2}\\times 4\\times 2=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52551286.png"
  },
  {
    "solution": "\\boxed{41^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle B=46^\\circ$, $\\angle C=52^\\circ$, $AD$ bisects $\\angle BAC$, intersects $BC$ at point $D$, $DE \\parallel AB$, and intersects $AC$ at point $E$, then what is the measure of $\\angle ADE$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle B=46^\\circ$ and $\\angle C=52^\\circ$,\\\\\nthus $\\angle BAC=180^\\circ-\\angle B-\\angle C=180^\\circ-46^\\circ-52^\\circ=82^\\circ$,\\\\\nand since AD bisects $\\angle BAC$,\\\\\nthus $\\angle BAD=\\frac{1}{2}\\times82^\\circ=41^\\circ$,\\\\\nsince DE$\\parallel$AB,\\\\\nthus $\\angle ADE=\\boxed{41^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52551715.png"
  },
  {
    "solution": "\\boxed{\\frac{4}{\\sqrt{5}}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, point $E$ is the midpoint of side $BC$ of a square $ABCD$ with side length 2, and $DF \\perp AE$ at $F$. What is the length of $DF$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect $DE$,\n\nsince point $E$ is the midpoint of side $BC$ of square $ABCD$ with a side length of 2,\n\ntherefore $BE=1$,\n\nthus, in $\\triangle ABE$ right-angled at $B$, $AE=\\sqrt{1^2+2^2}=\\sqrt{5}$,\n\nsince $DF\\perp AE$,\n\ntherefore $\\frac{1}{2}\\times AE\\times DF=\\frac{1}{2}\\times AD\\times AB$,\n\nthus, $DF=\\frac{AD\\times AB}{AE}=\\frac{2\\times 2}{\\sqrt{5}}=\\boxed{\\frac{4}{\\sqrt{5}}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52354693_83.png"
  },
  {
    "solution": "\\boxed{90^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB \\perp BC$, $EC \\perp BC$, point $D$ is on $BC$, $AB=1$, $BD=2$, $CD=3$, $CE=6$. $\\triangle ABD \\sim \\triangle DCE$; what is the measure of $\\angle ADE$ in degrees?\n",
    "completion": "Solution: From (1), we know that $\\triangle ABD\\sim \\triangle DCE$, thus $\\angle BAD=\\angle EDC$.\\\\\nSince $\\angle BAD+\\angle ADB=90^\\circ$,\\\\\nit follows that $\\angle ADB+\\angle EDC=90^\\circ$.\\\\\nHence, $\\angle ADE=180^\\circ-\\angle ADB-\\angle EDC=90^\\circ$.\\\\\nTherefore, the measure of $\\angle ADE$ is $\\boxed{90^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52916848_73.png"
  },
  {
    "solution": "\\boxed{-\\frac{41}{16}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure (1), it is known that the line $y=\\frac{4}{3}x+4$ intersects the x-axis at point A and the y-axis at point C. The parabola $y=ax^2+bx+c$ passes through points A and C, and has another intersection with the x-axis at point B. The axis of symmetry is the line $x=-1$. D is a moving point on the parabola in the second quadrant, and the x-coordinate of point D is denoted by m. A perpendicular line is drawn from point D to the y-axis, creating a perpendicular foot at point E, as shown in Figure (2). Is there a point D that makes $\\triangle CDE$ similar to $\\triangle AOC$? If so, find the value of the x-coordinate m of point D; if not, please explain why.\n",
    "completion": "\\textbf{Solution:} Existence proven. From the given information, we have: $OA=3$, $OC=4$, $DE=-m$,\\\\\nsince $\\triangle CDE$ is similar to $\\triangle AOC$, $\\angle AOC=\\angle CED=90^\\circ$\\\\\ntherefore $\\triangle CDE\\sim \\triangle ACO$ or $\\triangle CDE\\sim \\triangle CAO$\\\\\nthus $\\frac{CE}{DE}=\\frac{OA}{OC}=\\frac{3}{4}$ or $\\frac{CE}{DE}=\\frac{OC}{OA}=\\frac{4}{3}$\\\\\ntherefore $CE=-\\frac{3}{4}m$ or $-\\frac{4}{3}m$\\\\\nhence, point $D$'s coordinates are $(m, -\\frac{3}{4}m+4)$ or $(m, \\frac{3}{4}m+4)$ or $(m, -\\frac{4}{3}m+4)$ or $(m, \\frac{4}{3}m+4)$\\\\\nSubstituting $(m, -\\frac{3}{4}m+4)$ into $y=-\\frac{4}{3}x^2-\\frac{8}{3}x+4$, yields: $-\\frac{3}{4}m+4=-\\frac{4}{3}m^2-\\frac{8}{3}m+4$\\\\\nSolving gives $m=\\boxed{-\\frac{23}{16}}$;\\\\\nSubstituting $(m, \\frac{3}{4}m+4)$ into $y=-\\frac{4}{3}x^2-\\frac{8}{3}x+4$, yields: $\\frac{3}{4}m+4=-\\frac{4}{3}m^2-\\frac{8}{3}m+4$\\\\\nSolving gives $m=\\boxed{-\\frac{41}{16}}$;\\\\\nSubstituting $(m, -\\frac{4}{3}m+4)$ into $y=-\\frac{4}{3}x^2-\\frac{8}{3}x+4$, yields: $-\\frac{4}{3}m+4=-\\frac{4}{3}m^2-\\frac{8}{3}m+4$\\\\\nSolving gives $m=\\boxed{-1}$;\\\\\nSubstituting $(m, \\frac{4}{3}m+4)$ into $y=-\\frac{4}{3}x^2-\\frac{8}{3}x+4$, yields: $\\frac{4}{3}m+4=-\\frac{4}{3}m^2-\\frac{8}{3}m+4$\\\\\nSolving gives $m=-3$, which is discarded;\\\\\ntherefore, the possible values for $m$ are $\\boxed{-\\frac{23}{16}}$, $\\boxed{-1}$, or $\\boxed{-\\frac{41}{16}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53043766_127.png"
  },
  {
    "solution": "\\boxed{4\\pi}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, AB is the diameter of $\\odot O$, and the chord CD $\\perp$ AB at point E. Connect AC and BC. $\\angle ACO = \\angle BCD$; If CD $= 6$, $\\angle A = 30^\\circ$, find the area of the shaded region.\n",
    "completion": "\\textbf{Solution:} From the given problem, \\\\\n$\\because \\angle A=30^\\circ$, \\\\\n$\\therefore \\angle BOC=60^\\circ$, \\\\\n$\\therefore \\angle AOC=120^\\circ$, \\\\\n$\\because$ AB is the diameter of circle $O$, and chord CD $\\perp$ AB, \\\\\n$\\therefore$ CE= $\\frac{1}{2}$ CD=3, \\\\\nIn $\\triangle COE$, OC= $\\frac{CE}{\\sin 60^\\circ}$ =2$\\sqrt{3}$, \\\\\n$\\therefore$ the area of sector OAC (shaded region) = $\\frac{120^\\circ \\times \\pi \\times (2\\sqrt{3})^2}{360^\\circ}$ =4$\\pi$, \\\\\nAnswer: The area of the shaded region is $\\boxed{4\\pi}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53124729_103.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point C is on $BE$, $\\angle B = \\angle E = \\angle ACF$, $AC = CF$, $AB = 4$, and $EF = 6$. What is the length of $BE$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle B+\\angle BAC=\\angle ACE=\\angle FCE+\\angle ACF$,\\\\\nand since $\\angle B=\\angle ACF$,\\\\\nit follows that $\\angle BAC=\\angle FCE$,\\\\\nSince $\\angle B=\\angle E$, $AC=CF$,\\\\\nit follows that $\\triangle ABC\\cong \\triangle CEF$,\\\\\ntherefore $AB=CE$, $BC=EF$,\\\\\nSince $AB=4$, $EF=6$,\\\\\nit follows that $BE=BC+CE=6+4=\\boxed{10}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52978720.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the diagonals $AC$ and $BD$ of the rhombus $ABCD$ intersect at point $O$. $P$ is a moving point on side $AB$ (not coinciding with points $A$ or $B$). $PE \\perp OA$ at point $E$, and $PF \\perp OB$ at point $F$. If $AB=8$ and $\\angle BAD=60^\\circ$, what is the minimum length of segment $EF$?\n",
    "completion": "\\textbf{Solution}: Connect OP, \\\\\n$\\because$ Quadrilateral ABCD is a rhombus, \\\\\n$\\therefore$ AC$\\perp$BD, $\\angle$CAB=$\\frac{1}{2}\\angle$DAB=$30^\\circ$, \\\\\n$\\because$ PE$\\perp$OA at point E, PF$\\perp$OB at point F, \\\\\n$\\therefore$ $\\angle$EOF=$\\angle$OEP=$\\angle$OFP=$90^\\circ$, \\\\\n$\\therefore$ Quadrilateral OEPF is a rectangle, \\\\\n$\\therefore$ EF=OP, \\\\\n$\\because$ When OP takes the minimum value, the value of EF is also minimum, \\\\\n$\\therefore$ When OP$\\perp$AB, OP is minimized, \\\\\n$\\because$ AB=8, \\\\\n$\\therefore$ OB=$\\frac{1}{2}$AB=4, OA=$\\sqrt{AB^2−OB^2}$=$4\\sqrt{3}$, \\\\\n$\\therefore$ Area of $\\triangle ABO$=$\\frac{1}{2}$OA$\\cdot$OB=$\\frac{1}{2}$AB$\\cdot$OP, \\\\\n$\\therefore$ OP=$\\frac{4\\times 4\\sqrt{3}}{8}=\\boxed{2\\sqrt{3}}$, \\\\\n$\\therefore$ The minimum value of EF is $2\\sqrt{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52188837.png"
  },
  {
    "solution": "\\boxed{58^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in square $ABCD$, $E$ lies on the extension of $BC$, $AE$ and $BD$ intersect at point $F$. Connect $FC$. Then, what is the degree measure of $\\angle BCF$?\n",
    "completion": "\\textbf{Solution:} Given square ABCD,\\\\\nit follows that AD$\\parallel$BC, $\\angle$DCB=$90^\\circ$, AD=DC, $\\angle$ADF=$\\angle$CDF=$45^\\circ$,\\\\\nthus $\\triangle$ADF$\\cong$ $\\triangle$CDF (by SAS),\\\\\nwhich means $\\angle$DAF=$\\angle$DCF,\\\\\nalso, since AD$\\parallel$BC, $\\angle$E=$32^\\circ$,\\\\\nit follows that $\\angle$DAF=$32^\\circ$,\\\\\ntherefore $\\angle$DCF=$32^\\circ$,\\\\\nthus $\\angle$BCF=$\\angle$DCB-$\\angle$DCF=$90^\\circ-32^\\circ=\\boxed{58^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52076992.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that points $D$ and $E$ are the midpoints of sides $BC$ and $AB$ in an equilateral triangle $ABC$, respectively, with $AD=6$. Point $F$ is a moving point on segment $AD$. What is the minimum value of $BF+EF$?\n",
    "completion": "\\textbf{Solution:} Draw line CE intersecting AD at point F, and connect BF. \\\\\n$\\because$ $\\triangle ABC$ is an equilateral triangle and D is the midpoint of BC, \\\\\n$\\therefore$ BF = CF, \\\\\n$\\therefore$ BF + EF = CF + EF = CE, \\\\\nAt this point, the value of BF + EF is minimal, and the minimum value is CE, \\\\\n$\\because$ D and E are respectively the midpoints of sides BC and AB of the equilateral $\\triangle ABC$, \\\\\n$\\therefore$ AD = CE, \\\\\n$\\because$ AD = 6, \\\\\n$\\therefore$ CE = 6, \\\\\n$\\therefore$ The minimum value of BF + EF is \\boxed{6},",
    "image_path": "dataset/mm_math/images/MM_Math/52655146.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\triangle ABC$, EF is the perpendicular bisector of AB, intersecting AB at point D. If BF = 6 and CF = 2, then what is the length of AC?\n",
    "completion": "\\textbf{Solution:} Since EF is the perpendicular bisector of AB, and BF = 6, \\\\\nwe have AF=BF=6, \\\\\nand since CF = 2, \\\\\nit follows that AC=AF+CF=\\boxed{8}.",
    "image_path": "dataset/mm_math/images/MM_Math/51434797.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the figure, within $\\triangle ABC$, $\\angle A=60^\\circ$, $BD\\perp AC$ at $D$, $CE\\perp AB$ at $E$, $BD$ and $CE$ intersect at point $H$. If $CE=4$ and $BD=5$, what is the value of $\\frac{DH}{HB}$?\n",
    "completion": "\\textbf{Solution:} Given $BD \\perp AC$, $CE \\perp AB$,\\\\\nthus $\\angle ADB = \\angle AEC = 90^\\circ$,\\\\\nIn $\\triangle BDA$ with $\\angle A = 60^\\circ$, and $BD = 5$,\\\\\nthus $\\angle ABD = 30^\\circ$,\\\\\nthus $AB = 2AD$,\\\\\nGiven $AB^{2} = AD^{2} + BD^{2}$, that is $4AD^{2} = AD^{2} + 25$,\\\\\nthus $AD = \\frac{5\\sqrt{3}}{3}$,\\\\\nIn $\\triangle ACE$ with $\\angle A = 60^\\circ$, and $CE = 4$,\\\\\nthus $\\angle ACE = 30^\\circ$,\\\\\nthus $AC = 2AE$,\\\\\nGiven $AC^{2} = AE^{2} + CE^{2}$, that is $4AE^{2} = AE^{2} + 16$,\\\\\nthus $AE = \\frac{4\\sqrt{3}}{3}$, and $AC = \\frac{8\\sqrt{3}}{3}$,\\\\\nthus $CD = AC - AD = \\frac{8\\sqrt{3}}{3} - \\frac{5\\sqrt{3}}{3} = \\sqrt{3}$,\\\\\nIn $\\triangle CDH$ with $\\angle DCH = 30^\\circ$, and $CD = \\sqrt{3}$,\\\\\nthus $CH = 2DH$,\\\\\nGiven $CH^{2} = DH^{2} + CD^{2}$, that is $4DH^{2} = DH^{2} + 3$,\\\\\nthus $DH = 1$,\\\\\nthus $BH = BD - DH = 5 - 1 = 4$,\\\\\nthus $\\frac{DH}{HB} = \\boxed{\\frac{1}{4}}$,",
    "image_path": "dataset/mm_math/images/MM_Math/51368137.png"
  },
  {
    "solution": "\\boxed{7}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point C is the midpoint of line segment AB, point D is a point on line segment BC, and point E is the midpoint of line segment CD. Given that $AB=20$ and $BD=4$, find the length of line segment BE.\n",
    "completion": "\\textbf{Solution:} Since $E$ is the midpoint of $CD$,\\\\\nthus, $ED = \\frac{1}{2}CD = 3$,\\\\\ntherefore, $BE = BD + ED = 4 + 3 = \\boxed{7}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51379914_96.png"
  },
  {
    "solution": "\\boxed{(0,1)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=2x+1$ intersects the x-axis at point $A$ and the y-axis at point $B$. What are the coordinates of points $A$ and $B$?\n",
    "completion": "\\textbf{Solution:} Substituting $x=0$ into $y=2x+1$, we get $y=1$,\\\\\nthus the coordinates of point $B$ are $(0,1)$;\\\\\nSubstituting $y=0$ into $y=2x+1$, we get $0=2x+1$,\\\\\nsolving for $x$, we find $x=-\\frac{1}{2}$,\\\\\nthus the coordinates of point $A$ are $\\left(-\\frac{1}{2},0\\right)$.\\\\\nHence, the coordinates of points $A$ and $B$ are $\\boxed{\\left(-\\frac{1}{2},0\\right)}$ and $\\boxed{(0,1)}$, respectively.",
    "image_path": "dataset/mm_math/images/MM_Math/50519365_191.png"
  },
  {
    "solution": "\\boxed{3\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the side length of rhombus ABCD is 6, with diagonals AC and BD intersecting at point O, and $ \\angle ABC = 120^\\circ$. Point P is a moving point on AC, and point E is the midpoint of AB. What is the minimum value of $PB + PE$?\n",
    "completion": "\\textbf{Solution:} Construct the line segment DE, intersecting AC at point P, and draw BP as shown in the figure: \\\\\nThen $PB+PE=DE$, at this point $PB+PE$ has its minimum value;\\\\\nSince in the rhombus ABCD, $\\angle ABC=120^\\circ$,\\\\\nHence, AD=BD, $\\angle DAB=60^\\circ$,\\\\\nThus, $\\triangle ABD$ is an equilateral triangle,\\\\\nSince point E is the midpoint of AB,\\\\\nHence, DE $\\perp$ AB,\\\\\nSince $AB=BD=6$, $BE=\\frac{1}{2}AB=3$, \\\\\nHence, $DE=\\sqrt{6^2-3^2}=\\boxed{3\\sqrt{3}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52251984.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC \\cong \\triangle DEF$, $BE=5$, $BF=1$, what is the length of $CF$?\n",
    "completion": "\\textbf{Solution:} Since $BE=5$, and $BF=1$,\\\\\nit follows that $EF=BE-BF=4$,\\\\\nSince $\\triangle ABC \\cong \\triangle DEF$,\\\\\nit follows that $BC=EF=4$,\\\\\nTherefore, $CF=BC-BF=4-1=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51435165.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $B$ and $C$ are two points on the line segment $AD$, and $AB:BC:CD = 2:3:4$. Let $M$ be the midpoint of $AD$, with $MC = 1$. If $N$ is a point on the line segment $AD$ such that $CN = \\frac{1}{7}DB$, find the length of the line segment $DN$.\n",
    "completion": "\\textbf{Solution:} From (1), we know that $DB=18\\times \\frac{3+4}{2+3+4}=14, CN=\\frac{1}{7}DB$,\\\\\n$\\therefore CN=\\frac{1}{7}\\times 14=2$.\\\\\nConsider the following two cases:\\\\\n(1) When point $N$ is on segment $CD$, $DN=CD-CN=8-2=6$;\\\\\n(2) When point $N$ is on segment $CB$, $DN=CD+CN=8+2=10$.\\\\\nIn conclusion, the length of segment $DN$ is either $6$ or $\\boxed{10}$.",
    "image_path": "dataset/mm_math/images/MM_Math/54661145_69.png"
  },
  {
    "solution": "\\boxed{(3,\\frac{7}{2})}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given that the graph of the function $y=-\\frac{1}{2}x+b$ intersects the x-axis and the y-axis at points A and B, respectively, and intersects the graph of the function $y=2x$ at point M (2,4); there is a moving point P on the x-axis, vertical lines are drawn from point P, intersecting the graphs of $y=-\\frac{1}{2}x+b$ and $y=2x$ at points C and D, respectively. Let point P be $(a,0)$, if $CD=\\frac{1}{2}OB$, find the value of a and the coordinates of point C?\n",
    "completion": "\\textbf{Solution:} Substitute $x=0$ into $y=-\\frac{1}{2}x+5$ to get $y=5$,\\\\\n$\\therefore$ The coordinates of point B are $(0,5)$, that is: $OB=5$,\\\\\nGiven $CD=\\frac{1}{2}OB$, we have $CD=\\frac{5}{2}$,\\\\\nSince PC$\\perp x$ axis, and point P is $(a,0)$,\\\\\n$\\therefore$ The coordinates of point C are $(a,-\\frac{1}{2}a+5)$, and the coordinates of point D are $(a,2a)$,\\\\\n$\\therefore$ $|2a-(-\\frac{1}{2}a+5)|=\\frac{5}{2}$, solving this yields: $a=\\boxed{3}$ or $1$,\\\\\nWhen $a=3$, $y=-\\frac{1}{2}x+5=\\frac{7}{2}$; When $a=1$, $y=-\\frac{1}{2}x+5=\\frac{9}{2}$;\\\\\n$\\therefore$ The coordinates of point C are $(1,\\frac{9}{2})$ or \\boxed{(3,\\frac{7}{2})}.",
    "image_path": "dataset/mm_math/images/MM_Math/52251693_82.png"
  },
  {
    "solution": "\\boxed{\\left(0,\\frac{3}{2}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the Cartesian coordinate system, the coordinate of point A is $(-3, 0)$, and the coordinate of point B is $(0, 4)$. Point C is a point on OB. If $\\triangle ABC$ is folded along AC, so that point B lands exactly on the x-axis at point $B'$, what is the coordinate of point C?\n",
    "completion": "\\textbf{Solution:} By the problem statement, we have $A{B}^{\\prime }=AB$ and $C{B}^{\\prime }=CB$,\\\\\nSince the coordinates of point A are $(-3,0)$ and the coordinates of point B are $(0,4)$,\\\\\nTherefore, OA=3 and OB=4,\\\\\nTherefore, $AB=\\sqrt{OA^{2}+OB^{2}}=5$,\\\\\nTherefore, $A{B}^{\\prime }=5$,\\\\\nTherefore, $O{B}^{\\prime }=2$,\\\\\nIn $\\triangle {B}^{\\prime }OC$ right-angled at $B'$, $O{C}^{2}+O{{B}^{\\prime }}^{2}={B}^{\\prime }{C}^{2}$,\\\\\nTherefore, $O{C}^{2}+2^{2}={(4-OC)}^{2}$,\\\\\nSolving, we get: $OC=\\frac{3}{2}$,\\\\\nTherefore, the point $C\\left(0,\\frac{3}{2}\\right)$.\\\\\nHence, the answer is: The coordinates of point C are $\\boxed{\\left(0,\\frac{3}{2}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52827761.png"
  },
  {
    "solution": "\\boxed{18}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AD$ and $CE$ are medians. If the area of the quadrilateral $BDFE$ is $6$, what is the area of $\\triangle ABC$?\n",
    "completion": "\\textbf{Solution:} Connect BF,\\\\\nsince AD and CE are the medians of $\\triangle ABC$,\\\\\ntherefore AE=BE, BD=CD,\\\\\ntherefore $S_{\\triangle CBE}=\\frac{1}{2}S_{\\triangle ABC}$, $S_{\\triangle ACD}=\\frac{1}{2}S_{\\triangle ABC}$,\\\\\ntherefore $S_{\\triangle CBE}=S_{\\triangle ACD}$,\\\\\ntherefore $S_{\\triangle ACF}=S_{\\text{quadrilateral } BDFE}=6$,\\\\\nsince AE=BE, BD=CD,\\\\\ntherefore $S_{\\triangle AEF}=S_{\\triangle BEF}$, $S_{\\triangle CDF}=S_{\\triangle BDF}$,\\\\\ntherefore $S_{\\triangle AEF}+S_{\\triangle CDF}=S_{\\text{quadrilateral } BDFE}=6$,\\\\\ntherefore $S_{\\triangle ABC}=3S_{\\text{quadrilateral } BDFE}=\\boxed{18}$",
    "image_path": "dataset/mm_math/images/MM_Math/52856058.png"
  },
  {
    "solution": "\\boxed{1\\colon 15}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $DE$ is the median of $\\triangle ABC$, $M$ is the midpoint of $DE$, and the extension line of $CM$ intersects $AB$ at point $N$. What is the value of $\\frac{{S}_{\\triangle DMN}}{{S}_{\\text{quadrilateral}DBCM}}$?\n",
    "completion": "\\textbf{Solution:} Given that $DE$ is the midline of $\\triangle ABC$,\\\\\nthus $\\frac{DE}{BC}=\\frac{1}{2}$, $DE\\parallel BC$\\\\\nFurthermore, since M is the midpoint of $DE$,\\\\\nthus $\\frac{DM}{DE}=\\frac{1}{2}$,\\\\\ntherefore $\\frac{DM}{BC}=\\frac{1}{4}$,\\\\\nsince $DE\\parallel BC$,\\\\\nthus $\\triangle DMN\\sim \\triangle BCN$,\\\\\ntherefore $\\frac{{S}_{\\triangle DMN}}{{S}_{\\triangle BCN}}=\\left(\\frac{DM}{BC}\\right)^2=\\frac{1}{16}$,\\\\\ntherefore ${S}_{\\triangle DMN}\\colon {S}_{\\text{quadrilateral} DBCM}=1\\colon 15$.\\\\\nHence the answer is: $\\boxed{1\\colon 15}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52917633.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $AB \\parallel CD$, $\\angle BAE = 30^\\circ$, $\\angle DCE = 60^\\circ$, and $EF$, $EG$ trisect $\\angle AEC$ (that is, $\\angle AEF = \\angle FEG = \\angle GEC$). What is the measure of $\\angle AEF$ in degrees?\n",
    "completion": "\\textbf{Solution:} Construct $EF' \\parallel AB$ through point $E$, then $\\angle BAE=\\angle F'EA=30^\\circ$,\\\\\nbecause $AB\\parallel CD$,\\\\\ntherefore $EF' \\parallel CD$,\\\\\nthus $\\angle DCE=\\angle F'EC=60^\\circ$\\\\\nAlso, because $\\angle AEC=\\angle F'EA+\\angle F'EC$,\\\\\nthus $\\angle AEC=90^\\circ$,\\\\\nbecause $EF, EG$ trisect $\\angle AEC$ (i.e., $\\angle AEF=\\angle FEG=\\angle GEC$)\\\\\ntherefore $\\angle AEF=\\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53205833_107.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $ \\triangle ABO \\sim \\triangle CDO$, if $ BO=8$, $ DO=4$, and $ CD=3$, then what is the length of $ AB$?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABO \\sim \\triangle CDO$, $BO=8$, $DO=4$, $CD=3$,\n\\[\n\\therefore \\frac{BO}{DO}=\\frac{AB}{CD}, \\text{ which means } \\frac{8}{4}=\\frac{AB}{3},\n\\]\n\\[\n\\therefore AB=\\boxed{6}.\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/51351726.png"
  },
  {
    "solution": "\\boxed{80^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $O$ is on the line $AC$, $OD$ bisects $\\angle AOB$, $\\angle BOE = \\frac{1}{2}\\angle EOC$, and $\\angle DOE = 70^\\circ$. Find the measure of $\\angle EOC$.\n",
    "completion": "\\textbf{Solution:} Let $\\angle EOC = \\alpha$ \\\\\nSince $\\angle BOE = \\frac{1}{2} \\angle EOC$ \\\\\nThus $\\angle BOE = \\frac{1}{2} \\alpha$ \\\\\nSince $\\angle DOE = 70^\\circ$ \\\\\nThus $\\angle BOD = \\angle DOE - \\angle BOE = 70^\\circ - \\frac{1}{2} \\alpha$ \\\\\nSince $OD$ bisects $\\angle AOB$ \\\\\nThus $\\angle AOD = \\angle BOD = 70^\\circ - \\frac{1}{2} \\alpha$ \\\\\nSince point O is on line AC, \\\\\nThus $\\angle AOC = 180^\\circ$ \\\\\nTherefore $\\angle EOC + \\angle DOE + \\angle AOD = 180^\\circ$ \\\\\nTherefore $\\alpha + 70^\\circ + 70^\\circ - \\frac{1}{2}\\alpha = 180^\\circ$ \\\\\nTherefore $\\alpha = 80^\\circ$ \\\\\nTherefore $\\angle EOC = \\boxed{80^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/50057191.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, given point $E$ is on the edge $AB$ of the rectangular sheet $ABCD$. The sheet is folded along $DE$ such that the corresponding point of $A$, denoted as $A'$, falls exactly on the line segment $CE$. If $AD=3$ and $AE=1$, then what is the length of $CD$?\n",
    "completion": "\\textbf{Solution:} Given that the quadrilateral ABCD is a rectangle,\\\\\nhence, $AB\\parallel CD$ ,\\\\\ntherefore, $\\angle CDE=\\angle AED$,\\\\\nbased on the property of folding, it is known that,\\\\\n$A'D=AD=3$, $AE=A'E=1$, $\\angle AED=\\angle A'ED$\\\\\ntherefore, $\\angle CDE=\\angle CED$\\\\\nthus, $CD=CE$, $CA'=BE$,\\\\\nsince $A'$ lies on $CE$,\\\\\ntherefore, $\\angle DA'C=\\angle B=90^\\circ$\\\\\nLet $CD=x$, then $AB=x$,\\\\\nthus $BE=x-1$,\\\\\ntherefore $x^2=3^2+(x-1)^2$,\\\\\nSolving for $x$ gives: $x=5$\\\\\ni.e., $CD=\\boxed{5}$",
    "image_path": "dataset/mm_math/images/MM_Math/50482002.png"
  },
  {
    "solution": "\\boxed{\\frac{\\alpha}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $O$ is a point on the line $AB$, $\\angle COD = 90^\\circ$, and $OE$ bisects $\\angle BOC$. If $\\angle AOC = \\alpha$, what is the degree measure of $\\angle DOE$ (expressed in an algebraic expression involving $\\alpha$)?\n",
    "completion": "\\textbf{Solution:} Given the problem, \\\\\n$\\because \\angle AOC=\\alpha$, \\\\\n$\\therefore \\angle BOC=180^\\circ-\\alpha$, \\\\\n$\\because OE$ bisects $\\angle BOC$, \\\\\n$\\therefore \\angle COE=\\frac{1}{2}\\angle BOC=90^\\circ-\\frac{\\alpha}{2}$, \\\\\n$\\because \\angle COD=90^\\circ$, \\\\\n$\\therefore \\angle DOE=90^\\circ-(90^\\circ-\\frac{\\alpha}{2})=\\boxed{\\frac{\\alpha}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53119355_128.png"
  },
  {
    "solution": "\\boxed{25^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of $\\odot O$, and the lines $BD$ and $CD$ are tangents to $\\odot O$ at points $B$ and $C$, respectively. If $\\angle BDC = 130^\\circ$, what is the degree measure of $\\angle A$?\n",
    "completion": "\\textbf{Solution:} Let's connect $OC$, $BC$.\\\\\nSince $BD$, $CD$ are tangents to the circle centered at $O$ through points $B$ and $C$ respectively,\\\\\nTherefore $OC\\perp CD$, $OB\\perp BD$,\\\\\nTherefore $\\angle OCD=\\angle OBD=90^\\circ$,\\\\\nSince $\\angle BDC=130^\\circ$,\\\\\nTherefore $\\angle BOC=360^\\circ-\\angle OCD-\\angle BDC-\\angle OBD=50^\\circ$,\\\\\nTherefore $\\angle A=\\frac{1}{2}\\angle BOC=\\boxed{25^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52551457_91.png"
  },
  {
    "solution": "\\boxed{\\frac{12}{13}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC=13$ and $BC=10$, what is the value of $\\sin B$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw AD $\\perp$ BC at point D,\\\\\n$\\therefore$ $\\angle$ADB$=90^\\circ$,\\\\\n$\\because$ AB=AC=13, BC=10,\\\\\n$\\therefore$ BD=5,\\\\\n$\\therefore$ AD=$\\sqrt{13^2-5^2}=12$,\\\\\n$\\therefore$ $\\sin B=\\frac{AD}{AB}=\\boxed{\\frac{12}{13}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51034127.png"
  },
  {
    "solution": "\\boxed{1.5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $C$ is on the line segment $AB$, and point $D$ is the midpoint of the line segment $AB$. If $AC=6\\,cm$ and $BC=3\\,cm$, what is the length of the line segment $CD$ in $cm$?\n",
    "completion": "\\textbf{Solution:} Since $AC=6\\,cm$, $BC=3\\,cm$,\n\nit follows that $AB=AC+BC=6+3=9\\,cm$,\n\nSince point D is the midpoint of segment AB,\n\nit follows that $AD=\\frac{1}{2}AB=\\frac{1}{2}\\times9=4.5\\,cm$,\n\nthus $CD=AC-AD=6-4.5=\\boxed{1.5}\\,cm$",
    "image_path": "dataset/mm_math/images/MM_Math/52795546.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{x}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the line $y=-x+m+1$ and the hyperbolic function $y=\\frac{m}{x}$ ($x>0$, $m>0$) intersect at points A and B respectively, and intersect with the x-axis at point C and with the y-axis at point D. The coordinates of point A are $(1, 3)$. What is the equation of line AB and the hyperbolic function $y=\\frac{m}{x}$?\n",
    "completion": "\\textbf{Solution:} Substitute $ \\left(1, 3\\right)$ to get $ 3 = m$,\\\\\n$\\therefore$ the equation of line $AB$ is $y = -x + 4$, $y = \\boxed{\\frac{3}{x}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53080174_33.png"
  },
  {
    "solution": "\\boxed{98^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $CD$ bisects $\\angle ACB$, $\\angle A=68^\\circ$, $\\angle ACD=30^\\circ$. Find: the measure of $\\angle BDC$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle BDC=\\angle A+\\angle ACD$, and $\\angle A=68^\\circ$, $\\angle ACD=30^\\circ$,\\\\\ntherefore, $\\angle BDC=68^\\circ+30^\\circ=\\boxed{98^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53106969_147.png"
  },
  {
    "solution": "\\boxed{\\frac{8}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, there is a semicircle $O$ with the diameter $AB$ of $\\triangle ABC$, and point $C$ is exactly on the semicircle. Draw $CD\\perp AB$ at $D$. Given that $\\cos\\angle ACD=\\frac{3}{5}$ and $BC=2$, what is the length of $AC$?\n",
    "completion": "\\textbf{Solution:} Given that point C is on the semicircle, and AB is the diameter\\\\\n\\(\\therefore\\angle ACB=90^\\circ\\)\\\\\nSince \\(CD\\perp AB\\), and \\(\\cos\\angle ACD= \\frac{3}{5}\\)\\\\\n\\(\\therefore\\sin\\angle CAB=\\cos\\angle ACD= \\frac{3}{5}\\)\\\\\n\\(\\therefore\\cos\\angle CAB= \\sqrt{1-\\sin^2\\angle CAB}=\\sqrt{1-\\left(\\frac{3}{5}\\right)^2}=\\frac{4}{5}\\)\\\\\n\\(\\therefore\\tan\\angle CAB= \\frac{\\sin\\angle CAB}{\\cos\\angle CAB}=\\frac{3}{4}\\)\\\\\n\\(\\therefore \\frac{BC}{AC}=\\frac{3}{4}\\)\\\\\n\\(\\therefore AC= \\frac{4}{3}BC= \\frac{4}{3}\\times 2=\\boxed{\\frac{8}{3}}\\)",
    "image_path": "dataset/mm_math/images/MM_Math/55497829.png"
  },
  {
    "solution": "\\boxed{}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated, the diagonals $AC$ and $BD$ of quadrilateral $\\square ABCD$ intersect at point O, with $AC \\perp BD$, $BD = 12\\,cm$, and $AC = 6\\,cm$. Point E moves from point B towards point O on segment $BO$ at a speed of $1\\,cm/s$, and point F moves from point O towards point D on segment $OD$ at a speed of $2\\,cm/s$. Assuming points E and F start moving at the same time, and defining the movement duration as t seconds, what are the lengths of $EO$ and $OF$ in cm, respectively?\n",
    "completion": "\\textbf{Solution:} Given the problem, the specific values of $EO$ and $OF$ cannot be directly obtained; more information is needed.\\\\\nTherefore, a concrete answer cannot be provided.",
    "image_path": "dataset/mm_math/images/MM_Math/52250892_119.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $ \\triangle ADE \\sim \\triangle ACB$. If $AD=6$, $DE=4$, and $BC=8$, then what is the length of AC?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ADE \\sim \\triangle ACB$,\\\\\nwe have $\\frac{AD}{AC}=\\frac{DE}{CB}$,\\\\\nthus $AC=\\frac{AD \\cdot CB}{DE}=\\boxed{12}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51322568.png"
  },
  {
    "solution": "\\boxed{140^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, $\\triangle ABC$ is folded along $EF$ such that point $A$ falls on point ${A}^{\\prime}$. $BP$ and $CP$ are respectively the angle bisectors of $\\angle ABD$ and $\\angle ACD$. If $\\angle P=30^\\circ$ and $\\angle {A}^{\\prime}EB=20^\\circ$, what is the degree measure of $\\angle {A}^{\\prime}FC$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure,\\\\\nsince $BP$ and $CP$ are the angle bisectors of $\\angle ABD$ and $\\angle ACD$ respectively,\\\\\nit follows that $\\angle PBD=\\frac{1}{2}\\angle ABD$ and $\\angle BCP=\\frac{1}{2}\\angle BCA$.\\\\\nAlso, since $\\angle PBD=\\angle P+\\angle PCB$,\\\\\nwe have $\\angle P=\\angle PBD−\\angle PCB=\\frac{1}{2}\\angle ABD−\\frac{1}{2}\\angle BCA=\\frac{1}{2}(\\angle ABD−\\angle ACB)$,\\\\\nand since $\\angle ABD=\\angle A+\\angle ACB$,\\\\\nit follows that $\\angle ABD−\\angle ACB=\\angle A$,\\\\\nthus $\\angle P=\\frac{1}{2}\\angle A$,\\\\\nand hence $\\angle A=2\\angle P=2\\times 30^\\circ=60^\\circ$,\\\\\nGiven that: $\\angle A\\prime =\\angle A=60^\\circ$,\\\\\nit follows that $\\angle 1=\\angle A\\prime +\\angle A\\prime EB=60^\\circ+20^\\circ=80^\\circ$,\\\\\nthus $\\angle A\\prime FC=\\angle A+\\angle 1=60^\\circ+80^\\circ=\\boxed{140^\\circ}$,\\\\",
    "image_path": "dataset/mm_math/images/MM_Math/53686575.png"
  },
  {
    "solution": "\\boxed{\\sqrt{10}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, quadrilateral $OABC$ is a rectangle with $OA=1$. What is the value represented by point $P$?\n",
    "completion": "\\textbf{Solution:} Given that $\\because$OA=1, OC=3,\\\\\n$\\therefore$OB=$\\sqrt{3^2+1^2}$=$\\sqrt{10}$,\\\\\nthus, the number represented by point P is $\\boxed{\\sqrt{10}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52826952.png"
  },
  {
    "solution": "\\boxed{72^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the isosceles $\\triangle ABC$, where $AB=AC$ and $\\angle A=36^\\circ$. $BD$ bisects $\\angle ABC$. What is the measure of $\\angle BDC$?\n",
    "completion": "\\textbf{Solution:} Given that $AB = AC$ and $\\angle A = 36^\\circ$,\\\\\nit follows that $\\angle ABC = \\frac{180^\\circ - 36^\\circ}{2} = 72^\\circ$,\\\\\nsince BD bisects $\\angle ABC$,\\\\\nwe have $\\angle ABD = \\frac{1}{2} \\angle ABC = 36^\\circ$,\\\\\ntherefore, $\\angle BDC = \\angle A + \\angle ABD = \\boxed{72^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52828906.png"
  },
  {
    "solution": "\\boxed{70^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, D is a point on the extension of BC, $\\angle B=50^\\circ$, $\\angle ACD=120^\\circ$. What is the value of $\\angle A$?\n",
    "completion": "\\textbf{Solution:} It is known from the properties of the exterior angles of a triangle that,\n\\[\n\\angle A = \\angle ACD - \\angle B = 70^\\circ,\n\\]\nthus, \\(\\angle A = \\boxed{70^\\circ}\\).",
    "image_path": "dataset/mm_math/images/MM_Math/52754259.png"
  },
  {
    "solution": "\\boxed{\\sqrt{3}-1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: It seems there's no content provided for translation. Please provide the specific text you want to be translated into English and in LaTeX format.\n",
    "completion": "\\textbf{Solution:} As shown in Figure 2, draw $BD$ intersecting $AC$ at point $O$,\\\\\nsince quadrilateral $ABCD$ is a rhombus with $AB=2$,\\\\\nit follows that $AB=AD$, $AO=OC$, $OB=OD=\\frac{1}{2}BD$, and $AC\\perp BD$,\\\\\ngiven $\\angle BAD=60^\\circ$,\\\\\ntriangle $ABD$ is an equilateral triangle,\\\\\nthus $BD=AB=2$, $OB=OD=1$, $OC=OA=\\sqrt{AB^{2}-OB^{2}}=\\sqrt{2^{2}-1^{2}}=\\sqrt{3}$,\\\\\nsince $BG\\perp DF$,\\\\\nwe have $OE=OB=OD=1$,\\\\\ntherefore, $CE=OC-OE=\\sqrt{3}-1=\\boxed{\\sqrt{3}-1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52160377_158.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of semicircle $O$, and chord $AC$ makes an angle of $30^\\circ$ with $AB$. Given that $AC=CD$ and $OA=2$, what is the length of $AC$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect $BC$, \\\\\nsince $AB$ is the diameter of $\\odot O$, \\\\\ntherefore, $\\angle ACB=90^\\circ$, \\\\\nsince $\\angle BAC=30^\\circ$ and $OA=OB=2$, \\\\\ntherefore $AC=AB\\cdot\\cos30^\\circ=4\\times \\frac{\\sqrt{3}}{2}=\\boxed{2\\sqrt{3}}.$",
    "image_path": "dataset/mm_math/images/MM_Math/52623378.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, where $AB = AC = 4$ and $BC = 5$, points D and E are on line segments $BC$ and $AC$ respectively, with $\\angle ADE = \\angle B$. When $BD = 2$, find the length of $EC$.\n",
    "completion": "\\textbf{Solution:} Given: $AB=AC$,\\\\\nthus $\\angle B=\\angle C$.\\\\\nSince $\\angle B+\\angle BAD+\\angle ADB=180^\\circ$,\\\\\nand $\\angle ADE+\\angle EDC+\\angle ADB=180^\\circ$,\\\\\n$\\angle ADE=\\angle B$,\\\\\ntherefore $\\angle BAD=\\angle EDC$,\\\\\nthus $\\triangle ABD\\sim \\triangle DCE$,\\\\\nhence $\\frac{EC}{BD}=\\frac{DC}{AB}$.\\\\\nGiven $AB=4, BC=5, BD=2$,\\\\\nthus $\\frac{EC}{2}=\\frac{5-2}{4}$,\\\\\ntherefore $EC=\\boxed{\\frac{3}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52991859.png"
  },
  {
    "solution": "\\boxed{y=\\frac{4}{x}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=ax+1$ intersects with the x-axis and y-axis at points $A$ and $B$, respectively, and intersects with the hyperbola $y=\\frac{k}{x}$ ($x>0$) at point $P$. $PC\\perp x$-axis at point $C$, and $PC=2$. The coordinates of point $A$ are $(-2, 0)$. Find the expressions of line $AP$ and the hyperbola?\n",
    "completion": "\\textbf{Solution:} Substitute $A(-2,0)$ into $y=\\frac{1}{2}x+1$ and obtain $a=\\frac{1}{2}$,\\\\\n$\\therefore y=\\frac{1}{2}x+1$,\\\\\nGiven $PC=2$, substitute $y=2$ into $y=\\frac{1}{2}x+1$ to find $x=2$,\\\\\nthat is, $P(2,2)$,\\\\\nSubstituting $P(2,2)$ into $y=\\frac{k}{x}$ yields $k=4$,\\\\\n$\\therefore$ the expression for the hyperbola is $\\boxed{y=\\frac{4}{x}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52189578_137.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, point $D$ is on side $AB$ of $\\triangle ABC$, and point $E$ is the midpoint of $CD$. Connect $AE$ and $BE$. If the area of $\\triangle ABC$ is 8, what is the area of the shaded region?\n",
    "completion": "\\textbf{Solution:} Since point E is the midpoint of CD, \\\\\nit follows that the area of $\\triangle ACE$ equals the area of $\\triangle ADE$, and the area of $\\triangle BCE$ equals the area of $\\triangle BDE$, \\\\\nhence, the area of the shaded region equals $\\frac{1}{2}$ the area of $\\triangle ABC$ = $\\frac{1}{2} \\times 8 = \\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52655206.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the chord $CD \\perp AB$ at point E, $AB$ passes through the center O, $BD=5$, $BE=3$, then what is the length of $CD$?\n",
    "completion": "\\textbf{Solution:} Since $CD \\perp AB$ and $AB$ passes through the center $O$,\\\\\nit follows that $DE = \\frac{1}{2}CD$,\\\\\nIn the right triangle $\\triangle DEB$, with $BD=5$ and $BE=3$,\\\\\nby the Pythagorean theorem, we have: $DE = \\sqrt{BD^{2}-BE^{2}} = 4$,\\\\\ntherefore, $CD = \\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53110792.png"
  },
  {
    "solution": "\\boxed{\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, on two sides of $\\triangle MON$, segments OA and OB are intercepted respectively, such that $OA=OB$; arcs are drawn with point A and B as centers and the length of OA as the radius, intersecting at point C; lines AC, BC, AB, and OC are drawn, intersecting at point D. If $AB=4\\,cm$ and the area of quadrilateral OACB is $16\\,cm^2$. Point E is the midpoint of CB, and line DE is drawn. What is the length of segment DE?\n",
    "completion": "\\textbf{Solution}: Based on the given information, we know that $AC=BC=OA$,\\\\\n$\\because$ $OA=OB$,\\\\\n$\\therefore$ $AC=BC=OA=OB$,\\\\\n$\\therefore$ the quadrilateral OACB is a rhombus,\\\\\n$\\therefore$ $OD=\\frac{1}{2}OC$, $BD=\\frac{1}{2}AB=2\\text{cm}$, $AB\\perp OC$,\\\\\n$\\because$ $AB=4\\text{cm}$, the area of the quadrilateral OACB is $16\\text{cm}^2$,\\\\\n$\\therefore$ $\\frac{1}{2}AB\\times OC=16$,\\\\\n$\\therefore$ $OC=8\\text{cm}$,\\\\\n$\\therefore$ $OD=\\frac{1}{2}OC=4\\text{cm}$,\\\\\n$\\because$ in $\\triangle ODB$, by the Pythagorean theorem, we have $OB^2=OD^2+BD^2$,\\\\\n$\\therefore$ $OB=\\sqrt{OD^2+BD^2}=\\sqrt{4^2+2^2}=2\\sqrt{5}\\text{cm}$,\\\\\n$\\because$ point E is the midpoint of CB, point D is the midpoint of OC,\\\\\n$\\therefore$ DE is the midline of $\\triangle OBC$,\\\\\n$\\therefore$ $DE=\\frac{1}{2}OB=\\boxed{\\sqrt{5}}\\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52189085.png"
  },
  {
    "solution": "\\boxed{\\frac{13}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, there exists a point $C$ on the circle $\\odot O$ with diameter $AB$. Connect $BC$, and draw the bisector $BD$ of $\\angle ABC$ that intersects $\\odot O$ at point $D$. Through point $D$, draw the extension line $DE \\perp BC$ at point $E$. $DE$ is the tangent to the circle $\\odot O$; if $BE=9$ and $DE=6$, find the radius of $\\odot O$.\n",
    "completion": "\\textbf{Solution:} In right triangle $\\triangle BDE$, by the Pythagorean theorem, we have $BD = \\sqrt{BE^2 + DE^2} = \\sqrt{9^2 + 6^2} = 3\\sqrt{13}$,\\\\\nsince $AB$ is the diameter,\\\\\nthen $\\angle ADB = 90^\\circ$,\\\\\nthus $\\angle ADB = \\angle E$,\\\\\nsince $\\angle ABD = \\angle DBE$,\\\\\nit follows that $\\triangle ABD \\sim \\triangle DBE$,\\\\\nhence $\\frac{AB}{BD} = \\frac{BD}{BE}$,\\\\\nthus $\\frac{AB}{3\\sqrt{13}} = \\frac{3\\sqrt{13}}{9}$,\\\\\ntherefore $AB = 13$,\\\\\nhence the radius of circle $O$ is $\\boxed{\\frac{13}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53147490_128.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $C$ is on the line segment $AB$, and point $D$ is the midpoint of the line segment $BC$. Given that $AB=10$ and $AC=6$, what is the length of the line segment $BD$?\n",
    "completion": "\\textbf{Solution.} Since point D is the midpoint of line segment $BC$, and $AB=10$, $AC=6$,\n\\[\\therefore BD = \\frac{1}{2}BC =\\frac{1}{2}\\left(AB-AC\\right) =\\frac{1}{2}\\left(10-6\\right)=\\boxed{2}.\\]",
    "image_path": "dataset/mm_math/images/MM_Math/53106927.png"
  },
  {
    "solution": "\\boxed{35^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, within $\\odot O$, $A$, $B$, and $C$ are three points on $\\odot O$. If $\\angle AOB=70^\\circ$, what is the degree measure of $\\angle C$?\n",
    "completion": "\\textbf{Solution:} Given: $\\angle AOB$ and $\\angle ACB$ both subtend $\\overset{\\frown}{AB}$, and $\\angle AOB=70^\\circ$, \\\\\n$\\therefore \\angle C=\\frac{1}{2}\\angle AOB=\\boxed{35^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51351122.png"
  },
  {
    "solution": "\\boxed{\\frac{9}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in $ \\triangle ABC$, $ \\angle ACB=90^\\circ$, $CD\\perp AB$ with the foot of the perpendicular at D, $AC=4$, $BC=3$. What is the length of $BD$?\n",
    "completion": "Solution: In right-angled triangle $ABC$, we have $\\angle ACB=90^\\circ$, $AC=4$, and $BC=3$,\\\\\ntherefore, $AB=\\sqrt{AC^2+BC^2}=5$,\\\\\nsince $CD\\perp AB$,\\\\\nand $\\cos B=\\frac{BD}{BC}=\\frac{BC}{AB}$,\\\\\nit follows that $\\frac{BD}{3}=\\frac{3}{5}$,\\\\\nthus, $BD=\\boxed{\\frac{9}{5}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53148902_119.png"
  },
  {
    "solution": "\\boxed{k^2(2+k)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, an isosceles triangle with a vertex angle of $36^\\circ$ and the ratio of its base to its legs equal to $k$ is known as a golden triangle. Given that the leg $AB=1$ in the $\\triangle ABC$ which is the first golden triangle, $\\triangle BCD$ is the second golden triangle, and $\\triangle CDE$ is the third golden triangle, and so on, what is the perimeter of the third golden triangle?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABC$ and $\\triangle BCD$ are both isosceles triangles with a vertex angle of $36^\\circ$,\n\\[\n\\therefore \\triangle ABC \\sim \\triangle BCD,\n\\]\n\\[\n\\therefore \\text{the perimeter of } \\triangle BCD : \\text{the perimeter of } \\triangle ABC = BC: AB,\n\\]\nGiven that $AB=AC=1$, and $\\frac{BC}{AB}=k$,\n\\[\n\\therefore BC=k,\n\\]\n\\[\n\\therefore \\text{the perimeter of } \\triangle ABC = AB+AC+BC=2+k,\n\\]\n\\[\n\\therefore \\text{the perimeter of } \\triangle BCD : \\text{the perimeter of } \\triangle ABC = k,\n\\]\n\\[\n\\therefore \\text{the perimeter of } \\triangle BCD = k(2+k),\n\\]\nBy the same reasoning, \\text{the perimeter of } $\\triangle CDE$ : \\text{the perimeter of } $\\triangle BCD = k$,\n\\[\n\\therefore \\text{the perimeter of } \\triangle CDE = k^2(2+k) = \\boxed{k^2(2+k)}.\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/51298868.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $l_{1}$ intersects the x-axis and y-axis at points A and B, respectively, with the coordinates of A and B being $(2, 0)$ and $(0, 3)$, respectively. The line $l_{2}\\colon y=\\frac{1}{2}x+3$ passes through point B and intersects the x-axis at point C. Point $D(n, 6)$ lies on the line $l_{1}$. Connect $CD$. What is the area of $\\triangle BCD$?\n",
    "completion": "\\textbf{Solution:} By the problem statement, we have\n\\[\nS_{\\triangle BCD} = S_{\\triangle DAC} - S_{\\triangle BAC} = \\frac{1}{2} \\times (2+6) \\times 6 - \\frac{1}{2} \\times (2+6) \\times 3 = \\boxed{12}.\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/51699679_177.png"
  },
  {
    "solution": "\\boxed{15^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given that $O$ is a point on line $AB$, place the right-angle vertex of the right-angled triangle $OMN$ at point $O$. Ray $OC$ bisects $\\angle MOB$. As shown in Figure 1, if $\\angle AOM = 30^\\circ$, what is the measure of $\\angle CON$?\n",
    "completion": "\\textbf{Solution:}\nGiven $\\angle BOM = 180^\\circ - \\angle AOM = 150^\\circ$,\\\\\nsince $\\angle MON$ is a right angle, and $OC$ bisects $\\angle BOM$,\\\\\nthus, $\\angle CON = \\angle MON - \\frac{1}{2} \\angle BOM = 90^\\circ - \\frac{1}{2} \\times 150^\\circ = \\boxed{15^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52169848_188.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the diagonals $AC$ and $BD$ of rectangle $ABCD$ intersect at point $O$, with $\\angle BOC=120^\\circ$ and $AC=6$. What is the length of $AB$?\n",
    "completion": "\\textbf{Solution:} Given: Quadrilateral ABCD is a rectangle,\\\\\ntherefore $OA=OC=\\frac{1}{2}AC=3$, $OB=OD=\\frac{1}{2}BD$, $AC=BD$,\\\\\ntherefore $OA=OB=3$,\\\\\nsince $\\angle BOC=120^\\circ$,\\\\\ntherefore $\\angle AOB=60^\\circ$,\\\\\nthus, triangle AOB is an equilateral triangle,\\\\\ntherefore $AB=OA=OB=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52302711.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $AB=60\\,cm$, point C is the midpoint of line segment AB, and point D is a point on line segment AB with the ratio of the lengths of AD to DB being $2\\colon 1$. What is the length of CD in $cm$?\n",
    "completion": "\\textbf{Solution:} Since $AB=60\\,cm$, and point C is the midpoint of line segment AB,\\\\\ntherefore, $BC=\\frac{1}{2}AB=30\\,cm$,\\\\\ntherefore, $CD=BC-BD=30-20=\\boxed{10}\\,cm$.",
    "image_path": "dataset/mm_math/images/MM_Math/47112699_209.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ and $CD$ are two chords of the circle $\\odot O$, and they intersect at point $P$. Connect $AD$ and $BD$. Given that $AD=BD=4$ and $PC=6$, what is the length of $CD$?\n",
    "completion": "\\textbf{Solution:} Connect AC, \\\\\naccording to the inscribed angle theorem, we have $\\angle C = \\angle B$, \\\\\nsince AD = BD, \\\\\nit follows that $\\angle B = \\angle DAB$, \\\\\nthus, $\\angle DAP = \\angle C$, \\\\\ntherefore $\\triangle DAP \\sim \\triangle DCA$, \\\\\nhence, AD : CD = DP : AD, \\\\\nwe obtain $AD^{2} = DP \\cdot CD = CD \\cdot (CD - PC)$, \\\\\nsubstituting AD = 4, PC = 6 into the equation, we get, \\\\\n$CD^{2} - 6CD - 16 = 0$, \\\\\nsolving this, we find CD = \\boxed{8} or CD = -2 (discard the latter).",
    "image_path": "dataset/mm_math/images/MM_Math/51458273.png"
  },
  {
    "solution": "\\boxed{3+2\\sqrt{21}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within the circle $\\odot O$, the arcs $\\overset{\\frown}{AB} = \\overset{\\frown}{AC}$, $AB=10$, $BC=12$, and $D$ is a point on the arc $\\overset{\\frown}{BC}$ with $CD=5$. What is the length of $AD$?\n",
    "completion": "\\textbf{Solution:} Draw $AE\\perp BC$ at $E$, and draw $CF\\perp AD$ at $F$, then $\\angle AEB=\\angle CFD=90^\\circ$,\n\n$\\because$ $\\stackrel{\\frown}{AB} = \\stackrel{\\frown}{AC}$, and $AB=10$,\n\n$\\therefore$ $\\angle ACB=\\angle B=\\angle D$, and $AB=AC=10$,\n\n$\\because$ $AE\\perp BC$ and $BC=12$,\n\n$\\therefore$ $BE=CE=6$,\n\n$\\therefore$ $AE=\\sqrt{AB^{2}-BE^{2}}=\\sqrt{10^{2}-6^{2}}=8$,\n\n$\\because$ $\\angle B=\\angle D$ and $\\angle AEB=\\angle CFD=90^\\circ$,\n\n$\\therefore$ $\\triangle ABE\\sim \\triangle CDF$,\n\n$\\therefore$ $\\frac{AB}{CD}=\\frac{BE}{DF}=\\frac{AE}{CF}$,\n\n$\\because$ $AB=10$, $CD=5$, $BE=6$, $AE=8$,\n\n$\\therefore$ $\\frac{10}{5}=\\frac{6}{DF}=\\frac{8}{CF}$,\n\nSolving gives: $DF=3$, $CF=4$,\n\nIn $\\triangle AFC$, $\\angle AFC=90^\\circ$, $AC=10$, $CF=4$,\n\nthus $AF=\\sqrt{AC^{2}-CF^{2}}=\\sqrt{10^{2}-4^{2}}=2\\sqrt{21}$,\n\n$\\therefore$ $AD=DF+AF=3+2\\sqrt{21}$.\n\nTherefore, the answer is: $AD=\\boxed{3+2\\sqrt{21}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51433280.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $D$ and $E$ are the midpoints of line segments $BC$ and $BA$ respectively. Let the area of $\\triangle ABC$ be $S_{1}$, and the area of $\\triangle EBD$ be $S_{2}$. What is the value of $\\frac{S_{2}}{S_{1}}$?\n",
    "completion": "\\textbf{Solution:} Since $D$ and $E$ are the midpoints of line segments $BC$ and $BA$ respectively, \\\\\nthus $DE$ is the midline of $\\triangle ABC$, \\\\\ntherefore $DE=\\frac{1}{2}AC$ and $DE\\parallel AC$, \\\\\ntherefore $\\triangle EBD\\sim \\triangle ABC$, \\\\\ntherefore $\\frac{S_{2}}{S_{1}}=\\frac{S_{\\triangle EBD}}{S_{\\triangle ABC}}=\\left(\\frac{DE}{AC}\\right)^{2}=\\left(\\frac{1}{2}\\right)^{2}=\\boxed{\\frac{1}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52917358.png"
  },
  {
    "solution": "\\boxed{Q(0, 13)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, a rectangular paper $OABC$ is placed in the Cartesian coordinate system, where $O$ is the origin, point $C$ is on the $x$-axis, and point $A$ is on the $y$-axis, with $OA=9$ and $OC=15$. Under the given conditions, if point $T$ has the coordinates $(6,\\frac{5}{2})$, and point $P$ is on the line $MN$, is there a point $Q$ on the coordinate axes such that the quadrilateral with vertices $M$, $D'$, $Q$, $P$ is a parallelogram? If such a point exists, please directly write the coordinates of point $Q$.\n",
    "completion": "\\textbf{Solution:} There exists such a point Q, $\\because$ in rectangle ABCO, $D'G \\perp OC$, $\\therefore$ quadrilateral $AD'GO$ is also a rectangle, $\\therefore GD'=AO=9$, $AD'=OG$, $\\because T(6, \\frac{5}{2})$, $\\therefore AD'=OG=6$, $TG=\\frac{5}{2}$, $D'(6, 9)$, $\\therefore TD'=GD'-TG=9-\\frac{5}{2}=\\frac{13}{2}$, $\\therefore$ the side length of the rhombus $MD'TO$ is $\\frac{13}{2}$, that is $OM=MD'=D'T=TO=\\frac{13}{2}$, $\\therefore$ the coordinates of point M are: $(0, \\frac{13}{2})$.\n\nSuppose the equation of the line MN is $y=kx+b$, with point T on line MN, substitute $M(0, \\frac{13}{2})$, $T(6, \\frac{5}{2})$ into it, yielding $\\left\\{\\begin{array}{l}6k+b=\\frac{5}{2}\\\\ b=\\frac{13}{2}\\end{array}\\right.$, solve to get $\\left\\{\\begin{array}{l}k=-\\frac{2}{3}\\\\ b=\\frac{13}{2}\\end{array}\\right.$, thus the equation of MN is $y=-\\frac{2}{3}x+\\frac{13}{2}$.\n\n$\\because$ Point P is on the line MN, and point Q is on the coordinate axis, $\\therefore$ assume the coordinates of point P are $(x, -\\frac{2}{3}x+\\frac{13}{2})$,\n\nWhen point Q is on the x-axis, let the coordinates of Q be $(a, 0)$, i.e.: $P(x, -\\frac{2}{3}x+\\frac{13}{2})$, $Q(a, 0)$, $M(0, \\frac{13}{2})$, $D'(6, 9)$, these four points can form a parallelogram, knowing that the diagonals of a parallelogram bisect each other, it is solved that $x=6$, $a=0$, $\\therefore$ at this time the coordinates of Q are $Q(0, 0)$; solved to get $x=\\frac{27}{2}$, $a=\\frac{39}{2}$, $\\therefore$ at this time the coordinates of Q are $Q(\\frac{39}{2}, 0)$; solved to get $x=-\\frac{27}{2}$, $a=\\frac{39}{2}$, $\\therefore$ at this time the coordinates of Q are $Q(\\frac{39}{2}, 0)$.\n\nWhen point Q is on the y-axis, let the coordinates of Q be $(0, a)$, knowing that the diagonals of a parallelogram bisect each other, it is solved that $x=6$, $a=0$, $\\therefore$ at this time the coordinates of Q are $Q(0, 0)$; solved to get $x=-6$, $a=13$, $\\therefore$ at this time the coordinates of Q are $Q(0, 13)$; solved to get $x=6$, $a=13$, $\\therefore$ at this time the coordinates of Q are $Q(0, 13)$.\n\nIn conclusion: The coordinates of the point Q that satisfy the conditions are: \\boxed{Q(0, 0)}, \\boxed{Q\\left(\\frac{39}{2}, 0\\right)}, \\boxed{Q(0, 13)}.",
    "image_path": "dataset/mm_math/images/MM_Math/52381479_146.png"
  },
  {
    "solution": "\\boxed{34^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, we have $AB=AD=DC$. If $\\angle BAC=78^\\circ$, then what is the measure of $\\angle C$ in degrees?\n",
    "completion": "\\textbf{Solution:} We have $\\angle C = \\boxed{34^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52845731_112.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated in the figure, the parabola $y=\\frac{3}{4}x^{2}-\\frac{15}{4}x+3$ intersects the x-axis at points $A$ and $B$, with point $A$ located to the left of point $B$, and intersects the y-axis at point $C$. Point $E$ is a moving point on line segment $BC$. A perpendicular line to the x-axis is drawn through point $E$ and it intersects the parabola at point $F$. When $EF=OC$, find the x-coordinate of point $E$.\n",
    "completion": "\\textbf{Solution:} Let the equation of the line $BC$ be $y=-\\frac{3}{4}x+3$.\\\\\nSince it passes through point $B(4, 0)$,\\\\\nthus we find $k=-\\frac{3}{4}$.\\\\\nTherefore, the equation of the line $BC$ is $y=-\\frac{3}{4}x+3$.\\\\\nSince point $E$ is a moving point on line $BC$, and a perpendicular to the $x$ axis through point $E$ intersects the parabola at point $F$,\\\\\nthus, let the coordinates of point $E$ be $\\left(m, -\\frac{3}{4}m+3\\right)$, then the coordinates of point $F$ are $\\left(m, \\frac{3}{4}m^2-\\frac{15}{4}m+3\\right)$.\\\\\nSince $EF=OC$, and the coordinates of point $C$ are $(0, 3)$,\\\\\ntherefore, $\\left|\\frac{3}{4}m^2-\\frac{15}{4}m+3-\\left(-\\frac{3}{4}m+3\\right)\\right|=3$.\\\\\nTherefore, $\\frac{3}{4}m^2-3m=3$, or $\\frac{3}{4}m^2-3m=-3$.\\\\\nSolving gives $m_1=2+2\\sqrt{2}$, $m_2=2-2\\sqrt{2}$, $m_3=m_4=2$.\\\\\nTherefore, the $x$-coordinates of point $E$ are $\\boxed{2+2\\sqrt{2}}$, $\\boxed{2-2\\sqrt{2}}$, and $\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52719985_72.png"
  },
  {
    "solution": "\\boxed{y=-x+3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=kx+b$ intersects the $x$-axis at point $B$ and the $y$-axis at point $C(0,3)$, and intersects with $y=2x$ at point $A(a,2)$. Point $M$ is on line $OA$. What is the equation of line $AB$?\n",
    "completion": "\\textbf{Solution:} Since point A$(1,2)$ lies on the line $y=2x$, it follows that $2=2 \\cdot 1$, hence $a=1$. Therefore, A$(1,2)$. Since the line $y=kx+b$ passes through $C(0,3)$ and $A(1,2)$, it follows that $\\begin{cases} 3=0\\cdot k+b \\\\ 2=k+b \\end{cases}$. \\\\\nSolving these equations, we get $\\begin{cases} k=-1 \\\\ b=3 \\end{cases}$. Therefore, the equation of line $AB$ is \\boxed{y=-x+3};",
    "image_path": "dataset/mm_math/images/MM_Math/52354898_182.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in circle $\\odot O$, $OA=2$, and $\\angle ACB=30^\\circ$, what is the length of chord $AB$?\n",
    "completion": "\\textbf{Solution:} Since $\\overset{\\frown}{AB}=\\overset{\\frown}{AB}$,\\\\\nit follows that $\\angle AOB=2\\angle ACB=2\\times 30^\\circ=60^\\circ$,\\\\\nsince $OA=OB$,\\\\\nthus $\\triangle AOB$ is an equilateral triangle,\\\\\ntherefore $OA=AB=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52733112.png"
  },
  {
    "solution": "\\boxed{-t^2-3t}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the Cartesian coordinate system, point O is the origin. The parabola $y = ax^2 + bx + 3$ passes through points $B(-3, 0)$ and $C(1, 0)$, and intersects the y-axis at point A. As shown in Figure 1, point P lies on the parabola. A line through point P parallel to the y-axis intersects the line segment $AB$ at point D. Denoting the x-coordinate of point P as t, and the length of segment $PD$ as d, find the functional expression of d in terms of t (it is not required to state the domain of the independent variable t).\n",
    "completion": "\\textbf{Solution:}\n\n\\[\n\\begin{aligned}\n&\\because a=-1, b=-2 \\\\\n&\\therefore y=x^2-2x+3 \\\\\n&\\text{when} x=0, y=3 \\\\\n&\\therefore A(0, 3) \\\\\n&\\text{Let the line} AB: y=kx+b' \\\\\n&\\because B(-3, 0) \\\\\n&\\therefore\n\\left\\{\n\\begin{array}{l}\nb'=3 \\\\\n-3k+b'=0\n\\end{array}\n\\right. \\\\\n&\\therefore\n\\left\\{\n\\begin{array}{l}\nk=1 \\\\\nb'=3\n\\end{array}\n\\right. \\\\\n&\\therefore \\text{The equation of line} AB \\text{is} y=x+3 \\\\\n&\\because DP \\parallel \\text{the y-axis}, \\\\\n&\\therefore P, O \\text{have the same x-coordinate} t \\\\\n&\\therefore p(t, -t^2-2t+3), O(t, t+3) \\\\\n&\\therefore d=-t^2-2t+3-(t+3) \\\\\n&=-t^2-3t \\\\\n&\\therefore d=\\boxed{-t^2-3t}\n\\end{aligned}\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/53019865_161.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $C$ is a point on the line segment $AB$ ($AC>BC$), $D$ is on the line segment $BC$, with $BD=2CD$, and point $E$ is the midpoint of $AB$. If $AD=10$ and $EC=3CD$, what is the length of line segment $CD$?\n",
    "completion": "\\textbf{Solution:} Let $CD=x$, then $BD=2CD=2x$,\\\\\nsince $AD=10$,\\\\\nthus $AB=10+2x$,\\\\\nsince point E is the midpoint of $AB$,\\\\\nthus $AE=\\frac{1}{2}AB=5+x=BE$,\\\\\nthus $EC=EB-CB=5+x-3x=5-2x$,\\\\\nthus $5-2x=3x$,\\\\\nsolving gives $x=\\boxed{1}$,\\\\\nhence $CD=\\boxed{1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53206007_56.png"
  },
  {
    "solution": "\\boxed{\\sqrt{15}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the four corners of the rectangular paper ABCD are folded inward to perfectly form a quadrilateral EFGH with no gaps and no overlaps. If $BC=8$ and $GH=7$, then what is the length of $EH$?\n",
    "completion": "\\textbf{Solution:} Given that folding the four corners of the rectangle ABCD inward perfectly forms a quadrilateral EFGH without any gaps or overlaps,\\\\\nit follows that $\\angle CFE = \\angle EFH = \\frac{1}{2} \\angle CFH$, $\\angle BFG = \\angle HFG = \\frac{1}{2} \\angle BFH$, FM = CF, FN = BF, $\\angle GNH = \\angle EMH = 90^\\circ$,\\\\\nhence $\\angle EFH + \\angle HFG = \\frac{1}{2} (\\angle CFH + \\angle BFH) = 90^\\circ$, meaning $\\angle EFG = 90^\\circ$,\\\\\nSimilarly, it can be derived that $\\angle FEH = \\angle GHE = 90^\\circ$,\\\\\nthus quadrilateral EFGH is a rectangle,\\\\\nhence $\\angle GHF = \\angle EFH$, EF = GH = 7,\\\\\nIn $\\triangle GHN$ and $\\triangle EFM$,  $\\left\\{\\begin{array}{l} \\angle GNH = \\angle EMF\\hfill \\\\ \\angle GHN = \\angle EFM\\hfill \\\\ GH = EF\\hfill \\end{array} \\right.$,\\\\\nthus $\\triangle GHN \\cong \\triangle EFM$,\\\\\nhence HN = FM,\\\\\nthus CF = HN,\\\\\nhence FH = NH + FN = CF + BF = BC = 8,\\\\\nthus EH = $\\sqrt{FH^2 - EF^2}$ = $\\sqrt{FH^2 - GH^2}$ = $\\sqrt{8^2 - 7^2}$ = $\\boxed{\\sqrt{15}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50496662.png"
  },
  {
    "solution": "\\boxed{35^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, $\\angle BAC=70^\\circ$. The triangle $\\triangle ABC$ is rotated $70^\\circ$ clockwise around point A, with the corresponding points of B and C after rotation being B' and C', respectively. Connecting BB', what is the degree measure of $\\angle BB'C'$?\n",
    "completion": "\\textbf{Solution:}\n\nGiven the properties of rotation, we know that: $AB=AB'$, $\\angle BAB'=70^\\circ$, and $\\angle ACB=\\angle AC'B'=90^\\circ$ \\\\\n$\\therefore \\angle ABB'=\\angle AB'B=\\frac{1}{2}\\times (180^\\circ-70^\\circ)=55^\\circ$ \\\\\nIn $\\triangle AB'C'$, $\\angle AB'C'=90^\\circ-\\angle B'AB=20^\\circ$ \\\\\n$\\because \\angle BB'C+\\angle AB'C'=\\angle AB'B$ \\\\\n$\\therefore \\angle BB'C'=\\angle AB'B-\\angle AB'C'=55^\\circ-20^\\circ=\\boxed{35^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/51367477.png"
  },
  {
    "solution": "\\boxed{x<4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the intersection point of the lines $y_1=k_1x+a$ and $y_2=k_2x+b$ is $(4, 8)$. What is the range of values of $x$ for which $y_1<y_2$?\n",
    "completion": "\\textbf{Solution:} From the graph, it can be observed that when $x<4$, the line $y_1$ is below the line $y_2$. \\\\\nTherefore, the range of $x$ for which $y_1<y_2$ is: $\\boxed{x<4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53182974.png"
  },
  {
    "solution": "\\boxed{48}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system, the rhombus OABC has side OA on the x-axis, with point $A(10, 0)$, and $\\sin\\angle COA=\\frac{4}{5}$. If the inverse proportion function $y=\\frac{k}{x}$ ($k>0, x>0$) passes through point C, what is the value of $k$?\n",
    "completion": "\\[ \\text{{\\bfseries Solution:}} \\text{ As shown in the figure, draw } CE \\perp OA \\text{ at point E,} \\]\n\\[ \\because \\text{ the side OA of the rhombus OABC lies on the x-axis, with point } A(10,0), \\]\n\\[ \\therefore OC = OA = 10, \\]\n\\[ \\because \\sin\\angle COA = \\frac{4}{5} = \\frac{CE}{OC}, \\]\n\\[ \\therefore CE = 8, \\]\n\\[ \\therefore OE = \\sqrt{CO^2 - CE^2} = 6, \\]\n\\[ \\therefore \\text{ the coordinates of point C are }(6, 8), \\]\n\\[ \\because \\text{ if the inverse ratio function } y = \\frac{k}{x} (k>0, x>0) \\text{ passes through point C,} \\]\n\\[ \\therefore k = 6 \\times 8 = \\boxed{48} \\]",
    "image_path": "dataset/mm_math/images/MM_Math/51174880.png"
  },
  {
    "solution": "\\boxed{\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the side length of the square is 1. A point E is chosen on ray AB, and DE is connected. Then, triangle ADE is rotated $90^\\circ$ around point D, and point E falls on point F. Connect EF, which intersects the line containing diagonal BD at point M, and intersects ray DC at point N. Connect AM, and let the line AM intersect line BC at point G. When $BG = \\frac{1}{3}$, find the value of AE.\n",
    "completion": "\\textbf{Solution:} For point G on segment BC, where $BG=\\frac{1}{3}$,\n\nsince quadrilateral ABCD is a square,\n\ntherefore AD$\\parallel$BG,\n\nthus $\\angle$DAM=$\\angle$BGM, $\\angle$ADM=$\\angle$GBM,\n\nwhich implies $\\triangle$BGM$\\sim$$\\triangle$DAM,\n\nhence $\\frac{BG}{DA}=\\frac{BM}{DM}=\\frac{\\frac{1}{3}}{1}=\\frac{1}{3}$,\n\ngiven from (2) that $\\triangle$BEM$\\sim$$\\triangle$FDM,\n\nthus $\\frac{BM}{MF}=\\frac{BE}{DF}$,\n\nsince DB=$\\sqrt{AB^{2}+AD^{2}}=\\sqrt{2}$,\n\nthen BM=$\\frac{\\sqrt{2}}{4}$,\n\nhence $\\frac{\\frac{\\sqrt{2}}{4}}{y}=\\frac{1-x}{\\sqrt{1+x^{2}}}$,\n\ngiven $y=\\frac{1}{2}(1+x)\\sqrt{2+2x^{2}}$,\n\nthus $\\frac{\\frac{\\sqrt{2}}{4}}{\\frac{1}{2}(1+x)\\sqrt{2+2x^{2}}}=\\frac{1-x}{\\sqrt{1+x^{2}}}$ results in $1-x^{2}=\\frac{1}{2}$,\n\nsolution $x_{1}=\\frac{\\sqrt{2}}{2}$, $x_{2}=-\\frac{\\sqrt{2}}{2}$ discard;\n\nFor point G on the extension of CB, where $BG=\\frac{1}{3}$, draw ML$\\perp$BC intersecting line BC at L,\n\nsince GB$\\parallel$AD,\n\nthus $\\angle$DAM=$\\angle$BGM, $\\angle$ADM=$\\angle$GBM,\n\nwhich implies $\\triangle$BGM$\\sim$$\\triangle$DAM,\n\nhence $\\frac{BG}{DA}=\\frac{BM}{DM}=\\frac{\\frac{1}{3}}{1}=\\frac{1}{3}$,\n\nthus BM=$\\frac{1}{3}DM$,\n\nhence BM=$\\frac{1}{2}BD$,\n\ngiven $\\angle$LBM=$\\angle$CBD=$45^\\circ$, and ML$\\perp$BC,\n\n$\\triangle$MLB forms an isosceles right triangle,\n\nsince ML$\\parallel$CD,\n\nthus $\\angle$LMB=$\\angle$CDB, $\\angle$L=$\\angle$DCB,\n\nwhich implies $\\triangle$MLB$\\sim$$\\triangle$DCB,\n\nthus $\\frac{BM}{BD}=\\frac{ML}{DC}=\\frac{1}{2}$, with CD=1,\n\nhence ML=$\\frac{1}{2}$,\n\nsince ML$\\parallel$BE,\n\nthus $\\angle$L=$\\angle$FBE, $\\angle$LMF=$\\angle$BEF,\n\nwhich implies $\\triangle$LMF$\\sim$$\\triangle$BEF,\n\nthus $\\frac{LM}{BE}=\\frac{LF}{BF}$,\n\ngiven BE=AE−AB=x−1, LF=LB+BC+CF=$\\frac{1}{2}+1+x=\\frac{3}{2}+x$, BF=BC+CF=1+x,\n\nthus $\\frac{\\frac{1}{2}}{x-1}=\\frac{\\frac{3}{2}+x}{1+x}$,\n\nrearranging gives: $2x^{2}=4$,\n\nsolution $x_{3}=\\sqrt{2}$, $x_{4}=-\\sqrt{2}$ discard,\n\ntherefore, the value of AE is $\\boxed{\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51403292_160.png"
  },
  {
    "solution": "\\boxed{3\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the Cartesian coordinate system $xOy$, the line $y=-2x+a$ intersects the $y$ axis at point $A$ and intersects the line $y=x+1$ at point $P(3,b)$. Point $B$ lies on the line $y=x+1$. Find a point $C$ on the $x$ axis such that the value of $AC-PC$ is maximized. Please write the coordinates of point $C$ and find the maximum value.\n",
    "completion": "\\textbf{Solution:} When points $A$, $P$, and $C$ are collinear, the maximum value of $AC-PC$ is equal to $AP$. Solving $0 = -2x + 10$ gives $x = 5$, thus $C(5,0)$.\\\\\nSince $P(3,4)$,\\\\\n$AP= \\sqrt{AH^{2}+PH^{2}}=\\sqrt{(10-4)^{2}+3^{2}}=\\boxed{3\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50677275_151.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=6\\,cm$, $BC=12\\,cm$, $\\angle B=90^\\circ$. Point P starts from point A and moves towards point B along side $AB$ at a speed of $1\\,cm/s$, and point Q starts from point B and moves towards point C along side $BC$ at a speed of $2\\,cm/s$. If P and Q start moving from A and B at the same time, respectively, and the moving time is denoted as $t\\,(s)$. At what value of $t$ is the area of $\\triangle PBQ$ equal to $9\\,cm^{2}$?\n",
    "completion": "\\textbf{Solution:} Given the problem, point $P$ starts moving from point $A$ towards point $B$ along side $AB$ at a speed of $1\\,cm/s$, and point $Q$ starts moving from point $B$ towards point $C$ along side $BC$ at a speed of $2\\,cm/s$,\\\\\n$\\therefore AP=t$, $BQ=2t$,\\\\\n$\\therefore BP=AB-AP=6-t$;\\\\\nSince the area of $\\triangle PBQ$ is $9\\,cm^{2}$,\\\\\n$\\therefore \\frac{1}{2}\\cdot 2t\\cdot (6-t)=9$,\\\\\nsolving yields, $t=\\boxed{3}$.\\\\\nThat is, when $t=3$, the area of $\\triangle PBQ$ is $9\\,cm^{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53106523_96.png"
  },
  {
    "solution": "\\boxed{x\\in(-\\infty,-1)\\cup(3,\\infty)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=ax^2+bx+2$ is symmetric about the line $x=1$, and the point $(3, 0)$ lies on the parabola. What is the range of $x$ for which $ax^2+bx+2<0$?\n",
    "completion": "\\textbf{Solution} \\\\\nSince the parabola $y=ax^2+bx+2$ is symmetrical about the line $x=1$, and it intersects the x-axis at $\\left(3,0\\right)$,\\\\\nit follows that the other intersection point with the x-axis is $\\left(-1,0\\right)$, \\\\\ntherefore, when $ax^2+bx+2<0$, the value range of $x$ is $x<-1$ or $x>3$, \\\\\nhence, the solution is: $\\boxed{x\\in(-\\infty,-1)\\cup(3,\\infty)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52865967.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points C and E are two points on the line segment AB, point D is the midpoint of line segment AB, and AB = 6, CD = 1. If AE : EC = 1 : 3, find the length of EC.\n",
    "completion": "\\textbf{Solution:} Since point D is the midpoint of line segment AB, and AB$=6$,\\\\\nthus AD$=\\frac{1}{2}$AB$=3$,\\\\\nsince CD$=1$, thus AC$=$AD$+$CD$=4$,\\\\\nsince AE:EC$=1:3$, thus EC$=\\frac{3}{1+3}\\times 4=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51653655_167.png"
  },
  {
    "solution": "\\boxed{-3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points A and B are on the graphs of the inverse proportion functions $y=\\frac{1}{x} (x>0)$ and $y=\\frac{k}{x} (x>0)$, respectively, and $\\angle AOB=90^\\circ$, $\\sin B=\\frac{1}{2}$. What is the value of $k$?\n",
    "completion": "\\textbf{Solution:} Construct $AC\\perp y$-axis through $A$, and $BD\\perp y$-axis through $B$, it can be obtained that $\\angle ACO=\\angle BDO=90^\\circ$, \\\\\n$\\therefore \\angle AOC+\\angle OAC=90^\\circ$, \\\\\n$\\because OA\\perp OB$, \\\\\n$\\therefore \\angle AOC+\\angle BOD=90^\\circ$, \\\\\n$\\therefore \\angle OAC=\\angle BOD$, \\\\\n$\\therefore \\triangle AOC\\sim \\triangle OBD$, \\\\\n$\\because$ points $A$ and $B$ lie on the graphs of the inverse proportion functions $y=\\frac{1}{x}(x>0)$ and $y=\\frac{k}{x}(x>0)$ respectively, \\\\\n$\\therefore S_{\\triangle AOC}=\\frac{1}{2}$, $S_{\\triangle OBD}=|\\frac{k}{2}|$, \\\\\n$\\therefore S_{\\triangle AOC}\\colon S_{\\triangle OBD}=1\\colon |k|$, \\\\\n$\\therefore \\left(\\frac{OA}{OB}\\right)^2=1\\colon |k|$, \\\\\nthen, in $\\triangle AOB$ with right angle, $\\because \\sin B=\\frac{1}{2}$, \\\\\n$\\therefore \\angle B=30^\\circ$, \\\\\n$\\therefore \\tan B=\\frac{OA}{OB}=\\frac{\\sqrt{3}}{3}$, \\\\\n$\\therefore 1\\colon |k|=1\\colon 3$, \\\\\n$\\therefore |k|=3$, \\\\\n$\\because$ the graph of $y=\\frac{k}{x}(x>0)$ is in the fourth quadrant, \\\\\n$\\therefore k=\\boxed{-3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53065562.png"
  },
  {
    "solution": "\\boxed{\\frac{51}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, it is known that in rectangle $ABCD$, $AC$ and $BD$ intersect at point $O$. Through point $O$, draw $EF\\perp BD$ intersecting $AB$ and $CD$ at $E$ and $F$, respectively. Quadrilateral $BEDF$ is a rhombus. If $BC=3$ and $CD=5$, find the area of the rhombus $BEDF$.\n",
    "completion": "\\textbf{Solution:} Let $CF = x$, then $DF = CD - CF = 5 - x$, \\\\\nsince quadrilateral $BEDF$ is a rhombus, \\\\\ntherefore $BF = DF = 5 - x$, \\\\\nin $\\triangle BFC$, $BF^2 = BC^2 + CF^2$, \\\\\ntherefore $(5-x)^2 = 3^2 + x^2$, \\\\\nsolving for $x$ yields $x = \\frac{8}{5}$, \\\\\nthus $DF = 5 - \\frac{8}{5} = \\frac{17}{5}$, \\\\\ntherefore the area of rhombus $BEDF$ is $DF \\cdot BC = \\frac{17}{5} \\times 3 = \\boxed{\\frac{51}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52189001_133.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the quadrilateral $ABCD$ is a rectangle, with two diagonals $AC$ and $BD$ intersecting at point $O$, and $AE \\parallel BD$, $DE \\parallel AC$. If $BC = 4$ and $AC = 6$, what is the area of the quadrilateral $AODE$?\n",
    "completion": "\\textbf{Solution:}\n\nSince $BC=4$ and $AC=6$, \n\nwe have $AB=\\sqrt{AC^{2}-BC^{2}}=\\sqrt{36-16}=2\\sqrt{5}$.\n\nGiven that quadrilateral ABCD is a rectangle,\n\nit follows that $OB=OD$,\n\nimplying $Area_{\\triangle AOD}=\\frac{1}{2}Area_{\\triangle ABD}$.\n\nSince quadrilateral AODE is a rhombus,\n\nit follows that $Area_{\\triangle AOD}=\\frac{1}{2}Area_{\\text{rhombus} AODE}$.\n\nTherefore, $Area_{\\text{rhombus} AODE}=Area_{\\triangle ABD}=\\frac{1}{2}AB\\cdot AD=\\frac{1}{2}\\times 2\\sqrt{5}\\times 4=\\boxed{4\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53313100_115.png"
  },
  {
    "solution": "\\boxed{y=\\frac{3}{4}x-\\frac{15}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given the graph of a linear function $y=kx+b$ intersects with the graph of the inverse proportional function $y=\\frac{m}{x}$ $(x>0)$ at point A, and intersects with the x-axis at point B $(5,0)$. If $OB=AB$ and the area of $\\triangle OAB$ is $\\frac{15}{2}$, find the expressions of the inverse proportional function and the linear function.\n",
    "completion": "\\textbf{Solution:} Let's consider the diagram, draw $AM\\perp x$-axis at M,\n\n$\\because B(5,0)$,\n\n$\\therefore OB=5$,\n\n$\\because S_{\\triangle OAB}=\\frac{15}{2}$,\n\n$\\therefore \\frac{1}{2} \\cdot OB \\cdot AM=\\frac{15}{2}$,\n\n$\\therefore AM=3$,\n\n$\\because OB=AB$,\n\n$\\therefore AB=5$,\n\nWithin $\\triangle AMB$, by applying the Pythagorean theorem, $BM=\\sqrt{AB^2-AM^2}=4$,\n\n$\\therefore OM=OB+BM=9$,\n\n$\\therefore A(9,3)$,\n\n$\\because$ point A is on the graph of the inverse proportion function $y=\\frac{m}{x}$,\n\n$\\therefore m=9\\times 3=27$,\n\n$\\therefore$ the expression for the inverse proportion function is $\\boxed{y=\\frac{27}{x}}$;\n\n$\\because$ points A$(9,3)$, B$(5,0)$ are on the graph of the linear function $y=kx+b$,\n\n$\\therefore \\left\\{\\begin{array}{l}9k+b=3\\\\ 5k+b=0\\end{array}\\right.$,\n\n$\\therefore \\left\\{\\begin{array}{c}k=\\frac{3}{4}\\\\ b=-\\frac{15}{4}\\end{array}\\right.$,\n\n$\\therefore$ the expression for the linear function is $\\boxed{y=\\frac{3}{4}x-\\frac{15}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50496638_60.png"
  },
  {
    "solution": "\\boxed{70^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that quadrilateral ABCD is a rhombus, AE $\\perp$ BC at point E, and AF $\\perp$ CD at point F. AE=AF. If $\\angle B=70^\\circ$, find the degree measure of $\\angle EAF$.\n",
    "completion": "\\textbf{Solution:} Since $AE\\perp BC$ at point $E$, and $\\angle B=70^\\circ$, it follows that $\\angle BAE=20^\\circ$. Because $\\triangle ABE\\cong \\triangle ADF$, it follows that $\\angle BAE=\\angle DAF=20^\\circ$, thus $\\angle EAF=180^\\circ-\\angle B-\\angle BAE-\\angle DAF=\\boxed{70^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51614858_79.png"
  },
  {
    "solution": "\\boxed{\\left(-1, 0\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the Cartesian coordinate system, let points $A(1, 0)$ and $B(0, 2)$. Construct $\\triangle BOC$ such that $\\triangle BOC$ is congruent to $\\triangle ABO$ (with point $C$ not coinciding with point $A$). What are the coordinates of point $C$?\n",
    "completion": "\\textbf{Solution:} According to the problem, we draw $\\triangle BOC$ as shown in the figure,\\\\\nthen $OC_1=BC_2=BC_3=OA=1$, $\\angle OBC_2=\\angle OBC_3=90^\\circ$,\\\\\nthus, the coordinates of point C are $\\boxed{\\left(-1, 0\\right)}$ or $\\left(-1, 2\\right)$ or $\\left(1, 2\\right)$.",
    "image_path": "dataset/mm_math/images/MM_Math/52950525.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: To further prevent the \"novel coronavirus\", a school in Xi'an has disinfected all classrooms through the method of \"fumigation\". It is known that during the combustion release process of the medicine, the concentration of the medicine in the air of the classrooms per cubic meter, \\(y\\) (mg), is related to the burning time, \\(x\\) (min), as shown in the graph. Here, when \\(x < 6\\), \\(y\\) is directly proportional to \\(x\\), and when \\(x \\ge 6\\), \\(y\\) is inversely proportional to \\(x\\). Based on the information provided by the graph, solve the following question: Find the coordinates of point A.\n",
    "completion": "\\textbf{Solution:} Solve: When $x=6$, we have $10y=60$. Solving for $y$, we get $y=6$. \\\\\n$\\therefore$, the coordinates of point A are $(6, \\boxed{6})$.",
    "image_path": "dataset/mm_math/images/MM_Math/52304593_212.png"
  },
  {
    "solution": "\\boxed{\\frac{7}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=\\frac{1}{4}x^{2}-4$ intersects the x-axis at points A and B. Let P be a moving point on the circle with center C(0,3) and radius 2, and let Q be the midpoint of line segment PA. Connect OQ. What is the maximum value of line segment OQ?\n",
    "completion": "\\textbf{Solution:} Let us connect BP as shown in the diagram.\\\\\nWhen $y=0$, we have $\\frac{1}{4}x^{2}-4=0$. Solving this, we find $x_{1}=4$ and $x_{2}=-4$, thus $A(-4,0)$ and $B(4,0)$.\\\\\nSince Q is the midpoint of segment PA,\\\\\nit follows that OQ is the median of $\\triangle ABP$,\\\\\ntherefore OQ$=\\frac{1}{2}BP$.\\\\\nWhen BP is at its maximum, OQ is also at its maximum.\\\\\nMoreover, when BP passes through the circle's center C, PB is maximized. As shown in the diagram, when point P moves to position P', BP is maximized.\\\\\nSince BC$=\\sqrt{3^{2}+4^{2}}=5$,\\\\\nit follows that BP'$=5+2=7$.\\\\\nTherefore, the maximum value of segment OQ is $\\boxed{\\frac{7}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53009781.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $BC=6\\,cm$, $BD=7\\,cm$, $D$ is the midpoint of $AC$, find the length of $AB$.\n",
    "completion": "\\textbf{Solution:} Since $BC=6\\,\\text{cm}$ and $BD=7\\,\\text{cm}$,\\\\\n$\\therefore CD=BD-BC=1\\,\\text{cm}$,\\\\\nSince point $D$ is the midpoint of $AC$,\\\\\n$\\therefore AD=CD=1\\,\\text{cm}$,\\\\\n$\\therefore AB=AD+BD=1+7=\\boxed{8}\\,\\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55175882.png"
  },
  {
    "solution": "\\boxed{2-\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, where $AB=2$ and $AD=1$, point $M$ is on the side $CD$. If $AM$ bisects $\\angle DMB$, what is the length of $DM$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral ABCD is a rectangle,\\\\\nhence $CD=AB=2$, and $AB\\parallel CD$, $BC=AD=1$, $\\angle C=90^\\circ$,\\\\\nthus $\\angle BAM=\\angle AMD$,\\\\\nsince $AM$ bisects $\\angle DMB$,\\\\\nthus $\\angle AMD=\\angle AMB$,\\\\\nhence $\\angle BAM=\\angle AMB$,\\\\\nthus $BM=AB=2$,\\\\\nhence $CM= \\sqrt{MB^{2}-BC^{2}}=\\sqrt{2^{2}-1^{2}}=\\sqrt{3}$,\\\\\nthus $DM=CD−CM=2−\\sqrt{3}=\\boxed{2-\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52354873.png"
  },
  {
    "solution": "\\boxed{-\\frac{4}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=ax^2-2ax-3a$ (where $a$ is a constant and $a<0$) intersects the x-axis at points A and B (with point A to the left of point B) and intersects the y-axis at point C, with OB=OC. Point D is the vertex of the parabola, and point P $(m, n)$ is a point on the parabola in the third quadrant. Lines BD, BC, CD, and BP are drawn. When $\\angle PBA=\\angle CBD$, find the value of $m$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $PE\\perp x$ axis at point $E$,\\\\\n$\\because y=-x^2+2x+3=-(x-1)^2+4$\\\\\n$\\therefore D(1, 4)$\\\\\n$\\because B(3, 0), C(3, 0)$\\\\\n$\\therefore CD=\\sqrt{1^2+1^2}=\\sqrt{2}, BC=\\sqrt{3^2+3^2}=3\\sqrt{2}, BD=\\sqrt{(3-1)^2+4^2}=2\\sqrt{5}$\\\\\n$\\therefore CD^2+BC^2=20, BD^2=20$\\\\\n$\\therefore CD^2+BC^2=BD^2$\\\\\n$\\therefore \\triangle BCD$ is a right triangle, with $\\angle BCD=90^\\circ$\\\\\n$\\because PE\\perp AB$\\\\\n$\\therefore \\angle PEB=\\angle PCD=90^\\circ$\\\\\nAlso, $\\because \\angle PBA=\\angle CBD$\\\\\n$\\therefore \\triangle BCD\\sim \\triangle BEP$\\\\\n$\\therefore \\frac{CD}{PE}=\\frac{BC}{BE}$\\\\\n$\\because P(m, n)$ lies on the parabola $y=-x^2+2x+3$,\\\\\n$\\therefore n=-m^2+2m+3$\\\\\n$\\therefore PE=-n=m^2-2m-3, BE=3-m$\\\\\n$\\therefore \\frac{\\sqrt{2}}{m^2-2m-3}=\\frac{3\\sqrt{2}}{3-m}$\\\\\nAfter rearranging, we get $(3m+4)(m-3)=0$\\\\\nSolving this, we find $m_1=-\\frac{4}{3}, m_2=3$ (discard)\\\\\n$\\because P(m, n)$ is in the third quadrant,\\\\\n$\\therefore m=\\boxed{-\\frac{4}{3}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51433217_216.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{3}{4}x^{2}+\\frac{9}{4}x+3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=ax^{2}+bx+c$ passes through three points $A(-1,0)$, $B(4,0)$, and $C(0,3)$. Point $D$ is a moving point on the parabola above line $BC$, and $DE\\perp BC$ at $E$. Find the expression of the parabola.\n",
    "completion": "\\textbf{Solution:} Given the problem, we have\n\\[\n\\begin{cases}\na-b+c=0\\\\\n16a+4b+c=0\\\\\nc=3\n\\end{cases}\n\\]\nSolving these, we find\n\\[\n\\begin{cases}\na=-\\frac{3}{4}\\\\\nb=\\frac{9}{4}\\\\\nc=3\n\\end{cases}\n\\]\nThus, the equation of the parabola is \\boxed{y=-\\frac{3}{4}x^{2}+\\frac{9}{4}x+3}.",
    "image_path": "dataset/mm_math/images/MM_Math/52551774_95.png"
  },
  {
    "solution": "\\boxed{7.5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y=k_1x+b$ intersects with the graph of the inverse proportion function $y=\\frac{k_2}{x}$ at points A and B, where the coordinates of point A are $(-1, 4)$, and the coordinates of point B are $(4, n)$. Connect OA and OB, what is the area of $\\triangle AOB$?\n",
    "completion": "\\[\n\\text{{Solution}}: \\text{{As shown in the figure,}} \\\\\n\\text{{let }} AB \\text{{ intersect the }} y \\text{{-axis at point }} C, \\\\\n\\text{{since }} C \\text{{ lies on the line }} y = -x + 3, \\\\\n\\text{{it follows that }} C(0, 3), \\\\\n\\text{{therefore, }} {S}_{\\triangle AOB} = \\frac{1}{2} OC \\cdot (\\left|{x}_{A}\\right| + \\left|{x}_{B}\\right|) = \\frac{1}{2} \\times 3 \\times (1+4) = \\boxed{7.5};\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/51446477_240.png"
  },
  {
    "solution": "\\boxed{\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figures, two triangles of different sizes are each chosen and overlaid as in Figure 1 (with the right-angled vertices coinciding). It is given that $ \\angle A=45^\\circ$, $ \\angle CDE=30^\\circ$, and $ CE:BC=1:\\sqrt{6}$. After rotating $ \\triangle DCE$ $45^\\circ$ clockwise around point $ C$, line $ ED$ intersects line $ AC$ at point $ O$. What is the ratio of $ OA:OC$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, draw OF $\\perp$ CE through O,\\\\\nby the property of rotation, we have $\\angle$OCE$=45^\\circ$,\\\\\nin $\\triangle OEF$,\\\\\nsince $\\angle$CDE$=30^\\circ$,\\\\\nthus $\\angle$OEF$=60^\\circ$,\\\\\ntherefore OF$=EF\\tan\\angle$OEF$= \\sqrt{3}EF$,\\\\\nthus OF$=CF= \\sqrt{3}EF$, and $OC=\\sqrt{OF^{2}+CF^{2}}=\\sqrt{6}EF$,\\\\\ntherefore CE$=EF+CF= (\\sqrt{3}+1)EF$,\\\\\ntherefore $\\frac{CE}{OC}=\\frac{(\\sqrt{3}+1)EF}{\\sqrt{6}EF}=\\frac{\\sqrt{3}+1}{\\sqrt{6}}$,\\\\\nsince CE$\\colon BC=1\\colon \\sqrt{6}$,\\\\\nthus $\\frac{BC}{OC}=\\frac{BC}{CE} \\cdot \\frac{CE}{OC}=\\sqrt{6} \\times \\frac{\\sqrt{3}+1}{\\sqrt{6}}=\\sqrt{3}+1$,\\\\\nthus $\\frac{OA}{OC}=\\frac{AC-OC}{OC}=\\frac{AC}{OC}-1=\\frac{BC}{OC}-1=\\boxed{\\sqrt{3}}$,\\\\\nthus $OA\\colon OC=\\sqrt{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50360712.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown, $\\triangle ABO \\sim \\triangle CDO$, if $BO = 10$, $DO = 5$, and $AC = 9$, then what is the length of $OC$?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABO\\sim \\triangle CDO$,\\\\\nit follows that $\\frac{OB}{OD}=\\frac{OA}{OC}$,\\\\\ntherefore $\\frac{10}{5}=\\frac{9-OC}{OC}$,\\\\\nsolving this yields $OC=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52623484.png"
  },
  {
    "solution": "\\boxed{\\sqrt{10}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $AD = 9$, $AB = 3$. The paper is folded to make point $D$ coincide with point $B$, and the crease is $EF$. What is the length of the crease $EF$?\n",
    "completion": "\\textbf{Solution:} Construct $FH \\perp AD$ meeting AD at point H.\\\\\nSince quadrilateral $EFCB$ is obtained by folding quadrilateral $EFCD$ along EF,\\\\\ntherefore ED=BE, CF=$CF$, $BC'=CD=3$\\\\\nSince ED=BE, DE=AD−AE=9−AE\\\\\ntherefore $BE=9−AE$\\\\\nSince in $\\triangle ABE$, AB=3, BE=9−AE\\\\\nthus $(9−AE)^{2}=3^{2}+AE^{2}$\\\\\ntherefore $AE=4$\\\\\ntherefore $DE=5$\\\\\ntherefore $CF=BC−BF=9−BF$\\\\\nSince in $\\triangle BCF$, BC=3, CF=9−BF\\\\\nthus $(9−BF)^{2}+3^{2}=BF^{2}$\\\\\ntherefore $BF=5$, EH=1\\\\\nSince in $\\triangle EFH$, HF=3, EH=1\\\\\ntherefore $EF=\\sqrt{EH^{2}+HF^{2}}=\\sqrt{3^{2}+1^{2}}=\\boxed{\\sqrt{10}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51212036.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the radius of the drainage pipe's cross-section is 5 decimeters, and the width of the water surface $AB=8$ decimeters, with $OC\\perp AB$. What is the maximum depth $CD$ of the water?\n",
    "completion": "Solution: Since the diameter of circle O is 10 dm,\\\\\nit follows that OA=5 (dm),\\\\\nand since OD$\\perp$AB, and AB=8 (dm),\\\\\nit follows that AD=BD=4,\\\\\nhence, OD$=\\sqrt{OA^{2}-AD^{2}}=3$,\\\\\ntherefore, the maximum depth of the water CD=OC-OD=5-3=\\boxed{2} (dm).",
    "image_path": "dataset/mm_math/images/MM_Math/52622479.png"
  },
  {
    "solution": "\\boxed{\\sqrt{17}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the book \"Gougu Juyu,\" the Qing Dynasty mathematician Mei Wending proved the Pythagorean theorem by using four congruent right triangles to piece together square ABCD, as shown in the figure. By connecting CE, if $CE=5$ and $BE=4$, what is the side length of square ABCD?\n",
    "completion": "As shown in the figure:\n\nFrom the four congruent right triangles, it is known that $BE=CF=4$,  and $AE=BF$,\n\nBy Pythagorean theorem, $EF=\\sqrt{CE^{2}-CF^{2}}=\\sqrt{5^{2}-4^{2}}=3$,\n\n$\\therefore BF=BE-EF=4-3=1$,\n\nBy Pythagorean theorem, $AB=\\sqrt{AE^{2}+BE^{2}}=\\sqrt{1^{2}+4^{2}}=\\boxed{\\sqrt{17}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52827844.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{1}{2}x+3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the equation of line $l_{1}$ is $y=kx+b$ ($k\\neq 0$), and the equation of line $l_{2}$ is $y=x$. Line $l_{1}$ passes through points $A(6, 0)$ and $B(0, 3)$, and line $l_{1}$ intersects with line $l_{2}$ at point C. What is the equation of line $l_{1}$?\n",
    "completion": "\\textbf{Solution}: Because point $A(6, 0)$ and $B(0, 3)$ are on the line $y=kx+b$,\\\\\ntherefore $\\begin{cases}6k+b=0\\\\ b=3\\end{cases}$,\\\\\nsolving we get: $\\begin{cases}k=-\\frac{1}{2}\\\\ b=3\\end{cases}$,\\\\\ntherefore the equation of line $l_1$ is: \\boxed{y=-\\frac{1}{2}x+3}.",
    "image_path": "dataset/mm_math/images/MM_Math/52386525_134.png"
  },
  {
    "solution": "\\boxed{144}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a high school in Liupanshui City has an irregular quadrilateral space ABCD. The school plans to lay a suspended floor on the space. After measurement, $\\angle ABC=90^\\circ$, $BC=6\\text{m}$, $AB=8\\text{m}$, $AD=26\\text{m}$, $CD=24\\text{m}$. What is the area of the space ABCD in square meters?\n",
    "completion": "\\textbf{Solution:} Connect AC,\\\\\nIn the right-angled triangle ABC,\\\\\n$\\because \\angle ABC=90^\\circ$, BC=6m, AB=8m,\\\\\n$\\therefore AC=\\sqrt{BC^2+AB^2}=10$m,\\\\\n$\\because AC^2+CD^2=10^2+24^2=676=AD^2$,\\\\\n$\\therefore \\angle ACD=90^\\circ$,\\\\\n$\\therefore S_{\\text{quadrilateral }ABCD}=S_{\\triangle ABC}+S_{\\triangle ACD}=\\frac{1}{2}\\times 6\\times 8+\\frac{1}{2}\\times 10\\times 24=\\boxed{144}$m$^2$.",
    "image_path": "dataset/mm_math/images/MM_Math/51367124_120.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Place a triangular board $ABC$ ($\\angle ACB=90^\\circ$, $\\angle A=30^\\circ$) inside the acute angle $\\angle POQ$ as shown in figure (1), with the right-angle side $BC$ on side $OQ$, and let $\\angle POQ=\\alpha$. Now, rotate the triangular board $ABC$ counterclockwise around point B at a speed of $m^\\circ$ per second for $t$ seconds (with the right-angle side $BC$ moving to the position shown in figure (2), and point A always within $\\angle POQ$). Draw $MN\\parallel OQ$ intersecting ray $OP$ at point M, and bisect $\\angle MAB$ with $AD$ intersecting ray $OQ$ at point D. What value of $m$ minimizes the value of the algebraic expression $\\left|10−m\\right|+3$? What is the value of $m$?\n",
    "completion": "\\textbf{Solution:} We have $m=\\boxed{10}$.",
    "image_path": "dataset/mm_math/images/MM_Math/54883358_109.png"
  },
  {
    "solution": "\\boxed{65^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, PA and PB are tangent lines to the circle $O$, with A and B being the points of tangency. AC is a diameter of the circle $O$. Given that $\\angle P = 50^\\circ$, what is the measure of $\\angle ACB$?\n",
    "completion": "\\textbf{Solution:} Connect OB, as shown in the diagram,\\\\\nsince PA and PB are tangents to the circle centered at O,\\\\\nit follows that OA $\\perp$ PA, and OB $\\perp$ PB,\\\\\nthus, $\\angle$OAP=$\\angle$OBP=90$^\\circ$,\\\\\nhence, $\\angle$AOB=360$^\\circ$-90$^\\circ$-90$^\\circ$-50$^\\circ$=130$^\\circ$,\\\\\nsince OB=OC,\\\\\nit follows that $\\angle$OCB=$\\angle$OBC,\\\\\nand since $\\angle$AOB=$\\angle$OCB+$\\angle$OBC,\\\\\nthus, $\\angle$OCB= $\\frac{1}{2} \\times 130^\\circ=65^\\circ$,\\\\\nthat is, $\\angle$ACB=$\\boxed{65^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51213310.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB = 90^\\circ$, $\\angle A = 30^\\circ$, $AC = 15\\, \\mathrm{cm}$, point $O$ is on the median $CD$, when a circle with radius $3\\, \\mathrm{cm}$ centered at $O$ is tangent to the sides of $\\triangle ABC$, what is the length of $OC$?\n",
    "completion": "\\textbf{Solution:} In $\\triangle ABC$, $\\angle ACB=90^\\circ$, $\\angle A=30^\\circ$,\\\\\n$\\therefore \\angle B=60^\\circ$,\\\\\n$\\because AC = 15\\, \\text{cm}$,\\\\\n$\\therefore \\cos A=\\frac{AC}{AB}$,\\\\\n$\\therefore AB=\\frac{AC}{\\cos A}=\\frac{15}{\\frac{\\sqrt{3}}{2}}=10\\sqrt{3}$,\\\\\n$\\because CD$ is the median to side $AB$,\\\\\n$\\therefore CD=AD=BD=\\frac{1}{2}AB=5\\sqrt{3}$,\\\\\n$\\therefore \\angle BDC=\\angle BCD=\\angle B=60^\\circ$, $\\angle ACD=\\angle A=30^\\circ$,\\\\\n(1) When circle $O$ is tangent to $AB$, draw $OE\\perp AB$ at $E$,\\\\\nIn $\\triangle ODE$, $\\angle BDC=60^\\circ$, $OE=3$,\\\\\n$\\therefore \\sin\\angle BDC=\\frac{OE}{OD}$,\\\\\n$\\therefore OD=\\frac{OE}{\\sin\\angle BDC}=\\frac{3}{\\frac{\\sqrt{3}}{2}}=2\\sqrt{3}$;\\\\\n$\\therefore OC=CD-OD=5\\sqrt{3}-2\\sqrt{3}=3\\sqrt{3}$;\\\\\n(2) When circle $O$ is tangent to $BC$, draw $OE\\perp BC$,\\\\\nIn $\\triangle OCE$, $\\angle BCD=60^\\circ$, $OE=3$,\\\\\n$\\therefore \\sin\\angle BCD=\\frac{OE}{OC}$,\\\\\n$\\therefore OC=\\frac{OE}{\\sin\\angle BCD}=\\frac{3}{\\frac{\\sqrt{3}}{2}}=2\\sqrt{3}$;\\\\\n(3) When circle $O$ is tangent to $AC$, draw $OE\\perp AC$ at $E$,\\\\\nIn $\\triangle OCE$, $\\angle ACD=30^\\circ$, $OE=3$,\\\\\n$\\therefore \\sin\\angle ACD=\\frac{OE}{OC}$,\\\\\n$\\therefore OC=\\frac{OE}{\\sin\\angle ACD}=\\frac{3}{\\frac{1}{2}}=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51322900.png"
  },
  {
    "solution": "\\boxed{32^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of $\\odot O$, point $C$ is on $\\odot O$, extend $BC$ to point $D$ such that $DC=CB$, extend $DA$ to meet $\\odot O$ at another point $E$, and connect $AC$, $CE$. If $\\angle E = 32^\\circ$, what is the measure of $\\angle D$ in degrees?\n",
    "completion": "\\textbf{Solution:} Given: $\\because$ AB is the diameter of $\\odot$O,\\\\\n$\\therefore$ $\\angle$ACB=$90^\\circ$,\\\\\n$\\because$ CD=CB,\\\\\n$\\therefore$ AC is the perpendicular bisector of BD.\\\\\n$\\therefore$ AB=AD,\\\\\n$\\therefore$ $\\angle$B=$\\angle$D,\\\\\n$\\because$ $\\angle$B=$\\angle$E,\\\\\n$\\therefore$ $\\angle$D=$\\angle$E,\\\\\n$\\because$ $\\angle$E=$32^\\circ$,\\\\\n$\\therefore$ $\\angle$D=$\\boxed{32^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52874653_101.png"
  },
  {
    "solution": "\\boxed{y=-2x+4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the line $AC$ intersects the x-axis at point $C$ and the y-axis at point $A$, the line $AB$ intersects the x-axis at point $B$ and the y-axis at point $A$, with $A(0,4)$, $B(2,0)$ known. What is the analytical expression of line $AB$?\n",
    "completion": "\\textbf{Solution:} Let the equation of line AB be $y=kx+b$.\\\\\nSince line AB passes through A$(0, 4)$ and B$(2, 0)$,\\\\\ntherefore,\n\\[\n\\left\\{\n\\begin{array}{l}\nb=4 \\\\\n2k+b=0\n\\end{array}\n\\right.\n\\]\nSolving this, we get $k=-2$ and $b=4$,\\\\\ntherefore, the equation of line AB is $y=-2x+4=\\boxed{y=-2x+4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50555230_98.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{5}}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the vertices of $\\triangle ABC$ are at the vertices of a square grid. What is the value of $\\sin A$?\n",
    "completion": "\\textbf{Solution:} Draw line segment DC, \\\\\nfrom the grid we know that $CD\\perp AB$, \\\\\nthus $DC = \\sqrt{2}$, $AC = \\sqrt{10}$, \\\\\nhence $\\sin A = \\frac{DC}{AC} = \\frac{\\sqrt{2}}{\\sqrt{10}} = \\boxed{\\frac{\\sqrt{5}}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53080924.png"
  },
  {
    "solution": "\\boxed{y=(x-2)^{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system, the vertex C of the parabola passing through points $A\\left(0, 4\\right)$ and $B\\left(5, 9\\right)$ is on the positive half of the x-axis. Find the equation of the parabola.\n",
    "completion": "\\textbf{Solution:} Since the vertex $C$ of the parabola is on the positive x-axis, \\\\\nlet the equation of the parabola be $y=a(x-h)^{2}$. \\\\\nSubstituting points $A(0, 4)$ and $B(5, 9)$ into $y=a(x-h)^{2}$, we get: $\\left\\{\\begin{array}{l}a(0-h)^{2}=4 \\\\ a(5-h)^{2}=9\\end{array}\\right.$, \\\\\nSolving this yields: $h=-10$ (discard) or $h=2$, \\\\\nthus $a=1$, \\\\\nTherefore, the equation of the parabola is: $\\boxed{y=(x-2)^{2}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52514156_165.png"
  },
  {
    "solution": "\\boxed{2<x<3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y=kx+b$ intersects the $x$ axis at point $A(3, 0)$ and intersects the $y$ axis at point $B(0, 4)$. It intersects the graph of the direct proportion function $y=ax$ at point $C$, and the $x$-coordinate of point $C$ is 2. What is the solution set of the inequality $0<kx+b<ax$?\n",
    "completion": "\\textbf{Solution:}\n\nSubstitute $A(3, 0)$ and $B(0, 4)$ into $y=kx+b$ to get:\n\\[\n\\begin{cases}\nb=4\\\\ \n3k+b=0\n\\end{cases}\n\\]\nSolving this yields:\n\\[\n\\begin{cases}\nb=4\\\\ \nk=-\\frac{4}{3}\n\\end{cases}\n\\]\nTherefore, the equation of the linear function is $y=-\\frac{4}{3}x+4$.\n\nSubstituting $x=2$ into $y=-\\frac{4}{3}x+4$ gives $y=\\frac{4}{3}$. Therefore, the coordinates of point $C$ are $\\left(2, \\frac{4}{3}\\right)$.\n\nSubstituting $C$ into $y=ax$ gives $a=\\frac{2}{3}$. Therefore, the equation of the direct proportion function is $y=\\frac{2}{3}x$.\n\nThe solution set of the inequality $0<kx+b<ax$ is the same as the solution set of $0<-\\frac{4}{3}x+4<\\frac{2}{3}x$. From the graph, it is found that the solution set of $0<-\\frac{4}{3}x+4<\\frac{2}{3}x$ is $2<x<3$. Hence, the answer is $\\boxed{2<x<3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50520424.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{2}\\le PD\\le \\sqrt{10}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in rectangle $ABCD$, $AB=2$, $AD=4$, and $E$ is the midpoint of $AD$, $F$ is a moving point on the line segment $EC$, $P$ is the midpoint of $BF$. When the line segment $PD$ is drawn, what is the range of possible values for $PD$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram:\\\\\nWhen point F coincides with point C, point P is at ${P}_{1}$, where $C{P}_{1}=B{P}_{1}$,\\\\\nWhen point F coincides with point E, point P is at ${P}_{2}$, where $E{P}_{2}=B{P}_{2}$,\\\\\n$\\therefore$ ${P}_{1}{P}_{2}\\parallel EC$ and ${P}_{1}{P}_{2}=\\frac{1}{2}CE$,\\\\\nWhen point F lies on EC excluding points C and E, we have BP=FP,\\\\\nBy the Midline Theorem, it is known that: ${P}_{1}P\\parallel CF$ and ${P}_{1}P=\\frac{1}{2}CF$,\\\\\n$\\therefore$ The trajectory of point P is the line segment ${P}_{1}{P}_{2}$,\\\\\n$\\because$ In rectangle ABCD, AB=2, AD=4, E is the midpoint of AD,\\\\\n$\\therefore$ $\\triangle ABE$, $\\triangle BEC$, $\\triangle DCP_{1}$ are isosceles right triangles,\\\\\n$\\therefore$ $\\angle ECB=45^\\circ$, $\\angle DP_{1}C=45^\\circ$,\\\\\n$\\because$ ${P}_{1}{P}_{2}\\parallel EC$,\\\\\n$\\therefore$ $\\angle P_{2}P_{1}B=\\angle ECB=45^\\circ$,\\\\\n$\\therefore$ $\\angle P_{2}P_{1}D=90^\\circ$,\\\\\n$\\therefore$ The length of DP is smallest at $D{P}_{1}$ and largest at $D{P}_{2}$,\\\\\n$\\because$ $CD=C{P}_{1}=DE=2$,\\\\\n$\\therefore$ $D{P}_{1}=2\\sqrt{2}$, $CE=2\\sqrt{2}$,\\\\\n$\\therefore$ ${P}_{1}{P}_{2}=\\sqrt{2}$,\\\\\n$\\therefore$ $D{P}_{2}=\\sqrt{(2\\sqrt{2})^2+(\\sqrt{2})^2}=\\sqrt{10}$,\\\\\nTherefore, the answer is: $\\boxed{2\\sqrt{2}\\le PD\\le \\sqrt{10}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52212059.png"
  },
  {
    "solution": "\\boxed{x=5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system, the line $y=x+2$ and the line $y=ax+b$ ($a \\neq 0$) intersect at point $P$. According to the graph, what is the value of $x$ that solves the equation $x+2=ax+b$?\n",
    "completion": "\\textbf{Solution:} Substituting $y=7$ into $y=x+2$ yields $7=x+2$, \\\\\nSolving this gives $x=5$, \\\\\n$\\therefore$ The x-coordinate of point P is 5, \\\\\n$\\because$ The line $y=x+2$ intersects with the line $y=ax+b$ ($a\\neq 0$) at point P, \\\\\n$\\therefore$ The solution to the equation $x+2=ax+b$ is $\\boxed{x=5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52342889.png"
  },
  {
    "solution": "\\boxed{x>2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y=kx+b$ is depicted. What is the range of $x$ when $y<0$?\n",
    "completion": "\\textbf{Solution:} From the graph, we can conclude: when $y<0$, the range of $x$ is $x>2$. \\\\\nTherefore, the answer is $\\boxed{x>2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52380774.png"
  },
  {
    "solution": "\\boxed{x\\leq -2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: The graph of the linear function $y=kx+b$ is shown in the figure. What is the solution to the inequality $kx+b\\leq 0$?\n",
    "completion": "\\textbf{Solution:} From the graph, we can see that\\\\\nthe graph of the linear function $y=kx+b$ intersects the $x$-axis at the point $(-2,0)$,\\\\\nthe line rises from left to right, meaning $y$ increases as $x$ increases,\\\\\n$\\therefore$ the solution set of the inequality $kx+b\\leq 0$ is $x\\leq -2$,\\\\\nhence, the answer is: $\\boxed{x\\leq -2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52386773.png"
  },
  {
    "solution": "\\boxed{130^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\angle AOB$ is a straight angle, and $OM$, $ON$ are the angle bisectors of $\\angle AOC$, $\\angle BOD$ respectively. Given that $\\angle AOC=40^\\circ$, $\\angle BOD=60^\\circ$, what is the degree measure of $\\angle MON$?\n",
    "completion": "Solution:Given that $\\angle AOB$ is a straight angle, $\\angle AOC=40^\\circ$, $\\angle BOD=60^\\circ$,\\\\\nthus, $\\angle COD=\\angle AOB-\\angle AOC-\\angle BOD=180^\\circ-40^\\circ-60^\\circ=80^\\circ$,\\\\\nsince $OM$ and $ON$ are the angle bisectors of $\\angle AOC$ and $\\angle BOD$ respectively,\\\\\nthus, $\\angle COM= \\frac{1}{2} \\angle AOC=20^\\circ$, $\\angle DON= \\frac{1}{2} \\angle BOD=30^\\circ$,\\\\\nsince $\\angle MON=\\angle COM+\\angle COD+\\angle DON$,\\\\\nthus, $\\angle MON=20^\\circ+80^\\circ+30^\\circ=\\boxed{130^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/8753334_44.png"
  },
  {
    "solution": "\\boxed{\\sqrt{13}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a large square ABCD is formed by four congruent right-angled triangles and a small square. A perpendicular line DF is drawn from point D to intersect the extension of the diagonal EF of the small square at point G. Connect BG. If the area of the large square is 5 times the area of the small square, what is the value of $ \\frac{BG}{BE}$?\n",
    "completion": "Solution:Let the extension lines of $BE$ and $GD$ intersect at point $M$, and $BM$ intersect $CF$ at point $N$, as shown in the diagram.\\\\\nSuppose the side length of the small square is $x$, then the side length of the large square is $\\sqrt{5}x$,\\\\\nLet $CN = t$, then $BE = t$,\\\\\nIn $\\triangle BCN$, since $CN = t$, $BN = t + x$, $BC = \\sqrt{5}x$,\\\\\nTherefore, $x^2 + (t + x)^2 = (\\sqrt{5}x)^2$,\\\\\nSolving gives $t = x$,\\\\\nTherefore, $BE = CN = x$,\\\\\nSince $ED$ is the diagonal of the small square,\\\\\nTherefore, $\\angle FEN = \\angle EFN = 45^\\circ$,\\\\\nTherefore, $\\angle GFD = 45^\\circ$,\\\\\nSince $GD \\perp DF$,\\\\\nTherefore, $\\triangle GDF$ is an isosceles right triangle,\\\\\nTherefore, $\\angle FGD = 45^\\circ$, $DG = DF = x$,\\\\\nSince $\\angle GEM = \\angle EGM = 45^\\circ$,\\\\\nTherefore, $\\triangle MGE$ is an isosceles right triangle,\\\\\nTherefore, $ME = MG = 2x$,\\\\\nIn $\\triangle BMG$, since $BM = 3x, GM = 2x$,\\\\\nTherefore, $BG= \\sqrt{(3x)^2 + (2x)^2} = \\sqrt{13}x$,\\\\\nTherefore, $\\frac{BG}{BE} = \\frac{\\sqrt{13}x}{x} = \\boxed{\\sqrt{13}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52991672.png"
  },
  {
    "solution": "\\boxed{\\frac{12}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $AD = 3$, $AB = 4$. Let $M$ be a moving point on line segment $BD$. Let $MP \\perp CD$ at point $P$, and $MQ \\perp BC$ at point $Q$. What is the minimum value of $PQ$?\n",
    "completion": "\\textbf{Solution:} Connect CM as shown in the diagram:\n\\[\n\\because \\text{MP} \\perp \\text{CD at point P, MQ} \\perp \\text{BC at point Q,}\n\\]\n\\[\n\\therefore \\angle \\text{CPM} = \\angle \\text{CQM} = 90^\\circ,\n\\]\n\\[\n\\because \\text{Quadrilateral ABCD is a rectangle,}\n\\]\n\\[\n\\therefore \\text{BC} = \\text{AD} = 3, \\text{CD} = \\text{AB} = 4, \\angle \\text{BCD} = 90^\\circ,\n\\]\n\\[\n\\therefore \\angle \\text{CPM} = \\angle \\text{CQM} = \\angle \\text{BCD} = 90^\\circ,\n\\]\n\\[\n\\therefore \\text{Quadrilateral PCQM is a rectangle,}\n\\]\n\\[\n\\therefore \\text{PQ} = \\text{CM},\n\\]\n\\[\n\\therefore \\text{When CM is at its minimum, PQ is also at its minimum,}\n\\]\n\\[\n\\because \\text{Point M moves along BD,}\n\\]\n\\[\n\\therefore \\text{When CM} \\perp \\text{BD, CM is at its minimum, then PQ is at its minimum,}\n\\]\nUsing the Pythagorean theorem, we get:\n\\[\n\\text{BD} = \\sqrt{\\text{DC}^2 + \\text{BC}^2} = \\sqrt{4^2 + 3^2} = 5,\n\\]\n\\[\n\\because {S}_{\\triangle \\text{BCD}} = \\frac{1}{2} \\text{BC} \\times \\text{CD} = \\frac{1}{2} \\text{BD} \\times \\text{CM},\n\\]\n\\[\n\\therefore \\text{At this moment, CM} = \\frac{\\text{BC} \\times \\text{CD}}{\\text{BD}} = \\frac{3 \\times 4}{5} = \\boxed{\\frac{12}{5}},\n\\]\n\\[\n\\therefore \\text{The minimum value of PQ is} \\frac{12}{5}.\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/52303920.png"
  },
  {
    "solution": "\\boxed{\\frac{5}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle B = 90^\\circ$, point $D$ is on line segment $AC$, and circle $\\odot O$ with diameter $CD$ intersects $AB$ at point $E$. Line segment $CE$ is drawn, and $CE$ bisects $\\angle ACB$. $AB$ is a tangent to circle $\\odot O$; line segment $DE$ is drawn. If $BE=2$ and $BC=4$, what is the radius of circle $\\odot O$?\n",
    "completion": "Solution: As shown in the diagram, connect $DE$, then $\\angle DEC=90^\\circ$,\\\\\n$\\therefore \\angle DEC=\\angle B$,\\\\\n$\\because \\angle DCE=\\angle ECB$,\\\\\n$\\therefore \\triangle DCE \\sim \\triangle ECB$,\\\\\n$\\therefore \\frac{CD}{CE}=\\frac{CE}{BC}$,\\\\\n$\\because \\angle B=90^\\circ$, $BE=2$, $BC=4$,\\\\\nin $Rt\\triangle BCE$, $CE=\\sqrt{BC^2+BE^2}=\\sqrt{4^2+2^2}=2\\sqrt{5}$,\\\\\n$\\therefore CD=\\frac{(CE)^2}{BC}=\\frac{(2\\sqrt{5})^2}{4}=5$,\\\\\n$\\therefore$ The radius of $\\odot O$ is $\\boxed{\\frac{5}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52977753_139.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC$, D is the midpoint of side $AB$, E is a point on side $AC$, and line $BF$ is drawn parallel to $AC$ through point B and intersects the extension of $ED$ at point F. If $AD=6$ and $BF=9$, find the length of $CE$.\n",
    "completion": "\\textbf{Solution:} Given that $BF\\parallel AC$, \\\\\nit follows that $\\angle F=\\angle AED$, \\\\\nsince $D$ is the midpoint of $AB$, \\\\\nit implies $AD=BD$, \\\\\nconsidering $\\triangle ADE$ and $\\triangle BDF$, \\\\\nwe have \n$\\left\\{\\begin{array}{l}\n\\angle AED=\\angle F \\\\\n\\angle ADE=\\angle BDF \\\\\nAD=BD\n\\end{array}\\right.$, \\\\\nthus, $\\triangle ADE\\cong \\triangle BDF$ (by AAS), \\\\\nwhich means $AE=BF=9$, \\\\\ngiven $AB=AC$, \\\\\nit results in $AC=2AD=12$, \\\\\ntherefore, $CE=AC-AE=12-9=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53107094.png"
  },
  {
    "solution": "\\boxed{\\frac{10}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=\\frac{3}{2}x-2$ intersects with $y=-x+3$ at point P, and the line $y=\\frac{3}{2}x-2$ intersects the x-axis and y-axis at points C and D respectively; the line $y=-x+3$ intersects the x-axis and y-axis at points A and B, respectively. Connect BC. What is the area of $\\triangle BCD$?\n",
    "completion": "\\textbf{Solution:} Substitute $x=0$ into the line $y=\\frac{3}{2}x-2$ to get: $y=-2$,\\\\\nthus $D(0, -2)$,\\\\\nSubstitute $x=0$ into the line $y=-x+3$ to get: $y=3$,\\\\\nthus $B(0, 3)$,\\\\\nSubstitute $y=0$ into the line $y=\\frac{3}{2}x-2$ to get: $x=\\frac{4}{3}$,\\\\\nthus $C\\left(\\frac{4}{3}, 0\\right)$,\\\\\n$\\therefore BD=5, OC=\\frac{4}{3}$,\\\\\n$\\therefore {S}_{\\triangle BCD}=\\frac{1}{2}BD\\cdot OC=\\boxed{\\frac{10}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52386556_128.png"
  },
  {
    "solution": "\\boxed{42^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, lines \\(m\\), \\(n\\), and \\(l\\) intersect, and \\(m \\perp n\\), \\(\\angle 1 = 48^\\circ\\). What is the degree measure of \\(\\angle 3\\)?\n",
    "completion": "\\textbf{Solution:} Since line m intersects with line n,\\\\\nTherefore $\\angle 1 = \\angle 2 = 48^\\circ$,\\\\\nSince m $\\perp$ n,\\\\\nTherefore $\\angle 2 + \\angle 3 = 90^\\circ$,\\\\\nTherefore $\\angle 3 = 90^\\circ - 48^\\circ = \\boxed{42^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/54927610.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within the rectangle $ABCO$, point $C$ is on the x-axis, and point $A$ is on the y-axis. The coordinates of point $B$ are $(-6, 8)$. The rectangle $ABCO$ is folded along the line $BD$ such that point $A$ falls on point $E$ which is on the diagonal $OB$. The line containing $BD$ intersects $OA$ and the x-axis at points $D$ and $F$, respectively. What is the length of the segment $BO$?\n",
    "completion": "\\textbf{Solution:} Since $B(-6,8)$, it follows that $OC=6$ and $BC=8$. Given that quadrilateral $ABCO$ is a rectangle, it implies $\\angle BCO=90^\\circ$. By the Pythagorean theorem, we have: $BO= \\sqrt{BC^{2}+OC^{2}}=\\sqrt{8^{2}+6^{2}}=\\boxed{10}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51864146_100.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the quadratic function $y=ax^{2}+bx-3$ intersects the x-axis at points $A(-1,0)$ and $B(3,0)$, and intersects the y-axis at point $C$. The vertex of the parabola is point $M$. Point $P$ is a point on the parabola, with the x-coordinate of point $P$ being $m$ ($m>3$). Point $Q$ is on the axis of symmetry, and $AQ \\perp PQ$. If $AQ=2PQ$, find the value of $m$.\n",
    "completion": "\\textbf{Solution:} Draw a line parallel to the x-axis passing through point Q, intersecting the line parallel to the y-axis passing through point P at point N, and intersecting the line parallel to the y-axis passing through point A at point M.\\\\\n$\\therefore \\angle AMQ=\\angle QNP=90^\\circ$\\\\\n$\\therefore \\angle QPN+\\angle PQN=90^\\circ$\\\\\n$\\because \\angle AQP=90^\\circ$,\\\\\n$\\therefore \\angle AQM+\\angle PQN=90^\\circ$,\\\\\n$\\therefore \\angle AQM=\\angle QPN$,\\\\\n$\\therefore \\triangle AMQ\\sim \\triangle QNP$.\\\\\n$\\therefore \\frac{AM}{QN}=\\frac{MQ}{NP}=\\frac{AQ}{QP}=2$.\\\\\nThe axis of symmetry of this parabola is the line $x=1$,\\\\\nthus let the coordinates of point Q be (1, t),\\\\\ngiven that the coordinates of point P are ($m$, $m^{2}-2m-3$),\\\\\nhence $AM=t$, $QN=m-1$, $MQ=2$,\\\\\n$NP=t-m^{2}+2m+3$.\\\\\n$\\therefore \\frac{t}{m-1}=\\frac{2}{t-m^{2}+2m+3}=2$.\\\\\nSolving, we get $m=0$ (discard) or $4$.\\\\\nTherefore, $m=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52536969_109.png"
  },
  {
    "solution": "\\boxed{0<x<12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: A triangular piece of material is shown in the figure, where $ \\angle A=30^\\circ$, $ \\angle C=90^\\circ$, and $AB=12$. Cut out a rectangle CDEF from this material, where points D, E, and F are respectively on AC, AB, and BC. Let the length of AE be x, and the area of rectangle CDEF be S. Write down the expression of S as a function of x, and state the domain of x.\n",
    "completion": "\\textbf{Solution:} We have $AB=12$, $AE=x$, and the point E does not coincide with points A or B,\\\\\ntherefore $0<x<12$,\\\\\nthe quadrilateral CDEF is a rectangle,\\\\\ntherefore $\\angle AFE=90^\\circ$,\\\\\nsince $\\angle A=30^\\circ$,\\\\\ntherefore $EF=\\frac{1}{2}AE$,\\\\\nin $\\triangle ABC$, $\\angle C=90^\\circ$, $\\angle A=30^\\circ$, $AB=12$,\\\\\ntherefore $BC=\\frac{1}{2}AB=6$,\\\\\naccording to the Pythagorean theorem: $AC=\\sqrt{12^2-6^2}=6\\sqrt{3}$,\\\\\ntherefore $CF=AC-AF=6\\sqrt{3}-\\frac{\\sqrt{3}}{2}AE$,\\\\\ntherefore the area $S_{\\text{rectangle CDEF}}=CF\\cdot EF=\\frac{1}{2}x\\left(6\\sqrt{3}-\\frac{\\sqrt{3}}{2}x\\right)=-\\frac{\\sqrt{3}}{4}x^2+3\\sqrt{3}x$;\\\\\nThe range of $x$ is \\boxed{0<x<12}.",
    "image_path": "dataset/mm_math/images/MM_Math/52602949_91.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $C$ is on the line segment $AB$. Through point $B$, a line $l \\perp AB$ is drawn, and point $P$ is a point on line $l$. Connect $AP$, and point $Q$ is the midpoint of $AP$. Draw $QR \\perp AB$ with the foot of the perpendicular being $R$. Connect $CQ$, with $AC=2$, $BC=1$, and $CQ=1$. What is the length of $CR$?\n",
    "completion": "\\textbf{Solution:} Given that, $QR\\perp AB$ and $l\\perp AB$, \\\\\n$\\therefore QR\\parallel BP$, \\\\\n$\\therefore \\frac{AQ}{AP}=\\frac{AR}{AB}$, \\\\\nAs point Q is the midpoint of AP, \\\\\n$\\therefore \\frac{AQ}{AP}=\\frac{AR}{AB}=\\frac{1}{2}$, \\\\\nAlso given that $AC=2$ and $BC=1$, \\\\\n$\\therefore AB=3$, \\\\\n$\\therefore AR=\\frac{3}{2}$, \\\\\n$\\therefore CR=AC-AR=\\boxed{\\frac{1}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51545341_176.png"
  },
  {
    "solution": "\\boxed{90^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is a straight line, $OD$ and $OE$ are respectively the bisectors of $\\angle AOC$ and $\\angle BOC$. Given $\\angle BOC=72^\\circ 20'$, find the measures of $\\angle 1$, $\\angle 2$, and $\\angle DOE$.\n",
    "completion": "\\textbf{Solution}: Since $AB$ is a straight line, and $OD, OE$ are the bisectors of $\\angle AOC, \\angle BOC$ respectively, and $\\angle BOC=72^\\circ 20'$,\\\\\nthus $\\angle 1=\\angle EOB=\\frac{1}{2}\\angle BOC=36^\\circ 10'$,\\\\\nthus $\\angle DOC=\\angle AOD=\\frac{1}{2}\\angle AOC=\\frac{1}{2}(180^\\circ-\\angle BOC)=\\frac{1}{2}(180^\\circ-72^\\circ 20')=53^\\circ 50'$\\\\\nthus $\\angle DOE=\\angle 1+\\angle 2=36^\\circ 10'+53^\\circ 50'=\\boxed{90^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53403641_155.png"
  },
  {
    "solution": "\\boxed{8\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the rhombus $ABCD$, it is known that $\\angle A : \\angle B = 1 : 2$, and the perimeter of the rhombus is $16$. What is the area of the rhombus?\n",
    "completion": "\\textbf{Solution:} As shown in the figure: \\\\\nDraw DE$\\perp$BC at point E through point D,\\\\\nSince in rhombus ABCD, the perimeter is 16,\\\\\nTherefore, AD=AB=4, $AB\\parallel CD$,\\\\\nTherefore, $\\angle A + \\angle B = 180^\\circ$,\\\\\nSince the ratio $\\angle A\\colon\\angle B=1\\colon2$,\\\\\nTherefore, $\\angle A = 60^\\circ$,\\\\\nTherefore, $\\angle ADE = 30^\\circ$,\\\\\nTherefore, AE=$\\frac{1}{2}AD=2$,\\\\\nTherefore, DE=$\\sqrt{AD^{2}-AE^{2}}=2\\sqrt{3}$,\\\\\nTherefore, the area $S$ of rhombus ABCD is $AB\\cdot DE=\\boxed{8\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53498119.png"
  },
  {
    "solution": "\\boxed{19^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given that $AB$ is the diameter of $\\odot O$, and the chord $CD$ intersects $AB$ at point $E$. A tangent to $\\odot O$ at point $C$ intersects the extension of $BA$ at point $P$, with $\\angle BPC=38^\\circ$. Connect $OD$. If $D$ is the midpoint of $\\overset{\\frown}{AB}$, what is the measure of $\\angle ODC$?\n",
    "completion": "\\textbf{Solution:} As shown in Figure 1, connect $OC$, \\\\\nsince $D$ is the midpoint of $\\overset{\\frown}{AB}$,\\\\\nit follows that $\\overset{\\frown}{AD}=\\overset{\\frown}{BD}$,\\\\\nthus $\\angle AOD=\\angle BOD=\\frac{1}{2}\\angle AOB=90^\\circ$,\\\\\nsince $PC$ is the tangent to the circle $\\odot O$ at point $C$,\\\\\nit follows that $\\angle PCO=90^\\circ$,\\\\\nthus $\\angle POC=90^\\circ-\\angle BPC=90^\\circ-38^\\circ=52^\\circ$,\\\\\ntherefore, $\\angle COD=\\angle POC+\\angle AOD=52^\\circ+90^\\circ=142^\\circ$,\\\\\nsince $OC=OD$,\\\\\nit follows that $\\angle OCD=\\angle ODC$,\\\\\nthus $\\angle OCD=\\angle ODC=\\frac{180^\\circ-\\angle COD}{2}=\\frac{180^\\circ-142^\\circ}{2}=19^\\circ$\\\\\nhence, $\\angle ODC=\\boxed{19^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52720239_162.png"
  },
  {
    "solution": "\\boxed{64^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within circle $\\odot O$, diameter $AB \\perp CD$, and $\\angle A=26^\\circ$, what is the degree measure of the inscribed angle subtended by chord $AC$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram: Connect OC, BC, take a point E on arc AC, and connect AE, CE,\\\\\n$\\because OA=OC$,\\\\\n$\\therefore \\angle AOC=180^\\circ-2\\times 26^\\circ=128^\\circ$,\\\\\n$\\therefore \\angle B=\\frac{1}{2}\\angle AOC=\\frac{1}{2}\\times 128^\\circ=64^\\circ$,\\\\\n$\\therefore \\angle E=180^\\circ-\\angle B=116^\\circ$,\\\\\n$\\therefore$ the measure of the inscribed angle facing chord AC is $\\boxed{64^\\circ}$ or $116^\\circ$.",
    "image_path": "dataset/mm_math/images/MM_Math/52871736.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{5}}{3}a}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, ABCD is a square with a side length of \\(a\\). Point \\(P\\) is on the diagonal \\(AC\\), and line \\(PB\\) is drawn. Line \\(PQ\\) is perpendicular to \\(BP\\) and intersects segment \\(CD\\) at point \\(Q\\). When \\(DQ = 2CQ\\), what is the length of \\(BP\\)?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw PF$\\perp$CD at point F, and PE$\\perp$BC at point E.\\\\\nSince square ABCD has a side length of $a$,\\\\\nit follows that CP bisects $\\angle$BCD, and $\\angle$PEC=$\\angle$ECF=90$^\\circ$,\\\\\ntherefore PE=PF,\\\\\ntherefore quadrilateral PECF is a square,\\\\\nhence EC=CF.\\\\\nSince $\\angle$BPE+$\\angle$EPQ=90$^\\circ$ and $\\angle$EPQ+$\\angle$QPF=90$^\\circ$,\\\\\nit follows that $\\angle$BPE=$\\angle$QPF.\\\\\nSince $\\angle$BPE=$\\angle$QPF and $\\angle$PEB=$\\angle$PFQ=90$^\\circ$ and PE=PF,\\\\\nit follows that $\\triangle$PEB$\\cong$$\\triangle$PFQ,\\\\\ntherefore PB=PQ, and BE=QF.\\\\\nSince DQ=2CQ and CD=$a$,\\\\\nit follows that QD=$\\frac{2}{3}a$ and CQ=$\\frac{1}{3}a$.\\\\\nLet BE=FQ=$m$, then CE=$a-m$, and CF=QC+FQ=$\\frac{1}{3}a+m$,\\\\\nthus $a-m=\\frac{1}{3}a+m$,\\\\\ntherefore $m=\\frac{1}{3}a$,\\\\\nhence BE=$\\frac{1}{3}a$ and PE=CF=$\\frac{2}{3}a$,\\\\\ntherefore BP=$\\sqrt{B{E}^{2}+P{E}^{2}}$=$\\sqrt{(\\frac{1}{3}a{)}^{2}+(\\frac{2}{3}a{)}^{2}}$=$\\boxed{\\frac{\\sqrt{5}}{3}a}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50475082.png"
  },
  {
    "solution": "\\boxed{2<DE<5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=3$, $AC=7$, and $D$, $E$ are the midpoints of $AB$, $AC$, respectively. What could be the length of $DE$?\n",
    "completion": "\\textbf{Solution:} Since $D$ and $E$ are the midpoints of $AB$ and $AC$ respectively, \\\\\nit follows that $DE$ is the midsegment of $\\triangle ABC$, \\\\\nhence $DE=\\frac{1}{2}BC$. \\\\\nIn $\\triangle ABC$, with $AB=3$ and $AC=7$, the triangle inequality theorem implies that $7-3<BC<7+3$, i.e., $4<BC<10$, \\\\\nthus $2<DE<5$. \\\\\nTherefore, the answer is: $\\boxed{2<DE<5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53498146.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the Cartesian coordinate system, the graph of the linear function $y=ax+b$ ($a\\neq 0$) intersects the graph of the inverse proportion function $y=\\frac{k}{x}$ ($k\\neq 0$) at points $A$ and $B$ within the second and fourth quadrants, and intersects the $y$-axis at point $C$. A perpendicular line is drawn from point $A$ to the $y$-axis, with $H$ as the foot, where $OH=3$, $\\tan\\angle AOH =\\frac{4}{3}$, and the coordinates of point $B$ are ($m$, $-2$). Calculate the perimeter of $\\triangle AHO$.\n",
    "completion": "\\textbf{Solution:} Given that $\\because$ OH $=$ 3, $\\tan\\angle AOH =\\frac{4}{3}$, and $AH\\perp$ y-axis\\\\\n$\\therefore$ $\\frac{AH}{OH}=\\frac{4}{3}$ \\\\\n$\\therefore$ $AH=\\frac{4OH}{3}=4$ \\\\\n$\\therefore$ $AO=\\sqrt{AH^{2}+OH^{2}}=5$ \\\\\n$\\therefore$ the perimeter of $\\triangle AHO$ is $AH+OH+AO=4+3+5=\\boxed{12}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51367530_122.png"
  },
  {
    "solution": "\\boxed{y=\\frac{1}{2}x+1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the straight line \\(l\\) is the graph of the linear function \\(y=kx+b\\).\n",
    "completion": "\\textbf{Solution:} Based on the problem, we know that the points $(-2,0)$ and $(2,2)$ lie on the graph of a linear function,\\\\\nSubstituting $(-2,0)$ and $(2,2)$ into $y=kx+b$, we get: $\\begin{cases}-2k+b=0 \\\\ 2k+b=2\\end{cases}$,\\\\\nSolving this, we find: $\\begin{cases}k=\\frac{1}{2} \\\\ b=1\\end{cases}$,\\\\\nTherefore, the equation of the linear function is \\boxed{y=\\frac{1}{2}x+1};",
    "image_path": "dataset/mm_math/images/MM_Math/52423839_226.png"
  },
  {
    "solution": "\\boxed{1.5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\angle AOP=\\angle BOP=15^\\circ$, $PC \\parallel OA$ intersects $OB$ at point C, $PD \\perp OA$ at D, if $PC=3$, then what is the value of $PD$?\n",
    "completion": "\\textbf{Solution:} Draw PE $\\perp$ OB at point E,\\\\\nsince PC $\\parallel$ OA,\\\\\nit follows that $\\angle$CPO = $\\angle$AOP,\\\\\nsince $\\angle$AOP = $\\angle$BOP = $15^\\circ$,\\\\\nit follows that $\\angle$BOP = $\\angle$CPO = $15^\\circ$,\\\\\nthus, $\\angle$BCP = $\\angle$BOP + $\\angle$CPO = $30^\\circ$,\\\\\ntherefore, PE = $\\frac{1}{2}$PC = $\\frac{3}{2}$,\\\\\nsince $\\angle$AOP = $\\angle$BOP, and PE $\\perp$ OB, PD $\\perp$ OA at D,\\\\\nthus, PD = PE = $\\frac{3}{2}$ = \\boxed{1.5}.",
    "image_path": "dataset/mm_math/images/MM_Math/53058544.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that point C is on line segment AB, and points M and N are on line segments AC and BC, respectively, with $AM=2MC$, $BN=2NC$. If $AC=9$ and $BC=6$, what is the length of line segment MN?\n",
    "completion": "\\textbf{Solution:} Given that $AC=9$ and $BC=6$, and also that $AM=2MC$ and $BN=2NC$.\\\\\nTherefore, $MC= \\frac{1}{3} AC=3$ and $NC= \\frac{1}{3} BC=2$,\\\\\nTherefore, $MN=MC+NC=3+2=\\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53206849_47.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, $AD\\parallel BC$, $\\angle B=90^\\circ$, $AB=8\\,cm$, $AD=12\\,cm$, $BC=18\\,cm$. Point $P$ starts moving from point $A$ towards point $D$ at a speed of $1\\,cm/s$; simultaneously, point $Q$ starts moving from point $C$ towards point $B$ at a speed of $2\\,cm/s$. When point $Q$ reaches point $B$, point $P$ also stops moving. Let the time taken for points $P$ and $Q$ to move be $t(s)$. There exists a value of $t$ during the entire motion such that quadrilateral $PQCD$ is a rhombus. Find the value of $t$.\n",
    "completion": "\\textbf{Solution:} Since quadrilateral PQCD is a rhombus, \\\\\ntherefore CQ=CD, \\\\\ntherefore $2t=10$, \\\\\nSolving this, we get $t=\\boxed{5}$. At this time, $DP=AD-AP=12-5=7$, \\\\\ntherefore $DP \\neq CD$, \\\\\ntherefore, the quadrilateral PQCD is not a rhombus.",
    "image_path": "dataset/mm_math/images/MM_Math/51614702_95.png"
  },
  {
    "solution": "\\boxed{\\frac{3+\\sqrt{3}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a set of triangular plates is placed as in Figure 1, with $AB=CD= \\sqrt{6}$, and vertex $E$ coinciding. When $\\triangle DEC$ is rotated around its vertex $E$, as in Figure 2, during the rotation, when $\\angle AED=75^\\circ$, and $AD$ and $BC$ are connected, what is the area of $\\triangle ADE$?\n",
    "completion": "\\textbf{Solution:} As shown in Figure 2, draw $EG \\parallel CD$ through point E, extend CE to meet AB at point H, \\\\\n$\\therefore \\angle EDC=\\angle DEG$, \\\\\nGiven that, we know: $\\angle EDC=30^\\circ$, $\\angle DEC=60^\\circ$, $\\angle ECD=\\angle AEB=90^\\circ$, $\\angle ABE=\\angle BAE=45^\\circ$, \\\\\n$\\because \\angle AED=75^\\circ$, \\\\\n$\\therefore \\angle AEG=\\angle AED-\\angle DEG=75^\\circ-30^\\circ=45^\\circ$, \\\\\n$\\therefore EG\\parallel AB\\parallel CD$, \\\\\n$\\therefore HC\\perp AB$, \\\\\n$\\therefore HE=BH=AH=\\frac{1}{2}AB$, \\\\\nAnd since $AB=CD$, \\\\\n$\\therefore$ Quadrilateral ABCD is a parallelogram, \\\\\n$\\because AB=CD=\\sqrt{6}$, \\\\\n$\\therefore HE=BH=AH=\\frac{\\sqrt{6}}{2}$, EC=$\\frac{CD}{\\sqrt{3}}=\\frac{\\sqrt{6}}{\\sqrt{3}}=\\sqrt{2}$, \\\\\nHC=HE+EC=$\\frac{\\sqrt{6}}{2}+\\sqrt{2}=\\frac{\\sqrt{6}+2\\sqrt{2}}{2}$, \\\\\n$\\therefore S_{\\triangle ABE}=\\frac{1}{2} \\times \\sqrt{6} \\times \\frac{\\sqrt{6}}{2}=\\frac{3}{2}$, $S_{\\triangle DEC}=\\frac{1}{2} \\times \\sqrt{6} \\times \\sqrt{2}=\\sqrt{3}$, $S_{\\triangle CEB}=\\frac{1}{2} \\times \\sqrt{2} \\times \\frac{\\sqrt{6}}{2}=\\frac{\\sqrt{3}}{2}$, \\\\\n$S_{\\square ABCD}=\\sqrt{6} \\times \\frac{\\sqrt{6}+2\\sqrt{2}}{2}=3+2\\sqrt{3}$, \\\\\n$\\therefore S_{\\triangle ADE}=S_{\\square ABCD}-S_{\\triangle ABE}-S_{\\triangle DEC}-S_{\\triangle CEB}=3+2\\sqrt{3}-\\frac{3}{2}-\\sqrt{3}-\\frac{\\sqrt{3}}{2}=\\boxed{\\frac{3+\\sqrt{3}}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51825223.png"
  },
  {
    "solution": "\\boxed{\\sqrt{2} - 1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within rectangle $ABCD$, point $E$ is on side $AB$, and triangles $\\triangle BEC$ and $\\triangle FEC$ are symmetrical about line $EC$. The symmetric point of point $B$, point $F$, is on side $AD$, with $G$ being the midpoint of $CD$. Connecting $BG$ intersects $CE$ and $CF$ at points $M$ and $N$ respectively. If $BM=BE$ and $MG=2$, what is the length of $BN$? What is the value of $\\sin\\angle AFE$?\n",
    "completion": "\\textbf{Solution:} Given that $BM = BE$, it follows that $\\angle BEM = \\angle BME$. Since $AB \\parallel CD$, we have $\\angle BEM = \\angle GCM$. Moreover, given $\\angle BME = \\angle GMC$, it leads to $\\angle GCM = \\angle GMC$, thus $MG = GC = 2$. Given $G$ is the midpoint of $CD$, it follows that $CD = AB = 4$. Connecting $BF$ and $FM$, by folding, we obtain $\\angle FEM = \\angle BEM$ and $BE = EF$, thus $BM = EF$. Since $\\angle BEM = \\angle BME$, it follows $\\angle FEM = \\angle BME$, thus $EF \\parallel BM$. Consequently, quadrilateral $BEFM$ is a parallelogram. Given $BM = BE$, quadrilateral $BEFM$ is a rhombus. Since $\\angle EBC = \\angle EFC = 90^\\circ$ and $EF \\parallel BG$, it follows $\\angle BNF = 90^\\circ$. Given $BF$ bisects $\\angle ABN$, it follows $FA = FN$. Hence, right triangle $ABF$ is congruent to right triangle $NBF$ (HL), leading to $BN = AB = \\boxed{4}$. Since $FE = FM$, $FA = FN$, and $\\angle A = \\angle BNF = 90^\\circ$, right triangle $AEF$ is congruent to right triangle $NMF$ (HL), leading to $AE = NM$. Let $AE = NM = x$, then $BE = FM = 4 - x$, and $NG = MG - NM = 2 - x$. Since $FM \\parallel GC$, triangles $FMN$ and $CGN$ are similar, leading to $\\frac{CG}{FM} = \\frac{GN}{NM}$, i.e., $\\frac{2}{4-x} = \\frac{2-x}{x}$. Solving this equation gives $x = 4 + 2\\sqrt{2}$ (discard) or $x = 4 - 2\\sqrt{2}$, thus $EF = BE = 4 - x = 2\\sqrt{2}$. Hence, $\\sin\\angle AFE = \\frac{AE}{EF} = \\frac{4 - 2\\sqrt{2}}{2\\sqrt{2}} = \\boxed{\\sqrt{2} - 1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51265327.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{2}−1}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle BAC=90^\\circ$, point $D$ is on line $AC$, connecting $BD$. If $\\angle DBC=45^\\circ$, what is the maximum value of $\\frac{AD}{CD}$?\n",
    "completion": "\\textbf{Solution:} As shown in Figure 1, construct the circumcircle of $\\triangle BCD$, denoted as $\\odot O$. Construct $OE \\perp AC$ at point E, and $OF \\perp AB$ at point F, and connect OB,\\\\\n$\\because \\angle AEO = \\angle AFO = \\angle BAC = 90^\\circ$,\\\\\n$\\therefore$ Quadrilateral OFAE is a rectangle,\\\\\n$\\therefore$ AE=OF,\\\\\n$\\because DE=CE= \\frac{1}{2}CD$,\\\\\n$\\therefore$ CD=2CE,\\\\\nThe lengths of CD, CE remain constant. When OF is at its maximum, then AE is at its maximum, and thus AD is at its maximum,\\\\\n$\\therefore$ the value of $\\frac{AD}{CD}$ is also at its maximum,\\\\\n$\\because OF \\leq OB$,\\\\\n$\\therefore$ when OF=OB, OF is at its maximum,\\\\\nAt this time, point B coincides with point F,\\\\\nAs shown in Figure 2, when point B coincides with point F, connect OD and OC,\\\\\nLet DE=CE=m, then CD=2m,\\\\\n$\\because \\angle DBC=45^\\circ$,\\\\\n$\\therefore \\angle DOC=2\\angle DBC=90^\\circ$,\\\\\n$\\because OD=OC$,\\\\\n$\\therefore OE= \\frac{1}{2}CD=DE=m$,\\\\\n$\\therefore AE=OB=OD= \\sqrt{DE^2+OE^2}=\\sqrt{m^2+m^2}=\\sqrt{2}m$,\\\\\n$\\therefore AD=AE−DE= \\sqrt{2}m−m$,\\\\\n$\\therefore \\frac{AD}{CD}=\\frac{\\sqrt{2}m−m}{2m}=\\boxed{\\frac{\\sqrt{2}−1}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52872337.png"
  },
  {
    "solution": "\\boxed{90^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is a straight line, and $OD$, $OE$ are the angle bisectors of $\\angle AOC$, $\\angle BOC$, respectively. If $\\angle BOC = \\alpha$, what is $\\angle DOE$?\n",
    "completion": "\\textbf{Solution:} Since $AB$ is a straight line, and $OD, OE$ are the bisectors of $\\angle AOC, \\angle BOC$ respectively,\\\\\nthus $\\angle 1=\\angle EOB=\\frac{1}{2}\\angle BOC$\\\\\nthus $\\angle DOC=\\angle AOD=\\frac{1}{2}\\angle AOC$,\\\\\nthus $\\angle DOE=\\angle 1+\\angle 2=\\frac{1}{2}\\angle AOC+\\frac{1}{2}\\angle BOC=90^\\circ=\\boxed{90^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53403641_156.png"
  },
  {
    "solution": "\\boxed{20^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ is rotated $40^\\circ$ clockwise around point C to obtain $\\triangle A'CB'$. If $AC \\perp A'B'$, and $AA'$ is connected, what is the measure of $\\angle AA'B'$?\n",
    "completion": "\\textbf{Solution:} Let the intersection point of $AC$ and ${A}^{\\prime}{B}^{\\prime}$ be O, \\\\\nfrom the property of rotation it follows that $AC={A}^{\\prime}C$, $\\angle {A}^{\\prime}CA=\\angle BC{B}^{\\prime}=40^\\circ$, \\\\\n$\\therefore$ $\\angle CA{A}^{\\prime}=\\angle C{A}^{\\prime}A=\\frac{180^\\circ-\\angle {A}^{\\prime}CA}{2}=70^\\circ$, \\\\\n$\\because$ $AC\\perp {A}^{\\prime}{B}^{\\prime}$, \\\\\n$\\therefore$ $\\angle {A}^{\\prime}OC=90^\\circ$, \\\\\n$\\therefore$ $\\angle C{A}^{\\prime}O=180^\\circ-\\angle {A}^{\\prime}CO-\\angle {A}^{\\prime}OC=50^\\circ$, \\\\\n$\\therefore$ $\\angle A{A}^{\\prime}{B}^{\\prime}=\\angle C{A}^{\\prime}A-\\angle C{A}^{\\prime}O=\\boxed{20^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51654675.png"
  },
  {
    "solution": "\\boxed{9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given that the length of segment \\(AB=a\\) (as shown in the diagram), where \\(C\\) is a point on the extension of \\(AB\\) in the opposite direction, and \\(AC=\\frac{1}{3}AB\\), \\(D\\) is the midpoint of segment \\(BC\\). If \\(AD=3\\text{cm}\\), find the value of \\(a\\).\n",
    "completion": "\\textbf{Solution}: Since $AC= \\frac{1}{3}a$ and $AD=3\\text{cm}$,\\\\\nit follows that $CD= \\frac{1}{3}a+3$,\\\\\nwhich implies $\\frac{1}{3}a+3= \\frac{2}{3}a$.\\\\\nSolving for $a$ yields: $a=\\boxed{9}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51495344_111.png"
  },
  {
    "solution": "\\boxed{\\sqrt{3}+1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, the line segment $AC$ is rotated $60^\\circ$ counterclockwise around point $C$ to the line segment $CD$, where point $D$ falls exactly on side $AB$. Point $E$ is a point on side $BC$, and $AE$ and $DE$ are connected, with $\\angle AEC=\\angle BED=60^\\circ$. Given that $AC=\\sqrt{6}$ and $\\angle BCD=15^\\circ$, what is the length of $AE$?\n",
    "completion": "\\textbf{Solution:} Construct $CH\\perp AE$ at point H,\n\n$\\because$ rotating line segment AC about point C by $60^\\circ$ counterclockwise to line segment CD,\n\n$\\therefore AC= CD$, $\\angle ACD=60^\\circ$,\n\n$\\therefore \\triangle ACD$ is an equilateral triangle, thus $\\angle CAD=\\angle CDA=60^\\circ$, $AD=AC=CD$,\n\n$\\because \\angle BCD=15^\\circ$, $\\angle AEC=\\angle BED=60^\\circ$,\n\n$\\therefore \\angle EAD=180^\\circ-\\angle CDA-\\angle AQD=180^\\circ-\\angle CEA-\\angle CQE=\\angle ECQ=15^\\circ$,\n\n$\\therefore \\angle CAH=\\angle CAD-\\angle EAD =45^\\circ$,\n\n$\\because AC= \\sqrt{6}$, $\\angle CAH=45^\\circ$, $CH\\perp AE$,\n\n$\\therefore CH=AH= \\frac{\\sqrt{2}}{2}AC= \\sqrt{3}$,\n\n$\\because \\angle CEH=60^\\circ$,\n\n$\\therefore EH= \\frac{CH}{\\tan 60^\\circ}=1$,\n\n$\\therefore AE= AH+ EH= \\sqrt{3}+1=\\boxed{\\sqrt{3}+1}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51737317_120.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, where $AB=4$ and $BC=5$, point $E$ is a point on $AB$. The rectangle is folded along $DE$ so that side $BC$ exactly passes through point $A$. What is the length of $BE$?\n",
    "completion": "\\textbf{Solution:} In rectangle $ABCD$, fold the rectangle along $DE$ such that side $BC$ exactly passes through point $A$,\\\\\n$\\therefore \\angle C'=\\angle C=90^\\circ$, $C'D=CD=AB=4$, $AD=BC=5$, $\\angle B=\\angle B'=90^\\circ$,\\\\\nIn $\\triangle AC'D$, we have $AC'=\\sqrt{AD^2-C'D^2}=\\sqrt{5^2-4^2}=3$,\\\\\nLet $BE=x$, then $AE=AB-BE=4-x$, $AB'=B'C'-AC'=5-3=2$,\\\\\nIn $\\triangle AB'E$, we have $AE^2=AB'^2+B'E^2$,\\\\\nwhich gives $(4-x)^2=2^2+x^2$,\\\\\nsolving for $x$ yields $x=\\boxed{\\frac{3}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52160165.png"
  },
  {
    "solution": "\\boxed{\\left(-3, -8\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: The line $y=-\\frac{1}{3}x+1$ intersects the x-axis and the y-axis at points A and B, respectively. Does there exist a point Q on the line BG such that the triangle formed by points A, B, and Q is similar to $\\triangle AOB$? If such a point exists, find the coordinates of point Q; if it does not exist, explain why.\n",
    "completion": "\\textbf{Solution:} Let the equation of line BG be $y=kx+b$, then\n\\[\n\\left\\{\n\\begin{array}{l}\n7=2k+b \\\\\nb=1\n\\end{array}\n\\right.\n\\]\nSolving gives\n\\[\n\\left\\{\n\\begin{array}{l}\nk=3 \\\\\nb=1\n\\end{array}\n\\right.\n\\]\nThus, the equation of line BG is $y=3x+1$. Let point Q be $(m,3m+1)$,\n\n(1) When $\\triangle ABQ \\sim \\triangle AOB$,\nthen $\\frac{AB}{AO}=\\frac{BQ}{OB}$, i.e., $\\frac{\\sqrt{10}}{3}=\\frac{\\sqrt{m^2+(3m+1-1)^2}}{1}$, solving gives $m=\\pm \\frac{1}{3}$,\n$\\therefore$ $Q_1\\left(\\frac{1}{3}, 2\\right)$, $Q_2\\left(-\\frac{1}{3}, 0\\right)$\n\n(2) When $\\triangle ABQ \\sim \\triangle BOA$,\n$\\frac{AB}{OB}=\\frac{BQ}{AO}$, i.e., $\\frac{\\sqrt{10}}{1}=\\frac{\\sqrt{m^2+(3m+1-1)^2}}{3}$, solving gives $m=\\pm3$,\n$\\therefore$ $Q_3\\left(3, 10\\right)$, $Q_4\\left(-3, -8\\right)$\n\nHence, the coordinates of point Q are $\\boxed{\\left(\\frac{1}{3}, 2\\right)}$ or $\\boxed{\\left(-\\frac{1}{3}, 0\\right)}$ or $\\boxed{\\left(3, 10\\right)}$ or $\\boxed{\\left(-3, -8\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51051573_104.png"
  },
  {
    "solution": "\\boxed{\\frac{16}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, the quadrilateral $ABCD$ is a square, and point $G$ is any point on side $BC$. $DE \\perp AG$ at point $E$, and $BF \\parallel DE$ intersecting $AG$ at point $F$. $DE = AF$; If $AB=4$ and $BG=3$, what is the length of $AF$?\n",
    "completion": "\\textbf{Solution:} In right triangle $\\text{Rt}\\triangle ABG$, $AB=4$, $BG=3$. According to the Pythagorean theorem, $AG=5$.\\\\\n$\\because \\frac{1}{2}AB \\cdot BG = \\frac{1}{2}AG \\cdot BF$,\\\\\n$\\therefore BF = \\frac{AB \\cdot BG}{AG} = \\frac{12}{5}$,\\\\\nIn right triangle $\\text{Rt}\\triangle ABF$, $AF = \\sqrt{AB^2 - BF^2} = \\boxed{\\frac{16}{5}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53313073_127.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point P is a point on the diagonal BD of the rhombus ABCD. Line CP is extended to intersect the extension of BA at point F and AD at point E. $\\angle DCP=\\angle DAP$; if $PE=4$ and $EF=5$, find the length of the line segment PA.\n",
    "completion": "\\textbf{Solution:} Since $\\triangle APD \\cong \\triangle CPD$,\\\\\n$\\therefore \\angle DAP = \\angle DCP$,\\\\\n$\\because CD \\parallel AB$,\\\\\n$\\therefore \\angle DCF = \\angle DAP = \\angle CFB$,\\\\\nAlso, $\\because \\angle FPA = \\angle FPA$,\\\\\n$\\therefore \\triangle APE \\sim \\triangle FPA$,\\\\\n$\\therefore \\frac{AP}{PF} = \\frac{PE}{PA}$,\\\\\n$\\therefore PA^{2} = PE \\cdot PF$,\\\\\n$\\because PE = 4, EF = 5$,\\\\\n$\\therefore PF = PE + EF = 9$,\\\\\n$\\therefore PA^{2} = PE \\cdot PF = 4 \\times 9 = 36$,\\\\\n$\\therefore PA = \\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51322304_125.png"
  },
  {
    "solution": "\\boxed{11}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given: In $\\triangle ABC$, $AB=AC$, the perpendicular bisector of $AB$, $DE$, intersects $AB$ and $AC$ at $D$ and $E$ respectively. If $AD=3$ and $BC=5$, then what is the perimeter of $\\triangle BEC$?\n",
    "completion": "\\textbf{Solution:} Since $AB=AC$, and $DE$ is the perpendicular bisector of segment $AB$,\\\\\nit follows that $AD=BD$, and $AE=BE$,\\\\\nsince $AD=3$,\\\\\nthen $AB=AC=2AD=6$,\\\\\nsince $BC=5$,\\\\\nthus the perimeter of $\\triangle BEC$ is $BC+BE+EC=BC+AE+EC=5+6=\\boxed{11}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52655046.png"
  },
  {
    "solution": "\\boxed{x>4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given the graph of the linear function $y=kx-b$ as shown in the figure, what is the solution set of the inequality $kx-1>b$ with respect to $x$?\n",
    "completion": "\\textbf{Solution:} Given $\\because kx-1>b$,\\\\\n$\\therefore kx-b>1$,\\\\\n$\\therefore$ the graph of the linear function lies above the line $y=1$. Observing the graph, we find that $x>4$,\\\\\n$\\therefore$ the solution set for the inequality $kx-1>b$ in terms of $x$ is $\\boxed{x>4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52344437.png"
  },
  {
    "solution": "\\boxed{13.5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given, as shown in the graph, the parabola $y=ax^{2}+3ax+c$ ($a>0$) intersects with the y-axis at point C, and intersects with the x-axis at points A and B, where point A is to the left of point B. The coordinates of point B are $(1, 0)$, and $OC=3OB$. If point D is a moving point on the parabola below line segment AC, what is the maximum area of quadrilateral AOCD?\n",
    "completion": "\\textbf{Solution:} As shown in Figure 1: draw line DE through point D parallel to y-axis, intersecting AC at point E.\n\nSince $x=-\\frac{b}{2a}=\\frac{-\\frac{9}{4}}{2\\times \\frac{3}{4}}=-\\frac{3}{2}$, B(1,0),\n\ntherefore, A(-4,0).\n\nThus, AB=5.\n\nTherefore, $S_{\\triangle ABC}=\\frac{1}{2}$AB$\\cdot$OC$=\\frac{1}{2}\\times 5\\times 3=7.5$.\n\nSuppose the equation of AC is $y=kx+b$.\n\nSince substituting A(-4,0) and C(0,-3) gives $\\begin{cases}-4k+b=0\\\\ b=-3\\end{cases}$, we solve to get $\\begin{cases}k=-\\frac{3}{4}\\\\ b=-3\\end{cases}$,\n\nTherefore, the equation of line AC is $y=-\\frac{3}{4}x-3$.\n\nLet D(a, $\\frac{3}{4}a^{2}+\\frac{9}{4}a-3$), then E(a, $-\\frac{3}{4}a-3$).\n\nSince DE$=-\\frac{3}{4}a-3-\\left(\\frac{3}{4}a^{2}+\\frac{9}{4}a-3\\right)=-\\frac{3}{4}(a+2)^{2}+3$,\n\nTherefore, when $a=-2$, DE reaches its maximum value of 3.\n\nTherefore, the maximum area of $\\triangle ADC =\\frac{1}{2}$DE$\\cdot$AO$=\\frac{1}{2}\\times 3\\times 4=6$.\n\nTherefore, $S_{\\text{quadrilateral }ABCD}$=$S_{\\triangle ABC}$+$S_{\\triangle ACD}$=7.5+6=\\boxed{13.5},\n\nThus, the maximum area of quadrilateral ABCD is \\boxed{13.5}.",
    "image_path": "dataset/mm_math/images/MM_Math/53080621_38.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{5}}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $O$ is the circumcircle of $\\triangle ABC$, and $AD$ is the diameter of $O$ with $CD$ connected. If $AD=3$ and $AC=2$, what is the value of $\\cos B$?\n",
    "completion": "\\textbf{Solution:} Given that $AD$ is the diameter of circle $O$,\\\\\nit follows that $\\angle ACD=90^\\circ$,\\\\\nand therefore $DC=\\sqrt{AD^{2}-CD^{2}}=\\sqrt{3^{2}-2^{2}}=\\sqrt{5}$,\\\\\nGiven that arc $AC$ equals arc $AC$,\\\\\nit implies that $\\angle B = \\angle D$,\\\\\nthus $\\cos B = \\cos D = \\frac{DC}{AD} = \\frac{\\sqrt{5}}{3}$.\\\\\nHence, $\\cos B = \\boxed{\\frac{\\sqrt{5}}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50951659.png"
  },
  {
    "solution": "\\boxed{P_4\\left(-1, -2\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the parabola $y=-x^2+bx+c$ passes through points A, B ($-3$, $0$), and C ($0$, $3$). Let point P be a moving point on the axis of symmetry of the parabola. Directly write the coordinates of point P that make $\\triangle BPC$ a right-angled triangle.\n",
    "completion": "\\textbf{Solution:} The coordinates of point P that make $\\triangle BPC$ a right-angled triangle are $\\boxed{P_1(-1, \\frac{3+\\sqrt{17}}{2})}$, $\\boxed{P_2\\left(-1, \\frac{3-\\sqrt{17}}{2}\\right)}$, $\\boxed{P_3\\left(-1, 4\\right)}$, and $\\boxed{P_4\\left(-1, -2\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52602225_174.png"
  },
  {
    "solution": "\\boxed{19}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AD\\perp BC$ and $CE\\perp AB$ with perpendicular feet at points D and E respectively, $AD$ and $CE$ intersect at point H, and $AE=CE$. Since $\\triangle BEC\\cong \\triangle HEA$, if $BE=8$ and $CH=3$, what is the length of the line segment $AB$?\n",
    "completion": "Solution: From Equation (1), we know that $\\triangle BEC \\cong \\triangle HEA$,\\\\\nhence, BE = EH = 8,\\\\\nsince CH = 3,\\\\\nthus, CE = AE = 3 + 8 = 11,\\\\\ntherefore, AB = AE + BE = 11 + 8 = \\boxed{19}.",
    "image_path": "dataset/mm_math/images/MM_Math/52905611_136.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the two diagonals of rectangle $ABCD$ intersect at point $O$. If $AD=\\sqrt{3}$ and $CD=1$, then what is the degree measure of $\\angle ADB$?\n",
    "completion": "\\textbf{Solution:} Given a rectangle ABCD,\\\\\nit follows that $\\angle BCD=90^\\circ$, AD$\\parallel$BC, and AD=BC.\\\\\nHence, $\\angle ADB=\\angle DBC$.\\\\\nThus, $BD=\\sqrt{BC^{2}+CD^{2}}=\\sqrt{(\\sqrt{3})^{2}+1^{2}}=2$.\\\\\nThis means BD=2CD,\\\\\nand so $\\angle ADB=\\angle DBC=30^\\circ$.\\\\\nTherefore, the answer is: $\\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52251402.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AC=5$, $\\cot A=2$, $\\cot B=3$. Point D is on side $AB$, and $\\angle BDC=45^\\circ$. What is the length of segment $BD$?\n",
    "completion": "\\textbf{Solution:}\n\nConstruct $CE\\perp AB$ at point E,\n\nsince $\\angle BDC=45^\\circ$,\n\ntherefore $\\angle DCE=90^\\circ-\\angle CDE=45^\\circ=\\angle CDE$,\n\ntherefore $DE=CE$,\n\nsince $AC=5$, and $\\cot A=2$,\n\nlet $CE=x$, and $\\cot A=\\frac{AE}{CE}=2$,\n\ntherefore, $AE=2CE=2x$,\n\nin $\\triangle ACE$, according to Pythagoras' theorem $AC^2=CE^2+AE^2$ hence $5^2=x^2+(2x)^2$,\n\nsolving for $x$ gives $x=\\sqrt{5}$,\n\ntherefore, $DE=CE=\\sqrt{5}$,\n\nin $\\triangle CEB$, $\\cot B=\\frac{EB}{CE}=3$,\n\ntherefore, $BE=3CE=3\\sqrt{5}$,\n\ntherefore, $BD=DE+BE=\\sqrt{5}+3\\sqrt{5}=4\\sqrt{5}$;\n\nHence, the length of segment $BD$ is $BD=\\boxed{4\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51353456_90.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AD \\parallel EB \\parallel FC$, $AB = 2BC$, $DF = 6$. What is the length of $DE$?\n",
    "completion": "\\textbf{Solution}: Since $AD\\parallel EB\\parallel FC$ and $AB=2BC$,\\\\\nit follows that $\\frac{DE}{EF}=\\frac{AB}{BC}$. Therefore, $\\frac{DE}{EF}=\\frac{2BC}{BC}=2$,\\\\\nGiven $DF=DE+EF=6$,\\\\\nit results in $DE=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52977733.png"
  },
  {
    "solution": "\\boxed{5\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the rhombus $ABCD$, $AB=BD=5$, find: What is the length of $AC$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rhombus,\\\\\nit follows that OA = OC, BO = OD = $\\frac{1}{2}$BD = $\\frac{5}{2}$, and BD$\\perp$AC,\\\\\nIn $\\triangle AOB$, OA = $\\sqrt{AB^{2}-BO^{2}}=\\frac{5\\sqrt{3}}{2}$,\\\\\ntherefore, AC = $2\\text{OA}=\\boxed{5\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53359704_117.png"
  },
  {
    "solution": "\\boxed{5\\sqrt{13}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in parallelogram $ABCD$, the parallelogram is folded such that point $D$ falls on point $F$ on side $BC$, with the crease being $CE$. Connect $EF$, $CE$, $DF$, where $CE$ and $DF$ intersect at point $G$, and connect $BG$. Quadrilateral $DEFC$ is a rhombus. If $AB=10$, $AD=20$, and $\\angle ABC=60^\\circ$, find the length of $BG$.\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, draw $GH\\perp BC$ at point $H$ passing through $G$,\\\\\nsince quadrilateral $ABCD$ is a parallelogram and $\\angle ABC=60^\\circ$,\\\\\ntherefore, $\\angle ADC=60^\\circ$.\\\\\nBecause quadrilateral $DEFC$ is a rhombus and $AB=10$,\\\\\nthus $AB=CD=10$, $DF\\perp CE$, $\\angle FDC=30^\\circ$, and $\\angle ECF=60^\\circ$,\\\\\nhence $CG=\\frac{1}{2}CD=5$, $GH=CG\\cdot \\sin 60^\\circ=\\frac{\\sqrt{3}}{2}\\times 5=\\frac{5\\sqrt{3}}{2}$,\\\\\ntherefore, $CH=\\frac{1}{2}CG=\\frac{5}{2}$.\\\\\nBecause quadrilateral $ABCD$ is a parallelogram,\\\\\ntherefore $BC=AD=20$,\\\\\nthus $BH=BC-CH=20-\\frac{5}{2}=\\frac{35}{2}$,\\\\\ntherefore, $BG=\\sqrt{GH^2+BH^2}=\\sqrt{\\left(\\frac{5\\sqrt{3}}{2}\\right)^2+\\left(\\frac{35}{2}\\right)^2}=\\boxed{5\\sqrt{13}}$",
    "image_path": "dataset/mm_math/images/MM_Math/53308271_110.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in rhombus ABCD with an area of 24, and $AC+BD=14$, what is the length of AB?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, AC intersects BD at point O, \\\\\n$\\because$ ABCD is a rhombus with an area of 24, \\\\\n$\\therefore$ AC $\\perp$ BD, and AC$\\cdot$BD = 48, \\\\\nAlso, since AC + BD = 14, \\\\\nLet AC = a, then BD = 14 − a, \\\\\n$\\therefore$ a(14−a) = 48, \\\\\nUpon rearrangement and solving, we get: a = 8 or a = 6 (which does not fit the diagram, hence discarded), \\\\\n$\\therefore$ AC = 8, BD = 6, \\\\\n$\\therefore$ AO = 4, BO = 3, \\\\\n$\\therefore$ AB = $\\sqrt{AO^{2}+BO^{2}}$ = $\\sqrt{4^{2}+3^{2}}$ = \\boxed{5}.",
    "image_path": "dataset/mm_math/images/MM_Math/52076985.png"
  },
  {
    "solution": "\\boxed{56^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, \\(AB\\) is the diameter of the circle \\(\\odot O\\), and \\(CD\\) is a chord. If \\(\\angle BCD=34^\\circ\\), then what is the measure of \\(\\angle ABD\\)?\n",
    "completion": "\\textbf{Solution:} Given that AB is the diameter of circle O, \\\\\nthus, $\\angle ADB=90^\\circ$, \\\\\ntherefore, $\\angle DAB + \\angle ABD = 90^\\circ$, \\\\\nsince $\\angle DAB = \\angle BCD = 34^\\circ$, \\\\\nthus, $\\angle ABD = 90^\\circ - 34^\\circ = \\boxed{56^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51402603.png"
  },
  {
    "solution": "\\boxed{3\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in square $ABCD$, points $E$ and $F$ are the midpoints of $AD$ and $DC$, respectively. Through point $C$, construct $CM\\perp AB$ to meet the extension of $AB$ at $M$. Connect $EF$. If $CD=4, BM=2, CM=6$, then what is the length of $EF$?\n",
    "completion": "\\textbf{Solution:} As illustrated in the diagram, connect AC,\n\n$\\because$ Quadrilateral ABCD is a parallelogram,\n\n$\\therefore$ AB=CD=4,\n\n$\\therefore$ AM=AB+BM=4+2=6,\n\n$\\because$ CM$\\perp$AB, CM=6,\n\n$\\therefore$ AC=$\\sqrt{AM^{2}+CM^{2}}=\\sqrt{6^{2}+6^{2}}=6\\sqrt{2}$,\n\n$\\because$ Points E and F are the midpoints of AD and CD, respectively,\n\n$\\therefore$ EF is the midline of $\\triangle ADC$,\n\n$\\therefore$ EF=$\\frac{1}{2}$AC=$\\boxed{3\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52284940.png"
  },
  {
    "solution": "\\boxed{\\left(-\\frac{3}{4}, \\frac{9}{4}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system $xOy$, $AB=2\\sqrt{10}$. Connect $AB$ and extend it to $C$, also connect $OC$. If it satisfies $OC^2=BC\\cdot AC$ and $\\tan\\alpha =3$, what are the coordinates of point $C$?\n",
    "completion": "\\textbf{Solution:}\n\nGiven that\n\\[ OC^2 = BC \\cdot AC, \\]\nit follows that\n\\[ \\frac{OC}{BC} = \\frac{AC}{OC}, \\]\nand given that\n\\[ \\angle C = \\angle C, \\]\nit implies\n\\[ \\triangle OBC \\sim \\triangle AOC, \\]\ntherefore\n\\[ \\angle A = \\angle COB, \\]\nGiven that\n\\[ \\alpha + \\angle COB = 90^\\circ, \\quad \\angle A + \\angle ABO = 90^\\circ, \\]\nit follows that\n\\[ \\angle ABO = \\alpha, \\]\nGiven that\n\\[ \\tan \\alpha = 3, \\]\nit follows that\n\\[ \\tan \\angle ABO = \\frac{OA}{OB} = 3, \\]\ntherefore\n\\[ OA = 3OB, \\]\nGiven that\n\\[ AB = 2\\sqrt{10}, \\]\nby the Pythagorean theorem, we have\n\\[ OA^2 + OB^2 = AB^2, \\]\nwhich implies\n\\[ (3OB)^2 + OB^2 = (2\\sqrt{10})^2, \\]\nSolving this, we get\n\\[ OB = 2, \\]\ntherefore\n\\[ OA = 6, \\]\ntherefore\n\\[ \\tan \\angle A = \\frac{OB}{OA} = \\frac{1}{3}, \\]\nAs shown in the figure, draw $CD \\perp x$-axis at point $D$.\n\nGiven that\n\\[ \\tan \\alpha = 3, \\]\nlet $C(-m, 3m)$, with $m > 0$.\n\nTherefore\n\\[ AD = 6 + m, \\]\nGiven that\n\\[ \\tan \\angle A = \\frac{1}{3}, \\]\nit follows that\n\\[ \\frac{CD}{AD} = \\frac{1}{3}, \\]\nleading to\n\\[ \\frac{3m}{6+m} = \\frac{1}{3}, \\]\nSolving this, we get\n\\[ m = \\frac{3}{4}, \\]\nUpon checking, $m = \\frac{3}{4}$ is a solution of the original equation.\n\nTherefore\n\\[ -m = -\\frac{3}{4}, \\quad 3m = \\frac{9}{4}, \\]\nThus, the coordinates of point $C$ are\n\\[ \\boxed{\\left(-\\frac{3}{4}, \\frac{9}{4}\\right)}. \\]",
    "image_path": "dataset/mm_math/images/MM_Math/52797163.png"
  },
  {
    "solution": "\\boxed{12^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $P$ is outside circle $\\odot O$, line $PA$ intersects circle $\\odot O$ at point $C$, line $PB$ intersects circle $\\odot O$ at point $D$, if $\\overset{\\frown}{AB}=80^\\circ$, $\\angle P=28^\\circ$, then what is the degree measure of $\\angle CAD$?\n",
    "completion": "\\textbf{Solution:} Connect OA and OB, as shown in the figure,\\\\\n$\\because$ The degree of $\\overset{\\frown}{AB}$ is $80^\\circ$,\\\\\n$\\therefore$ $\\angle AOB=80^\\circ$,\\\\\n$\\therefore$ $\\angle ADB= \\frac{1}{2}\\angle AOB=40^\\circ$,\\\\\n$\\because$ $\\angle ADB=\\angle PAD+\\angle P$,\\\\\n$\\therefore$ $\\angle PAD=40^\\circ-28^\\circ=12^\\circ$,\\\\\nthat is, the degree of $\\angle CAD$ is $\\boxed{12^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52872370.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AC$ and $BD$ are the diagonals of the quadrilateral $ABCD$, points $E$ and $F$ are the midpoints of $AD$ and $BC$ respectively, and points $M$ and $N$ are the midpoints of $AC$ and $BD$ respectively. Connecting $EM$, $MF$, $FN$, $NE$ in sequence, if $AB=CD=2$, then what is the perimeter of the quadrilateral $ENFM$?\n",
    "completion": "\\textbf{Solution:} Given that points E and F are the midpoints of $AD$ and $BC$ respectively, and points M and N are the midpoints of $AC$ and $BD$ respectively, \\\\\n$\\therefore EN$, $MF$, $NF$, $ME$ are the medians of $\\triangle ABD$, $\\triangle ABC$, $\\triangle BCD$, $\\triangle ADC$ respectively, \\\\\n$\\because AD=CD=2$, \\\\\n$\\therefore EN=MF=\\frac{1}{2}AB=1$, $NF=ME=\\frac{1}{2}CD=1$, \\\\\nThe perimeter of quadrilateral $ENFM$ equals $EN+MF+NF+ME=1+1+1+1=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52395430.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB \\parallel CD$, and $AC$ and $BD$ intersect at point $O$. If $AB = 6$, $CD = 3$, and $DO = 4$, what is the length of $BO$?\n",
    "completion": "\\textbf{Solution:} Since $AB\\parallel CD$, \\\\\nit follows that $\\triangle ABO\\sim \\triangle CDO$, \\\\\nwhich implies $\\frac{OB}{OD}=\\frac{AB}{CD}$, \\\\\nthus $\\frac{OB}{4}=\\frac{6}{3}$, \\\\\nsolving yields: $OB=\\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52948354.png"
  },
  {
    "solution": "\\boxed{50}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points E and F are inside the square ABCD, with AE=CF=4, EF=6, and $\\angle E=\\angle F=90^\\circ$. What is the area of square ABCD?\n",
    "completion": "\\textbf{Solution:} Join AC and intersect EF at point O, as shown in the diagram: \\\\\n$\\because$ $\\angle E=\\angle F=90^\\circ$ and $\\angle AOE=\\angle COF$, AE=CF, \\\\\n$\\therefore$ $\\triangle AEO \\cong \\triangle CFO$, \\\\\n$\\therefore$ AO=OC, EO=OF, \\\\\n$\\because$ EF=6, \\\\\n$\\therefore$ EO=OF=3, \\\\\n$\\because$ AE=CF=4, \\\\\nIn $\\triangle AOE$, according to the Pythagorean theorem, we get OA=5, \\\\\n$\\therefore$ AC=10, \\\\\nIn the square ABCD, AB=BC and $\\angle ABC=90^\\circ$, \\\\\nAccording to the Pythagorean theorem, we have $AB^2+BC^2=AC^2$, \\\\\nSolving this gives $AB^2=50$, \\\\\n$\\therefore$ The area of the square ABCD is $\\boxed{50}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53359743.png"
  },
  {
    "solution": "\\boxed{\\sqrt{5}−1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $E$ and $F$ are two movable points on the side $AD$ of the square $ABCD$, satisfying $AE=DF$. Connect $CF$ and intersect $BD$ at point $G$, connect $BE$ and intersect $AG$ at point $H$. If the side length of the square is $2$, what is the minimum value of the length of segment $DH$?\n",
    "completion": "\\textbf{Solution:} Take the midpoint O of AB, and connect OH and OD,\\\\\nsince it is a square ABCD,\\\\\ntherefore AB＝AD＝CD, $\\angle BAD＝\\angle CDA$, $\\angle ADG＝\\angle CDG$,\\\\\nin $\\triangle ABE$ and $\\triangle DCF$,\\\\\n\\[\n\\left\\{\n\\begin{array}{l}\nAB=CD\\\\\n\\angle BAD=\\angle CDA\\\\\nAE=DF\n\\end{array}\n\\right.\n\\]\\\\\ntherefore $\\triangle ABE \\cong \\triangle DCF$ (SAS),\\\\\ntherefore $\\angle 1＝\\angle 2$,\\\\\nin $\\triangle ADG$ and $\\triangle CDG$,\\\\\n\\[\n\\left\\{\n\\begin{array}{l}\nAD=CD\\\\\n\\angle ADG=\\angle CDG\\\\\nDG=DG\n\\end{array}\n\\right.\n\\]\\\\\ntherefore $\\triangle ADG \\cong \\triangle CDG$ (SAS),\\\\\ntherefore $\\angle 2＝\\angle 3=\\angle 1$,\\\\\nsince $\\angle BAH+\\angle 3＝\\angle BAD＝90^\\circ$,\\\\\ntherefore $\\angle 1+\\angle BAH＝90^\\circ$,\\\\\ntherefore $\\angle AHB＝180^\\circ−90^\\circ＝90^\\circ$,\\\\\nsince OH is the median of Rt$\\triangle ABH$,\\\\\ntherefore OH＝AO＝$\\frac{1}{2}$AB＝1,\\\\\nin Rt$\\triangle AOD$,\\\\\n$OD=\\sqrt{AO^{2}＋AD^{2}}=\\sqrt{1^{2}+2^{2}}=\\sqrt{5}$\\\\\nsince OH＋DH＞OD,\\\\\ntherefore when points O, D, and H are collinear, the length of DH is minimal,\\\\\nThe minimum value of DH ＝ OD−OH ＝ $\\boxed{\\sqrt{5}−1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51721905.png"
  },
  {
    "solution": "\\boxed{(3,15)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, in a Cartesian coordinate system, the linear function $y=-\\frac{2}{3}x+4$ intersects the $x$-axis at point $B$ and the $y$-axis at point $A$. Point $C$ is the midpoint of line segment $AB$. A line is drawn through point $C$ such that $DC\\perp x$-axis, with the foot of the perpendicular being $D$. As shown in Figure 2, does there exist a point $Q$ on the line $CD$ such that $\\angle ABQ=45^\\circ$? Find the coordinates of point $Q$.\n",
    "completion": "Solution:When point $Q$ is above the x-axis, $\\angle ABQ=45^\\circ$. Draw $AM\\perp AB$ at point $A$, intersecting $BQ$ at point $M$. Draw $MH\\perp y$-axis at point $H$,\\\\\n$\\therefore \\angle BAM=90^\\circ$,\\\\\n$\\therefore \\triangle BAM$ is an isosceles right triangle,\\\\\n$\\therefore AM=AB$,\\\\\n$\\because \\angle HAM+\\angle OAB=90^\\circ$, $\\angle OAB+\\angle ABO=90^\\circ$,\\\\\n$\\therefore \\angle HAM=\\angle ABO$,\\\\\nIn $\\triangle AMH$ and $\\triangle ABO$\n\\[\n\\left\\{\\begin{array}{l}\n\\angle AHM=\\angle AOB\\\\\n\\angle HAM=\\angle ABO\\\\\nAM=AB\n\\end{array}\\right.\n\\]\n$\\therefore \\triangle AMH\\cong \\triangle ABO$ (AAS),\\\\\n$\\therefore MH=AO=4$, $AH=BO=6$,\\\\\n$\\therefore OH=AH+AO=4+6=10$,\\\\\n$\\therefore$ the coordinates of point $M$ are $(4,10)$,\\\\\nLet the equation of the line $MB$ be $y=mx+n$ \n\\[\n\\left\\{\\begin{array}{l}\n4m+n=10\\\\\n6m+n=0\n\\end{array}\\right.\n\\]\nSolving it, we get: \n\\[\n\\left\\{\\begin{array}{l}\nm=-5\\\\\nn=30\n\\end{array}\\right.\n\\]\n$\\therefore y=-5x+30$,\\\\\nWhen $x=3$, $y=-15+30=15$,\\\\\n$\\therefore$ the coordinates of point $Q$ are $(3,15)$;\\\\\nWhen point $Q$ is below the x-axis, $\\angle ABQ=45^\\circ$. Draw $AN\\perp AB$ at point $A$, intersecting the extended line of $BQ$ at point $N$. Draw $NG\\perp y$-axis at point $G$,\\\\\nSimilarly, it can be proved that $\\triangle ANG\\cong \\triangle ABO$,\\\\\n$\\therefore NG=AO=4$, $AG=BO=6$,\\\\\n$\\therefore OG=6-4=2$,\\\\\n$\\therefore$ the coordinates of point $N$ are $(-4,-2)$,\\\\\nLet the equation of $BN$ be $y=ax+c$,\n\\[\n\\left\\{\\begin{array}{l}\n-4a+c=-2\\\\\n6a+c=0\n\\end{array}\\right.\n\\]\nSolving it, we get:\n\\[\n\\left\\{\\begin{array}{l}\na=\\frac{1}{5}\\\\\nc=-\\frac{6}{5}\n\\end{array}\\right.\n\\]\n$\\therefore y=\\frac{1}{5}x-\\frac{6}{5}$,\\\\\nWhen $x=3$, $y=\\frac{1}{5}\\times 3-\\frac{6}{5}=-\\frac{3}{5}$,\\\\\n$\\therefore$ the coordinates of point $Q$ are $\\left(3, -\\frac{3}{5}\\right)$ or $(3,15)$.\\\\\nHence, the coordinates of point $Q$ are $\\boxed{\\left(3, -\\frac{3}{5}\\right)}$ or $\\boxed{(3,15)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52815665_106.png"
  },
  {
    "solution": "\\boxed{\\frac{15}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $D$ is the midpoint of the hypotenuse $AB$, and $DE \\perp AB$ intersects $AC$ at point $E$. If $AC=8$ and $BC=6$, then what is the length of $DE$?\n",
    "completion": "\\textbf{Solution:} In right triangle $\\triangle ABC$, we have\n\\[AB=\\sqrt{AC^{2}+BC^{2}}=\\sqrt{8^{2}+6^{2}}=10,\\]\nsince point D is the midpoint of AB,\n\\[AD=\\frac{1}{2}AB=5;\\]\nsince DE$\\perp$AB,\n\\[\\angle ADE=\\angle C=90^\\circ,\\]\nand since $\\angle A=\\angle A$,\n\\[\\triangle ADE\\sim \\triangle ACB,\\]\nhence\n\\[\\frac{DE}{BC}=\\frac{AD}{AC},\\]\nwhich gives\n\\[\\frac{DE}{6}=\\frac{5}{8}.\\]\nSolving this, we find $DE=\\boxed{\\frac{15}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52661220.png"
  },
  {
    "solution": "\\boxed{3\\colon 1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within square $ABCD$, point $E$ is on side $BC$, and $AB = 3BE$. Draw $BF\\perp AE$ through point $B$, intersecting with side $CD$ at point $F$. With $C$ as the center and $CF$ as the radius, draw a circle intersecting side $BC$ at point $G$. Connect $DG$, intersecting $BF$ at point $H$. What is the ratio of $DH$ to $HG$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, connect AH and CH. Let AE and BF intersect at M,\\\\\n$\\because BF \\perp AE$,\\\\\n$\\therefore \\angle AMB = 90^\\circ$,\\\\\n$\\therefore \\angle BAM + \\angle ABM = 90^\\circ$,\\\\\n$\\because$ Quadrilateral ABCD is a square,\\\\\n$\\therefore AB = BC = CD = AD$, $\\angle ABE = \\angle BCF = 90^\\circ$,\\\\\n$\\therefore \\angle ABM + \\angle CBF = 90^\\circ$,\\\\\n$\\therefore \\angle BAE = \\angle CBF$,\\\\\n$\\therefore \\triangle ABE \\cong \\triangle BCF$ (ASA),\\\\\n$\\therefore BE = CF$,\\\\\n$\\therefore BF = DF$,\\\\\n$\\because CG = CF$, $\\angle DCG = \\angle BCF$, DC = BC,\\\\\n$\\therefore \\triangle BCF \\cong \\triangle DCG$ (SAS),\\\\\n$\\therefore \\angle CBF = \\angle CDG$,\\\\\nAlso, $\\because \\angle BHG = \\angle DHF$,\\\\\n$\\therefore \\triangle BHG \\cong \\triangle DHF$ (AAS),\\\\\n$\\therefore HG = HF$,\\\\\nAlso, $\\because HC = HC$, CG = CF,\\\\\n$\\therefore \\triangle HCG \\cong \\triangle HCF$ (SSS),\\\\\n$\\therefore \\angle HCG = \\angle HCF = 45^\\circ$,\\\\\n$\\therefore$ Points A, H, and C are collinear,\\\\\n$\\because AD \\parallel CG$,\\\\\n$\\therefore \\triangle ADH \\sim \\triangle CGH$,\\\\\n$\\therefore \\frac{DH}{HG} = \\frac{AD}{CG} = \\frac{AD}{BE} = \\frac{AB}{BE} = \\boxed{3\\colon 1}$,\\\\",
    "image_path": "dataset/mm_math/images/MM_Math/53110836.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC=4$ and $\\angle B=\\angle C=40^\\circ$. Point $D$ moves on line segment $BC$ (not coinciding with $B$ or $C$), and $AD$ is connected. Let $\\angle ADE=40^\\circ$, and let $DE$ intersect line segment $AC$ at $E$. When does $DC$ equal such that $\\triangle ABD\\cong \\triangle DCE$? Please state your reasons.\n",
    "completion": "\\textbf{Solution:} When $DC=4$, $\\triangle ABD\\cong \\triangle DCE$ for the following reason:\\\\\n$\\because \\angle C=40^\\circ$,\\\\\n$\\therefore \\angle DEC+\\angle EDC=140^\\circ$,\\\\\nAlso, $\\because \\angle ADE=40^\\circ$,\\\\\n$\\therefore \\angle ADB+\\angle EDC=140^\\circ$,\\\\\n$\\therefore \\angle ADB=\\angle DEC$,\\\\\nAlso, $\\because AB=DC=4$,\\\\\nIn $\\triangle ABD$ and $\\triangle DCE$, we have $\\left\\{\\begin{array}{l}\\angle ADB=\\angle DEC \\\\ \\angle B=\\angle C \\\\ AB=DC\\end{array}\\right.$,\\\\\n$\\therefore \\triangle ABD\\cong \\triangle DCE$ (AAS);\\\\\nThus, $DC=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52913913_103.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle BAC = 90^\\circ$, extend the hypotenuse $BC$ to point $D$ so that $CD = \\frac{1}{2}BC$. Connect $AD$. If $\\tan B = \\frac{4}{3}$, what is the value of $\\tan \\angle CAD$?\n",
    "completion": "\\textbf{Solution:} Construct line $CF \\perp AC$ intersecting $AD$ at point $F$, \\\\\nit follows that $\\angle ACF=90^\\circ$, \\\\\nsince $\\angle BAC=90^\\circ$, \\\\\nit implies $AB\\parallel CF$, \\\\\nhence $\\angle DCF=\\angle DBA$, \\\\\nthus $\\triangle DCF\\sim \\triangle DBA$, \\\\\nimplying $\\frac{CF}{AB}=\\frac{CD}{BD}$, \\\\\nsince $CD=\\frac{1}{2}BC$, \\\\\nthus $CD=\\frac{1}{3}BD$, \\\\\nthat is $\\frac{CD}{BD}=\\frac{1}{3}$, \\\\\nthus $\\frac{CF}{AB}=\\frac{1}{3}$, \\\\\nhence $CF=\\frac{1}{3}AB$, \\\\\nIn right $\\triangle ABC$, \\\\\nsince $\\tan B=\\frac{4}{3}$, that is $\\frac{AC}{AB}=\\frac{4}{3}$, \\\\\nlet $AC=4x$, then $AB=3x$, \\\\\nthus $CF=\\frac{1}{3} \\times 3x=x$, \\\\\ntherefore $\\tan\\angle CAD=\\frac{CF}{AC}=\\frac{x}{4x}=\\boxed{\\frac{1}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51231580.png"
  },
  {
    "solution": "\\boxed{\\sqrt{6}+\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $C$ and $D$ are points on the circle $\\odot O$, and they are located on both sides of the diameter $AB$, with $BC=2\\sqrt{3}$, and $\\angle ABC=30^\\circ$. When $D$ is the midpoint of arc $AB$, what is the length of $CD$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, construct $BF\\perp CD$ at point F,\\\\\nsince D is the midpoint of $\\overset{\\frown}{AB}$, and $BC=2\\sqrt{3}$,\\\\\nthus $\\angle BCD=45^\\circ$,\\\\\ntherefore, $\\triangle BCF$ is an isosceles right triangle,\\\\\nthus $BF=CF=\\frac{\\sqrt{2}BC}{2}=\\sqrt{6}$,\\\\\nalso, since $\\angle BDC=60^\\circ$,\\\\\nthus $DF=\\frac{\\sqrt{3}BF}{3}=\\sqrt{2}$,\\\\\ntherefore, $CD=\\sqrt{6}+\\sqrt{2}=\\boxed{\\sqrt{6}+\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52991389_145.png"
  },
  {
    "solution": "\\boxed{\\frac{17}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the parabola $y=ax^2+bx+8\\left(a\\ne 0\\right)$ passes through points $A(-3, 5)$ and $B(5, 3)$, and intersects the y-axis at point C. The circle with diameter AB passes through points O and C, and intersects the x-axis at point D. Lines AO and AC are drawn. Point E is on the x-axis. Lines BD and BE are drawn. When $\\triangle BDE$ is similar to $\\triangle OAC$, find the length of OE that satisfies this condition.\n",
    "completion": "\\textbf{Solution:} From the given information, \\\\\n$\\because y=-\\frac{1}{4}x^2+\\frac{1}{4}x+8$ intersects the y-axis at point C, \\\\\n$\\therefore$ C(0, 8), \\\\\nDraw AF $\\perp$ y-axis through point A and BG $\\perp$ x-axis through point B, then F(0, 5), G(5, 0), \\\\\n$\\therefore \\tan\\angle ACF=\\frac{AF}{CF}=\\frac{3}{8-5}=1$, $\\tan\\angle BDG=\\frac{BG}{DG}=\\frac{3}{5-2}=1$, \\\\\n$\\therefore \\angle ACF=\\angle BDG=45^\\circ$, \\\\\n(1) When $\\triangle BDE\\sim \\triangle ACO$, \\\\\nthen $\\frac{AC}{BD}=\\frac{AO}{BE}=\\frac{CO}{DE}$, \\\\\n$\\because AC=\\sqrt{3^2+3^2}=3\\sqrt{2}$, $BD=\\sqrt{3^2+3^2}=3\\sqrt{2}$, \\\\\n$\\therefore \\frac{CO}{DE}=\\frac{3\\sqrt{2}}{3\\sqrt{2}}=1$, \\\\\n$\\therefore DE=CO=8$, \\\\\n$\\therefore OE=OD+DE=2+8=\\boxed{10}$; \\\\\n(2) When $\\triangle EBD\\sim \\triangle AOC$, \\\\\nthen $\\frac{DE}{CA}=\\frac{BD}{OC}=\\frac{BE}{OA}$, \\\\\n$\\because AC=\\sqrt{3^2+3^2}=3\\sqrt{2}$, $BD=\\sqrt{3^2+3^2}=3\\sqrt{2}$, OC=8, \\\\\n$\\therefore \\frac{DE}{3\\sqrt{2}}=\\frac{3\\sqrt{2}}{8}$, \\\\\n$\\therefore DE=\\frac{9}{4}$, \\\\\n$\\therefore OE=OD+DE=2+\\frac{9}{4}=\\boxed{\\frac{17}{4}}$; \\\\\nIn summary, the length of OE is \\boxed{10} or \\boxed{\\frac{17}{4}}.",
    "image_path": "dataset/mm_math/images/MM_Math/52872196_81.png"
  },
  {
    "solution": "\\boxed{14\\sqrt{2}+4\\sqrt{17}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system $xOy$, point $A$ is on the positive half of the $y$-axis, and point $B$ is on the positive half of the $x$-axis, with $OA=OB=10$. Line $AB$ is translated downwards by 10 units to obtain line $l$, and points $M$ and $N$ are moving points on line $l$ (the abscissas of $M$ and $N$ are $x_{M}$ and $x_{N}$, respectively, and $x_{M}<x_{N}$). Given $MN=4\\sqrt{2}$, find the minimum perimeter of quadrilateral $ABNM$ and explain the reasoning.\n",
    "completion": "\\textbf{Solution:} From the given problem, we can construct as shown in the figure: Draw point C as the symmetric point of A with respect to MN, draw MD $\\parallel$ BN, and connect MC and CD,\\\\\n$\\therefore$ $AM=MC$, $AC\\perp MN$,\\\\\n$\\because$ MN $\\parallel$ AB,\\\\\n$\\therefore$ quadrilateral $MNBD$ is a parallelogram, $\\angle CAD=90^\\circ$,\\\\\n$\\therefore$ $MN=BD$, $MD=BN$,\\\\\n$\\because$ MN $=4\\sqrt{2}$,\\\\\n$\\therefore$ $BD=4\\sqrt{2}$,\\\\\n$\\because$ $OA=OB=10$,\\\\\n$\\therefore$ $\\triangle AOB$ is an isosceles right-angled triangle,\\\\\n$\\therefore$ $AB=\\sqrt{2}OA=10\\sqrt{2}$, the distance from line l to AB is $\\frac{1}{2}AB=5\\sqrt{2}$,\\\\\n$\\therefore$ based on the property of axis symmetry, $AC=10\\sqrt{2}$,\\\\\n$\\therefore$ the perimeter of quadrilateral $ABNM$ is $AB+BN+MN+AM=AB+MD+MN+CM=14\\sqrt{2}+MD+MC$,\\\\\nTo minimize the perimeter of quadrilateral $ABNM$, it is necessary to make $MC+MD$ as small as possible, which is achieved when points C, M, D are collinear. At this time, $MC+MD=CD$,\\\\\n$\\therefore$ in $\\triangle ADC$, $AC=10\\sqrt{2}$, $AD=AB−BD=6\\sqrt{2}$, by Pythagoras' theorem, we have:\\\\\n$CD=\\sqrt{AC^2+AD^2}=4\\sqrt{17}$,\\\\\n$\\therefore$ $MC+MD=4\\sqrt{17}$,\\\\\n$\\therefore$ The minimum perimeter of quadrilateral $ABNM$ is $14\\sqrt{2}+4\\sqrt{17}=\\boxed{14\\sqrt{2}+4\\sqrt{17}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50493321_221.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the line $l_1: y=kx+b$ intersects with the line $l_2: y=mx+n$ at point A$(1,2)$, the line $l_2$ intersects with the $y$-axis at point B$(0,3)$, and the line $l_1$ intersects with the $x$-axis at point C$(-1,0)$. Connect $BC$. What is the area of triangle ABC?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram: \\\\\nWhen $y=0$, we have $0=-x+3$, which means $x=3$, \\\\\n$\\therefore$ the coordinates of point D are: $(3, 0)$, \\\\\n$\\therefore$ CD=4, \\\\\n$\\therefore$ the area of the triangle ABC is $S_{\\triangle ABC}=S_{\\triangle BCD}-S_{\\triangle ACD}=\\frac{1}{2}\\times 4\\times 3-\\frac{1}{2}\\times 4\\times 2=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52188903_158.png"
  },
  {
    "solution": "\\boxed{25^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points $A$, $B$, $P$ are three points on the circle $\\odot O$. If $\\angle AOB = 50^\\circ$, what is the degree measure of $\\angle APB$?\n",
    "completion": "Given that points A, B, and P lie on the circle centered at O, and $\\angle AOB = 50^\\circ$, \\\\\nit follows that $\\angle APB = \\frac{1}{2} \\angle AOB = \\boxed{25^\\circ}.$",
    "image_path": "dataset/mm_math/images/MM_Math/51402444.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: After shifting the parabola \\(y=x^{2}-4x+1\\) to the left so that its vertex falls on the y-axis, as shown in the figure, what is the area \\(S\\) of the shape formed by the two parabolas, the line \\(y=-3\\), and the x-axis (the shaded area in the figure)?\n",
    "completion": "\\textbf{Solution:} The parabola $y=x^{2}-4x+1=(x-2)^{2}-3$ has its vertex at $C(2,-3)$. Shifting it leftwards to have the vertex lie on the y-axis, we get the vertex $B(0,-3)$. Point $A$ is one of the intersections of the parabola with the $x$-axis. Connecting $OC$, $AB$,\\\\\nas shown in the figure,\\\\\nthe shaded area equals the area of $\\triangle ABCO$, with $S=2\\times 3=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52536743.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the perimeter of the rhombus $ABCD$ is 20, and $ \\angle A=60^\\circ$. What is the length of the diagonal $BD$?\n",
    "completion": "\\textbf{Solution:}\n\nSince quadrilateral ABCD is a rhombus and its perimeter is 20,\\\\\nit follows that AB=AD=5.\\\\\nSince $\\angle$A=60$^\\circ$,\\\\\nit follows that $\\triangle$ABD is an equilateral triangle,\\\\\ntherefore, AB=AD=BD=5.\\\\\nHence, the answer is: BD=\\boxed{5}.",
    "image_path": "dataset/mm_math/images/MM_Math/52250623.png"
  },
  {
    "solution": "\\boxed{\\frac{48}{5}cm}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, with the side $AB$ of $\\triangle ABC$ as the diameter, the semicircle $O$ intersects sides $AC$ and $BC$ at points $E$ and $D$, respectively, and $D$ is the midpoint of the arc $\\overset{\\frown}{BE}$. Given $AB=AC$, if the radius of $\\odot O$ is $5cm$ and $BC=12cm$, what is the length of segment $BE$ in $cm$?\n",
    "completion": "\\textbf{Solution:} Connect $AC$,\\\\\n$\\because AB=AC$,\\\\\n$\\therefore \\triangle ABC$ is an isosceles triangle,\\\\\n$\\because AB$ is the diameter of $\\odot O$,\\\\\n$\\therefore \\angle ADB=90^\\circ$, which means $AD\\perp BC$,\\\\\n$\\because BC=12cm$,\\\\\n$\\therefore CD=BD=\\frac{1}{2}BC=6cm$,\\\\\nAlso, $AC=2OD=10cm$,\\\\\nBy the Pythagorean theorem, $AD=\\sqrt{AC^{2}-CD^{2}}=\\sqrt{10^{2}-6^{2}}=8cm$,\\\\\n$\\therefore \\sin C=\\frac{AD}{AC}=\\frac{8}{10}=\\frac{4}{5}$,\\\\\n$\\therefore \\sin\\angle ODB=\\sin C=\\frac{4}{5}$,\\\\\n$\\therefore \\frac{BH}{BD}=\\frac{4}{5}$,\\\\\n$\\therefore BH=\\frac{4}{5}BD=\\frac{24}{5}$,\\\\\n$\\therefore BE=2BH=\\boxed{\\frac{48}{5}cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51265419_143.png"
  },
  {
    "solution": "\\boxed{25^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, it is known that in rhombus ABCD, one interior angle $\\angle BAD=80^\\circ$, the diagonals AC and BD intersect at point O, and point E is on AB, with BE=BO. What is the measure of $\\angle EOA$ in degrees?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rhombus,\\\\\nit follows that AC$\\perp$BD, $\\angle$BAO= $\\frac{1}{2}\\angle$BAD=$40^\\circ$,\\\\\nhence $\\angle$ABO=$90^\\circ$−$\\angle$BAO=$50^\\circ$,\\\\\nsince BE=BO,\\\\\nit follows that $\\angle$BOE= $\\frac{180^\\circ−\\angle ABO}{2}$=$65^\\circ$,\\\\\ntherefore $\\angle$EOA=$\\angle$AOB−$\\angle$BOE=$90^\\circ−65^\\circ=\\boxed{25^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51614464.png"
  },
  {
    "solution": "\\boxed{x>1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the Cartesian coordinate system $xOy$, the graphs of the linear functions ${y}_{1}=kx$ and ${y}_{2}=-x+b$ intersect at point $A(1, 2)$. What is the solution set of the inequality $kx>-x+b$ with respect to $x$?\n",
    "completion": "\\textbf{Solution:} Since the graph of linear functions $y_{1}=kx$ and $y_{2}=-x+b$ intersects at point $A(1,2)$,\\\\\nit follows from the graph that the solution set of the inequality $kx>-x+b$ with respect to $x$ is $x>1$.\\\\\nTherefore, the answer is: $\\boxed{x>1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52380621.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $ \\angle AOB : \\angle BOC : \\angle COD = 2 : 3 : 4$, the rays $OM$ and $ON$ bisect $ \\angle AOB$ and $ \\angle COD$ respectively. If $ \\angle MON$ is a right angle, what is the measure of $ \\angle COD$ in degrees?\n",
    "completion": "\\textbf{Solution:} Let $\\angle AOB=2x^\\circ$, then $\\angle BOC=3x^\\circ$, $\\angle COD=4x^\\circ$,\\\\\nsince ray OM and ON bisect $\\angle AOB$ and $\\angle COD$ respectively,\\\\\ntherefore $\\angle BOM= \\frac{1}{2} \\angle AOB=x^\\circ$\\\\\n$\\angle CON= \\frac{1}{2} \\angle COD=2x^\\circ$\\\\\nsince $\\angle MON=90^\\circ$\\\\\ntherefore $\\angle CON+\\angle BOC+\\angle BOM=90^\\circ$\\\\\ntherefore $2x+3x+x=90$\\\\\nSolving for x, we get: $x=15$\\\\\ntherefore $\\angle COD=4x =15^\\circ \\times 4=\\boxed{60^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51323803.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, points D, E, and F are the midpoints of the sides, respectively. If the area of $\\triangle ABC$ is $16\\,cm^2$, then what is the area of $\\triangle DEF$ in $cm^2$?\n",
    "completion": "\\textbf{Solution:}\n\nSince points D and F are the midpoints of AB and AC, respectively, \\\\\n$\\therefore$DF$\\parallel $BC and DF$ = \\frac{1}{2}$BC, \\\\\n$\\therefore$DF$\\parallel $BE, \\\\\nSince E is the midpoint of BC, \\\\\n$\\therefore$BE$ = \\frac{1}{2}$BC, \\\\\n$\\therefore$DF$ = $BE, \\\\\n$\\therefore$Quadrilateral BEFD is a parallelogram, \\\\\n$\\therefore$BD$ = $EF, \\\\\nIn $\\triangle $BDE and $\\triangle $FED, \\\\\n$ \\left\\{\\begin{array}{l}DF=BE \\\\ BD=EF \\\\ DE=DE \\end{array}\\right.$ \\\\\n$\\therefore$$\\triangle $BDE$\\cong $$\\triangle $FED \\text{ (SSS)}, \\\\\nSimilarly, it can be proven that $\\triangle $DAF$\\cong $$\\triangle $FED and $\\triangle $EFC$\\cong $$\\triangle $FED, \\\\\ni.e., $\\triangle $BDE$\\cong $$\\triangle $DAF$\\cong $$\\triangle $EFC$\\cong $$\\triangle $FED, \\\\\n$\\therefore$Area of $\\,\\triangle $DEF$ = \\frac{1}{4}$Area of $\\,\\triangle $ABC$ = \\frac{1}{4} \\times 16 = \\boxed{4} \\text{ (cm}^2\\text{)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51534333.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{6}+\\sqrt{2}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within the square $ABCD$, $P$ is a point on the diagonal $BD$, and point $E$ is on the extension of $AB$, such that $PC=PE$ and $PA=PE$; $AE=\\sqrt{2}DP$; given that the side length of the square is $2$, what is the minimum value of $CP+\\frac{1}{2}BP$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, rotate segment BD $30^\\circ$ counterclockwise around point B, intersecting AD at G. Draw $PH\\perp BG$ at H through point P, and connect AC, intersecting BD at O,\\\\\n$\\because \\angle PBH=30^\\circ$,\\\\\n$\\therefore PH=\\frac{1}{2}PB$,\\\\\n$\\therefore$ when points C, P, H are collinear, $PC+\\frac{1}{2}PB$ has the minimum value CH,\\\\\n$\\because$ the side length of the square is 2,\\\\\n$\\therefore OC=OB=\\sqrt{2}$,\\\\\n$\\because \\angle PBH+\\angle BPH=90^\\circ$, $\\angle OCP+\\angle OPC=90^\\circ$, $\\angle OPC=\\angle BPH$,\\\\\n$\\therefore \\angle OCP=\\angle PBH=30^\\circ$,\\\\\n$\\therefore OP=\\frac{1}{2}CP$,\\\\\nIn $\\triangle OPC$, $PC^2=OP^2+OC^2$, that is $4OP^2=OP^2+2$,\\\\\n$\\therefore OP=\\frac{\\sqrt{6}}{3}$ (discard the negative value),\\\\\n$\\therefore CP=\\frac{2\\sqrt{6}}{3}$,\\\\\n$\\therefore PB=\\sqrt{2}-\\frac{\\sqrt{6}}{3}$, $PH=\\frac{1}{2}PB=\\frac{\\sqrt{2}}{2}-\\frac{\\sqrt{6}}{6}$,\\\\\n$\\therefore CH=CP+PH=\\frac{2\\sqrt{6}}{3}+\\frac{\\sqrt{2}}{2}-\\frac{\\sqrt{6}}{6}=\\boxed{\\frac{\\sqrt{6}+\\sqrt{2}}{2}}$,\\\\\n$\\therefore CP+\\frac{1}{2}BP$'s minimum value is $\\boxed{\\frac{\\sqrt{6}+\\sqrt{2}}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53313110_132.png"
  },
  {
    "solution": "\\boxed{\\left(0, -6\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the graph, the parabola $y=2x^{2}-4x-6$ intersects the x-axis at points A and B, with point A being to the left of point B, and intersects the y-axis at point C. Point D is a moving point on the parabola. Find the coordinates of points A, B, and C.\n",
    "completion": "\\textbf{Solution:} Substitute $y=0$ into $y=2x^{2}-4x-6$,\\\\\nwe get $2x^{2}-4x-6=0$.\\\\\nSolving it, we find $x_{1}=-1$, $x_{2}=3$.\\\\\nTherefore, the coordinates of point A are $\\boxed{\\left(-1, 0\\right)}$, and the coordinates of point B are $\\boxed{\\left(3, 0\\right)}$.\\\\\nSubstitute $x=0$ into $y=2x^{2}-4x-6$, we get $y=-6$.\\\\\nTherefore, the coordinates of point C are $\\boxed{\\left(0, -6\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52977819_205.png"
  },
  {
    "solution": "\\boxed{24}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, C is a point on the line segment AB, with M and N being the midpoints of AC and BC, respectively. If the ratio of the lengths of segments CM to CN is 2:1, and the length of segment CN is 4 cm, find the length of segment AB.\n",
    "completion": "\\textbf{Solution:} Given that the ratio of the lengths of segments CM to CN is $2\\colon 1$ and $CN=4\\,\\text{cm}$, \\\\\n$\\therefore$ the length of segment CM is $8\\,\\text{cm}$.\\\\\nSince M and N are the midpoints of AC and BC, respectively, \\\\\n$\\therefore$ AC$=2\\cdot CM=16\\,\\text{cm}$ and BC$=2\\cdot CN=8\\,\\text{cm}$,\\\\\n$\\therefore$ AB$=AC+BC=16+8=\\boxed{24}\\,\\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53274798_99.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{3}}{6}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ and $\\triangle ADE$ are both equilateral triangles, where point $E$ is the incenter of $\\triangle ABC$. Taking $E$ as the center and the length of $DE$ as the radius, an arc is drawn to intersect $BC$ at point $B$. Then, the arc $DB$ is rotated $60^\\circ$ counterclockwise around point $A$ to the position of arc $EC$. Given $AB=1$, what is the area of the shaded region in the figure?\n",
    "completion": "\\textbf{Solution:}\n\nConnect BD and CE as shown in the diagram.\\\\\nDue to the properties of rotation: BD=CE, $\\overset{\\frown}{BD}=\\overset{\\frown}{CE}$\\\\\n$\\therefore$ The area of the arc with BD as its chord is equal to the area of the arc with CE as its chord.\\\\\n$\\because$ $\\triangle ABC$ is an equilateral triangle, and point $E$ is the incenter of $\\triangle ABC$\\\\\n$\\therefore$ $\\angle$BAE=$\\angle$CAE= $\\frac{1}{2}$ $\\angle$BAC=$30^\\circ$, AE=CE\\\\\n$\\therefore$ $\\angle$ECA=$\\angle$CAE=$30^\\circ$\\\\\n$\\therefore$ $\\angle$AEC=$180^\\circ$−($\\angle$ECA+$\\angle$CAE)=$120^\\circ$\\\\\n$\\because$ $\\triangle ADE$ is an equilateral triangle\\\\\n$\\therefore$ AE=DE, $\\angle$AED=$60^\\circ$\\\\\n$\\therefore$ $\\angle$AED+$\\angle$AEC=$180^\\circ$, $\\angle$BAE+$\\angle$AED=$90^\\circ$\\\\\nThat is, DE and CE are collinear, and DE=CE=AE, AB$\\perp$DE, let the foot of the perpendicular be F\\\\\n$\\therefore$ Point F is the midpoint of AB\\\\\n$\\therefore$ $AF=BF=\\frac{1}{2}AB=\\frac{1}{2}$\\\\\n$\\therefore$ $AE=AF÷\\cos30^\\circ=\\frac{\\sqrt{3}}{3}$\\\\\n$\\therefore$ $CD=DE+CD=\\frac{2\\sqrt{3}}{3}$\\\\\n$\\therefore$ ${S}_{\\triangle DBC}=\\frac{1}{2}CD·BF=\\frac{1}{2}\\times \\frac{2\\sqrt{3}}{3}\\times \\frac{1}{2}=\\frac{\\sqrt{3}}{6}$\\\\\nTherefore, the area of the shaded part is $\\boxed{\\frac{\\sqrt{3}}{6}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51497073.png"
  },
  {
    "solution": "\\boxed{y=-(x-1)^2+4=-x^2+2x+3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the vertex of the parabola is $P(1, 4)$, it intersects with the $y$-axis at point $C(0, 3)$, and intersects with the $x$-axis at points $A$ and $B$. Find the equation of the parabola.\n",
    "completion": "\\textbf{Solution:} Let $y=-a(x-1)^2+4$, where $a\\neq 0$.\n\nSubstituting point $C(0, 3)$, we get: $-a+4=3$, which means $a=-1$.\n\n$\\therefore$ The equation of the parabola is $\\boxed{y=-(x-1)^2+4=-x^2+2x+3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52049799_105.png"
  },
  {
    "solution": "\\boxed{\\left(3+\\sqrt{6}, 3-\\sqrt{6}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the line $y=-x+m+1$ and the inverse proportional function $y=\\frac{m}{x}$ (where $x>0$, $m>0$) intersect at points A and B, respectively, and intersect the $x$-axis at point C and the $y$-axis at point D, and the coordinate of point A is $(1, 3)$. If point $P$ is a point on the $y$-axis, when the area of $\\triangle ABP$ is 6, find the coordinates of point $P$.\n",
    "completion": "\\textbf{Solution:} Solve the system \n\\[\n\\begin{cases}\ny=\\frac{3}{x}\\\\\ny=-x+4\n\\end{cases},\n\\]\nto obtain\n\\[\n\\begin{cases}\nx_1=1\\\\\ny_1=3\n\\end{cases}, \n\\begin{cases}\nx_2=3\\\\\ny_2=1\n\\end{cases},\n\\]\n\\(\\therefore\\) the coordinates of point \\(B\\) are \\((3, 1)\\),\n\n\\(\\therefore AB=\\sqrt{(3-1)^2+(1-3)^2}=2\\sqrt{2}\\),\n\nsubstitute \\(x=0\\) into \\(y=-x+4\\) to get \\(y=4\\),\n\nsubstitute \\(y=0\\) into \\(y=-x+4\\) to get \\(x=4\\),\n\n\\(\\therefore\\) the coordinates of point \\(D\\) are \\((0, 4)\\), and the coordinates of point \\(C\\) are \\((4, 0)\\), \\(\\triangle DOC\\) is an isosceles right triangle, construct a line parallel to \\(AB\\) through point \\(P\\) intersecting the \\(x\\)-axis at point \\(M\\), construct \\(CF\\perp PM\\) at point \\(F\\),\n\nthen \\({S}_{\\triangle ABP}=\\frac{1}{2}AB\\cdot CF=2\\),\n\n\\(\\therefore CF=\\sqrt{2}\\)\n\n\\(\\because \\angle DCO=45^\\circ, \\angle BCF=90^\\circ\\), \\(PM\\parallel AB\\),\n\n\\(\\therefore \\angle FCM=\\angle FMC=45^\\circ\\)\n\n\\(\\therefore \\triangle CFM\\) is an isosceles right triangle,\n\n\\(\\therefore CM=\\sqrt{2}CF=2\\),\n\n\\(\\therefore\\) the coordinates of point \\(M\\) are \\((6, 0)\\),\n\nlet the equation of \\(PM\\) be \\(y=-x+b\\),\n\nsubstitute \\((6, 0)\\) into \\(y=-x+b\\): \\(0=-6+b\\),\n\nsolve to get \\(b=6\\),\n\nthe equation of line \\(PM\\) is \\(y=-x+6\\),\n\nlet \\(\\frac{3}{x}=-x+6\\)\n\nsolve to get \\({x}_{1}=3-\\sqrt{6}, {x}_{2}=3+\\sqrt{6}\\),\n\nsubstitute \\(x=3-\\sqrt{6}\\) into \\(y=-x+6\\) to get \\(y=3+\\sqrt{6}\\),\n\nsubstitute \\(x=3+\\sqrt{6}\\) into \\(y=-x+6\\) to get \\(y=3-\\sqrt{6}\\),\n\n\\(\\therefore\\) the coordinates of point \\(P\\) are \\(\\boxed{\\left(3-\\sqrt{6}, 3+\\sqrt{6}\\right)}\\) or \\(\\boxed{\\left(3+\\sqrt{6}, 3-\\sqrt{6}\\right)}\\).",
    "image_path": "dataset/mm_math/images/MM_Math/53080174_34.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle B=90^\\circ$, $AB=12\\,cm$, $BC=16\\,cm$. Point $P$ starts from point $A$ and moves towards point $B$ along edge $AB$ at a speed of $1\\,cm/s$, and point $Q$ starts from point $B$ and moves towards point $C$ along edge $BC$ at a speed of $2\\,cm/s$. If $P$ and $Q$ start simultaneously from $A$ and $B$, respectively, and one of them stops moving when it reaches its endpoint, the other stops as well. Let the time of movement be $t$ seconds. For what value of $t$ is the length of $PQ$ equal to $8\\sqrt{2}\\,cm$?\n",
    "completion": "\\textbf{Solution:} Let $t$ seconds later, the length of $PQ$ be $8\\sqrt{2}\\,cm$. According to the Pythagorean theorem, we have\n\\[ PQ^{2}=BP^{2}+BQ^{2}=(12-t)^{2}+(2t)^{2}=128, \\]\n\\[ 5t^{2}-24t+144-128=0, \\]\nSolving this, we obtain \\[ t_{1}=\\frac{4}{5},\\] \\[t_{2}=4.\\]\nTherefore, when $t$ is \\boxed{\\frac{4}{5}} or \\boxed{4}, the length of $PQ$ equals $8\\sqrt{2}\\,cm$.",
    "image_path": "dataset/mm_math/images/MM_Math/52622502_115.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, port A is located to the true east of observation point O, with OA = 4km. A ship sets sail from port A and arrives at point B after traveling in a direction of 15° north of east for a certain distance. At this moment, it is measured from observation point O that the ship is in a direction of 60° north of east. The question is, what is the distance the ship has traveled (i.e., the length of AB)?\n",
    "completion": "\\textbf{Solution:} Construct $AD \\perp OB$ through point A, so that $AD= \\frac{1}{2}OA=2\\,\\text{km}$. According to the problem statement, $\\triangle ABD$ is an isosceles right triangle, hence $AB=2\\sqrt{2}\\,\\text{km}$. Therefore, the distance traveled by the ship (i.e., the length of $AB$) is $AB=\\boxed{2\\sqrt{2}}\\,\\text{km}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51403117.png"
  },
  {
    "solution": "\\boxed{105^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ is inscribed in circle $\\odot O$, and $AB=AC$. Point $P$ is on arc $\\overset{\\frown}{AC}$, and $\\angle BAC=30^\\circ$. What is the degree measure of $\\angle APC$?\n",
    "completion": "\\textbf{Solution:} Given that $AB=AC$, $\\angle BAC=30^\\circ$,\\\\\nhence, $\\angle ABC=\\angle ACB= \\frac{1}{2}\\left(180^\\circ-\\angle BAC\\right)=\\frac{1}{2}\\left(180^\\circ-30^\\circ\\right)=75^\\circ$,\\\\\nsince quadrilateral ABCP is a cyclic quadrilateral,\\\\\nit follows that $\\angle ABC+\\angle APC=180^\\circ$,\\\\\nthus, $\\angle APC=180^\\circ-\\angle ABC=180^\\circ-75^\\circ=\\boxed{105^\\circ}$,",
    "image_path": "dataset/mm_math/images/MM_Math/51458286_106.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{1}{2}{x}^{2}+2x+\\frac{5}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y = ax^2 + bx + \\frac{5}{2}$ intersects the line $AB$ at points $A(-1, 0)$ and $B(4, \\frac{5}{2})$. Point $D$ is a moving point on the parabola between points $A$ and $B$ (not coinciding with points $A$ and $B$). The line $CD$ is parallel to the y-axis and intersects the line $AB$ at point $C$. Connect $AD$ and $BD$. What is the analytic expression of the parabola?\n",
    "completion": "\\textbf{Solution:} Solve: The parabola $y=a{x}^{2}+bx+\\frac{5}{2}$ intersects the line $AB$ at point $A\\left(-1, 0\\right)$ and $B\\left(4, \\frac{5}{2}\\right)$,\\\\\n$\\therefore \\left\\{\\begin{array}{l}\n0=a(-1)^{2}-b+\\frac{5}{2}\\\\\n\\frac{5}{2}=16a+4b+\\frac{5}{2}\n\\end{array}\\right.$,\\\\\nSolving, we get: $\\left\\{\\begin{array}{l}\na=-\\frac{1}{2}\\\\\nb=2\n\\end{array}\\right.$,\\\\\n$\\therefore$ the equation of the parabola is $\\boxed{y=-\\frac{1}{2}{x}^{2}+2x+\\frac{5}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53043850_140.png"
  },
  {
    "solution": "\\boxed{9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC$, $\\angle BAC=120^\\circ$, $AD\\perp BC$ at $D$, $DE\\perp AC$ at $E$. If $AE=3$, what is the length of $CE$?\n",
    "completion": "\\textbf{Solution:} Given: In $\\triangle ABC$, $AB=AC$ and $\\angle BAC=120^\\circ$, \\\\\n$\\therefore$ $\\angle B=\\angle C=30^\\circ$, \\\\\nSince $AD\\perp BC$ at $D$, \\\\\n$\\therefore$ $\\angle ADC=90^\\circ$, \\\\\n$\\therefore$ $\\angle DAC=60^\\circ$, \\\\\nSince $DE\\perp AC$, \\\\\n$\\therefore$ $\\angle AED=90^\\circ$, \\\\\n$\\therefore$ $\\angle ADE=30^\\circ$, \\\\\nGiven $AE=3$, \\\\\n$\\therefore$ $AD=2AE=6$, \\\\\n$\\therefore$ $AC=2AD=12$, \\\\\n$\\therefore$ $CE=AC-AE=\\boxed{9}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52603537.png"
  },
  {
    "solution": "\\boxed{\\frac{4a}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $C$ is a point on line $AB$, and point $M$ is the midpoint of segment $AC$. If $AB=a$, and $AB-AC=\\frac{a}{3}$, find the length of $BM$ (expressed in an algebraic expression with $a$).\n",
    "completion": "\\textbf{Solution:}\n\n(1) When point $C$ lies between points $A$ and $B$, \\\\\n$\\because BC=AB-AC=\\frac{a}{3}$, \\\\\n$\\therefore AC=AB-BC=a-\\frac{a}{3}=\\frac{2a}{3}$, \\\\\n$\\because$ point $M$ is the midpoint of $AC$, \\\\\n$\\therefore CM=\\frac{1}{2}AC=\\frac{1}{2}\\times \\frac{2a}{3}=\\frac{a}{3}$, \\\\\n$\\therefore BM=CM+BC=\\frac{a}{3}+\\frac{a}{3}=\\boxed{\\frac{2a}{3}}$, \\\\\n(2) When point $C$ is to the left of point $A$, \\\\\n$\\because AC=AB-\\frac{a}{3}=\\frac{2a}{3}$, \\\\\nAdditionally, $\\because$ point $M$ is the midpoint of $AC$, \\\\\n$\\therefore AM=\\frac{1}{2}AC=\\frac{1}{2}\\times \\frac{2a}{3}=\\frac{a}{3}$, \\\\\n$\\therefore BM=AM+AB=\\frac{a}{3}+a=\\boxed{\\frac{4a}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53364701_168.png"
  },
  {
    "solution": "\\boxed{5cm}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line segments $AB$ and $CD$ are on the same straight line, where $M$ and $N$ are the midpoints of $AB$ and $CD$, respectively. Given that $AB = 6\\,\\text{cm}$ and $CD = 8\\,\\text{cm}$. When point $C$ is on line segment $AB$, if the common part $BC$ of the line segments $AB$ and $CD$ is $2\\,\\text{cm}$, then what is the length of $MN$ in cm?\n",
    "completion": "\\textbf{Solution:} Since M and N are the midpoints of the line segments AB and CD, respectively, and AB=6cm, CD=8cm,\\\\\ntherefore, AM=3cm, DN=4cm.\\\\\nSince the common part of the line segments AB and CD is BC=2cm,\\\\\ntherefore, the length of AD is AB+CD−BC=6+8−2=12cm,\\\\\ntherefore, the length of MN is AD−AM−DN=12−3−4=\\boxed{5cm};",
    "image_path": "dataset/mm_math/images/MM_Math/52916287_138.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\angle AOP = \\angle BOP = 15^\\circ$, $PC \\parallel OA$, $PD \\perp OA$, $PE \\perp OB$. If $PC = 4$, what is the length of $PD$?\n",
    "completion": "\\textbf{Solution:} Since $PC \\parallel OA$,\\\\\nit follows that $\\angle AOP = \\angle CPO = 15^\\circ$,\\\\\nthus $\\angle PCE = \\angle BOP + \\angle CPO = 15^\\circ + 15^\\circ = 30^\\circ$,\\\\\nand since $PC = 4$,\\\\\nit follows that $PE = \\frac{1}{2}PC = \\frac{1}{2} \\times 4 = 2$,\\\\\nsince $\\angle AOP = \\angle BOP$, $PD \\perp OA$, $PE \\perp OB$,\\\\\nit follows that $PD = PE = \\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52827547.png"
  },
  {
    "solution": "\\boxed{\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the circle $O$, the radius $OC$ is perpendicular to the chord $AB$ at $D$, point $E$ is on the circle $O$, $\\angle E=22.5^\\circ$, and $AB=2$. What is the length of the radius $OB$?\n",
    "completion": "\\textbf{Solution:} By the problem statement, radius $OC$ is perpendicular to chord $AB$ at point $D$,\\\\\n$\\therefore \\overset{\\frown}{AC}=\\overset{\\frown}{BC}$,\\\\\n$\\therefore \\angle E=\\frac{1}{2}\\angle BOC=22.5^\\circ$,\\\\\n$\\therefore \\angle BOD=45^\\circ$,\\\\\n$\\therefore \\triangle ODB$ is an isosceles right triangle,\\\\\n$\\because AB=2$,\\\\\n$\\therefore DB=OD=1$,\\\\\n$\\therefore OB=\\sqrt{1^2+1^2}=\\boxed{\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52622748.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, it is known that $\\angle ABC=90^\\circ$. With $AC$ as the right angle side, $\\triangle ACD$ is constructed outside ($\\angle CAD=90^\\circ$). Semicircles are drawn outside with diameters $AB$, $BC$, $CD$, $DA$ respectively, and their areas are denoted as $S_1$, $S_2$, $S_3$, and $S_4$. It is known that $S_1=3$, $S_2=1$, and $S_3=7$. What is the value of $S_4$?\n",
    "completion": "\\textbf{Solution:} From the given problem, we have\n\\begin{align*}\nS_1 &= \\frac{1}{8}AB^2\\pi = 3, \\\\\nS_2 &= \\frac{1}{8}BC^2\\pi = 1, \\\\\nS_3 &= \\frac{1}{8}CD^2\\pi = 7, \\\\\nS_4 &= \\frac{1}{8}AD^2\\pi,\n\\end{align*}\nUsing the Pythagorean theorem, we can derive:\n\\begin{align*}\nAC^2 &= BC^2 + AB^2, \\\\\nAD^2 &= CD^2 - AC^2,\n\\end{align*}\nHence,\n\\begin{align*}\nAD^2 &= CD^2 - BC^2 - AB^2, \\\\\n\\end{align*}\nFrom the above equations, we reach:\n\\begin{align*}\nAB^2 &= \\frac{24}{\\pi}, \\\\\nBC^2 &= \\frac{8}{\\pi}, \\\\\nCD^2 &= \\frac{56}{\\pi},\n\\end{align*}\nTherefore,\n\\begin{align*}\nAD^2 &= CD^2 - BC^2 - AB^2 = \\frac{24}{\\pi},\n\\end{align*}\nThus,\n\\begin{align*}\nS_4 &= \\frac{1}{8}AD^2\\pi = \\frac{1}{8} \\times \\frac{24}{\\pi} \\times \\pi = \\boxed{3}.\n\\end{align*}",
    "image_path": "dataset/mm_math/images/MM_Math/55403712.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, fold a rectangular piece of paper \\textit{ABCD} with sides of lengths 8 and 16 respectively, such that point \\textit{C} coincides with point \\textit{A}. What is the length of the crease \\textit{EF}?\n",
    "completion": "\\textbf{Solution:} Given the properties of folding, we know that $EC=AE$, and $\\angle AEF=\\angle CEF$, \\\\\nsince $AD\\parallel BC$, \\\\\nit follows that $\\angle AFE=\\angle CEF$, \\\\\nthus, $\\angle AEF=\\angle AFE$, \\\\\nwhich means $AE=AF=CE$, \\\\\nusing the Pythagorean theorem, we find $AB^2+BE^2=AE^2$ that is $8^2+(16-AE)^2=AE^2$, \\\\\nsolving this gives us $AE=AF=10$, and $BE=6$, \\\\\nlet $EG\\perp AF$ at point G, \\\\\nthen the quadrilateral $AGEB$ is a rectangle, with $AG=6$, $GF=4$, $GE=AB=8$, and by the Pythagorean theorem, $EF=\\boxed{4\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52423904.png"
  },
  {
    "solution": "\\boxed{\\left(\\frac{1}{2}, -\\frac{3}{2}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=ax^2+bx-3$ intersects the $x$-axis at points $A(-1,0)$ and $B(3,0)$. A line $l$ passing through point $A$ intersects the parabola at point $C(2,n)$. Point $P$ is a moving point on segment $AC$, and a perpendicular line to the $x$-axis passing through point $P$ intersects the parabola at point $E$. Find the coordinates of point $P$ when the length of segment $PE$ is maximal.\n",
    "completion": "Solution: To solve this problem, substitute the x-coordinate $x=2$ into $y=x^2-2x-3$,\\\\\nwe get $y=-3$,\\\\\n$\\therefore C(2, -3)$,\\\\\nSubstituting $A(-1, 0)$ and $C(2, -3)$ into $y=kx+b$ yields $\\left\\{\\begin{array}{l}-k+b=0 \\\\ 2k+b=-3\\end{array}\\right.$,\\\\\nfrom which we solve $\\left\\{\\begin{array}{l}k=-1 \\\\ b=-1\\end{array}\\right.$,\\\\\n$\\therefore$ the equation of line AC is $y=-x-1$,\\\\\nLet the x-coordinate of point P be $m(-1\\leq m\\leq 2)$, then, the coordinates of P and E are: $P\\left(m, -m-1\\right)$, $E\\left(m, m^2-2m-3\\right)$;\\\\\n$\\therefore PE=\\left(-m-1\\right)-\\left(m^2-2m-3\\right)$\\\\\n$=-m^2+m+2$\\\\\n$=-\\left(m-\\frac{1}{2}\\right)^2+\\frac{9}{4}$,\\\\\n$\\because -1<0$,\\\\\n$\\therefore$ when $m=\\frac{1}{2}$, the maximum value of PE $=\\frac{9}{4}$,\\\\\n$\\therefore -m-1=-\\frac{1}{2}-1=-\\frac{3}{2}$,\\\\\nat this time $P\\left(\\frac{1}{2}, -\\frac{3}{2}\\right)$,\\\\\n$\\therefore$ when the segment PE is at its maximum, the coordinates of point P are $\\boxed{\\left(\\frac{1}{2}, -\\frac{3}{2}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51446075_227.png"
  },
  {
    "solution": "\\boxed{(1,0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=ax^2+bx+c$ passes through the points $A(-1,0)$, $B(3,0)$, and $C(0,3)$, where $D$ is the vertex and the axis of symmetry is the line $DE$. Find a point $P$ on the axis of symmetry of the parabola such that $\\triangle PAC$ is an isosceles triangle. Please write down the coordinates of point $P$ directly.\n",
    "completion": "\\textbf{Solution:} The coordinates of point $P$ are $\\boxed{(1,1)}$, $\\boxed{\\left(1, \\sqrt{6}\\right)}$, $\\boxed{\\left(1, -\\sqrt{6}\\right)}$, or $\\boxed{(1,0)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52699702_88.png"
  },
  {
    "solution": "\\boxed{\\frac{49}{8}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a small ball M is thrown from point O on slope $OA$. The trajectory of the ball is a part of a parabola. A Cartesian coordinate system is established as shown in the figure, and the slope can be characterized by the linear function $y=\\frac{1}{2}x$. If the highest point reached by the ball is at coordinates $(4, 8)$, find the maximum height of ball M from the slope $OA$ during its flight.\n",
    "completion": "\\textbf{Solution:} Solve the system of equations $\\begin{cases}y=-\\frac{1}{2}x^2+4x\\\\ y=\\frac{1}{2}x\\end{cases}$, and we find: $\\begin{cases}x_1=0\\\\ y_1=0\\end{cases}$, $\\begin{cases}x_2=7\\\\ y_2=\\frac{7}{2}\\end{cases}$. Therefore, $0\\le x\\le 7$. As shown in the figure, draw $MF\\perp x$-axis at point F, intersecting $OA$ at point E. Let $E(t, \\frac{1}{2}t)(0\\le t\\le 7)$, then $M(t, -\\frac{1}{2}t^2+4t)$, hence $EM=y_M-y_E=-\\frac{1}{2}t^2+4t-\\frac{1}{2}t=-\\frac{1}{2}(t-\\frac{7}{2})^2+\\frac{49}{8}$. Since $-\\frac{1}{2}<0$ and $0\\le t\\le 7$, when $t=\\frac{7}{2}$, $EM$ reaches its maximum value, which is \\boxed{\\frac{49}{8}} meters.",
    "image_path": "dataset/mm_math/images/MM_Math/52934216_236.png"
  },
  {
    "solution": "\\boxed{65^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, $AB$ is the diameter of $\\odot O$, $AC$ is a chord, and $\\angle BAC=25^\\circ$. A point D is arbitrarily chosen on $\\odot O$, and point D and point C are located on opposite sides of diameter $AB$. Connecting $AD$ and $DC$, what is the measure of $\\angle D$?\n",
    "completion": "\\textbf{Solution:} Draw line $BC$, as shown in the diagram,\\\\\nsince $AB$ is the diameter of $\\odot O$,\\\\\nthus $\\angle ACB=90^\\circ$,\\\\\nsince $\\angle BAC=25^\\circ$,\\\\\nthus $\\angle B=65^\\circ$,\\\\\nsince $\\overset{\\frown}{AC}=\\overset{\\frown}{AC}$,\\\\\ntherefore $\\angle D=\\angle B=\\boxed{65^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52917391.png"
  },
  {
    "solution": "\\boxed{18}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in parallelogram $ABCD$, point $E$ is on side $AD$ and $AE=2ED$. $EC$ intersects the diagonal $BD$ at point $F$ and the area of $\\triangle DEF$ is $2$. What is the area of $\\triangle BCF$?\n",
    "completion": "\\textbf{Solution:} Since $AE=2ED$, and $AD=AE+DE=3DE$,\\\\\nthus $\\frac{ED}{AD}=\\frac{1}{3}$,\\\\\nSince quadrilateral ABCD is a parallelogram,\\\\\nthus $AD\\parallel BC$, and $BC=AD$,\\\\\nthus $\\triangle EDF\\sim \\triangle CBF$,\\\\\nthus $\\frac{ED}{BC}=\\frac{EF}{CF}=\\frac{ED}{AD}=\\frac{1}{3}$,\\\\\nthus $\\frac{{S}_{\\triangle EDF}}{{S}_{\\triangle BCF}}=\\frac{ED^{2}}{BC^{2}}=\\left(\\frac{1}{3}\\right)^{2}=\\frac{1}{9}$,\\\\\nSince ${S}_{\\triangle EDF}=2$,\\\\\nthus ${S}_{\\triangle BCF}=9\\times 2=\\boxed{18}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51322317.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the side length of square $ABCD$ is 4. Point $E$ is a moving point on line segment $BC$. Connect $AE$, and rotate $AE$ clockwise by $90^\\circ$ around point $E$ to $EF$. Connect $BF$, and take the midpoint $M$ of $BF$. If point $E$ moves from point $B$ to point $C$, what is the length of the path traveled by point $M$?\n",
    "completion": "\\textbf{Solution:} Because rotating line AE clockwise by $90^\\circ$ around point E brings it to line EF,\\\\\ntherefore EF$\\perp$AE.\\\\\nWhen point E is located at point B, point M is at the midpoint G of BC,\\\\\nwhen point E is at point C, point M is at the midpoint H of CD, the path traveled by point M is the length of GH\\\\\nBecause the side length of square ABCD is 4,\\\\\ntherefore $GH=2\\sqrt{2}$,\\\\\nthus, the length of the path traveled by point M is $\\boxed{2\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51510349.png"
  },
  {
    "solution": "\\boxed{35}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, where $AB=AC=20cm$, $DE$ is perpendicular to and bisects $AB$ at point $E$, and intersects $AC$ at $D$. If $BC=15cm$, what is the perimeter of $\\triangle DBC$?\n",
    "completion": "\\textbf{Solution:} Since $DE$ is the perpendicular bisector of $AB$,\\\\\n$\\therefore AD=BD$,\\\\\n$\\therefore$ the perimeter of $\\triangle BCD = BD + CD + BC = AD + CD + BC = AC + BC$\\\\\nAlso, $\\because AB = AC = 20\\,\\text{cm}$, $BC = 15\\,\\text{cm}$,\\\\\n$\\therefore$ the perimeter of $\\triangle BCD = 20 + 15 = \\boxed{35}\\,\\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53148324.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in circle $O$, $\\angle ACB=30^\\circ$, what is the measure of $\\angle AOB$ in degrees?\n",
    "completion": "\\textbf{Solution:} Given that in circle $\\odot O$, $\\angle ACB=30^\\circ$,\\\\\ntherefore, $\\angle AOB=2\\angle ACB=60^\\circ$,\\\\\nhence, the answer is: $\\boxed{60^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53080394.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the diameter of $O$ is $20$, the length of chord $AB$ is $16$, $ON\\perp AB$ with the foot of the perpendicular being $N$. What is the length of $ON$?\n",
    "completion": "\\textbf{Solution:} From the given information, we have \\\\\nOA=10, $\\angle$ONA=$90^\\circ$, AB=16, \\\\\n$\\therefore$AN=8, \\\\\n$\\therefore$ON= $ \\sqrt{OA^{2}-AN^{2}}=\\sqrt{10^{2}-8^{2}}=\\boxed{6}$,",
    "image_path": "dataset/mm_math/images/MM_Math/51368175.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{7}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in an equilateral $\\triangle ABC$ with side length 6, $AD$ is the median on side $BC$, and $M$ is a moving point on $AD$, $E$ is a point on side $AC$. If $AE=2$, what is the minimum value of $EM+CM$?\n",
    "completion": "\\textbf{Solution:} Draw line BE intersecting AD at point M, and draw CM. Construct EH$\\perp$BC at point H,\\\\\n$\\therefore$ $\\angle EHC=\\angle BHE=90^\\circ$,\\\\\n$\\because$ $\\triangle ABC$ is an equilateral triangle, and AD is the median to side BC,\\\\\n$\\therefore$ AD$\\perp$BC, BD=CD,\\\\\n$\\therefore$ Points B and C are symmetric about AD,\\\\\n$\\therefore$ BM=CM,\\\\\n$\\therefore$ EM+CM=BM+ME=BE,\\\\\nThe shortest distance between two points is along the straight line between them, the minimum value of EM+CM is the length of BE,\\\\\n$\\because$ $\\triangle ABC$ is an equilateral triangle, AE=2, BC=AC=6,\\\\\n$\\therefore$ CE=4, $\\angle ECH=60^\\circ$, $\\angle CEH=90^\\circ-60^\\circ=30^\\circ$,\\\\\n$\\therefore$ $CH=\\frac{1}{2}CE=2$,\\\\\n$\\therefore$ BH=6-2=4,\\\\\n$\\therefore$ $EH=\\sqrt{CE^2-CH^2}=\\sqrt{4^2-2^2}=2\\sqrt{3}$;\\\\\nIn $\\triangle BEH$,\\\\\n$BE=\\sqrt{BH^2+EH^2}=\\sqrt{4^2+(2\\sqrt{3})^2}=2\\sqrt{7}$.\\\\\nTherefore, the minimum value of EM+CM is $\\boxed{2\\sqrt{7}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52815743.png"
  },
  {
    "solution": "\\boxed{x > -4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the point $P(-4, 3)$ is on the graph of the linear function $y = kx + b$ (where $k \\neq 0$). Then, what is the solution set of the inequality $kx + b < 3$ with respect to $x$?\n",
    "completion": "\\textbf{Solution:} When $x > -4$, $y < 3$, i.e., $kx + b < 3$,\\\\\nthus the solution set of the inequality $kx + b < 3$ with respect to $x$ is $x > -4$.\\\\\nTherefore, the answer is: $\\boxed{x > -4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50682791.png"
  },
  {
    "solution": "\\boxed{\\frac{3\\sqrt{3}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, the vertices $A$ and $C$ of the rectangle $OABC$ are located on the positive semi-axes of the $x$ and $y$ axes, respectively, and the coordinate of point $B$ is $(2\\sqrt{3}, 4)$. The graph of the linear function $y=-\\frac{\\sqrt{3}}{3}x+b$ intersects the sides $OC$, $AB$, and the $x$ axis at points $D$, $E$, and $F$, respectively, with $\\angle DFO=30^\\circ$, and it satisfies $OD=BE$. Point $M$ is a movable point on segment $DF$. Find the minimum value of $OM+\\frac{1}{2}MF$.\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, draw $MN\\perp x$ axis through point $M$, intersecting at point $N$,\\\\\n$\\because \\angle DFO=30^\\circ$,\\\\\n$\\therefore MN=\\frac{1}{2}MF$,\\\\\n$\\therefore OM+\\frac{1}{2}MF=OM+MN$,\\\\\nConstruct the symmetric point ${O}^{\\prime}$ of point $O$ with respect to the linear function, and let $OO’$ intersect line $DF$ at point $Q$. Draw a perpendicular to the $x$ axis from point ${O}^{\\prime}$, intersecting the $x$ axis at point ${N}^{\\prime}$,\\\\\n$\\therefore OM={O}^{\\prime}M$,\\\\\n$\\therefore OM+\\frac{1}{2}MF=OM+MN={O}^{\\prime}M+MN$,\\\\\nWhen ${O}^{\\prime}$, $M$, and $N$ are on the same line, ${O}^{\\prime}M+MN$ is minimized,\\\\\nThat is, the minimum value of $OM+\\frac{1}{2}MF=OM+MN={O}^{\\prime}M+MN$ is ${O}^{\\prime}{N}^{\\prime}$,\\\\\n$\\because \\angle DFO=30^\\circ$,\\\\\n$\\therefore \\angle ODF=60^\\circ$, $\\angle DOQ=30^\\circ$, $\\angle {O}^{\\prime}O{N}^{\\prime}=60^\\circ$,\\\\\nIn $Rt\\triangle ODQ$, $OQ=OD\\cdot \\sin60^\\circ=3\\times \\frac{\\sqrt{3}}{2}=\\frac{3\\sqrt{3}}{2}$,\\\\\n$\\therefore OO^{\\prime}=2OQ=3$,\\\\\nIn $Rt\\triangle O{N}^{\\prime}{O}^{\\prime}$, ${O}^{\\prime}{N}^{\\prime}=OO^{\\prime}\\sin60^\\circ=3\\times \\frac{\\sqrt{3}}{2}=\\frac{3\\sqrt{3}}{2}$,\\\\\nTherefore, the minimum value of $OM+\\frac{1}{2}MF$ is $\\boxed{\\frac{3\\sqrt{3}}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51152292_114.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the graph of the linear function $y=kx+b$ passes through points $A(-2, -2)$, $B(1, 4)$, and intersects the x-axis at point C, and the y-axis at point D. What is the area of $\\triangle AOB$?\n",
    "completion": "\\textbf{Solution:} When $x=0$, $y=2x+2=2$,\\\\\n$\\therefore$ the coordinates of point D are $\\left(0, 2\\right)$;\\\\\n$\\therefore S_{\\triangle AOB}=\\frac{1}{2}\\times 2\\times 1+\\frac{1}{2}\\times 2\\times 2=\\boxed{3}$\\\\\n$\\therefore$ The area of $\\triangle AOB$ is \\boxed{3}.",
    "image_path": "dataset/mm_math/images/MM_Math/51663193_197.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that points $B$ and $C$ divide the line segment $AD$ into three parts in the ratio $2:4:3$, with $M$ being the midpoint of $AD$. If $CD=6$, find the length of line segment $MC$.\n",
    "completion": "\\textbf{Solution:} Given that points B and C divide the line segment AD into three parts in the ratio 2:4:3, that is, 2+4+3=9,\\\\\nit follows that $CD=\\frac{1}{3}AD$,\\\\\nSince CD equals 6,\\\\\nit follows that AD equals 18,\\\\\nGiven that M is the midpoint of AD,\\\\\nit follows that $MD=\\frac{1}{2}AD=9$,\\\\\ntherefore, $MC=MD-CD=9-6=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52794817_104.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{13}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rhombus ABCD, the diagonals AC and BD intersect at point O, with $BD = 6$ and $AC = 4$. Find the perimeter of rhombus ABCD.\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rhombus,\\\\\nit follows that $\\angle AOB = 90^\\circ$,\\\\\n$BO = \\frac{1}{2}BD = \\frac{1}{2} \\times 6 = 3$,\\\\\n$AO = \\frac{1}{2}AC = \\frac{1}{2} \\times 4 = 2$,\\\\\n$AB = BC = CD = AD$.\\\\\nIn $\\triangle AOB$, by the Pythagorean theorem, we have\\\\\n$AB = \\sqrt{AO^{2} + BO^{2}} = \\sqrt{13}$,\\\\\nthus, the perimeter ${C}_{\\text{rhombus} ABCD} = 4AB = 4\\sqrt{13} = \\boxed{4\\sqrt{13}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52303706.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, it is known that point $D$ is the midpoint of line segment $AB$, and point $C$ is on line segment $AB$. If $AC=8\\,\\text{cm}$ and $BC=6\\,\\text{cm}$, what is the length of line segment $CD$ in cm?\n",
    "completion": "\\textbf{Solution:} Given that $AC=8\\text{cm}, BC=6\\text{cm}$,\\\\\nit follows that $AB=AC+BC=8+6=14\\text{cm}$,\\\\\nsince point D is the midpoint of the line segment $AB$,\\\\\nwe have $BD=\\frac{1}{2}AB=\\frac{1}{2}\\times 14=7\\text{cm}$.\\\\\nSince $CD=BD-BC$,\\\\\nwe conclude that $CD=7-6=\\boxed{1}\\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51000645_151.png"
  },
  {
    "solution": "\\boxed{FE}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $E$ is a point on $BC$, and $DF \\perp AE$ at point $F$, with $\\frac{AD}{AE} = \\lambda$ ($\\lambda > 0$). If $\\lambda = 1$, prove that $CE=FE$.\n",
    "completion": "\\textbf{Solution.} Proof: Draw $DE$ as shown in the figure:\\\\\nSince quadrilateral $ABCD$ is a rectangle,\\\\\nit follows that $\\angle C=90^\\circ$, and $AD\\parallel BC$,\\\\\nthus, $\\angle ADE=\\angle CED$,\\\\\nsince $DF\\perp AE$,\\\\\nit follows that $\\angle DFE=90^\\circ$,\\\\\nthus, $\\angle DFE=\\angle C$,\\\\\nsince $\\frac{AD}{AE}=\\lambda=1$,\\\\\nit follows that $AD=AE$,\\\\\nthus, $\\angle ADE=\\angle FED$,\\\\\nthus, $\\angle FED=\\angle CED$,\\\\\nin $\\triangle DFE$ and $\\triangle DCE$,\\\\\n$\\left\\{\\begin{array}{l}\n\\angle DFE=\\angle C \\\\\n\\angle FED=\\angle CED \\\\\nDE=DE\n\\end{array}\\right.$,\\\\\nit follows that $\\triangle DFE\\cong \\triangle DCE$ (AAS),\\\\\nthus, $CE=FE=\\boxed{FE}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53080796_9.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\triangle ABC$, $\\angle BAC=120^\\circ$, point D is the midpoint of BC, and point E is a point on AC, such that $ AB+AE=EC$. If $ DE=2$, what is the length of AB?\n",
    "completion": "\\textbf{Solution:} Extend $CA$ to point $F$ such that $AF=AB$, and connect $BF$.\\\\\nSince $\\angle BAC=120^\\circ$,\\\\\nit follows that $\\angle BAF=180^\\circ-\\angle BAC=180^\\circ-120^\\circ=60^\\circ$,\\\\\nand since $AF=AB$,\\\\\ntriangle $ABF$ is equilateral,\\\\\nthus $AB=AF=BF$,\\\\\nGiven $AB+AE=EC$,\\\\\nit follows that $AF+AE=EC$, namely $EF=EC$,\\\\\nhence $E$ is the midpoint of $FC$,\\\\\nGiven $D$ is the midpoint of $BC$,\\\\\nit follows that $DE$ is the midline of triangle $BCF$,\\\\\nthus $BF=2DE$,\\\\\nGiven $DE=2$,\\\\\nthus $AB=BF=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53359453.png"
  },
  {
    "solution": "\\boxed{-3 < x < 1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the parabola $y=x^2+bx+c$ passes through points A($-3$, $0$) and C($0$, $-3$). Determine the coordinates of another intersection point B of the parabola with the x-axis, and based on the graph, write the range of values for $x$ when $y<0$.\n",
    "completion": "\\textbf{Solution:} From (1) we know the equation of the parabola is $y = x^2 + 2x - 3$,\\\\\nWhen $y = 0$, $x^2 + 2x - 3 = 0$,\\\\\nWe find: $x_1 = 1$, $x_2 = -3$,\\\\\n$\\therefore$ the other intersection point B of the parabola and the x-axis is at $\\boxed{(1,0)}$,\\\\\nFrom the graph, we know that when $y < 0$, $\\boxed{-3 < x < 1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52622522_219.png"
  },
  {
    "solution": "\\boxed{\\left(-\\frac{1}{2},0\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the graph, the line $y=-2x-4$ intersects the x-axis and y-axis at points A and B, respectively. Points C and D are the midpoints of line segment $AB$ and $OB$, respectively. Point M is a moving point on $OA$. What are the coordinates of point M when the value of $MC+MD$ is minimized?\n",
    "completion": "Solution:Given the problem, we know $A(-2,0)$, $B(0,-4)$,\\\\\nthen $C(-1,-2)$, $D(0,-2)$,\\\\\nConstruct the symmetric point of D with respect to the x-axis, $E(0,2)$, as shown below:\\\\\nFrom this, we know: $MD=ME$,\\\\\n$MC+MD=MC+ME$,\\\\\n$\\therefore$When points $C$, $M$, and $E$ are collinear, $MC+MD$ is minimized,\\\\\nLet the line through $CE$ be $y=kx+b$, substituting $C(-1,-2)$, $E(0,2)$ into the equation, we get\\\\\n$\\left\\{\\begin{array}{l}\nb=2 \\\\\n-k+b=-2\n\\end{array}\\right.$, solving this yields $\\left\\{\\begin{array}{l}\nb=2 \\\\\nk=4\n\\end{array}\\right.$,\\\\\n$y=4x+2$,\\\\\nSetting $y=0$, we solve to get $x=-\\frac{1}{2}$, hence $M\\left(-\\frac{1}{2},0\\right)$,\\\\\nTherefore, the answer is $M\\left(-\\frac{1}{2},0\\right)=\\boxed{\\left(-\\frac{1}{2},0\\right)}$",
    "image_path": "dataset/mm_math/images/MM_Math/50481996.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{3}-3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in parallelogram $ABCD$, $BG$ is perpendicular to $CD$, and $AB=BG=BE$, $AE$ intersects $BG$ at point $F$. If $AB=3$ and $\\angle BAD=60^\\circ$, what is the length of $CE$?\n",
    "completion": "Solution: As shown in the figure, in the parallelogram $ABCD$, $\\angle BAD=\\angle C=60^\\circ$.\\\\\nSince $BG$ is perpendicular to $CD$,\\\\\n$\\therefore \\angle BGC=90^\\circ$,\\\\\n$\\therefore BC=\\frac{BG}{\\sin 60^\\circ}$.\\\\\nAlso, since $AB=BG=BE=3$,\\\\\n$\\therefore BC=\\frac{3}{\\frac{\\sqrt{3}}{2}}=2\\sqrt{3}$,\\\\\n$\\therefore CE=BC-BE=BC-BG=2\\sqrt{3}-3$;\\\\\nThus, the length of $CE$ is $\\boxed{2\\sqrt{3}-3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52551191_85.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, a rectangular piece of paper $ABCD$, with points $E$ and $F$ located on $AB$ and $CD$ respectively, and points $G$, $H$ located on $AF$ and $EC$ respectively. Now, cut the paper along $AF$, $GH$, $EC$, and reassemble it into the rectangle shown in Figure 2. Given that the ratio $DF:AD=5:12$ and $GH=6$, what is the length of $AD$?\n",
    "completion": "\\[ \\textbf{Solution:} \\text{ As shown in the figure, let } DF = 5x, \\text{ according to the problem, we have } AD = 12x, \\]\n\\[ AF = \\sqrt{AD^{2} + DF^{2}} = 13x, \\]\n\\[ \\text{In figure 2, since } \\angle CHA = \\angle FDA = 90^\\circ \\text{ and } \\angle CAH = \\angle FAD, \\]\n\\[ \\therefore \\triangle ADF \\sim \\triangle AHC, \\]\n\\[ \\therefore \\frac{AD}{AH} = \\frac{DF}{HC} = \\frac{AF}{AC}, \\]\n\\[ \\therefore \\frac{12x}{6+12x} = \\frac{5x}{HC} = \\frac{13x}{FC+13x}, \\]\n\\[ \\therefore HC = 5x + \\frac{5}{2}, FC = \\frac{13}{2}, \\]\n\\[ \\therefore \\text{ The area of the rectangle formed as shown in figure 2 } = AH \\times HC = \\left(12x + 6\\right) \\left(5x + \\frac{5}{2} \\right) = 60\\left(x + \\frac{1}{2}\\right)^{2}, \\]\n\\[ \\text{In figure 1,} \\]\n\\[ CD = DF + FC = 5x + \\frac{13}{2}, \\]\n\\[ \\text{The area of the original rectangle } = AD \\times DC = 12x\\left(5x + \\frac{13}{2}\\right), \\]\n\\[ \\therefore 60\\left(x + \\frac{1}{2}\\right)^{2} = 12x\\left(5x + \\frac{13}{2}\\right), \\]\n\\[ \\text{Solving yields } x = \\frac{5}{6}, \\]\n\\[ \\therefore AD = 12x = 12 \\times \\frac{5}{6} = \\boxed{10}. \\]",
    "image_path": "dataset/mm_math/images/MM_Math/52514018.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC=4$, $D$ and $E$ are the midpoints of $AB$ and $AC$ respectively. Line $CD$ is drawn, and through point $E$, line $EF\\parallel DC$ intersects the extension of $BC$ at point $F$. $DE=CF$, if $\\angle B=60^\\circ$, what is the length of $EF$?\n",
    "completion": "\\textbf{Solution:} Since $AB=AC$ and $\\angle B=60^\\circ$ \\\\\nit follows that $\\triangle ABC$ is an equilateral triangle \\\\\nHence, $BC=AB=4$ \\\\\nGiven $D$ is the midpoint of $AB$ \\\\\nit follows $\\angle BDC=90^\\circ$ and $BD=\\frac{1}{2}AB$ \\\\\nThus, $BD=2$ \\\\\nTherefore, $CD=\\sqrt{BC^{2}-BD^{2}}=\\sqrt{4^{2}-2^{2}}=2\\sqrt{3}$ \\\\\nSince quadrilateral $DCFE$ is a parallelogram \\\\\nit follows that $EF=CD=2\\sqrt{3}$ \\\\\nThus, the length of $EF$ is $\\boxed{2\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51708898_149.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points A, B, C, and D are on the circle, where $\\overset{\\frown}{AB}=\\overset{\\frown}{AD}=\\overset{\\frown}{CD}$. The lines BD and AC intersect at point E. As shown in figure 1, if AE=4 and CE=8, what is the length of AB?\n",
    "completion": "\\textbf{Solution:} Given that $\\because \\overset{\\frown}{AB}=\\overset{\\frown}{AD}=\\overset{\\frown}{CD}$, $\\therefore \\angle ABD=\\angle CBD=\\angle ACD=\\angle ACB$, \\\\\nFurthermore, $\\because \\angle BAE=\\angle CAB$, \\\\\n$\\therefore \\triangle ABE \\sim \\triangle ACB$, \\\\\n$\\therefore \\frac{AE}{AB}=\\frac{AB}{AC}$, that is, $\\frac{4}{AB}=\\frac{AB}{12}$ \\\\\n$\\therefore AB=\\boxed{4\\sqrt{3}}$",
    "image_path": "dataset/mm_math/images/MM_Math/55452515_65.png"
  },
  {
    "solution": "\\boxed{y=x^2-2x-3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated, the parabola $y=ax^2+bx+c$ intersects the x-axis at points $A(-1, 0)$ and $B(3, 0)$, and intersects the y-axis at point C, also passing through point $D(2, -3)$. Points P and Q are moving points on the parabola $y=ax^2+bx+c$. What is the equation of the parabola?\n",
    "completion": "\\textbf{Solution:} The expression for the function is: $y=a(x+1)(x-3)$. Substituting the coordinates of point D into the above equation and solving gives: $a=1$, \\\\\nHence, the equation of the parabola is: $y=x^2-2x-3$. \\boxed{y=x^2-2x-3}",
    "image_path": "dataset/mm_math/images/MM_Math/51284522_192.png"
  },
  {
    "solution": "\\boxed{65^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AD$ is the altitude on side $BC$, and $CE$ is an angle bisector that intersects with $AD$ at point $P$. Given that $\\angle APE=55^\\circ$ and $\\angle AEP=80^\\circ$, find the measure of $\\angle BAC$.\n",
    "completion": "\\textbf{Solution:} Since $AD\\perp BC$,\\\\\nit follows that $\\angle PDC=90^\\circ$,\\\\\nGiven $\\angle CPD=\\angle APE=55^\\circ$,\\\\\nit implies $\\angle PCD=90^\\circ-55^\\circ=35^\\circ$,\\\\\nSince $CE$ is an angle bisector,\\\\\nit follows $\\angle ACE=\\angle PCD=35^\\circ$,\\\\\nHence, $\\angle BAC=180^\\circ-35^\\circ-80^\\circ=\\boxed{65^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52991413.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in parallelogram $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$, with $OA=3$. To make parallelogram $ABCD$ a rectangle, what should the length of $BD$ be?\n",
    "completion": "\\textbf{Solution:} Suppose parallelogram \\(ABCD\\) is a rectangle, \\\\\n\\(\\therefore\\) \\(OA=OC=OB=OD\\), \\\\\n\\(\\because\\) \\(OA=3\\), \\\\\n\\(\\therefore\\) \\(BD=2OB=\\boxed{6}\\).",
    "image_path": "dataset/mm_math/images/MM_Math/53313461.png"
  },
  {
    "solution": "\\boxed{\\frac{9}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $AB=3\\text{cm}$, and $BC=6\\text{cm}$. Point $P$ starts moving from point $D$ towards point $A$ and stops upon reaching point $A$; simultaneously, point $Q$ starts moving from point $B$ towards point $C$ and stops upon reaching point $C$. The speeds of points $P$ and $Q$ are both $1\\text{cm/s}$. Connect $PQ$, $AQ$, $CP$. Let the time during which points $P$ and $Q$ move be $t$ seconds. When $t$ equals what value, is quadrilateral $AQCP$ a rhombus?\n",
    "completion": "\\textbf{Solution:} Since $AP=CQ$ and $AP\\parallel CQ$, \\\\\n$\\therefore$ quadrilateral $AQCP$ is a parallelogram, \\\\\n$\\therefore$ when $AQ=CQ$, quadrilateral $AQCP$ becomes a rhombus, \\\\\ni.e., when $\\sqrt{3^2+t^2}=6-t$, quadrilateral $AQCP$ is a rhombus, solving this gives $t=\\boxed{\\frac{9}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53313476_111.png"
  },
  {
    "solution": "\\boxed{110^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point O is on line AB, and ray OC bisects $\\angle AOD$. If $\\angle AOC = 35^\\circ$, then what is the measure of $\\angle BOD$ in degrees?\n",
    "completion": "\\textbf{Solution:} Since line OC bisects $\\angle$DOA, \\\\\n$\\therefore$ $\\angle$AOD = 2$\\angle$AOC, \\\\\nSince $\\angle$COA = $35^\\circ$, \\\\\n$\\therefore$ $\\angle$DOA = $70^\\circ$, \\\\\n$\\therefore$ $\\angle$BOD = $180^\\circ - 70^\\circ = \\boxed{110^\\circ}$,",
    "image_path": "dataset/mm_math/images/MM_Math/52795759.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, isosceles right triangles $S_1$ and $S_2$ are constructed on hypotenuses $AC$ and $BC$, respectively, and a square $S$ is constructed on side $AB$. If the sum of the areas of $S_1$ and $S_2$ equals $9$, what is the length of the side of square $S$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure: draw DE $\\perp$ AC at E. In the isosceles right triangle $\\triangle$ACD, we have AD=CD, and $\\angle$ADC=$90^\\circ$,\\\\\nthus, DE=AE=CE=$\\frac{1}{2}$AC,\\\\\ntherefore, ${S}_{1}=\\frac{1}{2}AC\\cdot \\frac{1}{2}AC=\\frac{1}{4}AC^{2}$,\\\\\nSimilarly, ${S}_{2}=\\frac{1}{4}BC^{2}$,\\\\\ntherefore, ${S}_{1}+{S}_{2}=\\frac{1}{4}AC^{2}+\\frac{1}{4}BC^{2}=9$,\\\\\ntherefore, $AC^{2}+BC^{2}=36$,\\\\\nIn $\\triangle$ABC, $\\angle$ACB=$90^\\circ$, $AC^{2}+BC^{2}=AB^{2}$,\\\\\ntherefore, S=$AB^{2}=36$,\\\\\ntherefore, the side length of the square S equals $\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51367277.png"
  },
  {
    "solution": "\\boxed{40^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $\\angle AOB=120^\\circ$. OC is a ray inside $\\angle AOB$, and $\\angle AOC: \\angle BOC=1: 2$. Draw ray OM to bisect $\\angle AOC$. Inside $\\angle BOC$, draw ray ON such that $\\angle CON: \\angle BON=1: 3$. What is the measure of $\\angle MON$ in degrees?\n",
    "completion": "\\textbf{Solution:} Given that OM bisects $\\angle$AOC,\\\\\n$\\therefore \\angle$COM$= \\frac{1}{2}\\angle$AOC$= \\frac{1}{2}\\times 40^\\circ=20^\\circ$,\\\\\nSince $\\angle$CON$\\colon \\angle$BON$=1\\colon 3$,\\\\\n$\\therefore \\angle$CON$= \\frac{1}{4}\\angle$BOC$= \\frac{1}{4}\\times 80^\\circ=20^\\circ$,\\\\\nHence, $\\angle$MON$=\\angle$COM$+\\angle$CON$=20^\\circ+20^\\circ=\\boxed{40^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51438377_169.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{2}x^2+2x}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the parabola $y=-x^2+2x+4$ intersects the $y$-axis at point $C$ and has the vertex at $D$. Connect $CD$. Point $Q$ is on the line $CD$ in the first quadrant. Draw $QM\\perp x$-axis, intersecting the $x$-axis at point $M$. If the abscissa of point $Q$ is $x$, and the area of $\\triangle QMO$ is $S$, what is the function expression of $S$ in terms of $x$?\n",
    "completion": "\\textbf{Solution:} Let the equation of line $CD$ be $y=ax+b$. Substituting points $D(1, 5)$ and $C(0, 4)$ into this, we get\n\\[\n\\left\\{\n\\begin{array}{l}\na+b=5 \\\\\nb=4\n\\end{array}\n\\right.\n\\]\nSolving this system yields\n\\[\n\\left\\{\n\\begin{array}{l}\na=1 \\\\\nb=4\n\\end{array}\n\\right.\n\\]\nTherefore, the equation of line $CD$ is $y=x+4$.\\\\\nSince the x-coordinate of point $Q$ is $x$,\\\\\nTherefore, $Q(x, x+4)$, $M(x, 0)$,\\\\\nSince point $Q$ is in the first quadrant,\\\\\nTherefore, $QM=x+4$, $MO=x$,\\\\\nTherefore, $S=\\frac{1}{2}QM\\cdot MO=\\frac{1}{2}x(x+4)=\\boxed{\\frac{1}{2}x^2+2x}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52660399_102.png"
  },
  {
    "solution": "\\boxed{20}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, D is the midpoint of AB, $AE\\parallel CD$, and $CE\\parallel AB$. Quadrilateral ADCE is a rhombus. If $BC=6$ and $\\tan B=\\frac{4}{3}$, what is the perimeter of quadrilateral ADCE?\n",
    "completion": "\\textbf{Solution:} In right-angled $\\triangle ABC$, where $BC=6$ and $\\tan B=\\frac{AC}{BC}=\\frac{4}{3}$,\\\\\n$\\therefore AC=\\frac{4}{3}BC=\\frac{4}{3}\\times 6=8$,\\\\\n$\\therefore AB=\\sqrt{AC^2+BC^2}=\\sqrt{8^2+6^2}=10$,\\\\\n$\\therefore CD=\\frac{1}{2}AB=5$,\\\\\n$\\because$ Quadrilateral $ADCE$ is a rhombus,\\\\\n$\\therefore CD=DA=AE=EC=5$,\\\\\n$\\therefore$ The perimeter of rhombus $ADCE$ is: $5\\times 4=\\boxed{20}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51445991_121.png"
  },
  {
    "solution": "\\boxed{y=-2x+6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y=kx+b$ intersects with the graph of the inverse proportion function $y=\\frac{4}{x}$ (where $x>0$) at points $A(m, 4)$ and $B(2, n)$, and intersects with the coordinate axes at points $M$ and $N$, respectively. What is the analytical expression of the linear function?\n",
    "completion": "\\textbf{Solution:} Given that A $(1, 4)$ and B $(2, 2)$ are on the graph of the inverse proportion function $y=\\frac{4}{x}$,\n\n\\[\n\\therefore\n\\begin{cases}\n4=\\frac{4}{m} \\\\\nn=\\frac{4}{2}\n\\end{cases}\n\\text{, which implies }\n\\begin{cases}\nm=1 \\\\\nn=2\n\\end{cases},\n\\]\nthus A $(1, 4)$ and B $(2, 2)$.\n\nSubstituting A $(1, 4)$ and B $(2, 2)$ into $y=kx+b$, we get:\n\\[\n\\begin{cases}\n4=k+b \\\\\n2=2k+b\n\\end{cases}\n\\]\nwhich solves to:\n\\[\n\\begin{cases}\nk=-2 \\\\\nb=6\n\\end{cases},\n\\]\ntherefore, the equation of the linear function is \\boxed{y=-2x+6}.",
    "image_path": "dataset/mm_math/images/MM_Math/52601975_176.png"
  },
  {
    "solution": "\\boxed{\\pi-2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $ \\triangle ABC$ is an inscribed triangle of $ \\odot O$, and $ AB=2\\sqrt{2}$. What is the area of the shaded region in the figure?\n",
    "completion": "\\textbf{Solution:} Connect OA and OB.\\\\\nIn $\\triangle OAB$, OA=OB=2, AB=2$\\sqrt{2}$,\\\\\n$\\therefore$ OA$^{2}$+OB$^{2}$=2$^{2}$+2$^{2}$=8, AB$^{2}$=(2$\\sqrt{2}$)$^{2}$=8,\\\\\n$\\therefore$ OA$^{2}$+OB$^{2}$=AB$^{2}$,\\\\\n$\\therefore$ $\\angle$AOB=$90^\\circ$,\\\\\n$\\therefore$ Area of the sector $OAB$= $\\frac{90}{360}\\times \\pi \\times 4=\\pi$, Area of $\\triangle OAB$= $\\frac{1}{2}$ ×2×2=2,\\\\\n$\\therefore$ Area of the shaded region=Area of the sector $OAB$−Area of $\\triangle OAB$=$\\pi$−2.\\\\\nTherefore, the area of the shaded region in the diagram is $\\boxed{\\pi-2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51213420.png"
  },
  {
    "solution": "\\boxed{56^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, AD is the angle bisector of $\\triangle ABC$, $\\angle C=28^\\circ$, and AB+BD=AC. When $\\triangle ABD$ is folded along the line containing AD, the point where B lands on side AC is marked as point E. What is the measure of $\\angle AED$?\n",
    "completion": "\\textbf{Solution:} According to the property of folding, we have \\(BD=DE\\), \\(AB=AE\\).\n\n\\(\\because AB+BD=AC\\), \\(AC=AE+EC\\),\n\n\\(\\therefore AB+BD=AE+EC\\),\n\n\\(\\therefore DE=EC\\),\n\n\\(\\therefore \\angle EDC=\\angle C=28^\\circ\\),\n\n\\(\\therefore \\angle AED=\\angle EDC+\\angle C=28^\\circ+28^\\circ=\\boxed{56^\\circ}\\).",
    "image_path": "dataset/mm_math/images/MM_Math/52603188.png"
  },
  {
    "solution": "\\boxed{(-2, 6)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the quadratic function $y=ax^2+bx+4$ ($a\\neq 0$) passes through points $A(-4,0)$, $B(1,0)$, and intersects the y-axis at point $C$. Point $P$ is a point on the parabola in the second quadrant. Connect $BP$ and $AC$, intersecting at point $Q$, and draw $PD\\perp x$ axis at point $D$. Determine whether $\\frac{PQ}{QB}$ has a maximum value? If so, find the coordinates of point $P$ when $\\frac{PQ}{QB}$ is maximized; if not, explain why.\n",
    "completion": "\\textbf{Solution}: Set point $N$ where line segment $PD$ intersects with line segment $AC$. Draw a line parallel to the y-axis through point $B$ that intersects line segment $AC$ at point $M$. Given that the coordinates of points $A$ and $C$ are $(-4,0)$ and $(0,4)$, respectively, the equation of line $AC$ is $y=x+4$. Therefore, the coordinates of point $M$ are $(1,5)$, and the length of $BM$ is $5$. Since $BM \\parallel PN$, triangles $\\triangle PNQ$ and $\\triangle BMQ$ are similar. Thus, $\\frac{PQ}{QB}=\\frac{PN}{BM}=\\frac{PN}{5}$. Let $P\\left(a_{0}, -a_{0}^{2}-3a_{0}+4\\right)\\left(-4<a_{0}<0\\right)$, then $N\\left(a_{0}, a_{0}+4\\right)$. Consequently, $\\frac{PQ}{QB}=\\frac{-a_{0}^{2}-3a_{0}+4-(a_{0}+4)}{5}=\\frac{-a_{0}^{2}-4a_{0}}{5}=\\frac{-\\left(a_{0}+2\\right)^{2}+4}{5}$. Therefore, when $a_{0}=-2$, $\\frac{PQ}{QB}$ reaches its maximum value of $0.8$. At this point, the coordinates of point $P$ are $(-2, 6)$. Hence, the maximum value of $\\frac{PQ}{QB}$ is $\\boxed{0.8}$, and the coordinates of point $P$ at this moment are $\\boxed{(-2, 6)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53059889_153.png"
  },
  {
    "solution": "\\boxed{\\sqrt{6}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given a rectangle $ABCD$ with a point $P$ satisfying $PA=1$, $PB=2$, $PC=3$, what is the value of $PD$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw GH$\\parallel$BC through point P intersecting AB and CD at G and H, respectively; draw EF$\\parallel$AB through point P intersecting AD and BC at E and F, respectively. Let AG=DH=x, GB=HC=y, AE=BF=m, and ED=FC=n.\\\\\nTherefore, AP$^{2}$=x$^{2}$+m$^{2}$, CP$^{2}$=y$^{2}$+n$^{2}$, BP$^{2}$=y$^{2}$+m$^{2}$, DP$^{2}$=n$^{2}$+x$^{2}$,\\\\\nThus, AP$^{2}$+CP$^{2}$=BP$^{2}$+DP$^{2}$=x$^{2}$+y$^{2}$+m$^{2}$+n$^{2}$,\\\\\nGiven that, PA=1, PB=2, PC=3,\\\\\nThis implies $1+9=4+DP^{2}$,\\\\\nSolving yields: $DP=\\boxed{\\sqrt{6}}$ or $-\\sqrt{6}$ (discard the latter).",
    "image_path": "dataset/mm_math/images/MM_Math/52225958.png"
  },
  {
    "solution": "\\boxed{25^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $\\angle AOB = 40^\\circ$, and $OD$ is the bisector of $\\angle BOC$. When $\\angle AOB$ and $\\angle BOC$ are complementary, what is the degree measure of $\\angle COD$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle AOB$ and $\\angle BOC$ are complementary,\\\\\nit follows that $\\angle AOB+\\angle BOC=90^\\circ$,\\\\\nhence $\\angle BOC=90^\\circ-40^\\circ=50^\\circ$,\\\\\nsince $OD$ bisects $\\angle BOC$,\\\\\nit results in $\\angle COD=\\frac{1}{2}\\angle BOC=\\boxed{25^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52916444_92.png"
  },
  {
    "solution": "\\boxed{105^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $OC$ bisects $\\angle AOD$, $\\angle BOC=30^\\circ$, and $\\angle BOC$ is complementary to $\\angle AOD$. Find the measures of $\\angle AOD$ and $\\angle BOD$.\n",
    "completion": "\\textbf{Solution:}\\\\\nSince $\\angle$BOC and $\\angle$AOD are complementary,\\\\\n$\\therefore$ $\\angle$BOC + $\\angle$AOD = 180$^\\circ$,\\\\\nSince $\\angle$BOC = 30$^\\circ$,\\\\\n$\\therefore$ $\\angle$AOD = 150$^\\circ$,\\\\\nSince OC bisects $\\angle$AOD,\\\\\n$\\therefore$ $\\angle$DOC = $\\frac{1}{2}\\angle$AOD = $\\frac{1}{2} \\times 150^\\circ = 75^\\circ$,\\\\\n$\\therefore$ $\\angle$BOD = $\\angle$DOC + $\\angle$BOC\\\\\n= 75$^\\circ$ + 30$^\\circ$\\\\\n= \\boxed{105^\\circ},\\\\\nthat is, $\\angle$AOD = \\boxed{150^\\circ}, $\\angle$BOD = \\boxed{105^\\circ}.",
    "image_path": "dataset/mm_math/images/MM_Math/52773591_162.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB=18\\,cm$, C is the midpoint of $AB$. Point D is on line segment $AC$, and $AD:CB=1:3$, what is the length of $DC$?\n",
    "completion": "\\textbf{Solution:} Given $\\because AB=18$cm, and $C$ is the midpoint of $AB$,\\\\\n$\\therefore AC=BC=\\frac{1}{2}AB=9$cm,\\\\\n$\\because AD:CB=1:3$,\\\\\n$\\therefore AD=3$cm,\\\\\n$\\therefore DC=AC-AD=9-3=\\boxed{6}$cm.",
    "image_path": "dataset/mm_math/images/MM_Math/52732598.png"
  },
  {
    "solution": "\\boxed{7}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, the traces of compass and straightedge constructions intersect sides $BC$ and $AB$ of $\\triangle ABC$ at points D and E, respectively. Construct $DF \\perp AB$ at point F. On $AC$, pick a point G such that $DE = DG$. If the area of $\\triangle ADG$ is $52$ and the area of $\\triangle AED$ is $38$, what is the area of $\\triangle DEF$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, construct $DH\\perp AC$ at H,\\\\\nfrom the construction traces, $AD$ bisects $\\angle BAC$,\\\\\n$\\because$ $AD$ bisects $\\angle BAC$, and $DF\\perp AB$, $DH\\perp AC$,\\\\\n$\\therefore$ $DF=DH$,\\\\\nIn right triangles $\\triangle DEF$ and $\\triangle DGH$,\\\\\n$\\left\\{\\begin{array}{l}DE=DG\\\\ DF=DH\\end{array}\\right.$,\\\\\n$\\therefore$ right $\\triangle DEF\\cong$ right $\\triangle DGH$ (HL),\\\\\nIn right triangles $\\triangle DAF$ and $\\triangle DAH$,\\\\\n$\\left\\{\\begin{array}{l}DF=DM\\\\ AD=AD\\end{array}\\right.$,\\\\\n$\\therefore$ right $\\triangle DAF\\cong$ right $\\triangle DAH$ (HL),\\\\\n$\\therefore$ area of $\\triangle DEF$ = area of $\\triangle DGH$, area of $\\triangle DAF$ = area of $\\triangle DAH$,\\\\\n$\\because$ area of $\\triangle ADG$ = area of $\\triangle AHD$ + area of $\\triangle DGH$, area of $\\triangle AHD$ = area of $\\triangle ADE$ + area of $\\triangle DEF$,\\\\\n$\\therefore$ area of $\\triangle ADE$ + area of $\\triangle DEF$ = area of $\\triangle ADG$ - area of $\\triangle DGH$,\\\\\ni.e., $38$ + area of $\\triangle DEF$ = $52$ - area of $\\triangle DEF$,\\\\\n$\\therefore$ area of $\\triangle DEF$ = $\\boxed{7}$,\\\\",
    "image_path": "dataset/mm_math/images/MM_Math/52950954.png"
  },
  {
    "solution": "\\boxed{110^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is a chord of the circle $\\odot O$, and point $C$ is a point on the circle $\\odot O$. Given that $\\angle ACB = 55^\\circ$, what is the measure of $\\angle AOB$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle ACB = 55^\\circ$, \\\\\n$\\therefore \\angle AOB = 2\\angle ACB = 110^\\circ$, \\\\\n$\\therefore \\angle AOB = \\boxed{110^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52796840.png"
  },
  {
    "solution": "\\boxed{80}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, $AE$ bisects $\\angle BAC$ and intersects $BC$ at point $E$, and point $D$ is on $AB$, with $DE\\perp AE$. The circumcircle $\\odot O$ of $\\triangle ADE$ intersects $AC$ at point $F$. The line $BC$ is a tangent to the circle $\\odot O$; if the radius of $\\odot O$ is $10$ and $AC=16$, find the area of $\\triangle ADE$.\n",
    "completion": "\\textbf{Solution:} Given that $AD$ is the diameter of $\\odot O$,\\\\\n$\\therefore \\angle AED=90^\\circ$,\\\\\n$\\therefore \\angle C=\\angle AED=90^\\circ$,\\\\\nGiven that $\\angle 1=\\angle 2$,\\\\\n$\\therefore \\triangle ACE\\sim \\triangle AED$,\\\\\n$\\therefore \\frac{AC}{AE}=\\frac{AE}{AD}$, that is $\\frac{16}{AE}=\\frac{AE}{20}$,\\\\\n$\\therefore AE=8\\sqrt{5}$ (negative values are discarded),\\\\\n$\\therefore DE=\\sqrt{AD^{2}-AE^{2}}=\\sqrt{20^{2}-\\left(8\\sqrt{5}\\right)^{2}}=4\\sqrt{5}$,\\\\\n$\\therefore S_{\\triangle ADE}=\\frac{1}{2}AE\\cdot DE=\\frac{1}{2}\\times 8\\sqrt{5}\\times 4\\sqrt{5}=\\boxed{80}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53010285_78.png"
  },
  {
    "solution": "\\boxed{10.5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y_{1}=kx+b$ intersects with the graph of the inverse proportion function $y_{2}=\\frac{m}{x}$ at points $A\\left(-2, -5\\right)$ and $C\\left(5, n\\right)$. It intersects the y-axis at point B and the x-axis at point D. Connect $OA$ and $OC$. What is the area of $\\triangle AOC$?\n",
    "completion": "Solution: Substituting $y=0$ into $y_1=x-3$ yields: $x=3$,\\\\\ntherefore, the coordinates of point D are $(3,0)$,\\\\\ntherefore, $OD=3$,\\\\\ntherefore, $S_{\\triangle AOC}=S_{\\triangle DOC}+S_{\\triangle AOD}=\\frac{1}{2}\\times 3\\times 2+\\frac{1}{2}\\times 3\\times |{-5}|=\\boxed{10.5}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52601939_122.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, in $\\triangle ABC$, $BO \\perp AC$ at point $O$, $AO=BO=3$, $OC=1$. A line is drawn through point $A$ such that $AH \\perp BC$ at point $H$, intersecting $BO$ at point $P$. What is the length of line segment $OP$?\n",
    "completion": "\\textbf{Solution:} Since $BO \\perp AC$ and $AH \\perp BC$,\\\\\n$\\therefore$ $\\angle AOP=\\angle BOC=\\angle AHC=90^\\circ$,\\\\\n$\\therefore$ $\\angle OAP+\\angle C=\\angle OBC+\\angle C=90^\\circ$,\\\\\n$\\therefore$ $\\angle OAP=\\angle OBC$,\\\\\nIn $\\triangle OAP$ and $\\triangle OBC$,\\\\\n$\\left\\{\\begin{array}{l}\n\\angle AOP=\\angle BOC \\\\\nAO=BO \\\\\n\\angle OAP=\\angle OBC\n\\end{array}\\right.$,\\\\\n$\\therefore$ $\\triangle OAP \\cong \\triangle OBC$ (ASA),\\\\\n$\\therefore$ OP=OC=\\boxed{1};",
    "image_path": "dataset/mm_math/images/MM_Math/52913848_121.png"
  },
  {
    "solution": "\\boxed{\\frac{8}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the vertices A and B of the square $ABCD$ both lie on the circle $\\odot O$, and side $CD$ is tangent to the circle $\\odot O$ at point E. If the radius of the circle $\\odot O$ is 1, what is the side length of the square $ABCD$?\n",
    "completion": "\\textbf{Solution:} Let's extend line $EO$ and have it intersect $AB$ at point $F$, and also connect $OA$,\\\\\n$\\because$ side $CD$ of square $ABCD$ is tangent to circle $O$ at point E,\\\\\n$\\therefore$ $OE\\perp CD$, $AB\\parallel CD$, $\\angle BAD=\\angle ADC=90^\\circ$,\\\\\n$\\therefore$ $OF\\perp AB$,\\\\\n$\\therefore$ quadrilateral $EFAD$ forms a rectangle,\\\\\n$\\therefore$ $EF=AD$,\\\\\n$\\because$ vertices A and B of square $ABCD$ are located on circle $O$,\\\\\n$\\therefore$ $AF=\\frac{1}{2}AB$,\\\\\n$\\because$ the radius of circle $O$ is 1,\\\\\n$\\therefore$ $OE=OA=1$,\\\\\nLet the side length of the square be $a$, then: $OF=EF-OE=a-1$, $AF=\\frac{a}{2}$,\\\\\nIn right triangle $\\triangle OFA$: $OA^{2}=OF^{2}+AF^{2}$,\\\\\nwhich gives: $1^{2}=(a-1)^{2}+\\left(\\frac{a}{2}\\right)^{2}$,\\\\\nSolving this equation results in: $a=0$ (discard) or $a=\\frac{8}{5}$;\\\\\n$\\therefore$ the side length of the square is: $\\boxed{\\frac{8}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53110761.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, point $D$ is the midpoint of segment $AC$, and $AB=\\frac{2}{3}BC$, $BC=6$. Find the length of segment $BD$.\n",
    "completion": "\\textbf{Solution:} By the given conditions, \\\\\nsince $AB=\\frac{2}{3}BC$ and $BC=6$, \\\\\nthen $AB=4$ and $AC=10$, \\\\\nsince point $D$ is the midpoint of line segment $AC$, \\\\\nthus $AD=\\frac{1}{2}AC=5$, \\\\\ntherefore $BD=5-4=\\boxed{1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51385662_124.png"
  },
  {
    "solution": "\\boxed{16^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rhombus $ABCD$, $AB=10$, diagonals $AC$ and $BD$ intersect at point $O$. If $E$ is the midpoint of side $AD$ and $\\angle AEO = 32^\\circ$, what is the length of $OE$? What is the degree measure of $\\angle ADO$?\n",
    "completion": "\\textbf{Solution:} Given: In diamond ABCD, AC and BD intersect at point O,\\\\\nit follows that OA=OC, $\\angle$ADB=$\\angle$CDB, and CD=AB=10,\\\\\nsince E is the midpoint of side AD,\\\\\nit follows AE=DE,\\\\\nthus OE is the median line of $\\triangle$ADC,\\\\\nhence OE $\\parallel$ CD, and OE=$\\frac{1}{2}$CD=$\\frac{1}{2} \\times 10=\\boxed{5}$,\\\\\ntherefore $\\angle$ADC=$\\angle$AEO=32$^\\circ$,\\\\\nhence $\\angle$ADO=$\\frac{1}{2}\\angle$ADC=$\\frac{1}{2} \\times 32^\\circ=\\boxed{16^\\circ}$,",
    "image_path": "dataset/mm_math/images/MM_Math/53333412.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $ \\triangle ABC$ is inscribed in a circle $\\odot O$ with a radius of 6. Given $\\sin\\angle A=\\frac{2}{3}$, what is the length of $BC$?\n",
    "completion": "\\textbf{Solution:} Construct the diameter $CD$ of $\\odot O$, and connect $BD$, hence $CD=2\\times 6=12$.\\\\\n$\\therefore$ $\\angle CBD=90^\\circ$, $\\angle D=\\angle A$,\\\\\n$\\therefore$ $BC=CD\\cdot \\sin D=CD\\cdot \\sin A=12\\times \\frac{2}{3}=\\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53066096.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $\\triangle ABC$ and $\\triangle EDC$ are similar figures with point $C$ as the center of similarity, and the similarity ratio between $\\triangle ABC$ and $\\triangle EDC$ is $1\\colon 2$. If the area of $\\triangle ABC$ is $2$, what is the area of $\\triangle EDC$?\n",
    "completion": "\\textbf{Solution:} Since triangles $\\triangle ABC$ and $\\triangle EDC$ are homothetic,\\\\\nit follows that $\\triangle ABC \\sim \\triangle EDC$,\\\\\nand since the homothety ratio is $1\\colon 2$,\\\\\nthe similarity ratio is $1\\colon 2$,\\\\\nthus, the area ratio of $\\triangle ABC$ to $\\triangle EDC$ is: $1\\colon 4$,\\\\\nsince the area of $\\triangle ABC$ is $2$,\\\\\nthe area of $\\triangle EDC$ is: $2 \\times 4 = \\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53136801.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $EG \\parallel BC$, if $\\frac{EG}{BC} = \\frac{1}{3}$, then what is the value of $\\frac{AF}{AD}$?\n",
    "completion": "\\textbf{Solution:} Given $EG \\parallel BC$,\\\\\nthus $\\triangle AEG \\sim \\triangle ABC$,\\\\\nso $\\frac{AE}{AB}=\\frac{EG}{BC}=\\frac{1}{3}$,\\\\\nGiven $EG \\parallel BC$,\\\\\nthus $\\triangle AEF \\sim \\triangle ABD$,\\\\\nso $\\frac{AF}{AD}=\\frac{AE}{AB}=\\boxed{\\frac{1}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51285006.png"
  },
  {
    "solution": "\\boxed{9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, points D and E are on AB and AC respectively, $\\angle AED=\\angle B$, $AD=\\frac{3}{4}AC$. If the area of the quadrilateral BCED is 7, what is the area of $\\triangle ADE$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle A = \\angle A$ and $\\angle AED = \\angle B$,\n\nit follows that $\\triangle ADE \\sim \\triangle ACB$.\n\nThus, $\\frac{AD}{AC} = \\frac{\\frac{3}{4}AC}{AC} = \\frac{3}{4}$,\n\nand $\\frac{S_{\\triangle ADE}}{S_{\\triangle ABC}} = \\left(\\frac{AD}{AC}\\right)^2 = \\left(\\frac{3}{4}\\right)^2 = \\frac{9}{16}$.\n\nLet $S_{\\triangle ADE} = 9x$, then $S_{\\triangle ABC} = 16x$.\n\nTherefore, $S_{BCED} = S_{\\triangle ABC} - S_{\\triangle ADE} = 16x - 9x = 7x = 7$,\n\nimplying $x = 1$,\n\nand therefore $S_{\\triangle ADE} = 9x = \\boxed{9}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51545217.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in squares $ABCD$ and $CEFG$, where point $D$ is on $CG$, given that $BC=1$ and $CE=7$, and point $H$ is the midpoint of $AF$, what is the length of $CH$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, extend $AD$ to meet $EF$ at $M$, and draw lines $AC$, $CF$.\n\nSince in squares $ABCD$ and $CEFG$, point $D$ lies on $CG$, given that $BC=1$, $CE=7$,\n\nHence, $AB=AD=CD=BC=1$, $EF=CE=7$,\n\n$\\angle GCE=\\angle E=90^\\circ$, $\\angle ADC=90^\\circ$,\n\nTherefore, $\\angle CDM=180^\\circ-\\angle ADC=90^\\circ$,\n\nTherefore, quadrilateral $DCEM$ is a rectangle,\n\nTherefore, $DM=CE=7$, $ME=CD=1$, $\\angle DME=90^\\circ$,\n\nTherefore, $AM=AD+DM=1+7=8$, $FM=EF-ME=7-1=6$,\n\n$\\angle AMF=180^\\circ-\\angle DME=180^\\circ-90^\\circ=90^\\circ$,\n\nTherefore, in $\\triangle AMF$ right-angled at $M$, $AF=\\sqrt{AM^2+FM^2}=\\sqrt{8^2+6^2}=10$,\n\nSince squares $ABCD$ and $CEFG$ are squares,\n\nTherefore, $\\angle ACD=\\angle GCF=45^\\circ$,\n\nTherefore, $\\angle ACF=90^\\circ$,\n\nSince point $H$ is the midpoint of $AF$,\n\nTherefore, $CH=\\frac{1}{2}AF=\\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52302533.png"
  },
  {
    "solution": "\\boxed{x>1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system xOy, if the line $y_1=-2x+a$ and the line $y_2=bx-4$ intersect at point $P(1, -3)$, what is the solution set of the inequality $-2x+a<bx-4$ with respect to x?\n",
    "completion": "\\textbf{Solution:} When $x>1$, the graph of the function $y_{1}=-2x+a$ is entirely below the graph of $y_{2}=bx-4$. Hence, the solution set of the inequality $-2x+a<bx-4$ is $x>1$;\\\\\nTherefore, the answer is: $\\boxed{x>1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52094910.png"
  },
  {
    "solution": "\\boxed{\\left(16, 8\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, within the Cartesian coordinate system, the line passing through point $C(0, 12)$ intersects with line OA at point $A(8, 4)$. Suppose a point M is moving along the ray OA. Is there any position of M that makes the area of $\\triangle OBM$ equal to the area of $\\triangle OAC$? If such a point exists, find the coordinates of M; if not, explain why.\n",
    "completion": "\\textbf{Solution:} Exists, \\\\\nBy substituting $y=0$ into $y=-x+12$, we get $x=12$, \\\\\n$\\therefore$ point $B(12, 0)$, \\\\\n$\\therefore OB=12$, \\\\\nLet the equation of line OA be $y=mx$, \\\\\nSubstituting $A(8, 4)$ gives $8m=4$, \\\\\n$\\therefore m=\\frac{1}{2}$, \\\\\nThe equation of line OA is $y=\\frac{1}{2}x$, \\\\\nLet the coordinates of point M be $\\left(x, \\frac{1}{2}x\\right)$, \\\\\n$\\frac{1}{2}\\times OB\\times \\frac{1}{2}x=48$, \\\\\n$\\frac{1}{2}\\times 12\\times \\frac{1}{2}x=48$, \\\\\n$x=16$, \\\\\nThus, the coordinates of point M are $\\boxed{\\left(16, 8\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52690876_66.png"
  },
  {
    "solution": "\\boxed{\\frac{5}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, rotate rectangle $ABCD$ clockwise about point $A$, causing point $E$ to fall on $BD$, thus forming rectangle $AEFG$. $EF$ intersects $AD$ at point $H$. Connect $AF$. $BD \\parallel AF$; If $AB=1$ and $BC=2$, find the length of $AH$.\n",
    "completion": "\\textbf{Solution:} Given that $BD\\parallel AF$, $\\therefore \\angle 3=\\angle EFA$. Since $\\angle 4=\\angle EFA$, $\\therefore \\angle 4=\\angle 3$. Hence, $EH=DH$. Given $AB=1$, $BC=2$, we have $AE=1$, $AD=2$. Let $EH=DH=x$, then $AH=2-x$. In $\\triangle AEH$, by Pythagoras' theorem, we have $AE^2+EH^2=AH^2$, which implies $1^2+x^2=(2-x)^2$. Solving this, we find $x=\\frac{3}{4}$, hence $AH=2-\\frac{3}{4}=\\boxed{\\frac{5}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51445148_143.png"
  },
  {
    "solution": "\\boxed{\\text{pentagonal prism}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: The figure on the right is the unfolded shape of a certain solid. What is this solid?\n",
    "completion": "\\textbf{Solution:} According to the net of the solid, we can conclude that the solid is a \\boxed{\\text{pentagonal prism}}.",
    "image_path": "dataset/mm_math/images/MM_Math/55300127.png"
  },
  {
    "solution": "\\boxed{15}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, points D and E are on sides $AB$ and $AC$, respectively. If $DE \\parallel BC$ and $\\frac{AD}{DB} = \\frac{2}{3}$, $DE = 6\\,\\text{cm}$, then what is the length of $BC$ in cm?\n",
    "completion": "\\textbf{Solution:} Since $\\frac{AD}{DB}=\\frac{2}{3}$,\\\\\nit follows that $\\frac{AD}{AB}=\\frac{2}{5}$,\\\\\nSince $DE\\parallel BC$,\\\\\nit implies $\\triangle ADE\\sim \\triangle ABC$,\\\\\nthus $\\frac{AD}{AB}=\\frac{DE}{BC}$, and given $DE=6$,\\\\\nit means $\\frac{6}{BC}=\\frac{2}{5}$,\\\\\ntherefore $BC=\\boxed{15}$, which is verified to be in accordance with the problem statement;",
    "image_path": "dataset/mm_math/images/MM_Math/53043255.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point C is the midpoint of segment $AB$, and point D is on segment $CB$. If $AB=10$ and $DB=2$, then what is the length of segment $CD$?\n",
    "completion": "\\textbf{Solution:} Since point $C$ is the midpoint of line segment $AB$ and $AB=10$, \\\\ \nit follows that $BC=\\frac{1}{2}AB=5$, \\\\\nFurthermore, since $DB=2$, \\\\\nit follows that $CD=BC-DB=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53253227.png"
  },
  {
    "solution": "\\boxed{8−4\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the length of segment $BC$ is $8$. With $B$ as the center and $4$ as the radius, a circle $\\odot B$ is drawn. Point $A$ is a moving point on $\\odot B$. Segment $AC$ is connected, and then segment $AC$ is rotated $90^\\circ$ counterclockwise about point $A$ to obtain segment $AD$. Connect $DB$. What is the minimum value of $DB$?\n",
    "completion": "\\textbf{Solution:} Connect DC, AB, and rotate segment BC clockwise around point B by $90^\\circ$ to get segment BF. Connect CF, DF,\\\\\n$\\therefore$ $\\triangle BCF$ is an isosceles right triangle,\\\\\n$\\therefore$ $CF=\\sqrt{BC^2+BF^2}=\\sqrt{8^2+8^2}=8\\sqrt{2}$;\\\\\n$\\because$ rotating segment AC around point A clockwise by $90^\\circ$ gets segment AD,\\\\\n$\\therefore$ $\\triangle ADC$ is an isosceles right triangle,\\\\\n$\\therefore$ $\\triangle ADC \\sim \\triangle CBF$,\\\\\n$\\therefore$ $\\frac{DC}{AC}=\\frac{CF}{AB}$,\\\\\n$\\because$ $\\angle BCF=\\angle ACD$,\\\\\n$\\therefore$ $\\angle DCF=\\angle ACB$,\\\\\n$\\therefore$ $\\triangle DCF \\sim \\triangle ACB$,\\\\\n$\\therefore$ $\\frac{AB}{DF}=\\frac{BC}{CF}$ thus $\\frac{4}{DF}=\\frac{8}{8\\sqrt{2}}$,\\\\\nSolving this, $DF=4\\sqrt{2}$;\\\\\nAs shown in the diagram, point D moves on circle F. If circle F intersects BF at point E, when point D moves to point E, the length of BD is the smallest,\\\\\n$\\therefore$ The minimal value is $BE=BF−EF=BF−DF=8−4\\sqrt{2}$,\\\\\n$\\therefore$ The minimal value of $DB$ is $\\boxed{8−4\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52861191.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point C is on the semicircle O with diameter AB, and $AB=2$. AD bisects $\\angle BAC$ and intersects BC at point D. CP bisects $\\angle BCA$ and intersects AD at point P. $PF\\perp AC$, $PE\\perp BC$. Quadrilateral CEPF is a square; what is the maximum value of $AC\\cdot BC$?\n",
    "completion": "\\textbf{Solution:} Let line $CG\\perp AB$ pass through point C. From the area formula of triangle $ABC$, $S_{\\triangle ABC}=\\frac{1}{2}AB\\cdot CG=\\frac{1}{2}AC\\cdot BC$, we know that when $CG$ is at its maximum, $AC\\cdot BC$ will also reach its maximum value, that is, $CG=\\frac{1}{2}AB=\\frac{1}{2}\\times 2=1$. Using the area formula of a triangle, $S_{\\triangle ABC}=\\frac{1}{2}AB\\cdot CG=\\frac{1}{2}AC\\cdot BC$, since $AB=2$, it follows that $\\frac{1}{2}\\times 2\\times 1=\\frac{1}{2}AC\\cdot BC$, therefore, $AC\\cdot BC=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51433177_152.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{10}-2\\sqrt{13}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, it is given that $AB=8$, $BC=12$. Points $O$ and $P$ are the midpoints of sides $AB$ and $AD$, respectively. Point $H$ is a moving point on side $CD$. Connect $OH$. Quadrilateral $OBCH$ is folded along $OH$ to obtain quadrilateral $OFEH$. Connect $PE$. What is the minimum value of the length of $PE$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect EO, PO, OC, \\\\\n$\\because$ Quadrilateral ABCD is a rectangle, \\\\\n$\\therefore \\angle B=\\angle OAP=90^\\circ$, \\\\\n$\\because$ Points O, P are the midpoints of sides AB, AD respectively, \\\\\n$\\therefore$ OB=AO=4, AP=DP=6, \\\\\nIn $\\triangle OBC$, BC=12, OB=4, \\\\\n$\\therefore$ OC=$\\sqrt{OB^2+BC^2}=\\sqrt{4^2+12^2}=4\\sqrt{10}$, \\\\\nIn $\\triangle AOP$, OA=4, PA=6, \\\\\n$\\therefore$ OP=$\\sqrt{OA^2+PA^2}=\\sqrt{4^2+6^2}=2\\sqrt{13}$, \\\\\nDue to folding, OE=OC=$4\\sqrt{10}$, \\\\\n$\\because$ PE$\\geq$ OE−OP, \\\\\n$\\therefore$ The minimum value of PE=OE−OP=$4\\sqrt{10}-2\\sqrt{13}$, \\\\\nThus, the answer is: $PE_{\\min} = \\boxed{4\\sqrt{10}-2\\sqrt{13}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52250424.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $C$ is the midpoint of segment $AB$, point $D$ is a point on segment $CB$, and point $E$ is the midpoint of segment $DB$. Given that $AB=20$ and $EB=3$, what is the length of segment $CD$?\n",
    "completion": "\\textbf{Solution:} Since point C is the midpoint of line segment AB,\\\\\nit follows that $CB=\\frac{1}{2}AB=\\frac{1}{2}\\times 20=10$\\\\\nGiven that $CD=CB-DB$,\\\\\nit then results in $CD=10-6=\\boxed{4}$",
    "image_path": "dataset/mm_math/images/MM_Math/46945019_173.png"
  },
  {
    "solution": "\\boxed{y=-x+4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=-x+m$ intersects the $x$ axis at point $B(4,0)$ and the $y$ axis at point $A$. Point $C$ is a point on the $x$ axis, and it is known that the area of $\\triangle ABC$ is $4$. What is the equation of line $AB$?\n",
    "completion": "The solution is as follows: Substitute point B$(4,0)$ into $y=-x+m$ to obtain $-4+m=0$,\\\\\nsolving for $m$ gives $m=4$,\\\\\n$\\therefore$ the equation of line $AB$ is $\\boxed{y=-x+4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50482083_79.png"
  },
  {
    "solution": "\\boxed{20^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a ray $OC$ is drawn from point $O$ on line $AB$, making an angle $ \\angle BOC=70^\\circ$. A right-angle triangle ruler is placed with its right-angle vertex at point $O$. (Note: $ \\angle DOE=90^\\circ$) As shown in figure 2, if $OC$ exactly bisects $ \\angle BOE$, what is the degree measure of $ \\angle COD$?\n",
    "completion": "\\textbf{Solution:} Since $OC$ bisects $\\angle EOB$ and $\\angle BOC=70^\\circ$,\\\\\nit follows that $\\angle EOB=2\\angle BOC=140^\\circ$.\\\\\nGiven $\\angle DOE=90^\\circ$,\\\\\nwe have $\\angle BOD=\\angle BOE-\\angle DOE=50^\\circ$,\\\\\ntherefore $\\angle COD=\\angle BOC-\\angle BOD=\\boxed{20^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55426249_5.png"
  },
  {
    "solution": "\\boxed{80^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle BAC=130^\\circ$, the perpendicular bisectors of AB and AC respectively intersect BC at points E and F, and intersect AB and AC at points D and G, respectively. What is the measure of $\\angle EAF$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle$BAC=$130^\\circ$,\\\\\nit follows that $\\angle$B+$\\angle$C=$180^\\circ-\\angle$BAC=$180^\\circ-130^\\circ$=$50^\\circ$;\\\\\nSince DE bisects AB perpendicularly, and FG bisects AC perpendicularly,\\\\\nit follows that BE=AE, and AF=CF,\\\\\nwhich implies that $\\angle$B=$\\angle$BAE, $\\angle$C=$\\angle$CAF,\\\\\nhence $\\angle$EAF=$\\angle$BAC-$\\angle$BAE-$\\angle$CAF=$\\angle$BAC-$\\angle$B-$\\angle$C=$130^\\circ-50^\\circ=\\boxed{80^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53059797.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle B=90^\\circ$, the perpendicular bisector $l$ of $AC$ intersects $BC$ at point $D$, and $AD$ is connected. If $AB=3$, $BC=9$, what is the length of $BD$?\n",
    "completion": "\\textbf{Solution:} Let BD $= x$, thus CD $= 9 - x$, \\\\\n$\\because$ line $l$ is the perpendicular bisector of AC, \\\\\n$\\therefore$ AD = DC = $9-x$, \\\\\nIn $\\triangle ABD$, $AB^{2} + BD^{2} = AD^{2}$, that is, $3^{2} + x^{2} = (9-x)^{2}$, \\\\\nSolving gives: $x = \\boxed{4}$, which means the length of BD is 4.",
    "image_path": "dataset/mm_math/images/MM_Math/52836849.png"
  },
  {
    "solution": "\\boxed{\\frac{49\\sqrt{3}}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, D is a point outside the equilateral triangle $\\triangle ABC$, with $AD=5$, $CD=3$. When the length of $BD$ is maximized, what is the area of $\\triangle ABC$?\n",
    "completion": "\\textbf{Solution:} Construct an equilateral triangle $DCE$ with $CD$ as a side. Join $AE$.\\\\\n$\\because$ $BC=AC$, $CD=CE$, $\\angle BCA=\\angle DCE=60^\\circ$,\\\\\n$\\therefore$ $\\angle BCD=\\angle ACE$,\\\\\nIn $\\triangle BCD$ and $\\triangle ACE$,\\\\\n$\\left\\{\\begin{array}{l}\nBC=AC\\\\\n\\angle BCD=\\angle ACE\\\\\nCD=CE\n\\end{array}\\right.$,\\\\\n$\\therefore$ $\\triangle BCD \\cong \\triangle ACE$ (SAS),\\\\\n$\\therefore$ $BD=AE$,\\\\\nIn $\\triangle ADE$, $\\because$ $AD=5$, $DE=CD=3$,\\\\\n$\\therefore$ $AE \\leq AD + DE$,\\\\\n$\\therefore$ The maximum value of $AE$ is 8,\\\\\n$\\therefore$ When points A, D, E are collinear, the value of $BD$ is maximum, and it is 8,\\\\\nAs shown in the diagram, draw $CF\\perp AD$ with the foot at F, and draw $BM\\perp AC$ intersecting AC at point M,\\\\\n$\\because$ $\\angle CDE=\\angle BDC=\\angle E=60^\\circ$,\\\\\n$\\therefore$ $\\angle DCF=30^\\circ$,\\\\\n$\\therefore$ $DF=\\frac{1}{2}CD=\\frac{3}{2}$,\\\\\n$\\therefore$ $CF=\\sqrt{3^2-\\left(\\frac{3}{2}\\right)^2}=\\frac{3\\sqrt{3}}{2}$, $AF=AD+DF=5+\\frac{3}{2}=\\frac{13}{2}$,\\\\\n$\\therefore$ $AC=\\sqrt{AF^2+CF^2}=7$,\\\\\nAccording to the properties of an equilateral triangle, $AM=CM=\\frac{1}{2}AC=\\frac{7}{2}$,\\\\\n$\\therefore$ The height $BM$ on the side $AC$ of $\\triangle ABC$ is $\\sqrt{7^2-\\left(\\frac{7}{2}\\right)^2}=\\frac{7\\sqrt{3}}{2}$,\\\\\n$\\therefore$ The area of $\\triangle ABC$ is $\\frac{1}{2} \\times \\frac{7\\sqrt{3}}{2} \\times 7=\\boxed{\\frac{49\\sqrt{3}}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53206622.png"
  },
  {
    "solution": "\\boxed{(6,3)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Synthesis and Exploration: As shown in the figure, in the Cartesian coordinate system, the line $l_{1}: y_{1}=-\\frac{1}{2}x+6$ intersects with the x-axis and y-axis at point B and C, respectively, and intersects with the line $l_{2}: y_{2}=\\frac{1}{2}x$ at point A. Determine the coordinates of point A.\n",
    "completion": "\\textbf{Solution:} Solve the system of equations for the two lines to get $\\begin{cases}y=-\\frac{1}{2}x+6\\\\ y=\\frac{1}{2}x\\end{cases}$, and we find: $\\begin{cases}x=6\\\\ y=3\\end{cases}$, $\\therefore$ the coordinates of point A are $\\boxed{(6,3)}$;",
    "image_path": "dataset/mm_math/images/MM_Math/50677413_139.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points $A$ and $D$ lie on the circle $\\odot O$, and $BC$ is a diameter. If $\\angle ADC=30^\\circ$, then what is the measure of $\\angle OCA$ in degrees?\n",
    "completion": "\\textbf{Solution:} Connect AB, as shown in the figure,\\\\\nsince BC is a diameter,\\\\\nhence $\\angle BAC=90^\\circ$,\\\\\nsince $\\angle B=\\angle ADC=30^\\circ$,\\\\\nthus $\\angle ACB=90^\\circ-\\angle B=90^\\circ-30^\\circ=60^\\circ$,\\\\\nthat is, $\\angle OCA=\\boxed{60^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52536747.png"
  },
  {
    "solution": "\\boxed{(1,\\frac{2}{3}) or (1, -2)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=-x^2+2x+c$ intersects the $x$-axis at points $A(-1,0)$ and $B(3,0)$, and intersects the $y$-axis at point $C$. $D$ is the vertex of the parabola. The point symmetric to $C$ with respect to the axis of symmetry of the parabola is point $E$. Point $M$ is a point on the axis of symmetry of the parabola, and $\\triangle DMB$ is similar to $\\triangle BCE$. What are the coordinates of point $M$?\n",
    "completion": "\\textbf{Solution:} The coordinates of point $M$ are $ \\boxed{(1,\\frac{2}{3}) or (1, -2)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51445487_37.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ABCD$, $E$ and $F$ are the midpoints of $AB$ and $BC$ respectively, $EH \\perp AC$ with the foot at $H$, and it intersects $AF$ at point $G$. If $AC = 24$ and $GF = 6\\sqrt{5}$, then what is the length of $EG$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, connect EF, \\\\\nbecause E and F are the midpoints of AB and BC respectively, \\\\\ntherefore EF is the midline of $\\triangle$BAC, \\\\\ntherefore $EF=\\frac{1}{2}AC=12$, $EF\\parallel AC$, \\\\\nbecause EH$\\perp$AC, \\\\\ntherefore EH$\\perp$EF, that is $\\angle$FEG$=90^\\circ$, \\\\\ntherefore $EG=\\sqrt{GF^{2}-EF^{2}}=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53359566.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the diagonals of square $ABCD$ intersect at point $O$, and point $O$ is also a vertex of square $A_1B_1C_1O$, with both squares having equal side lengths. No matter how square $A_1B_1C_1O$ is rotated around point $O$, the area of the overlapping part of the two squares is always equal to what fraction of the area of a square?\n",
    "completion": "\\textbf{Solution}: As shown in the figure, construct OG$\\perp$AB at G, and OH$\\perp$BC at H. Assume AB intersects A$_{1}$O at point M.\\\\\nSince quadrilateral ABCD is a square,\\\\\n$\\therefore \\angle$OBH=$\\angle$OBG=$45^\\circ$,\\\\\n$\\therefore$ quadrilateral OGBH is a square,\\\\\n$\\therefore$OG=OH,\\\\\nSince $\\angle$GOM+$\\angle$FOG=$\\angle$FOH+$\\angle$FOG=$90^\\circ$,\\\\\n$\\therefore \\angle$GOM=$\\angle$FOH,\\\\\n$\\therefore \\triangle FOH \\cong \\triangle MOG$ (ASA),\\\\\n$\\therefore$Area of quadrilateral OFBM=Area of quadrilateral OGBH=$\\boxed{\\frac{1}{4}}$ Area of square ABCD.",
    "image_path": "dataset/mm_math/images/MM_Math/51770518.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, points D and E are on AC and AB respectively, BD bisects $\\angle ABC$, $DE\\perp AB$ at point $E$, $AE=6$, $\\cos A=\\frac{3}{5}$. What is the value of $\\tan \\angle DBC$?\n",
    "completion": "\\textbf{Solution:} Given $AD=10$, $DC=8$,\n\n$\\therefore AC=AD+DC=18$,\n\nIn triangles $\\triangle ADE$ and $\\triangle ABC$,\n\n$\\because \\angle A=\\angle A$, $\\angle AED=\\angle ACB$,\n\n$\\therefore \\triangle ADE \\sim \\triangle ABC$,\n\n$\\therefore \\frac{DE}{BC}=\\frac{AE}{AC}$, which means $\\frac{8}{BC}=\\frac{6}{18}$, thus $BC=24$,\n\n$\\therefore \\tan\\angle DBC=\\frac{CD}{BC}=\\frac{8}{24}=\\boxed{\\frac{1}{3}}.$",
    "image_path": "dataset/mm_math/images/MM_Math/51299773_166.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, points D and E are respectively the midpoints of AB and AC. If the area of $\\triangle ABC$ is 16, what is the area of quadrilateral BCED?\n",
    "completion": "\\textbf{Solution:} Given that in $\\triangle ABC$, points D and E are the midpoints of AB and AC, respectively,\\\\\nthen $DE \\parallel BC$ and $DE= \\frac{1}{2}BC$,\\\\\nthus $\\triangle ADE \\sim \\triangle ABC$,\\\\\nsince $\\frac{DE}{BC}=\\frac{1}{2}$,\\\\\nit follows that $\\frac{S_{\\triangle ADE}}{S_{\\triangle ABC}}=\\frac{1}{4}$,\\\\\ngiven that $S_{\\triangle ABC}=16$,\\\\\nthen $S_{\\triangle ADE}=\\frac{1}{4}S_{\\triangle ABC}=\\frac{1}{4}\\times 16=4$,\\\\\ntherefore the area of quadrilateral $BCED= S_{\\triangle ABC}-S_{\\triangle ADE}=16-4=\\boxed{12}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51321575.png"
  },
  {
    "solution": "\\boxed{y=-x^2+4x+5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=-x^2+bx+c$ intersects the x-axis at points $A(-1,0)$ and $B(5,0)$, and intersects the y-axis at point $C$. The line $y=-x+n$ passes through point $B$. Point $P$ is a moving point on the parabola above the line $BC$. A perpendicular line from point $P$ to the x-axis meets point $F$, and intersects the line $BC$ at point $E$. Let the x-coordinate of point $P$ be $m$. What is the equation of the parabola?\n",
    "completion": "\\textbf{Solution:} \\\\\nSince the parabola $y=-x^2+bx+c$ intersects the x-axis at points A ($-1$, $0$) and B ($5$, $0$), \\\\\ntherefore, we can set the equation of the function as $y=-\\left(x+1\\right)\\left(x-5\\right)$ \\\\\nthus, $y=-x^2+4x+5$ \\\\\nHence, the equation of the parabola is $\\boxed{y=-x^2+4x+5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52756020_82.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{3}}{2}-\\frac{\\pi}{6}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, it is known that $AB$ is the diameter of $\\odot O$, $AD$ is a chord of $\\odot O$, $BC$ is a tangent of $\\odot O$ with the point of tangency at B, and points D, F are the trisection points of the arc $\\overset{\\frown}{ADB}$. The extensions of $BA$ and $CD$ intersect at point E. $DC$ is a tangent of $\\odot O$. If the radius of $\\odot O$ is 1, calculate the area of the shaded region.\n",
    "completion": "Solution: From the construction, we have\n\\[\n\\because OD\\perp EC,\n\\]\n\\[\n\\therefore \\angle EDO=90^\\circ,\n\\]\n\\[\n\\therefore \\text{in} \\triangle DOE, \\angle OED + \\angle DOE = 90^\\circ,\n\\]\n\\[\n\\because \\angle DOE=60^\\circ,\n\\]\n\\[\n\\therefore \\angle DEO=30^\\circ,\n\\]\n\\[\n\\therefore \\text{in} \\triangle DOE, OE=2OD=2,\n\\]\n\\[\n\\therefore \\text{in} \\triangle DOE, \\text{by the Pythagorean theorem}: DE=\\sqrt{OE^2-OD^2}=\\sqrt{2^2-1^2}=\\sqrt{3},\n\\]\n\\[\n\\therefore S_{\\triangle DOE}=\\frac{1}{2}\\times 1\\times \\sqrt{3}=\\frac{\\sqrt{3}}{2},\n\\]\n\\[\nS_{\\text{sector} DOE}=\\frac{60^\\circ \\pi \\times 1^2}{360}=\\frac{\\pi}{6},\n\\]\n\\[\n\\therefore S_{\\text{shaded}}=S_{\\triangle DOE}-S_{\\triangle AOD}=\\boxed{\\frac{\\sqrt{3}}{2}-\\frac{\\pi}{6}}.\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/52602698_157.png"
  },
  {
    "solution": "\\boxed{\\frac{4}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the figure, within $\\triangle ABC$, $\\angle C=90^\\circ$, point $D$ is on $BC$, $AD=BC=5$, and $\\cos\\angle ADC=\\frac{3}{5}$. What is the value of $\\tan B$?\n",
    "completion": "\\textbf{Solution:} Given: $AD=BC=5$, \\\\\n$\\cos\\angle ADC=\\frac{3}{5}$, \\\\\nHence, $\\frac{CD}{AD}=\\frac{3}{5}$, \\\\\nTherefore, $CD=3$, \\\\\nBy the Pythagorean theorem, we have: $AC=\\sqrt{AD^{2}-CD^{2}}=\\sqrt{5^{2}-3^{2}}=4$, \\\\\nHence, $\\tan B=\\frac{AC}{BC}=\\boxed{\\frac{4}{5}}$",
    "image_path": "dataset/mm_math/images/MM_Math/53043752.png"
  },
  {
    "solution": "\\boxed{\\frac{16}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the direct proportion function $y = kx \\left(k \\neq 0\\right)$ passes through point $A\\left(2, -4\\right)$. The line $y = x + b$ is translated parallel to the y-axis, intersects the x-axis and y-axis at points $B$ and $C$, respectively, and intersects line $OA$ at point $D$. When the symmetric point $A'$ of point $A$ with respect to line $BC$ happens to fall on the y-axis, what is the area of $\\triangle OBD$?\n",
    "completion": "\\textbf{Solution:} Substitute point A(2, -4) into $y=kx$, we get $-4=2k$, thus $k=-2$,\\\\\n$\\therefore$ the equation of line OA is $y=-2x$,\\\\\nIn $y=x+b$, let $y=0$, then $x=-b$,\\\\\n$\\therefore$ B$(-b, 0)$, which means $OB=|b|$,\\\\\nIn $y=x+b$, let $x=0$, then $y=b$,\\\\\n$\\therefore$ C$(0, b)$, which means $OC=|b|$,\\\\\n$\\therefore$ OB=OC,\\\\\nAlso, $\\because$ $\\angle$BOC$=90^\\circ$,\\\\\n$\\therefore$ $\\angle$OCB=$\\angle$OBC$=45^\\circ$,\\\\\n$\\because$ the symmetric point of point A with respect to line BC precisely falls on the y-axis,\\\\\n$\\therefore$ CD perpendicularly bisects AA$^{\\prime }$,\\\\\n$\\therefore$ CA=CA$^{\\prime }$,\\\\\n$\\therefore$ $\\angle$ACD=$\\angle$OCB$=45^\\circ$,\\\\\n$\\therefore$ $\\angle$ACO$=90^\\circ$,\\\\\n$\\therefore$ OC$=|y_{A}|=4$,\\\\\n$\\therefore$ OB=OC$=4$, that is, C$(0, -4)$,\\\\\nSubstitute point C$(0, -4)$ into $y=x+b$, we get $-4=0+b$,\\\\\n$\\therefore$ b$=-4$,\\\\\n$\\therefore$ the equation of line BC is $y=x-4$,\\\\\nFrom $\\begin{cases}y=-2x\\\\ y=x-4\\end{cases}$, we obtain $\\begin{cases}x=\\frac{4}{3}\\\\ y=-\\frac{8}{3}\\end{cases}$,\\\\\n$\\therefore$ point D$\\left(\\frac{4}{3}, -\\frac{8}{3}\\right)$,\\\\\n$\\therefore$ the area of $\\triangle OBD=\\frac{1}{2}OB\\cdot \\left|y_{D}\\right|=\\frac{1}{2}\\times 4\\times \\frac{8}{3}=\\boxed{\\frac{16}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52094911.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AD$ is the median of $ \\triangle ABC$, $CE$ is the median of $ \\triangle ACD$. If the area of $ \\triangle ABC$ is $12\\text{cm}^2$, what is the area of $ \\triangle CDE$?\n",
    "completion": "\\textbf{Solution:} Since $AD$ is the median of side $BC$ of $\\triangle ABC$, and the area of $\\triangle ABD$ is $12\\text{cm}^2$,\\\\\ntherefore, the area of $\\triangle ADC$ is: $\\frac{1}{2}\\times 12=6\\text{cm}^2$,\\\\\nSince $CE$ is the median of side $AD$ of $\\triangle ACD$,\\\\\ntherefore, the area of $\\triangle CDE$ is: $\\frac{1}{2}\\times 6=\\boxed{3}\\text{cm}^2$.",
    "image_path": "dataset/mm_math/images/MM_Math/52950675.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\mathrm{Rt}\\triangle ABC$ with $\\angle A=90^\\circ$, $O$ is a point on side $BC$. A semicircle centered at $O$ is tangent to side $AB$ at point $D$ and intersects sides $AC$ and $BC$ at points $E$, $F$, and $G$, respectively. Line $OD$ is drawn, given that $BD = 2$, $AE = 3$, and $\\tan\\angle BOD= \\frac{2}{3}$. What is the radius of the circle centered at $O$, $OD$?\n",
    "completion": "\\textbf{Solution:} Since AB is tangent to circle O,\\\\\nit follows that OD$\\perp$AB. In $\\triangle BDO$, BD=2, $\\tan\\angle BOD= \\frac{BD}{OD} = \\frac{2}{OD}$,\\\\\nhence, OD=\\boxed{3}.",
    "image_path": "dataset/mm_math/images/MM_Math/51317358_158.png"
  },
  {
    "solution": "\\boxed{-4<y<-1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the inverse proportion function $y=\\frac{k}{x} (k\\ne 0)$ passes through the points $(-2, 4)$ and point $A(a, -2)$. If the point $C(x, y)$ is also on the graph of the inverse proportion function $y=\\frac{k}{x} (k\\ne 0)$, when $2<x<8$, find the range of values for the function $y$.\n",
    "completion": "\\textbf{Solution:} Given that point $C(x, y)$ is also on the graph of the inverse proportion function $y=-\\frac{8}{x}$, \\\\\n$\\therefore$ when $2<x<8$, $y$ increases as $x$ increases, \\\\\nwhen $x=2$, we have $y=-4,$ \\\\\nwhen $x=8$, we have $y=-1,$ \\\\\n$\\therefore -4<y<-1.$ \\\\\nTherefore, the range of function $y$ is $\\boxed{-4<y<-1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52720153_116.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the small tree AB casts a shadow BC under the illumination of the street light O. If the height of the tree AB is 2m, the length of the shadow BC is 3m, and the horizontal distance between the tree and the street light BP is 4.5m, what is the height of the street light OP?\n",
    "completion": "Solution: Under the same lighting conditions, the ratio of the height of any object to its shadow remains constant:\\\\\nSince the height of the tree, AB = 2m, and the shadow of the tree, BC = 3m, and BP = 4.5m\\\\\nTherefore, $\\frac{OP}{PC}=\\frac{AB}{BC}$, substituting the values gives: $\\frac{OP}{7.5}=\\frac{2}{3}$\\\\\nTherefore, $OP=\\boxed{5}$m",
    "image_path": "dataset/mm_math/images/MM_Math/51212846.png"
  },
  {
    "solution": "\\boxed{20^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within $\\triangle ABC$, $AD$ is the height of $\\triangle ABC$, $AE$ and $BF$ are the angle bisectors of $\\triangle ABC$, and $AE$ intersects $BF$ at point $O$. Given that $\\angle BOA = 125^\\circ$, what is the measure of $\\angle DAC$ in degrees?\n",
    "completion": "\\textbf{Solution:} Since $\\angle OAB + \\angle OBA + \\angle AOB = 180^\\circ$ and $\\angle AOB = 125^\\circ$,\n\\[\n\\therefore \\angle OAB + \\angle OBA = 180^\\circ - 125^\\circ = 55^\\circ,\n\\]\nSince AE and BF are the angle bisectors of $\\triangle ABC$,\n\\[\n\\therefore \\angle OAB = \\frac{1}{2} \\angle BAC, \\angle OBA = \\frac{1}{2} \\angle ABC,\n\\]\n\\[\n\\therefore \\frac{1}{2} \\angle BAC + \\frac{1}{2} \\angle ABC = 55^\\circ,\n\\]\n\\[\n\\therefore \\angle BAC + \\angle ABC = 110^\\circ,\n\\]\nSince $\\angle BAC + \\angle ABC + \\angle ACB = 180^\\circ$,\n\\[\n\\therefore \\angle ACB = 70^\\circ,\n\\]\nSince AD is the altitude of $\\triangle ABC$,\n\\[\n\\therefore \\angle ADC = 90^\\circ,\n\\]\n\\[\n\\therefore \\angle DAC = 90^\\circ - 70^\\circ = \\boxed{20^\\circ}.\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/53066357.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, there is a street lamp pole $AB$ on the opposite side of the river. Under the illumination of the lamp, Xiao Ming is at point $D$, and his shadow is $DF=4m$ long. He moves in the direction of $BD$ to reach point $F$, and then measures his shadow to be $FG=5m$ long. If Xiao Ming's height is $1.6m$, calculate the height of the street lamp pole $AB$.\n",
    "completion": "\\textbf{Solution:} Given that $CD\\parallel EF\\parallel AB$,\\\\\nwe can infer that $\\triangle ABF\\sim \\triangle CDF$ and $\\triangle ABG\\sim \\triangle EFG$,\\\\\nhence, $\\frac{AB}{CD}=\\frac{BF}{DF}$ and $\\frac{AB}{EF}=\\frac{BG}{FG}$,\\\\\nsince $CD=EF$,\\\\\nit follows that $\\frac{BF}{DF}=\\frac{BG}{FG}$\\\\\nGiven $DF=4$, $FG=5$, $BF=BD+DF=BD+4$, $BG=BD+DF+FG=BD+9$,\\\\\nwe get $\\frac{4+BD}{4}=\\frac{9+BD}{5}$,\\\\\nsolving for $BD$ yields $BD=16, BF=16+4=20$,\\\\\ntherefore $\\frac{AB}{1.6}=\\frac{20}{4}$,\\\\\nsolving gives $AB=\\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53010292.png"
  },
  {
    "solution": "\\boxed{9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the right-angled triangle $ABC$ where $\\angle C=90^\\circ$, $D$ is the midpoint of $BC$, and $\\angle CAD=30^\\circ$, with $BC=6$, what is the length of $AD+DB$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle C=90^\\circ$ and D is the midpoint of BC, with $\\angle CAD=30^\\circ$ and BC=6,\\\\\nit follows that $AD=2CD$, BD=CD=$\\frac{1}{2}$BC=3,\\\\\ntherefore $AD+BD=3CD=\\boxed{9}.$",
    "image_path": "dataset/mm_math/images/MM_Math/52095948.png"
  },
  {
    "solution": "\\boxed{Q\\left(-2, 5\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the axis of symmetry of the parabola is the line $x=1$, and it intersects the x-axis at points A and B, and intersects the y-axis at point $C\\left(0, -3\\right)$, where $OB=OC$. Is there a point Q on the parabola such that $\\triangle BCQ$ is a right-angled triangle with BC as its right angle side? If it exists, find the coordinates of point Q; if not, please explain why.\n",
    "completion": "\\textbf{Solution:} There exists a point $Q$ such that $ \\triangle BCQ$ is a right triangle with $BC$ as the right angle side, for the reasons below:\\\\\n(1) When $ \\angle BCQ=90^\\circ$, as shown in the figure:\\\\\nDraw a line from point Q to the y-axis at point H, perpendicular to it, since $ OB=OC$ and $ \\angle BOC=90^\\circ$, thus $\\triangle BOC$ is an isosceles right triangle, hence $ \\angle OCB=45^\\circ$; since $ \\angle BCQ=90^\\circ$, thus $ \\angle HCQ=45^\\circ$, hence $\\triangle HCQ$ is also an isosceles right triangle, thus $ HC=HQ$; consider point $ Q\\left(t, {t}^{2}-2t-3\\right)$, then $ HQ=t$ and $ CH=-3-{t}^{2}+2t+3=-{t}^{2}+2t$, thus $ t=-{t}^{2}+2t$, solving this yields: ${t}_{1}=1$, ${t}_{2}=0$ (discarded as it does not meet the question's requirement), hence point $ Q\\left(1, -4\\right)$;\\\\\n(2) When $ \\angle CBQ=90^\\circ$, as shown in the figure:\\\\\nDraw a perpendicular line from point B to the x-axis, then from points Q and C draw lines QE$\\perp$BE at point E, and CF$\\perp$BE at point F, thus $ CF=BF=3$, hence $\\triangle BFC$ is an isosceles right triangle, thus $ \\angle CBF=45^\\circ$; since $ \\angle CBQ=90^\\circ$, thus $ \\angle EBQ=45^\\circ$, hence $\\triangle QEB$ is also an isosceles right triangle, thus $ EQ=EB$; consider point $ Q\\left(t, {t}^{2}-2t-3\\right)$, then $ EQ=3-t$ and $ EB={t}^{2}-2t-3$, thus $ 3-t={t}^{2}-2t-3$, solving this yields: ${t}_{1}=-2$, ${t}_{2}=3$ (discarded as it does not meet the question's requirement), hence point $ Q\\left(-2, 5\\right)$;\\\\\nIn conclusion: When $ \\triangle BCQ$ is a right triangle with $BC$ as the right angle side, point $\\boxed{Q\\left(1, -4\\right)}$ or $\\boxed{Q\\left(-2, 5\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51433738_208.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in $ \\triangle ABC$, with $ \\angle C=90^\\circ$, AC and BC are tangent to circle $ \\odot O$ at points E and F, respectively. AB passes through point M on circle $ \\odot O$, and $ AE=AM$. AB is a tangent to circle $ \\odot O$; if $ AC=8$ and $ BC=6$, what is the radius of circle $ \\odot O$?\n",
    "completion": "\\textbf{Solution:} Connect OF. Let the radius of $\\odot O$ be $r$.\\\\\nSince BC is tangent to $\\odot O$ at point F,\\\\\nthus OF$\\perp$BC,\\\\\ntherefore $\\angle$OFC$=90^\\circ$,\\\\\nAlso, since $\\angle$C$=90^\\circ$, $\\angle$OEC$=90^\\circ$, and OF=OE,\\\\\ntherefore, quadrilateral OFCE is a square,\\\\\nthus CF=CE=OE$=r$,\\\\\nSince AB, BC, AC are all tangent to $\\odot O$,\\\\\ntherefore BM=BF$=6-r$, AM=AE$=8-r$,\\\\\nIn $\\triangle ABC$, $AB=\\sqrt{BC^2+AC^2}=10$,\\\\\nSince BM+AM=AB,\\\\\ntherefore $6-r+8-r=10$,\\\\\ntherefore $r=\\boxed{2}$\\\\\nthus the radius of $\\odot O$ is \\boxed{2}.",
    "image_path": "dataset/mm_math/images/MM_Math/52602632_145.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, line AC intersects the graph of the inverse proportion function $y=\\frac{k}{x}$ ($x>0$) at points A and C (with point A to the left of point C), and intersects the x-axis at point B. A rectangle ADMN is drawn downwards with point A as the vertex, and its diagonals intersect at point O, with AD bisecting $\\angle OAB$. Given that AC=CB and line CD is drawn, if the area of $\\triangle ACD$ is 6, what is the value of $k$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw AN $\\perp$ x-axis at point N through point A, and draw CM $\\perp$ x-axis at point M through point C, connect CO, \\\\\n$\\because$ it's a rectangle ADMN, \\\\\n$\\therefore$ OA=OD, \\\\\n$\\therefore$ $\\angle$OAD=$\\angle$ODA, \\\\\nAlso, $\\because$ AD bisects $\\angle$OAB, \\\\\n$\\therefore$ $\\angle$OAD=$\\angle$DAB, \\\\\n$\\therefore$ $\\angle$ODA=$\\angle$DAB, \\\\\n$\\therefore$ OD$\\parallel$AB, \\\\\n$\\because$ the area of $\\triangle$ACD is 6, and AC=CB, \\\\\n$\\therefore$ $S_{\\triangle AOC}$=$S_{\\triangle OCB}$=6, CM is the median line of $\\triangle$ANB, \\\\\n$\\therefore$ AN=2CM, NM=MB, \\\\\nLet point A be (x, $\\frac{k}{x}$), then ON=x, AN=$\\frac{k}{x}$, \\\\\n$\\therefore$ CM=$\\frac{k}{2x}$, \\\\\n$\\therefore$ C(2x, $\\frac{k}{2x}$), \\\\\n$\\therefore$ MB=x, OB=3x, \\\\\n$\\therefore$ $S_{\\triangle OCB}$=6=$\\frac{1}{2}$OB·CM=$\\frac{1}{2}$×3x·$\\frac{k}{2x}$=$\\frac{3k}{4}$, \\\\\n$\\therefore$ k=\\boxed{8}.",
    "image_path": "dataset/mm_math/images/MM_Math/52082364.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, points D and E are the midpoints of $AB$ and $AC$ respectively. Connect $BE$. If $AE=6$, $DE=5$, and $\\angle BEC=90^\\circ$, what is the length of $BE$?\n",
    "completion": "Given that points D and E are the midpoints of AB and AC respectively, with AE=6 and DE=5,\n\nit follows that EC=AE=6 and BC=$2\\cdot DE$=10,\n\nin the right triangle $\\triangle BEC$, we have $BE= \\sqrt{BC^{2}-EC^{2}} = \\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50481447.png"
  },
  {
    "solution": "\\boxed{20^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle B = 40^\\circ$, $\\angle C = 80^\\circ$. $AE$ bisects $\\angle BAC$ and intersects $BC$ at $E$, $AD \\perp BC$ at $D$. Find the degree measure of $\\angle EAD$.\n",
    "completion": "\\textbf{Solution:}\\\\\nSince $AD \\perp BC$,\\\\\nit follows that $\\angle ADC = 90^\\circ$,\\\\\nSince $\\angle DAC = 180^\\circ - \\angle ADC - \\angle C$,\\\\\nit follows that $\\angle DAC = 10^\\circ$,\\\\\nSince $AE$ bisects $\\angle BAC$,\\\\\nit follows that $\\angle BAE = \\angle CAE = 30^\\circ$,\\\\\nTherefore, $\\angle EAD = \\angle CAE - \\angle DAC = \\boxed{20^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52807557_124.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of $\\odot O$. Points $F$ and $C$ are on $\\odot O$, and arc $AF = \\text{arc} FC = \\text{arc} BC$. Connect $AC$ and $AF$. Through point $C$, draw $CD \\perp AF$ to intersect the extended line of $AF$ at point $D$, with the foot of the perpendicular being $D$. $CD$ is a tangent of $\\odot O$; If $CD = 2\\sqrt{3}$, what is the radius of $\\odot O$?\n",
    "completion": "\\textbf{Solution:} Connect BC as shown in the figure. \\\\\nSince AB is the diameter, it follows that $\\angle ACB=90^\\circ$. Since $\\overset{\\frown}{AF} = \\overset{\\frown}{FC} = \\overset{\\frown}{BC}$, it follows that $\\angle BOC = \\frac{1}{3} \\times 180^\\circ = 60^\\circ$.\\\\\nTherefore, $\\angle BAC = 30^\\circ$, thus $\\angle DAC = 30^\\circ$. In $\\triangle ADC$, given $CD = 2\\sqrt{3}$, it follows that $AC = 2CD = 4\\sqrt{3}$.\\\\\nIn $\\triangle ACB$, $BC = \\frac{\\sqrt{3}}{3}AC = \\frac{\\sqrt{3}}{3} \\times 4\\sqrt{3} = 4$, therefore $AB = 2BC = 8$.\\\\\nHence, the radius of the circle $O$ is $\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51299533_129.png"
  },
  {
    "solution": "\\boxed{24\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the radius of the circle is 4. What is the perimeter of the shaded area in the figure?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, connect OA and OB, and draw OC $\\perp$ AB at point C\\\\\n$\\angle OCB=90^\\circ$, $\\angle OBA=30^\\circ$, with the radius of the circle being 4,\\\\\n$\\therefore$OC=2,\\\\\n$\\therefore$BC=$2\\sqrt{3}$,\\\\\n$\\therefore$AB=2BC=$4\\sqrt{3}$,\\\\\n$\\therefore$the perimeter of the shaded area is: $6\\times 4\\sqrt{3}=\\boxed{24\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51545212.png"
  },
  {
    "solution": "\\boxed{(-10, 0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system $xOy$, points $A$ and $B$ lie on the line $y=\\frac{1}{2}x$. The graph of the quadratic function $y=ax^2+bx-2$ also passes through points $A$ and $B$, and intersects the $y$-axis at point $C$. If $BC$ is parallel to the $x$-axis, and the $x$-coordinate of point $A$ is $2$. Suppose the axis of symmetry of the graph of this quadratic function intersects $BC$ at point $D$, and point $E$ is on the negative $x$-axis. If the triangle formed by points $E$, $O$, and $B$ is similar to $\\triangle OBD$, and the ratio of similarity is not $1$, what are the coordinates of point $E$?\n",
    "completion": "\\textbf{Solution:} From (1), we get that the axis of symmetry of the quadratic function $y=\\frac{1}{4}x^{2}+x-2$ is the line $x=-2$, \\\\\n$\\therefore$ the coordinates of point $D$ are $(-2, -2)$, \\\\\n$\\therefore$ $OB=2\\sqrt{5}$, $BD=2$, \\\\\n$\\because$ $BC\\parallel x$-axis, \\\\\n$\\therefore$ $\\angle OBD=\\angle BOE$, \\\\\n$\\therefore$ The triangle formed by points $E$, $O$, and $B$ is similar to $\\triangle OBD$, there are two possibilities: \\\\\nIf $\\frac{BO}{OB}=\\frac{BD}{OE}$, then $\\triangle BOD\\sim \\triangle OBE$, \\\\\nObviously, these two similar triangles have a similarity ratio of $1$, \\\\\nThis contradicts the given information that the similarity ratio is not $1$; thus, this case should be discarded; \\\\\nWhen $\\frac{BO}{OE}=\\frac{BD}{OB}$, then $\\triangle BOD\\sim \\triangle OEB$, \\\\\n$\\therefore$ $\\frac{2\\sqrt{5}}{OE}=\\frac{2}{2\\sqrt{5}}$, \\\\\n$\\therefore$ $OE=10$, \\\\\nSince point $E$ is on the negative half of the x-axis, \\\\\n$\\therefore$ the coordinates of point $E$ are $\\boxed{(-10, 0)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51379457_208.png"
  },
  {
    "solution": "\\boxed{\\left(\\frac{3}{2}, 0\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the graph, the line $y=-\\frac{3}{4}x+3$ intersects the x-axis at point A and the y-axis at point B. The $\\triangle AOB$ is folded along a certain line passing through point B, such that point A falls on point D located on the negative y-axis. The fold line intersects the x-axis at point C. Please find the coordinates of points A, B, and C?\n",
    "completion": "\\textbf{Solution:} Let's solve it. When $x=0$, then $y=3$; \\\\\nwhen $y=0$, then $x=4$, \\\\\ntherefore, the coordinates of point A are $\\boxed{(4,0)}$, and the coordinates of point B are $\\boxed{(0,3)}$, \\\\\ntherefore, $AB=\\sqrt{3^2+4^2}=5$, \\\\\nsince BD=AB=5, \\\\\ntherefore, the coordinates of point D are $\\boxed{(0,-2)}$, \\\\\nlet the coordinates of point C be $(a,0)$, according to the problem, we have CD=AC, \\\\\ntherefore, $\\sqrt{4+a^2}=4-a$, \\\\\nsolving it we find $a=\\frac{3}{2}$, \\\\\ntherefore, the coordinates of point C are $\\boxed{\\left(\\frac{3}{2}, 0\\right)}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51894652_191.png"
  },
  {
    "solution": "\\boxed{2.8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=kx+b$ (where $k$ and $b$ are constants) intersects the $x$ axis and $y$ axis at points $A(-4,0)$ and $B(0,3)$, respectively. The parabola $y=-x^2+2x+1$ intersects the $y$ axis at point $C$. Point $E$ moves on the axis of symmetry of the parabola $y=-x^2+2x+1$, and point $F$ moves on the line $AB$. What is the minimum value of $CE+EF$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, let the point $C^{\\prime}$ be the symmetric point of C with respect to the axis of symmetry of the parabola, by the property of symmetry we have $CE = C^{\\prime}E$, \\\\\n$\\therefore CE + EF = C^{\\prime}E + EF$, \\\\\n$\\therefore$ when points F, E, $C^{\\prime}$ are collinear and $C^{\\prime}F$ is perpendicular to AB, $CE + EF$ is minimized, \\\\\nAccording to the problem we have $\\left\\{\\begin{array}{l}-4k+b=0\\\\ b=3\\end{array}\\right.$, solving this gives $\\left\\{\\begin{array}{l}k=\\frac{3}{4}\\\\ b=3\\end{array}\\right.$, \\\\\n$\\therefore$ the equation of the line is $y= \\frac{3}{4}x+3$; \\\\\n$\\because C(0,1)$, $\\therefore C^{\\prime}(2,1)$, \\\\\n$\\therefore$ the equation of line $C^{\\prime}F$ is $y=-\\frac{3}{4}x+\\frac{11}{3}$, \\\\\nFrom $\\left\\{\\begin{array}{l}y=\\frac{3}{4}x+3\\\\ y=−\\frac{3}{4}x+\\frac{11}{3}\\end{array}\\right.$ we solve $\\left\\{\\begin{array}{l}x=\\frac{8}{25}\\\\ y=\\frac{81}{25}\\end{array}\\right.$, \\\\\n$\\therefore F\\left(\\frac{8}{25}, \\frac{81}{25}\\right)$, \\\\\n$\\therefore CF = \\sqrt{{\\left(\\frac{8}{25}−2\\right)}^{2}+{\\left(\\frac{81}{25}−1\\right)}^{2}} = \\frac{14}{5} = 2.8$, \\\\\nhence, the minimum of $CE+EF$ is $\\boxed{2.8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51368325.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{3}-1}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown, in $\\triangle ABC$, $\\angle ACB=75^\\circ$, $AE\\perp BC$ at point E, $CD\\perp AB$ at point D, $AE$ and $CD$ intersect at point F, and $\\angle AFC=120^\\circ$. If $BD=1$, $AD=\\sqrt{3}$, then how long is $EF$?\n",
    "completion": "\\textbf{Solution:} We have $EF=\\boxed{\\frac{\\sqrt{3}-1}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53066385_62.png"
  },
  {
    "solution": "\\boxed{70^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $ \\angle B=35^\\circ$, and $CD$ is the perpendicular bisector of $AB$, then what is the value of $ \\angle ACE$?\n",
    "completion": "\\textbf{Solution:} Since CD is the perpendicular bisector of AB,\\\\\nit follows that $BC=AC$,\\\\\nhence $\\angle A=\\angle B=35^\\circ$,\\\\\ntherefore, $\\angle ACE=\\angle A+\\angle B=2\\times 35^\\circ=70^\\circ$,\\\\\nthus, $\\angle ACE = \\boxed{70^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/52793322.png"
  },
  {
    "solution": "\\boxed{4\\text{cm}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC$, $\\angle A=120^\\circ$, $BC=12\\,cm$. The perpendicular bisector of $AB$ intersects $BC$ at point M and intersects $AB$ at point E. The perpendicular bisector of $AC$ intersects $BC$ at point N and intersects $AC$ at point F. What is the length of $MN$?\n",
    "completion": "\\textbf{Solution:} Draw $AM$ and $AN$, \\\\\nsince the perpendicular bisector of $AB$ intersects $BC$ at point $M$, and $AB$ at point $E$, and the perpendicular bisector of $AC$ intersects $BC$ at point $N$, and $AC$ at point $F$, \\\\\nthus $AM=BM$, $AN=CN$, \\\\\ntherefore, $\\angle BAM=\\angle B$, $\\angle CAN=\\angle C$, \\\\\nsince $AB=AC$ and $\\angle A=120^\\circ$, \\\\\ntherefore, $\\angle B=\\angle C=30^\\circ$, \\\\\nthus, $\\angle AMN=\\angle ANM=60^\\circ$, \\\\\ntherefore, $\\triangle AMN$ is an equilateral triangle, \\\\\nthus, $AM=MN=AN$, \\\\\nthus, $BM=MN=BN$, \\\\\nsince $BC=12\\text{cm}$, \\\\\ntherefore, $MN=\\frac{1}{3}BC=\\boxed{4\\text{cm}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53044174.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the diagonals $AC$ and $BD$ of quadrilateral $ABCD$ intersect at point $O$. It is given that $AC\\perp BD$, $BD=12\\,cm$, and $AC=6\\,cm$. Point $E$ moves towards point $O$ at a speed of $1\\,cm/s$ from point $B$ along the segment $BO$, and point $F$ moves towards point $D$ at a speed of $2\\,cm/s$ from point $O$ along the segment $OD$. If points $E$ and $F$ start moving at the same time and the moving time is $t$ seconds, for what value of $t$ is the quadrilateral $AECF$ a parallelogram?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral $AECF$ is a parallelogram,\\\\\nit follows that $AO=OC$ and $EO=OF$,\\\\\nwhere $EO=6-t$ and $OF=2t$,\\\\\nthus $6-t=2t$,\\\\\ntherefore $t=\\boxed{2}$,\\\\\nhence, when $t=2$, quadrilateral $AECF$ is a parallelogram.",
    "image_path": "dataset/mm_math/images/MM_Math/52250892_120.png"
  },
  {
    "solution": "\\boxed{40^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB \\parallel CD$, $DE$ bisects $\\angle BDC$, $\\angle AED = 110^\\circ$. What is the degree measure of $\\angle B$?\n",
    "completion": "\\textbf{Solution:} Given that $AB\\parallel CD$,\\\\\nit follows that $\\angle AED + \\angle CDE = 180^\\circ$, $\\angle B + \\angle BDC = 180^\\circ$,\\\\\nsince $\\angle AED = 110^\\circ$,\\\\\nit follows that $\\angle CDE = 70^\\circ$,\\\\\nsince DE bisects $\\angle BDC$,\\\\\nit follows that $\\angle BDC = 2\\angle CDE = 140^\\circ$,\\\\\ntherefore $\\angle B = \\boxed{40^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/52916008.png"
  },
  {
    "solution": "\\boxed{(-2, 2)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the line \\(y=\\frac{4}{3}x+8\\) intersects the x-axis and y-axis at points A and B, respectively. Point C is on the positive half of the x-axis, and \\(AB=AC\\). Line \\(BC\\) is drawn. Given condition (2), point E is on \\(AF\\), point G is on \\(CD\\), and \\(CG=AE\\). Line \\(EG\\) is drawn. Triangle \\(\\triangle EGM\\) is an isosceles right triangle with \\(EG\\) as the hypotenuse. Point M is located above line \\(AD\\). The symmetric point of M with respect to line \\(EG\\) is marked as N. Lines \\(NC\\), \\(NE\\), and \\(NG\\) are drawn. \\(MG\\) intersects \\(AD\\) at point H. When \\(\\angle NCD=90^\\circ\\), find the coordinates of point H.\n",
    "completion": "Solution: Construct $MP\\perp MD$ through point M, meeting $AD$ at point P,\\\\\nsince $\\angle PMD=\\angle EMG=90^\\circ$, $\\angle EMP+\\angle PMG=\\angle PMG+\\angle DMG$,\\\\\nthus, $\\angle EMP=\\angle DMG$.\\\\\nSince $\\angle MHE=\\angle DHG$, $\\angle EGM=\\angle ADC=90^\\circ$,\\\\\nthus, $\\angle MEP=\\angle MGD$,\\\\\nsince $ME=MG$,\\\\\nthus $\\triangle MPE\\cong \\triangle MDG$,\\\\\nthus $DG=PE$.\\\\\nSince $D\\left(2, 4\\right)$,\\\\\nthus $DQ=4, AQ=OA+OQ=8, CQ=2$,\\\\\nthus $AD=\\sqrt{DQ^{2}+AQ^{2}}=4\\sqrt{5}$, $CD=\\sqrt{CQ^{2}+DQ^{2}}=2\\sqrt{5}$.\\\\\nSince $AE=CG$, $PE=DG$,\\\\\nthus $DG=PE$,\\\\\nthus $AP=AE+PE=CG+DG=CD=2\\sqrt{5}$,\\\\\nthus $PD=AD−AP=2\\sqrt{5}$.\\\\\nThus $PD=AP$,\\\\\nTake the midpoint V of $AQ$, connect $PV$,\\\\\nthus $PV$ is the median of $\\triangle ADQ$,\\\\\nsince $DQ=4$,\\\\\nthus $PV=\\frac{1}{2}DQ=2$,\\\\\nsince $A(-6, 0)$, $Q\\left(2, 0\\right)$,\\\\\nthus $V\\left(-2, 0\\right)$\\\\\nthus $P\\left(-2, 2\\right)$.\\\\\n$\\triangle PMD$ is an isosceles right triangle, construct $MS\\perp PV$ at point S through M, construct $DT\\perp MS$ at point T through D,\\\\\nsince $\\angle MPS=\\angle DMT, \\angle MSP=\\angle MTD, PM=MD$,\\\\\nthus $\\triangle MSP\\cong \\triangle DTM$,\\\\\nsince $D\\left(2, 4\\right)$, $P\\left(-2, 2\\right)$,\\\\\nthus $MS+MT=4, PS-DT=2$,\\\\\nthus $MT=3, DT=1$,\\\\\nthus $M\\left(-1, 5\\right)$.\\\\\nConstruct $MR\\perp DP$ at R through M, construct $MK\\perp BD$ at point K through M,\\\\\nsince $MP=MD, \\angle PMD=90^\\circ$\\\\\nthus $\\triangle DMP$ is an isosceles right triangle.\\\\\nthus $MR=RD=RP=\\sqrt{5}$,\\\\\nsince $\\angle MDP=\\angle MDK=90^\\circ$,\\\\\nthus $MK=MR=\\sqrt{5}$.\\\\\nPoints M, N are symmetric with respect to line $EG$,\\\\\nthus quadrilateral $MENG$ is a rhombus,\\\\\nsince $\\angle EMG=90^\\circ$,\\\\\nthus rhombus $MENG$ is a square.\\\\\nSince $\\angle MGK+\\angle KMG=\\angle MGK+\\angle NGC=90^\\circ$,\\\\\nthus $\\angle KMG=\\angle NGC$.\\\\\nSince $\\angle MKG=\\angle NCG=90^\\circ, GM=GN$,\\\\\nthus $\\triangle MKG\\cong \\triangle GCN$,\\\\\nthus $CG=MK=\\sqrt{5}$.\\\\\nSince $CD=2\\sqrt{5}$,\\\\\nthus $CD=DG$, take the midpoint W of $CQ$, connect $GW$,\\\\\nthus $GW$ intersects $AD$ at point H,\\\\\nthus the coordinates of point H are \\boxed{(-2, 2)}.\\\\",
    "image_path": "dataset/mm_math/images/MM_Math/53728859_93.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=-\\frac{3}{4}x^{2}+bx+c$ intersects the x-axis at point $A(4,0)$ and the y-axis at point $B(0,3)$. Point $M(m,0)$ is a moving point on line segment $OA$, and the line passing through point $M$ and perpendicular to the x-axis intersects line $AB$ and the parabola at points $P$ and $N$, respectively. If the quadrilateral formed by points $P$, $N$, $B$, and $O$ is a parallelogram, find the value of $m$?\n",
    "completion": "\\textbf{Solution:} Let the equation of line AB be $y=px+q$,\\\\\nSubstituting A(4, 0) and B(0, 3) gives $\\left\\{\\begin{array}{l}4p+q=0 \\\\ q=3\\end{array}\\right.$,\\\\\nSolving for $p$ and $q$ yields $\\left\\{\\begin{array}{l}p=-\\frac{3}{4} \\\\ q=3\\end{array}\\right.$,\\\\\n$\\therefore$ The equation of line AB is $y=-\\frac{3}{4}x+3$,\\\\\n$\\because$ M(m, 0), and MN$\\perp$x-axis,\\\\\n$\\therefore$ N(m, $-\\frac{3}{4}m^{2}+\\frac{9}{4}m+3$), and P(m, $-\\frac{3}{4}m+3$),\\\\\n$\\therefore$ NP= $-\\frac{3}{4}m^{2}+3m$, OB=3,\\\\\n$\\because$ NP$\\parallel$OB, and the quadrilateral formed by vertices P, N, B, and O is a parallelogram,\\\\\n$\\therefore$ NP= OB, i.e., $-\\frac{3}{4}m^{2}+3m=3$,\\\\\nUpon rearrangement: $m^{2}-4m+4=0$,\\\\\nSolving yields: $m=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51403288_130.png"
  },
  {
    "solution": "\\boxed{\\frac{9}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that quadrilateral $ABFE \\sim$ quadrilateral $EFCD$, $AB=2$, and $EF=3$. What is the length of $DC$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral $ABFE\\sim$ quadrilateral $EFCD$,\\\\\nwe have $\\frac{AB}{EF}=\\frac{EF}{CD}$,\\\\\nConsidering $AB=2$ and $EF=3$,\\\\\nwe obtain $CD=\\frac{EF^2}{AB}=\\boxed{\\frac{9}{2}}$",
    "image_path": "dataset/mm_math/images/MM_Math/53043599.png"
  },
  {
    "solution": "\\boxed{(-4, -2)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a linear function $y_{1}=k_{1}x+b$ intersects with the hyperbolic function $y_{2}=\\frac{k_{2}}{x}\\left(x<0\\right)$ at points $A$ and $B$, and intersects with the coordinate axes at $\\left(-6, 0\\right)$ and $\\left(0, 6\\right)$. The x-coordinate of point $B$ is $-4$. Directly write the solution to the inequality $k_{1}x+b>\\frac{k_{2}}{x}$.\n",
    "completion": "\\textbf{Solution:} Solve: $-4<x<-2$.\n\nDirectly write the solution of the inequality $k_{1}x+b>\\frac{k_{2}}{x}$ as $x \\in \\boxed{(-4, -2)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52622495_250.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ is rotated $60^\\circ$ clockwise around point $A$ to obtain $\\triangle AED$. If the length of segment $AB=4$, what is the length of $BE$?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABC$ is rotated $60^\\circ$ clockwise around point A to get $\\triangle AED$,\\\\\nit follows that $AB=AE$ and $\\angle BAE=60^\\circ$.\\\\\nThus, $\\triangle AEB$ is an equilateral triangle,\\\\\nwhich means $BE=AB$.\\\\\nGiven $AB=4$,\\\\\nit follows that $BE=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53136539.png"
  },
  {
    "solution": "\\boxed{20}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $AB=8$ and $BC=4$. A line passing through the midpoint $O$ of diagonal $BD$ intersects $AB$ and $CD$ at points $E$ and $F$, respectively. When quadrilateral $BEDF$ is a rhombus, find the area of the rhombus $BEDF$.\n",
    "completion": "\\textbf{Solution:} When the quadrilateral $BEDF$ is a rhombus, let $BE=x$, then $DE=x$, $AE=8-x$. In $\\triangle ADE$, we have $DE^2=AD^2+AE^2$, therefore\n\\[\nAE=AB-BE=8-x, \\quad AD=4,\n\\]\n\\[\n\\therefore (8-x)^2+16=x^2\n\\]\nSolving gives: $x=5$, that is $BE=5$,\n\\[\n\\therefore S_{\\text{rhombus}BEDF}=5\\times 4=\\boxed{20}.\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/50496842_89.png"
  },
  {
    "solution": "\\boxed{105^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, extend $BC$ to point $D$ and connect $AD$. Let $E$ be a point on $AD$. It is known that $\\angle B=50^\\circ$, $\\angle CAE=\\angle D$, $\\angle DCE=\\angle BAC=20^\\circ$. What is the degree measure of $\\angle CED$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle ACD = \\angle B + \\angle BAC$, and $\\angle DCE = \\angle BAC = 20^\\circ$, \\\\\n$\\therefore \\angle ACD = 50^\\circ + 20^\\circ = 70^\\circ$, \\\\\n$\\therefore \\angle CAE = \\angle D = \\frac{180^\\circ - \\angle ACD}{2} = 55^\\circ$, \\\\\n$\\therefore \\angle CED = 180^\\circ - \\angle DCE - \\angle D = 180^\\circ - 20^\\circ - 55^\\circ = \\boxed{105^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/53066432.png"
  },
  {
    "solution": "\\boxed{8\\times \\left(\\frac{5}{4}\\right)^{2020}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in rectangle $ABCD$, with $AB=2$ and $CB=4$, connect $AC$. Form rectangle $ACC_1B_1$ with diagonal $AC$ as a side, rotating in a counterclockwise direction, such that rectangle $ACC_1B_1\\sim$ rectangle $ADCB$. Then, connect $AC_1$ and form rectangle $AC_1C_2B_2$ with diagonal $AC_1$ as a side, rotating in a counterclockwise direction, so that rectangle $AC_1C_2B_2\\sim$ rectangle $ACC_1B_1$, and so on, following this pattern. Let the area of rectangle $ABCD$ be denoted as $S_1$, the area of rectangle $ACC_1B_1$ as $S_2$, the area of rectangle $AC_1C_2B_2$ as $S_3$, ..., then what is the value of $S_{2021}$?\n",
    "completion": "\\textbf{Solution:}\\\\\n$\\because$ Quadrilateral $ABCD$ is a rectangle,\\\\\n$\\therefore AB\\perp BC$,\\\\\n$\\therefore AC=\\sqrt{AB^{2}+CB^{2}}=\\sqrt{2^{2}+4^{2}}=2\\sqrt{5}$,\\\\\n$\\because$ Constructing a similar rectangle $AB_{1}C_{1}C$ to rectangle $ABCD$ in a counterclockwise direction,\\\\\n$\\therefore$ The ratio of the side lengths of rectangle $AB_{1}C_{1}C$ to rectangle $ABCD$ is $\\sqrt{5}:2$,\\\\\n$\\therefore$ The ratio of the area of rectangle $AB_{1}C_{1}C$ to the area of rectangle $ABCD$ is $5:4$,\\\\\n$\\because S_{1}=2\\times 4=8$,\\\\\n$S_{2}=8\\times \\frac{5}{4}$,\\\\\n$S_{3}=8\\times \\left(\\frac{5}{4}\\right)^{2}$, $\\dots$\\\\\n$\\therefore S_{2020}=8\\times \\left(\\frac{5}{4}\\right)^{2019}$,\\\\\n$S_{2021}=8\\times \\left(\\frac{5}{4}\\right)^{2020}=\\boxed{8\\times \\left(\\frac{5}{4}\\right)^{2020}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51433786.png"
  },
  {
    "solution": "\\boxed{75^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle A=32^\\circ$, and E is a point on side AC. First, fold $\\triangle ABE$ along BE, after the fold, side AB of $\\triangle ABE$ intersects AC at point D. Then fold $\\triangle BCD$ along BD, such that point C precisely lands on BE. If at this moment $\\angle CDB=82^\\circ$, what is the degree measure of $\\angle B$ in the original triangle?\n",
    "completion": "\\textbf{Solution}: As shown in the figure,\\\\\ndue to the properties of folding, we have: $\\angle GDB = \\angle CDB = 82^\\circ$, $\\angle FBE = \\angle ABE = \\angle ABG$,\\\\\nsince $\\angle GDB$ is an external angle of $\\triangle BDF$,\\\\\ntherefore, $\\angle DBF = \\angle GDB - \\angle F = 82^\\circ - 32^\\circ = 50^\\circ$,\\\\\ntherefore, $\\angle FBE = \\angle ABE = 25^\\circ$,\\\\\ntherefore, $\\angle FBG = 3 \\times 25^\\circ = 75^\\circ$,\\\\\nthus, the angle $B$ of the original triangle is $\\boxed{75^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52551645.png"
  },
  {
    "solution": "\\boxed{(5,3)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the graph, in the Cartesian coordinate system, $\\triangle OAB$ is translated rightward along the x-axis to obtain $\\triangle O'A'B'$. The coordinates of point A are $(0, 4)$, and the corresponding point A' lies on the line $y =\\frac{5}{4}x-1$. Point B is on the angle bisector of $\\angle A'AO$. If the area of quadrilateral $AA'B'B$ is 4, what are the coordinates of point B?\n",
    "completion": "\\textbf{Solution:} According to the properties of translation, we have: $AA'=BB'$, $AA'\\parallel BB'$, and the ordinate of $A'$ is 4,\\\\\n$\\because$ point $A'$ lies on the line $y=\\frac{5}{4}x-1$,\\\\\n$\\therefore$ 4=$\\frac{5}{4}x-1$,\\\\\n$\\therefore$ x=4,\\\\\n$\\therefore$ AA'=BB'=4,\\\\\n$\\because$ the area of the quadrilateral $AA'B'B$ is 4,\\\\\n$\\therefore$ the distance from point B to $AA'$ is 1,\\\\\n$\\because$ point B is on the angle bisector of $\\angle A'AO$,\\\\\n$\\therefore$ the distance from point B to OA is 1,\\\\\n$\\therefore$ the coordinates of point B' are $\\boxed{(5,3)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51419801.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within the Cartesian coordinate system, the line $y=\\frac{1}{2}x+b$ intersects the x-axis at point A, and the line $y=-x+2$ intersects the x-axis at point B. The two lines intersect at point $C(-2, a)$. A point $P(m, 0)$ is to the right of point A, and segment $PC$ is drawn. When $\\triangle ACP$ is an isosceles triangle, find the value of $m$.\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, draw $CD\\perp AB$ and connect $PC$,\\\\\nwe have $CD=4$, $AD=8$, $PA=m+10$,\\\\\n$\\therefore AC=\\sqrt{8^2+4^2}=4\\sqrt{5}$.\\\\\nWhen $PC=PA$, let $PC=PA=m+10$, then $PD=-2-m$,\\\\\nin $\\triangle PCD$, $PC^2=PD^2+CD^2$,\\\\\n$\\therefore (m+10)^2=(-2-m)^2+4^2$, solving this yields: $m=\\boxed{-5}$;\\\\\nWhen $AC=AP$, we have $AP=4\\sqrt{5}$,\\\\\n$\\therefore m=\\boxed{-10+4\\sqrt{5}}$;\\\\\nWhen $CA=CP$, point D is the midpoint of $AP$,\\\\\n$\\therefore AP=2AD=16$,\\\\\n$\\therefore m=\\boxed{6}$.\\\\\nIn summary, when $\\triangle ACP$ is an isosceles triangle, the value of $m$ is \\boxed{-5}, \\boxed{-10+4\\sqrt{5}}, or \\boxed{6}.",
    "image_path": "dataset/mm_math/images/MM_Math/52423819_218.png"
  },
  {
    "solution": "\\boxed{\\frac{26}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, $AC=5$, and $BC=12$. If $\\triangle ABC$ is folded along $AD$ such that point $C$ coincides with point $E$ on $AB$, what is the length of $DB$?\n",
    "completion": "\\textbf{Solution}: By the property of folding, we have: $AE=AC=5$, $DC=DE$, and $\\angle AED=\\angle BED=\\angle C=90^\\circ$.\\\\\nSince $\\angle C=90^\\circ$ and $AC=5$, $BC=12$,\\\\\nthus $AB=\\sqrt{AC^2+BC^2}=13$,\\\\\nand hence $BE=AB-AE=8$.\\\\\nLet $DC=x$, then $DE=x$, and $BD=BC-DC=12-x$.\\\\\nIn $\\triangle BDE$, by the Pythagorean theorem, we have $DE^2+BE^2=BD^2$,\\\\\nthus $x^2+8^2=(12-x)^2$,\\\\\nsolving this, we find $x=\\frac{10}{3}$.\\\\\nTherefore, $BD=12-x=\\boxed{\\frac{26}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51367879.png"
  },
  {
    "solution": "\\boxed{8+6\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=30^\\circ$, $AC=12$, $\\sin B=\\frac{3}{5}$, find the length of $BC$.\n",
    "completion": "\\textbf{Solution:} Construct $AD\\perp BC$ at point A, with D as the foot\\\\\nIn $\\triangle ADC$, $AC=12$, $\\angle C=30^\\circ$\\\\\n$\\therefore AD=\\frac{1}{2}AC=6$, $CD=\\sqrt{3}AD=6\\sqrt{3}$\\\\\nIn $\\triangle ABD$, $\\sin B=\\frac{3}{5}$\\\\\n$\\therefore AB=10$\\\\\n$\\therefore BD=8$\\\\\n$\\therefore BC=BD+CD=8+6\\sqrt{3}$\\\\\n$\\therefore$ The length of $BC$ is $\\boxed{8+6\\sqrt{3}}$",
    "image_path": "dataset/mm_math/images/MM_Math/53043266.png"
  },
  {
    "solution": "\\boxed{(-1, 0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the line $y=2x+6$ intersects the x-axis at point A and the y-axis at point C. The parabola $y=-2x^{2}+bx+c$ passes through points A and C, and intersects the x-axis at another point B. If point E is located to the left of the axis of symmetry, point M is a point on the axis of symmetry of the parabola, and point N is a point on the parabola, when the quadrilateral formed by points M, N, E, B is a rhombus, directly write down the coordinates of point M.\n",
    "completion": "\\textbf{Solution:} The coordinates of point M are \\boxed{(-1, -4)}, \\boxed{(-1, -16)}, or \\boxed{(-1, 0)}.",
    "image_path": "dataset/mm_math/images/MM_Math/53066295_95.png"
  },
  {
    "solution": "\\boxed{(-4, \\frac{84}{3})}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given the parabola $y=\\frac{4}{3}x^{2}+bx+c$ intersects the x-axis at points $A(-1,0)$ and $B$ (with point $B$ to the right of point $A$) and intersects the y-axis at point $C(0,-4)$. As shown in the figure, point $H$ is a moving point on the parabola, and the area of $\\triangle OCH$ is $8$. What are the coordinates of point $H$?\n",
    "completion": "\\textbf{Solution:} Let the x-coordinate of point H be $m$, \n\n$\\therefore S_{\\triangle OCH}=\\frac{1}{2}\\text{OC} \\cdot |x_H|$, \n\n$\\therefore 2|m|=8$, \n\nSolving this gives: $m=4$ or $m=-4$, \n\nWhen $m=4$, $y=\\frac{4}{3}(4-1)^2-\\frac{16}{3}=\\frac{20}{3}$, \n\nWhen $m=-4$, $y=\\frac{4}{3}(-4-1)^2-\\frac{16}{3}=\\frac{84}{3}$, \n\n$\\therefore$ The coordinates of point H are $\\boxed{(4, \\frac{20}{3})}$ or $\\boxed{(-4, \\frac{84}{3})}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52719833_112.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{10}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point A is on the line segment BG. The quadrilateral ABCD and the quadrilateral DEFG are both squares with areas of 10 and 19 respectively. What is the area of $\\triangle CDE$?\n",
    "completion": "\\textbf{Solution:} Draw EH$\\perp$CD at point H.\\\\\n$\\because \\angle$ADG+$\\angle$GDH=$\\angle$EDH+$\\angle$GDH,\\\\\n$\\therefore \\angle$ADG=$\\angle$EDH.\\\\\nAlso, $\\because$DG=DE, $\\angle$DAG=$\\angle$DHE.\\\\\n$\\therefore \\triangle ADG\\cong \\triangle HDE$.\\\\\n$\\therefore$HE=AG.\\\\\n$\\because$Both quadrilaterals ABCD and DEFG are squares, with areas 5 and 9, respectively. That is, AD$^{2}$=5, DG$^{2}$=9.\\\\\n$\\therefore$In the right-angled $\\triangle ADG$,\\\\\nAG=$\\sqrt{DG^{2}-AD^{2}}=\\sqrt{9-5}=2$,\\\\\n$\\therefore$EH=AG=2.\\\\\n$\\therefore$The area of $\\triangle CDE$ is $\\frac{1}{2}$CD$\\cdot$EH=$\\frac{1}{2}\\times\\sqrt{10}\\times2=\\boxed{\\frac{\\sqrt{10}}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52416366.png"
  },
  {
    "solution": "\\boxed{15}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\triangle ABC$, $ \\angle C=90^\\circ$, and $ AD$ bisects $ \\angle BAC$. Given $ AB=10$ and $ CD=3$, what is the area of $ \\triangle ABD$?\n",
    "completion": "\\textbf{Solution}: As shown in the figure, draw DE$\\perp$AB at point E through point D,\\\\\nsince $\\angle C=90^\\circ$ and AD bisects $\\angle$BAC,\\\\\nit follows that $DE=CD=3$,\\\\\nthus, the area of $\\triangle ABD$ $=\\frac{1}{2}AB\\cdot DE=\\frac{1}{2}\\times 10\\times 3=\\boxed{15}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52861522.png"
  },
  {
    "solution": "\\boxed{AECD}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, $AC$ intersects $BD$ at point $O$, and $AO=CO$. Point $E$ is on segment $BO$, and $\\angle CEO=\\angle ADO$. Prove that quadrilateral $AECD$ is a parallelogram.\n",
    "completion": "\\textbf{Solution.} Proof:\n\nSince $\\angle CEO = \\angle ADO$,\n\nit follows that $CE \\parallel AD$.\n\nGiven $\\angle CEO = \\angle ADO$, $\\angle COE = \\angle AOD$, and $OC = OA$,\n\nwe have $\\triangle COE \\cong \\triangle AOD$,\n\nwhich implies $CO = AO$, $OE = OD$.\n\nThus, quadrilateral \\boxed{AECD} is a parallelogram.",
    "image_path": "dataset/mm_math/images/MM_Math/52424185_125.png"
  },
  {
    "solution": "\\boxed{\\left(2^{2022}-1,2^{2021}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Consider placing the squares $A_1B_1C_1O$, $A_2B_2C_2C_1$, $A_3B_3C_3C_2$ as shown in the figure, where the points $A_1$, $A_2$, $A_3,\\dots$ and the points $C_1$, $C_2$, $C_3,\\dots$ lie on the line $y=x+1$ and the x-axis, respectively. What are the coordinates of point $B_3$? What are the coordinates of point $B_{2022}$?\n",
    "completion": "\\textbf{Solution:} When $x=0$, $y=x+1=1$, \\\\\n$\\therefore$ the coordinates of point $A_1$ are $(0,1)$, \\\\\n$\\therefore$ $OA_1=1$. \\\\\n$\\because$ Quadrilateral $A_1B_1C_1O$ is a square, \\\\\n$\\therefore$ $OC_1=B_1C_1=OA_1=1$, \\\\\n$\\therefore$ the coordinates of point $B_1$ are $(1,1)$, and the coordinates of point $C_1$ are $(1,0)$. \\\\\nWhen $x=1$, $y=x+1=2$, \\\\\n$\\therefore$ the coordinates of point $A_2$ are $(1,2)$, \\\\\n$\\therefore$ $A_2C_1=2$. \\\\\n$\\because$ $A_2B_2C_2C_1$ is a square, \\\\\n$\\therefore$ $C_1C_2=B_2C_2=A_2C_1=2$, \\\\\n$\\therefore$ the coordinates of point $B_2$ are $(3,2)$, and the coordinates of point $C_2$ are $(3,0)$. \\\\\nSimilarly, we can obtain that the coordinates of point $B_3$ are $\\boxed{(7,4)}$, \\\\\n$\\therefore$ the coordinates of point $B_n$ are $\\left(2^n-1,2^{n-1}\\right)$. \\\\\n$\\therefore$ the coordinates of point $B_{2022}$ are $\\boxed{\\left(2^{2022}-1,2^{2021}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52395011.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, $AD \\parallel BC$, $\\angle B = \\angle ACD = 90^\\circ$, $\\triangle ABC \\sim \\triangle DCA$. If $BC=1$, $AC=2$, find the length of $AD$.\n",
    "completion": "Solution: Since $\\triangle ABC \\sim \\triangle DCA$, \\\\\n$\\therefore \\frac{BC}{AC}=\\frac{AC}{AD}$, which means $\\frac{1}{2}=\\frac{2}{AD}$, \\\\\n$\\therefore AD=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51299801_113.png"
  },
  {
    "solution": "\\boxed{50^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point E is a point on side AD of quadrilateral ABCD. Extend line EB such that $BF = BE$, and extend line EC such that $CG = CE$. Connect FG. H is the midpoint of FG. Connect DH and AF. If $\\angle BAE = 70^\\circ$ and $\\angle DCE = 20^\\circ$, what is the measure of $\\angle DEC$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral ABCD is a parallelogram, \\\\\n$\\therefore \\angle BAE=\\angle BCD=70^\\circ$, with AD$\\parallel$BC, \\\\\nsince $\\angle DCE=20^\\circ$, and given $AB\\parallel CD$, \\\\\n$\\therefore \\angle CDE=180^\\circ-\\angle BAE=110^\\circ$, \\\\\n$\\therefore \\angle DEC=180^\\circ-\\angle DCE-\\angle CDE=\\boxed{50^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52354319_91.png"
  },
  {
    "solution": "\\boxed{130^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, line AB and CD intersect at point O, $ \\angle DOE = 90^\\circ$, OD bisects $ \\angle BOF$, $ \\angle BOE = 50^\\circ$. What is the degree measure of $ \\angle EOF$?\n",
    "completion": "\\textbf{Solution:} Since $OD$ bisects $\\angle BOF$, \\\\\n$\\therefore \\angle DOF=\\angle BOD=40^\\circ$, \\\\\n$\\therefore \\angle EOF=\\angle DOF+\\angle DOE=40^\\circ+90^\\circ=\\boxed{130^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52795107_101.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB = 90^\\circ$, and $D$, $E$, and $F$ are the midpoints of $AB$, $BC$, and $CA$ respectively. If $CD = 4$, what is the length of $EF$?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABC$ is a right-angled triangle, and $CD$ is the median to the hypotenuse,\\\\\ntherefore $CD=\\frac{1}{2}AB$,\\\\\nthus $AB=2CD=8$,\\\\\nsince $EF$ is the median of $\\triangle ABC$,\\\\\ntherefore $EF=\\frac{1}{2}AB= \\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52225662.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ is an inscribed triangle of circle $O$ with $OB$ and $OC$ connected. If the length of chord $BC$ is $2\\sqrt{3}$, what is the measure of $\\angle BAC$ in degrees?\n",
    "completion": "\\textbf{Solution:} Construct OE $\\perp$ BC at E. \\\\\nSince OE $\\perp$ BC, \\\\\nthen BE=EC= $\\sqrt{3}$, and $\\angle$BOE=$\\angle$COE, \\\\\nthus OE= $\\sqrt{OB^{2}-BE^{2}}=\\sqrt{2^{2}-(\\sqrt{3})^{2}}=1$, \\\\\ntherefore OB=2OE, \\\\\nhence $\\angle$OBE=30$^\\circ$, \\\\\nthus $\\angle$BOE=$\\angle$COE=60$^\\circ$, \\\\\ntherefore $\\angle$BOC=120$^\\circ$, \\\\\nwhich implies $\\angle$BAC=\\boxed{60^\\circ}.",
    "image_path": "dataset/mm_math/images/MM_Math/51433195.png"
  },
  {
    "solution": "\\boxed{\\frac{4}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=\\frac{1}{2}x+2$ intersects the $x$-axis and $y$-axis at points $A$ and $C$, respectively. The parabola $y=-\\frac{1}{2}x^2+bx+c$ passes through points $A$ and $C$, and its other intersection point with the $x$-axis is $B$. Point $D$ is a moving point on the parabola. When point $D$ is above line $AC$, by connecting $BC$, $CD$, $BD$, and with $BD$ intersecting $AC$ at point $E$, let the area of $\\triangle CDE$ be $S_1$ and the area of $\\triangle BCE$ be $S_2$. What is the maximum value of $\\frac{S_1}{S_2}$?\n",
    "completion": "\\textbf{Solution:} Let $y=0$,\n\n$\\therefore -\\frac{1}{2}x^2-\\frac{3}{2}x+2=0$,\n\n$\\therefore x_1=-4$, $x_2=1$,\n\n$\\therefore B(1,0)$,\n\nDraw $DM\\perp x$ axis from $D$ meeting $AC$ at point $M$, draw $BN\\perp x$ axis from $B$ meeting $AC$ at $N$,\n\n$\\therefore DM\\parallel BN$,\n\n$\\therefore \\triangle DME\\sim \\triangle BNE$,\n\n$\\therefore \\frac{S_1}{S_2}=\\frac{DE}{BE}=\\frac{DM}{BN}$,\n\nLet $D\\left(a, -\\frac{1}{2}a^2-\\frac{3}{2}a+2\\right)$,\n\n$\\therefore M\\left(a, \\frac{1}{2}a+2\\right)$,\n\n$\\because B(1,0)$,\n\n$\\therefore N\\left(1, \\frac{5}{2}\\right)$,\n\n$\\therefore \\frac{S_1}{S_2}=\\frac{DM}{BN}=-\\frac{1}{5}(a+2)^2+\\frac{4}{5}$,\n\n$\\therefore$ When $a=-2$, the maximum value of $\\frac{S_1}{S_2}$ is $\\boxed{\\frac{4}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52622506_111.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in parallelogram $ABCD$, $AC$ and $BD$ intersect at point $O$, and $E$ is the midpoint of $AB$. If $OE = 6$, what is the length of $AD$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a parallelogram, \\\\\nit follows that $BO=DO$, \\\\\nsince point E is the midpoint of AB, \\\\\nit follows that $OE$ is the median line of $\\triangle ABD$, \\\\\nhence $AD=2OE$, \\\\\nsince $OE=6$, \\\\\nit follows that $AD=\\boxed{12}$",
    "image_path": "dataset/mm_math/images/MM_Math/52251026.png"
  },
  {
    "solution": "\\boxed{3\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the rhombus $ABCD$, we have $AB=6$ and $\\angle CAB=30^\\circ$. If we extend $AB$ and $CD$ to form the rectangle $AECF$, what is the length of the side $CE$ of the rectangle?\n",
    "completion": "Solution: Since in rhombus ABCD, AB=6, and quadrilateral AECF is a rectangle, it follows that $CB=AB=6$, $\\angle E=90^\\circ$. Since $\\angle CAB=30^\\circ$, it follows that $\\angle ACB=\\angle CAB=30^\\circ$, $\\angle ACE=60^\\circ$, hence $\\angle BCE=30^\\circ$. Therefore, $BE=\\frac{1}{2}CB=3$. In right triangle $BCE$, $CE=\\sqrt{BC^{2}-BE^{2}}=\\sqrt{6^{2}-3^{2}}=\\boxed{3\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52303698.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: The three views of a geometric body are interrelated. Given the three views of a right triangular prism as shown in the figure, in $\\triangle PMN$, $\\angle MPN=90^\\circ$, $PN=4$, and $\\sin\\angle PMN=\\frac{4}{5}$. If the rectangles in the front view and the left view are similar, what is the length of $AB$?\n",
    "completion": "\\textbf{Solution:} Since rectangle ABCD is similar to rectangle EFGH, and AB=EF, \\\\\nit follows that $\\frac{AB}{FG} = \\frac{BC}{EF}$, which means $\\frac{AB}{\\frac{12}{5}} = \\frac{5}{EF}$. Solving this gives: $AB = \\boxed{2\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51246271_90.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB \\parallel DE$, and points $B$, $C$, $D$ are on the same line. If $\\angle BCE = 55^\\circ$ and $\\angle E = 25^\\circ$, then what is the measure of $\\angle B$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle BCE=55^\\circ$ and $\\angle E=25^\\circ$, we have $\\angle BCE=\\angle E+\\angle D$. \\\\\nTherefore, $\\angle D=\\angle BCE-\\angle E=55^\\circ-25^\\circ=30^\\circ$. \\\\\nThus, $AB\\parallel DE$. \\\\\nConsequently, $\\angle B=\\angle D$. \\\\\nHence, $\\angle B=\\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52950404.png"
  },
  {
    "solution": "\\boxed{20}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points $A, B, C$ are on the circle $\\odot O$, $\\angle B=55^\\circ$, and $\\angle BOC=70^\\circ$. What is the measure of $\\angle C$ in degrees?\n",
    "completion": "\\textbf{Solution:} Since $\\angle BOC=70^\\circ$ and it subtends the same arc $\\overset{\\frown}{BC}$,\\\\\nit follows that $\\angle A=\\frac{1}{2}\\angle BOC=35^\\circ$,\\\\\nSince $\\angle B+\\angle A+\\angle BDA=180^\\circ$,\\\\\nit follows that $\\angle ADB=90^\\circ$,\\\\\nand hence $\\angle ODC=90^\\circ$,\\\\\ntherefore $\\angle C=180^\\circ-\\angle ODC-\\angle BOC=20^\\circ$.\\\\\nThus, the answer is: $\\boxed{20}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53147978.png"
  },
  {
    "solution": "\\boxed{25^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of $\\odot O$. Points $D$ and $C$ are on $\\odot O$. Connect $AD$, $DC$, $AC$. If $\\angle C=65^\\circ$, then what is the degree measure of $\\angle BAD$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect BD.\\\\\nSince $AB$ is the diameter of $\\odot O$,\\\\\n$\\therefore$ $\\angle ADB=90^\\circ$.\\\\\nSince $\\angle C=65^\\circ$,\\\\\n$\\therefore$ $\\angle ABD=65^\\circ$,\\\\\n$\\therefore$ $\\angle BAD=90^\\circ-65^\\circ=\\boxed{25^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52871899.png"
  },
  {
    "solution": "\\boxed{1 : 16}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $D,E$ are points on the sides $AB$ and $BC$ of $\\triangle ABC$, respectively, with $DE\\parallel AC$. If the ratio of the areas ${S}_{\\triangle BDE}:{S}_{\\triangle ADE}=1:3$, what is the value of ${S}_{\\triangle DOE}:{S}_{\\triangle AOC}$?\n",
    "completion": "\\textbf{Solution:} From the construction, we have\n\\[\n\\because S_{\\triangle BDE} : S_{\\triangle ADE} = 1 : 3,\n\\]\n\\[\n\\therefore BD : AD = 1 : 3,\n\\]\n\\[\n\\therefore BD : BA = 1 : 4,\n\\]\n\\[\n\\because DE \\parallel AC,\n\\]\n\\[\n\\therefore \\triangle DOE \\sim \\triangle AOC,\n\\]\n\\[\n\\therefore \\frac{DE}{AC} = \\frac{BD}{BA} = \\frac{1}{4},\n\\]\n\\[\n\\therefore S_{\\triangle DOE} : S_{\\triangle AOC} = \\boxed{1 : 16}.\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/53065543.png"
  },
  {
    "solution": "\\boxed{94^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, $\\angle DAC=\\angle ACB$, $\\angle D=86^\\circ$, what is the measure of $\\angle BCD$ in degrees?\n",
    "completion": "\\textbf{Solution:} Since $\\angle DAC = \\angle ACB$, \\\\\nit follows that $AD \\parallel BC$, \\\\\nhence $\\angle D + \\angle BCD = 180^\\circ$, \\\\\nsince $\\angle D = 86^\\circ$, \\\\\nit follows that $\\angle BCD = 180^\\circ - \\angle D = 180^\\circ - 86^\\circ = \\boxed{94^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53404559.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, it is known that point $D$ is on segment $AB$, and point $C$ is the midpoint of segment $AB$. If $AB=8\\,\\text{cm}$ and $BD=3\\,\\text{cm}$, what is the length of segment $CD$ in cm?\n",
    "completion": "\\textbf{Solution:} Given that point $C$ is the midpoint of line segment $AB$,\\\\\nhence, $CB = \\frac{1}{2}AB = 4\\,\\text{cm}$,\\\\\nthus, $CD = CB - BD = 4 - 3 = 1\\,\\text{cm}$,\\\\\nmeaning the length of line segment $CD$ is \\boxed{1}\\,\\text{cm}.",
    "image_path": "dataset/mm_math/images/MM_Math/47110741_184.png"
  },
  {
    "solution": "\\boxed{24}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\triangle ABC$, $\\angle C=90^\\circ$, D is a point on $AC$. If $DA=DB=15$ and the area of $\\triangle ABD$ is $90$, what is the length of AC?\n",
    "completion": "\\textbf{Solution:} Since the area of $\\triangle ABD$ is $90$, and $\\angle C=90^\\circ$\\\\\nTherefore, $\\frac{1}{2}AD\\cdot BC=90$\\\\\nTherefore, $BC=\\frac{90\\times 2}{AD}=12$\\\\\nIn $Rt\\triangle ABC$, $CD=\\sqrt{BD^{2}-BC^{2}}=\\sqrt{15^{2}-12^{2}}=9$\\\\\nTherefore, $AC=AD+CD=24$\\\\\nHence, the answer is: $AC=\\boxed{24}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52827971.png"
  },
  {
    "solution": "\\boxed{20^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $O$ is a point on the line $AB$, $\\angle COD$ is a right angle, $OE$ bisects $\\angle BOC$, $\\angle AOC=40^\\circ$, then what is the degree measure of $\\angle DOE$?\n",
    "completion": "\\textbf{Solution:} Given $\\because$ $\\angle AOC = 40^\\circ$,\\\\\n$\\therefore$ $\\angle BOC = 180^\\circ - \\angle AOC = 140^\\circ$,\\\\\n$\\because$ OE bisects $\\angle BOC$,\\\\\n$\\therefore$ $\\angle COE = \\angle BOE = 70^\\circ$,\\\\\n$\\because$ $\\angle COD$ is a right angle, that is $\\angle COD = 90^\\circ$,\\\\\n$\\therefore$ $\\angle DOE = \\angle COD - \\angle COE = \\boxed{20^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53010644.png"
  },
  {
    "solution": "\\boxed{x > -3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=kx+b$ intersects with the x-axis and y-axis at points $A(-3, 0)$ and $B(0, 2)$ respectively. What is the solution set of the inequality $kx+b>0$?\n",
    "completion": "\\[ \\text{\\textbf{Solution:}} \\quad \\text{According to the graph, it is known that the function values of y to the right of point A are all greater than 0. Therefore, when } x > -3, \\text{ the function values are greater than 0, that is, } kx + b > 0; \\]\n\\[ \\text{Thus, the answer is: } \\boxed{x > -3}. \\]",
    "image_path": "dataset/mm_math/images/MM_Math/52189353.png"
  },
  {
    "solution": "\\boxed{\\left(\\frac{3}{2}, 0\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the line $y=4x-6$ intersects the $x$-axis at point $A$, and the line $y=-2x-2$ intersects the $x$-axis at point $B$, and these two lines intersect at point $C$. What are the coordinates of point $A$? What are the coordinates of point $B$?\n",
    "completion": "\\textbf{Solution:} The coordinates of point $A$ are $\\boxed{\\left(\\frac{3}{2}, 0\\right)}$.\\\\\nThe coordinate information of point $B$ is not provided, so an answer cannot be given.",
    "image_path": "dataset/mm_math/images/MM_Math/52423437_60.png"
  },
  {
    "solution": "\\boxed{1.4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ is inscribed in $\\odot O$, with $AB$ being the diameter of $\\odot O$, $AB=10$, $AC=6$. Connect $OC$, and the chord $AD$ intersects $OC$, $BC$ at points $E$, $F$, where point $E$ is the midpoint of $AD$. $\\angle CAD=\\angle CBA$. What is the length of $OE$?\n",
    "completion": "\\textbf{Solution:} Since $AB$ is a diameter, \\\\\n$\\therefore \\angle ACB=90^\\circ$, \\\\\nBecause $AE=DE$, \\\\\n$\\therefore OC\\perp AD$, \\\\\n$\\therefore \\angle AEC=90^\\circ$, \\\\\n$\\therefore \\angle AEC=\\angle ACB$, \\\\\n$\\therefore \\triangle AEC\\sim \\triangle BCA$, \\\\\n$\\therefore \\frac{CE}{AC} = \\frac{AC}{AB}$, \\\\\n$\\therefore \\frac{CE}{6} = \\frac{6}{10}$, \\\\\n$\\therefore CE=3.6$, \\\\\nSince $OC= \\frac{1}{2} AB=5$, \\\\\n$\\therefore OE=OC-EC=5-3.6=\\boxed{1.4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51179503_163.png"
  },
  {
    "solution": "\\boxed{AE=3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rhombus $ABCD$, $AC$ is a diagonal, and $AB=AC=6$. Points $E$ and $F$ are on sides $AB$ and $BC$ respectively, satisfying $AE=BF$. The segments $AF$ and $CE$ intersect at point $G$. Point $O$ is the midpoint of segment $CE$, and segment $EO$ is rotated clockwise by $60^\\circ$ around point $E$ to obtain segment $EM$. When $\\triangle MAC$ forms an isosceles triangle, please directly state the length of $AE$.\n",
    "completion": "Solution:\n\n(1) If $AM=MC$, then $\\triangle MAC$ is an isosceles triangle. In this case, let point $P$ be the midpoint of $AC$, and connect $OP$, $OM$, and $BM$,\n\n$\\because \\angle MEO=60^\\circ$, $EO=EM$,\n\n$\\therefore \\triangle OEM$ is an equilateral triangle,\n\n$\\because \\angle FGC=60^\\circ$,\n\n$\\therefore \\angle MEO=\\angle FGC$,\n\n$\\therefore ME\\parallel AF$,\n\n$\\because O$ is the midpoint of $CE$, $P$ is the midpoint of $AC$,\n\n$\\therefore OP$ is the midline of $\\triangle AEC$, $OP\\parallel AB$,\n\n$\\because \\triangle ABC$ is an equilateral triangle, $\\triangle MAC$ is an isosceles triangle, $P$ is the midpoint of $AC$,\n\n$\\therefore from$ \"three lines coincide,\" points $B$, $M$, $P$ are collinear, and $BP\\perp AC$, $AP=PC= \\frac{1}{2}AC=3$, $\\angle ABP= \\frac{1}{2} \\angle ABC=30^\\circ$,\n\n$\\because \\triangle OEM$ is an equilateral triangle,\n\n$\\therefore OE=OM$, $\\angle OEM=\\angle OME$,\n\n$\\because OE=OC$,\n\n$\\therefore OM=OC$, $\\angle OMC=\\angle OCM$,\n\n$\\therefore \\angle OEM+\\angle OCM=\\angle OME+\\angle OMC$,\n\nthat is: $\\angle OEM+\\angle OCM=\\angle EMC$,\n\n$\\therefore \\angle EMC=90^\\circ$, $CM\\perp EM$,\n\n$\\therefore in \\triangle CEM$, $\\angle ECM=90^\\circ-60^\\circ=30^\\circ$,\n\nAt this point, as shown in the figure, rotate $\\triangle AEC$ counterclockwise around point $C$ through $60^\\circ$ until $\\triangle BNC$, connect $MN$,\n\nthen $\\angle ACE=\\angle BCN$, $\\angle NBC=60^\\circ$,\n\n$\\because \\angle ECM=30^\\circ$,\n\n$\\therefore \\angle ACE+\\angle MCB=30^\\circ$,\n\n$\\therefore \\angle BCN+\\angle MCB=\\angle MCN=30^\\circ$,\n\n$\\therefore \\angle MCN=\\angle MCE=30^\\circ$,\n\n$\\because CE=CN$, $\\angle MCN=\\angle MCE$, $CM=CM$,\n\n$\\therefore \\triangle MCE\\cong \\triangle MCN$,\n\n$\\therefore \\angle CMN=\\angle CME=90^\\circ$,\n\n$\\therefore points $E, $M$, $N$ are collinear,\n\n$\\therefore \\triangle ECN$ is an equilateral triangle,\n\n$\\because \\angle NBC=\\angle ACB=60^\\circ$,\n\n$\\therefore BN\\parallel AC$,\n\n$\\because \\angle BPC=90^\\circ$,\n\n$\\therefore \\angle NBM=90^\\circ$,\n\n$\\because \\angle CMN=90^\\circ$,\n\n$\\therefore \\angle BMN+\\angle CMP=90^\\circ$,\n\n$\\because \\angle BMN+\\angle BNM=90^\\circ$,\n\n$\\therefore \\angle BNM=\\angle CMP$,\n\n$\\therefore \\triangle BMN\\sim \\triangle PCM$,\n\n$\\therefore \\frac{BM}{PC}=\\frac{NM}{MC}$,\n\n$\\because \\frac{NM}{MC}=\\tan\\angle MCN=\\tan30^\\circ$,\n\n$\\therefore \\frac{BM}{PC}=\\tan30^\\circ=\\frac{\\sqrt{3}}{3}$,\n\n$\\because PC=3$,\n\n$\\therefore BM= \\sqrt{3}$,\n\nIn $\\triangle ABP$, $AP=3$, $\\angle ABP=30^\\circ$,\n\n$\\therefore BP= 3\\sqrt{3}$,\n\n$\\therefore PM=BP-BM= 2\\sqrt{3}$,\n\n$\\because \\angle MBC=30^\\circ$, $\\angle OMC=90^\\circ-\\angle OME=30^\\circ$,\n\n$\\therefore \\angle MBC+\\angle MCB=\\angle OMC+\\angle OMP$,\n\n$\\therefore \\boxed{AE=3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51152250_121.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\triangle ABC$, D, E, and F are the midpoints of the three sides respectively. Given $ AB=6$ and $AC=4$, what is the perimeter of the quadrilateral AEDF?\n",
    "completion": "\\textbf{Solution:}\nGiven that in $\\triangle ABC$, E, D, and F are the midpoints of AB, BC, and CA, respectively,\n$\\therefore$ DE=AF=$\\frac{1}{2}$AC=2, DF=AE=$\\frac{1}{2}$AB=3,\n$\\therefore$ the perimeter of quadrilateral AEDF is $(2+3)\\times 2=\\boxed{10}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52065420.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $D$ and $E$ are the midpoints of $AB$ and $AC$, respectively. Point $F$ is on segment $DE$, and $AF\\perp BF$. If $AB=4$ and $BC=7$, what is the length of $EF$?\n",
    "completion": "\\textbf{Solution:} Since points D and E are the midpoints of sides AB and AC, respectively,\n\nit follows that DE is the midline of $\\triangle ABC$.\n\nSince BC = 7,\n\nwe have DE = $\\frac{1}{2}$BC = $\\frac{1}{2} \\times 7 = \\frac{7}{2}$.\n\nSince AF$\\perp$BF and D is the midpoint of AB, with AB = 4,\n\nit follows that DF = $\\frac{1}{2}$AB = $\\frac{1}{2} \\times 4 = 2$.\n\nTherefore, EF = DE - DF = $\\frac{7}{2} - 2 = \\boxed{\\frac{3}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52189050.png"
  },
  {
    "solution": "\\boxed{x \\in (-\\infty, -1) \\cup (0, 2)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y=kx+b$ intersects with the graph of the inverse proportion function $y=\\frac{m}{x}$ at points $P(2, a)$ and $Q(-1, -4)$. Based on the graph, directly write the solution set for the inequality $\\frac{m}{x}>kx+b$ with respect to $x$.\n",
    "completion": "\\textbf{Solution:} The solution set for $x<-1$ or $0<x<2$ is $\\boxed{x \\in (-\\infty, -1) \\cup (0, 2)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52797108_43.png"
  },
  {
    "solution": "\\boxed{C(1,2)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the graph of the linear function $y=kx+b$ intersects the $x$ axis and the $y$ axis at points $A$ and $B$ respectively, and intersects the graph of the inverse proportion function $y= \\frac{m}{x}$ in the first quadrant at point $C$, with $CD$ perpendicular to the $x$ axis, and the foot of the perpendicular being $D$. If $OA=OB=OD=1$, find: the coordinates of points $A$, $B$, and $C$?\n",
    "completion": "\\textbf{Solution:} Since $OA=OB=OD=1$, \\\\\nit follows that $A(-1,0)$, $B(0,1)$, substituting into the linear equation of line AB $y=kx+b$, \\\\\nwe obtain $\\begin{cases}0=-k+b \\\\ 1=0+b\\end{cases}$, solving this gives $\\begin{cases}k=1 \\\\ b=1\\end{cases}$, \\\\\nhence, the equation of line AB is $y=x+1$. Substituting $x=1$ into $y=x+1$ yields $y=2$, \\\\\nthus $C(1,2)$.\\\\\nTherefore, the coordinates of points $A$, $B$, and $C$ are $\\boxed{A(-1,0)}$, $\\boxed{B(0,1)}$, and $\\boxed{C(1,2)}$ respectively.",
    "image_path": "dataset/mm_math/images/MM_Math/51546388_67.png"
  },
  {
    "solution": "\\boxed{110^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, quadrilateral $ABCD$ is inscribed in circle $\\odot O$, $\\angle BAD=70^\\circ$, then what is the degree measure of $\\angle BCD$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral ABCD is inscribed in circle O, $\\angle$BAD=$70^\\circ$,\\\\\ntherefore, $\\angle$BAD+$\\angle$BCD=$180^\\circ$,\\\\\ntherefore, $\\angle$BCD=$\\boxed{110^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52814896.png"
  },
  {
    "solution": "\\boxed{y=-x^{2}-3x+4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: The graph of the quadratic function $y=ax^2+bx+4$ (where $a\\neq 0$) passes through points $A(-4,0)$ and $B(1,0)$, and intersects the y-axis at point $C$. Point $P$ is a point on the parabola in the second quadrant. Lines $BP$ and $AC$ intersect at point $Q$. A line is drawn from point $P$ perpendicular to the x-axis, meeting it at point $D$. Find the expression of the quadratic function.\n",
    "completion": "\\textbf{Solution:} From the given conditions, we have:\n\\[\n\\left\\{\n\\begin{array}{l}\na\\cdot (-4)^{2}+b\\cdot (-4)+4=0\\\\\na+b+4=0\n\\end{array}\n\\right.\n\\]\nSolving the system, we get:\n\\[\n\\left\\{\n\\begin{array}{l}\na=-1\\\\\nb=-3\n\\end{array}\n\\right.\n\\]\nTherefore, the expression for the quadratic function is \\boxed{y=-x^{2}-3x+4}.",
    "image_path": "dataset/mm_math/images/MM_Math/53059889_151.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, the diagonals $AC$ and $BD$ intersect at point $O$, and $AE$ bisects $\\angle BAD$ and intersects $BC$ at point $E$. Connect $OE$. If $\\angle EAO = 15^\\circ$, what is the degree measure of $\\angle COE$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rectangle, \\\\\n$\\therefore$ AD $\\parallel$ BC, AC = BD, OA = OC, OB = OD, $\\angle$BAD = $90^\\circ$, \\\\\n$\\therefore$ OA = OB, $\\angle$DAE = $\\angle$AEB, \\\\\nSince AE bisects $\\angle$BAD, \\\\\n$\\therefore$ $\\angle$BAE = $\\angle$DAE = $45^\\circ$ = $\\angle$AEB, \\\\\n$\\therefore$ AB = BE, \\\\\nSince $\\angle$EAO = $15^\\circ$, \\\\\n$\\therefore$ $\\angle$BAC = $45^\\circ + 15^\\circ$ = $60^\\circ$, \\\\\n$\\therefore$ $\\triangle$BAO is an equilateral triangle, \\\\\n$\\therefore$ AB = OB, $\\angle$ABO = $\\angle$AOB = $60^\\circ$, \\\\\n$\\therefore$ $\\angle$OBC = $90^\\circ - 60^\\circ$ = $30^\\circ$, $\\angle$BOC = $180^\\circ - 60^\\circ$ = $120^\\circ$, \\\\\nSince AB = OB = BE, \\\\\n$\\therefore$ $\\angle$BOE = $\\angle$BEO = $\\frac{1}{2}$($180^\\circ - 30^\\circ$) = $75^\\circ$, \\\\\n$\\therefore$ $\\angle$COE = $120^\\circ - 75^\\circ$ = $\\boxed{45^\\circ}$, \\\\",
    "image_path": "dataset/mm_math/images/MM_Math/53359603.png"
  },
  {
    "solution": "\\boxed{B\\left(3, \\sqrt{3}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system, the vertices of rectangle $OABC$ are such that vertex $A$ is on the positive $y$-axis, and point $C$ is on the positive $x$-axis, with $AO=\\sqrt{3}$, and $\\angle BAC=30^\\circ$. Point $D$ is a point on segment $OC$, and triangle $AOD$ is folded along line $AD$ such that point $O$ coincides with point $E$ on the diagonal $AC$ of the rectangle. What are the coordinates of point $B$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral $OABC$ is a rectangle, it follows that $BC=AO=\\sqrt{3}$. In $\\triangle ACB$, with $\\angle BAC=30^\\circ$, we have $AC=2BC=2\\sqrt{3}$. Thus, $AB=\\sqrt{AC^{2}-BC^{2}}=3$. Therefore, the coordinates of point $B$ are $\\boxed{B\\left(3, \\sqrt{3}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50519550_187.png"
  },
  {
    "solution": "\\boxed{\\sqrt{19}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $ABCD$ is a parallelogram. Extend $AB$ to $E$ such that $BE=AB$, and connect $BD$, $ED$, $EC$. If $ED=AD$, $CD=BE$; quadrilateral $BECD$ is a rectangle; connect $AC$, if $AD=\\sqrt{7}$, $CD=2$, find the length of $AC$.\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, connect $AC$,\n\nsince quadrilateral $ABCD$ is a parallelogram,\n\ntherefore $CB=AD=\\sqrt{7}$, $AB=CD=2$,\n\nsince quadrilateral $BECD$ is a rectangle,\n\ntherefore $BE=CD=2$, $\\angle BEC=90^\\circ$,\n\ntherefore, in $\\triangle BEC$,\n\n$CE=\\sqrt{BC^2-BE^2}=\\sqrt{(\\sqrt{7})^2-2^2}=\\sqrt{3}$,\n\nin $\\triangle AEC$,\n\n$AC=\\sqrt{AE^2+CE^2}=\\sqrt{(AB+BE)^2+CE^2}=\\sqrt{4^2+(\\sqrt{3})^2}=\\boxed{\\sqrt{19}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52252006_117.png"
  },
  {
    "solution": "\\boxed{G\\left(-5, -\\frac{5}{2}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, in the Cartesian coordinate system $xOy$, the line $l_1$ intersects the $x$-axis at point $A(m, 0)$ and the $y$-axis at point $B(0, mn)$ ($m<0$, $n>0$). As shown in Figure 2, under the condition of (1), the line $l_2: y=\\frac{1}{2}x$ intersects the line $l_1$ at point $C$, and let's consider point $D(0, 2)$. Does there exist a point $G$ on the line $l_2$ such that $\\frac{S_{\\triangle ACD}}{S_{\\triangle CDG}}=\\frac{4}{3}$? If it exists, find the coordinates of point $G$.\n",
    "completion": "\\textbf{Solution:} Solve the system $\\begin{cases}y=-\\frac{1}{2}x-2\\\\ y=\\frac{1}{2}x\\end{cases}$. We find $\\begin{cases}x=-2\\\\ y=-1\\end{cases}$, thus $C(-2,-1)$.\n\nLet the equation of line CD be: $y=mx+n$.\n\nHence $\\begin{cases}n=2\\\\ -2m+n=-1\\end{cases}$. Solving this, we find $\\begin{cases}n=2\\\\ m=\\frac{3}{2}\\end{cases}$, thus $y=\\frac{3}{2}x+2$.\n\nSetting $y=0$, $0=\\frac{3}{2}x+2$ leads to $x=-\\frac{4}{3}$, thus $E\\left(-\\frac{4}{3},0\\right)$.\n\nTherefore, $AE=4-\\frac{4}{3}=\\frac{8}{3}$.\n\nTherefore, the area of $\\triangle ACD$, $S_{\\triangle ACD}=\\frac{1}{2}AE\\cdot DF=\\frac{1}{2}\\times\\frac{8}{3}\\times3=4$.\n\nSince $\\frac{S_{\\triangle ACD}}{S_{\\triangle CDG}}=\\frac{4}{3}$,\n\nThus, $S_{\\triangle CDG}=3$.\n\nLet $G(x, \\frac{1}{2}x)$.\n\nThus, $\\frac{1}{2}OD\\cdot |x+2|=3$,\n\nor $\\frac{1}{2}\\times2\\cdot |x+2|=3$,\n\nThus, $x_{1}=1, x_{2}=-5$,\n\nTherefore, \\boxed{G(1, \\frac{1}{2})} or \\boxed{G\\left(-5, -\\frac{5}{2}\\right)}.",
    "image_path": "dataset/mm_math/images/MM_Math/51983386_180.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the graph of the linear function $y=kx+k+1$ intersects the graph of the linear function $y=-x+4$ at point $A(1, a)$. What are the values of $a$ and $k$?\n",
    "completion": "\\textbf{Solution:} From the given, we have $a=-1+4=3$,\\\\\n$\\therefore A(1,3)$,\\\\\nSubstituting the coordinates of point A into the function, we get $3=2k+1$,\\\\\nSolving for $k$ yields $k=\\boxed{1}$,\\\\\n$\\therefore a=\\boxed{3}$, $k=1$.",
    "image_path": "dataset/mm_math/images/MM_Math/52345530_206.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within $\\triangle ABC$, $AB=AC=10\\text{cm}$, $DE$ is a perpendicular bisector of $AB$ with the foot of the perpendicular being at $E$, and it intersects $AC$ at $D$. If the perimeter of $\\triangle DBC$ is $18\\text{cm}$, what is the length of $BC$ in $\\text{cm}$?\n",
    "completion": "\\textbf{Solution:} Since the perimeter of $\\triangle DBC$ equals BC + BD + CD $= 18\\,\\text{cm}$,\\\\\nand since DE perpendicularly bisects AB,\\\\\nthus AD $=$ BD,\\\\\nhence BC + AD + CD $= 18\\,\\text{cm}$,\\\\\nsince AC $=$ AD + DC $= 10\\,\\text{cm}$,\\\\\ntherefore BC $= 18 - 10 = \\boxed{8}\\,\\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52978929.png"
  },
  {
    "solution": "\\boxed{\\left(1, -3\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the vertices A and B of the square ABCD are on the x-axis, and the parabola $y=-x^2+bx+c$ passes through points A and $C(4, -5)$, and intersects with line DC at another point E. Let N be a point in the plane. Is there a point M on the axis of symmetry of the parabola such that the quadrilateral formed by points M, N, E, A is a rhombus? If such a point exists, please write down the coordinates of point M directly; if it does not exist, please explain why.\n",
    "completion": "\\textbf{Solution:} Exist, the coordinates of point M are $\\boxed{\\left(1, -3\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52132998_103.png"
  },
  {
    "solution": "\\boxed{48}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: After filling a cylindrical oil tank with a diameter of 52 cm with some oil, the cross-section is shown in the figure. If the maximum depth CD of the oil is 16 cm, calculate the width AB of the oil surface.\n",
    "completion": "\\textbf{Solution:} From the problem statement, we have: $OC \\perp AB$ at point D,\\\\\nAccording to the perpendicular diameter theorem, point D is the midpoint of AB, $AB = 2BD$,\\\\\n$\\because$ the diameter is $52\\,\\text{cm}$,\\\\\n$\\therefore$OB = $26\\,\\text{cm}$,\\\\\n$\\therefore$OD = OC - CD = $26 - 16 = 10\\,\\text{cm}$,\\\\\nBy the Pythagorean theorem,\\\\\n$BD = \\sqrt{BO^2 - OD^2} = 24\\,\\text{cm}$,\\\\\n$\\therefore$AB = \\boxed{48}\\,\\text{cm}.",
    "image_path": "dataset/mm_math/images/MM_Math/53080751.png"
  },
  {
    "solution": "\\boxed{16 \\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, points $A_{1}$, $A_{2}$, $A_{3}$, ..., are all on the positive $x$-axis, and points $B_{1}$, $B_{2}$, $B_{3}$, ..., are all on the line $y=\\frac{\\sqrt{3}}{3}x$. The triangles $\\triangle A_{1}B_{1}A_{2}$, $\\triangle A_{2}B_{2}A_{3}$, $\\triangle A_{3}B_{3}A_{4}$, ..., are all equilateral triangles, and $OA_{1}=1$. What is the $y$-coordinate of point $B_{6}$?\n",
    "completion": "\\textbf{Solution:} Let the side length of $\\triangle B_{n}A_{n}A_{n+1}$ be $a_{n}$,\\\\\nsince points $B_{1}$, $B_{2}$, $B_{3}$, ... are on the line $y = \\frac{\\sqrt{3}}{3}x$ within the first quadrant,\\\\\ndraw $A_{1}N \\perp x$ axis from $A_{1}$ to intersect the line $OB_{1}$ at point $N$,\\\\\nsince $OA_{1} = 1$,\\\\\ntherefore, the $x$-coordinate of point $N$ is $1$, substituting $x=1$ into $y= \\frac{\\sqrt{3}}{3}x$, we get $y= \\frac{\\sqrt{3}}{3}$,\\\\\ntherefore, the coordinates of point $N$ are $(1, \\frac{\\sqrt{3}}{3})$,\\\\\ntherefore $A_{1}N= \\frac{\\sqrt{3}}{3}$,\\\\\nin $\\triangle NOA_{1}$, $\\tan\\angle A_{1}ON= \\frac{A_{1}N}{A_{1}O} = \\frac{\\sqrt{3}}{3}$,\\\\\ntherefore $\\angle A_{1}OB_{1} = 30^\\circ$,\\\\\nalso, since $\\triangle B_{n}A_{n}A_{n+1}$ is an equilateral triangle,\\\\\ntherefore $\\angle B_{n}A_{n}A_{n+1} = 60^\\circ$,\\\\\ntherefore $\\angle OB_{n}A_{n} = 30^\\circ$, $A_{n}B_{n} = OA_{n}$,\\\\\nsince $OA_{1}=1$, $a_{1} =1$, $a_{2}=1+1=2= 2a_{1}$, $a_{3}= 1+a_{1} +a_{2}=4= 2a_{2}$, $a_{4}= 1+a_{1}+a_{2}+a_{3}=8= 2a_{3}$, $a_{n+1}= 2a_{n}$, $a_{5}=2a_{4}= 16$, $a_{6}= 2a_{5} = 32$, $a_{7}= 2a_{6}= 64$,\\\\\n$\\triangle A_{6}B_{6}A_{7}$ is an equilateral triangle,\\\\\nthe coordinates of point $B_{6}$ are $\\left(a_{7}- \\frac{1}{2} a_{6}, \\frac{\\sqrt{3}}{3} (a_{7}- \\frac{1}{2} a_{6})\\right)$,\\\\\ntherefore, the coordinates of point $B_{6}$ are $(48,16 \\sqrt{3})$,\\\\\nthus, the $y$-coordinate of point $B_{6}$ is $\\boxed{16 \\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50560242.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\triangle ABC$, we have $AB=AC$ and $AD$ is the median to side $BC$. Point $E$ is on side $AB$, and $DE \\parallel AC$. If $DE=3$, what is the value of $AB$?\n",
    "completion": "\\textbf{Solution:} Given that $AB=AC$, and $AD$ is the median to side $BC$,\\\\\nhence $\\angle EAD=\\angle CAD$, $\\angle B=\\angle C$,\\\\\nGiven $DE\\parallel AC$,\\\\\nhence $\\angle EDA=\\angle CAD$, $\\angle EDB=\\angle C$,\\\\\nthus $\\angle EAD=\\angle EDA$, $\\angle EDB=\\angle B$,\\\\\nhence $DE=AE$, $DE=BE$,\\\\\nthus $AB=2DE=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52828915.png"
  },
  {
    "solution": "\\boxed{P\\left(-\\frac{4}{3}, \\frac{1}{3}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the square OABC, point O is the origin, point $C(-3, 0)$, point A is on the positive y-axis, points E and F are respectively on BC and CO, with $CE=CF=2$. The graph of the linear function $y=kx+b\\ (k\\ne 0)$ passes through points E and F, intersecting the y-axis at point G. The graph of the inverse proportion function $y=\\frac{m}{x}\\ (m\\ne 0)$ through point E intersects AB at point D. Is there a point P on segment EF such that $S_{\\triangle ADP}=S_{\\triangle APG}$? If so, find the coordinates of point P; if not, please explain why.\n",
    "completion": "\\textbf{Solution:} Suppose there exists a point $P$ such that $S_{\\triangle ADP}=S_{\\triangle APG}$. In the equation $y=-x-1$, let $x=0$ to obtain $y=-1$,\\\\\ntherefore $G(0, -1)$,\\\\\nhence $AG=4$,\\\\\nin the equation $y=-\\frac{6}{x}$, let $y=3$ to obtain $x=-2$,\\\\\ntherefore $D(-2, 3)$,\\\\\nthus $AD=2$,\\\\\nlet $P(t, -t-1)$,\\\\\nsince $S_{\\triangle ADP}=S_{\\triangle APG}$,\\\\\nit follows that $\\frac{1}{2}\\times 2\\cdot [3-(-t-1)]=\\frac{1}{2}\\times 4\\times (-t)$,\\\\\nsolving this yields $t=-\\frac{4}{3}$,\\\\\ntherefore \\boxed{P\\left(-\\frac{4}{3}, \\frac{1}{3}\\right)}.",
    "image_path": "dataset/mm_math/images/MM_Math/52949741_253.png"
  },
  {
    "solution": "\\boxed{\\frac{15}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, a right-angled triangle paper has its two right-angled sides measuring $6$ and $8$, respectively. Now, it is folded such that point $A$ coincides with point $B$, with the crease being $DE$. What is the value of $DE$?\n",
    "completion": "Solution: According to the properties of the fold transformation, we know that $\\triangle DBE \\cong \\triangle DAE$, $\\therefore$AD=BD,\\\\\n$\\because$ the lengths of the two right-angled sides of the rectangular paper are 6 and 8, respectively, $\\therefore$AB=10, $\\tan A = \\frac{3}{4}$,\\\\\nIn $\\triangle ADE$, $DE = \\tan A \\times AD = \\boxed{\\frac{15}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50583927.png"
  },
  {
    "solution": "\\boxed{\\frac{2\\sqrt{5}}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the Cartesian coordinate system, the parabola $y=x^{2}+(k-1)x-k$ intersects with the line $y=kx+1$ at points $A$ and $B$, where point $A$ is to the left of point $B$. As shown in the figure, the parabola $y=x^{2}+(k-1)x-k$ ($k>0$) intersects with the x-axis at points $C$ and $D$, with point $C$ being to the left of point $D$. There exists a unique point $Q$ on the line $y=kx+1$ such that $\\angle OQC=90^\\circ$. What is the value of $k$ in this case?\n",
    "completion": "\\textbf{Solution:} Method 1:\\\\\nLet line AB: $y=kx+1$ intersect the x-axis and y-axis at point E and F, respectively,\\\\\nthen E$\\left(-\\frac{1}{k},0\\right)$, F$\\left(0,1\\right)$, OE$=\\frac{1}{k}$, OF$=1$.\\\\\nIn $\\triangle EOF$, by the Pythagorean theorem, we have: $EF=\\sqrt{\\left(\\frac{1}{k}\\right)^2+1}=\\frac{\\sqrt{1+k^2}}{k}$\\\\\nLet $y=x^2+(k-1)x-k=0$, that is, $(x+k)(x-1)=0$, solving this yields: $x=-k$ or $x=1$.\\\\\n$\\therefore$C$\\left(-k,0\\right)$, OC$=k$.\\\\\nI. Assume there exists a unique point Q such that $\\angle OQC=90^\\circ$,\\\\\nthen the circle with OC as its diameter touches line AB at point Q, according to the inscribed angle theorem, this implies $\\angle OQC=90^\\circ$.\\\\\nLet point N be the midpoint of OC, and connect NQ, then NQ$\\perp$EF, NQ$=CN=ON=\\frac{k}{2}$.\\\\\n$\\therefore$EN$=OE-ON=\\frac{1}{k}-\\frac{k}{2}$.\\\\\n$\\because \\angle NEQ=\\angle FEO$, $\\angle EQN=\\angle EOF=90^\\circ$,\\\\\n$\\therefore \\triangle EQN\\sim \\triangle EOF$,\\\\\n$\\therefore \\frac{NQ}{OF}=\\frac{EN}{EF}$, that is: $\\frac{\\frac{k}{2}}{1}=\\frac{\\frac{1}{k}-\\frac{k}{2}}{\\frac{\\sqrt{1+k^2}}{k}}$\\\\\nSolving this yields: $k=\\pm\\frac{2\\sqrt{5}}{5}$,\\\\\n$\\because k>0$,\\\\\n$\\therefore k=\\boxed{\\frac{2\\sqrt{5}}{5}}$.\\\\\nII. If line AB passes through point C, at that point the line and circle have only another point Q as their intersection, hence there exists a unique point Q such that $\\angle OQC=90^\\circ$,\\\\\nSubstituting C$\\left(-k,0\\right)$ into $y=kx+1$,\\\\\nwe get $k=1$, $k=-1$ (discard),\\\\\nthus, there exists a unique point Q such that $\\angle OQC=90^\\circ$, at this time $k=1$.\\\\\nTo sum up, when $k=\\frac{2\\sqrt{5}}{5}$ or $1$, there exists a unique point Q such that $\\angle OQC=90^\\circ$.",
    "image_path": "dataset/mm_math/images/MM_Math/53009969_94.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In quadrilateral \\(ABCD\\), triangles \\(OAB\\) and \\(OCD\\) share the common vertex \\(O\\), and \\( \\triangle OAB \\cong \\triangle OCD\\). As shown in Figure 2, points \\(B\\), \\(O\\), and \\(C\\) are not on the same line, and \\(\\angle AOB=90^\\circ\\), given that \\(AD^2+BC^2=50\\), \\(AO=3\\), find the area of \\(\\triangle OAB\\).\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect $AC, BD$ and they intersect at point Q, \\\\\n$\\because$ $\\triangle OAB \\cong \\triangle OCD$, \\\\\n$\\therefore$ $\\angle COD = \\angle AOB = 90^\\circ$, $OA = OC, OB = OD$, $AB = CD$, \\\\\n$\\therefore$ $\\angle AOC = \\angle BOD$, \\\\\n$\\therefore$ $\\angle OAC = \\angle OCA = \\angle OBD = \\angle ODB$. \\\\\n$\\because$ $\\angle OBC + \\angle OCB = 180^\\circ - \\angle BOC$, $\\angle AOD = 180^\\circ - \\angle BOC$, \\\\\n$\\therefore$ $\\angle OBC + \\angle OCB = \\angle AOD$, \\\\\n$\\therefore$ $\\angle DQC = \\angle QBO + \\angle OBC + \\angle QCO + \\angle OCB$ \\\\\n$= \\angle OAC + \\angle OCA + \\angle AOD$ \\\\\n$= 180^\\circ - 90^\\circ = 90^\\circ$. \\\\\n$\\therefore$ $AD^2 = AQ^2 + DQ^2$, $BC^2 = BQ^2 + CQ^2$, \\\\\n$\\therefore$ $AD^2 + BC^2 = AQ^2 + DQ^2 + BQ^2 + CQ^2$ \\\\\n$= AB^2 + CD^2$, \\\\\n$\\because$ $AB = CD$, \\\\\n$\\therefore$ $AD^2 + BC^2 = 2AB^2 = 50$, \\\\\n$\\therefore$ $AB = 5$, \\\\\n$\\because$ $AO = 3$, \\\\\n$\\therefore$ $BO = 4$, \\\\\n$\\therefore$ the area of $\\triangle OAB = \\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53033486_103.png"
  },
  {
    "solution": "\\boxed{\\frac{10}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, $\\angle B=90^\\circ$, $AB=8$, $BC=6$. Point $M$ is the midpoint of diagonal $AC$, and point $N$ is the midpoint of side $AD$. Connecting $BM$, $MN$, if $BM = 3MN$, then what is the length of the line segment $CD$?\n",
    "completion": "\\textbf{Solution:} Given $\\angle B=90^\\circ$ and lengths AB=8, BC=6, \\\\\n$\\therefore$ the length of AC is given by $\\sqrt{8^2+6^2}=10$, \\\\\nSince M is the midpoint of AC, \\\\\n$\\therefore$ BM= $\\frac{1}{2}$AC=5, \\\\\nGiven that BM is three times MN, \\\\\n$\\therefore$ MN=$\\frac{5}{3}$, \\\\\nAlso, since N is the midpoint of AD, \\\\\n$\\therefore$ MN is the midsegment of $\\triangle ACD$, \\\\\n$\\therefore$ CD is twice MN, which is $2 \\times \\frac{5}{3} = \\boxed{\\frac{10}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51942329.png"
  },
  {
    "solution": "\\boxed{B}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, this is the net of a solid. What is the solid?\n",
    "completion": "\\textbf{Solution:} From the unfolded figure, it can be seen that the solid is composed of three rectangular faces and two triangular faces,\\\\\ntherefore, the solid is a triangular prism.\\\\\nHence, the answer is: \\boxed{B}.",
    "image_path": "dataset/mm_math/images/MM_Math/51379704.png"
  },
  {
    "solution": "\\boxed{78^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle BAC = \\angle ABC = 42^\\circ$. Through point C, draw $CD \\perp AB$ at point D. Let point E be a point on $CD$. If $\\triangle ACE$ is folded along $AE$ to obtain $\\triangle AFE$ and points E, F, B are exactly on the same line, connect $CF$. What is the measure of $\\angle CFA$?\n",
    "completion": "\\textbf{Solution:} \\\\\nSince $CD \\perp AB$\\\\\n$\\therefore \\angle ADC = 90^\\circ$ \\\\\nSince $\\angle BAC = \\angle ABC = 42^\\circ$\\\\\n$\\therefore \\angle ACD = 90^\\circ - \\angle BAC = 48^\\circ$, $AC = BC$\\\\\nSince $CD \\perp AB$\\\\\n$\\therefore \\angle BCD = \\angle ACD = 48^\\circ$ \\\\\nIn $\\triangle ACE$ and $\\triangle BCE$\\\\\n$\\left\\{\\begin{array}{l}\nAC = BC\\\\\n\\angle ACD = \\angle BCD\\\\\nCE = CE\n\\end{array}\\right.$ \\\\\n$\\therefore \\triangle ACE \\cong \\triangle BCE$\\\\\n$\\therefore \\angle CAE = \\angle CBE$ \\\\\nSince folding $\\triangle ACE$ along $AE$ yields $\\triangle AFE$,\\\\\n$\\therefore \\angle AFE = \\angle ACD = 48^\\circ$, $\\angle CAE = \\angle FAE$, $AC=AF$\\\\\n$\\therefore \\angle CAE = \\angle FAE = \\angle CBE$\\\\\nLet $\\angle CAE = \\angle FAE = \\angle CBE = x$\\\\\n$\\therefore \\angle BAF = \\angle BAC - \\angle CAE - \\angle FAE = 42^\\circ - 2x$, $\\angle ABF = \\angle ABC - \\angle CBE = 42^\\circ - x$ \\\\\nSince $\\angle AFE = \\angle BAF + \\angle ABF = 42^\\circ - 2x + (42^\\circ - x) = 84^\\circ - 3x$ \\\\\n$\\therefore 48^\\circ = 84^\\circ - 3x$\\\\\n$\\therefore x = 12^\\circ$ \\\\\n$\\therefore \\angle CAF = \\angle CAE + \\angle FAE = 24^\\circ$ \\\\\nSince $AC=AF$\\\\\n$\\therefore \\angle CFA = \\angle FCA = \\frac{180^\\circ - \\angle CAF}{2} = \\boxed{78^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/52836686.png"
  },
  {
    "solution": "\\boxed{36^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the regular pentagon ABCDE is inscribed in circle $O$. What is the measure of $\\angle CAD$ in degrees?\n",
    "completion": "\\textbf{Solution}: Since pentagon ABCDE is a regular pentagon,\\\\\ntherefore, \\(\\angle BAE=108^\\circ\\), and \\(BC=CD=DE\\),\\\\\ntherefore, \\(\\angle BAC=\\angle CAD=\\angle DAE=\\frac{1}{3}\\angle BAE=36^\\circ\\),\\\\\nthus, the answer is: \\(\\boxed{36^\\circ}\\).",
    "image_path": "dataset/mm_math/images/MM_Math/52874521.png"
  },
  {
    "solution": "\\boxed{\\frac{7\\sqrt{2}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated in the figure, $\\triangle ABC$ is inscribed in $\\odot O$, with $AB$ being the diameter of $\\odot O$, and $AB=5$, $AC=3$. If $D$ is the midpoint of $\\overset{\\frown}{AB}$, and $CD$ and $BD$ are connected, what is the length of chord $CD$?\n",
    "completion": "\\textbf{Solution:} Draw $BE\\perp CD$ at E passing through point B, \\\\\nsince D is the midpoint of $\\overset{\\frown}{AB}$, \\\\\nit follows that $\\overset{\\frown}{AD}=\\overset{\\frown}{BD}$, \\\\\nthus, $\\angle ACD=\\angle BCD=45^\\circ$, \\\\\ngiven that BC=4, \\\\\nin $\\triangle BCE$, we have $BE=CE=2\\sqrt{2}$, \\\\\nsince $\\angle A=\\angle D$, \\\\\nit follows that $\\tan D=\\tan A=\\frac{4}{3}$, \\\\\nin $\\triangle BDE$, \\\\\n$DE=\\frac{BE}{\\tan D}=\\frac{2\\sqrt{2}}{\\frac{4}{3}}=\\frac{3\\sqrt{2}}{2}$, \\\\\ntherefore, $CD=CE+DE=\\boxed{\\frac{7\\sqrt{2}}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51379192_90.png"
  },
  {
    "solution": "\\boxed{7}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $C$ is a point on line segment $AD$, $B$ is the midpoint of $CD$, and $AD=13\\, \\text{cm}$, $BC=3\\, \\text{cm}$. What is the length of $AC$ in centimeters?\n",
    "completion": "\\textbf{Solution:} Since $B$ is the midpoint of $CD$, and $BC=3\\text{cm}$, \\\\\nthen $CD=2BC=6\\text{cm}$. \\\\\nSince $AD=13\\text{cm}$, \\\\\nthen $AC=AD-CD=13-6=\\boxed{7}\\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/54962520_33.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, DE is the median of $\\triangle ABC$, the angle bisector of $\\angle ABC$ intersects DE at point F, $AB=8$, $BC=12$, then what is the length of $EF$?\n",
    "completion": "\\textbf{Solution:} Given that $DE$ is the median line of $\\triangle ABC$,\\\\\nit follows that $DE \\parallel BC$ and $DE = \\frac{1}{2}BC = 6$, $BD = AD = \\frac{1}{2}AB = 4$,\\\\\nthus $\\angle DFB = \\angle FBC$,\\\\\nsince $BF$ bisects $\\angle ABC$,\\\\\nit follows that $\\angle DBF = \\angle FBC$,\\\\\nthus $\\angle DFB = \\angle DBF$,\\\\\ntherefore $DF = BD = 4$,\\\\\nhence $EF = DE - DF = 6 - 4 = \\boxed{2}$",
    "image_path": "dataset/mm_math/images/MM_Math/51534395.png"
  },
  {
    "solution": "\\boxed{Q\\left(1+\\sqrt{3}, 2\\sqrt{3}-2\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the straight line $y=ax+1$ intersects the $x$-axis and the $y$-axis at points $A$ and $B$ respectively, and intersects the hyperbola $y=\\frac{k}{x}$ (where $x>0$) at point $P$, $PC\\perp x$-axis at point $C$, and $PC=2$, the coordinates of point $A$ are $(-2, 0)$. If point $Q$ is a point on the right side of point $P$ on the hyperbola, and $QH\\perp x$-axis at $H$, when the triangle formed by points $Q$, $C$, and $H$ is similar to $\\triangle AOB$, what are the coordinates of point $Q$?\n",
    "completion": "\\textbf{Solution:} Given the diagram, draw $QH\\perp$ x-axis and let $Q(m,n)$, \\\\\nsince $Q(m,n)$ lies on $y= \\frac{4}{x}$, \\\\\nthus, $n= \\frac{4}{m}$.\\\\\nWhen $\\triangle QCH\\sim \\triangle BAO$, it follows that $\\frac{CH}{AO}=\\frac{QH}{BO}$,\\\\\nwhich means $\\frac{m-2}{2}=\\frac{n}{1}$,\\\\\ntherefore $m-2=2n$,\\\\\nwhich implies $m-2=\\frac{8}{m}$,\\\\\nleading to $m^{2}-2m-8=0$,\\\\\nsolving this, we find $m=4$ or $m=-2$ (discard),\\\\\nhence, $Q\\left(4, 1\\right)$,\\\\\nWhen $\\triangle QCH\\sim \\triangle ABO$, it follows that $\\frac{CH}{BO}=\\frac{QH}{AO}$,\\\\\nwhich means $\\frac{m-2}{1}=\\frac{n}{2}$,\\\\\nrearranging gives $2m-4=\\frac{8}{m}$,\\\\\nleading to $2m^{2}-4m-8=0$,\\\\\nsolving this, we find $m=1+\\sqrt{3}$ or $m=1-\\sqrt{3}$ (discard),\\\\\nhence, $Q\\left(1+\\sqrt{3}, 2\\sqrt{3}-2\\right)$,\\\\\nIn conclusion, $Q\\left(4, 1\\right)$ or $Q\\left(1+\\sqrt{3}, 2\\sqrt{3}-2\\right)$.\\\\\nTherefore, the coordinates of point $Q$ are $\\boxed{Q\\left(4, 1\\right)}$ or $\\boxed{Q\\left(1+\\sqrt{3}, 2\\sqrt{3}-2\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52189578_138.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $ OB$ is a ray inside $ \\angle AOC$, $ OE$ bisects $ \\angle AOC$, and $ OF$ bisects $ \\angle BOC$. If $ AO\\perp BO$ and $ \\angle BOC=60^\\circ$, find the measure of $ \\angle EOF$.\n",
    "completion": "\\textbf{Solution:} Given $AO\\perp BO$, $\\angle BOC=60^\\circ$\\\\\n$\\therefore \\angle AOB=90^\\circ, \\angle AOC=\\angle AOB+\\angle BOC=150^\\circ$\\\\\nSince OE bisects $\\angle AOC$, and OF bisects $\\angle BOC$\\\\\n$\\therefore \\angle EOC=\\frac{1}{2}\\angle AOC=75^\\circ, \\angle COF=\\frac{1}{2}\\angle BOC=30^\\circ$\\\\\n$\\therefore \\angle EOF=\\angle EOC-\\angle COF=75^\\circ-30^\\circ=\\boxed{45^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/52792815_139.png"
  },
  {
    "solution": "\\boxed{\\frac{75}{8}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, the perpendicular bisector of AB intersects AB and AC at points D and E, respectively. If $AC=8$ and $BD=5$, what is the area of $\\triangle ADE$?\n",
    "completion": "\\textbf{Solution:} Connect BE,\\\\\nsince DE is the perpendicular bisector of AB,\\\\\nit follows that EA=EB, AD=DB=5,\\\\\nsince $\\angle C=90^\\circ$, AC=8, BD=5,\\\\\nit follows that AB=2BD=10,\\\\\nby the Pythagorean theorem, we have BC=$\\sqrt{AB^{2}-AC^{2}}=\\sqrt{10^{2}-8^{2}}=6$,\\\\\nthus CE=8−AE=8−EB,\\\\\nin $\\triangle CBE$, we have $BE^{2}=CE^{2}+BC^{2}$, that is $BE^{2}=(8-BE)^{2}+36$,\\\\\nsolving this, we get $BE=\\frac{25}{4}$, therefore, AE=$\\frac{25}{4}$,\\\\\nthus the area of $\\triangle ABE$ is $\\frac{1}{2}AE\\times BC=\\frac{1}{2}\\times\\frac{25}{4}\\times6=\\frac{75}{4}$,\\\\\nthus the area of $\\triangle ADE$ is $\\frac{1}{2}$ area of $\\triangle ABE$=$\\boxed{\\frac{75}{8}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51366593.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $OP$ bisects $\\angle AOB$, $\\angle AOP=15^\\circ$, $PC\\parallel OA$, $PC=4$, $PD\\perp OA$, with the foot of the perpendicular being $D$. What is the length of $PD$?\n",
    "completion": "\\textbf{Solution:} Let PE$\\perp$OB at E,\\\\\nsince $\\angle$BOP=$\\angle$AOP, PD$\\perp$OA, PE$\\perp$OB,\\\\\ntherefore PE=PD (The distances from a point on the angle bisector to the two sides of the angle are equal),\\\\\nsince $\\angle$BOP=$\\angle$AOP=15$^\\circ$,\\\\\ntherefore $\\angle$AOB=30$^\\circ$,\\\\\nsince PC$\\parallel$OA,\\\\\ntherefore $\\angle$BCP=$\\angle$AOB=30$^\\circ$,\\\\\ntherefore, in Rt$\\triangle$PCE, PE=$\\frac{1}{2}$PC=$\\frac{1}{2}\\times 4=2 $(In a right triangle, the leg opposite the 30$^\\circ$ angle is half the length of the hypotenuse),\\\\\ntherefore, PD=PE=\\boxed{2},",
    "image_path": "dataset/mm_math/images/MM_Math/52096487.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, divide the circle $ \\odot O$ into six equal arcs and connect the division points sequentially to obtain the regular hexagon ABCDEF. If the circumference of $ \\odot O$ is $ 12\\pi $, what is the side length of this regular hexagon?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect OA, OB, OC, OD, OE, OF. \\\\\n$\\because$ the hexagon ABCDEF is regular, \\\\\n$\\therefore$ AB = BC = CD = DE = EF = FA, $\\angle$AOB = $\\angle$BOC = $\\angle$COD = $\\angle$DOE = $\\angle$EOF = $\\angle$FOA = $60^\\circ$, \\\\\n$\\therefore$ $\\triangle$AOB, $\\triangle$BOC, $\\triangle$DOC, $\\triangle$EOD, $\\triangle$EOF, $\\triangle$AOF are all equilateral triangles, \\\\\n$\\because$ the circumference of $\\odot O$ is $12\\pi$, \\\\\n$\\therefore$ the radius of $\\odot O$ is $\\frac{12\\pi}{2\\pi}=6$, \\\\\nthus, the side length of the regular hexagon is $\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51402227.png"
  },
  {
    "solution": "\\boxed{\\frac{48}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C = 90^\\circ$, $AC = 6$, $BC = 8$, $D$ is the midpoint of side $BC$, $E$ is a moving point on side $AB$. Construct $\\angle DEF = 90^\\circ$, and let $EF$ intersect the extension of line $BC$ at point $F$. Let $BE = x$, and the area of $\\triangle BED$ be $y$. If the triangle formed by vertices $B$, $E$, $F$ is similar to $\\triangle BED$, find the area of $\\triangle BED$.\n",
    "completion": "\\textbf{Solution}: From the problem, we know that $ \\angle BEF\\ne 90^\\circ$, so we can consider two cases.\\\\\n(1) As shown in Figure 2, when $ \\angle BEF$ is acute,\\\\\nfrom the known fact that the triangle with vertices $B, E, F$ is similar to $\\triangle BED$ and knowing that $\\angle EBF=\\angle DBE$ and $\\angle BEF<\\angle BED$, it follows that $\\angle BEF=\\angle BDE$.\\\\\nDraw $DM\\perp BA$ at $M$ through point $D$, and draw $EH\\perp BC$ at $H$ through $E$.\\\\\n$\\therefore$ $ \\angle MDE=\\angle HDE$,\\\\\n$\\therefore$ $ EM=EH$.\\\\\nSince $ EM=MB-EB=\\frac{16}{5}-x$\\\\\nand $\\because$ $ EH=\\frac{3}{5}x$\\\\\n$\\therefore$ $ \\frac{16}{5}-x=\\frac{3}{5}x$,\\\\\nwe find $ x=2$\\\\\n$\\therefore$ $ y=\\frac{6}{5}\\times 2=\\frac{12}{5}$\\\\\n(2) As shown in Figure 3, when $\\angle BEF$ is obtuse,\\\\\nsimilarly, we find $ x-\\frac{16}{5}=\\frac{3}{5}x$,\\\\\n$\\therefore$ $ x=8$.\\\\\n$\\therefore$ $ y=\\frac{6}{5}\\times 8=\\frac{48}{5}$\\\\\nIn summary, the area of $\\triangle BED$ is $\\boxed{\\frac{12}{5}}$ or $\\boxed{\\frac{48}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53078900_116.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, extend the line segment $AB$ to $C$ so that $BC=4AB$. Point $D$ is the midpoint of the line segment $BC$. If $CD=4\\,cm$, find the length of $AC$.\n",
    "completion": "\\textbf{Solution:} Let $AB=x$,\\\\\n$\\therefore BC=4AB=4x$, $AC=AB+BC=5x$\\\\\n$\\because$ point $D$ is the midpoint of line segment $BC$,\\\\\n$\\therefore CD=BD=\\frac{1}{2}BC=2x$,\\\\\nthen $2x=4$, solving this yields: $x=2$,\\\\\n$CD=BD=2x=4\\text{cm}$,\\\\\n$\\therefore AC=5x=\\boxed{10}\\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53106951_62.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the quadrilateral $ABCD$ is inscribed in $\\odot O$. If the quadrilateral $AOCD$ is a rhombus, what is the degree measure of $\\angle B$? .\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is inscribed in circle O,\\\\\nit follows that $\\angle B+\\angle D=180^\\circ$,\\\\\nSince quadrilateral OACD is a rhombus,\\\\\nit follows that $\\angle AOC=\\angle D$,\\\\\nBy the inscribed angle theorem, we have $\\angle B= \\frac{1}{2}\\angle AOC$,\\\\\nThus $\\angle B+2\\angle B=180^\\circ$,\\\\\nSolving, we find $\\angle B=\\boxed{60^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52622680.png"
  },
  {
    "solution": "\\boxed{6.5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $C$, $D$, $E$ are points on the line segment $AB$, with $AC=5$, $DB=3$. Points $C$ and $E$ are the midpoints of line segments $AD$ and $BD$, respectively. Find the length of $CE$.\n",
    "completion": "\\textbf{Solution:} Since points C and E are the midpoints of segments $AD$ and $BD$, respectively, \\\\\ntherefore $CD=AC=5$, $DE=\\frac{1}{2}BD=\\frac{1}{2}\\times 3=1.5$, \\\\\ntherefore $CE=CD+DE=5+1.5=\\boxed{6.5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53119205.png"
  },
  {
    "solution": "\\boxed{\\sqrt{10}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure (1), the quadrilateral ABCD and the quadrilateral AEFG are both squares. Rotate the square AEFG around point A and connect BE and CF. When points G, F, and C are collinear, as shown in Figure (2), if $AB=5$ and $AE=\\sqrt{5}$, then what is the length of $BE$?\n",
    "completion": "\\textbf{Solution:} We have $BE=\\boxed{\\sqrt{10}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51445446_40.png"
  },
  {
    "solution": "\\boxed{\\sqrt{5}+1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in square $ABCD$ with $AB=4$, $BE=CE$, and $F$ being a moving point on side $AB$, connect $EF$. Fold $\\triangle BEF$ over to $\\triangle GEF$ such that point $B$ falls on point $G$, and connect $DG$. When the perimeter of quadrilateral $AFGD$ reaches its minimum value, what is the length of $BF$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect ED, FG, DF,\\\\\nsince square ABCD, with AB=4,\\\\\ntherefore $\\angle ABC = \\angle DCB = 90^\\circ$, and DC=AD=BC=4,\\\\\nsince the triangle BEF is folded onto triangle GEF, we have BE=EC,\\\\\ntherefore BE=EG=EC=2,\\\\\ntherefore point G moves on a circle with E as its center and a radius of 2,\\\\\ntherefore, when points E, G, and D are collinear, the length of GD is the shortest, thus the length of FG=BF is the maximum,\\\\\ntherefore, the length of AF is the shortest,\\\\\ntherefore, the perimeter of quadrilateral AFGD is the minimum,\\\\\nin $\\triangle DCE$, ED=$\\sqrt{2^2+4^2}=2\\sqrt{5}$,\\\\\ntherefore GD=$2\\sqrt{5}-2$,\\\\\nlet FG=BF=x, AF=4-x,\\\\\nin triangles $\\triangle FAD$ and $\\triangle FGD$, we have DF$^2$=AF$^2$+AD$^2$=FG$^2$+GD$^2$,\\\\\ntherefore $(4-x)^2+16=x^2+(2\\sqrt{5}-2)^2$,\\\\\nsolving this, we get $x=\\sqrt{5}+1$,\\\\\ntherefore BF=$\\boxed{\\sqrt{5}+1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52050077.png"
  },
  {
    "solution": "\\boxed{7}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the figure, if $\\triangle ACB$ and $\\triangle DCE$ are both isosceles right triangles with $\\angle ACB = \\angle DCE = 90^\\circ$, points A, D, and E lie on the same straight line, and $CM$ is the altitude from $C$ to the side $DE$ in $\\triangle DCE$, with $BE$ connected. $\\triangle ACD \\cong \\triangle BCE$; if $CM = 2$ and $BE = 3$, find the length of $AE$. The core problem of this LaTeX is {to find the length of $AE$.}\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ACD\\cong \\triangle BCE$, it follows that $AD=BE=3$. Since $CD=CE$ and $CM\\perp DE$, it follows that $DM=ME$, and $\\angle MCD=\\angle MCE=45^\\circ$. Since $\\angle CDE=\\angle CED=45^\\circ$, it follows that $DM=ME=CM$. Therefore, $DE=2CM=4$. Thus, $AE=AD+DE=\\boxed{7}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52950975_84.png"
  },
  {
    "solution": "\\boxed{\\left(\\frac{3}{2}, -\\frac{15}{2}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=2x^2-4x-6$ intersects the x-axis at points $A$ and $B$, with point $A$ to the left of point $B$, and intersects the y-axis at point $C$. Point $D$ is a moving point on the parabola. As shown in figure 2, when point $D$ is in the fourth quadrant, connect $BD$, $CD$, and $BC$ to form $\\triangle BCD$. When the area of $\\triangle BCD$ is maximized, what are the coordinates of point $D$?\n",
    "completion": "\\textbf{Solution:} Let the coordinates of point D be $\\left(m, 2m^{2}-4m-6\\right)$.\n\nAs shown in the figure, draw $DH\\perp x$ axis at point H and $DG\\perp y$ axis at point G, then connect $OD$.\n\nTherefore, $DG=m$, $DH=-2m^{2}+4m+6$.\n\nBecause the coordinates of point B are $\\left(3, 0\\right)$ and the coordinates of point C are $\\left(0, -6\\right)$.\n\nTherefore, $OB=3$, $OC=6$.\n\nBecause $S_{\\triangle BCD}=S_{\\triangle OCD}+S_{\\triangle OBD}-S_{\\triangle OBC}$,\n\nTherefore, $S_{\\triangle BCD}=\\frac{OC\\cdot DG}{2}+\\frac{OB\\cdot DH}{2}-\\frac{OC\\cdot OB}{2}$\n\n$S_{\\triangle BCD}=\\frac{6m}{2}+\\frac{3(-2m^{2}+4m+6)}{2}-\\frac{6\\times 3}{2}$\n\nUpon simplification, we get $S_{\\triangle BCD}=-3\\left(m-\\frac{3}{2}\\right)^{2}+\\frac{27}{4}$.\n\nBecause $-3<0$,\n\nTherefore, when $m=\\frac{3}{2}$, the maximum area of $\\triangle BCD$ is $\\frac{27}{4}$.\n\nTherefore, $2m^{2}-4m-6=2\\times \\left(\\frac{3}{2}\\right)^{2}-4\\times \\frac{3}{2}-6=2\\times \\frac{9}{4}-6-6=-\\frac{15}{2}$.\n\nTherefore, the coordinates of point D are $\\boxed{\\left(\\frac{3}{2}, -\\frac{15}{2}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52917427_261.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point A is a moving point on the parabola \\(y=\\frac{1}{8}x^{2}\\) that does not coincide with the origin O. \\(AB\\perp x\\) axis at point B, draw a perpendicular line from point B to \\(OA\\) and extend it to intersect the y-axis at point C. Connect \\(AC\\), then what is the length of segment \\(OC\\)? What is the minimum value of \\(AC\\)?\n",
    "completion": "\\textbf{Solution}: Let point A be $(a, \\frac{1}{8}a^{2})$, then the coordinates of point B are $(a, 0)$,\n\n$\\therefore$ OB = $|a|$, AB = $\\frac{1}{8}a^{2}$,\n\n$\\because$ $\\angle$ABO = $\\angle$BOC = $90^\\circ$,\n\n$\\therefore$ $\\angle$AOB + $\\angle$OBC = $90^\\circ$, $\\angle$OBC + $\\angle$BCO = $90^\\circ$,\n\n$\\therefore$ $\\angle$AOB = $\\angle$BCO,\n\n$\\therefore$ $\\triangle$AOB $\\sim$ $\\triangle$BCO,\n\n$\\therefore$ $\\frac{OB}{CO}=\\frac{AB}{BO}$,\n\n$\\therefore$ OB$^{2}$ = CO$\\cdot$AB, i.e., a$^{2}$ = $\\frac{1}{8}a^{2}\\cdot$CO,\n\nSolving for CO gives CO = 8,\n\n$\\therefore$ C(0, 8),\n\n$\\because$ AC$^{2}$ = (xC - xA)$^{2}$ + (yC - yA)$^{2}$ = a$^{2}$ + $\\frac{1}{64}a^{4}$ - 2a$^{2}$ + 64 = $\\frac{1}{64}(a^{4} - 64a^{2}) + 64 = \\frac{1}{64}(a^{2} - 32)^{2} + 48$,\n\n$\\therefore$ when a$^{2}$ = 32, AC$^{2}$ = 48 is the minimum value, hence AC = $4\\sqrt{3}$.\n\nThus, the answer is: CO = \\boxed{8}, and the minimum value of AC is \\boxed{4\\sqrt{3}}.",
    "image_path": "dataset/mm_math/images/MM_Math/51458317.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{5}}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the vertices of $\\triangle ABC$ are all grid points in a square grid. What is the value of $\\sin\\angle BAC$?\n",
    "completion": "\\textbf{Solution:} Construct $CD\\perp AB$ at D,\\\\\nFrom the figure, it is known that $BC=2$,\\\\\nBy the Pythagorean theorem, $AC=\\sqrt{3^2+1^2}=\\sqrt{10}$, $AB=\\sqrt{3^2+3^2}=3\\sqrt{2}$,\\\\\nBy the formula for the area of a triangle, $\\frac{1}{2}\\times 2\\times 3=\\frac{1}{2}\\times 3\\sqrt{2}\\times CD$,\\\\\nSolving, we find $CD=\\sqrt{2}$,\\\\\n$\\therefore \\sin\\angle BAC=\\frac{CD}{AC}=\\frac{\\sqrt{2}}{\\sqrt{10}}=\\frac{\\sqrt{5}}{5}$,\\\\\nTherefore, the answer is: $\\boxed{\\frac{\\sqrt{5}}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51367511.png"
  },
  {
    "solution": "\\boxed{(0,-9)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: It is known that, as shown in the diagram, the graph of the linear function $y=kx+b$ intersects the $x$-axis and $y$-axis at points $A(3,0)$ and $B(0,-4)$ respectively. Point $M$ is on the $y$-axis, and the area of $\\triangle ABM$ is $7.5$. Write down the coordinates of point $M$ directly.\n",
    "completion": "\\textbf{Solution:} The coordinates of point $M$ are $\\boxed{(0,1)}$ or $\\boxed{(0,-9)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52423792_224.png"
  },
  {
    "solution": "\\boxed{\\left(-1, 1\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line BC intersects the x-axis at point C, the y-axis at point B, and intersects the line $y=ax$ at point A. The x-coordinate of point A is 2, $\\angle ACO = 45^\\circ$, and the area of $\\triangle ABO$ is 1. Suppose point P is a point on the line BC, and point Q is a point in the coordinate plane. There exist points P and Q that satisfy the conditions making the quadrilateral OAPQ a rhombus. Write down the coordinates of point Q.\n",
    "completion": "\\textbf{Solution:} The coordinates of point $Q$ are $\\boxed{\\left(-1, 1\\right)}$ or $\\left(\\frac{\\sqrt{26}}{2}, \\frac{\\sqrt{26}}{2}\\right)$ or $\\left(-\\frac{\\sqrt{26}}{2}, -\\frac{\\sqrt{26}}{2}\\right)$ or $\\left(\\frac{13}{10}, \\frac{13}{10}\\right)$.",
    "image_path": "dataset/mm_math/images/MM_Math/52056412_202.png"
  },
  {
    "solution": "\\boxed{\\frac{n}{OA}<1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, in the Cartesian coordinate system $xOy$, the line ${l}_{1}$ intersects the $x$-axis at point $A(m, 0)$ and the $y$-axis at point $B(0, mn)$ ($m<0$, $n>0$). There is a point $P$ below the line ${l}_{1}$, whose abscissa is $m+n$. Connect $PB$. If $\\angle PBA=2\\angle BAO$, find the range of values for $\\frac{n}{OA}$.\n",
    "completion": "Solution: As shown in Figure 2,\n(1) When $m+n<0$, that is $-\\frac{n}{m}<1$,\\\\\nextend line $AO$ and take $OC=OA$,\\\\\nsince $OB\\perp AC$,\\\\\nit follows that $AB=BC$,\\\\\ntherefore $\\angle BCO=\\angle BAO$,\\\\\nhence $\\angle APB=\\angle BAO+\\angle BCO=2\\angle BAO$,\\\\\nhence point $P$ lies on the extension of line $CB$,\\\\\nthus, there exists a point $P$ below $l_1$, satisfying $\\angle PBA=2\\angle BAO$,\\\\\nas shown in Figure 3,\\\\\n(2) Extend line $AO$ and take $OC=OA$,\\\\\nwhen $m+n>0$, that is $-\\frac{n}{m}>1$,\\\\\nfrom (1) it is known that: $\\angle ABE=2\\angle BAO$,\\\\\nthus $\\angle PBA=\\angle ABE+\\angle PBE$,\\\\\ntherefore $\\angle PBA>\\angle ABE$,\\\\\ntherefore $\\angle PBA\\neq 2\\angle BAO$,\\\\\nIn summary: $\\boxed{\\frac{n}{OA}<1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51983386_181.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{5}-2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As depicted in the figure, in square $ABCD$, $E$ is a point on $AB$. Line $DE$ is drawn, and $\\triangle DAE$ is rotated $90^\\circ$ counterclockwise around point $A$ to $\\triangle BAF$ (with $E$ corresponding to $F$ and $D$ to $B$). Lines $FE$ and $FC$ are drawn. If $AB=4$ and $FE$ bisects $\\angle BFC$, what is the length of $AE$?\n",
    "completion": "\\textbf{Solution:} Let DE and CF intersect at O, as shown in the diagram: \\\\\n$\\because$ Rotating $\\triangle DAE$ $90^\\circ$ counterclockwise around point A to $\\triangle BAF$, \\\\\n$\\therefore AF=AE$, $\\angle ADE=\\angle ABF$ \\\\\n$\\therefore \\angle AFE=\\angle AEF=45^\\circ$, \\\\\n$\\because FE$ bisects $\\angle BFC$, \\\\\n$\\therefore \\angle EFB=\\angle EFC$, \\\\\n$\\therefore \\angle OFD=45^\\circ-\\angle EFC$, $\\angle ADE=\\angle ABF=45^\\circ-\\angle EFB$, \\\\\n$\\therefore \\angle OFD=\\angle ADE$, that is $\\angle CFD=\\angle ADE$, \\\\\n$\\therefore \\tan\\angle CFD=\\tan\\angle ADE$, that is $\\frac{CD}{DF}=\\frac{AE}{AD}$, \\\\\nLet $AE=AF=x$, \\\\\n$\\therefore \\frac{4}{4+x}=\\frac{x}{4}$, \\\\\n$\\therefore x=\\boxed{2\\sqrt{5}-2}$ or $-2\\sqrt{5}-2$ (discard the latter), \\\\\n$\\therefore AE=2\\sqrt{5}-2$.",
    "image_path": "dataset/mm_math/images/MM_Math/52457426.png"
  },
  {
    "solution": "\\boxed{18^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC$, and through point $A$, $DA\\perp AC$ intersects $BC$ at point $D$. If $\\angle B=2\\angle BAD$, what is the degree measure of $\\angle BAD$?\n",
    "completion": "\\textbf{Solution:} Since AB = AC, \\\\\nit follows that $\\angle B = \\angle C = 2\\angle BAD$, \\\\\nhence $\\angle ADC = \\angle BAD + \\angle B$, \\\\\nsince $\\angle B = 2\\angle BAD$, \\\\\nit implies $\\angle ADC = 3\\angle BAD$, \\\\\ngiven DA $\\perp$ AC, \\\\\nthus $\\angle DAC = 90^\\circ$, \\\\\ntherefore $\\angle ADC + \\angle C = 90^\\circ$, \\\\\nit follows that $3\\angle BAD + 2\\angle BAD = 90^\\circ$, \\\\\nhence $\\angle BAD = \\boxed{18^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52836886.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle BAC=90^\\circ$, AD is the altitude from A to side BC, E and F are the midpoints of sides AB and AC, respectively. If AB = 8 and AC = 6, what is the perimeter of $\\triangle DEF$?\n",
    "completion": "\\textbf{Solution:} Given $AD$ is the altitude from $A$ to side $BC$,\\\\\nthus $\\angle ADB=\\angle ADC=90^\\circ$,\\\\\nsince $E,F$ are the midpoints of sides $AB,AC$ respectively, and $AB=8,AC=6$,\\\\\nthen $DE=\\frac{1}{2}AB=4,DF=\\frac{1}{2}AC=3$,\\\\\nsince in $\\triangle ABC$, $\\angle BAC=90^\\circ,AB=8,AC=6$,\\\\\nthen $BC=\\sqrt{AB^{2}+AC^{2}}=10$,\\\\\nalso since $E,F$ are the midpoints of sides $AB,AC$ respectively,\\\\\nthen $EF=\\frac{1}{2}BC=5$,\\\\\ntherefore, the perimeter of $\\triangle DEF$ is $DE+DF+EF=4+3+5=\\boxed{12}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52188862.png"
  },
  {
    "solution": "\\boxed{\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, $\\angle B=30^\\circ$. Point $D$ is on side $AB$, and point $E$ is on side $BC$. After folding $\\triangle BDE$ along line $DE$, point $B$ falls exactly on point $P$, which is on the extension of line segment $AC$. If $\\angle APE=2\\angle B$, what is the value of $\\frac{BD}{AD}$?\n",
    "completion": "\\textbf{Solution:} Let's consider $Rt\\triangle ABC$ where $\\angle C=90^\\circ$,\\\\\n$\\because \\angle B=30^\\circ$,\\\\\n$\\therefore \\angle A=60^\\circ$,\\\\\n$\\therefore \\angle APE=2\\angle B=60^\\circ$,\\\\\nBy folding, it is known that: $PD=BD\\text{, }\\angle DPE=\\angle B=30^\\circ$,\\\\\n$\\therefore \\angle APD=\\angle APE−\\angle DPE=30^\\circ$,\\\\\n$\\therefore \\angle ADP=60^\\circ+30^\\circ=90^\\circ$,\\\\\n$\\therefore \\frac{PD}{AD}=\\sqrt{3}$,\\\\\n$\\therefore \\frac{BD}{AD}=\\boxed{\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52977934.png"
  },
  {
    "solution": "\\boxed{x=-1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, parabola $P_{1}\\colon y=x^{2}+2x+m$ is translated to get parabola $P_{2}\\colon y=x^{2}-5x+n$. The two parabolas intersect the $y$ axis at points $C$ and $D$ respectively. The intersection point $E$ of parabolas $P_{1}$ and $P_{2}$ has an abscissa of $1$. A line parallel to the $x$ axis is drawn through point $E$, intersecting parabolas $P_{1}$ and $P_{2}$ at points $A$ and $B$, respectively. Find the axis of symmetry of parabola $P_{1}$ and the abscissa of point $A$.\n",
    "completion": "\\textbf{Solution:} To solve, the parabola $P_{1}\\colon y=x^{2}+2x+m$ has its axis of symmetry as the line $x=-1$,\\\\\n$\\because AB\\parallel x$ axis,\\\\\nand $\\because$ Points A and E are symmetric about the axis $x=-1$, and the x-coordinate of point E is 1,\\\\\n$\\therefore$ the x-coordinate of point A is $\\boxed{-3}$;\\\\\nThus, the axis of symmetry of the parabola $P_{1}$ is the line $\\boxed{x=-1}$ and the x-coordinate of point A is $-3$.",
    "image_path": "dataset/mm_math/images/MM_Math/52731763_79.png"
  },
  {
    "solution": "\\boxed{\\frac{3\\sqrt{85}}{10}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in rhombus $ABCD$, $\\tan \\angle DAB = \\frac{4}{3}$, $AB=3$, and point $P$ is a moving point on side $AB$. Extend $BA$ to point $Q$ such that $AQ=2AP$, and $CQ$, $DP$ intersect at point $T$. When point $P$ moves from point $A$ to the right towards point $B$, what is the length of the path traced by point $T$?\n",
    "completion": "\\textbf{Solution:} Extend $AT$ until it intersects $CD$ at $N$, as shown in the diagram: \\\\\nSince $CD\\parallel BQ$,\\\\\nit follows that $ \\frac{AP}{DN} = \\frac{AT}{NT} = \\frac{AQ}{CN}$,\\\\\ntherefore, $ \\frac{AP}{AQ} = \\frac{1}{2} = \\frac{DN}{CN}$,\\\\\nthus, point $N$ is the trisection point on $CD$ closer to $D$,\\\\\nhence, point $T$ moves along segment $AN$,\\\\\nwhen $P$ moves from point $A$ to the right until it coincides with $B$, as depicted in the diagram: \\\\\nthe path of point $T$ is along $AT$, draw $DH\\perp AB$ at $H$, and draw $TM\\perp AB$ at $M$,\\\\\nin $\\triangle ADH$, $\\tan \\angle DAB = \\frac{4}{3}$,\\\\\nlet $DH = 4k$, then $AH = 3k$, $AD = 5k$,\\\\\nsince $AD = AB = 3$,\\\\\nit follows that $5k = 3$,\\\\\nthus, $k = \\frac{3}{5}$,\\\\\ntherefore, $DH = \\frac{12}{5}$, $AH = \\frac{9}{5}$,\\\\\nthus, $BH = AB - AH = \\frac{6}{5}$,\\\\\nsince $ \\frac{DT}{PT} = \\frac{CD}{PQ} = \\frac{AP}{AP+AQ} = \\frac{1}{3}$,\\\\\nit follows that $ \\frac{PT}{PD} = \\frac{3}{4}$,\\\\\nsince $DH\\perp AB$, $TM\\perp AB$,\\\\\nit follows that $TM\\parallel DH$,\\\\\nhence, $ \\frac{PT}{PD} = \\frac{TM}{DH} = \\frac{BM}{BH}$, namely $ \\frac{3}{4} = \\frac{TM}{\\frac{12}{5}} = \\frac{BM}{\\frac{6}{5}}$,\\\\\nthus, $TM = \\frac{9}{5}$, $BM = \\frac{9}{10}$,\\\\\ntherefore, $AM = AB - BM = \\frac{21}{10}$,\\\\\nin $\\triangle ATM$, $AT = \\sqrt{AM^{2}+TM^{2}} = \\boxed{\\frac{3\\sqrt{85}}{10}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52483257.png"
  },
  {
    "solution": "\\boxed{\\left(\\frac{3}{2},-3\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the straight line $y=-2x+7$ intersects the x-axis and y-axis at points C and B, respectively, and intersects the straight line $y=\\frac{3}{2}x$ at point A. In the Cartesian coordinate system xOy, is there a point M such that the quadrilateral with vertices O, A, M, and C is a parallelogram? Write down the coordinates of all the points M that meet the conditions; if not, please explain the reason;\n",
    "completion": "\\textbf{Solution:} The solution exists. Let $y=0$ and substitute it into $y=-2x+7$, we get $0=-2x+7$. Solving this we find $x=\\frac{7}{2}$, thus $C\\left(\\frac{7}{2},0\\right)$. Let $M(x,y)$ be as shown in the figure:\n\\begin{enumerate}\n    \\item When $AC$ is the diagonal, $x=2+\\frac{7}{2}-0=\\frac{7}{2}$, $y=3$. Therefore, the coordinates of point $M$ are $\\left(\\frac{11}{2},3\\right)$;\n    \\item When $AO$ is the diagonal, $x=2+0-\\frac{7}{2}=-\\frac{3}{2}$, $y=3$. Therefore, the coordinates of point $M$ are $\\left(-\\frac{3}{2},3\\right)$;\n    \\item When $CO$ is the diagonal, $x=0+\\frac{7}{2}-2=\\frac{3}{2}$, $y=-3$. Therefore, the coordinates of point $M$ are $\\left(\\frac{3}{2},-3\\right)$,\n\\end{enumerate}\nIn summary: the coordinates of point $M$ are $\\boxed{\\left(\\frac{11}{2},3\\right)}$, $\\boxed{\\left(-\\frac{3}{2},3\\right)}$, $\\boxed{\\left(\\frac{3}{2},-3\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52342217_128.png"
  },
  {
    "solution": "\\boxed{42^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, fold the triangular paper $ABC$ along $DE$. When point $A$ falls outside of the quadrilateral $BCED$, the measurements show that $\\angle 1 = 70^\\circ$ and $\\angle 2 = 152^\\circ$. What is the measure of $\\angle A$?\n",
    "completion": "\\[\n\\text{Solution:} \\\\\n\\text{Based on the fold, we know } \\angle A'=\\angle A, \\\\\n\\because \\angle 1=70^\\circ, \\\\\n\\therefore \\angle A'DA=180^\\circ-\\angle 1=110^\\circ, \\\\\n\\therefore \\text{according to the external angle of a triangle } \\angle A'=\\angle 2-\\angle A'DA=152^\\circ-110^\\circ=42^\\circ, \\\\\n\\therefore \\angle A=\\boxed{42^\\circ}.\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/52096886.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{8}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the lines $l_1 \\parallel l_2 \\parallel l_3$, and the lines $a$ and $b$ intersect with these three lines at points $A$, $B$, $C$ and $E$, $F$, $F$, respectively. If $AB:BC = 5:3$, then what is the ratio of $EF:DF$?\n",
    "completion": "\\textbf{Solution:} Since ${l}_{1}\\parallel {l}_{2}\\parallel {l}_{3}$\\\\\n$\\therefore \\frac{AB}{BC}=\\frac{DE}{EF}=\\frac{5}{3}$\\\\\n$\\therefore \\frac{EF}{DE}=\\frac{3}{5}$\\\\\n$\\therefore \\frac{EF}{DF}=\\boxed{\\frac{3}{8}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51379316.png"
  },
  {
    "solution": "\\boxed{40^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of the circle $\\odot O$, with chord $CD \\perp AB$. Connecting $OD$ and $BC$, if $\\angle CBA = 70^\\circ$, what is the measure of $\\angle BOD$?\n",
    "completion": "\\textbf{Solution:} Since $CD\\perp AB$ and $\\angle CBA=70^\\circ$, \\\\\n$\\therefore \\angle BCD=20^\\circ$, \\\\\n$\\therefore \\angle BOD=2\\angle BCD=40^\\circ$, \\\\\nHence, the answer is: $\\boxed{40^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53106831.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: The graph of the function $y=kx$ and $y=6-x$ is shown in the figure below. What is the value of $k$?\n",
    "completion": "\\textbf{Solution:} Since the abscissa of the point of intersection between the linear function $y=6-x$ and $y=kx$ is 2,\\\\\n$\\therefore y=6-2=4$,\\\\\n$\\therefore$ the coordinates of the intersection point are $(2,4)$,\\\\\nSubstituting $(2,4)$ into $y=kx$, we get $2k=4$, which gives: $k=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52342520.png"
  },
  {
    "solution": "\\boxed{quadrilateral\\ BCFD\\ is\\ a\\ rhombus}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral ABCD, $AD\\parallel BC$, $\\angle A=90^\\circ$, $BD=BC$, point E is the midpoint of CD, ray BE intersects the extension of AD at point F, connect CF. Prove that the quadrilateral BCFD is a rhombus.\n",
    "completion": "\\textbf{Solution.} Proof: Since AF$\\parallel$BC,\\\\\nit follows that $\\angle$DCB=$\\angle$CDF, $\\angle$FBC=$\\angle$BFD,\\\\\nsince point E is the midpoint of CD,\\\\\nit follows that DE=EC,\\\\\nin $\\triangle$ BCE and $\\triangle$ FDE,\\\\\nsince $\\left\\{\\begin{array}{l}\n\\angle FBC=\\angle BFD \\\\\n\\angle DCB=\\angle CDF \\\\\nDE=EC\n\\end{array}\\right.$,\\\\\nit follows that $\\triangle$ BCE$\\cong$ $\\triangle$ FDE,\\\\\nthus, DF=BC,\\\\\nand since DF$\\parallel$BC,\\\\\nit follows that quadrilateral BCFD is a parallelogram,\\\\\nsince BD=BC,\\\\\nit follows that \\boxed{quadrilateral\\ BCFD\\ is\\ a\\ rhombus};",
    "image_path": "dataset/mm_math/images/MM_Math/53599308_94.png"
  },
  {
    "solution": "\\boxed{90^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $\\angle ADC=90^\\circ$, $DC\\parallel AB$, $BA=BC$, $AE\\perp BC$ with the foot of the perpendicular being point E, and point F is the midpoint of $AC$. Is $\\angle AFB=90^\\circ$?\n",
    "completion": "\\textbf{Solution:} Proof: \\\\\nSince $BA = BC$, and $F$ is the midpoint of $AC$, \\\\\ntherefore $BF \\perp AC$. \\\\\nTherefore, $\\angle AFB = 90^\\circ$. \\\\\nHence, the answer is $\\boxed{90^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53044604_146.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\square ABCD $, with $ AD=6 $, E is a moving point on AD, and M, N are the midpoints of BE and CE, respectively. What is the length of MN?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, in parallelogram $ABCD$, we have $BC=AD=6$.\\\\\nSince $M$, $N$ are the midpoints of $BE$, $CE$, respectively,\\\\\ntherefore $MN$ is the midline of $\\triangle EBC$,\\\\\ntherefore $MN=\\frac{1}{2}BC=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52211052.png"
  },
  {
    "solution": "\\boxed{105^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point O is on line DB. Given that $\\angle 1=15^\\circ$ and $\\angle AOC=90^\\circ$, what is the measure of $\\angle 2$ in degrees?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle 1=15^\\circ$ and $\\angle AOC=90^\\circ$, \\\\\nit follows that $\\angle BOC=75^\\circ$, \\\\\ntherefore, $\\angle 2=180^\\circ-75^\\circ=\\boxed{105^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55190294.png"
  },
  {
    "solution": "\\boxed{-1<x<3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the quadratic function $y=x^2+bx+c$ intersects the $x$-axis at points $A$ and $B$, and intersects the $y$-axis at point $C$. It is symmetric about the line $x=1$, and the coordinates of point $A$ are $(-1,0)$. When $y<0$, write down the range of values for $x$.\n",
    "completion": "\\textbf{Solution:} When $y<0$, the range of $x$ is \\boxed{-1<x<3}.",
    "image_path": "dataset/mm_math/images/MM_Math/51369956_189.png"
  },
  {
    "solution": "\\boxed{24}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $ \\triangle ABC$ and $\\triangle A'B'C'$ are similar figures with the coordinate origin $O$ as the center of similarity. If point $A$ is the midpoint of $OA'$, and the area of $\\triangle ABC$ is $6$, what is the area of $\\triangle A'B'C'$?\n",
    "completion": "\\textbf{Solution:} Given the problem, we know the similarity ratio between triangles $ \\triangle ABC$ and $ \\triangle A'B'C'$ is $ \\frac{1}{2}$.\\\\\n$\\therefore$ $ \\frac{{S}_{\\triangle ABC}}{{S}_{\\triangle A'B'C'}}={\\left(\\frac{1}{2}\\right)}^{2}$\\\\\n$\\because$ $ {S}_{\\triangle ABC}=6$\\\\\n$\\therefore$ $ {S}_{\\triangle A'B'C'}=\\boxed{24}$",
    "image_path": "dataset/mm_math/images/MM_Math/51496703.png"
  },
  {
    "solution": "\\boxed{y=-x+1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the graph, in the Cartesian coordinate system $xOy$, the line $y=mx$ intersects the hyperbola $y=\\frac{n}{x}$ at points $A(-1, a)$ and $B$, with $BC\\perp x$-axis and the foot of the perpendicular being $C$. The area of $\\triangle AOC$ is $1$. Find the analytical expression of line $AC$.\n",
    "completion": "\\textbf{Solution:} Let the analytical expression of line $AC$ be $y=kx+b$,\\\\\nsince $y=kx+b$ passes through points $A(-1,2)$ and $C(1,0)$,\\\\\nwe have $\\begin{cases} -k+b=2 \\\\ k+b=0 \\end{cases}$,\\\\\nsolving which gives $k=-1$, $b=1$,\\\\\ntherefore, the analytical expression of line $AC$ is $\\boxed{y=-x+1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52551767_143.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated, a square OABC with a side length of 3 rotates counterclockwise by $n^\\circ$ ($0<n<90$) around point O to form square ODEF. DE intersects BC at point P, and the extension of ED intersects AB at point Q and the extension of OA at point M. If $AQ=1$, what is the value of $PD$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect OQ and OP,\n\nIn $\\triangle OPD$ and $\\triangle OPC$,\n\n$\\because \\angle PDO=\\angle C=90^\\circ$,\n\nthen $\\left\\{\\begin{array}{l}\nOP=OP\\\\\nOD=OC\n\\end{array}\\right.$,\n\n$\\therefore \\triangle OPD\\cong \\triangle OPC$,\n\n$\\therefore PD=PC$,\n\nSimilarly, $\\triangle OQA\\cong \\triangle OQD$, $QA=QD$,\n\nLet $PD=x$, then $PC=x$, $BP=3-x$, $PQ=1+x$,\n\nIn $Rt\\triangle BPQ$, $BP^2+BQ^2=PQ^2$,\n\ni.e., $(3-x)^2+(3-1)^2=(1+x)^2$,\n\nyielding $x=\\boxed{\\frac{3}{2}}$,\n\nthus $PD=\\frac{3}{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51708611.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{1}{2}x+4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, place the rectangular paper $OABC$ into the Cartesian coordinate system such that $OA$ and $OC$ lie on the positive x-axis and y-axis, respectively. Connect $AC$, and let $AC=4\\sqrt{5}$ and $OA=2CO$. What is the equation of the line on which $AC$ lies?\n",
    "completion": "\\textbf{Solution:} Let $\\because$ OA=2CO, assume OC=x, then OA=2x. In $\\triangle AOC$, by Pythagorean theorem we have $OC^2+OA^2=AC^2$,\n$\\therefore x^2+(2x)^2=(4\\sqrt{5})^2$. Solving this, we get $x=4$ (discarding $x=-4$).\n$\\therefore$ OC=4, OA=8.\n$\\therefore$ A(8,0), C(0,4).\nLet the equation of line AC be $y=kx+b$,\n$\\therefore \\left\\{\\begin{array}{l}8k+b=0\\\\ b=4\\end{array}\\right.$, solving this we get $\\left\\{\\begin{array}{l}k=-\\frac{1}{2}\\\\ b=4\\end{array}\\right.$,\n$\\therefore$ the equation of line AC is $\\boxed{y=-\\frac{1}{2}x+4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52342628_122.png"
  },
  {
    "solution": "\\boxed{-\\frac{2}{x}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y_{1}=-x-1$ intersects the x-axis at point A and the y-axis at point B. It intersects the graph of the inverse proportion function $y_{2}= \\frac{k}{x}$ at point M($-2$, $m$). What is the analytic expression of the inverse proportion function?\n",
    "completion": "\\textbf{Solution:} Since point M($-2$, $m$) lies on the graph of the linear function $y_{1}=-x-1$,\\\\\nby substitution we get: $m=-(-2)-1=1$,\\\\\nthus, the coordinates of M are ($-2$, $1$),\\\\\nsubstituting M's coordinates into $y_{2}=\\frac{k}{x}$ gives: $k=-2$,\\\\\ntherefore, the equation of the inverse proportion function is: $y_{2}=\\boxed{-\\frac{2}{x}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51457952_272.png"
  },
  {
    "solution": "\\boxed{1 \\text{ or } 4 \\text{ or } 2.5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, it is known that: $AB=3$, $AD=5$, and point $P$ is a point on $BC$, and $\\triangle PAD$ is an isosceles triangle. What is the length of $BP$?\n",
    "completion": "\\textbf{Solution:}\n(1) When $DP=AD$,\\\\\nsince rectangle $ABCD$,\\\\\nthus $DC=AB=3$, $AD=BC=5$,\\\\\nsince $\\triangle PAD$ is an isosceles triangle,\\\\\nthus $DP=AD=5$,\\\\\nin right $\\triangle PCD$,\\\\\n$PC=\\sqrt{PD^2-DC^2}=4$,\\\\\nthus $BP=BC-CP=5-4=1$.\\\\\n(2) When $AD=AP$,\\\\\nthus $AP=AD=5$,\\\\\nin right $\\triangle ABP$,\\\\\nby the Pythagorean theorem,\\\\\n$BP=\\sqrt{PA^2-AB^2}=4$,\\\\\n(3) When $AP=DP$,\\\\\ndraw $PE\\perp AD$ at point $E$,\\\\\nthus $AE=\\frac{1}{2}AD=2.5$,\\\\\nsince $\\angle B=\\angle BAE=\\angle AEP=90^\\circ$,\\\\\nthus quadrilateral $ABPE$ is a rectangle,\\\\\nthus $BP=AE=2.5$.\\\\\nIn summary, $BP=1$ or $4$ or $2.5$.\\\\\nTherefore, the answer is: \\boxed{1 \\text{ or } 4 \\text{ or } 2.5}.",
    "image_path": "dataset/mm_math/images/MM_Math/52423830.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, two right-angled triangles with angles $45^\\circ$ and $30^\\circ$ are placed on the same plane. If $AB=2\\sqrt{2}$, what is the length of $BD$?\n",
    "completion": "\\textbf{Solution:} Given that $\\triangle ABC$ is a right-angled triangle with a $45^\\circ$ angle, and $AB=2\\sqrt{2}$, \\\\\ntherefore $AC^2+BC^2=AB^2$, \\\\\nthus $AC=BC=2$, \\\\\nsince $\\triangle DBC$ is a right-angled triangle with a $30^\\circ$ angle, and $\\angle D=30^\\circ$, $BC=2$, \\\\\ntherefore $BD=2BC=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52828913.png"
  },
  {
    "solution": "\\boxed{x\\leq -2 or 0<x\\leq 2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the graph of the direct proportion function $y_1=-\\frac{3}{2}x$ intersects with the graph of the inverse proportion function $y_2=\\frac{k}{x}$ at point A in the second quadrant, and the x-coordinate of point A is $-2$. Write directly the range of values for $x$ when $y_1>y_2$.\n",
    "completion": "\\textbf{Solution:} When $y_1>y_2$, the range of values for $x$ is $\\boxed{x\\leq -2 or 0<x\\leq 2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52602886_84.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AD \\perp BC$ at point $D$, $AD$ and $BE$ intersect at point $F$, and $AC=BF$, $DF=DC$. If $\\angle ABE=15^\\circ$, what is the degree measure of $\\angle DBF$?\n",
    "completion": "\\textbf{Solution:} Since $AD \\perp BC$, \\\\\ntherefore $\\angle BDF = \\angle ADC = 90^\\circ$, \\\\\nIn right triangles $\\triangle BDF$ and $\\triangle ADC$, \\\\\n$\\left\\{\\begin{array}{l}AC=BF\\\\ DC=DF\\end{array}\\right.$, \\\\\ntherefore right $\\triangle BDF \\cong \\triangle ADC$ (HL), \\\\\ntherefore $AD = BD$, \\\\\ntherefore $\\angle ABD = \\angle DAB = 45^\\circ$, \\\\\nsince $\\angle ABE = 15^\\circ$, \\\\\ntherefore $\\angle DBF = \\angle ABD - \\angle ABE = 45^\\circ - 15^\\circ = \\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52836660.png"
  },
  {
    "solution": "\\boxed{56^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the isosceles $\\triangle ABC$ with $AB=AC$, $\\angle BAC=56^\\circ$. The bisector of $\\angle BAC$ intersects the perpendicular bisector of $AB$ at point $O$. If point $C$ is folded along $EF$ to coincide with point $O$, what is the measure of $\\angle CEF$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect BO,\\\\\n$\\because$ in the isosceles $\\triangle ABC$, $AB=AC$, $\\angle BAC=56^\\circ$,\\\\\n$\\therefore \\angle ABC=\\angle ACB=(180^\\circ-56^\\circ)÷2=62^\\circ$,\\\\\n$\\because AO$ is the bisector of $\\angle BAC$,\\\\\n$\\therefore \\angle BAO=28^\\circ$,\\\\\nAlso $\\because OD$ is the perpendicular bisector of AB,\\\\\n$\\therefore$ OA=OB,\\\\\n$\\therefore \\angle OBA=\\angle OAB=28^\\circ$,\\\\\n$\\therefore \\angle OBC=\\angle OCB=62^\\circ-28^\\circ=34^\\circ$,\\\\\n$\\because EF$ is the perpendicular bisector of the segment OC,\\\\\n$\\therefore \\angle CEF=90^\\circ-34^\\circ=\\boxed{56^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52906068.png"
  },
  {
    "solution": "\\boxed{\\frac{7}{8}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AD \\perp BC$ at point $D$, $E$ is a point on $AD$, and $AE:ED = 7:5$. Connect $CE$ and extend it to intersect side $AB$ at point $F$, $AC=13$, $BC=8$, $\\cos\\angle ACB = \\frac{5}{13}$. What is the value of $\\frac{AF}{BF}$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw line $DG\\parallel CF$ passing through point $D$ and intersecting $AB$ at point $G$, \\\\\nsince $BC=8, CD=5$, \\\\\nthus $BD=BC-CD=3$. \\\\\nSince $DG\\parallel CF$, it follows that $\\frac{BD}{CD}=\\frac{BG}{FG}=\\frac{3}{5}$, $\\frac{AF}{FG}=\\frac{AE}{DE}=\\frac{7}{5}$, \\\\\ntherefore $AF=\\frac{7}{5}FG$. Let $BG=3x$, then $FG=5x, AF=7x$. \\\\\nHence $BF=FG+BG=8x$, therefore $\\frac{AF}{BF}=\\boxed{\\frac{7}{8}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55595013_156.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, $AD \\parallel BC$, and $\\angle B + \\angle C = 90^\\circ$. Squares $ABEF$, $ADHG$, and $DCJI$ are constructed externally on sides $AB$, $AD$, and $DC$, respectively, and their areas are denoted as $S_{1}$, $S_{2}$, and $S_{3}$ in sequence. If $S_{1} + S_{3} = 4S_{2}$, then what is the value of $\\frac{BC}{AD}$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, draw $AQ\\parallel CD$ through point A, and let it intersect $BC$ at point Q,\\\\\n$\\therefore$ $\\angle AQB=\\angle QCD$,\\\\\n$\\because$ $AD\\parallel BC$,\\\\\n$\\therefore$ Quadrilateral $ADCQ$ is a parallelogram,\\\\\n$\\therefore$ $AD=QC$, $AQ=DC$, $\\angle BCD=\\angle AEB$,\\\\\n$\\because$ $\\angle ABC+\\angle BCD=90^\\circ$,\\\\\n$\\therefore$ $\\angle ABC+\\angle QCD=90^\\circ$,\\\\\n$\\therefore$ $\\angle BAQ=90^\\circ$,\\\\\n$\\therefore$ $AB^{2}+AQ^{2}=BQ^{2}$, that is, $AB^{2}+DC^{2}=BQ^{2}$,\\\\\n$\\therefore$ $S_{1}+S_{3}=BQ^{2}$,\\\\\nFurthermore, $\\because$ $S_{1}+S_{3}=4S_{2}$,\\\\\n$\\therefore$ $4S_{2}=BQ^{2}=4AD^{2}$,\\\\\n$\\therefore$ $BQ=2AD$,\\\\\n$\\therefore$ $BC=BQ+QC=2AD+AD=3AD$,\\\\\n$\\therefore$ $\\frac{BC}{AD}=\\frac{3AD}{AD}=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52065936.png"
  },
  {
    "solution": "\\boxed{x>4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the line $y_1=k_1x+b$ intersects with the hyperbola $y_2=\\frac{k_2}{x}$ at points $A(4, 2)$ and $B(m, -4)$. Based on the graph of the function, directly write down the solution set of the inequality $k_1x+b>\\frac{k_2}{x}$ in terms of $x$.\n",
    "completion": "\\textbf{Solution:} Observing the graph, the solution set of the inequality ${k}_{1}x+b>\\frac{{k}_{2}}{x}$ with respect to $x$ is \\boxed{-2<x<0} or \\boxed{x>4}.",
    "image_path": "dataset/mm_math/images/MM_Math/52381008_158.png"
  },
  {
    "solution": "\\boxed{P\\left(12, 7\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system \\(xOy\\), the line \\(AB: y=-\\frac{1}{2}x+3\\) and the line \\(CD: y=kx-2\\) intersect at point \\(M(4, a)\\), intersecting the coordinate axes at points A, B, C, and D, respectively. As depicted, point \\(P\\) is a moving point on line \\(CD\\). When the area of triangle \\(PBM\\) is \\(20\\), what are the coordinates of point \\(P\\)?\n",
    "completion": "\\textbf{Solution:} Let $P\\left(m, \\frac{3}{4}m-2\\right)$. The equation of line CD is: $y=\\frac{3}{4}x-2$. Setting $x=0$, we find $y=-2$, thus point $D\\left(0, -2\\right)$. For line AB: $y=-\\frac{1}{2}x+3$, setting $x=0$, we obtain $y=3$, thus point $B\\left(0, 3\\right)$, hence $BD=5$. Therefore, the area of $\\triangle PBM$ equals $\\frac{1}{2}\\times BD\\times \\left|{x}_{M}-m\\right|=\\frac{1}{2}\\times 5\\times \\left|4-m\\right|=20$, therefore $\\left|4-m\\right|=8$. Solving this results in: $m=-4$ or $m=12$. In conclusion, the points are $P\\left(-4, -5\\right)$ or $P\\left(12, 7\\right)$; therefore, the answer is $\\boxed{P\\left(-4, -5\\right)}$ or $\\boxed{P\\left(12, 7\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52381257_179.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point E is on the bisector of $\\angle BAC$ on the line AP, with AB = 4, and the area of $\\triangle ABE$ is 12. What is the distance from point E to the line AC?\n",
    "completion": "\\textbf{Solution:}\nGiven that $AB = 4$ and the area of $\\triangle ABE$ is $12$,\\\\\nit follows that the distance from point E to line AB $= \\frac{2 \\times 12}{4} = 6$,\\\\\nsince E is a point on the bisector of $\\angle BAC$,\\\\\ntherefore, the distance from point E to line AC $=$ \\boxed{6}.",
    "image_path": "dataset/mm_math/images/MM_Math/52793324.png"
  },
  {
    "solution": "\\boxed{(6, 4)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the quadrilateral $ABCD$ is a square. The coordinate of point $A$ is $(0,2)$, and the coordinate of point $B$ is $(4,0)$. What is the coordinate of point $C$?\n",
    "completion": "\\textbf{Solution:} As illustrated, draw CE$\\perp$x-axis at point C, with E being the foot of the perpendicular.\\\\\nSince quadrilateral ABCD is a square, and points A(0, 2), B(4, 0),\\\\\nthen AB=BC, $\\angle$ABC=90$^\\circ$, AO=2, OB=4,\\\\\nthus $\\angle$AOB=$\\angle$BEC= 90$^\\circ$,  $\\angle$ABO=$\\angle$BCE=90$^\\circ$−$\\angle$CBE,\\\\\ntherefore $\\triangle$AOB$\\cong$$\\triangle$BEC,\\\\\nthus BE=AO=2, EC=OB=4,\\\\\ntherefore OE=OB+BE=2+4=6,\\\\\nhence, the coordinates of point C are $\\boxed{(6, 4)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52422876.png"
  },
  {
    "solution": "\\boxed{M_4(-1, -3)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the figure, the line $y=kx$ intersects with the parabola $y=x^2+c$ at points $A$ and $B$, where the coordinates of point $B$ are $(2, 2)$. Point $P$ is on the parabola, point $Q$ is on line $AB$, and point $M$ is within the coordinate plane. When the quadrilateral with vertices $A$, $P$, $Q$, $M$ forms a square, what are the coordinates of point $M$?\n",
    "completion": "\\textbf{Solution:} Let $M_1(1, -1)$, $M_2(0, -2)$, $M_3(-1, 1)$, and $M_4(-1, -3)$. $\\boxed{M_1(1, -1)}$, $\\boxed{M_2(0, -2)}$, $\\boxed{M_3(-1, 1)}$, and $\\boxed{M_4(-1, -3)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53066184_91.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{5}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in square ABCD, AB=2, E is the midpoint of CD, and $\\triangle BCE$ is folded along BE to $\\triangle BFE$. G is the midpoint of BE. Connect FG. What is the length of FG?\n",
    "completion": "\\textbf{Solution:}\nSince $\\triangle BCE$ is folded along $BE$ to form $\\triangle BFE$,\n\nthus $\\angle C=\\angle BFE=90^\\circ$,\n\ntherefore $\\triangle BFE$ is a right-angled triangle.\n\nHence, $FG=\\frac{1}{2}BE=\\frac{1}{2}\\sqrt{1^2+2^2}=\\boxed{\\frac{\\sqrt{5}}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52250970.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given points $A$, $B$, $C$, and $D$ are on the same line, where point $C$ is the midpoint of segment $AB$, and point $D$ is on segment $AB$. If $AB=6$ and $BD=\\frac{1}{3}BC$, what is the length of segment $CD$?\n",
    "completion": "\\textbf{Solution:} Given that point C is the midpoint of segment AB, and $AB=6$,\\\\\n$\\therefore$ $BC=\\frac{1}{2}AB=3$,\\\\\n$\\therefore$ $BD=\\frac{1}{3}BC$,\\\\\n$\\therefore$ $BD=\\frac{1}{3}\\times 3=1$,\\\\\n$\\therefore$ $CD=BC-BD=3-1=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53195484_120.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $BC=4\\,\\text{cm}$, $BD=7\\,\\text{cm}$, and point $D$ is the midpoint of $AC$. What is the length of $AC$ in centimeters?\n",
    "completion": "\\textbf{Solution:} Given that $CD = BD - BC = 7 - 4 = 3\\,(\\text{cm})$\\\\\nSince point D is the midpoint of AC, it follows that $AC = 2CD = \\boxed{6}\\,(\\text{cm})$",
    "image_path": "dataset/mm_math/images/MM_Math/55204863.png"
  },
  {
    "solution": "\\boxed{\\frac{2}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, Xiaohui used two right-angled triangle paper pieces to form a quadrilateral, where $\\angle A=90^\\circ$, $AD=2\\text{cm}$, $AB=3\\text{cm}$, and the hypotenuses overlap to form a quadrilateral. Then, points $E$ and $F$ are chosen on $CB$ and $CD$ respectively, and $AE$ and $BF$ are connected such that $AE \\perp BF$. If the quadrilateral is formed as shown in the figure, what is the value of $\\frac{BF}{AE}$?\n",
    "completion": "\\textbf{Solution:} Thus, $\\frac{BF}{AE} = \\boxed{\\frac{2}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51545297_61.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, $\\triangle ABC$ is inscribed in circle $\\odot O$, with $D$ being a point on the extension of diameter $AB$ of $\\odot O$, and $\\angle DCB = \\angle OAC$. A line parallel to $BC$ passing through the center $O$ of the circle intersects the extension of $DC$ at point $E$. $CD$ is a tangent to circle $\\odot O$; if $CD = 4$ and $CE = 6$, find the radius of circle $\\odot O$ and the value of $\\tan \\angle OCB$.\n",
    "completion": "\\textbf{Solution:} Since OE $\\parallel$ BC, \\\\\n$\\therefore \\frac{BD}{OB}=\\frac{CD}{CE}$, \\\\\nGiven that CD = 4, CE = 6, \\\\\n$\\therefore \\frac{BD}{OB}=\\frac{4}{6}=\\frac{2}{3}$, \\\\\nLet BD = 2x, then OB = OC = 3x, OD = OB + BD = 5x, \\\\\nSince OC $\\perp$ DC, \\\\\n$\\therefore \\triangle OCD$ is a right-angled triangle, \\\\\nIn $\\triangle OCD$, $OC^{2}+CD^{2}=OD^{2}$, \\\\\n$\\therefore (3x)^{2}+4^{2}=(5x)^{2}$, \\\\\nSolving this, we find x = 1, \\\\\n$\\therefore OC=3x=3$, i.e., the radius of $\\odot O$ is $\\boxed{3}$, \\\\\nSince BC $\\parallel$ OE, \\\\\n$\\therefore \\angle OCB=\\angle EOC$, \\\\\nIn $\\triangle OCE$, $\\tan \\angle EOC= \\frac{EC}{OC}=\\frac{6}{3}=2$, \\\\\n$\\therefore \\tan \\angle OCB=\\tan \\angle EOC=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51379304_88.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, points D and E are on the sides AB and AC respectively, and $\\frac{AE}{AB} = \\frac{AD}{AC} = \\frac{1}{2}$, then what is the ratio of the perimeter of $\\triangle ADE$ to the perimeter of $\\triangle ABC$?\n",
    "completion": "\\textbf{Solution:} Since $\\frac{AE}{AB} = \\frac{AD}{AC}$, and $\\angle DAE = \\angle CAB$,\\\\\ntherefore $\\triangle ADE \\sim \\triangle ACB$,\\\\\nthus, the ratio of the perimeter of $\\triangle ADE$ to that of $\\triangle ABC$ is $\\boxed{\\frac{1}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51179480.png"
  },
  {
    "solution": "\\boxed{80^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $EF \\parallel CD$, $\\angle 1 + \\angle 2 = 180^\\circ$. If $CD$ bisects $\\angle ACB$, $DG$ bisects $\\angle CDB$, and $\\angle A = 40^\\circ$, what is the measure of $\\angle ACB$?\n",
    "completion": "\\textbf{Solution:} Given from the previous question, we know that $GD\\parallel CA$,\\\\\n$\\therefore \\angle GDB = \\angle A = 40^\\circ$,\\\\\n$\\because DG$ bisects $\\angle CDB$, and $\\angle CDB$ is the exterior angle of triangle $ADC$,\\\\\n$\\therefore \\angle CDB = \\angle A + \\angle ACD = 40^\\circ + \\angle ACD = 80^\\circ$,\\\\\n$\\therefore \\angle ACD = 40^\\circ$,\\\\\nMoreover, since $CD$ bisects $\\angle ACB$,\\\\\n$\\therefore \\angle ACB = 40^\\circ \\times 2 = \\boxed{80^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/9349800_146.png"
  },
  {
    "solution": "\\boxed{x>5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, it is known that point $M$ is the vertex of the parabola $y=-\\left(x-b\\right)^2+4b+1$, and the line $y=mx+5$ intersects with the positive x-axis and y-axis at points $A$ and $B$ respectively. If the parabolic graph also passes through points $A$ and $B$, and if $mx+5>-\\left(x-b\\right)^2+4b+1$, according to the graph, write down the range of values for $x$.\n",
    "completion": "\\textbf{Solution:} As shown in Figure 1,\\\\\nthe line $y=mx+5$ intersects the y-axis at point $B$,\\\\\n$\\therefore$ the coordinates of point $B$ are $(0,5)$,\\\\\nSince $B$ lies on the parabola,\\\\\n$\\therefore 5 = -\\left(0-b\\right)^2 + 4b + 1 = 5$,\\\\\nsolving for $b$ yields $b=2$,\\\\\nThe analytical expression for the quadratic function is $y=-\\left(x-2\\right)^2+9$,\\\\\nWhen $y=0$, $-\\left(x-2\\right)^2+9=0$,\\\\\nsolving for $x$ yields $x_1=5$, $x_2=-1$,\\\\\n$\\therefore A(5,0)$,\\\\\nFrom the graph, we find that when $mx+5 > -\\left(x-b\\right)^2 + 4b + 1$, the range of $x$ is $x<0$ or $x>5$;\\\\\nTherefore, the range of $x$ is $\\boxed{x<0}$ or $\\boxed{x>5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53009807_168.png"
  },
  {
    "solution": "\\boxed{\\frac{21}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the length of segment $AB=15\\text{cm}$, point $C$ is on segment $AB$, and $AC:CB=3:2$. If point $P$ is the midpoint of segment $AB$, and point $M$ is the midpoint of segment $AP$, find the length of segment $MC$.\n",
    "completion": "\\textbf{Solution:} Since point P is the midpoint of line segment AB,\\\\\nit follows that $AP=BP=\\frac{1}{2}AB=\\frac{15}{2}$,\\\\\nFurthermore, since point M is the midpoint of line segment AP,\\\\\nit follows that $MP=\\frac{1}{2}AP=\\frac{15}{4}$,\\\\\nAlso, since $BC=6$,\\\\\nit follows that $PC=PB−BC=\\frac{15}{2}−6=\\frac{3}{2}$,\\\\\nTherefore, $MC=MP+PC=\\frac{15}{4}+\\frac{3}{2}=\\boxed{\\frac{21}{4}}cm$.",
    "image_path": "dataset/mm_math/images/MM_Math/51366955_124.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{3}{4}x+4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the line $y=-2x+4$ intersects the $x$-axis and $y$-axis at points $A$ and $C$ respectively. A rectangle $OABC$ is constructed in the first quadrant with $OA$ and $OC$ as sides. $\\triangle ABC$ is folded so that point $A$ coincides with point $C$, and the fold intersects $AB$ at point $D$. Find the equation of line $CD$.\n",
    "completion": "\\textbf{Solution:} Given from the fold: $CD=AD$. Let $AD=x$, then $CD=x$, $BD=4-x$. According to the given conditions, we have: $(4-x)^{2}+2^{2}=x^{2}$. Solving this gives: $x=\\frac{5}{2}$. At this point, $AD=\\frac{5}{2}$, $D\\left(2, \\frac{5}{2}\\right)$. Let the equation of line $CD$ be $y=kx+b$, substituting $D\\left(2, \\frac{5}{2}\\right)$ into it yields $\\frac{5}{2}=2k+b$. Solving this gives: $k=-\\frac{3}{4}$, $b=4$. $\\therefore$ The equation of line $CD$ is $\\boxed{y=-\\frac{3}{4}x+4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51419687_120.png"
  },
  {
    "solution": "\\boxed{\\sqrt{7}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, AB is the diameter of semicircle O, D is the midpoint of BC, extending OD intersects the arc $BC$ at point E, and point F is a point on the extension of OD such that $\\angle B = \\angle F$. CF is a tangent to the circle $O$; if AB = 4, $\\angle B = 30^\\circ$, connect AD, find the length of AD.\n",
    "completion": "\\textbf{Solution:} Join AD,\\\\\n$\\because$ AB is the diameter of semicircle $O$,\\\\\n$\\therefore \\angle ACB=90^\\circ$.\\\\\n$\\because$ AB$=4$, and $\\angle B=30^\\circ$,\\\\\n$\\therefore$ AC$=2$, and $BC=\\sqrt{4^2-2^2}=2\\sqrt{3}$,\\\\\n$\\therefore$ CD$=\\sqrt{3}$,\\\\\nIn $\\triangle ACD$, $AD=\\sqrt{2^2+(\\sqrt{3})^2}=\\boxed{\\sqrt{7}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52765493_175.png"
  },
  {
    "solution": "\\boxed{130^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $BD$ is the angle bisector of $\\triangle ABC$, points $E$ and $F$ are respectively on sides $AB$ and $BC$, and $BE=CF$, $ED\\parallel BC$. Quadrilateral $EFCD$ is a parallelogram; $\\angle ABC=60^\\circ$, $\\angle ADB=100^\\circ$, what is the degree measure of $\\angle AEF$?\n",
    "completion": "\\textbf{Solution:} Given that $BD$ is the angle bisector of $\\triangle ABC$, and $\\angle ABC=60^\\circ$, \\\\\nit follows that $\\angle ABD=\\frac{1}{2}\\angle ABC=30^\\circ$, \\\\\nsince $\\angle ADB=100^\\circ$, \\\\\nthen $\\angle A=180^\\circ-\\angle ABD-\\angle ADB=50^\\circ$, \\\\\nGiven that quadrilateral EFCD is a parallelogram, \\\\\nit implies $EF\\parallel AC$, \\\\\nthus, $\\angle AEF=180^\\circ-\\angle A=\\boxed{130^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52211459_90.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{5}}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the diagram, it is known that in $\\triangle ABC$, $\\angle C=90^\\circ$, $\\tan A= \\frac{1}{2}$, D is a point on AC, and $\\angle CBD=\\angle A$. Then, what is the value of $\\cos \\angle CDB$?\n",
    "completion": "\\[\n\\text{Solution:} \\quad \\because \\angle CBD=\\angle A,\n\\]\n\\[\n\\therefore \\tan\\angle CBD=\\tan A=\\frac{1}{2},\n\\]\n\\[\n\\text{Let } CD=a,\n\\]\n\\[\n\\therefore \\tan\\angle CBD=\\frac{CD}{BC}=\\frac{1}{2},\n\\]\n\\[\n\\therefore BC=2a,\n\\]\n\\[\n\\text{In } \\triangle CBD,\n\\]\n\\[\nBD=\\sqrt{CD^2+BC^2}=\\sqrt{a^2+(2a)^2}=\\sqrt{5}a,\n\\]\n\\[\n\\therefore \\cos\\angle CDB=\\frac{CD}{BD}=\\frac{a}{\\sqrt{5}a}=\\boxed{\\frac{\\sqrt{5}}{5}}.\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/52977394.png"
  },
  {
    "solution": "\\boxed{15^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $D$ is a point on $AB$, $E$ is the midpoint of $AC$, $DE$ is extended to point $F$ such that $EF=ED$, and $CF$ is drawn. $CF\\parallel AB$. If $\\angle A=70^\\circ$, $\\angle F=35^\\circ$, and $BE\\perp AC$, what is the degree measure of $\\angle BED$?\n",
    "completion": "Solution: From (1) we know that $\\angle A = \\angle ACF$,\\\\\n$\\therefore \\angle ACF = \\angle A = 70^\\circ$,\\\\\nAdditionally, since $\\angle F = 35^\\circ$,\\\\\n$\\therefore \\angle DEC = \\angle F + \\angle ACF = 70^\\circ + 35^\\circ = 105^\\circ$\\\\\nAlso, since $BE \\perp AC$\\\\\n$\\therefore \\angle BEC = 90^\\circ$\\\\\nThus, $\\angle BED = \\angle DEC - \\angle BEC = 105^\\circ - 90^\\circ = \\boxed{15^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53078535_116.png"
  },
  {
    "solution": "\\boxed{-m^2+3m}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the parabola $y=ax^2+2x+c$ intersects the x-axis at points $A(-1,0)$ and $B$, and intersects the y-axis at point $C(0,3)$. Point $M$ is a point on line segment $BC$ (not coinciding with $B$ or $C$), and a line through $M$ is drawn parallel to the y-axis intersecting the parabola at $N$. If the abscissa of point $M$ is $m$, please express the length of $MN$ using an algebraic expression that includes $m$.\n",
    "completion": "\\textbf{Solution}: From (1) it is known that $B(3, 0)$. Assume the equation of line $BC$ is: $y=kx+b$. Substituting $B(3, 0)$, $C(0, 3)$, we get: \\[\n\\begin{cases}\n3k+b=0 \\\\\nb=3\n\\end{cases}\n\\], solving this system gives: \\[\n\\begin{cases}\nk=-1 \\\\\nb=3\n\\end{cases}\n\\]. Therefore, the equation of line $BC$ is $y=-x+3$, thus $M(m, -m+3)$. Furthermore, since $MN \\perp x$ axis, hence $N(m, -m^2+2m+3)$, therefore $MN=\\boxed{-m^2+3m}$ $(0<m<3)$.",
    "image_path": "dataset/mm_math/images/MM_Math/52766412_104.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function \\(y=-2x+8\\) intersects the graph of the function \\(y=\\frac{k}{x}\\) (\\(x>0\\)) at points \\(A(m, 6)\\) and \\(B(n, 2)\\). The line \\(AC\\) is perpendicular to the \\(y\\)-axis at point \\(C\\), and the line \\(BD\\) is perpendicular to the \\(x\\)-axis at point \\(D\\). What is the value of \\(k\\)?\n",
    "completion": "\\textbf{Solution:} Since the graph of the linear function $y=-2x+8$ passes through the points $A(m,6)$ and $B(n,2)$,\\\\\nit follows that $-2m+8=6$ and $-2n+8=2$,\\\\\nfrom which we find $m=1$ and $n=3$.\\\\\nSince the graph of the function $y=\\frac{k}{x}$ (for $x>0$) passes through the points $A(1,6)$ and $B(3,2)$,\\\\\nit follows that $k=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53078921_53.png"
  },
  {
    "solution": "\\boxed{x<2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a linear function, $y=-x-2$, intersects with the graph of $y=kx+b$ at point $P\\left(2, n\\right)$. What then is the solution set of the inequality $kx+b<-x-2$ in terms of $x$?\n",
    "completion": "\\textbf{Solution:} From the graph, it is known that when $x<2$, the graph of the line $y=kx+b$ is below the graph of the line $y=-x-2$. Therefore, the solution set of the inequality $kx+b<-x-2$ with respect to $x$ is $\\boxed{x<2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52211441.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given that $AB$ is the diameter of the circle $\\odot O$, and point $C$ is a point on the circle $\\odot O$. A tangent line to $\\odot O$ passing through point $C$ intersects the extension line of $AB$ at point $P$. Let $D$ be a point on arc $AC$, and connect $BD$, $BC$, and $DC$. If quadrilateral $CDBP$ is a parallelogram, find the measures of $\\angle PCB$ and $\\angle ADC$?\n",
    "completion": "\\textbf{Solution:} Connect CO, AC,\n\nsince quadrilateral CDBP is a parallelogram,\n\nit follows that $\\angle P=\\angle CDB$,\n\nsince AB is the diameter of $\\odot O$,\n\nit follows that $\\angle ACB=\\angle ADB=90^\\circ$, i.e., $\\angle ACO+\\angle OCB=90^\\circ$,\n\nsince PC is tangent to $\\odot O$,\n\nit follows that $\\angle PCO=90^\\circ$, i.e., $\\angle BCP+\\angle OCB=90^\\circ$,\n\ntherefore, $\\angle ACO=\\angle BCP$,\n\nsince $\\angle OAC=\\angle BDC$,\n\nit follows that $\\angle OAC=\\angle OCA=\\angle BCP=\\angle P=\\angle CDB$,\n\nhence $\\angle COB=2\\angle OAC$, $\\angle OBC=2\\angle P$,\n\ntherefore, $\\angle COB=\\angle OBC$,\n\nsince $OB=OC$,\n\nit follows that $\\angle OCB=\\angle OBC$,\n\ntherefore, $\\angle OCB=\\angle OBC=\\angle COB=60^\\circ$,\n\ntherefore, $\\angle PCB=\\frac{1}{2}\\angle CBO=30^\\circ$,\n\ntherefore, $\\angle OAC=\\angle OCA=\\angle BCP=\\angle P=\\angle CDB=30^\\circ$,\n\ntherefore, $\\angle ADC=\\angle ADB+\\angle BDC=90^\\circ+30^\\circ=\\boxed{120^\\circ}$;\n\ntherefore, $\\angle PCB=\\boxed{30^\\circ}$, $\\angle ADC=120^\\circ$.",
    "image_path": "dataset/mm_math/images/MM_Math/51319037_128.png"
  },
  {
    "solution": "\\boxed{(-2, \\infty) \\cup (0, 1)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the Cartesian coordinate system, suppose the inverse proportion function $y_{1}=\\frac{k_{1}}{x}\\left(k_{1}\\ne 0\\right)$ and the linear function $y_{2}=k_{2}x+b$ ($k_{2}\\neq 0$) have graphs passing through points $A$ and $B$, where the coordinate of point $A$ is $(1, m)$ and the coordinate of point $B$ is $(-2, -2)$. When $y_{1}>y_{2}$, directly write the range of $x$.\n",
    "completion": "\\textbf{Solution:} From the given, we have $x > -2$ or $0 < x < 1$,\n\n$\\therefore$ the value range of $x$ is \\boxed{(-2, \\infty) \\cup (0, 1)}.",
    "image_path": "dataset/mm_math/images/MM_Math/52457688_182.png"
  },
  {
    "solution": "\\boxed{140^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ is inscribed in $\\odot O$, with $CA=CB$. If $\\angle C=40^\\circ$, what is the degree measure of $\\overset{\\frown}{BC}$?\n",
    "completion": "\\textbf{Solution:} Given $\\because$CA＝CB, $\\angle$C＝$40^\\circ$,\\\\\n$\\therefore$$\\angle$A＝$\\angle$B＝$\\frac{1}{2}\\times(180^\\circ-40^\\circ)＝70^\\circ$,\\\\\n$\\therefore$ the degree of arc $\\overset{\\frown}{BC}$ is $\\boxed{140^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52755170.png"
  },
  {
    "solution": "\\boxed{}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, $\\angle B=30^\\circ$, point $D$ is on the extension of $BA$, $\\angle CDB=45^\\circ$, point $M$ is on side $BC$, and quadrilateral $ACMN$ is a square. What is the degree measure of $\\angle ACD$?\n",
    "completion": "\\textbf{Solution:} Since the information provided by the problem is insufficient, a direct solution cannot be offered. Therefore, a specific mathematical solution process cannot be provided. \\boxed{}",
    "image_path": "dataset/mm_math/images/MM_Math/53182827_160.png"
  },
  {
    "solution": "\\boxed{24}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the goal is to draw a painting with a rectangular frame on the gable roof of a building to enhance its beauty. It is known that $BC = 12\\,m$ and the height $AD = 8\\,m$. If the painting's area is to be maximized, and one side of the rectangle is on $BC$, with the other two vertices on $AB$ and $AC$ respectively, find the dimensions of the frame's length and width as well as the maximum area.\n",
    "completion": "\\textbf{Solution:} Let the length of the rectangle be $PN=x$ and $PQ=y$. Then $AE=8-y$. \\\\\nSince $\\triangle APN\\sim \\triangle ABC$, \\\\\nit follows that $\\frac{PN}{BC}=\\frac{AE}{AD}$, \\\\\ni.e., $\\frac{x}{12}=\\frac{8-y}{8}$, \\\\\nyielding $y=8-\\frac{2}{3}x$, \\\\\nTherefore, $S=xy=x\\left(8-\\frac{2}{3}x\\right)=-\\frac{2}{3}x^2+8x=-\\frac{2}{3}\\left(x-6\\right)^2+24$ \\\\\nTherefore, when the length is $6$ meters and the width is $4$ meters, the maximum area of the rectangle that can be drawn is $\\boxed{24}$ square meters.",
    "image_path": "dataset/mm_math/images/MM_Math/53080410_113.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, lines AB and CD intersect at point O, with OE bisecting $\\angle AOC$. If $\\angle BOD=60^\\circ$, what is the measure of $\\angle AOE$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle BOD=60^\\circ$,\\\\\nit follows that $\\angle AOC=\\angle BOD=60^\\circ$,\\\\\nsince OE bisects $\\angle AOC$,\\\\\nthus $\\angle AOE= \\frac{1}{2}\\angle AOC=30^\\circ$,\\\\\ntherefore, the answer is: $\\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55190265.png"
  },
  {
    "solution": "\\boxed{-\\frac{1}{2}x^2+\\frac{5}{2}x-2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola passes through points \\(A(4, 0)\\), \\(B(1, 0)\\), and \\(C(0, -2)\\). Find the analytical expression of this parabola.\n",
    "completion": "\\textbf{Solution:} Given that the parabola passes through point $C(0, -2)$,\\\\\nwe can assume the equation of the parabola to be $y=ax^2+bx-2$.\\\\\nSubstituting points $A(4, 0)$, $B(1, 0)$ into it,\\\\\nwe get $\\begin{cases}16a+4b-2=0\\\\ a+b-2=0\\end{cases}$,\\\\\nfrom which we find $\\begin{cases}a=-\\frac{1}{2}\\\\ b=\\frac{5}{2}\\end{cases}$,\\\\\ntherefore, the equation of this parabola is $y=-\\frac{1}{2}x^2+\\frac{5}{2}x-2=\\boxed{-\\frac{1}{2}x^2+\\frac{5}{2}x-2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52513841_161.png"
  },
  {
    "solution": "\\boxed{100^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $OE$ is the angle bisector of $\\angle AOC$, and $OD$ is the angle bisector of $\\angle BOC$. If $\\angle AOE=70^\\circ$ and $\\angle COD=20^\\circ$, what is the degree measure of $\\angle AOB$?\n",
    "completion": "\\textbf{Solution:} Since $OD$ is the angle bisector of $\\angle BOC$ and $\\angle COD=20^\\circ$,\\\\\n$\\therefore \\angle BOC=2\\angle COD=40^\\circ$.\\\\\nSince $OE$ is the angle bisector of $\\angle AOC$,\\\\\n$\\therefore \\angle AOC=2\\angle AOE=140^\\circ$.\\\\\n$\\therefore \\angle AOB=\\angle AOC-\\angle BOC=140-40^\\circ=\\boxed{100^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/47111555_199.png"
  },
  {
    "solution": "\\boxed{105^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points A, B, and C are on the same straight line, and BD bisects $\\angle ABE$, $\\angle EBC=30^\\circ$. What is the degree measure of $\\angle DBC$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle EBC=30^\\circ$,\\\\\ntherefore $\\angle ABE=180^\\circ-30^\\circ=150^\\circ$,\\\\\nsince BD bisects $\\angle ABE$,\\\\\nthus $\\angle DBE= \\frac{1}{2}\\angle ABE=75^\\circ$,\\\\\ntherefore $\\angle DBC=\\angle DBE+\\angle EBC=\\boxed{105^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55190283.png"
  },
  {
    "solution": "\\boxed{\\frac{2}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in parallelogram $ABCD$, point $E$ is on side $BC$, and $AE$ intersects diagonal $BD$ at point $F$. If $BE:BC=2:3$, then what is the value of $\\frac{EF}{AE}$?\n",
    "completion": "\\textbf{Solution}: Since quadrilateral ABCD is a parallelogram, \\\\\ntherefore BC$\\parallel$AD and BC=AD, \\\\\ntherefore $\\frac{BE}{BC} = \\frac{BE}{AD}$, $\\triangle BEF\\sim \\triangle DAF$, \\\\\ntherefore $\\frac{BE}{AD}=\\frac{BF}{FD}=\\frac{EF}{AF}=\\frac{2}{3}$, \\\\\ntherefore $AF=\\frac{3}{2}EF$, \\\\\ntherefore $AE=AF+EF=\\frac{5}{2}EF$, \\\\\ntherefore $\\frac{EF}{AE}=\\boxed{\\frac{2}{5}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51368290.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point \\(A\\) is on the graph of the inverse proportion function \\(y=\\frac{1}{x}\\left(x>0\\right)\\), point \\(B\\) is on the graph of the inverse proportion function \\(y=\\frac{3}{x}\\left(x>0\\right)\\), and \\(AB\\) is parallel to the \\(x\\) axis. Points \\(C\\) and \\(D\\) are on the \\(x\\) axis. If the quadrilateral \\(ABCD\\) is a rectangle, what is its area?\n",
    "completion": "\\textbf{Solution:} Extend BA to intersect the y-axis at point E, \\\\\n$\\because$ Quadrilateral ABCD is a rectangle, and AB$\\parallel$x-axis, with points C, D on the x-axis, \\\\\n$\\therefore$ AE$\\perp$y-axis, \\\\\n$\\therefore$ Quadrilaterals ADOE and BCOE are rectangles, \\\\\n$\\because$ point A is on the graph of the inverse proportion function y=$\\frac{1}{x}$, and point B is on the graph of the inverse proportion function y=$\\frac{3}{x}$, \\\\\n$\\therefore$ S$_{\\text{rectangle}}$ADOE=1, S$_{\\text{rectangle}}$BCOE=3, \\\\\n$\\therefore$ S$_{\\text{rectangle}}$ABCD=S$_{\\text{rectangle}}$BCOE−S$_{\\text{rectangle}}$ADOE=3−1=\\boxed{2}.",
    "image_path": "dataset/mm_math/images/MM_Math/51458242.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{5}}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in a $6\\times6$ square grid, where each small square has a side length of $1$, points $A$, $B$, and $C$ are all located on the grid intersections, and $\\odot O$ is the circumcircle of $\\triangle ABC$, what is the sine value of $\\angle BAC$?\n",
    "completion": "\\textbf{Solution:} Let $E$ be the midpoint of $BC$, and connect $BO$, $CO$, and $OE$. \\\\\n$\\because$ $\\angle BAC = \\frac{1}{2}\\angle BOC$\\\\\n$\\therefore$ $\\angle BAC = \\angle BOE$\\\\\n$\\because$ $BE=1$, $EO=2$\\\\\n$\\therefore$ $BO=\\sqrt{BE^2+EO^2}=\\sqrt{5}$\\\\\n$\\therefore$ $\\sin\\angle BAC=\\sin\\angle BOE=\\frac{BE}{BO}=\\frac{1}{\\sqrt{5}}=\\boxed{\\frac{\\sqrt{5}}{5}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51358246.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of $\\odot O$, and $CD$ is a chord of $O$ with $CD\\perp AB$ at point $E$. $\\angle BCO = \\angle D$; If $BE=9\\text{cm}$ and $CD=6\\text{cm}$, what is the radius of $\\odot O$?\n",
    "completion": "\\textbf{Solution:} Let the radius be $x$, then $OE=9-x$.\\\\\nSince diameter $AB \\perp CD$ and $CD= 6$,\\\\\nit follows that $CE=3$.\\\\\nIn $\\triangle OCE$, we have $OC^{2}=OE^{2}+CE^{2}$,\\\\\ntherefore $x^{2}=(9-x)^{2}+3^{2}$,\\\\\nhence $x=\\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52808502_125.png"
  },
  {
    "solution": "\\boxed{x>4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown, rectangle $OCBA$ is positioned in the Cartesian coordinate system, where point $C$ is on the positive half of the $x$-axis, and point $A$ is on the positive half of the $y$-axis. Given that the coordinates of point $B$ are $(4,2)$, the graph of the inverse proportion function $y=\\frac{k}{x}$ passes through the midpoint $D$ of $AB$ and intersects $BC$ at point $E$. Let the equation of line $DE$ be $y=mx+n$. Connecting $OD$ and $OE$, directly write out the solution set for the inequality $\\frac{k}{x}>mx+n$.\n",
    "completion": "\\textbf{Solution:} The solution set is \\boxed{0<x<2} or \\boxed{x>4}.",
    "image_path": "dataset/mm_math/images/MM_Math/52536619_157.png"
  },
  {
    "solution": "\\boxed{\\frac{12}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the rhombus $ABCD$, the diagonals $AC$ and $BD$ intersect at point $O$, with $AC=8$, $BD=6$, and $OE\\perp BC$ with the foot of the perpendicular being point $E$. Find the length of $OE$.\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rhombus,\\\\\nit follows that AC $\\perp$ BD, OB = OD = $\\frac{1}{2}$BD = 3, OA = OC = $\\frac{1}{2}$AC = 4,\\\\\nin $\\triangle OBC$, since OB = 3, OC = 4,\\\\\nthus BC = $\\sqrt{3^2 + 4^2} = 5$,\\\\\nsince OE $\\perp$ BC,\\\\\nit follows that $\\frac{1}{2}$OE $\\cdot$ BC = $\\frac{1}{2}$ OB $\\cdot$ OC,\\\\\nthus OE = $\\frac{3 \\times 4}{5} = \\boxed{\\frac{12}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51684389.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, point $O$ is the midpoint of diagonal $AC$, and a line $EF\\perp AC$ is drawn through point $O$, intersecting $BC$ at point $E$ and $AD$ at point $F$. Lines $AE$ and $CF$ are connected. The quadrilateral $AECF$ forms a rhombus. When $OB$ is connected, given $AB=4$ and $AF=5$, find the length of $OB$.\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect $BO$, \\\\\n$\\because AB=4$, $AF=AE=EC=5$, \\\\\n$\\therefore BE= \\sqrt{AE^{2}-AB^{2}}=\\sqrt{25-16}=3$, \\\\\n$\\therefore BC=8$, \\\\\n$\\therefore AC= \\sqrt{AB^{2}+BC^{2}}=\\sqrt{16+64}=4\\sqrt{5}$, \\\\\n$\\because AO=CO$, $\\angle ABC=90^\\circ$, \\\\\n$\\therefore BO= \\frac{1}{2} AC= 2\\sqrt{5}$, hence the length of $OB$ is $\\boxed{2\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50496635_104.png"
  },
  {
    "solution": "\\boxed{20^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, AB is the diameter of the circle $O$, and point C is a point on circle $O$, $\\angle ABC=70^\\circ$. What is the measure of $\\angle BAC$?\n",
    "completion": "\\textbf{Solution:} Since AB is the diameter of circle O, and point C is a point on circle O,\\\\\nit follows that $\\angle ACB=90^\\circ$,\\\\\nSince $\\angle ABC=70^\\circ$,\\\\\nit follows that $\\angle BAC=180^\\circ-\\left(\\angle ACB+\\angle ABC\\right)=180^\\circ-90^\\circ-70^\\circ=\\boxed{20^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52796833.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ABC=90^\\circ$, $\\angle A=60^\\circ$, D is a point on side $AC$, $\\angle DBC=30^\\circ$, $AC=12\\, \\text{cm}$, what is the perimeter of $\\triangle ABD$?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABC\\cong \\triangle ADE$,\\\\\nit follows that $\\angle E=\\angle C$,\\\\\nIn $\\triangle ABC$, $\\angle BAC=80^\\circ$, $\\angle B=40^\\circ$,\\\\\ntherefore, $\\angle E=\\angle C=180^\\circ-80^\\circ-40^\\circ=\\boxed{60^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52826729.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system \\(xOy\\), the graph of the linear function \\(y=x\\) intersects the graph of the quadratic function \\(y=-x^2+bx\\) (where \\(b\\) is a constant) at points \\(O\\) and \\(A\\). The coordinates of point \\(A\\) are \\((3, m)\\). What is the value of \\(m\\) and the expression of the quadratic function?\n",
    "completion": "\\textbf{Solution:} \n\nSubstitute the coordinates of point A, $(3,m)$, into the linear function $y=x$ to get:\\\\\n$m=3$,\\\\\n$\\therefore A(3,3)$,\\\\\nSubstituting the coordinates of point A, $(3,3)$, into the quadratic function $y=-x^{2}+bx$ gives:\\\\\n$3=-9+3b$,\\\\\nSolving for $b$ yields: $b=4$,\\\\\n$\\therefore y=-x^{2}+4x$,\\\\\nAnswer: The value of $m$ is $\\boxed{3}$, and the expression for the quadratic function is: $y=-x^{2}+4x$;",
    "image_path": "dataset/mm_math/images/MM_Math/52872343_271.png"
  },
  {
    "solution": "\\boxed{4\\,cm}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $C$ is on line segment $AD$, point $B$ is the midpoint of $CD$, and $AD=9\\,cm$, $BC=2\\,cm$. If point $E$ is on line segment $AD$ and $AE=3\\,cm$, what is the length of $BE$?\n",
    "completion": "\\textbf{Solution:} \\\\\nSince $AB = AC + BC = 7\\,cm$, \\\\\n$AE = 3\\,cm$, \\\\\nTherefore $BE = AB - AE = 4\\,cm$, \\\\\nhence the length of $BE$ is \\boxed{4\\,cm}.",
    "image_path": "dataset/mm_math/images/MM_Math/53066707_137.png"
  },
  {
    "solution": "\\boxed{y_1=x-4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the graph of the linear function $y_1=kx+b$ passes through the point $(-1, -5)$ and intersects the $y$-axis at point $B$, and it intersects the graph of the direct proportion function $y_2=\\frac{1}{3}x$ at point $A(6, a)$. Find the equation of the line $y_1=kx+b$.\n",
    "completion": "\\textbf{Solution:} Substitute $A(6, 2)$ into $y_2=\\frac{1}{3}x$ to get $a=6\\times \\frac{1}{3}=2$,\\\\\n$\\therefore A(6, 2)$.\\\\\n$\\because$ Points $(-1, -5)$ and $A(6, 2)$ are on the graph of the linear function $y_1=kx+b$,\\\\\n$\\therefore \\left\\{\\begin{array}{l}\n6k+b=2 \\\\\n-k+b=-5\n\\end{array}\\right.$,\\\\\nsolving this, we get $\\left\\{\\begin{array}{l}\nk=1 \\\\\nb=-4\n\\end{array}\\right.$,\\\\\n$\\therefore \\boxed{y_1=x-4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50595463_134.png"
  },
  {
    "solution": "\\boxed{\\frac{4\\sqrt{42}}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, with C as the common vertex, in $\\triangle ABC$ and $\\triangle CED$, $\\angle ACB=\\angle CDE=90^\\circ$ and $\\angle A=\\angle DCE=30^\\circ$, and with point D on line segment $AB$, what is $\\angle ABE$? If $AC=10$ and $CD=9$, what is the length of $BE$?\n",
    "completion": "\\textbf{Solution:}\n\nGiven that $\\angle ACB=\\angle CDE=90^\\circ$ and $\\angle A=\\angle DCE=30^\\circ$,\\\\\nit follows that $\\angle DBC=\\angle DEC=60^\\circ$,\\\\\nhence, points B, C, D, and E are concyclic,\\\\\ntherefore $\\angle DBE=\\angle DCE=30^\\circ$,\\\\\nwhich leads us to conclude that $\\angle ABE=\\boxed{30^\\circ}$,\\\\\nconsidering that in right triangle $ABC$, $\\tan\\angle BAC=\\frac{BC}{AC}$,\\\\\nthus $BC=AC\\cdot\\tan30^\\circ=10\\times \\frac{\\sqrt{3}}{3}=\\frac{10\\sqrt{3}}{3}$,\\\\\nsimilarly, in right triangle $CED$, $\\cos\\angle DCE=\\frac{CD}{CE}$,\\\\\ntherefore $CE=\\frac{CD}{\\cos30^\\circ}=\\frac{9}{\\frac{\\sqrt{3}}{2}}=6\\sqrt{3}$,\\\\\ngiven $\\angle ABC=60^\\circ$ and $\\angle ABE=30^\\circ$,\\\\\nit follows that $\\angle CBE=\\angle ABC+\\angle ABE=90^\\circ$,\\\\\nconsequently, $BE=\\sqrt{CE^2-BC^2}=\\sqrt{108-\\frac{300}{9}}=\\boxed{\\frac{4\\sqrt{42}}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53102155.png"
  },
  {
    "solution": "\\boxed{13}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $C$ is a point on the semicircle $O$ with $AB$ as its diameter. Connect $AC$ and $BC$. Construct squares $ACDE$ and $BCFG$ externally on sides $AC$ and $BC$, respectively. Let $M$, $N$, $P$, and $Q$ be the midpoints of $DE$, $FG$, arc $AC$, and arc $BC$, respectively. If $MP + NQ = 14$ and $AC + BC = 18$, what is the length of $AB$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect OP and OQ which intersect AC and BC at points I and H respectively,\\\\\n$\\because$ The midpoints of DE, FG, $\\overset{\\frown}{AC}$, and $\\overset{\\frown}{BC}$ are M, N, P, and Q respectively,\\\\\n$\\therefore$ OP $\\perp$ AC, OQ $\\perp$ BC,\\\\\n$\\therefore$ H and I are the midpoints of AC and BD,\\\\\n$\\therefore$ OH+OI=$\\frac{1}{2}$(AC+BC)=9,\\\\\n$\\therefore$ MH+NI=AC+BC=18, MP+NQ=14,\\\\\n$\\therefore$ PH+QI=18−14=4,\\\\\n$\\therefore$ AB=OP+OQ=OH+OI+PH+QI=9+4=\\boxed{13},",
    "image_path": "dataset/mm_math/images/MM_Math/52871878.png"
  },
  {
    "solution": "\\boxed{4.8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, quadrilateral ABCD is a rhombus with diagonal AC = 8 and DB = 6. If DH $\\perp$ AB at point H, what is the length of DH?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rhombus,\\\\\nit follows that AC$\\perp$BD, and OA＝OC＝ $\\frac{1}{2}$ AC＝4, OB＝OD＝3,\\\\\nthus $AB=\\sqrt{OA^{2}+OB^{2}}=\\sqrt{3^{2}+4^{2}}=5$,\\\\\nthus $S_{\\text{rhombus} ABCD}$＝ $\\frac{1}{2}$ AC$\\cdot$BD＝AB$\\cdot$DH,\\\\\nthus DH＝ $\\frac{AC\\cdot BD}{2AB}$ ＝$\\boxed{4.8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52303916.png"
  },
  {
    "solution": "\\boxed{\\frac{15}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AF$ and $AG$ are the altitudes of $\\triangle ABC$ and $\\triangle ADE$, respectively, with $\\angle BAF = \\angle DAG$. $\\triangle ABC \\sim \\triangle ADE$; If $DE = 3$ and $\\frac{AD}{AB} = \\frac{2}{5}$, find the length of $BC$.\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABC\\sim \\triangle ADE$,\\\\\n$\\therefore \\frac{AD}{AB}=\\frac{DE}{BC}$,\\\\\nand given $\\frac{AD}{AB}=\\frac{2}{5}$, $DE=3$,\\\\\n$\\therefore \\frac{2}{5}=\\frac{3}{BC}$,\\\\\n$\\therefore BC=\\boxed{\\frac{15}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51212669_77.png"
  },
  {
    "solution": "\\boxed{\\left(\\frac{4}{3}, 0\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the coordinate of point $A$ is $(5, 0)$. Point $B$ is located on the positive half of the $y$-axis ($OB<OA$). Line segment $AB$ is rotated clockwise around point $A$ by $90^\\circ$ to obtain line segment $AC$. Perpendiculars are drawn from point $C$ to the $x$-axis and the $y$-axis, with the feet of the perpendiculars being $D$ and $E$ respectively. Under the conditions of (2), point $P$ is a point on the extension of $OE$. Line $PC$ is connected, and the bisector of $\\angle PCD$ intersects the $x$-axis at point $F$. If $\\triangle PCF$ is an isosceles triangle, what are the coordinates of point $F$?\n",
    "completion": "\\textbf{Solution:} Let the coordinates of point $F$ be $\\left(m, 0\\right)$,\n\n$\\because$ $CF$ bisects $\\angle PCD$,\n\n$\\therefore$ $\\angle PCF=\\angle DCF$,\n\n(1) When $PF=PC$, then $\\angle PFC=\\angle PCF=\\angle DCF$,\n\n$\\therefore$ $PF\\parallel CD$,\n\n$\\therefore$ $F$ coincides with $O$,\n\n$\\therefore$ $F\\left(0, 0\\right)$;\n\n(2) When $PF=CF$, draw $FM\\perp CP$ at point $F$ with the foot $M$,\n\nthen $PM=CM$, $\\angle CMF=\\angle CDF=90^\\circ$,\n\nAlso $\\because$ $\\angle PCF=\\angle DCF$, $CF=CF$,\n\n$\\therefore$ $\\triangle CMF\\cong \\triangle CDF$,\n\n$\\therefore$ $CM=CD=5$,\n\n$\\therefore$ $PC=10$, in $Rt\\triangle PCE$, by the Pythagorean theorem, $PE=6$,\n\n$\\therefore$ $OP=11$, in $Rt\\triangle POF$, $PF^{2}=OF^{2}+OP^{2}$,\n\nin $Rt\\triangle CDF$, $CF^{2}=CD^{2}+DF^{2}$,\n\n$\\therefore$ $OF^{2}+OP^{2}=CD^{2}+DF^{2}$,\n\n$\\therefore$ $\\left(-m\\right)^{2}+11^{2}=5^{2}+\\left(8-m\\right)^{2}$,\n\nSolving gives $m=-2$,\n\n$\\therefore$ $F\\left(-2, 0\\right)$;\n\n(3) When $CP=CF$, extend $CF$ to intersect the $y$-axis at point $N$,\n\n$\\because$ $CD\\parallel PN$, and $\\angle PCF=\\angle DCF$,\n\n$\\therefore$ $\\angle CNP=\\angle PCF=\\angle DCF$,\n\n$\\therefore$ $PN=PC$, draw $PQ\\perp CF$ at point $P$ with the foot $Q$,\n\nthen $CQ=QN$, $\\angle PQC=\\angle CDF=90^\\circ$,\n\n$\\therefore$ $\\triangle PCQ\\cong \\triangle FCD$,\n\n$\\therefore$ $CQ=CD=5$,\n\n$\\therefore$ $CN=10$, in $Rt\\triangle CEN$, by the Pythagorean theorem, $EN=6$,\n\n$\\therefore$ $ON=1$,\n\n$\\therefore$ $N\\left(0, -1\\right)$,\n\n$\\because$ $C\\left(8, 5\\right)$, let the equation of line $CN$ be $y=kx+b$,\n\nthen $\\begin{cases}8k+b=5\\\\ b=-1\\end{cases}$, solving gives $\\begin{cases}k=\\frac{3}{4}\\\\ b=-1\\end{cases}$,\n\n$\\therefore$ the equation of line $CN$ is $y=\\frac{3}{4}x-1$,\n\nWhen $y=0$, solving gives $x=\\frac{4}{3}$,\n\n$\\therefore$ $F\\left(\\frac{4}{3}, 0\\right)$.\n\nIn summary, when $\\triangle PCF$ is an isosceles triangle, the coordinates of point $F$ are $\\boxed{\\left(0, 0\\right)}$, or $\\boxed{\\left(-2, 0\\right)}$, or $\\boxed{\\left(\\frac{4}{3}, 0\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50545780_116.png"
  },
  {
    "solution": "\\boxed{9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the perimeter of $\\triangle ABC$ is 18. D and E are the midpoints of sides AB and BC, respectively. What is the perimeter of $\\triangle BDE$?\n",
    "completion": "\\textbf{Solution:} Since the perimeter of $\\triangle ABC$ is 18,\\\\\nwe have AB+AC+BC=18,\\\\\nSince points D and E are the midpoints of sides AB and BC respectively,\\\\\nthen DE is the midsegment of $\\triangle ABC$, BD=$\\frac{1}{2}$AB, BE=$\\frac{1}{2}$BC,\\\\\nthus, DE=$\\frac{1}{2}$AC,\\\\\ntherefore, the perimeter of $\\triangle DBE$=BD+BE+DE=$\\frac{1}{2}$(AB+AC+BC)=$\\boxed{9}$,",
    "image_path": "dataset/mm_math/images/MM_Math/52386771.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that quadrilateral $ABCD$ is a rhombus, $E$ is a point moving on diagonal $AC$, $DE$ is extended to intersect the extension line of $AB$ at point $F$, and $BE$ is connected. $\\angle AFD = \\angle EBC$. If $DE = EC$ and $BE \\perp AF$, find the degree measure of $\\angle DAB$?\n",
    "completion": "\\textbf{Solution:} Given rhombus ABCD, it follows that $\\angle DAB = \\angle DCB$ and DC$\\parallel$AB implies $\\angle CBF = \\angle DCB$. From (1) it has been proven that $\\triangle DCE \\cong \\triangle BCE$, thus $\\angle EDC = \\angle EBC$ and $\\angle DCE = \\angle BCE$. Since DE=EC, it results in $\\angle EDC = \\angle ECD$, hence $\\angle EDC = \\angle ECD = \\angle EBC = \\angle BCE$. Let $\\angle EDC = \\angle ECD = \\angle EBC = \\angle BCE = x$, thus $\\angle CBF = \\angle DCB = 2x$. Given BE$\\perp$AF, it leads to $\\angle CBF + \\angle EBC = 2x + x = 90^\\circ$, consequently $x = 30^\\circ$, which means $\\angle DAB = \\angle DCB = 2x = \\boxed{60^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51510300_134.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $OB$, $OC$, and $OD$ are three rays within $\\angle AOE$, $OB$ bisects $\\angle AOE$, $OD$ bisects $\\angle COE$. If $\\angle AOE=136^\\circ$ and $AO\\perp CO$, what is the degree measure of $\\angle BOD$?\n",
    "completion": "\\textbf{Solution:} Given the construction,\\\\\n$\\because$ OB bisects $\\angle AOE$, OD bisects $\\angle COE$,\\\\\n$\\therefore \\angle BOE=\\frac{1}{2}\\angle AOE$, $\\angle DOE=\\frac{1}{2}\\angle COE$.\\\\\n$\\because \\angle BOD=\\angle BOE-\\angle DOE$,\\\\\n$\\therefore \\angle BOD=\\angle BOE-\\angle DOE=\\frac{1}{2}\\left(\\angle AOE-\\angle COE\\right)=\\frac{1}{2}\\angle AOC$,\\\\\n$\\because$ AO$\\perp$CO,\\\\\n$\\therefore \\angle AOC=90^\\circ$,\\\\\n$\\therefore \\angle BOD=\\boxed{45^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/54927250_123.png"
  },
  {
    "solution": "\\boxed{108^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AD$ and $BE$ are the altitude and angle bisector of $\\triangle ABC$, respectively, with $\\angle BAC=86^\\circ$ and $\\angle C=58^\\circ$. Find the measure of $\\angle AOB$?\n",
    "completion": "\\textbf{Solution:} Given $\\angle BAC=86^\\circ$, $\\angle C=58^\\circ$\\\\\nTherefore, $\\angle ABC=180^\\circ-86^\\circ-58^\\circ=36^\\circ$\\\\\nSince $AD$ and $BE$ are the altitude and angle bisector of $\\triangle ABC$ respectively,\\\\\nTherefore, $\\angle OBD=\\frac{1}{2}\\angle ABC=18^\\circ$, $\\angle ADB=90^\\circ$\\\\\nHence, $\\angle AOB=\\angle OBD+\\angle ADB=18^\\circ+90^\\circ=\\boxed{108^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/52950323.png"
  },
  {
    "solution": "\\boxed{99^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, quadrilateral $ABCD$ is inscribed in $\\odot O$, with $AD\\parallel BC$. $BD$ bisects $\\angle ABC$, and $\\angle A=126^\\circ$. What is the degree measure of $\\angle BDC$?\n",
    "completion": "\\textbf{Solution:} Since $AD \\parallel BC$ and $\\angle A=126^\\circ$,\\\\\nit follows that $\\angle ABC=180^\\circ-\\angle A=180^\\circ-126^\\circ=54^\\circ$,\\\\\nGiven that $BD$ bisects $\\angle ABC$,\\\\\nwe have $\\angle DBC=54^\\circ\\div 2=27^\\circ$,\\\\\nSince $AD \\parallel BC$,\\\\\nit follows that $\\angle ADB=\\angle DBC=27^\\circ$,\\\\\nGiven that quadrilateral $ABCD$ is circumscribed around circle $O$,\\\\\nit follows that $\\angle ADC=180^\\circ-54^\\circ=126^\\circ$,\\\\\nTherefore, $\\angle BDC=\\angle ADC-\\angle ADB=126^\\circ-27^\\circ=\\boxed{99^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51402196.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, it is known that $AB=3\\sqrt{2}$, $\\angle B=45^\\circ$, $\\angle C=30^\\circ$, then what is the length of $AC$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $AD \\perp BC$ at point D.\\\\\n$\\therefore$ $\\angle ADB = \\angle ADC = 90^\\circ$,\\\\\n$\\because$ $\\angle B = 45^\\circ$,\\\\\n$\\therefore$ $\\angle BAD = \\angle B = 45^\\circ$,\\\\\n$\\therefore$ AD = BD,\\\\\n$\\because$ $AB^{2} = AD^{2} + BD^{2}$, AB = $3\\sqrt{2}$,\\\\\n$\\therefore$ AD = BD = 3,\\\\\nAlso, $\\because$ $\\angle C = 30^\\circ$,\\\\\n$\\therefore$ AC = 2AD = \\boxed{6}",
    "image_path": "dataset/mm_math/images/MM_Math/51213066.png"
  },
  {
    "solution": "\\boxed{\\frac{8}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the length of the line segment $AB=8$, point $C$ is the midpoint of line segment $AB$, and point $D$ is the midpoint of line segment $BC$. There is a point $E$ on the line segment $AC$, where $CE=\\frac{1}{3}BC$. Find the length of $AE$.\n",
    "completion": "\\textbf{Solution:} \\\\\nSince $CE = \\frac{1}{3}BC$ and $BC = 4$, \\\\\nit follows that $CE = \\frac{4}{3}$, \\\\\ntherefore $AE = AC - CE = 4 - \\frac{4}{3} = \\boxed{\\frac{8}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50733799_146.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a rectangular sheet $OABC$ is placed in the Cartesian coordinate system, with $OA$ and $OC$ lying on the positive halves of the x-axis and y-axis, respectively. Connect $AC$, and let $AC=4\\sqrt{5}$ and $OA=2CO$. Fold the sheet $OABC$ so that point A coincides with point C (the fold line is $EF$). Find the area of the overlapping part of the sheet after folding.\n",
    "completion": "\\textbf{Solution:} Given that $AE=CE$ from the fold, let $AE=CE=y$, then $OE=8-y$. In $\\triangle OCE$, by the Pythagorean theorem, we have $OE^2+OC^2=CE^2$, \\\\\n$\\therefore$ $(8-y)^2+4^2=y^2$, solving this yields $y=5$\\\\\n$\\therefore$ $AE=CE=5$\\\\\nIn the rectangle $OABC$,\\\\\n$\\because$ $BC \\parallel OA$,\\\\\n$\\therefore$ $\\angle CFE=\\angle AEF$, and from the fold $\\angle AEF=\\angle CEF$,\\\\\n$\\therefore$ $\\angle CFE=\\angle CEF$\\\\\n$\\therefore$ $CF=CE=5$\\\\\n$\\therefore$ $Area_{\\triangle CEF}=\\frac{1}{2}CF \\cdot OC=\\frac{1}{2} \\times 5 \\times 4=\\boxed{10}$\\\\\nTherefore, the area of the overlapping section is \\boxed{10}.",
    "image_path": "dataset/mm_math/images/MM_Math/52342628_123.png"
  },
  {
    "solution": "\\boxed{\\sqrt{305}\\pi}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points A, C, D are all on the circle $\\odot O$, and point B is inside $\\odot O$. Furthermore, $AB \\perp BC$ at point B, and $BC \\perp CD$ at point C. Given that $AB = 6$, $BC = 12$, and $CD = 4$, what is the circumference of $\\odot O$?\n",
    "completion": "\\textbf{Solution:} Extend $CB$ to meet $\\odot O$ at point $E$, connect $AE$, $DE$, $AC$, and have $AD$ intersect $CB$ at point $F$.\\\\\nTherefore, $AC=\\sqrt{AB^{2}+BC^{2}}=6\\sqrt{5}$.\\\\\nSince $AB\\perp BC$ and $BC\\perp CD$,\\\\\nTherefore, $AB\\parallel CD$,\\\\\nTherefore, $\\triangle ABF\\sim \\triangle DCF$,\\\\\nTherefore, $\\frac{AB}{CD}=\\frac{BF}{CF}$,\\\\\nTherefore, $\\frac{BF}{CF}=\\frac{6}{4}=\\frac{3}{2}$.\\\\\nSince $BC=BF+CF=12$,\\\\\nTherefore, $BF=\\frac{3}{5}\\times 12=\\frac{36}{5}$, $CF=\\frac{2}{5}\\times 12=\\frac{24}{5}$,\\\\\nTherefore, $AF=\\sqrt{AB^{2}+BF^{2}}=\\sqrt{6^{2}+\\left(\\frac{36}{5}\\right)^{2}}=\\frac{6}{5}\\sqrt{61}$, $DF=\\sqrt{CF^{2}+CD^{2}}=\\sqrt{\\left(\\frac{24}{5}\\right)^{2}+4^{2}}=\\frac{4}{5}\\sqrt{61}$,\\\\\nTherefore, $AD=AF+DF=2\\sqrt{61}$.\\\\\nSince $\\angle DCE=90^\\circ$,\\\\\nTherefore, DE is the diameter,\\\\\nTherefore, $\\angle EAD=\\angle ABC=90^\\circ$.\\\\\nSince $\\angle EDA=\\angle ACB$,\\\\\nTherefore, $\\triangle EDA\\sim \\triangle ACB$,\\\\\nTherefore, $\\frac{AD}{BC}=\\frac{DE}{AC}$, that is, $\\frac{2\\sqrt{61}}{12}=\\frac{DE}{6\\sqrt{5}}$,\\\\\nSolving gives: $DE=\\sqrt{305}$, that is, the diameter of $\\odot O$ is $\\sqrt{305}$,\\\\\nTherefore, the circumference of $\\odot O$ is $\\boxed{\\sqrt{305}\\pi}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53147979.png"
  },
  {
    "solution": "\\boxed{(-1, 4)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated, it is known that the graph of the quadratic function $y=-x^2+bx+c$ passes through points $A(-2, -1)$ and $B(0, 7)$. On the upper side of the x-axis, a line $l$ is drawn parallel to the x-axis, intersecting the parabola at points $C$ and $D$ (with point $C$ being on the left side of the axis of symmetry). Perpendiculars are drawn from points $C$ and $D$ to the x-axis, with foot points at $F$ and $E$, respectively. When the rectangle $CDEF$ becomes a square, what are the coordinates of point $C$?\n",
    "completion": "\\textbf{Solution:} When the rectangle $CDEF$ is a square,\\\\\nlet the coordinates of point $C$ be $(x, -x^2+2x+7)$,\\\\\n$\\therefore$ the coordinates of point $D$ are $(-x^2+3x+7, -x^2+2x+7)$,\\\\\n$\\because$ the axis of symmetry is the line $x=1$, and the distance from $D$ to the axis of symmetry is equal to the distance from $C$ to the axis of symmetry,\\\\\n$\\therefore -x^2+3x+7-1=-x+1$,\\\\\nsolving this gives $x_1=-1$, $x_2=5$ (discard as it does not meet the condition),\\\\\nwhen $x=-1$, $-x^2+2x+7=4$,\\\\\n$\\therefore$ the coordinates of point $C$ are: $\\boxed{(-1, 4)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52916890_266.png"
  },
  {
    "solution": "\\boxed{P_{2}\\left(\\frac{1}{2}, \\frac{13}{2}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the Cartesian coordinate system, the graph of the linear function \\(y = x + b\\) intersects the x-axis and the y-axis at points \\(A\\) and \\(B\\) respectively, and intersects the graph of the direct proportionality function \\(y = -2x\\) at point \\(C\\), where the y-coordinate of point \\(C\\) is \\(4\\). If a moving point \\(P\\) is on the ray \\(CA\\), and the area of triangle \\(OAP\\) is \\(\\frac{1}{8}\\) of the area of triangle \\(OBC\\), what are the coordinates of point \\(P\\)?\n",
    "completion": "\\textbf{Solution:} Let $P\\left(x, x+6\\right)$,\\\\\nwhen $S_{\\triangle OAP}=\\frac{1}{8}S_{\\triangle OBC}$, then $\\frac{1}{2}AO\\cdot |x|=\\frac{1}{8}\\times \\frac{1}{2}\\times BO\\cdot |y_c|$,\\\\\n$\\therefore \\frac{1}{2}\\times 6|x|=\\frac{1}{8}\\times \\frac{1}{2}\\times 6\\times 4$, hence $x=\\pm \\frac{1}{2}$,\\\\\nMoreover, since the moving point P is on the ray $CA$,\\\\\n$\\therefore x\\ge -2$,\\\\\n$\\therefore$ the coordinates of point P are $\\boxed{P_{1}\\left(-\\frac{1}{2}, \\frac{11}{2}\\right)}$, $\\boxed{P_{2}\\left(\\frac{1}{2}, \\frac{13}{2}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52190033_151.png"
  },
  {
    "solution": "\\boxed{\\frac{5}{{2}^{n-3}}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the rhombus $ABCD$ has a side length of $10$, and $\\angle A=60^\\circ$. By successively connecting the midpoints of each side of the rhombus $ABCD$, quadrilateral ${A}_{1}{B}_{1}{C}_{1}{D}_{1}$ is obtained; by successively connecting the midpoints of each side of quadrilateral ${A}_{1}{B}_{1}{C}_{1}{D}_{1}$, quadrilateral ${A}_{2}{B}_{2}{C}_{2}{D}_{2}$ is obtained; by successively connecting the midpoints of each side of quadrilateral ${A}_{2}{B}_{2}{C}_{2}{D}_{2}$, quadrilateral ${A}_{3}{B}_{3}{C}_{3}{D}_{3}$ is obtained; and so on. Following this pattern, what is the perimeter of the quadrilateral ${A}_{2n}{B}_{2n}{C}_{2n}{D}_{2n}$?\n",
    "completion": "\\textbf{Solution:} By utilizing the properties of mid-quadrilaterals, we know quadrilaterals $A_{1}B_{1}C_{1}D_{1}$ and $A_{3}B_{3}C_{3}D_{3}$ are rectangles, while $A_{2}B_{2}C_{2}D_{2}$ and $A_{4}B_{4}C_{4}D_{4}$ are rhombuses.\\\\\nSince the side length of rhombus ABCD is 10, and $\\angle A=60^\\circ$,\\\\\ntherefore, $\\triangle ADB$ and $\\triangle AA_{1}D_{1}$ are equilateral triangles, and the perimeter of rhombus ABCD is $10\\times 4=40$;\\\\\ntherefore, the perimeter of rhombus $A_{2}B_{2}C_{2}D_{2}$ is $\\frac{1}{2}\\times 40=20$;\\\\\nThe perimeter of rhombus $A_{4}B_{4}C_{4}D_{4}$ is $\\frac{1}{{2}^{2}}\\times 40$;\\\\\n$ \\dots $\\\\\ntherefore, the perimeter of quadrilateral ${A}_{2n}{B}_{2n}{C}_{2n}{D}_{2n}$ is $40\\times \\frac{1}{{2}^{n}}=5\\times 8\\times \\frac{1}{{2}^{n}}=\\boxed{\\frac{5}{{2}^{n-3}}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51614697.png"
  },
  {
    "solution": "\\boxed{110^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, place the right angle vertex $O$ of the right-angled triangle $OMN$ on the line $AB$, with the ray $OC$ bisecting the angle $\\angle AON$. Rotate the right-angled triangle $OMN$ counterclockwise about vertex $O$. During the rotation process:\n",
    "completion": "\\textbf{Solution:}\n\n(1) When ON is above line AB,\n\n$\\because \\angle BON=140^\\circ$,\n\n$\\therefore \\angle AON=40^\\circ$,\n\n$\\because OC$ bisects $\\angle AON$,\n\n$\\therefore \\angle CON=20^\\circ$,\n\n$\\because \\angle MON=90^\\circ$,\n\n$\\therefore \\angle COM=70^\\circ$.\n\nWhen ON is below line AB,\n\n$\\because \\angle BON=140^\\circ$,\n\n$\\therefore \\angle AON=40^\\circ$,\n\n$\\because OC$ bisects $\\angle AON$,\n\n$\\therefore \\angle CON=20^\\circ$,\n\n$\\because \\angle MON=90^\\circ$,\n\n$\\therefore \\angle COM=\\boxed{110^\\circ}$.\n\n(2) $\\angle BON=2\\angle COM$ or $\\angle BON+2\\angle COM=360^\\circ$",
    "image_path": "dataset/mm_math/images/MM_Math/51385746_143.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within $\\triangle ABC$, points D, E, and F are located on the sides AB, AC, and BC, respectively, and lines DE and EF are connected. It is known that quadrilateral BFED is a parallelogram, and $\\frac{DE}{BC} = \\frac{1}{4}$. If AB = 8, find the length of the line segment AD.\n",
    "completion": "\\textbf{Solution}: Since quadrilateral BFED is a parallelogram,\\\\\nit implies that DE$\\parallel$BF, and hence DE$\\parallel$BC, which further implies $\\triangle$ADE$\\sim$$\\triangle$ABC,\\\\\nthus $\\frac{AD}{AB} = \\frac{DE}{BC} = \\frac{1}{4}$,\\\\\ngiven that AB$=8$,\\\\\nit follows that AD$=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52694105_74.png"
  },
  {
    "solution": "\\boxed{5\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In right triangle $ABC$, $\\angle C = 90^\\circ$, and $AB = 10$. If a circle centered at point $C$ with radius $CB$ exactly passes through the midpoint $D$ of $AB$, what is the length of $AC$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect CD,\\\\\n$\\because$ $\\angle C=90^\\circ$ and D is the midpoint of AB,\\\\\n$\\therefore$ CD=DA=DB.\\\\\nAnd CD=CB,\\\\\n$\\therefore$ CD=CB=DB, thus $\\triangle CDB$ is an equilateral triangle.\\\\\n$\\therefore$ $\\angle B=60^\\circ$.\\\\\n$\\because$ AB=10,\\\\\n$\\therefore$ $AC=AB\\cdot \\sin\\angle B=10\\times \\frac{\\sqrt{3}}{2}=\\boxed{5\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53065816.png"
  },
  {
    "solution": "\\boxed{\\frac{2\\sqrt{2}}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $AD=1, AB=2\\sqrt{2}$, $M$ is a moving point on line segment $BD$, $MP\\perp CD$ at point $P$, $MQ\\perp BC$ at point $Q$, what is the minimum value of $PQ$?\n",
    "completion": "\\textbf{Solution:} Connect CM as shown in the figure,\\\\\n$ \\because MP\\perp CD$ at point P, $MQ\\perp BC$ at point Q,\\\\\n$ \\therefore \\angle CPM=\\angle CQM=90^\\circ$,\\\\\n$ \\because $ quadrilateral ABCD is a rectangle,\\\\\n$ \\therefore BC=AD=1, CD=AB=2\\sqrt{2}$, $ \\angle BCD=90^\\circ$,\\\\\n$ \\therefore $ quadrilateral PCQM is a rectangle,\\\\\n$ \\therefore PQ=CM$,\\\\\nIn $ \\triangle BCD$, $BD=\\sqrt{BC^{2}+CD^{2}}=\\sqrt{1^{2}+(2\\sqrt{2})^{2}}=3$,\\\\\nWhen CM $ \\perp $BD, CM is minimized, then PQ is minimized,\\\\\nAt this time, $ S_{\\triangle BCD}=\\frac{1}{2}BD\\cdot CM=\\frac{1}{2}BC\\cdot CD$,\\\\\n$ \\therefore CM=\\frac{BC\\cdot CD}{BD}=\\frac{1\\times 2\\sqrt{2}}{3}=\\frac{2\\sqrt{2}}{3}$,\\\\\n$ \\therefore PQ$'s minimum value is $ \\boxed{\\frac{2\\sqrt{2}}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53728548.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, point $D$ is the midpoint of the hypotenuse $AB$, $DE$ bisects $\\angle ADC$, and $BC=4$, what is the length of $DE$?\n",
    "completion": "\\textbf{Solution:} Let point $D$ be the midpoint of the hypotenuse $AB$, and $\\angle ACB=90^\\circ$, \\\\\n$\\therefore$AD=CD=BD,\\\\\n$\\because$ $DE$ bisects $\\angle ADC$,\\\\\n$\\therefore$DE$\\perp$AC, and AE=CE,\\\\\n$\\therefore$DE is the midline of $\\triangle ABC$,\\\\\n$\\therefore$DE= $\\frac{1}{2}BC=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52076824.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points E, F are the midpoints of the sides AB and BC of the rhombus ABCD, respectively, with $AB=2$ and $EF=\\sqrt{3}$. What is the area of the rhombus ABCD?\n",
    "completion": "\\textbf{Solution:} Let the intersection of AC and BD be point P. Since points E and F are the midpoints of AB and BC respectively, it follows that $AC=2\\cdot EF=2\\sqrt{3}$, and $AP=\\sqrt{3}$. In $\\triangle ABP$, we have $BP=\\sqrt{AB^{2}-AP^{2}}=\\sqrt{2^{2}-(\\sqrt{3})^{2}}=1$, and $BD=2\\cdot BP=2$. Therefore, the area of rhombus ABCD is $\\frac{1}{2}\\cdot AC\\cdot BD=\\frac{1}{2}\\times 2\\sqrt{3}\\times 2=\\boxed{2\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52284598.png"
  },
  {
    "solution": "\\boxed{95^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the circumcircle of $\\triangle ABC$, with point D on the arc $\\overset{\\frown}{BC}$, connect AD and CD. If $\\angle CAD=25^\\circ$, then what is the measure of $\\angle ACD$?\n",
    "completion": "\\textbf{Solution:} Given $\\triangle ABC$ is an equilateral triangle,\\\\\n$\\therefore$ $\\angle ACB = \\angle ABC = \\angle BAC = 60^\\circ$,\\\\\nGiven $\\angle CAD = 25^\\circ$,\\\\\n$\\therefore$ $\\angle BAD = \\angle BAC - \\angle CAD = 35^\\circ$,\\\\\nSince arc $\\overset{\\frown}{BD} = \\overset{\\frown}{BD}$,\\\\\n$\\therefore$ $\\angle BCD = \\angle BAD = 35^\\circ$,\\\\\n$\\therefore$ $\\angle ACD = \\angle ACB + \\angle BCD = \\boxed{95^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/52720283.png"
  },
  {
    "solution": "\\boxed{y=\\frac{4}{x}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=ax^2+bx+3$ intersects the x-axis at points $A$ and $B$ (with $A$ to the left of $B$) and intersects the y-axis at point $C$, with $\\tan\\angle CAB=3$. The hyperbola $y=\\frac{k}{x}$ (with $k\\neq 0$) passes through the vertex $D$ of the parabola $y=ax^2+bx+3$, where the x-coordinate of point $D$ is 1. Find the equations of the parabola and the hyperbola.\n",
    "completion": "\\textbf{Solution:} Let us set $x=0$, which yields: $y=3$,\\\\\nTherefore, the coordinates of point C are $\\left(0, 3\\right)$,\\\\\nTherefore, $OC=3$.\\\\\nSince $\\tan\\angle CAB=3$,\\\\\nTherefore, $\\frac{OC}{OA}=3$, that is $\\frac{3}{OA}=3$,\\\\\nTherefore, $OA=1$.\\\\\nTherefore, the coordinates of point A are $\\left(-1, 0\\right)$.\\\\\nSince the axis of symmetry for the parabola is $x=1$,\\\\\nTherefore, the coordinates of point B are $\\left(3, 0\\right)$.\\\\\nSubstituting $A\\left(-1, 0\\right)$ and $B\\left(3, 0\\right)$ yields: $\\left\\{\\begin{array}{l}a-b+3=0 \\\\ 9a+3b+3=0\\end{array}\\right.$,\\\\\nSolving this, we find: $a=-1, b=2$,\\\\\nTherefore, the equation of the parabola is $y=-x^{2}+2x+3$.\\\\\nSubstituting $x=1$ yields: $y=-1+2+3=4$.\\\\\nTherefore, the coordinates of point D are $\\left(1, 4\\right)$.\\\\\nSubstituting $\\left(1, 4\\right)$ into the equation of the inverse proportion function yields: $4=\\frac{k}{1}$,\\\\\nSolving this, we find: $k=4$.\\\\\nTherefore, the equation of the inverse proportion function is $y=\\frac{4}{x}$.\\\\\n$\\therefore \\boxed{y=-x^{2}+2x+3}$ and $\\boxed{y=\\frac{4}{x}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53043696_144.png"
  },
  {
    "solution": "\\boxed{5\\,\\text{cm}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given that point $C$ is a point on segment $AB$, point $M$ is the midpoint of segment $AC$, and point $N$ is the midpoint of segment $BC$, if $AB=10\\,\\text{cm}$, what is the length of $MN$?\n",
    "completion": "\\textbf{Solution:} We have $MN=CM+CN=\\frac{1}{2}AC+\\frac{1}{2}BC=\\frac{1}{2}AB=5\\,\\text{cm}$,\\\\\ntherefore $MN=\\boxed{5\\,\\text{cm}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/43067255_148.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in rhombus $ABCD$, $E$ is the midpoint of $AB$, and $DE \\perp AB$, with $AE=2$. What are the lengths of the diagonals $AC$ and $BD$?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABD$ is an equilateral triangle and $AE=2$,\n\n$\\therefore AB=BD=AD=4$,\n\n$\\therefore DE=AO=\\sqrt{AD^{2}-AE^{2}}=\\sqrt{4^{2}-2^{2}}=2\\sqrt{3}$,\n\n$\\therefore AC=\\boxed{4\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51614472_113.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: It is known that there are three rays inside $\\angle AOB$. Among them, $OE$ bisects $\\angle BOC$, and $OF$ bisects $\\angle AOC$. As shown in Figure 1, if $\\angle AOB=90^\\circ$ and $\\angle AOC=30^\\circ$, what is the measure of $\\angle EOF$ in degrees?\n",
    "completion": "\\textbf{Solution:}\n\nGiven: $\\angle BOC=\\angle AOB-\\angle AOC=90^\\circ-30^\\circ=60^\\circ$,\\\\\nSince OE bisects $\\angle BOC$, and OF bisects $\\angle AOC$,\\\\\nThen $\\angle EOC=\\frac{1}{2}\\angle BOC=\\frac{1}{2}\\times 60^\\circ=30^\\circ$ and $\\angle COF= \\frac{1}{2}\\angle AOC=\\frac{1}{2}\\times 30^\\circ=15^\\circ$,\\\\\nTherefore $\\angle EOF=\\angle EOC+\\angle COF=30^\\circ+15^\\circ=\\boxed{45^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/52795749_120.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{1}{2}x^2+2x+\\frac{5}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the vertices of rectangle $OABC$ are $A(0, 3)$ and $C(-1, 0)$. Now, rectangle $OABC$ is rotated $90^\\circ$ clockwise around the origin $O$ to obtain rectangle $O{A}^{\\prime }{B}^{\\prime }{C}^{\\prime }$. The line $B{B}^{\\prime }$ intersects the x-axis at point $M$ and the y-axis at point $N$. The graph of the parabola $y=ax^{2}+bx+c$ passes through points $C$, $M$, and $N$. Find the equation of the parabola.\n",
    "completion": "\\textbf{Solution:} Let the equation of line BB' be $y=kx+b$. Substituting the points B $(-1, 3)$, B' $(3, 1)$ into it, we obtain\n\\[\n\\begin{cases}\n-k+b=3\\\\\n3k+b=1\n\\end{cases},\n\\]\nfrom which it follows that\n\\[\n\\begin{cases}\nk=-\\frac{1}{2}\\\\\nb=\\frac{5}{2}\n\\end{cases}\n\\]\n$\\therefore$ The equation of line BB' is $y=-\\frac{1}{2}x+\\frac{5}{2}$; setting $x=0$, we find $y=\\frac{5}{2}$, and setting $y=0$, we find $x=5$, $\\therefore$ the intersection points of line BB' with the x-axis and y-axis are M $(5, 0)$ and N $(0, \\frac{5}{2})$, respectively. $\\because$ C $(-1, 0)$, $\\therefore$ let the equation of the parabola be $y=a(x-5)(x+1)$, $\\because$ the parabola passes through point N, $\\therefore$ $\\frac{5}{2}=a\\times (-5)\\times 1$, $\\therefore$ a $=-\\frac{1}{2}$, $\\therefore$ the equation of the parabola is $y=-\\frac{1}{2}(x-5)(x+1)=-\\frac{1}{2}x^2+2x+\\frac{5}{2}$; hence, the equation of the parabola is $\\boxed{y=-\\frac{1}{2}x^2+2x+\\frac{5}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52602539_163.png"
  },
  {
    "solution": "\\boxed{P(-5, 0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the graph of the inverse proportion function \\(y = \\frac{k}{x}\\) intersects with the line \\(y = ax + b\\) at points \\(A\\left(\\frac{1}{3}, m + 9\\right)\\) and \\(B(1, m)\\). There is a point \\(P\\) on the x-axis such that the area of \\(\\triangle PAB\\) is \\(18\\). Find the coordinates of point \\(P\\).\n",
    "completion": "\\textbf{Solution:} Let us draw line $AB$ and construct $AC\\perp x$-axis at point $C$ and $BD\\perp x$-axis at point $D$. Let $AB$ intersect the $x$-axis at point $E$. When $y=-3x-3=0$, we solve and find $x=-1$. That is, $E(-1, 0)$. The area of $\\triangle PAB$ is $\\frac{1}{2}PE\\cdot AC+\\frac{1}{2}PE\\cdot DB=\\frac{3}{2}PE+\\frac{6}{2}PE=\\frac{9}{2}PE$, and since the area of $\\triangle PAB=18$, we have $\\frac{9}{2}PE=18$, $\\therefore PE=4$. Thus, when point $P$ is to the right of the origin, $P(3, 0)$; when point $P$ is to the left of the origin, $P(-5, 0)$.\n\n$\\therefore \\boxed{P(3, 0)}$ and $\\boxed{P(-5, 0)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51368308_107.png"
  },
  {
    "solution": "\\boxed{PD\\text{ is the tangent to circle }O}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $AB$ is the diameter of semicircle $O$, $C$ is a point on the semicircle, $P$ is the midpoint of $BC$, and $PD\\parallel BC$ intersects the extension of $AB$ at point $D$. Prove that $PD$ is the tangent to $\\odot O$.\n",
    "completion": "\\textbf{Solution.} Proof: Connect $OP$,\\\\\n$\\because$ Point $P$ is the midpoint of arc $BC$,\\\\\n$\\therefore$ $OP\\perp BC$,\\\\\n$\\because$ $PD\\parallel BC$,\\\\\n$\\therefore$ $PD\\perp OP$,\\\\\n$\\therefore$ \\boxed{PD\\text{ is the tangent to circle }O}.",
    "image_path": "dataset/mm_math/images/MM_Math/51369905_94.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given the figure, in the Cartesian coordinate system, the midpoint of diagonal $AC$ of rectangle $ABCD$ coincides with the origin. Point $E$ is a point on the $x$-axis, and $AE$ is connected. If $AD$ bisects $\\angle OAE$, the graph of the inverse proportion function $y=\\frac{k}{x}$ ($k>0$, $x>0$) passes through two points $A$ and $F$ on $AE$, and $AF=EF$, the area of $\\triangle ABE$ is $18$, then what is the value of $k$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect BD and AC at point O. Draw AG $\\perp$ x-axis at point G from point A, and draw FH $\\perp$ x-axis at point H from point F. Connect OF, \\\\\n$\\therefore$ FH $\\parallel$ AG, \\\\\n$\\because$ AE=EF, \\\\\n$\\therefore$ FH is the median line of $\\triangle$ AGE, \\\\\n$\\therefore$ GH=HE, AG=2FH \\\\\n$\\because$ points A and F are on the graph of the inverse proportion function y=$\\frac{k}{x}$ (k>0, x>0), \\\\\n$\\therefore$ S$_{\\triangle AOG}$=S$_{\\triangle FOH}$=$\\frac{k}{2}$, \\\\\n$\\therefore$ OG·AG=OH·FH, \\\\\n$\\therefore$ OH=2OG, \\\\\n$\\therefore$ OG=GH=HE, \\\\\n$\\because$ of the rectangle ABCD, \\\\\n$\\therefore$ OA=OD, \\\\\n$\\therefore$ $\\angle$OAD=$\\angle$ODA, \\\\\nAdditionally, $\\because$ AD bisects $\\angle$OAE, \\\\\n$\\therefore$ $\\angle$OAD=$\\angle$EAD, \\\\\n$\\therefore$ $\\angle$ODA=$\\angle$EAD, \\\\\n$\\therefore$ AE $\\parallel$ BD, \\\\\n$\\therefore$ S$_{\\triangle AOE}$=S$_{\\triangle ABE}$=18, \\\\\n$\\therefore$ S$_{\\triangle AOG}$=$\\frac{1}{3}$S$_{\\triangle AOE}$=6, \\\\\n$\\therefore$ $\\frac{k}{2}$=6, \\\\\n$\\therefore$ k=\\boxed{12}.",
    "image_path": "dataset/mm_math/images/MM_Math/52205064.png"
  },
  {
    "solution": "\\boxed{10^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in $ \\triangle ABC$, $AD$ bisects $ \\angle BAC$ and $AE$ is the altitude. If $ \\angle B=40^\\circ$ and $ \\angle C=60^\\circ$, what is the measure of $ \\angle EAD$?\n",
    "completion": "\\textbf{Solution:} Given that in $\\triangle ABC$, $\\angle B=40^\\circ$ and $\\angle C=60^\\circ$, \\\\\ntherefore $\\angle BAC=180^\\circ-\\angle B-\\angle C=180^\\circ-40^\\circ-60^\\circ=80^\\circ$ \\\\\nSince $AD$ bisects $\\angle BAC$, \\\\\nthus $\\angle BAD=\\angle CAD=40^\\circ$ \\\\\nAlso, since $AE$ is the altitude, \\\\\nthus $\\angle BAE=90^\\circ-\\angle B=90^\\circ-40^\\circ=50^\\circ$ \\\\\nTherefore, $\\angle EAD=\\angle BAE-\\angle BAD=50^\\circ-40^\\circ=\\boxed{10^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/53042516.png"
  },
  {
    "solution": "\\boxed{B(-1, -2)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the inverse proportion function $y=\\frac{k}{x} (k\\neq 0)$ intersects with the graph of the direct proportion function $y=2x$ at points $A(1, a)$ and $B$. Point $C$ is located in the fourth quadrant, with $CA\\parallel y$-axis and $AB\\perp BC$. What is the equation of the inverse proportion function and the coordinates of point $B$?\n",
    "completion": "\\textbf{Solution:} Since point A$(1, a)$ is on $y=2x$,\\\\\nit follows that $a=2$,\\\\\nhence $A(1, 2)$.\\\\\nSubstituting A$(1, 2)$ into $y=\\frac{k}{x}$ yields $k=2$,\\\\\ntherefore, the equation of the inverse proportion function is $y=\\frac{2}{x}$.\\\\\nSince points A and B are centrally symmetric with respect to the origin,\\\\\nit follows that \\boxed{B(-1, -2)}.",
    "image_path": "dataset/mm_math/images/MM_Math/52765898_183.png"
  },
  {
    "solution": "\\boxed{15}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\triangle ABC$, the perpendicular bisector of side AB intersects BC and AB at points D and E, respectively, with AE = 3cm and the perimeter of $ \\triangle ADC$ being 9cm. What is the perimeter of $ \\triangle ABC$ in cm?\n",
    "completion": "\\textbf{Solution:} Given that in triangle $ABC$, the perpendicular bisector of side $AB$ intersects $BC$ and $AB$ at points $D$ and $E$, respectively, with $AE=3cm$,\\\\\nit follows that $BD=AD$ and $AB=2AE=6cm$,\\\\\ngiven that the perimeter of $\\triangle ADC$ is $9cm$,\\\\\nthen, $AC+AD+CD=AC+BD+CD=AC+BC=9cm$,\\\\\nthus, the perimeter of $\\triangle ABC$ is given by: AB+AC+BC=\\boxed{15}cm.",
    "image_path": "dataset/mm_math/images/MM_Math/51213252.png"
  },
  {
    "solution": "\\boxed{y=x^{2}-4x+3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the parabola $y=ax^{2}+bx+c\\ (a\\ne 0)$ has its vertex at point $Q(2, -1)$, it intersects the y-axis at point $C(0, 3)$, and intersects the x-axis at points $A$ and $B$ (with point $A$ to the right of point $B$). Point $P$ is a moving point on this parabola, moving from point $C$ along the parabola towards point $A$ (point $P$ does not coincide with $A$). A line is drawn through point $P$ parallel to the y-axis, intersecting $AC$ at point $D$. What is the equation of this parabola?\n",
    "completion": "\\textbf{Solution:} Since the vertex of the parabola is Q(2,−1),\\\\\nhence, let $y=a(x-2)^{2}-1$,\\\\\nSubstituting C(0, 3) into the above equation, we get $3=a(0-2)^{2}-1$,\\\\\nSolving for $a$, we find that $a=1$,\\\\\ntherefore $y=(x-2)^{2}-1$,\\\\\nwhich simplifies to $\\boxed{y=x^{2}-4x+3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52622423_221.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{2}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $AB$ is the diameter of $\\odot O$, $BC \\perp AB$, connect $OC$, the chord $AD \\parallel OC$, and the line $CD$ intersects the extension line of $BA$ at point $E$. The line $CD$ is the tangent line of $\\odot O$; if $BC=1$, $DE=2BC$, what is the radius of $\\odot O$?\n",
    "completion": "\\textbf{Solution:} In $\\triangle OEC$, since $AD \\parallel OC$,\\\\\n$\\therefore \\frac{ED}{CD} = \\frac{EA}{OA}$.\\\\\nAdditionally, since $\\triangle OBC \\cong \\triangle ODC$,\\\\\n$\\therefore CB = CD$,\\\\\nGiven that $BC = 1$, $DE = 2BC$,\\\\\n$\\therefore DE = 2BC = 2CD = 2$,\\\\\n$\\therefore \\frac{ED}{CD} = \\frac{EA}{OA} = 2$,\\\\\n$\\therefore EA = 2OA$.\\\\\nLet the radius of $\\odot O$ be $x$, then $OA = OD = x$, $EA = 2x$,\\\\\n$\\therefore OE = OA + EA = x + 2x = 3x$,\\\\\nSince the line $CD$ is a tangent to $\\odot O$,\\\\\n$\\therefore \\angle ODE = 90^\\circ$,\\\\\nBy the Pythagorean theorem, $OD^2 + DE^2 = OE^2$,\\\\\n$\\therefore x^2 + 2^2 = (3x)^2$,\\\\\nSolving yields $x = \\frac{\\sqrt{2}}{2}$,\\\\\nHence, the radius of $\\odot O$ is $\\boxed{\\frac{\\sqrt{2}}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53106853_103.png"
  },
  {
    "solution": "\\boxed{x<2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: If the graph of the function $y=kx+b$ is shown in the figure, what is the solution set of the inequality $kx+b>0$?\n",
    "completion": "\\textbf{Solution:} From the graph, we can determine that the function $y=kx+b$ intersects the x-axis at $(2,0)$. Since $y$ decreases as $x$ increases, \\\\\n$\\therefore$ the solution set of the inequality $kx+b>0$ is $\\boxed{x<2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52381338.png"
  },
  {
    "solution": "\\boxed{B\\left(4, 0\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system, the parabola $y=\\frac{1}{2}x^{2}+bx+c$ (where $b$ and $c$ are constants) has its vertex at $\\left(\\frac{3}{2}, -\\frac{25}{8}\\right)$, intersects the x-axis at points $A$ and $B$ (with point $A$ to the left of point $B$), and intersects the y-axis at point $C$. Point $C$ and point $D$ are symmetric about the x-axis. Line $AD$ is drawn, and line $BD$ is constructed. What are the coordinates of points $A$ and $B$?\n",
    "completion": "\\textbf{Solution:} Let $\\frac{1}{2}x^{2}-\\frac{3}{2}x-2=0$,\\\\\nwe rearrange it to get: $x^{2}-3x-4=0$,\\\\\nfrom which we find $x_{1}=-1$ and $x_{2}=4$,\\\\\n$\\therefore \\boxed{A\\left(-1, 0\\right)}$ and $\\boxed{B\\left(4, 0\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53079018_49.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in square $ABCD$, $AB=6$, $E$ is a point on the diagonal $AC$, and line $BE$ is drawn. Through point $E$, line $EF\\perp BE$ intersects $AD$ at point $F$. The areas of $\\triangle BCE$ and $\\triangle AEF$ are ${S}_{1}$ and ${S}_{2}$, respectively. If $2{S}_{1}=3{S}_{2}$, then what is the length of $CE$?\n",
    "completion": "\\textbf{Solution:} Connect ED, and through E draw $MN \\perp BC$ at N intersecting AD at M,\\\\\nsince quadrilateral ABCD is a square,\\\\\ntherefore $AB=BC=CD=AD=6$, $\\angle ABC=\\angle BCD=\\angle ADC=\\angle DAB=90^\\circ$,\\\\\ntherefore $\\angle 1=\\angle 2=45^\\circ$,\\\\\nsince $MN \\perp BC$,\\\\\ntherefore $\\angle ENC=\\angle ENB=90^\\circ$,\\\\\ntherefore quadrilateral MNCD is a rectangle,\\\\\ntherefore $MN=CD=6$, $DM=CN$, $\\angle DME=90^\\circ$,\\\\\nin $\\triangle CDE$ and $\\triangle CBE$,\\\\\n$\\left\\{ \\begin{array}{l} CD=CB \\\\ \\angle 2=\\angle 1 \\\\ CE=CE \\end{array} \\right.$,\\\\\ntherefore $\\triangle CDE \\cong \\triangle CBE$ (SAS),\\\\\ntherefore $ED=EB$, $\\angle EDC=\\angle EBC$,\\\\\nsince $\\angle CDA=\\angle CBA=90^\\circ$,\\\\\ntherefore $\\angle CDA-\\angle EDC=\\angle CBA-\\angle EBC$,\\\\\nthat is $\\angle ADE=\\angle ABE$,\\\\\nsince $EF \\perp BE$,\\\\\ntherefore $\\angle FEB=90^\\circ$,\\\\\nsince $\\angle FEB+\\angle DAB+\\angle AFE+\\angle ABE=360^\\circ$,\\\\\ntherefore $\\angle AFE+\\angle ABE=360^\\circ-\\angle FEB-\\angle DAB=180^\\circ$,\\\\\nsince $\\angle AFE+\\angle EFD=180^\\circ$,\\\\\ntherefore $\\angle ABE=\\angle EFD$,\\\\\ntherefore $\\angle ADE=\\angle EFD$,\\\\\ntherefore $ED=EF$,\\\\\nsince $\\angle DME=90^\\circ$,\\\\\ntherefore $EM \\perp DF$,\\\\\ntherefore $DM=MF$,\\\\\nin $\\triangle NEC$, $\\angle 1=45^\\circ$,\\\\\ntherefore $\\triangle NEC$ is an isosceles right triangle,\\\\\nlet $NE=NC=x$,\\\\\nthen $CE= \\sqrt{2}x$, $DM=MF=CN=x$,\\\\\ntherefore $AF=AD-DM-MF=6-2x$,\\\\\n$ME=MN-EN=6-x$,\\\\\ntherefore $S_{1}=\\frac{1}{2}BC\\cdot EN=\\frac{1}{2}\\times 6\\times x=3x$, $S_{2}=\\frac{1}{2}(6-2x)(6-x)=x^{2}-9x+18$,\\\\\nsince $2S_{1}=3S_{2}$,\\\\\ntherefore $2\\cdot 3x=3(x^{2}-9x+18)$,\\\\\nsolving, we find: $x_{1}=2$, $x_{2}=9$ (discard),\\\\\ntherefore $CE= \\sqrt{2}x=2\\sqrt{2}$,\\\\\nhence the answer is: $CE = \\boxed{2\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50474806.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that in $\\triangle ABC$, with side $BC=6$ and height $AD=3$, the vertices $E$ and $F$ of the square $EFGH$ lie on side $BC$, while vertices $H$ and $G$ lie on sides $AB$ and $AC$, respectively. What is the length of the side of this square?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram:\\\\\n$ \\because $ Quadrilateral $EFMN$ is a square,\\\\\n$ \\therefore HG\\parallel BC$, $HG=EF$,\\\\\n$ \\therefore \\angle AHG=\\angle B$,\\\\\n$ \\because \\angle BAC=\\angle BAC$,\\\\\n$ \\therefore \\triangle AHG\\sim \\triangle ABC$,\\\\\n$ \\therefore \\frac{AH}{AB}=\\frac{HG}{BC}$,\\\\\nAlso $ \\because AD\\perp BC$,\\\\\n$ \\therefore AD\\perp HG$, $HG=EF=MD$,\\\\\n$ \\because HG\\parallel BC$,\\\\\n$ \\therefore \\frac{AM}{AD}=\\frac{AH}{AB}$, that is $\\frac{AM}{AD}=\\frac{HG}{BC}$,\\\\\nLet $HG=x$, then $AM=AD-MD=3-x$,\\\\\n$ \\therefore \\frac{3-x}{3}=\\frac{x}{6}$, solving this gives: $x=\\boxed{2}$,\\\\\n$ \\therefore $ The side length of this square is $2$.",
    "image_path": "dataset/mm_math/images/MM_Math/52917671.png"
  },
  {
    "solution": "\\boxed{\\frac{35}{9}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In quadrilateral $ABCD$, $E$ is the midpoint of $DC$. Line $AE$ is extended to meet the extended line of $BC$ at point $F$, and $BC = CF$. Let $G$ be a point on $CF$, and line $AG$ intersects $CD$ at point $H$, where $\\angle DAF = \\angle GAF$. Given $CG = 2$ and $GF = 5$, find the length of $AH$.\n",
    "completion": "\\textbf{Solution:} Given that $AD\\parallel BC$,\\\\\nit follows that $\\angle DAF=\\angle F$,\\\\\nsince $\\angle GAF=\\angle DAF$,\\\\\nwe have $\\angle GAF=\\angle F$,\\\\\nthus $AG=GF=5$,\\\\\ngiven that $CG=2$, $GF=5$,\\\\\nit follows that $AD=CF=7$,\\\\\nsince $AD\\parallel BC$,\\\\\nit implies $\\triangle AHD\\sim \\triangle GHC$,\\\\\ntherefore $\\frac{AH}{GH}=\\frac{AD}{GC}=\\frac{7}{2}$,\\\\\nwhich means $\\frac{AH}{AG}=\\frac{7}{9}$, i.e., $\\frac{AH}{5}=\\frac{7}{9}$,\\\\\n$AH=\\boxed{\\frac{35}{9}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51407093_86.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC$, $D$ and $E$ are two points inside $\\triangle ABC$, $AE$ bisects $\\angle BAC$, $\\angle D=\\angle DBC=60^\\circ$. If $BD=5\\text{cm}$ and $DE=3\\text{cm}$, what is the length of $BC$ in $\\text{cm}$?\n",
    "completion": "\\textbf{Solution:} Extend DE to meet BC at M, extend AE to meet BC at N, construct EF $\\parallel$ BC at F, \\\\\n$\\because$ AB=AC, AE bisects $\\angle$BAC, \\\\\n$\\therefore$ AN$\\perp$BC, BN=CN, \\\\\n$\\because$ $\\angle$DBC=$\\angle$D=$60^\\circ$, \\\\\n$\\therefore$ $\\triangle$BDM is an equilateral triangle, \\\\\n$\\therefore$ $\\triangle$EFD is an equilateral triangle, \\\\\n$\\because$ BD=5, DE=3, \\\\\n$\\therefore$ EM=2, \\\\\n$\\because$ $\\triangle$BDM is an equilateral triangle, \\\\\n$\\therefore$ $\\angle$DMB=$60^\\circ$, \\\\\n$\\because$ AN$\\perp$BC, \\\\\n$\\therefore$ $\\angle$ENM=$90^\\circ$, \\\\\n$\\therefore$ $\\angle$NEM=$30^\\circ$, \\\\\n$\\therefore$ NM=1, \\\\\n$\\therefore$ BN=4, \\\\\n$\\therefore$ BC=2BN=\\boxed{8} (cm),",
    "image_path": "dataset/mm_math/images/MM_Math/52905602.png"
  },
  {
    "solution": "\\boxed{x>-2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given the linear function $y=kx+b$ whose graph passes through points $A(0, 2)$ and point $B(-2, 0)$, for what values of $x$ is $y > 0$?\n",
    "completion": "\\textbf{Solution:} From the graph, it is known that when $x>-2$, $y>0$. Therefore, $\\boxed{x>-2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50676718_162.png"
  },
  {
    "solution": "\\boxed{70^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, quadrilateral ABCD is a rhombus. Extend BC to E so that BD = BE. Connect DE. If $\\angle ABC = 80^\\circ$, what is the degree measure of $\\angle E$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rhombus,\\\\\nit follows that $BD$ bisects $\\angle ABC$,\\\\\nand since $\\angle ABC=80^\\circ$,\\\\\nit follows that $\\angle DBC=\\frac{1}{2}\\angle ABC=40^\\circ$,\\\\\nand since $BD=BE$,\\\\\nit follows that $\\angle BDE=\\angle BED=\\frac{1}{2}\\left(180^\\circ-40^\\circ\\right)=\\boxed{70^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52160164.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the quadrilateral $ABCD$ is inscribed in circle $\\odot O$. If the quadrilateral $ABCO$ is a rhombus, then what is the measure of $\\angle D$ in degrees?\n",
    "completion": "\\textbf{Solution:} Let $\\angle ADC=\\alpha$ and $\\angle ABC=\\beta$;\\\\\nSince quadrilateral ABCO is a rhombus,\\\\\nit follows that $\\angle ABC=\\angle AOC=\\beta$;\\\\\nTherefore $\\angle ADC=\\frac{1}{2}\\beta$;\\\\\nSince quadrilateral $ABCD$ is circumscribed within a circle,\\\\\nit follows that $\\alpha +\\beta =180^\\circ$,\\\\\ntherefore $\\left\\{\\begin{array}{l}\n\\alpha +\\beta =180^\\circ\\\\\n\\alpha =\\frac{1}{2}\\beta\n\\end{array}\\right.$,\\\\\nFrom which we find: $\\beta =120^\\circ$, $\\alpha =60^\\circ$, thus $\\angle ADC=\\boxed{60^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51351172.png"
  },
  {
    "solution": "\\boxed{24}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the two medians $AD$ and $BE$ of $\\triangle ABC$ intersect at point $F$. Connecting $CF$, if the area of $\\triangle ABF$ is $8$, what is the area of $\\triangle ABC$?\n",
    "completion": "\\textbf{Solution:} Given that the two medians $AD$ and $BE$ of $\\triangle ABC$ intersect at point $F$,\\\\\n$\\therefore 2EF=BF$,\\\\\n$\\because$ the area of $\\triangle ABF$ is $8$,\\\\\n$\\therefore$ the area of $\\triangle AEF$ is $4$,\\\\\n$\\therefore$ the area of $\\triangle ABE$ is $12$.\\\\\n$\\because AE$ is the median of $\\triangle ABC$,\\\\\n$\\therefore$ the area of $\\triangle ABC$ is $2$ times the area of $\\triangle ABE$ which equals $\\boxed{24}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52653366.png"
  },
  {
    "solution": "\\boxed{x^2-2x-3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=ax^{2}+bx-3 \\ (a \\neq 0)$ intersects the x-axis at points $A(-1, 0)$ and $B(3, 0)$, and intersects the y-axis at point C. What is the expression of the parabola?\n",
    "completion": "\\textbf{Solution}: Substitute point $A(-1, 0)$ and point $B(3, 0)$ into $y=ax^2+bx-3$, \\\\\n$\\therefore \\begin{cases}\na-b-3=0 \\\\\n9a+3b-3=0\n\\end{cases}$, \\\\\nUpon solving, we get $\\begin{cases}\na=1 \\\\\nb=-2\n\\end{cases}$, \\\\\n$\\therefore y=x^2-2x-3=\\boxed{x^2-2x-3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52796987_45.png"
  },
  {
    "solution": "\\boxed{\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of $\\odot O$, with $AB=2$. $OC$ is the radius of $\\odot O$, and $OC\\perp AB$. Point $D$ lies on $\\overset{\\frown}{AC}$, and $\\overset{\\frown}{AD}=2\\overset{\\frown}{CD}$. Point $P$ is a moving point on the radius $OC$. What is the minimum value of $AP+PD$?\n",
    "completion": "\\textbf{Solution:} Construct line BD, intersecting OC at point P. Draw lines OD, AP, AD.\\\\\nSince $OC\\perp AB$ and point $P$ is a movable point on the radius $OC$,\\\\\nit follows that $AP=PB$\\\\\nSince $DP+AP=DP+PB\\ge DB$\\\\\nthe minimum value of $AP+PD$ is the length of $DB$,\\\\\nhence, $\\overset{\\frown}{AC}=\\overset{\\frown}{CB}$, $\\angle AOC=90^\\circ$\\\\\nSince $\\overset{\\frown}{AD}=2\\overset{\\frown}{CD}$\\\\\nit follows that $\\overset{\\frown}{AD}=\\frac{2}{3}\\overset{\\frown}{AC}$\\\\\nTherefore, $\\angle AOD=\\frac{2}{3}\\angle AOC=60^\\circ$\\\\\nSince $\\overset{\\frown}{AD}=\\overset{\\frown}{AD}$\\\\\nit follows that $\\angle ABD=30^\\circ$\\\\\nIn right triangle $\\triangle ADC$\\\\\n$AB=2$, $AD=\\frac{1}{2}AB=1$\\\\\nTherefore, $BD=\\sqrt{3}$\\\\\nHence, the minimum value of $AP+PD$ is $\\boxed{\\sqrt{3}}$",
    "image_path": "dataset/mm_math/images/MM_Math/53080600.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, the quadrilateral $ABCD$ is a rhombus, $\\angle DAB=120^\\circ$, $M$ is a moving point on side $BC$, and $AM$ intersects side $BD$ at point $N$. When the length of segment $AM$ is minimal, $NM=1$. At this point, what is the distance from point $N$ to line $CD$?\n",
    "completion": "\\textbf{Solution}: Since quadrilateral ABCD is a rhombus and $\\angle DAB=120^\\circ$,\nit follows that $AD\\parallel BC$, and BD bisects $\\angle ADC$ and $\\angle ABC$.\nTherefore, $\\angle ABC=60^\\circ$,\nwhich means $\\angle DBC=30^\\circ$.\nWhen AM$\\perp BC$ and AM is shortest, then $\\angle BAM=30^\\circ$.\nGiven $NM=1$,\nwe find $BN=2$ and $BM=\\sqrt{2^2-1^2}=\\sqrt{3}$.\nThus, $AB=2\\sqrt{3}$ and $AM=\\sqrt{AB^2-BM^2}=3$,\nleading to $AN=AM-NM=3-1=2$.\nSince $AD\\parallel BC$ and AM$\\perp BC$,\nit follows that AM$\\perp AD$.\nBecause BD bisects $\\angle ADC$,\nthe distance from point N to line CD is equal to its distance to line AD, with NA=2.\nTherefore, at this point, the distance from point N to the line CD is $\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52211841.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the figure, suppose $\\triangle AED$ where $\\angle AED=90^\\circ$, $AB=AC=AD$, $EC=2$, and $BE=8$. If $AB=\\sqrt{41}$, what is the value of $AE$?\n",
    "completion": "\\textbf{Solution:} We have $AE = \\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52867068_60.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in circle $\\odot O$, through the midpoint $C$ of the radius $OD$, draw $AB\\perp OD$ intersecting $\\odot O$ at points $A$ and $B$, and $AB=2\\sqrt{3}$. What is the length of $OD$?\n",
    "completion": "\\textbf{Solution:} Since $AB \\perp OD$,\\\\\n$\\therefore \\angle OCB=90^\\circ$,\\\\\n$AC=BC= \\frac{1}{2} AB= \\sqrt{3}$,\\\\\nSince point C is the midpoint of OD,\\\\\n$\\therefore OC= \\frac{1}{2} OB$,\\\\\nSince $\\cos\\angle COB= \\frac{OC}{OB} = \\frac{1}{2}$,\\\\\n$\\therefore \\angle COB=60^\\circ$,\\\\\n$\\therefore OC= \\frac{\\sqrt{3}}{3} BC= \\frac{\\sqrt{3}}{3} \\times \\sqrt{3} =1$,\\\\\n$\\therefore OB=2OC=2$,\\\\\n$\\therefore OD=OB=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52754810_76.png"
  },
  {
    "solution": "\\boxed{-1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=ax+6$ intersects with the line $y=2x$ at point A $(m, 4)$, and intersects the x-axis at point B and the y-axis at point C. Find the values of $a$ and $m$?\n",
    "completion": "\\textbf{Solution:} Substitute point A$(m,4)$ into $y=2x$, we get $4=2m$\\\\\n$\\therefore m=2$\\\\\n$\\therefore$ the coordinates of point A are $(2,4)$,\\\\\nSubstituting point A$(2,4)$ into $y=ax+6$, we get $4=2a+6$\\\\\nSolving this, we find $a=-1$,\\\\\n$\\therefore m=\\boxed{2}$, $a=\\boxed{-1}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52189216_154.png"
  },
  {
    "solution": "\\boxed{\\frac{2\\sqrt{10}}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, let's consider a square $ABCD$ with a side length of $2$, where points $E$ and $F$ are located on sides $AD$ and $BC$, respectively. Points $G$ and $H$ are located on the sides of the square $ABCD$. Connect $GH$. If the angle between line segments $GH$ and $EF$ is $45^\\circ$ and $GH=\\sqrt{5}$, what is the length of $EF$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw BK $\\parallel$ EF intersecting AD at K, and BM $\\parallel$ GH intersecting CD at M,\\\\\nthen BK $=$ EF, BM $=$ GH $=$ $\\sqrt{5}$,\\\\\nsince the angle between line segments GH and EF is $45^\\circ$,\\\\\n$\\therefore \\angle KBM=45^\\circ$,\\\\\n$\\therefore \\angle ABK+\\angle CBM=90^\\circ-45^\\circ=45^\\circ$,\\\\\ndraw $\\angle KBN=45^\\circ$ intersecting the extension of AD at N,\\\\\nthen $\\angle ABN+\\angle ABK=45^\\circ$,\\\\\n$\\therefore \\angle ABN=\\angle CBM$,\\\\\nin $\\triangle ABN$ and $\\triangle CBM$,\\\\\n$\\left\\{\\begin{array}{l}\n\\angle ABN=\\angle CBM \\\\\nAB=BC \\\\\n\\angle BAN=\\angle C=90^\\circ\n\\end{array}\\right.$,\\\\\n$\\therefore \\triangle ABN \\cong \\triangle CBM$ (ASA),\\\\\n$\\therefore BN=BM, AN=CM$,\\\\\nin $\\triangle BCM$, CM $=$ $\\sqrt{BM^2-BC^2}=\\sqrt{5-4}=1$,\\\\\ndraw KP$\\perp$BN at point K through point P,\\\\\nsince $\\angle KBN=45^\\circ$,\\\\\n$\\therefore \\triangle BKP$ is an isosceles right-angled triangle,\\\\\nlet EF $=$ BK $=$ x, then BP $=$ KP $=$ $\\frac{\\sqrt{2}}{2}x$,\\\\\n$\\therefore NP=\\sqrt{5}-\\frac{\\sqrt{2}}{2}x$, $AK=\\sqrt{x^2-4}$,\\\\\nsince $\\frac{1}{2}BN \\cdot KP=\\frac{1}{2}KN \\cdot AB$,\\\\\n$\\therefore \\sqrt{5} \\times \\frac{\\sqrt{2}}{2}x=2(1+\\sqrt{x^2-4})$,\\\\\n$\\therefore 3x^2+4\\sqrt{10}x-40=0$,\\\\\n$\\therefore (3x-2\\sqrt{10})(x+2\\sqrt{10})=0$,\\\\\nsolving, we get $x=\\frac{2\\sqrt{10}}{3}$, $x=-2\\sqrt{10}$,\\\\\nupon verification: $x=\\frac{2\\sqrt{10}}{3}$ is a root of the original equation, $x=-2\\sqrt{10}$ is an extraneous root, discard it,\\\\\nthus, EF $=$ $\\boxed{\\frac{2\\sqrt{10}}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50482133.png"
  },
  {
    "solution": "\\boxed{22}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, within quadrilateral $ABCD$, points $E$ and $F$ are on the diagonal $AC$ such that $AE=EF=FC$. Quadrilateral $DEBF$ is a parallelogram; if $\\angle CDE=90^\\circ$, $DC=8$, and $DE=6$, what is the perimeter of quadrilateral $DEBF$?\n",
    "completion": "\\textbf{Solution:} Given, \\\\\n$\\because \\angle CDE=90^\\circ$, $DC=8$, $ED=6$, \\\\\n$\\therefore CE= \\sqrt{CD^2+DE^2}=\\sqrt{64+36}=10$, \\\\\n$\\because EF=CF$, \\\\\n$\\therefore DF= \\frac{1}{2}CE=5$, \\\\\n$\\therefore$ the perimeter of $\\square DEBF$ $=2(DF+DE)=2\\times(5+6)=\\boxed{22}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53490642_84.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In $\\triangle ABC$, points D and E are on AB and AC respectively, and $DE \\parallel BC$. If $AD=2$, $AB=3$, and $DE=4$, what is the length of $BC$?\n",
    "completion": "\\textbf{Solution:} Since $DE \\parallel BC$, \\\\\nit follows that $\\angle ADE = \\angle ABC$, $\\angle AED = \\angle ACB$, \\\\\ntherefore, $\\triangle ADE \\sim \\triangle ABC$, \\\\\nthus, $\\frac{BC}{DE} = \\frac{AB}{AD}$, that is $\\frac{BC}{4} = \\frac{3}{2}$, \\\\\nhence, $BC = \\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52796878.png"
  },
  {
    "solution": "\\boxed{\\left(5, -\\frac{27}{28}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, point $B$ is in the first quadrant. $BA\\perp x$-axis at point $A$, $BC\\perp y$-axis at point $C$, with $BA=3$ and $BC=5$. There is an inverse proportion function graph that exactly passes through point $B$. A moving point $P$ is on the ray $CA$ (excluding point $C$), and a line $l\\perp x$-axis is drawn through point $P$, intersecting the graph of the inverse proportion function at point $D$. Is there such a point $Q$, such that the quadrilateral with vertices $B$, $D$, $P$, and $Q$ forms a rhombus? Determine the coordinates of point $Q$.\n",
    "completion": "\\textbf{Solution:} Let point P be $ \\left(m, -\\frac{3}{5}m+3\\right)$ and point D be $ \\left(m, \\frac{15}{m}\\right)$.\\\\\n$\\therefore$PD$= \\frac{15}{m}-\\left(-\\frac{3}{5}m+3\\right)$,\\\\\n$\\because$The quadrilateral formed by points B, D, P, Q is a rhombus, and when point Q lies on line BA and PD=BD=BQ,\\\\\n$ \\frac{15}{m}-\\left(-\\frac{3}{5}m+3\\right)=\\sqrt{(m-5)^{2}+\\left(3-\\frac{15}{m}\\right)^{2}}$\\\\\nSimplifying gives $ \\frac{16}{25}m^{2}-\\frac{32}{5}m+7=0$\\\\\nSolving this yields $ m_{1}=\\frac{5}{4}, m_{2}=\\frac{35}{4}$.\\\\\nWhen $ m=\\frac{5}{4}$, $ PD=BQ=\\frac{15}{m}-\\left(-\\frac{3}{5}m+3\\right)=\\frac{39}{4}$,\\\\\n$\\because$ BA＝3,\\\\\n$\\therefore$the y-coordinate of point Q is $ 3-\\frac{39}{4}=-\\frac{27}{4}$,\\\\\n$\\therefore$Point Q is $ \\left(5, -\\frac{27}{4}\\right)$;\\\\\nWhen $ m=\\frac{35}{4}$, $ PD=BQ=\\frac{15}{m}-\\left(-\\frac{3}{5}m+3\\right)=\\frac{111}{28}$,\\\\\n$\\because$ BA＝3,\\\\\n$\\therefore$the y-coordinate of point Q is $ 3-\\frac{111}{28}=-\\frac{27}{28}$,\\\\\n$\\therefore$Point Q is $ \\left(5, -\\frac{27}{28}\\right)$;\\\\\nWhen point Q lies on line AC, PD and BQ bisect each other perpendicularly,\\\\\n$\\therefore$BQ$\\parallel$x-axis,\\\\\n$\\therefore$the y-coordinate of point Q is 3,\\\\\n$\\therefore$ $ \\frac{\\frac{3}{5}m+3+\\frac{15}{m}}{2}=3$\\\\\nSolving this yields $ m=\\frac{-5\\pm 5\\sqrt{3}}{2}$\\\\\n$\\because$m＞0,\\\\\n$\\therefore$ $ m=\\frac{-5+5\\sqrt{3}}{2}$\\\\\n$\\therefore$Point Q is $ \\left(\\frac{-5+5\\sqrt{3}}{2}, 3\\right)$,\\\\\n$\\therefore$The coordinates of point Q are $\\boxed{\\left(5, -\\frac{27}{4}\\right)}$ or $\\boxed{\\left(\\frac{-5+5\\sqrt{3}}{2}, 3\\right)}$ or $\\boxed{\\left(5, -\\frac{27}{28}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52160945_161.png"
  },
  {
    "solution": "\\boxed{(2, 3)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=ax^2-3ax-4a$ passes through point $C(0, 2)$ and intersects the x-axis at points $A$ and $B$ (point $A$ is to the left of point $B$). Line $BC$ is drawn, and the line $y=kx+1$ ($k>0$) intersects the y-axis at point $D$ and intersects the parabola above line $BC$ at point $E$, and intersect line $BC$ at point $F$. Does $\\frac{EF}{DF}$ have a maximum value? If it exists, please find the maximum value and the coordinates of point $E$ at that time; if it does not exist, please explain why.\n",
    "completion": "\\textbf{Solution:} Existence is shown as follows,\\\\\nAs illustrated, according to the problem, point $E$ is on the right side of the $y$-axis. Construct $EG\\parallel y$-axis, intersecting $BC$ at point $G$\\\\\n$\\therefore CD\\parallel EG$\\\\\n$\\therefore \\frac{EF}{DF}=\\frac{EG}{CD}$\\\\\n$\\because$ The line $y=kx+1\\,(k>0)$ intersects the $y$-axis at point $D$\\\\\n$\\therefore D(0, 1),$\\\\\n$\\therefore CD=2-1=1,$\\\\\n$\\therefore \\frac{EF}{DF}=EG$\\\\\nAssume the equation of the line where $BC$ lies is $y=mx+n\\,(m\\ne 0)$,\\\\\nSubstituting $B(4, 0), C(0, 2)$ into the above equation yields: $\\left\\{\\begin{array}{l} 0=4m+n \\\\ 2=n \\end{array}\\right.$\\\\\nSolving gives: $\\left\\{\\begin{array}{l} m=-\\frac{1}{2} \\\\ n=2 \\end{array}\\right.$\\\\\n$\\therefore$ The equation of line $BC$ is $y=-\\frac{1}{2}x+2$\\\\\nLet $E(t, -\\frac{1}{2}t^2+\\frac{3}{2}t+2)$\\\\\nThen $G(t, -\\frac{1}{2}t+2)$, where $0<t<4$.\\\\\n$\\therefore EG=-\\frac{1}{2}t^2+\\frac{3}{2}t+2-\\left(-\\frac{1}{2}t+2\\right)=-\\frac{1}{2}(t-2)^2+2$\\\\\n$\\therefore \\frac{EF}{DF}=-\\frac{1}{2}(t-2)^2+2,$\\\\\n$\\because -\\frac{1}{2}<0,$\\\\\n$\\therefore$ The parabola opens downwards\\\\\n$\\therefore$ At $t=2$, it has the maximum value, which is $\\boxed{2}$.\\\\\nSubstituting $t=2$ into $-\\frac{1}{2}t^2+\\frac{3}{2}t+2=-2+3+2=3$\\\\\n$\\therefore$ The coordinates of point $E$ are $\\boxed{(2, 3)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52977788_221.png"
  },
  {
    "solution": "\\boxed{y=-x+1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=2x+1$ intersects the $x$ axis at point $A$, and the $y$ axis at point $B$. Draw line $BP$ from point $B$ to intersect the $x$ axis at point $P$, and let $OP=2OA$, find the equation of the line $BP$.\n",
    "completion": "\\textbf{Solution:}\n\nGiven that the coordinates of point A is $\\left(-\\frac{1}{2}, 0\\right)$, \\\\\ntherefore, OA $= \\frac{1}{2}$, \\\\\nthus, OP $= 2\\text{OA} = 1$, \\\\\nwhen point P is on the positive half of the x-axis, then the coordinates of P is $(1,0)$, \\\\\nlet the equation of line BP be: $y=kx+b$, \\\\\nsubstituting P $(1,0)$ and B $(0,1)$ into the equation, we get\n\\[\n\\begin{cases}\nk+b=0\\\\\nb=1\n\\end{cases}\n\\]\nSolving this we get:\n\\[\n\\begin{cases}\nk=-1\\\\\nb=1\n\\end{cases}\n\\]\ntherefore, the equation of line BP is: $y=-x+1$; \\\\\nwhen point P is on the negative half of the x-axis, then the coordinates of P is $(-1,0)$, \\\\\nlet the equation of line BP be $y=mx+n$, \\\\\nsubstituting P $(-1, 0)$ and B $(0, 1)$, we get\n\\[\n\\begin{cases}\n-m+n=0\\\\\nn=1\n\\end{cases}\n\\]\nSolving this we get\n\\[\n\\begin{cases}\nm=1\\\\\nn=1\n\\end{cases}\n\\]\ntherefore, the equation of line BP is: $y=x+1$; \\\\\nIn summary, the equation of line BP is either \\boxed{y=x+1} or \\boxed{y=-x+1}.",
    "image_path": "dataset/mm_math/images/MM_Math/50519365_192.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y_1=k_1x+b$ intersects with the graph of the inverse proportion function $y_2=\\frac{k_2}{x}$ (where $x<0$) at points $A$ and $B$. The graphs also intersect the coordinate axes at $(-6, 0)$ and $(0, 6)$. The x-coordinate of point $B$ is $-4$. What is the area of $\\triangle AOB$?\n",
    "completion": "Solution: Since point A and point B are the intersection points of the inverse proportion function and the linear function,\n\nit follows that: $x+6=-\\frac{8}{x}$,\n\nSolving this, we get $x=-2$ or $x=-4$,\n\nhence $A(-2, 4)$,\n\ntherefore, the area of $\\triangle AOB$ is $6\\times 6\\div 2-6\\times 2=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52622495_249.png"
  },
  {
    "solution": "\\boxed{y=x^{2}-4x+3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, the parabola $y=x^2+bx+c$ intersects the $x$-axis at points $A$ and $B(3,0)$, and intersects the $y$-axis at point $C(0,3)$. Find the analytical expression of the parabola.\n",
    "completion": "\\textbf{Solution:} Substitute points B$(3,0)$, C$(0,3)$ into the parabola $y=x^{2}+bx+c$, we get:\n\\[\n\\left\\{\n\\begin{array}{l}\n0=9+3b+c\\\\\n3=c\n\\end{array}\n\\right.\n\\]\nSolving this, we obtain:\n\\[\n\\left\\{\n\\begin{array}{l}\nb=-4\\\\\nc=3\n\\end{array}\n\\right.\n\\]\n$\\therefore$ The equation of the parabola is \\boxed{y=x^{2}-4x+3}.",
    "image_path": "dataset/mm_math/images/MM_Math/53124516_61.png"
  },
  {
    "solution": "\\boxed{\\frac{13}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of the semicircle $\\odot O$, and $C,D$ are two points on the circle, with $OD\\parallel AC$ and $OD$ intersecting $BC$ at point $E$. $E$ is the midpoint of $BC$. If $BC=6$ and $DE=2$, what is the length of $AB$?\n",
    "completion": "\\textbf{Solution:} From (1), we have $BE=3$. Let $OB=OD=x$, then $OE=x-2$. In right triangle $OEB$, we have $BE^2+OE^2=OB^2$, that is $3^2+(x-2)^2=x^2$. Solving this, we get: $x=\\frac{13}{4}$, so $OB=\\frac{13}{4}$, therefore $AB=2OB=\\boxed{\\frac{13}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52747354_134.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that in quadrilateral $ABCD$, $AB \\parallel CD$, $BC=AD=4$, $AB=CD=10$, $\\angle DCB=90^\\circ$. $E$ is a point on the side $CD$, with $DE=7$. A moving point $P$ starts from point $A$ and moves towards point $B$ along side $AB$ at a speed of 1 unit per second. Line $PE$ is drawn, and the time taken for point $P$ to move is $t$ seconds. What is the length of $BE$?\n",
    "completion": "\\textbf{Solution:} Since $CD=10$ and $DE=7$, \\\\\nthus $CE=10-7=3$, \\\\\nSince $\\angle DCB=90^\\circ$, \\\\\nhence, in $\\triangle CBE$, $BE=\\sqrt{BC^2+CE^2}=\\sqrt{4^2+3^2}=\\boxed{5}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53339149_121.png"
  },
  {
    "solution": "\\boxed{\\frac{13}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the equilateral $\\triangle ABC$, $AB=4$, and $D$ and $P$ are points on the sides $BC$ and $AC$, respectively, with $BP=1$ and $\\angle APD=60^\\circ$. What is the length of $AD$?\n",
    "completion": "\\textbf{Solution:} Given that $\\because \\triangle ABC$ is an equilateral triangle,\\\\\n$\\therefore \\angle B=\\angle C=60^\\circ$,\\\\\nIn $\\triangle ABP$, $\\angle B+\\angle APB+\\angle BAP=180^\\circ$,\\\\\n$\\therefore \\angle BAP+\\angle APB=120^\\circ$,\\\\\n$\\because \\angle APB+\\angle CPD=180^\\circ-\\angle APD=120^\\circ$,\\\\\n$\\therefore \\angle BAP=\\angle CPD$,\\\\\n$\\therefore \\triangle ABP\\sim \\triangle PCD$,\\\\\n$\\therefore \\frac{AB}{PC}=\\frac{BP}{CD}$, which gives $\\frac{4}{4-1}=\\frac{1}{CD}$,\\\\\n$\\therefore CD=\\frac{3}{4}$,\\\\\n$\\therefore AD=AC-CD=\\boxed{\\frac{13}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52872013.png"
  },
  {
    "solution": "\\boxed{14}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle B=90^\\circ$, arcs are drawn with centers at $A$ and $C$ and radii greater than the length of $AC$, intersecting at points $M$ and $N$. Line $MN$ is drawn, intersecting $AC$ and $BC$ at $D$ and $E$ respectively. Connect $AE$. If $AB=6$ and $AC=10$, what is the perimeter of $\\triangle ABE$?\n",
    "completion": "\\textbf{Solution:} From the construction, we have $ED$ perpendicular bisector of $AC$,\\\\\n$\\therefore EA=EC$,\\\\\nIn $\\triangle ABC$, $BC= \\sqrt{AC^{2}-AB^{2}}=\\sqrt{10^{2}-6^{2}}=8$,\\\\\n$\\therefore$ The perimeter of $\\triangle ABE$ $=AB+BE+AE=AB+BE+CE=AB+BC=6+8=\\boxed{14}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52837588.png"
  },
  {
    "solution": "\\boxed{4\\text{cm}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, \\(CM\\) is the median of \\(\\triangle ABC\\), and \\(AM=4\\,cm\\), what is the length of \\(BM\\)?\n",
    "completion": "Solution: Since $CM$ is the median of $\\triangle ABC$ and $AM=4\\text{cm}$, \\\\\ntherefore $BM=AM=4\\text{cm}$,\\\\\nso $BM=\\boxed{4\\text{cm}}$",
    "image_path": "dataset/mm_math/images/MM_Math/53044131.png"
  },
  {
    "solution": "\\boxed{80^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle BAC=130^\\circ$, the perpendicular bisectors of $AB$ and $AC$ intersect $BC$ at points $E$ and $F$, and intersect $AB$ and $AC$ at points $D$ and $G$, respectively. What is the measure of $\\angle EAF$?\n",
    "completion": "\\textbf{Solution:} \\\\\nSince DE bisects AB perpendicularly, and FG bisects AC perpendicularly, \\\\\nit follows that EB=EA, and FA=FC, \\\\\nthus, $\\angle$BAE=$\\angle$B, $\\angle$FAC=$\\angle$C, \\\\\nSince in $\\triangle ABC$, $\\angle$BAC=$130^\\circ$, \\\\\nit follows that $\\angle$B+$\\angle$C=$50^\\circ$, \\\\\nthus, $\\angle$BAE+$\\angle$FAC=$50^\\circ$, \\\\\ntherefore, $\\angle$EAF=$\\angle$BAC−($\\angle$BAE+$\\angle$FAC)=$\\boxed{80^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51367760.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In rectangle $ABCD$, where $AB=6, BC=8$, the diagonals intersect at point $O$. What is the length of $BO$?\n",
    "completion": "\\textbf{Solution:} \nSince quadrilateral ABCD is a rectangle, and $AB=6$, $BC=8$, \\\\\nit follows that $\\angle ABC=90^\\circ$, \\\\\ntherefore $BD=AC=\\sqrt{AB^{2}+BC^{2}}=10$, \\\\\nsince $OB=\\frac{1}{2}BD$, \\\\\nit follows that $OB=\\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52422309.png"
  },
  {
    "solution": "\\boxed{2-\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated, in square $ABCD$ with side length $AB=2$, point $P$ is located on side $AB$ (not coinciding with $A$ or $B$). A square $PBEF$ is constructed inside square $ABCD$ with $PB$ as a side, intersecting side $BC$ at point $E$. Lines $DF$ and $CF$ are drawn. When $\\triangle CDF$ is an isosceles triangle with $D$ as the apex, what is the length of $PB$?\n",
    "completion": "\\textbf{Solution:} Connect BF,\\\\\nQuadrilaterals ABCD, PBEF are squares, AB=2,\\\\\n$\\therefore$ Points B, F, D are collinear, CD=AB=2, $\\angle A=90^\\circ$\\\\\n$\\therefore$ $BD=\\sqrt{AD^{2}+AB^{2}}=\\sqrt{2^{2}+2^{2}}=2\\sqrt{2}$\\\\\n$\\because$ $\\triangle CDF$ is an isosceles triangle with D as the vertex,\\\\\n$\\therefore$FD=CD=2, BF=BD-FC=$2\\sqrt{2}-2$,\\\\\n$\\because$2PB$^{2}$=BF$^{2}$\\\\\n$\\therefore$ $PB=\\frac{\\sqrt{2}}{2}BF=\\frac{\\sqrt{2}}{2}\\left(2\\sqrt{2}-2\\right)=\\boxed{2-\\sqrt{2}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52251415.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system, the graph of the linear function $y=kx+b$ passes through point $A(-2, 6)$, intersects the $x$ axis at point $B$, and intersects the graph of the direct proportion function $y=3x$ at point $C$, where the $x$-coordinate of point $C$ is $1$. Find the values of $k$ and $b$?\n",
    "completion": "\\textbf{Solution:} When $x=1$, $y=3x=3$. Therefore, the coordinates of point $C$ are $(1, 3)$. Since the line $y=kx+b$ passes through $(-2, 6)$ and $(1, 3)$, we have\n\\[\n\\begin{cases}\n6=-2k+b \\\\\n3=k+b\n\\end{cases}\n\\]\nSolving this system of equations, we find\n\\[\n\\begin{cases}\nk=\\boxed{-1} \\\\\nb=\\boxed{4}\n\\end{cases}\n\\]\nTherefore, the explicit formula for the linear function is $y=-x+4$, which means $k=-1$, $b=4$.",
    "image_path": "dataset/mm_math/images/MM_Math/52381468_174.png"
  },
  {
    "solution": "\\boxed{55^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the quadrilateral $ABCD$ is an inscribed quadrilateral of the circle, with $AB$ as the diameter. It is given that $\\overset{\\frown}{DC} = \\overset{\\frown}{CB}$, and if $\\angle DCB = 110^\\circ$, what is the degree measure of $\\angle ABC$?\n",
    "completion": "\\textbf{Solution:} Connect AC, \\\\\n$\\because$ Quadrilateral ABCD is inscribed in circle O, \\\\\n$\\therefore$ $\\angle$DAB + $\\angle$DCB = 180$^\\circ$, \\\\\n$\\therefore$ $\\angle$DAB = 180$^\\circ$ − 110$^\\circ$ = 70$^\\circ$, \\\\\n$\\because$ $ \\overset{\\frown}{DC} = \\overset{\\frown}{CB}$, \\\\\n$\\therefore$ $\\angle$DAC = $\\angle$CAB = $\\frac{1}{2}$ $\\angle$DAB = $\\frac{1}{2}$ × 70$^\\circ$ = 35$^\\circ$; \\\\\n$\\because$ AB is the diameter, \\\\\n$\\therefore$ $\\angle$ACB = 90$^\\circ$, \\\\\n$\\therefore$ $\\angle$ABC = 90$^\\circ$ − $\\angle$CAB = 90$^\\circ$ − 35$^\\circ$ = \\boxed{55^\\circ}.",
    "image_path": "dataset/mm_math/images/MM_Math/52861148.png"
  },
  {
    "solution": "\\boxed{\\frac{7}{6}\\,\\text{cm}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, there is a rectangular piece of paper $ABCD$, where $AD=8\\text{cm}$ and $AB=6\\text{cm}$. First, fold along the diagonal $BD$, so that point $C$ falls on the position of point $C'$, and $BC'$ intersects $AD$ at point $G$. It is given that $BG=DG$; as shown in Figure 2, fold it again such that point $D$ coincides with point $A$. The crease $EN$ intersects $AD$ at $M$. Find the length of $EM$ in cm.\n",
    "completion": "\\textbf{Solution:} Since point $D$ coincides with point $A$, we get the crease $EN$,\\\\\n$\\therefore DM=4\\,\\text{cm}$,\\\\\n$\\because AD=8\\,\\text{cm}$, $AB=6\\,\\text{cm}$,\\\\\n$\\therefore$ in $\\triangle ABD$, $BD=10\\,\\text{cm}$,\\\\\n$\\because EN\\perp AD$, $AB\\perp AD$,\\\\\n$\\therefore EN\\parallel AB$,\\\\\n$\\therefore MN$ is the midline of $\\triangle ABD$,\\\\\n$\\therefore DN=\\frac{1}{2}BD=5\\,\\text{cm}$,\\\\\nIn $\\triangle MND$, $MN=\\sqrt{5^2-4^2}=3\\,\\text{cm}$,\\\\\nBy the properties of folding, it is known that $\\angle NDE=\\angle NDC$,\\\\\n$\\because EN\\parallel CD$,\\\\\n$\\therefore \\angle END=\\angle NDC$,\\\\\n$\\therefore \\angle END=\\angle NDE$,\\\\\n$\\therefore EN=ED$,\\\\\nLet $EM=x$, then $ED=EN=x+3$,\\\\\nBy the Pythagorean theorem, $ED^2=EM^2+DM^2$, that is, $(x+3)^2=x^2+4^2$,\\\\\nSolving for $x$, we get $x=\\frac{7}{6}$, thus $EM=\\boxed{\\frac{7}{6}\\,\\text{cm}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51920381_134.png"
  },
  {
    "solution": "\\boxed{\\frac{50\\sqrt{2}}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $AB$ is the diameter of circle $O$, and chord $CD$ intersects diameter $AB$ at point $E$. Given $AC=CD$, connect $CO$ and extend it to intersect chord $AD$ at point $F$. Connect $AC$, $BC$, $BD$, and given $AB=10$, $BE=2$. What is the area of $\\triangle ABC$?\n",
    "completion": "\\textbf{Solution:} In $\\triangle \\mathrm{ACF}$, $\\angle \\mathrm{AFC} = 90^\\circ$, $\\mathrm{AF} = \\frac{10\\sqrt{2}}{3}$, $\\mathrm{CF} = \\mathrm{OC} + \\mathrm{OF} = 5 + \\frac{5}{3} = \\frac{20}{3}$, \\\\\n$\\therefore \\mathrm{AC} = \\sqrt{\\mathrm{AF}^2 + \\mathrm{CF}^2} = \\sqrt{\\left(\\frac{10\\sqrt{2}}{3}\\right)^2 + \\left(\\frac{20}{3}\\right)^2} = \\frac{10\\sqrt{6}}{3}$, \\\\\nIn $\\triangle \\mathrm{ABC}$, $\\mathrm{BC} = \\sqrt{\\mathrm{AB}^2 - \\mathrm{AC}^2} = \\sqrt{10^2 - \\left(\\frac{10\\sqrt{6}}{3}\\right)^2} = \\frac{10\\sqrt{3}}{3}$, \\\\\n$\\therefore S_{\\triangle \\mathrm{ABC}} = \\frac{1}{2} \\cdot \\mathrm{AC} \\cdot \\mathrm{BC} = \\frac{1}{2} \\times \\frac{10\\sqrt{6}}{3} \\times \\frac{10\\sqrt{3}}{3} = \\boxed{\\frac{50\\sqrt{2}}{3}}.$",
    "image_path": "dataset/mm_math/images/MM_Math/51317366_156.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\angle ACD=90^\\circ$, $\\angle D=15^\\circ$, and point $B$ is on the perpendicular bisector of $AD$. If $AC=4$, what is the length of $AB$?\n",
    "completion": "\\textbf{Solution:} Since point B is on the perpendicular bisector of AD,\\\\\nit follows that $BA=BD$,\\\\\nwhich means $\\angle D=\\angle BAD=15^\\circ$\\\\\nThus, $\\angle ABC=\\angle D+\\angle BAD=30^\\circ$\\\\\nSince $\\angle ACD=90^\\circ$ and $AC=4$,\\\\\nit follows that $AB=2AC=\\boxed{8}$",
    "image_path": "dataset/mm_math/images/MM_Math/53148185.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{15}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\triangle ABC$, M and N are points on the sides AB and AC, respectively, where AM = $ \\frac{1}{4}$AB and AN = $ \\frac{1}{4}$AC. What is the ratio of the area of $ \\triangle AMN$ to the area of the quadrilateral MBCN?\n",
    "completion": "\\textbf{Solution:} Since $AM = \\frac{1}{4}AB$ and $AN = \\frac{1}{4}AC$, and $\\angle MAN = \\angle BAC$,\n\n$\\therefore \\triangle AMN \\sim \\triangle ABC$, \n\n$\\therefore \\frac{S_{\\triangle AMN}}{S_{\\triangle ABC}} = \\left(\\frac{AM}{AB}\\right)^2 = \\frac{1}{16}$,\n\n$\\therefore S_{\\triangle AMN} = \\frac{1}{16}S_{\\triangle ABC}$,\n\n$\\therefore \\frac{S_{\\triangle AMN}}{S_{\\text{quadrilateral }MBCN}} = \\frac{S_{\\triangle AMN}}{S_{\\triangle ABC} - S_{\\triangle AMN}} = \\boxed{\\frac{1}{15}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52745998.png"
  },
  {
    "solution": "\\boxed{70^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the direction of ray $OA$ is $15^\\circ$ east of north, and the direction of ray $OB$ is $40^\\circ$ west of north. $OA$ is the angle bisector of $\\angle BOC$. The ray $OF$ is the extension in the opposite direction of $OB$. Find the degree measure of $\\angle COF$.\n",
    "completion": "\\textbf{Solution:} Since $\\angle BOC = \\angle AOB + \\angle AOC$\\\\\n$=55^\\circ \\times 2 = 110^\\circ$,\\\\\nthen $\\angle COF = 180^\\circ - \\angle BOC$\\\\\n$=180^\\circ - 110^\\circ$\\\\\n$=70^\\circ$.\\\\\nThus, the measure of $\\angle COF$ is $\\boxed{70^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53404231_160.png"
  },
  {
    "solution": "\\boxed{(1, -3)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given: As shown in the figure, the linear function $y_{1}=kx-2$ intersects with the x-axis at point $B(-2, 0)$, $y_{2}=x+b$ intersects with the x-axis at point $C(4, 0)$, and the graphs of these two functions intersect at point $A$. Find the values of $k$ and $b$ and the coordinates of point $A$.\n",
    "completion": "\\textbf{Solution:} Substitute $B(-2, 0)$ into $y_1=kx-2$ to get, $0=-2k-2$,\\\\\nthus, $k=-1$;\\\\\nSubstitute $C(4, 0)$ into $y_2=x+b$ to get, $0=4+b$,\\\\\nthus, $b=-4$;\\\\\nSolving the system of equations yields, $\\left\\{\\begin{array}{l}y=-x-2 \\\\ y=x-4\\end{array}\\right.$,\\\\\nwhich gives, $\\left\\{\\begin{array}{l}x=1 \\\\ y=-3\\end{array}\\right.$,\\\\\nThe coordinates of point A are: $A(1, -3)$.\\\\\nTherefore, $k=\\boxed{-1}$, $b=\\boxed{-4}$, and the coordinates of point $A$ are $\\boxed{(1, -3)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52381120_148.png"
  },
  {
    "solution": "\\boxed{y=-x^{2}-2x+3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=ax^{2}+bx+3$ ($a\\neq 0$) intersects with the x-axis at points $A(1,0)$ and $B(-3,0)$ and with the y-axis at point $C$. Line segment $BC$ intersects with the axis of symmetry of the parabola at point $E$, with the vertex being point $D$. Find the equation of the parabola.\n",
    "completion": "\\textbf{Solution:} Given that the parabola $y=ax^{2}+bx+3$ passes through points $A(1,0)$ and $B(-3,0)$,\\\\\nwe have $\\begin{cases} a+b+3=0 \\\\ 9a-3b+3=0 \\end{cases}$,\\\\\nsolving these equations yields $\\begin{cases} a=-1 \\\\ b=-2 \\end{cases}$,\\\\\ntherefore, the equation of the parabola is: \\boxed{y=-x^{2}-2x+3}.",
    "image_path": "dataset/mm_math/images/MM_Math/52602919_158.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that in $\\triangle ABC$ and $\\triangle CDB$, $\\angle ACB=\\angle CDB=90^\\circ$, $AC=BC$, $CD=BD$. Connect $AD$ and intersect $BC$ at point $P$. What is the value of $\\tan\\angle CAP$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle ACB=\\angle CDB=90^\\circ$ and $AC=BC$, $CD=BD$,\\\\\nit follows that $\\triangle ACB$ and $\\triangle CDB$ are isosceles right triangles,\\\\\ntherefore $\\angle ABC+\\angle CBD=90^\\circ$\\\\\nthus $\\angle ABD+\\angle CDB=180^\\circ$\\\\\nwhich means $CD\\parallel AB$\\\\\nhence $\\triangle CDP\\sim \\triangle BAP$,\\\\\nlet $CD=DB=a$, then $CB=\\sqrt{2}a$, $AB=2a$\\\\\ntherefore, $\\frac{CD}{AB}=\\frac{CP}{BP}=\\frac{1}{2}$\\\\\nthus, $\\frac{CP}{BC}=\\frac{1}{3}$\\\\\nSince $AC=BC$\\\\\nit follows that $\\tan\\angle CAP=\\boxed{\\frac{1}{3}}$",
    "image_path": "dataset/mm_math/images/MM_Math/53079124.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Consider a point $E$ on the side of rectangle $ABCD$. Fold $\\triangle BCE$ along $BE$ such that point $C$ exactly aligns with point $F$ on side $AD$. As shown in the diagram, extend $EF$ and let it intersect the angle bisector of $\\angle ABF$ at point $M$. Let $BM$ intersect $AD$ at point $N$. Given that $NF = AN + FD$, find the value of $\\frac{AB}{BC}$.\n",
    "completion": "\\textbf{Solution:} Draw $NG \\perp BF$ at point G,\n\nsince $NF=AN+FD$,\n\nthus $NF= \\frac{1}{2}AD= \\frac{1}{2}BC$,\n\nsince $BC=BF$,\n\nthus $NF= \\frac{1}{2}BF$,\n\nsince $\\angle NFG=\\angle AFB$, $\\angle NGF=\\angle BAF=90^\\circ$,\n\nthus $\\triangle NFG \\sim \\triangle BFA$,\n\nthus $\\frac{NG}{AB}=\\frac{FG}{FA}=\\frac{NF}{BF}=\\frac{1}{2}$,\n\nLet $AN=x$,\n\nsince $BN$ bisects $\\angle ABF$, $AN \\perp AB$, $NG \\perp BF$,\n\nthus $AN=NG=x$, $AB=BG=2x$,\n\nLet $FG=y$, then $AF=2y$,\n\nsince $AB^2+AF^2=BF^2$,\n\nthus $(2x)^2+(2y)^2=(2x+y)^2$,\n\nsolving yields $y=\\frac{4}{3}x$,\n\nthus $BF=BG+GF=2x+\\frac{4}{3}x=\\frac{10}{3}x$,\n\nthus $\\frac{AB}{BC}=\\frac{AB}{BF}=\\frac{2x}{\\frac{10}{3}x}=\\boxed{\\frac{3}{5}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51402971_127.png"
  },
  {
    "solution": "\\boxed{120^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $\\angle AOB= 40^\\circ$. As shown in figure 2, $\\angle AOC= 30^\\circ$, $OM$ is a straight line inside $\\angle AOB$, $ON$ is the quadrisection line of $\\angle MOC$, and $3\\angle CON=\\angle NOM$, what is the value of $4\\angle AON+\\angle COM$?\n",
    "completion": "\\textbf{Solution:} Let $\\angle CON = \\alpha$, \\\\\nsince ON is the angle bisector of $\\angle MOC$, \\\\\nthus $\\angle MOC = 4\\alpha$. \\\\\nSince $3\\angle CON = \\angle NOM$, \\\\\nthus $\\angle MON = 3\\angle CON = 3\\alpha$ \\\\\nSince $\\angle AOC = 30^\\circ$, \\\\\nthus $\\angle AON = \\angle AOC - \\angle CON = 30^\\circ - \\alpha$, \\\\\ntherefore $4\\angle AON + \\angle COM = 4(30^\\circ-\\alpha) + 4\\alpha = \\boxed{120^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55192442_108.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is a chord of the circle $\\odot O$, $OP\\perp OA$ intersects $AB$ at point $P$, and the line passing through point $B$ intersects the extension of $OP$ at point $C$, and $BC$ is a tangent to $\\odot O$. If $OA=6$, $OP=2$, find the length of $CB$.\n",
    "completion": "\\textbf{Solution}: Let $BC=x$, then $PC=x$,\\\\\nin $\\triangle OBC$, $OB=OA=6$, $OC=CP+OP=x+2$,\\\\\nsince $OB^2+BC^2=OC^2$,\\\\\nthus $6^2+x^2=(x+2)^2$,\\\\\nwe find $x=8$,\\\\\nthat is, the length of $BC$ is $\\boxed{8}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51322517_139.png"
  },
  {
    "solution": "\\boxed{75^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $ \\angle AOB = 120^\\circ$ and $ \\angle BOC = 30^\\circ$. If $OD$ is the angle bisector of $ \\angle AOC$, find the measure of $ \\angle BOD$.\n",
    "completion": "\\textbf{Solution:} Given that $\\because \\angle AOB=120^\\circ$ and $\\angle BOC=30^\\circ$,\\\\\n$\\therefore \\angle AOC=\\angle AOB - \\angle BOC=90^\\circ$,\\\\\nAlso, $\\because$ OD is the bisector of $\\angle AOC$,\\\\\n$\\therefore \\angle COD=\\frac{1}{2}\\angle AOC=45^\\circ$,\\\\\n$\\therefore \\angle BOD=\\angle COD+\\angle BOC =45^\\circ+30^\\circ=\\boxed{75^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/47019465.png"
  },
  {
    "solution": "\\boxed{(2\\sqrt{3},1)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, points B and C are on the y-axis, and $\\triangle ABC$ is an equilateral triangle with $AB=4$. The intersection point D of $AC$ and the x-axis has coordinates $(\\sqrt{3}, 0)$. What are the coordinates of point A?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw AE$\\perp$BC through point A, thus AE$\\parallel$OD, \\\\\nbecause $\\triangle ABC$ is an equilateral triangle, \\\\\ntherefore $BC=AB=4$, $BE=CE=\\frac{1}{2}BC=2$, \\\\\ntherefore $AE=\\sqrt{AB^{2}-BE^{2}}=\\sqrt{4^{2}-2^{2}}=2\\sqrt{3}$, \\\\\nbecause $D(\\sqrt{3},0)$, \\\\\ntherefore $OD=\\sqrt{3}$, \\\\\ntherefore point O is the midpoint of CE, \\\\\ntherefore $OE=\\frac{1}{2}CE=1$, \\\\\ntherefore the coordinates of point A are $\\boxed{(2\\sqrt{3},1)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51366536.png"
  },
  {
    "solution": "\\boxed{9 \\text{ or } 18}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the rectangle $ABCD$, it is known that $AD=18$ and $AB=24$. Point $E$ is a moving point on side $DC$, and $\\triangle AD'E$ and $\\triangle ADE$ are symmetrical with respect to line $AE$. When $\\triangle CD'E$ is a right-angled triangle, what is the length of $DE$?\n",
    "completion": "\\textbf{Solution:}\n\n(1) When $\\angle CED'=90^\\circ$, as shown in Figure (1),\\\\\n$\\because \\angle CED'=90^\\circ$,\\\\\naccording to the property of axial symmetry, $\\angle AED=\\angle AED'=\\frac{1}{2}\\times90^\\circ=45^\\circ$,\\\\\n$\\because \\angle D=90^\\circ$,\\\\\n$\\therefore \\triangle ADE$ is an isosceles right triangle,\\\\\n$\\therefore DE=AD=18$;\\\\\n(2) When $\\angle ED'A=90^\\circ$, as shown in Figure (2),\\\\\naccording to the property of axial symmetry, $\\angle AD'E=\\angle D=90^\\circ$, $AD'=AD$, $DE=D'E$,\\\\\n$\\triangle CD'E$ is a right triangle,\\\\\nthat is $\\angle CD'E=90^\\circ$,\\\\\n$\\therefore \\angle AD'E+\\angle CD'E=180^\\circ$,\\\\\n$\\therefore A$, $D'$, $C$ are collinear,\\\\\naccording to the Pythagorean theorem, $AC=\\sqrt{AD^{2}+CD^{2}}=30$,\\\\\n$\\therefore CD'=30-18=12$,\\\\\nlet $DE=D'E=x$, then $EC=CD-DE=24-x$,\\\\\nin $\\triangle D'EC$, $D'E^{2}+D'C^{2}=EC^{2}$,\\\\\nthat is $x^{2}+144=(24-x)^{2}$,\\\\\nsolving for $x$ gives $x=9$,\\\\\nthat is $DE=9$;\\\\\nIn summary: The length of $DE$ is either 9 or 18;\\\\\nHence, the answer is: \\boxed{9 \\text{ or } 18}.",
    "image_path": "dataset/mm_math/images/MM_Math/53578729.png"
  },
  {
    "solution": "\\boxed{24}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given rectangle $ABCD$, with point $E$ on side $AD$, $BE$ and $BD$ are connected, $\\angle BED=2\\angle BDC$, $BE=25$, $BC=32$, what is the length of $CD$?\n",
    "completion": "\\textbf{Solution:} Construct $EF\\perp BD$ at point $F$,\\\\\nsince quadrilateral $ABCD$ is a rectangle,\\\\\nthus $\\angle C=\\angle ADC=90^\\circ$,\\\\\nsince $\\angle BED=2\\angle BDC$,\\\\\nthus $180^\\circ-\\angle AEB=2\\left(90^\\circ-\\angle ADB\\right)$,\\\\\nthus $\\angle AEB=2\\angle EDB$,\\\\\nsince $\\angle AEB=\\angle ADB+\\angle EBD$,\\\\\nthus $\\angle EDB=\\angle EBD$,\\\\\nthus $BE=DE$,\\\\\nthus $BF=\\frac{1}{2}BD$,\\\\\nsince $AD\\parallel BC$,\\\\\nthus $\\angle ADB=\\angle DBC$,\\\\\nthus $\\angle EBF=\\angle DBC$,\\\\\nthus $\\triangle EBF\\sim \\triangle DBC$,\\\\\nthus $\\frac{BE}{BD}=\\frac{BF}{BC}$,\\\\\nthus $BD^2=2BC\\cdot BE=2\\times 25\\times 32=40^2$,\\\\\nthus $BD=40$,\\\\\nthus $CD=\\sqrt{BD^2-BC^2}=\\boxed{24}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52623151.png"
  },
  {
    "solution": "\\boxed{(10,3)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the edges $OB$ and $OA$ of the rectangle $AOBC$ are on the $x$-axis and $y$-axis, respectively, and point $D$ is on edge $BC$. When the rectangle is folded along $AD$, point $C$ falls exactly on point $E$ on edge $OB$. If point $A$ is at $(0, 8)$ and point $B$ is at $(10, 0)$, what are the coordinates of point $D$?\n",
    "completion": "\\textbf{Solution:} Given $A(0,8)$ and point $B(10,0)$, $\\therefore$ $OA=BC=8$, $OB=AC=10$. Let $BD=a$, then $CD=8-a$. According to the problem, $CD=DE=8-a$. Knowing from folding, $AE=AC=10$, $\\therefore$ $OE=\\sqrt{AE^{2}-AO^{2}}=\\sqrt{10^{2}-8^{2}}=6$, $\\therefore$ $BE=OB-OE=10-6=4$. Since $\\angle DBE=90^\\circ$, $\\therefore$ $a^{2}+4^{2}=(8-a)^{2}$. Solving this, we find $a=3$. Therefore, the coordinates of point D are $(10,3)$. Hence, the answer is: $\\boxed{(10,3)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51434867.png"
  },
  {
    "solution": "\\boxed{y=\\frac{1}{2}x+2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system $xOy$, line $AB$ intersects the $x$-axis and $y$-axis at point $A$ and point $B$, respectively, and line $CD$ intersects the $x$-axis and $y$-axis at point $C$ and point $D$, respectively. The equation of line $AB$ is $y=-\\frac{5}{8}x+5$, and the equation of line $CD$ is $y=kx+b\\ (k\\neq 0)$. The two lines intersect at point $E\\left(m, \\frac{10}{3}\\right)$, and $OB:OC=5:4$. What is the equation of line $CD$?\n",
    "completion": "\\textbf{Solution:} Let's substitute point $E\\left(\\frac{8}{3}, \\frac{10}{3}\\right)$ into $y=-\\frac{5}{8}x+5$, then we get $m=\\frac{8}{3}$, hence the coordinates of point $E$ are $\\left(\\frac{8}{3}, \\frac{10}{3}\\right)$. Setting $x=0$ in $y=-\\frac{5}{8}x+5$, we find $y=5$, hence the coordinates of point $B$ are $(0, 5)$, thus $OB=5$. Since $OB:OC=5:4$, it follows that $OC=4$, hence the coordinates of point $C$ are $(-4, 0)$. Substituting $E\\left(\\frac{8}{3}, \\frac{10}{3}\\right)$, $C(-4, 0)$ into $y=kx+b$ (where $k\\neq 0$), we obtain\n\\[\n\\left\\{\n\\begin{array}{l}\n\\frac{8}{3}k+b=\\frac{10}{3}\\\\\n-4k+b=0\n\\end{array}\n\\right.\n\\]\nSolving the system, we find\n\\[\n\\left\\{\n\\begin{array}{l}\nk=\\frac{1}{2}\\\\\nb=2\n\\end{array}\n\\right.\n\\]\nTherefore, the equation of line $CD$ is $\\boxed{y=\\frac{1}{2}x+2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52304489_229.png"
  },
  {
    "solution": "\\boxed{40}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the distance from the midpoint \\(M\\) of one side of the rhombus \\(ABCD\\) to the intersection point \\(O\\) of the diagonals is \\(5 \\, \\text{cm}\\). What is the perimeter of the rhombus \\(ABCD\\)?\n",
    "completion": "Solution: Since quadrilateral ABCD is a rhombus,\\\\\nit follows that $\\angle AOB=90^\\circ$,\\\\\nsince M is the midpoint of side AB,\\\\\nit follows that AB=2OM=10,\\\\\ntherefore, the perimeter of rhombus ABCD is $10\\times 4=\\boxed{40}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52354153.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $P$ is any point on the diagonal $AC$ of the square $ABCD$, with $PE \\perp AB$ at $E$, and $PF \\perp BC$ at $F$. If $AC= \\sqrt{2}$, then what is the value of $PE+PF$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral $ABCD$ is a square,\\\\\nthus $\\angle A=90^\\circ$, and $AB=BC$,\\\\\nthus $AB^2+BC^2=AC^2$,\\\\\nsince $AC=\\sqrt{2}$,\\\\\nthus $AB=BC=1$,\\\\\nsince quadrilateral $ABCD$ is a square,\\\\\nthus $\\angle BAC=\\angle BCA=45^\\circ$,\\\\\nsince $PE\\perp AB$ at $E$, and $PF\\perp BC$ at $F$,\\\\\nthus quadrilateral $PEBF$ is a rectangle, and $\\triangle AEP$ and $\\triangle PFC$ are isosceles right triangles,\\\\\nthus $PF=BE$, $PE=AE$,\\\\\nthus $PE+PF=AB=\\boxed{1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51863531.png"
  },
  {
    "solution": "\\boxed{\\frac{20}{9}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $O$ is a point on side $AB$ of $\\triangle ABC$, where $\\angle ACB = 90^\\circ$. A circle $\\odot O$ with radius OB is drawn and tangent to AC at point D. If $BC = 4$ and $\\sin A = \\frac{4}{5}$, what is the length of the radius of $\\odot O$?\n",
    "completion": "\\textbf{Solution:} Solve: In $\\triangle ABC$ with a right angle at $B$, we have $BC=4$, and $\\sin A=\\frac{4}{5}$,\\\\\n\\(\\therefore \\frac{BC}{AB}=\\frac{4}{5}\\), that is, \\(\\frac{4}{AB}=\\frac{4}{5}\\),\\\\\n\\(\\therefore AB=5\\),\\\\\nConnect $OD$,\\\\\n\\(\\because AC\\) is a tangent to circle $O$,\\\\\n\\(\\therefore OD \\perp AC\\),\\\\\nLet the radius of circle $O$ be $r$, then $OD = OB = r$,\\\\\n\\(\\therefore AO=5-r\\),\\\\\nIn $\\triangle AOD$ with a right angle at $O$, and \\(\\sin A = \\frac{4}{5}\\),\\\\\n\\(\\therefore \\frac{OD}{AO} = \\frac{4}{5}\\), that is, \\(\\frac{r}{5-r} = \\frac{4}{5}\\),\\\\\n\\(\\therefore r=\\boxed{\\frac{20}{9}}\\).",
    "image_path": "dataset/mm_math/images/MM_Math/51353198.png"
  },
  {
    "solution": "\\boxed{\\frac{25}{12}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, it is known that the graph of the quadratic function $y=-\\frac{4}{3}(x+1)^2+\\frac{16}{3}$ intersects the x-axis at points A and B (with point A to the left of B) and intersects the y-axis at point C. Point D is the vertex of the parabola. As shown in Figure 2, connect AC and DC, and draw $CE\\parallel AB$ through point C to intersect the parabola at point E. As shown in Figure 3, $\\angle DCE = \\angle CAO$; connect BC, and on the extension of EC there is a point P such that the triangle with vertices D, E, P is similar to $\\triangle ABC$. What is the length of EP?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect DE,\\\\\n$\\because A(-3,0), B(1,0), C(0,4), D(-1,\\frac{16}{3})$,\\\\\n$\\therefore AC=\\sqrt{3^2+4^2}=5$, AB=4, $DC=\\sqrt{1^2+\\left(\\frac{4}{3}\\right)^2}=\\frac{5}{3}$,\\\\\nUsing the symmetry property of the parabola, we have: $DE=DC=\\frac{5}{3}$, CE=2,\\\\\n$\\therefore \\angle ECD=\\angle DEC$,\\\\\nFrom (2) we get $\\angle ECD=\\angle CAO$,\\\\\n$\\therefore \\angle DEC=\\angle CAB$,\\\\\n$\\because \\triangle DEP$ and $\\triangle ABC$ are similar,\\\\\n(1) When $\\triangle DEP \\sim \\triangle CAB$, we have $\\frac{DE}{AC}=\\frac{EP}{AB}$,\\\\\n$\\therefore \\frac{\\frac{5}{3}}{5}=\\frac{EP}{4}$,\\\\\n$\\therefore EP=\\boxed{\\frac{4}{3}}$.\\\\\n(2) When $\\triangle DEP \\sim \\triangle BAC$, we have $\\frac{DE}{AB}=\\frac{EP}{AC}$,\\\\\n$\\therefore \\frac{\\frac{5}{3}}{4}=\\frac{EP}{5}$,\\\\\n$\\therefore EP=\\frac{25}{12}$.\\\\\nIn conclusion, EP=$\\boxed{\\frac{4}{3}}$ or $\\boxed{\\frac{25}{12}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52872397_97.png"
  },
  {
    "solution": "\\boxed{(4,4)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, the quadrilateral $ABCD$ is a rhombus. The coordinates of vertices $A$ and $C$ are $(0, 2)$ and $(8, 2)$ respectively. Given that point $D$ is on the x-axis, what are the coordinates of vertex $B$?\n",
    "completion": "\\textbf{Solution:} Connect AC and BD, where AC and BD intersect at point N, as shown in the figure,\\\\\n$\\because$A(0,2), C(8,2),\\\\\n$\\therefore$Points A and C lie on the line $y=2$, and AC=8, OA=2,\\\\\n$\\therefore$AC$\\perp$y-axis,\\\\\n$\\because$In the rhombus ABCD, AC$\\perp$BD, and AC, BD bisect each other,\\\\\n$\\therefore$BD$\\parallel$y-axis, BN=ND=$\\frac{1}{2}BD$, AN=NC=$\\frac{1}{2}AC=4$,\\\\\n$\\therefore$BD$\\perp$x-axis,\\\\\n$\\because$BD$\\perp$x-axis, AC$\\perp$y-axis,\\\\\n$\\therefore$$\\angle$AOD=$\\angle$AND=$\\angle$NAO=$\\angle$ODN=$90^\\circ$,\\\\\n$\\therefore$Quadrilateral ANDO is a rectangle,\\\\\n$\\therefore$OA=DN=2, OD=AN=4,\\\\\n$\\therefore$BD=2DN=4,\\\\\n$\\therefore$The coordinates of point B are $\\boxed{(4,4)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53706850.png"
  },
  {
    "solution": "\\boxed{\\frac{9}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the rectangle $ABCD$, where $AB=12$ and $AD=16$, $E$ and $F$ are the midpoints of $AD$ and $BC$, respectively. Points $G$ and $H$ are two moving points on the diagonal $AC$, moving towards each other from points $A$ and $C$ at a speed of $1$ unit per second, for a duration of $t$ seconds, with $0\\le t<10$. If point $E'$ starts from point $E$ and moves to the right along line $AD$, and point $F'$ starts from point $F$ and moves to the left along line $CB$, both at the same speed as points $G$ and $H$ starting simultaneously, find the value of $t$ when the quadrilateral $E'G{F'}H$ forms a rhombus.\n",
    "completion": "\\textbf{Solution:} Let's assume at time $t$ seconds, the quadrilateral $E'GF'H$ forms a rhombus, with point E and F moved to positions $E'$ and $F'$ respectively (as shown in Figure 2). Connect $AF'$, $CE'$, $E'F'$, where $E'F'$ intersects AC at point O.\\\\\n$\\because$ Quadrilateral $E'GF'H$ is a rhombus, $E'F' \\perp AC$, $OG=OH$, $OE'=OF'$,\\\\\n$\\therefore OA=OC$, $AE'=CF'$, $\\therefore$ Quadrilateral $AE'CF'$ is a rhombus, $\\therefore CE'=AE'=AE+EE'=8+t$, $DE'=ED-EE'=8-t$. In $\\triangle CDE'$,\\\\\nby Pythagoras' theorem, we have $CD^2+DE'^2=CE'^2$, that is $12^2+(8-t)^2=(8+t)^2$. Solving this gives $t=\\boxed{\\frac{9}{2}}$, $\\therefore$ when $t=\\frac{9}{2}$, the quadrilateral $E'GF'H$ forms a rhombus.",
    "image_path": "dataset/mm_math/images/MM_Math/52416356_99.png"
  },
  {
    "solution": "\\boxed{-x+4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system, the graph of the linear function \\(y=kx+b\\) intersects the x-axis and y-axis at points A and B, respectively, and it passes through the point \\(D(-2, 6)\\). It intersects the graph of the direct proportion function \\(y=3x\\) at point C, where the \\(x\\)-coordinate of point C is 1. What is the equation of the linear function \\(y=kx+b\\)?\n",
    "completion": "\\textbf{Solution:} Solve: When $x=1$, $y=3$. Let the line $y=kx+b$ pass through $(1, 3)$ and $(-2, 6)$\\\\\n$\\therefore$ $\\begin{cases}6=-2k+b\\\\ 3=k+b\\end{cases}$\\\\\nSolving these, we get: $\\begin{cases}k=-1\\\\ b=4\\end{cases}$\\\\\n$\\therefore$ The equation of the function is $y=-x+4=\\boxed{-x+4}$",
    "image_path": "dataset/mm_math/images/MM_Math/50520008_183.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: It is known that the acute triangle $ABC$ is inscribed in $\\odot O$. As shown in Figure 1, when point $A$ is the midpoint of $\\overset{\\frown}{BAC}$,\n",
    "completion": "\\textbf{Solution:}\n\nLet us connect OB and OC, and assume AB and BC intersect at point P,\n\n$\\because$ point A is the midpoint of $\\overset{\\frown}{BAC}$,\n\n$\\therefore \\overset{\\frown}{AB}=\\overset{\\frown}{AC}$,\n\n$\\therefore AB=AC$,\n\nFurthermore, $\\because OB=OC$,\n\n$\\therefore AO$ is the perpendicular bisector of BC,\n\n$\\therefore AO\\perp BC$;\n\n(2) $\\because AB=AC$ and $AP\\perp BC$,\n\n$\\therefore BP=CP=4$,\n\n$\\therefore AP=\\sqrt{AB^{2}-BP^{2}}=8$,\n\n$\\because BO^{2}=OP^{2}+BP^{2}$,\n\n$\\therefore BO^{2}=(8-OB)^{2}+16$,\n\n$\\therefore BO=\\boxed{5}$,\n\n$\\therefore$ the radius of the circle $O$ is \\boxed{5};",
    "image_path": "dataset/mm_math/images/MM_Math/52872084_83.png"
  },
  {
    "solution": "\\boxed{y=x+2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the line $l_{1}\\colon y=3x$ intersects with the line $l_{2}\\colon y=kx+b$ at point $A(a, 3)$, and point $B(2, 4)$ lies on the line $l_{2}$. Find the value of $a$ and the explicit equation of the line $l_{2}$.\n",
    "completion": "\\textbf{Solution}: The line $l_{1}\\colon y=3x$ intersects with the line $l_{2}\\colon y=kx+b$ at the point $A(a,3)$, which yields $3a=3$. \\\\\nSolving this gives $a=\\boxed{1}$. \\\\\n$\\therefore$ For the point $A(1,3)$, \\\\\nthe line $l_{2}\\colon y=kx+b$ passes through the point $A(1,3)$, and point $B(2,4)$, \\\\\nhence $\\begin{cases}k+b=3 \\\\ 2k+b=4 \\end{cases}$, \\\\\nsolving this leads to $\\begin{cases}k=1 \\\\ b=2 \\end{cases}$, \\\\\nthus, the equation of the line $l_{2}$ is $\\boxed{y=x+2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51983442_183.png"
  },
  {
    "solution": "\\boxed{\\sqrt{13}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system $xOy$, rectangle $OABC$ has its sides $OA$ and $OC$ along the $x$-axis and $y$-axis, respectively, and the coordinates of point $B$ are $(3, 2)$. Let $D$ be a moving point on side $OA$, and a rectangle $BDEF$ is constructed to the right side with $BD$ as one of its sides, where the ratio of $DE$ to $BD$ is $2:3$. Connect $CD$ and $AF$. The range of motion of point $E$ is a line segment, and the length of this line segment is to be found.\n",
    "completion": "\\textbf{Solution:} Join $OB$ and $BH$.\\\\\nSince $\\frac{AB}{AH}=\\frac{AO}{AB}=\\frac{2}{3}$, and $\\angle BAH=\\angle OAB=90^\\circ$.\\\\\nTherefore, $\\triangle BAH\\sim \\triangle OAB$.\\\\\nThus, $\\angle ABH=\\angle AOB$,\\\\\nTherefore, $\\angle OBA+\\angle ABH=\\angle OBA+\\angle AOB=90^\\circ$.\\\\\nThat is, $OB\\perp BH$, and $\\frac{BH}{OB}=\\frac{2}{3}$.\\\\\nConstruct rectangle $OBHG$ with $OB$ and $BH$ as adjacent sides.\\\\\nThen, when point $D$ coincides with $O$, the position of point $E$ is point $G$; when point $D$ coincides with $A$, the position of point $E$ is point $H$.\\\\\nJoin $BE$.\\\\\nSince $\\frac{BA}{AH}=\\frac{BD}{DE}$, and $\\angle BAH=\\angle BDE=90^\\circ$.\\\\\nTherefore, $\\triangle BAH\\sim \\triangle BDE$.\\\\\nThus, $\\frac{BH}{BA}=\\frac{BE}{BD}$, and $\\angle ABH=\\angle DBE$.\\\\\nTherefore, $\\angle EBH=\\angle DBA$.\\\\\nThus, $\\triangle BHE\\sim \\triangle BAD$.\\\\\nTherefore, $\\angle BHE=\\angle BAD=90^\\circ$, that is, $EH\\perp BH$.\\\\\nTherefore, the range of point $E$ is the line segment $GH$.\\\\\n$GH=OB=\\sqrt{OA^2+AB^2}=\\boxed{\\sqrt{13}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53058741_160.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB = 90^\\circ$, $AC=BC$, $BE\\perp CE$ at $E$, and $AD\\perp CE$ at $D$. We have $\\triangle ADC \\cong \\triangle CEB$. Given $AD=5\\text{cm}$ and $DE=3\\text{cm}$, find the length of $BE$ in cm.\n",
    "completion": "\\textbf{Solution:} \\\\\nFrom (1), we know that $\\triangle ADC \\cong \\triangle CEB$,\\\\\nhence $AD = CE = 5\\,\\text{cm}$, and $CD = BE$.\\\\\nSince $CD = CE - DE$,\\\\\nit follows that $BE = AD - DE = 5 - 3 = \\boxed{2}\\,\\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52978934_107.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points $D$ and $E$ are the midpoints of $AB$ and $AC$, respectively. What is the ratio of the area of $\\triangle ADE$ to the area of the quadrilateral $BCED$?\n",
    "completion": "Solution: Since points $D$ and $E$ are the midpoints of $AB$ and $AC$ respectively, \\\\\n$\\therefore$ $DE=\\frac{1}{2}BC$, i.e., $DE$ is the midline of $\\triangle ABC$, \\\\\nand $\\triangle ADE \\sim \\triangle ABC$, \\\\\n$\\therefore$ $\\frac{AD}{AB}=\\frac{1}{2}$, \\\\\nbased on the fact that the area ratio of similar triangles equals the square of the similarity ratio, \\\\\n$\\therefore$ $S_{\\triangle ADE} : S_{\\triangle ABC} = 1 : 4$, thus $S_{\\text{quadrilateral} BCED} = S_{\\triangle ABC} - S_{\\triangle ADE}$, \\\\\n$\\therefore$ $S_{\\triangle ADE} : S_{\\text{quadrilateral} BCED} = 1 : (4-1) = 1 : 3 = \\boxed{\\frac{1}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52977964.png"
  },
  {
    "solution": "\\boxed{\\frac{2}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in square $ABCD$, point $E$ is on $BC$, and $CE=\\frac{1}{2}BC$. Draw $BG\\perp DE$ at $G$, and extend $DC$ to intersect with $BG$ at $H$. Connect $AG$ and extend it to intersect the extended line of $DC$ at $F$. Then, what is the value of $\\tan\\angle CBH$? What is the value of $\\frac{CF}{DF}$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a square,\\\\\n$\\therefore \\angle BCD = \\angle BCH = 90^\\circ$, and CD=CB,\\\\\nsince $BG \\perp DG$,\\\\\n$\\therefore \\angle BGE = \\angle DCE = 90^\\circ$,\\\\\nsince $\\angle BEG = \\angle CED$,\\\\\n$\\therefore \\angle CDE = \\angle CBH$,\\\\\n$\\therefore \\triangle DCE \\cong \\triangle BCH$ (ASA),\\\\\n$\\therefore \\angle CBH = \\angle CDE$,\\\\\nsince CE = $\\frac{1}{2}BC$,\\\\\n$\\therefore \\tan\\angle CBH = \\tan\\angle CDE = \\frac{CE}{CD} = \\boxed{\\frac{1}{2}}$,\\\\\nLet BE=EC=m, then BC=CD=2m,\\\\\n$\\therefore EG = \\frac{\\sqrt{5}}{5}m$, $BG = \\frac{2\\sqrt{5}}{5}m$, $DE = \\sqrt{CD^2 + EC^2} = \\sqrt{5}m$,\\\\\n$\\therefore DH = DC + CH = 3m$, $DG = DE + EG = \\frac{6\\sqrt{5}}{5}m$,\\\\\n$\\therefore GH = \\sqrt{DH^2 - DG^2} = \\sqrt{(3m)^2 - \\left(\\frac{6\\sqrt{5}}{5}m\\right)^2} = \\frac{3\\sqrt{5}}{5}m$,\\\\\nsince AB $\\parallel$ FH,\\\\\n$\\therefore \\triangle FHG \\sim \\triangle ABG$,\\\\\n$\\therefore \\frac{FH}{2m} = \\frac{\\frac{3\\sqrt{5}}{5}m}{\\frac{2\\sqrt{5}}{5}m}$,\\\\\n$\\therefore FH = 3m$,\\\\\n$\\therefore DF = 6m$, CF = 4m,\\\\\n$\\therefore \\frac{CF}{DF} = \\frac{4m}{6m} = \\boxed{\\frac{2}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51106900.png"
  },
  {
    "solution": "\\boxed{18}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, all the shaded quadrilaterals are squares, and all triangles are right-angled triangles. If the areas of squares $A$, $B$, and $C$ are $4$, $8$, and $6$ respectively, what is the area of square $D$?\n",
    "completion": "Solution: As shown in the diagram,\\\\\nsince all the shaded quadrilaterals are squares and all the triangles are right-angled triangles, the areas of squares A, B, C are 4, 8, 6, respectively,\\\\\ntherefore, EF$^{2}$ = 4 + 8 = 12, GH$^{2}$ = area of square D - 6, with EF$^{2}$ = GH$^{2}$,\\\\\ntherefore, GH$^{2}$ = area of square D - 6 = 12\\\\\nUpon solving: the area of square D = $\\boxed{18}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52991885.png"
  },
  {
    "solution": "\\boxed{64}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, $AD \\parallel BC$, $\\angle ABC + \\angle DCB = 90^\\circ$, and $BC = 2AD$. Squares are constructed externally on the sides $AB$, $BC$, and $DC$, with areas denoted as $S_{1}$, $S_{2}$, and $S_{3}$ respectively. Given that $S_{1} = 4$ and $S_{3} = 12$, what is the value of $S_{2}$?\n",
    "completion": "\\textbf{Solution:} Given $S_{1}=4$ and $S_{3}=12$, \\\\\nit follows that $AB=2$ and $CD=2\\sqrt{3}$. \\\\\nDraw AE $\\parallel$ CD intersecting BC at E, \\\\\nhence $\\angle AEB=\\angle DCB$, \\\\\nsince $AD\\parallel BC$, \\\\\nthe quadrilateral AECD is a parallelogram, \\\\\nthus $CE=AD$ and $AE=CD=2\\sqrt{3}$, \\\\\ngiven $\\angle ABC+\\angle DCB=90^\\circ$, \\\\\nit implies $\\angle AEB+\\angle ABC=90^\\circ$, \\\\\nthus $\\angle BAE=90^\\circ$, \\\\\ntherefore $BE=\\sqrt{AB^{2}+AE^{2}}=\\sqrt{2^{2}+12}=4$, \\\\\nsince $BC=2AD$, \\\\\nit follows that $BC=2BE=8$, \\\\\ntherefore $S_{2}=(8)^{2}=\\boxed{64}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51534410.png"
  },
  {
    "solution": "\\boxed{28}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AD$ bisects $\\angle BAC$, $DE \\perp AB$ at point $E$. If $AB=8$, $AC=6$, and $DE=4$, what is the area of $\\triangle ABC$?\n",
    "completion": "\\textbf{Solution:} Draw $DF \\perp AC$ at $F$, as shown in the figure.\\\\\nSince $AD$ is the angle bisector of $\\angle BAC$ in $\\triangle ABC$, $DE \\perp AB$, $DF \\perp AC$,\\\\\ntherefore $DE=DF=4$,\\\\\nSince $AB=8$, $AC=6$,\\\\\ntherefore the area of $\\triangle ABC$ = area of $\\triangle ADB$ + area of $\\triangle ADC$\\\\\n= $\\frac{1}{2}$AB$\\cdot$DE + $\\frac{1}{2}$AC$\\cdot$DF\\\\\n= $\\frac{1}{2}\\times 8 \\times 4 + \\frac{1}{2}\\times 6 \\times 4\\\\\n= \\boxed{28}$,",
    "image_path": "dataset/mm_math/images/MM_Math/52855867.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $OABC$, $OA=3$, $OC=4$. The lines containing $OA$ and $OC$ are taken as the $x$-axis and the $y$-axis, respectively, to establish a Cartesian coordinate system. Point $D$ is a moving point on side $CB$ (not coinciding with $C$ or $B$), and the graph of the inverse proportion function $y=\\frac{k}{x}$ ($k>0$) passes through point $D$ and intersects side $BA$ at point $E$. Line $DE$ is drawn. Connect $DA$. When the area of $\\triangle DAE$ is $\\frac{4}{3}$, what is the value of $k$?\n",
    "completion": "\\textbf{Solution:} From equation (2), we know that $AE=\\frac{k}{3}$ and $BD=3-\\frac{k}{4}$,\\\\\ntherefore, the area of $\\triangle DAE$ is ${S}_{\\triangle DAE}=\\frac{1}{2}AE\\cdot BD=\\frac{1}{2}\\times \\frac{k}{3}\\times \\left(3-\\frac{k}{4}\\right)=\\frac{4}{3}$,\\\\\nSimplifying, we get: ${k}^{2}-12k+32=0$,\\\\\nSolving, we find ${k}_{1}=\\boxed{4}$ and ${k}_{2}=\\boxed{8}$, therefore, when the area of $\\triangle DAE$ is $\\frac{4}{3}$, the value of $k$ is \\boxed{4} or \\boxed{8}.",
    "image_path": "dataset/mm_math/images/MM_Math/52934017_240.png"
  },
  {
    "solution": "\\boxed{24}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that points $E$ and $F$ are the midpoints of sides $BC$ and $AD$ of quadrilateral $ABCD$, respectively. Connect $AC$, $AE$, and $CF$. Quadrilateral $AECF$ is a parallelogram; If $BC=12$ and $\\angle BAC=90^\\circ$, what is the perimeter of parallelogram $AECF$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle BAC=90^\\circ$ in $\\triangle ABC$, and point E is the midpoint of BC,\\\\\n$\\therefore AE=CE=\\frac{1}{2}BC=6$,\\\\\n$\\therefore$ Quadrilateral AECF is a rhombus,\\\\\n$\\therefore$ The perimeter of quadrilateral AECF is $6\\times 4=\\boxed{24}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52160932_149.png"
  },
  {
    "solution": "\\boxed{1\\text{cm}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given point $B$ is on line segment $AC$, and point $D$ is on line segment $AB$. If $AB=8\\text{cm}$, $BC=6\\text{cm}$, and $D$ is the midpoint of line segment $AC$, what is the length of line segment $DB$ in $\\text{cm}$?\n",
    "completion": "\\textbf{Solution:} As shown in Figure 1:\\\\\n$\\because AB=8\\text{cm}$, $BC=6\\text{cm}$\\\\\n$\\therefore AC=AB+BC=8+6=14\\text{cm}$\\\\\nFurthermore, $\\because D$ is the midpoint of segment $AC$\\\\\n$\\therefore DC=\\frac{1}{2}AC=\\frac{1}{2}\\times14=7\\text{cm}$\\\\\n$\\therefore DB=DC-BC=7-6=\\boxed{1\\text{cm}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51237402_169.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, AC intersects BD at point E, with AD $\\parallel$ BC. If $AE : AC = 1 : 3$, what is the ratio ${S}_{\\triangle AED} : {S}_{\\triangle CEB}$?\n",
    "completion": "\\textbf{Solution:} Given $\\because AE: AC = 1: 3$,\\\\\n$\\therefore AE: EC = 1: 2$,\\\\\n$\\because AD \\parallel BC$,\\\\\n$\\therefore \\triangle ADE \\sim \\triangle CBE$,\\\\\n$\\therefore \\frac{S_{\\triangle ADE}}{S_{\\triangle CBE}} = \\left(\\frac{AE}{EC}\\right)^2 = \\left(\\frac{1}{2}\\right)^2 = \\boxed{\\frac{1}{4}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51367380.png"
  },
  {
    "solution": "\\boxed{\\sqrt{13}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, $AD\\parallel BC$, $AB=BC$, and the diagonals $AC$ and $BD$ intersect at point $O$, where $BD$ bisects $\\angle ABC$. Quadrilateral $ABCD$ is a rhombus; through point $D$, draw $DE\\perp BC$ and extend it to meet the extension of $BC$ at point $E$, then connect $OE$. Given that $OE=3$ and $AC=4$, what is the length of the sides of the rhombus?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral ABCD is a rhombus, \\\\\nthus AC$\\perp$BD, OB=OD=$\\frac{1}{2}$BD, OC=$\\frac{1}{2}$AC=2, \\\\\nsince DE$\\perp$BC, \\\\\nthus $\\angle$DEB$=90^\\circ$, \\\\\nsince OB=OD, \\\\\nthus OE=$\\frac{1}{2}$BD, \\\\\nsince OD=$\\frac{1}{2}$BD, \\\\\nthus OD=OE=3, \\\\\nin $\\triangle$OCD, by the Pythagorean theorem, $CD=\\sqrt{OD^2+OC^2}=\\sqrt{3^2+2^2}=\\boxed{\\sqrt{13}},$ \\\\\ntherefore, the side length of the rhombus is $\\boxed{\\sqrt{13}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52423280_91.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the diameter $CD$ of $O$ is perpendicular to the chord $AB$ at point $E$, and $CE=2$, $OB=4$. What is the length of $AB$?\n",
    "completion": "\\textbf{Solution:} Because the diameter CD of the circle $\\odot$O is perpendicular to the chord AB at point E, therefore $AB=2BE$. Because $CE=2$, and $OB=4$, therefore $OE=4-2=2$, hence $BE= \\sqrt{OB^{2}-OE^{2}}=\\sqrt{4^{2}-2^{2}}=2\\sqrt{3}$, therefore $AB=\\boxed{4\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52536668.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, there are four congruent squares placed inside the parallelogram ABCD, all arranged parallely. The top-left vertices of each square, B, E, F, and G, all lie on the line AB, with $BE = EF = FG$. If the line PQ passes exactly through the point D, and $AB = 14$, $CH = 5$, with $\\angle A = 60^\\circ$, then what is the area of each square?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, extend EP to meet CD at K,\\\\\ngiven $BH\\parallel EP$ and ABCD is a parallelogram,\\\\\n$\\therefore AB\\parallel CD$, $AB=CD=14$, and $\\angle A=\\angle C=60^\\circ$,\\\\\n$\\therefore$ quadrilateral BCKE is a parallelogram, and $\\angle EKD=\\angle C=60^\\circ$,\\\\\n$\\therefore BC=EK$, $BE=CK$,\\\\\nby the property of congruent squares we get EP=BH, $\\angle KPD=90^\\circ$,\\\\\n$\\therefore BC-BH=EK-EP$, that is $PK=CH=5$, $\\angle PDK=30^\\circ$,\\\\\n$\\therefore DK=2PK=10$,\\\\\n$\\therefore CK=CD-DK=4$,\\\\\n$\\therefore BE=4$,\\\\\nAlso, since $BE=EF=FG$,\\\\\n$\\therefore AG=AB-BE-EF-FG=2$,\\\\\nSimilarly, it is found $AG=2AM$,\\\\\n$\\therefore AM=1$,\\\\\n$\\therefore GM=\\sqrt{AG^2-AM^2}=\\sqrt{3}$,\\\\\n$\\therefore$ the area of each small square is $\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53498558.png"
  },
  {
    "solution": "\\boxed{18^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, line AD intersects with BE at point O, angles $\\angle DOE$ and $\\angle COE$ are complementary. If $\\angle COE=72^\\circ$, what is the measure of $\\angle AOB$?\n",
    "completion": "Solution: Since $\\angle DOE$ and $\\angle COE$ are complementary,\\\\\ntherefore $\\angle DOE + \\angle COE = 90^\\circ$,\\\\\nsince $\\angle COE = 72^\\circ$,\\\\\ntherefore $\\angle DOE = 18^\\circ$,\\\\\nthus $\\angle AOB = \\angle DOE = \\boxed{18^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/55191143.png"
  },
  {
    "solution": "\\boxed{11}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in $\\triangle ABC$, $DE$ is the perpendicular bisector of $AC$, $AB=4$, and $BC=7$. What is the perimeter of $\\triangle ABD$?\n",
    "completion": "\\textbf{Solution:} Given that $DE$ is the perpendicular bisector of $AC$,\\\\\n$\\therefore AD=CD$,\\\\\n$\\therefore BC=BD+CD=BD+AD$,\\\\\n$\\therefore$ the perimeter of $\\triangle ABD$ $=AB+BD+AD=AB+BC=4+7=\\boxed{11}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52867065.png"
  },
  {
    "solution": "\\boxed{\\frac{39-9\\pi }{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle A = 90^\\circ$, O is a point on side BC. A semicircle centered at O is tangent to side AB at point D and intersects sides AC and BC at points E, F, and G, respectively. Segment OD is drawn. It is known that BD = 2, AE = 3, and $\\tan \\angle BOD = \\frac{2}{3}$. AE is a tangent to circle O; what is the sum of the areas of the two shaded regions?\n",
    "completion": "\\textbf{Solution:} Since $OD \\parallel AC$, it follows that $\\frac{BD}{AB} = \\frac{OD}{AC}$, that is, $\\frac{2}{2+3} = \\frac{3}{AC}$. Thus, $AC = 7.5$, and therefore $EC = AC - AE = 7.5 - 3 = 4.5$. Hence, the area of the shaded region $S_{\\text{shaded}} = S_{\\triangle BDO} + S_{\\triangle OEC} - S_{\\text{sector} FOD} - S_{\\text{sector} EOG}$ \\\\\n$= \\frac{1}{2} \\times 2 \\times 3 + \\frac{1}{2} \\times 3 \\times 4.5 - \\frac{90^\\circ \\pi \\times {3}^{2}}{360}$ \\\\\n$= 3 + \\frac{27}{4} - \\frac{9\\pi }{4}$ \\\\\n$= \\boxed{\\frac{39-9\\pi }{4}}.$",
    "image_path": "dataset/mm_math/images/MM_Math/51317358_160.png"
  },
  {
    "solution": "\\boxed{20}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: The cross-section of the arch of Zhaozhou Bridge is approximately parabolic, as illustrated in the figure, with the equation given by $y=-\\frac{1}{25}x^{2}$. When the height $DO$ from the water surface to the top of the arch is $4m$, what is the width $AB$ of the water surface in meters?\n",
    "completion": "\\textbf{Solution:} According to the problem, we have $-4 = -\\frac{1}{25}x^{2}$. Solving this yields $x = \\pm 10$. Therefore, the coordinates of point A are $(-10, -4)$ and the coordinates of point B are $(10, -4)$. At this moment, the width of the water surface $AB$ is $\\boxed{20}$ meters.",
    "image_path": "dataset/mm_math/images/MM_Math/52949319.png"
  },
  {
    "solution": "\\boxed{32}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=ax^2+3x+c\\ (a\\neq 0)$ intersects the x-axis at points $A(-2, 0)$ and $B$, and intersects the y-axis at point $C(0, 8)$. Point $P$ is a moving point on the parabola above the line $BC$. Connect $CP$ and $PB$. The line $BC$ intersects the axis of symmetry $l$ of the parabola at point $E$. What is the maximum area of $\\triangle BCP$?\n",
    "completion": "\\textbf{Solution:} Let $y=0$, then $-\\frac{1}{2}x^2+3x+8=0$,\\\\\nwe get $x=-2$ or $x=8$,\\\\\n$\\therefore B(8, 0)$,\\\\\nLet the equation of line $BC$ be $y=kx+b$,\\\\\n$\\therefore \\left\\{\\begin{array}{l} b=8 \\\\ 8k+b=0 \\end{array}\\right.$,\\\\\nSolving, we get $\\left\\{\\begin{array}{l} k=-1 \\\\ b=8 \\end{array}\\right.$,\\\\\n$\\therefore y=-x+8$,\\\\\nDraw a line through $P$ parallel to the $y$ axis intersecting $BC$ at $G$,\\\\\nLet $P(t, -\\frac{1}{2}t^2+3t+8)$, then $G(t, -t+8)$,\\\\\n$\\therefore PG=-\\frac{1}{2}t^2+3t+8+t-8=-\\frac{1}{2}t^2+4t$,\\\\\n$\\therefore S_{\\triangle CBP}=\\frac{1}{2}\\times 8\\times \\left(-\\frac{1}{2}t^2+4t\\right)=-2t^2+16t=-2(t-4)^2+32$,\\\\\n$\\therefore$ when $t=4$, the area of $\\triangle BCP$ reaches its maximum value, which is $\\boxed{32}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53058938_102.png"
  },
  {
    "solution": "\\boxed{\\sqrt{13}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a large square ABCD is composed of four congruent right-angled triangles and a small square. Points P and Q are the midpoints of AB and GH, respectively. If PQ happens to pass through point F, what is the value of $\\frac{AB}{EF}$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw line PM$\\perp$BE at point M,\\\\\nLet BE=a, GH=b,\\\\\nsince $\\triangle AHD \\cong \\triangle BEA$,\\\\\ntherefore AH=BE=a,\\\\\nsince quadrilateral EFGH is a square,\\\\\ntherefore FG=GH=EH=EF=b,\\\\\nsince $\\angle AEB = \\angle PMB = 90^\\circ$,\\\\\ntherefore PM$\\parallel$AE,\\\\\ntherefore BP:AP=BM:ME,\\\\\nsince point P is the midpoint of AB,\\\\\ntherefore AP=BP,\\\\\ntherefore BM=EM=$\\frac{1}{2}$a,\\\\\ntherefore PM is the median of $\\triangle ABE$,\\\\\ntherefore PM=$\\frac{1}{2}$AE=$\\frac{1}{2}$(a−b),\\\\\nfurthermore, since point Q is the midpoint of GH,\\\\\ntherefore GQ=$\\frac{1}{2}$GH=$\\frac{1}{2}$b,\\\\\nsince $\\angle PMF = \\angle BFC = 90^\\circ$,\\\\\ntherefore $\\triangle PMF \\sim \\triangle FGQ$,\\\\\ntherefore PM:FG=MF:GQ,\\\\\ntherefore PM$\\times$GQ=FG$\\times$MF,\\\\\nsince MF=EM−EF=$\\frac{1}{2}$a−b,\\\\\ntherefore $\\frac{1}{2}$(a−b)$\\times\\frac{1}{2}$b=b($\\frac{1}{2}$a−b),\\\\\nUpon simplification, we get $3b^2−ab=0$, i.e., $b(3b−a)=0$,\\\\\nsince b$\\neq$0,\\\\\ntherefore 3b−a=0,\\\\\ntherefore a=3b,\\\\\ntherefore AE=AH−EH=a−b=2b,\\\\\ntherefore $AB=\\sqrt{AE^2+BE^2}=\\sqrt{(2b)^2+(3b)^2}=\\sqrt{13}b$,\\\\\ntherefore $AB:EF = \\boxed{\\sqrt{13}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52935302.png"
  },
  {
    "solution": "\\boxed{x\\leq 1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, it is known that the graphs of the functions $y=2x$ and $y=ax+4$ ($a$ is a constant, and $a\\neq 0$) intersect at point $A(1, 2)$. What is the solution set of the inequality $ax+4\\ge 2x$ with respect to $x$?\n",
    "completion": "\\textbf{Solution:} The graphs of the functions $y=2x$ and $y=ax+4$ intersect at point $A\\left(1, 2\\right)$, which means when $x=1$, $2x=ax+4$. \\\\\nBy observing the graph, we can conclude that when $x\\le 1$, $2x\\le ax+4$\\\\\nTherefore, the answer is $\\boxed{x\\leq 1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50560440.png"
  },
  {
    "solution": "\\boxed{135^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, on the square paper $ABCD$, there is a point $P$ such that $PA=1$, $PD=2$, and $PC=3$. Now, cut $\\triangle PCD$ and paste it to the position as shown in the figure (where $C$ coincides with $A$, $P$ coincides with $G$, and $D$ coincides with $D$). Find: What is the degree measure of $\\angle APD$?\n",
    "completion": "\\textbf{Solution:} Given equation (1), we know that $\\angle GPD=45^\\circ$ and $PG=\\sqrt{2}PD=2\\sqrt{2}$,\\\\\nbecause $AG=PC=3$ and $AP=1$,\\\\\ntherefore, $1^2+(2\\sqrt{2})^2=3^2$,\\\\\nthus, $AP^2+PG^2=AG^2$,\\\\\nwhich means $\\angle GPA=90^\\circ$,\\\\\nhence $\\angle APD=90^\\circ+45^\\circ=\\boxed{135^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52188913_142.png"
  },
  {
    "solution": "\\boxed{P\\left(-2, 3\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the line $y=\\frac{1}{2}x+2$ intersects the x-axis at point A and the y-axis at point C. The parabola $y=ax^2+bx+c$ has its axis of symmetry at $x=-\\frac{3}{2}$, and it passes through points A and C, with another intersection with the x-axis at point B. Suppose point P is a point on the parabola above line AC, and let $PQ\\perp x$ axis, intersecting AC at point Q. Find the maximum value of the line segment PQ and the coordinates of point P at this maximum value.\n",
    "completion": "\\textbf{Solution:} Let $P\\left(m, -\\frac{1}{2}m^{2}-\\frac{3}{2}m+2\\right)$, and $PQ\\perp x$ axis, intersecting line AC at point Q, \\\\\n$\\therefore Q\\left(m, \\frac{1}{2}m+2\\right)$, \\\\\n$\\therefore PQ=\\left(-\\frac{1}{2}m^{2}-\\frac{3}{2}m+2\\right)-\\left(\\frac{1}{2}m+2\\right)=-\\frac{1}{2}m^{2}-2m$, \\\\\n$PQ=-\\frac{1}{2}(m+2)^{2}+2$, \\\\\n$\\therefore$ when $m=-2$, the maximum value of $PQ$ is $\\boxed{2}$, at this time $\\boxed{P\\left(-2, 3\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51457966_270.png"
  },
  {
    "solution": "\\boxed{7.2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, Xiaoxin stands under the light, casting a shadow $AB=2.4\\,m$ on the ground. When she squats down, the shadow becomes $AC=1.05\\,m$. Given that Xiaoming's height $AD=1.6\\,m$ and the height while squatting is half the standing height, what is the height of the lamp above the ground $PH$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure,\n\n$\\because$ AD $\\parallel$ PH, $\\therefore \\triangle ADB \\sim \\triangle HPB$; $\\triangle AMC \\sim \\triangle HPC$ (M is the midpoint of AD),\n\n$\\therefore$ AB : HB = AD : PH, AC : AM = HC : PH, that is $2.4 : (2.4 + AH) = 1.6 : PH$, $1.05 : 0.8 = (1.05 + HA) : PH$,\n\nSolving, we get: $AH = 8.4$, $PH = \\boxed{7.2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52796970.png"
  },
  {
    "solution": "\\boxed{25^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given point O on line AB, draw ray OC through point O such that $\\angle BOC=65^\\circ$. Place the right angle vertex of a right-angled triangle ruler at point O. As shown in the diagram, rotate the triangle ruler MON about point O anticlockwise by a certain angle, such that OC is the angle bisector of $\\angle MOB$. Find the degrees of $\\angle BON$ and $\\angle CON$?\n",
    "completion": "\\textbf{Solution:} \\\\\nSince $\\angle BOC=65^\\circ$ and OC bisects $\\angle MOB$, \\\\\nit follows that $\\angle MOB=2\\angle BOC=130^\\circ$. \\\\\nHence, $\\angle BON=\\angle MOB-\\angle MON=130^\\circ-90^\\circ=\\boxed{40^\\circ}$. \\\\\nTherefore, $\\angle CON=\\angle COB-\\angle BON=65^\\circ-40^\\circ=\\boxed{25^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52169374_193.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C = 90^\\circ$, $\\angle B = 30^\\circ$, and $AB = 8$, what is the length of $BC$?\n",
    "completion": "\\textbf{Solution:} In right triangle $\\triangle ABC$, we have $\\cos B = \\frac{BC}{AB}$. \\\\\nThus, $BC = AB \\cdot \\cos B = 8 \\times \\cos 30^\\circ = 8 \\times \\frac{\\sqrt{3}}{2} = \\boxed{4\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52537139.png"
  },
  {
    "solution": "\\boxed{80^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, points $M$ and $N$ are located on $AB$ and $BC$, respectively. Triangle $\\triangle BMN$ is folded along $MN$ to obtain $\\triangle FMN$. If $MF\\parallel AD$ and $FN\\parallel DC$, what is the measure of $\\angle B$?\n",
    "completion": "\\[\n\\text{Solution: Since } MF \\parallel AD, \\text{ and } FN \\parallel DC,\n\\]\n\\[\n\\therefore \\angle FMB = \\angle A = 110^\\circ, \\angle FNB = \\angle C = 90^\\circ,\n\\]\n\\[\n\\text{Since } \\triangle FMN \\text{ and } \\triangle BMN \\text{ are symmetric about } MN,\n\\]\n\\[\n\\therefore \\angle BMN = \\angle FMN = 55^\\circ, \\angle BNM = \\angle FNM = 45^\\circ,\n\\]\n\\[\n\\therefore \\angle B = 180^\\circ - \\angle BMN - \\angle BNM = \\boxed{80^\\circ}.\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/52754267.png"
  },
  {
    "solution": "\\boxed{1:3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ is similar to $\\triangle DEF$ with the center of similarity at point O. If $OC:OF=1:3$, then what is the ratio of the perimeters of $\\triangle ABC$ to $\\triangle DEF$?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABC$ is similar to $\\triangle DEF$ with OC:OF=1:3,\\\\\nit follows that $\\triangle ABC\\sim \\triangle DEF$, and AB:DE=1:3,\\\\\nthus, the ratio of the perimeters of $\\triangle ABC$ to $\\triangle DEF$ is: $1:3$,\\\\\nTherefore, the ratio of the perimeters is $\\boxed{1:3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51299753.png"
  },
  {
    "solution": "\\boxed{9-4\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, there is a right-angled triangle with one of its legs measuring 1 unit and the hypotenuse measuring 3 units. By arranging them as illustrated in Figure 2 to form a square, and ensuring that the pieces do not overlap or leave gaps at the joints, what is the area of the quadrilateral ABCD, which is the blank space in the middle of Figure 2?\n",
    "completion": "\\textbf{Solution:} From Figure 2, we have $AB=BC=CD=DA$ and $\\angle BAD=90^\\circ$, \\\\\n$\\therefore$ quadrilateral ABCD is a square, \\\\\n$\\because$ one of the legs of the right-angled triangle paper is 1 and the hypotenuse is 3, \\\\\naccording to the Pythagorean theorem, we can find that the other leg of the right-angled triangle is $\\sqrt{3^2-1^2}=2\\sqrt{2}$, \\\\\n$\\therefore$ the area of square ABCD is $(2\\sqrt{2}-1)^2=9-4\\sqrt{2}$, \\\\\nthus, the answer is: $\\boxed{9-4\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52422706.png"
  },
  {
    "solution": "\\boxed{\\frac{30}{13}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the line $y=-\\frac{4}{3}x+8$ intersects the x-axis at point A and the y-axis at point B. A moving point C starts from point B and moves at a uniform speed of 1 unit per second in the direction of $BA$. Simultaneously, a moving point D starts from point A and moves at a uniform speed of 2 units per second in the direction of $AO$ towards point O. When one point stops moving, the other point will also stop moving. Let the time of movement be $t$ seconds ($t>0$). Directly state the value of $t$ when $\\triangle ACD$ is similar to $\\triangle AOB$.\n",
    "completion": "\\textbf{Solution:} From the problem, we know that when $\\triangle ACD$ is similar to $\\triangle AOB$, the value of $t$ is $\\boxed{\\frac{30}{13}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52917352_81.png"
  },
  {
    "solution": "\\boxed{\\left(\\frac{\\sqrt{3}}{2}, \\frac{1}{2}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle AOB$ in the Cartesian coordinate system is rotated $90^\\circ$ clockwise around point $O$ to get $\\triangle A^\\prime OB^\\prime$. Given that $\\angle AOB=60^\\circ$, $\\angle B=90^\\circ$, and $AB= \\sqrt{3}$, what are the coordinates of point $B^\\prime$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw line $B^{\\prime}C \\perp x$-axis at point C, \\\\\nsince $\\triangle AOB$ rotates $90^\\circ$ clockwise around point O to become $\\triangle A^{\\prime}OB^{\\prime}$,\\\\\ntherefore, $OB^{\\prime} = OB$, and $\\angle BOB^{\\prime} = 90^\\circ$,\\\\\nsince $\\angle AOB = 60^\\circ$, and $AB= \\sqrt{3}$,\\\\\ntherefore, $OB = 1$,\\\\\ntherefore, $OB^{\\prime} = 1$,\\\\\n$\\angle B^{\\prime}OC = 180^\\circ - \\angle AOB - \\angle BOB^{\\prime} = 180^\\circ - 60^\\circ - 90^\\circ = 30^\\circ$,\\\\\ntherefore, $OC = OB^{\\prime}\\cos30^\\circ = 1 \\times \\frac{\\sqrt{3}}{2} = \\frac{\\sqrt{3}}{2}$,\\\\\n$B^{\\prime}C = OB^{\\prime}\\sin30^\\circ = 1 \\times \\frac{1}{2} = \\frac{1}{2}$,\\\\\ntherefore, the coordinates of $B^{\\prime}$ are $\\boxed{\\left(\\frac{\\sqrt{3}}{2}, \\frac{1}{2}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51083013.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=14$, $BC=8$, D and E are the midpoints of side AC and BC, respectively, point F is on DE, and $\\angle BFC=90^\\circ$, then what is the length of DF?\n",
    "completion": "\\textbf{Solution:} Since points D and E are the midpoints of sides AB and AC, respectively,\\\\\nthus, DE is the midline of $\\triangle ABC$,\\\\\nsince AB=14,\\\\\nthus, DE= $\\frac{1}{2}AB = 7$,\\\\\nsince $\\angle$BFC=$90^\\circ$ and BC=8,\\\\\nthus, EF= $\\frac{1}{2}BC = 4$,\\\\\ntherefore, DF=DE−EF=7−4=\\boxed{3}",
    "image_path": "dataset/mm_math/images/MM_Math/52050157.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points $D$ and $E$ are the midpoints of sides $BC$ and $AB$ of the equilateral triangle $\\triangle ABC$, respectively, with $AD=6$. If $F$ is a moving point on $AD$, what is the minimum value of $BF+EF$?\n",
    "completion": "\\textbf{Solution:} Connect CE, intersecting AD at F, and connect BF, then BF+EF is the minimum (according to the fact that the shortest distance between two points is a straight line; the perpendicular distance from a point to a line is the shortest),\\\\\n$\\because$E is the midpoint of AB, $\\triangle ABC$ is an equilateral triangle,\\\\\n$\\therefore CE\\perp AB$\\\\\nSince C and B are symmetric about AD, then BF+EF=CF,\\\\\n$\\because$In the equilateral $\\triangle ABC$, BD=CD,\\\\\n$\\therefore$AD$\\perp$BC,\\\\\n$\\therefore$AD is the perpendicular bisector of BC (three lines in one),\\\\\n$\\therefore$C and B are symmetric about the line AD,\\\\\n$\\therefore$CF=BF,\\\\\nThat is, BF+EF=CF+EF=CE,\\\\\n$\\because$AD$\\perp$BC, CE$\\perp$AB,\\\\\n$\\therefore$$\\angle$ADB=$\\angle$CEB=$90^\\circ$,\\\\\nIn $\\triangle ADB$ and $\\triangle CEB$,\\\\\n$\\left\\{\\begin{array}{l}\n\\angle ADB=\\angle CEB\\\\\n\\angle ABD=\\angle CBE\\\\\nAB=CB\n\\end{array}\\right.$ ,\\\\\n$\\therefore$$\\triangle ADB\\cong \\triangle CEB$ (AAS),\\\\\n$\\therefore$CE=AD=6,\\\\\nThat is, BF+EF=\\boxed{6}.",
    "image_path": "dataset/mm_math/images/MM_Math/51434998.png"
  },
  {
    "solution": "\\boxed{\\frac{25}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, it is known that $AB=6$, $BC=8$. The perpendicular bisector of $BD$ intersects $AD$ at point $E$ and intersects $BC$ at point $F$. What is the length of $BF$?\n",
    "completion": "\\textbf{Solution:}\nGiven that quadrilateral ABCD is a rectangle,\\\\\n$\\therefore \\angle A=90^\\circ$, AB=6, AD=BC=8,\\\\\n$\\therefore$BD=$\\sqrt{AB^{2}+AD^{2}}=10$,\\\\\nAdditionally, since EF is the perpendicular bisector of BD,\\\\\n$\\therefore$OB=OD=5, $\\angle$BOF=$90^\\circ$,\\\\\nAlso, $\\because \\angle C=90^\\circ$,\\\\\n$\\therefore \\triangle BOF\\sim \\triangle BCD$,\\\\\n$\\therefore \\frac{BO}{BC}=\\frac{BF}{BD}$, that is: $\\frac{5}{8}=\\frac{BF}{10}$, solving this yields: $BF=\\boxed{\\frac{25}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53599255.png"
  },
  {
    "solution": "\\boxed{120^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given that quadrilateral $ABCD$ is a rhombus, $E$ is a moving point on the diagonal $AC$. Extend line $DE$ to intersect the extension of line $AB$ at point $F$, and connect $BE$. If $\\angle DAB=90^\\circ$ and $\\triangle BEF$ is an isosceles triangle, find the measure of $\\angle EFB$.\n",
    "completion": "\\textbf{Solution:}\n\n(1) When $F$ is on the extension line of $AB$, as shown in the figure, if $\\angle DAB=90^\\circ$, the rhombus $ABCD$ is a square, $\\therefore \\angle DAB=\\angle ADC=90^\\circ$, $DC\\parallel AB$, $\\therefore \\angle CDF=\\angle EFB$, $\\because \\triangle BEF$ is an isosceles triangle, $\\therefore \\angle EFB=\\angle FEB=\\angle CDF$, let $\\angle EFB=\\angle FEB=\\angle CDF=y$, $\\therefore \\angle ABE=2y$, $\\angle CDF=y$. Combining the conclusions from (1), it is easy to prove that $\\triangle ADE\\cong \\triangle ABE$, $\\therefore \\angle ADE=\\angle ABE$, $\\therefore 90^\\circ-y=2y$, $\\therefore \\boxed{y=30^\\circ}$, $\\therefore \\angle EFB=30^\\circ$;\n\n(2) When $F$ is on the line segment $AB$, as shown in the figure, if $\\angle DAB=90^\\circ$, the rhombus $ABCD$ is a square, $\\therefore \\angle DAB=\\angle ADC=90^\\circ$, $DC\\parallel AB$, $\\therefore \\angle CDF=\\angle AFD$, $\\because \\triangle BEF$ is an isosceles triangle, $\\therefore \\angle EBF=\\angle BEF$, let $\\angle EFB=\\angle FEB=\\angle CDF=y$, $\\therefore \\angle CBE=90^\\circ-y$, $\\angle CDF=\\angle AFD=2y$. Combining the conclusions from (1), $\\triangle CDE\\cong \\triangle CBE$, $\\therefore \\angle CDE=\\angle CBE=90^\\circ-y$, $\\therefore 2y=90^\\circ-y$, $\\therefore y=30^\\circ$, $\\therefore \\angle EFB=180^\\circ-30^\\circ-30^\\circ=\\boxed{120^\\circ}$,\n\nIn summary, $\\angle EFB=30^\\circ$ or $\\boxed{120^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51510300_135.png"
  },
  {
    "solution": "\\boxed{\\frac{9}{2}\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $BD$ is the angle bisector of $\\triangle ABC$, with $AB=15$, $BC=9$, and $AC=12$. What is the length of $BD$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, draw $DE\\perp AB$ at point $E$,\\\\\nbecause $AB=15$, $BC=9$, and $AC=12$,\\\\\ntherefore, $AB^{2}=225$, and $BC^{2}+AC^{2}=9^{2}+12^{2}=225$\\\\\ntherefore, $AB^{2}=BC^{2}+AC^{2}$\\\\\ntherefore, $\\triangle ABC$ is a right-angled triangle\\\\\ntherefore, $\\angle C=90^\\circ$\\\\\ntherefore, $DC\\perp BC$\\\\\nbecause $BD$ is the angle bisector of $\\triangle ABC$,\\\\\ntherefore, $DE=DC$\\\\\nIn $Rt\\triangle DEB$ and $Rt\\triangle DCB$\\\\\n$\\left\\{\\begin{array}{l}DB=DB\\\\ DC=DE\\end{array}\\right.$\\\\\ntherefore, $Rt\\triangle DEB \\cong Rt\\triangle DCB$\\\\\ntherefore, $BE=BC=9$\\\\\ntherefore, $AE=AB-BE=15-9=6$\\\\\nLet $DC=DE =x$, then $AD=AC-DC=12-x$\\\\\nIn $Rt\\triangle ADE$, $AD^{2}=AE^{2}+DE^{2}$\\\\\nthat is $\\left(12-x\\right)^{2}=x^{2}+6^{2}$\\\\\nSolving for $x$ yields $x=\\frac{9}{2}$\\\\\nIn $Rt\\triangle BDC$\\\\\n$BD=\\sqrt{DC^{2}+BC^{2}}=\\sqrt{\\left(\\frac{9}{2}\\right)^{2}+9^{2}}=\\boxed{\\frac{9}{2}\\sqrt{5}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51378985.png"
  },
  {
    "solution": "\\boxed{48}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the line AC passing through the point $C(0, 12)$ intersects with the line OA at the point $A(8, 4)$. What is the area of $\\triangle OAC$?\n",
    "completion": "\\textbf{Solution:} Construct $AH \\perp OC$ at H, as shown in the diagram:\\\\\n$\\because A\\left(8, 4\\right)$ and $AH \\perp OC$,\\\\\n$\\therefore AH=8$,\\\\\n$\\because C\\left(0, 12\\right)$,\\\\\n$\\therefore OC=12$,\\\\\n$\\therefore {S}_{\\triangle OAC}=\\frac{1}{2}OC \\cdot AH=\\frac{1}{2} \\times 12 \\times 8=\\boxed{48}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52690876_65.png"
  },
  {
    "solution": "\\boxed{x>6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the figure, in the Cartesian coordinate system, the line $y=-\\frac{1}{2}x$ intersects with the graph of the inverse proportion function $y=\\frac{k}{x}$ at points $A$ and $B$, which are symmetric about the origin. It is known that the ordinate of point $A$ is 3. Based on the graph, directly write the solution set of $-\\frac{1}{2}x<\\frac{k}{x}$.\n",
    "completion": "\\textbf{Solution:} Solve the system of equations:\n\\[\n\\begin{cases}\ny=-\\frac{1}{2}x \\\\\ny=\\frac{-18}{x}\n\\end{cases}\n\\]\nThe solutions are:\n\\[\n\\begin{cases}\nx=-6 \\\\\ny=3\n\\end{cases}\n\\]\nor\n\\[\n\\begin{cases}\nx=6 \\\\\ny=-3\n\\end{cases}\n\\]\nTherefore, for point $B(6, -3)$, when $-6<x<0$ or $x>6$, we have $-\\frac{1}{2}x<\\frac{k}{x}$. Hence, the solution set for $-\\frac{1}{2}x<\\frac{k}{x}$ is \\boxed{-6<x<0} or \\boxed{x>6}.",
    "image_path": "dataset/mm_math/images/MM_Math/50496356_65.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the graph of the linear function $y = kx + b$ intersects with the graph of the inverse proportional function $y = -\\frac{8}{x}$ at points $A$ and $B$. Moreover, the x-coordinate of point $A$ and the y-coordinate of point $B$ are both $-2$. Find: What is the area of $\\triangle AOB$?\n",
    "completion": "\\textbf{Solution:} Let $y=-x+2$ and $x=0$, then $y=2$,\\\\\n$\\therefore$ the coordinates of point N are $\\left(0, 2\\right)$,\\\\\n$\\therefore$ the area of $\\triangle AOB$ is $\\frac{1}{2}ON\\cdot \\left({x}_{B}-{x}_{A}\\right)=\\frac{1}{2}\\times 2\\times \\left[4-\\left(-2\\right)\\right]=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52765753_96.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rhombus $ABCD$, the diagonals $AC$ and $BD$ intersect at point $O$. Point $E$ is the midpoint of side $AB$. If $OE=4$, what is the length of $BC$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rhombus,\\\\\nit follows that AB = BC = CD = AD, and BD$\\perp$AC,\\\\\nFurthermore, since point E is the midpoint of side AB,\\\\\nit follows that OE = AE = EB $ =\\frac{1}{2}$ AB,\\\\\nTherefore, BC = AB = 2OE = $4\\times 2$ = \\boxed{8}",
    "image_path": "dataset/mm_math/images/MM_Math/52381893.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $OC$ is the bisector of $\\angle AOB$. Given the condition (1), that $\\angle EOC = 90^\\circ$, please complete the figure in the diagram and find the measure of $\\angle AOE$?\n",
    "completion": "\\textbf{Solution:} Since OE $\\perp$ OC, it follows that $\\angle$EOC$=90^\\circ$, as shown in Figure 1, $\\angle$AOE = $\\angle$COE + $\\angle$COA = $90^\\circ + 30^\\circ = \\boxed{120^\\circ}$.\n\nAs shown in Figure 2, $\\angle$AOE = $\\angle$COE - $\\angle$COA = $90^\\circ - 30^\\circ = \\boxed{60^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52795995_114.png"
  },
  {
    "solution": "\\boxed{105^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $E$ is a point outside the square $ABCD$, and $ED=CD$, connect $AE$ and intersect $BD$ at point $F$. If $\\angle CDE=30^\\circ$, then what is the measure of $\\angle DFC$?\n",
    "completion": "\\textbf{Solution}:  \nSince quadrilateral $ABCD$ is a square,\\\\ \nthus $AD=DC, \\angle ADC=90^\\circ$,\\\\ \ntherefore $\\angle ADB=\\angle BDC=45^\\circ$,\\\\ \nsince $DC=DE$,\\\\ \nthus $AD=DE$,\\\\ \ntherefore $\\angle DAE=\\angle DEA$,\\\\ \nsince $\\angle ADE=90^\\circ+30^\\circ=120^\\circ$,\\\\ \nthus $\\angle DAE=\\frac{180^\\circ−120^\\circ}{2}=30^\\circ$,\\\\ \ntherefore $\\angle AFD=180^\\circ−30^\\circ−45^\\circ=105^\\circ$,\\\\ \nIn $\\triangle ADF$ and $\\triangle CDF$,\\\\ \nsince $\\left\\{\\begin{array}{c}AD=DC\\\\ \\angle ADB=\\angle BDC\\\\ DF=DF\\end{array}\\right.$,\\\\ \nthus $\\triangle ADF\\cong \\triangle CDF$,\\\\ \ntherefore $\\angle DFC=\\angle AFD=\\boxed{105^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50496865.png"
  },
  {
    "solution": "\\boxed{(0, -4)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Integrated Exploration\n\nAs shown in the figure, in the Cartesian coordinate system, the graph of the linear function $y=kx+b$ passes through point $A(-2,6)$, intersects with the $x$-axis at point $B$, and intersects with the graph of the direct proportion function $y=3x$ at point $C$, where the $x$-coordinate of $C$ is $1$. If point $D$ is on the $y$-axis and satisfies $S_{\\triangle COD} = \\frac{1}{3} S_{\\triangle BOC}$, what are the coordinates of point $D$?\n",
    "completion": "\\textbf{Solution:} From the problem statement, we have $y=-x+4$,\\\\\nthus $B(4, 0)$,\\\\\ntherefore $OB=4$,\\\\\nthus $\\frac{1}{3}S_{\\triangle BOC}=\\frac{1}{3}\\times \\frac{1}{2}\\times 4\\times 3=2$,\\\\\nsince $S_{\\triangle COD}=\\frac{1}{3}S_{\\triangle BOC}$,\\\\\nthus $S_{\\triangle COD}=2=\\frac{1}{2}\\times 1\\times |OD|$,\\\\\ntherefore $OD=4$,\\\\\nhence the coordinates of $D$ are $\\boxed{(0, 4)}$ or $\\boxed{(0, -4)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53706908_199.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=ax^2+bx+6$ passes through points $A(-2, 0)$, $B(4, 0)$ and intersects the $y$-axis at point $C$. Let point $D$ be a moving point on the parabola with the $x$-coordinate of $m(1<m<4)$. Connect $AC$, $BC$, $DB$, and $DC$. When the area of $\\triangle BCD$ is $\\frac{3}{4}$ of the area of $\\triangle AOC$, what is the value of $m$?\n",
    "completion": "\\textbf{Solution:} Given the expression for the parabola, point $C(0, 6)$ is known,\\\\\nfrom the coordinates of points $B$ and $C$, the equation of line $BC$ can be obtained: $y=-\\frac{3}{2}x+6$,\\\\\nas shown in the figure, a line parallel to the $y$-axis through point $D$ intersects line $BC$ at point $H$,\\\\\nlet point $D$ be $(m, -\\frac{3}{4}m^2+\\frac{3}{2}m+6)$, then point $H$ is $(m, -\\frac{3}{2}m+6)$,\\\\\nthus the area of $\\triangle BDC$ is $\\frac{1}{2}HD\\times OB=2\\left(-\\frac{3}{4}m^2+\\frac{3}{2}m+6+\\frac{3}{2}m-6\\right)=2\\left(-\\frac{3}{4}m^2+3m\\right)$,\\\\\n$\\therefore \\frac{3}{4}{S}_{\\triangle ACO}=\\frac{3}{4}\\times \\frac{1}{2}\\times 6\\times 2=\\frac{9}{2}$,\\\\\ni.e., $2\\left(-\\frac{3}{4}m^2+3m\\right)=\\frac{9}{2}$,\\\\\nSolving yields: $m=1$ or $3$ (discard $1$),\\\\\nhence $m=\\boxed{3}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52458461_178.png"
  },
  {
    "solution": "\\boxed{\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, AB is the diameter of the circle $\\odot O$, and the chord CD is perpendicular to AB at point E. Connect AD and BD, and $\\angle ADC = \\angle ABD$. Construct OF $\\perp$ AD at point F. If the radius of $\\odot O$ is 5 and OE = 3, find the length of OF.\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, connect OD.\\\\\nIn $\\triangle OED$, $DE = \\sqrt{OD^2 - OE^2} = \\sqrt{5^2 - 3^2} = 4$,\\\\\nIn $\\triangle ADE$, $AD = \\sqrt{AE^2 + DE^2} = \\sqrt{8^2 + 4^2} = 4\\sqrt{5}$,\\\\\n$\\because \\sin\\angle A = \\frac{OF}{OA} = \\frac{DE}{AD}$,\\\\\n$\\therefore \\frac{OF}{5} = \\frac{4}{4\\sqrt{5}}$,\\\\\n$\\therefore OF = \\boxed{\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52754813_80.png"
  },
  {
    "solution": "\\boxed{\\frac{4}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, quadrilateral $ABCD$ is a diamond, $AC$ and $BD$ intersect at point $O$, $E$ is the midpoint of $OA$, connect $BE$ and extend it to intersect $AD$ at point $F$. Given $BE=4$, what is the length of $EF$?\n",
    "completion": "\\textbf{Solution:} Given that point $E$ is the midpoint of $OA$,\\\\\nthus $AO=2AE$,\\\\\nsince $ABCD$ is a rhombus,\\\\\nthus $AD \\parallel BC$, and $AO=OC$,\\\\\ntherefore, $\\frac{AE}{EC}=\\frac{EF}{BE}$, and $EC=3AE$,\\\\\nthus, $\\frac{AE}{3AE}=\\frac{EF}{4}$\\\\\nSolving yields: $EF=\\boxed{\\frac{4}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52991845.png"
  },
  {
    "solution": "\\boxed{73^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a right-angled triangle ruler with an angle of $45^\\circ$ is placed on two parallel lines $m$ and $n$. If $\\angle \\alpha = 118^\\circ$, then what is the degree measure of $\\angle \\beta$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, \\\\\n$\\because m\\parallel n$, \\\\\n$\\therefore \\angle 1=\\angle 2$, \\\\\n$\\because \\angle \\alpha =\\angle 2+\\angle A$, \\\\\nand $\\angle A=45^\\circ$, $\\angle \\alpha =118^\\circ$, \\\\\n$\\therefore \\angle 2=118^\\circ-45^\\circ=73^\\circ$, \\\\\n$\\therefore \\angle 1=73^\\circ$, \\\\\n$\\therefore \\angle \\beta =\\boxed{73^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51367416.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{\\sqrt{3}}{3}x+\\frac{4\\sqrt{3}}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated, point $A$ in the Cartesian coordinate system has coordinates $(1, \\sqrt{3})$. Line $AB$ passes through point $A$, and point $B$ is the intersection of the line with the $x$-axis. A parallelogram $AOBC$ is formed with $OA$ and $OB$ as adjacent sides. If $OD$ is the bisector of $\\angle AOB$, and $D$ is the midpoint of $AC$. What is the equation of line $AB$?\n",
    "completion": "\\textbf{Solution:} Let the equation of the line AB be $y=kx+b$. Given that $A(1, \\sqrt{3})$ and $B(4,0)$, according to the problem, we get\n\\[\n\\left\\{\n\\begin{array}{l}\n\\sqrt{3}=k+b \\\\\n0=4k+b\n\\end{array}\n\\right.\n\\]\nSolving this, we find\n\\[\n\\left\\{\n\\begin{array}{l}\nk=-\\frac{\\sqrt{3}}{3} \\\\\nb=\\frac{4\\sqrt{3}}{3}\n\\end{array}\n\\right.\n\\]\n$\\therefore$ The equation of the line AB is $\\boxed{y=-\\frac{\\sqrt{3}}{3}x+\\frac{4\\sqrt{3}}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50613254_111.png"
  },
  {
    "solution": "\\boxed{\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, $BC=\\sqrt{5}$, and point D is on segment $AC$, with $BD$ connected. If $\\tan A=\\frac{1}{2}$ and $\\tan \\angle ABD=\\frac{1}{3}$, what is the length of $CD$?\n",
    "completion": "\\textbf{Solution:} Draw $DE \\perp AB$ at $E$ from point $D$, \\\\\n$\\because \\tan A = \\frac{1}{2}, \\tan \\angle ABD = \\frac{1}{3}$, \\\\\n$\\therefore \\frac{DE}{AE} = \\frac{1}{2}, \\frac{DE}{BE} = \\frac{1}{3}$, \\\\\n$\\therefore AE = 2DE, BE = 3DE$, \\\\\n$\\therefore AE + BE = 2DE + 3DE = 5DE = AB$, \\\\\n$\\because$ In $\\triangle ABC$, $\\angle C = 90^\\circ$, $BC = \\sqrt{5}$, $\\tan A = \\frac{1}{2}$, \\\\\n$\\therefore \\frac{BC}{AC} = \\frac{\\sqrt{5}}{AC} = \\frac{1}{2}$, \\\\\n$\\therefore AC = 2\\sqrt{5}$, \\\\\n$\\therefore AB = \\sqrt{AC^2 + BC^2} = 5$, \\\\\n$\\therefore DE = 1$, \\\\\n$\\therefore AE = 2$, \\\\\n$\\therefore AD = \\sqrt{AE^2 + DE^2} = \\sqrt{5}$, \\\\\n$\\therefore CD = AC - AD = \\boxed{\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52976996.png"
  },
  {
    "solution": "\\boxed{\\sqrt{10}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the rectangle $AOBD$, the coordinate of point $D$ is $(1,3)$. What is the length of $AB$?\n",
    "completion": "\\textbf{Solution:} Since the quadrilateral AOBD is a rectangle,\\\\\nit follows that AB=OD,\\\\\nand since the coordinates of point D are $(1,3)$,\\\\\nit then follows that $OD=\\sqrt{1^2+3^2}=\\sqrt{10}$,\\\\\ntherefore, $AB=OD=\\boxed{\\sqrt{10}}$",
    "image_path": "dataset/mm_math/images/MM_Math/52302569.png"
  },
  {
    "solution": "\\boxed{\\sqrt{41}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, point $P$ is a moving point on the edge of rectangle $ABCD$. Point $P$ starts from $A$ and moves along the four edges of the rectangle, finally returning to $A$. Suppose the distance traveled by point $P$ is $x$, and the area of $\\triangle ABP$ is $y$. Figure 2 is the graph of function $y$ as $x$ varies. Then, what is the length of the diagonal $BD$ of rectangle $ABCD$?\n",
    "completion": "\\textbf{Solution:} According to Figure 2, it is known that $AB=5$,\\\\\nwhen $P$ moves to point $C$,\\\\\n$y=\\frac{1}{2}AB\\cdot BC=10$,\\\\\n$\\therefore \\frac{1}{2}\\times 5\\cdot BC=10$,\\\\\n$\\therefore BC=4$,\\\\\n$\\because$ the diagonals of a rectangle are equal,\\\\\n$\\therefore BD=AC=\\sqrt{5^{2}+4^{2}}=\\sqrt{41}$.\\\\\nHence, the answer is: $BD=\\boxed{\\sqrt{41}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52345207.png"
  },
  {
    "solution": "\\boxed{5.4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $l_{1}\\parallel l_{2}\\parallel l_{3}$. The line AB intersects $l_{1}$, $l_{2}$, and $l_{3}$ at A, M, and B, respectively. Similarly, the line CD intersects $l_{1}$, $l_{2}$, and $l_{3}$ at C, N, and D, respectively. Given that AM = 4, MB = 6, and CD = 9, what is the length of ND?\n",
    "completion": "\\textbf{Solution:} Since $l_1 \\parallel l_2 \\parallel l_3$,\\\\\nit follows that $\\frac{AM}{AB} = \\frac{CN}{CD}$,\\\\\nthus $\\frac{4}{4+6} = \\frac{CN}{9}$,\\\\\nsolving this gives $CN = 3.6$,\\\\\nhence $ND = CD - CN = 9 - 3.6 = \\boxed{5.4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51402841.png"
  },
  {
    "solution": "\\boxed{70^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in rhombus ABCD, E and F are on BC and DC, respectively, with BE=DF and AE=AB. If $\\angle EAF=30^\\circ$, what is the degree measure of $\\angle D$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral ABCD is a rhombus, \\\\\n$\\therefore$AB=AD, $\\angle B=\\angle D$, \\\\\n$\\because$BE=FD, \\\\\n$\\therefore \\triangle ABE \\cong \\triangle ADF$ (SAS), \\\\\n$\\therefore$AE=AF, $\\angle BAE=\\angle DAF$, \\\\\nLet $\\angle B=x$, $\\angle BAE=y$, \\\\\n$\\because \\angle BAE+\\angle B+\\angle AEB=2x+y=180^\\circ$, $\\angle BAD+\\angle B=2y+x+30^\\circ=180^\\circ$, \\\\\n$\\therefore 4x+2y-2y-x-30^\\circ=180^\\circ$, \\\\\nSimplifying, we get $3x=210^\\circ$, \\\\\n$\\therefore x=\\boxed{70^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50475214.png"
  },
  {
    "solution": "\\boxed{\\frac{5}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $AD$ is the angle bisector of $\\triangle ABC$, and $DE \\parallel AB$ intersects $AC$ at $E$. If $\\frac{AE}{EC}=\\frac{3}{5}$, then what is the value of $\\frac{AC}{AB}$?\n",
    "completion": "\\textbf{Solution:} Since $DE\\parallel AB$,\\\\\n$\\therefore$ $\\angle ADE=\\angle BAD$,\\\\\nSince $AD$ is the bisector of $\\angle A$ in $\\triangle ABC$,\\\\\n$\\therefore$ $\\angle BAD=\\angle EAD$,\\\\\n$\\therefore$ $\\angle EAD=\\angle ADE$,\\\\\n$\\therefore$ $AE=DE$,\\\\\nSince $\\frac{AE}{EC}=\\frac{3}{5}$,\\\\\n$\\therefore$ $\\frac{EC}{DE}=\\frac{5}{3}$,\\\\\nSince $DE\\parallel AB$,\\\\\n$\\therefore$ $\\triangle CDE\\sim \\triangle CBA$,\\\\\n$\\therefore$ $\\frac{DE}{AB}=\\frac{EC}{AC}$,\\\\\n$\\therefore$ $\\frac{AC}{AB}=\\frac{EC}{DE}=\\boxed{\\frac{5}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51402152.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated in the diagram, within rectangle ABCD, point E is the midpoint of BC. Line AE intersects BD at point F. If $BF = 2$, what is the length of BD?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral ABCD is a rectangle,\\\\\nit follows that $AD \\parallel BC$,\\\\\nwhich implies $\\triangle ADF \\sim \\triangle EBF$,\\\\\ntherefore, $\\frac{BF}{DF} = \\frac{BE}{DA}$,\\\\\nsince $BF = 2$ and point E is the midpoint of BC,\\\\\nit leads to $\\frac{2}{DF} = \\frac{1}{2}$,\\\\\nthus, $DF = 4$,\\\\\nhence, $BD = BF + DF = 2 + 4 = \\boxed{6}$,\\\\",
    "image_path": "dataset/mm_math/images/MM_Math/52303693.png"
  },
  {
    "solution": "\\boxed{y=x-2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: It seems like there was a mistake in your question. You've asked for a translation of a text, but you haven't provided any text to translate. Could you please provide the text you'd like translated into English?\n",
    "completion": "\\textbf{Solution:} Since the graph of the linear function $y=kx+b$ ($k\\neq 0$) passes through the points $A(-2,-4)$ and $B(2,0)$,\\\\\nwe have $\\begin{cases} -2k+b=-4\\\\ 2k+b=0 \\end{cases}$,\\\\\nsolving this system of equations gives $k=1$ and $b=-2$,\\\\\ntherefore, the expression of the function is: \\boxed{y=x-2}.",
    "image_path": "dataset/mm_math/images/MM_Math/51919603_188.png"
  },
  {
    "solution": "\\boxed{-\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the number represented by point $A$ is $x$, then what is the value of $x$?\n",
    "completion": "\\textbf{Solution:} According to the question, as illustrated,\n\nsince $Rt\\triangle BCD$ is an isosceles right triangle, and $BC=BD=1$,\n\ntherefore, $CD=\\sqrt{BC^2+BD^2}=\\sqrt{1^2+1^2}=\\sqrt{2}$,\n\nalso, since arc $\\overset{\\frown}{AD}$ is part of a circle with $CD$ as its radius,\n\ntherefore, $CA=CD=\\sqrt{2}$,\n\nsince it is at the origin of the coordinate axis,\n\ntherefore, the number represented by point $A$ is $-\\sqrt{2}$, that is, $x=\\boxed{-\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53044048.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the lines $AB, IL, JK, DC$ are parallel to each other, and the lines $AD, IJ, LK, BC$ are also parallel to each other. The area of the quadrilateral $ABCD$ is 60, and the area of quadrilateral $EFGH$ is 35. What is the area of quadrilateral $IJKL$?\n",
    "completion": "\\textbf{Solution:} Since lines $AB$, $IL$, $JK$, $DC$ are parallel to each other, and lines $AD$, $IJ$, $LK$, $BC$ are parallel to each other, \\\\\ntherefore, quadrilaterals AFJE, DGKF, CHLG, BEIH are parallelograms, \\\\\ntherefore, $S_{\\triangle AEF}=S_{\\triangle EFJ}$, $S_{\\triangle BEH}=S_{\\triangle EHI}$, $S_{\\triangle CHG}=S_{\\triangle HGL}$, $S_{\\triangle DGF}=S_{\\triangle GFK}$, \\\\\nsince the area of quadrilateral $ABCD$ is 60, and the area of quadrilateral $EFGH$ is 35, \\\\\ntherefore, $S_{\\triangle AEF}+S_{\\triangle BEH}+S_{\\triangle CHG}+S_{\\triangle DGF}=25$, \\\\\ntherefore, $S_{\\triangle EFJ}+S_{\\triangle EHI}+S_{\\triangle HGL}+S_{\\triangle GFK}=25$, \\\\\ntherefore, $S_{\\text{quadrilateral }IJKL}=S_{\\text{quadrilateral }EFGH}-(S_{\\triangle EFJ}+S_{\\triangle EHI}+S_{\\triangle HGL}+S_{\\triangle GFK})=35-25=\\boxed{10}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53538187.png"
  },
  {
    "solution": "\\boxed{50^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points A, B, C, and D are all on the circle \\( \\odot O\\), with \\( OA\\perp BC\\), and \\( \\angle CDA=25^\\circ\\). What is the degree measure of \\( \\angle AOB\\)?\n",
    "completion": "\\textbf{Solution:}\\\\\nSince $ AO\\perp BC$\\\\\nIt follows that $\\overset{\\frown}{AC}=\\overset{\\frown}{AB}$\\\\\nSince $\\angle CDA=25^\\circ$\\\\\nThus, $\\angle AOB=2\\angle CDA=50^\\circ$\\\\\nTherefore, the answer is: $\\boxed{50^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/52602647.png"
  },
  {
    "solution": "\\boxed{x \\ge 3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the graph, the intersection point of the line $y = -x + 2$ and $y = kx + b$ ($k \\neq 0$ and $k$, $b$ are constants) is at the coordinates $(3, -1)$. What is the solution set of the inequality $kx + b \\geq -x + 2$ with respect to $x$?\n",
    "completion": "\\textbf{Solution:} From the graph, we obtain that when $x \\ge 3$, the graph corresponding to the points of $y = -x + 2$ lies below the graph of the function $y = kx + b$ (where $k \\ne 0$ and $k$, $b$ are constants).\\\\\nTherefore, the solution set of the inequality $kx + b \\ge -x + 2$ is $\\boxed{x \\ge 3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51662198.png"
  },
  {
    "solution": "\\boxed{(1,-1)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown, it is known that the radius of $\\odot P$ is 1, and the center P moves on the parabola $y=\\frac{1}{2}x^{2}-x-\\frac{1}{2}$. When $\\odot P$ is tangent to the x-axis, what are the coordinates of the center P?\n",
    "completion": "\\textbf{Solution:} Let point P$(x, y)$, where $y=\\frac{1}{2}x^{2}-x-\\frac{1}{2}$,\\\\\nsince $\\odot P$ is tangent to the x-axis,\\\\\nthus $|y|=1$,\\\\\ntherefore $y=\\pm 1$.\\\\\n(1) When $y=1$, $1=\\frac{1}{2}x^{2}-x-\\frac{1}{2}$,\\\\\nwe get: $x_{1}=3$, $x_{2}=-1$\\\\\nthus, points P$(3,1)$, $(-1,1)$\\\\\n(2) When $y=-1$, $-1=\\frac{1}{2}x^{2}-x-\\frac{1}{2}$,\\\\\nwe get: $x=1$\\\\\nthus, point P$(1,-1)$\\\\\nTherefore, the answers are $\\boxed{(3,1)}$ or $\\boxed{(-1,1)}$ or $\\boxed{(1,-1)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52765623.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the right-angled triangle $\\triangle OAB$ has its right-angle vertex $A$ on the $y$-axis. The graph of the inverse proportional function $y=\\frac{k}{x}$ (where $k>0$, $x>0$) passes through the midpoint $D$ of line segment $OB$ and intersects line segment $AB$ at point $C$. Connect $CD$. If the area of $\\triangle BCD$ is $3$, what is the value of $k$?\n",
    "completion": "\\textbf{Solution:} Join OC, and draw a perpendicular line from D to the x-axis, intersecting AB at point E and the x-axis at point F,\\\\\n\\(\\because S_{\\triangle CAO} = S_{\\triangle DOF} = \\frac{1}{2}|k|\\),\\\\\n\\(\\because\\) D is the midpoint of the line segment OB,\\\\\n\\(\\therefore S_{\\triangle BDE} = S_{\\triangle DOF} = \\frac{1}{2}|k|\\),\\\\\n\\(\\because\\) DE is the median of \\(\\triangle AOB\\),\\\\\n\\(\\therefore DE : OA = 1 : 2\\),\\\\\n\\(\\therefore S_{\\triangle BDE} = \\frac{1}{4}S_{\\triangle AOB} = S_{\\triangle CAO}\\),\\\\\n\\(\\therefore S_{\\triangle OBC} = S_{\\triangle AOB} - S_{\\triangle CAO} = \\frac{3}{4}S_{\\triangle AOB}\\),\\\\\n\\(\\because\\) D is the midpoint of the line segment OB,\\\\\n\\(\\therefore S_{\\triangle OBC} = 2S_{\\triangle BCD} = 6 = \\frac{3}{4}S_{\\triangle AOB}\\),\\\\\n\\(\\therefore S_{\\triangle AOB} = 8\\),\\\\\n\\(\\therefore S_{\\triangle AOC} = \\frac{1}{4}S_{\\triangle AOB} = 2 = \\frac{1}{2}|k|\\),\\\\\n\\(\\because k > 0\\),\\\\\n\\(\\therefore k = \\boxed{4}\\).",
    "image_path": "dataset/mm_math/images/MM_Math/52082526.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, line AB and line CD intersect at point O, with OM$\\perp$AB at point O. If $\\angle BOC=2\\angle AOC$, what is the degree measure of $\\angle BOD$?\n",
    "completion": "\\textbf{Solution}:\nGiven that $\\because \\angle BOC = 2\\angle AOC$, and $\\angle BOC + \\angle AOC = 180^\\circ$,\\\\\n$\\therefore 2\\angle AOC + \\angle AOC = 180^\\circ$,\\\\\n$\\therefore 3\\angle AOC = 180^\\circ$,\\\\\n$\\therefore \\angle AOC = 60^\\circ$,\\\\\n$\\therefore \\angle BOD = \\angle AOC = \\boxed{60^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/54962548_116.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the equilateral triangle $\\triangle ABC$, $P$ is a point on $BC$, and $D$ is a point on $AC$, with $\\angle APD=60^\\circ$, $BP=4$, and $CD=2$. What is the side length of $\\triangle ABC$?\n",
    "completion": "\\textbf{Solution:} Given that $\\triangle ABC$ is an equilateral triangle,\\\\\nit follows that AB = BC = AC, and $\\angle B = \\angle C = 60^\\circ$,\\\\\nthus $\\angle BAP + \\angle APB = 180^\\circ - 60^\\circ = 120^\\circ$,\\\\\nsince $\\angle APD = 60^\\circ$,\\\\\nit follows that $\\angle APB + \\angle DPC = 180^\\circ - 60^\\circ = 120^\\circ$,\\\\\ntherefore, $\\angle BAP = \\angle DPC$,\\\\\nwhich means $\\angle B = \\angle C$ and $\\angle BAP = \\angle DPC$,\\\\\nhence $\\triangle ABP \\sim \\triangle PCD$;\\\\\nthus $\\frac{AB}{PC} = \\frac{BP}{CD}$,\\\\\ngiven BP = 4 and CD = 2,\\\\\nit follows that $\\frac{AB}{AB - 4} = \\frac{4}{2}$, solving gives AB = \\boxed{8},\\\\\ntherefore, the side length of $\\triangle ABC$ is 8.",
    "image_path": "dataset/mm_math/images/MM_Math/52720073.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, $D$ is a point inside $\\triangle ABC$, where $CD$ bisects $\\angle ACB$, and $BE\\perp CD$ with the foot of the perpendicular being $D$, intersecting $AC$ at point $E$. It is given that $\\angle A = \\angle ABE$. If $AC=7$ and $BC=4$, what is the length of $BD$?\n",
    "completion": "\\textbf{Solution:} Given that $CD$ bisects $\\angle ACB$,\\\\\nthus $\\angle BCD=\\angle ECD$,\\\\\nsince $BE\\perp CD$,\\\\\nthus $\\angle BDC=\\angle EDC=90^\\circ$,\\\\\nsince $CD=CD$,\\\\\nthus $\\triangle BDC\\cong \\triangle EDC$ (by ASA),\\\\\ntherefore $BC=CE=4$, $BD=DE$,\\\\\nalso since $\\angle A=\\angle ABE$,\\\\\nthus $AE=BE$,\\\\\nsince $AC=7$, $BC=4$,\\\\\nthus $AE=AC-CE=3$,\\\\\ntherefore $BE=AE=3$,\\\\\ntherefore $BD=\\frac{1}{2}BE=\\boxed{\\frac{3}{2}}$,",
    "image_path": "dataset/mm_math/images/MM_Math/53044234.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the vertex \\(A\\) of \\(\\square OABC\\) is on the inverse proportion function \\(y=\\frac{1}{x}\\), the vertex \\(B\\) is on the inverse proportion function \\(y=\\frac{5}{x}\\), and the point \\(C\\) is on the positive half of the x-axis. What is the area of \\(\\square OABC\\)?\n",
    "completion": "\\textbf{Solution:} Construct line AF $\\perp$ x-axis at point F through point A, and construct line BE $\\perp$ x-axis at point E through point B. Extend BA to intersect the y-axis at point G. Thus, both quadrilaterals OFAG and OEBG are rectangles. \\\\\n$\\because$ Point A lies on the inverse proportional function $y=\\frac{1}{x}$, and vertex B lies on the inverse proportional function $y=\\frac{5}{x}$, \\\\\n$\\therefore$ Area of rectangle OFAG = 1, Area of rectangle OEBG = 5, \\\\\n$\\therefore$Area of $\\square OABC$ = Area of rectangle ABEF = Area of rectangle OEBG - Area of rectangle OFAG = \\boxed{4}.",
    "image_path": "dataset/mm_math/images/MM_Math/52514129.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, the vertices A and C of $\\triangle ABC$ are located on circle O, and circle O intersects with side AB at point D. Through point D, $DF \\parallel BC$ is drawn, intersecting AC at point E and circle O at point F. DC is connected, and point C is the midpoint of arc DF. BC is the tangent to circle O; if the radius of circle O is 3 and $DF=4\\sqrt{2}$, find the value of $CE\\cdot CA$.\n",
    "completion": "\\textbf{Solution:} Connect CO and extend it to meet $\\odot O$ at G, DF and CG intersect at point M, connect DO, as shown in the figure:\\\\\nSince $OC\\perp BC$ and $DF\\parallel BC$,\\\\\nit follows that $OM\\perp DF$\\\\\nTherefore, $DM=FM=\\frac{1}{2}DF=2\\sqrt{2}$\\\\\nSince the radius of $\\odot O$ is 3,\\\\\nit follows that $OD=OC=OG=3$\\\\\nTherefore, $OM=\\sqrt{OD^{2}-DM^{2}}=1$, and $CG=OC+OG=6$\\\\\nTherefore, $MC=OC-OM=2$\\\\\nSince CG is a diameter,\\\\\nit follows that $\\angle CAG=90^\\circ$\\\\\nSince $\\angle MCE=\\angle ACG$\\\\\nit follows that $\\triangle MCE\\sim \\triangle ACG$\\\\\nTherefore, $\\frac{MC}{CA}=\\frac{CE}{CG}$\\\\\nTherefore, $\\frac{2}{CA}=\\frac{CE}{6}$\\\\\nTherefore, $CE\\cdot CA=\\boxed{12}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51367439_117.png"
  },
  {
    "solution": "\\boxed{3\\sqrt{10}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, extend the side $BC$ of the rectangle $ABCD$ to the point $E$ such that $CE = BD$, and connect $AE$. If $AB = 3$ and $CE = 5$, what is the length of $AE$?\n",
    "completion": "\\textbf{Solution}: From (1), we know $AC=CE=5$, \\\\\n$\\therefore BC= \\sqrt{AC^{2}-AB^{2}} = \\sqrt{5^{2}-3^{2}}=4$, \\\\\n$\\therefore BE=BC+CE=4+5=9$, \\\\\n$\\because \\angle ABE=90^\\circ$, \\\\\n$\\therefore AE=\\sqrt{AB^{2}+BE^{2}}=\\sqrt{3^{2}+9^{2}}=\\boxed{3\\sqrt{10}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51770540_140.png"
  },
  {
    "solution": "\\boxed{15^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $\\triangle ABC$ is an equilateral triangle, and points $B$, $C$, $D$, $E$ are on the same line, with $CG=CD$, $DF=DE$. What is the measure of $\\angle E$ in degrees?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABC$ is an equilateral triangle,\\\\\ntherefore, $\\angle ACB=60^\\circ$,\\\\\nsince CG=CD,\\\\\ntherefore, $\\angle CDG=\\angle CGD=30^\\circ$,\\\\\nsince DF=DE,\\\\\ntherefore, $\\angle E=\\angle DFE=\\boxed{15^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52604175.png"
  },
  {
    "solution": "\\boxed{3-\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $E$ is a moving point on the side $CD$ of square $ABCD$, with $AB=2$. Connect $BE$, and draw $AF \\perp BE$, which intersects $BC$ at $F$ and $BE$ at $G$. Connect $CG$. When $CG$ is at its minimum value, what is the length of $CF$?\n",
    "completion": "\\textbf{Solution:} As shown in Figure 1, take the midpoint O of AB, and connect OG and OC.\\\\\n$\\because$ Quadrilateral ABCD is a square,\\\\\n$\\therefore \\angle ABC=90^\\circ$,\\\\\n$\\because$AB=2,\\\\\n$\\therefore$OB=OA=1,\\\\\n$\\therefore$OC=$\\sqrt{1^2+2^2}=\\sqrt{5}$,\\\\\n$\\because$AF$\\perp$BE,\\\\\n$\\therefore \\angle AGB=90^\\circ$,\\\\\n$\\therefore$ point G lies on the circle $\\odot$O with AB as the diameter,\\\\\n$\\because$AO=OB,\\\\\n$\\therefore$OG=$\\frac{1}{2}$AB=1,\\\\\n$\\therefore$ when O, G, and C are collinear, the value of CG is minimum, and the minimum value=$\\sqrt{5}-1$ (as shown in Figure 2),\\\\\n$\\because$OB=OG=1,\\\\\n$\\therefore \\angle OBG=\\angle OGB$,\\\\\n$\\because$AB$\\parallel$CD,\\\\\n$\\therefore \\angle OBG=\\angle CEG$,\\\\\n$\\because \\angle OGB=\\angle CGE$,\\\\\n$\\therefore \\angle CGE=\\angle CEG$,\\\\\n$\\therefore$CE=CG=$\\sqrt{5}-1$,\\\\\n$\\because \\angle ABF=\\angle BCE=\\angle AGB=90^\\circ$,\\\\\n$\\therefore \\angle BAF+\\angle ABG=90^\\circ$, $\\angle ABG+\\angle CBE=90^\\circ$,\\\\\n$\\therefore \\angle BAF=\\angle CBE$,\\\\\n$\\because$AB=BC,\\\\\n$\\therefore \\triangle ABF \\cong \\triangle BCE$ (ASA),\\\\\n$\\therefore$BF=CE=$\\sqrt{5}-1$,\\\\\n$\\therefore$CF=BC−BF=2−($\\sqrt{5}-1$)=3−$\\sqrt{5}$,\\\\\nTherefore, $CF = \\boxed{3-\\sqrt{5}}$",
    "image_path": "dataset/mm_math/images/MM_Math/52623034.png"
  },
  {
    "solution": "\\boxed{\\frac{4}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within rectangle $ABCD$, $AB=2$, $BC=4$, point $D$ is the midpoint of side $AB$, the graph of the inverse proportional function $y_1=\\frac{k}{x}$ ($x>0$) passes through point $D$ and intersects $BC$ at point $E$. What is the area of $\\triangle PDE$?\n",
    "completion": "\\textbf{Solution:} From (1), we know the equation of line $DE$ is $y_2=-\\frac{1}{2}x+3$. Let its intersection with the $x$-axis be $Q$,\\\\\nthen it is known that the coordinates of $Q$ are $Q(6, 0)$,\\\\\n$\\because P\\left(\\frac{10}{3}, 0\\right)$, then $PQ=6-\\frac{10}{3}=\\frac{8}{3}$,\\\\\nand $D(4, 1)$, $y_E=2$,\\\\\n$\\therefore {S}_{\\triangle PQE}=\\frac{1}{2}PQ\\cdot y_E=\\frac{1}{2}\\times \\frac{8}{3}\\times 2=\\frac{8}{3}$\\\\\n$\\therefore {S}_{\\triangle PQD}=\\frac{1}{2}PQ\\cdot y_D=\\frac{1}{2}\\times \\frac{8}{3}\\times 1=\\frac{4}{3}$\\\\\n$\\therefore {S}_{\\triangle PDE}={S}_{\\triangle PQE}-{S}_{\\triangle PQD}=\\frac{8}{3}-\\frac{4}{3}=\\boxed{\\frac{4}{3}}$",
    "image_path": "dataset/mm_math/images/MM_Math/52251436_129.png"
  },
  {
    "solution": "\\boxed{P(1, \\frac{8\\sqrt{3}}{3})}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that in $\\triangle ABC$, $\\angle BAC = 30^\\circ$, $\\angle C = 90^\\circ$, the coordinate of point A is $(-1,0)$, and the coordinate of point B is $(3,0)$. The vertex of parabola $y_1$ is denoted as Q, and it passes through the three vertices A, B, and C of $\\triangle ABC$ (with A to the left of B, and C below the x-axis). Parabola $y_2$ also intersects the x-axis at points A and B, with its vertex denoted as P. Suppose point M is a moving point on parabola $y_1$, located to the right of its axis of symmetry. If $\\triangle PQM$ is similar to $\\triangle ABC$, determine the coordinates of vertex P of parabola $y_2$.\n",
    "completion": "\\textbf{Solution:} Since point M is on the right side of the symmetry axis of the parabola $y_{1}$, it follows that point Q is not the right-angle vertex. Let point M be $(a, \\frac{\\sqrt{3}}{3}a^{2}- \\frac{2\\sqrt{3}}{3}a- \\sqrt{3})$, thus, the distance of point M from the symmetry axis is $a-1$, and $y_{M}-y_{Q} = \\frac{\\sqrt{3}}{3}a^{2}- \\frac{2\\sqrt{3}}{3}a- \\sqrt{3} -(- \\frac{4\\sqrt{3}}{3}) = \\frac{\\sqrt{3}}{3}a^{2}- \\frac{2\\sqrt{3}}{3}a+ \\frac{\\sqrt{3}}{3}$. Since triangle $PQM$ is similar to triangle $ABC$ with $\\angle ACB = 90^\\circ$.\n\n(1) When $\\angle PMQ = 90^\\circ$,\n\n(i) If $\\angle PQM = 30^\\circ$ and $\\angle QPM = 60^\\circ$, then $y_{M}-y_{Q} = \\sqrt{3}(a-1)$ and $y_{P}-y_{M} = \\frac{\\sqrt{3}}{3}(a-1)$. Therefore, $\\frac{\\sqrt{3}}{3}a^{2}- \\frac{2\\sqrt{3}}{3}a+ \\frac{\\sqrt{3}}{3} = \\sqrt{3}(a-1)$. Solving, we find $a=1$ (discard) or $a=4$. Therefore, $M(4, \\frac{5\\sqrt{3}}{3})$, and $y_{P}-y_{M} = \\frac{\\sqrt{3}}{3}(4-1) = \\sqrt{3}$. Hence, the y-coordinate of point P is $\\frac{5\\sqrt{3}}{3} + \\sqrt{3} = \\frac{8\\sqrt{3}}{3}$. Therefore, $P(1, \\frac{8\\sqrt{3}}{3})$.\n\n(ii) If $\\angle PQM = 60^\\circ$ and $\\angle QPM = 30^\\circ$, then $\\sqrt{3}(y_{M}-y_{Q}) = a-1$, and $y_{P}-y_{M} = \\sqrt{3}(a-1)$. Therefore, $\\sqrt{3}(\\frac{\\sqrt{3}}{3}a^{2}- \\frac{2\\sqrt{3}}{3}a+ \\frac{\\sqrt{3}}{3}) = a-1$. Solving, we find $a=1$ (discard) or $a=2$. Therefore, $M(2, -\\sqrt{3})$, and $y_{P}-y_{M} = \\sqrt{3} \\times (2-1) = \\sqrt{3}$. Hence, the y-coordinate of point P is $-\\sqrt{3} + \\sqrt{3} = 0$. Therefore, $P(1, 0)$.\n\n(2) When $\\angle MPQ = 90^\\circ$,\n\n(i) If $\\angle PQM = 30^\\circ$, then $y_{M}-y_{Q} = \\sqrt{3}(a-1)$. Therefore, $\\frac{\\sqrt{3}}{3}a^{2}- \\frac{2\\sqrt{3}}{3}a- \\frac{\\sqrt{3}}{3} = \\sqrt{3}(a-1)$. Solving, we find $a=1$ (discard) or $a=4$. Therefore, $M(4, \\frac{5\\sqrt{3}}{3})$. Hence, $P(1, \\frac{5\\sqrt{3}}{3})$.\n\n(ii) If $\\angle PQM = 60^\\circ$, then $\\sqrt{3}(y_{M}-y_{Q}) = a-1$. Therefore, $\\sqrt{3}(\\frac{\\sqrt{3}}{3}a^{2}- \\frac{2\\sqrt{3}}{3}a- \\frac{\\sqrt{3}}{3}) = a-1$. Solving, we find $a=1$ (discard) or $a=?$.\n\n\\boxed{P(1, \\frac{8\\sqrt{3}}{3})}",
    "image_path": "dataset/mm_math/images/MM_Math/51317409_152.png"
  },
  {
    "solution": "\\boxed{40^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the lines $AB$ and $CD$ intersect at point $O$, with $OE\\perp CD$. If $\\angle AOE=50^\\circ$, what is the degree measure of $\\angle BOD$?\n",
    "completion": "\\textbf{Solution:} Since $OE \\perp CD$,\\\\\nit follows that $\\angle EOC=90^\\circ$,\\\\\nwhich means $\\angle EOA+\\angle AOC=90^\\circ$,\\\\\nSince $\\angle AOE=50^\\circ$,\\\\\nit follows that $\\angle AOC=90^\\circ-50^\\circ=40^\\circ$,\\\\\nTherefore, $\\angle BOD=\\angle AOC=\\boxed{40^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53414108.png"
  },
  {
    "solution": "\\boxed{\\left(0, -6\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=2x^2-4x-6$ intersects the x-axis at points A and B, with point A to the left of point B, and intersects the y-axis at point C. Point D is a moving point on the parabola. Find the coordinates of points A, B, and C.\n",
    "completion": "\\textbf{Solution:} Substitute $y=0$ into $y=2x^{2}-4x-6$, \\\\\nwe get $2x^{2}-4x-6=0$. \\\\\nSolving this, we find $x_{1}=-1$, $x_{2}=3$. \\\\\nTherefore, the coordinates of point A are $\\boxed{\\left(-1, 0\\right)}$, and the coordinates of point B are $\\boxed{\\left(3, 0\\right)}$. \\\\\nSubstitute $x=0$ into $y=2x^{2}-4x-6$, we get $y=-6$. \\\\\nTherefore, the coordinates of point C are $\\boxed{\\left(0, -6\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52917427_260.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $C$ is on the circle $\\odot O$, point $P$ is on the extension of the diameter $AB$ of $\\odot O$. Connect $AC$, $OC$, $BC$, $CP$, and it is given that $BO=BC=BP$. $CP$ is the tangent to the circle $\\odot O$; If $OP=4$, find the length of $AC$.\n",
    "completion": "\\textbf{Solution:} Since $OP=4$, and $BO=BC=BP$.\\\\\nTherefore, $OB=OC=OA=BC=BP=2$, and $AB=4$,\\\\\nSince $AB$ is the diameter,\\\\\nTherefore, $\\angle ACB=90^\\circ$,\\\\\nThus, $AC=\\sqrt{AB^{2}-BC^{2}}=\\sqrt{4^{2}-2^{2}}=\\boxed{2\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52765452_169.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in parallelogram $ABCD$, point $M$ is the midpoint of $CD$, and $AM$ intersects $BD$ at point $N$. If it is known that the area of $\\triangle DMN$ is 3, then what is the area of $\\triangle ADN$?\n",
    "completion": "\\textbf{Solution:} Since point $M$ is the midpoint of $CD$, \\\\\n$\\therefore$ DM=$\\frac{1}{2}CD$, \\\\\nIn parallelogram $ABCD$ where AB$\\parallel$CD, AB=CD, \\\\\n$\\therefore$ $\\angle$ABN=$\\angle$MDN, $\\angle$BAN=$\\angle$DMN, \\\\\n$\\therefore$ $\\triangle ABN\\sim \\triangle MDN$, \\\\\n$\\therefore$ $\\frac{AB}{MD}=\\frac{AN}{MN}=\\frac{CD}{DM}=2$, \\\\\n$\\therefore$ AN=2MN, \\\\\n$\\therefore$ $\\frac{{S}_{\\triangle ADN}}{{S}_{\\triangle DMN}}=AN:MN=2:1$, \\\\\nSince ${S}_{\\triangle DMN}=3$, \\\\\n$\\therefore$ ${S}_{\\triangle ADN}=2{S}_{\\triangle DMN}=2\\times 3=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51352002.png"
  },
  {
    "solution": "\\boxed{3\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within a square $ABCD$ with a side length of $4\\text{cm}$, point $P$ starts moving from point $A$, travels uniformly along the path $AB \\rightarrow BC$, and stops at point $C$. Through point $P$, line $PQ$ is drawn parallel to $BD$, where $PQ$ intersects with side $AD$ (or side $CD$) at point $Q$. The length of $PQ$, $y$ (in centimeters), varies as point $P$ moves, as illustrated in the figure. What is the length of $PQ$ when point $P$ has moved for $2.5s$?\n",
    "completion": "\\textbf{Solution:} Given from (2), when point P moves for 2s, PQ overlaps with BD, and the side length of quadrilateral ABCD is 4cm.\\\\\n$\\therefore$ The speed of point P is 2cm/s.\\\\\n$\\therefore$ When P moves for 2.5s, the distance covered by point P is 5cm, and at this moment, P is on side BC.\\\\\n$\\therefore$ CP$_{1}$=3cm.\\\\\nAlso, since PQ$\\parallel$BD,\\\\\n$\\therefore$CQ$_{1}$=3cm.\\\\\n$\\therefore$${P}_{1}{Q}_{1}=\\sqrt{{3}^{2}+{3}^{2}}=\\boxed{3\\sqrt{2}}$cm.",
    "image_path": "dataset/mm_math/images/MM_Math/52211279.png"
  },
  {
    "solution": "\\boxed{1\\colon 9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the regular hexagon $ABCDEF$ is obtained through a similarity transformation from the regular hexagon $A^{\\prime}B^{\\prime}C^{\\prime}D^{\\prime}E^{\\prime}F^{\\prime}$. Given that $AB=3$ and $B^{\\prime}C^{\\prime}=1$, what is the ratio of the area of the regular hexagon $A^{\\prime}B^{\\prime}C^{\\prime}D^{\\prime}E^{\\prime}F^{\\prime}$ to that of the regular hexagon $ABCDEF$?\n",
    "completion": "\\textbf{Solution:} Given that the hexagon $ABCDEF$ is obtained from the hexagon $A'B'C'D'E'F'$ through a similarity transformation,\\\\\nthus hexagon $ABCDEF \\sim$ hexagon $A'B'C'D'E'F'$,\\\\\nhence, the ratio of the areas of the hexagons $A'B'C'D'E'F'$ to $ABCDEF$ is equal to $(1\\colon 3)^{2}=\\boxed{1\\colon 9}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51654922.png"
  },
  {
    "solution": "\\boxed{\\frac{5}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, $AC=3$, $BC=4$. Point $D$ is on side $AB$, with $AD=AC$, and $AE \\perp CD$ at foot $F$, intersecting $BC$ at point $E$. What is the length of $BE$?\n",
    "completion": "\\textbf{Solution}: As shown in the diagram, connect DE, \\\\\n$\\because$AD＝AC=3, AE$\\perp$CD, \\\\\n$\\therefore$AE is the perpendicular bisector of CD, \\\\\n$\\therefore$CE＝DE, \\\\\n$\\therefore$$\\angle$ADE＝$\\angle$ACB＝$90^\\circ$, \\\\\nIn $\\triangle ABC$, where AC=3, BC=4, \\\\\n$\\therefore$AB＝$\\sqrt{AC^{2}+BC^{2}}$=5, \\\\\n$\\therefore$BD＝AB−AD＝2, \\\\\n$\\therefore$Area of $\\triangle ABC$＝Area of $\\triangle ACE$+Area of $\\triangle ABE$, \\\\\n$\\therefore$AC·BC＝AC·CE+AB·DE, \\\\\n$\\therefore$3×4＝3CE+5DE=8DE, \\\\\n$\\therefore$DE＝$\\frac{3}{2}$, \\\\\nIn $\\triangle BDE$, BE＝$\\sqrt{DE^{2}+BD^{2}}$＝$\\sqrt{\\left(\\frac{3}{2}\\right)^{2}+2^{2}}$=$\\boxed{\\frac{5}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52935543.png"
  },
  {
    "solution": "\\boxed{A}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ABC=90^\\circ$, $\\angle A=60^\\circ$, D is a point on side $AC$, $\\angle DBC=30^\\circ$, $AC=12\\,cm$, then what is the perimeter of $\\triangle ABD$?\n",
    "completion": "\\textbf{Solution:} The net of this solid consists entirely of triangles,\\\\\n$\\therefore$ this solid is a triangular pyramid.\\\\\nHence, the answer is: \\boxed{A}",
    "image_path": "dataset/mm_math/images/MM_Math/53206750.png"
  },
  {
    "solution": "\\boxed{27^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\angle AOB$ is a right angle, $\\angle BOC = 36^\\circ$, and $OD$ bisects $\\angle AOC$. What is the measure of $\\angle BOD$ in degrees?\n",
    "completion": "\\textbf{Solution:} Since $\\angle AOB$ is a right angle, and $\\angle BOC=36^\\circ$, \\\\\nthen $\\angle AOC=90^\\circ+36^\\circ=126^\\circ$ \\\\\nSince $OD$ bisects $\\angle AOC$, \\\\\nthen $\\angle COD=\\frac{1}{2}\\angle AOC=63^\\circ$, \\\\\ntherefore, $\\angle BOD=\\angle COD-\\angle BOC=63^\\circ-36^\\circ=\\boxed{27^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/53211124.png"
  },
  {
    "solution": "\\boxed{105^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown, it is known that $ \\triangle ABD \\cong \\triangle ACE$, $ \\angle A=53^\\circ$, $ \\angle B=22^\\circ$. Then, how many degrees is $ \\angle AEC$?\n",
    "completion": "\\textbf{Solution}: Since $\\triangle ABD \\cong \\triangle ACE$,\\\\\nit follows that $\\angle C = \\angle B = 22^\\circ$,\\\\\ntherefore, in $\\triangle AEC$, $\\angle AEC = 180^\\circ - \\angle A - \\angle C = 180^\\circ - 53^\\circ - 22^\\circ = \\boxed{105^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52603218.png"
  },
  {
    "solution": "\\boxed{\\frac{16}{3}\\pi}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within circle $\\odot O$, chord $BC$ is perpendicular to radius $OA$, point $D$ is a point on the major arc $BC$, connecting $BD$, $AD$, $OC$, with $\\angle ADB=30^\\circ$. If chord $BC=8\\sqrt{3}$ cm, what is the length of the arc corresponding to chord $BC$ in the figure in cm?\n",
    "completion": "\\textbf{Solution:} Draw line OB. Since OA$\\perp$BC, it follows that $\\overset{\\frown}{AB}=\\overset{\\frown}{AC}$, and therefore $\\angle$AOC=$\\angle$AOB. According to the inscribed angle theorem, $\\angle$AOB=$2\\angle$ADB=$60^\\circ$, hence $\\angle$AOC=$\\angle$AOB=$60^\\circ$. Since OA$\\perp$BC, it follows that BE= $\\frac{1}{2}$ BC=$4\\sqrt{3}$ cm. In right triangle BOE, with $\\angle$AOB=$60^\\circ$, it follows that $OB=\\frac{BE}{\\sin 60^\\circ}=8$ cm. Therefore, the length of the minor arc BC is $\\frac{120\\pi \\times 8}{180}=\\boxed{\\frac{16}{3}\\pi}$ cm.",
    "image_path": "dataset/mm_math/images/MM_Math/51008731.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $CE$ bisects $\\angle ACB$, and $CF$ bisects the supplementary angle of $\\angle ACD$. If $EF \\parallel BC$ and intersects $AC$ at $M$, with $CE=8\\, cm$ and $CF=6\\, cm$, what is the length of $EM$ in $cm$?\n",
    "completion": "\\textbf{Solution:} Since CE bisects $\\angle ACB$, and CF bisects the supplementary angle $\\angle ACD$, \\\\\nit follows that $\\angle MCE = \\angle BCE = \\frac{1}{2}\\angle ACB$, $\\angle FCM = \\angle FCD = \\frac{1}{2}\\angle ACD$, \\\\\nhence $\\angle ECF = \\frac{1}{2}\\angle ACB + \\frac{1}{2}\\angle ACD = \\frac{1}{2}(\\angle ACD + \\angle ACD) = 90^\\circ$, \\\\\naccording to the Pythagorean theorem, it can be found that $EF = \\sqrt{EC^{2} + CF^{2}} = 10$cm, \\\\\nsince $EF\\parallel BC$, \\\\\nit is inferred that $\\angle MEC = \\angle BCE$, $\\angle MFC = \\angle FCD$, \\\\\nwhich implies $MEC = \\angle MCE$, $\\angle MFC = \\angle MCF$, \\\\\nthus $EM = CM = FM$, \\\\\nhence $EM = \\frac{1}{2}EF = \\boxed{5}$cm,",
    "image_path": "dataset/mm_math/images/MM_Math/52699331.png"
  },
  {
    "solution": "\\boxed{\\frac{24}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given the rhombus $ABCD$ as shown in the figure, where the diagonals $AC$ and $BD$ intersect at point $O$, with $BD=8$ and $BC=5$, and $AE \\perp BC$ at point $E$, what is the length of $AE$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral $ABCD$ is a rhombus with $BD=8$, \\\\\n$\\therefore BO=DO=4$, $\\angle BOC=90^\\circ$, \\\\\nIn $\\triangle OBC$, $OC=\\sqrt{BC^2-OB^2}=\\sqrt{5^2-4^2}=3$, \\\\\n$\\therefore AC=2OC=6$, \\\\\nThus, $S_{\\triangle ABC}=\\frac{1}{2}\\times AE\\times BC=\\frac{1}{2}\\times BO\\times AC$, \\\\\nHence, $5AE=24$, \\\\\nSolving gives: $AE=\\boxed{\\frac{24}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52188921.png"
  },
  {
    "solution": "\\boxed{y=-2x-2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, the graph of the linear function $y=mx+n$ and the graph of the inverse proportion function $y=\\frac{k}{x}$ intersect at point $A(-2,2)$ and point $B(1,a)$. What are the expressions of these two functions?\n",
    "completion": "\\textbf{Solution:} Let's substitute $A(-2,2)$ into $y=\\frac{k}{x}$ to obtain $k=-2\\times 2=-4$,\\\\\n$\\therefore$ the equation for the inverse proportion function is $y=-\\frac{4}{x}$,\\\\\nSubstituting $B(1,a)$ into $y=-\\frac{4}{x}$ gives $a=-4$,\\\\\n$\\therefore \\boxed{B(1,-4)}$,\\\\\nSubstituting $A(-2,2)$ and $B(1,-4)$ into $y=mx+n$ we get $\\left\\{\\begin{array}{l}-2m+n=2 \\\\ m+n=-4\\end{array}\\right.$,\\\\\nSolving this system gives $\\left\\{\\begin{array}{l}m=-2 \\\\ n=-2\\end{array}\\right.$,\\\\\n$\\therefore$ the equation for the linear function is $\\boxed{y=-2x-2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53065859_80.png"
  },
  {
    "solution": "\\boxed{130^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the quadrilateral $ABCD$ is inscribed in circle $\\odot O$. If $\\angle BOD = 100^\\circ$, then what is the measure of $\\angle DCB$?\n",
    "completion": "\\textbf{Solution:} Given that $\\because \\angle BOD=100^\\circ$,\\\\\n$\\therefore \\angle BAD=\\frac{1}{2}\\angle BOD=50^\\circ$;\\\\\nSince $\\because \\angle BAD+\\angle BCD=180^\\circ$,\\\\\n$\\therefore \\angle BCD=180^\\circ-\\angle BAD=180^\\circ-50^\\circ=\\boxed{130^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53066050.png"
  },
  {
    "solution": "\\boxed{\\sqrt{13}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, squares are constructed externally on each side of the right-angled triangle $ABC$. Line segment $CG$ is drawn, where square $ACDE$ and square $ABGF$ have areas of 1 and 5, respectively. What is the length of $CG$?\n",
    "completion": "\\textbf{Solution:} Draw line AH, \\\\\nsince the areas of square ACDE and square ABGF are 1 and 5, respectively, \\\\\ntherefore, AC$^{2}=1$, AB$^{2}=5$, GB=AB, \\\\\nthus, AC=1, \\\\\nsince $\\angle$ACB$=90^\\circ$, \\\\\ntherefore, BC$^{2}=AB^{2}-AC^{2}=5-1=4$, \\\\\nhence, BC=2, \\\\\nsince quadrilateral BDIH is a square, \\\\\ntherefore, CI=HI=BC=2, $\\angle$ICB$=90^\\circ$, CB=HB, \\\\\nthus, $\\angle$ACB+$\\angle$ICB$=180^\\circ$, \\\\\nhence, points A, C, and I are on the same line, \\\\\nthus, AI=AC+CI=1+2=3, \\\\\ntherefore, HA=$\\sqrt{HI^{2}+AI^{2}}=\\sqrt{2^{2}+3^{2}}=\\sqrt{13}$, \\\\\nsince $\\angle$ABG=$\\angle$CBH$=90^\\circ$, \\\\\ntherefore, $\\angle$GBC=$\\angle$ABH$=90^\\circ+\\angle$ABC, \\\\\nin $\\triangle GBC$ and $\\triangle ABH$, \\\\\n$\\left\\{\\begin{array}{l}GB=AB \\\\ \\angle GBC=\\angle ABH \\\\ CB=HB \\end{array}\\right.$, \\\\\nthus, $\\triangle GBC\\cong \\triangle ABH$ (SAS), \\\\\ntherefore, CG=HA=$\\sqrt{13}$, \\\\\nhence, the length of CG is $\\boxed{\\sqrt{13}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52861959.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, it is known that $AB=6$, $AC=10$, and $AD$ bisects $\\angle BAC$, with $BD\\perp AD$ at point $D$, and $E$ is the midpoint of $BC$. Find the length of $DE$.\n",
    "completion": "\\textbf{Solution:} As shown in the figure, extend $BD$ to meet $AC$ at point $F$.\\\\\nSince $AD$ bisects $\\angle BAC$,\\\\\nit follows that $\\angle BAD=\\angle FAD$.\\\\\nSince $BD\\perp AD$,\\\\\nit follows that $\\angle ADB=\\angle ADF=90^\\circ$.\\\\\nHence, $\\angle ABD=\\angle AFD$.\\\\\nTherefore, $AB=AF$ and $BD=DF$.\\\\\nHence, $D$ is the midpoint of $BF$.\\\\\nSince $AC=10$ and $AB=6$,\\\\\nit follows that $FC=AC-AF=AC-AB=10-6=4$.\\\\\nSince $E$ is the midpoint of $BC$,\\\\\nthus $DE$ is the midline of $\\triangle BFC$.\\\\\nTherefore, $DE=\\frac{1}{2}FC=\\frac{1}{2}\\times 4=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52423436.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated in the figure, point $C$ is a point on line segment $AB$, point $M$ is the midpoint of line segment $AC$, and point $N$ is the midpoint of line segment $BC$. If $AB = 12\\,cm$ and $AM = 5\\,cm$, what is the length of $BC$ in centimeters?\n",
    "completion": "\\textbf{Solution:} Since point M is the midpoint of line segment AC,\\\\\nit follows that AC $= 2$AM,\\\\\nsince AM $= 5$cm,\\\\\nit follows that AC $= 10$cm,\\\\\nsince AB $= 12$cm,\\\\\nit follows that BC $=$ AB $-$ AC $= \\boxed{2}$cm;",
    "image_path": "dataset/mm_math/images/MM_Math/52916470_77.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, $\\cos B=\\frac{4}{5}$, $AB=10$, and point $D$ is on side $BC$ with $AC=DC$. What is the length of $BD$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle C=90^\\circ$ and $\\cos B=\\frac{4}{5}$,\\\\\nit follows that $\\frac{BC}{AB}=\\frac{4}{5}$, hence $\\frac{BC}{10}=\\frac{4}{5}$,\\\\\nsolving this gives $BC=8$,\\\\\nthus, $AC=\\sqrt{AB^{2}-BC^{2}}=\\sqrt{10^{2}-8^{2}}=6$,\\\\\nwhich means $CD=AC=6$,\\\\\ntherefore, $BD=BC-CD=8-6=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52977974_146.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a set of triangles is placed on a table (ignoring thickness), with the right angle $(\\angle DAE)$ and the $60^\\circ$ angle $(\\angle BAC)$ having coincident vertices. The triangle $ADE$ can rotate around point $A$. During the rotation process, keeping $AD$ inside $\\angle BAC$, what is the degree measure of $\\angle 1 - \\angle 2$?\n",
    "completion": "\\textbf{Solution:} During the rotation process, when $AD$ is inside $\\angle BAC$, we have $\\angle 1 = \\angle DAE - \\angle DAC = 90^\\circ - \\angle DAC$, and $\\angle 2 = \\angle BAC - \\angle DAC = 60^\\circ - \\angle DAC$.\n\n$\\therefore \\angle 1 - \\angle 2 = (90^\\circ - \\angle DAC) - (60^\\circ - \\angle DAC) = \\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51321764_136.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the line segment $AD=30\\,cm$, with points C and B both on the line segment $AD$ and point E being the midpoint of $AB$. If $AC=\\frac{1}{3}AD$, and point F is the midpoint of line segment $CD$, what is the length of line segment $EF$ in $cm$?\n",
    "completion": "\\textbf{Solution:} Given that $\\because AC= \\frac{1}{3} AD=10\\,cm$,\\\\\n$\\therefore CD=AD-AC=20\\,cm$,\\\\\nsince $F$ is the midpoint of $CD$,\\\\\n$\\therefore DF= \\frac{1}{2} CD=10\\,cm$,\\\\\n$\\therefore BF=DF-BD=4\\,cm$,\\\\\n$\\therefore EF=BE-BF=AE-BF=\\boxed{8}\\,cm$.",
    "image_path": "dataset/mm_math/images/MM_Math/51482139_105.png"
  },
  {
    "solution": "\\boxed{y=\\frac{2}{3}x+2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the graph of the linear function $y=kx+b$ intersects with the x-axis at point $A(-3, 0)$ and with the y-axis at point B. It also intersects the graph of the directly proportional function $y=\\frac{4}{3}x$ at point $C(a, 4)$. Find the value of $a$ and the equation of the linear function $y=kx+b$?\n",
    "completion": "\\textbf{Solution:} Since point C lies on the graph of a direct proportion function, $\\therefore \\frac{4}{3}a=4$. Solving this gives: $a=\\boxed{3}$. Since points C(3, 4) and A(-3, 0) lie on the graph of a linear function, $\\therefore \\left\\{\\begin{array}{l} -3k+b=0 \\\\ 3k+b=4 \\end{array}\\right.$. Solving this system of equations yields $\\left\\{\\begin{array}{l} k=\\frac{2}{3} \\\\ b=2 \\end{array}\\right.$, therefore the equation of the linear function is $\\boxed{y=\\frac{2}{3}x+2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52423287_74.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y=-\\frac{1}{2}x+1$ intersects the graph of the hyperbolic function $y=\\frac{k}{x}$ at points $A$ and $B$, with the $x$-coordinate of point $A$ being $-2$ and the $x$-coordinate of point $B$ being $4$. The graph of the linear function intersects the $y$-axis at point $C$. If point $P$ is on the $y$-axis, and the area of $\\triangle ABP$ is $6$, find the coordinates of point $P$.\n",
    "completion": "\\textbf{Solution:}\n\nGiven that the graph of the inverse proportion function $y=-\\frac{4}{x}$ passes through point B, and the x-coordinate of point B is 4,\n\n$\\therefore y=-1$, i.e., B(4, -1),\n\nLet the coordinates of point P be (0, $a$),\n\nSince the graph of the linear function $y=-\\frac{1}{2}x+1$ intersects the y-axis at point C,\n\n$\\therefore$ C(0, 1),\n\n$\\therefore$ PC=$|1-a|$,\n\nSince the area of $\\triangle ABP$ is 6,\n\n$\\therefore$ $\\frac{1}{2}\\times |1-a|\\times 6=6$,\n\nSolving gives $a=-1$ or $a=3$,\n\n$\\therefore$ the coordinates of point P are (0, $-1$) or (0, $\\boxed{3}$).",
    "image_path": "dataset/mm_math/images/MM_Math/52602046_156.png"
  },
  {
    "solution": "\\boxed{6\\text{cm}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given that the length of segment $AB=15\\text{cm}$, and point $C$ is on segment $AB$ with $AC:CB=3:2$. Find the lengths of segments $AC$ and $CB$.\n",
    "completion": "\\textbf{Solution:} Given that $AB=15\\text{cm}$, and the ratio of $AC$ to $CB$ is $3:2$,\n\nit follows that $AC=\\frac{3}{3+2}AB=\\frac{3}{5}\\times 15=\\boxed{9\\text{cm}}$, and $CB=\\frac{2}{3+2}AB=\\frac{2}{5}\\times 15=\\boxed{6\\text{cm}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51366955_123.png"
  },
  {
    "solution": "\\boxed{40^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $BD$ is the diameter of $\\odot O$, with $A$ and $C$ on the circle. If $\\angle A=50^\\circ$, what is the measure of $\\angle DBC$?\n",
    "completion": "\\textbf{Solution:} Given that BD is the diameter of circle $O$,\\\\\nit follows that $\\angle BCD = 90^\\circ$,\\\\\nwhich implies $\\angle D = \\angle A = 50^\\circ$,\\\\\nthus, $\\angle DBC = 90^\\circ - \\angle D = \\boxed{40^\\circ}$,",
    "image_path": "dataset/mm_math/images/MM_Math/53106899.png"
  },
  {
    "solution": "\\boxed{P\\left(2, 1\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y = \\frac{3}{2}x - 2$ intersects with the line $y = -x + 3$ at point P. The line $y = \\frac{3}{2}x - 2$ intersects the x-axis and y-axis at points C and D, respectively; the line $y = -x + 3$ intersects the x-axis and y-axis at points A and B, respectively. What are the coordinates of the intersection point P?\n",
    "completion": "\\textbf{Solution}: According to the problem, we have:\\\\\n\\[\n\\begin{cases}\ny=\\frac{3}{2}x-2 \\\\\ny=-x+3\n\\end{cases}\n\\], solving this we get: \\[\n\\begin{cases}\nx=2 \\\\\ny=1\n\\end{cases}\n\\], \\\\\n$\\therefore \\boxed{P\\left(2, 1\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52386556_127.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, there is a square $ABCD$ with a side length of 10. Let $E$ be a moving point on side $BC$ (not coinciding with $B$ or $C$), connect $AE$. Let $G$ be a point on the extension line of $BC$. A perpendicular line is drawn from point $E$ to the angle bisector of $\\angle DCG$, intersecting at point $F$. If $FG \\perp BG$, please directly write down the value of $EC$ for which the area of $\\triangle CEF$ is maximized.\n",
    "completion": "\\textbf{Solution:} Let $CE=x$, thus $BE=10-x$, \\\\\n$\\therefore EG=CE+CG=x+FG$, \\\\\nAccording to (1), $\\triangle BAE \\sim \\triangle GEF$, \\\\\n$\\therefore \\frac{AB}{EG}=\\frac{BE}{FG}$, \\\\\n$\\therefore \\frac{10}{x+FG}=\\frac{10-x}{FG}$, \\\\\n$\\therefore FG=10-x$, \\\\\n$\\therefore \\text{Area of }\\triangle ECF=\\frac{1}{2}\\times CE\\times FG=\\frac{1}{2}\\cdot x\\cdot (10-x)=-\\frac{1}{2}(x^2-10x)=-\\frac{1}{2}(x-5)^2+\\frac{25}{2}$, \\\\\nWhen $x=\\boxed{5}$, $\\text{Area of }\\triangle ECF$ is maximal.",
    "image_path": "dataset/mm_math/images/MM_Math/52550935_93.png"
  },
  {
    "solution": "\\boxed{x>2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, line $l_1\\colon y=2x-1$ intersects with line $l_2\\colon y=kx+b\\ (k\\neq 0)$ at point $P(2, 3)$. What is the solution set of the inequality $2x-1>kx+b$ with respect to $x$?\n",
    "completion": "\\textbf{Solution:} According to the graph, the solution set for the inequality $2x-1>kx+b$ is: $\\boxed{x>2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52354905.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a regular hexagon $ABCDEF$ is inscribed in circle $\\odot O$. If the distance from the center to a side $OH=\\sqrt{3}$, then what is the radius of circle $\\odot O$?\n",
    "completion": "\\textbf{Solution:} Let's connect OD. Since ABCDEF is a regular hexagon and OH is the apothem, it follows that $\\angle DOH = \\frac{360^\\circ}{12} = 30^\\circ$ and $\\angle OHD = 90^\\circ$. Therefore, OD equals twice DH. In the right triangle $ODH$, we have $OH^{2}$ + $DH^{2}$ = $OD^{2}$. Hence, $3 + DH^{2} = 4DH^{2}$. Solving this gives: DH=1 (considering positive value). Therefore, DH=\\boxed{2}.",
    "image_path": "dataset/mm_math/images/MM_Math/53081474.png"
  },
  {
    "solution": "\\boxed{\\left(\\frac{7}{3}, \\frac{20}{3}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the line $y=2x+10$ intersects the $x$-axis at point $A$ and the $y$-axis at point $B$. Another line passing through point $B$ intersects the positive half of the $x$-axis at point $C$, and the area of $\\triangle ABC$ is $60$. If $M$ is a point on segment $BC$ and the line $AM$ divides the area of $\\triangle ABC$ into two parts with an area ratio of $1\\colon 2$, what are the coordinates of $M$?\n",
    "completion": "\\textbf{Solution:} Let $M(m,-\\frac{10}{7}m+10)$,\\\\\n(1) When $\\frac{S_{\\Delta ACM}}{S_{\\Delta ABM}} = \\frac{1}{2}$, that is $S_{\\Delta ACM} = \\frac{1}{3}S_{\\Delta ABC}$,\\\\\n$\\therefore \\frac{1}{2}AC \\cdot y_{M} = \\frac{1}{3} \\times 60 = 20$,\\\\\n$\\because AC = 12$,\\\\\n$\\therefore y_{M} = \\frac{10}{3}$,\\\\\n$\\therefore \\frac{10}{3} = -\\frac{10}{7}m + 10$,\\\\\nSolving, we get $m = \\frac{14}{3}$,\\\\\n$\\therefore M\\left(\\frac{14}{3}, \\frac{10}{3}\\right)$;\\\\\n(2) When $\\frac{S_{\\Delta ABM}}{S_{\\Delta ACM}} = \\frac{1}{2}$, that is $S_{\\Delta ACM} = \\frac{2}{3}S_{\\Delta ABC}$,\\\\\n$\\therefore \\frac{1}{2}AC \\cdot y_{M} = \\frac{2}{3} \\times 60 = 40$,\\\\\n$\\because AC = 12$,\\\\\n$\\therefore y_{M} = \\frac{20}{3}$,\\\\\n$\\therefore \\frac{20}{3} = -\\frac{10}{7}m + 10$,\\\\\nSolving, we get $m = \\frac{7}{3}$,\\\\\n$\\therefore M\\left(\\frac{7}{3}, \\frac{20}{3}\\right)$,\\\\\nIn conclusion, the coordinates of point M are $\\boxed{\\left(\\frac{14}{3}, \\frac{10}{3}\\right)}$, $\\boxed{\\left(\\frac{7}{3}, \\frac{20}{3}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52205244_124.png"
  },
  {
    "solution": "\\boxed{x > -2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the equation of line $l_1$ is $y_1=k_1x+b_1$, and the equation of line $l_2$ is $y_2=k_2x+b_2$. What is the solution set of the inequality $k_1x+b_1<k_2x+b_2$?\n",
    "completion": "\\textbf{Solution:} Solving the inequality $k_1x + b_1 < k_2x + b_2$ implies that the line $l_1$ is below the line $l_2$.\\\\\nFrom the graph of the function, we know that the range of $x$ is $x > -2$,\\\\\nmeaning the solution set for the inequality $k_1x + b_1 < k_2x + b_2$ is $x > -2$,\\\\\nthus the answer is: $\\boxed{x > -2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52212386.png"
  },
  {
    "solution": "\\boxed{-\\frac{15}{8}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=-2x^{2}+8x-6$ intersects the x-axis at points $A$ and $B$. Let the part of the parabola above and on the x-axis be denoted as $C_{1}$. After shifting $C_{1}$ to the right, we obtain $C_{2}$, which intersects the x-axis at points $B$ and $D$. If the line $y=x+m$ has 3 different intersection points with $C_{1}$ and $C_{2}$, what is the range of values for $m$?\n",
    "completion": "\\textbf{Solution:} Let $y=-2x^{2}+8x-6=0$, \\\\\nthat is, $x^{2}-4x+3=0$, \\\\\nsolving yields $x=1$ or $3$, \\\\\nthus, points $A(1,0)$, $B(3,0)$, \\\\\nsince translating $C_{1}$ rightward by 2 units gives $C_{2}$, \\\\\nthen, the equation for $C_{2}$ is $y=-2(x-4)^{2}+2$, $3\\leq x\\leq 5$, \\\\\nwhen $y=x+m_{1}$ is tangent to $C_{2}$, \\\\\nlet $y=x+m_{1}=-2(x-4)^{2}+2$, \\\\\nthat is, $2x^{2}-15x+30+m_{1}=0$, \\\\\n$\\Delta=-8m_{1}-15=0$, \\\\\nsolving yields $m_{1}=\\boxed{-\\frac{15}{8}}$, \\\\\nwhen $y=x+m_{2}$ passes through point $B$, \\\\\nthat is, $0=3+m_{2}$, \\\\\n$m_{2}=-3$, \\\\\nwhen $-3<m<-\\frac{15}{8}$, the line $y=x+m$ intersects with $C_{1}$ and $C_{2}$ at 3 distinct points.",
    "image_path": "dataset/mm_math/images/MM_Math/52731465.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, extend the line segment AB to C such that $BC = \\frac{1}{2}AB$. Let D be the midpoint of the line segment AC. If $DC = 3$, then what is the length of AB?\n",
    "completion": "\\textbf{Solution:}\n\nSince $D$ is the midpoint of $AC$, we have\n\n\\[AC = 2DC\\]\n\\[= 2 \\times 3\\]\n\\[= 6\\]\n\nAlso, since $BC = \\frac{1}{2}AB$ and $AC = AB + BC$, \n\n\\[\\therefore BC = \\frac{1}{3}AC\\]\n\\[= \\frac{1}{3} \\times 6\\]\n\\[= 2\\]\n\nFrom the relationship between the sums and differences of line segments, we have\n\n\\[AB = AC - BC\\]\n\\[= 6 - 2\\]\n\\[= AB = \\boxed{4}\\]",
    "image_path": "dataset/mm_math/images/MM_Math/51379682.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, given $\\triangle ABC$, $\\angle CAB=45^\\circ$, $AB=7$, $AC=3\\sqrt{2}$, and $CD\\perp AB$ at point D. E is a moving point on side BC. A circle $\\odot O$ is drawn with DE as the diameter, intersecting BC at point F and AB at point G. Lines DF and FG are drawn. $\\angle BCD=\\angle FDB$. When point E is on segment BF, and $\\triangle DFG$ is an isosceles triangle, what is the length of DG?\n",
    "completion": "\\textbf{Solution:}\n\n(i) When $DF=DG$, as shown in the figure:\n\nSince $\\angle CAB=45^\\circ$, $CD\\perp AB$, $AC=3\\sqrt{2}$\n\nTherefore $AD=CD=3$\n\nSince $AB=7$\n\nTherefore $BD=7-3=4$\n\nTherefore $BC=\\sqrt{3^2+4^2}=5$\n\nTherefore $DF=\\frac{3\\times 4}{5}=\\frac{12}{5}$\n\nTherefore $DG=\\boxed{\\frac{12}{5}}$\n\n(ii) As shown in the figure:\n\nWhen $DF=FG$, draw $FH\\perp BD$ intersecting $BD$ at point $H$,\n\nTherefore $\\triangle DFH\\sim \\triangle CBD$\n\nTherefore $\\frac{DH}{CD}=\\frac{DF}{CB}$\n\nTherefore $DH=DF\\times \\frac{3}{5}=\\frac{12}{5}\\times \\frac{3}{5}=\\frac{36}{25}$\n\nTherefore $DG=2DH=\\boxed{\\frac{72}{25}}$\n\n(iii) As shown in the figure:\n\nWhen $FG=DG$, $\\angle 1=\\angle 2$\n\nSince $\\angle 1+\\angle 3=\\angle 2+\\angle 4=90^\\circ$\n\nTherefore $\\angle 3=\\angle 4$\n\nTherefore $FG=GB=DG$\n\nTherefore $DG=\\frac{1}{2}BD=\\boxed{2}$",
    "image_path": "dataset/mm_math/images/MM_Math/51322957_132.png"
  },
  {
    "solution": "\\boxed{\\frac{7}{2}\\leq S\\leq 8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=x^2+bx+c$ (where $b$ and $c$ are constants) intersects the x-axis at points $A(-1,0)$ and $B(3,0)$, and the y-axis at point $C$. Let $P$ be a point on the parabola with the x-coordinate $m$. Let the area of $\\triangle ABP$ be denoted as $S$. Find the range of possible values for $S$ when $0 \\leq m \\leq \\frac{5}{2}$.\n",
    "completion": "\\textbf{Solution:} Draw $PE\\perp AB$ at point E through point P, as shown in the figure,\\\\\nsince $0\\leq m\\leq \\frac{5}{2}$,\\\\\nthus P($m$, $m^{2}-2m-3$) is in the fourth quadrant,\\\\\ntherefore $PE=-m^{2}+2m+3$.\\\\\nSince A($-1$,0), B(3,0),\\\\\nthus $OA=1$, $OB=3$,\\\\\ntherefore $AB=OA+OB=4$.\\\\\nTherefore, the area of $\\triangle PAB=\\frac{1}{2}AB\\cdot PE$\\\\\n$=\\frac{1}{2}\\times 4\\times (-m^{2}+2m+3)$\\\\\n$=-2m^{2}+4m+6$\\\\\n$=-2(m-1)^{2}+8$.\\\\\nThus, when $m=1$, the area of $\\triangle PAB$ reaches its maximum value of $8$.\\\\\nSince $0\\leq m\\leq \\frac{5}{2}$,\\\\\nthus, when $m=\\frac{5}{2}$, the area of $\\triangle PAB$ reaches its minimum value of $\\frac{7}{2}$.\\\\\nTherefore, the range of values for $S$ is: $\\boxed{\\frac{7}{2}\\leq S\\leq 8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52623662_242.png"
  },
  {
    "solution": "\\boxed{22^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ABC=90^\\circ$, $\\angle A=60^\\circ$, $D$ is a point on side $AC$, $\\angle DBC=30^\\circ$, and $AC=12\\,cm$. What is the perimeter of $\\triangle ABD$?\n",
    "completion": "\\textbf{Solution:} Given that $\\triangle ABC \\cong \\triangle DBE$ and $\\angle ABD = 23^\\circ$, $\\angle DBE = 45^\\circ$,\\\\\nit follows that $\\angle ABC = \\angle DBE = 45^\\circ$,\\\\\ntherefore, $\\angle DBC = \\angle ABC - \\angle ABD = \\boxed{22^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53066376.png"
  },
  {
    "solution": "\\boxed{9\\sqrt{3}\\,\\text{cm}^2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, there is a paper circle $\\odot O$ with a radius of 6cm. $\\triangle ABC$ is an inscribed triangle in $\\odot O$. The paper circle $\\odot O$ is folded along the straight lines $AB$ and $AC$, and both arcs $\\overset{\\frown}{AB}$ and $\\overset{\\frown}{AC}$ pass through the center $O$. What is the area of the shaded region in the figure?\n",
    "completion": "\\textbf{Solution:} Draw lines $AO, BO, CO$, and extend $AO$ to intersect $BC$ at point D, as shown in the diagram:\\\\\nSince $\\triangle ABC$ is an inscribed triangle of circle $O$ with a radius of $6\\,\\text{cm}$,\\\\\ntherefore, $AD\\perp BC$ and $AO=6\\,\\text{cm}$,\\\\\ntherefore, $OD=3\\,\\text{cm}$,\\\\\ntherefore, $CD=\\sqrt{OC^{2}-OD^{2}}=3\\sqrt{3}$,\\\\\ntherefore, $BC=2CD=6\\sqrt{3}\\,\\text{cm}$,\\\\\nFrom the diagram, it can be seen that the area of the shaded part is the same as the area of $\\triangle OBC$,\\\\\ntherefore, $S_{\\text{shaded}}=S_{\\triangle COB}=\\frac{1}{2}\\times BC\\times OD=\\frac{1}{2}\\times 6\\sqrt{3}\\times 3=\\boxed{9\\sqrt{3}\\,\\text{cm}^2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52917397.png"
  },
  {
    "solution": "\\boxed{9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points A, B, C, and D are four points on the circle $\\odot O$. AC bisects $\\angle BAD$, AC intersects BD at point E, $CE=4$, $CD=6$. What is the length of AC?\n",
    "completion": "\\textbf{Solution:} Given that $AC$ bisects $\\angle BAD$, \\\\\nit follows that $\\overset{\\frown}{BC}=\\overset{\\frown}{CD}$, \\\\\nhence $\\angle BDC=\\angle CAD$, \\\\\ngiven that $\\angle ACD=\\angle DCE$, \\\\\nit follows that $\\triangle CDE\\sim \\triangle CAD$, \\\\\ntherefore, $CD: AC=CE: CD$, \\\\\nwhich implies $CD^{2}=AC\\cdot CE$, \\\\\nhence $6^{2}=4(4+AE)$, \\\\\nthus $AE=5$, \\\\\ntherefore, $AC=AE+CE=9$, \\\\\nthus, the length of $AC$ is $\\boxed{9}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52991552.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in a triangular piece of paper $ABC$, $\\angle B = 90^\\circ$, $AB = 3$. Point $D$ is on side $BC$. When $\\triangle ABD$ is folded along the straight line $AD$, point $B$ exactly falls on point $E$ on the side $AC$. If $AD = CD$, what is the length of $AC$?\n",
    "completion": "\\textbf{Solution:} It follows from the property of folding that $AE=AB=3$,\\\\\n$\\angle AED=90^\\circ$ \\\\\n$\\because AD=CD$ \\\\\n$\\therefore \\triangle ADC$ is an isosceles triangle \\\\\n$\\because DE\\perp AC$ \\\\\n$\\therefore EC=AE$ \\\\\n$\\therefore AC=\\boxed{6}$",
    "image_path": "dataset/mm_math/images/MM_Math/52720674.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown, both $\\triangle ABC$ and $\\triangle ADE$ are isosceles right-angled triangles, with $\\angle BAC = \\angle DAE = 90^\\circ$, and $AB = 4$. Point $F$ is the midpoint of $DE$. If point $E$ is a moving point on line $BC$, and $BF$ is connected, what is the minimum value of $BF$?\n",
    "completion": "\\textbf{Solution:} Since both $\\triangle ABC$ and $\\triangle ADE$ are isosceles right-angled triangles,\\\\\nit follows that $\\triangle ABC\\sim \\triangle ADE$,\\\\\nhence $\\angle ADE = \\angle ABE$,\\\\\ntherefore, points A, D, B, E are concyclic,\\\\\nsince $\\angle DAE = 90^\\circ$,\\\\\nit follows that $\\angle DBE = 90^\\circ$,\\\\\nsince F is the midpoint of DE,\\\\\nit follows that BF = $\\frac{1}{2}$DE,\\\\\ntherefore, when DE is at its minimum, the value of BF is also minimal,\\\\\nsince if point E is a moving point on line BC,\\\\\nit follows that when AE$\\perp$BC, AE is at its minimum, at this time, DE is also minimal,\\\\\nsince $\\angle BAC = 90^\\circ$, AB = 4 = AC,\\\\\nit follows that BC = $4\\sqrt{2}$,\\\\\ntherefore, AE = $\\frac{AB \\cdot AC}{BC} = \\frac{4 \\times 4}{4\\sqrt{2}} = 2\\sqrt{2}$,\\\\\ntherefore, DE = 4,\\\\\ntherefore, BF = $\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52550647.png"
  },
  {
    "solution": "\\boxed{(-2\\sqrt{5},0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the direct proportion function $y_1=kx$ intersects with the graph of the inverse proportion function $y_2=\\frac{m}{x}$ at points $A(2,4)$ and point $B$. The coordinates of point $C$ are $(4,0)$, and point $D$ is on the graph of $y_2=\\frac{m}{x}$. Suppose point $E$ is on the x-axis, and $\\angle AEB = 90^\\circ$, find the coordinates of point $E$.\n",
    "completion": "\\textbf{Solution:} As shown in Figure 1, draw $AG\\perp x$-axis at $G$, and $BH\\perp x$-axis at $H$,\\\\\nsince the graph of the direct proportion function $y_{1}=kx$ intersects with the graph of the inverse proportion function $y_{2}=\\frac{8}{x}$ at points $A(2,4)$ and $B$,\\\\\ntherefore $B(-2,-4)$,\\\\\nhence $OG=OH=2$, $AG=BH=4$,\\\\\nlet the coordinate of $E$ be $(x,0)$ (where $|x|>2$), then $EG=|2-x|$, $EH=|x+2|$,\\\\\nsince $\\angle AEG+\\angle BEH=\\angle AEB=90^\\circ$, $\\angle BEH+\\angle EBH=90^\\circ$,\\\\\ntherefore $\\angle AEG=\\angle EBH$,\\\\\nsince $\\angle AGE=\\angle EHB=90^\\circ$,\\\\\ntherefore $\\triangle EBH\\sim \\triangle AEG$,\\\\\ntherefore $\\frac{EH}{AG}=\\frac{BH}{EG}$, that is $\\frac{|x+2|}{4}=\\frac{4}{|x-2|}$,\\\\\nrewriting, we get $x^2-4=16$,\\\\\nsolving, $x=\\pm 2\\sqrt{5}$,\\\\\ntherefore the coordinates of point $E$ are $\\boxed{(2\\sqrt{5},0)}$ or $\\boxed{(-2\\sqrt{5},0)}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51403218_43.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $OC \\perp AB$ at point O, $OD$ bisects $\\angle BOC$, $OE$ bisects $\\angle AOD$. Find the degree measure of $\\angle BOD$.\n",
    "completion": "\\textbf{Solution:} Since $OC \\perp AB$,\\\\\nit follows that $\\angle BOC = \\angle AOC = 90^\\circ$,\\\\\nsince $OD$ bisects $\\angle BOC$,\\\\\nit follows that $\\angle BOD = \\angle COD = \\frac{1}{2}\\angle BOC = 45^\\circ = \\boxed{45^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53364075_164.png"
  },
  {
    "solution": "\\boxed{9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the equation of line $AB$ is $y=x+2$, and the equation of line $AC$ is $y=-x+4$. The two lines intersect at point $A$ and intersect the $x$-axis at points $B$ and $C$, respectively. What is the area of $\\triangle ABC$?\n",
    "completion": "\\textbf{Solution:} Draw $AE \\perp x$ axis at point $E$ through point A. By solving the system of equations given by the problem statement\n\\[\n\\left\\{\n\\begin{array}{l}\ny=x+2 \\\\\ny=-x+4\n\\end{array}\n\\right.\n\\]\nwe find:\n\\[\n\\left\\{\n\\begin{array}{l}\nx=1 \\\\\ny=3\n\\end{array}\n\\right.\n\\]\nTherefore, $A(1, 3)$ and thus $AE=3$. When $y=0$, we have $x+2=0$, therefore $x=-2$, leading to $B(-2, 0)$. When $y=0$, $-x+4=0$, therefore $x=4$, leading to $C(4, 0)$. Therefore, $BC=4-(-2)=6$, therefore the area of $\\triangle ABC = \\frac{1}{2}BC\\cdot AE = \\frac{1}{2} \\times 6 \\times 3 = \\boxed{9}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50590416_149.png"
  },
  {
    "solution": "\\boxed{\\sqrt{2}+\\sqrt{6}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in parallelogram $ABCD$, $\\angle ABC = 45^\\circ$. Points $E$ and $F$ are located on the extensions of $CD$ and $BC$ respectively, with $AE \\parallel BD$ and $\\angle EFC = 30^\\circ$. Given $AB = 1$, what is the length of $CF$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw line EH $\\perp$ BF at point H,\\\\\n$\\because$ Quadrilateral ABCD is a parallelogram,\\\\\n$\\therefore AB \\parallel CD$, AB=DC,\\\\\n$\\because AE \\parallel DB$,\\\\\n$\\therefore$ Quadrilateral ABDE is a parallelogram,\\\\\n$\\therefore AB=DE=CD$, thus D is the midpoint of CE.\\\\\n$\\because AB=1$,\\\\\n$\\therefore CE=2$,\\\\\n$\\therefore AB \\parallel CD$,\\\\\n$\\therefore \\angle ECF=\\angle ABC=45^\\circ$,\\\\\n$\\because CE=8$, $\\angle ECF=45^\\circ$,\\\\\n$\\therefore EH=CH=\\sqrt{2}$,\\\\\n$\\because \\angle EFC=30^\\circ$,\\\\\n$\\therefore FH=\\sqrt{3}EH=\\sqrt{6}$,\\\\\n$\\therefore CF=\\sqrt{2}+\\sqrt{6}=\\boxed{\\sqrt{2}+\\sqrt{6}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51684372.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ is an equilateral triangle with a side length of 4. Point $P$ is on $AB$. Through point $P$, draw $PE\\perp AC$ with the foot of the perpendicular as $E$. Extend $BC$ to point $Q$ such that $CQ=PA$. Connect $PQ$ and let it intersect $AC$ at point $D$. Find the length of $DE$.\n",
    "completion": "\\textbf{Solution:} Draw $PF \\parallel BC$ through point P intersecting $AC$ at F, then $\\angle FPD = \\angle Q$, $\\angle APF = \\angle B$, $\\angle AFP = \\angle ACB$,\n\n$\\because$ $\\triangle ABC$ is an equilateral triangle,\n\n$\\therefore$ $\\angle B = \\angle A = \\angle ACB = 60^\\circ$,\n\n$\\therefore$ $\\angle APF = \\angle AFP = \\angle A = 60^\\circ$,\n\n$\\therefore$ $\\triangle APF$ is an equilateral triangle, and $PE \\perp AC$,\n\n$\\therefore$ AP=PF, AE=EF, and AP=CQ,\n\n$\\therefore$ PF=CQ,\n\nIn $\\triangle PDF$ and $\\triangle QDC$,\n\n$\\left\\{\\begin{array}{l}\n\\angle FPD = \\angle Q \\\\\n\\angle PDF = \\angle QDC \\\\\nPF = CQ\n\\end{array}\\right.$,\n\n$\\therefore$ $\\triangle PDF \\cong \\triangle QDC$ (AAS),\n\n$\\therefore$ FD=CD,\n\n$\\therefore$ EF + FD = AE + CD,\n\n$\\therefore$ DE = $\\frac{1}{2}$AC, and AC=4,\n\n$\\therefore$ DE = \\boxed{2}.",
    "image_path": "dataset/mm_math/images/MM_Math/52856072.png"
  },
  {
    "solution": "\\boxed{\\frac{7\\sqrt{3}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In rhombus $ABCD$, $AB=AC=10$, diagonals $AC$ and $BD$ intersect at point $O$. Point $M$ is on line segment $AC$, and $AM=3$. Point $P$ is a moving point on line segment $BD$. What is the minimum value of $MP+\\frac{1}{2}PB$?\n",
    "completion": "\\textbf{Solution:} Solve: Draw $PE\\perp BC$ at E through P, and draw $MF\\perp BC$ at F through M, \\\\\nsince quadrilateral ABCD is a rhombus, \\\\\ntherefore AB=BC, \\\\\nsince AB=AC=10, \\\\\ntherefore AB=BC=AB=AD=CD=10, \\\\\ntherefore $\\triangle ABC$ and $\\triangle ACD$ are both equilateral triangles, \\\\\ntherefore $\\angle ABC=\\angle ACB=\\angle ADC=60^\\circ$, \\\\\nsince BD is the diagonal of the rhombus ABCD, \\\\\ntherefore BD bisects $\\angle ABC$, \\\\\ntherefore $\\angle CBO= \\frac{1}{2}\\angle ABC=30^\\circ$, \\\\\ntherefore PE=$ \\frac{1}{2}$BP, \\\\\ntherefore PM+$ \\frac{1}{2}$BP=PM+PE$\\leq$ MF, \\\\\nsince point P is on BD, \\\\\ntherefore the shortest distance is when points M, P, E are collinear, $MP+\\frac{1}{2}PB_{\\text{min}}=MF$, \\\\\nsince AM=3, \\\\\ntherefore CM=AC−AM=10−3=7, \\\\\nin $\\triangle MFC$, $\\angle MCF=60^\\circ$, \\\\\ntherefore $\\sin\\angle MCF= \\frac{MF}{MC}$, \\\\\ntherefore MF=$MC\\sin60^\\circ= 7\\times \\frac{\\sqrt{3}}{2}=\\boxed{\\frac{7\\sqrt{3}}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52302717.png"
  },
  {
    "solution": "\\boxed{8\\sqrt{5} - 3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the square $ABCD$ with side length 8, $E$ and $F$ are moving points on sides $AB$ and $BC$, respectively, and $EF=6$. Let $M$ be the midpoint of $EF$, and $P$ be a moving point on side $AD$. What is the minimum value of $CP+PM$?\n",
    "completion": "\\textbf{Solution:} Extend CD to $C'$, making $C'D = CD$,\\\\\n$CP + PM = C'P + PM$,\\\\\nWhen $C'$, P, and M are collinear, the value of $C'P + PM$ is minimized,\\\\\nAccording to the problem, the trajectory of point M is on the circular arc with B as the center and radius 3,\\\\\nThe minimum distance from a point outside a circle $C'$ to a point M on the circle, $C'M = C'B - 3$,\\\\\nsince $BC = CD = 8$,\\\\\nthus $CC' = 16$,\\\\\nthus $C'B = \\sqrt{CC'^2 + BC^2} = \\sqrt{16^2 + 8^2} = 8\\sqrt{5}$,\\\\\ntherefore, the minimum value of $CP + PM$ is $8\\sqrt{5} - 3 = \\boxed{8\\sqrt{5} - 3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52354875.png"
  },
  {
    "solution": "\\boxed{x>-2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the graphs of the functions $y=3x+b$ and $y=ax-3$ intersect at point $P(-2, -5)$. Based on the graph, what is the solution set of the inequality $3x+b>ax-3$?\n",
    "completion": "\\textbf{Solution:} Since the graphs of the functions $y=3x+b$ and $y=ax-3$ intersect at point $P(-2,-5)$,\\\\\nit follows that the solution set of the inequality $3x+b>ax-3$ is $x>-2$,\\\\\ntherefore, the answer is: $\\boxed{x>-2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50555655.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{3}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the regular hexagon ABCDEF is inscribed in $\\odot O$, where the radius of $\\odot O$ is 1. What is the length of the apothem $OM$?\n",
    "completion": "\\textbf{Solution:} Connect OB, \\\\\n$\\because$ hexagon ABCDEF is a regular hexagon inscribed in $\\odot$ O, \\\\\n$\\therefore \\angle$BOM = $\\frac{360^\\circ}{6\\times 2}$ = $30^\\circ$, \\\\\n$\\therefore$ OM = OB $\\cdot \\cos \\angle$BOM = 1 $\\times \\frac{\\sqrt{3}}{2}$ = $\\boxed{\\frac{\\sqrt{3}}{2}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52913781.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{1}{3}x+5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Place a rectangular sheet $OABC$ in the Cartesian coordinate system, where $O$ is the origin, point $C$ is on the $x$-axis, and point $A$ is on the $y$-axis, with $OA=9$ and $OC=15$. As shown in Figure 1, take a point $E$ on $OA$ and fold $\\triangle EOC$ along $EC$, such that point $O$ falls on point $D$ on side $AB$. Find the equation of the line $EC$.\n",
    "completion": "\\textbf{Solution:} In rectangle $ABCO$, we have $AB=OC$, $OA=BC$, $\\angle B=\\angle BAO=\\angle BCO=\\angle EOC=90^\\circ$, since $OA=9$, $OC=15$, therefore $AB=15$, $BC=9$. The coordinates of point $C$ are $(15,0)$. Based on the property of folding, we have: $DC=OC$, $EO=ED$, therefore $DC=15$. In $\\triangle DBC$, $DB=\\sqrt{DC^{2}-BC^{2}}=\\sqrt{15^{2}-9^{2}}=12$. Therefore, $AD=AB-DB=15-12=3$. Since $OA=9$, $EO=ED$, thus $AE=OA-EO=9-ED$. In $\\triangle AED$, we have $AE^{2}+AD^{2}=ED^{2}$, with $AD=3$, we get $(9-ED)^{2}+3^{2}=ED^{2}$. Solving this gives $DE=5$, therefore $EO=ED=5$. Thus, the coordinates of point $E$ are $(0,5)$. Let the equation of line $EC$ be $y=kx+b$. Substituting the coordinates of point $C$ as $(15,0)$ and point $E$ as $(0,5)$, we obtain $\\begin{cases} 15k+b=0 \\\\ b=5 \\end{cases}$. Solving this, we get $\\begin{cases} k=-\\frac{1}{3} \\\\ b=5 \\end{cases}$, hence the equation of line $EC$ is \\boxed{y=-\\frac{1}{3}x+5}.",
    "image_path": "dataset/mm_math/images/MM_Math/52381479_144.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{6}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $BC$ bisects $\\angle ABD$, $\\angle A = \\angle BCD$. If $AB = 4$ and $BD = 6$, what is the length of $BC$?\n",
    "completion": "\\textbf{Solution:} Since $BC$ bisects $\\angle ABD$, \\\\\nit follows that $\\angle ABC = \\angle CBD$, \\\\\nMoreover, since $\\angle A = \\angle BCD$, \\\\\nit follows that $\\triangle ABC \\sim \\triangle CBD$,\\\\\nthus, $\\frac{AB}{BC} = \\frac{BC}{BD}$, \\\\\nGiven $AB = 4$ and $BD = 6$, \\\\\nit follows that $BC^{2} = AB \\times BD = 4 \\times 6 = 24$, \\\\\nhence, $BC = \\boxed{2\\sqrt{6}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52977734.png"
  },
  {
    "solution": "\\boxed{y=\\frac{1}{4}x^2+\\frac{3}{2}x-4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the parabola $y=ax^2+bx-4$ passes through points $A(-8, 0)$ and $B(2, 0)$, and the line $x=-4$ intersects the $x$-axis at point $C$ and intersects the parabola at point $D$. What is the equation of this parabola?\n",
    "completion": "\\textbf{Solution:} Since the parabola $y=ax^2+bx-4$ passes through points $A(-8, 0)$ and $B(2, 0)$,\\\\\nlet the equation of the parabola be $y=a(x+8)(x-2)=ax^2+6ax-16a$,\\\\\nthus $-16a=-4$, solving this gives $a=\\frac{1}{4}$,\\\\\ntherefore, the equation of the parabola is $\\boxed{y=\\frac{1}{4}x^2+\\frac{3}{2}x-4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52936367_230.png"
  },
  {
    "solution": "\\boxed{120^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, place the right-angled vertex $O$ of the right triangle $OMN$ on the line $AB$, and the ray $OC$ bisects $\\angle AON$. Rotate the right triangle $OMN$ counterclockwise around vertex $O$. During the rotation process: when $\\angle BON = 120^\\circ$, what is the measure of $\\angle COM$?\n",
    "completion": "\\textbf{Solution:} When $NO$ is above the line $AB$,\n\n$\\because \\angle AON = 180^\\circ - \\angle BON$,\n\n$\\therefore \\angle AON = 60^\\circ$,\n\n$\\because OC$ bisects $\\angle AON$,\n\n$\\therefore \\angle CON = \\frac{1}{2}\\angle AON = 30^\\circ$,\n\n$\\because \\angle MON = 90^\\circ$,\n\n$\\therefore \\angle COM = \\angle MON - \\angle CON = 90^\\circ - 30^\\circ = 60^\\circ$;\n\nWhen $ON$ is below the line $AB$,\n\n$\\because \\angle AON = 180^\\circ - \\angle BON$,\n\n$\\therefore \\angle AON = 60^\\circ$,\n\n$\\because OC$ bisects $\\angle AON$,\n\n$\\therefore \\angle CON = 30^\\circ$,\n\n$\\therefore \\angle COM = \\angle MON + \\angle CON = 90^\\circ + 30^\\circ = 120^\\circ$;\n\n$\\therefore$ The measure of $\\angle COM$ is either $\\boxed{60^\\circ}$ or $\\boxed{120^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53520190_89.png"
  },
  {
    "solution": "\\boxed{\\sqrt{7}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the diagonals $AC$ and $BD$ of rhombus $ABCD$ intersect at point $O$. Through point $D$, draw $DE \\parallel AC$ and $DE = \\frac{1}{2}AC$. Connect $CE$ and $OE$, and connect $AE$ which intersects $OD$ at point $F$. $OE = CD$; If the side length of rhombus $ABCD$ is $2$, and $\\angle ABC=60^\\circ$, what is the length of $AE$?\n",
    "completion": "\\textbf{Solution:} Given that in diamond ABCD, $\\angle ABC=60^\\circ$,\\\\\ntherefore, $\\triangle ABC$ is an equilateral triangle,\\\\\ntherefore, $AC=AB=2$,\\\\\nsince $OA=\\frac{1}{2}AC=1$ and $AC\\perp BD$, $AD=2$,\\\\\nthus, $OD=\\sqrt{AD^{2}-OA^{2}}=\\sqrt{3}$,\\\\\ntherefore, in rectangle OCED, $CE=OD=\\sqrt{AD^{2}-AO^{2}}=\\sqrt{3}$,\\\\\ntherefore, in $\\triangle ACE$, $AE=\\sqrt{AC^{2}+CE^{2}}=\\boxed{\\sqrt{7}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53639911_178.png"
  },
  {
    "solution": "\\boxed{36}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, $AC \\perp BC$, $AB=13$, $BC=12$, $CD=3$, $AD=4$. What is the area of quadrilateral $ABCD$?\n",
    "completion": "\\textbf{Solution:} Given that $\\because A{D}^{2}+C{D}^{2}=16+9=25=A{C}^{2}$,\\\\\n$\\therefore \\angle D=90^\\circ$,\\\\\n$\\therefore {S}_{\\text{quadrilateral }ABCD}={S}_{\\triangle ABC}+{S}_{\\triangle ACD}=\\frac{12\\times 5}{2}+\\frac{4\\times 3}{2}=\\boxed{36}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52890848_108.png"
  },
  {
    "solution": "\\boxed{\\frac{1+\\sqrt{13}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, the parabola $y=-x^2+bx+c$ intersects the x-axis at points A and B, and intersects the y-axis at point C (0, 3). The axis of symmetry is the line $x=1$, which intersects the x-axis at point D, and the vertex is point E. As shown in Figure 2, point F is on the y-axis, and $OF=\\sqrt{2}$, point N is a moving point on the parabola within the first quadrant, and is to the right of the parabola's axis of symmetry. The line ON intersects the axis of symmetry at point G, and line GF is drawn. If GF bisects $\\angle OGE$, find the coordinates of point N.\n",
    "completion": "\\textbf{Solution:} Let $FP \\perp DE$ at point P through point F, thus $FP=1$. Draw $FQ \\perp ON$ at Q through point F, \\\\\nsince $GF$ bisects $\\angle OGE$, \\\\\ntherefore $FQ=FP=1$, \\\\\nin $\\triangle FQO$, $OF= \\sqrt{2}$, \\\\\naccording to the Pythagorean theorem, we get $OQ=\\sqrt{OF^{2}-FQ^{2}}=1$, \\\\\nthus $OQ=FQ$, \\\\\nthus $\\angle FOQ=45^\\circ$, \\\\\nthus $\\angle BON=90^\\circ-45^\\circ=45^\\circ$, \\\\\ndraw $QM \\perp OB$ at M through point Q, $OM=QM$ \\\\\ntherefore, the equation of $ON$ is $y=x$, \\\\\nsince point N lies on the parabola $y=-x^{2}+2x+3$, \\\\\nby solving the system $\\begin{cases}y=x\\\\ y=-x^{2}+2x+3\\end{cases}$, \\\\\nwe obtain $\\left\\{\\begin{array}{l}x=\\frac{1+\\sqrt{13}}{2}\\\\ y=\\frac{1+\\sqrt{13}}{2}\\end{array}\\right\\}$ or $\\left\\{\\begin{array}{l}x=\\frac{1-\\sqrt{13}}{2}\\\\ y=\\frac{1-\\sqrt{13}}{2}\\end{array}\\right\\}$ (since point N is to the right of the axis of symmetry $x=1$, the latter is discarded), \\\\\nthus, the coordinates of point N are: $\\left(\\boxed{\\frac{1+\\sqrt{13}}{2}}, \\boxed{\\frac{1+\\sqrt{13}}{2}}\\right)$.",
    "image_path": "dataset/mm_math/images/MM_Math/53080571_43.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the figure, in $\\triangle AED$, $\\angle AED=90^\\circ$, $AB=AC=AD$, $EC=2$, $BE=8$, then what is the value of $DE$?\n",
    "completion": "\\textbf{Solution:} We have $DE=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52867068_61.png"
  },
  {
    "solution": "\\boxed{3.6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, two straight lines are intersected by three parallel lines, where $AB=4$, $BC=6$, and $EF=5.4$. What is the length of $DE$?\n",
    "completion": "\\textbf{Solution:} Given $\\because AD\\parallel BE\\parallel CF$,\\\\\n$\\therefore \\frac{AB}{AC}=\\frac{DE}{DF}$,\\\\\n$\\because AB=4, BC=6, EF=5.4$,\\\\\n$\\therefore AC=AB+BC=4+6=10, DF=5.4+DE$,\\\\\n$\\therefore \\frac{4}{10}=\\frac{DE}{5.4+DE}$,\\\\\n$\\therefore DE=\\boxed{3.6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53102142.png"
  },
  {
    "solution": "\\boxed{P(-2, -5)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the parabola $y=-x^2+bx+c$ passes through points $A(-1,0)$ and $B(3,0)$. Let point $P$ be a point on the parabola. If the area of $\\triangle PAB$ is $10$, find the coordinates of point $P$ at this time.\n",
    "completion": "\\textbf{Solution:} Let $P(x, y)$,\\\\\nthus, the height of $\\triangle PAB$ is $|y|$,\\\\\nsince $A(-1, 0)$, $B(3, 0)$,\\\\\nthus, $AB=4$,\\\\\ntherefore, the area of $\\triangle ABP$ is $\\frac{1}{2} \\times 4 \\times |y| = 10$,\\\\\nsolving gives $y=\\pm 5$,\\\\\nwhen $y=5$,\\\\\n$5=-x^2+2x+3$,\\\\\nno solution for this equation,\\\\\nwhen $y=-5$,\\\\\n$-5=-x^2+2x+3$,\\\\\nsolving gives $x_1=4, x_2=-2$,\\\\\nthus, the coordinates of $P$ are $\\boxed{P(4, -5)}$ or $\\boxed{P(-2, -5)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52977350_189.png"
  },
  {
    "solution": "\\boxed{4−2\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $AB=2$, point $E$ is on side $AD$, $EB$ bisects $\\angle AEC$, and $\\angle DEC=30^\\circ$. What is the length of $AE$?\n",
    "completion": "\\textbf{Solution:} In rectangle ABCD, we have AD=BC, CD=AB=2, AD$\\parallel$BC,\\\\\n$\\because$ $\\angle$DEC＝$30^\\circ$,\\\\\n$\\therefore$ $\\angle$BCE=$\\angle$DEC=$30^\\circ$, $\\angle$AEC=$150^\\circ$, CE=2CD=4,\\\\\n$\\therefore$ $DE=\\sqrt{CE^{2}−CD^{2}}=2\\sqrt{3}$,\\\\\n$\\because$ EB bisects $\\angle$AEC,\\\\\n$\\therefore$ $\\angle BEC=\\frac{1}{2}\\angle AEC=75^\\circ$,\\\\\n$\\therefore$ $\\angle$CBE=$75^\\circ$,\\\\\n$\\therefore$ $\\angle$CBE=$\\angle$BEC,\\\\\n$\\therefore$ CE=BC=AD=4,\\\\\n$\\therefore$ $AE=AD−DE=4−2\\sqrt{3}$.\\\\\nThus, the answer is: $AE=\\boxed{4−2\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52354640.png"
  },
  {
    "solution": "\\boxed{15^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $O$ is a point on line $AB$, and the right-angled vertex of a right-angled triangle $OMN$ is placed at point $O$. The ray $OC$ bisects $\\angle MOB$. As shown in Figure 1, if $\\angle AOM = 30^\\circ$, what is the degree measure of $\\angle CON$?\n",
    "completion": "\\textbf{Solution}: From the given, we find that \\angle BOM=180^\\circ-\\angle AOM=150^\\circ,\\\\\nFurthermore, since \\angle MON is a right angle and $OC$ bisects \\angle BOM,\\\\\nit follows that \\angle CON=\\angle MON-\\frac{1}{2}\\angle BOM \\\\\n=90^\\circ-\\frac{1}{2}\\times 150^\\circ,\\\\\n= \\boxed{15^\\circ}",
    "image_path": "dataset/mm_math/images/MM_Math/52794314_145.png"
  },
  {
    "solution": "\\boxed{50^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that in $\\triangle ABC$, $AD$ bisects $\\angle BAC$ and intersects $BC$ at $D$, $AE \\perp BC$ at $E$. If $\\angle ADE = 80^\\circ$, $\\angle EAC = 20^\\circ$, what is the degree measure of $\\angle B$?\n",
    "completion": "\\textbf{Solution:} Since $AE \\perp BC$ and $\\angle EAC = 20^\\circ$,\\\\\nit follows that $\\angle C = 70^\\circ$,\\\\\nhence $\\angle BAC + \\angle B = 110^\\circ$.\\\\\nSince $\\angle ADE = \\angle B + \\angle BAD = \\frac{1}{2}(\\angle BAC + \\angle B) + \\frac{1}{2}\\angle B$,\\\\\nit follows that $\\angle B = \\boxed{50^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52977143.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $C$ is a point on line segment $AB$, and $D$ and $E$ are the midpoints of line segments $AC$ and $BC$, respectively. If $AB = 10$ and $AD = 2$, find the length of $CE$.\n",
    "completion": "\\textbf{Solution:} Since $D$ is the midpoint of line segment $AC$,\\\\\n$AD=2$,\\\\\ntherefore $AC=2AD=4$,\\\\\nsince $AB=10$,\\\\\ntherefore $BC=10-4=6$,\\\\\nsince $E$ is the midpoint of line segment $BC$,\\\\\ntherefore $CE=\\frac{1}{2}BC=\\boxed{3}$",
    "image_path": "dataset/mm_math/images/MM_Math/55190302.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within rectangle $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$, $AE$ perpendicular bisects $BO$, with $AE = \\sqrt{3}\\,\\text{cm}$, what is the length of $OD$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rectangle,\\\\\nit follows that OB=OD, OA=OC, AC=BD,\\\\\nwhich implies OA=OB,\\\\\nsince AE perpendicularly bisects OB,\\\\\nwe have AB=AO,\\\\\ntherefore, OA=AB=OB,\\\\\ngiven that AE=$\\sqrt{3}$cm,\\\\\nit results in OB=OD=\\boxed{2};",
    "image_path": "dataset/mm_math/images/MM_Math/53232772.png"
  },
  {
    "solution": "\\boxed{Q\\left(\\frac{16}{3}, 0\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, $AB\\parallel OC$, $A(0, 12)$, $B(a, c)$, $C(b, 0)$, and $a, b$ satisfy $\\left|a-21\\right|+\\left(b-16\\right)^2=0$. A moving point $P$ starts from point $A$ and moves towards point $B$ along the segment $AB$ at a speed of 2 units per second; another moving point $Q$ starts from point $O$ and moves towards point $C$ along the segment $OC$ at a speed of 1 unit per second. Points $P, Q$ start simultaneously from points $A, O$ respectively, and the movement of point $P$ stops when it reaches point $B$, and consequently, point $Q$ stops moving. Suppose the movement duration is $t$ seconds. At what value of $t$ is $\\triangle PQC$ an isosceles triangle with $PQ$ as its equal side? Determine the coordinates of points $P$ and $Q$ at this time.\n",
    "completion": "\\textbf{Solution:} Given: $Q(t,0)$, $P(2t,12)$,\n\n(1) When $PQ=PC$, constructing $PM \\perp x$-axis at point $M$, it follows that: $QM=t$, $CM=16-2t$,\n\n$\\therefore QM=CM$ (coincidence of three lines),\n\n$\\therefore t=16-2t$,\n\nSolving for $t$ gives $t=\\frac{16}{3}$,\n\n$\\therefore AP=2t=\\frac{32}{3}$, $OQ=\\frac{16}{3}$,\n\n$\\therefore P\\left(\\frac{32}{3}, 12\\right), Q\\left(\\frac{16}{3}, 0\\right)$;\n\n(2) When $PQ=CQ$, constructing $QN \\perp AB$ at point $N$, it follows that: $QN=12$, $PN=2t-t=t$, $CQ=16-t$,\n\n$\\because PQ^2=CQ^2$,\n\n$\\therefore 12^2+(2t-t)^2=(16-t)^2$,\n\nSolving for $t$ gives $t=\\frac{7}{2}$,\n\n$\\therefore AP=2t=7$, $OQ=\\frac{7}{2}$,\n\n$\\therefore P(7, 12), Q\\left(\\frac{7}{2}, 0\\right)$;\n\nIn summary, the coordinates of points $P$ and $Q$ are $\\boxed{P(7, 12)}, \\boxed{Q\\left(\\frac{7}{2}, 0\\right)}$ or $\\boxed{P\\left(\\frac{32}{3}, 12\\right)}, \\boxed{Q\\left(\\frac{16}{3}, 0\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52345544_117.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $ \\angle AOB=90^\\circ$, $ \\angle AOC$ is an acute angle, $ ON$ bisects $ \\angle AOC$, and the ray $ OM$ is inside $ \\angle AOB$. How many angles in the figure are less than a straight angle?\n",
    "completion": "\\textbf{Solution:} Let's consider angles that start from edge $OC$: there are $\\angle CON, \\angle COA, \\angle COM, \\angle COB$,\\\\\nAngles that start from edge $ON$: $\\angle NOA, \\angle NOM, \\angle NOB$,\\\\\nAngles that start from edge $OA$: $\\angle AOM, \\angle AOB$,\\\\\nAngles that start from edge $OM$: $\\angle MOB$,\\\\\n$\\therefore$ In the figure, there are a total of $4+3+2+1=\\boxed{10}$ angles that are less than a straight angle.",
    "image_path": "dataset/mm_math/images/MM_Math/53403787_53.png"
  },
  {
    "solution": "\\boxed{2.4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the quadrilateral ADEF is a rhombus, and $AB=6$, $AC=4$. Then, what is the value of $DE$?\n",
    "completion": "\\[ \\textbf{Solution:} \\]\n\\[ \\because \\text{ Quadrilateral ADEF is a rhombus,} \\]\n\\[ \\therefore AD=DE, DE\\parallel AC, \\]\n\\[ \\therefore \\triangle BDE \\sim \\triangle BAC, \\]\n\\[ \\therefore \\frac{DE}{AC}=\\frac{BD}{AB}, \\]\n\\[ \\therefore \\frac{DE}{4}=\\frac{AB-AD}{AB}, \\]\n\\[ \\therefore \\frac{DE}{4}=\\frac{6-DE}{6}, \\]\n\\[ \\text{Solving gives } DE=\\boxed{2.4}. \\]",
    "image_path": "dataset/mm_math/images/MM_Math/51457948.png"
  },
  {
    "solution": "\\boxed{75^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, B is on AC, D is on CE, $AD=BD=BC$, $\\angle ACE=25^\\circ$, what is the degree measure of $\\angle ADE$?\n",
    "completion": "\\textbf{Solution}: Since $BD=BC$, $\\angle ACE=25^\\circ$,\\\\\nhence $\\angle BDC=\\angle C=25^\\circ$,\\\\\nthus $\\angle ABD=50^\\circ$,\\\\\nsince $AD=BD$,\\\\\nthus $\\angle A=\\angle ABD=50^\\circ$,\\\\\ntherefore $\\angle ADE=\\angle A+\\angle C=\\boxed{75^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51368721.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ABC = \\angle BAC = 45^\\circ$, point $P$ lies on $AB$, $AD \\perp CP$, $BE \\perp CP$, with the feet of the perpendiculars being $D$ and $E$ respectively. Given that $DC = 2$, what is the length of $BE$?\n",
    "completion": "\\textbf{Solution:} \\\\\nGiven that $\\angle ABC=\\angle BAC=45^\\circ$, \\\\\nthus, $\\angle ACB=90^\\circ$ and AC=BC, \\\\\nsince $\\angle DAC+\\angle ACD=90^\\circ$ and $\\angle BCE+\\angle ACD=90^\\circ$, \\\\\nit follows that $\\angle DAC=\\angle BCE$, \\\\\nIn $\\triangle ACD$ and $\\triangle CEB$, \\\\\nhence, $\\triangle ACD \\cong \\triangle CEB$ (AAS), \\\\\ntherefore, $BE=CD=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/2283164.png"
  },
  {
    "solution": "\\boxed{\\frac{15}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AD \\parallel EF \\parallel BC$, point $G$ is the midpoint of $EF$, $\\frac{EF}{BC} = \\frac{3}{5}$. If $EF = 6$, what is the length of $AD$?\n",
    "completion": "\\textbf{Solution:} Given: $EF\\parallel BC$\\\\\n$\\therefore$ $\\triangle AEF\\sim \\triangle ABC$\\\\\n$\\therefore$ $\\frac{AE}{AB}=\\frac{EF}{BC}=\\frac{3}{5}$\\\\\n$\\therefore$ $\\frac{BE}{BA}=\\frac{2}{5}$\\\\\nGiven: $EF\\parallel AD$\\\\\n$\\therefore$ $\\triangle BEG\\sim \\triangle BAD$\\\\\n$\\therefore$ $\\frac{EG}{AD}=\\frac{BE}{BA}=\\frac{2}{5}$\\\\\nGiven: Point G is the midpoint of EF, and EF=6\\\\\n$\\therefore$ EG=3\\\\\n$\\therefore$ $AD=\\frac{5}{2}EG=\\boxed{\\frac{15}{2}}$",
    "image_path": "dataset/mm_math/images/MM_Math/52601722.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, DE is the perpendicular bisector of the line segment AB. If the perimeter of $\\triangle ABC$ is 18, and the length of the line segment AE is 4, then what is the perimeter of $\\triangle BCD$?\n",
    "completion": "\\textbf{Solution:} Since the perimeter of $\\triangle ABC$ is $18$, \\\\\nit follows that $AC+BC+AB=18$, \\\\\nSince $DE$ is the perpendicular bisector of segment $AB$, and $AE=4$, \\\\\nit follows that $AB=2AE=8$, and $DA=DB$, \\\\\nthus, $AC+BC=10$, \\\\\ntherefore, the perimeter of $\\triangle BCD$ = $BD+CD+BC=AD+CD+BC=AC+BC=\\boxed{10}$",
    "image_path": "dataset/mm_math/images/MM_Math/53107076.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point C is on the line segment BG. With BC and CG as sides, squares are constructed on both sides, with areas denoted by $S_{1}$ and $S_{2}$ respectively. The sum of the areas of the two squares is $S_{1}+S_{2}=40$. Given that BG=8, what is the area of the shaded part in the figure?\n",
    "completion": "\\textbf{Solution:} Let $BC = a$ and $CG = b$. Then, $S_{1} = a^{2}$ and $S_{2} = b^{2}$, with $a + b = BG = 8$.\\\\\nHence, $a^{2} + b^{2} = 40$.\\\\\nSince $(a + b)^{2} = a^{2} + b^{2} + 2ab = 64$,\\\\\nit follows that $2ab = 64 - 40 = 24$,\\\\\nand thus, $ab = 12$.\\\\\nTherefore, the area of the shaded part is equal to $\\frac{1}{2} ab = \\frac{1}{2} \\times 12 = \\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52836781.png"
  },
  {
    "solution": "\\boxed{\\left(\\frac{3}{2}, -\\frac{15}{2}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Comprehensive and Practical\n\nAs shown in the figure, the parabola $y=2x^2-4x-6$ intersects the x-axis at points A and B, with point A on the left of point B, and intersects the y-axis at point C. Point D is a moving point on the parabola. As shown in Figure 2, when point D is in the fourth quadrant, connect $BD$, $CD$, and $BC$ to form $\\triangle BCD$. What are the coordinates of point D when the area of $\\triangle BCD$ is maximized?\n",
    "completion": "\\textbf{Solution:} Let the coordinates of point D be $\\left(\\frac{3}{2}, 2m^2-4m-6\\right)$.\n\nDraw $DH \\perp x$-axis at point H, and draw $DG \\perp y$-axis at point G, connecting $OD$.\n\n$\\therefore DG=m$, $DH=-2m^2+4m+6$.\n\n$\\because$ The coordinates of point B are $(3, 0)$ and the coordinates of point C are $(0, -6)$.\n\n$\\therefore OB=3$, $OC=6$.\n\n$\\because S_{\\triangle BCD}=S_{\\triangle OCD}+S_{\\triangle OBD}-S_{\\triangle OBC}$,\n\n$\\therefore S_{\\triangle BCD}=\\frac{OC \\cdot DG}{2}+\\frac{OB \\cdot DH}{2}-\\frac{OC \\cdot OB}{2}$\n\n$S_{\\triangle BCD}=\\frac{6m}{2}+\\frac{3(-2m^2+4m+6)}{2}-\\frac{6 \\times 3}{2}$\n\nSimplifying, we get $S_{\\triangle BCD}=-3\\left(m-\\frac{3}{2}\\right)^2+\\frac{27}{4}$.\n\n$\\because -3<0$,\n\n$\\therefore$ When $m=\\frac{3}{2}$, the area of $\\triangle BCD$ is maximized at $\\frac{27}{4}$.\n\n$\\therefore 2m^2-4m-6=2\\times \\left(\\frac{3}{2}\\right)^2-4\\times \\frac{3}{2}-6=2\\times \\frac{9}{4}-6-6=-\\frac{15}{2}$.\n\n$\\therefore$ The coordinates of point D are $\\boxed{\\left(\\frac{3}{2}, -\\frac{15}{2}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52977885_210.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=14$, $AC=10$, point $D$ is on $BC$, and point $M$ is on the extension of $BA$. Given that $\\tan\\angle CAM=\\frac{4}{3}$ and $\\angle DAB=45^\\circ$, what is the length of $AD$?\n",
    "completion": "\\textbf{Solution:} Draw $CE\\parallel AD$ through $C$ intersecting ray $AM$ at point $E$, and draw $CF\\perp AM$ with the foot at $F$, \\\\\nsince $AC=10$ and $\\tan\\angle CAM=\\frac{4}{3}$, $CF\\perp AM$, \\\\\ntherefore, $\\tan\\angle CAM=\\frac{CF}{AF}=\\frac{4}{3}$, \\\\\nlet $CF=4m$ and $AF=3m$, \\\\\naccording to the Pythagorean theorem: $CF^2+AF^2=AC^2$, that is $16m^2+9m^2=100$, \\\\\nwe find: $m=2$ (discard $m=-2$), \\\\\nthus, $CF=8$ and $AF=6$, \\\\\nsince $CE\\parallel AD$ and $\\angle DAB=45^\\circ$, \\\\\nthus, $\\angle CEF=\\angle DAB=45^\\circ$, \\\\\nthus, $EF=CF=8$, $CE=8\\sqrt{2}$, \\\\\nsince $\\angle CEF=\\angle DAB=45^\\circ$ and $\\angle B=\\angle B$, \\\\\nthus, $\\triangle DBA\\sim \\triangle CBE$, \\\\\nthus, $\\frac{AB}{BE}=\\frac{AD}{CE}$, that is $\\frac{14}{AF+EF+AB}=\\frac{14}{28}=\\frac{AD}{8\\sqrt{2}}$, \\\\\ntherefore, $AD=\\boxed{4\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51379325.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $E$ is a moving point on $BC$, and $\\triangle ABE$ is folded along $AE$ to obtain $\\triangle AFE$. Point $F$ is inside the rectangle $ABCD$, and extending $AF$ intersects $CD$ at point $G$, with $AB=3$ and $AD=4$. As shown in figure 3, when the perimeter of $\\triangle CFE$ is at its minimum during the movement of point $E$, write down the length of $BE$.\n",
    "completion": "\\textbf{Solution:} We have $BE = \\boxed{\\frac{3}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51684393_169.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in square $ABCD$, $E$ is the midpoint of $BC$, and $F$ is a point on $CD$ with $CF=\\frac{1}{4}CD$. If the length of $CF$ is 1, what are the lengths of $EF$ and $AE$ respectively?\n",
    "completion": "\\textbf{Solution}: Given by the problem, $ABCD$ is a square, \\\\\n$\\therefore AB=BC=CD=AD$. \\\\\n$\\because E$ is the midpoint of $BC$, \\\\\n$\\therefore BE=EC$. \\\\\nAlso, $\\because CF=\\frac{1}{4}CD$, \\\\\n$\\therefore AB=2BE=2EC=4FC$. \\\\\n$\\therefore AB=4$, $BE=EC=2$. \\\\\nIn $\\triangle ABE$, \\\\\n$AE=\\sqrt{AB^2+BE^2}=\\sqrt{4^2+2^2}=2\\sqrt{5}$. \\\\\n$\\therefore$ Similarly, $EF=\\sqrt{EC^2+CF^2}=\\sqrt{2^2+1^2}=\\sqrt{5}$. \\\\\n$\\therefore$ The lengths of $EF$ and $AE$ are $\\boxed{\\sqrt{5}}$ and $\\boxed{2\\sqrt{5}}$, respectively.",
    "image_path": "dataset/mm_math/images/MM_Math/51798762_141.png"
  },
  {
    "solution": "\\boxed{\\frac{20}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within rectangle $ABCD$, point $E$ is on side $CD$. When $\\triangle BCE$ is folded along $BE$ such that point $C$ coincides with point $F$ on side $AD$, and a line is drawn from point $F$ such that $FG \\parallel CD$ and intersects $BE$ at point $G$, and $CG$ is connected. Quadrilateral $CEFG$ is a rhombus. Given that $AB=6$ and $AD=10$, what is the area of quadrilateral $CEFG$?\n",
    "completion": "\\textbf{Solution:} Given rectangle ABCD, where AD=10,\\\\\nit follows that BC=10.\\\\\nGiven that $\\triangle$BCE is folded along BE, such that point C lands on point F on side AD,\\\\\nit then follows that BF=BC=10.\\\\\nIn $\\triangle$ABF, where AB=6, AF=$\\sqrt{BF^{2}-AB^{2}}$=8,\\\\\nthus, DF=AD−AF=2.\\\\\nLet EF=x, then CE=x, and DE=6−x.\\\\\nIn $\\triangle$DEF, we have $DF^{2}+DE^{2}=EF^{2}$,\\\\\nthus, $2^{2}+(6-x)^{2}=x^{2}$. Solving this, we find $x=\\frac{10}{3}$,\\\\\ntherefore, CE=$\\frac{10}{3}$.\\\\\nHence, the area of quadrilateral CEFG is: CE$\\cdot$DF=$\\frac{10}{3}\\times 2=\\boxed{\\frac{20}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51807002_129.png"
  },
  {
    "solution": "\\boxed{24}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that in $\\triangle ABC$, $BC=18$, $E$ and $F$ are the trisection points of $BC$, $AE=10$, $AF=8$, $G$ and $H$ are the midpoints of $AC$ and $AB$ respectively. What is the perimeter of the quadrilateral $EFGH$?\n",
    "completion": "\\textbf{Solution:} Since $E$ and $F$ are the trisection points of $BC$,\\\\\nwe have $EF=\\frac{1}{3}BC=6$, and $BE=EF=FC$.\\\\\nSince $G$ and $H$ are the midpoints of $AC$ and $AB$ respectively,\\\\\n$GH$ is the midline of $\\triangle ABC$, and $GH=\\frac{1}{2}BC=9$.\\\\\n$HE$ is the midline of $\\triangle ABF$, and $HE=\\frac{1}{2}AF=4$.\\\\\n$GF$ is the midline of $\\triangle ACE$, and $GF=\\frac{1}{2}AE=5$.\\\\\nTherefore, the perimeter of the quadrilateral $EFGH=6+9+4+5=\\boxed{24}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53454092.png"
  },
  {
    "solution": "\\boxed{\\left(\\frac{5}{2}, \\frac{1}{2}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the rectangle $ABCD$ has its side $AB$ on the $x$-axis, with $AB=3$ and $AD=2$. The line passing through point $C$ and described by $y=x-2$ intersects the $x$-axis and $y$-axis at points $E$ and $F$, respectively. The question is whether there exists a point $P$ on the line $y=x-2$ such that $\\triangle PDC$ is an isosceles right-angled triangle. Determine the coordinates of point $P$.\n",
    "completion": "\\textbf{Solution:} There exists.\n\\\\\n$\\because \\triangle EBC$ is an isosceles right-angled triangle,\\\\\n$\\therefore \\angle CEB=\\angle ECB=45^\\circ$,\\\\\nAlso, $\\because DC\\parallel AB$,\\\\\n$\\therefore \\angle DCE=\\angle CEB=45^\\circ$,\\\\\n$\\therefore \\triangle PDC$ can only be an isosceles right-angled triangle with $P$ and $D$ as the right-angled vertices.\\\\\nAs shown in the figure, (1) when $\\angle D=90^\\circ$, extend $DA$ to intersect the line $y=x-2$ at point $P_1$,\\\\\n$\\because$ the coordinates of point $D$ are $(1, 2)$,\\\\\n$\\therefore$ the $x$-coordinate of point $P_1$ is 1,\\\\\nSubstituting $x=1$ into $y=x-2$ gives, $y=-1$,\\\\\n$\\therefore$ the point $P_1(1, -1)$;\\\\\n(2) when $\\angle DPC=90^\\circ$, draw the perpendicular bisector of $DC$ to intersect the line $y=x-2$ at point $P_2$,\\\\\nTherefore, the $x$-coordinate of point $P_2$ is $\\frac{1+4}{2}=\\frac{5}{2}$,\\\\\nSubstituting $x=\\frac{5}{2}$ into $y=x-2$ gives, $y=\\frac{1}{2}$,\\\\\nTherefore, the point $P_2\\left(\\frac{5}{2}, \\frac{1}{2}\\right)$,\\\\\nIn summary, the coordinates of the suitable point $P$ are \\boxed{(1, -1)} or \\boxed{\\left(\\frac{5}{2}, \\frac{1}{2}\\right)}.",
    "image_path": "dataset/mm_math/images/MM_Math/50561201_58.png"
  },
  {
    "solution": "\\boxed{(0, 2\\sqrt{3})}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the vertices of the equilateral triangle \\( ABC \\) are on the coordinate axes, and the side length is \\( 4 \\). What are the coordinates of point \\( A \\)?\n",
    "completion": "\\textbf{Solution:} Given that it’s an equilateral triangle, it follows from the rule that all medians intersect at one point that: $OC = \\frac{1}{2}BC = 2$,\\\\\nApplying Pythagoras’ theorem, we find: $OA = \\sqrt{AC^2 - CO^2} = 2\\sqrt{3}$,\\\\\n$\\therefore$The coordinates of point A are $\\boxed{(0, 2\\sqrt{3})}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51212038.png"
  },
  {
    "solution": "\\boxed{\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, within rectangle ABCD, E is a movable point on BC. After folding $\\triangle ABE$ along AE, we obtain $\\triangle AFE$, where point F is within rectangle ABCD. Extend AF to intersect CD at point G, with AB = 3 and AD = 4. As shown in Figure 1, when $\\angle DAG=30^\\circ$, find the length of BE.\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rectangle,\\\\\nit follows that $\\angle BAD=90^\\circ$,\\\\\nand since $\\angle DAG=30^\\circ$,\\\\\nit follows that $\\angle BAG=60^\\circ$.\\\\\nFrom the folding, we know that $\\angle BAE=\\frac{1}{2}\\angle BAG=30^\\circ$,\\\\\nIn $\\triangle BAE$, where $\\angle BAE=30^\\circ$ and $AB=3$,\\\\\nit follows that $BE=\\boxed{\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51684393_167.png"
  },
  {
    "solution": "\\boxed{\\frac{-3+\\sqrt{19}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that point $D$ lies on the extension of the diameter $AB$ of circle $O$, and point $C$ lies on circle $O$. A line is drawn through $D$ perpendicular to $AD$ and intersects with the extension of $AC$ at $E$. $CD$ is a tangent to circle $O$; $AB=2$ and $AE=3$. $CD=DE$; find the length of $BD$?\n",
    "completion": "\\textbf{Solution:} Method One: \\\\\nSince $AB=2$, \\\\\nit follows that $OA=OB=OC=1$, \\\\\nSince $OC\\perp CD$, \\\\\nby the Pythagorean theorem, we have $CD^{2}=(1+BD)^{2}-1^{2}$, \\\\\nSince $ED\\perp AD$, \\\\\nby the Pythagorean theorem, we have $DE^{2}=3^{2}-(2+BD)^{2}$, \\\\\nSince $CD=DE$, \\\\\nit follows that $(1+BD)^{2}-1^{2}=3^{2}-(2+BD)^{2}$, \\\\\nSolving this gives: $BD=\\frac{-3+\\sqrt{19}}{2}$ or $\\frac{-3-\\sqrt{19}}{2}$ (discard this result),\\\\\nthus $BD=\\boxed{\\frac{-3+\\sqrt{19}}{2}}$. \\\\\n\nMethod Two: \\\\\nBy the chord tangent theorem, we have $\\angle DCB=\\angle DAC$, \\\\\nSince $\\angle CDB=\\angle ADC$, \\\\\nit follows that $\\triangle CDB\\sim \\triangle ADC$, \\\\\nhence $\\frac{CD}{AD}=\\frac{BD}{CD}$, that is, $CD^{2}=AD\\cdot BD=(2+BD)\\cdot BD$, \\\\\nSince $ED\\perp AD$, \\\\\nby the Pythagorean theorem, we have $DE^{2}=3^{2}-(2+BD)^{2}$, \\\\\nSince $CD=DE$, \\\\\nit follows that $(2+BD)\\times BD=3^{2}-(2+BD)^{2}$, \\\\\nSolving this gives $BD=\\frac{-3+\\sqrt{19}}{2}$ or $\\frac{-3-\\sqrt{19}}{2}$ (discard this result),\\\\\nthus $BD=\\boxed{\\frac{-3+\\sqrt{19}}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53147522_132.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure: AB is the diameter of $\\odot O$, chord CD $\\perp$ AB at E. If AB $= 20$ and CD $= 16$, what is the length of the line segment BE?\n",
    "completion": "\\textbf{Solution:} Draw line OC, \\\\\n$\\because$AB=20, \\\\\n$\\therefore$OC=OA=OB=10, \\\\\n$\\because$AB$\\perp$CD and AB passes through O, \\\\\n$\\therefore$CE=DE=$\\frac{1}{2}$CD=8, \\\\\nIn $\\triangle OCE$, by the Pythagorean theorem, we have: OE=$\\sqrt{10^{2}-8^{2}}$=6, \\\\\n$\\therefore$BE=10-6=\\boxed{4}.",
    "image_path": "dataset/mm_math/images/MM_Math/52991432.png"
  },
  {
    "solution": "\\boxed{3:5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that points $A$, $B$, $C$, $D$ are on the same line, with point $C$ being the midpoint of segment $AB$, and point $D$ on segment $AB$. If point $E$ is a point on segment $AB$, and $AE=2BE$, when $AD:BD=2:3$, what is the value of $CD:CE$?\n",
    "completion": "\\textbf{Solution:} Let $AD=2x$ and $BD=3x$, hence $AB=5x$, \\\\\nsince point C is the midpoint of segment AB, \\\\\ntherefore $AC=\\frac{1}{2}AB=\\frac{5}{2}x$, \\\\\ntherefore $CD=AC-AD=\\frac{1}{2}x$, \\\\\nsince $AE=2BE$, \\\\\ntherefore $AE=\\frac{2}{3}AB=\\frac{10}{3}x$, \\\\\ntherefore $CE=AE-AC=\\frac{10}{3}x-\\frac{5}{2}x=\\frac{5}{6}x$, \\\\\ntherefore $CD: CE=\\frac{1}{2}x: \\frac{5}{6}x=3:5$, \\\\\nthus, the ratio of segment $CD:CE$ is $\\boxed{3:5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53195484_121.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AD$ is the perpendicular bisector of $BC$ with the foot of the perpendicular as $D$, $\\angle BAC=45^\\circ$, $CE\\perp AB$ at $E$, intersecting $AD$ at $F$, and $BD=2$. What is the length of $AF$?\n",
    "completion": "\\textbf{Solution:} Since $AD$ is the perpendicular bisector of $BC$, and $BD=2$, \\\\\nthus, $BC=2BD=4$, \\\\\nSince $CE\\perp AB$, \\\\\nthus, $\\angle AEC=90^\\circ$, \\\\\nSince $\\angle BAC=45^\\circ$, \\\\\nthus, $\\angle ACE=90^\\circ-45^\\circ=45^\\circ$, \\\\\nthus, $\\angle EAC=\\angle ACE$, \\\\\nthus, $AE=CE$, \\\\\nSince $AB=AC$, and point $D$ is the midpoint of $BC$, \\\\\nthus, $AD\\perp BC$, \\\\\nthus, $\\angle ADB=90^\\circ$, \\\\\nthus, $\\angle B+\\angle BAD=90^\\circ$, \\\\\nSince $\\angle B+\\angle BCE=90^\\circ$, \\\\\nthus, $\\angle BAD=\\angle BCE$, \\\\\nIn $\\triangle AEF$ and $\\triangle CEB$, \\\\\n$\\left\\{\\begin{array}{l} \\angle AEF=\\angle CEB \\\\ AE=CE \\\\ \\angle EAF=\\angle BCE \\end{array}\\right.$, \\\\\nthus, $\\triangle AEF\\cong \\triangle CEB$ (ASA), \\\\\nthus, $AF=BC=\\boxed{4}$,",
    "image_path": "dataset/mm_math/images/MM_Math/52653754.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, it is known that $\\angle AOB=90^\\circ$, $\\angle BOC=30^\\circ$, $OM$ bisects $\\angle AOC$, and $ON$ bisects $\\angle BOC$. What is the measure of $\\angle MON$ in degrees?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle AOB = 90^\\circ$ and $\\angle BOC = 30^\\circ$,\\\\\nit follows that $\\angle AOC = 90^\\circ + 30^\\circ = 120^\\circ$,\\\\\nsince OM bisects $\\angle AOC$,\\\\\nwe have $\\angle AOM = \\frac{1}{2} \\angle AOC = \\frac{1}{2} (\\angle AOB + \\angle BOC) = \\frac{1}{2} \\times 120^\\circ = 60^\\circ$,\\\\\nsince ON bisects $\\angle BOC$,\\\\\nit follows that $\\angle CON = \\frac{1}{2} \\angle BOC = \\frac{1}{2} \\times 30^\\circ = 15^\\circ$,\\\\\nthus, $\\angle MON = \\angle AOC - \\angle AOM - \\angle CON = 120^\\circ - 60^\\circ - 15^\\circ = \\boxed{45^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52580669.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $A$ is a point on the graph of the inverse proportion function $y = \\frac{k}{x}$, and $AB$ is perpendicular to the $x$-axis at $B$. If the area of $\\triangle AOB = 2$, what is the value of $k$?\n",
    "completion": "\\textbf{Solution:} Since ${S}_{\\triangle AOB}=2$,\\\\\ntherefore $\\frac{k}{2}=2$,\\\\\nthus $k=\\boxed{4}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53080709.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, $\\cos A=\\frac{1}{3}$, then what is the value of $\\sin B$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle C=90^\\circ$,\\\\\nit follows that $\\angle A+\\angle B=90^\\circ$,\\\\\nhence $\\sin B=\\cos A=\\boxed{\\frac{1}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52872325.png"
  },
  {
    "solution": "\\boxed{15^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC$, $\\angle A=50^\\circ$, and the perpendicular bisector of $AB$, $DE$, intersects $AC$ and $AB$ at points $D$ and $E$, respectively. What is the degree measure of $\\angle CBD$?\n",
    "completion": "\\textbf{Solution:} Given that in $\\triangle ABC$, $AB=AC$ and $\\angle A=50^\\circ$ \\\\\n$\\therefore$ $\\angle ABC=\\angle ACB=(180^\\circ-50^\\circ)÷2=65^\\circ$ \\\\\nSince $DE$ is the perpendicular bisector of $AB$ \\\\\n$\\therefore$ $DA=DB$ \\\\\n$\\therefore$ $\\angle ABD=\\angle A=50^\\circ$ \\\\\n$\\therefore$ $\\angle CBD=\\angle ABC-\\angle ABD=65^\\circ-50^\\circ=\\boxed{15^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/10253872.png"
  },
  {
    "solution": "\\boxed{B}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a cut is made on a cubic box, with the intersections of the cut surface and the edges being points $A$, $B$, and $C$ respectively. After cutting off the corner and unfolding the box into a plane, what are the impossible configurations for the unfolded figure?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, the square's net is as follows:\n\n\\begin{itemize}\n  \\item \"141\" type:\n  \\item \"132\" type:\n  \\item \"222\" type:\n  \\item \"33\" type:\n\\end{itemize}\n\nConsidering the characteristics of the square's net, after cutting off corners and unfolding the box into a plane, we know that the faces with corners cut off are three adjacent faces; it cannot be that the faces with corners cut off are opposite faces.\n\nHence, the answer is: \\boxed{B}.",
    "image_path": "dataset/mm_math/images/MM_Math/53042405.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{3}}{2}-\\frac{\\pi}{6}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$ and $\\angle B=30^\\circ$. Point $D$ is the midpoint of side $AB$, and point $O$ is on side $BC$. A circle with center $O$ passes through vertex $C$ and intersects side $AB$ at point $D$. The line $AB$ is a tangent to circle $O$; if $AC=\\sqrt{3}$, what is the area of the shaded region?\n",
    "completion": "\\textbf{Solution:} From (1) we know: $AC=AD=BD=\\frac{1}{2}AB$,\\\\\nand since $AC=\\sqrt{3}$,\\\\\ntherefore $BD=AC=\\sqrt{3}$,\\\\\nsince $\\angle B=30^\\circ$, $\\angle BDO=\\angle ADO=90^\\circ$,\\\\\ntherefore $\\angle BOD=60^\\circ$, $BO=2DO$,\\\\\nby the Pythagorean theorem we have: $BO^2=OD^2+BD^2$,\\\\\nwhich gives $(2OD)^2=OD^2+(\\sqrt{3})^2$,\\\\\nsolving this yields: $OD=1$ (negatives disregarded),\\\\\nthus the area of the shaded region $S=S_{\\triangle BDO}-S_{\\text{sector} DOE}=\\frac{1}{2}\\times 1\\times \\sqrt{3}-\\frac{60^\\circ\\pi \\times 1^2}{360}=\\boxed{\\frac{\\sqrt{3}}{2}-\\frac{\\pi}{6}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52512571_103.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in $\\triangle ABC$, $\\angle B=\\angle C$, $AB=3$, what is the length of $AC$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle B = \\angle C$ and AB=3, \\\\\nit follows that AB=AC=\\boxed{3}.",
    "image_path": "dataset/mm_math/images/MM_Math/52978549.png"
  },
  {
    "solution": "\\boxed{120^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $BD$ and $CE$ are the medians of the equilateral triangle $ABC$, and $BD$ and $CE$ intersect at point $F$. What is the measure of $\\angle BFC$ in degrees?\n",
    "completion": "\\textbf{Solution:} Given that $\\triangle ABC$ is an equilateral triangle\\\\\nHence, $\\angle ABC=\\angle ACB=60^\\circ$\\\\\nSince BD, CE are medians of the equilateral triangle ABC\\\\\nTherefore, $\\angle DBC=\\frac{1}{2}\\angle ABC=30^\\circ$, $\\angle ECB=\\frac{1}{2}\\angle ACB=30^\\circ$\\\\\nAdditionally, $\\angle BFC=180^\\circ-\\angle DBC-\\angle ECB$\\\\\nTherefore, $\\angle BFC=180^\\circ-30^\\circ-30^\\circ=\\boxed{120^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/51378737.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AD$ is the angle bisector of $\\triangle ABC$, $\\angle ADE = \\angle B$, $AE=3$, and $CE=1$. What is the length of $CD$?\n",
    "completion": "\\textbf{Solution:} \\\\\nSince $AD$ is the angle bisector of $\\triangle ABC$, \\\\\nthen $\\angle BAD=\\angle DAC$, \\\\\nand since $\\angle ADE=\\angle B$, \\\\\nthus $\\angle AED=\\angle ADB$, \\\\\nthus $\\angle CED=\\angle CDA$, \\\\\nand since $\\angle C=\\angle C$, \\\\\nthus $\\triangle ADC\\sim \\triangle DEC$, \\\\\ntherefore $\\frac{CE}{CD}=\\frac{CD}{AC}$, \\\\\nGiven that $AE=3$ and $CE=1$, \\\\\nthus $AC=AE+EC=3+1=4$, \\\\\ntherefore $\\frac{1}{CD}=\\frac{CD}{4}$, i.e., $CD^2=4$, \\\\\nSolving gives $CD=2$ or $CD=-2$ (discard this), \\\\\nthus $CD=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52872414.png"
  },
  {
    "solution": "\\boxed{y=-x^{2}+2x+3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, the parabola $y=ax^2+bx+3$ intersects the x-axis at points $A(-1, 0)$ and $B(3, 0)$, and intersects the y-axis at point C. Connect $BC$. Find the analytical expression of the parabola.\n",
    "completion": "\\textbf{Solution:} Substitute points $A(-1, 0)$ and $B(3, 0)$ into $y=ax^{2}+bx+3$,\\\\\nwe get $\\begin{cases}a-b+3=0\\\\ 9a+3b+3=0\\end{cases}$,\\\\\nsolving this yields $\\begin{cases}a=-1\\\\ b=2\\end{cases}$,\\\\\n$\\therefore \\boxed{y=-x^{2}+2x+3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53107063_278.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $ \\triangle ACB \\cong \\triangle ACB^{\\prime }$, $ \\angle BCB^{\\prime } = 30^{\\circ}$, what is the measure of $ \\angle ACA^{\\prime }$?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ACB\\cong \\triangle A'C'B'$,\\\\\nit follows that $\\angle ACB=\\angle A'C'B'$, which means $\\angle ACA'+\\angle A'CB=\\angle B'CB+\\angle A'CB$,\\\\\ntherefore, $\\angle ACA'=\\angle B'CB$,\\\\\nand given that $\\angle B'CB=30^\\circ$,\\\\\nit follows that $\\angle ACA'=\\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52653482.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC$, $AD$ is the altitude, $E$ is the midpoint of $AB$. Given that the area of $\\triangle ABC$ is $8$, what is the area of $\\triangle ADE$?\n",
    "completion": "\\textbf{Solution:} Given that $AB=AC$ and $AD$ is the altitude, \\\\\nthus $BD=CD$, $AD\\perp BC$, \\\\\nthus $S_{\\triangle ABD}=S_{\\triangle ACD}=\\frac{1}{2} S_{\\triangle ABC}=4$, \\\\\ngiven that $E$ is the midpoint of $AB$, \\\\\nthus $AE=BE$, \\\\\nthus $S_{\\triangle ADE}=S_{\\triangle BDE}=\\frac{1}{2} S_{\\triangle ABD}=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52991698.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{2}-\\frac{\\pi }{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of $\\odot O$. Point $D$ is on $\\odot O$, and $\\angle DAB=45^\\circ$, $BC\\parallel AD$, and $CD\\parallel AB$. If the radius of $\\odot O$ is 1, what is the area of the shaded region? (Keep $\\pi$ in the result).\n",
    "completion": "\\textbf{Solution:} Connect $DO$,\\\\\n$\\because \\angle DAB=45^\\circ$, $\\overset{\\frown}{DB}=\\overset{\\frown}{DB}$\\\\\n$\\therefore \\angle DOB=90^\\circ$ meaning $DO\\perp AB$\\\\\n$\\because$ the radius of $\\odot O$ is 1\\\\\n$\\therefore AB=2$\\\\\n$\\therefore {S}_{\\square ABCD}=AB\\cdot OD=2\\times 1=2$\\\\\n${S}_{\\triangle ADO}=\\frac{1}{2}AO\\cdot OD=\\frac{1}{2}\\times 1\\times 1=\\frac{1}{2}$\\\\\n$S_{\\text{sector}}{OBD}=\\frac{90^\\circ}{360^\\circ}\\pi \\cdot OA^{2}=\\frac{\\pi }{4}$\\\\\n$\\therefore S_{\\text{shaded area}}={S}_{\\square ABCD}-{S}_{\\triangle ADO}-S_{\\text{sector}}{OBD}$\\\\\n$=2-\\frac{1}{2}-\\frac{\\pi }{4}$\\\\\n$=\\boxed{\\frac{3}{2}-\\frac{\\pi }{4}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51446659.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, C is a point on the line segment AB, M is the midpoint of the line segment AC, N is the midpoint of the line segment BC. If $AC=x$ cm, and $BC= (10-x)$ cm, then what is the length of $MN$ in cm?\n",
    "completion": "\\textbf{Solution:} Since $M$ is the midpoint of line segment $AC$ and $N$ is the midpoint of line segment $BC$, \\\\\nit follows that $CM=\\frac{1}{2}AC=\\frac{1}{2}x$ and $CN=\\frac{1}{2}BC=\\frac{1}{2}\\left(10-x\\right)$, \\\\\ntherefore, $MN=CM+CN=\\boxed{5}$ cm.",
    "image_path": "dataset/mm_math/images/MM_Math/54962545_12.png"
  },
  {
    "solution": "\\boxed{y_2=-\\frac{3}{4}x+\\frac{9}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y_1=3x+6$ intersects the x-axis and y-axis respectively at points A and B. The line $y_2=kx+b$ passes through point D(3,0) and intersects with the line $y_1=3x+6$ at point C(m,3). What is the analytical equation of line CD?\n",
    "completion": "\\textbf{Solution:} \nSince line $y_2$ intersects line $y_1$ at point C ($m$, 3), \\\\\nsubstituting point C ($m$, 3) into $y_1=3x+6$, we get $3m+6=3$, and solve for $m=-1$, \\\\\nthus, point C is ($-1$, 3). Substituting points C ($-1$, 3) and D (3, 0) into $y_2=kx+b$, \\\\\nwe get $\\begin{cases}\n-k+b=3 \\\\\n3k+b=0\n\\end{cases}$, \\\\\nsolving which we find $\\begin{cases}\nk=-\\frac{3}{4} \\\\\nb=\\frac{9}{4}\n\\end{cases}$, \\\\\ntherefore, the equation of line CD is $\\boxed{y_2=-\\frac{3}{4}x+\\frac{9}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52159946_163.png"
  },
  {
    "solution": "\\boxed{y=x^2-4x+3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y_1=-x+3$ intersects the x-axis and y-axis at points $B$ and $C$ respectively. The parabola $y_2=ax^2+bx+c$ passing through points $B$ and $C$ has another intersection point with the x-axis at point $A$, and its axis of symmetry is the line $x=2$. Find the analytical expression of the parabola.\n",
    "completion": "Solution: Let $x=0$, we obtain $y=-x+3=3$,\\\\\nLet $y=0$, we obtain $-x+3=0$,\\\\\nSolving yields $x=3$,\\\\\n$\\therefore B(3, 0)$ and $C(0, 3)$,\\\\\n$\\because$ The axis of symmetry of the parabola is the line $x=2$, and the other intersection point of the parabola $y=ax^2+bx+c$ with the x-axis is A,\\\\\n$\\therefore A(1, 0)$,\\\\\nSuppose the equation of the parabola is $y=a(x-1)(x-3)$,\\\\\nSubstituting $C(0, 3)$ into it, we get $3a=3$,\\\\\n$\\therefore a=1$,\\\\\n$\\therefore$ The equation of the parabola is $y=x^2-4x+3$; hence, the equation of the parabola is $\\boxed{y=x^2-4x+3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52871717_293.png"
  },
  {
    "solution": "\\boxed{(3, -\\frac{25}{4})}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given the parabola: $y=ax^2-6ax-16a$ ($a>0$) intersects the $x$-axis at points $A$ and $B$ ($A$ is to the left of $B$), and intersects the $y$-axis at point $C$. The midpoint of $AC$ is point $G$. It is known that point $P$ is a moving point in the fourth quadrant on the parabola, and a perpendicular line to the $x$-axis through point $P$ intersects $BC$ at point $E$ and the $x$-axis at point $F$. If the triangle formed by points $C$, $P$, $E$ is similar to $\\triangle AOG$, determine the coordinates of point $P$.\n",
    "completion": "\\[ \\textbf{Solution:} \\]\nGiven \\(y=\\frac{1}{4}x^{2}-\\frac{3}{2}x-4\\), we know \\(C(0, -4)\\),\n\nSuppose the equation of line BC is \\(y=kx+b\\),\n\n\\[\n\\therefore \\left\\{\\begin{array}{l}\nb=-4\\\\\n8k+b=0\n\\end{array}\\right.,\n\\]\n\n\\[\n\\therefore \\left\\{\\begin{array}{l}\nk=\\frac{1}{2}\\\\\nb=-4\n\\end{array}\\right.,\n\\]\n\n\\[\n\\therefore y=\\frac{1}{2}x-4,\n\\]\n\nLet \\(P(t, \\frac{1}{4}t^{2}-\\frac{3}{2}t-4)\\), then \\(E(t, \\frac{1}{2}t-4)\\),\n\n\\[\n\\therefore PE=-\\frac{1}{4}t^{2}+2t,\n\\]\n\n\\[\n\\therefore CE=\\sqrt{t^{2}+(\\frac{1}{2}t-4)^{2}}, \\quad CP=\\sqrt{t^{2}+\\frac{1}{16}t^{2}(t-6)^{2}},\n\\]\n\nSince point G is the midpoint of AC,\n\n\\[\n\\therefore G(-1, -2),\n\\]\n\n\\[\n\\therefore AG=\\sqrt{5}, \\quad GO=\\sqrt{5},\n\\]\n\n\\[\n\\therefore \\triangle AOG \\text{ is an isosceles triangle},\n\\]\n\nSince OA=2, OC=4,\n\n\\[\n\\therefore \\tan\\angle OAC=2,\n\\]\n\nSince OB=8,\n\n\\[\n\\therefore \\tan\\angle OCB=2,\n\\]\n\n\\[\n\\therefore \\angle CAO=\\angle OCB,\n\\]\n\nSince \\(PE\\parallel OC\\),\n\n\\[\n\\therefore \\angle FEB=\\angle OCB,\n\\]\n\n\\[\n\\therefore \\angle FEB=\\angle CAO,\n\\]\n\n(1) When \\(CP=PE\\), \\(\\triangle AOG\\sim \\triangle CEP\\),\n\n\\[\n\\therefore PE=PC,\n\\]\n\n\\[\n\\therefore \\sqrt{t^{2}+\\frac{1}{16}t^{2}(t-6)^{2}}=-\\frac{1}{4}t^{2}+2t,\n\\]\n\nSolving gives \\(t=3\\), \\(t=0\\),\n\nSince point P is in the fourth quadrant,\n\n\\[\n\\therefore t>0,\n\\]\n\n\\[\n\\therefore P(3, -\\frac{25}{4});\n\\]\n\n(2) When \\(PC=CE\\), \\(\\triangle AOG\\sim \\triangle PEC\\),\n\n\\[\n\\therefore \\sqrt{t^{2}+\\frac{1}{16}t^{2}(t-6)^{2}}=\\sqrt{t^{2}+(\\frac{1}{2}t-4)^{2}},\n\\]\n\nSolving gives \\(t=4\\),\n\n\\[\n\\therefore P(4, -6).\n\\]\n\nIn conclusion, the coordinates of point P are \\(\\boxed{(4, -6)}\\) or \\(\\boxed{(3, -\\frac{25}{4})}\\).",
    "image_path": "dataset/mm_math/images/MM_Math/51545315_152.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the diagonals $AC$ and $BD$ of rhombus $ABCD$ intersect at point $O$, and $E$ is the midpoint of $DC$. If the perimeter of the rhombus is 16, what is the length of $OE$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rhombus, \\\\\nit follows that AB=BC=CD=AD, and BO=DO.\\\\\nSince the perimeter of rhombus ABCD is 16, \\\\\nit follows that BC=4.\\\\\nSince point E is the midpoint of CD, \\\\\nit follows that OE is the midsegment of $\\triangle BCD$, \\\\\nhence OE=$\\frac{1}{2}$BC= \\boxed{2}.",
    "image_path": "dataset/mm_math/images/MM_Math/52386683.png"
  },
  {
    "solution": "\\boxed{144^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $O$ is on line $AB$. Place the right-angled vertex of the right-angled triangle $OMN$ at point $O$. The ray $OC$ bisects $\\angle MOB$. Rotate the right-angled triangle $OMN$ in Figure 1 clockwise around the vertex $O$ to the position in Figure 2, with one side $OM$ above the ray $OB$ and the other side $ON$ below line $AB$.\n",
    "completion": "\\textbf{Solution:} Let $ \\angle AOM=\\alpha $, then $ \\angle BOM=180^\\circ-\\alpha $,\n\n(1) $\\angle AOM=2\\angle CON$, the reason is as follows:\n\nSince $OC$ bisects $ \\angle BOM$,\n\n$\\therefore \\angle MOC=\\frac{1}{2}\\angle BOM=\\frac{1}{2}\\left(180^\\circ-\\alpha \\right)$\n\n$=90^\\circ-\\frac{1}{2}\\alpha $,\n\nSince $\\angle MON=90^\\circ$,\n\n$\\therefore \\angle CON=\\angle MON-\\angle MOC=90^\\circ-\\left(90^\\circ-\\frac{1}{2}\\alpha \\right)=\\frac{1}{2}\\alpha $,\n\n$\\therefore \\angle AOM=2\\angle CON$,\n\n(2) From (1), we know $ \\angle BON=\\angle MON-\\angle BOM=90^\\circ-\\left(180^\\circ-\\alpha \\right)=\\alpha -90^\\circ$,\n\n$ \\angle AOC=\\angle AOM+\\angle MOC=\\alpha +90^\\circ-\\frac{1}{2}\\alpha =90^\\circ+\\frac{1}{2}\\alpha $,\n\nSince $\\angle AOC=3\\angle BON$,\n\n$\\therefore 90^\\circ+\\frac{1}{2}\\alpha =3\\left(\\alpha -90^\\circ\\right)$, solving this gives $ \\alpha =\\boxed{144^\\circ}$,\n\nThus, $\\angle AOM=144^\\circ$.",
    "image_path": "dataset/mm_math/images/MM_Math/52794314_147.png"
  },
  {
    "solution": "\\boxed{120^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, one corner of a rectangular piece of paper is folded so that the vertex \\(A\\) falls on \\(A'\\), with \\(EF\\) as the crease. If \\(EA'\\) exactly bisects \\(\\angle FEB\\), what is the degree measure of \\(\\angle FEB\\)?\n",
    "completion": "\\textbf{Solution:}\\\\\nSince $EA'$ exactly bisects $\\angle FEB$,\\\\\nit follows that $\\angle FEA'=\\angle A'EB$,\\\\\nFrom the folding, we get: $\\angle FEA=\\angle FEA'$,\\\\\nthus $\\angle FEA'=\\angle A'EB=\\angle FEA$,\\\\\nSince $\\angle FEA'+\\angle A'EB+\\angle FEA=180^\\circ$,\\\\\nit follows that $\\angle FEA'=\\angle A'EB=\\angle FEA=60^\\circ$,\\\\\nTherefore, $\\angle FEB=\\angle FEA'+\\angle A'EB=\\boxed{120^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51482123.png"
  },
  {
    "solution": "\\boxed{55^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the tangent to the circle $\\odot O$ with $A$ being the point of tangency. The line segments $OA$ and $OB$ are connected. If $\\angle B = 35^\\circ$, what is the measure of $\\angle AOB$?\n",
    "completion": "\\textbf{Solution:} Since AB is the tangent line of the circle O, \\\\\nit follows that $\\angle$OAB=$90^\\circ$, \\\\\nSince $\\angle$B=$35^\\circ$, \\\\\nit follows that $\\angle$AOB=$90^\\circ-\\angle$B=\\boxed{55^\\circ}.",
    "image_path": "dataset/mm_math/images/MM_Math/51446300.png"
  },
  {
    "solution": "\\boxed{15^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, there are two points A and B on the outer edge of the protractor, and their readings are $70^\\circ$ and $40^\\circ$ respectively. What is the degree measure of $\\angle 1$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle$AOB $= 70^\\circ - 40^\\circ = 30^\\circ$, \\\\\ntherefore $\\angle 1 = \\frac{1}{2} \\angle$AOB $= \\boxed{15^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/51261696.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, $AB=10\\,\\text{cm}$, $BC=8\\,\\text{cm}$. Point $D$ is on side $AB$ with $AD=AC$, $AE\\perp CD$ at foot $E$. Point $F$ is the midpoint of $BC$. Then, what is the length of $EF$ in $\\text{cm}$?\n",
    "completion": "\\textbf{Solution}: In $\\triangle ABC$, where $\\angle ACB=90^\\circ$, $AB=10\\text{cm}$, $BC=8\\text{cm}$,\\\\\nwe have $AC=\\sqrt{AB^{2}-BC^{2}}=\\sqrt{10^{2}-8^{2}}=6\\text{cm}$.\\\\\nThus, $AD=AC=6\\text{cm}$,\\\\\nand $BD=AB-AD=10-6=4\\text{cm}$.\\\\\nSince $AD=AC$ and $AE\\perp CD$,\\\\\nit follows that $CE=DE$, meaning point $E$ is the midpoint of $CD$,\\\\\nand given that point $F$ is the midpoint of $BC$,\\\\\n$EF$ serves as the midsegment of $\\triangle CBD$,\\\\\nthus $EF=\\frac{1}{2}BD=\\boxed{2}\\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50496612.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{26}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the sides of square $ABCD$ and square $DEFG$ are respectively $3$ and $2$, with points $E$ and $G$ located on sides $AD$ and $CD$, respectively. Point $H$ is the midpoint of $BF$. Connect $HG$. What is the length of $HG$?\n",
    "completion": "\\textbf{Solution:} Extend $GF$ to intersect $AB$ at $M$, and draw $HN \\perp GM$ at $N$,\\\\\nsince squares $ABCD$ and $DEFG$,\\\\\nthus $GM \\perp AB$, $FM=3-2=1$, $BM=3-2=1$,\\\\\nthus $FM=BM$, $NH\\parallel BM$,\\\\\nsince $H$ is the midpoint of $BF$,\\\\\nthus $NH=FN=\\frac{1}{2}BM=\\frac{1}{2}$,\\\\\nthus $GN=GF+FN=\\frac{5}{2}$,\\\\\nthus $GH=\\sqrt{GN^{2}+NH^{2}}=\\sqrt{\\left(\\frac{5}{2}\\right)^{2}+\\left(\\frac{1}{2}\\right)^{2}}=\\boxed{\\frac{\\sqrt{26}}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53729180.png"
  },
  {
    "solution": "\\boxed{\\frac{72}{25}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, $\\triangle ABC$ is an isosceles triangle with $AB=BC=5$ and $AC=6$. Point $O$ is the midpoint of $AC$, and a line is drawn from $O$ perpendicular to $AB$ at point $D$. Point $P$ is located on ray $BO$, and point $Q$ is located on line segment $BC$ (not coinciding with points $B$ or $C$). Line $PQ$ is drawn. When point $P$ coincides with point $O$, a point $F$ is chosen on line segment $AB$ such that points $C$ and $F$ are symmetrical about $PQ$. What is the distance $h$ from point $F$ to $AC$?\n",
    "completion": "\\textbf{Solution:}\n\nLet $O$ be the origin, the line where $AC$ lies be the $x$-axis, and the line where $OB$ lies be the $y$-axis to establish a Cartesian coordinate system. Connect $CF$,\\\\\nsince $C(3, 0)$, $B(0, 4)$, $A(3, 0)$,\\\\\nLet the equation of line $BC$ be $y=kx+b$,\\\\\nthus $\\left\\{\\begin{array}{l}b=4 \\\\ -3k+b=0\\end{array}\\right.$,\\\\\nsolving yields $\\left\\{\\begin{array}{l}b=4 \\\\ k=\\frac{4}{3}\\end{array}\\right.$,\\\\\ntherefore $y=\\frac{4}{3}x+4$,\\\\\nSimilarly, the equation of line $AB$ can be found as $y=-\\frac{4}{3}x+4$,\\\\\nLet $Q\\left(m, \\frac{4}{3}m+4\\right)$, $F\\left(n, -\\frac{4}{3}n+4\\right)$,\\\\\nThe equation of line $OQ$ can be found as $y=\\left(\\frac{4}{3}+\\frac{4}{m}\\right)x$,\\\\\nsince points $C$ and $F$ are symmetric about $PQ$,\\\\\nthus the midpoint $K$ of $CF$ is $\\left(\\frac{n-3}{2}, -\\frac{2}{3}n+2\\right)$,\\\\\nSubstituting point $K$ into $y=\\left(\\frac{4}{3}+\\frac{4}{m}\\right)x$,\\\\\nyields $-\\frac{2}{3}n+2=\\left(\\frac{4}{3}+\\frac{4}{m}\\right)\\cdot \\left(\\frac{n-3}{2}\\right)$,\\\\\nsolving for $m$ gives $m=-\\frac{3}{2}$,\\\\\nthus $Q\\left(-\\frac{3}{2}, 2\\right)$,\\\\\nsince $CQ=QF$,\\\\\nthus $\\left(\\frac{3}{2}\\right)^{2}+4=\\left(n+\\frac{3}{2}\\right)^{2}+\\left(2-\\frac{4}{3}n\\right)^{2}$,\\\\\nsolving yields $n=0$ (discarded) or $n=\\frac{21}{25}$,\\\\\nthus $F\\left(\\frac{21}{25}, \\frac{72}{25}\\right)$,\\\\\ntherefore, the distance $h$ from point $F$ to $AC$ is $\\boxed{\\frac{72}{25}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52914212_130.png"
  },
  {
    "solution": "\\boxed{y=-2x+6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, the analytical expression of line $AB$ is $y=kx+6$. The coordinates of point $D$ are $\\left(8, 0\\right)$. The symmetric point $C$ of point $O$ with respect to line $AB$ lies on line $AD$. What are the analytical expressions of lines $AD$ and $AB$?\n",
    "completion": "\\textbf{Solution:} From the problem, we have\n\\[\n\\because y=kx+6,\n\\]\n\\[\n\\therefore A\\left(0, 6\\right),\n\\]\nthat is, $OA=6$. Also,\n\\[\n\\because D\\left(8, 0\\right),\n\\]\n\\[\n\\therefore OD=8,\n\\]\nSuppose the equation of line $AD$ is $y=nx+6$. Substituting point $D\\left(8, 0\\right)$ into it, we get the equation of line $AD$ as $y=-\\frac{3}{4}x+6$.\n\nIn $\\triangle ABD$, $AD=\\sqrt{6^2+8^2}=10$,\n\n\\[\n\\because \\text{Point $O$ and point $C$ are symmetric about line $AB$},\n\\]\n\\[\n\\therefore \\text{Let $OB=BC=a$, $OA=AC=6$, $CD=4$},\n\\]\n\\[\n\\therefore BD=8-a,\n\\]\nIn $\\triangle BCD$, $a^2+4^2=(8-a)^2$,\n\n\\[\n\\therefore a=3 \\text{(area method is also applicable)}\n\\]\n\\[\n\\therefore B\\left(3, 0\\right),\n\\]\n\\[\n\\therefore \\text{The equation of line $AB$ is } y=-2x+6.\n\\]\n\nTherefore, the equation of line $AD$ is $\\boxed{y=-\\frac{3}{4}x+6}$, and the equation of line $AB$ is $\\boxed{y=-2x+6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50560467_65.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{3}-2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, $AD \\parallel BC$, $\\angle ABC = \\angle BCD = 90^\\circ$. The diagonals $AC$ and $BD$ intersect at point $O$. Line $DE$ bisects $\\angle ADC$ and meets $BC$ at point $E$. Connect $OE$. Quadrilateral $ABCD$ is a rectangle, if $CD = 2$ and $\\angle DBC = 30^\\circ$, find the area of $\\triangle BED$.\n",
    "completion": "Solution: In the right-angled triangle $BCD$, we have $\\angle BCD=90^\\circ$, $\\angle DBC=30^\\circ$, and $CD=2$. \\\\\nTherefore, $BD=2CD=4$, $BC=\\sqrt{BD^2-CD^2}=2\\sqrt{3}$, \\\\\nAs proved in (1), the quadrilateral $ABCD$ is a rectangle, \\\\\nTherefore, $\\angle ADC=90^\\circ$, \\\\\nBecause $DE$ bisects $\\angle ADC$, \\\\\nTherefore, $\\angle CDE=\\frac{1}{2}\\angle ADC=45^\\circ$, \\\\\nTherefore, $\\angle CED=90^\\circ-\\angle CDE=45^\\circ=\\angle CDE$, \\\\\nTherefore, $CE=CD=2$, \\\\\nTherefore, $BE=BC-CE=2\\sqrt{3}-2$, \\\\\nThus, the area of $\\triangle BED$ is $\\frac{1}{2}BE\\cdot CD=\\frac{1}{2}\\times (2\\sqrt{3}-2)\\times 2=\\boxed{2\\sqrt{3}-2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52204235_110.png"
  },
  {
    "solution": "\\boxed{\\sqrt{97}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In rhombus $ABCD$, the diagonals $AC$ and $BD$ intersect at point $O$. Rectangle $AEFO$ is constructed with $OA$ as its side, where point $E$ lies exactly on the perpendicular bisector of segment $AB$, and point $F$ is on $OB$. Let $G$ be a point on $OD$ such that $OG=BF$. Connect $CG$ and $BF$. As illustrated, given that $BO=2AO=8$ and $\\angle AMB=45^\\circ+\\frac{1}{2}\\angle MAC$, connect $MG$. Determine the length of segment $MG$.\n",
    "completion": "\\textbf{Solution:} As shown in figure 3, draw $GH\\perp MC$ through point $G$ meeting the extended line of $MC$ at point $H$.\\\\\nFrom (1) and (2), it is known that $AC\\perp BD$, $\\angle AMB=BAC$, $\\angle AMC=\\angle EBC+\\angle ABO$, $\\angle EBA=\\angle ABO$,\\\\\n$\\therefore OB\\perp AC$, $\\angle AOB=\\angle COG=90^\\circ$,\\\\\n$\\therefore \\angle BAO+\\angle ABO=90^\\circ$, $\\angle AMC=2\\angle ABO$,\\\\\n$\\therefore \\angle AMB+\\frac{1}{2}\\angle AMC=90^\\circ$,\\\\\n$\\because \\angle AMB=45^\\circ+\\frac{1}{2}\\angle MAC$,\\\\\n$\\therefore 45^\\circ+\\frac{1}{2}\\angle MAC+\\frac{1}{2}\\angle AMC=90^\\circ$,\\\\\n$\\therefore \\angle MAC+\\angle AMC=90^\\circ$,\\\\\n$\\therefore \\angle ACM=90^\\circ$,\\\\\n$\\because \\angle COG=\\angle OCH=\\angle H=90^\\circ$,\\\\\n$\\therefore$ quadrilateral $OCGH$ is a rectangle,\\\\\n$\\because \\angle BAM+\\angle MAC=\\angle BAC=\\angle AMB=45^\\circ+\\frac{1}{2}\\angle MAC$,\\\\\n$\\therefore \\angle BAM=45^\\circ-\\frac{1}{2}\\angle MAC$,\\\\\n$\\because \\angle MAC+\\angle AMC=90^\\circ$, $\\angle AMC=2\\angle ABO$,\\\\\n$\\therefore \\angle MAC+2\\angle ABO=90^\\circ$,\\\\\n$\\therefore \\angle ABO=45^\\circ-\\frac{1}{2}\\angle MAC$,\\\\\n$\\therefore \\angle BAM=\\angle ABO$,\\\\\n$\\therefore BN=AN$,\\\\\n$\\therefore \\angle EBA=\\angle EAB=\\angle ABN=\\angle BAN$,\\\\\n$\\because AB=AB$,\\\\\n$\\therefore \\triangle ABN\\cong \\triangle ABE(ASA)$,\\\\\n$\\therefore BE=BN$,\\\\\n$\\therefore AE=BE=BN=AN$,\\\\\n$\\because BO=2AO=8$,\\\\\n$\\therefore OA=4$,\\\\\n$\\therefore OC=OA=4$, $NO=BO-BN=BO-AN=8-AN$,\\\\\n$\\because AN^2=NO^2+OA^2$,\\\\\n$\\therefore AN^2=(8-AN)^2+4^2$,\\\\\n$\\therefore AN=5$, $ON=8-5=3$,\\\\\n$\\therefore AN=AE=BE=5$,\\\\\n$\\therefore CG=BE=5$, $OF=5$, $BF=BO-OF=3$,\\\\\n$\\therefore OG=BG=3$,\\\\\n$\\therefore OG=CH=3$, $GH=OC=5$,\\\\\n$\\because MC\\parallel OB$,\\\\\n$\\therefore \\triangle ANO\\sim \\triangle AMC$,\\\\\n$\\therefore \\frac{NO}{MC}=\\frac{AO}{AC}$, which means $MC=\\frac{NO\\cdot (AO+OC)}{AO}=\\frac{3\\times (4+4)}{4}=6$,\\\\\n$\\therefore MG=\\sqrt{MH^2+GH^2}=\\sqrt{9^2+4^2}=\\boxed{\\sqrt{97}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53728960_104.png"
  },
  {
    "solution": "\\boxed{y=3x-3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, two lines \\( {l}_{1} \\) and \\( {l}_{2} \\) passing through point \\(A(1, 0)\\) intersect the y-axis at points \\(B\\) and \\(C\\) respectively, where point \\(B\\) is above the origin and point \\(C\\) is below the origin. It is known that \\( AB = \\sqrt{10} \\). If the area of \\( \\triangle ABC \\) is 3, what is the equation of line \\( {l}_{2} \\)?\n",
    "completion": "\\textbf{Solution:} Given that the area of $\\triangle ABC$ is 3, it follows that $\\frac{1}{2}BC \\cdot AO = 3$. Hence, $\\frac{1}{2}BC \\times 1 = 3$, which implies $BC=6$. Since $BO=3$, it follows that $CO=3$. Therefore, the coordinates of point C are $(0, -3)$. Let the equation of line $l_{2}$ be $y= kx + b$ ($k \\neq 0$). We then have \n\\[\n\\left\\{\n\\begin{array}{l}\nb=-3 \\\\\nk+b=0\n\\end{array}\n\\right.\n\\]\nSolving this system of equations, we obtain\n\\[\n\\left\\{\n\\begin{array}{l}\nk=3 \\\\\nb=-3\n\\end{array}\n\\right.\n\\]\nThus, the equation of the line $l_{2}$ is $y=3x-3$. Therefore, the answer is \\boxed{y=3x-3}.",
    "image_path": "dataset/mm_math/images/MM_Math/52690999_58.png"
  },
  {
    "solution": "\\boxed{9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the square $ABCD$, $M$ is a point on $BC$, $ME\\perp AM$, $ME$ intersects $CD$ at $F$, and intersects the extension of $AD$ at point $E$. $\\triangle ABM\\sim \\triangle MCF$; if $AB=4$ and $BM=2$, what is the area of $\\triangle DEF$?\n",
    "completion": "\\textbf{Solution:}\n\nSince $\\triangle ABM\\sim \\triangle MCF$,\n\nit follows that $\\frac{AB}{MC}=\\frac{BM}{CF}$,\n\nSince quadrilateral $ABCD$ is a square, $AB=4$, $BM=2$,\n\ntherefore $BC\\parallel AD$, $BC=CD=4$, $MC=BC-BM=4-2=2$,\n\n$\\angle FDE=\\angle CDA=90^\\circ$,\n\nthus $CF=\\frac{MC\\cdot BM}{AB}=\\frac{2\\times 2}{4}=1$,\n\nthus $DF=DC-CF=4-1=3$,\n\nSince $BC\\parallel AD$,\n\nit follows that $\\triangle MCF\\sim \\triangle EDF$,\n\nthus $\\frac{MC}{DE}=\\frac{CF}{DF}=\\frac{1}{3}$,\n\nthus $DE=6$,\n\nthus $S_{\\triangle DEF}=\\frac{1}{2}DF\\times DE=\\frac{1}{2}\\times 3\\times 6=\\boxed{9}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51434139_134.png"
  },
  {
    "solution": "\\boxed{(-\\frac{3}{2}, 5)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure (1), it's known that the line $y=\\frac{4}{3}x+4$ intersects the x-axis at point A and the y-axis at point C. The parabola $y=ax^2+bx+c$ passes through points A and C, also intersects the x-axis at another point B, with its axis of symmetry being the line $x=-1$. Point D is a moving point on the parabola in the second quadrant, and the x-coordinate of D is denoted as $m$. What is the maximum value of the area $S$ of quadrilateral $ABCD$ and the coordinates of D at this maximum?\n",
    "completion": "\\textbf{Solution:} Connect $OD$,\\\\\nsince point B is symmetric to point $A(-3,0)$ with respect to the line $x=-1$,\\\\\ntherefore, the coordinates of point B are $(1,0)$.\\\\\nAccording to the problem, the coordinates of point D are $(m, -\\frac{4}{3}m^{2}-\\frac{8}{3}m+4)$.\\\\\nTherefore, the area $S_{\\text{quadrilateral} ABCD}$ of quadrilateral $ABCD$ equals $S_{\\triangle OAD}+S_{\\triangle OCD}+S_{\\triangle OBC}$\\\\\n$=\\frac{1}{2}\\times3\\times(-\\frac{4}{3}m^{2}-\\frac{8}{3}m+4)+\\frac{1}{2}\\times4\\times(-m)+\\frac{1}{2}\\times1\\times4$\\\\\n$=-2m^{2}-6m+8=-2(m+\\frac{3}{2})^{2}+\\frac{25}{2}$\\\\\nSince $-2<0$,\\\\\ntherefore, when $m=-\\frac{3}{2}$, the maximum area of quadrilateral $ABCD$ is $\\boxed{\\frac{25}{2}}$.\\\\\nAt this point, the coordinates of point D are $\\boxed{(-\\frac{3}{2}, 5)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53043766_126.png"
  },
  {
    "solution": "\\boxed{(2,0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a circular arc in the Cartesian coordinate system passes through the grid points $A(0, 4)$, $B(4, 4)$, and $C(6, 2)$. What are the coordinates of the center of the circle containing this arc?\n",
    "completion": "\\textbf{Solution:} The center of the circle containing the arc has coordinates $\\boxed{(2,0)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51402739_127.png"
  },
  {
    "solution": "\\boxed{66^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle B=48^\\circ$, the angle bisectors of the exterior angles $\\angle DAC$ and $\\angle ACF$ intersect at point E. What is the measure of $\\angle AEC$ in degrees?\n",
    "completion": "Solution: Since the bisectors of exterior angles $\\angle DAC$ and $\\angle ACF$ of the triangle intersect at point E, \\\\\nit follows that $\\angle CAE + \\angle ACE = \\frac{1}{2}(\\angle B + \\angle ACB) + \\frac{1}{2}(\\angle B + \\angle BAC)\\\\\n= \\frac{1}{2}(\\angle BAC + \\angle B + \\angle ACB + \\angle B)\\\\\n= \\frac{1}{2}(180^\\circ + 48^\\circ)\\\\\n= 114^\\circ$\\\\\nIn $\\triangle ACE$, $\\angle AEC = 180^\\circ - (\\angle CAE + \\angle ACE)\\\\\n= 180^\\circ - 114^\\circ\\\\\n= \\boxed{66^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/52604050.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, line segments AB and CD intersect at point E, with AC $\\parallel$ BD. If AC = 3, BD = 1.5, and BE = 2, what is the length of AE?\n",
    "completion": "\\textbf{Solution:} Given that $AC \\parallel BD$,\\\\\nit follows that $\\angle A = \\angle B$ and $\\angle C = \\angle D$,\\\\\ntherefore $\\triangle AEC \\sim \\triangle BED$,\\\\\nwhich leads to $\\frac{AC}{BD} = \\frac{AE}{BE}$,\\\\\ngiven that $AC = 3$, $BD = 1.5$, and $BE = 2$,\\\\\nit follows that $\\frac{3}{1.5} = \\frac{AE}{2}$,\\\\\nsolving this, we find $AE = \\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52602898.png"
  },
  {
    "solution": "\\boxed{40}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $BC$ is the diameter of $\\odot O$, points $D$ and $F$ are on $\\odot O$. Connect $CD$ and $BF$ and extend them to intersect at point $A$, where $AD=2CD=8$, $AF=3BF$. If point $E$ is the midpoint of arc $BC$, connect $BE$ and $DE$. Find the value of $EG\\cdot EF$?\n",
    "completion": "\\textbf{Solution:} \\\\\nSince $E$ is the midpoint of the arc $BC$, \\\\\nhence $\\overset{\\frown}{BE}=\\overset{\\frown}{CE}$, \\\\\ntherefore $\\angle BFE=\\angle CBE$, \\\\\nalso since $\\angle GEB=\\angle BEF$, \\\\\nhence $\\triangle EBG\\sim \\triangle EFB$, \\\\\nthus $\\frac{BE}{EF}=\\frac{EG}{BE}$, \\\\\nthat is $EG\\cdot EF=BE^{2}$, \\\\\nsince $\\overset{\\frown}{BE}=\\overset{\\frown}{CE}$ again, \\\\\nhence $BE=CE$, \\\\\nsince $BE^{2}+CE^{2}=BC^{2}$, \\\\\nthat is $2BE^{2}=BC^{2}$, \\\\\ntherefore $BE^{2}=\\frac{1}{2}BC^{2}=\\frac{1}{2}\\times \\left(4\\sqrt{5}\\right)^{2}=40$, \\\\\ntherefore $EG\\cdot EF=\\boxed{40}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53148615_113.png"
  },
  {
    "solution": "\\boxed{\\frac{7}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in rectangle $ABCD$, $AB=4$, $BC=5$, P is a moving point on the ray $BC$, construct $PE\\perp AP$, with $PE$ intersecting the ray $DC$ at point E, and the ray $AE$ intersecting the ray $BC$ at point F. Let $BP=x$, $CF=y$. When $\\frac{PE}{AP}=\\frac{1}{2}$, find the length of $CF$.\n",
    "completion": "\\textbf{Solution:} (1) When point $P$ lies on segment $BC$, and $E$ lies on segment $CD$,\n\\\\\n$\\because \\triangle PCE\\sim \\triangle ABP$,\n\\\\\n$\\therefore \\frac{PC}{AB}=\\frac{CE}{BP}=\\frac{PE}{AP}=\\frac{1}{2}$,\n\\\\\n$\\therefore PC=\\frac{1}{2}AB=2$, $CE=\\frac{1}{2}BP$,\n\\\\\n$\\therefore BP=BC-PC=5-2=3$,\n\\\\\n$\\therefore CE=\\frac{3}{2}$,\n\\\\\n$\\because AB\\parallel CD$,\n\\\\\n$\\therefore \\triangle CEF\\sim \\triangle BAF$,\n\\\\\n$\\therefore \\frac{CF}{BF}=\\frac{CE}{AB}$,\n\\\\\nwhich gives $\\frac{CF}{5+CF}=\\frac{\\frac{3}{2}}{4}$,\n\\\\\nSolving this we get: $CF=\\boxed{3}$;\n\\\\\n(2) When $P$ lies on the extension of $BC$, as shown in the diagram:\n\\\\\n$\\because \\angle EPC+\\angle BPA=90^\\circ$, $\\angle BPA+\\angle BAP=90^\\circ$,\n\\\\\n$\\therefore \\angle EPC=\\angle BAP$,\n\\\\\n$\\because \\angle B=\\angle PCE$,\n\\\\\n$\\therefore \\triangle ABP\\sim \\triangle PCE$,\n\\\\\n$\\therefore \\frac{PC}{AB}=\\frac{CE}{BP}=\\frac{PE}{AP}=\\frac{1}{2}$,\n\\\\\n$\\therefore PC=\\frac{1}{2}AB=2$, $CE=\\frac{1}{2}BP$,\n\\\\\n$\\therefore BP=BC+PC=7$, $CE=\\frac{7}{2}$,\n\\\\\n$\\because \\angle AFB=\\angle CFE$, $\\angle B=\\angle FCE$,\n\\\\\n$\\therefore \\triangle ABF\\sim \\triangle ECF$,\n\\\\\n$\\therefore \\frac{AB}{EC}=\\frac{BF}{CF}$,\n\\\\\nwhich gives $\\frac{4}{\\frac{7}{2}}=\\frac{5-CF}{CF}$,\n\\\\\nSolving this we get: $CF=\\boxed{\\frac{7}{3}}$;\n\\\\\nIn summary, the length of $CF$ can be $3$ or $\\boxed{\\frac{7}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52917616_88.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $DE\\perp AB$ at $E$, $DF\\perp AC$ at $F$. Given that $BD=CD$ and $BE=CF$, $\\triangle BDE \\cong \\triangle CDF$. It's known that $AC=12$ and $BE=2$, find the length of $AB$.\n",
    "completion": "\\textbf{Solution:} From (1) it is known that: $Rt\\triangle BDE \\cong Rt\\triangle CDF$,\\\\\n$\\therefore DE=DF$, $BE=CF=2$,\\\\\nAlso, since $AC=12$, $\\therefore AF=10$,\\\\\nBecause $\\angle E=\\angle DFA=90^\\circ$ and $AD=AD$,\\\\\n$\\therefore$ $Rt\\triangle AED \\cong Rt\\triangle AFD$ (HL),\\\\\n$\\therefore AE=AF=10$,\\\\\n$\\therefore AB=AE-BE=10-2=\\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52935555_93.png"
  },
  {
    "solution": "\\boxed{\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the side length of rhombus $ABCD$ is 2, and $\\angle ADC=120^\\circ$. Point $E$ is the midpoint of side $AB$, and point $P$ is a moving point on the diagonal $AC$. What is the minimum value of $PB+PE$?\n",
    "completion": "\\textbf{Solution:} Connect BD, DE, and DE intersects AC at point P$^{\\prime}$. Connect BP$^{\\prime}$,\n\n$\\because$ rhombus ABCD,\n\n$\\therefore$ point B and point D are symmetrical about AC,\n\n$\\therefore$ DP$^{\\prime}$=BP$^{\\prime}$,\n\n$\\therefore$ P$^{\\prime}$B + P$^{\\prime}$E = DP$^{\\prime}$ + P$^{\\prime}$E = DE,\n\n$\\because$ the line segment between two points is the shortest,\n\n$\\therefore$ the minimum value of P$^{\\prime}$B + P$^{\\prime}$E is equal to the length of DE;\n\n$\\because$ rhombus ABCD,\n\n$\\therefore$ DC$\\parallel$AB, AD=AB,\n\n$\\therefore$ $\\angle$ADC + $\\angle$DAB = 180$^\\circ$,\n\n$\\therefore$ $\\angle$DAB = 60$^\\circ$,\n\n$\\therefore$ $\\triangle$ADB is an equilateral triangle,\n\n$\\because$ point E is the midpoint of AB,\n\n$\\therefore$ DE $\\perp$ AB, BE=$\\frac{1}{2}$AB=1, BD=AB=2,\n\nIn $\\triangle$BDE, $BE=\\sqrt{BD^{2}-BE^{2}}=\\sqrt{2^{2}-1^{2}}=\\sqrt{3}$,\n\n$\\therefore$ the minimum value of PB+PE is $\\boxed{\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52251409.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within a square grid with a side length of 1, where points A, B, C, and D are all located at the grid points, and line segments $AB$ and $CD$ intersect at point E, what is the value of $\\frac{AE}{BE}$?\n",
    "completion": "\\textbf{Solution:} We have $AD=\\sqrt{1+4}=\\sqrt{5}$, $BC=\\sqrt{16+4}=2\\sqrt{5}$,\\\\\nsince it is known from the lattice points that $AD \\parallel BC$,\\\\\nit follows that $\\triangle ADE \\sim \\triangle BCE$,\\\\\ntherefore $\\frac{AE}{BE}=\\frac{AD}{BC}=\\frac{\\sqrt{5}}{2\\sqrt{5}}=\\boxed{\\frac{1}{2}}$",
    "image_path": "dataset/mm_math/images/MM_Math/52991847.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the vertex B of the rhombus OABC is on the y-axis, and the coordinates of vertex C are $(-3, 2)$. If the graph of the inverse proportion function $y = \\frac{k}{x}$ $(x>0)$ passes through point A, what is the value of $k$?\n",
    "completion": "\\textbf{Solution:} Since diamond OABC is axially symmetrical, \\\\\n$\\therefore$ A and C are symmetrical with respect to the y-axis, \\\\\nthen A(3, 2), \\\\\n$\\because$ A lies on the graph of y=$\\frac{k}{x}$, \\\\\n$\\therefore k=3\\times 2=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52623483.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the graph of the quadratic function $y=-x^2+bx+c$ passes through two points $A(-2, -1)$ and $B(0, 7)$. What is the analytic expression of the parabola and its axis of symmetry?\n",
    "completion": "\\textbf{Solution:} Let the graph of the quadratic function $y=-x^2+bx+c$ pass through points $A(-2, -1)$, and $B(0, 7)$.\n\nHence, $\\left\\{\\begin{array}{l}\n-1=-4-2b+c \\\\\nc=7\n\\end{array}\\right.$,\n\nSolving this system, we get: $\\left\\{\\begin{array}{l}\nb=2 \\\\\nc=7\n\\end{array}\\right.$,\n\nTherefore, $y=-x^2+2x+7$,\n\n$=-\\left(x^2-2x\\right)+7$,\n\n$=-\\left[\\left(x^2-2x+1\\right)-1\\right]+7$,\n\n$=-(x-1)^2+8$,\n\nHence, the axis of symmetry is the line $x=\\boxed{1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52916890_264.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $AB=8\\,\\text{cm}$, $BC=10\\,\\text{cm}$. A point $E$ is taken on side $CD$, and $\\triangle ADE$ is folded such that point $D$ aligns exactly with point $F$ on side $BC$. What is the length of $CE$ in $\\text{cm}$?\n",
    "completion": "\\textbf{Solution:} \nGiven: rectangle ABCD, \\\\\nhence AD=BC=10cm, and CD=AB=8cm, \\\\\nBy the problem statement: Rt$\\triangle$ADE$\\cong$Rt$\\triangle$AFE,\\\\\nthus $\\angle$AFE=90$^\\circ$, AF=10cm, and EF=DE,\\\\\nLet CE=xcm, then DE=EF=CD−CE=$(8−x)$cm,\\\\\nIn Rt$\\triangle$ABF, by the Pythagorean theorem: AB$^{2}$+BF$^{2}$=AF$^{2}$,\\\\\nthat is $8^{2}$+BF$^{2}$=$10^{2}$,\\\\\nthus BF=6cm,\\\\\nthus CF=BC−BF=10−6=4(cm),\\\\\nIn Rt$\\triangle$ECF, by the Pythagorean theorem: EF$^{2}$=CE$^{2}$+CF$^{2}$,\\\\\nthat is $(8−x)^{2}$=x$^{2}$+4$^{2}$,\\\\\nthus 64−16x+x$^{2}$=x$^{2}$+16,\\\\\nthus x=\\boxed{3},\\\\\nwhich means CE=3cm.",
    "image_path": "dataset/mm_math/images/MM_Math/52827074.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point E is on $AC$, $\\triangle ABC\\cong \\triangle DAE$, $BC=3$, $DE=7$, then what is the length of $CE$?\n",
    "completion": "\\textbf{Solution:} Since it follows from the problem statement that $\\triangle ABC\\cong \\triangle DAE$,\n\nwe have $AE=BC=3$, $AC=DE=7$,\n\nthus $CE=AC-AE=7-3=\\boxed{4}$",
    "image_path": "dataset/mm_math/images/MM_Math/52866931.png"
  },
  {
    "solution": "\\boxed{130^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB \\parallel CD$, $\\angle FGB = 155^\\circ$, $FG$ bisects $\\angle EFD$, what is the size of $\\angle BEF$?\n",
    "completion": "\\textbf{Solution:} Since $AB\\parallel CD$ and $\\angle FGB=155^\\circ$,\\\\\nit follows that $\\angle BEF+\\angle EFD=180^\\circ$,\\\\\nhence $\\angle GFD=180^\\circ-\\angle FGB=180^\\circ-155^\\circ=25^\\circ$,\\\\\nGiven that $FG$ bisects $\\angle EFD$,\\\\\nit implies $\\angle EFD=2\\angle GFD=2\\times 25^\\circ=50^\\circ$,\\\\\nthus $\\angle BEF=180^\\circ-\\angle EFD=180^\\circ-50^\\circ=\\boxed{130^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53119447.png"
  },
  {
    "solution": "\\boxed{70^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in $\\triangle ABC$, $AB=AC$, a semicircle with diameter $AB$ intersects $AC$ and $BC$ at points $E$ and $D$, respectively. Connect $ED$. If $\\angle BAC=55^\\circ$, find the degree measure of arc $AE$.\n",
    "completion": "\\textbf{Solution:} Connect $BE$, $OE$\\\\\nSince $AB$ is a diameter\\\\\nTherefore $\\angle AEB=90^\\circ$\\\\\nSince $\\angle BAC=55^\\circ$\\\\\nTherefore $\\angle ABE=90^\\circ-\\angle BAC=35^\\circ$\\\\\nTherefore $\\angle AOE=2\\angle ABE=70^\\circ$;\\\\\nTherefore, the degree measure of arc $AE$ is $\\boxed{70^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53080407_96.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in a grid composed of small squares with side length of 1, points A, B, and C are all on the grid points. If a circle with AB as its diameter passes through points C and D, then what is the value of $\\tan\\angle ADC$?\n",
    "completion": "\\textbf{Solution:} Since $AB$ is the diameter,\\\\\n$\\therefore\\angle ACB=90^\\circ$,\\\\\nIn $\\triangle ABC$, $\\tan\\angle ABC= \\frac{AC}{BC}=\\frac{3}{2}$,\\\\\nSince $\\angle ADC=\\angle ABC$,\\\\\n$\\therefore\\tan\\angle ADC= \\frac{3}{2}$,\\\\\nTherefore $\\tan\\angle ADC = \\boxed{\\frac{3}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51545288.png"
  },
  {
    "solution": "\\boxed{3\\sqrt{7}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rhombus $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$. Line $DE$ is drawn through point $D$ such that $DE \\parallel AC$ and $DE = OC$, and $CE$ is then connected. Quadrilateral $OCED$ is a rectangle. Line $AE$ is drawn and intersects $OD$ at point $F$. If the side length of rhombus $ABCD$ is $6$ and $\\angle ABC = 60^\\circ$, find the length of $AE$.\n",
    "completion": "\\textbf{Solution:}\\\\\n$\\because$ The side length of rhombus ABCD is 6,\\\\\n$\\therefore AB=BC=CD=AD=6$, $BD\\perp AC$, $AO=CO=\\frac{1}{2}AC$\\\\\n$\\because \\angle ABC=60^\\circ$, $AB=BC$\\\\\n$\\therefore \\triangle ABC$ is an equilateral triangle\\\\\n$\\therefore AC=AB=6$\\\\\n$\\because$ In $\\triangle AOD$, $BD\\perp AC$, $AD=6$, $AO=3$,\\\\\n$\\therefore OD=\\sqrt{AD^2-AO^2}=3\\sqrt{3}$\\\\\n$\\because$ Quadrilateral OCED is a rectangle,\\\\\n$\\therefore CE=OD=3\\sqrt{3}$\\\\\n$\\because$ In $\\triangle ACE$, $AC=6$, $CE=3\\sqrt{3}$,\\\\\n$\\therefore AE=\\sqrt{AC^2+CE^2}=\\sqrt{6^2+(3\\sqrt{3})^2}=\\boxed{3\\sqrt{7}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51435842_144.png"
  },
  {
    "solution": "\\boxed{16\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, rectangle $ABCD$ is folded along $EF$, with point B exactly landing on point $B'$ on side $AD$. If $AE=2$, $DE=6$, and $\\angle EFB=60^\\circ$, what is the area of rectangle $ABCD$? The core question is: What is the area of rectangle $ABCD$?\n",
    "completion": "\\textbf{Solution:} In rectangle $ABCD$, since $AD\\parallel BC$, it follows that $\\angle DEF=\\angle EFB=60^\\circ$,\\\\\nsince folding rectangle $ABCD$ along $EF$ precisely positions point $B$ at $B^{\\prime}$ on side $AD$,\\\\\ntherefore $\\angle EFB=\\angle EFB=60^\\circ$, $\\angle B=\\angle A^{\\prime}B^{\\prime}F=90^\\circ$, $\\angle A=\\angle A^{\\prime}=90^\\circ$,\\\\\n$AE=A^{\\prime}E=2$, $AB=A^{\\prime}B^{\\prime}$,\\\\\nin $\\triangle EFB^{\\prime}$, since $\\angle DEF=\\angle EFB=\\angle EB^{\\prime}F=60^\\circ$, it follows that $\\triangle EFB^{\\prime}$ is an equilateral triangle,\\\\\nin $\\triangle A^{\\prime}EB^{\\prime}$, since $\\angle A^{\\prime}B^{\\prime}E=90^\\circ-60^\\circ=30^\\circ$, it follows that $B^{\\prime}E=2A^{\\prime}E$, and since $A^{\\prime}E=2$,\\\\\nthus $B^{\\prime}E=4$, therefore $A^{\\prime}B^{\\prime}=2\\sqrt{3}$, that is $AB=2\\sqrt{3}$, since $AE=2$, $DE=6$,\\\\\nthus $AD=AE+DE=2+6=8$, therefore the area of rectangle $ABCD$ = $AB\\cdot AD=2\\sqrt{3}\\times 8=\\boxed{16\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53360048.png"
  },
  {
    "solution": "\\boxed{90^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In square $ABCD$ and square $AEFG$, connect $AC$, $AF$, and $CF$ respectively, with point $H$ being the midpoint of $CF$. As shown in Figure 1, when points $F$ and $G$ are respectively on line segments $AB$ and $AC$, what is the measure of $\\angle BHG$?\n",
    "completion": "Solution: Since squares $ABCD$ and $AEFG$, where $AC$ and $AF$ are diagonals,\\\\\n$\\therefore$ $\\angle AFG=\\angle ACB=45^\\circ$, $\\angle AGF=\\angle ABC=90^\\circ$,\\\\\nSince point $F$ is on segment $AB$,\\\\\n$\\therefore$ $\\angle GFB=180^\\circ-\\angle ACB=135^\\circ$.\\\\\nSince in $\\triangle CGF$ and $\\triangle CBF$, point $H$ is the midpoint of hypotenuse $CF$,\\\\\n$\\therefore$ $GH=\\frac{1}{2}CF=HF$, $HB=\\frac{1}{2}CF=HF$,\\\\\n$\\therefore$ $\\angle HGF=\\angle HFG$, $\\angle HFB=\\angle HBF$,\\\\\n$\\therefore$ $\\angle HGF+\\angle HBF=\\angle HFG+\\angle HFB=\\angle GFB=135^\\circ$,\\\\\n$\\therefore$ $\\angle BHG=360^\\circ-\\angle HGF-\\angle HBF-\\angle GFB=360^\\circ-135^\\circ-135^\\circ=\\boxed{90^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53728774_87.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\triangle ABC$, $ \\angle ACB=90^\\circ$, CD is the median to side AB, and $ CD+AB=12$, what is the length of AB?\n",
    "completion": "\\textbf{Solution:} Since $\\angle ACB=90^\\circ$ and D is the midpoint of side AB, \\\\\nit follows that $CD=\\frac{1}{2}AB$, \\\\\nand since $CD+AB=12$, \\\\\nwe have $AB=\\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51545519.png"
  },
  {
    "solution": "\\boxed{\\frac{16}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC=10$ and $AD\\perp BC$ at point $D$, $AD=8$. If point $E$ is the centroid of $\\triangle ABC$, and point $F$ is the centroid of $\\triangle ACD$, then what is the area of $\\triangle AEF$?\n",
    "completion": "Solution:As shown in the figure, extend AF to meet BC at point M, \\\\\n$\\because AB=AC=10, AD\\perp BC, AD=8$, \\\\\n$\\therefore CD=BD=\\sqrt{AB^2-AD^2}=6$, \\\\\n$\\because$ point E is the centroid of $\\triangle ABC$, point F is the centroid of $\\triangle ACD$, \\\\\n$\\therefore \\frac{AE}{DE}=2, \\frac{AF}{FM}=2, DM=\\frac{1}{2}CD=3$, \\\\\n$\\therefore \\frac{AE}{AD}=\\frac{AF}{AM}=\\frac{2}{3}, AE=\\frac{2}{3}AD=\\frac{16}{3}$, \\\\\nFurthermore, $\\because \\angle EAF=\\angle DAM$, \\\\\n$\\therefore \\triangle AEF\\sim \\triangle ADM$, \\\\\n$\\therefore \\frac{EF}{DM}=\\frac{AE}{AD}=\\frac{2}{3}, \\angle AEF=\\angle ADM=90^\\circ$, \\\\\n$\\therefore \\frac{EF}{3}=\\frac{2}{3}$, \\\\\nSolving gives $EF=2$, \\\\\nHence, the area of $\\triangle AEF$ is $\\frac{1}{2}AE\\cdot EF=\\frac{1}{2}\\times \\frac{16}{3}\\times 2=\\boxed{\\frac{16}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51322975.png"
  },
  {
    "solution": "\\boxed{115^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points A, B, and C are on the circle $\\odot O$. If $\\angle AOB=130^\\circ$, what is the measure of $\\angle ACB$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, take a point \\(D\\) on the major arc \\(AB\\), and connect \\(AD\\), \\(DB\\).\\\\\nSince \\(\\angle ADB = \\frac{1}{2} \\angle AOB\\), \\(\\angle AOB = 130^\\circ\\),\\\\\nTherefore, \\(\\angle ADB = 65^\\circ\\),\\\\\nSince \\(\\angle ACB + \\angle ADB = 180^\\circ\\),\\\\\nTherefore, \\(\\angle ACB = \\boxed{115^\\circ}\\).",
    "image_path": "dataset/mm_math/images/MM_Math/52766573.png"
  },
  {
    "solution": "\\boxed{\\frac{2\\sqrt{29}}{29}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that in $\\triangle ABC$, $CD\\perp AB$ with the foot of the perpendicular at point $D, AD=2, BD=6, \\tan\\angle B=\\frac{2}{3}$. Point $E$ is the midpoint of side $BC$. Find the sine of $\\angle EAB$.\n",
    "completion": "\\textbf{Solution:} Construct $EF\\perp AB$ at point F through point $E$, as shown in the figure,\\\\\nsince $CD\\perp AB$,\\\\\ntherefore $EF\\parallel CD$\\\\\ntherefore $\\triangle BEF\\sim \\triangle BCD$\\\\\ntherefore $\\frac{BE}{BC}=\\frac{BF}{BD}=\\frac{EF}{CD}$\\\\\nsince point $E$ is the midpoint of side $BC$\\\\\ntherefore $BE=CE=\\frac{1}{2}BC$\\\\\ntherefore $\\frac{BF}{BD}=\\frac{EF}{CD}=\\frac{1}{2}$\\\\\nsince $CD=4, BD=6$\\\\\ntherefore $EF=2, BF=3$\\\\\ntherefore $DF=3$\\\\\ntherefore $AF=DF+DF=2+3=5$\\\\\nIn $\\triangle EAF$, since $AF^{2}+EF^{2}=AE^{2}$\\\\\ntherefore $AE=\\sqrt{AF^{2}+EF^{2}}=\\sqrt{5^{2}+2^{2}}=\\sqrt{29}$\\\\\ntherefore $\\sin\\angle EAB=\\frac{EF}{AE}=\\frac{2}{\\sqrt{29}}=\\boxed{\\frac{2\\sqrt{29}}{29}}",
    "image_path": "dataset/mm_math/images/MM_Math/51379561_98.png"
  },
  {
    "solution": "\\boxed{\\triangle ABC \\sim \\triangle AED}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in quadrilateral $ABCD$, $AC$ and $BD$ intersect at point $F$. Point $E$ is on $BD$, and $\\angle BAE = \\angle CAD$, $\\frac{AB}{AE} = \\frac{AC}{AD}$. Prove that: $\\triangle ABC \\sim \\triangle AED$.\n",
    "completion": "\\textbf{Solution:} Since $\\angle BAE = \\angle CAD$, \\\\\nit follows that $\\angle BAE + \\angle EAF = \\angle CAD + \\angle EAF$, \\\\\nwhich implies $\\angle BAC = \\angle DAE$, \\\\\nand thus $\\frac{AB}{AE}=\\frac{AC}{AD}$. \\\\\nConsidering triangles $\\triangle ABC$ and $\\triangle AED$, where \\\\\n$\\left\\{\\begin{array}{l}\n\\frac{AB}{AE}=\\frac{AC}{AD} \\\\\n\\angle BAC = \\angle DAE\n\\end{array}\\right.$, \\\\\nit can be concluded that \\boxed{\\triangle ABC \\sim \\triangle AED}.",
    "image_path": "dataset/mm_math/images/MM_Math/52872215_89.png"
  },
  {
    "solution": "\\boxed{70^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, rectangle $ABCD$ is folded along crease $EF$. The corresponding edge of $BC$, denoted as $B'C'$, intersects $CD$ at point $M$. If $\\angle BMD = 50^\\circ$, then what is the measure of $\\angle BEF$ in degrees?\n",
    "completion": "\\textbf{Solution:} Since we fold the rectangle ABCD along the crease EF,\\\\\nit follows that $\\angle C^{\\prime} = \\angle A = 90^\\circ$, $\\angle DMB^{\\prime} = \\angle C^{\\prime}MF = 50^\\circ$, $\\angle EFC = \\angle C^{\\prime}FE$, and $DC \\parallel AB$\\\\\ntherefore, $\\angle C^{\\prime}FM = 90^\\circ - 50^\\circ = 40^\\circ$, $\\angle MFE = \\angle FEB$, and $\\angle CFE + \\angle FEB = 180^\\circ$,\\\\\nthus, $\\angle C^{\\prime}FE = 40^\\circ + \\angle MFE$, $\\angle EFC = 180^\\circ - \\angle MFE$,\\\\\nit leads to $40^\\circ + \\angle MFE = 180^\\circ - \\angle MFE$,\\\\\nSolving gives: $\\angle MFE = 70^\\circ$,\\\\\nhence, $\\angle BEF = \\boxed{70^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51614711.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the rhombus $ABCD$, the heights $CE$ and $CF$ are drawn from point $C$ onto sides $AB$ and $AD$ respectively. Line $BF$ is drawn and intersects $CE$ at point $G$. If point $E$ is the midpoint of $AB$, what is the value of $\\frac{EG}{CG}$?\n",
    "completion": "\\textbf{Solution:} As the diagram shows, construct $FH \\perp CE$ intersecting $CE$ at $H$,\\\\\n$\\because \\angle D = \\angle EBC, \\angle DFC = \\angle BEC = 90^\\circ, CD = BC$,\\\\\n$\\therefore \\triangle DFC \\cong \\triangle BEC \\left( AAS \\right)$,\\\\\n$\\therefore BE = DF, CF = CE, \\angle DCF = \\angle BCE$,\\\\\n$\\because AD \\parallel BC, CF \\perp AD$,\\\\\n$\\therefore \\angle FCB = 90^\\circ$,\\\\\nSimilarly, $\\angle DCE = 90^\\circ$,\\\\\n$\\because \\angle FCE + \\angle DCF = 90^\\circ, \\angle D + \\angle DCF = 90^\\circ$,\\\\\n$\\therefore \\angle D = \\angle FCE$,\\\\\n$\\because FH \\perp CE$,\\\\\n$\\therefore \\triangle DFC \\sim \\triangle CHF$,\\\\\n$\\therefore \\frac{CD}{CF} = \\frac{CF}{FH}$,\\\\\nLet $BE = x$, then $DF = BE = x, CD = AB = BC = 2x, CF = CE = \\sqrt{3}x$,\\\\\n$\\therefore \\frac{2x}{\\sqrt{3}x} = \\frac{\\sqrt{3}x}{FH}$,\\\\\n$\\therefore FH = \\frac{3}{2}x$,\\\\\n$\\therefore CH = \\frac{\\sqrt{3}}{2}x$,\\\\\n$\\because FH \\perp CE, CE \\perp AB$,\\\\\n$\\therefore \\triangle FHG \\sim \\triangle BEG$,\\\\\n$\\therefore \\frac{FH}{BE} = \\frac{HG}{EG} = \\frac{\\frac{3}{2}x}{x}$,\\\\\n$\\because HE = CE - CH = \\frac{\\sqrt{3}}{2}x$,\\\\\n$\\therefore HG = \\frac{3\\sqrt{3}}{10}x, GE = \\frac{\\sqrt{3}}{5}x$,\\\\\n$\\therefore \\frac{EG}{CG} = \\frac{\\frac{\\sqrt{3}}{5}x}{\\frac{8\\sqrt{3}}{10}x} = \\boxed{\\frac{1}{4}}$,",
    "image_path": "dataset/mm_math/images/MM_Math/53110864.png"
  },
  {
    "solution": "\\boxed{9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=ax^{2}+bx-4a$ (where $a\\neq 0$) passes through points $A(-1, 0)$ and $C(0, 4)$, and intersects with the x-axis at another point $B$. Lines $AC$ and $BC$ are drawn. A line parallel to the x-axis, $y=-14$, intersects the parabola at points $D$ and $E$. Find the length of the line segment $DE$.\n",
    "completion": "\\textbf{Solution:} When $y=-14$, $-x^2+3x+4=-14$,\\\\\nrearrange to get: $x^2-3x-18=0$,\\\\\nsolving gives $x_1=-3$, $x_2=6$,\\\\\n$\\therefore DE=\\left|x_1-x_2\\right|=\\left|-3-6\\right|=\\boxed{9}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52536660_153.png"
  },
  {
    "solution": "\\boxed{x\\leq 3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: The graph of the linear function $y=kx+b$ (where $k$ and $b$ are constants) is shown in the figure. What is the solution set of the linear inequality in one variable $kx+b\\ge 0$ with respect to $x$?\n",
    "completion": "\\textbf{Solution:} When $x\\leq 3$, $y\\geq 0$, \\\\\nhence the solution set of the linear inequality in one variable $kx+b\\geq 0$ with respect to $x$ is $\\boxed{x\\leq 3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50447097.png"
  },
  {
    "solution": "\\boxed{(2,0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure: It is known that the line $y_{1}=kx+b$ passes through points $A(5,0)$ and $B(1,4)$, and intersects with line $y_{2}=2x-4$ at point $C$. Find the equation of line $y_{1}$ as well as the coordinates of point $D$, the intersection point of $y_{2}$ and the $x$-axis.\n",
    "completion": "\\textbf{Solution:} Based on the given problem, we have:\n\\[\n\\left\\{\n\\begin{array}{l}\n5k+b=0 \\\\\nk+b=4\n\\end{array}\n\\right.\n\\]\nSolving this, we find:\n\\[\n\\left\\{\n\\begin{array}{l}\nk=-1 \\\\\nb=5\n\\end{array}\n\\right.\n\\]\nThus, the equation for $y_{1}$ is $y_{1}=-x+5$. Setting $y_{2}=0$ in $y_{2}=2x-4$, we get:\n$2x-4=0$,\nhence $x=2$,\nwhich means the coordinates of point D are $\\boxed{(2,0)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50561230_53.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=x+b$ intersects the $x$-axis at point $C(4,0)$ and the $y$-axis at point $B$, and intersects the hyperbola $y=\\frac{m}{x}$ (where $x<0$) at point $A(-1,n)$. Connect $OA$. What is the area of $\\triangle OAC$?\n",
    "completion": "\\textbf{Solution:} Since point $B$ passes through the y-axis,\\\\\n$ \\therefore x=0$,\\\\\n$ \\therefore y=-4$,\\\\\n$ \\therefore {S}_{\\triangle AOC}={S}_{\\triangle AOB}+{S}_{\\triangle BOC}$,\\\\\n$ =\\frac{1}{2}\\times 4\\times 1+\\frac{1}{2}\\times 4\\times 4$,\\\\\n$ =\\boxed{10}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51379219_204.png"
  },
  {
    "solution": "\\boxed{8\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the rhombus $ABCD$, $AB=8$, $\\angle ADC=120^\\circ$. What is the length of $AC$?\n",
    "completion": "\\textbf{Solution:} Connect AC and BD, where AC intersects BD at point O. \\\\\nSince quadrilateral ABCD is a rhombus, \\\\\nit follows that AB $\\parallel$ CD, AB = AD, AC$\\perp$BD, and AO = OC = $\\frac{1}{2}$AC, \\\\\nSince $\\angle$ADC = $120^\\circ$, \\\\\nit follows that $\\angle$DAB = $180^\\circ - 120^\\circ = 60^\\circ$, \\\\\nSince AB = AD, \\\\\nit follows that $\\triangle$ABD is an equilateral triangle, \\\\\nSince AC$\\perp$BD, \\\\\nit follows that $\\angle$OAB = $\\frac{1}{2}$$\\angle$BAD = $30^\\circ$, $\\angle$AOB = $90^\\circ$, \\\\\nthus OB = $\\frac{1}{2}$AB = 4, \\\\\nthus AO = $\\sqrt{AB^{2} - OB^{2}} = \\sqrt{8^{2} - 4^{2}} = 4\\sqrt{3}$, \\\\\nthus AC = 2AO = $\\boxed{8\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52381887.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $\\triangle ABC$ is an equilateral triangle with a side length of 6, and point D is on side $BC$ with $BD=2$. Construct an equilateral triangle $ADE$ with $AD$ as its side. Through point E, draw $EF\\parallel BC$ intersecting $AC$ at point F. Connect $BF$. What is the area $S_{\\text{quadrilateral }BDEF}$?\n",
    "completion": "\\textbf{Solution:} Draw $EC$, and construct $CH\\perp EF$ at $H$,\\\\\nsince $\\triangle ABC$ and $\\triangle ADE$ are both equilateral triangles,\\\\\ntherefore $AB=AC$, $AD=AE$, $\\angle BAC=\\angle DAE=\\angle ABC=\\angle ACB=60^\\circ$,\\\\\nthus $\\angle BAD=\\angle CAE$,\\\\\nin $\\triangle BAD$ and $\\triangle CAE$,\\\\\n$\\left\\{\\begin{array}{l}\nAB=AC\\\\\n\\angle BAD=\\angle CAE\\\\\nAD=AE\n\\end{array}\\right.$,\\\\\nthus $\\triangle BAD \\cong \\triangle CAE(SAS)$,\\\\\nthus $BD=EC=2$, $\\angle ACE=\\angle ABD=60^\\circ$,\\\\\nsince $EF\\parallel BC$,\\\\\nthus $\\angle EFC=\\angle ACB=60^\\circ$,\\\\\nthus $\\triangle EFC$ is an equilateral triangle,\\\\\ntherefore $CH=\\sqrt{3}$, $EF=EC=BD=2$,\\\\\nsince $EF\\parallel BD$,\\\\\nthus quadrilateral $BDEF$ is a parallelogram,\\\\\nthus the area of quadrilateral $BDEF=BD\\cdot CH=2\\sqrt{3}$.\\\\\nHence, the answer is: $S_{\\text{quadrilateral}BDEF}=\\boxed{2\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53538342.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, D and E are points on sides AB and BC of $\\triangle ABC$, respectively, and $DE\\parallel AC$. AE and CD intersect at point O. If $\\frac{S_{\\triangle DOE}}{S_{\\triangle COA}}=\\frac{1}{25}$, what is the value of $\\frac{S_{\\triangle BDE}}{S_{\\triangle CDE}}$?\n",
    "completion": "\\textbf{Solution:} Given that $DE\\parallel AC$,\\\\\n$\\therefore \\triangle DEO\\sim \\triangle CAO$,\\\\\n$\\therefore \\frac{{S}_{\\triangle DEO}}{{S}_{\\triangle CAO}}=\\left(\\frac{DE}{AC}\\right)^{2}=\\frac{1}{25}$,\\\\\n$\\therefore \\frac{DE}{AC}=\\frac{BE}{BC}=\\frac{1}{5}$,\\\\\n$\\therefore \\frac{BE}{EC}=\\frac{1}{4}$,\\\\\n$\\therefore \\frac{{S}_{\\triangle BED}}{{S}_{\\triangle DEC}}=\\boxed{\\frac{1}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51298872.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within a grid composed of small squares with side length 1, there is a circle $\\odot O$ with radius 1 centered on a grid point. What is the value of the tangent of $\\angle AED$?\n",
    "completion": "\\textbf{Solution:} According to the inscribed angle theorem, we have $\\angle AED = \\angle ABC$, thus $\\tan\\angle AED = \\tan\\angle ABC = \\frac{AC}{AB} = \\boxed{\\frac{1}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51212618.png"
  },
  {
    "solution": "\\boxed{75}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, there is a slope AB with the height of the top B from the ground BC being 30m. The gradient of slope AB, \\(i = 1 \\colon 2.5\\), what is the horizontal distance AC of this slope?\n",
    "completion": "\\textbf{Solution:} Since the slope $AB$ has a gradient $i = 1:2.5$, \\\\\nhence $BC:AC = 1:2.5$, \\\\\nsince $BC = 30m$, \\\\\ntherefore $AC = 30 \\times 2.5 = \\boxed{75}m$",
    "image_path": "dataset/mm_math/images/MM_Math/52537007.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $AB\\perp BD$, $ED\\perp BD$, $C$ is the midpoint of line segment $BD$, and $AC\\perp CE$, with $ED=1$, $BD=4$. What is the length of $AB$?\n",
    "completion": "\\textbf{Solution:} Since $AB\\perp BD$ and $ED\\perp BD$,\\\\\nit follows that $\\angle B=\\angle D=90^\\circ$,\\\\\nwhich means $\\angle A+\\angle ACB=90^\\circ$,\\\\\nSince $AC\\perp CE$,\\\\\nwe have $\\angle ACE=90^\\circ$,\\\\\ntherefore $\\angle ECD+\\angle ACB=90^\\circ$,\\\\\nthus $\\angle A=\\angle ECD$,\\\\\nwhich implies $\\triangle ABC\\sim \\triangle CDE$,\\\\\nhence $\\frac{AB}{CD}=\\frac{BC}{DE}$\\\\\nGiven $C$ is the midpoint of segment $BD$, $ED=1$, and $BD=4$,\\\\\nwe find $BC=CD=2$,\\\\\ntherefore $\\frac{AB}{2}=\\frac{2}{1}$\\\\\nleading to $AB=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53009947.png"
  },
  {
    "solution": "\\boxed{20^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\square ABCD$, $ \\angle D=100^\\circ$. AC is the diagonal. If $ \\triangle ACD$ is rotated clockwise by a certain angle around point A to obtain $ \\triangle AFE$, such that the corresponding point E of point D falls on side AB, and the corresponding point F of point C falls on the extension of side CB, then what is the degree measure of $ \\angle EFB$?\n",
    "completion": "\\textbf{Solution:} In parallelogram $ABCD$, $\\angle D=100^\\circ$,\\\\\n$\\therefore \\angle DAB=180^\\circ - \\angle D=80^\\circ$,\\\\\n$\\because \\triangle ACD$ is rotated clockwise around point $A$ by a certain angle to obtain $\\triangle AEF$,\\\\\n$\\therefore AF=AC$, $\\angle FAE=\\angle CAD$, $\\angle AFE=\\angle ACD$,\\\\\n$\\therefore \\angle FAC=\\angle FAE+\\angle BAC=\\angle CAD+\\angle BAC=\\angle BAD=80^\\circ$\\\\\n$\\therefore \\angle AFC=\\angle ACF=\\frac{1}{2}\\left(180^\\circ-\\angle FAC\\right)=50^\\circ$\\\\\n$\\because AD \\parallel BC$,\\\\\n$\\therefore \\angle DAC=\\angle ACF=50^\\circ$,\\\\\n$\\therefore \\angle ACD=180^\\circ-\\angle D-\\angle CAD=180^\\circ-100^\\circ-50^\\circ=30^\\circ$,\\\\\n$\\therefore \\angle AFE=\\angle ACD=30^\\circ$,\\\\\n$\\therefore \\angle EFB=\\angle AFC-\\angle AFE=50^\\circ-30^\\circ=\\boxed{20^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/55595071.png"
  },
  {
    "solution": "\\boxed{80^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, the perpendicular bisectors of sides $AB$ and $AC$ respectively meet $BC$ at $D$ and $E$. If $\\angle BAC=130^\\circ$, find the measure of $\\angle DAE$.\n",
    "completion": "\\textbf{Solution:} Since $\\angle BAC=130^\\circ$,\\\\\nthen $\\angle B+\\angle C=50^\\circ$,\\\\\nSince $AD=BD$, $CE=AE$,\\\\\nthen $\\angle BAD=\\angle B$, $\\angle EAC=\\angle C$,\\\\\nthus $\\angle BAD+\\angle EAC=50^\\circ$,\\\\\nhence $\\angle DAE=\\angle BAC-(\\angle BAD+\\angle EAC)=130^\\circ-50^\\circ=\\boxed{80^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53044271_86.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated, quadrilateral $ABCD$ is inscribed in circle $\\odot O$, with $AB$ being the diameter of circle $\\odot O$. Let $DE \\perp AB$ at point $E$, intersecting $AC$ at point $F$. It is given that arc $\\overset{\\frown}{BC} = \\overset{\\frown}{AD}$. We have $\\triangle ABC \\sim \\triangle DAE$. When $AD = \\sqrt{5}$ and $BE = 4$, what is the length of $CD$?\n",
    "completion": "\\textbf{Solution:} Draw $CH\\perp AB$ at point H.\\\\\nSince $\\triangle ABC\\sim \\triangle DAE$,\\\\\nit follows that $\\frac{AE}{BC}=\\frac{AD}{AB}$, $\\angle BAC=\\angle ADE$,\\\\\nhence $\\frac{AE}{\\sqrt{5}}=\\frac{\\sqrt{5}}{4+AE}$,\\\\\nthus $AE=1$ (negative values are discarded),\\\\\ntherefore $AB=AE+BE=5$,\\\\\n$DE=\\sqrt{AD^{2}-AE^{2}}=\\sqrt{5-1}=2$,\\\\\ntherefore $AC=\\sqrt{AB^{2}-BC^{2}}=\\sqrt{25-5}=2\\sqrt{5}$,\\\\\nsince $\\overset{\\frown}{BC}=\\overset{\\frown}{AD}$,\\\\\nit follows that $\\angle ACD=\\angle BAC$, hence $\\angle ADE=\\angle ACD$,\\\\\nalso, since $\\angle DAF=\\angle DAC$,\\\\\nit follows that $\\triangle DAF\\sim \\triangle CAD$,\\\\\nthus $\\frac{AD}{AC}=\\frac{AF}{AD}=\\frac{DF}{DC}$,\\\\\nhence $AF=\\frac{\\sqrt{5}}{2}$, $DC=2DF$,\\\\\nthus $EF=\\sqrt{AF^{2}-AE^{2}}=\\frac{1}{2}$,\\\\\ntherefore $DF=\\frac{3}{2}$,\\\\\ntherefore $CD=2DF=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52935324_172.png"
  },
  {
    "solution": "\\boxed{\\sqrt{10}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, $A(0,4)$, $B(2,4)$, $C(4,2)$. What is the radius of this circle?\n",
    "completion": "\\textbf{Solution:} The radius of the circle is $\\boxed{\\sqrt{10}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52623631_56.png"
  },
  {
    "solution": "\\boxed{50^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, PA and PB are tangents to the circle $\\odot O$ with points A and B as the points of tangency. AC is a diameter of $\\odot O$. If $\\angle BAC=25^\\circ$, what is the degree measure of $\\angle P$?\n",
    "completion": "\\textbf{Solution:} Connect OB, \\\\\n$\\because$ PA, PB are the tangents to circle $O$ with A and B as the points of tangency, \\\\\n$\\therefore$ $\\angle OAP = \\angle OBP = 90^\\circ$, \\\\\n$\\because$ $\\angle BAC = 25^\\circ$, \\\\\n$\\therefore$ $\\angle BOC = 50^\\circ$, \\\\\n$\\therefore$ $\\angle AOB = 180^\\circ - 50^\\circ = 130^\\circ$, \\\\\n$\\therefore$ $\\angle P = 360^\\circ - 90^\\circ - 90^\\circ - 130^\\circ = \\boxed{50^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52966971.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $CD$ is the diameter of $\\odot O$, $AB$ is a chord, $CD\\perp AB$ at $E$, $DE=2$, and $AB=8$. What is the length of $AC$?\n",
    "completion": "\\textbf{Solution:} Draw line OA as shown in the diagram:\\\\\nSince $CD \\perp AB$,\\\\\nthus $AE=BE=\\frac{1}{2}AB=4$,\\\\\nLet $OA=r$, then $OE=OD-DE=r-2$,\\\\\nIn $\\triangle OAE$, $OA^2=OE^2+AE^2$,\\\\\nwhich gives $r^2=(r-2)^2+4^2$,\\\\\nSolving this, we find: $r=5$,\\\\\nthus $CE=2r-DE=10-2=8$,\\\\\nthus $AC=\\sqrt{CE^2+AE^2}=\\sqrt{8^2+4^2}=\\boxed{4\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53043251.png"
  },
  {
    "solution": "\\boxed{y = -x^2 + x}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the Cartesian coordinate system $xOy$, it is known that the parabola $y=ax^{2}+bx$ passes through the points $A(4,0)$ and $B(1,4)$. Point $P$ lies on the parabola and above the line $AB$. What is the equation of the parabola?\n",
    "completion": "\\textbf{Solution:} Substituting $A(4, 0)$ and $B(1, 4)$ into $y = ax^2 + bx$,\\\\\n$\\therefore$ we find the solution.\\\\\n$\\therefore$ The equation of the parabola is: \\boxed{y = -x^2 + x}.",
    "image_path": "dataset/mm_math/images/MM_Math/55533039_81.png"
  },
  {
    "solution": "\\boxed{\\sqrt{2}-1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system \\(xOy\\), given points \\(A(1,0)\\), \\(B(3,0)\\), and \\(C\\) as a moving point in the plane satisfying \\(\\angle ACB=90^\\circ\\), and \\(D\\) as a moving point on the line \\(y=x\\), what is the minimum length of the segment \\(CD\\)?\n",
    "completion": "\\textbf{Solution:} Let E be the midpoint of AB. Draw a perpendicular line from point E to the line y=x, and let D be the foot of the perpendicular. \\\\\n$\\because$ Point A is (1,0), and point B is (3,0), \\\\\n$\\therefore$ OA=1, OB=3, \\\\\n$\\therefore$ OE=2, \\\\\n$\\therefore$ ED=2$\\times\\frac{\\sqrt{2}}{2}=\\sqrt{2}$, \\\\\n$\\because$ $\\angle$ACB$=90^\\circ$, \\\\\n$\\therefore$ point C lies on the circle with diameter AB, \\\\\n$\\therefore$ the minimum length of segment CD is $\\boxed{\\sqrt{2}-1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51433567.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $CD$ is the altitude on side $AB$, $BE$ bisects $\\angle ABC$ and intersects $CD$ at point $E$, $BC=5$. If the area of $\\triangle BCE$ is $5$, what is the length of $ED$?\n",
    "completion": "\\textbf{Solution:} Draw EF $\\perp$ BC at F through E, \\\\\nsince CD is the altitude from AB, and BE bisects $\\angle ABC$, intersecting CD at point E, \\\\\nthus DE=EF, \\\\\nsince the area of $\\triangle BCE= \\frac{1}{2}\\times$BC$\\times$EF=5, \\\\\nthus $\\frac{1}{2}\\times$5$\\times$EF=5, \\\\\ntherefore, EF=DE=\\boxed{2}",
    "image_path": "dataset/mm_math/images/MM_Math/51545444.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the figure, point $A(-1, m)$ lies on the line $y=2x+3$. When connecting $OA$ and $\\angle AOB=90^\\circ$, point $B$ is on the line $y=-x+b$ and $OA=OB$. What is the value of $b$?\n",
    "completion": "\\textbf{Solution:} Substituting point A ($-1$, $m$) into the line equation $y=2x+3$, we get: $m=-2+3=1$,\\\\\nsince $\\angle AOB=90^\\circ$ and $OA=OB$,\\\\\ntherefore, rotating line segment $OA$ around point $O$ clockwise by $90^\\circ$, we obtain line segment $OB$. Hence, the coordinates of point $B$ are ($1$, $1$),\\\\\nsubstituting point $B$ into the line $y=-x+b$, we get: $1=-1+b$,\\\\\ntherefore $b=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52386067.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of $\\odot O$, with chord $CD \\perp AB$ at point $E$. Connect $AC$, $BC$, $BD$, and $OF \\perp AC$ at point $F$, where $OF = 1$. What is the length of $BD$?\n",
    "completion": "\\textbf{Solution:} Since $OF \\perp AC$,\\\\\nit follows that $AF=FC$,\\\\\nSince $OA=OB$,\\\\\nit follows that $OF$ is the median of $\\triangle ABC$,\\\\\nTherefore, $BC=2OF=2$,\\\\\nSince $AB \\perp CD$,\\\\\nit follows that $\\overset{\\frown}{BC}=\\overset{\\frown}{BD}$,\\\\\nTherefore, $BD=BC=2$;\\\\\nHence, the length of $BD$ is $\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51262145_130.png"
  },
  {
    "solution": "\\boxed{24}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in rhombus $ABCD$, the diagonals $AC$ and $BD$ intersect at point $O$. Given $AB=5$ and $AC=8$, what is the area of rhombus $ABCD$?\n",
    "completion": "\\textbf{Solution:} Given that rhombus $ABCD$,\\\\\ntherefore $AC\\perp BD$, $AC=2OA$, $BD=2BO$,\\\\\nthus $OA=4$, $\\angle AOB=90^\\circ$,\\\\\ntherefore $BO=\\sqrt{AB^{2}-OA^{2}}=\\sqrt{5^{2}-4^{2}}=3$,\\\\\nthus $BD=2\\times 3=6$,\\\\\ntherefore the area of the rhombus $ABCD$ is $\\frac{1}{2}AC\\cdot BD=\\frac{1}{2}\\times 8\\times 6=\\boxed{24}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52345439.png"
  },
  {
    "solution": "\\boxed{3\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: The line $l_1$ passing through point $B(2, 0)$ intersects with the line $l_2: y=2x+8$ at point $P(-1, n)$. Try to find the length of $PB$.\n",
    "completion": "\\textbf{Solution:} Construct $PI\\perp x$-axis at point $I$,\n\nsince the coordinates of point $P$ are $(-1, 6)$,\n\nand point $B(2, 0)$,\n\ntherefore $BP=\\sqrt{PI^{2}+BI^{2}}=\\sqrt{6^{2}+3^{2}}=\\boxed{3\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50553735_105.png"
  },
  {
    "solution": "\\boxed{100m}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB \\parallel PQ$, $AB = 80 \\, \\text{m}$, $PQ = 100 \\, \\text{m}$, and points $P$, $A$, $C$ are on the same line, as are points $Q$, $B$, $C$. If the distance between $AB$ and $PQ$ is $20 \\, \\text{m}$, what is the distance from point $C$ to line $PQ$ in meters?\n",
    "completion": "\\textbf{Solution:} Given that $AB\\parallel PQ$ and $AB=80m$, $PQ=100m$, \\\\\nthus $\\triangle CAB\\sim \\triangle CPQ$, \\\\\nhence, $\\frac{AB}{PQ}=\\frac{CB}{CQ}$, that is $\\frac{CB}{CQ}=\\frac{80}{100}=\\frac{4}{5}$, \\\\\nas shown in the figure, draw a line through point $C$ perpendicular to the extension of $PQ$, with the foot at point $D$, intersecting the extension of $AB$ at point $E$, \\\\\nsince the distance between $AB$ and $PQ$ is $20m$, that is $DE=20m$, and according to the drawing, $CE\\perp AB$, $CE\\perp PD$, $AB\\parallel PQ$, thus $BE\\parallel PD$, \\\\\ntherefore $\\triangle CBE\\sim \\triangle CQD$, \\\\\ntherefore $\\frac{CB}{CQ}=\\frac{CE}{CD}=\\frac{4}{5}$, \\\\\nsince $CD=CE+ED=CE+20$, \\\\\nthus $\\frac{CE}{CE+20}=\\frac{4}{5}$, \\\\\ntherefore $CE=80$, then $CD=20+80=\\boxed{100m}$, which means the distance from point $C$ to line $PQ$ is $100m$.",
    "image_path": "dataset/mm_math/images/MM_Math/52827684.png"
  },
  {
    "solution": "\\boxed{y=\\frac{1}{2}x^{2}-x+2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the parabola $y=\\frac{1}{2}x^{2}+bx+c$ passes through points $A\\left(0, 2\\right)$ and $B\\left(1, \\frac{3}{2}\\right)$. What is the equation of the parabola?\n",
    "completion": "\\textbf{Solution:} Let us substitute points $A\\left(0, 2\\right)$ and $B\\left(1, \\frac{3}{2}\\right)$ into $y=\\frac{1}{2}x^{2}+bx+c$,\\\\\nwe get: $\\begin{cases}c=2\\\\ \\frac{1}{2}+b+c=\\frac{3}{2}\\end{cases}$,\\\\\nSolving these, we find: $\\begin{cases}b=-1\\\\ c=2\\end{cases}$,\\\\\nTherefore, the equation of the parabola is $\\boxed{y=\\frac{1}{2}x^{2}-x+2}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53033041_156.png"
  },
  {
    "solution": "\\boxed{B(2,2)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, line $l_1$ passes through points $A(0, 4)$ and $D(4, 0)$, line $l_2\\colon y=\\frac{1}{2}x+1$ intersects the $x$-axis at point $C$, and the two lines $l_1$ and $l_2$ intersect at point $B$. Find the equation of line $l_1$ and the coordinates of point $B$.\n",
    "completion": "Solution: Let the equation of line $l_1$ be $y=kx+b$,\\\\\nAccording to the problem, we have $\\begin{cases}b=4 \\\\ 4k+b=0\\end{cases}$,\\\\\nSolving these equations gives: $\\begin{cases}k=-1 \\\\ b=4\\end{cases}$,\\\\\nTherefore, the equation of line $l_1$ is $y=-x+4$;\\\\\nBy combining $\\begin{cases}y=-x+4 \\\\ y=\\frac{1}{2}x+1\\end{cases}$,\\\\\nWe find: $\\begin{cases}x=2 \\\\ y=2\\end{cases}$,\\\\\n$\\therefore$The coordinates of point \\boxed{B(2,2)};",
    "image_path": "dataset/mm_math/images/MM_Math/52056352_194.png"
  },
  {
    "solution": "\\boxed{\\frac{m+n}{n}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, fold the rectangular paper ABCD (AD > DC) at point A along the line passing through point D, making point A land on the side BC, with the landing point being E, and the fold line intersecting side AB at point F. If $BE: EC = m: n$, find $AF: FB$ (expressed as an algebraic expression involving $m$ and $n$).\n",
    "completion": "\\textbf{Solution:} From the given conditions, we can derive: $\\angle EFD=\\angle A=\\angle B=\\angle C=90^\\circ$\\\\\n$\\therefore$ $\\angle BEF+\\angle DEC=90^\\circ$, $\\angle DEC+\\angle EDC=90^\\circ$\\\\\n$\\therefore$ $\\angle BEF=\\angle EDC$\\\\\n$\\therefore$ $\\triangle BEF\\sim \\triangle CDE$\\\\\n$\\therefore$ $\\frac{EF}{DE}=\\frac{BF}{CE}$, that is $\\frac{EF}{BF}=\\frac{DE}{CE}$\\\\\nSince $BE\\colon EC=m\\colon n$\\\\\nLet $BE=mk$, $EC=nk$\\\\\nThen $DE=AD=BC=BE+CE=mk+nk$\\\\\n$\\therefore$ $\\frac{EF}{BF}=\\frac{DE}{CE}=\\frac{mk+nk}{nk}=\\frac{m+n}{n}$\\\\\nFrom the property of folding, we obtain: $AF=EF$\\\\\n$\\therefore$ $AF\\colon FB=\\boxed{\\frac{m+n}{n}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51155328_178.png"
  },
  {
    "solution": "\\boxed{\\frac{n^\\circ}{{2}^{2023}}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, given that $\\angle A=n^\\circ$, if point $P_{1}$ is the intersection of the angle bisector of $\\angle ABC$ and the external angle bisector of $\\angle ACE$, point $P_{2}$ is the intersection of the angle bisector of $\\angle P_{1}BC$ and the external angle bisector of $\\angle P_{1}CE$, point $P_{3}$ is the intersection of the angle bisector of $\\angle P_{2}BC$ and the external angle bisector of $\\angle P_{2}CE$, and so on, then what is the degree measure of $\\angle P_{2023}$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle ACE = \\angle ABC + \\angle A$, $\\angle P_{1}CE = \\angle P_{1}BC + \\angle P_{1}$, and point $P_{1}$ is the intersection of the angle bisector of $\\angle ABC$ and the exterior angle $\\angle ACE$,\\\\\n$\\therefore \\angle ACE = 2\\angle P_{1}CE$, $\\angle ABC = 2\\angle P_{1}BC$,\\\\\n$\\therefore \\angle ABC + \\angle A = 2(\\angle P_{1}BC + \\angle P_{1})$,\\\\\n$\\therefore \\angle P_{1} = \\frac{1}{2}\\angle A = \\frac{1}{2}n^\\circ$; By the same reasoning: $\\angle P_{2} = \\frac{1}{2}\\angle P_{1} = \\frac{1}{2} \\times \\frac{1}{2}n^\\circ = \\frac{n^\\circ}{{2}^{2}}$, ..., $\\angle P_{m} = \\frac{n^\\circ}{{2}^{m}}$;\\\\\n$\\therefore \\angle P_{2023} = \\frac{n^\\circ}{{2}^{2023}}$.\\\\\nHence, the answer is: $\\boxed{\\frac{n^\\circ}{{2}^{2023}}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55426603.png"
  },
  {
    "solution": "\\boxed{\\frac{8}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, DP is the tangent of the circle $\\odot O$, with D being the point of tangency. The chord AB is parallel to DP. Extend BO to intersect $\\odot O$ at point C and DP at point E. Extend AC to intersect DP at point F. Connect OD. AF is parallel to OD; If OD = 5 and AB = 8, find the length of the line segment EF.\n",
    "completion": "\\textbf{Solution:}\n\nGiven that OH $\\perp$ AB, and AB = 8, \\\\\n$\\therefore$ BH = AH = 4, \\\\\n$\\therefore$ OH = $\\sqrt{OB^{2} - BH^{2}}$ = $\\sqrt{5^{2} - 4^{2}}$ = 3, \\\\\nGiven that BH $\\parallel$ ED, \\\\\n$\\therefore$ $\\triangle$ BOH $\\sim$ $\\triangle$ EOD, \\\\\n$\\therefore$ $\\frac{BH}{ED}$ = $\\frac{OH}{OD}$, that is, $\\frac{4}{ED}$ = $\\frac{3}{5}$, \\\\\nSolving this yields: ED = $\\frac{20}{3}$, \\\\\nGiven that $\\angle$BAC = $90^\\circ$, DH $\\perp$ AB, DH $\\perp$ DP, \\\\\n$\\therefore$ Quadrilateral AFDH is a rectangle, \\\\\n$\\therefore$ DF = AH = 4, \\\\\n$\\therefore$ EF = ED - DF = $\\frac{20}{3} - 4$ = $\\boxed{\\frac{8}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51402858_156.png"
  },
  {
    "solution": "\\boxed{18}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $C$ is a moving point on line segment $AB$. Isoceles right triangles $\\triangle ACD$ and $\\triangle CBE$ are constructed on the same side of $AB$ with $AC$ and $BC$ as their hypotenuses, respectively. Line segment $DE$ is drawn, and point $F$ is located on $DE$. Line segments $CF$ and $FD=FC$ are drawn. Given that $AB=12$, and the sum of the areas of triangles $\\triangle ACD$ and $\\triangle CEB$ is denoted as $S$, find the minimum value of $S$.\n",
    "completion": "\\textbf{Solution:} Let $\\because AB=12$,\\\\\n$\\therefore$let $AC=x$, then $BC=12-x$,\\\\\n$\\because DM\\perp AB$, $EN\\perp AB$,\\\\\n$\\therefore DM=\\frac{1}{2}AC$, $EN=\\frac{1}{2}BC$,\\\\\n$\\therefore S=\\frac{1}{2}AC\\cdot DM+\\frac{1}{2}BC\\cdot EN=\\frac{1}{4}AC^{2}+\\frac{1}{4}BC^{2}=\\frac{1}{4}x^{2}+\\frac{1}{4}(12-x)^{2}$,\\\\\ni.e., $S=\\frac{1}{2}x^{2}-6x+36=\\frac{1}{2}(x-6)^{2}+18$\\\\\n$\\therefore S$'s minimum value is $\\boxed{18}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55588163_86.png"
  },
  {
    "solution": "\\boxed{x < -1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y_{1}=x+b$ intersects with the line $y_{2}=kx-1$ at point $P$, where the $x$-coordinate of point $P$ is $-1$. What is the solution set of the inequality $x+b<kx-1$ in terms of $x$?\n",
    "completion": "\\textbf{Solution:} By observing the graph of the function, it is found that when $x < -1$, the graph of the function $y = x + b$ is below the graph of $y = kx - 1$,\\\\\ntherefore, the solution set of the inequality $x + b < kx - 1$ is $x < -1$.\\\\\nHence, the answer is: $\\boxed{x < -1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50554003.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the quadrilateral ABCD is a rhombus, with diagonals AC and BD intersecting at point O. We have $AC=4\\sqrt{3}$ and $BD=4$. Point P is a moving point on AC, and point E is the midpoint of AB. What is the minimum value of $PD+PE$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect DE,\\\\\nsince quadrilateral ABCD is a rhombus, with diagonals AC and BD intersecting at point O, and AC = $4\\sqrt{3}$, BD = 4,\\\\\nthus AO = $\\frac{1}{2}$AC = $2\\sqrt{3}$, BO = $\\frac{1}{2}$BD = 2, and AC $\\perp$ BD,\\\\\nthus AB = $\\sqrt{AO^{2}+BO^{2}} = \\sqrt{(2\\sqrt{3})^{2}+2^{2}} = 4$,\\\\\nthus AB = AD = BD, meaning $\\triangle ABD$ is an equilateral triangle,\\\\\nsince point E is the midpoint of AB,\\\\\nthus DE $\\perp$ AB,\\\\\nthus DE = $\\sqrt{AD^{2}-AE^{2}} = \\sqrt{4^{2}-2^{2}} = 2\\sqrt{3}$,\\\\\nsince DP+PE $\\geq$ DE,\\\\\nthus the minimum value of PD+PE is the length of DE,\\\\\ni.e., the minimum value of PD+PE is $2\\sqrt{3}$,\\\\\nhence, the answer is: the minimum value of $PD+PE$ is $\\boxed{2\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53599256.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the length of line segment AB is 12. Point C is on line segment AB with BC being $\\frac{1}{3}$ of AB. Point D is the midpoint of AB. What is the length of CD?\n",
    "completion": "\\textbf{Solution:} Since BC = $\\frac{1}{3}$ AB = 4, \\\\\nSince point D is the midpoint of AB, \\\\\nTherefore, BD = 6, \\\\\nTherefore, CD = BD - BC = \\boxed{2}.",
    "image_path": "dataset/mm_math/images/MM_Math/51385719.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In $\\triangle ABC$, $\\angle C=90^\\circ$, $\\tan A=\\frac{3}{4}$, $BC=9$, find the length of $AC$ and the value of $\\sin A$.\n",
    "completion": "\\textbf{Solution}: Given that in $\\triangle ABC$,\\\\\n$\\tan A = \\frac{3}{4}$ and $BC = 9$,\\\\\nthen $\\frac{BC}{AC} = \\frac{3}{4}$,\\\\\nthus $AC = \\boxed{12}$,\\\\\nand $AB = \\sqrt{AC^2 + BC^2} = \\sqrt{9^2 + 12^2} = 15$,\\\\\ntherefore $\\sin A = \\frac{BC}{AB} = \\frac{9}{15} = \\boxed{\\frac{3}{5}}$",
    "image_path": "dataset/mm_math/images/MM_Math/53043063.png"
  },
  {
    "solution": "\\boxed{65^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $ \\triangle ABC \\cong \\triangle DEF$. If $ \\angle A = 65^\\circ$, what is the measure of $ \\angle EDC$?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABC \\cong \\triangle DEF$, \\\\\nit follows that $\\angle EDC = \\angle A = 65^\\circ$, \\\\\ntherefore, $\\angle EDC = \\boxed{65^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51385796.png"
  },
  {
    "solution": "\\boxed{-\\frac{3}{4}x^2+\\frac{3}{2}x+6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, the parabola $y=ax^2+bx+6$ passes through points $A(-2, 0)$ and $B(4, 0)$, and intersects the $y$-axis at point $C$. Point $D$ is a moving point on the parabola with the $x$-coordinate $m(1<m<4)$. Connect $AC$, $BC$, $DB$, and $DC$. Find the equation of the parabola.\n",
    "completion": "\\textbf{Solution:} \nGiven the equation of the intersection of the parabola as: $y=a(x+2)(x-4)=a(x^2-2x-8)=ax^2-2ax-8a$,\\\\\nwe have $-8a=6$. Solving this gives: $a=-\\frac{3}{4}$,\\\\\nthus the equation of the parabola is: $y=-\\frac{3}{4}x^2+\\frac{3}{2}x+6=\\boxed{-\\frac{3}{4}x^2+\\frac{3}{2}x+6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52458461_177.png"
  },
  {
    "solution": "\\boxed{\\left(\\frac{5}{2}, -1\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the vertices A and C of the parallelogram OABC have coordinates (2, 0) and (1, 2), respectively. If $D$ is a moving point on the line segment $AC$ (with $D$ not coinciding with $A$ or $C$), and the area of $\\triangle DBC$ is 3, determine the coordinates of point $D$.\n",
    "completion": "\\textbf{Solution:} Given that $BC=OA=2$, and the area of $\\triangle DBC$ is 3,\\\\\n$\\therefore$ Let the height from $D$ to $BC$ in $\\triangle DBC$ be $h$,\\\\\nthen $\\frac{1}{2}BC\\cdot h=3$,\\\\\nSolving gives: $h=3$,\\\\\n$\\therefore$ The y-coordinate of point $D$ is $2+3=5$ or $2-3=-1$,\\\\\nAlso, since $D$ is a point on line $AC$,\\\\\n$\\therefore$ In the equation $y=-2x+4$,\\\\\nwhen $y=5$, $-2x+4=5$ gives $x=-\\frac{1}{2}$,\\\\\nwhen $y=-1$, $-2x+4=-1$ gives $x=\\frac{5}{2}$,\\\\\n$\\therefore$ The coordinates of point $D$ are $\\boxed{\\left(-\\frac{1}{2}, 5\\right)}$ or $\\boxed{\\left(\\frac{5}{2}, -1\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50682377_129.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y=-x+4$ intersects with the graph of the inverse proportion function $y=\\frac{k}{x}$ at points $A(3, a)$ and $B$, and intersects with the y-axis at point $C$. Find the coordinates of point $B$, and directly write out the area of $\\triangle AOB$.\n",
    "completion": "\\textbf{Solution:} Substitute $(3,1)$ into the inverse proportion function $y=\\frac{3}{x}$, to get $k=3$,\\\\\ntherefore, the analytical expression of the inverse proportion function is: $y=\\frac{3}{x}$,\\\\\nBy solving the system of equations $\\left\\{\\begin{array}{l}y=-x+4 \\\\ y=\\frac{3}{x}\\end{array}\\right.$, we find $\\left\\{\\begin{array}{l}x_1=1 \\\\ y_1=3\\end{array}\\right.\\left\\{\\begin{array}{l}x_2=3 \\\\ y_2=1\\end{array}\\right.$,\\\\\ntherefore, the coordinates of point B are $\\boxed{(3,1)}$,\\\\\nThe area of $\\triangle AOB$ is $\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51353246_280.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point A is on the hyperbola $y=\\frac{4}{x}$, point B is on the hyperbola $y=\\frac{k}{x}$, and $AB\\parallel x$-axis, intersecting the y-axis at point C. If $AB=2AC$, what is the value of $k$?\n",
    "completion": "\\textbf{Solution:} Let us draw AD $\\perp$ x-axis at D through point A, and draw BE $\\perp$ x-axis at E through point B,\\\\\n$\\because$ AB $\\parallel$ x-axis,\\\\\n$\\therefore$ the quadrilateral ACOD is a rectangle, and the quadrilateral BCOE is a rectangle,\\\\\n$\\because$ AB = 2AC,\\\\\n$\\therefore$ BC = 3AC,\\\\\n$\\because$ point A is on the hyperbola $y = \\frac{4}{x}$,\\\\\n$\\therefore$ the area of the rectangle ACOD = 4,\\\\\nSimilarly, the area of the rectangle BCOE = k,\\\\\n$\\therefore$ the area of the rectangle BCOE = $3 \\times$ the area of the rectangle ACOD = 12,\\\\\n$\\therefore$ k = \\boxed{12}",
    "image_path": "dataset/mm_math/images/MM_Math/52601797.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown, $\\triangle ABE\\cong \\triangle ADC\\cong \\triangle ABC$, if $\\angle 1=150^\\circ$, what is the measure of $\\angle \\alpha$?\n",
    "completion": "\\textbf{Solution:} Given that $\\triangle ABE \\cong \\triangle ADC \\cong \\triangle ABC$,\\\\\nit follows that $\\angle E = \\angle DCA$, and $\\angle BAE = \\angle 1 = 150^\\circ$,\\\\\nsince $\\angle EFD = \\angle CFA$, and $\\angle \\alpha + \\angle E + \\angle EFD = \\angle EAC + \\angle DCA + \\angle CFA$\\\\\nit implies that $\\angle EAC = \\angle \\alpha$;\\\\\nGiven that $\\angle EAC = 360^\\circ - \\angle BAE - \\angle 1 = 360^\\circ - 150^\\circ - 150^\\circ = 60^\\circ$,\\\\\nthus, $\\angle \\alpha = \\boxed{60^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53078592.png"
  },
  {
    "solution": "\\boxed{180^\\circ-\\frac{1}{2}\\alpha}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, OC is outside of $\\angle AOB$, and OM, ON are the angle bisectors of $\\angle AOC$ and $\\angle BOC$, respectively. If $\\angle AOB = \\alpha$ and $\\angle BOC = 40^\\circ$, with all other conditions remaining the same, find the value of $\\angle MON$ (expressed in terms of $\\alpha$).\n",
    "completion": "\\textbf{Solution:} When $OM$ and $ON$ are on the same side of $OC$,\\\\\n$\\because$ $\\angle AOB=\\alpha$, $\\angle BOC=40^\\circ$,\\\\\n$\\therefore$ $\\angle AOC=\\angle AOB+\\angle BOC=40^\\circ+\\alpha$,\\\\\n$\\because$ $OM$ bisects $\\angle AOC$,\\\\\n$\\therefore$ $\\angle MOC=\\angle MOA=\\frac{1}{2}\\angle AOC=\\frac{1}{2}\\alpha +20^\\circ$,\\\\\n$\\because$ $ON$ bisects $\\angle BOC$,\\\\\n$\\therefore$ $\\angle BON=\\angle CON=20^\\circ$,\\\\\n$\\therefore$ $\\angle MON=\\angle COM-\\angle CON=\\frac{1}{2}\\alpha +20^\\circ-20^\\circ=\\frac{1}{2}\\alpha$; When $OM$ and $ON$ are on opposite sides of $OC$,\\\\\n$\\because$ $\\angle AOB=\\alpha$, $\\angle BOC=40^\\circ$,\\\\\n$\\therefore$ $\\angle AOC=360^\\circ-\\angle AOB-\\angle BOC=320^\\circ-\\alpha$,\\\\\n$\\because$ $OM$ bisects $\\angle AOC$,\\\\\n$\\therefore$ $\\angle MOC=\\angle MOA=\\frac{1}{2}\\angle AOC=160^\\circ-\\frac{1}{2}\\alpha$,\\\\\n$\\because$ $ON$ bisects $\\angle BOC$,\\\\\n$\\therefore$ $\\angle BON=\\angle CON=20^\\circ$,\\\\\n$\\therefore$ $\\angle MON=\\angle COM+\\angle CON=160^\\circ-\\frac{1}{2}\\alpha +20^\\circ=180^\\circ-\\frac{1}{2}\\alpha$; In summary, the value of $\\angle MON$ is either $\\boxed{\\frac{1}{2}\\alpha}$ or $\\boxed{180^\\circ-\\frac{1}{2}\\alpha}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53010670_128.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{5}}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, $AC=6$, $BC=8$. $\\triangle ABC$ is rotated counterclockwise about point $A$ to obtain $\\triangle AB'C'$, such that point $C'$ falls on side $AB$. When connecting $BB'$, what is the value of $\\sin\\angle BB'C'$?\n",
    "completion": "\\textbf{Solution:} Because $\\angle C=90^\\circ$, $AC=6$, and $BC=8$, \\\\\ntherefore $AB=\\sqrt{AC^{2}+BC^{2}}=10$, \\\\\nAlso, because $AC'=AC=6$, \\\\\ntherefore $BC'=AB-AC'=4$, \\\\\nBecause $B'C'=BC=8$, $\\angle B'C'B=90^\\circ$, \\\\\ntherefore $BB'=\\sqrt{B'C'^{2}+BC'^{2}}=4\\sqrt{5}$, \\\\\ntherefore $\\sin\\angle BB'C'=\\frac{BC'}{BB'}=\\boxed{\\frac{\\sqrt{5}}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51918111.png"
  },
  {
    "solution": "\\boxed{180^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure below, what is the sum of $ \\angle A+\\angle B+\\angle C+\\angle D+\\angle E$?\n",
    "completion": "Extend $BE$ to meet $AC$ at point $G$, as shown in the figure,\\\\\n$\\because \\angle BGC=\\angle A+\\angle B$ and $\\angle GFC=\\angle D+\\angle E$,\\\\\n$\\therefore \\angle A+\\angle B+\\angle C+\\angle D+\\angle E=\\angle BGC+\\angle GFC+\\angle C=180^\\circ$.\\\\\nHence, the answer is: $\\boxed{180^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52827839.png"
  },
  {
    "solution": "\\boxed{Q(6,0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, the graph of the linear function $y=-\\frac{1}{2}x+4$ intersects the $x$-axis and $y$-axis at points $A$ and $B$, respectively, and intersects the graph of the direct proportionality function $y=\\frac{3}{2}x$ at point $C$. Moving point $C$ 1 unit to the right and then 6 units downward yields point $D$. If point $Q$ is a moving point on the $x$-axis, and $\\triangle OQD$ forms an isosceles triangle, directly write the coordinates of point $Q$.\n",
    "completion": "\\textbf{Solution:} Let point Q be $(x,0)$. Since D is $(3,-3)$ and O is $(0,0)$, it follows that $OD^2=(3-0)^2+(-3-0)^2=18$, and $OQ^2=(x-0)^2+(0-0)^2=x^2$. The distance $DQ^2=(x-3)^2+(0+3)^2=x^2-6x+18$. For $\\triangle OQD$ to be an isosceles triangle, there are three cases:\n- When $OD=OQ$, from $18=x^2$, we have $x=\\pm 3\\sqrt{2}$. Hence, $Q(-3\\sqrt{2},0)$ or $Q(3\\sqrt{2},0)$,\n- When $OD=DQ$, from $18=x^2-6x+18$, we get $x=6$ or $x=0$ (coinciding with O, which is disregarded). Therefore, $Q(6,0)$,\n- When $OQ=DQ$, from $x^2=x^2-6x+18$, we obtain $x=3$. Therefore, $Q(3,0)$,\n\nIn summary, when $\\triangle OQD$ is an isosceles triangle, the coordinates of point Q are $\\boxed{Q(-3\\sqrt{2},0)}$, or $\\boxed{Q(3\\sqrt{2},0)}$, or $\\boxed{Q(3,0)}$, or $\\boxed{Q(6,0)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50481662_84.png"
  },
  {
    "solution": "\\boxed{\\frac{24}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, quadrilateral ABCD is a rhombus, with $AC=8$ and $\\tan \\angle DAC= \\frac{3}{4}$. If $DH \\perp AB$ at H, what is the distance from point D to side AB?\n",
    "completion": "\\textbf{Solution:} Let the intersection of AC and BD be O,\\\\\n$\\because$ Quadrilateral ABCD is a rhombus,\\\\\n$\\therefore$ AC$\\perp$BD, $AO=\\frac{1}{2}AC=4$, $BD=2OD$, AB=AD,\\\\\n$\\therefore$ $\\tan\\angle DAC=\\frac{OD}{OA}=\\frac{3}{4}$,\\\\\n$\\therefore$ $OD=3$,\\\\\n$\\therefore$ $AD=\\sqrt{AO^{2}+DO^{2}}=5$, $BD=6$,\\\\\n$\\therefore$ $AB=AD=5$,\\\\\n$\\because$ $S_{\\text{rhombus}ABCD}=\\frac{1}{2}AC\\cdot BD=AB\\cdot DH$,\\\\\n$\\therefore$ $DH=\\frac{AC\\cdot BD}{2AB}=\\boxed{\\frac{24}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51285366.png"
  },
  {
    "solution": "\\boxed{(-2, 2)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the coordinates of points $A$ and $B$ are $(2,0)$ and $(0,-4)$, respectively. If segment $AB$ is rotated $90^\\circ$ clockwise around point $A$ to obtain segment $AC$, then what are the coordinates of point $C$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw CD $\\perp$ x-axis at point D,\\\\\n$\\because$ The coordinates of points A and B are (2, 0) and (0, -4) respectively,\\\\\n$\\therefore$ OA=2, OB=4,\\\\\nFrom the property of rotation, we get: $\\angle BAC=90^\\circ$, AC=AB,\\\\\n$\\because$ $\\angle ADC=\\angle AOB=90^\\circ$,\\\\\n$\\therefore$ $\\angle CAD=\\angle B$,\\\\\n$\\therefore$ $\\triangle ACD \\cong \\triangle BAO$,\\\\\n$\\therefore$ CD=OA=2, AD=BO=4,\\\\\n$\\therefore$ OD=AD−OA=2,\\\\\n$\\therefore$ The coordinates of point C are $\\boxed{(-2, 2)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51628060.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle AOB$, $\\angle AOB=90^\\circ$, with vertex $A$ on the graph of the inverse proportion function $y=\\frac{1}{x}$ $(x>0)$ and vertex $B$ on the graph of $y=\\frac{-3}{x}$ $(x<0)$, then what is the degree measure of $\\angle BAO$?\n",
    "completion": "\\textbf{Solution:} Draw $AC \\perp x$-axis at $C$, and draw $BD \\perp x$-axis at $D$,\\\\\nthen $\\angle BDO = \\angle ACO = 90^\\circ$,\\\\\nsince the vertex $A$ is on the graph of the inverse proportion function $y=\\frac{1}{x}\\left(x>0\\right)$ and vertex $B$ is on the graph of $y=\\frac{-3}{x}\\left(x<0\\right)$,\\\\\nthus, the area ${S}_{\\triangle AOC} = \\frac{1}{2} \\cdot {x}_{A} \\cdot {y}_{A} = \\frac{1}{2} \\times 1 = \\frac{1}{2}$, and the area ${S}_{\\triangle BDO} = \\frac{1}{2} \\cdot |\\,{x}_{B} \\cdot {y}_{B}| = \\frac{1}{2} \\times |-3| = \\frac{3}{2}$,\\\\\nsince $\\angle AOB=90^\\circ$,\\\\\nthus $\\angle BOD + \\angle DBO = \\angle BOD + \\angle AOC = 90^\\circ$,\\\\\ntherefore, $\\angle DBO = \\angle AOC$,\\\\\ntherefore, $\\triangle BDO \\sim \\triangle OCA$,\\\\\ntherefore, $\\frac{{S}_{\\triangle BOD}}{{S}_{\\triangle OAC}} = \\left(\\frac{OB}{OA}\\right)^2 = \\frac{\\frac{3}{2}}{\\frac{1}{2}} = 3$,\\\\\ntherefore, $\\frac{OB}{OA} = \\sqrt{3}$,\\\\\ntherefore, $\\tan\\angle BAO = \\frac{OB}{OA} = \\sqrt{3}$,\\\\\ntherefore, $\\angle BAO = \\boxed{60^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53043676.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, point O is the origin. The line $y=2x-1$ and the line $y=\\frac{3}{4}x+\\frac{3}{2}$ intersect at point A. A perpendicular line to the x-axis is drawn from point A, and point B is the foot of the perpendicular. The x-coordinate of point C is $-1$, and point C lies on the line $y=2x-1$. Connect points B and C. What is the degree measure of $\\angle CBO$?\n",
    "completion": "\\textbf{Solution:} Construct $CD\\perp x$-axis at D, \\\\\n$\\because A(2,3)$, \\\\\n$\\therefore B(2,0)$, \\\\\n$\\because$ point C has an x-coordinate of $-1$ and lies on the line $y=2x-1$, \\\\\n$\\therefore y=2\\times(-1)-1=-3$, \\\\\n$\\therefore C(-1,-3)$, \\\\\n$\\therefore BD=3$, $CD=3$, \\\\\n$\\therefore$ triangle $CBD$ is an isosceles right triangle, \\\\\n$\\therefore \\angle CBO=\\boxed{45^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50676594_165.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the same plane, if one side of a regular hexagon and a square, both with equal side lengths, coincide, then what is the measure of $\\angle 1$?\n",
    "completion": "\\textbf{Solution:} To find: $\\angle 1=\\frac{(6-2)\\times 180^\\circ}{6}-\\frac{(4-2)\\times 180^\\circ}{4}$\\\\\n$=120^\\circ-90^\\circ$\\\\\n$=\\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52082439.png"
  },
  {
    "solution": "\\boxed{\\frac{4039}{4040}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Arrange 2020 squares with side length 1 as shown in the figure. The points A, $A_{1}$, $A_{2}$, $A_{3}$, \\ldots, $A_{2020}$ and points M, $M_{1}$, $M_{2}$, \\ldots, $M_{2019}$ are the vertices of the squares. The lines connecting AM$_{1}$, AM$_{2}$, AM$_{3}$, \\ldots, AM$_{2019}$ intersect the sides of the squares $A_{1}M$, $A_{2}M_{1}$, $A_{3}M_{2}$, \\ldots, $A_{2019}M_{2018}$ at the points $N_{1}$, $N_{2}$, $N_{3}$, \\ldots, $N_{2019}$, respectively. The area of the quadrilateral $M_{1}N_{1}A_{1}A_{2}$ is denoted as $S_{1}$, the area of the quadrilateral $M_{2}N_{2}A_{2}A_{3}$ is denoted as $S_{2}$, \\ldots, then what is the area of $S_{2019}$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, assume the upper left vertex of the leftmost square is denoted as O.\\\\\n$\\because$ The 2019 squares, each with a side length of 1, are arranged as illustrated.\\\\\n$\\therefore$ OA$\\parallel$MA$_{1}$$\\parallel$M$_{1}$A$_{2}$$\\parallel$M$_{2}$A$_{3}$$\\parallel$…$\\parallel$M$_{2018}$A$_{2019}$.\\\\\n$\\therefore$ $\\triangle$M$_{1}$MN$_{1}$$\\sim$$\\triangle$M$_{1}$OA.\\\\\n$\\therefore$ $\\frac{MM_{1}}{OM_{1}}=\\frac{MN_{1}}{OA}=\\frac{1}{2}$.\\\\\n$\\therefore$MN$_{1}$=$\\frac{1}{2}$,\\\\\n$\\therefore$The area of quadrilateral M$_{1}$N$_{1}$A$_{1}$A$_{2}$, S$_{1}$=$1-\\frac{1}{2}\\times1\\times\\frac{1}{2}=\\frac{3}{4}$;\\\\\nSimilarly, it follows that: $ \\frac{M_{1}M_{2}}{OM_{2}}=\\frac{M_{1}N_{2}}{OA}=\\frac{1}{3}$.\\\\\n$\\therefore$The area of quadrilateral M$_{2}$N$_{2}$A$_{2}$A$_{3}$, S$_{2}$=$1-\\frac{1}{2}\\times1\\times\\frac{1}{3}=\\frac{5}{6}$;\\\\\n…\\\\\n$\\therefore$The area of quadrilateral MnNnAnAn$_{+1}$, Sn=$1-\\frac{1}{2(n+1)}=\\frac{2n+1}{2n+2}$.\\\\\n$\\therefore$S$_{2019}$=$\\boxed{\\frac{4039}{4040}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51322836.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $P$ is on the side $AB$ of the square $ABCD$ (not coinciding with points $A$, $B$). Connect $PD$ and rotate the line segment $PD$ $90^\\circ$ clockwise around point $P$ to obtain the line segment $PE$, where $PE$ intersects side $BC$ at point $F$. Connect $BE$, $DF$. If $\\triangle PFD\\sim \\triangle BFP$, find $\\frac{AP}{AB}$?\n",
    "completion": "\\textbf{Solution:} From the given information,\n\n$\\because \\triangle PFD \\sim \\triangle BFP$,\n\n$\\therefore \\frac{PB}{BF} = \\frac{PD}{PF}$,\n\n$\\therefore PD \\cdot BF = PB \\cdot PF$,\n\n$\\because \\angle ADP = \\angle EPB, \\angle CBP = \\angle A = 90^\\circ$,\n\n$\\therefore \\triangle DAP \\sim \\triangle PBF$,\n\n$\\therefore \\frac{PD}{PF} = \\frac{PA}{BF}$,\n\n$\\therefore PD \\cdot BF = AP \\cdot PF$,\n\n$\\therefore PB \\cdot PF = AP \\cdot PF$,\n\n$\\therefore PA = PB$,\n\n$\\because AB = PA + PB = 2PA$,\n\n$\\therefore \\frac{AP}{AB} = \\boxed{\\frac{1}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51322751_129.png"
  },
  {
    "solution": "\\boxed{\\frac{81}{8}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle BAC=120^\\circ$, $AB=AC=6$, $D$ is a moving point on side $AB$ (not coinciding with point $B$). Connect $CD$, and rotate line segment $CD$ $90^\\circ$ counterclockwise around point $D$ to obtain $DE$. Connect $BE$. What is the maximum value of $S_{\\triangle BDE}$?\n",
    "completion": "Solution: Draw a line segment CM$\\perp$AB at M through point C, and draw EN$\\perp$AB at N through point E, \\\\\ntherefore, $\\angle END=\\angle DMC=90^\\circ$, \\\\\ntherefore, $\\angle EDN+\\angle DEN=90^\\circ$, \\\\\nsince $\\angle EDC=90^\\circ$, \\\\\ntherefore, $\\angle EDN+\\angle CDM=90^\\circ$, \\\\\ntherefore, $\\angle DEN=\\angle CDM$, \\\\\nin $\\triangle EDN$ and $\\triangle DCM$, \\\\\n$\\left\\{\\begin{array}{l}\n\\angle DEN=\\angle CDM\\\\\n\\angle END=\\angle DMC\\\\\nED=DC\n\\end{array}\\right.$, \\\\\ntherefore, $\\triangle EDN \\cong \\triangle DCM$ (AAS), \\\\\ntherefore, $EN=DM$, \\\\\nsince $\\angle BAC=120^\\circ$, \\\\\ntherefore, $\\angle MAC=60^\\circ$, \\\\\ntherefore, $\\angle ACM=30^\\circ$, \\\\\ntherefore, $AM=\\frac{1}{2}AC=\\frac{1}{2}\\times 6=3$, \\\\\ntherefore, $BM=AB+AM=6+3=9$, \\\\\nlet $BD=x$, then $EN=DM=9-x$, \\\\\ntherefore, $S_{\\triangle BDE}=\\frac{1}{2}BD\\cdot EN=\\frac{1}{2}x(9-x)=-\\frac{1}{2}(x-4.5)^2+\\frac{81}{8}$, \\\\\ntherefore, when $BD=4.5$, $S_{\\triangle BDE}$ reaches its maximum value of $\\boxed{\\frac{81}{8}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52808497.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the side length of each small square is 1. What is the degree measure of $\\angle ABC$?\n",
    "completion": "\\textbf{Solution:} Connect AC,\\\\\nby the Pythagorean theorem, we have $AC^{2}=2^{2}+1^{2}=5$,\\\\\n$BC^{2}=2^{2}+1^{2}=5$,\\\\\n$AB^{2}=1^{2}+3^{2}=10$,\\\\\n$\\therefore AC^{2}+BC^{2}=5+5=10=AB^{2}$,\\\\\n$\\therefore \\triangle ABC$ is an isosceles right triangle, $\\angle ACB=90^\\circ$,\\\\\n$\\therefore \\angle ABC=\\boxed{45^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53043939.png"
  },
  {
    "solution": "\\boxed{135^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the right-angle vertex $P$ of the set square $ABP$ is on the line $CD$, and points $A$ and $B$ are on the same side of line $CD$. As shown in Figure 2, if $PM$ bisects $\\angle APC$ and $PN$ bisects $\\angle BPD$, find the degree measurement of $\\angle MPN$.\n",
    "completion": "\\textbf{Solution:} Given that $PM$ bisects $\\angle APC$, and $PN$ bisects $\\angle BPD$,\\\\\nit follows that $\\angle APM=\\angle CPM$ and $\\angle BPN=\\angle DPN$.\\\\\nSince $\\angle APB=90^\\circ$,\\\\\nwe have $2\\angle APM+2\\angle BPN=90^\\circ$,\\\\\nwhich leads to $\\angle APM+\\angle BPN=45^\\circ$,\\\\\nthus, $\\angle MPN=\\angle APM+\\angle APB+\\angle BPN=45^\\circ+90^\\circ=\\boxed{135^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53413592_117.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\square ABCD$, E is the midpoint of $AB$, F is on $AD$, and $AF:AD=1:3$, $EF$ intersects $AC$ at G. If $AC=40$, then what is the value of $AG$?\n",
    "completion": "\\textbf{Solution:} Let the midpoint of AC be O, and connect EO, \\\\\nsince E is the midpoint of AB, \\\\\n$\\therefore$EO$\\parallel$BC and EO= $\\frac{1}{2}$ BC, \\\\\nalso AD$\\parallel$BC, \\\\\n$\\therefore$AF$\\parallel$EO, \\\\\n$\\therefore$$\\triangle$AFG$\\sim$$\\triangle$OEG, \\\\\n$\\therefore$ $\\frac{AG}{GO}=\\frac{AF}{OE}$.\\\\\nSince AC=40 and O is the midpoint, \\\\\n$\\therefore$OA=20, then GO=20−AG\\\\\nSince AF$\\colon$ AD=1$\\colon$ 3 and AD=BC, \\\\\n$\\therefore$ $\\frac{AG}{GO}=\\frac{AF}{EO}=\\frac{2}{3}$, \\\\\n$\\therefore$ $\\frac{AG}{OG}=\\frac{AG}{AO−AG}=\\frac{2}{3}$.\\\\\n$\\therefore$ $\\frac{AG}{20−AG}=\\frac{2}{3}$, \\\\\nSolving gives $AG=\\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51458243.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that points B and C divide the line segment AD from left to right into three parts in the ratio $2:4:3$. If M is the midpoint of AD and BM $= 5\\,\\text{cm}$, then what is the length of the line segment MC?\n",
    "completion": "\\textbf{Solution:} Given, let $AB=2x$, $BC=4x$, $CD=3x$\\\\\nSince M is the midpoint of AD,\\\\\nit follows that $AM=MD=\\frac{1}{2}AD=\\frac{1}{2}(2x+4x+3x)=\\frac{9}{2}x$\\\\\nSince $BM=5cm$,\\\\\nit follows that $AD-AB-MC-CD=5$\\\\\nThus, $9x-2x-MC-3x=5$\\\\\nTherefore, $MC=4x-5$\\\\\nSince $MC=MD-CD=\\frac{9}{2}x-3x=\\frac{3}{2}x$,\\\\\nit follows that $4x-5=\\frac{3}{2}x$\\\\\nThus, $\\frac{3}{2}x=4x-5$\\\\\nThus, $x=2$\\\\\nTherefore, $MC=\\frac{3}{2}\\times 2=\\boxed{3}$",
    "image_path": "dataset/mm_math/images/MM_Math/51495484.png"
  },
  {
    "solution": "\\boxed{36}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point B is a point on line segment AC, and \\(AB = 27cm\\), \\(BC = \\frac{1}{3}AB\\). What is the length of line segment \\(AC\\)?\n",
    "completion": "\\textbf{Solution:} Solution: According to the given problem, $AB=27\\text{cm}$, $BC=\\frac{1}{3}AB$,\\\\\n$\\therefore BC=9\\text{cm}$,\\\\\n$\\therefore AC=AB+BC=27+9=\\boxed{36}\\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53206769_125.png"
  },
  {
    "solution": "\\boxed{y=-x+12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the line AC passing through the point $C(0, 12)$ intersects with the line OA at the point $A(8, 4)$. What is the equation of line AC?\n",
    "completion": "\\textbf{Solution:} Let the equation of the line $AC$ be $y=kx+b$, \\\\\nSubstituting $C(0, 12)$ and $A(8, 4)$ into it yields: $\\begin{cases}12=b\\\\ 4=8k+b\\end{cases}$, \\\\\nSolving these equations, we get $\\begin{cases}k=-1\\\\ b=12\\end{cases}$, \\\\\n$\\therefore$ the equation of the line $AC$ is \\boxed{y=-x+12}.",
    "image_path": "dataset/mm_math/images/MM_Math/52690876_64.png"
  },
  {
    "solution": "\\boxed{50^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $\\angle AOE = \\angle COD$, and the ray $OC$ bisects $\\angle BOE$, $\\angle EOD = 30^\\circ$. What is the degree measure of $\\angle AOD$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle AOB=180^\\circ$ and $\\angle EOD=30^\\circ$,\\\\\nit follows that $\\angle AOD+\\angle EOC+\\angle COB=150^\\circ$.\\\\\nSince $\\angle AOE=\\angle COD$,\\\\\nit follows that $\\angle AOD=\\angle EOC$.\\\\\nSince $OC$ bisects $\\angle EOB$,\\\\\nit follows that $\\angle EOC=\\angle COB$,\\\\\nhence $\\angle EOC=\\angle COB=\\angle AOD=\\boxed{50^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51284764.png"
  },
  {
    "solution": "\\boxed{\\frac{6\\sqrt{5}}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ is inscribed in a semicircle $O$ with radius $\\sqrt{5}$, where $AB$ is the diameter. Point $M$ is the midpoint of the arc $\\overset{\\frown}{AC}$. Connecting $BM$ intersects $AC$ at point $E$. $AD$ bisects $\\angle CAB$ and intersects $BM$ at point $D$, and $D$ is the midpoint of $BM$. What is the length of $BC$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, construct $MH\\perp AB$ at $H$, connect $AM$, $OM$, where $OM$ intersects $AC$ at $F$. \\\\\n$\\because AB$ is the diameter,\\\\\n$\\therefore \\angle AMB=90^\\circ$, $\\angle ACB=90^\\circ$,\\\\\n$\\therefore \\angle CAB+\\angle CBA=90^\\circ$,\\\\\n$\\because$ point $M$ is the midpoint of $\\overset{\\frown}{AC}$,\\\\\n$\\therefore \\overset{\\frown}{CM}=\\overset{\\frown}{AM}$\\\\\n$\\therefore \\angle CBM=\\angle ABM$,\\\\\n$\\because \\angle CAD=\\angle BAD$,\\\\\n$\\therefore \\angle DAB+\\angle DBA=\\frac{1}{2}(\\angle CAB+\\angle CBA)=45^\\circ$,\\\\\n$\\therefore \\angle ADB=180^\\circ-(\\angle DAB+\\angle DBA)=135^\\circ$,\\\\\n$\\because \\angle ADM=180^\\circ-\\angle ADB=45^\\circ$,\\\\\n$\\therefore MA=MD$,\\\\\n$\\because DM=DB$,\\\\\n$\\therefore BM=2AM$, let $AM=x$, then $BM=2x$,\\\\\n$\\because AB=2\\sqrt{5}$,\\\\\n$\\therefore x^2+4x^2=20$,\\\\\n$\\therefore x=2$ (the negative root has been discarded),\\\\\n$\\therefore AM=2$, $BM=4$,\\\\\n$\\because \\frac{1}{2}\\cdot AM\\cdot BM=\\frac{1}{2}\\cdot AB\\cdot MH$,\\\\\n$\\therefore MH=\\frac{2\\times 4}{2\\sqrt{5}}=\\frac{4\\sqrt{5}}{5}$,\\\\\n$\\therefore OH=\\sqrt{OM^2-MH^2}=\\sqrt{(\\sqrt{5})^2-\\left(\\frac{4\\sqrt{5}}{5}\\right)^2}=\\frac{3\\sqrt{5}}{5}$,\\\\\n$\\because \\overset{\\frown}{CM}=\\overset{\\frown}{AM}$,\\\\\n$\\therefore OM\\perp AC$,\\\\\n$\\therefore AF=FC$,\\\\\n$\\because OA=OB$,\\\\\n$\\therefore BC=2OF$,\\\\\n$\\because \\angle OHM=\\angle OFA=90^\\circ$, $\\angle AOF=\\angle MOH$, $OA=OM$,\\\\\n$\\therefore \\triangle OAF\\cong \\triangle OMH\\left(AAS\\right)$,\\\\\n$\\therefore OF=OH=\\frac{3\\sqrt{5}}{5}$,\\\\\n$\\therefore BC=2OF=\\boxed{\\frac{6\\sqrt{5}}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53080331.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in circle $O$, chord $AB=8$, point $C$ is on circle $O$ ($C$ does not coincide with $A$ and $B$). Connect $CA$ and $CB$, and draw $OD\\perp AC$ and $OE\\perp BC$ through point $O$, where the feet of the perpendiculars are points $D$ and $E$, respectively. Find the length of line segment $DE$.\n",
    "completion": "\\textbf{Solution:} Since $OD$ passes through the center $O$, and $OD\\perp AC$,\\\\\n$\\therefore AD=DC$,\\\\\nSimilarly, $CE=EB$,\\\\\n$\\therefore DE$ is the median of $\\triangle ABC$,\\\\\n$\\therefore DE=\\frac{1}{2}AB$,\\\\\nSince $AB=8$,\\\\\n$\\therefore DE=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53080238_90.png"
  },
  {
    "solution": "\\boxed{-\\frac{5}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=\\frac{1}{2}x^2-2x+c$ intersects the x-axis at points A and B, and intersects the negative half of the y-axis at point C. The vertex of the parabola is M, and D and E are the midpoints of AB and BM, respectively. If the area ratio of $\\triangle DEB$ to $\\triangle ACD$ is $9:10$, what is the value of $c$?\n",
    "completion": "\\textbf{Solution:} Let $S_{\\triangle DEB} = \\frac{1}{2} \\text{DB} \\cdot |y_E|$, and $S_{\\triangle ADC} = \\frac{1}{2} \\text{AD} \\cdot |y_C|$,\n\\\\\nSince D is the midpoint of AB,\n\\\\\nTherefore, AD = DB,\n\\\\\nAlso, since the area ratio of $\\triangle DEB$ to $\\triangle ACD$ is $9:10$,\n\\\\\nTherefore, $\\frac{|y_E|}{|y_C|} = \\frac{9}{10}$,\n\\\\\nAlso, since E is the midpoint of BM,\n\\\\\nTherefore, $|y_E| = \\frac{1}{2} |y_M|$,\n\\\\\nSubstituting $x = 0$ into the equation yields $y_C = c$,\n\\\\\nTherefore, $|y_E| = \\left| \\frac{9}{10} c \\right|$,\n\\\\\nTherefore, $|y_M| = \\left| \\frac{9}{5} c \\right|$,\n\\\\\nSince $y_M < 0$ and $c < 0$,\n\\\\\nTherefore, $y_M = \\frac{9}{5} c$,\n\\\\\nSince M is the vertex of the parabola,\n\\\\\nTherefore, $x_M = -\\frac{b}{2a} = -\\frac{-2}{2 \\times \\frac{1}{2}} = 2$,\n\\\\\nSubstituting $x = 2$ into the equation yields: $y_M = \\frac{1}{2} \\times 2^2 - 2 \\times 2 + c = c - 2 = \\frac{9}{5} c$,\n\\\\\nSolving this gives $c = \\boxed{-\\frac{5}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52755076.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the circle $\\odot O$, the chord $AC=\\sqrt{15}$. When the minor arc $\\stackrel{\\frown}{AC}$ is folded along $AC$, it intersects the diameter $AB$ at $D$, where $DB=\\frac{1}{2}$. What is the length of the diameter $AB$?\n",
    "completion": "\\textbf{Solution:} Draw $CE\\perp AB$ at point $E$, and connect $CB$, $CD$,\n\\[\n\\because \\overset{\\frown}{AC}=\\overset{\\frown}{ADC},\n\\]\n\\[\n\\therefore \\angle B=\\angle A+\\angle ACD=\\angle CDB,\n\\]\n\\[\n\\therefore CD=CB,\n\\]\n\\[\n\\therefore DE=BE=\\frac{1}{2}BD=\\frac{1}{4};\n\\]\n\\[\n\\because AB \\text{ is the diameter},\n\\]\n\\[\n\\therefore \\angle ACB=\\angle BEC=90^\\circ,\n\\]\n\\[\n\\because \\angle B=\\angle B,\n\\]\n\\[\n\\therefore \\triangle ACE\\sim \\triangle CBE,\n\\]\n\\[\n\\therefore \\frac{CE}{AE}=\\frac{BE}{CE},\n\\]\n\\[\n\\therefore CE^2=AE\\cdot BE=\\frac{1}{4}AE;\n\\]\n\\[\n\\because AE^2=AC^2−AE^2=15−\\frac{1}{4}AE\n\\]\nSolving this, we find $AE=\\frac{15}{4},$\n\\[\n\\therefore AB=AE+BE=\\frac{15}{4}+\\frac{1}{4}=\\boxed{4}.\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/52861190.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{13} - 2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the side length of square $ABCD$ is $4$, and point $E$ is a point inside the square. Connect $BE$ and $CE$, and $\\angle ABE = \\angle BCE$. Point $F$ is a moving point on side $AB$, connecting $FD$ and $FE$. What is the minimum value of the length of $FD+FE$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a square, \\\\\n$\\therefore \\angle ABC = 90^\\circ$, \\\\\n$\\therefore \\angle ABE + \\angle CBE = 90^\\circ$, \\\\\nSince $\\angle ABE = \\angle BCE$, \\\\\n$\\therefore \\angle BCE + \\angle CBE = 90^\\circ$, \\\\\n$\\therefore \\angle BEC = 90^\\circ$, \\\\\n$\\therefore$ point E moves on the semicircle with BC as the diameter, \\\\\nAs shown in the diagram, let the midpoint of BC be O, and construct the square APGB symmetric to square ABCD with respect to line AB, then the corresponding point of D is P, \\\\\nConnect PO, intersecting AB at F and the semicircle O at E, then the length of segment EP is the minimum value of $FD+FE$, OE $= 4$, \\\\\nSince $\\angle G = 90^\\circ$, PG $=$ BG $=$ AB $= 4$, \\\\\n$\\therefore$ OG $= 6$, \\\\\n$\\therefore$ OP $= \\sqrt{PG^2 + OG^2} = \\sqrt{4^2 + 6^2} = 2\\sqrt{13}$, \\\\\n$\\therefore$ EP $= 2\\sqrt{13} - 2$, \\\\\n$\\therefore$ The minimum value of the length of FD+FE is $\\boxed{2\\sqrt{13} - 2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51684446.png"
  },
  {
    "solution": "\\boxed{2^{2021}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: The squares $A_1B_1C_1O$, $A_2B_2C_2C_1$, $A_3B_3C_3C_2$, … are placed as shown in the figure, where points $A_1, A_2, A_3, $… and points $C_1, C_2, C_3, $… are on the line $y=x+1$ and the x-axis, respectively. What is the y-coordinate of point $B_{2022}$?\n",
    "completion": "\\textbf{Solution:} Since point $A_{1}$ is the intersection point of the line $y=x+1$ and the y-axis,\\\\\n$\\therefore A_{1}(0,1)$,\\\\\nbecause the quadrilateral $A_{1}B_{1}C_{1}O$ is a square,\\\\\n$\\therefore B_{1}(1,1)$,\\\\\nbecause point $A_{2}$ is on the line $y=x+1$,\\\\\n$\\therefore A_{2}(1,2)$,\\\\\nSimilarly, we can derive that $A_{3}(3,4)$, $B_{2}(3,2)$, $B_{3}(7,4)$,\\\\\n$\\therefore$ the length of the sides of the first three squares $=1+2+4=7$,\\\\\n$\\therefore A_{4}(7,8)$,\\\\\nbecause $B_{1}(1,1)$, $B_{2}(3,2)$, $B_{3}(7,4)$,\\\\\n$\\therefore$ the y-coordinate of $B_{n}$ is $2^{n-1}$,\\\\\nwhen $n=2022$, $2^{n-1}=2^{2022-1}=2^{2021}$.\\\\\nThus, the answer is: the y-coordinate of $B_{2022}$ is $\\boxed{2^{2021}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52225656.png"
  },
  {
    "solution": "\\boxed{\\left(-2, -1\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the two diagonals $AC$ and $BD$ of the quadrilateral $\\square ABCD$ intersect at the origin $O$ of the Cartesian coordinate system. The coordinates of point $D$ are $\\left(2, 1\\right)$. What are the coordinates of point $B$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral $ABCD$ is a parallelogram, and $O$ is the intersection point of diagonals $AC$ and $BD$,\\\\\nit follows that points $B$ and $D$ are symmetrical with respect to the origin $O$,\\\\\nsince the coordinates of point $D$ are $\\left(2, 1\\right)$,\\\\\nthus, the coordinates of point $B$ are $\\boxed{\\left(-2, -1\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52204756.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABD$, C is a point on $AD$, and $AD = 4CD$, $\\angle CBD = \\angle A$. A line is drawn through point D such that $DH \\parallel AB$, intersecting the extension of $BC$ at point H. We have $\\triangle HCD \\sim \\triangle HDB$. If $BC = 6$, find the length of $DH$.\n",
    "completion": "\\textbf{Solution:} Since $AD=4CD$,\\\\\nit follows that $\\frac{CD}{AC}=\\frac{1}{3}$,\\\\\nGiven $DH\\parallel AB$,\\\\\nit leads to $\\triangle HCD\\sim \\triangle BCA$,\\\\\nwhich implies $\\frac{HC}{BC}=\\frac{CD}{AC}=\\frac{1}{3}$,\\\\\nSince $BC=6$,\\\\\nwe find that $HC=2$ and $HB=HC+BC=8$,\\\\\nBecause $\\triangle HCD\\sim \\triangle HDB$,\\\\\nit can be deduced that $\\frac{HC}{HD}=\\frac{HD}{HB}$, thus $HD^2=HC\\cdot HB=2\\times 8=16$,\\\\\nTherefore, $HD=\\boxed{4}$ (the negative value has been discarded).",
    "image_path": "dataset/mm_math/images/MM_Math/53019683_87.png"
  },
  {
    "solution": "\\boxed{70^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given $\\angle AOB=40^\\circ$, $OD$ is the bisector of $\\angle BOC$. As shown in Figure 1, when $\\angle AOB$ and $\\angle BOC$ are complementary, what is the degree measure of $\\angle COD$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle AOB$ and $\\angle BOC$ are complementary,\\\\\nit follows that $\\angle AOB+\\angle BOC=180^\\circ$,\\\\\ntherefore $\\angle BOC=180^\\circ-40^\\circ=140^\\circ$,\\\\\nsince $OD$ is the bisector of $\\angle BOC$,\\\\\nit results in $\\angle COD=\\frac{1}{2}\\angle BOC=\\boxed{70^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52916444_91.png"
  },
  {
    "solution": "\\boxed{}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the diamond $ABCD$, $AD\\parallel x$ axis, the coordinate of point $A$ is $\\left(0, 4\\right)$, and the coordinate of point $B$ is $\\left(3, 0\\right)$. The line $y_1=mx+n$ on which side $CD$ lies intersects the $x$ axis at point $C$, and intersects the hyperbola $y_2=\\frac{k}{x}\\left(x<0\\right)$ at point $D$. Directly write down the range of the independent variable $x$ for which $y_1\\ge y_2$.\n",
    "completion": "\\textbf{Solution:} The range of the independent variable $x$, which satisfies $y_1 \\geq y_2$, is $x \\leq -5$.",
    "image_path": "dataset/mm_math/images/MM_Math/52250758_125.png"
  },
  {
    "solution": "\\boxed{1\\colon 4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle A'B'C'$ and $\\triangle ABC$ are similar figures with $O$ as the center of similarity. If $OA'=A'A$, then what is the ratio of the areas of $\\triangle A'B'C'$ to $\\triangle ABC$?\n",
    "completion": "\\textbf{Solution:} Given that, $\\triangle ABC$ and $\\triangle A'B'C'$ are similar figures,\\\\\n$\\therefore \\triangle ABC \\sim \\triangle A'B'C'$, and the ratio AB$\\colon$ A'B'=OA$\\colon$ AA'=1$\\colon$ 2,\\\\\n$\\therefore$ the ratio of the areas of $\\triangle A'B'C'$ to $\\triangle ABC$ is: $1\\colon 4$.\\\\\nTherefore, the ratio of the areas is $\\boxed{1\\colon 4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51299759.png"
  },
  {
    "solution": "\\boxed{130^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points A, B, and C are all on circle $O$. When $\\angle OAC = 40^\\circ$, what is the measure of $\\angle B$?\n",
    "completion": "\\textbf{Solution:} Let's select any point D on the major arc AB, and connect AD and CD.\\\\\nSince OA=OC, $\\angle$OAC=40$^\\circ$,\\\\\nit follows that $\\angle$OAC=$\\angle$OCA=40$^\\circ$,\\\\\nwhich implies $\\angle$AOC=180$^\\circ$-40$^\\circ$-40$^\\circ$=100$^\\circ$,\\\\\nthus $\\angle$D=$\\frac{1}{2}$$\\angle$AOC=50$^\\circ$,\\\\\nSince quadrilateral ABCD is circumscribed around $\\odot$O,\\\\\nit follows that $\\angle$D+$\\angle$B=180$^\\circ$,\\\\\ntherefore $\\angle$B=180$^\\circ$-50$^\\circ$=\\boxed{130^\\circ},",
    "image_path": "dataset/mm_math/images/MM_Math/52536701.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{13}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, rectangle $ABCD$, $BF \\perp AC$ intersects $CD$ at point $E$, and intersects the extension line of $AD$ at point $F$. $AB^{2} = BC \\cdot AF$. When $\\frac{BC}{AB} = \\frac{2}{3}$ and $DF = 5$, find the length of $AC$.\n",
    "completion": "\\textbf{Solution:}\nGiven that,\n\\begin{align*}\n\\because \\frac{BC}{AB} &= \\frac{2}{3}, \\\\\n\\therefore \\text{let} BC &= 2x, AB = 3x, \\\\\n\\because AB^2 &= BC \\cdot AF, \\\\\n\\therefore (3x)^2 &= 2x(2x + 5), \\\\\n\\therefore x_1 &= 0, x_2 = 2, \\\\\n\\therefore BC &= 4, AB = 6, \\\\\n\\therefore AC &= \\sqrt{6^2 + 4^2} = 2\\sqrt{13}.\n\\end{align*}\nThus, the length of $AC$ is $\\boxed{2\\sqrt{13}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51209236_87.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $D, E$ are two points on sides $AB, AC$ of $\\triangle ABC$, respectively, and $DE \\parallel BC$, $DE:BC = 1:3$. What is the ratio of $AD:AB$?\n",
    "completion": "\\textbf{Solution:} Since $DE \\parallel BC$, \\\\\nthen $\\triangle ADE \\sim \\triangle ABC$, \\\\\ntherefore $\\frac{AD}{AB} = \\frac{DE}{BC} = \\boxed{\\frac{1}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52977928.png"
  },
  {
    "solution": "\\boxed{16}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, we have $AB=AC$, $AD=BC=8$. Points $E$ and $F$ are on the median $AD$. What is the area of the shaded region?\n",
    "completion": "\\textbf{Solution:} Given $\\because$ AB = AC, AD = BC = 8, AD$\\perp$BC at point D,\\\\\n$\\therefore$ BD=CD=$\\frac{1}{2}$BC=4,\\\\\n$\\therefore$ $\\triangle ABC$ is symmetrical about AD,\\\\\n$\\therefore$ S$_{\\triangle BEF}$=S$_{\\triangle CEF}$,\\\\\n$\\therefore$ the area of the shaded part in the figure = $\\frac{1}{2}$S$_{\\triangle BEF}$=$\\frac{1}{2} \\times \\frac{1}{2} \\times 8 \\times 8 = \\boxed{16}$",
    "image_path": "dataset/mm_math/images/MM_Math/52733745.png"
  },
  {
    "solution": "\\boxed{-3<x<0}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, the line $y=ax+b$ intersects the x-axis at point $A(1, 0)$ and intersects the graph of the function $y=\\frac{k}{x}$ at points B and C. The x-coordinate of point B is 4, and the x-coordinate of point C is $-3$. What is the solution to the system of inequalities $0<ax+b<\\frac{k}{x}$?\n",
    "completion": "\\textbf{Solution:} Observing the graph yields: when $-3<x<0$, the line $y=ax+b$ is above the x-axis and below the graph of the inverse proportion function $y=\\frac{k}{x}$,\\\\\n$\\therefore$ the solution to the system of inequalities $0<ax+b<\\frac{k}{x}$ is $-3<x<0$.\\\\\nThus, the answer is: $\\boxed{-3<x<0}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53102105.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within a square grid, the quadrilateral ABCD is a rhombus. What is the value of $\\tan\\frac{\\angle BAD}{2}$?\n",
    "completion": "\\textbf{Solution}: From the properties of a rhombus, we know that $\\frac{\\angle BAD}{2}=\\angle BAC$\\\\\nFrom the figure, we have $\\tan\\angle BAC=\\frac{3}{4}$\\\\\n$\\therefore \\tan\\frac{\\angle BAD}{2}=\\tan\\angle BAC=\\boxed{\\frac{3}{4}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51446460.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $D$ and $E$ are points on the sides $AB$ and $AC$ of $\\triangle ABC$ respectively, with $\\angle AED=\\angle B$, $AD=3$, $AB=8$, and $AE=4$. What is the length of $AC$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle AED = \\angle B$, and $\\angle A = \\angle A$,\\\\\n$\\therefore \\triangle ADE \\sim \\triangle ACB$,\\\\\n$\\therefore \\frac{AD}{AC} = \\frac{AE}{AB}$,\\\\\nGiven that $AD = 3$, $AB = 8$, $AE = 4$,\\\\\n$\\therefore \\frac{3}{AC} = \\frac{4}{8}$,\\\\\n$\\therefore AC = \\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51446580.png"
  },
  {
    "solution": "\\boxed{130^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within $\\odot O$, points B and C divide $\\overset{\\frown}{AD}$ into three equal parts. The chords $AC$ and $BD$ intersect at point E. $AC=BD$; $AB$ is connected. If $\\angle BAC=25^\\circ$, what is the degree measure of $\\angle BEC$?\n",
    "completion": "\\textbf{Solution:} Given $\\because \\angle BAC=25^\\circ$, $\\overset{\\frown}{AB}=\\overset{\\frown}{BC}=\\overset{\\frown}{CD}$, \\\\\n$\\therefore \\angle BAC=\\angle CAD=\\angle BDA=25^\\circ$, \\\\\n$\\because \\angle AED+\\angle CAD+\\angle BDA=180^\\circ$, \\\\\n$\\therefore \\angle AED=180^\\circ-\\angle CAD-\\angle BDA=130^\\circ$, \\\\\n$\\therefore \\angle BEC=\\angle AED=\\boxed{130^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52916773_86.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure (1), two triangular boards, OAB and OCD, are placed with $\\angle ABO=30^\\circ$ and $\\angle ODC=45^\\circ$. If OM and ON bisect $\\angle AOD$ and $\\angle BOC$ respectively, rotate the triangular board OCD clockwise around point O as shown in Figure 2. If OD bisects $\\angle AOB$, what is the rotation angle $\\alpha$?\n",
    "completion": "\\textbf{Solution:} Since OD bisects $\\angle AOB$,\n\n$\\therefore \\angle BOD = \\frac{1}{2}\\angle AOB = 30^\\circ$,\n\n$\\therefore$ The angle of rotation $\\angle BOD = \\alpha = \\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/8928654_72.png"
  },
  {
    "solution": "\\boxed{52.5^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Join a set of triangles together as shown in the figure, where points B, C, and D lie on the same straight line, $\\angle ACB=45^\\circ$, and $\\angle DCE=60^\\circ$. If CM bisects $\\angle BCE$ and CN bisects $\\angle DCA$, as shown in Figure 2, what is the measure of $\\angle MCN$?\n",
    "completion": "\\textbf{Solution:} We have $\\angle MCN = \\boxed{52.5^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51438029_48.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given, as shown in the figure, line MN intersects $\\odot O$ at points A and B, AC is a diameter, DE is tangent to $\\odot O$ at point D, and DE is drawn perpendicular to MN at point E. AD bisects $\\angle CAE$; If AE＝2 and AD＝4, what is the radius of $\\odot O$?\n",
    "completion": "\\textbf{Solution:} Draw OF$\\perp$AB through point O, it's clear that the quadrilateral ODEF is a rectangle,\\\\\nhence, OF=DE, OD=EF,\\\\\nLet the radius of the circle OD=EF=OA=$r$,\\\\\nsince AE=2, AD=4, and $\\angle$AED=90$^\\circ$,\\\\\ntherefore, DE=$\\sqrt{AD^2-AE^2}=2\\sqrt{3}$,\\\\\nthus, OF=DE=$2\\sqrt{3}$, AF=EF−AE=$r−2$,\\\\\nIn $\\triangle$AOF, according to the Pythagorean theorem, we have: OA$^2$=AF$^2$+OF$^2$, i.e., $r^2$=$(r−2)^2+(2\\sqrt{3})^2$,\\\\\nSolving this, we find: $r=\\boxed{4}$,\\\\\nHence, the radius of circle O is \\boxed{4}.",
    "image_path": "dataset/mm_math/images/MM_Math/51446547_115.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point D is the midpoint of line segment AC, $BC = \\frac{1}{2}AB$, and $BD = 1$cm. What is the length of AB?\n",
    "completion": "\\textbf{Solution:} Let $BC=x$ and $BC=\\frac{1}{2}AB$,\\\\\nthus $AB=2BC=2x$, $AC=AB+BC=3x$,\\\\\nsince point D is the midpoint of line segment AC,\\\\\nthus $AD=CD=\\frac{3}{2}x$ and $DC=DB+BC$, $DB=1$,\\\\\nhence $\\frac{3}{2}x=1+x$,\\\\\nSolving for $x$ gives: $x=2$,\\\\\nthus $AB=2x=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51284809.png"
  },
  {
    "solution": "\\boxed{-\\frac{20}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, point B lies on the hyperbolic function $y=\\frac{k}{x}$ (where $k<0$, $x<0$), and point A lies on the negative half of the x-axis. Connect points A and B, and rotate the line segment AB $90^\\circ$ counterclockwise around point A. The corresponding point to B after the rotation is point C. Suppose point C also lies on the graph of the hyperbolic function $y=\\frac{k}{x}$. Given that the y-coordinates of points B and C are 4 and 1 respectively, what is the value of $k$?\n",
    "completion": "\\textbf{Solution}: As illustrated, draw BM perpendicular to the x-axis at point M through point B; draw CN perpendicular to the x-axis at point N through point C. \\\\\nHence, $BM=4$, $CN=1$, $\\angle AMB=\\angle CNA=90^\\circ$, \\\\\n$\\therefore \\angle BAM+\\angle ABM=90^\\circ$, \\\\\nDue to the property of rotation: $\\angle BAC=90^\\circ$, $AB=CA$, \\\\\n$\\therefore \\angle BAM+\\angle CAN=90^\\circ$,\\\\\n$\\therefore \\angle ABM=\\angle CAN$,\\\\\nIn $\\triangle ABM$ and $\\triangle CAN$, we have $\\left\\{\\begin{array}{l}\\angle AMB=\\angle CNA=90^\\circ \\\\ \\angle ABM=\\angle CAN \\\\ AB=CA \\end{array}\\right.$,\\\\\n$\\therefore \\triangle ABM\\cong \\triangle CAN(AAS)$, \\\\\n$\\therefore BM=AN=4$, $AM=CN=1$, \\\\\nLet $OA=a(a>0)$, hence $OM=OA-AM=a-1$, $ON=AN+OA=4+a$, \\\\\n$\\therefore B(1-a, 4)$, $C(-4-a, 1)$, \\\\\nSubstituting points $B(1-a, 4)$, $C(-4-a, 1)$ into the inverse proportion function $y=\\frac{k}{x}$ yields: $4(1-a)=k=-4-a$, \\\\\nSolving, we get $a=\\frac{8}{3}$, \\\\\nThus, $k=-4-a=-4-\\frac{8}{3}=\\boxed{-\\frac{20}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51496547.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system, the vertex $C$ of the parabola passing through points $A(0, 4)$ and $B(5, 9)$ is located on the positive half of the $x$-axis. Let $P(x, y)$ be a point on segment $AB$, with $1\\leq x\\leq 4$. Draw $PM\\parallel y$-axis to intersect the parabola at point $M$. Find the maximum and minimum values of $PM$.\n",
    "completion": "\\textbf{Solution:} Let the equation of AB be: $y=kx+b$. Substituting points A$(0,4)$ and B$(5,9)$ into $y=kx+b$, we obtain:\n\\[\n\\left\\{\n\\begin{array}{l}\nb=4\\\\\n5k+b=9\n\\end{array}\n\\right.\n\\]\nSolving this gives:\n\\[\n\\left\\{\n\\begin{array}{l}\nk=1\\\\\nb=4\n\\end{array}\n\\right.\n\\]\n$\\therefore$The equation of AB is: $y=x+4$,\n\n$\\because$Point P is a point on line segment AB, and point M is a point on the parabola $y=(x-2)^2$, $PM\\parallel y$-axis,\n\nLet $P(t, t+4)$, then $M(t, t^2-4t+4)$,\n\n$\\therefore PM=(t+4)-(t^2-4t+4)=-t^2+5t=-(t-\\frac{5}{2})^2+\\frac{25}{4}$,\n\nAdditionally, $\\because 1\\leq t\\leq 4$,\n\n$\\therefore$When $1\\le t\\le \\frac{5}{2}$, the value of PM increases with increasing t; when $\\frac{5}{2}\\le t\\le 4$, the value of PM decreases with increasing t;\n\n$\\therefore$When $t=\\frac{5}{2}$, $PM_{\\max}=\\boxed{\\frac{25}{4}}$;\n\nSince $\\because 4-\\frac{5}{2}=\\frac{5}{2}-1=\\frac{3}{2}$,\n\n$\\therefore$When $t=1$ and $t=4$, the minimum value of PM is reached, $PM_{\\min}=\\boxed{4}$.\n\nIn conclusion, the maximum value of PM is $\\boxed{\\frac{25}{4}}$, and the minimum value of PM is $\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51369908_194.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Based on the diagram, the triangle \\(ABC\\) is placed inside a grid with each small square having a side length of 1. The points \\(A\\), \\(B\\), \\(C\\) all lie on the grid points. How do we determine the value of \\(\\tan A\\)?\n",
    "completion": "\\textbf{Solution:}\n\nConsider $BD \\perp AC$ at point $D$, \\\\\n$\\because BC = 2$, $AC = \\sqrt{3^{2}+3^{2}} = 3\\sqrt{2}$, the distance from point $A$ to $BC$ is 3, $AB=\\sqrt{3^{2}+1^{2}}=\\sqrt{10}$,\\\\\n$\\therefore \\frac{AC \\cdot BD}{2}=\\frac{BC\\cdot 3}{2}$, that is $\\frac{3\\sqrt{2}\\times BD}{2}=\\frac{2\\times 3}{2}$,\\\\\nSolving for it, we get: $BD=\\sqrt{2}$,\\\\\n$\\therefore AD= \\sqrt{AB^{2}-BD^{2}}=\\sqrt{(\\sqrt{10})^{2}-(\\sqrt{2})^{2}}=2\\sqrt{2}$,\\\\\n$\\therefore \\tan A=\\frac{BD}{AD}=\\frac{\\sqrt{2}}{2\\sqrt{2}}=\\boxed{\\frac{1}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52797160.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given that $\\angle AOB = 90^\\circ$, as shown in figure 2, $OE$ and $OD$ bisect $\\angle AOC$ and $\\angle BOC$ respectively. If $\\angle BOC = 40^\\circ$, find the measure of $\\angle EOD$.\n",
    "completion": "\\textbf{Solution:} Given that $\\angle EOD = \\angle EOC - \\angle COD$ \\\\\n= $\\frac{1}{2}\\angle AOC - \\frac{1}{2}\\angle BOC$ \\\\\n= $\\frac{1}{2}(\\angle AOB + \\angle BOC) - \\frac{1}{2}\\angle BOC$ \\\\\n= $\\frac{1}{2}(90^\\circ + 40^\\circ) - \\frac{1}{2} \\times 40^\\circ$ \\\\\n= $65^\\circ - 20^\\circ$ \\\\\n= $\\boxed{45^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/51439127_176.png"
  },
  {
    "solution": "\\boxed{176^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $\\angle AOB=120^\\circ$, and $OC$ is a ray inside $\\angle AOB$, with $\\angle AOC : \\angle BOC = 1 : 2$. A ray $OD$ passes through point $O$. If $2\\angle AOD = 3\\angle BOD$, what is the measure of $\\angle COD$ in degrees?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, when OD is inside $\\angle AOB$,\\\\\nlet $\\angle BOD=x^\\circ$,\\\\\nsince $2\\angle AOD=3\\angle BOD$,\\\\\nthen $\\angle AOD= \\frac{3}{2}x^\\circ$,\\\\\nsince $\\angle AOB=120^\\circ$,\\\\\nthen $x+ \\frac{3}{2}x=120$,\\\\\nsolving this gives: $x=48$,\\\\\nthus $\\angle BOD=48^\\circ$,\\\\\ntherefore $\\angle COD=\\angle BOC-\\angle BOD=80^\\circ-48^\\circ=32^\\circ$,\\\\\nAs shown in the figure, when OD is outside $\\angle AOB$,\\\\\nlet $\\angle BOD=y^\\circ$,\\\\\nsince $2\\angle AOD=3\\angle BOD$,\\\\\nthen $\\angle AOD= \\frac{3}{2}y^\\circ$,\\\\\nsince $\\angle AOB=120^\\circ$,\\\\\nthen $\\frac{3}{2}y+y+120^\\circ=360^\\circ$,\\\\\nsolving this gives: $y=96^\\circ$,\\\\\nthus $\\angle COD=\\angle BOD+\\angle BOC=96^\\circ+80^\\circ=176^\\circ$,\\\\\nIn conclusion, the degree measure of $\\angle COD$ is $\\boxed{32^\\circ}$ or $\\boxed{176^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51438377_170.png"
  },
  {
    "solution": "\\boxed{130^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, $AO \\perp BO$, $CO \\perp DO$, where $O$ is the foot of the perpendiculars, $\\angle BOC = 50^\\circ$. What is the degree measure of $\\angle AOD$?\n",
    "completion": "\\textbf{Solution:} Given $AO\\perp BO$, $CO\\perp DO$,\\\\\ntherefore, $\\angle AOB = \\angle COD = 90^\\circ$,\\\\\nsince $\\angle AOB + \\angle BOC + \\angle COD + \\angle AOD = 360^\\circ$, and $\\angle BOC = 50^\\circ$,\\\\\nthus, $90^\\circ + 50^\\circ + 90^\\circ + \\angle AOD = 360^\\circ$,\\\\\nwe find that $\\angle AOD = \\boxed{130^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55191220.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, in $\\triangle ACB$, $\\angle ACB=90^\\circ, AC=4, BC=3$. Point $D$ is a moving point on the hypotenuse $AB$ ($0<AD<3.2$). Taking point $A$ as the center and $AD$ as the radius, draw circle $A$ intersecting $AC$ at point $F$. Extend $CD$ to intersect circle $A$ at point $E$, and connect $AE$ and $DF$. $\\angle FAE=2\\angle FDC$. As shown in Figure 2, if $AE\\parallel CB$, find the length of $EC$.\n",
    "completion": "\\textbf{Solution:} Given that $AE\\parallel CB$,\\\\\n$\\therefore \\angle EAC+\\angle ACB=180^\\circ$, $\\angle E=\\angle BCD$,\\\\\nSince $\\angle ACB=90^\\circ$, $AE=AD$,\\\\\n$\\therefore \\angle EAC=90^\\circ$, $\\angle E=\\angle ADE$,\\\\\nSince $\\angle ADE=\\angle CDB$,\\\\\n$\\therefore \\angle BCD=\\angle CDB$,\\\\\n$\\therefore BD=BC=3$,\\\\\nSince $\\angle ACB=90^\\circ, AC=4, BC=3$,\\\\\n$\\therefore AB=\\sqrt{AC^{2}+BC^{2}}=5$,\\\\\n$\\therefore AE=AD=2$,\\\\\n$\\therefore CE=\\sqrt{AC^{2}+AE^{2}}=\\boxed{2\\sqrt{5}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52991468_108.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle B=90^\\circ$, $AB=12\\,cm$, $BC=16\\,cm$. Point $P$ starts from point $A$ and moves towards point $B$ along side $AB$ at a speed of $1\\,cm/s$, and point $Q$ starts from point $B$ and moves towards point $C$ along side $BC$ at a speed of $2\\,cm/s$. If $P$ and $Q$ depart from $A$ and $B$ simultaneously, when one point reaches its destination, the other point also stops moving. Let the time of movement be $t$ seconds. What value of $t$ makes the area of $\\triangle PBQ$ equal to $35\\,cm^{2}$?\n",
    "completion": "\\textbf{Solution:} According to the problem, we know that $BQ=2t$, $AP=t$, \\\\\n$\\therefore QC=BC-BQ=16-2t$, $BP=AB-AP=12-t$. \\\\\nUsing the formula for the area of a triangle, we have $\\frac{1}{2}BQ\\cdot BP=35$, \\\\\n$\\frac{1}{2}\\times 2t(12-t)=35$, \\\\\n$t^{2}-12t+35=0$, \\\\\nSolving this, we get $t_{1}=\\boxed{5}$, $t_{2}=7$. \\\\\nTherefore, when $t$ is 5 or 7, the area of $\\triangle PBQ$ equals $35\\,cm^{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52622502_114.png"
  },
  {
    "solution": "\\boxed{80^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, $OB$ and $OE$ are two rays within $\\angle AOC$, $OD$ bisects $\\angle AOB$, $\\angle BOE=\\frac{1}{2}\\angle EOC$. Given $\\angle DOE=55^\\circ$ and $\\angle AOC=150^\\circ$, calculate the degree measure of $\\angle EOC$.\n",
    "completion": "\\textbf{Solution:} Let $\\angle BOE = x^\\circ$, then $\\angle DOB = 55^\\circ - x^\\circ$,\\\\\nsince $\\angle BOE = \\frac{1}{2} \\angle EOC$, we have $\\angle EOC = 2x^\\circ$,\\\\\nas $OD$ bisects $\\angle AOB$, we get $\\angle AOB = 2\\angle DOB$,\\\\\nthus $2x + x + 2(55 - x) = 150$,\\\\\nsolving the equation gives $x = 40$,\\\\\nhence $\\angle EOC = 2x = \\boxed{80^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51481943.png"
  },
  {
    "solution": "\\boxed{\\begin{cases}x=1\\\\ y=3\\end{cases}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y=x+2$ intersects with the graph of $y=kx+b$ $(k\\ne 0)$ at point $P$. What is the solution to the system of linear equations in two variables $\\begin{cases}y=x+2\\\\ y=kx+b\\end{cases}$ for $x$ and $y$?\n",
    "completion": "\\textbf{Solution:} Observing the graph, we find: The graph of the linear function $y=x+2$ intersects with the graph of $y=kx+b \\ (k \\neq 0)$ at point P(1,3), \\\\\n$\\therefore$ the solution to the system of linear equations $\\begin{cases}y=x+2\\\\ y=kx+b\\end{cases}$ is $\\boxed{\\begin{cases}x=1\\\\ y=3\\end{cases}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53706360.png"
  },
  {
    "solution": "\\boxed{75^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\angle AOB = 120^\\circ$, $\\angle AOC = \\frac{1}{3}\\angle BOC$, $OM$ bisects $\\angle BOC$. What is the degree measure of $\\angle AOM$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle AOB=120^\\circ$, and $\\angle AOC= \\frac{1}{3} \\angle BOC$,\\\\\nthus $\\angle AOC= \\frac{1}{4} \\angle AOB=30^\\circ$, and $\\angle BOC= \\frac{3}{4} \\angle AOB=90^\\circ$,\\\\\ngiven that OM bisects $\\angle BOC$,\\\\\nthus $\\angle BOM=\\angle COM= \\frac{1}{2} \\angle BOC=45^\\circ$,\\\\\ntherefore $\\angle AOM=\\angle AOC+\\angle COM=\\boxed{75^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/51367684.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $AB = 6$, and $E$ is the midpoint of $BC$. $AE$ and $BD$ intersect at point $F$. Connect $CF$. If $AE \\perp BD$, what is the length of $CF$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, draw $FM\\perp BC$ through point F, with FM intersecting BC at point M\\\\\n$\\because$ rectangle ABCD\\\\\n$\\therefore$ $\\angle ABE=\\angle BAD=\\angle ACD=90^\\circ$, $BC=AD$\\\\\n$\\because$AE$\\perp$BD\\\\\n$\\therefore$ $\\angle BAF+\\angle ABF=\\angle ADB+\\angle ABF=90^\\circ$\\\\\n$\\therefore$ $\\angle BAF=\\angle ADB$\\\\\n$\\therefore$ $\\triangle BAD\\sim \\triangle EBA$\\\\\n$\\therefore$ $\\frac{AD}{AB}=\\frac{AB}{BE}$\\\\\n$\\because$E is the midpoint of BC\\\\\n$\\therefore$ $AD=BC=2BE$\\\\\n$\\therefore$ $\\frac{2BE}{6}=\\frac{6}{BE}$\\\\\n$\\therefore$ $2BE^2=36$\\\\\n$\\therefore$ $BE=\\sqrt{18}=3\\sqrt{2}$\\\\\nUpon verification, $BE\\ne 0$, it is a solution to the original equation $\\frac{2BE}{6}=\\frac{6}{BE}$\\\\\n$\\therefore$ $AD=BC=2BE=6\\sqrt{2}$\\\\\n$\\therefore$ $BD=\\sqrt{AB^2+AD^2}=6\\sqrt{3}$, $AE=\\sqrt{AB^2+BE^2}=3\\sqrt{6}$\\\\\n$\\therefore$ $BF=\\frac{AB\\times BE}{AE}=\\frac{6\\times 3\\sqrt{2}}{3\\sqrt{6}}=2\\sqrt{3}$\\\\\n$\\because$ $\\angle FBE=\\angle CBD$, $\\angle BFE=\\angle ACD=90^\\circ$\\\\\n$\\therefore$ $\\triangle BFE\\sim \\triangle BCD$\\\\\n$\\therefore$ $\\frac{FM}{CD}=\\frac{BF}{BD}$\\\\\n$\\therefore$ $FM=\\frac{CD\\times BF}{BD}=\\frac{6\\times 2\\sqrt{3}}{6\\sqrt{3}}=2$\\\\\n$\\therefore$ $BM=\\sqrt{BF^2−FM^2}=\\sqrt{12−4}=2\\sqrt{2}$\\\\\n$\\therefore$ $CM=BC−BM=6\\sqrt{2}−2\\sqrt{2}=4\\sqrt{2}$\\\\\n$\\therefore$ $FC=\\sqrt{FM^2+CM^2}=\\sqrt{4+32}=\\boxed{6}$",
    "image_path": "dataset/mm_math/images/MM_Math/51402956.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in square $ABCD$, circle $O$ with diameter $AB$ intersects the midpoint of side $BC$ at point $E$ and intersects diagonal $AC$ at point $F$. Draw $EG \\perp AC$ with the foot of the perpendicular at $M$. If $FA=3CF$, what is the value of $\\frac{CG}{GD}$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect BF and extend GE, where AB intersects at point H, let AC intersect GE at point M,\\\\\nsince AB is the diameter of circle O,\\\\\ntherefore $\\angle AFB=90^\\circ$,\\\\\nsince quadrilateral ABCD is a parallelogram,\\\\\ntherefore $DC\\parallel AB$, $DC=AB$,\\\\\ntherefore $\\angle DCE=\\angle CBH$, $\\angle CGE=\\angle EHB$,\\\\\nsince E is the midpoint of BC,\\\\\ntherefore $CE=BE=\\frac{1}{2}CB$,\\\\\ntherefore $\\triangle GEC\\cong \\triangle HEB$ (AAS),\\\\\ntherefore $GC=BH$,\\\\\nsince $EG\\perp AC$, $BF\\perp AC$,\\\\\ntherefore $ME\\parallel BF$,\\\\\ntherefore $CM:CF=CE:CB=\\frac{1}{2}$,\\\\\ntherefore let $CM=MF=a$, $CF=2a$,\\\\\ntherefore $FA=3CF=6a$,\\\\\ntherefore $CM:MA=a:7a=\\frac{1}{7}$,\\\\\nsince $CD\\parallel AB$\\\\\ntherefore $\\triangle GMC\\sim \\triangle HMA$,\\\\\ntherefore $GC:AH=MC:MA=\\frac{1}{7}$,\\\\\ntherefore $GC:AB=\\frac{1}{6}$,\\\\\ntherefore $GC:CD=\\frac{1}{6}$,\\\\\ntherefore $GC:DG=\\boxed{\\frac{1}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52774551.png"
  },
  {
    "solution": "\\boxed{40^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the lines AB and CD intersect at point O. If $\\angle BOD = 40^\\circ$ and OA bisects $\\angle COE$, what is the measure of $\\angle AOE$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle BOD = 40^\\circ$,\\\\\ntherefore $\\angle AOC = \\angle BOD = 40^\\circ$,\\\\\nsince OA bisects $\\angle COE$,\\\\\ntherefore $\\angle AOE = \\angle AOC = \\boxed{40^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/6776773.png"
  },
  {
    "solution": "\\boxed{7}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle BAD = \\angle CAD$, $BE = CE$, $AD \\perp BD$, $DE = \\frac{3}{2}$, $AB = 4$, what is the value of $AC$?\n",
    "completion": "\\textbf{Solution:} Extend $BD$ to intersect $AC$ at $F$, as shown in the diagram:\\\\\n$\\because$ $\\angle BAD=\\angle CAD$ and $AD \\perp BD$,\\\\\n$\\therefore$ $AF=AB=4$, point $D$ is the midpoint of segment $BF$,\\\\\nFurther, $\\because$ $BE=CE$,\\\\\n$\\therefore$ point $E$ is the midpoint of segment $BC$,\\\\\n$\\therefore$ $DE$ is the midline of $\\triangle BCF$,\\\\\n$\\therefore$ $FC=2DE=3$,\\\\\n$\\therefore$ $AC=AF+FC=4+3=\\boxed{7}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52211273.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AD$ perpendicularly bisects $BC$ at $D$, $\\angle BAC=45^\\circ$, $CE \\perp AB$ at $E$, intersecting $AD$ at $F$, and $BD=2$. What is the length of $AF$?\n",
    "completion": "\\textbf{Solution:} Given that $AD$ vertically bisects $BC$, and $BD=2$, \\\\\nit follows that $BC=2BD=4$, \\\\\nsince $CE\\perp AB$, \\\\\nit follows that $\\angle CEB=90^\\circ$, \\\\\nthus, $\\angle AEC=90^\\circ$, \\\\\ngiven that $\\angle BAC=45^\\circ$, \\\\\nit follows that $\\angle ACE=90^\\circ-45^\\circ=45^\\circ$, \\\\\nhence, $\\angle EAC=\\angle ACE$, \\\\\nthus, $AE=CE$, \\\\\nsince $AD$ vertically bisects $BC$, \\\\\nit follows that $\\angle ADB=90^\\circ$, \\\\\nthus, $\\angle B+\\angle BAD=90^\\circ$, \\\\\ngiven that $\\angle B+\\angle BCE=90^\\circ$, \\\\\nit follows that $\\angle BAD=\\angle BCE$, \\\\\nin $\\triangle AEF$ and $\\triangle CEB$, we have $\\left\\{\\begin{array}{l}\\angle AEF=\\angle CEB \\\\ AE=CE \\\\ \\angle EAF=\\angle BCE\\end{array}\\right.$, \\\\\ntherefore, $\\triangle AEF\\cong \\triangle CEB\\left(ASA\\right)$, \\\\\nhence, $AF=BC= \\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52793334.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within a $5\\times4$ square grid, where each small square has a side length of $1$, the vertices of $\\triangle ABC$ are all located at the vertices of these small squares. What is the value of $\\tan\\angle ABC$?\n",
    "completion": "\\textbf{Solution:} Let point $D$ be chosen on the grid, such that $\\angle BDC = 90^\\circ$,\\\\\nsince $CD = 4$, and $BD = 1$\\\\\ntherefore, $\\tan\\angle ABC = \\frac{CD}{BD} = \\frac{4}{1} = \\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51212854.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{4}\\sqrt{17}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the figure, $AB$ is the diameter of $\\odot O$, and the chord $CD\\perp AB$. Point $E$ is on the extension of $CA$. Connect $DE$ intersecting $\\odot O$ at point $F$, and connect $AF, CF$. Given the conditions in (3), connect $AD$, $EH$, intersecting at point $G$. What is the length of $GH$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram below, connect $AD$ and $EH$, intersecting at point G,\\\\\n$\\because DE^2+DC^2=CE^2$,\\\\\n$\\therefore DE=\\sqrt{CE^2-DC^2}=8$,\\\\\n$\\because DE^2+DH^2=HE^2$,\\\\\n$\\therefore HE=2\\sqrt{17}$,\\\\\n$\\because AH\\parallel DE$,\\\\\n$\\therefore \\triangle AHG\\sim \\triangle DGE$,\\\\\n$\\therefore \\frac{AH}{DE}=\\frac{HG}{GE}$,\\\\\n$\\therefore \\frac{AH}{DE}=\\frac{HG}{HE-HG}$,\\\\\n$\\therefore \\frac{HG}{2\\sqrt{17}-HG}=\\frac{4}{8}$,\\\\\n$\\therefore HG=\\boxed{\\frac{3}{4}\\sqrt{17}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52991580_102.png"
  },
  {
    "solution": "\\boxed{\\left(\\frac{2}{3}, -\\frac{10}{3}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the line $y=4x-6$ intersects the $x$-axis at point $A$, and the line $y=-2x-2$ intersects the $x$-axis at point $B$, and these two lines intersect at point $C$. What are the coordinates of the intersection point $C$ of these two lines?\n",
    "completion": "\\textbf{Solution:} The coordinates of the intersection point $C$ of these two lines is $\\boxed{\\left(\\frac{2}{3}, -\\frac{10}{3}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52423437_61.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\triangle ABC$, $ \\angle A=90^\\circ$, $\\sin\\angle ABC=\\frac{1}{2}$. $\\odot O$ is the inscribed circle of $ \\triangle ABC$, which is tangent to $ AC$, $ AB$, $ BC$ at points $ F$, $ P$, and $ E$ respectively. What is the value of $\\angle EPF$ in degrees?\n",
    "completion": "\\textbf{Solution:} We have $\\angle EPF = \\boxed{60^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55502222_58.png"
  },
  {
    "solution": "\\boxed{y=\\frac{4}{x}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the inverse proportion function $y=\\frac{k_1}{x}$ intersects with the graph of the direct proportion function $y=k_2x$ at points $A(a, 1)$ and B. Point $M(a-3, a)$ lies on the graph of the inverse proportion function. Connect $OM$ and $BM$, where $BM$ intersects the y-axis at point N. Determine the analytic expression of the inverse proportion function.\n",
    "completion": "\\textbf{Solution:} Since points A$(4,1)$ and M$(1,4)$ lie on the graph of an inverse variation function, \\\\\nit follows that the equation of the inverse variation function is $\\boxed{y=\\frac{4}{x}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53043068_129.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the figure, point D is on side AB of $\\triangle ABC$, with $\\angle ABC = \\angle ACD$. $\\triangle ABC \\sim \\triangle ACD$; if $AD = 2$ and $AB = 8$, find the length of $AC$.\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABC\\sim \\triangle ACD$,\\\\\nit follows that $\\frac{AC}{AD}=\\frac{AB}{AC}$,\\\\\nGiven $AD=2$, $AB=8$,\\\\\nwe have $\\frac{AC}{2}=\\frac{8}{AC}$,\\\\\ntherefore $AC= \\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52719745_181.png"
  },
  {
    "solution": "\\boxed{-4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the inverse proportion function $y=\\frac{k}{x}$ is depicted. Point A is on the graph of this function, and AB is perpendicular to the x-axis with B being the foot of the perpendicular. If the area of $\\triangle AOB$ is 2, what is the value of $k$?\n",
    "completion": "\\textbf{Solution:} Let $A\\left(x_{0},y_{0}\\right)$. As $S_{\\triangle AOB}=2$ we have\\\\\n$\\frac{1}{2}\\left|x_{0}\\right|\\left|y_{0}\\right|=2$,\\\\\nthus $\\left|x_{0}\\right|\\left|y_{0}\\right|=4$.\\\\\nSince point A is in the second quadrant,\\\\\nwe have $x_{0}+y_{0}=-4$,\\\\\n$\\because$ point A lies on the graph of the function,\\\\\nthus $y_{0}=\\frac{k}{x_{0}}$,\\\\\nwhich leads to $k=x_{0}y_{0}=\\boxed{-4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52797102.png"
  },
  {
    "solution": "\\boxed{CP \\text{ is the tangent to the circle } O}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the circumcircle of $\\triangle ABC$ has its center at $O$, with $AB$ being the diameter. $AB \\perp CD$ at point $E$, and $P$ is a point on the extension of $AB$, and $\\angle BCP = \\angle BCD$. Prove that $CP$ is a tangent to $\\odot O$.\n",
    "completion": "\\textbf{Solution:} Proof: Connect OC.\\\\\n$\\because$OB=OC,\\\\\n$\\therefore \\angle OBC=\\angle OCB$,\\\\\n$\\because$AB$\\perp$CD at point E,\\\\\n$\\therefore \\angle CEB=90^\\circ$,\\\\\n$\\therefore \\angle OBC+\\angle BCD=90^\\circ$,\\\\\n$\\therefore \\angle OCB+\\angle BCD=90^\\circ$,\\\\\n$\\because \\angle BCP=\\angle BCD$,\\\\\n$\\therefore \\angle OCB+\\angle BCP=90^\\circ$,\\\\\n$\\therefore$OC$\\perp$CP,\\\\\n$\\therefore \\boxed{CP \\text{ is the tangent to the circle } O}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51350909_109.png"
  },
  {
    "solution": "\\boxed{14}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, if $AB=7$ and $\\angle DBC=30^\\circ$, what is the length of $AC$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral $ABCD$ is a rectangle, and $AB=7$,\n\nhence $CD=AB=7$, $AC=BD$, and $\\angle BCD=90^\\circ$,\n\nsince in $\\triangle BCD$, $\\angle DBC=30^\\circ$,\n\nit follows that $BD=2CD=14$,\n\ntherefore $AC=BD=\\boxed{14}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52204025.png"
  },
  {
    "solution": "\\boxed{\\frac{2\\sqrt{5}}{15}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC=6$, $\\sin B=\\frac{2}{3}$. Point D is on the extension line of side BC, and $CD=BC$. What is the value of $\\tan D$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $AE\\perp BD$ at point E through point $A$.\\\\\n$\\therefore$ $\\sin B=\\frac{AE}{AB}=\\frac{2}{3}$.\\\\\n$\\because$ $AB=AC=6$,\\\\\n$\\therefore$ $AE=4$,\\\\\n$\\therefore$ $BE=\\sqrt{AB^{2}-AE^{2}}=2\\sqrt{5}$.\\\\\n$\\because$ $AE\\perp BD$, $AB=AC=6$,\\\\\n$\\therefore$ $CE=BE=2\\sqrt{5}$,\\\\\n$\\therefore$ $CD=BC=BE+CE=4\\sqrt{5}$,\\\\\n$\\therefore$ $DE=CD+CE=6\\sqrt{5}$,\\\\\n$\\therefore$ $\\tan D=\\frac{AE}{DE}=\\frac{4}{6\\sqrt{5}}=\\boxed{\\frac{2\\sqrt{5}}{15}}$",
    "image_path": "dataset/mm_math/images/MM_Math/52977742.png"
  },
  {
    "solution": "\\boxed{50}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, quadrilateral $ABCD$ is a circumscribed quadrilateral of circle $\\odot O$, and $AB=10$, $CD=15$. What is the perimeter of quadrilateral $ABCD$?\n",
    "completion": "\\textbf{Solution:}\\\\\n$\\because$ Quadrilateral ABCD is a circumscribed quadrilateral of $\\odot$O,\\\\\n$\\therefore$AE = AH, BE = BF, CF = CG, DH = DG,\\\\\n$\\therefore$AD + BC = AB + CD = 25,\\\\\n$\\therefore$The perimeter of quadrilateral ABCD = AD + BC + AB + CD = 25 + 25 = \\boxed{50}",
    "image_path": "dataset/mm_math/images/MM_Math/51213350.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given: As shown in the figure, in $\\triangle ABC$, $AB=4$, $BC=8$. $D$ is a point on side $BC$, and $\\angle BDA = \\angle BAC$. Find the length of $BD$.\n",
    "completion": "\\textbf{Solution:} In $\\triangle BDA$ and $\\triangle BAC$,\\\\\n$\\angle B=\\angle B$, $\\angle BDA=\\angle BAC$,\\\\\n$\\therefore \\triangle BDA\\sim \\triangle BAC$,\\\\\n$\\therefore \\frac{AB}{BC}=\\frac{BD}{AB}$,\\\\\n$\\because AB=4$, $BC=8$,\\\\\n$\\therefore \\frac{4}{8}=\\frac{BD}{4}$,\\\\\n$\\therefore BD=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52866412.png"
  },
  {
    "solution": "\\boxed{\\sqrt{6}-\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points $A$, $P$, and $T$ lie on the circle $\\odot O$, $OT\\perp TB$, the arc $\\overset{\\frown}{AT}$ measures $60^\\circ$, the radius of $\\odot O$ is 2, $TB=2$, point $B$ and point $A$ are on opposite sides of line $OT$, $PB$ intersects $\\odot O$ at point $C$. When $\\angle APB=60^\\circ$, what is the value of $\\angle PBT$? What is the length of $BC$?\n",
    "completion": "\\textbf{Solution}: Connect $OA$, $OC$, $TC$, and draw $CD\\perp TB$ at point $C$ with foot $D$,\\\\\n$\\because \\angle APB=60^\\circ$,\\\\\n$\\therefore \\angle AOC=2\\angle APC=120^\\circ$,\\\\\n$\\because$ the degree of $\\overset{\\frown}{AT}$ is $60^\\circ$,\\\\\n$\\therefore \\angle AOT=60^\\circ$,\\\\\n$\\therefore \\angle TOC=\\angle AOC-\\angle AOT=60^\\circ$,\\\\\n$\\because OT=OC$,\\\\\n$\\therefore \\triangle OTC$ is an equilateral triangle,\\\\\n$\\therefore TC=OT=OC=2$, and $\\angle OTC=60^\\circ$,\\\\\n$\\because OT\\perp TB$,\\\\\n$\\therefore \\angle OTB=90^\\circ$,\\\\\n$\\therefore \\angle CTB=\\angle OTB-\\angle OTC=30^\\circ$,\\\\\n$\\because TC=TB=2$,\\\\\n$\\therefore \\angle PBT=\\angle TCB=\\frac{1}{2}\\left(180^\\circ-\\angle CTB\\right)=75^\\circ$,\\\\\nIn $\\triangle TCD$ right-angled at $D$, $\\angle CTB=30^\\circ$, $TC=2$,\\\\\n$\\therefore CD=\\frac{1}{2}TC=1$, $TD=\\sqrt{3}CD=\\sqrt{3}$,\\\\\n$\\therefore BD=TB-TD=2-\\sqrt{3}$,\\\\\n$\\therefore CB=\\sqrt{CD^{2}+DB^{2}}=\\sqrt{1^{2}+(2-\\sqrt{3})^{2}}=\\sqrt{8-4\\sqrt{3}}=\\sqrt{8-2\\sqrt{12}}=\\sqrt{(\\sqrt{6}-\\sqrt{2})^{2}}=\\boxed{\\sqrt{6}-\\sqrt{2}}$,\\\\",
    "image_path": "dataset/mm_math/images/MM_Math/53009650.png"
  },
  {
    "solution": "\\boxed{8-2t}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, $BC\\parallel AD$, $\\angle ADC=90^\\circ$, $BC=8\\,\\text{cm}$, $AD=CD=10\\,\\text{cm}$. Point $E$ starts from point $B$ and moves along direction $BC$ at a uniform speed of $2\\,\\text{cm/s}$; at the same time, point $F$ starts from point $D$ and moves along direction $DA$ at a uniform speed of $3\\,\\text{cm/s}$. Draw $EH\\perp AD$ at point $E$ with the perpendicular foot being $H$, and $EH$ intersects $AC$ at point $G$. Connect $FG$. Let the movement time be $t\\,\\text{s}$ $\\left(0<t<\\frac{10}{3}\\right)$. Solve the following problem: Find the length of $DH$ (expressed in an algebraic expression involving $t$).\n",
    "completion": "\\textbf{Solution:} From the given, we have $BE=2t$, $DF=3t$, \\\\\n$\\therefore$ $CE=8-2t$, $AF=10-3t$, \\\\\n$\\because$ $CD\\perp AD$, $EH\\perp AD$, \\\\\n$\\therefore$ $\\angle D=\\angle EHA=90^\\circ$, \\\\\n$\\therefore$ $CD\\parallel EH$, \\\\\nAdditionally, $\\because$ $CB\\parallel AD$, \\\\\n$\\therefore$ the quadrilateral $CEHD$ is a parallelogram, \\\\\n$\\therefore$ the quadrilateral $CEHD$ is a rectangle, \\\\\n$\\therefore$ $DH=CE=8-2t$; \\\\\nThus, the length of $DH$ is $\\boxed{8-2t}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52211373_84.png"
  },
  {
    "solution": "\\boxed{\\frac{5}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\triangle ABC$, $\\angle BCA=90^\\circ$, $AC=3$, $BC=4$. Point $P$ is a point on hypotenuse $AB$. A ray $PD$ is drawn perpendicular to $PE$, intersecting $AC$ and $BC$ at points $D$ and $E$, respectively. $DE$ is connected. If $\\triangle PDE$ is similar to $\\triangle ABC$, what is the value of $BP$?\n",
    "completion": "\\textbf{Solution:} Construct $PM\\perp AC$, $PN\\perp BC$, intersecting $AC$, $BC$ at points $M$, $N$ respectively.\\\\\nSince $\\angle BCA=90^\\circ$\\\\\nTherefore, quadrilateral $PMCN$ is a rectangle,\\\\\nSince $\\angle DPE=90^\\circ$\\\\\nTherefore $\\angle DPM=\\angle EPN$\\\\\nSince $PM\\perp AC$, $PN\\perp BC$,\\\\\nTherefore $\\angle DMP=\\angle ENP=90^\\circ$\\\\\nTherefore $\\triangle DPM\\sim \\triangle EPN$\\\\\n(1) If $\\triangle DPE\\sim \\triangle ACB$, it can be obtained that $\\frac{PD}{PE}=\\frac{AC}{BC}=\\frac{3}{4}$\\\\\nSince $\\triangle DPM\\sim \\triangle EPN$\\\\\nTherefore $\\frac{PM}{PN}=\\frac{PD}{PE}=\\frac{3}{4}$\\\\\nLet $PM=3m$, then $PN=4m$\\\\\nSince quadrilateral $PMCN$ is a rectangle, $CN=3m$\\\\\nIn $\\triangle ACB$, with $\\angle C=90^\\circ$, it can be obtained that $\\tan B=\\frac{AC}{BC}=\\frac{3}{4}$\\\\\nIn $\\triangle PNB$, $\\tan B=\\frac{PN}{BN}=\\frac{4m}{BN}=\\frac{3}{4}$, solving gives $BN=\\frac{16}{3}m$\\\\\nSince $BC=CN+BN$\\\\\nTherefore $4=3m+\\frac{16}{3}m$, solving gives $m=\\frac{12}{25}$\\\\\nTherefore $PN=\\frac{48}{25}$\\\\\nSince $\\angle B=\\angle B$, $\\angle PNB=\\angle C=90^\\circ$\\\\\nTherefore $\\triangle BPN\\sim \\triangle BAC$\\\\\nTherefore $\\frac{BP}{BA}=\\frac{PN}{AC}$, that is $\\frac{BP}{5}=\\frac{\\frac{48}{25}}{3}$, solving gives $BP=\\boxed{\\frac{16}{5}}$\\\\\n(2) If $\\triangle DPE\\sim \\triangle BCA$, it can be obtained that $\\frac{PD}{PE}=\\frac{BC}{AC}=\\frac{4}{3}$\\\\\nSince $\\triangle DPM\\sim \\triangle EPN$\\\\\nTherefore $\\frac{PM}{PN}=\\frac{PD}{PE}=\\frac{4}{3}$\\\\\nLet $PM=4m$, then $PN=3m$,\\\\\nSince quadrilateral $PMCN$ is a rectangle, $CN=4m$\\\\\nIn $\\triangle ACB$, with $\\angle C=90^\\circ$, it can be obtained that $\\tan B=\\frac{AC}{BC}=\\frac{3}{4}$\\\\\nIn $\\triangle PNB$, $\\tan B=\\frac{PN}{BN}=\\frac{3m}{BN}=\\frac{3}{4}$, solving gives $BN=4m$\\\\\nSince $BC=CN+BN$\\\\\nTherefore $4=4m+4m$, solving gives $m=\\frac{1}{2}$\\\\\nTherefore $PN=\\frac{3}{2}$\\\\\nSince $\\angle B=\\angle B$, $\\angle PNB=\\angle C=90^\\circ$\\\\\nTherefore $\\triangle BPN\\sim \\triangle BAC$\\\\\nTherefore $\\frac{BP}{BA}=\\frac{PN}{AC}$, that is $\\frac{BP}{5}=\\frac{\\frac{3}{2}}{3}$,\\\\\nSolving gives $BP=\\boxed{\\frac{5}{2}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51446440_77.png"
  },
  {
    "solution": "\\boxed{17^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AD$ and $AE$ are respectively the altitude and the angle bisector, $\\angle B=33^\\circ$, $\\angle C=67^\\circ$. What is the measure of $\\angle EAD$ in degrees?\n",
    "completion": "\\textbf{Solution:} Given $\\angle B=33^\\circ$ and $\\angle C=67^\\circ$,\\\\\nit follows that $\\angle BAC=80^\\circ$. Since AE is the bisector of $\\angle BAC$,\\\\\nit follows that $\\angle BAE=40^\\circ$ and thus $\\angle AED=\\angle BAE+\\angle B=33^\\circ+40^\\circ=73^\\circ$.\\\\\nSince AD is the altitude of the triangle,\\\\\nit follows that $\\angle EAD=90^\\circ-73^\\circ=\\boxed{17^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/40626917.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $D$ is the midpoint of segment $AC$, and $AB = \\frac{2}{3} BC$, $BC = 6$. Find the length of segment $BD$.\n",
    "completion": "\\textbf{Solution:} Given that $AB= \\frac{2}{3} BC$ and $BC=6$,\n\\[\n\\therefore AB= \\frac{2}{3} \\times 6=4,\n\\]\n\\[\n\\therefore AC=AB+BC=10,\n\\]\nSince point D is the midpoint of segment AC,\n\\[\n\\therefore AD= \\frac{1}{2} AC=5,\n\\]\n\\[\n\\therefore BD=AD-AB=5-4=\\boxed{1}.\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/51482322_177.png"
  },
  {
    "solution": "\\boxed{y=-x+2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the graph of the linear function $y=kx+b$ intersects with the graph of the inverse proportion function $y=-\\frac{8}{x}$ at points $A$ and $B$, and both the $x$-coordinate of point $A$ and the $y$-coordinate of point $B$ are $-2$. Find the analytical expression of the linear function.\n",
    "completion": "\\textbf{Solution:} Let's consider the inverse proportion function $y=-\\frac{8}{x}$. When $x=-2$, we have $y=4$,\n\n$\\therefore$ the coordinates of point A are $\\left(-2, 4\\right)$;\n\nFor the inverse proportion function $y=-\\frac{8}{x}$, when $y=-2$, we have $-2=-\\frac{8}{x}$,\n\nSolving this gives: $x=4$,\n\n$\\therefore$ the coordinates of point B are $\\left(4, -2\\right)$.\n\n$\\because$ A linear function passes through points A and B,\n\n$\\therefore$\n$\\begin{cases}\n-2k+b=4\\\\\n4k+b=-2\n\\end{cases}$, solving these equations yields:\n$\\begin{cases}\nk=-1\\\\\nb=2\n\\end{cases}$,\n\n$\\therefore$ the equation of the linear function is $\\boxed{y=-x+2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52765753_95.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point C is a point on line segment AB, point D is the midpoint of line segment BC, and $AB=16\\,cm$, $CD=3\\,cm$. Find the length of line segment AC.\n",
    "completion": "\\textbf{Solution:} Since point D is the midpoint of line segment BC, and $CD=3\\,\\text{cm}$,\\\\\nit follows that $BC=2CD=6\\,\\text{cm}$,\\\\\ntherefore, $AC=AB-BC=16-6=\\boxed{10}\\,\\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51495383_119.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Fold a parallelogram sheet $ABCD$ as shown in Diagram 1, so that vertices $A$ and $B$ coincide at point $M$ on segment $HF$, and vertices $C$ and $D$ coincide at point $N$ on segment $HF$, where the length of $AD$ is $6$, and the length of $AE$ is $x$. Quadrilateral $EFGH$ is a rectangle; please express the length of $HF$ with an algebraic expression involving $x$; if you can't, simply write out the length of $HF$.\n",
    "completion": "\\textbf{Solution:} By folding, we get: $\\angle B=\\angle FME$, $\\angle D=\\angle HNG$, $AH=MH$, $FC=FN$, $BF=FM$, $HD=HN$,\\\\\n$\\therefore HF=MH+FM=AH+BF$,\\\\\n$\\because$ Rectangle EFGH,\\\\\n$\\therefore EH=FG$, $EH\\parallel FG$,\\\\\n$\\therefore \\angle EHM=\\angle NFG$,\\\\\nFurthermore, $\\because$ parallelogram ABCD,\\\\\n$\\therefore \\angle B=\\angle D$,\\\\\n$\\therefore \\angle FME=\\angle HNG$,\\\\\n$\\therefore \\angle EMH=\\angle GNF$,\\\\\n$\\therefore \\triangle EMH \\cong \\triangle GNF$,\\\\\n$\\therefore MH=FN$,\\\\\n$\\therefore BF=MF=NH=DH$,\\\\\n$\\therefore HF=AH+DH=AD$,\\\\\n$\\because AD=6$,\\\\\n$\\therefore HF=\\boxed{6}$,\\\\\nHence, the length of $HF$ remains unchanged.",
    "image_path": "dataset/mm_math/images/MM_Math/52077012_87.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=\\frac{1}{2}x-1$ intersects the x-axis at point B, and the hyperbola $y=\\frac{k}{x}$ (where $x>0$) intersects at point A. A perpendicular line from point B to the x-axis intersects the hyperbola $y=\\frac{k}{x}$ at point C, and $AB=AC$. What is the value of $k$?\n",
    "completion": "\\textbf{Solution:} The line $y=\\frac{1}{2}x-1$ intersects the x-axis at point B, thus: B(2,0),\\\\\nSince AB=AC and BC is perpendicular to the x-axis, point A lies on the perpendicular bisector of BC, hence: C(2, $\\frac{k}{2}$), A(4, $\\frac{k}{4}$),\\\\\nSubstituting point A into the line equation $y=x-1$ yields: $k=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51446264.png"
  },
  {
    "solution": "\\boxed{x>3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the line $y=kx+5$ intersects the $x$ axis at point $A$ and the $y$ axis at point $B$, and the coordinate of point $A$ is $(5,0)$. The line $y=2x-4$ intersects the $x$ axis at point $D$, and intersects the line $AB$ at point $C$. Based on the graph, what is the solution set of the inequality $2x-4>kx+5$ with respect to $x$?\n",
    "completion": "\\textbf{Solution:} Observing the graph of the function, we see that when $x>3$, the line $y=2x-4$ is above the line $y=kx+5$. \\\\\n$\\therefore$ The solution set of the inequality $2x-4>kx+5$ is $\\boxed{x>3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52354168_187.png"
  },
  {
    "solution": "\\boxed{\\frac{60}{17}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, $AC=12$, $BC=5$. A square $CDEF$ is cut out from the triangle. What is the length of the side of square $CDEF$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral CDEF is a square, \\\\\nit follows that EF $\\parallel$ CD, and EF = FC = CD = DE. Let EF = x, then AF = 12 - x, \\\\\ntherefore, $\\triangle$ AFE $\\sim$ $\\triangle$ ACB, \\\\\nthus, $\\frac{EF}{BC} = \\frac{AF}{AC}$, \\\\\nwhich leads to $\\frac{x}{5} = \\frac{12 - x}{12}$, \\\\\nsolving this gives x = $\\boxed{\\frac{60}{17}}$, \\\\\nmeaning the side length of square CDEF is $\\frac{60}{17}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52866013.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system, the line $y =\\frac{3}{4}x +\\frac{3}{2}$ intersects the x-axis at point A and passes through point B $(2, a)$. There is a moving point P on the y-axis, and a moving point M on line BC, where point C is $(3, 0)$. What is the value of $a$?\n",
    "completion": "\\textbf{Solution:} From the given problem, we have \\(a=\\boxed{3}\\).",
    "image_path": "dataset/mm_math/images/MM_Math/52837765_67.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $D$ is a point on $AC$, and $DE \\parallel AB$, with $\\angle B = \\angle DAE$. $\\triangle ABC \\sim \\triangle DAE$; If $AB = 4$, $AD = 3$, and $AE = 6$, find the length of $BC$.\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABC\\sim \\triangle DAE$,\\\\\nit follows that $\\frac{BC}{AE}=\\frac{BA}{AD}$,\\\\\ngiven $AB=4$, $AD=3$, and $AE=6$,\\\\\nwe have $\\frac{BC}{6}=\\frac{4}{3}$,\\\\\nthus, $BC=\\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53078854_119.png"
  },
  {
    "solution": "\\boxed{48^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, BD is the diagonal of the rhombus ABCD, and $\\angle CBD = 76^\\circ$. The perpendicular bisector of segment AB, with the foot of the perpendicular being point E, intersects AD at point F. Connect BF. Find the degree of $\\angle DBF$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rhombus, \\\\\nit follows that CD=CB, $AB\\parallel CD$, and $\\angle A=\\angle C$, \\\\\nBecause $\\angle CBD=76^\\circ$, \\\\\nit follows that $\\angle CDB=\\angle CBD=76^\\circ$, \\\\\nhence $\\angle ABD=\\angle CDB=76^\\circ$, \\\\\nthus $\\angle A=\\angle C=180^\\circ-\\angle CDB-\\angle CBD=28^\\circ$, \\\\\nBecause FE is the perpendicular bisector of segment AB, \\\\\nit follows that AF=BF, \\\\\nhence $\\angle FBA=\\angle A=28^\\circ$, \\\\\nthus $\\angle DBF=\\angle ABD-\\angle FBA=\\boxed{48^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52160216_152.png"
  },
  {
    "solution": "\\boxed{\\frac{81}{64}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the parabola $y=ax^2+bx+3$ ($a\\neq 0$) intersects with the x-axis at points $A(-4,0)$ and $B(6,0)$, and intersects with the y-axis at point $C$. If $G$ is a moving point between $A$ and $C$ on the parabola, draw line $GD\\parallel x$ axis through point $G$, intersecting the parabola at point $D$. Draw perpendiculars to the x-axis through points $D$ and $G$, with feet at $E$ and $F$ respectively, to form rectangle $DEFG$. If line $BC$ intersects $DG$ and $DE$ at points $M$ and $N$ respectively, what is the maximum area of $\\triangle DMN$?\n",
    "completion": "\\textbf{Solution:} Let the equation of line BC be $y=kx+m$ ($k\\neq 0$). By substituting B(6,0) and C(0,3) into the equation, we get\n\\[\n\\left\\{\n\\begin{array}{l}\n6k+m=0\\\\\nm=3\n\\end{array}\n\\right.\n\\]\nFrom which we solve\n\\[\n\\left\\{\n\\begin{array}{l}\nk=-\\frac{1}{2}\\\\\nm=3\n\\end{array}\n\\right.\n\\]\nTherefore, the expression for line BC is $y=-\\frac{1}{2}x+3$.\n\nAssume the x-coordinate of point D is n, due to symmetry, $2\\leq n\\leq 6$,\n\nTherefore, the coordinates of points D and N are respectively D$\\left(n, -\\frac{1}{8}n^{2}+\\frac{1}{4}n+3\\right)$, N$\\left(n, -\\frac{1}{2}n+3\\right)$.\n\nTherefore $DN=-\\frac{1}{8}n^{2}+\\frac{1}{4}n+3-\\left(-\\frac{1}{2}n+3\\right)=-\\frac{1}{8}\\left(n-3\\right)^{2}+\\frac{9}{8}$.\n\nTherefore, when $n=3$, DN reaches its maximum value of $\\frac{9}{8}$.\n\nSince DG is parallel to the x-axis,\n\nTherefore, $\\angle DMN=\\angle OBC$.\n\nAlso, since $\\angle MDN=\\angle BOC=90^\\circ$\n\nTherefore, $\\triangle DMN\\sim \\triangle OBC$.\n\nTherefore, $\\frac{S_{\\triangle DMN}}{S_{\\triangle OBC}}=\\left(\\frac{DN}{OC}\\right)^{2}$.\n\nTherefore, when DN is at its maximum, the area of $\\triangle DMN$ is also at its maximum.\n\nSince $S_{\\triangle OBC}=3\\times 6\\times \\frac{1}{2}=9$,\n\nTherefore $S_{\\triangle DMN}=S_{\\triangle OBC}\\times \\left(\\frac{DN}{OC}\\right)^{2}=9\\times \\left(\\frac{9}{8}\\div 3\\right)^{2}=\\boxed{\\frac{81}{64}}$.\n\nTherefore, the maximum area of $\\triangle DMN$ is $\\boxed{\\frac{81}{64}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52550670_89.png"
  },
  {
    "solution": "\\boxed{9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $OP$ bisects $\\angle MON$, and point $C$ is any point on $OP$ with $CA\\perp ON$ at foot $A$. The perpendicular bisector of line segment $OA$, denoted as $BG$, intersects $OM$ at point $B$ and intersects $OA$ at point $G$. Given $AB=6$ and $AC=3$, what is the area of $\\triangle OBC$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw line $CE\\perp OM$ at point E. Since $OP$ bisects $\\angle MON$, $CA\\perp ON$, and $CE\\perp OM$, it follows that $CE=CA=3$. Since $BG$ is the perpendicular bisector of $OA$, we have $BA=BO=6$. The area of $\\triangle OBC$ is given by $S_{\\triangle OBC}=\\frac{1}{2}\\times OB\\times CE=\\frac{1}{2}\\times 6\\times 3=\\boxed{9}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55553380.png"
  },
  {
    "solution": "\\boxed{18m}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the famous scenic spot Sheng'an Pavilion in the Western Hills of Kunming is founded on a regular hexagon with a radius of 3m. What is the perimeter of the regular hexagon in meters?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect $OA$ and $OB$,\\\\\nsince the radius of the regular hexagon is $3m$,\\\\\ntherefore $OA=OB=3m$,\\\\\nsince the figure is a regular hexagon,\\\\\ntherefore $\\angle AOB=60^\\circ$,\\\\\ntherefore $\\triangle AOB$ is an equilateral triangle,\\\\\ntherefore $AB=3m$,\\\\\ntherefore the perimeter of the regular hexagon is: $3\\times 6=\\boxed{18m}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52602674.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, by changing the interior angles of the square $ABCD$, the square is transformed into a rhombus $AB{C}^{\\prime }{D}^{\\prime }$. If $\\angle {D}^{\\prime }AB=30^\\circ$, what is the ratio of the area of the rhombus $AB{C}^{\\prime }{D}^{\\prime }$ to the area of the square $ABCD$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $D'M\\perp AB$ at point M through point $D'$,\\\\\n$\\because \\angle D'AB=30^\\circ$,\\\\\n$\\therefore D'M=\\frac{1}{2}AD'$,\\\\\n$\\because$ Quadrilateral $ABC'D'$ is a rhombus,\\\\\n$\\therefore AD'=AB$,\\\\\n$\\therefore$ The area of rhombus $ABC'D'$ is $AB\\times D'M= AB\\times \\frac{1}{2}AD'=AB\\times \\frac{1}{2}AB=\\frac{1}{2}AB^2$,\\\\\n$\\because$ The area of square $ABCD=AB^2$,\\\\\n$\\therefore$ The ratio of the area of rhombus $ABC'D'$ to the area of square $ABCD$ is $\\boxed{\\frac{1}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51510273.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, given that $\\angle AOB=90^\\circ$, draw rays $OC$, and $OD$ is the angle bisector of $\\angle AOC$, $OE$ is the angle bisector of $\\angle BOC$. As shown in Figure (1), when $\\angle BOC=70^\\circ$, find the degree measure of $\\angle DOE$.\n",
    "completion": "\\textbf{Solution:} Since $\\angle AOB=90^\\circ$ and $\\angle BOC=70^\\circ$,\\\\\nit follows that $\\angle AOC=90^\\circ-\\angle BOC=20^\\circ$,\\\\\nSince lines $OD$ and $OE$ bisect $\\angle AOC$ and $\\angle BOC$ respectively,\\\\\nit follows that $\\angle COD=\\frac{1}{2}\\angle AOC=10^\\circ$ and $\\angle COE=\\frac{1}{2}\\angle BOC=35^\\circ$,\\\\\ntherefore $\\angle DOE=\\angle COD+\\angle COE=\\boxed{45^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/47364521_202.png"
  },
  {
    "solution": "\\boxed{y=3x^2+2x-3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a parabola $y=ax^2+bx+c$ ($a\\neq 0$) with the axis of symmetry $x=-1$ intersects the x-axis at points $A$ and $B$ $(-3, 0)$, and the point $(2, 5)$ lies on the parabola $y=ax^2+bx+c$. Determine the equation of the parabola.\n",
    "completion": "\\textbf{Solution}: By the symmetry axis of the parabola being $x=-1$ and point $A(-3, 6)$ being on the parabola, we get the following system of equations:\n\\[\n\\begin{cases}\n9a-3b+c=6\\\\\n4a+2b+c=5\\\\\n-\\frac{b}{2a}=-1\n\\end{cases}\n\\]\nSolving this, we obtain:\n\\[\n\\begin{cases}\na=3\\\\\nb=2\\\\\nc=-3\n\\end{cases}\n\\]\n$\\therefore$ The equation of the parabola is: \\boxed{y=3x^2+2x-3}.",
    "image_path": "dataset/mm_math/images/MM_Math/55138456_268.png"
  },
  {
    "solution": "\\boxed{7.5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in rectangle $ABCD$, $AB=3\\text{cm}$, $AD=9\\text{cm}$. If this rectangle is folded such that point $B$ coincides with point $D$, with the crease being $EF$, what is the area of $\\triangle BEF$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, the point corresponding to point C after folding is G,\\\\\nIn the rectangle ABCD, AB=CD=3cm, AD=9cm, $\\angle C=90^\\circ$,\\\\\nBased on the folding, it can be concluded that: $\\angle G=\\angle C=90^\\circ$, BG=CD=3cm, GF=CF,\\\\\nLet BF=xcm, then GF=CF=$(9-x)$cm,\\\\\nIn $\\triangle BGF$, according to the Pythagorean theorem: \\\\\n$3^2+(9-x)^2=x^2$, solving for x gives $x=5$,\\\\\n$\\therefore S_{\\triangle BEF}=\\frac{1}{2}\\times BF\\times AB=\\frac{1}{2}\\times 5\\times 3=\\boxed{7.5}\\text{(cm}^2\\text{)}$",
    "image_path": "dataset/mm_math/images/MM_Math/53437522.png"
  },
  {
    "solution": "\\boxed{-\\frac{4}{7}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the Cartesian coordinate system, the coordinate of point $A$ is $(6, 4)$. Line segments $AB$ and $AC$ are drawn, perpendicular to the x-axis at point $B$ and to the y-axis at point $C$, respectively. The graph of the linear function $y=kx+b$ passes through point $P(2, 3)$. When the line $y=kx+b$ ($k<0$) divides the rectangle $OBAC$ into two parts with an area ratio of $3:5$, directly write down the value of $k$.\n",
    "completion": "\\textbf{Solution:} Solve: $k=-4, k=-\\frac{4}{7}$.\n\nTherefore, the value of $k$ is $k=\\boxed{-4}$ or $k=\\boxed{-\\frac{4}{7}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52342947_117.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $\\angle AOB = 80^\\circ$, $OC$ is the bisector of $\\angle AOB$, and $OD$ is the bisector of $\\angle BOC$. What is the measure of $\\angle AOD$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle AOB=80^\\circ$ and OC is the bisector of $\\angle AOB$, \\\\\nit follows that $\\angle AOC=\\angle BOC=0.5\\times 80^\\circ=40^\\circ$. \\\\\nAlso, since OD is the bisector of $\\angle BOC$, \\\\\nit follows that $\\angle COD=\\angle BOD=0.5\\times 40^\\circ=20^\\circ$. \\\\\nTherefore, $\\angle AOD=\\angle AOC+\\angle COD=40^\\circ+20^\\circ=\\boxed{60^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52796437_117.png"
  },
  {
    "solution": "\\boxed{5\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the perimeter of square $ABCD$ is $40$. Point $P$ is a moving point on the diagonal $AC$ of square $ABCD$. Perpendiculars are drawn from point $P$ to $AB$ and $BC$ respectively, with the feet of the perpendiculars being $E$ and $F$, respectively. Quadrilateral $PEBF$ is a rectangle, and $PD=EF$. As point $P$ moves, the length of $EF$ also changes. What is the minimum value of $EF$?\n",
    "completion": "\\textbf{Solution:} Given equation (2), we have $DP=EF$. Thus, the minimum value of $EF$, which is also the minimum value of $DP$, occurs when $DP\\perp AC$, meaning $DP$ is perpendicular to $AC$.\n\nSince the perimeter of square ABCD is 40,\n\\[AD=CD=10.\\]\n\nGiven $AD=CD$ and $\\angle ADC=90^\\circ$,\n\\[AC=\\sqrt{10^2+10^2}=10\\sqrt{2}.\\]\n\nSince $DP\\perp AC$,\n\\[DP=\\frac{1}{2}AC=5\\sqrt{2}.\\]\n\nTherefore, the minimum value of $EF$ is \\boxed{5\\sqrt{2}}.",
    "image_path": "dataset/mm_math/images/MM_Math/52422339_95.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle B=45^\\circ$, $\\tan C=\\frac{2}{3}$, and $AC=2\\sqrt{13}$. Find the length of $BC$.\n",
    "completion": "\\textbf{Solution:} According to the problem, draw $AD \\perp BC$ through point A, as shown:\n\n$\\therefore$ $\\triangle ABD$ and $\\triangle ACD$ are both right-angled triangles,\n\n$\\because$ $\\tan C = \\frac{AD}{CD} = \\frac{2}{3}$,\n\nLet $AD = 2x$, $CD = 3x$,\n\n$\\therefore$ $AC = \\sqrt{(2x)^2 + (3x)^2} = 2\\sqrt{13}$,\n\nSolving, we get: $x = 2$ (negative value is discarded),\n\n$\\therefore$ $AD = 4$, $CD = 6$,\n\n$\\because$ $\\angle B = 45^\\circ$,\n\n$\\therefore$ $BD = AD = 4$,\n\n$\\therefore$ $BC = 4 + 6 = \\boxed{10}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51351195.png"
  },
  {
    "solution": "\\boxed{\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the side length of square $ABCD$ and square $EFCG$ are respectively 3 and 1, with points $F$ and $G$ respectively on sides $BC$ and $CD$. Let $P$ be the midpoint of $AE$. When connecting $PG$, what is the length of $PG$?\n",
    "completion": "\\textbf{Solution:} Extend $GE$ to meet $AB$ at point $O$, and draw $PH \\perp OE$ at point $H$,\\\\\nsince square $ABCD$, and square $EFCG$,\\\\\ntherefore, $EF=1$, $AB=3$, quadrilateral $EFBO$, $ODG$ are rectangles,\\\\\ntherefore, $EF=OB=1$ therefore $AO=BF=AB-OB=3-1=2$;\\\\\nsince $PH\\parallel AB$, and since $P$ is the midpoint of $AE$,\\\\\ntherefore, $PH$ is the midline of $\\triangle AOE$,\\\\\ntherefore, $PH=\\frac{1}{2}OA=\\frac{1}{2}\\times 2=1$,\\\\\nsince in $\\triangle AOE$, $\\angle OAE=45^\\circ$,\\\\\ntherefore, $\\triangle AOE$ is an isosceles right triangle,\\\\\ntherefore, $OA=OE=2$, similarly, it is known that $HE=PH=1$.\\\\\ntherefore, $HG=HE+EG=1+1=2$.\\\\\nIn $\\triangle PHG$,\\\\\n$PG=\\sqrt{PH^{2}+HG^{2}}=\\sqrt{1^{2}+2^{2}}=\\boxed{\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52188275.png"
  },
  {
    "solution": "\\boxed{\\frac{76}{17}\\le t \\leq \\frac{80}{17}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the line $y=-\\frac{3}{4}x+m$ intersects the x-axis at point B and the y-axis at point A, where the coordinate of point B is (12, 0). The line $y=\\frac{3}{8}x$ intersects line AB at point C. Point D is a moving point on line AB, and a perpendicular line is drawn from point D to the x-axis, intersecting line OC at point E. Let the x-coordinate of point D be $t$, and the length of segment DE be $d$. Find the functional formula of $d$ in terms of $t$ (including the range of the independent variable).\n",
    "completion": "\\textbf{Solution:}\n\n(1) As shown in the figure,\\\\\nsince the x-coordinate of point D is $t$, hence point D$(t, -\\frac{3}{4}t+9)$, point E$(t, \\frac{3}{8}t)$,\\\\\nwhen $t<8$, $d= -\\frac{3}{4}t+9-\\frac{3}{8}t=-\\frac{9}{8}t+9$,\\\\\nwhen $t>8$, $d= \\frac{3}{8}t+\\frac{3}{4}t-9=\\frac{9}{8}t-9$;\\\\\n(2) since the line segment with endpoints H$(\\frac{1}{2}, t)$ and G$(1, t)$ intersects with the edge of the square DEPQ at only one point,\\\\\nhence $\\frac{1}{2}\\leq t \\leq 1$ or $\\left\\{\\begin{array}{l}\n-\\frac{9}{8}t+9\\le t-\\frac{1}{2} \\\\\n-\\frac{9}{8}t+9\\ge t-1\n\\end{array}\\right.$,\\\\\nhence $\\frac{1}{2}\\leq t \\leq 1$ or \\boxed{\\frac{76}{17}\\le t \\leq \\frac{80}{17}}.",
    "image_path": "dataset/mm_math/images/MM_Math/52381062_172.png"
  },
  {
    "solution": "\\boxed{D\\left(2, 3\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Synthesis and Exploration\\\\ As shown in Figure 1, the parabola $y=-\\frac{1}{2}x^{2}+\\frac{3}{2}x+2$ intersects the x-axis at points A and B (point A is to the left of point B) and intersects the y-axis at point C. Point D lies on the parabola in the first quadrant.\\\\ If $ S_{\\triangle COD}=\\frac{1}{3}S_{\\triangle OBD}$, find the coordinates of point D.\n",
    "completion": "\\textbf{Solution:} Let $DG \\perp x$-axis and $DH \\perp y$-axis through the point D, with the feet of the perpendiculars being G and H, respectively.\n\nSuppose $D\\left(m, -\\frac{1}{2}m^{2}+\\frac{3}{2}m+2\\right)$.\n\nSince the area of $\\triangle DCO = \\frac{1}{2}OC \\cdot DH = \\frac{1}{2} \\times 2 \\times m$.\n\nThe area of $\\triangle BOD = \\frac{1}{2}OB \\cdot DG = \\frac{1}{2} \\times 4 \\times \\left(-\\frac{1}{2}m^{2}+\\frac{3}{2}m+2\\right)$\n\nSince the area of $\\triangle COD = \\frac{1}{3}$ the area of $\\triangle CBD$.\n\nTherefore, $\\frac{1}{2} \\times 2 \\times m = \\frac{1}{3} \\times \\frac{1}{2} \\times 4 \\times \\left(-\\frac{1}{2}m^{2}+\\frac{3}{2}m+2\\right)$\n\nSolving this, we get $m_{1}=2$, $m_{2}=-2$ (discard).\n\nWhen $m=2$, $-\\frac{1}{2}m^{2}+\\frac{3}{2}m+2=3$.\n\nTherefore, $D\\left(2, 3\\right)$.\n\nHence, the coordinates of point D are $\\boxed{D\\left(2, 3\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52866560_297.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $AB$ is the diameter of the circle $\\odot O$, $\\angle ACD$ is the angle subtended by arc $AD$ on the circumference of the circle, and $\\angle ACD=30^\\circ$. A line is drawn through point $D$ such that $DE\\perp AB$ with $E$ being the foot of the perpendicular, and the extension of $DE$ intersects $\\odot O$ at point $F$. If $AB=4$, what is the length of $DF$?\n",
    "completion": "\\textbf{Solution:} Extend line OD, \\\\\n$\\because \\angle DAB=60^\\circ$ and OD=OA=2,\\\\\n$\\therefore \\triangle AOD$ is an equilateral triangle,\\\\\n$\\therefore AD=2$,\\\\\n$\\because DE \\perp AB$,\\\\\n$\\therefore \\angle ADE=30^\\circ$ and DF=2DE,\\\\\n$\\therefore AE=\\frac{1}{2}AD=1$,\\\\\nIn $\\triangle ADE$, $DE=\\sqrt{AD^{2}-AE^{2}}=\\sqrt{3}$,\\\\\n$\\therefore DF=2\\sqrt{3}=\\boxed{2\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52874717_96.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in rhombus $ABCD$, point $P$ is a moving point on the diagonal $AC$, through point $P$, draw $PE \\perp AB$ at point $E$, and $PF \\perp BC$ at point $F$. If the perimeter of the rhombus $ABCD$ is 48 and its area is 48, what is the value of $PE+PF$?\n",
    "completion": "\\textbf{Solution:} Extend FP to meet AD at point G, as shown in the figure:\\\\\nIn rhombus ABCD, AD$\\parallel$BC, $\\angle$DAC=$\\angle$BAC,\\\\\nAlso, $\\because$PF$\\perp$BC,\\\\\n$\\therefore$PF$\\perp$AD,\\\\\n$\\therefore$$\\angle$AGP=$90^\\circ$,\\\\\n$\\because$PE$\\perp$AB,\\\\\n$\\therefore$$\\angle$AEP=$90^\\circ$,\\\\\n$\\therefore$$\\angle$AEP=$\\angle$AGP,\\\\\nAlso, $\\because$$\\angle$EAP=$\\angle$GAP, AP=AP,\\\\\n$\\therefore$$\\triangle $EAP$\\cong $$\\triangle $GAP (AAS),\\\\\n$\\therefore$GP=EP,\\\\\n$\\because$The perimeter of rhombus ABCD is 48,\\\\\n$\\therefore$BC=12,\\\\\n$\\because$The area of rhombus ABCD is 48, i.e., $BC\\cdot FG$=48,\\\\\n$\\therefore$FG=48÷12=4,\\\\\n$\\therefore$PE+PF=GF=\\boxed{4},",
    "image_path": "dataset/mm_math/images/MM_Math/52250418.png"
  },
  {
    "solution": "\\boxed{13}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in $\\triangle ABC$, $AB=AC$, $AD$ bisects $\\angle BAC$. Given $BC=10$, $AD=12$, find the length of $AC$.\n",
    "completion": "\\textbf{Solution:} Since $AB=AC$ and $AD$ bisects $\\angle BAC$,\\\\\nit follows that $AD\\perp BC$, $BD=CD=\\frac{1}{2}BC=5$,\\\\\nGiven that $AD=12$,\\\\\nwe have $AC=\\sqrt{AD^{2}+CD^{2}}=\\sqrt{12^{2}+5^{2}}=\\boxed{13}$",
    "image_path": "dataset/mm_math/images/MM_Math/52914200.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated, $E$ is a point within $\\triangle ABC$, $BE$ bisects $\\angle ABC$, $AD \\perp BE$ at point $E$, intersecting $BC$ at point $D$, where point $D$ is precisely on the perpendicular bisector of side $AC$, $BC=10$, and $AB=6$. What is the length of $AE$?\n",
    "completion": "\\textbf{Solution:} Given that BE bisects $\\angle$ABC, and AD$\\perp$BE,\\\\\nit follows that $\\triangle ABD$ is an isosceles triangle,\\\\\nhence, AB=BD=6, and AE=ED,\\\\\ngiven that BC=10,\\\\\nthus, DC=BC-BD=10-6=4,\\\\\nsince point D lies exactly on the perpendicular bisector of side AC,\\\\\nthus, DA=DC=4,\\\\\ntherefore, AE=\\boxed{2},",
    "image_path": "dataset/mm_math/images/MM_Math/52603530.png"
  },
  {
    "solution": "\\boxed{22.5^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $OC \\perp AB$ at point O, $OD$ bisects $\\angle BOC$, and $OE$ bisects $\\angle AOD$. Find the degree measure of $\\angle COE$.\n",
    "completion": "\\textbf{Solution:} From (1), we have\n\\[ \\angle AOD = \\angle AOC + \\angle COD = 90^\\circ + 45^\\circ = 135^\\circ, \\]\n\\[ \\because OE \\text{ bisects } \\angle AOD, \\]\n\\[ \\therefore \\angle AOE = \\frac{1}{2}\\angle AOD = 67.5^\\circ \\]\n\\[ \\therefore \\angle COE = 90^\\circ - 67.5^\\circ = \\boxed{22.5^\\circ} \\]",
    "image_path": "dataset/mm_math/images/MM_Math/53364075_165.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $l: y=k_{1}x+b$ intersects the hyperbola $y=\\frac{k_{2}}{x}$ (for $x>0$) at points $A$ and $B$, and intersects the $x$-axis at point $C$. If the abscissas of points $A$ and $B$ are $1$ and $2$, respectively, and the area of $\\triangle AOB$ is $3$, what is the value of $k_{2}$?\n",
    "completion": "\\textbf{Solution:} Let $AM\\perp x$-axis at $M$, and $BN\\perp x$-axis at $N$. Hence, the area of $\\triangle AOM$ and the area of $\\triangle BON$ are both equal to $\\frac{1}{2}|k_{2}|$, \\\\\nLet $A(1,k_{2})$, and $B(2,\\frac{k_{2}}{2})$, \\\\\nSince the area of $\\triangle AOB$ is $3$, \\\\\nTherefore, $S_{\\triangle AOB}=S_{\\triangle AOM}+S_{\\text{trapezoid} AMNB}-S_{\\triangle BON}=S_{\\text{trapezoid} AMNB}=\\frac{1}{2}\\left(k_{2}+\\frac{k_{2}}{2}\\right)\\times(2-1)=3$, \\\\\nHence, $k_{2}=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52601705_137.png"
  },
  {
    "solution": "\\boxed{\\frac{2\\sqrt{5}}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, square $ABCD$ has a side length of $2$. Points $E$ and $F$ are the midpoints of $AB$ and $CD$, respectively. Connect $CE$ and $AF$. Draw $DG \\perp AF$ through point $D$ with $G$ as the foot of the perpendicular. Extend $DG$ to intersect $CE$ at point $H$. Find the length of $GH$.\n",
    "completion": "\\textbf{Solution:} Since points E and F are the midpoints of AB and CD respectively, \\\\\ntherefore $AE=\\frac{1}{2}AB$, $CF=\\frac{1}{2}CD$, \\\\\nhence $AE=CF$, \\\\\nsince $AE\\parallel CF$, \\\\\nthus quadrilateral AFCE is a parallelogram, \\\\\ntherefore $AF\\parallel CE$, \\\\\nsince point F is the midpoint of DC, \\\\\nthus point G is the midpoint of DH, \\\\\ntherefore $DG=GH=\\boxed{\\frac{2\\sqrt{5}}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52160938_145.png"
  },
  {
    "solution": "\\boxed{90^\\circ+\\frac{1}{2}\\alpha}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in $\\triangle ABC$, if $\\triangle ABD$ and $\\triangle ACE$ are constructed with $AB$ and $AC$ as their respective sides, and $\\angle DAB = \\angle CAE = \\alpha$, $AD = AB$, $AC = AE$, and $DC$ and $BE$ intersect at point $P$, with $AP$ connected, then what is the measure of $\\angle APC$?\n",
    "completion": "\\textbf{Solution:} As illustrated, construct $AF\\perp CD$ at point F, and $AG\\perp BE$ at point G, then $\\angle AFC=\\angle AGE=90^\\circ$,\\\\\n$\\because \\angle DAB=\\angle CAE=\\alpha$,\\\\\n$\\therefore \\angle DAC=\\angle BAE=\\alpha +\\angle BAC$,\\\\\nIn $\\triangle DAC$ and $\\triangle BAE$,\\\\\n$ \\left\\{\\begin{array}{l}AD=AB\\\\ \\angle DAC=\\angle BAE\\\\ AC=AE\\end{array}\\right.$,\\\\\n$\\therefore \\triangle DAC\\cong \\triangle BAE$ (SAS),\\\\\n$\\therefore \\angle ACF=\\angle AEG$,\\\\\nIn $\\triangle ACF$ and $\\triangle AEG$\\\\\n$ \\left\\{\\begin{array}{l}\\angle AFC=\\angle AGE\\\\ \\angle ACF=\\angle AEG\\\\ AC=AE\\end{array}\\right.$,\\\\\n$\\therefore \\triangle ACF\\cong \\triangle AEG$ (AAS),\\\\\n$\\therefore AF=AG$,\\\\\n$\\therefore$ point A is on the bisector of $\\angle DPE$,\\\\\n$\\therefore \\angle APE=\\angle APD=\\frac{1}{2}\\angle DPE$,\\\\\n$\\because \\angle CPE+\\angle ACF=\\angle CAE+\\angle AEG=\\angle AHP$,\\\\\n$\\therefore \\angle CPE=\\angle CAE=\\alpha$,\\\\\n$\\therefore \\angle APE=\\frac{1}{2}\\angle DPE=\\frac{1}{2}(180^\\circ-\\angle CPE)=90^\\circ-\\frac{1}{2}\\alpha$,\\\\\n$\\therefore \\angle APC=\\angle APE+\\angle CPE=90^\\circ-\\frac{1}{2}\\alpha+\\alpha=90^\\circ+\\frac{1}{2}\\alpha$,\\\\\n$\\therefore$ the measure of $\\angle APC$ is $90^\\circ+\\frac{1}{2}\\alpha$.\\\\\nTherefore, the measure of $\\angle APC$ is $\\boxed{90^\\circ+\\frac{1}{2}\\alpha}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52551635.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB = 90^\\circ$, $CD \\perp AB$ with the foot of the perpendicular as D. If $AD = 2$ and $BD = 4$, then what is the length of $CD$?\n",
    "completion": "\\textbf{Solution:} In the right $\\triangle ACB$, $\\angle ACB=90^\\circ$, $CD\\perp AB$,\\\\\n$\\therefore \\angle ACD=\\angle B=90^\\circ-\\angle A$,\\\\\nAlso, $\\because \\angle ACD=\\angle CDB=90^\\circ$,\\\\\n$\\therefore \\triangle ACD\\sim \\triangle CBD$,\\\\\n$\\therefore CD^2=AD\\cdot BD=8$, that is $CD=\\boxed{2\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/9061650.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in a square grid, the three vertices of triangle $ABC$ are all on the grid points. What is the value of $\\tan\\angle B$?\n",
    "completion": "\\textbf{Solution:} To find point $B$, move point $C$ down by 1 unit and then left by 3 units. Applying the same procedure to point $A$ gives us point $D$. Connect $AD$, $BD$, $CD$, $AB$ with $AB$ and $CD$ intersecting at point $O$, as shown in the figure: \\\\\nTherefore, quadrilateral $ACBD$ is a parallelogram. \\\\\nSince $AC=BC=\\sqrt{3^2+1^2}=\\sqrt{10}$ and $AB=\\sqrt{4^2+4^2}=4\\sqrt{2}$, $CD=\\sqrt{2^2+2^2}=2\\sqrt{2}$, \\\\\nthus, parallelogram $ACBD$ is a rhombus. \\\\\nTherefore, $AB\\perp CD$, $OB=\\frac{1}{2}AB=2\\sqrt{2}$, $OC=\\frac{1}{2}CD=\\sqrt{2}$, \\\\\nin $\\triangle BOC$, $\\tan\\angle ABC=\\frac{OC}{OB}=\\frac{\\sqrt{2}}{2\\sqrt{2}}=\\boxed{\\frac{1}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52511772.png"
  },
  {
    "solution": "\\boxed{2.5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within rectangle $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$. Points $E$ and $F$ are respectively the midpoints of $AO$ and $AD$. Connecting $EF$, if $AB = 6\\,cm$ and $BC = 8\\,cm$, then what is the length of $EF$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rectangle,\\\\\nit follows that $\\angle$ABC$=90^\\circ$, BD = AC, and BO = OD,\\\\\ngiven that AB = 6 cm, and BC = 8 cm,\\\\\nby the Pythagorean theorem, we find: AC$= \\sqrt{AB^{2}+BC^{2}} = \\sqrt{6^{2}+8^{2}} = 10$ (cm),\\\\\nhence, BD = 10 cm, and DO = 5 cm,\\\\\ngiven that points E and F are the midpoints of AO and AD respectively,\\\\\nit follows that EF is the midline of $\\triangle AOD$,\\\\\ntherefore, EF$= \\frac{1}{2} \\times$OD = \\boxed{2.5} cm.",
    "image_path": "dataset/mm_math/images/MM_Math/51919732.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{2}a}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\angle AOB$ is a right angle and $\\angle BOC$ is an acute angle. Moreover, $OM$ bisects $\\angle AOC$ and $ON$ bisects $\\angle BOC$. Given $\\angle AOB = 90^\\circ$ and $\\angle BOC = \\beta$, with $\\alpha$ and $\\beta$ being acute angles and all other conditions unchanged, what is the degree measure of $\\angle MON$?\n",
    "completion": "\\textbf{Solution:} \nGiven: $\\angle AOB=a$ and $\\angle BOC=\\beta$,\\\\\nThus, $\\angle AOC=\\angle AOB+\\angle BOC= a+\\beta$,\\\\\nSince OM bisects $\\angle AOC$ and ON bisects $\\angle BOC$,\\\\\nThus $\\angle MOC= \\frac{1}{2}\\angle AOC= \\frac{1}{2}(a+\\beta)$ and $\\angle NOC= \\frac{1}{2}\\angle BOC= \\frac{1}{2}\\beta$,\\\\\nTherefore, $\\angle MON=\\angle MOC-\\angle NOC= \\frac{1}{2}(a+\\beta)- \\frac{1}{2}\\beta= \\frac{1}{2}a$. Hence, the measure of $\\angle MON$ is $\\boxed{\\frac{1}{2}a}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52794787_108.png"
  },
  {
    "solution": "\\boxed{1.5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in parallelogram $ABCD$, $AB = 3.5\\,cm$, $BC = 5\\,cm$, $AE$ bisects $\\angle BAD$, $CF\\parallel AE$, what is the length of $AF$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a parallelogram, \\\\\n$\\therefore$ AD $\\parallel$ BC, \\\\\n$\\therefore$ $\\angle$DAE = $\\angle$AEB, \\\\\nSince AE $\\parallel$ CF, \\\\\n$\\therefore$ quadrilateral AECF is a parallelogram, \\\\\n$\\therefore$ AF = CE, \\\\\nSince AE bisects $\\angle$BAD, \\\\\n$\\therefore$ $\\angle$BAE = $\\angle$EAD, \\\\\n$\\therefore$ $\\angle$BAE = $\\angle$AEB, \\\\\n$\\therefore$ AB = BE = 3.5\\text{cm}, \\\\\n$\\therefore$ EC = BC - BE = 5 - 3.5 = 1.5\\text{cm}, \\\\\n$\\therefore$ AF = \\boxed{1.5}\\text{cm}",
    "image_path": "dataset/mm_math/images/MM_Math/52386169.png"
  },
  {
    "solution": "\\boxed{\\sqrt{13}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the diagonals $AC, BD$ of the rhombus $ABCD$ intersect at point $O$. If $AC=4, BD=6$, what is the length of $AB$?\n",
    "completion": "\\textbf{Solution:} Since rhombus $ABCD$ has diagonals $AC, BD$ intersecting at point $O$, and $AC=4, BD=6$,\\\\\nit follows that $AC\\perp BD$, $AO=\\frac{1}{2}AC=2$, $BO=\\frac{1}{2}BD=3$,\\\\\nIn $\\triangle AOB$, $AB^2=AO^2+BO^2$,\\\\\nthus $AB=\\sqrt{AO^2+BO^2}=\\sqrt{2^2+3^2}=\\boxed{\\sqrt{13}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52065356.png"
  },
  {
    "solution": "\\boxed{y=-x^2-3x+4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: The parabola $y=ax^2+bx+4$ intersects the x-axis at points $A(-4, 0)$ and $B(1, 0)$, and intersects the y-axis at point C. Point P is a moving point on the parabola above the line $AC$. Find the equation of the parabola.\n",
    "completion": "\\textbf{Solution:} Since the parabola $y=ax^2+bx+4$ intersects the x-axis at points $A(-4, 0)$ and $B(1, 0)$, \\\\\nwe have $\\left\\{\\begin{array}{l}\n16a+4b+4=0 \\\\\na+b+4=0\n\\end{array}\\right.$. \\\\\nSolving this system, we find $\\left\\{\\begin{array}{l}\na=-1 \\\\\nb=-3\n\\end{array}\\right.$. \\\\\nTherefore, the quadratic function is $\\boxed{y=-x^2-3x+4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53107240_275.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of $\\odot O$, and chord $CD$ is perpendicular to $AB$ at point $E$, $OC=5\\,cm$, $CD=8\\,cm$. What is the length of $OE$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram:\\\\\nSince $AB$ is the diameter of the circle, and $CD \\perp AB$,\\\\\nTherefore, $\\angle OEC = 90^\\circ$, $CE = \\frac{1}{2}CD = 4\\,\\text{cm}$,\\\\\nIn $\\triangle OCE$ right-angled at $E$, $OE = \\sqrt{OC^{2} - CE^{2}} = \\sqrt{5^{2} - 4^{2}} = \\boxed{3}\\,\\text{cm}$",
    "image_path": "dataset/mm_math/images/MM_Math/52874699.png"
  },
  {
    "solution": "\\boxed{Q_3\\left(\\frac{3+\\sqrt{17}}{2}, \\frac{-1-\\sqrt{17}}{2}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the vertex of the parabola is $P(1, 4)$, it intersects the y-axis at point $C(0, 3)$, and intersects the x-axis at points $A$ and $B$. Let $Q$ be a point on the parabola other than point $P$, and the area of $\\triangle BCQ$ is equal to the area of $\\triangle BCP$, find the coordinates of point $Q$.\n",
    "completion": "\\textbf{Solution:} Given the parabola's equation $y=-x^2+2x+3$, by setting $y=0$, we find $x_1=-1, x_2=3$,\\\\\n$\\therefore$ points $A(-1, 0), B(3, 0)$,\\\\\nLet the equation of line BC be $y=kx+b$, then:\\\\\n$\\begin{cases}\n3k+b=0\\\\\nb=3\n\\end{cases}$, solving gives: $\\begin{cases}\nk=-1\\\\\nb=3\n\\end{cases}$,\\\\\n$\\therefore$ the equation of line BC is $y=-x+3$,\\\\\nDraw $PQ_1 \\parallel BC$ through $P$, intersecting the parabola at point $Q_1$,\\\\\nLet the equation of line $PQ_1$ be $y=-x+n$,\\\\\n$\\because P(1, 4)$,\\\\\n$\\therefore$ equation of line $PQ_1$ is $y=-x+5$,\\\\\nBy solving: $\\begin{cases}\ny=-x+5\\\\\ny=-x^2+2x+3\n\\end{cases}$,\\\\\nWe get: $\\begin{cases}\nx=1\\\\\ny=4\n\\end{cases}$ or $\\begin{cases}\nx=2\\\\\ny=3\n\\end{cases}$, that is, $(1, 4)$ coincides with $P$, $Q_1(2, 3)$;\\\\\nThrough point $P$, draw a perpendicular to the $x$-axis intersecting BC at $G$, take $PG=GH$ on line $PG$,\\\\\n$\\because$ the equation of line BC is $y=-x+3$, $P(1, 4)$,\\\\\n$\\therefore G(1, 2)$,\\\\\n$\\therefore PG=GH=2$,\\\\\n$\\therefore H(1, 0)$,\\\\\nDraw line $Q_2Q_3 \\parallel BC$ through $H$, intersecting the parabola at points $Q_2, Q_3$,\\\\\nSimilarly, we find the equation of line $Q_2Q_3$ to be $y=-x+1$,\\\\\nBy solving: $\\begin{cases}\ny=-x+1\\\\\ny=-x^2+2x+3\n\\end{cases}$,\\\\\nWe get: $\\begin{cases}\nx=\\frac{3+\\sqrt{17}}{2}\\\\\ny=\\frac{-1-\\sqrt{17}}{2}\n\\end{cases}$ or $\\begin{cases}\nx=\\frac{3-\\sqrt{17}}{2}\\\\\ny=\\frac{-1+\\sqrt{17}}{2}\n\\end{cases}$,\\\\\n$\\therefore Q_2\\left(\\frac{3-\\sqrt{17}}{2}, \\frac{-1+\\sqrt{17}}{2}\\right)$, $Q_3\\left(\\frac{3+\\sqrt{17}}{2}, \\frac{-1-\\sqrt{17}}{2}\\right)$;\\\\\nTherefore, the coordinates of point $Q$ are $\\boxed{Q_2\\left(\\frac{3-\\sqrt{17}}{2}, \\frac{-1+\\sqrt{17}}{2}\\right)}$ or $\\boxed{Q_3\\left(\\frac{3+\\sqrt{17}}{2}, \\frac{-1-\\sqrt{17}}{2}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52049799_107.png"
  },
  {
    "solution": "\\boxed{25^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, the quadrilateral ABCD is inscribed in circle $\\odot O$, with point C being the midpoint of arc $\\overset{\\frown}{BD}$. If $\\angle A=50^\\circ$, what is the measure of $\\angle CBD$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral ABCD is inscribed in circle O, $\\angle A=50^\\circ$, \\\\\n$\\therefore \\angle C=180^\\circ-\\angle A=180^\\circ-50^\\circ=130^\\circ$, \\\\\nsince point C is the midpoint of arc BD, \\\\\n$\\therefore CD=CB$, \\\\\n$\\therefore \\angle CBD=\\angle CDB$, \\\\\n$\\therefore \\angle CBD=\\frac{1}{2}\\times(180^\\circ-130^\\circ)=\\boxed{25^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52874668.png"
  },
  {
    "solution": "\\boxed{40^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, if $\\angle AOC = 110^\\circ$, what is the measure of $\\angle B$?\n",
    "completion": "\\textbf{Solution}: Since circle $O$ is inscribed in $\\triangle ABC$ and $\\angle AOC=110^\\circ$,\\\\\nit follows that $\\angle OAC+\\angle OCA=180^\\circ-\\angle AOC=70^\\circ$,\\\\\nand $\\angle OAC+\\angle OCA=\\frac{1}{2}(\\angle BAC+\\angle ACB)$,\\\\\ntherefore $\\angle BAC+\\angle ACB=140^\\circ$,\\\\\nthus $\\angle B=180^\\circ-(\\angle BAC+\\angle ACB)=180^\\circ-140^\\circ=\\boxed{40^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52720287.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the line \\( y=\\frac{3}{4}x+6\\) intersects the x-axis at point A and the y-axis at point B. Triangle \\(AOB\\) is folded along BC, with point O exactly falling on point D on side AB, where BC is the fold line. What is the length of segment AB?\n",
    "completion": "\\textbf{Solution:} Let $x=0$, then $y=6$,\\\\\nLet $y=0$, then $\\frac{3}{4}x+6=0$, solving this gives $x=-8$,\\\\\n$\\therefore$ the coordinates of point A are $(-8,0)$, and the coordinates of point B are $(0,6)$,\\\\\n$\\therefore$ OA=8, OB=6,\\\\\n$\\therefore$ $AB=\\sqrt{OA^2+OB^2}=\\boxed{10}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52212500_113.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{5}}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points $M$ and $N$ are respectively on the side $AD$ and diagonal $BD$ of the rectangle $ABCD$. By connecting $BM$, $MN$, we have $AB=1$, $BC=2$. What is the value of $\\sin \\angle ADB$?\n",
    "completion": "\\textbf{Solution}: Thus, $\\sin\\angle ADB = \\boxed{\\frac{\\sqrt{5}}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53151360_74.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Place the right-angled vertex O of the right-angled triangle OMN on the line AB, and let the ray OC bisect $\\angle AON$. As shown in the figure, if $\\angle BON = 60^\\circ$, what is the degree measure of $\\angle AOM$?\n",
    "completion": "\\textbf{Solution}: Since $\\angle MON=90^\\circ$ and $\\angle BON=60^\\circ$, \\\\\nit follows that $\\angle AOM=180^\\circ-\\angle MON-\\angle BON=180^\\circ-90^\\circ-60^\\circ=\\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53520190_87.png"
  },
  {
    "solution": "\\boxed{\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, $\\angle A > \\angle B$, the $\\triangle ABC$ is first folded along the crease $CE$ so that point $A$ can fall on $BC$. After flattening, $\\angle B$ is folded along the crease $GF$ making point $B$ coincide with point $A$, where $FG$ intersects sides $BC$ and $AB$ at points $F$ and $G$, respectively. $CD$ is the height from the hypotenuse. If $\\angle DCE = \\angle B$, then what is $\\frac{BF}{CE}$?\n",
    "completion": "\\textbf{Solution:} Connect $AF$, \\\\\n$\\because$ when $\\triangle ABC$ is folded for the first time along crease $CE$, making point $A$ fall onto $BC$, \\\\\n$\\therefore$ $\\angle ACE=\\angle BCE=45^\\circ$, \\\\\n$\\because$ by folding $\\angle B$ along crease $GF$, making point $B$ coincide with point $A$, \\\\\n$\\therefore$ $\\angle B=\\angle FAB$, $FA=FB$, \\\\\n$\\because$ $\\angle ACD+\\angle DCB=\\angle B+\\angle DCB=90^\\circ$, \\\\\n$\\therefore$ $\\angle ACD=\\angle B$, \\\\\n$\\because$ $\\angle DCE=\\angle B$, \\\\\n$\\therefore$ $\\angle ACD=\\angle DCE=\\angle B=\\frac{1}{2}\\angle ACE=22.5^\\circ$, \\\\\n$\\therefore$ $\\angle AFC=\\angle B+\\angle FAB=2\\angle B=45^\\circ$, \\\\\n$\\therefore$ $\\triangle$ AFC is an isosceles right triangle, \\\\\nLet $AC=CF=a$, then $AF=\\sqrt{a^2+a^2}=\\sqrt{2}a$, \\\\\n$\\because$ $\\angle CAB=90^\\circ-\\angle B=67.5^\\circ$, $\\angle CEA=\\angle B+\\angle BCE=67.5^\\circ$, \\\\\nthat is $\\angle CAE=\\angle CEA$, \\\\\n$\\therefore$ CA=CE, \\\\\n$\\therefore$ $\\frac{BF}{CE}=\\frac{AF}{CA}=\\frac{\\sqrt{2}a}{a}=\\boxed{\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52828076.png"
  },
  {
    "solution": "\\boxed{-4<x<3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: The graph of the linear function $y_1=mx+n$ ($m\\neq 0$) and the quadratic function $y_2=ax^2+bx+c$ ($a\\neq 0$) are shown in the figure. What is the solution set of the inequality $ax^2+bx+c>mx+n$?\n",
    "completion": "\\textbf{Solution}: From the graph, we know that the quadratic function $y_{2}=ax^{2}+bx+c$ ($a\\neq 0$) and the linear function $y_{1}=mx+n$ ($m\\neq 0$) intersect at the x-coordinates $-4$ and $3$, \\\\\n$\\because$ the inequality $ax^{2}+bx+c>mx+n$ implies that the graph of the quadratic function is above that of the linear function, \\\\\n$\\therefore$ the solution set of the inequality $ax^{2}+bx+c>mx+n$ is $-4<x<3$; \\\\\nHence, the answer is: $\\boxed{-4<x<3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53078759.png"
  },
  {
    "solution": "\\boxed{\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in parallelogram $ABCD$, $\\angle B=60^\\circ$, $AB=4$, point $H$ and point $G$ are moving points on side $DC$ and $BC$, respectively, where point $H$ does not coincide with point $C$. Connect $AH$ and $HG$. Point $E$ is the midpoint of $AH$ and point $F$ is the midpoint of $GH$. Connect $EF$. What is the minimum value of $EF$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect AG,\\\\\nbecause point E is the midpoint of AH, and point F is the midpoint of GH,\\\\\ntherefore EF=$\\frac{1}{2}AG$, thus the minimum value of EF,\\\\\ncan only be achieved when AG is at its minimum value, which is the vertical segment AG,\\\\\ndraw AM$\\perp$BC through point A, with the foot at M,\\\\\nbecause $\\angle B=60^\\circ$, $AB=4$,\\\\\ntherefore BM=2,\\\\\nAM=$\\sqrt{4^2-2^2}=2\\sqrt{3}$,\\\\\ntherefore the minimum value of EF is $\\frac{1}{2}AG=\\boxed{\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53599304.png"
  },
  {
    "solution": "\\boxed{36^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC$ and $\\angle A=36^\\circ$. An arc is drawn with $B$ as the center and the length of $BC$ as the radius, intersecting $AC$ at point $D$. When connecting $BD$, what is the measure of $\\angle ABD$?\n",
    "completion": "\\textbf{Solution:} Given $\\because$ AB=AC, $\\angle A=36^\\circ$,\\\\\n$\\therefore \\angle C= \\frac{180^\\circ-\\angle A}{2}=72^\\circ$,\\\\\n$\\because$ BC=BD,\\\\\n$\\therefore \\angle BDC=\\angle C=72^\\circ$,\\\\\n$\\therefore \\angle ABD=\\angle BDC-\\angle A=\\boxed{36^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51385932.png"
  },
  {
    "solution": "\\boxed{1.5\\pi}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $\\triangle ABC$ is an inscribed triangle of $\\odot O$, and $AD$ is the diameter of $\\odot O$. Connect $BD$, and $BC$ bisects $\\angle ABD$. If $AD=6$, then what is the length of $\\overset{\\frown}{CD}$?\n",
    "completion": "\\textbf{Solution:} Connect OC, \\\\\n$\\because$AD is the diameter, and AD=6 \\\\\n$\\therefore$$\\angle$ABD=90$^\\circ$, OD=3 \\\\\n$\\because$BC bisects $\\angle$ABD, \\\\\n$\\therefore$$\\angle$DBC= $ \\frac{1}{2}$$\\angle$ABD=45$^\\circ$, \\\\\n$\\because$$ \\overset{\\frown}{CD}=\\overset{\\frown}{CD}$, \\\\\n$\\therefore$$\\angle$DOC=2$\\angle$DBC=90$^\\circ$, \\\\\n$\\therefore$ the length of $ \\overset{\\frown}{CD}$ is $ \\frac{90^\\circ\\pi \\times 3}{180}=\\boxed{1.5\\pi} $.",
    "image_path": "dataset/mm_math/images/MM_Math/52861155.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $AE \\perp BD$ at point $E$, and $\\angle DAE = 2\\angle BAE$, what is the measure of $\\angle EAC$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral ABCD is a rectangle,\\\\\nit follows that OA=OC=OB=OD, and $\\angle$BAD=$90^\\circ$,\\\\\ntherefore, $\\angle$ABO=$\\angle$BAO,\\\\\ngiven that $\\angle$DAE=2$\\angle$BAE, and $\\angle$DAE+$\\angle$BAE=$90^\\circ$,\\\\\nit follows that $\\angle$BAE=$30^\\circ$,\\\\\ngiven that AE$\\perp$BD, that is $\\angle$AEB=$90^\\circ$,\\\\\nit follows that $\\angle$ABO=$60^\\circ$,\\\\\ntherefore, $\\angle$BAO=$60^\\circ$,\\\\\ntherefore, $\\angle$EAC=$\\angle$BAO−$\\angle$BAE=$60^\\circ−30^\\circ=\\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51614398.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{7}}{3}\\pi}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, points $A$, $B$, and $C$ are three points on the circle $\\odot O$, with $AC=BC$. Point $M$ is another point on $\\odot O$, and $CE$ is perpendicular to $AM$, with the foot of the perpendicular being point $E$. Given that $AE=2\\sqrt{3}$, $BM=\\sqrt{3}$, and $CM=\\sqrt{7}$, what is the length of arc $\\overset{\\frown}{CM}$?\n",
    "completion": "\\textbf{Solution:} Let's cut $AG=BM$ on $AE$, and connect $CG$,\\\\\nsince $AC=BC$ and $\\angle A=\\angle B$,\\\\\nthus $\\triangle ACG \\cong \\triangle BCM$ (SAS),\\\\\nhence $CG=CM=\\sqrt{7}$,\\\\\nsince $AE=2\\sqrt{3}$ and $AG=BM=\\sqrt{3}$,\\\\\nthus $GE=\\sqrt{3}$,\\\\\nsince $CE\\perp AM$,\\\\\nthus $CE = \\sqrt{CG^{2} - GE^{2}} = \\sqrt{(\\sqrt{7})^2 - (\\sqrt{3})^2} = 2$,\\\\\nthus $\\tan\\angle A = \\frac{CE}{AE} = \\frac{\\sqrt{3}}{3}$,\\\\\nthus $\\angle A = 30^\\circ$,\\\\\nthus $\\angle COM = 60^\\circ$,\\\\\nconnect $OM$ and $CO$,\\\\\nsince $OC=OM$,\\\\\nthus $\\triangle COM$ is an equilateral triangle,\\\\\nthus $OC=\\sqrt{7}$,\\\\\nthus the length of $\\overset{\\frown}{CM} = \\frac{60\\pi \\times \\sqrt{7}}{180} = \\boxed{\\frac{\\sqrt{7}}{3}\\pi}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51130768.png"
  },
  {
    "solution": "\\boxed{(1,3]}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the parabola $y=\\frac{1}{2}x^{2}+bx+c$ passes through points $A(0, 2)$ and $B(1, \\frac{3}{2})$. Point $D$ is on the parabola with an x-coordinate of $4$. Denote the segment of the parabola between points $A$ and $D$ (including points $A$ and $D$) as graph $G$. If graph $G$ is translated downwards by $t$ units ($t>0$) and intersects line $BC$ at only one point, find the range of values for $t$.\n",
    "completion": "\\textbf{Solution:} Let the equation of line BC be $y=mx+n$,\\\\\nsubstituting $B\\left(1, \\frac{3}{2}\\right)$ and $C\\left(2, 2\\right)$ into $y=mx+n$,\\\\\nwe get: $\\left\\{\\begin{array}{l}\nm+n=\\frac{3}{2}\\\\\n2m+n=2\n\\end{array}\\right.$,\\\\\nsolving this, we find: $\\left\\{\\begin{array}{l}\nm=\\frac{1}{2}\\\\\nn=1\n\\end{array}\\right.$,\\\\\n$\\therefore$ the equation of line BC is $y=\\frac{1}{2}x+1$,\\\\\nwhen $x=0$, $y=\\frac{1}{2}x+1=1$,\\\\\n$\\therefore$ when graph G is translated downwards by 1 unit, point A lies on line BC,\\\\\nwhen $x=4$, $y=\\frac{1}{2}x+1=3$,\\\\\nsince for $x=4$, $y=\\frac{1}{2}x^2-x+2=6$,\\\\\n$\\therefore$ when graph G is translated downwards by 3 units, point D lies on line BC,\\\\\n$\\therefore$ when $1<t\\le 3$, after graph G is translated downwards by t ($t>0$) units, it intersects line BC at only one point,\\\\\nthe desired range of values for $t$ is $t\\in\\boxed{(1,3]}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53033041_158.png"
  },
  {
    "solution": "\\boxed{\\frac{5}{6}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the parallelogram $ABCD$, $DE \\perp AB$ at point $E$, $DF \\perp BC$ at point $F$, the perimeter of the parallelogram $ABCD$ is 28, and the area is 40, with $AB:AD = 4:3$. Find the value of $\\sin\\angle EDF$.\n",
    "completion": "\\textbf{Solution:} In quadrilateral $DEBF$, we have $\\angle EDF+\\angle DEB+\\angle B+\\angle BFD=360^\\circ$,\\\\\nsince $DE\\perp AB$ and $DF\\perp BC$,\\\\\n$\\therefore \\angle EDF+\\angle B=180^\\circ$,\\\\\nMoreover, in parallelogram $ABCD$, $\\angle A+\\angle B=180^\\circ$,\\\\\n$\\therefore \\angle EDF=\\angle A$,\\\\\nIn $\\triangle ADE$, $\\sin A=\\frac{DE}{AD}=\\frac{5}{6}$,\\\\\n$\\therefore \\sin\\angle EDF=\\sin A=\\boxed{\\frac{5}{6}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52486708_92.png"
  },
  {
    "solution": "\\boxed{25^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, point $O$ is located on line segment $AB$. A ray $OC$ is drawn from point $O$ such that $\\angle BOC = 65^\\circ$. A right-angled triangle ruler is placed with its right angle vertex at point $O$. As shown in Diagram 2, the triangle ruler $MON$ is rotated counterclockwise around point $O$ by a certain angle, such that ray $OC$ bisects angle $\\angle MOB$. What are the measures of $\\angle BON$ and $\\angle CON$?\n",
    "completion": "\\textbf{Solution}: Since $\\angle BOC=65^\\circ$ and $OC$ is the angle bisector of $\\angle MOB$,\\\\\nit follows that $\\angle MOB=2\\angle BOC=130^\\circ.$\\\\\nHence, $\\angle BON=\\angle MOB-\\angle MON=130^\\circ-90^\\circ=40^\\circ.$\\\\\n$\\angle CON=\\angle COB-\\angle BON=65^\\circ-40^\\circ=25^\\circ.$\\\\\nThus, $\\angle BON=\\boxed{40^\\circ}, \\angle CON=\\boxed{25^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55503444_56.png"
  },
  {
    "solution": "\\boxed{16}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that points $B$ and $D$ are on the line segment $AC$. If $BD=\\frac{1}{3}AB=\\frac{1}{4}CD$, $AE=BE$, and $EC=13$, find the length of the line segment $AC$.\n",
    "completion": "\\textbf{Solution:} Given that $BD=\\frac{1}{3}AB=\\frac{1}{4}CD$,\\\\\nit follows that $AB=3BD, CD=4BD$,\\\\\nGiven $AE=BE$,\\\\\nit then follows $AE=BE=\\frac{1}{2}AB=\\frac{3}{2}BD$,\\\\\nSince $EC=BE+BD+DC=\\frac{3}{2}BD+BD+4BD=\\frac{13}{2}BD=13$,\\\\\nthen $BD=2$,\\\\\nThus, $AE=\\frac{3}{2}BD=\\frac{3}{2}\\times 2=3$,\\\\\nTherefore, $AC=AE+EC=3+13=\\boxed{16}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50009912_221.png"
  },
  {
    "solution": "\\boxed{(−6, 3)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the Cartesian coordinate system, the line $l_{1}\\colon y_{1}=-\\frac{1}{2}x+6$ intersects the x-axis and y-axis at points B and C, respectively, and intersects with the line $l_{2}\\colon y_{2}=\\frac{1}{2}x$ at point A. Is there a point Q in the plane such that the quadrilateral with vertices O, C, A, and Q is a parallelogram? Find the coordinates of point Q.\n",
    "completion": "\\textbf{Solution:} There exists a point Q such that the quadrilateral with vertices O, C, A, Q is a parallelogram; as shown in the diagram, consider three cases:\n- $\\because C(0, 6)$, A(6,3)\n  - (1) When the quadrilateral OAQ$_{1}$C is a parallelogram, $OC=AQ_{1}$, \\\\\n    $\\therefore$ the x-coordinate of point Q$_{1}$ is 6, and the y-coordinate is $3+6=9$,\\\\\n    $\\therefore$ the coordinates of Q$_{1}$ are (6,9),\n  - (2) When the quadrilateral OQ$_{2}$AC is a parallelogram, $OC=AQ_{2}$,\\\\\n    the x-coordinate of point Q$_{2}$ is 6, and the y-coordinate is $3-6=-3$,\\\\\n    $\\therefore$ the coordinates of Q$_{2}$ are (6, −3),\n  - (3) When the quadrilateral OACQ$_{3}$ is a parallelogram,\\\\\n    $\\because AC=\\sqrt{6^{2}+(6-3)^{2}}=3\\sqrt{5}$, $AO=\\sqrt{3^{2}+6^{2}}=3\\sqrt{5}$,\\\\\n    $\\therefore AC=AO$,\\\\\n    $\\therefore$ the quadrilateral OACQ$_{3}$ is a rhombus,\\\\\n    Given that a rhombus is an axis-symmetric figure, with $OC$ as a diagonal,\\\\\n    point Q$_{3}$ is symmetric to point A with respect to OC,\\\\\n    $\\therefore$ the coordinates of Q$_{3}$ are (−6, 3),\\\\\nSummarizing, the coordinates of point Q are: $\\boxed{(6,9)}$ or $\\boxed{(6, −3)}$ or $\\boxed{(−6, 3)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50677413_141.png"
  },
  {
    "solution": "\\boxed{(2,0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the parabola $y=ax^{2}+bx+8\\left(a\\neq 0\\right)$ passes through point $A\\left(-3, 5\\right)$ and $B\\left(5, 3\\right)$, and intersects the y-axis at point C. The circle with diameter AB passes through points O and C, and intersects the x-axis at point D. Connect AO and AC. What are the coordinates of point D?\n",
    "completion": "\\textbf{Solution:} Let's connect AD, and assume the coordinates of point D are $(x,0)$,\\\\\nsince AB is the diameter of the circle,\\\\\nit follows that $\\angle ADB=90^\\circ$,\\\\\nhence $AB^2=AD^2+BD^2$,\\\\\ntherefore $\\left(5+3\\right)^2+\\left(3-5\\right)^2=\\left(-3-x\\right)^2+5^2+\\left(5-x\\right)^2+3^2$,\\\\\nsolving this gives $x=2$ or $x=0$ (discard),\\\\\nthus, the coordinates of point D are $\\boxed{(2,0)}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52872196_80.png"
  },
  {
    "solution": "\\boxed{54^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $\\angle AOE$ is a straight angle, $OD$ bisects $\\angle COE$, $OB$ bisects $\\angle AOC$, and the ratio $\\angle COD : \\angle BOC = 2 : 3$. Find the measures of $\\angle COD$ and $\\angle BOC$ in degrees.\n",
    "completion": "\\textbf{Solution:} Since OD bisects $\\angle COE$, \\\\\nthen $\\angle COD= \\frac{1}{2} \\angle COE$, \\\\\nSince OB bisects $\\angle AOC$, \\\\\nthen $\\angle BOC= \\frac{1}{2} \\angle AOC$, \\\\\ntherefore $\\angle BOD=\\angle COD+\\angle BOC= \\frac{1}{2} (\\angle COE+\\angle AOC)=90^\\circ$, \\\\\nSince the ratio $\\angle COD\\colon \\angle BOC=2\\colon 3$, \\\\\nthus $\\angle COD= \\frac{2}{5}\\times 90^\\circ =\\boxed{36^\\circ}$, $\\angle BOC= \\frac{3}{5}\\times 90^\\circ =\\boxed{54^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50810088.png"
  },
  {
    "solution": "\\boxed{x<1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the graph of the linear function $y=kx+k+1$ intersects with the graph of the linear function $y=-x+4$ at point $A(1, a)$. Based on the graph, write the solution set of the inequality $-x+4>kx+k+1$.\n",
    "completion": "\\textbf{Solution:} Since the intersection point of the two lines is A$(1,3)$, when $x<1$, the graph of the linear function $y=-x+4$ lies above the graph of the linear function $y=kx+k+1$, \\\\\nthus, the solution set of the inequality $-x+4>kx+k+1$ is $\\boxed{x<1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52345530_207.png"
  },
  {
    "solution": "\\boxed{y=\\frac{1}{4}x}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system, perpendicular lines are drawn through points $C(1, 4)$ and $D(4, 1)$ to the x-axis, meeting the x-axis at points A and B, respectively. Find the analytical expressions for the lines CD and OD.\n",
    "completion": "\\textbf{Solution:} Let the equation of line CD be $y=kx+b$,\\\\\nSubstituting $C\\left(1, 4\\right)$ and $D\\left(4, 1\\right)$ into the equation,\\\\\nwe get $\\begin{cases}k+b=4\\\\ 4k+b=1\\end{cases}$,\\\\\nSolving this system of equations, we find $\\begin{cases}k=-1\\\\ b=5\\end{cases}$,\\\\\nTherefore, the equation of line CD is $\\boxed{y=-x+5}$,\\\\\nLet the equation of line OD be $y=mx$,\\\\\nSubstituting $D\\left(4, 1\\right)$ into the equation, we get $m=\\frac{1}{4}$,\\\\\nHence, the equation of line OD is $\\boxed{y=\\frac{1}{4}x}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52394997_105.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{13}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in parallelogram $ABCD$, the diagonals $AC$ and $BD$ intersect at point $O$. $E$ and $F$ are respectively the midpoints of $OB$ and $OD$. Connect $AE$, $AF$, $CE$, and $CF$. Quadrilateral $AECF$ is a parallelogram; if $AB \\perp AC$ and $AB = 3$, $BC = 5$. What is the length of $BD$?\n",
    "completion": "\\textbf{Solution:} Since AB$\\perp$AC,\\\\\nit follows that $\\angle$BAC$=90^\\circ$,\\\\\nhence, AC$= \\sqrt{BC^{2}-AB^{2}}=\\sqrt{5^{2}-3^{2}}=4$,\\\\\nthus, OA$=\\frac{1}{2}AC=2$,\\\\\nIn right triangle $\\triangle$AOB, by the Pythagorean theorem, we get OB$=\\sqrt{AB^{2}+OA^{2}}=\\sqrt{3^{2}+2^{2}}=\\sqrt{13}$,\\\\\ntherefore, BD$=2OB=2\\sqrt{13}$. Therefore, the length of BD is $\\boxed{2\\sqrt{13}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52425278_123.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, $\\angle A=30^\\circ$. The perpendicular bisector of AC intersects AC and AB at points D and F, respectively. A perpendicular line is drawn from point B to DF, and the foot of the perpendicular is E. If BC=2, what is the area of the quadrilateral BCDE?\n",
    "completion": "\\textbf{Solution:} Given that in $\\triangle ABC$, $\\angle C=90^\\circ$ and $\\angle A=30^\\circ$, $BC=2$,\\\\\nhence $AC= \\sqrt{3}BC=2\\sqrt{3}$,\\\\\nsince DE is perpendicular bisector of AC, with $\\angle CDE=90^\\circ$,\\\\\nthen $AD=DC= \\frac{1}{2}AC=\\sqrt{3}$,\\\\\nsince BE$\\perp$ED,\\\\\nthus $\\angle C=\\angle CDE=\\angle E=90^\\circ$,\\\\\ntherefore, quadrilateral BCDE is a rectangle,\\\\\nthus the area of quadrilateral BCDE=BC$\\cdot$DC=$2\\times\\sqrt{3}=\\boxed{2\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51614426.png"
  },
  {
    "solution": "\\boxed{y=\\frac{4}{x}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, the vertices A and B of the rhombus ABCD are located on the y-axis and the positive half of the x-axis, respectively, while C and D are in the first quadrant, with $AC \\parallel x$-axis. The graph of the inverse proportion function $y=\\frac{k}{x}$ passes through vertex D. Given $A(0, 2)$ and $B(1, 0)$, find the equation of the inverse proportion function.\n",
    "completion": "**Solution:**\n\n(1) Draw a perpendicular line from point D to the y-axis, intersecting at point F,\n\n$\\because ABCD$ is a rhombus,\n\n$\\therefore BD\\perp AC$, $BE=DE$, $AE=EC$\n\nIt is easy to prove that quadrilaterals AOBE and AEDF are rectangles.\n\n$\\therefore AO=BE=DE=2$, $\\therefore D\\left(1, 4\\right)$, $\\therefore y=\\frac{4}{x}$\n\n(2) Draw a perpendicular line from point C to the x-axis, intersecting at point G,\n\nIt is easy to prove that quadrilaterals AEBO and ACGO are rectangles.\n\n$\\therefore AO=CG=2$, $AC=2AE=2$ $\\therefore C\\left(2, 2\\right)$, $\\therefore$C lies on the graph of the inverse proportional function $\\boxed{y=\\frac{4}{x}}$",
    "image_path": "dataset/mm_math/images/MM_Math/52077009_83.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{5}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC\\sim \\triangle EDC$, given that $\\angle ABC=90^\\circ$, $\\frac{AB}{BC}=\\frac{1}{2}$, what is the value of $\\frac{AE}{BD}$?\n",
    "completion": "\\textbf{Solution:} Given that $\\triangle ABC \\sim \\triangle EDC$,\\\\\nit follows that $\\frac{AC}{EC}=\\frac{BC}{DC}$ and $\\angle ACB=\\angle ECD$,\\\\\nhence $\\frac{AC}{EC}=\\frac{BC}{DC}$ and $\\angle ACE=\\angle BCD$,\\\\\nimplying $\\triangle AEC \\sim \\triangle BDC$,\\\\\nwhich leads to $\\frac{AE}{BD}=\\frac{AC}{BC}$,\\\\\nsince $\\angle ABC=90^\\circ$ and $\\frac{AB}{BC}=\\frac{1}{2}$,\\\\\nit results in $\\frac{AB^2}{BC^2}=\\frac{1}{4}$,\\\\\nthus $\\frac{AB^2+BC^2}{BC^2}=\\frac{5}{4}$, which means $\\frac{AC^2}{BC^2}=\\frac{5}{4}$,\\\\\ntherefore $\\frac{AC}{BC}=\\frac{\\sqrt{5}}{2}$,\\\\\nconcluding that $\\frac{AE}{BD}=\\boxed{\\frac{\\sqrt{5}}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52602351.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{5}x - 2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, it is known that the line $l$: $y=-x+6$ intersects with the $x$ axis at point $A$, and with the $y$ axis at point $B$. The line $m$ intersects with the $y$ axis at point $C(0, -2)$, and intersects with the line $l$ at point $D(t, 1)$. What is the analytical expression of line $m$?\n",
    "completion": "\\textbf{Solution:} Let $\\because D(5, 1)$ lies on the line $l: y = -x + 6$,\\\\\n$\\therefore 1 = -5 + 6$,\\\\\n$\\therefore t = 5$,\\\\\n$\\therefore D(5, 1)$,\\\\\nLet the equation of line $m$ be $y = kx + b$,\\\\\nSubstituting points C and D, we get,\\\\\n$\\begin{cases}\n5k + b = 1\\\\\nb = -2\n\\end{cases}$,\\\\\nSolving this, we find,\\\\\n$\\begin{cases}\nk = \\frac{3}{5}\\\\\nb = -2\n\\end{cases}$,\\\\\nTherefore, the equation of line $m$ is $y = \\boxed{\\frac{3}{5}x - 2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50496938_56.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{1}{2}x+5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the graph of the linear function $y_1=kx+b$ ($k\\ne 0$, $k$, $b$ are constants), denoted as $l_1$, intersects with the $x$ axis and $y$ axis at points $A$ and $B$ respectively, where the coordinates of point $B$ are (0,5). The graph of the direct proportion function $y_2=2x$, denoted as $l_2$, intersects with $l_1$ at point $C(m, 4)$. Find the expression of the linear function?\n",
    "completion": "\\textbf{Solution:} Substitute point C$(2, 4)$ into the direct proportion function $y=2x$, and we find $m=2$.\\\\\n$\\therefore$C$(2, 4)$.\\\\\nSubstitute points C$(2, 4)$ and B$(0, 5)$ into the linear function $y=kx+b$, yielding the system of equations\n\\[\n\\left\\{\n\\begin{array}{l}\n2k+b=4 \\\\\nb=5\n\\end{array}\n\\right.\n\\]\nSolving this, we find $k=-\\frac{1}{2}$ and $b=5$.\\\\\n$\\therefore$The equation of the linear function is $y=-\\frac{1}{2}x+5$, thus the equation is $\\boxed{y=-\\frac{1}{2}x+5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50677475_136.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the graph, the function \n\n\\[\ny = \n\\begin{cases} \n3-x & (0 \\le x \\le 2)\\\\ \nx-1 & (2 \\le x \\le 4)\n\\end{cases}\n\\]\n\nis illustrated. What is the minimum value of this function?\n",
    "completion": "\\textbf{Solution:} Since the minimum value of the function is $(2,1)$, \\\\\ntherefore, the minimum value of the function is $\\boxed{1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52423578.png"
  },
  {
    "solution": "\\boxed{58}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, this is a beautiful Pythagorean tree, in which all the quadrilaterals are squares, and all the triangles are right-angled triangles. If the side lengths of squares $A$, $B$, $C$, and $D$ are $3$, $6$, $2$, and $3$ respectively, what is the area of the largest square $E$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure,\\\\\nby the Pythagorean theorem, the area of square $F$ is equal to the area of square $A$ plus the area of square $B$ = $3^{2}+6^{2}=45$,\\\\\nSimilarly, the area of square $G$ is equal to the area of square $C$ plus the area of square $D$ = $2^{2}+3^{2}=13$,\\\\\n$\\therefore$ the area of square $E$ is equal to the area of square $F$ plus the area of square $G$ = $45+13=\\boxed{58}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52160204.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{1}{2}x-1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system (with $O$ as the origin), it is known that the line $y=kx+b$ intersects the $x$-axis and the $y$-axis at points $A(-2,0)$ and $B(0,-1)$ respectively, and the coordinates of point $C$ are $(0,2)$. What is the equation of line $AB$?\n",
    "completion": "\\textbf{Solution:} Since the line $y=kx+b$ intersects the x-axis and y-axis at points $A(-2,0)$ and $B(0,-1)$ respectively,\n\n$\\therefore$ $\\begin{cases} -2k+b=0\\\\ b=-1 \\end{cases}$,\n\nSolving this yields $\\begin{cases} k=-\\frac{1}{2}\\\\ b=-1 \\end{cases}$,\n\n$\\therefore$ the equation of line $AB$ is: $\\boxed{y=-\\frac{1}{2}x-1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55498635_91.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\angle AOB = 30^\\circ$, point $P$ is a fixed point inside $\\angle AOB$ and $OP=3$. If points $M$ and $N$ are moving points on rays $OA$ and $OB$ different from point $O$, respectively, what is the minimum perimeter of $\\triangle PMN$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, let $E$ and $D$ be the symmetric points of point $P$ with respect to $OA$ and $OB$, respectively. Draw lines $DN$, $EM$, $DE$, $OD$, and $OE$,\\\\\nthen $OA$ perpendicularly bisects $PE$, and $OB$ perpendicularly bisects $PD$,\\\\\n$\\therefore EM=PM$, $DN=PN$, $OD=OP=OE=3$,\\\\\n$\\therefore$ the perimeter of $\\triangle PMN$ is $PM+MN+PN=EM+MN+DN$,\\\\\nAs the shortest distance between two points is a straight line, when points $E$, $M$, $N$, $D$ are collinear, the value of $EM+MN+DN$ is minimized, and the minimum value is the length of $DE$,\\\\\n$\\because OE=OP$ and $OA\\perp PE$,\\\\\n$\\therefore \\angle AOE=\\angle AOP$ (angle-side-angle congruence in isosceles triangle),\\\\\nSimilarly, it can be obtained: $\\angle BOD=\\angle BOP$,\\\\\n$\\because \\angle AOB=30^\\circ$,\\\\\n$\\therefore \\angle BOP+\\angle AOP=30^\\circ$,\\\\\n$\\therefore \\angle DOE=\\angle BOD+\\angle BOP+\\angle AOP+\\angle AOE=2\\left(\\angle BOP+\\angle AOP\\right)=60^\\circ$,\\\\\nAnd since $OD=OE=3$,\\\\\n$\\therefore \\triangle DOE$ is an equilateral triangle,\\\\\n$\\therefore DE=OD=3$,\\\\\n$\\therefore$ the minimum value of $EM+MN+DN$ is 3,\\\\\n$\\therefore$ the minimum perimeter of $\\triangle PMN$ is $\\boxed{3}$,",
    "image_path": "dataset/mm_math/images/MM_Math/52653459.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{4}a}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the length of line segment AB is $a$. Extend BA to point C such that $AC = \\frac{1}{2} AB$. Point D is the midpoint of line segment BC. What is the length of CD?\n",
    "completion": "\\textbf{Solution:} Since $AB = a$, \\\\\ntherefore $AC = \\frac{1}{2}AB = \\frac{1}{2}a$, \\\\\nthus $BC = AC + AB = a + \\frac{1}{2}a = \\frac{3}{2}a$, \\\\\nsince D is the midpoint of the line segment BC, \\\\\nthus $CD = BD = \\frac{1}{2}BC = \\frac{3}{4}a = \\boxed{\\frac{3}{4}a}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55327759_85.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{1}{12}(x-6)^2+4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: \"The Joy of Winter Olympics: Mingxi Enthusiastically Welcomes the Basketball Match\" has begun with athletes attacking and defending, every second counts. In this event, the basketball player Xiaoming performs a long throw from position O. The ball is thrown from point A, which is 1 meter above the ground. It reaches its highest point at B, 6 meters away from O, with the highest point C being 4 meters above the ground. The ball bounces again and lands at point E, with the distance between E and F being 2 meters. According to experiments, the trajectory of the basketball's second bounce on the court has the same shape as the first but its maximum height is reduced to half of the original. Taking Xiaoming's standing position O as the origin, a Cartesian coordinate system is established as shown in the figure (all calculated results are to be rounded to the nearest integer, $ \\sqrt{3} \\approx 1.75$; $ \\sqrt{6} \\approx 2.5$). What is the equation of the parabola ACD?\n",
    "completion": "\\textbf{Solution:} Let us assume that the equation of the parabola describing the basketball's trajectory from its initial launch to its first landing is $y=a(x-h)^2+k$,\\\\\nwhere $h=6$, $k=4$,\\\\\nthus $y=a(x-6)^2+4$.\\\\\nFrom the given information, when $x=0$, $y=1$,\\\\\nwhich implies $1=36a+4$,\\\\\ntherefore $a=-\\frac{1}{12}$,\\\\\nhence the equation is $\\boxed{y=-\\frac{1}{12}(x-6)^2+4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51403162_46.png"
  },
  {
    "solution": "\\boxed{116^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points $A$, $B$, $C$, $D$, $E$ are all on the circle $\\odot O$, with $AC=AE$, and $\\angle D = 128^\\circ$. Then, what is the degree measure of $\\angle B$?\n",
    "completion": "\\textbf{Solution:} Connect AC and CE, \\\\\n$\\because$ Quadrilateral ACDE is inscribed in circle O, \\\\\n$\\therefore$ $\\angle$CAE + $\\angle$C = 180$^\\circ$, \\\\\n$\\therefore$ $\\angle$CAE = 180$^\\circ$ − 128$^\\circ$ = 52$^\\circ$, \\\\\n$\\because$ AC = CE, \\\\\n$\\therefore$ $\\angle$ACE = $\\angle$AEC = $ \\frac{1}{2}$(180$^\\circ$ − 52$^\\circ$) = 64$^\\circ$, \\\\\n$\\because$ Quadrilateral ABCE is inscribed in circle O, \\\\\n$\\therefore$ $\\angle$B + $\\angle$AEC = 180$^\\circ$, \\\\\n$\\therefore$ $\\angle$B = 180$^\\circ$ − 64$^\\circ$ = \\boxed{116^\\circ}.",
    "image_path": "dataset/mm_math/images/MM_Math/52815682.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point \\(C\\) is a point on line segment \\(AB\\), and points \\(M\\), \\(N\\), and \\(P\\) are the midpoints of line segments \\(AC\\), \\(BC\\), and \\(AB\\), respectively. If \\(AC = 3\\, \\text{cm}\\) and \\(CP = 1\\, \\text{cm}\\), what is the length of line segment \\(PN\\)?\n",
    "completion": "\\textbf{Solution:} \\\\ \nSince $AC=3$ and $CP=1$, \\\\\nthus $AP=AC+CP=4$, \\\\\nSince point P is the midpoint of segment AB, \\\\\nthus $AB=2AP=8$, $CB=AB-AC=5$, \\\\\nSince point N is the midpoint of segment CB, \\\\\nthus $CN=\\frac{1}{2}CB=\\frac{5}{2}$, \\\\\ntherefore $PN=CN-CP=\\frac{5}{2}-1=\\boxed{\\frac{3}{2}}$ (cm).",
    "image_path": "dataset/mm_math/images/MM_Math/53848489_45.png"
  },
  {
    "solution": "\\boxed{\\frac{2.2}{\\tan 80^\\circ}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, the sunlight strikes the ground at an angle of $80^\\circ$. Given that window $AB$ is 2 meters high, to install a horizontal sunshade $DC$ at point $D$, which is 0.2 meters above the window, such that the sunlight cannot directly enter the room, what is the minimum length of sunshade $DC$?\n",
    "completion": "\\textbf{Solution}: Given that $DA=0.2$ meters, $AB=2$ meters,\\\\\nthus $DB=DA+AB=2.2$ meters,\\\\\nsince the angle between the ray and the ground is $80^\\circ$, thus $\\angle BCD=80^\\circ$.\\\\\nAlso, since $\\tan \\angle BCD= \\frac{DB}{DC}$,\\\\\nthus $DC= \\frac{DB}{\\tan \\angle BCD}=\\frac{2.2}{\\tan 80^\\circ}=\\boxed{\\frac{2.2}{\\tan 80^\\circ}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51174756.png"
  },
  {
    "solution": "\\boxed{-x^{2}-2x+3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=ax^{2}+2ax+c$ intersects the x-axis at points $A$ and $B$ (where point $A$ is to the left of point $B$), and intersects the y-axis at point $C(0, 3)$. The y-coordinate of its vertex $D$ is $4$. What is the expression of this parabola?\n",
    "completion": "\\textbf{Solution:} Substitute the point $C(0, 3)$ into $y=ax^{2}+2ax+c$ to get: $c=3$\\\\\n$\\therefore y=ax^{2}+2ax+3$\\\\\n$\\therefore -\\frac{b}{2a}=-\\frac{2a}{2a}=-1$\\\\\nWhen $x=-1$, $\\therefore y=a-2a=-a+3$\\\\\nSince the ordinate of the vertex D is $4$,\\\\\n$\\therefore -a+3=4$\\\\\n$a=-1$\\\\\n$\\therefore D(-1, 4)$\\\\\nTherefore, the equation of the parabola is $y=-x^{2}-2x+3=\\boxed{-x^{2}-2x+3}$",
    "image_path": "dataset/mm_math/images/MM_Math/51319005_104.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that points $A$, $B$, $C$, and $D$ lie on the same straight line, point $C$ is the midpoint of segment $AB$, and point $D$ is on segment $AB$. If $AB=6$ and $BD=\\frac{1}{3}BC$, what is the length of segment $CD$?\n",
    "completion": "\\textbf{Solution:} Since point C is the midpoint of segment AB, and $AB=6$, \\\\\ntherefore $BC= \\frac{1}{2}AB=3$, \\\\\nsince $BD=\\frac{1}{3}BC$, \\\\\nthus $BD=\\frac{1}{3} \\times 3=1$, \\\\\ntherefore $CD=BC-BD=3-1=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50810294_199.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, line segment $AB=24\\,cm$, C is a point on $AB$, and $AC=15\\,cm$, O is the midpoint of $AB$. What is the length of line segment $OC$?\n",
    "completion": "\\textbf{Solution:} Since AB=24cm, and point O is the midpoint of AB,\\\\\nthen OB=$\\frac{1}{2}$AB=$\\frac{1}{2}\\times 24=12cm$,\\\\\nSince BC=AB−AC,\\\\\nthen BC=24−15=9;\\\\\nSince OC=OB−BC,\\\\\nthen OC=12−9=$\\boxed{3}$cm.",
    "image_path": "dataset/mm_math/images/MM_Math/53124986.png"
  },
  {
    "solution": "\\boxed{\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown, for the square $ABCD$ with side length $2$, where point $E$ is the midpoint of side $BC$, and point $P$ moves along the diagonal $AC$, what is the minimum value of $PE+PB$?\n",
    "completion": "\\textbf{Solution:} Draw line DE, intersecting AC at point P, and draw BD. \\\\\n$\\because$ Point B and point D are symmetric about AC, \\\\\n$\\therefore$ DP=PB, \\\\\n$\\therefore$ PB+PE=PD+PE=DE \\\\\n$\\therefore$ The length of DE is the minimum value of PE+PB, \\\\\n$\\because$ In the square ABCD, AB=2, and E is the midpoint of BC, \\\\\n$\\therefore$ CE=1, \\\\\nIn $\\triangle CDE$, \\\\\nDE=$\\sqrt{CD^{2}+CE^{2}}$ \\\\\n=$\\sqrt{2^{2}+1^{2}}$ \\\\\n=$\\sqrt{5}$ \\\\\nHence, the answer is: $\\boxed{\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52225663.png"
  },
  {
    "solution": "\\boxed{\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ACB$, $\\angle ACB=90^\\circ$, point M is the midpoint of side $AB$, point E is on segment $AM$, $EF\\perp AC$ at point F, connect $CM$, $CE$. It is known that $\\angle A=50^\\circ$, $\\angle ACE=30^\\circ$, and $CE=CM$. If $AB=4$, find the length of segment $FC$.\n",
    "completion": "\\textbf{Solution:} Since $AB=4$,\\\\\nit follows that $CE=CM=\\frac{1}{2}AB=2$,\\\\\nsince $EF\\perp AC$ and $\\angle ACE=30^\\circ$,\\\\\nit follows that $EF=\\frac{1}{2}CE=1$,\\\\\ntherefore $FC=\\sqrt{CE^{2}-EF^{2}}=\\sqrt{2^{2}-1^{2}}=\\boxed{\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52866951_161.png"
  },
  {
    "solution": "\\boxed{\\frac{4}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ is inscribed in $\\odot O$, with $AB$ being the diameter of $\\odot O$ and $AB=5$, $AC=3$. What is the value of $\\tan A$?\n",
    "completion": "\\textbf{Solution:} Since AB is the diameter of the circle $\\odot$O,\\\\\ntherefore $\\angle ACB=90^\\circ$,\\\\\nsince AB=5, and AC=3,\\\\\ntherefore BC=4,\\\\\nhence $\\tan A=\\boxed{\\frac{4}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51379192_89.png"
  },
  {
    "solution": "\\boxed{\\sqrt{2}-1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, $O$ is the intersection point of the diagonals of square $ABCD$, and $E$ is a point on line segment $AO$. If $AB = 1$, and $\\angle BED = 135^\\circ$, what is the length of $AE$?\n",
    "completion": "\\textbf{Solution:} Since it is a square ABCD,\\\\\nit follows that $\\angle BCA=\\angle BAC=45^\\circ$, $\\angle ABC=90^\\circ$, and BC=AB=1,\\\\\nthus, AC=$\\sqrt{2}$AB=$\\sqrt{2}$,\\\\\nsince E is a point on the diagonal AC of the square, and $\\angle DEB=135^\\circ$,\\\\\nit is easy to derive that $\\angle CEB=\\angle CED=67.5^\\circ$,\\\\\nthus, $\\angle EBC=180^\\circ-67.5^\\circ-45^\\circ=67.5^\\circ$,\\\\\ntherefore, $\\angle EBC=CEB=67.5^\\circ$,\\\\\nhence, BC=EC=1,\\\\\nthus, AE=AC−EC=$\\sqrt{2}-1$.\\\\\nTherefore, the answer is: $AE=\\boxed{\\sqrt{2}-1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52082490.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the circle $\\odot O$ with radius $r=10$, and a chord $AB=16$, $P$ is a moving point on chord $AB$. What is the minimum length of segment $OP$?\n",
    "completion": "\\textbf{Solution:} Draw $OH \\perp AB$ at point H, and connect OB.\\\\\nTherefore, $AH=BH=\\frac{1}{2}AB=\\frac{1}{2}\\times 16=8$,\\\\\nIn $\\triangle BOH$, $OH=\\sqrt{OB^{2}-BH^{2}}=\\sqrt{10^{2}-8^{2}}=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52458117.png"
  },
  {
    "solution": "\\boxed{4\\,cm}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point C is a point on the line segment AB, with $AC=12\\, \\text{cm}$ and $AC=\\frac{3}{2}CB$. D and E are the midpoints of AC and AB, respectively. Find the length of DE.\n",
    "completion": "\\textbf{Solution:} Given that $AC=12\\,cm$ and $AC=\\frac{3}{2}CB$,\\\\\nit follows that $CB=\\frac{2}{3}AC=\\frac{2}{3}\\times 12=8\\,cm$,\\\\\nthus $AB=AC+CB=12+8=20\\,cm$,\\\\\nsince $D$ and $E$ are the midpoints of $AC$ and $AB$ respectively,\\\\\nwe have $AD=\\frac{1}{2}AC=6\\,cm$ and $AE=\\frac{1}{2}AB=10\\,cm$\\\\\ntherefore $DE=AE-AD=10-6=\\boxed{4\\,cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53027074.png"
  },
  {
    "solution": "\\boxed{1-2\\sqrt{2}<x<1+2\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the graph of the quadratic function $y=-x^2+bx+c$ passes through points $A(-2, -1)$ and $B(0, 7)$. For what values of $x$ is $y>0$?\n",
    "completion": "\\textbf{Solution}: When $y=0$,\\\\\n$0=-(x-1)^2+8$,\\\\\n$\\therefore x-1=\\pm 2\\sqrt{2}$,\\\\\n${x}_{1}=1+2\\sqrt{2}$, ${x}_{2}=1-2\\sqrt{2}$,\\\\\n$\\therefore$ the intersection points of the parabola and the $x$-axis are: $(1-2\\sqrt{2}, 0)$, $(1+2\\sqrt{2}, 0)$,\\\\\n$\\therefore$ when \\boxed{1-2\\sqrt{2}<x<1+2\\sqrt{2}}, $y>0$.",
    "image_path": "dataset/mm_math/images/MM_Math/52916890_265.png"
  },
  {
    "solution": "\\boxed{\\frac{10}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the square $ABCD$, $A(-10,0)$, $B(-2,0)$, $C(0,4)$. Points $P$, $Q$ are respectively on the line segments $OA$, $CD$, with $OP=2DQ$. Connect $PQ$, $PC$, and let $DQ=x$. For what value of $x$ is $PQ \\parallel AD$?\n",
    "completion": "\\textbf{Solution:} Given, $A(-10,0)$, $B(-2,0)$, $C(0,4)$,\\\\\nthus $OA=10$, $OB=2$, $OC=4$,\\\\\nthus $AB=OA-OB=10-2=8$,\\\\\nsince quadrilateral $ABCD$ is a parallelogram,\\\\\nthus $CD=AB=8$, $AB \\parallel CD$,\\\\\nsince $OP=2DQ$, $DQ=x$,\\\\\nthus $OP=2x$,\\\\\nthus $AP=AO-OP=10-2x$,\\\\\nsince $PQ \\parallel AD$, $AP \\parallel DQ$,\\\\\nthus quadrilateral $APQD$ is a parallelogram,\\\\\nthus $AP=DQ$,\\\\\nthus $10-2x=x$,\\\\\nthus $x=\\boxed{\\frac{10}{3}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53498608_79.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that in an equilateral triangle $ABC$, $D$ is the midpoint of $AC$, $E$ is a point on the extension line of $BC$, and $CD = CE$, $M$ is the midpoint of $BE$. What is the measure of $\\angle E$ in degrees?\n",
    "completion": "\\textbf{Solution:} Given that $\\triangle ABC$ is an equilateral triangle, \\\\\n$\\therefore \\angle ACB=60^\\circ$, \\\\\nsince $CD=CE$, \\\\\n$\\therefore \\angle CDE=\\angle E$, \\\\\nAlso, since $\\angle ACB=\\angle CDE+\\angle E$, \\\\\n$\\therefore \\angle E=\\boxed{30^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52950696_89.png"
  },
  {
    "solution": "\\boxed{240^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a triangular paper piece with a $60^\\circ$ angle is cut to remove this $60^\\circ$ angle, resulting in a quadrilateral. What is the degree measurement of $\\angle 1+\\angle 2$?\n",
    "completion": "\\textbf{Solution}: As shown in the figure, \\\\\nsince $\\angle C=60^\\circ$, \\\\\ntherefore $\\angle A+\\angle B=180^\\circ-\\angle C=180^\\circ-60^\\circ=120^\\circ$, \\\\\ntherefore $\\angle 1+\\angle 2=360^\\circ-(\\angle A+\\angle B)=360^\\circ-120^\\circ=\\boxed{240^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52755868.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, it is known that the graph of the linear function $y=\\frac{1}{2}x+1$ intersects the x-axis and y-axis at points $A$ and $B$, respectively. A square $ABCD$ is constructed inside the second quadrant with $AB$ as its side. What is the area of square $ABCD$?\n",
    "completion": "\\textbf{Solution:} For the line $y=\\frac{1}{2}x+1$, let $x=0$, then we obtain $y=1$; let $y=0$, then we obtain $x=-2$,\\\\\n$\\therefore A(-2, 0)$, $B(0, 1)$,\\\\\nIn the right triangle $Rt\\triangle AOB$, $OA=2$, $OB=1$,\\\\\nAccording to the Pythagorean theorem: $AB=\\sqrt{2^2+1^2}=\\sqrt{5}$;\\\\\nTherefore, the area of square $ABCD$ is $\\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52386761_123.png"
  },
  {
    "solution": "\\boxed{100}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the detector on the hot air balloon indicates that the angle of depression from the bottom of the hot air balloon at point $A$ to the top of a building is $30^\\circ$, and the angle of depression to the bottom of the building is $60^\\circ$. The distance from point $A$ on the hot air balloon to the ground is $150m$. What is the height of the building in meters?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, draw $AH\\perp BC$, extending through point $CB$ to $H$,\\\\\nIn right $\\triangle ACD$,\\\\\n$\\because \\angle CAD=30^\\circ$, $AD=150m$\\\\\n$\\therefore CD=AD\\cdot \\tan 30^\\circ=150\\times \\frac{\\sqrt{3}}{3}=50\\sqrt{3}(m)$,\\\\\n$\\therefore AH=CD=50\\sqrt{3}m$\\\\\nIn right $\\triangle ABH$,\\\\\n$\\because \\angle BAH=30^\\circ$, $AE=50\\sqrt{3}m$\\\\\n$\\therefore BH=AH\\cdot \\tan 30^\\circ=50\\sqrt{3}\\times \\frac{\\sqrt{3}}{3}=50(m)$,\\\\\n$\\therefore BC=AD−BH=150−50=\\boxed{100}(m)$",
    "image_path": "dataset/mm_math/images/MM_Math/51299763.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in quadrilateral \\(ABCD\\), \\(\\angle ABC=90^\\circ\\), points \\(E\\) and \\(F\\) are the midpoints of \\(AC\\) and \\(AD\\) respectively, and \\(BE=EF\\). If \\(AB=8\\) and \\(BC=4\\), what is the length of \\(CD\\)?\n",
    "completion": "\\textbf{Solution:} Let $\\angle ABC=90^\\circ$ and $AB=8$, $BC=4$,\\\\\ntherefore $AC=\\sqrt{AB^2+BC^2}=\\sqrt{8^2+4^2}=4\\sqrt{5}$.\\\\\nSince point E is the midpoint of AC,\\\\\ntherefore $BE=\\frac{1}{2}AC=2\\sqrt{5}$,\\\\\nand since $BE=EF$,\\\\\nthus $EF=2\\sqrt{5}$.\\\\\nSince points E and F are respectively the midpoints of AC and AD,\\\\\ntherefore $EF=\\frac{1}{2}DC$.\\\\\nThus $DC=2EF=4\\sqrt{5}$.\\\\\nHence, the length of $DC$ is $\\boxed{4\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52386625.png"
  },
  {
    "solution": "\\boxed{150^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that lines $AB$ and $CD$ intersect at point $O$, and $\\angle COE=90^\\circ$. If $\\angle BOD:\\angle BOC = 3:6$, what is the degree measure of $\\angle AOE$?\n",
    "completion": "\\textbf{Solution:}\n\nGiven that $\\angle BOD : \\angle BOC = 3 : 6$,\\\\\nand $\\angle BOD + \\angle BOC = 180^\\circ$,\\\\\nhence, $\\angle BOD = 60^\\circ$,\\\\\nSince $\\angle BOD = \\angle AOC$,\\\\\nit follows that $\\angle AOC = 60^\\circ$,\\\\\nGiven $\\angle COE = 90^\\circ$,\\\\\nthus, $\\angle AOE = \\angle COE + \\angle AOC = 90^\\circ + 60^\\circ = \\boxed{150^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51420172_72.png"
  },
  {
    "solution": "\\boxed{30}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given point $B$ is on line segment $AC$, and point $D$ is on line segment $AB$. As shown in the figure, if $BD = \\frac{1}{4}AB = \\frac{1}{3}CD$, $E$ is the midpoint of line segment $AB$, $EC=20\\,cm$, find the length of line segment $AC$.\n",
    "completion": "\\textbf{Solution:} Let $BD=x$ cm, \\\\\nsince $BD= \\frac{1}{4}AB= \\frac{1}{3}CD$, \\\\\nthus $AB=4BD=4x$ cm, $CD=3BD=3x$ cm, \\\\\nhence $BC=DC-DB=3x-x=2x$, \\\\\nthus $AC=AB+BC=4x+2x=6x$ cm, \\\\\nsince $E$ is the midpoint of segment $AB$, \\\\\nthus $BE= \\frac{1}{2}AB= \\frac{1}{2}\\times4x=2x$ cm, \\\\\nthus $EC=BE+BC=2x+2x=4x$ cm, \\\\\nalso, since $EC=20$ cm, \\\\\nthus $4x=20$, \\\\\nsolving gives: $x=5$, \\\\\nhence $AC=6x=6\\times5=\\boxed{30}$ cm.",
    "image_path": "dataset/mm_math/images/MM_Math/51237402_170.png"
  },
  {
    "solution": "\\boxed{2^{2019}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\angle MON=30^\\circ$, points $A_1$, $A_2$, $A_3$, ... are on ray ON, and points $B_1$, $B_2$, $B_3$, ... are on ray OM. The triangles $\\triangle A_1B_1A_2$, $\\triangle A_2B_2A_3$, $\\triangle A_3B_3A_4$, ... are all equilateral triangles, with the side length of the first equilateral triangle from the left being $a_1$, the second being $a_2$, and so on. If $OA_1=1$, what is the value of $a_{2020}$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure,\\\\\n$\\because$ $\\triangle A_1B_1A_2$ is an equilateral triangle,\\\\\n$\\therefore$ $A_1B_1=A_2B_1$, $\\angle 3=\\angle 4=\\angle 1+\\angle O=60^\\circ$,\\\\\n$\\therefore$ $\\angle 2=120^\\circ$,\\\\\n$\\because$ $\\angle MON=30^\\circ$,\\\\\n$\\therefore$ $\\angle 1=180^\\circ-120^\\circ-30^\\circ=30^\\circ$,\\\\\nFurthermore, $\\because$ $\\angle 3=60^\\circ$,\\\\\n$\\therefore$ $\\angle 5=180^\\circ-60^\\circ-30^\\circ=90^\\circ$,\\\\\n$\\because$ $\\angle MON=\\angle 1=30^\\circ$,\\\\\n$\\therefore$ $OA_1=A_1B_1=1$,\\\\\n$\\therefore$ $A_2B_1=1$,\\\\\n$\\because$ $\\triangle A_2B_2A_3$, $\\triangle A_3B_3A_4$ are equilateral triangles,\\\\\n$\\therefore$ $\\angle 11=\\angle 10=60^\\circ$, $\\angle 13=60^\\circ$,\\\\\n$\\because$ $\\angle 4=\\angle 12=60^\\circ$,\\\\\n$\\therefore$ $A_1B_1 \\parallel A_2B_2 \\parallel A_3B_3$, $B_1A_2 \\parallel B_2A_3$,\\\\\n$\\therefore$ $\\angle 1=\\angle 6=\\angle 7=30^\\circ$, $\\angle 5=\\angle 8=90^\\circ$,\\\\\n$\\therefore$ $a_2=2a_1$, $a_3=4a_1=4$, $a_4=8a_1=8$, $a_5=16a_1$,\\\\\nBy analogy: $a_{2020}=2^{2019}$.\\\\\nTherefore, the answer is: $a_{2020}=\\boxed{2^{2019}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51366817.png"
  },
  {
    "solution": "\\boxed{(3,1)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, in the Cartesian coordinate system, the linear function $y=-\\frac{2}{3}x+4$ intersects the x-axis at point $B$ and the y-axis at point $A$. Point $C$ is the midpoint of segment $AB$. Through point $C$, $DC\\perp x$ axis, and the foot of the perpendicular is $D$. Assume point $E$ is on the negative half of the y-axis, and connecting $CE$ intersects the x-axis at point $F$, with $CF=FE$. On the line $CD$, there exists a point $P$ such that $AP+EP$ is minimized. What are the coordinates of point $P$?\n",
    "completion": "\\textbf{Solution:} Since point C is the midpoint of AB, and given points A(0,4) and B(6,0), it follows that point C(3,2), therefore CD=2, and OD=3; since CD is perpendicular to the x-axis, it follows that $\\angle EOF=\\angle CDF=90^\\circ$. In $\\triangle EOF$ and $\\triangle CDF$, we have\n\\[\n\\left\\{\n\\begin{array}{l}\n\\angle EOF=\\angle CDF\\\\\n\\angle OFE=\\angle CFD\\\\\nEF=CF\n\\end{array}\n\\right.\n\\]\ntherefore, $\\triangle EOF\\cong \\triangle CDF$ (by AAS), hence OE=CD=2, and OF=FD=$\\frac{1}{2}$OD=$\\frac{3}{2}$, therefore point F($\\frac{3}{2}$,0), and point E(0,-2); Construct the symmetric point A$^{\\prime}$ of point A with respect to line CD, and draw line A$^{\\prime}$E intersecting CD at point F, draw line AP, thus point A$^{\\prime}$(6,4). Suppose the equation of line A$^{\\prime}$E is y=kx+b, then\n\\[\n\\left\\{\n\\begin{array}{l}\n6k+b=4\\\\\nb=-2\n\\end{array}\n\\right.\n\\]\nSolving this yields:\n\\[\n\\left\\{\n\\begin{array}{l}\nk=1\\\\\nb=-2\n\\end{array}\n\\right.\n\\]\ntherefore, the equation of A$^{\\prime}$E is y=x-2, when x=3, y=1, therefore point P(3,1) is $\\boxed{(3,1)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52815665_105.png"
  },
  {
    "solution": "\\boxed{x>1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y=k_1x+b$ intersects with the graph of the inverse proportion function $y=\\frac{k_2}{x}$ at points $A(1, 2)$ and $B(-2, n)$. Based on the graph, directly write the range of values of $x$ for which $k_1x+b>\\frac{k_2}{x}$.\n",
    "completion": "\\textbf{Solution:} Observing the graph, the range of $x$ for $k_1x+b>\\frac{k_2}{x}$ is \\boxed{-2<x<0} or \\boxed{x>1}.",
    "image_path": "dataset/mm_math/images/MM_Math/51402407_31.png"
  },
  {
    "solution": "\\boxed{\\frac{13}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that point $E$ is located on the side $AB$ of the square $ABCD$. A square $BEFG$ is constructed outside square $ABCD$ with $E$ as one of its sides. Line segments $DF$ and $MN$ are drawn, where $M$ and $N$ are the midpoints of $DC$ and $DF$, respectively. If $AB=7$ and $BE=5$, what is the length of $MN$?\n",
    "completion": "\\textbf{Solution:} Draw line CF, \\\\\n$\\because$ In square ABCD and square BEFG, we have AB=7, BE=5, \\\\\n$\\therefore$ GF=GB=5, BC=7, \\\\\n$\\therefore$ GC=GB+BC=5+7=12, \\\\\n$\\therefore$ CF= $\\sqrt{GF^{2}+GC^{2}}=\\sqrt{5^{2}+12^{2}}=13$, \\\\\n$\\because$ M and N are the midpoints of DC and DF respectively, \\\\\n$\\therefore$ MN= $\\frac{1}{2}$ CF= $\\boxed{\\frac{13}{2}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/50496336.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated, point C is the midpoint of segment AB, point D is a point on segment BC, and point E is the midpoint of segment CD. Given $AB=20$ and $BD=4$, find the length of segment CD.\n",
    "completion": "\\textbf{Solution:} Because $AB=20$ and $BD=4$,\\\\\nit follows that $AD=AB-BD=20-4=16$,\\\\\nsince point C is the midpoint of line segment AB,\\\\\nit follows that $AC=BC=10$,\\\\\ntherefore $CD=AD-AC=16-10=\\boxed{6};$",
    "image_path": "dataset/mm_math/images/MM_Math/51379914_95.png"
  },
  {
    "solution": "\\boxed{y=x-6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $AB$ intersects the x-axis at point A and the y-axis at point B, where $\\angle OAB=30^\\circ$, and the coordinates of point B are $(0, 2\\sqrt{3})$. The line $y=kx+b$ passes through point A and intersects the y-axis at point C, and $OC=OA$. What is the equation of the line $AC$?\n",
    "completion": "\\textbf{Solution:} Given $B(0,2\\sqrt{3})$, we have $OB=2\\sqrt{3}$. In $\\triangle AOB$, $\\angle OAB=30^\\circ$, hence $AB=2OB=4\\sqrt{3}$. Therefore, $AO=\\sqrt{AB^{2}-OB^{2}}=6$, resulting in $A(6,0)$. Since $OA=6$ and $OA=OC$, we get $OC=6$ and consequently $C(0,-6)$. Assuming the equation of line AC is $y=kx-6$ and substituting point A into it gives $k=1$. Therefore, the equation of line AC is $y=x-6$. Hence, the answer is $\\boxed{y=x-6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50482334_72.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, ship A leaves the port at a speed of 16 nautical miles per hour, heading southeast. At the same time and from the same location, ship B heads southwest. It is known that one and a half hours after leaving the port, they reach points B and A respectively. Given that the distance between points A and B is 30 nautical miles, how many nautical miles does ship B travel per hour?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram: $OA = 16 \\times 1.5 = 24$ (nautical miles),\\\\\n$AB = 30$ (nautical miles),\\\\\nBy the Pythagorean theorem, we have: $OB = \\sqrt{30^2 - 24^2} = 18$ (nautical miles),\\\\\n$18 \\div 1.5 = \\boxed{12}$ (nautical miles/hour),\\\\\nThus, the ship B travels $12$ nautical miles per hour.",
    "image_path": "dataset/mm_math/images/MM_Math/52814879.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in a $4\\times4$ square grid, each small square has a side length of $1$. Points $A$, $B$, and $C$ are all on the grid points, and $AD\\perp BC$ at point $D$. What is the length of $AD$?\n",
    "completion": "\\textbf{Solution:} From the Pythagorean theorem, we have: $AB =\\sqrt{2^{2}+4^{2}}=2\\sqrt{5}$, $AC =\\sqrt{1^{2}+2^{2}}=\\sqrt{5}$, and $BC =\\sqrt{3^{2}+4^{2}}=5$,\\\\\n$\\because AB^{2}+AC^{2}=25$ and $BC^{2}=25$,\\\\\n$\\therefore AB^{2}+AC^{2}=BC^{2}$,\\\\\n$\\therefore \\angle BAC=90^\\circ$,\\\\\n$\\therefore S_{\\triangle ABC} =\\frac{1}{2}AC\\cdot AB=\\frac{1}{2}BC\\cdot AD$,\\\\\n$\\therefore \\sqrt{5}\\times 2\\sqrt{5}=5\\times AD$,\\\\\n$\\therefore AD=\\boxed{2}$,",
    "image_path": "dataset/mm_math/images/MM_Math/51367108.png"
  },
  {
    "solution": "\\boxed{-6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system, the line $y=-\\frac{2}{3}x$ and the line $y=-\\frac{2}{3}x+2$ intersect with the graph of the function $y=\\frac{k}{x}$ $(x<0)$ at points $A$ and $B$, respectively. Connecting $AB$ and $OB$, if the area of $\\triangle OAB$ is $3$, what is the value of $k$?\n",
    "completion": "\\textbf{Solution:} Let the line $y=-\\frac{2}{3}x+2$ intersect the $y$ axis at point $C$, and connect $AC$, \\\\\n$\\therefore$ the point $C(0,2)$, \\\\\n$\\because$ $OA\\parallel CB$, \\\\\n$\\therefore$ ${S}_{\\triangle AOC}={S}_{\\triangle AOB}=3$, \\\\\n$\\therefore$ ${S}_{\\triangle AOC}=\\frac{1}{2}\\times OC\\times |{x}_{A}|=3$, \\\\\n$\\therefore$ $|{x}_{A}|=3$, \\\\\n$\\therefore$ ${x}_{A}=\\pm 3$, \\\\\n$\\because$ point $A$ is in the second quadrant, \\\\\n$\\therefore$ ${x}_{A}=-3$, \\\\\n$\\because$ point $A$ is on the line $y=-\\frac{2}{3}x$, \\\\\n$\\therefore$ $y=-\\frac{2}{3}\\times (-3)=2$, \\\\\n$\\therefore$ the coordinates of point $A$ are: $A(-3,2)$, \\\\\n$\\therefore$ substituting point $A(-3,2)$ into $y=\\frac{k}{x}\\ (x<0)$, \\\\\n$\\therefore$ $2=\\frac{k}{-3}$, \\\\\n$\\therefore$ $k=\\boxed{-6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53078880.png"
  },
  {
    "solution": "\\boxed{7}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, $\\cos B=\\frac{4}{5}$, $AB=10$, and point D is a point on side $BC$, with $AC=DC$. Find the value of $\\cot \\angle BAD$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $DE \\perp AB$ at E,\\\\\nin $\\triangle BED$, $BD=2$, $\\cos B=\\frac{4}{5}$,\\\\\n$\\therefore \\frac{BE}{BD}=\\frac{4}{5}$, which implies $\\frac{BE}{2}=\\frac{4}{5}$,\\\\\nsolving this gives $BE=\\frac{8}{5}$,\\\\\n$\\therefore AE=10-\\frac{8}{5}=\\frac{42}{5}$, and $DE=\\sqrt{BD^2-BE^2}=\\sqrt{2^2-\\left(\\frac{8}{5}\\right)^2}=\\frac{6}{5}$,\\\\\nin $\\triangle ADE$, $\\cot \\angle BAD=\\frac{AE}{DE}=\\frac{\\frac{42}{5}}{\\frac{6}{5}}=\\boxed{7}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52977974_147.png"
  },
  {
    "solution": "\\boxed{\\sqrt{7}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the Cartesian coordinate system $xOy$, the coordinates of point $P$ are $(m-1, -m+5)$. The radius of the circle $\\odot O$ is 1. Point $Q$ lies on $\\odot O$, and line segment $PQ$ is drawn. If $PQ$ is tangent to $\\odot O$, what is the minimum value of line segment $PQ$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, draw tangent PQ from point P to the circle, and connect OQ;\\\\\n$\\because$ The coordinates of point P are $(m−1, −m+5)$\\\\\n$\\therefore$ $O{P}^{2}={\\left(−m+5\\right)}^{2}+{\\left(m−1\\right)}^{2}=2{m}^{2}−12m+26$\\\\\n$\\because$ The radius of circle $O$ is 1, that is, OQ=1\\\\\n$\\therefore$ PQ$^{2}$= OP$^{2}$−OQ$^{2}$= $2{m}^{2}−12m+25=2{\\left(m−3\\right)}^{2}+7$\\\\\nHence, when m=3, PQ$^{2}$ attains its minimum value\\\\\n$\\therefore$ When m=3, PQ attains its minimum value $\\boxed{\\sqrt{7}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51368215.png"
  },
  {
    "solution": "\\boxed{\\frac{4}{5}a+\\frac{1}{5}b}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AC=5$, $\\cot A=2$, $\\cot B=3$. Point $D$ lies on side $AB$, and $\\angle BDC=45^\\circ$. If we let $\\overrightarrow{CA}=a$ and $\\overrightarrow{CB}=b$, then $\\overrightarrow{AB}=\\square$, $\\overrightarrow{AD}=\\square$, $\\overrightarrow{CD}=\\square$ (expressed in terms of $a$ and $b$).\n",
    "completion": "\\textbf{Solution:} From the given information, we have $ \\overrightarrow{CA}=a, \\overrightarrow{CB}=b$,\n\nsince $ \\overrightarrow{AB}=\\overrightarrow{AC}+\\overrightarrow{CB}$ and $ \\overrightarrow{AC}=-\\overrightarrow{CA}$,\n\nthus $ \\overrightarrow{AB}=\\overrightarrow{CB}-\\overrightarrow{CA}=b-a$,\n\nsince AD=AE−DE=2DE−DE=DE and BD=4DE,\n\nthus AB=AD+BD=5AD,\n\nthus AD=$ \\frac{1}{5}AB$,\n\nthus $ \\overrightarrow{AD}=\\frac{1}{5}\\overrightarrow{AB}=\\frac{1}{5}(b-a)=\\frac{1}{5}b-\\frac{1}{5}a$,\n\nthus $ \\overrightarrow{CD}=\\overrightarrow{CA}+\\overrightarrow{AD}=a+\\frac{1}{5}b-\\frac{1}{5}a=\\frac{4}{5}a+\\frac{1}{5}b$.\n\nTherefore, the answers are: $ \\boxed{b-a}$, $ \\boxed{\\frac{1}{5}b-\\frac{1}{5}a}$, $ \\boxed{\\frac{4}{5}a+\\frac{1}{5}b}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51353456_91.png"
  },
  {
    "solution": "\\boxed{\\left(\\frac{3}{2}, -4\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the graph of the linear function $y_1=kx+b$ intersects the x-axis at point A and the y-axis at point B. It intersects with the graph of the hyperbolic function $y_2=\\frac{m}{x}$ at points C and D, with point C being (-2, 3) and the ordinate of point D being -1. If point E is a point on the graph of the hyperbolic function in the fourth quadrant, and a line is drawn from point E perpendicular to the y-axis to point F, and lines OE and AF are connected, if the area of $\\triangle BAF$ is 4 times that of $\\triangle EFO$, determine the coordinates of point E.\n",
    "completion": "\\textbf{Solution:} From the problem, since the graph of the linear function $y_{1}=-\\frac{1}{2}x+2$ intersects the x-axis at point A and the y-axis at point B, \\\\\n$\\therefore A(4, 0)$, $B(0, 2)$, that is $OA=4$, $OB=2$, \\\\\nLet $E(x, -\\frac{6}{x})$, \\\\\nsince $E$ is in the fourth quadrant, \\\\\n$\\therefore EF=x$, $OF=\\frac{6}{x}$, \\\\\n$\\therefore$ The area of $\\triangle EFO=\\frac{1}{2}EF\\cdot OF=\\frac{1}{2}x\\cdot \\frac{6}{x}=3$, \\\\\nsince the area of $\\triangle BAF=4$ times the area of $\\triangle EFO$, \\\\\n$\\therefore$ The area of $\\triangle BAF=12$, \\\\\n$\\therefore$ The area of $\\triangle BAF=\\frac{1}{2}BF\\cdot OA=\\frac{1}{2}BF\\cdot 4=12$, \\\\\n$\\therefore BF=6$, \\\\\n$\\therefore OF=6-2=4$, \\\\\n$\\therefore F(0, -4)$, \\\\\nWhen $y=-4$, substitute into $y=-\\frac{6}{x}$, and solve to obtain $x=\\frac{3}{2}$, \\\\\n$\\therefore E\\left(\\frac{3}{2}, -4\\right)$．Therefore, the coordinates of point E are $\\boxed{\\left(\\frac{3}{2}, -4\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52765598_143.png"
  },
  {
    "solution": "\\boxed{25^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $O$ is a point on line $AB$, and ray $OC$ passes through point $O$ such that $\\angle BOC=65^\\circ$. A right-angled triangle ruler is placed with its right-angled vertex at point $O$. As shown in Figure 1, when one side $ON$ of the triangle ruler $MON$ coincides with ray $OB$, then what is the degree measure of $\\angle MOC$?\n",
    "completion": "\\textbf{Solution:} We have $\\angle MOC = \\boxed{25^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55503444_55.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within the circle $\\odot O$, $CD$ is the diameter of $\\odot O$, and $AB \\perp CD$ at point $E$. If $AB=8$ and $CE=2$, what is the radius of $\\odot O$?\n",
    "completion": "\\textbf{Solution:} Let the radius of $\\odot O$ be $r$, \\\\\nsince CD is the diameter of $\\odot O$ and AB $\\perp$ CD, with AB=8, \\\\\ntherefore AE=$\\frac{1}{2}$AB=4, \\\\\nin $\\triangle OAE$, by the Pythagorean theorem we have: $AE^{2}+OE^{2}=OA^{2}$, \\\\\nthat is $4^{2}+(r-2)^{2}=r^{2}$, \\\\\nsolving this gives: $r=\\boxed{5}$, \\\\\nthus, the radius of $\\odot O$ is 5.",
    "image_path": "dataset/mm_math/images/MM_Math/51445770.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, point $E$ is a point on the diagonal $BD$ of the square $ABCD$, $EF\\perp BC$, $EG\\perp CD$, with the feet of the perpendiculars being $F$ and $G$, respectively. If $CG=6$ and $CF=8$, what is the length of $AE$?\n",
    "completion": "\\textbf{Solution:} Connect CE, as shown in the diagram: \\\\\n$\\because$ Quadrilateral ABCD is a square, \\\\\n$\\therefore AD=CD$, $\\angle ADE=\\angle CDE$, $\\angle C=90^\\circ$, \\\\\nIn $\\triangle ADE$ and $\\triangle CDE$, \\\\\n$\\left\\{\\begin{array}{l}\nAD=CD\\\\\n\\angle ADE=\\angle CDE\\\\\nDE=DE\n\\end{array}\\right.$\\\\\n$\\therefore \\triangle ADE\\cong \\triangle CDE$, \\\\\n$\\therefore AE=CE$, \\\\\n$\\because$ EF$\\perp$BC, EG$\\perp$CD, \\\\\n$\\therefore \\angle EGC=\\angle EFC=\\angle C=90^\\circ$, \\\\\n$\\therefore$ Quadrilateral EFCG is a rectangle, \\\\\n$\\therefore$ EF=CG, \\\\\n$\\because CG=6$, $CF=8$, \\\\\n$\\therefore AE=CE=\\sqrt{6^2+8^2}=\\boxed{10}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53333445.png"
  },
  {
    "solution": "\\boxed{\\frac{5}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\sin\\angle BAC=\\frac{5}{13}$, $AB=13$, $AC=7.2$, $BD\\perp AC$ with the foot of the perpendicular at point $D$, point $E$ is the midpoint of $BD$, and $AE$ intersects $BC$ at point $F$. Find the value of $\\frac{BF}{CF}$.\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, draw $FH\\parallel BD$ through point F and intersect AD at H, \\\\\n$\\therefore \\angle FHC=\\angle EDC=90^\\circ$, $\\angle HFC=\\angle DBC$, \\\\\n$\\therefore \\triangle CHF\\sim \\triangle CDE$, $\\triangle AHF\\sim \\triangle ADE$, $\\tan\\angle HFC=\\tan\\angle CBD=\\frac{CH}{FH}=\\frac{24}{25}$, \\\\\n$\\therefore \\frac{HF}{BD}=\\frac{CF}{CB}$, $CH=\\frac{24}{25}FH$, $\\frac{AH}{AD}=\\frac{HF}{DE}$, \\\\\nLet $FH=x$, then $CH=\\frac{24}{25}x$, $AH=AC+CH=7.2+\\frac{24}{25}x$, \\\\\nSince E is the midpoint of BD, \\\\\n$\\therefore DE=\\frac{1}{2}BD=\\frac{5}{2}$, \\\\\n$\\therefore \\frac{7.2+\\frac{24}{25}x}{12}=\\frac{x}{\\frac{5}{2}}$, \\\\\n$\\therefore$ solving yields $x=\\frac{15}{8}$, \\\\\n$\\therefore FH=\\frac{15}{8}$, \\\\\n$\\therefore \\frac{\\frac{15}{8}}{5}=\\frac{CF}{CB}$, that is $\\frac{CF}{CB}=\\frac{3}{8}$, \\\\\n$\\therefore CF=\\frac{3}{8}CB$, \\\\\n$\\therefore BF=\\frac{5}{8}CB$, \\\\\n$\\therefore \\frac{BF}{CF}=\\boxed{\\frac{5}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52746343_94.png"
  },
  {
    "solution": "\\boxed{F_2(2+\\sqrt{2}, 1)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the parabola $y=ax^2+bx+c$ ($a\\ne 0$) has its vertex at point Q$(2, -1)$ and intersects the $y$-axis at point C$(0, 3)$ and intersects the $x$-axis at points A and B (with point A to the right of point B). Point P is a moving point on the parabola, moving from point C towards point A (point P does not coincide with A). A perpendicular line $PD\\perp$ to line $AC$ is drawn at point P. If point P coincides with point Q, point E is on the $x$-axis, and point F is on the parabola, is there a parallelogram with vertices A, P, E, F? If it exists, write down the coordinates of point F directly; if not, please explain the reason.\n",
    "completion": "\\textbf{Solution:} Solution: Exists, \\boxed{F_1(2-\\sqrt{2}, 1)}, \\boxed{F_2(2+\\sqrt{2}, 1)}.",
    "image_path": "dataset/mm_math/images/MM_Math/52977854_225.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the side length of square $ABCD$ is 8, and point $E$ is on $AB$ with $BE=2$. Point $F$ is a moving point on the diagonal $AC$. What is the minimum perimeter of $\\triangle BFE$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect $DB$, $DE$, $DF$, \\\\\nsince the quadrilateral $ABCD$ is a square with side length equal to 8, \\\\\ntherefore $AB=AD=8, \\angle BAD=90^\\circ$, and $AC$ is the perpendicular bisector of $BD$, \\\\\nthus $BF=DF$, \\\\\nsince $BE=2$, \\\\\ntherefore $AE=AB-BE=6$, the perimeter of $\\triangle BFE$ is $BE+BF+EF=2+DF+EF$, \\\\\naccording to the principle that the length of the segment between two points is shortest, when points $D$, $F$, $E$ are collinear, the sum of $DF+EF$ is minimal, with the minimum being the length of $DE$, \\\\\nin $\\triangle ADE$, $DE=\\sqrt{AD^2+AE^2}=10$, \\\\\nthus, the minimum value of the perimeter of $\\triangle BFE$ is $2+DE=2+10=\\boxed{12}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52386464.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point D lies on the circle $\\odot O$, and point C is on the extension line of diameter BA. Connect CD, and $\\angle CDA=\\angle CBD$. CD is the tangent of $\\odot O$; if $DC=4$ and $AC=2$, what is the length of OC?\n",
    "completion": "\\textbf{Solution:} Since $\\angle CDA=\\angle CBD$ and $\\angle ACD=\\angle DCB$,\\\\\nit follows that $\\triangle ACD\\sim \\triangle DCB$,\\\\\nthus $\\frac{CD}{CB}=\\frac{AC}{DC}$,\\\\\ni.e., $\\frac{4}{CB}=\\frac{2}{4}$,\\\\\nhence $CB=8$,\\\\\ntherefore $OA=\\frac{CB-AC}{2}=\\frac{8-2}{2}=3$,\\\\\nthus $OC=OA+AC=3+2=\\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51445791_100.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{3}{4}x+\\frac{3}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, the line $l: y = \\frac{4}{3}x + b$ passes through the point $A(-3, 0)$ and intersects the $y$-axis at point $B$. The bisector of $\\angle OAB$ intersects the $y$-axis at point $C$. A perpendicular line to $AB$ is drawn through point $C$, intersecting the $x$-axis at point $E$, with the foot of the perpendicular being point $D$. What is the equation of the line $DE$?\n",
    "completion": "\\textbf{Solution:} Since $CD\\perp AB$ and $CO\\perp OA$, \\\\\nit follows that $\\angle CDB = \\angle COE = 90^\\circ$, \\\\\nSince $CD=CO$ and $\\angle BCD = \\angle ECO$, \\\\\nit follows that $\\triangle BCD \\cong \\triangle ECO$ (ASA), \\\\\nOE=BD=2, \\\\\nThus, the coordinates of E are $(2,0)$, \\\\\nGiven that $C(0, \\frac{3}{2})$, \\\\\nlet the equation of line DE be $y=kx+\\frac{3}{2}$, \\\\\nThus, $0=2k+\\frac{3}{2}$, solving this gives $k=-\\frac{3}{4}$, \\\\\nTherefore, the equation of line DE is $\\boxed{y=-\\frac{3}{4}x+\\frac{3}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52189067_134.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, two strips of paper, each with a width of 2, are overlapped at an angle of $30^\\circ$. What is the area of the quadrilateral $ABCD$ formed by the overlapping part?\n",
    "completion": "\\textbf{Solution:} Since $AB \\parallel CD$, \\\\\nit follows that $\\angle ABC = 30^\\circ$, \\\\\nAlso, since the widths of both strips are 2, \\\\\nit follows that $AB = 4$, \\\\\nSimilarly, it can be concluded that $AD = 4$, \\\\\nAgain, because $AB \\parallel CD$ and $AD \\parallel BC$, \\\\\nit follows that quadrilateral $ABCD$ is a parallelogram, \\\\\nAlso, since $AB = AD = 4$, \\\\\nit follows that quadrilateral $ABCD$ is a rhombus, \\\\\nTherefore, $S_{\\text{rhombus}}ABCD = 4 \\times 2 = \\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52302538.png"
  },
  {
    "solution": "\\boxed{x \\le 2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=kx+b$ ($k\\neq 0$) intersects the x-axis at point $A(6,0)$ and the y-axis at point $B(0,3)$, and intersects the line $y=x$ at point $C$. Directly write down the solution to the inequality $kx+b\\geq x$.\n",
    "completion": "\\textbf{Solution:} Upon observing the graph, in conjunction with $C(2, 2)$,\n\n$\\therefore$ the solution set for the inequality $kx+b\\geq x$ is: $x\\boxed{x \\le 2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50677506_134.png"
  },
  {
    "solution": "\\boxed{\\frac{24}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, for the rhombus $ABCD$ with diagonals $AC=6\\,cm$ and $BD=8\\,cm$, and $AH \\perp BC$ at point $H$, what is the length of $AH$ in $cm$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral ABCD is a rhombus, \\\\\n$\\therefore$ CO = $\\frac{1}{2}$AC = 3cm, BO = $\\frac{1}{2}$BD = 4cm, AO$\\perp$BO, \\\\\n$\\therefore$ BC=$\\sqrt{OC^{2}+BO^{2}}$=5cm, \\\\\n$\\therefore$ Area of rhombus ABCD = $\\frac{BD\\cdot AC}{2}$ = $\\frac{1}{2}\\times 6\\times 8=24\\text{cm}^{2}$, \\\\\nSince the area of rhombus ABCD = BC$\\times$AH, \\\\\n$\\therefore$ BC$\\times$AH = 24, \\\\\n$\\therefore$ AH=$\\boxed{\\frac{24}{5}}$cm.",
    "image_path": "dataset/mm_math/images/MM_Math/52344754.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the graph of the function $y=kx+2$ passes through point $A(3,0)$. After the graph is translated upwards by $2$ units, it intersects the $x$ axis at point $B$ and the $y$ axis at point $C$. Find the area of $\\triangle OBC$.\n",
    "completion": "\\textbf{Solution:} In the line $y=-\\frac{2}{3}x+4$, when $x=0$, then $y=4$; when $y=0$, then $x=6$,\\\\\n$\\therefore$ the coordinates of $B$ are $(6, 0)$, and the coordinates of $C$ are $(0, 4)$\\\\\n$\\therefore$ $OB=6, OC=4$,\\\\\n$\\therefore$ the area of $\\triangle OBC$ is $\\frac{1}{2}\\times 6\\times 4=\\boxed{12}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52250984_103.png"
  },
  {
    "solution": "\\boxed{y=-2x+20}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, point O is the origin, and the vertex A of the rhombus OABC has coordinates (3, 4). Connect OB and AC to intersect at point H. Draw line $BD \\parallel AC$ passing through point B to intersect the x-axis at point D. Find the equation of line BD.\n",
    "completion": "\\textbf{Solution:} Since quadrilateral OABC is a rhombus, \\\\\nit follows that OB$\\perp$AC, \\\\\nand since BD$\\parallel$AC, \\\\\nit also follows that OB$\\perp$BD, that is $\\angle$OBD$=90^\\circ$, \\\\\ntherefore, $\\angle$OBF+$\\angle$DBF$=90^\\circ$, \\\\\nand since $\\angle$DBF+$\\angle$BDF$=90^\\circ$, \\\\\nit follows that $\\angle$OBF$=\\angle$BDF, \\\\\nfurthermore, since $\\angle$OFB$=\\angle$BFD$=90^\\circ$, \\\\\nit follows that $\\triangle$OBF$\\sim$$\\triangle$BDF, \\\\\nthus, $\\frac{OF}{BF}=\\frac{BF}{DF}$, \\\\\nleading to $\\frac{8}{4}=\\frac{4}{DF}$, \\\\\nfrom which we solve that DF$=2$, \\\\\ntherefore, OD$=$OF$+$DF$=8+2=10$, \\\\\nthus, D(10,0). \\\\\nSuppose the equation of the line on which BD lies is $y=k_{1}x+b$, \\\\\nsubstituting B(8,4), D(10,0) into the equation gives $\\begin{cases}8k_{1}+b=4\\\\ 10k_{1}+b=0\\end{cases}$, \\\\\nsolving this yields $\\begin{cases}k_{1}=-2\\\\ b=20\\end{cases}$. \\\\\nTherefore, the equation of line BD is $\\boxed{y=-2x+20}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51402702_65.png"
  },
  {
    "solution": "\\boxed{76^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the perpendicular bisectors $l_{1}$ and $l_{2}$ of line segments AB and BC intersect at point O. If $\\angle 1=37^\\circ$, what is the degree measure of $\\angle AOC$?\n",
    "completion": "\\textbf{Solution}: Draw and extend line BO to P,\\\\\nsince the perpendicular bisectors $l_{1}$ and $l_{2}$ of line segments AB and BC intersect at point O,\\\\\ntherefore, AO = OB = OC, $\\angle$BDO = $\\angle$BEO = $90^\\circ$,\\\\\ntherefore, $\\angle$DOE + $\\angle$ABC = $180^\\circ$,\\\\\nsince $\\angle$DOE + $\\angle 1$ = $180^\\circ$,\\\\\ntherefore, $\\angle$ABC = $\\angle 1$ = $38^\\circ$,\\\\\nsince OA = OB = OC,\\\\\ntherefore, $\\angle A$ = $\\angle$ABO, $\\angle$OBC = $\\angle C$,\\\\\nsince $\\angle$AOP = $\\angle A$ + $\\angle$ABO, $\\angle$COP = $\\angle C$ + $\\angle$OBC,\\\\\ntherefore, $\\angle$AOC = $\\angle$AOP + $\\angle$COP = $\\angle A$ + $\\angle$ABC + $\\angle C$ = $2\\times38^\\circ$ = $\\boxed{76^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51378679.png"
  },
  {
    "solution": "\\boxed{625}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, the quadrilaterals with sides AB, BC, AC in the diagram are all squares, and through measurement, the areas of the three squares are 225, 400, and \\(S\\) respectively. What is the value of \\(S\\)?\n",
    "completion": "\\textbf{Solution:} Since $\\angle ACB=90^\\circ$,\\\\\nit follows that $AB^{2}=BC^{2}+AC^{2}=400+225=625$,\\\\\nand the value of S is exactly $AB^{2}$,\\\\\nhence, the value of S is $\\boxed{625}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51737142.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{3}{5}x-3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, the straight line $AB$ intersects the coordinate axes at points $A(0, -3)$ and $B(-5, 0)$, respectively. Point $C$ is the midpoint of line segment $AB$, and point $P$ is a point on the y-axis. Connecting $CP$, and drawing a perpendicular line from point $C$ to line segment $BO$ intersecting at point $Q$. What is the analytic equation of line $AB$?\n",
    "completion": "\\textbf{Solution:} Let the equation of line AB be $y=kx+b$ ($k\\neq 0$). Given $A(0, -3)$, $B(-5, 0)$, we have $\\begin{cases}b=-3\\\\ -5k+b=0\\end{cases}$. Solving these equations yields $\\begin{cases}k=-\\frac{3}{5}\\\\ b=-3\\end{cases}$. Therefore, the equation of line AB is $\\boxed{y=-\\frac{3}{5}x-3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51419593_123.png"
  },
  {
    "solution": "\\boxed{18\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the regular hexagon $ABCDEF$ inscribed in circle $\\odot O$ has a side length of $2\\sqrt{3}$ cm. What is the area of this regular hexagon in cm$^{2}$?\n",
    "completion": "\\textbf{Solution:} Construct \\(OH \\perp AB\\) at point H, and connect OA and OB.\\\\\nSince the side length of the inscribed regular hexagon ABCDEF in circle \\(O\\) is \\(2\\sqrt{3}\\) cm,\\\\\nTherefore, OA = OB = AB = \\(2\\sqrt{3}\\) cm,\\\\\nTherefore, OH = OA\\(\\cdot\\cos 30^\\circ\\) = \\(2\\sqrt{3} \\cdot \\frac{\\sqrt{3}}{2}\\) = 3 (cm),\\\\\nTherefore, the area of the regular hexagon ABCDEF = 6\\(\\cdot\\) Area of \\(\\triangle OAB\\) = 6\\(\\cdot \\frac{1}{2} \\cdot 2\\sqrt{3}\\cdot 3\\) = \\(\\boxed{18\\sqrt{3}}\\) (cm)\\(^{2}\\).",
    "image_path": "dataset/mm_math/images/MM_Math/51005030.png"
  },
  {
    "solution": "\\boxed{-1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the Cartesian coordinate system $xOy$, A and B are two points on the x-axis, and C and D are two points on the y-axis. A part of the parabola passing through points A, C, and B, denoted as $C_{1}$, and a part of the parabola passing through points A, D, and B, denoted as $C_{2}$, combine to form a closed curve. We call this closed curve an \"egg curve\". Given that the coordinates of point C are (0,), and point M is the vertex of the parabola $C_{2}\\colon y=mx^{2}-2mx-3m$ ($m<0$). When $\\triangle BDM$ is a right-angled triangle, what is the value of $m$?\n",
    "completion": "\\textbf{Solution:} From $C_{2}$, it is known that: $B(3,0)$, $D(0,-3m)$, $M(1,-4m)$, \\\\\n$\\therefore BD^{2}=9m^{2}+9$, $BM^{2}=16m^{2}+4$, $DM^{2}=m^{2}+1$. \\\\\n$\\because$ $\\angle MBD<90^\\circ$, $\\therefore$ we discuss the cases of $\\angle BMD=90^\\circ$ and $\\angle BDM=90^\\circ$: \\\\\nWhen $\\angle BMD=90^\\circ$, $BM^{2}+DM^{2}=BD^{2}$, i.e., $16m^{2}+4+m^{2}+1=9m^{2}+9$, \\\\\nwe obtain: $m_{1}=-\\frac{\\sqrt{2}}{2}$, $m_{2}=\\frac{\\sqrt{2}}{2}$ (discard). \\\\\nWhen $\\angle BDM=90^\\circ$, $BD^{2}+DM^{2}=BM^{2}$, i.e., $9m^{2}+9+m^{2}+1=16m^{2}+4$, \\\\\nwe obtain: $m_{1}=-1$, $m_{2}=1$ (discard). \\\\\nIn summary, when $m=-\\frac{\\sqrt{2}}{2}$ or $m=-1$, $\\triangle BDM$ is a right-angled triangle. \\\\\nTherefore, the value of $m$ is $\\boxed{-\\frac{\\sqrt{2}}{2}}$ or $\\boxed{-1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52977501_203.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, E is a point on side BC of the equilateral $\\triangle ABC$ (not coinciding with points B and C). Draw AE, and construct an equilateral $\\triangle AEF$ to the right side using AE as its side. Draw CF. It is known that the function relationship between the area ($S$) of $\\triangle ECF$ and the length ($x$) of BE is shown in Figure 2 (P is the vertex of the parabola). What is the side length of the equilateral $\\triangle ABC$?\n",
    "completion": "\\textbf{Solution:} It is known from the diagram that the maximum value of $S_{\\triangle ECF}$ is $2\\sqrt{3}$,\\\\\n$\\therefore \\frac{\\sqrt{3}}{16}a^{2}=2\\sqrt{3}$\\\\\nSolving gives $a=4\\sqrt{2}$ or $a=-4\\sqrt{2}$ (discard)\\\\\n$\\therefore$ The side length of equilateral $\\triangle ABC$ is $\\boxed{4\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51212719_88.png"
  },
  {
    "solution": "\\boxed{x < -1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $l_{1}\\colon y=ax-b$ and the line $l_{2}\\colon y=kx$ are in the same Cartesian coordinate system. What is the solution set for the inequality $ax-b>kx$ in terms of $x$?\n",
    "completion": "\\textbf{Solution:} When $x < -1$, we have $ax - b > kx$. \\\\\nTherefore, the solution set of the inequality $ax - b > kx$ with respect to $x$ is $x < -1$, \\\\\nthus the answer is: $\\boxed{x < -1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52049852.png"
  },
  {
    "solution": "\\boxed{(\\sqrt{3}, 3)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the coordinates of vertex $A$ of the rhombus $OABC$ are $(0, 2)$. The vertices $B$ and $C$ are in the first quadrant, and the ordinate of point $C$ is $1$. What are the coordinates of point $B$?\n",
    "completion": "\\textbf{Solution:} Extend BC to intersect the x-axis at D,\\\\\n$\\because$ the coordinates of point A are (0, 2),\\\\\n$\\therefore$ OA = 2,\\\\\n$\\because$ quadrilateral OABC is a rhombus,\\\\\n$\\therefore$ AO = OC = BC = 2,\\\\\n$\\because$ BC $\\parallel$ y-axis,\\\\\n$\\therefore$ BD $\\perp$ x-axis,\\\\\nIn $\\triangle$OCD,\\\\\n$\\because$ the ordinate of point C is 1,\\\\\n$\\therefore$ CD = 1,\\\\\n$\\therefore$ OD = $\\sqrt{OC^{2}-CD^{2}} = \\sqrt{2^{2}-1^{2}} = \\sqrt{3}$,\\\\\n$\\because$ BD = BC + CD = 2 + 1 = 3,\\\\\n$\\therefore$ the coordinates of point B are $\\boxed{(\\sqrt{3}, 3)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51662303.png"
  },
  {
    "solution": "\\boxed{10^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AD$ and $AE$ are the median and altitude on side $BC$, respectively. If $\\angle B=40^\\circ$ and $\\angle C=50^\\circ$, what is the measure of $\\angle DAE$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle B=40^\\circ$ and $\\angle C=50^\\circ$, \\\\\ntherefore, $\\angle BAC=90^\\circ$, \\\\\nand since AD is the median, \\\\\nthus AD$= \\frac{1}{2}$ BC$=$BD, \\\\\nhence, $\\angle ADE=2\\angle B=80^\\circ$, \\\\\nand since AE$\\perp$BC, \\\\\ntherefore, $\\angle DAE=\\boxed{10^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52861353_109.png"
  },
  {
    "solution": "\\boxed{\\frac{5}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the direct proportion function $y=kx$ intersects with the graph of the inverse proportion function $y=\\frac{m}{x}$ (where $x>0$) at point $A(2, 2)$. By translating the line $y=kx$ downward, we obtain line $l$. If line $l$ intersects with the graph of the inverse proportion function at point $B(3, n)$, connect $AB$ and $OB$. What is the area of $\\triangle AOB$?\n",
    "completion": "\\textbf{Solution:} Start by substituting point A(2,2) into $y=kx$, we get: $2=2k$, thus $k=1$, and therefore $y=x$. Since the line $y=kx$ is translated downwards to form line $l$, we set the equation of line $l$ as $y=x+b$. Given it intersects the x-axis at point C, substituting point B(3,$\\frac{4}{3}$) yields: $b=-\\frac{5}{3}$. Hence, the equation of line $l$ is $y=x-\\frac{5}{3}$. When $y=0$, we have $x=\\frac{5}{3}$, thus $OC=\\frac{5}{3}$. Connecting AC and knowing that $OA\\parallel BC$, we conclude that ${S}_{\\triangle AOB}={S}_{\\triangle AOC}=\\frac{1}{2}\\times \\frac{5}{3}\\times 2=\\boxed{\\frac{5}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52204038_128.png"
  },
  {
    "solution": "\\boxed{18}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the diagonals AC and BD of the square ABCD intersect at point O, and OA = 3. What is the area of this square?\n",
    "completion": "\\textbf{Solution:} Given the square ABCD, with OA=3,\\\\\nit follows that AC=BD=6, and AO$\\perp$BO,\\\\\nhence the area of the square is: $\\frac{1}{2}$AC$\\times$BD=$\\boxed{18}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51435943.png"
  },
  {
    "solution": "\\boxed{\\frac{9}{10}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\cos\\angle BAC=\\frac{12}{13}$, $AB=26$, $AC=15$, $BD\\perp AC$ with the foot of the perpendicular being D, E is the midpoint of BD, and AE intersects BC at point F. What is the tangent value of $\\angle CBD$?\n",
    "completion": "\\textbf{Solution:} In right-angled $\\triangle ADB$, where $AB=26$ and $\\cos\\angle BAC= \\frac{12}{13}$, \\\\\n$\\therefore AD=24$, \\\\\nBy the Pythagorean theorem, we have $BD= \\sqrt{AB^{2}-AD^{2}}=10$, \\\\\n$\\because AC=15$, \\\\\n$\\therefore CD=9$, \\\\\n$\\therefore \\tan\\angle CBD=\\frac{DC}{BD}=\\frac{9}{10}$, \\\\\n$\\therefore$ the tangent of $\\angle CBD$ is $\\boxed{\\frac{9}{10}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52602500_138.png"
  },
  {
    "solution": "\\boxed{\\frac{4}{7}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $AD$ is the angle bisector of $\\triangle ABC$, and $DE \\parallel AB$. If $\\frac{AE}{EC} = \\frac{3}{4}$, then what is the value of $\\frac{AE}{AB}$?\n",
    "completion": "\\textbf{Solution:} Since $DE \\parallel AB$, \\\\\nit follows that $\\triangle CDE \\sim \\triangle CBA$, \\\\\nhence $\\frac{DE}{AB}=\\frac{CE}{AC}$, \\\\\nGiven $\\frac{AE}{EC} = \\frac{3}{4}$, \\\\\nit then follows $\\frac{DE}{AB}=\\frac{CE}{AC} = \\frac{4}{7}$, \\\\\nSince $AD$ is the angle bisector of $\\triangle ABC$ and $DE \\parallel AB$, \\\\\nit follows that $\\angle ADE = \\angle BAD = \\angle DAE$, \\\\\nconsequently, $AE = DE$, \\\\\nthus, $\\frac{AE}{AB}=\\frac{DE}{AB}=\\frac{CE}{AC} = \\boxed{\\frac{4}{7}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52917594.png"
  },
  {
    "solution": "\\boxed{12+2\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the side length of the square $ABCD$ is $8$, with point $E$ on $AB$ and $BE=2$. Points $M$ and $N$ are on $AC$ such that $MN=2\\sqrt{2}$. Connecting $BN$ and $EM$, what is the minimum perimeter of the quadrilateral $BEMN$?\n",
    "completion": "\\textbf{Solution:} Connect BD and DN, construct point E's symmetric point F with respect to BD, thus BE=BF=2,\\\\\nconnect NF and DF,\\\\\n$\\because$ Quadrilateral ABCD is a square, $\\therefore \\angle ABC=\\angle BCD=90^\\circ$, DB$\\perp$AC, BN=DN, point F is on BC,\\\\\n$\\therefore$EF$\\parallel$AC, EF= $\\sqrt{BE^2+BF^2}=2\\sqrt{2}=MN$,\\\\\n$\\therefore$ Quadrilateral MEFN is a parallelogram,\\\\\n$\\therefore$ME=NF,\\\\\n$\\therefore$ME+BN=NF+DN$\\geq$DF (equality holds when D, N, F are collinear),\\\\\nIn $\\triangle DCF$, CD=8, CF=8-2=6, then DF= $\\sqrt{CD^2+CF^2}=10$,\\\\\n$\\therefore$ME+BN$\\geq 10$,\\\\\n$\\therefore$MN+BE+ME+BN$\\geq 2\\sqrt{2}+2+10=12+2\\sqrt{2}$,\\\\\nthus, the minimum perimeter of quadrilateral $BEMN$ is $\\boxed{12+2\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52188954.png"
  },
  {
    "solution": "\\boxed{16}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is a chord of $\\odot O$, $OC\\perp AB$ at point $M$, and intersects $\\odot O$ at point $C$. If the radius of $\\odot O$ is $10$, and the ratio $OM:MC = 3:2$, what is the length of $AB$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, connect OA. \\\\\nSince OM:MC=3:2, and OC=10, \\\\\nthus OM=$\\frac{3}{5}OC=\\frac{3}{5}\\times 10=6$. \\\\\nSince OC$\\perp$AB, \\\\\ntherefore $\\angle$OMA$=90^\\circ$, and AB=2AM. \\\\\nIn $\\triangle$AOM, with AO=10, and OM=6, \\\\\ntherefore AM$=\\sqrt{AO^{2}-OM^{2}}=\\sqrt{10^{2}-6^{2}}=8$. \\\\\nThus, AB=2AM$=\\boxed{16}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51351028.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated, the radius $OA$ of the arc $\\overset{\\frown}{AB}$ is 2, $OC \\perp AB$ at point $C$, and $\\angle AOC = 60^\\circ$. What is the length of the chord $AB$?\n",
    "completion": "\\textbf{Solution:} \\\\\n$\\because$ The radius $OA=2$ of arc $\\widehat{AB}$, and $OC\\perp AB$ at point C, $\\angle AOC=60^\\circ$,\\\\\n$\\therefore AC=OA\\cdot\\sin60^\\circ=2\\times \\frac{\\sqrt{3}}{2}=\\sqrt{3}$,\\\\\n$\\therefore AB=2AC=2\\sqrt{3}$;\\\\\nThus, the length of chord AB is $AB=\\boxed{2\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52765293_135.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ABC = 45^\\circ$. $F$ is the intersection point of altitudes $AD$ and $BE$, $AC = \\sqrt{5}$, $BD = 2$. What is the length of segment $DF$?\n",
    "completion": "\\textbf{Solution:} Given that $AD$ and $BE$ are altitudes of $\\triangle ABC$,\\\\\nit follows that $\\angle ADB = \\angle ADC = \\angle BEC = 90^\\circ$.\\\\\nThus, $\\angle C + \\angle DAC = 90^\\circ$; $\\angle C + \\angle DBF = 90^\\circ$.\\\\\nThis means $\\angle DAC = \\angle DBF$.\\\\\nSince $\\angle ABC = 45^\\circ$,\\\\\nit implies $\\angle DAB = 45^\\circ$.\\\\\nHence, $\\angle ABC = \\angle DAB$.\\\\\nTherefore, $DA = DB$.\\\\\nIn $\\triangle ADC$ and $\\triangle BDF$,\\\\\n\\[\\left\\{ \\begin{array}{l} \\angle ADC = \\angle BDF \\\\ DA = DB \\\\ \\angle DAC = \\angle DBF \\end{array} \\right.\\]\\\\\nThus, $\\triangle ADC \\cong \\triangle BDF$ (ASA).\\\\\nTherefore, $AC = BF = \\sqrt{5}$.\\\\\nIn right $\\triangle BDF$, with $\\angle BDF = 90^\\circ$,\\\\\nit follows that $BD^2 + DF^2 = BF^2$.\\\\\nGiven $BD = 2$, and $BF = \\sqrt{5}$,\\\\\nit concludes $DF = \\boxed{1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51378906.png"
  },
  {
    "solution": "\\boxed{-6<x<-3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the figure, the graph of the linear function $y=-x+b$ intersects with the graph of the hyperbolic function $y=\\frac{k}{x}$ ($x<0$) at point $A(-6,m)$, and intersects with the x-axis at point $B(-4,0)$. If the line $y=4$ intersects line $AB$ at point $C$, and intersects the hyperbola at point $D$, based on the graph, directly write out the solution set for the inequality $-x+b<\\frac{k}{x}<4$.\n",
    "completion": "\\textbf{Solution:} Solving directly, the solution set of the inequality $-x+b<\\frac{k}{x}<4$ is $\\boxed{-6<x<-3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52602187_170.png"
  },
  {
    "solution": "\\boxed{y=-x+6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in $\\triangle AOB$, $\\angle ABO=90^\\circ$, $OB=4$, and $AB=8$. The line $y=-x+b$ intersects $OA$ and $AB$ at points $C$ and $D$, respectively, and the area of $\\triangle BOD$ is $4$. Find the equation of line $CD$.\n",
    "completion": "\\textbf{Solution:} Given that $OB=4$ and $\\angle ABO=90^\\circ$, the area of $\\triangle BOD$ is $S_{\\triangle BOD} = \\frac{1}{2} \\times OB \\times BD = 4$, \\\\\nthus $DB=2$, \\\\\ntherefore, the coordinates of point D are $(4,2)$, \\\\\nsubstituting D$(4,2)$ into $y=-x+b$ yields: $b=6$, \\\\\ntherefore, the equation of line CD is $\\boxed{y=-x+6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52076885_175.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ABC=90^\\circ$, $\\angle A=60^\\circ$, D is a point on side $AC$, $\\angle DBC=30^\\circ$, $AC=12\\,cm$, then what is the perimeter of $\\triangle ABD$?\n",
    "completion": "\\textbf{Solution:} Connect $CO$, $AO$, passing through point $O$, construct $OM\\perp AB$ at $M$, $ON\\perp CD$ at $N$,\\\\\nthen quadrilateral $OMEN$ is a rectangle,\\\\\nsince $OM\\perp AB$, $ON\\perp CD$,\\\\\nand $AB=CD$,\\\\\nit follows that $CN=\\frac{1}{2}CD$, $AM=\\frac{1}{2}AB$,\\\\\nsince $CO=AO$,\\\\\nit follows that right triangle $CNO\\cong$ right triangle $AMO$ (HL),\\\\\nthus $NO=MO$,\\\\\nhence $NO=MO=EM$,\\\\\nseeing that $OM\\perp AB$,\\\\\nit follows that $AM=MB=\\frac{1}{2}AB=\\frac{1}{2}(AE+BE)=5$,\\\\\ntherefore, $EM=AM-AE=5-3=2$,\\\\\nthus $ON=EM=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52623527.png"
  },
  {
    "solution": "\\boxed{20^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC$, and the bisectors of $\\angle BAC$ and $\\angle ACB$ intersect at point $D$. Given that $\\angle ADC=130^\\circ$, find the measure of $\\angle BAC$?\n",
    "completion": "\\textbf{Solution:} Given that $AB=AC$, and $AE$ bisects $\\angle BAC$,\\\\\nit follows that $AE\\perp BC$ (for an isosceles triangle, three lines coincide),\\\\\nsince $\\angle ADC=130^\\circ$,\\\\\nit follows $\\angle CDE=50^\\circ$,\\\\\nthus $\\angle DCE=90^\\circ-\\angle CDE=40^\\circ$, and since $CD$ bisects $\\angle ACB$,\\\\\nit follows $\\angle ACB=2\\angle DCE=80^\\circ$.\\\\\nSince $AB=AC$,\\\\\nit follows $\\angle B=\\angle ACB=80^\\circ$,\\\\\nthus $\\angle BAC=180^\\circ-(\\angle B+\\angle ACB)=\\boxed{20^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53147949.png"
  },
  {
    "solution": "\\boxed{\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $E$ and $F$ are two points on $CD$, $EG \\perp AC$, and $FH \\perp AC$, with the feet of the perpendiculars being $G$ and $H$ respectively. If $AD=2$, $DE=1$, $CF=2$, and $AG=CH$, what is the sum of $EG+FH$?\n",
    "completion": "\\textbf{Solution:} Draw line $EM \\perp AB$ at $M$, and extend $EG$ to intersect $AB$ at $Q$, then $\\triangle EQM$ is a right-angled triangle.\\\\\nSince $EG \\perp AC$ and $FH \\perp AC$,\\\\\n$\\therefore \\angle CHF=\\angle AGQ=90^\\circ$,\\\\\nSince in rectangle $ABCD$, $CD\\parallel AB$,\\\\\n$\\therefore \\angle FCH=\\angle QAG$,\\\\\nIn $\\triangle FCH$ and $\\triangle QAG$,\\\\\n$\\left\\{\\begin{array}{l}\n\\angle CHF=\\angle AGQ\\\\\nCH=AG\\\\\n\\angle FCH=\\angle QAG\n\\end{array}\\right.$,\\\\\n$\\therefore \\triangle FCH\\cong \\triangle QAG$ (ASA),\\\\\n$\\therefore AQ=CF=2$, FH=QG,\\\\\nSince $\\angle D=\\angle DAM=\\angle AME=90^\\circ$,\\\\\n$\\therefore$ quadrilateral $ADEM$ is a rectangle,\\\\\n$\\therefore AM=DE=1$, EM=AD=2,\\\\\n$\\therefore MQ=2-1=1$,\\\\\n$\\therefore$ in $\\triangle EMQ$,\\\\\n$EQ= \\sqrt{EM^2+QM^2}=\\sqrt{5}$,\\\\\ni.e., $EG+QG=EG+FH= \\boxed{\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50481443.png"
  },
  {
    "solution": "\\boxed{3\\sqrt{10}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in square $ABCD$, $AB=9$, point $E$ is on side $CD$, and $DE=2CE$. Point $P$ is a moving point on diagonal $AC$. What is the minimum value of $PE+PD$?\n",
    "completion": "\\textbf{Solution:} Draw BP and BE,\\\\\nsince quadrilateral ABCD is a square,\\\\\nthus AB=AD, and $\\angle$BAP=$\\angle$DAP.\\\\\nsince AP=AP,\\\\\nthus $\\triangle$ABP$\\cong$$\\triangle$ADP,\\\\\nthus BP=DP,\\\\\nthus PE+PD=PE+PB$\\geq$BE,\\\\\nmeaning the minimum value of PE+PD is the length of BE,\\\\\nsince AB=9, and DE=2CE,\\\\\nthus CE=$\\frac{1}{3}CD$=$\\frac{1}{3}\\times 9$=3,\\\\\nthus BE=$\\sqrt{BC^{2}+CE^{2}}$=$\\sqrt{9^{2}+3^{2}}$=$3\\sqrt{10}$,\\\\\nthus the minimum value of PE+PD is \\boxed{3\\sqrt{10}}.",
    "image_path": "dataset/mm_math/images/MM_Math/53182977.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system $xOy$, a point light source is located at $P(2, 2)$. The coordinates of the ends of the rod $AB$ are respectively $(0, 1)$ and $(3, 1)$. What is the length of the shadow $CD$ of the rod $AB$ on the $x$-axis?\n",
    "completion": "\\textbf{Solution:} Construct PE$\\perp$x-axis at E, intersecting AB at M, as shown in the figure,\\\\\nsince P(2, 2), A(0, 1), B(3, 1).\\\\\nTherefore, PM = 1, PE = 2, AB = 3,\\\\\nsince AB$\\parallel$CD\\\\\nTherefore $\\frac{AB}{CD}=\\frac{PM}{PE}$ \\\\\nTherefore $\\frac{3}{CD}=\\frac{1}{2}$ \\\\\nTherefore CD= \\boxed{6},",
    "image_path": "dataset/mm_math/images/MM_Math/52536529.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in parallelogram $ABCD$, draw $DE\\perp AB$ at point $E$, with point $F$ on side $CD$ and $FC=AE$. Connect $AF$ and $BF$. Quadrilateral $DEBF$ is a rectangle; if $AF$ bisects $\\angle DAB$, $FC=3$, and $DF=5$, find the length of $BF$.\n",
    "completion": "\\textbf{Solution:} Given that $\\because AF$ bisects $\\angle DAB$,\\\\\n$\\therefore \\angle DAF=\\angle BAF$,\\\\\n$\\because CD\\parallel AB$,\\\\\n$\\therefore \\angle DFA=\\angle BAF$,\\\\\n$\\therefore \\angle DFA=\\angle DAF$,\\\\\n$\\therefore AD=DF=5$,\\\\\nIn $\\triangle AED$, $AE=FC=3$,\\\\\nBy the Pythagorean theorem, we have $DE=\\sqrt{AD^{2}-AE^{2}}=\\sqrt{5^{2}-3^{2}}=4$,\\\\\nFrom (1), it is deduced that the quadrilateral DEBF is a rectangle,\\\\\n$\\therefore BF=DE=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52189671_139.png"
  },
  {
    "solution": "\\boxed{(13, 0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: The straight line $y=kx+b$ and the inverse proportion function $y=\\frac{8}{x}$ $(x>0)$ intersect at points $A(m, 4)$ and $B(8, n)$, respectively, and intersect the coordinate axes at points $C$ and $D$, respectively. If $P$ is a moving point on the $x$-axis, and the area of $\\triangle ADP$ is $6$, find the coordinates of point $P$.\n",
    "completion": "\\[ \\text{Solution: From (1), the equation of line AB is } y=-\\frac{1}{2}x+5, \\]\n\\[ \\text{when } y=0, \\text{ we have } x=10, \\]\n\\[ \\therefore \\text{ the coordinates of point } D \\text{ are } (10, 0) \\]\n\\[ \\text{Let the coordinates of point } P \\text{ be } (a, 0), \\text{ then } PD=|10-a| \\]\n\\[ \\because \\text{ the area of } \\triangle ADP \\text{ is } 6 \\]\n\\[ \\therefore \\frac{1}{2} \\times 4 \\times PD = 6 \\]\n\\[ \\therefore PD = 3 \\]\n\\[ \\therefore |10-a| = 3 \\]\n\\[ \\text{Solving this gives } a=7 \\text{ or } 13 \\]\n\\[ \\therefore \\text{The coordinates of } P \\text{ are } (\\boxed{7}, 0) \\text{ or } \\boxed{(13, 0)} \\]",
    "image_path": "dataset/mm_math/images/MM_Math/51403084_62.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, AB is the diameter of the circle $\\odot$O, and points F, C are on circle $\\odot$O, with $\\overset{\\frown}{AF}=\\overset{\\frown}{FC}=\\overset{\\frown}{CB}$. Connect AC and AF, and draw $CD\\perp AF$ extending through point C to meet the extension of AF at point D, with D being the foot of the perpendicular. CD is the tangent of circle $\\odot$O; If CD=$2\\sqrt{3}$, what is the radius of $\\odot$O?\n",
    "completion": "\\textbf{Solution:} Connect BC as shown in the figure, \\\\\nsince AB is the diameter, \\\\\ntherefore, $\\angle ACB=90^\\circ$, \\\\\nsince $\\overset{\\frown}{AF}=\\overset{\\frown}{FC}=\\overset{\\frown}{CB}$, \\\\\ntherefore, $\\angle BOC= \\frac{1}{3}\\times180^\\circ=60^\\circ$, \\\\\ntherefore, $\\angle BAC=30^\\circ$, \\\\\ntherefore, $\\angle DAC=30^\\circ$, \\\\\nin $\\triangle ADC$, where CD=$2\\sqrt{3}$, \\\\\ntherefore, AC=2CD=$4\\sqrt{3}$, \\\\\nin $\\triangle ACB$, where BC$^{2}$+AC$^{2}$=AB$^{2}$, \\\\\nthat is, ($4\\sqrt{3}$)$^{2}$+($\\frac{1}{2}$AB)$^{2}$=AB$^{2}$, \\\\\ntherefore, AB=8, \\\\\ntherefore, the radius of $\\odot O$ is $\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52536794_97.png"
  },
  {
    "solution": "\\boxed{25}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, respectively along the diagonals AC of the rectangular paper piece ABCD and EG of the square paper piece EFGH, if cut and assembled as shown in Figure 2 into $\\square KLMN$, and if the quadrilateral OPQR in the middle is precisely a square, and the area of $\\square KLMN$ is 50, then what is the area of the square EFGH?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, let $PM = PL = NR = KR = a$, and the side length of square $ORQP$ be $b$.\\\\\nAccording to the problem: $a^{2} + b^{2} + (a + b)(a - b) = 50$,\\\\\n$\\therefore a^{2} = 25$,\\\\\n$\\therefore$ the area of square $EFGH$ = $a^{2} = \\boxed{25}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52210860.png"
  },
  {
    "solution": "\\boxed{4\\,cm}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given, as shown in the figure, points $B$ and $C$ divide the line segment $AD$ into three parts in the ratio $2:5:3$, $M$ is the midpoint of $AD$, and $BM=6\\,\\text{cm}$. What is the length of $CM$ in centimeters?\n",
    "completion": "\\textbf{Solution:} Let $AB=2x\\,cm$, $BC=5x\\,cm$, $CD=3x\\,cm$,\\\\\nthus, $AD=AB+BC+CD=10x\\,cm$\\\\\nGiven that $M$ is the midpoint of $AD$,\\\\\nthus, $AM=MD=\\frac{1}{2}AD=\\frac{1}{2}\\times 10x=5x$, and $AB=2x\\,cm$,\\\\\nthus, $BM=AM−AB=5x−2x=3x\\,cm$,\\\\\nGiven that $BM=6\\,cm$,\\\\\nthus, $3x=6$, $x=2$,\\\\\nGiven that $CM=MD−CD$,\\\\\nthus, $CM=5x−3x=2x=2\\times 2=\\boxed{4\\,cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53078633.png"
  },
  {
    "solution": "\\boxed{y = -x + 1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given the figure, it is known that the line \\(l_1\\) passing through the point \\(B(1, 0)\\) intersects with the line \\(l_2: y=2x+4\\) at the point \\(P(-1, a)\\). Find the equation of the line \\(l_1\\).\n",
    "completion": "\\textbf{Solution:} Since point P$(-1, a)$ lies on the line $y=2x+4$, we have $2 \\times (-1) + 4 = a$, which means $a=2$. Therefore, the coordinates of P are $(-1, 2)$. Let the equation of line $l_1$ be $y=kx+b$ (where $k \\neq 0$). Substituting points B$(1, 0)$ and P$(-1, 2)$ into this equation, we get $\\begin{cases} k + b = 0 \\\\ -k + b = 2 \\end{cases}$. Solving this system of equations, we find $\\begin{cases} k = -1 \\\\ b = 1 \\end{cases}$. Hence, the equation of line $l_1$ is \\boxed{y = -x + 1}.",
    "image_path": "dataset/mm_math/images/MM_Math/50431079_189.png"
  },
  {
    "solution": "\\boxed{\\frac{81}{16}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the right triangle $ABC$, both quadrilaterals $DBFE$ and $GFIH$ are squares. Given that $AD=1\\text{cm}$ and $DB=3\\text{cm}$, what is the area of the shaded part in $\\text{cm}^2$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect $BE$, $FH$,\\\\\n$\\because$ $S_{\\triangle ABC}=\\frac{1}{2}AB\\cdot BC=\\frac{1}{2}AB\\cdot DE+\\frac{1}{2}BC\\cdot EF$,\\\\\n$\\therefore$ $4BC=12+3BC$,\\\\\n$\\therefore$ $BC=12$,\\\\\n$\\therefore$ $FC=BC-BF=12-3=9$,\\\\\n$\\because$ $S_{\\triangle EFC}=\\frac{1}{2}EF\\cdot FC=\\frac{1}{2}EF\\cdot GH+\\frac{1}{2}FC\\cdot HI$,\\\\\n$\\therefore$ $27=3GH+9HI$,\\\\\n$\\therefore$ $GH=\\frac{9}{4}$,\\\\\n$\\therefore$ the area of square GFIH is: $\\boxed{\\frac{81}{16}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52076904.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, we know for the rectangle $ABCD$, where $AD=10$ and $AB=6$. Now, the side $AD$ is rotated about one of its endpoints, and when the other endpoint falls exactly at point E on the line where side $BC$ is located, what is the length of the segment $DE$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure,\\\\\n$\\because$ The quadrilateral ABCD is a rectangle,\\\\\n$\\therefore$ AB=CD=6, AD=BC=10, $\\angle$ABC=$\\angle$DCB=$90^\\circ$,\\\\\nWhen AD=AE=10, BE= $\\sqrt{AE^{2}-AB^{2}}=\\sqrt{10^{2}-6^{2}}=8$,\\\\\n$\\therefore$ DE$_{1}$= $\\sqrt{CD^{2}+EC^{2}}=\\sqrt{6^{2}+2^{2}}=2\\sqrt{10}$, DE$_{2}$= $\\sqrt{CD^{2}+EC^{2}}=\\sqrt{6^{2}+18^{2}}=6\\sqrt{10}$,\\\\\nWhen DE=DA=10, DE=10,\\\\\nIn summary, the values of DE that meet the conditions are $\\boxed{2\\sqrt{10}}$, $\\boxed{6\\sqrt{10}}$, or $\\boxed{10}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51213352.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the midpoints of the sides of the quadrilateral $ABCD$ are $E$, $F$, $G$, and $H$ respectively. The diagonals $AC$ and $BD$ intersect at point $O$. If the perimeter of the quadrilateral $EFGH$ is $3$, what is the length of $AC + BD$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure,\\\\\nsince E, F, G, H are the midpoints of line segments AB, BC, CD, AD, respectively,\\\\\ntherefore, EH and FG are the midlines of $\\triangle ABD$ and $\\triangle BCD$, respectively, EF and HG are the midlines of $\\triangle ACD$ and $\\triangle ABC$, respectively,\\\\\nbased on the properties of a triangle's midline, we know:\\\\\nEF$\\parallel$AC, GH$\\parallel$AC and EF=GH=$\\frac{1}{2}$AC, EH=FG=$\\frac{1}{2}$BD,\\\\\ntherefore, quadrilateral EFGH is a parallelogram,\\\\\nsince the perimeter of quadrilateral EFGH is 3, i.e., EF+GH+EH+FG=3,\\\\\ntherefore, AC+BD=\\boxed{3},",
    "image_path": "dataset/mm_math/images/MM_Math/53313177.png"
  },
  {
    "solution": "\\boxed{105^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Place a set of set squares as shown in the figure, what is the degree measure of $\\angle AOB$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure below: \\\\\nFrom the diagram, we can see that $\\angle 1=30^\\circ$, $\\angle 2=45^\\circ$, \\\\\n$\\therefore$ $\\angle 3=\\angle 1+\\angle 2=75^\\circ$, \\\\\n$\\therefore$ $\\angle AOB=180^\\circ-\\angle 3=\\boxed{105^\\circ}$,",
    "image_path": "dataset/mm_math/images/MM_Math/52580110.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the graph of the linear function $y=kx+b$ passes through point $A(0, 2)$ and point $B(-2, 0)$. What is the area of the triangle formed by the linear function and the coordinate axes?\n",
    "completion": "\\textbf{Solution:} Given that $A(0, 2)$, $B(-2, 0)$,\\\\\nit follows that $OB=OA=2$,\\\\\nthus, the area of $\\triangle AOB$ is given by ${S}_{\\triangle AOB}=\\frac{1}{2}OA\\cdot OB=\\frac{1}{2}\\times 2\\times 2=\\boxed{2}$,\\\\\nwhich means the area of the triangle formed by the linear function and the coordinate axes is \\boxed{2}.",
    "image_path": "dataset/mm_math/images/MM_Math/50676718_161.png"
  },
  {
    "solution": "\\boxed{50^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ is rotated $40^\\circ$ clockwise around point $A$ to get $\\triangle ADE$. If $AC \\perp DE$, what is the measure of $\\angle C$?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABC$ is rotated $40^\\circ$ clockwise around point $A$ to get $\\triangle ADE$\\\\\nthen $\\angle CAE=40^\\circ$, and $\\angle C = \\angle E$\\\\\nSince $AC \\perp DE$\\\\\nthen $\\angle E=50^\\circ$\\\\\nthus $\\angle C=\\boxed{50^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/53764975.png"
  },
  {
    "solution": "\\boxed{42^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within circle $O$, $AB$ is a diameter, and point $C$ is a point on the circle. The minor arc $AC$ is folded along the chord $AC$ to intersect $AB$ at point $D$, and $CD$ is connected. If point $D$ does not coincide with the center $O$ and $\\angle BAC=24^\\circ$, what is the degree measure of $\\angle DCA$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect BC,\\\\\n$\\because$ AB is the diameter of $ \\odot O$\\\\\n$\\therefore$ $\\angle ACB=90^\\circ$\\\\\n$\\therefore$ $\\angle B=180^\\circ-\\angle BAC-\\angle B=180^\\circ-90^\\circ-24^\\circ=66^\\circ$\\\\\nAccording to the reflection property, $\\angle B$ is the inscribed angle corresponding to arc $\\overset{\\frown}{AC}$, $\\angle ADC$ is the inscribed angle corresponding to arc $\\overset{\\frown}{ABC}$\\\\\n$\\therefore$ $\\angle B+\\angle ADC=180^\\circ$\\\\\n$\\therefore$ $\\angle ADC=114^\\circ$\\\\\n$\\therefore$ $\\angle DCA=180^\\circ-\\angle ADC-\\angle BAC=180^\\circ-114^\\circ-24^\\circ=\\boxed{42^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/52793776.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points E and F are the midpoints of $BD$ and $CE$ respectively. If the area of $\\triangle ABC$ is 32, then what is the area of the shaded part $\\triangle AEF$?\n",
    "completion": "\\textbf{Solution:} \\\\\nSince point $E$ is the midpoint of $BD$, \\\\\nit follows that $AE$ is the median of $\\triangle ABD$, and $CE$ is the median of $\\triangle CBD$, \\\\\nthus, ${S}_{\\triangle ADE}=\\frac{1}{2}{S}_{\\triangle ABD}$ and ${S}_{\\triangle CDE}=\\frac{1}{2}{S}_{\\triangle CBD}$, \\\\\nhence, ${S}_{\\triangle ACE}={S}_{\\triangle ADE}+{S}_{\\triangle CDE}=\\frac{1}{2}{S}_{\\triangle ABD}+\\frac{1}{2}{S}_{\\triangle CBD}=\\frac{1}{2}{S}_{\\triangle ABC}=\\frac{1}{2}\\times 32=16$, \\\\\nSince point $F$ is the midpoint of $CE$, \\\\\nit follows that $AF$ is the median of $\\triangle ACE$, \\\\\nthus, ${S}_{\\triangle AEF}=\\frac{1}{2}{S}_{\\triangle ACE}=\\frac{1}{2}\\times 16=\\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52950314.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the edge $DE$ of the rectangle $DEFG$ lies on the edge $BC$ of $\\triangle ABC$, with vertices $G$ and $F$ located on edges $AB$ and $AC$, respectively. Given that $BC=6\\,cm$, $DE=3\\,cm$, and $EF=2\\,cm$, what is the area of $\\triangle ABC$ in $cm^{2}$?\n",
    "completion": "\\textbf{Solution:} Let $AH \\perp BC$ at $H$, and intersect $GF$ at $M$, as shown in the figure,\\\\\nthen $MH=EF=2\\,cm$,\\\\\nsince quadrilateral $DEFG$ is a rectangle,\\\\\nthus $GF\\parallel BC$, $DG=EF=2\\,cm$, $GF=DE=3\\,cm$,\\\\\nsince $GF\\parallel BC$,\\\\\nthus $\\triangle AGF\\sim \\triangle ABC$,\\\\\ntherefore $\\frac{AM}{AH}=\\frac{GF}{BC}$,\\\\\nthat is $\\frac{AM}{AM+2}=\\frac{3}{6}$,\\\\\nsolving this, we get: $AM=2\\,cm$,\\\\\nthus $AH=AM+MH=4\\,cm$,\\\\\ntherefore, the area of $\\triangle ABC$ $=\\frac{1}{2}BC \\cdot AH=\\frac{1}{2} \\times 6 \\times 4 = \\boxed{12}\\,\\left(cm^{2}\\right)$.",
    "image_path": "dataset/mm_math/images/MM_Math/52977898.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the isosceles right triangle $\\triangle ABC$ with leg length of $4$, $D$ and $E$ are two moving points on sides $AB$ and $AC$ respectively, and $AE=BD$. What is the minimum value of $CD+BE$?\n",
    "completion": "\\textbf{Solution:} Construct CF $\\perp$ AC such that CF=AC=4. Connect EF and BF, intersecting AC at point O.\n\n$\\because$ AB=AC, AE=BD,\n\n$\\therefore$ AD=CE,\n\nIn $\\triangle ACD$ and $\\triangle CEF$,\n\n$\\left\\{\\begin{array}{l}\nAD=CE\\\\\n\\angle DAC=\\angle ECF\\\\\nAC=CF\n\\end{array}\\right.$\n\n$\\therefore$ $\\triangle ACD \\cong \\triangle CEF$ (SAS),\n\n$\\therefore$ CD=FE,\n\n$\\therefore$ BE+CD=BE+EF$\\geq$BF,\n\nWhen points B, E, and F are collinear, the value of BE+CD=BF is minimal,\n\n$\\because$ $\\angle$BAC=$\\angle$ACF=90$^\\circ$, AB=AC, AC=CF,\n\n$\\therefore$ AB$\\parallel$CF,\n\n$\\therefore$ quadrilateral ABCF is a parallelogram,\n\n$\\therefore$ BF=2OB, AO=OC=2,\n\nIn Rt $\\triangle ABO$, $BO=\\sqrt{AB^{2}+AO^{2}}=\\sqrt{4^{2}+2^{2}}=2\\sqrt{5}$\n\n$\\therefore$ BF=$\\boxed{4\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52861347.png"
  },
  {
    "solution": "\\boxed{2.4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the right-angled triangle $ABC$, where $\\angle ACB = 90^\\circ$, $AC = 3$, and $BC = 4$, point $M$ is a point on side $AB$ (not coinciding with points $A$ or $B$). Construct $ME \\perp AC$ at point $E$ and $MF \\perp BC$ at point $F$. What is the minimum value of $EF$?\n",
    "completion": "\\textbf{Solution:} Connect CM, \\\\\n$\\because$ ME $\\perp$ AC, and MF $\\perp$ BC, \\\\\n$\\therefore \\angle$ MEC = $\\angle$ MFC = 90$^\\circ$, \\\\\n$\\because \\angle$ C = 90$^\\circ$, \\\\\n$\\therefore$ Quadrilateral ECFM is a rectangle, \\\\\n$\\therefore$ EF = CM, \\\\\nWhen CM $\\perp$ AB, CM is at its shortest, as shown in the figure: \\\\\nWhen CM $\\perp$ AB, \\\\\n$S_{\\triangle ABC} = \\frac{1}{2}AC \\cdot BC = \\frac{1}{2}AB \\cdot CM$, \\\\\n$\\therefore \\frac{1}{2} \\times 3 \\times 4 = \\frac{1}{2}CM \\cdot AB$, \\\\\n$\\because$ In Rt $\\triangle$ ABC, \\\\\n$AB = \\sqrt{AC^2 + BC^2} = \\sqrt{3^2 + 4^2} = 5$, \\\\\n$\\therefore \\frac{1}{2} \\times 3 \\times 4 = \\frac{1}{2}CM \\times 5$, \\\\\n$\\therefore$ CM = 2.4, \\\\\n$\\therefore$ The minimum value of CM is 2.4, \\\\\n$\\therefore$ EF = CM = 2.4, \\\\\n$\\therefore$ The minimum value of EF is $\\boxed{2.4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53359734.png"
  },
  {
    "solution": "\\boxed{\\sqrt{21}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the rhombus $ABCD$, the diagonals $AC$ and $BD$ intersect at point $O$, with $DE \\parallel AC$ and $AE \\parallel BD$. Quadrilateral $AODE$ is a rectangle; if $AB=2\\sqrt{3}$ and $\\angle BCD=120^\\circ$, connect $CE$. What is the length of $CE$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle BCD=120^\\circ$ and quadrilateral ABCD is a rhombus,\\\\\n$\\therefore \\angle BAD=\\angle BCD=120^\\circ$, $\\angle CAB=\\angle CAD=60^\\circ$, AB=BC,\\\\\n$\\therefore \\triangle ABC$ is an equilateral triangle,\\\\\n$\\therefore AC=AB=2\\sqrt{3}$, OB=OD=AE=3,\\\\\nIn $\\triangle AEC$, $EC=\\sqrt{AC^2+AE^2}=\\sqrt{(2\\sqrt{3})^2+3^2}=\\boxed{\\sqrt{21}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51817965_92.png"
  },
  {
    "solution": "\\boxed{(4,5)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the parabola $y=x^{2}-2x+c$ intersects with the line $y=x+1$ at points $A$ and $C$, and the coordinates of point $A$ are $(-1,0)$. What are the coordinates of point $C$?\n",
    "completion": "Solution: Since point A $(-1,0)$ lies on the parabola $y=x^{2}-2x+c$, \\\\\n$\\therefore$ $0=-1^{2}-2(-1)+c$, \\\\\n$\\therefore$ $c=-3$, \\\\\n$\\therefore$ the equation of the parabola is $y=x^{2}-2x-3$, \\\\\nBy solving the system with the line $y=x+1$, we get $\\left\\{\\begin{array}{l}y=x^{2}-2x-3 \\\\ y=x+1\\end{array}\\right.$, \\\\\nwhich yields $\\left\\{\\begin{array}{l}x=4\\\\ y=5\\end{array}\\right.$ or $\\left\\{\\begin{array}{l}x=-1\\\\ y=0\\end{array}\\right.$, \\\\\n$\\therefore$ the coordinates of point C are $\\boxed{(4,5)}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52605441_116.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $D$ is a point on side $BC$, and $AB=AD=DC$. If $\\angle B=60^\\circ$, what is the degree measure of $\\angle C$?\n",
    "completion": "\\textbf{Solution:} Since $AB=AD$, and $\\angle B=60^\\circ$,\\\\\nthus, $\\triangle ABD$ is an equilateral triangle,\\\\\ntherefore, $\\angle ADB=60^\\circ$,\\\\\nSince $AD=CD$,\\\\\nthus $\\angle C=\\angle DAC=\\frac{1}{2}\\angle ADB= \\boxed{30^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/53044239_140.png"
  },
  {
    "solution": "\\boxed{M(0,\\frac{3}{2})}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, rectangle $OCBA$ is located in the Cartesian coordinate system. Point $C$ is on the positive x-axis, and point $A$ is on the positive y-axis. It is known that the coordinates of point $B$ are $(4,2)$. The graph of the inverse proportion function $y=\\frac{k}{x}$ passes through the midpoint $D$ of $AB$ and intersects $BC$ at point $E$. Assume the equation of line $DE$ is $y=mx+n$. Connect points $OD$ and $OE$. Point $M$ is on the positive y-axis. If the area of $\\triangle MBO$ is equal to the area of $\\triangle ODE$, what are the coordinates of point $M$?\n",
    "completion": "\\textbf{Solution}: Given, $D(2,2)$, $E(4,1)$,\\\\\n$\\therefore$ the area of $\\triangle ODE$ is $2\\times 4-\\frac{1}{2}\\times 2\\times 2-\\frac{1}{2}\\times 2\\times 1-\\frac{1}{2}\\times 4\\times 1=3$,\\\\\nLet $M(0,m)$, and the area of $\\triangle MBO$ $=\\frac{1}{2}|m|\\times 4=3$,\\\\\n$\\therefore m=\\pm\\frac{3}{2}$,\\\\\n$\\therefore \\boxed{M(0,\\frac{3}{2})}$, and $(0,-\\frac{3}{2})$ (discard).",
    "image_path": "dataset/mm_math/images/MM_Math/52536619_158.png"
  },
  {
    "solution": "\\boxed{135^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the right-angled vertex $O$ of the right triangle $OMN$ is placed on line $AB$, and ray $OC$ bisects $\\angle AON$. If $\\angle AOM = 2\\angle COM$, what is the measure of $\\angle AON$ in degrees?\n",
    "completion": "\\textbf{Solution:} Since line OC bisects $\\angle AON$,\n\\[\n\\therefore \\angle AON=2\\angle AOC,\n\\]\nLet $\\angle COM=x$, then $\\angle AOM=2x$,\n\\[\n\\therefore \\angle CON=\\angle AOC=\\angle COM+\\angle AOM=x+2x=3x,\n\\]\nSince $\\angle COM+\\angle CON=90^\\circ$,\n\\[\n\\therefore x+3x=90^\\circ,\n\\]\nSolving this, we find $x=22.5^\\circ$;\n\\[\n\\therefore \\angle AON=6x=6\\times22.5^\\circ=\\boxed{135^\\circ}.\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/53520190_88.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that lines $AB$ and $CD$ intersect at point $O$. $OE$ and $OF$ are rays, $\\angle AOE = 90^\\circ$, and $OF$ bisects $\\angle BOC$. If $\\angle EOF = 30^\\circ$, what is the degree measure of $\\angle BOD$?\n",
    "completion": "\\textbf{Solution:} Given the construction, we have\\\\\n$\\because \\angle AOE= 90^\\circ$,\\\\\n$\\therefore \\angle EOB= 180^\\circ-\\angle AOE= 90^\\circ$.\\\\\n$\\because \\angle EOF= 30^\\circ$,\\\\\n$\\therefore \\angle FOB=\\angle EOB-\\angle EOF= 60^\\circ$.\\\\\n$\\because$ OF bisects $\\angle BOC$,\\\\\n$\\therefore \\angle BOC=2\\angle FOB= 120^\\circ$,\\\\\n$\\therefore \\angle BOD=180^\\circ-\\angle BOC=\\boxed{60^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/54927213_126.png"
  },
  {
    "solution": "\\boxed{1:9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $\\triangle ABO \\sim \\triangle DEO$, and $BO: EO = 1: 3$. What is the ratio of the areas of $\\triangle ABO$ to $\\triangle DEO$?\n",
    "completion": "\\textbf{Solution:} Given $\\because \\triangle ABO \\sim \\triangle DEO$ and $BO:EO=1:3$, \\\\\n$\\therefore$ the ratio of the areas of $\\triangle ABO$ and $\\triangle DEO$ is $1:9$, \\\\\nthus the answer is: $\\boxed{1:9}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51445764.png"
  },
  {
    "solution": "\\boxed{30}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, points $D$, $E$, and $F$ are located on the three sides respectively, $E$ is the midpoint of $AC$, $AD$, $BE$, $CF$ intersect at point $G$, $BD=2DC$, ${S}_{\\triangle GEC}=3$, ${S}_{\\triangle GDC}=4$, then what is the area of $\\triangle ABC$?\n",
    "completion": "\\textbf{Solution:} $BD=2DC$,\\\\\n$\\therefore {S}_{\\triangle ABD}=2{S}_{\\triangle ACD}$,\\\\\n$\\therefore {S}_{\\triangle ABC}=3{S}_{\\triangle ACD}$,\\\\\n$\\because E$ is the midpoint of $AC$,\\\\\n$\\therefore {S}_{\\triangle AGE}={S}_{\\triangle CGE}$,\\\\\nAlso, $\\because {S}_{\\triangle GEC}=3$, ${S}_{\\triangle GDC}=4$,\\\\\n$\\therefore {S}_{\\triangle ACD}={S}_{\\triangle AGE}+{S}_{\\triangle CGE}+{S}_{\\triangle CGD}=3+3+4=10$,\\\\\n$\\therefore {S}_{\\triangle ABC}=3{S}_{\\triangle ACD}=3\\times 10=\\boxed{30}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52977136.png"
  },
  {
    "solution": "\\boxed{50^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In circle $ \\odot O$, $ AB$ is a diameter, and point $C$ is a point on the circle. Fold the minor arc along chord $ AC$ to intersect $ AB$ at point $D$. Connect $ CD$. As shown in Figure 2, if point $D$ and the center $O$ do not coincide, and $ \\angle BAC=20^\\circ$, find the degree measure of $ \\angle DCA$.\n",
    "completion": "\\textbf{Solution:} Let the symmetrical point of point D with respect to chord $AC$ be F, and connect $AF$ and $CB$, \\\\\nAccording to the problem, we get $\\angle BAC=\\angle FAC=20^\\circ$, $CD=CF$, \\\\\nHence, $CB=CF=CD$, \\\\\nThus, $\\angle B=\\angle CDB$; \\\\\nSince $AB$ is the diameter, \\\\\nTherefore, $\\angle ACB=90^\\circ, \\angle B=\\angle CDB=70^\\circ$, \\\\\nTherefore, $\\angle DCA=\\angle CDB-\\angle BAC=70^\\circ-20^\\circ=\\boxed{50^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52871917_108.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $AB$ is the diameter of the circle $\\odot O$ and $CD$ is a chord, with $AB \\perp CD$ at point $E$. $G$ is any point on the arc $AC$ (not coinciding with $A$ or $C$), and segments $AD$ and $GD$ are connected. When $G$ moves to the position where points $O$, $D$, and $G$ are collinear, find the length of $AG$ at this moment.\n",
    "completion": "\\textbf{Solution:} Given from (2), we know that the radius of $ \\odot O$ is 10cm,\\\\\ntherefore $GD=20cm$, $AE=16cm$, $DE=8cm$,\\\\\ntherefore $AD^{2}=DE^{2}+AE^{2}=320cm^2$,\\\\\nsince $GD$ is the diameter of $ \\odot O$,\\\\\ntherefore $\\angle GAD=90^\\circ$,\\\\\ntherefore $AG=\\sqrt{GD^{2}-AD^{2}}=\\boxed{4\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53149307_138.png"
  },
  {
    "solution": "\\boxed{5\\, cm}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the angle bisectors of $ \\angle ABC$ and $\\angle ACB$ intersect at point F. A line through F is drawn parallel to $BC$, intersecting $AB$ at point D and $AC$ at point E. Given that $BD = 3\\,cm$ and $EC = 2\\,cm$, what is the length of $DE$ in centimeters?\n",
    "completion": "\\textbf{Solution:}\\\\\nSince $BF$ is the bisector of $\\angle ABC$,\\\\\nit follows that $\\angle DBF=\\angle CBF$,\\\\\nSince $DE \\parallel BC$,\\\\\nit then implies $\\angle DFB=\\angle CBF$,\\\\\nThus, $\\angle DBF=\\angle DFB$,\\\\\nHence, $DF=BD=3\\, cm$,\\\\\nSimilarly, it can be concluded that $EF=EC=2\\, cm$,\\\\\nTherefore, $DE=DF+EF=\\boxed{5\\, cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51368723.png"
  },
  {
    "solution": "\\boxed{(0, \\frac{13}{5})}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y = x + 1$ intersects with the parabola $y = x^2 - 4x + 5$ at points $A$ and $B$. Point $P$ is a moving point on the y-axis. When the perimeter of $\\triangle PAB$ is minimized, what are the coordinates of point $P$?\n",
    "completion": "\\textbf{Solution:}\\\\\nFrom $\\begin{cases}y=x+1\\\\ y=x^2-4x+5\\end{cases}$, we can obtain $\\begin{cases}x=1\\\\ y=2\\end{cases}$ or $\\begin{cases}x=4\\\\ y=5\\end{cases}$,\\\\\n$\\therefore$ the coordinates of point A are (1, 2) and the coordinates of point B are (4, 5),\\\\\n$\\therefore$ AB = $\\sqrt{(5-2)^2+(4-1)^2}=3\\sqrt{2}$,\\\\\nConstruct the reflection of point A across the y-axis to get point A$^{\\prime}$, and connect A$^{\\prime}$B to intersect the y-axis at point P, then in this case, the perimeter of $\\triangle PAB$ is minimized,\\\\\n$\\because$ the coordinates of point A$^{\\prime}$ are ($-1$, 2) and the coordinates of point B are (4, 5),\\\\\n$\\therefore$, let the equation of line A$^{\\prime}$B be $y=kx+b$,\\\\\nthen from $\\begin{cases}-k+b=2\\\\ 4k+b=5\\end{cases}$, we can obtain $\\begin{cases}k=\\frac{3}{5}\\\\ b=\\frac{13}{5}\\end{cases}$,\\\\\n$\\therefore$ the equation of line A$^{\\prime}$B is $y= \\frac{3}{5}x+\\frac{13}{5}$,\\\\\n$\\therefore$ when $x=0$, $y= \\frac{13}{5}$,\\\\\nthus, the coordinates of point P are $\\boxed{(0, \\frac{13}{5})}$,",
    "image_path": "dataset/mm_math/images/MM_Math/51213422.png"
  },
  {
    "solution": "\\boxed{\\frac{15}{6}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known in $\\triangle ABC$ that $\\angle C=90^\\circ$, $AC=8$, and $BC=15$. Then, the value of $\\tan A$ is (\\quad)\n",
    "completion": "\\textbf{Solution:} Given that in the right triangle $\\text{Rt}\\triangle ABC$, $\\angle C=90^\\circ$, $AC=6$, $BC=15$, \\\\\nthus $\\tan A= \\frac{BC}{AC}=\\boxed{\\frac{15}{6}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51032231.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{2}+2\\sqrt{6}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a protractor is combined with a triangular plate with an angle of $30^\\circ$ ($\\angle CAB=30^\\circ$). The hypotenuse AB of the triangular plate coincides with the diameter MN of the circle on which the protractor lies. Now, point P is exactly the midpoint of the semicircular arc of the protractor, and we connect CP. If BC = 4, what is the length of CP?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, let the intersection of CP and AB be H, construct $HI\\perp BC$ at I, and $HJ\\perp AP$ at J,\\\\\n$\\because$ AB is the diameter of the semicircle, $\\angle ACB=90^\\circ$,\\\\\n$\\therefore$ C lies on the semicircle Q where the protractor is located, and P is the midpoint of $\\overset{\\frown}{APB}$,\\\\\n$\\therefore \\angle BCP=\\angle ACP=45^\\circ$,\\\\\n$\\therefore$ $CI=HI$,\\\\\n$\\because$ $\\angle CAB=30^\\circ$, $\\angle ACB=90^\\circ$,\\\\\n$\\therefore$ $\\angle IBH=60^\\circ$, $\\angle IHB=30^\\circ$,\\\\\nLet $BI=x$, then $BH=2x$, $HI=\\sqrt{3}x$, $CI=4−x$,\\\\\n$\\therefore$ $4−x=\\sqrt{3}x$,\\\\\nSolving yields: $x=2\\sqrt{3}−2$,\\\\\n$\\therefore$ $BH=2(2\\sqrt{3}-2)=4\\sqrt{3}-4$, $CH=\\sqrt{2} \\times \\sqrt{3}(2\\sqrt{3}-2)=6\\sqrt{2}-2\\sqrt{6}$,\\\\\nSimilarly, it can be obtained that: $AB=2BC=8$,\\\\\n$\\therefore$ $AH=8−BH=12-4\\sqrt{3}$,\\\\\nSimilarly, it can be obtained that: $\\angle JAH=45^\\circ$, $HJ=AJ=\\frac{\\sqrt{2}}{2}AH=6\\sqrt{2}-2\\sqrt{6}$,\\\\\nAnd $\\angle HPJ=\\angle ABC=60^\\circ$,\\\\\n$\\therefore$ $HP=\\frac{HJ}{\\sin 60^\\circ}=4\\sqrt{6}-4\\sqrt{2}$,\\\\\n$\\therefore$ $CP=4\\sqrt{6}-4\\sqrt{2}+6\\sqrt{2}-2\\sqrt{6}=\\boxed{2\\sqrt{2}+2\\sqrt{6}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52766425.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, there are points A and B on the graphs of the inverse proportion functions $y=-\\frac{a}{x}$ and $y=\\frac{6}{x}$, respectively, and $AB\\parallel x$-axis, $AD\\perp x$-axis at D, $BC\\perp x$-axis at C. If the area of rectangle ABCD is 8, what is the value of $a$?\n",
    "completion": "\\textbf{Solution:} According to the problem, the area of rectangle ABCD is $a+6=8$,\n\\[\\therefore a=\\boxed{2}.\\]",
    "image_path": "dataset/mm_math/images/MM_Math/51351897.png"
  },
  {
    "solution": "\\boxed{\\frac{3\\sqrt{21}}{7}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, D is the midpoint on side AC. Connect BD and fold $\\triangle BDC$ along BD to obtain $\\triangle BDC'$, where DC' intersects AB at point E. Connect AC'. If AD = AC' = 2 and BD = 3, what is the distance from point D to line BC?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, connect $CC'$ and let it intersect $BD$ at point $M$. Draw $DH \\perp BC'$ at point $H$, \\\\\nsince $AD=AC'=2$ and $D$ is the midpoint of side $AC$, \\\\\ntherefore $DC=AD=2$. By folding, $\\triangle BDC\\cong \\triangle BDC'$, and $BD$ perpendicularly bisects $CC'$, \\\\\ntherefore $DC=DC'=2$, $BC=BC'$, $CM=C'M$, \\\\\ntherefore $AD=AC'=DC'=2$, \\\\\nthus $\\triangle ADC'$ is an equilateral triangle, \\\\\ntherefore $\\angle ADC'=\\angle AC'D=\\angle C'AC=60^\\circ$, \\\\\nsince $DC=DC'$, \\\\\ntherefore $\\angle DCC'=\\angle DC'C= \\frac{1}{2} \\times 60^\\circ=30^\\circ$, \\\\\nIn $\\triangle C'DM$, with $\\angle DC'C=30^\\circ$ and $DC'=2$, \\\\\nthus $DM=1$, $C'M= \\sqrt{3}DM= \\sqrt{3}$, \\\\\ntherefore $BM=BD−DM=3−1=2$, \\\\\nIn $\\triangle BMC'$, $BC'= \\sqrt{BM^2+C'M^2}=\\sqrt{2^2+(\\sqrt{3})^2}=\\sqrt{7}$, \\\\\nsince the area of $\\triangle BDC'$ is $\\frac{1}{2} BC'\\cdot DH= \\frac{1}{2} BD\\cdot CM$, \\\\\nthus $\\sqrt{7}DH=3\\times \\sqrt{3}$, \\\\\ntherefore $DH= \\frac{3\\sqrt{21}}{7}$, \\\\\nsince $\\angle DBC=\\angle DBC'$, \\\\\ntherefore the distance from point $D$ to $BC$ is $\\boxed{\\frac{3\\sqrt{21}}{7}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51368376.png"
  },
  {
    "solution": "\\boxed{9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, AB is the diameter of $\\odot O$, points C and D are on $\\odot O$, $CE \\perp AB$ and $OF \\perp AD$, with the feet of the perpendiculars being points E and F, respectively, and $\\angle COF = 90^\\circ$. If $BE=3$ and $CE=6$, what is the length of $AD$?\n",
    "completion": "\\textbf{Solution:} Since $CE \\perp AB$ and $OF \\perp AD$, it follows that $\\angle OEC = \\angle AFO = 90^\\circ$. Hence, $\\angle COE + \\angle C = 90^\\circ$. Let the radius of the circle be $r$, then $OE = r - 3$. Therefore, $CE^{2} + OE^{2} = CO^{2}$, i.e., $6^{2} + (r - 3)^{2} = r^{2}$. Solving this yields $r = 7.5$, therefore $OE = 7.5 - 3 = 4.5$. Since $\\angle COF = 90^\\circ$, it follows that $\\angle COE + \\angle AOF = 90^\\circ$. Therefore, $\\angle C = \\angle AOF$, which means $\\triangle AOF \\cong \\triangle OCE$ (AAS). Hence, $OE = AF = 4.5$. Since $OF \\perp AD$, it follows that $AD = 2AF = 2 \\times 4.5 = \\boxed{9}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55452503.png"
  },
  {
    "solution": "\\boxed{36}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given the figure, $\\angle BAC=30^\\circ$, P is a point on the bisector of $\\angle BAC$, PM$\\parallel$AC, PD$\\perp$AC, and PD$=6$. What is the area of $\\triangle AMP$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw PE$\\perp$AB at point E through point P,\\\\\nsince P is a point on the bisector of $\\angle$BAC, and PD$\\perp$AC, PE$\\perp$AB,\\\\\nit follows that PE=PD=6,\\\\\nsince $\\angle$BAC=30$^\\circ$, and PM$\\parallel$AC,\\\\\nit follows that $\\angle$PME=$\\angle$BAC=30$^\\circ$, and $\\angle$APM=$\\angle$PAD,\\\\\ntherefore PM=2PE=12,\\\\\nsince $\\angle$BAP=$\\angle$PAD,\\\\\nit follows that $\\angle$BAP=$\\angle$APM,\\\\\nthus AM=PM=12.\\\\\nTherefore, the area of $\\triangle AMP$=  $\\frac{1}{2}\\times AM\\times PE=\\frac{1}{2}\\times 12\\times 6= \\boxed{36}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51367326.png"
  },
  {
    "solution": "\\boxed{P(2, -3)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the axis of symmetry of the parabola is the line $x=1$, and it intersects the x-axis at points A and B, and intersects the y-axis at point $C\\left(0, -3\\right)$, with $OB=OC$. Let the x-coordinate of a point $P$ on the parabola be $m$, and $P$ is below the line BC. Find the coordinates of point P when the area of $\\triangle BCP$ is the largest integer.\n",
    "completion": "\\textbf{Solution:} From equation (1), we know that $B(3, 0)$, $C(0, -3)$. Assume the equation of line BC is $y=kx+b$, then we have:\n\\[\n\\begin{cases}\n3k+b=0\\\\\nb=-3\n\\end{cases}\n\\]\nSolving these, we obtain:\n\\[\n\\begin{cases}\nk=1\\\\\nb=-3\n\\end{cases}\n\\]\nThus, the equation of line BC is $y=x-3$. Draw PM from point P perpendicular to the x-axis, intersecting BC at point M, as shown in the diagram. Therefore, $P(m, m^{2}-2m-3)$, thus $M(m, m-3)$, therefore $PM=m-3-m^{2}+2m+3=-m^{2}+3m$. Then, the area of $\\triangle BCP$, ${S}_{\\triangle BCP}={S}_{\\triangle CPM}+{S}_{\\triangle BPM}=\\frac{1}{2}PM\\cdot (x_{B}-x_{C})=\\frac{3}{2}(-m^{2}+3m)=-\\frac{3}{2}\\left(m-\\frac{3}{2}\\right)^{2}+\\frac{27}{8}$. Since $-\\frac{3}{2}<0$, the parabola opens downwards, therefore ${S}_{\\triangle BCP}\\le \\frac{27}{8}$. Hence, when the area of $\\triangle BCP$ is the largest integer, it equals $\\boxed{3}$. Thus, $-\\frac{3}{2}m^{2}+\\frac{9}{2}m=3$, solving for $m$ yields $m_{1}=1$, $m_{2}=2$. Therefore, point $\\boxed{P(1, -4)}$ or $\\boxed{P(2, -3)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51433738_209.png"
  },
  {
    "solution": "\\boxed{\\frac{36}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system, perpendicular lines to the $x$-axis are drawn through points $C(1, 4)$ and $D(4, 1)$, with the feet of the perpendiculars being $A$ and $B$, respectively. Point $M$ is a moving point on line $OD$, and a perpendicular line to the $x$-axis is drawn through point $M$, intersecting the $x$-axis at point $P$ and line $CD$ at point $N$.\n",
    "completion": "Solution:\n(1) Since PM is the midline of $\\triangle OBD$ with $D(4, 1)$,\nit follows that $M\\left(2, \\frac{1}{2}\\right)$.\nSubstituting $x=2$ into $y=-x+5$ yields $N\\left(2, 3\\right)$,\ntherefore $MN=\\frac{5}{2}$.\n\n(2) Possible;\nReason: Assume the coordinates of point M are $\\left(a, \\frac{1}{4}a\\right)$, and the coordinates of point N are $\\left(a, -a+5\\right)$.\nWhen $AC=MN$, the quadrilateral formed by the vertices A, C, N, M is a parallelogram.\nThat is, $\\left|-a+5-\\frac{1}{4}a\\right|=4$, solving this gives $a=\\frac{4}{5}$ or $\\frac{36}{5}$. Thus, the satisfying x-coordinate of point M is $\\boxed{\\frac{4}{5}}$ or $\\boxed{\\frac{36}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52394997_106.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, $AB$ is the diameter of $\\odot O$, the chord $AC = BC$, and $E$ is the midpoint of $OB$. Extend line $CE$ to point $F$ such that $EF=CE$, and draw line $AF$ intersecting $\\odot O$ at point $D$. Draw lines $BD$ and $BF$. The line $BF$ is tangent to $\\odot O$; if the length of $AF$ is $5\\sqrt{2}$, find the radius of $\\odot O$ and the length of $BD$.\n",
    "completion": "\\textbf{Solution:} Let the radius of $\\odot O$ be $r$, hence $AB=2r$, $BF=OC=r$. In $\\triangle ABF$, we have: ${\\left(2r\\right)}^{2}+{r}^{2}={\\left(5\\sqrt{2}\\right)}^{2}$. \n\n$\\therefore$ We only take $r=\\sqrt{10}$, thus the radius of $\\odot O$ is $\\boxed{\\sqrt{10}}$.\n\n$\\because$ $AB$ is the diameter of $\\odot O$, thus $AB=2\\sqrt{10}$,\n\n$\\therefore$ $BD\\perp AF$,\n\n$\\therefore$ $BF=\\sqrt{AF^{2}-AB^{2}}=\\sqrt{{\\left(5\\sqrt{2}\\right)}^{2}-{\\left(2\\sqrt{10}\\right)}^{2}}=\\sqrt{10}$,\n\n$\\because$ AB is a diameter,\n\n$\\therefore$ $\\angle ADB=90^\\circ$,\n\n$\\therefore$ ${s}_{\\triangle ABF}=\\frac{1}{2}AB\\times BF=\\frac{1}{2}AF\\times BD$,\n\n$\\therefore$ $\\frac{1}{2}\\times 2\\sqrt{10}\\times \\sqrt{10}=\\frac{1}{2}\\times 5\\sqrt{2}\\times BD$, solving this yields $BD=\\boxed{2\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51323934_106.png"
  },
  {
    "solution": "\\boxed{66^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, AD is the altitude from A to side BC, AE bisects $\\angle BAC$, $\\angle C=46^\\circ$, $\\angle DAE=10^\\circ$, what is the measure of $\\angle B$?\n",
    "completion": "\\textbf{Solution:} Given that $AD$ is the perpendicular from $A$ to the side $BC$,\\\\\nit follows that $\\angle ADC=90^\\circ$,\\\\\nsince $\\angle C=46^\\circ$,\\\\\nwe have $\\angle CAD=90^\\circ-\\angle C=44^\\circ$,\\\\\ngiven $\\angle DAE=10^\\circ$,\\\\\nit follows that $\\angle CAE=\\angle CAD-\\angle DAE=34^\\circ$,\\\\\nsince $AE$ bisects $\\angle BAC$,\\\\\nwe have $\\angle BAC=2\\angle CAE=68^\\circ$,\\\\\nconsequently, $\\angle B=180^\\circ-\\angle BAC-\\angle C=180^\\circ-68^\\circ-46^\\circ=\\boxed{66^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51212563.png"
  },
  {
    "solution": "\\boxed{(-1,-4)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is the graph of the quadratic function $y=ax^2+bx+c$, which intersects the x-axis at points B and C, and intersects the y-axis at point A. Point P is on the parabola, and the area of $\\triangle BCP$ is $\\frac{4}{3}$ times the area of $\\triangle ABC$. What are the coordinates of point P?\n",
    "completion": "\\textbf{Solution:} Because $S_{\\triangle BCP} = \\frac{4}{3}S_{\\triangle ABC}$, \\\\\nit follows that $|y_P| = \\frac{4}{3}|y_C|$, which means $y_P = \\pm 4$, \\\\\ni.e. $y = x^2 + 2x - 3 = \\pm 4$, \\\\\nSolving this gives: $x= -1 \\pm 2\\sqrt{2}$ or $-1$, \\\\\nHence, the coordinates of point P are $\\boxed{(-1+2\\sqrt{2},4)}$ or $\\boxed{(-1-2\\sqrt{2},4)}$ or $\\boxed{(-1,-4)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52977484_199.png"
  },
  {
    "solution": "\\boxed{36}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given the diagram, in $\\triangle ABC$, $AB=AC$. Point $D$ is on $AC$, and satisfies $AD^{2}=CD\\cdot AC$. If $AD=BC$, by connecting $BD$, what is the measure of $\\angle ABD$ in degrees?\n",
    "completion": "\\textbf{Solution:} Since $AB=AC$,\\\\\nit follows that $\\angle ABC=\\angle ACB$,\\\\\nSince $AD^{2}=CD\\cdot AC$, $AD=BC$,\\\\\nit follows that $BC^{2}=CD\\cdot AC$,\\\\\nwhich means $\\frac{CB}{CD}=\\frac{AC}{CB}$,\\\\\nAlso, since $\\angle C=\\angle C$,\\\\\nit follows that $\\triangle CBA\\sim \\triangle CDB$,\\\\\nthus, $\\angle BDC=\\angle C=\\angle ABC$, $\\angle A=\\angle DBC$,\\\\\nit follows that $BD=CB$,\\\\\nthus, $DB=DA$,\\\\\nit follows that $\\angle A=\\angle ABD$,\\\\\nLet $\\angle A=\\angle ABD =x^\\circ$,\\\\\nit follows that $\\angle BDC=\\angle A+\\angle ABD=2x^\\circ$,\\\\\nit follows that $\\angle ABC=\\angle C=2\\angle ABD=2x^\\circ$,\\\\\nIn $\\triangle ABC$, by the triangle angle sum theorem,\\\\\n$x+2x+2x=180$,\\\\\nSolving, we find $x=\\boxed{36}$,\\\\\nthus, $\\angle ABD=36^\\circ.$",
    "image_path": "dataset/mm_math/images/MM_Math/52601704.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, AE is the bisector of the external angle $\\angle BAD$ of $\\triangle ABC$, BF bisects $\\angle ABC$ and intersects the extended line of AE at point F. What is the measure of $\\angle BFE$?\n",
    "completion": "\\textbf{Solution:} Let $\\angle ABF = x$, \\\\\nsince $BF$ bisects $\\angle ABC$, \\\\\nthen $\\angle ABC = 2\\angle ABF = 2x$, \\\\\nsince $\\angle C = 90^\\circ$, \\\\\nthen $\\angle DAB = \\angle C + \\angle ABC = 90^\\circ+2x$, \\\\\nsince $AE$ is the bisector of $\\angle BAD$, \\\\\nthen $\\angle EAB = 45^\\circ+x$, \\\\\nsince $\\angle EAB = \\angle ABF + \\angle BFE$, \\\\\nthen $\\angle BFE = \\boxed{45^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/51434956.png"
  },
  {
    "solution": "\\boxed{25^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: The point O is on the line segment AB. A ray OC is drawn passing through point O such that $\\angle BOC=65^\\circ$. Place the right-angle vertex of a right-angled triangle ruler at point O. As shown in figure (1), when one side ON of the triangle ruler MON coincides with the ray OB, what is the degree measure of $\\angle MOC$?\n",
    "completion": "\\textbf{Solution:} We have $\\angle MOC = \\boxed{25^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52169374_192.png"
  },
  {
    "solution": "\\boxed{0 < x < 2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the line $y=kx+b$ ($k\\neq 0$) intersects with the hyperbola $y=\\frac{m}{x}$ ($m\\neq 0$) at points $A(2,-3)$ and $B(n,2)$. Directly write out the solution set of the inequality $kx+b>\\frac{m}{x}$ in terms of $x$.\n",
    "completion": "\\textbf{Solution:} Solve: \\boxed{x < -3} or \\boxed{0 < x < 2}.",
    "image_path": "dataset/mm_math/images/MM_Math/52623645_228.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{5}-1}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, fold the rectangular paper ABCD, first creating a crease through BD, then fold it again so that AD falls on the diagonal BD, with A corresponding to A$^{\\prime}$, creating the fold crease DG. If AB = 2 and BC = 1, what is the length of AG?\n",
    "completion": "\\textbf{Solution:} Given the conditions, we find that $AB=2$ and $AD=BC=1$.\\\\\nIn $\\triangle ABD$, according to the Pythagorean theorem, $BD=\\sqrt{AB^2+AD^2}=\\sqrt{5}$.\\\\\nBy the property of folding, $AG=A'G$, $DA=DA'=1$. Let $AG=x$, $A'G=AG=x$, $BG=2-x$, $BA'=\\sqrt{5}-1$,\\\\\nIn $\\triangle BGA'$, by the Pythagorean theorem,\\\\\n$BG^2=BA'^2+A'G^2$,\\\\\n$(2-x)^2=(\\sqrt{5}-1)^2+x^2$,\\\\\nSolving, we find $x=\\frac{\\sqrt{5}-1}{2}$,\\\\\nHence, the length of $AG$ is: $\\boxed{\\frac{\\sqrt{5}-1}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52251043.png"
  },
  {
    "solution": "\\boxed{2:3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown, $\\triangle ABC$ and $\\triangle A'B'C'$ are similar figures with point O as the center of similitude. If $OA: AA'=2:5$, then what is the ratio of the perimeters of $\\triangle ABC$ and $\\triangle A'B'C'$?\n",
    "completion": "\\textbf{Solution:} Since $OA:AA'=2:5$, \\\\\ntherefore, $OA:OA'=2:3$, \\\\\nthus, the ratio of similarity between $\\triangle ABC$ and $\\triangle A'B'C'$ is $2:3$, \\\\\nhence, the ratio of the perimeters of $\\triangle ABC$ and $\\triangle A'B'C'$ is $\\boxed{2:3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51807213.png"
  },
  {
    "solution": "\\boxed{\\left(\\sqrt{10}-1, \\frac{\\sqrt{10}+1}{3}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the linear function $y=x+b$ and the hyperbolic function $y=\\frac{k}{x}$ (where $k \\neq 0$) intersect at points $A$ and $B$. The coordinates of point $A$ are (1, 3). The linear function $y=x+b$ intersects the $x$-axis at point $C$. Lines $OA$ and $OB$ are drawn. A perpendicular line to the $y$-axis is drawn through point $A$, and its foot is point $D$. Point $M$ is a moving point on the graph of the hyperbolic function $y=\\frac{k}{x}$ in the first quadrant. A perpendicular line to the $x$-axis is drawn through point $M$, intersecting the $x$-axis at point $N$, and line $CM$ is drawn. When $\\triangle ADO$ is similar to $\\triangle CNM$, find the coordinates of point $M$.\n",
    "completion": "\\textbf{Solution:}\n\n(1) When $\\frac{MN}{OD}=\\frac{CN}{AD}$, $\\triangle ADO\\sim \\triangle CNM$,\\\\\n$MN=\\frac{3}{x}$, $OD=3$, $AD=1$, $CN=2+x$,\\\\\n$\\therefore \\frac{\\frac{3}{x}}{3}=\\frac{2+x}{1}$, i.e., $x^{2}+2x-1=0$,\\\\\nSolving for $x$, we get: $x_{1}=\\sqrt{2}-1$, $x_{2}=-\\sqrt{2}-1$,\\\\\nSince $M$ is in the first quadrant,\\\\\n$\\therefore x=\\sqrt{2}-1$, $y=\\frac{3}{\\sqrt{2}-1}=3\\sqrt{2}+3$,\\\\\n$\\therefore M(\\sqrt{2}-1, 3\\sqrt{2}+3)$,\\\\\n(2) When $\\frac{MN}{AD}=\\frac{CN}{OD}$, $\\triangle ADO\\sim \\triangle CNM$,\\\\\n$\\therefore \\frac{\\frac{3}{x}}{1}=\\frac{2+x}{3}$, i.e., $x^{2}+2x-9=0$,\\\\\nSolving for $x$, we get: $x_{1}=\\sqrt{10}-1$, $x_{2}=-\\sqrt{10}-1$,\\\\\nSince $M$ is in the first quadrant,\\\\\n$\\therefore x=\\sqrt{10}-1$, $y=\\frac{3}{\\sqrt{10}-1}=\\frac{\\sqrt{10}+1}{3}$,\\\\\n$\\therefore M\\left(\\sqrt{10}-1, \\frac{\\sqrt{10}+1}{3}\\right)$,\\\\\nIn summary, when $\\triangle ADO$ and $\\triangle CNM$ are similar, the coordinates of point $M$ are $\\boxed{(\\sqrt{2}-1, 3\\sqrt{2}+3)}$ or $\\boxed{\\left(\\sqrt{10}-1, \\frac{\\sqrt{10}+1}{3}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51322344_129.png"
  },
  {
    "solution": "\\boxed{y=x^{2}+2x-3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=ax^2+bx-3$ passes through three points $A$, $B$, and $C$, where point $A(-3,0)$, point $C(1,0)$, and point $B$ is on the $y$-axis. Point $P$ is a moving point on the parabola below the line $AB$ (not coinciding with $A$ or $B$). What is the analytical expression of this parabola?\n",
    "completion": "\\textbf{Solution:} Let's substitute points A $(-3,0)$ and C $(1,0)$ into $y=ax^{2}+bx-3$, to get\n\\[\n\\left\\{\n\\begin{array}{l}\n0=9a-3b-3 \\\\\n0=a+b-3\n\\end{array}\n\\right.\n\\]\nSolving this system, we find\n\\[\n\\left\\{\n\\begin{array}{l}\na=1 \\\\\nb=2\n\\end{array}\n\\right.\n\\]\n$\\therefore$ The equation of the parabola is $\\boxed{y=x^{2}+2x-3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51402562_50.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, $\\angle ABC=30^\\circ$, $AC=\\sqrt{5}$. By rotating $\\triangle ABC$ counterclockwise around point A, we get $\\triangle AB'C'$, making point C fall on the side AB. Connect $BB'$. What is the length of $BB'$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle C=90^\\circ$,\\\\\n$\\angle ABC=30^\\circ$,\\\\\n$AC=\\sqrt{5}$,\\\\\ntherefore $AB=2\\sqrt{5}$,\\\\\n$BC=\\sqrt{15}$,\\\\\nsince right triangle $ABC$ is rotated counterclockwise around point $A$ to get $\\triangle AB'C'$,\\\\\nthus $AC'=AC=\\sqrt{5}$,\\\\\n$B'C'=BC=\\sqrt{15}$,\\\\\ntherefore $BC'=\\sqrt{5}$,\\\\\nsince $\\angle BC'B'=90^\\circ$,\\\\\nthus $BB'=\\sqrt{B'C'^2+BC'^2}=\\sqrt{(\\sqrt{15})^2+(\\sqrt{5})^2}=\\boxed{2\\sqrt{5}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51654847.png"
  },
  {
    "solution": "\\boxed{(8,4)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the vertex $C$ of the rhombus $OABC$ is on the positive half of the $x$-axis. If the coordinates of point $A$ are $(3, 4)$, what are the coordinates of point $B$?\n",
    "completion": "\\textbf{Solution:} Since the coordinates of point $A$ are $(3,4)$, \\\\\n$\\therefore OA=\\sqrt{(3-0)^{2}+(4-0)^{2}}=5$, \\\\\nSince quadrilateral $OABC$ is a rhombus, \\\\\n$\\therefore AB\\parallel OC$ and $AB=OA=5$, \\\\\n$\\therefore$ the x-coordinate of point $B$ is $3+5=8$, the y-coordinate of point $B$ is the same as that of point $A$, which is 4, \\\\\nThus, we have $B(8,4)$, \\\\\nHence, the answer is: $B=\\boxed{(8,4)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53333402.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $P$ is the midpoint of side $AC$ of the equilateral triangle $\\triangle ABC$, $E$ is a point on the extension line of side $BC$, and $PE = PB$, what is the degree measure of $\\angle CPE$?\n",
    "completion": "\\textbf{Solution:} Since $P$ is the midpoint of side $AC$ of the equilateral triangle $\\triangle ABC$,\\\\\nit follows that $BP$ bisects $\\angle ABC$, and $\\angle ABC=60^\\circ=\\angle ACB$,\\\\\nthus, $\\angle PBC=30^\\circ$,\\\\\nsince $PE=PB$,\\\\\nit follows that $\\angle PBC=\\angle E=30^\\circ$,\\\\\nhence, $\\angle CPE=\\angle ACB-\\angle E=\\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52978709.png"
  },
  {
    "solution": "\\boxed{65^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in triangle $ABC$, $AB=AC$, $\\angle C=25^\\circ$, point $D$ is on the extension line of segment $CA$, and $DA=AC$. What is the measure of $\\angle ABD$ in degrees?\n",
    "completion": "\\textbf{Solution:} Since AB = AC, and $\\angle C = 25^\\circ$, \\\\\nhence $\\angle ABC = 25^\\circ$, \\\\\ntherefore $\\angle BAD = \\angle C + \\angle ABC = 50^\\circ$, \\\\\nsince DA = AC, \\\\\nhence AB = AD, \\\\\ntherefore $\\angle ABD =  \\frac{1}{2} (180^\\circ - \\angle BAD) = \\boxed{65^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52991712.png"
  },
  {
    "solution": "\\boxed{k_1x+b>\\frac{k_2}{x}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y=k_1x+b$ intersects with the graph of the inverse proportion function $y=\\frac{k_2}{x}$ at points A and B, where the coordinates of point A are $(-1, 4)$, and the coordinates of point B are $(4, n)$. Based on the graph, directly write down the range of values of $x$ for which $k_1x+b>\\frac{k_2}{x}$.\n",
    "completion": "\\textbf{Solution:} By observing the graph, it is known that when $x<-1$ or $0<x<4$, it satisfies \\boxed{k_1x+b>\\frac{k_2}{x}}.",
    "image_path": "dataset/mm_math/images/MM_Math/51446477_239.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system \\(xOy\\), it is known that the chord of \\(\\odot D\\) intersecting the \\(y\\)-axis has a length of 6, and the center of the circle is \\(D(2, 4)\\). What is the length of the shortest chord passing through point \\(B(2, 3)\\)?\n",
    "completion": "\\textbf{Solution:} Let the intersection points of circle $D$ with the y-axis be $E$ and $A$. Connect $DE$. Draw $DC \\perp y$-axis at $C$,\n\n$\\because$ The chord length of circle $D$ intersecting with the y-axis is 6,\n\n$\\therefore$ $AE = 6$,\n\n$\\therefore$ $CE = 3$,\n\n$\\because$ $D(2, 4)$,\n\n$\\therefore$ $CD = 2$,\n\n$\\therefore$ $DE= \\sqrt{CD^{2}+CE^{2}}=\\sqrt{2^{2}+3^{2}}=\\sqrt{13}$,\n\n$\\because$ $D(2, 4)$, $B(2, 3)$,\n\n$\\therefore$ $DB \\parallel y$-axis,\n\n$\\therefore$ Among all the chords through point $B(2, 3)$, the shortest chord is perpendicular to $DB$,\n\nDraw $MN \\perp DB$intersecting circle $D$ at $M$ and $N$. Connect $DN$,\n\nIn $\\triangle DBN$, $\\angle DBN=90^\\circ$, $DB=1$, $DN= \\sqrt{13}$,\n\n$\\therefore$ $BN= \\sqrt{DN^{2}-BD^{2}}=2\\sqrt{3}$,\n\n$\\therefore$ $MN=2BN= 4\\sqrt{3}$,\n\nHence, among all the chords through point $B(2, 3)$, the shortest chord length is $\\boxed{4\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52949726.png"
  },
  {
    "solution": "\\boxed{6.5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the quadrilateral ABCD, $\\angle A=90^\\circ$, AB = 12, and AD = 5. Points M and N are the moving points (including endpoints, but point M does not coincide with point B) on the line segments BC and AB, respectively. Points E and F are the midpoints of DM and MN, respectively. What is the maximum length of EF?\n",
    "completion": "\\textbf{Solution:} Connect DN and DB\\\\\n$\\because$ Points E and F are the midpoints of DM and MN respectively,\\\\\n$\\therefore$ EF is the midline of $\\triangle MDN$\\\\\n$\\therefore$ EF$=\\frac{1}{2}DN$\\\\\nAccording to the problem, when N coincides with point B, DN is at its maximum, at this time, the value of EF is maximal\\\\\nBy the Pythagorean theorem, DB$= \\sqrt{AD^{2}+AB^{2}}=13$\\\\\n$\\therefore$ The maximum value of EF is $\\boxed{6.5}$",
    "image_path": "dataset/mm_math/images/MM_Math/52387523.png"
  },
  {
    "solution": "\\boxed{130^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points $A$, $O$, and $B$ lie on the same straight line, and $\\angle AOC = \\angle BOD$. Lines $OM$ and $ON$ are the angle bisectors of $\\angle AOC$ and $\\angle BOD$, respectively. If $\\angle COD = 80^\\circ$, what is the measure of $\\angle MON$?\n",
    "completion": "\\textbf{Solution:} Since $OM$ and $ON$ are the angle bisectors of $\\angle AOC$ and $\\angle BOD$ respectively, \\\\\nhence, $\\angle MOC = \\frac{1}{2}\\angle AOC$ and $\\angle NOD = \\frac{1}{2}\\angle BOD$, \\\\\ntherefore, $\\angle MON = \\frac{1}{2}(\\angle AOC + \\angle BOD) + \\angle COD$ \\\\\n$= \\frac{1}{2}(180^\\circ - \\angle COD) + \\angle COD$ \\\\\n$= 90^\\circ + \\frac{1}{2}\\angle COD$, \\\\\nsince $\\angle COD = 80^\\circ$, \\\\\ntherefore, $\\angle MON = 90^\\circ + \\frac{1}{2} \\times 80^\\circ = \\boxed{130^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53404119_161.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the figure, in triangle $ABC$, $\\angle C=90^\\circ$, the angle bisector of $\\angle CAB$, $AD$, intersects $BC$ at $D$, $DE\\perp AB$ at $E$, $DE=2\\,\\text{cm}$, and $DB=4\\,\\text{cm}$. What is the length of $BC$?\n",
    "completion": "\\textbf{Solution:} Given that $\\because$AD bisects $\\angle$CAB, and DE$\\perp$AB at E, with $\\angle$C=$90^\\circ$,\\\\\n$\\therefore$CD=DE=2,\\\\\nAlso $\\because$ $DB=4\\text{cm}$,\\\\\n$\\therefore$BC=BD+CD=4+2=\\boxed{6}\\text{(cm)};",
    "image_path": "dataset/mm_math/images/MM_Math/51368566.png"
  },
  {
    "solution": "\\boxed{\\sqrt{5}-1 < DP < \\frac{2\\sqrt{5}}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $BC=1$, $AB=2$. A perpendicular line to $AB$ through a point $P$ on the diagonal $BD$ intersects $AB$ at point $F$ and $CD$ at point $E$. A line through $E$ such that $EG \\parallel BD$ intersects $BC$ at point $G$. Connect $FG$ and $BD$ at point $H$, and connect $DF$. The symmetrical point of $B$ with respect to $FG$ is denoted as $B'$. If $B'$ lies inside $\\triangle EFG$ (excluding the boundary), find the range of values for the length of $DP$.\n",
    "completion": "\\textbf{Solution:} \n\nWhen point $B^{\\prime}$ is on the side $EG$, let $BG = x$ and $B^{\\prime}F = BF = CE = 2-2x$,\\\\\n$\\because$ $\\angle FB^{\\prime}G = \\angle FBG = 90^\\circ$,\\\\\n$\\therefore$ $\\angle EFB^{\\prime} = \\angle CEG = \\angle CDB$, $\\angle C = \\angle EB^{\\prime}F = 90^\\circ$,\\\\\n$\\therefore$ $\\triangle EFB^{\\prime} \\sim \\triangle BDC$,\\\\\n$\\therefore$ $\\frac{B^{\\prime}F}{EF} = \\frac{DC}{BD}$,\\\\\n$\\therefore$ $\\frac{2-2x}{1} = \\frac{2}{\\sqrt{5}}$,\\\\\nSolving this, we get $x = 1-\\frac{\\sqrt{5}}{5}$,\\\\\n$\\therefore$ $DP = \\sqrt{5}$ $EP = \\sqrt{5}-1$\\\\\nWhen point $B^{\\prime}$ is on the side $EF$,\\\\\n$\\because$ $BG = B^{\\prime}G = CE$,\\\\\n$\\therefore$ $x = 2-2x$,\\\\\nSolving this, we get $x = \\frac{2}{3}$,\\\\\n$\\therefore$ $DP = \\sqrt{5}x = \\frac{2\\sqrt{5}}{3}$,\\\\\nIn summary, $\\boxed{\\sqrt{5}-1 < DP < \\frac{2\\sqrt{5}}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51722675_126.png"
  },
  {
    "solution": "\\boxed{(-2, 2)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a Cartesian coordinate system is established with the center of the square $ABCD$ as the origin. The coordinates of point $A$ are $(2,2)$, what are the coordinates of point $D$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure:\n\nSince the coordinate system is established with the center O of the square ABCD as the origin, and the coordinates of point A are (2, 2), and point D is symmetrical to point A with respect to the y-axis,\n\nTherefore, the coordinates of point D are: $\\boxed{(-2, 2)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52422698.png"
  },
  {
    "solution": "\\boxed{57^\\circ40'}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Place a set of triangle rulers as shown in the diagram. If $\\angle 1 = 27^\\circ 40'$, what is the measure of $\\angle 2$?\n",
    "completion": "\\textbf{Solution:} Given, $\\because$ $\\angle 1=27^\\circ40'$ and $\\angle$BAC=$60^\\circ$, \\\\\n$\\therefore$ $\\angle$EAC=$\\angle$BAC−$\\angle 1 = 60^\\circ−27^\\circ40'=32^\\circ20'$, \\\\\n$\\because$ $\\angle$DAE=$90^\\circ$, \\\\\n$\\therefore$ $\\angle 2=\\angle$DAE−$\\angle$EAC=$90^\\circ−32^\\circ20'=\\boxed{57^\\circ40'}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55503421.png"
  },
  {
    "solution": "\\boxed{(\\sqrt{5}+1)a}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the side length of the square $ABCD$ is $a$, and point $E$ is a moving point on side $AB$. Connect $ED$, and rotate $ED$ clockwise by $90^\\circ$ around point $E$ to $EF$. Connect $DF$ and $CF$. Then, when the sum of $DF+CF$ takes the minimum value, what is the perimeter of $\\triangle DCF$? (Express in an algebraic equation involving $a$)\n",
    "completion": "\\textbf{Solution:} Draw BF, and through F, draw FG $\\perp$ to the extension of AB at point G, \\\\\nsince rotating ED clockwise by $90^\\circ$ around point E gives EF, \\\\\ntherefore EF$\\perp$DE, EF=DE, \\\\\ntherefore $\\angle$DEA + $\\angle$FEG = $\\angle$DEA + $\\angle$ADE = $90^\\circ$, \\\\\ntherefore $\\angle$ADE = $\\angle$FEG, \\\\\ntherefore $\\triangle$AED $\\cong$ $\\triangle$GFE (AAS), \\\\\ntherefore FG=AE, AD=EG, \\\\\ntherefore AD=EG=AB, \\\\\ntherefore BG=AE=FG, \\\\\ntherefore $\\angle$CBF=$\\angle$GBF+$45^\\circ$, i.e., point F moves along the angle bisector of $\\angle$CBG. \\\\\nConstruct the symmetric point C$^{\\prime }$ of C with respect to BF, then point C$^{\\prime }$ lies on the extension of AB, hence when D, F, and C are collinear, DF+CF=DC is minimal. \\\\\nsince AD=a, AC$^{\\prime }$=2a, \\\\\ntherefore DC$^{\\prime }$= $\\sqrt{5}$a, \\\\\ntherefore the minimum value of DF+CF is $\\sqrt{5}$a, \\\\\ntherefore the perimeter of $\\triangle$DCF is DF+CF+CD=$\\sqrt{5}$a+a=($\\sqrt{5}$+1)a. \\\\\nThus, the answer is: $\\boxed{(\\sqrt{5}+1)a}$.",
    "image_path": "dataset/mm_math/images/MM_Math/54345169.png"
  },
  {
    "solution": "\\boxed{29^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, with vertex B of $\\triangle ABC$ as the center and the length of BA as the radius, an arc is drawn intersecting side BC at point D, and AD is connected. If $\\angle B=50^\\circ$ and $\\angle C=36^\\circ$, what is the degree measure of $\\angle DAC$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle B=50^\\circ$ and $\\angle C=36^\\circ$,\\\\\nhence $\\angle BAC=180^\\circ-(\\angle B+\\angle C)=180^\\circ-(50^\\circ+36^\\circ)=94^\\circ$,\\\\\nsince BA=BD,\\\\\nthus $\\angle BAD=\\angle BDA$,\\\\\ntherefore $\\angle BAD=\\frac{1}{2}(180^\\circ-\\angle B)=\\frac{1}{2}(180^\\circ-50^\\circ)=65^\\circ$,\\\\\nthus $\\angle DAC=\\angle BAC-\\angle BAD=94^\\circ-65^\\circ=\\boxed{29^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52856088.png"
  },
  {
    "solution": "\\boxed{15^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a right-angled triangle is placed with point C located on the extension of $FD$, $AB \\parallel CF$, $\\angle F = \\angle ACB = 90^\\circ$, $\\angle E = 30^\\circ$, and $\\angle A = 45^\\circ$. What is the measure of $\\angle CBD$?\n",
    "completion": "\\textbf{Solution:} Given: $\\because$ $\\angle A=45^\\circ$,\\\\\n$\\therefore$ $\\angle ABC=45^\\circ$,\\\\\n$\\because$ $AB\\parallel CF$,\\\\\n$\\therefore$ $\\angle BCD=45^\\circ$,\\\\\n$\\because$ $\\angle E=30^\\circ$,\\\\\n$\\therefore$ $\\angle EDF=60^\\circ$,\\\\\n$\\therefore$ $\\angle BDC=180^\\circ-\\angle EDF=120^\\circ$\\\\\nIn $\\triangle BCD$, $\\angle BCD=45^\\circ$, $\\angle BDC=120^\\circ$,\\\\\n$\\therefore$ $\\angle CBD=180^\\circ-\\angle BDC-\\angle BCD=15^\\circ$,\\\\\nTherefore, the answer is: $\\boxed{15^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53206218.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\triangle ABC$, $ \\angle C=90^\\circ$, AD bisects $ \\angle BAC$, $ DE\\perp AB$ with foot E, $ AB=12$, $ AC=8$, what is the length of BE?\n",
    "completion": "\\textbf{Solution:} Given $\\because$ AD bisects $\\angle$BAC,\\\\\n$\\therefore$$\\angle$DAE=$\\angle$DAC\\\\\n$\\because$$\\angle$C=90$^\\circ$, DE$\\perp$AB,\\\\\n$\\therefore$$\\angle$AED=$\\angle$ACD=90$^\\circ$,\\\\\nAlso, AD=AD,\\\\\n$\\therefore$ $\\triangle$ADE$\\cong$ $\\triangle$ADC,\\\\\n$\\therefore$AE=AC,\\\\\n$\\because$AB=12, AC=8,\\\\\n$\\therefore$BE=AB−AE=AB−AC=12−8=\\boxed{4}.",
    "image_path": "dataset/mm_math/images/MM_Math/52654988.png"
  },
  {
    "solution": "\\boxed{50^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is a chord of the circle $\\odot O$, $OD\\perp AB$ with the foot of the perpendicular at $E$, intersecting $\\odot O$ at points $C$ and $D$. If $\\angle AOD=50^\\circ$, what is the degree measure of $\\angle DOB$?\n",
    "completion": "\\textbf{Solution:} Since $OD \\perp AB$,\\\\\nit follows that $\\overset{\\frown}{AD} = \\overset{\\frown}{BD}$,\\\\\ntherefore, $\\angle DOB = \\angle AOD = \\boxed{50^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53081489_81.png"
  },
  {
    "solution": "\\boxed{(-5,2)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, $A\\left(1, 0\\right)$, $B\\left(0, 2\\right)$, and $\\triangle ABC$ is an isosceles right triangle. By using A as the center of dilation and enlarging $\\triangle ABC$ by a similarity ratio of $1:2$, the enlarged figure is denoted as $\\triangle AB'C'$. What are the coordinates of $C'$?\n",
    "completion": "\\textbf{Solution:}\\\\\nSince $A$ is the center of similitude, if we enlarge $\\triangle ABC$ by the similarity ratio $1\\colon 2$, the resulting figure is denoted as $\\triangle AB'C'$,\\\\\ntherefore $AB=\\frac{1}{2}AB'$,\\\\\nthus point $B$ is the midpoint of segment $AB'$,\\\\\nsince $A(1,0)$ and $B(0,2)$,\\\\\nby the Pythagorean theorem, $AB=\\sqrt{4+1}=\\sqrt{5}$,\\\\\nthus $AB' =2AB=2\\sqrt{5}$,\\\\\nbased on similarity, $\\triangle AB'C'$ is an isosceles right triangle,\\\\\n$AC'=\\sqrt{AB'^{2}+B'C'^{2}}=2\\sqrt{10}$,\\\\\nA, when $C'(-6,-2)$, by the Pythagorean theorem, $AC'=\\sqrt{(1+6)^{2}+(0+2)^{2}}=\\sqrt{53}\\ne 2\\sqrt{10}$, the option is incorrect and does not match the condition;\\\\\nB, when $C'(-5,2)$, by the Pythagorean theorem, $AC'=\\sqrt{(1+5)^{2}+(0-2)^{2}}=\\sqrt{40}=2\\sqrt{10}$, the option is correct and matches the condition;\\\\\nC, when $C'(-4,-2)$, by the Pythagorean theorem, $AC'=\\sqrt{(1+4)^{2}+(0+2)^{2}}=\\sqrt{29}\\ne 2\\sqrt{10}$, the option is incorrect and does not match the condition;\\\\\nD, when $C'(-4,2)$, by the Pythagorean theorem, $AC'=\\sqrt{(1+4)^{2}+(0-2)^{2}}=\\sqrt{29}\\ne 2\\sqrt{10}$, the option is incorrect and does not match the condition;\\\\\nTherefore, the answer is $C'(-5,2)$, that is, the coordinates of $C'$ are $\\boxed{(-5,2)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51496823.png"
  },
  {
    "solution": "\\boxed{\\frac{2\\sqrt{5}}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\triangle ABC$, $ \\angle ACB=90^\\circ$, $AC=\\frac{1}{2}BC=2$. Point D is a moving point on $ AB$. Connect $ CD$, rotate the line segment $ CD$ counterclockwise by $90^\\circ$ around point C to obtain line segment $ CE$, and connect $ DE$ and $ BE$. When the area of $ \\triangle BED$ is maximized, what is the length of $ AD$?\n",
    "completion": "\\textbf{Solution:} Let's connect point A and the midpoint F of $BC$, draw $FH\\perp AB$ at point H, and connect $EF$,\\\\\n$\\because$ point F is the midpoint of $BC$,\\\\\n$\\therefore$ $CF=\\frac{1}{2}BC=1$,\\\\\n$\\because$ $\\angle ACB=\\angle DCE=90^\\circ$,\\\\\n$\\therefore$ $\\angle ACB-\\angle DCF=\\angle DCE-\\angle DCF$, which means $\\angle ACD=\\angle FCE$,\\\\\nIn $\\triangle ACD$ and $\\triangle FCE$,\\\\\n$\\left\\{\\begin{array}{l}\nAC=CF\\\\\n\\angle ACD=\\angle FCE\\\\\nDC=CE\n\\end{array}\\right.$,\\\\\n$\\therefore$ $\\triangle ACD\\cong \\triangle FCE(SAS)$,\\\\\n$\\therefore$ $\\angle CFE=\\angle CAD$,\\\\\n$\\because$ $FH\\perp AB$,\\\\\n$\\therefore$ $\\angle ABC+\\angle BFH=\\angle ABC+\\angle CAD=90^\\circ$,\\\\\n$\\therefore$ $\\angle BFH=\\angle CAD$,\\\\\n$\\therefore$ $\\angle BFH=\\angle CFE$,\\\\\n$\\because$ $\\angle CFE+\\angle BFE=180^\\circ$,\\\\\n$\\therefore$ $\\angle BFH+\\angle BFE=180^\\circ$, thus points H, F, and E are collinear,\\\\\nTherefore, $EH$ is the height of $\\triangle BED$,\\\\\nLet $AD=x$, then $EF=x$,\\\\\nAccording to the Pythagorean theorem: $AB=\\sqrt{AC^2+BC^2}=\\sqrt{5}$,\\\\\n$\\therefore$ $BD=\\sqrt{5}-x$,\\\\\n$\\because$ $\\angle BFH=\\angle CAD$,\\\\\n$\\therefore$ $\\triangle BFH\\sim \\triangle ABC$,\\\\\n$\\therefore$ $\\frac{BF}{AB}=\\frac{FH}{AC}$, that is $\\frac{1}{\\sqrt{5}}=\\frac{FH}{1}$, solving gives: $FH=\\frac{\\sqrt{5}}{5}$,\\\\\n$\\therefore$ $EH=FH+EF=\\frac{\\sqrt{5}}{5}+x$,\\\\\n$S_{\\triangle BDE}=\\frac{1}{2}BD\\cdot EH=\\frac{1}{2}\\left(\\sqrt{5}-x\\right)\\left(\\frac{\\sqrt{5}}{5}+x\\right)=-\\frac{1}{2}\\left(x-\\sqrt{5}\\right)\\left(x+\\frac{\\sqrt{5}}{5}\\right)$,\\\\\n$\\therefore$ when the area is maximized, $x=\\frac{\\sqrt{5}-\\frac{\\sqrt{5}}{5}}{2}=\\boxed{\\frac{2\\sqrt{5}}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52933339.png"
  },
  {
    "solution": "\\boxed{\\frac{4}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that in the parallelogram $ABCD$, $G$ is a point on the extension line of $AB$. Connect $DG$, and let it intersect $AC$ and $BC$ at point $E$ and point $F$, respectively, and $AE: EC = 3: 2$. Find the value of $\\frac{EF}{FG}$.\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a parallelogram, \\\\\nit follows that AD $\\parallel$ BC, which means AD $\\parallel$ CF, \\\\\nhence, $\\angle ADE = \\angle CFE$, and $\\angle DAE = \\angle FCE$, \\\\\nthus, $\\triangle ADE \\sim \\triangle CFE$, \\\\\ntherefore, $\\frac{DE}{EF} = \\frac{AE}{EC} = \\frac{3}{2}$, \\\\\nFurthermore, since $\\triangle AGE \\sim \\triangle CDE$, \\\\\nit follows that $\\frac{DE}{GE} = \\frac{EC}{AE} = \\frac{2}{3}$, \\\\\ntherefore, $\\frac{EF}{GE} = \\frac{EF}{DE} \\times \\frac{DE}{GE} = \\frac{2}{3} \\times \\frac{2}{3} = \\frac{4}{9}$, \\\\\nhence, $\\frac{EF}{FG} = \\frac{EF}{GE-EF} = \\boxed{\\frac{4}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51379520_101.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $ \\triangle ABC \\cong \\triangle DEF$. If $AC = 5$ and $CF = 2$, what is the length of $CD$?\n",
    "completion": "\\textbf{Solution:} Given that $\\triangle ABC\\cong \\triangle DEF$, $\\angle A$ and $\\angle D$ are corresponding angles, and $AB$ and $DE$ are corresponding sides,\\\\\n$\\therefore AC=DF=5$,\\\\\nAlso, since $CF=2$,\\\\\n$\\therefore CD=DF-CF=5-2=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53066301.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=8$, $\\angle B=90^\\circ$, point $P$ is a moving point on side $AB$, draw $PD\\parallel AC$ intersecting $BC$ at $D$, and draw $PE\\parallel BC$ intersecting $AC$ at $E$. When point $P$ is the midpoint of $AB$, $PE=3$. What is the length of side $BC$?\n",
    "completion": "\\textbf{Solution:} Given $\\because PE\\parallel BC$, and point P is the midpoint of AB\\\\\n$\\therefore \\frac{AE}{EC}=\\frac{AP}{PB}=1$\\\\\n$\\therefore AE=EC$\\\\\nThis means E is the midpoint of AC\\\\\n$\\therefore$ PE is the median\\\\\n$\\therefore BC=2PE=2\\times 3=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51212675_79.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, where $AB=4$ and $AD=2$, $E$ is the midpoint of $AB$, $F$ is a moving point on $EC$, and $P$ is the midpoint of $DF$. Connect $PB$. What is the minimum value of $PB$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure:\n\nWhen point $F$ coincides with point $C$, point $P$ is at ${P}_{1}$, and $C{P}_{1}=D{P}_{1}$,\n\nWhen point $F$ coincides with point $E$, point $P$ is at ${P}_{2}$, and $E{P}_{2}=D{P}_{2}$,\n\n$\\therefore {P}_{1}{P}_{2}\\parallel CE$ and ${P}_{1}{P}_{2}=\\frac{1}{2}CE$.\n\nWhen point $F$ is on segment $EC$ except at points $C$ and $E$, $DP=FP$.\n\nAccording to the Midline Theorem: ${P}_{1}P\\parallel CE$ and ${P}_{1}P=\\frac{1}{2}CF$.\n\n$\\therefore$ The trajectory of point $P$ is the line segment ${P}_{1}{P}_{2}$,\n\n$\\therefore$ When $BP\\perp {P}_{1}{P}_{2}$, $PB$ reaches its minimum value.\n\n$\\because$ In rectangle $ABCD$, $AB=4$, $AD=2$, $E$ is the midpoint of $AB$,\n\n$\\therefore \\triangle CBE$, $\\triangle ADE$, $\\triangle BC{P}_{1}$ are isosceles right triangles, and $C{P}_{1}=2$.\n\n$\\therefore \\angle ADE=\\angle CDE=\\angle C{P}_{1}B=45^\\circ$, $\\angle DEC=90^\\circ$.\n\n$\\therefore \\angle D{P}_{2}{P}_{1}=90^\\circ$.\n\n$\\therefore \\angle D{P}_{1}{P}_{2}=45^\\circ$.\n\n$\\therefore \\angle {P}_{2}{P}_{1}B=90^\\circ$, i.e., $B{P}_{1}\\perp {P}_{1}{P}_{2}$,\n\n$\\therefore$ The minimum value of $BP$ is the length of $B{P}_{1}$.\n\nIn the isosceles right triangle $BC{P}_{1}$, $C{P}_{1}=BC=2$,\n\n$\\therefore B{P}_{1}=2\\sqrt{2}$\n\n$\\therefore$ The minimum value of $PB$ is $\\boxed{2\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50482309.png"
  },
  {
    "solution": "\\boxed{(-1, 0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the parabola $y = x^{2} + bx + c$ intersects the $x$-axis at points $A$ and $B$, where the coordinates of point $A$ are $(-3, 0)$. It intersects the $y$-axis at point $C$, and point $D(-2, -3)$ lies on the parabola. There is a moving point $M$ on the axis of symmetry of the parabola. When $\\triangle MAD$ is an isosceles triangle, directly write the coordinates of point $M$.\n",
    "completion": "\\textbf{Solution:} Let point M be ($-1$, $n$),\n\nsince $A (-3,0)$ and $D (-2,-3)$,\n\nthus, $AM^{2} = (-1+3)^{2}+n^{2} = n^{2}+4$,\n\n$AD^{2} = (-3+2)^{2}+9 = 10$,\n\n$MD^{2} = (-1+2)^{2} + (n+3)^{2} = n^{2} + 6n + 10$,\n\nWhen $AM=AD$, $n^{2}+4=10$, solving yields:\n\n$n_{1}=\\sqrt{6}$, $n_{2}=-\\sqrt{6}$,\n\nthus, the coordinates of point M are $(-1, \\sqrt{6})$ or $(-1, -\\sqrt{6})$;\n\nWhen $AM=MD$, $n^{2}+4 = n^{2}+6n+10$, solving yields:\n\n$n = -1$, thus, the coordinates of point M are $(-1, -1)$;\n\nWhen $AD=DM$, $n^{2}+6n+10=10$, solving yields:\n\n$n_{1}=0$, $n_{2}=-6$;\n\nthus, the coordinates of point P are $(-1,0)$ and $(-1,-6)$,\n\nLet the equation of line AD be $y=kx+b$,\n\nthus, $\\left\\{\\begin{array}{l}\n-3k+b=0 \\\\\n-2k+b=-3\n\\end{array}\\right.$, solving yields:\n\n$\\left\\{\\begin{array}{l}\nk=-3 \\\\\nb=-9\n\\end{array}\\right.$,\n\nthus, the equation of line AD is $y=-3x-9$,\n\nwhen $x=-1$, $y=3-9=-6$,\n\nthus, point $(-1, -6)$ is on line AD,\n\nthus, point $(-1, -6)$ does not meet the problem requirement, discard it,\n\nthus, when $\\triangle MAD$ is an isosceles triangle, the coordinates of point M are \\boxed{(-1, \\sqrt{6})} or \\boxed{(-1, -\\sqrt{6})} or \\boxed{(-1, -1)} or \\boxed{(-1, 0)}.",
    "image_path": "dataset/mm_math/images/MM_Math/52586455_122.png"
  },
  {
    "solution": "\\boxed{-20}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rhombus $ABCD$, $AD\\parallel x$-axis. The coordinate of point $A$ is $\\left(0, 4\\right)$, and the coordinate of point $B$ is $\\left(3, 0\\right)$. The line containing side $CD$, described by $y_{1}=mx+n$, intersects the $x$-axis at point $C$ and intersects the hyperbola $y_{2}=\\frac{k}{x}$ $(x<0)$ at point $D$. Find the equation of line $CD$ and the value of $k$?\n",
    "completion": "\\textbf{Solution:} Since the coordinates of point A are $(0,4)$ and the coordinates of point B are $(3,0)$,\n\\[AB= \\sqrt{3^2+4^2}=5,\\]\nsince quadrilateral ABCD is a rhombus,\n\\[AD=BC=AB=5,\\]\nthus $D(-5,4)$ and $C(-2,0)$. Plugging the coordinates of points C and D into the equation of the line yields\n\\[\\begin{cases}\n-5m+n=4 \\\\\n-2m+n=0\n\\end{cases},\\]\nfrom which we find\n\\[\\begin{cases}\nm=-\\frac{4}{3} \\\\\nn=-\\frac{8}{3}\n\\end{cases},\\]\ntherefore, the equation of line CD is $y_1=-\\frac{4}{3}x-\\frac{8}{3}$. Given that point D lies on the graph of an inverse ratio function,\n\\[4=\\frac{k}{-5},\\]\ntherefore $k=\\boxed{-20}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52250758_123.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB = 60^\\circ$, $DE$ is the perpendicular bisector of the hypotenuse $AC$, intersecting $AB$ and $AC$ at points $D$ and $E$, respectively. If $BD = 2$, what is the length of $AD$?\n",
    "completion": "\\textbf{Solution:} Given $\\triangle ABC$, $\\angle ACB=60^\\circ$, \\\\\n$\\therefore \\angle A=30^\\circ$. \\\\\nSince DE is the perpendicular bisector of AC, \\\\\n$\\therefore$ DA=DC, \\\\\nhence $\\angle DCA=\\angle A=30^\\circ$, \\\\\n$\\therefore \\angle DCB=\\angle ACB-\\angle DCA$, \\\\\n$=60^\\circ-30^\\circ$ \\\\\n$=30^\\circ$ \\\\\nIn $\\triangle BCD$, $\\angle DCB=30^\\circ$, \\\\\n$\\therefore BD= \\frac{1}{2}CD$, \\\\\n$\\therefore CD=2BD=4$, \\\\\n$\\therefore AD=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52655468.png"
  },
  {
    "solution": "\\boxed{90^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, CE is the bisector of the exterior angle $\\angle ACD$ of $\\triangle ABC$. If $\\angle B = 30^\\circ$ and $\\angle ACE = 60^\\circ$, what is the measure of $\\angle A$ in degrees?\n",
    "completion": "\\textbf{Solution:} Since $CE$ bisects $\\angle ACD$ and $\\angle ACE=60^\\circ$,\n\nit follows that $\\angle ACD=2\\angle ACE=120^\\circ$.\n\nSince $\\angle ACD$ is an exterior angle of $\\triangle ABC$ and $\\angle B=30^\\circ$,\n\nit follows that $\\angle ACD=\\angle B+\\angle A$,\n\nthus $\\angle A=\\angle ACD-\\angle B$\n\n$=120^\\circ-30^\\circ$\n\n$=\\boxed{90^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53102446.png"
  },
  {
    "solution": "\\boxed{-3 < x < 1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: The partial graph of the quadratic function is shown in the figure, with the axis of symmetry being the line $x=-1$. By observing the graph, find the range of the independent variable $x$ when $y>0$.\n",
    "completion": "\\textbf{Solution:} Since the parabola passes through the point ($-3$, $0$) and its axis of symmetry is the line $x = -1$, \\\\\nit follows that the parabola also passes through the point ($1$, $0$), \\\\\ntherefore, when \\boxed{-3 < x < 1}, we have $y > 0$.",
    "image_path": "dataset/mm_math/images/MM_Math/52605434_114.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, if $AC=48$ and M is the midpoint of $AC$, with $AB=\\frac{1}{3}AC$, what is the length of $BM$?\n",
    "completion": "\\textbf{Solution:} Given that $AC=48$ and M is the midpoint of $AC$, $AB=\\frac{1}{3}AC$\\\\\nTherefore, $AM=\\frac{1}{2}AC=\\frac{1}{2}\\times 48=24$, $AB=\\frac{1}{3}\\times 48=16$\\\\\nHence, $BM=AM-AB=24-16=\\boxed{8}$",
    "image_path": "dataset/mm_math/images/MM_Math/53211085.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In squares $ABCD$ and $AEFG$, connect $AC$, $AF$, and $CF$ respectively. Let point $H$ be the midpoint of $CF$. As shown in Figure 3, when point $G$ is on segment $AB$ and $AB=2AE=2$, what is the area of $\\triangle BGH$?\n",
    "completion": "\\textbf{Solution:} Given point $G$ on segment $AB$, if $AB=2AE=2$, then $BC=AB=2$, $AG=FG=AE=1$, \\\\\n$\\therefore BG=AB-AG=1$, \\\\\nExtend $GH$ to intersect $BC$ at point $Q$, as shown in the diagram: \\\\\n$\\because$ Point $H$ is the midpoint of $CF$, \\\\\n$\\therefore HC=HF$, \\\\\nIn squares $ABCD$ and $AEFG$, $\\angle ABC=\\angle AGF=90^\\circ$, \\\\\n$\\therefore \\angle BGF=90^\\circ=\\angle ABC$, \\\\\n$\\therefore FG\\parallel BC$, \\\\\n$\\therefore \\angle GFH=\\angle QCH$, \\\\\nIn $\\triangle GFH$ and $\\triangle QCH$, \\\\\n$\\left\\{\\begin{array}{l}\n\\angle GFH=\\angle QCH \\\\\nFH=CH \\\\\n\\angle GHF=\\angle QHC\n\\end{array}\\right.$, \\\\\n$\\therefore \\triangle GFH \\cong \\triangle QCH\\left(ASA\\right)$, \\\\\n$\\therefore CQ=FG=1$, $QH=GH$, \\\\\n$\\therefore BQ=BC-CQ=1$, $S_{\\triangle BGH}=S_{\\triangle BQH}=\\frac{1}{2}S_{\\triangle BGQ}$, \\\\\n$\\because S_{\\triangle BGQ}=\\frac{1}{2}BG\\cdot BQ=\\frac{1}{2}\\times 1\\times 1=\\frac{1}{2}$, \\\\\n$\\therefore S_{\\triangle BGH}=\\frac{1}{2}S_{\\triangle BGQ}=\\frac{1}{2}\\times \\frac{1}{2}=\\boxed{\\frac{1}{4}}$, \\\\\ni.e., the area of $\\triangle BGH$ is $\\frac{1}{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53728774_89.png"
  },
  {
    "solution": "\\boxed{-6<n<6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system, the line $l_{1}$ passes through point $A(1,0)$ and point $B(0,2)$. The moving point $P(m,n)$ is on the line $l_{1}$. When $-2<m<4$, what is the range of values for $n$?\n",
    "completion": "\\textbf{Solution:} Since $-2<0$, it follows that $y$ decreases as $x$ increases. When $x=m=-2$, we have $n=y=-2x+2=6$; when $x=m=4$, we find $n=y=-2x+2=-6$. Therefore, for $-2<m<4$, it holds that $\\boxed{-6<n<6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50561277_50.png"
  },
  {
    "solution": "\\boxed{40^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of $\\odot O$, $CD$ is a chord, and $\\angle CAB=50^\\circ$, then what is the degree measure of $\\angle D$?\n",
    "completion": "\\textbf{Solution:} Since AB is a diameter,\\\\\nit follows that $\\angle ACN=90^\\circ$,\\\\\ntherefore, $\\angle B=90^\\circ-\\angle CAB=90^\\circ-50^\\circ=40^\\circ$;\\\\\nSince arc AC equals arc AC,\\\\\nit follows that $\\angle B=\\angle D=\\boxed{40^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51369879.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the Cartesian coordinate system, for the right triangle $\\triangle ABC$ with vertices $A$ and $C$ having coordinates $(0,2)$ and $(2,0)$ respectively, and $\\angle ACB = 90^\\circ$, where $AC = 2BC$. If the graph of the function $y = \\frac{k}{x}$ ($k>0$, $x>0$) passes through point $B$, what is the value of $k$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $BE\\perp OC$ at $E$, \\\\\nsince the coordinates of A and C are respectively $(0, 2)$ and $(2, 0)$, \\\\\ntherefore, $OA=2$, $OC=2$, \\\\\ntherefore, $\\angle OAC=\\angle ACO=45^\\circ$, and $AC=2\\sqrt{2}$, \\\\\nsince $AC=2BC$ and $\\angle ACB=90^\\circ$, \\\\\ntherefore, $BC=\\sqrt{2}$, and $\\angle BCE=45^\\circ$, \\\\\ntherefore, $\\angle BCE=\\angle CBE=45^\\circ$, and $BE=CE=1$, \\\\\ntherefore, $B(3,1)$, \\\\\ntherefore, $k=xy=3\\times 1=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51402513.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, we have $AB=10$ and $AD=6$. Let point E be on side $BC$, and fold $\\triangle CDE$ along $DE$ such that point C falls onto point F on side $AB$. What is the length of $AF$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rectangle,\\\\\nit follows that $\\angle A=90^\\circ$, and AB=CD=10,\\\\\nby the properties of folding, we know that DF=CD=10,\\\\\ntherefore in $\\triangle ADF$, by the Pythagorean theorem, we have $AF=\\sqrt{DF^{2}-AD^{2}}=\\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52978448_134.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: It is known that the side length of square $ABCD$ is 4. Square $OMNE$ (with $OM > \\frac{1}{2}AC$) rotates around the center of symmetry $O$ of square $ABCD$. What is the area of the overlapping region of the two squares?\n",
    "completion": "\\textbf{Solution}: As shown in the figure, connect AC and BD. Let OE intersect CD at point Q, and OM intersect BC at point P,\n\n$\\because$ Point O is the center of the square ABCD,\n\n$\\therefore$ The diagonals of square ABCD intersect at point O,\n\n$\\therefore \\angle$ODQ=$\\angle$OCP=$45^\\circ$, OD=OC, $\\angle$COD=$90^\\circ$,\n\n$\\because$ Quadrilateral OMNE is a square,\n\n$\\therefore \\angle$EOM=$90^\\circ$,\n\n$\\because \\angle$DOC=$\\angle$DOQ+$\\angle$COQ=$90^\\circ$, $\\angle$EOM=$\\angle$COQ+$\\angle$COP=$90^\\circ$,\n\n$\\therefore \\angle$DOQ=$\\angle$COP,\n\nIn $\\triangle DOE$ and $\\triangle AOF$,\n\n$\\left\\{ \\begin{array}{l} \\angle DOQ=\\angle COP \\\\ OD=OC \\\\ \\angle ODQ=\\angle OCP \\end{array} \\right.$,\n\n$\\therefore \\triangle DOQ \\cong \\triangle COP$ (ASA),\n\n$\\therefore S_{\\triangle DOQ}=S_{\\triangle COP}$,\n\n$\\therefore$ The area of quadrilateral OPCQ=$S_{\\triangle OCP}+S_{\\triangle QOC}=S_{\\triangle ODQ}+S_{\\triangle QOC}=S_{\\triangle COD}$,\n\n$\\because S_{\\triangle COD}=\\frac{1}{4}S_{\\text{square ABCD}}=\\frac{1}{4}\\times 4\\times 4=4$,\n\n$\\therefore$ The area of the overlapping part of the two squares is $\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52225526.png"
  },
  {
    "solution": "\\boxed{140^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points A, B, and C are on the circle $\\odot O$. If $\\angle CAB=70^\\circ$, then what is the measure of $\\angle BOC$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle CAB=70^\\circ$,\\\\\nit follows that $\\angle BOC=2\\angle CAB=\\boxed{140^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/51446299.png"
  },
  {
    "solution": "\\boxed{20}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: The vertex $A$ and $C$ of the rectangle $OABC$ are respectively on the positive half-axis of $x$ and $y$. Point $F$ is a moving point on side $BC$ (not coinciding with points $B$ and $C$). The inverse proportion function $y=\\frac{k}{x}$ $(x>0)$ passing through point $F$ intersects with side $AB$ at point $E(8,m)$, where $AB=4$. As shown in the figure, connect $OF$ and $EF$. Please use an equation involving $m$ to express the area of quadrilateral $OAEF$, and find the maximum value of the area of $OAEF$.\n",
    "completion": "\\textbf{Solution:} Since points E and F are on the graph of the inverse proportion function $y=\\frac{k}{x}$ $(x>0)$,\\\\\nthus $CF \\times 4 = 8m$,\\\\\ntherefore $CF = 2m$,\\\\\ntherefore, the area of quadrilateral $OAEF$ is $8 \\times 4 - \\frac{1}{2} \\times 4 \\times 2m - \\frac{1}{2} \\times (8 - 2m) \\times (4 - m)$\\\\\n$= -m^{2} + 4m + 16 = -\\left(m - 2\\right)^{2} + 20$,\\\\\nsince $0 < m < 4$,\\\\\ntherefore, when $m = 2$, the maximum area of quadrilateral $OAEF$ is $\\boxed{20}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52457729_186.png"
  },
  {
    "solution": "\\boxed{x>1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given the graph, the linear function $y_{1}=x+b$ and the linear function $y_{2}=kx+5$ intersect at point $P(1, 4)$. What is the solution set of the inequality $x+b>kx+5$ in terms of $x$?\n",
    "completion": "\\textbf{Solution:} From the graph, we can see that $P(1,4)$, \\\\\nhence when $x>1$, $x+b>kx+5$, \\\\\ntherefore, the solution set is $\\boxed{x>1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52386462.png"
  },
  {
    "solution": "\\boxed{-x+1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the line $y_1=k_1x+b$ intersects with the hyperbola $y_2=\\frac{k_2}{x}$ at points $A(-2, 3)$ and $B(m, -2)$. What are the expressions for the functions corresponding to $y_1$ and $y_2$?\n",
    "completion": "\\textbf{Solution:} Let point $A\\left(-2, 3\\right)$ be substituted into the equation of the inverse proportion function to find: $k=-6$,\\\\\n$\\therefore y_{2}=-\\frac{6}{x}$,\\\\\n$\\because$ point B is on the graph of the inverse proportion function,\\\\\n$\\therefore -2m=-6$, which gives: $m=3$,\\\\\n$\\therefore B\\left(3, -2\\right)$,\\\\\nSubstituting points A and B into the equation of the line gives: $\\left\\{\\begin{array}{l}\n-2k_{1}+b=3 \\\\\n3k_{1}+b=-2\n\\end{array}\\right.$, solving this yields: $\\left\\{\\begin{array}{l}\nk_{1}=-1 \\\\\nb=1\n\\end{array}\\right.$,\\\\\n$\\therefore y_{1}=\\boxed{-x+1}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53043125_136.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points A and B are on the graphs of the inverse proportion functions $y=\\frac{2}{x}$ ($x>0$) and $y=\\frac{4}{x}$ ($x>0$) respectively, and AB is parallel to the x-axis. If C is any point on the x-axis, then what is the area of $\\triangle ABC$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, extend BA to intersect the y-axis at point M, and connect OA, OB,\\\\\nsince line AB is parallel to the x-axis,\\\\\nsince $S_{\\triangle AOM}=1$, $S_{\\triangle BOM}=2$,\\\\\ntherefore $S_{\\triangle ABC}=S_{\\triangle ABO}=S_{\\triangle BOM}-S_{\\triangle AOM}=2-1=\\boxed{1}$",
    "image_path": "dataset/mm_math/images/MM_Math/52765812.png"
  },
  {
    "solution": "\\boxed{13.5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, line $l_{1}$ intersects the y-axis at point A in the Cartesian coordinate system, and point B$(-3, 3)$ is also on line $l_{1}$. Point B is translated 1 unit to the right and then 2 units down to get point C, which is also on line $l_{1}$. It is known that line $l_{2}: y=x+b$ passes through point B and intersects the y-axis at point E. Find the area of $\\triangle ABE$.\n",
    "completion": "\\textbf{Solution:} Substitute the coordinates of point B into $y=x+b$ to get,\\\\\n$3=-3+b$, solving for $b$ gives $b=6$,\\\\\n$\\therefore y=x+6$.\\\\\nWhen $x=0$, $y=6$,\\\\\n$\\therefore$ the coordinates of point E are $(0,6)$.\\\\\nWhen $x=0$, $y=-3$,\\\\\n$\\therefore$ the coordinates of point A are $(0,-3)$,\\\\\n$\\therefore AE=6+3=9$,\\\\\n$\\therefore$ the area of $\\triangle ABE$ is $\\frac{1}{2} \\times 9 \\times |-3| = \\boxed{13.5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52019425_105.png"
  },
  {
    "solution": "\\boxed{38^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the lines AB and CD intersect at point O. OE is the bisector of $\\angle AOC$, $OF\\perp CD$, $OG\\perp OE$, and $\\angle BOD=52^\\circ$. What is the measure of $\\angle AOF$?\n",
    "completion": "\\textbf{Solution:} Since OF$\\perp$CD, \\\\\n$\\therefore$ $\\angle$COF$=90^\\circ$, \\\\\nAlso, since $\\angle$AOC and $\\angle$BOD are vertical angles, \\\\\n$\\therefore$ $\\angle$AOC$=\\angle$BOD$=52^\\circ$, \\\\\n$\\therefore$ $\\angle$AOF=$\\angle$COF-$\\angle$AOC=$90^\\circ-52^\\circ$=$\\boxed{38^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/55426370_88.png"
  },
  {
    "solution": "\\boxed{30}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points $D$ and $E$ are on the sides of $\\triangle ABC$. The angle bisector of $\\angle ABC$ is perpendicular to $AE$ with the foot of the perpendicular being $Q$, and the angle bisector of $\\angle ACB$ is perpendicular to $AD$ with the foot of the perpendicular being $P$. If $BC=12$ and $PQ=3$, what is the perimeter of $\\triangle ABC$?\n",
    "completion": "Solution: Since line BQ bisects $\\angle ABC$ and BQ $\\perp$ AE,\n\nit follows that $\\angle ABQ = \\angle EBQ$,\n\nSince $\\angle ABQ + \\angle BAQ = 90^\\circ$ and $\\angle EBQ + \\angle BEQ = 90^\\circ$,\n\nit follows that $\\angle BAQ = \\angle BEQ$,\n\nTherefore, $AB = BE$, similarly, $CA = CD$,\n\nThus, point Q is the midpoint of AE, and point P is the midpoint of AD (triangle median theorem),\n\nTherefore, PQ is the midline of $\\triangle ADE$,\n\nSince $PQ = 3$,\n\nit follows that $DE = 2 \\times PQ = 6$,\n\nThe perimeter of $\\triangle ABC$ is: $AB + AC + BC = BE + CD + BC = DE + BC + BC = 6 + 12 + 12 = \\boxed{30}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52019336.png"
  },
  {
    "solution": "\\boxed{40^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, A, B, and C are three points on the circle $\\odot O$. Given that $\\angle ABC=20^\\circ$, what is the measure of $\\angle AOC$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle ABC=20^\\circ$, \\\\\nthen $\\angle AOC= 2\\angle ABC = \\boxed{40^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52623055.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the parabola $y=ax^2+bx+3$ ($a \\neq 0$) intersects the x-axis at points $A(-4,0)$ and $B(6,0)$, and intersects the y-axis at point $C$. If $G$ is a moving point on the parabola between points $A$ and $C$, draw line $GD$ parallel to the x-axis, intersecting the parabola at point $D$. Draw vertical lines from points $D$ and $G$ to the x-axis, with the feet of the perpendiculars being $E$ and $F$, respectively, forming the rectangle $DEFG$. When point $G$ coincides with point $C$, what is the area of rectangle $DEFG$?\n",
    "completion": "\\textbf{Solution:} When point $G$ coincides with $C$, the coordinates of point $G$ are $(0,3)$.\\\\\nSubstituting $y=3$ into $y=-\\frac{1}{18}x^{2}+\\frac{1}{4}x+3$,\\\\\nwe get $-\\frac{1}{18}x^{2}+\\frac{1}{4}x+3=3$.\\\\\nSolving this equation, we find $x_{1}=0$ and $x_{2}=2$.\\\\\n$\\therefore$ The coordinates of point $D$ are $(2,3)$.\\\\\n$\\therefore GD=2$, $DE=3$.\\\\\n$\\therefore$ The area of rectangle DEFG is $S_{\\text{rectangle} DEFG}=GD \\cdot DE=2 \\times 3=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52550670_88.png"
  },
  {
    "solution": "\\boxed{(-2,-5)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the function expression of line $l_{1}$ is: $y=x-3$, which intersects the x-axis at point B. Line $l_{2}$ passes through point $A(-2, 0)$ and intersects line $l_{1}$ at point $C(-1, a)$. Point P is on line $l_{1}$, and point Q is on line $l_{2}$, where $PQ \\parallel y$-axis. If $PQ=AB$, find the coordinates of point P.\n",
    "completion": "\\textbf{Solution:} Solve: In the line $l_1\\colon y=x-3$, let $y=0$, we get $x=3$,\\\\\n$\\therefore B(3,0)$,\\\\\n$\\therefore AB=3-(-2)=5$,\\\\\nLet $P(b,b-3)$, since $PQ\\parallel x$-axis, we have $Q(b,-4b-8)$,\\\\\n$PQ=|b-3-(-4b-8)|=AB=5$,\\\\\nSolving this, we get $b=0$ or $b=-2$,\\\\\n$\\therefore P(0,-3)$ or $P(-2,-5)$,\\\\\nHence, the coordinates of point $P$ are $\\boxed{(0,-3)}$ or $\\boxed{(-2,-5)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52395325_103.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure of quadrilateral $ABCD$, it is known that $AB=3$, $BC=4$, $CD=12$, $AD=13$, and $\\angle B=90^\\circ$. Find: the length of $AC$?\n",
    "completion": "\\textbf{Solution:} Given that in $\\triangle ABC$, $AB=3$, $BC=4$, and $\\angle B=90^\\circ$,\\\\\nit follows that $AC=\\sqrt{AB^2+BC^2}=\\boxed{5}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53033473_98.png"
  },
  {
    "solution": "\\boxed{\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, it is known that point $D$ lies on the extension of the diameter $AB$ of circle $\\odot O$, point $C$ is on circle $\\odot O$. A line through $D$ is drawn perpendicular to $AD$ at $E$, intersecting the extension of $AC$ at $E$, and $CD$ is a tangent to circle $\\odot O$. It is given that $AB=2$ and $AE=3$. If the bisector of $\\angle ACB$ intersects circle $\\odot O$ at point $F$ and $P$ is the incenter of $\\triangle ABC$, determine the length of $PF$.\n",
    "completion": "\\textbf{Solution:} Draw $BF$, $PB$, and $AF$, \\\\\nsince $CF$ bisects $\\angle ACB$, \\\\\nthus $\\overset{\\frown}{A}F=\\overset{\\frown}{B}F$, \\\\\nhence $AF=BF$, \\\\\nsince $AB$ is the diameter and $AB=2$, \\\\\nthus $AF=BF=\\sqrt{2}$, \\\\\nsince $P$ is the incenter of $\\triangle ABC$, \\\\\nthus $\\angle 1=\\angle 2$, $\\angle CBP=\\angle ABP$, \\\\\nsince $\\angle 1=\\angle 3$, \\\\\nthus $\\angle 2=\\angle 3$, \\\\\nhence $\\angle 2+\\angle CBP=\\angle 3+\\angle ABP$, \\\\\nthus $\\angle FPB=\\angle FBP$, \\\\\nhence $FP=FB=\\boxed{\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53147522_133.png"
  },
  {
    "solution": "\\boxed{9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the length of segment $AB=23$, $BC=15$, and $M$ is the midpoint of $AC$. Take a point $N$ on $CB$ such that the ratio $CN: NB=1:2$. Find the length of segment $MN$.\n",
    "completion": "\\textbf{Solution:} Given that $BC=15$ and $CN:NB=1:2$,\\\\\nit follows that $CN=\\frac{1}{3}BC=\\frac{1}{3}\\times 15=5$.\\\\\nSince $M$ is the midpoint of $AC$,\\\\\nthen $CM=\\frac{1}{2}AC=4$.\\\\\nTherefore, $MN=CM+CN=4+5=\\boxed{9}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53363922_84.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the side DC of the rectangle ABCD is on the x-axis, and point B is on the graph of the hyperbola \\(y=\\frac{3}{x}\\). Point E is the trisection point of side AD closer to point A. Connect CE and extend it to intersect the y-axis at point F. What is the area of \\(\\triangle CDF\\)?\n",
    "completion": "\\textbf{Solution:} Let the lengths of the rectangle's sides be AB=CD=a, AD=BC=b,\\\\\nsince point B is on the graph of the inverse proportion function $y=\\frac{3}{x}$,\\\\\ntherefore B$\\left(\\frac{3}{b}, b\\right)$,\\\\\ntherefore OC=$\\frac{3}{b}$,\\\\\nsince point E is the trisection point on side AD closer to point A,\\\\\ntherefore DE=$\\frac{2}{3}b$,\\\\\nsince AD$\\parallel$y-axis,\\\\\ntherefore $\\triangle FOC\\sim \\triangle EDC$,\\\\\ntherefore $\\frac{OF}{ED}=\\frac{OC}{CD}$, that is, $\\frac{OF}{\\frac{2}{3}b}=\\frac{\\frac{3}{b}}{a}$,\\\\\ntherefore OF=$\\frac{2}{a}$,\\\\\ntherefore the area $S_{\\triangle CDF}=\\frac{1}{2}CD\\cdot OF=\\frac{1}{2}a\\cdot\\frac{2}{a}=\\boxed{1}$",
    "image_path": "dataset/mm_math/images/MM_Math/51445671.png"
  },
  {
    "solution": "\\boxed{\\left(-\\frac{3}{2}, -\\frac{15}{4}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=ax^2+bx-3$ passes through three points $A$, $B$, and $C$, with points $A(-3,0)$, $C(1,0)$, and point $B$ on the $y$-axis. Point $P$ is a moving point on the parabola below line $AB$ (not coinciding with $A$, $B$). A perpendicular line is drawn from point $P$ to the $x$-axis, the foot of the perpendicular is $D$, and it intersects line $AB$ at point $E$. When is $PE$ maximum, and what is the coordinate of point $P$ at this time?\n",
    "completion": "\\textbf{Solution:} Let $P(x, x^2+2x-3)$, and the equation of line AB is $y=kx+b$,\\\\\nFrom the parabolic equation $y=x^2+2x-3$,\\\\\nLet $x=0$, then $y=-3$,\\\\\n$\\therefore B(0, -3)$,\\\\\nSubstituting $A(-3, 0)$ and $B(0, -3)$ into $y=kx+b$, we get\n\\[\n\\left\\{\n\\begin{array}{l}\n0=-3k+b \\\\\n-3=b\n\\end{array}\n\\right.\n\\]\nSolving, we obtain\n\\[\n\\left\\{\n\\begin{array}{l}\nk=-1 \\\\\nb=-3\n\\end{array}\n\\right.\n\\]\n$\\therefore$ The equation of line AB is $y=-x-3$,\\\\\n$\\because PE\\perp x$-axis,\\\\\n$\\therefore E(x, -x-3)$,\\\\\n$\\because P$ is below the line AB,\\\\\n$\\therefore PE=-x-3-(x^2+2x-3)=-x^2-3x=-\\left(x+\\frac{3}{2}\\right)^2+\\frac{9}{4}$,\\\\\nWhen $x=-\\frac{3}{2}$, $y=x^2+2x-3=-\\frac{15}{4}$,\\\\\n$\\therefore$ When $PE$ is maximum, the coordinates of point $P$ are $\\boxed{\\left(-\\frac{3}{2}, -\\frac{15}{4}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51402562_51.png"
  },
  {
    "solution": "\\boxed{\\left(\\frac{2\\sqrt{6}}{3}, 2\\sqrt{6}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given: As shown in Figure 1, the point $A(4, n)$ lies on the graph of the inverse proportion function $y=\\frac{8}{x}(x>0)$. As shown in Figure 3, rotate line $OA$ counterclockwise by $45^\\circ$ around the origin $O$, and it intersects the graph of the inverse proportion function $y=\\frac{8}{x}(x>0)$ at point $B$. Find the coordinates of point $B$.\n",
    "completion": "\\textbf{Solution:} As shown in Figure 2, construct the symmetric point $A_1$ of point A with respect to the line OB, \\\\\nthen $OA=OA_1$, $AA_1 \\perp OB$, \\\\\nconstruct $A_1C \\perp y$-axis at point C, construct $AD \\perp x$-axis at point D, \\\\\nit is easy to prove $\\triangle A_1OC \\cong \\triangle AOD$, \\\\\n$\\therefore OC=OD$, $A_1C=AD$, \\\\\n$\\because$ The coordinates of A are $(4,2)$, \\\\\n$\\therefore$ The coordinates of $A_1$ are $(-2, 4)$, \\\\\n$\\therefore$ The equation of line $AA_1$ is: $y=-\\frac{1}{3}x+\\frac{10}{3}$, \\\\\n$\\therefore$ The equation of line $OB$ is: $y=3x$, \\\\\nFrom $\\begin{cases}y=\\frac{9}{x}\\\\ y=3x\\end{cases}$, we get $\\begin{cases}x=\\frac{2\\sqrt{6}}{3}\\\\ y=2\\sqrt{6}\\end{cases}$ or $\\begin{cases}x=-\\frac{2\\sqrt{6}}{3}\\\\ y=-2\\sqrt{6}\\end{cases}$ (negative solution is discarded) \\\\\n$\\therefore$ The coordinates of point $B\\left(\\frac{2\\sqrt{6}}{3}, 2\\sqrt{6}\\right)$ are $\\boxed{\\left(\\frac{2\\sqrt{6}}{3}, 2\\sqrt{6}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52250480_93.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{20}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in $\\triangle ABC$, $D$ and $E$ are points on $AB$ and $AC$ respectively, and $DE \\parallel BC$. If $AE:EC = 1:4$, then what is the value of $\\frac{S_{\\triangle ADE}}{S_{\\triangle BEC}}$?\n",
    "completion": "\\textbf{Solution:} Since $\\frac{AE}{EC}=\\frac{1}{4}$,\\\\\nthus $\\frac{{S}_{\\triangle ABE}}{{S}_{\\triangle EBC}}=\\frac{1}{4}$,\\\\\nthus ${S}_{\\triangle ABE}=\\frac{1}{4}{S}_{\\triangle EBC}$,\\\\\nSince DE$\\parallel$BC,\\\\\nthus $\\frac{AD}{DB}=\\frac{AE}{EC}=\\frac{1}{4}$,\\\\\nthus $\\frac{{S}_{\\triangle ADE}}{{S}_{\\triangle BDE}}=\\frac{1}{4}$,\\\\\nthus ${S}_{\\triangle BDE}=4{S}_{\\triangle ADE}$,\\\\\nAlso, since ${S}_{\\triangle BDE}={S}_{\\triangle ABE}-{S}_{\\triangle ADE}$,\\\\\nthus $4{S}_{\\triangle ADE}=\\frac{1}{4}{S}_{\\triangle EBC}-{S}_{\\triangle ADE}$,\\\\\nthus $\\frac{{S}_{\\triangle ADE}}{{S}_{\\triangle EBC}}=\\boxed{\\frac{1}{20}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52623826.png"
  },
  {
    "solution": "\\boxed{135^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\angle AOB$ is a straight angle, and $OM$ and $ON$ are the angle bisectors of $\\angle AOC$ and $\\angle BOD$ respectively. Given that $\\angle COD=90^\\circ$, find the measure of $\\angle MON$?\n",
    "completion": "Solution: Since $\\angle COD=90^\\circ$, it follows that $\\angle AOC+\\angle BOD =90^\\circ$. Furthermore, since OM and ON are the bisectors of $\\angle AOC$ and $\\angle BOD$ respectively, we have $\\angle MOC= \\frac{1}{2} \\angle AOC$ and $\\angle NOD= \\frac{1}{2} \\angle BOD$. Thus, $\\angle MOC+\\angle NOD= \\frac{1}{2} (\\angle AOC+\\angle BOD)=45^\\circ$. Also, $\\angle MON=\\angle MOC+\\angle NOD+\\angle COD$, yielding $\\angle MON=45^\\circ+90^\\circ=\\boxed{135^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/8753334_45.png"
  },
  {
    "solution": "\\boxed{70^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\triangle ABC$, we have $AB=AC$ and $\\angle BAC=110^\\circ$. Let $D$ be a point on $BC$. Connect $AD$ and construct $\\angle ADE=35^\\circ$, with $DE$ intersecting line segment $AC$ at point $E$. If $AE=DE$, what is the size of $\\angle BDA$?\n",
    "completion": "Solution: Since $AE=DE$,\\\\\nit follows that $\\angle DAE=\\angle ADE=35^\\circ$,\\\\\nSince $\\angle C=35^\\circ$,\\\\\nit follows that $\\angle BDA=\\angle DAE+\\angle C=35^\\circ+35^\\circ=\\boxed{70^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53009760_109.png"
  },
  {
    "solution": "\\boxed{\\left(0, -2\\sqrt{6}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the quadrilateral $ABCD$ is a rhombus with a perimeter of 24, and the coordinates of point $A$ are $\\left(4, 0\\right)$. What are the coordinates of point $D$?\n",
    "completion": "\\textbf{Solution}: Since quadrilateral $ABCD$ is a rhombus with a perimeter of 24,\\\\\nit follows that AB$=24\\div4=6$,\\\\\nsince the coordinates of point A are $\\left(4, 0\\right)$,\\\\\nit follows that OA$=4$,\\\\\nsince the diagonals of a rhombus are perpendicular bisectors of each other,\\\\\nit follows that in right-angled triangle AOB, by Pythagoras' theorem, OB$=\\sqrt{6^2-4^2}=2\\sqrt{6}$,\\\\\nhence OD=OB$=2\\sqrt{6}$,\\\\\ntherefore, the coordinates of D are $\\boxed{\\left(0, -2\\sqrt{6}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50482766.png"
  },
  {
    "solution": "\\boxed{25}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the coordinates of point $A$ are $\\left(5, 0\\right)$, and point $B$ is on the positive semi-axis of the $y$-axis ($OB<OA$). The segment $AB$ is rotated clockwise by $90^\\circ$ around point $A$ to obtain the segment $AC$. Perpendiculars are drawn from point $C$ to the $x$-axis and the $y$-axis, with the feet of the perpendiculars being $D$ and $E$, respectively. What is the area of the quadrilateral $ABEC$?\n",
    "completion": "\\textbf{Solution:} Given that $A\\left(5, 0\\right)$, it follows that $OA=5$. Connect $AE$ and construct $\\angle EAG=90^\\circ$, with $AG$ intersecting the extension of $EC$ at point $G$, as shown in the diagram. Consequently, $\\angle BAC=\\angle EAG=90^\\circ$, which implies $\\angle CAG=\\angle BAE$. Since $\\angle CEO+\\angle BAC=180^\\circ$, it means that $\\angle AEB+\\angle AEC+\\angle BAE+\\angle CAE=180^\\circ$. In $\\triangle ACE$, we have $\\angle AEC+\\angle CAE+\\angle ACE=180^\\circ$. Given that $\\angle AEB+\\angle BAE=180^\\circ-\\angle ABE$, it follows that $\\angle ABE+\\angle ACE=180^\\circ$. Additionally, since $\\angle ACG+\\angle ACE=180^\\circ$, it implies $\\angle ABE=\\angle ACG$. Given $AB=AC$, it follows that $\\triangle ABE\\cong \\triangle ACG$. Since $AE=AG$, we conclude $\\angle AEG=\\angle AGE=45^\\circ$ and therefore $\\angle AEO=\\angle EAO=45^\\circ$. Consequently, $OE=OA=5$, thus $AE=5\\sqrt{2}$ and the area of quadrilateral $ABEC$, $S_{\\text{quadrilateral }ABEC}=S_{\\triangle AEG}=\\frac{1}{2}\\times 5\\sqrt{2}\\times 5\\sqrt{2}=\\boxed{25}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50545780_114.png"
  },
  {
    "solution": "\\boxed{80^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, with point O as the endpoint, rays OA, OB, OC, OD, and OE are drawn successively in clockwise direction. Moreover, OB is the bisector of $\\angle AOC$, and OD is the bisector of $\\angle COE$. If $\\angle AOB=50^\\circ$ and $\\angle DOE=30^\\circ$, what is the degree measure of $\\angle BOD$?\n",
    "completion": "\\textbf{Solution:} Given that $OB$ is the bisector of $\\angle AOC$,\\\\\nhence $\\angle BOC = \\angle AOB = 50^\\circ$;\\\\\nsince $OD$ is the bisector of $\\angle COE$,\\\\\nthus $\\angle COD = \\angle DOE = 30^\\circ$,\\\\\ntherefore $\\angle BOD = \\angle BOC + \\angle COD = 50^\\circ + 30^\\circ = \\boxed{80^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52579788_199.png"
  },
  {
    "solution": "\\boxed{-1<x<3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the parabola $y=x^2+bx+c$ intersects the x-axis at one point $A(-1,0)$ and the y-axis at point $C(0,3)$. Based on the graph, for what values of $x$ is $y<0$?\n",
    "completion": "\\textbf{Solution:} It can be concluded from the graph that when $-1<x<3$, $y<0$. Therefore, when $x$ takes $\\boxed{-1<x<3}$, $y<0$.",
    "image_path": "dataset/mm_math/images/MM_Math/52874723_283.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola intersects the x-axis at $A(5,0)$ and $B(-1,0)$, and intersects the positive half-axis of the y-axis at point $C$. Lines $BC$ and $AC$ are drawn. It is known that $\\sin\\angle BAC=\\frac{\\sqrt{2}}{2}$. The line $y=kx$ ($k>0$) intersects segment $AC$ at point $M$. When the triangle with vertices $A$, $O$, and $M$ is similar to $\\triangle ABC$, find the value of $k$ and then find the coordinates of point $M$.\n",
    "completion": "\\textbf{Solution:} From (1) we obtain: $C(0,5)$, $A(5,0)$. Assume the equation of line AC is $y=k_1x+b$. Substituting the coordinates of points A and C, we get:\n\\[\n\\begin{cases}\nb=5\\\\\n5k_1+b=0\n\\end{cases}\n\\]\nSolving, we find:\n\\[\n\\begin{cases}\nb=5\\\\\nk_1=-1\n\\end{cases}\n\\]\n$\\therefore$ The equation of line AC is $y=-x+5$,\n\n$\\because$ Line $y=kx$ ($k>0$) intersects segment AC at point M, then let $M\\left(m, -m+5\\right)$,\n\n$\\therefore k=\\frac{-m+5}{m}$,\n\nFrom (1) it is known $OA=OC=5$, $OB=1$,\n\n$\\therefore AC=\\sqrt{(0-5)^2+(5-0)^2}=5\\sqrt{2}$, $AB=6$,\n\nAccording to the problem, we can divide into two cases:\n\n(1) When $\\triangle AOM\\sim \\triangle ABC$,\n\n$\\therefore \\frac{AO}{AB}=\\frac{AM}{AC}=\\frac{5}{6}$,\n\n$\\therefore AM=\\frac{5}{6}AC=\\frac{25\\sqrt{2}}{6}$,\n\n$\\therefore$ By the distance formula between two points:\n\n$(m-5)^2+(m-5)^2=\\frac{625}{18}$,\n\nSolving, we find: $m_1=\\frac{5}{6}$, $m_2=\\frac{55}{6}$,\n\n$\\because 0\\le m\\le 5$,\n\n$\\therefore m=\\frac{5}{6}$,\n\n$\\therefore M\\left(\\frac{5}{6}, \\frac{25}{6}\\right)$, $k=\\boxed{5}$;\n\n(2) When $\\triangle AOM\\sim \\triangle ACB$,\n\n$\\therefore \\frac{AO}{AC}=\\frac{AM}{AB}=\\frac{5}{5\\sqrt{2}}=\\frac{\\sqrt{2}}{2}$,\n\n$\\therefore AM=\\frac{\\sqrt{2}}{2}AB=3\\sqrt{2}$,\n\n$\\therefore$ By the distance formula between two points:\n\n$(m-5)^2+(m-5)^2=18$,\n\nSolving, we find: $m_1=2$, $m_2=8$ (discarding as it does not meet the problem criteria),\n\n$\\therefore M\\left(2, 3\\right)$, $k=\\boxed{\\frac{3}{2}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51368197_113.png"
  },
  {
    "solution": "\\boxed{65^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in $\\triangle ABC$, $D$ is a point on $AB$, and $E$ is the midpoint of $AC$. Connect $DE$ and extend it to point $F$ such that $EF=ED$, and connect $CF$. Given $CF\\parallel AB$ and $\\angle ABC=50^\\circ$, connect $BE$. $BE$ bisects $\\angle ABC$, and $AC$ bisects $\\angle BCF$. Find the measure of $\\angle A$.\n",
    "completion": "\\textbf{Solution:} Given $AC$ bisects $\\angle BCF$,\\\\\nthus, $\\angle ACB = \\angle ACF$,\\\\\nsince $\\angle A = \\angle ACF$,\\\\\nthus, $\\angle A = \\angle ACB$,\\\\\nsince $\\angle A + \\angle ABC + \\angle ACB = 180^\\circ$, $\\angle ABC = 50^\\circ$,\\\\\nthus, $2\\angle A = 130^\\circ$,\\\\\ntherefore, $\\angle A = \\boxed{65^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52913839_107.png"
  },
  {
    "solution": "\\boxed{6.5\\,\\text{cm}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that point $C$ is on the straight line $AB$, and points $M$ and $N$ are the midpoints of $AC$ and $BC$, respectively. If $C$ is on the line segment $AB$ with $AC=6$ cm and $MB=10$ cm, find the lengths of line segments $BC$ and $MN$.\n",
    "completion": "\\textbf{Solution:} Given that $M$ is the midpoint of $AC$,\\\\\nit follows that $MC=\\frac{1}{2}AC=3\\,\\text{cm}$,\\\\\nthus $BC=MB-MC=7\\,\\text{cm}$,\\\\\nand since $N$ is the midpoint of $BC$,\\\\\nit then follows that $CN=\\frac{1}{2}BC=3.5\\,\\text{cm}$,\\\\\ntherefore $MN=MC+NC=\\boxed{6.5\\,\\text{cm}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52872897_139.png"
  },
  {
    "solution": "\\boxed{130^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: An equilateral triangle, a right-angled triangle, and an isosceles triangle are placed as shown in the figure. The base angle of the isosceles triangle $\\angle 3=80^\\circ$, then what is the sum of $\\angle 1 + \\angle 2$?\n",
    "completion": "\\textbf{Solution}: As shown in the figure,\\\\\nfrom the equilateral triangle and the right triangle we can obtain $\\angle 1+\\alpha =120^\\circ$, $\\angle 2+\\beta =90^\\circ$,\\\\\n$\\therefore \\angle 1+\\angle 2+\\alpha +\\beta =90^\\circ+120^\\circ=210^\\circ$,\\\\\nand $\\angle 3=\\alpha +\\beta$,\\\\\n$\\therefore \\alpha +\\beta =80^\\circ$,\\\\\n$\\therefore \\angle 1+\\angle 2=210^\\circ-80^\\circ=\\boxed{130^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52720427.png"
  },
  {
    "solution": "\\boxed{\\left(\\frac{10}{3}, 0\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, where $AB=2$ and $BC=4$, point $D$ is the midpoint of side $AB$, and the graph of the inverse proportion function $y_1=\\frac{k}{x}(x>0)$ passes through point $D$ and intersects $BC$ at point $E$. Find a point $P$ on the $x$-axis such that the perimeter of $\\triangle PDE$ is minimized. What are the coordinates of point $P$ in this case?\n",
    "completion": "\\textbf{Solution:} Let $D(4, 1)$ be the point symmetrical about the $x$-axis as $D^{\\prime}(4, -1)$. Let the equation of line $D^{\\prime}E$ be $y=kx+b$. We have\n\\[\n\\left\\{\n\\begin{array}{l}\n2=2k+b \\\\\n-1=4k+b\n\\end{array}\n\\right.\n\\]\nSolving this, we get\n\\[\n\\left\\{\n\\begin{array}{l}\nk=-\\frac{3}{2} \\\\\nb=5\n\\end{array}\n\\right.\n\\]\nTherefore, the equation of line $D^{\\prime}E$ is $y=-\\frac{3}{2}x+5$ \\\\\nTherefore, the intersection of line $D^{\\prime}E$ with the $x$-axis is $P\\left(\\frac{10}{3}, 0\\right)$, \\\\\nTherefore, when the perimeter of $\\triangle PDE$ is minimized, the coordinates of $P$ are $\\boxed{\\left(\\frac{10}{3}, 0\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52251436_128.png"
  },
  {
    "solution": "\\boxed{(-8, -8)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated, the line $y = \\frac{2}{3}x + 4$ intersects the $x$-axis at point $B$ and the $y$-axis at point $A$. Translate line $AB$ to intersect the positive half of the $x$-axis at point $E$ and the $y$-axis at point $F$. Let $P$ be a point in the third quadrant on line $EF$. Draw $PG \\perp x$-axis at point $G$. Given that the area of $\\triangle PAB = 20$ and $GE = 12$, find the coordinates of point $P$.\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, since $PE \\parallel AB$,\\\\\nit follows that $S_{\\triangle PAB} = S_{\\triangle EAB} = 20$,\\\\\nhence $\\frac{1}{2} AO \\times BE = 20$, that is $4 \\times (6 + OE) = 40$,\\\\\ntherefore $OE = 4$.\\\\\nThus, $E(4, 0)$.\\\\\nSince $GE = 12$,\\\\\nit follows that $GO = 8$.\\\\\nThus, $G(-8, 0)$.\\\\\nLet the equation of EF be $y = \\frac{2}{3}x + b$. Substituting the coordinates of E yields $0 = \\frac{2}{3} \\times 4 + b$,\\\\\nsolving for $b$ gives $b = -\\frac{8}{3}$.\\\\\nTherefore, the equation of line EF is $y = \\frac{2}{3}x - \\frac{8}{3}$.\\\\\nSubstituting $x = -8$ results in $y = \\frac{2}{3} \\times (-8) - \\frac{8}{3} = -8$,\\\\\nthus, $P(-8, -8)$.\\\\\nHence, the coordinates of point $P$ are $P = \\boxed{(-8, -8)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50682487_126.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $l_{1}\\colon y_{1}=-2x+6$ intersects the x-axis at point A and the y-axis at point B. The line $l_{2}$ passes through point C ($-5$, $0$) and intersects line $l_{1}$ at point D ($a$, $8$) and the y-axis at point E. What is the area of $\\triangle BDE$?\n",
    "completion": "Solution: In $y=-2x+6$, let $x=0$, then $y=6$,\\\\\n$\\therefore B(0,6)$,\\\\\nIn $y=2x+10$, let $x=0$, then $y=10$,\\\\\n$\\therefore E(0,10)$,\\\\\n$\\therefore BE=10-6=4$,\\\\\n$\\therefore$ The area of $\\triangle BDE$ is $\\frac{1}{2} \\times 4 \\times 1 = \\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52381542_161.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the vertices A and B of rhombus ABCD are located on the positive y-axis and positive x-axis, respectively, while vertices C and D are located in the first quadrant. $AC$ is parallel to the x-axis, and the graph of the inverse proportion function $y=\\frac{k}{x}$ passes through vertex D. Given $k=18\\sqrt{3}$ and $\\angle ABD=30^\\circ$, what is the length of the sides of rhombus ABCD?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle ABD=30^\\circ$ and $AC\\perp BD$. Let $AE=a$, $BE=\\sqrt{3}a$, and $AB=2a$.\\\\\nThus, $D\\left(a, 2\\sqrt{3}a\\right)$. Since D lies on the inverse proportional function,\\\\\nwe have $2\\sqrt{3}a^2=18\\sqrt{3}$, thus $a=3\\left(a>0\\right)$.\\\\\nTherefore, $AB=2a=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52077009_84.png"
  },
  {
    "solution": "\\boxed{m<\\frac{1}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Synthesis and Exploration: As shown in the figure, in the Cartesian coordinate system, the graph of the quadratic function $y=x^2+bx+c$ passes through point $A\\left(0, -\\frac{7}{4}\\right)$ and point $B\\left(1, \\frac{1}{4}\\right)$. Point $P$ is an arbitrary point on the graph of this function, with the x-coordinate $m$. A line segment $PQ$ is drawn parallel to the x-axis from point $P$, and the x-coordinate of point $Q$ is $-2m+1$. It is known that point $P$ and point $Q$ do not coincide, and the length of segment $PQ$ decreases as $m$ increases. What is the range of values for $m$?\n",
    "completion": "Solution: $PQ=\\left|-3m+1\\right|$, \\\\\nwhen $-3m+1>0$, $PQ=-3m+1$, the length of $PQ$ decreases as $m$ increases, \\\\\nwhen $-3m+1<0$, $PQ=3m-1$, the length of $PQ$ increases as $m$ increases, \\\\\n$\\therefore$ $-3m+1>0$ satisfies the condition, \\\\\nsolving gives $\\boxed{m<\\frac{1}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52719950_150.png"
  },
  {
    "solution": "\\boxed{48^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that lines AB and CD intersect at point O, with OE $\\perp$ CD. If $\\angle AOC = 42^\\circ$, what is the degree measure of $\\angle BOE$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle AOC=42^\\circ$,\\\\\nit follows that $\\angle BOD=42^\\circ$,\\\\\nsince $OE \\perp CD$,\\\\\nit follows that $\\angle BOE=90^\\circ-42^\\circ=\\boxed{48^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52792885_151.png"
  },
  {
    "solution": "\\boxed{63^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, point $D$ is on the extension line of $BC$. If $\\angle ACD = 108^\\circ$ and $\\angle B = 45^\\circ$, then what is the degree measure of $\\angle A$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle ACD$ is an exterior angle of $\\triangle ABC$,\\\\\nit follows that $\\angle ACD=\\angle A+\\angle B$,\\\\\nGiven that $\\angle ACD=108^\\circ$, $\\angle B=45^\\circ$,\\\\\nthus $\\angle A=\\angle ACD−\\angle B=108^\\circ−45^\\circ=\\boxed{63^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53066374.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a rectangular piece of paper ABCD is folded along AE, where the width AB of the paper is 8 cm and the length BC is 10 cm. When folded along AE, the vertex D falls on point F on side BC. What is the length of CE in cm?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rectangle, \\\\\nhence AD = BC = 10cm, DC = AB = 8cm, $\\angle$B = $\\angle$D = $\\angle$C = $90^\\circ$, \\\\\nsince when folded along AE, vertex D falls on point F on side BC, \\\\\nthus AF = AD = 10cm, DE = EF, \\\\\nin $\\triangle ABF$, BF = $\\sqrt{AF^{2}-AB^{2}}$ = $\\sqrt{10^{2}-8^{2}}$ = 6 (cm), \\\\\nhence CF = BC - BF = 10 - 6 = 4 (cm), \\\\\nlet CE = x cm, then DE = EF = (8-x) cm, \\\\\nin $\\triangle CEF$, since $CF^{2}$ + $CE^{2}$ = $EF^{2}$, \\\\\nthus $4^{2}$ + $x^{2}$ = $(8-x)^{2}$, \\\\\nsolving gives x = $\\boxed{3}$, \\\\\nthat is, the length of CE is 3cm.",
    "image_path": "dataset/mm_math/images/MM_Math/53232615.png"
  },
  {
    "solution": "\\boxed{3.5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\triangle ABC$, $ \\angle ACB=90^\\circ$, and $ \\square AEDB$ is constructed outside the triangle with AB as a side, whose diagonals intersect at point F. Given that AE=2 and AB=5, what is the maximum value of CF?\n",
    "completion": "\\textbf{Solution:} As depicted, let O be the midpoint of AB, and draw lines FO and CO.\\\\\n$\\because$ Quadrilateral $AEDB$, AE=2, AB=5,\\\\\n$\\therefore$ BD=2, AO=BO=2.5, AF=DF,\\\\\n$\\therefore$ OF is the median line of $\\triangle ABD$,\\\\\n$\\therefore$ FO=1,\\\\\nAlso, since $\\angle ACB=90^\\circ$,\\\\\n$\\therefore$ OC=2.5,\\\\\nIn $\\triangle FOC$, CF $<$ FO+OC,\\\\\n$\\therefore$ when points F, O, and C are collinear, CF is maximized,\\\\\n$\\therefore$ CFmax=FO+OC=1+2.5=\\boxed{3.5}.",
    "image_path": "dataset/mm_math/images/MM_Math/52076995.png"
  },
  {
    "solution": "\\boxed{3\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In $\\triangle ABC$, $AB=6$, $BC=4$, and $\\angle B$ is an acute angle with $\\cos B=\\frac{1}{2}$. Find the tangent of $\\angle C$.\n",
    "completion": "\\textbf{Solution:} Let $AD \\perp BC$ at $D$, as shown in the figure:\\\\\nSince $\\cos B = \\frac{1}{2}$,\\\\\nit follows that $\\frac{BD}{AB} = \\frac{1}{2}$,\\\\\nGiven $AB = 6$,\\\\\nthen $BD = \\frac{1}{2}AB = 3$,\\\\\nthus $AD = \\sqrt{AB^{2} - BD^{2}} = 3\\sqrt{3}$,\\\\\nGiven $BC = 4$, and $BD = 3$,\\\\\nthen $CD = BC - BD = 1$,\\\\\ntherefore, $\\tan C = \\frac{AD}{CD} = \\frac{3\\sqrt{3}}{1} = \\boxed{3\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51322742_129.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, points $A,B$ lie on circle $O$, and the bisector of $\\angle BAO$ intersects circle $O$ at point $D$. Point $C$ is on the extension of $OA$ such that $\\angle CBA=\\angle D$. $CB$ is a tangent to circle $O$; If $DB\\parallel OA$ and $BD=3$, what is the radius of circle $O$?\n",
    "completion": "\\textbf{Solution:} Method 1:\\\\\nSince $DB\\parallel OA$,\\\\\nit follows that $\\angle OAD=\\angle D$,\\\\\nSince $AD$ is the bisector of $\\angle BAO$,\\\\\nit follows that $\\angle OAD=\\angle BAD$,\\\\\nTherefore $\\angle BAD=\\angle D$, hence $BD=AB=3$,\\\\\nSince $\\angle AOB=2\\angle E=2\\angle D=2\\angle BAD=\\angle OAB$, and $OA=OB$,\\\\\nit follows that $\\triangle AOB$ is an equilateral triangle, hence $OA=AB$,\\\\\nTherefore, $OA=\\boxed{3}$, which means the radius of circle O is 3.\\\\\nMethod 2:\\\\\nConnect $DO$, where $OD=OA$,\\\\\nTherefore $\\angle ODA=\\angle OAD$,\\\\\nMoreover, since $\\angle OAD=\\angle BAD$,\\\\\nTherefore $\\angle ODA=\\angle BAD$,\\\\\nTherefore $OD\\parallel AB$ (since the alternate interior angles are equal, the lines are parallel),\\\\\nAlso, since $DB\\parallel OA$,\\\\\nTherefore the quadrilateral $OABD$ is a parallelogram,\\\\\nHence, $OA=BD=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53019750_85.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, $O$ is the intersection point of the diagonals of $\\square ABCD$, $BE\\perp OC$ at point $E$, $BE$ is extended to point $F$ such that $EF=BE$, and $DF$ is connected. $\\angle F=90^\\circ$. When $\\square ABCD$ is a rectangle, $AC=6$, and $BF= 2\\sqrt{5}$, find the lengths of $DF$ and $CE$.\n",
    "completion": "\\textbf{Solution:} Since $ABCD$ is a rectangle, $BD=AC=6$, thus $OB=\\frac{1}{2}BD=3$. Given $EF=BE$ and $BF=2\\sqrt{5}$, it follows that $BE=\\sqrt{5}$. Since $BE\\perp OC$, $\\angle BEO=90^\\circ$. In $\\triangle BOE$, $OE=\\sqrt{BO^{2}-BE^{2}}=2$,\\\\\nthus $DF=2OE=\\boxed{4}$,\\\\\nand since $CO=\\frac{1}{2}AC=3$,\\\\\nit follows that $CE=CO-OE=3-2=\\boxed{1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52082505_81.png"
  },
  {
    "solution": "\\boxed{3\\sqrt{2}-\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, it is known that square $ABCD$ and isosceles triangle $\\triangle EFG$, with $\\angle EGF=90^\\circ$, points $E$ and $F$ slide along sides $AB$ and $BC$ respectively, and point $G$ is inside the square. The distances from point $G$ to $AB$ and $BC$ are equal. If $AB=4$ and $EF=\\sqrt{10}$, as shown in Figure 2, when point $F$ is the midpoint of side $BC$, what is the length of $DG$?\n",
    "completion": "\\textbf{Solution} Extend $MG$ to intersect $CD$ at point $P$, and connect $DG$,\n\n$\\because$ $GM=GN$,\n\n$\\therefore$ Rectangle $BNGM$ is a square,\n\n$\\therefore MG=GN=BN$,\n\n$\\because$ Point $F$ is the midpoint of side $BC$, $AB=4$,\n\n$\\therefore BF=2$,\n\nLet $FN=t$, then $EM=t$, $MG=GN=2+t$,\n\nIn $\\triangle GFN$, by the Pythagorean theorem, we have $GF^2=t^2+(t+2)^2$.\n\nIn $\\triangle EFG$, by the Pythagorean theorem, we have $EF^2=EG^2+GF^2$,\n\n$\\therefore t^2+(t+2)^2+t^2+(t+2)^2=10$,\n\nSolving this yields $t=-1+\\frac{\\sqrt{6}}{2}$ or $t=-1-\\frac{\\sqrt{6}}{2}$ (discard this solution),\n\n$\\therefore MG=GN=2+t=1+\\frac{\\sqrt{6}}{2}$, $GP=4-MG=3-\\frac{\\sqrt{6}}{2}$,\n\n$\\therefore DP=3-\\frac{\\sqrt{6}}{2}$,\n\nIn $\\triangle DGP$, by the Pythagorean theorem, we get $DG=\\sqrt{2}GP=3\\sqrt{2}-\\sqrt{3}$.\n\nHence, the length of $DG$ is $\\boxed{3\\sqrt{2}-\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52284026_76.png"
  },
  {
    "solution": "\\boxed{y=\\frac{1}{2}x+\\frac{3}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=\\frac{4}{3}x+4$ intersects the $x$-axis and $y$-axis at points $A$ and $B$, respectively. When $\\triangle AOB$ is folded along a line through point $A$, such that point $B$ falls on point $C$ on the positive half of the $x$-axis, and the fold intersects the $y$-axis at point $D$, what is the equation of the line where the fold lies?\n",
    "completion": "\\textbf{Solution:} Let $y=0$, then $\\frac{4}{3}x+4=0$, solving this we get $x=-3$,\\\\\n$\\therefore$ $A(-3, 0)$, $OA=3$,\\\\\nLet $x=0$, then $y=\\frac{4}{3}\\times 0+4=4$,\\\\\n$\\therefore$ $B(0, 4)$, $OB=4$,\\\\\nIn $\\triangle AOB$, $AB=\\sqrt{OA^2+OB^2}=\\sqrt{3^2+4^2}=5$,\\\\\nBy folding, we get: $AC=AB=5$ and $CD=DB$,\\\\\n$\\therefore$ $OC=AC-OA=5-3=2$, $CD=BD=OB-OD=4-OD$,\\\\\nIn $\\triangle COD$, $OD^2+OC^2=CD^2$, that is $OD^2+2^2=(4-OD)^2$,\\\\\nSolving this gives $OD=\\frac{3}{2}$,\\\\\n$\\therefore$ $D\\left(0,\\frac{3}{2}\\right)$,\\\\\nLet the equation of the line containing the crease $AD$ be $y=kx+b$,\\\\\n$\\because$ $A(-3, 0)$, $D\\left(0,\\frac{3}{2}\\right)$,\\\\\n$\\therefore$ $\\begin{cases}-3k+b=0\\\\ b=\\frac{3}{2}\\end{cases}$,\\\\\nSolving gives: $\\begin{cases}k=\\frac{1}{2}\\\\ b=\\frac{3}{2}\\end{cases}$,\\\\\n$\\therefore$ $y=\\frac{1}{2}x+\\frac{3}{2}$\\\\\nHence, the answer is $\\boxed{y=\\frac{1}{2}x+\\frac{3}{2}}$",
    "image_path": "dataset/mm_math/images/MM_Math/50589022.png"
  },
  {
    "solution": "\\boxed{x > 0}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: The graph of the linear function $y=kx+b$ ($k$ and $b$ are constants, and $k\\neq 0$) is shown in the figure. Using the graph, directly write down the solution set of the inequality $kx+b<2$ with respect to $x$.\n",
    "completion": "\\textbf{Solution:} Let \\boxed{x > 0}",
    "image_path": "dataset/mm_math/images/MM_Math/53183204_211.png"
  },
  {
    "solution": "\\boxed{\\frac{13}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, there is a cross-section of a tunnel. If its shape is part of a circle with center $O$, and the road surface $AB=12$ meters, the clear height $CD=9$ meters, then what is the radius $OA$ of this circle?\n",
    "completion": "\\textbf{Solution:} Let the radius of the circle be $r$, thus we have $OA=r$, and $OD=9-r$,\\\\\nsince $AB=12$ meters, and $CD\\perp AB$,\\\\\ntherefore $AD=\\frac{1}{2}AB=\\frac{1}{2}\\times 12=6$ meters,\\\\\nin $\\triangle AOD$,\\\\\nsince $OA=r$, $OD=9-r$, and $AD=6$ meters,\\\\\ntherefore $OA^{2}=OD^{2}+AD^{2}$, which means $r^{2}=(9-r)^{2}+6^{2}$,\\\\\nsolving for $r$ yields $r=\\boxed{\\frac{13}{2}}$ meters.",
    "image_path": "dataset/mm_math/images/MM_Math/52978021.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of $\\odot O$, and chord $CD\\perp AB$ at point $E$. A perpendicular line to $DB$ is drawn through point $C$, intersecting the extension of $AB$ at point $G$, with the foot of the perpendicular being point $F$. Connect $AC$, where $\\angle A=\\angle D$. $AC=CG$; given $CD=EG=8$, what is the radius of $\\odot O$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect $OC$,\\\\\nlet the radius of $\\odot O$ be $r$, thus $OA=OC=r$,\\\\\nsince $CA=CG$, $CD\\perp AB$, $CD=EG=8$,\\\\\ntherefore $AE=EG=8$, $EC=ED=\\frac{1}{2}CD=4$,\\\\\nhence $OE=AE−OA=8−r,$\\\\\nin $\\triangle OEC$, by the Pythagorean theorem, we have $OC^{2}=OE^{2}+EC^{2}$, which is $r^{2}=(8−r)^{2}+4^{2}$,\\\\\nsolving it, we get $r=\\boxed{5}$,\\\\\ntherefore, the radius of $\\odot O$ is $5$.",
    "image_path": "dataset/mm_math/images/MM_Math/53043302_118.png"
  },
  {
    "solution": "\\boxed{y=\\frac{1}{3}x-1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the line $y=2x-1$ intersects the x-axis and y-axis at point A and B, respectively. Given that $\\angle ABC=45^\\circ$, what is the equation of line BC?\n",
    "completion": "\\textbf{Solution:} Since the graph of the linear function $y=2x-1$ intersects the $x$ and $y$ axes at points $A$ and $B$, respectively, \\\\\nlet $x=0$, we get $y=-1$, and let $y=0$, we get $x=\\frac{1}{2}$, \\\\\nthus, point $A\\left(\\frac{1}{2},0\\right)$, point $B\\left(0,-1\\right)$, \\\\\nhence, OA$=\\frac{1}{2}$, OB$=1$, \\\\\nAs shown in the figure, draw $AF\\perp AB$ meeting $BC$ at $F$, and draw $FE\\perp x$ axis at $E$, \\\\\nsince $\\angle ABC=45^\\circ$, \\\\\nthus, $\\triangle ABF$ is an isosceles right triangle, \\\\\ntherefore, AB$=$AF, \\\\\nsince $\\angle OAB+\\angle ABO=\\angle OAB+\\angle EAF=90^\\circ$, \\\\\nthus, $\\angle ABO=\\angle EAF$, \\\\\ntherefore, $\\triangle ABO\\cong\\triangle FAE$ (AAS), \\\\\ntherefore, AE$=$OB$=1$, EF$=$OA$=\\frac{1}{2}$, \\\\\nthus, point $F\\left(\\frac{3}{2},-\\frac{1}{2}\\right)$, \\\\\nLet the equation of the line $BC$ be: $y=kx+b$, \\\\\n$\\left\\{\\begin{array}{l} \\frac{3}{2}k+b=-\\frac{1}{2}\\\\ b=-1\\end{array}\\right.$, solving, we get: $\\left\\{\\begin{array}{l}k=\\frac{1}{3}\\\\ b=-1\\end{array}\\right.$, \\\\\ntherefore, the equation of the line $BC$ is: $\\boxed{y=\\frac{1}{3}x-1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50446779.png"
  },
  {
    "solution": "\\boxed{M\\left(\\frac{10}{3},0\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola intersects the $x$-axis at points $A$ and $B(3,0)$, and intersects the $y$-axis at point $C(0,-3)$, with the vertex of the parabola being $D$. It is known that point $M$ is a moving point on the $x$-axis, and a perpendicular line is drawn from point $M$ to the parabola at point $G$. Is there such a point $M$ that makes the triangle with vertices $A$, $M$, and $G$ similar to $\\triangle BCD$? If such a point exists, find the coordinates of point $M$; if it does not exist, please explain why.\n",
    "completion": "\\textbf{Solution:} Let $y=0$, then $x^{2}-2x-3=0$. Solving this, we find $x_{1}=-1$ and $x_{2}=3$. Therefore, $A(-1,0)$. Since $y=x^{2}-2x-3=(x-1)^{2}-4$, the vertex $D$ of the parabola has coordinates $(1,-4)$. Also, given $C(0,-3)$, $B(3,0)$, and $D(1,-4)$, we have $BD^{2}=2^{2}+4^{2}=20$, $CD^{2}=1^{2}+1^{2}=2$, $BC^{2}=3^{2}+3^{2}=18$. Therefore, $BD^{2}=CD^{2}+BC^{2}$, implying that $\\triangle BDC$ is a right-angled triangle with $\\angle BCD=90^\\circ$. Let the coordinate of point $M$ be $(m,0)$, then the coordinate of point $G$ is $(m,m^{2}-2m-3)$. As shown in the figure, a perpendicular line to the $x$-axis through point $M$ intersects the parabola at point $G$. Therefore, $\\angle AMG=\\angle BCD=90^\\circ$. Thus, for the triangle with vertices $A$, $M$, $G$ to be similar to $\\triangle BCD$, it must satisfy the condition: $\\frac{AM}{MG}=\\frac{BC}{CD}$ or $\\frac{AM}{MG}=\\frac{CD}{BC}$. Since $AM=|-1-m|$ and $MG=|m^{2}-2m-3|$, with $CD=\\sqrt{2}$ and $BC=3\\sqrt{2}$, we have $\\frac{BC}{CD}=3$ and $\\frac{CD}{BC}=\\frac{1}{3}$. Therefore, we get $\\frac{|-1-m|}{|m^{2}-2m-3|}=3$ (1) and $\\frac{|-1-m|}{|m^{2}-2m-3|}=\\frac{1}{3}$ (2). Solving (1), we find $m=\\boxed{\\frac{8}{3}}$, or $m=\\boxed{\\frac{10}{3}}$, or $m=-1$ (discard). Solving (2) yields $m=0$, or $m=6$, or $m=-1$ (discard). Therefore, \\boxed{M(0,0)}, \\boxed{M(6,0)}, \\boxed{M\\left(\\frac{8}{3},0\\right)}, or \\boxed{M\\left(\\frac{10}{3},0\\right)}.",
    "image_path": "dataset/mm_math/images/MM_Math/52733221_72.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, $AC<BC$, and $CD$ bisects $\\angle ACB$ and intersects $AB$ at point $D$. A circle $\\odot O$ is drawn with $DB$ as the diameter, which intersects $CD$ and $BC$ at points $E$ and $F$ respectively. Connect $BE$ and $EF$. What is the degree measure of $\\angle EBF$? If $DE=DC$ and $BC=8$, what is the length of $EF$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect DF and construct EG$\\perp$BF at point G,\\\\\n$\\because$ $\\angle ACB=90^\\circ$ and CD bisects $\\angle ACB$,\\\\\n$\\therefore$ $\\angle DCB= \\frac{1}{2} \\angle ACB=45^\\circ$,\\\\\n$\\because$ BD is the diameter of circle $O$,\\\\\n$\\therefore$ $\\angle DFB=\\angle DEB=90^\\circ$,\\\\\n$\\because$ $\\angle CFD=180^\\circ-\\angle DFB=90^\\circ$,\\\\\n$\\therefore$ $\\angle FDC=\\angle FCD=45^\\circ$,\\\\\n$\\therefore$ $\\angle EBF=180^\\circ-\\angle EDF=\\angle FDC=45^\\circ$;\\\\\n$\\because$ $\\angle CEB=90^\\circ$ and $\\angle EBC=\\angle ECB=45^\\circ$, BC=8, DE=DC,\\\\\n$\\therefore$ CE=BE,\\\\\n$\\therefore$ CG=BG= $\\frac{1}{2}$ BC=4,\\\\\n$\\therefore$ EG= $\\frac{1}{2}$ BC=4,\\\\\n$\\because$ $\\angle CFD=\\angle CGE=90^\\circ$,\\\\\n$\\therefore$ DF$\\parallel$EG,\\\\\n$\\therefore$ $\\frac{FG}{FC}=\\frac{DE}{DC} =1$,\\\\\n$\\therefore$ FG=FC= $\\frac{1}{2}$ CG=2,\\\\\n$\\therefore$ EF= $\\sqrt{EG^2+FG^2}=\\sqrt{4^2+2^2}=\\boxed{2\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52755081.png"
  },
  {
    "solution": "\\boxed{15^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $O$ is a point on the line $AB$, $\\angle COD$ is a right angle, $OE$ bisects $\\angle BOC$. If $\\angle AOC = 30^\\circ$, what is the degree measure of $\\angle DOE$?\n",
    "completion": "\\textbf{Solution:}\\\\\n$\\because \\angle AOC=30^\\circ$,\\\\\n$\\therefore \\angle BOC=180^\\circ-30^\\circ=150^\\circ$,\\\\\n$\\because$ OE bisects $\\angle BOC$,\\\\\n$\\therefore \\angle COE= \\frac{1}{2} \\angle BOC=75^\\circ$,\\\\\n$\\because \\angle COD$ is a right angle,\\\\\n$\\therefore \\angle COD=90^\\circ$,\\\\\n$\\therefore \\angle DOE=90^\\circ-75^\\circ=\\boxed{15^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51317331_60.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the length of segment AB is 12, and point C is the midpoint of segment AB, while point D is the midpoint of segment BC. If there is a point E on segment AB such that $CE = \\frac{1}{3}BC$, calculate the length of AE.\n",
    "completion": "\\textbf{Solution:} Given that $BC=6$, and $CE=\\frac{1}{3}BC$, \\\\\nthus $CE=\\frac{1}{3}\\times 6=2$, \\\\\nwhen E is to the left of C, $AE=AC-CE=6-2=4$; \\\\\nwhen E is to the right of C, $AE=AC+CE=6+2=8$. \\\\\nTherefore, the length of $AE$ is $\\boxed{4}$ or $\\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52732986_66.png"
  },
  {
    "solution": "\\boxed{85^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $OB$ is the bisector of $\\angle AOC$, and $OD$ is the bisector of $\\angle COE$. If $\\angle AOB=50^\\circ$ and $\\angle DOE=35^\\circ$, then what is the degree measure of $\\angle BOD$?\n",
    "completion": "\\textbf{Solution:} Since OB is the bisector of $\\angle AOC$, and OD is the bisector of $\\angle COE$, \\\\\ntherefore $\\angle COD = \\angle DOE = 35^\\circ$, $\\angle COB = \\angle BOA = 50^\\circ$. \\\\\nThus, $\\angle BOD = \\angle COD + \\angle COB = \\boxed{85^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53119839_133.png"
  },
  {
    "solution": "\\boxed{80^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in acute-angled triangle $ABC$, we have $AB=AC$. Circle $O$ with diameter $BC$ intersects $AB$ and $AC$ at points $D$ and $E$ respectively. Connect $OD, OE$, and $DE$. If $\\angle A=50^\\circ$, what is the degree measure of $\\overset{\\frown}{DE}$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $BE$.\\\\\nSince $BC$ is the diameter,\\\\\nit follows that $\\angle BEC=90^\\circ$,\\\\\nthus $\\angle AEB=90^\\circ$,\\\\\nSince $\\angle BAC=50^\\circ$,\\\\\nit follows that $\\angle ABE=90^\\circ-50^\\circ=40^\\circ$,\\\\\nthus $\\angle DOE=2\\angle DBE=80^\\circ$,\\\\\nTherefore, the degree of $\\overset{\\frown}{DE}$ is $\\boxed{80^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52872091_112.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, on a billiard table, a ball is at point A and is to be shot from point A, reflected off the side cushion CD to hit ball B. Draw AC$\\perp$ CD at point C, and BD$\\perp$ CD at point D. Given that $\\angle AEC=\\angle BED$, with AC = 10cm, BD = 15cm, CD = 20cm, if the player is able to hit ball B, what is the length of DE?\n",
    "completion": "\\textbf{Solution}: Since $AC\\perp CD$ and $BD\\perp CD$,\\\\\nit follows that $\\angle ACE=\\angle BDE=90^\\circ$,\\\\\nSince $\\angle AEC=\\angle BED$,\\\\\nit implies that $\\triangle AEC\\sim\\triangle BED$,\\\\\nTherefore, $\\frac{AC}{BD}=\\frac{CE}{DE}$,\\\\\nGiven that $AC=10\\text{cm}$, $BD=15\\text{cm}$,\\\\\nit leads to $\\frac{10}{15}=\\frac{20-DE}{DE}$,\\\\\nwhich results in $DE=\\boxed{12}\\text{cm}$,\\\\\nHence, the length of $DE$ is $12\\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52514043.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In triangles $AOB$ and $COD$, $\\angle AOB = \\angle COD$ and it is known that $\\angle AOB = 90^\\circ$. When the two triangles are put together as shown in the figure, and $\\angle AOC = 2\\angle BOD$, what is the degree measure of $\\angle BOD$?\n",
    "completion": "\\textbf{Solution:}\\\\\n$\\because \\angle AOC = \\angle AOB + \\angle BOC = 90^\\circ + \\angle BOC$,\\\\\n$\\angle DOB = \\angle DOC - \\angle BOC = 90^\\circ - \\angle BOC$,\\\\\n$\\therefore \\angle AOC + \\angle DOB = 90^\\circ + \\angle BOC + 90^\\circ - \\angle BOC = 180^\\circ$,\\\\\n$\\because \\angle AOC = 2\\angle BOD$,\\\\\n$\\therefore \\angle BOD = \\boxed{60^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/17008253_58.png"
  },
  {
    "solution": "\\boxed{\\left(3,4\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the rectangle $OABC$, point $A$ is on the y-axis, and point $C$ is on the x-axis. We have $OA=BC=4$, and $AB=OC=8$. Point $D$ is on side $AB$, and point $E$ is on side $OC$. The rectangle is folded along the straight line $DE$ so that point $B$ coincides with point $O$. What are the coordinates of point $D$?\n",
    "completion": "\\textbf{Solution:} Let AD = x. By the property of folding, we know that OD = BD = 8 - x,\\\\\nIn right triangle $\\triangle OAD$,\\\\\n$\\because$ OA$^{2}$ + AD$^{2}$ = OD$^{2}$,\\\\\n$\\therefore$ $4^{2}$ + x$^{2}$ = (8 − x)$^{2}$,\\\\\n$\\therefore$ x = 3,\\\\\n$\\therefore$ D$(3,4)$,\\\\\nTherefore, the answer is: D$\\boxed{\\left(3,4\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51378936.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rhombus $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$. $E$ is the midpoint of $AD$, and points $F$, $G$ are on $CD$, with $EF\\perp CD$ and $OG\\parallel EF$. Quadrilateral $OEFG$ is a rectangle; if $AD=10$ and $EF=3$, find the lengths of $OE$ and $CG$.\n",
    "completion": "Solution: Since quadrilateral ABCD is a rhombus, \\\\\n$\\therefore$ $\\angle AOD=90^\\circ$, AD=CD=10,\\\\\nalso $\\because$ point E is the midpoint of AD,\\\\\n$\\therefore$ $OE=DE=\\frac{1}{2}AD=5$,\\\\\n$\\therefore$ $GF=OE=5$,\\\\\nIn $\\triangle DEF$, $\\angle EFD=90^\\circ$,\\\\\n$\\therefore$ $DF=\\sqrt{DE^2-EF^2}=\\sqrt{5^2-3^2}=4$,\\\\\n$\\therefore$ $CG=CD-GF-DF=10-5-4=\\boxed{1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52386487_121.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a hexagon $ABCDEF$ has equal interior angles, and $AD\\parallel BC$. What is the measure of $\\angle DAB$ in degrees?\n",
    "completion": "\\textbf{Solution}: Since the internal angles of the hexagon ABCDEF are all equal,\\\\\neach internal angle is: $(6-2) \\times 180^\\circ \\div 6 = 120^\\circ$,\\\\\nthat is, $\\angle ABC = 120^\\circ$,\\\\\nsince BC$\\parallel$AD,\\\\\nthus $\\angle DAB = 180^\\circ - \\angle ABC = \\boxed{60^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52345519.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that line $l_{1} \\parallel l_{2} \\parallel l_{3}$. Line AC intersects lines $l_{1}$, $l_{2}$, and $l_{3}$ at points A, B, and C, respectively. Line DF intersects lines $l_{1}$, $l_{2}$, and $l_{3}$ at points D, E, and F, respectively. Lines AC and DF intersect at point O. If BC = 2AO = 2OB and OD = 1, then what is the length of OF?\n",
    "completion": "Solution: $\\because BC = 2AO = 2OB$,\\\\\n$\\therefore OC = 3AO$,\\\\\n$\\because$ line $l_1 \\parallel l_2 \\parallel l_3$,\\\\\n$\\therefore \\frac{AO}{OC} = \\frac{OD}{OF}$,\\\\\n$\\therefore \\frac{AO}{OC} = \\frac{OD}{OF} = \\frac{1}{3}$,\\\\\n$\\because OD = 1$,\\\\\n$\\therefore OF = \\boxed{3}$",
    "image_path": "dataset/mm_math/images/MM_Math/53110858.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{4}{3}x^2-\\frac{8}{3}x+4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure (1), it is known that the line $y = \\frac{4}{3}x + 4$ intersects the x-axis at point A and the y-axis at point C. A parabola $y = ax^2 + bx + c$ passes through points A and C, and intersects the x-axis at another point, B, with its axis of symmetry being the line $x = -1$. Let D be a moving point on the parabola in the second quadrant with its x-coordinate being $m$. What is the equation of the parabola?\n",
    "completion": "\\textbf{Solution:} Substituting $x=0$ into $y=\\frac{4}{3}x+4$, we get $y=4$.\\\\\n$\\therefore$ The coordinates of point C are $(0,4)$.\\\\\nSubstituting $y=0$ into $y=\\frac{4}{3}x+4$, we get $x=-3$.\\\\\n$\\therefore$ The coordinates of point A are $(-3,0)$.\\\\\n$\\because$ The axis of symmetry for the parabola is the line $x=-1$,\\\\\n$\\therefore$ $-\\frac{b}{2a}=-1$, that is, $b=2a$.\\\\\nForming a system of equations from the given information, we get\\\\\n$\\begin{cases}\n9a-3b+c=0,\\\\\nb=2a,\\\\\nc=4.\n\\end{cases}$ Solving this system, we find $\\begin{cases}\na=-\\frac{4}{3},\\\\\nb=-\\frac{8}{3},\\\\\nc=4.\n\\end{cases}$\\\\\n$\\therefore$ The equation of the parabola is $y=-\\frac{4}{3}x^2-\\frac{8}{3}x+4$.\\\\\nHence, the expression for the parabola is $\\boxed{y=-\\frac{4}{3}x^2-\\frac{8}{3}x+4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53043766_125.png"
  },
  {
    "solution": "\\boxed{y=-2x+6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line ${l}_{1}\\colon y=x+3$ intersects with the line ${l}_{2}$, which passes through point $A(3, 0)$, at point $C(1, m)$, and intersects with the x-axis at point B. What is the equation of line ${l}_{2}$?\n",
    "completion": "\\textbf{Solution:} Let's substitute the point $C\\left(1, m\\right)$ into the line $l_{1}: y=x+3$, we get: $m=1+3=4$,\n\n$\\therefore$ the point $C\\left(1, 4\\right)$,\n\nLet's assume the equation of the line $l_{2}$ is: $y=kx+b\\left(k\\ne 0\\right)$,\n\nSubstituting points $A\\left(3, 0\\right)$ and $C\\left(1, 4\\right)$,\n\nWe obtain $\\begin{cases}3k+b=0\\\\ k+b=4\\end{cases}$,\n\nSolving these, we find $\\begin{cases}k=-2\\\\ b=6\\end{cases}$,\n\n$\\therefore$ the equation of $l_{2}$: \\boxed{y=-2x+6}.",
    "image_path": "dataset/mm_math/images/MM_Math/52425281_213.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{2}-2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, E is a point on the semicircle with center O, C is the midpoint of the arc $\\overset{\\frown}{BE}$, and the diameter $AB$ is parallel to chord $DC$, intersecting $AE$ at point $F$. It is given that $CF=AF$. When $AB=4$ and $OE \\perp CD$, find the value of $EF$.\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect OC, OE.\\\\\n$\\because$ OE$\\perp$CD, $\\therefore$ $\\overset{\\frown}{AE}=\\overset{\\frown}{BE}$, $\\therefore$ $\\angle AOE=\\angle BOE=90^\\circ$.\\\\\n$\\because$ AB=4, $\\therefore$ AO=EO=2, $\\therefore$ AE= $2\\sqrt{2}$, $\\angle OAC=\\angle OCA$. By (1) we have $\\angle DCA=\\angle EAC=\\angle CAB$, $\\therefore$ $\\angle FCA=\\angle FAC=\\angle OAC=\\angle OCA$. Also, $\\because$ AC=AC, $\\therefore$ $\\triangle AOC\\cong \\triangle AFC$, $\\therefore$ AF=AO=2, $\\therefore$ EF=AE−AF= $2\\sqrt{2}-2=\\boxed{2\\sqrt{2}-2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52733131_140.png"
  },
  {
    "solution": "\\boxed{20}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point \\(C\\) divides the line segment \\(AB\\) into two parts in the left-to-right order, in the ratio \\(2:3\\). Point \\(D\\) is the midpoint of \\(AB\\). If \\(CD=2\\), what is the length of the line segment \\(AB\\)?\n",
    "completion": "\\textbf{Solution}: Let $AC = 2x$, then $BC = 3x$, \\\\\n$\\therefore AB = AC + BC = 5x$, \\\\\n$\\because$ point D is the midpoint of AB, \\\\\n$\\therefore AD = \\frac{1}{2}AB = 2.5x$, \\\\\n$\\therefore CD = AD - AC = 2.5x - 2x = 0.5x$, \\\\\n$\\because CD = 2$, \\\\\n$\\therefore 0.5x = 2$, \\\\\n$\\therefore x = 4$, \\\\\n$\\therefore AB = 5x = \\boxed{20}$,",
    "image_path": "dataset/mm_math/images/MM_Math/51438558.png"
  },
  {
    "solution": "\\boxed{\\frac{7}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated, four small squares with side length $b$ are placed around a square with side length $a$ named $EFGH$, forming a larger square $ABCD$. The shaded parts of the figure form an icon resembling a \"red badge\". Let the area of the shaded part be denoted as $S_1$, and the area of each small square with side length $b$ be denoted as $S_2$. If $S_1 = 6S_2$, what is the value of $\\frac{a}{b}$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, we label the required intersections with letters:\\\\\n$S_{\\triangle DGI}=\\frac{1}{2}(a+b)\\times b=\\frac{1}{2}ab+\\frac{1}{2}b^{2}$,\\\\\n$S_{\\triangle KMD}=S_{ABCD}-S_{\\triangle DMC}-S_{\\triangle DKA}-S_{\\triangle KBM}$\\\\\n$=(a+2b)^{2}-\\frac{1}{2}(a+b)(a+2b)-\\frac{1}{2}(a+2b)(a+b)-\\frac{1}{2}b^{2}$\\\\\n$=ab+\\frac{3}{2}b^{2}$,\\\\\n$S_{\\triangle MNC}=\\frac{1}{2}(a+b)\\times b=\\frac{1}{2}ab+\\frac{1}{2}b^{2}$,\\\\\nTherefore, $S_{1}=S_{\\triangle DGI}+S_{\\triangle KMD}+S_{\\triangle MNC}=2ab+\\frac{5}{2}b^{2}$,\\\\\n$S_{2}=b^{2}$,\\\\\nSince $S_{1}=6S_{2}$,\\\\\ntherefore $2ab+\\frac{5}{2}b^{2}=6b^{2}$,\\\\\nSimplifying yields: $2a=\\frac{7}{2}b$,\\\\\nTherefore, $\\frac{a}{b}=\\boxed{\\frac{7}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51534305.png"
  },
  {
    "solution": "\\boxed{40^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, it is known that $\\angle AOB = 120^\\circ$, ray $OC$ is the angle bisector of $\\angle AOB$, point $D$ is a point inside $\\angle BOC$, and point $D$ does not lie on the bisector of $\\angle BOC$. When $\\angle BOD = 20^\\circ$, calculate the degree measure of $\\angle COD$?\n",
    "completion": "\\textbf{Solution:} Since the ray $OC$ is the angle bisector of $\\angle AOB$ and $\\angle AOB=120^\\circ$,\\\\\nit follows that $\\angle BOC=\\frac{1}{2}\\angle AOB=\\frac{1}{2}\\times 120^\\circ=60^\\circ$,\\\\\ntherefore $\\angle COD=\\angle BOC-\\angle BOD=60^\\circ-20^\\circ=\\boxed{40^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53253309_169.png"
  },
  {
    "solution": "\\boxed{\\frac{5\\sqrt{2}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in a parallelogram $ABCD$ made of paper, $\\angle BAD=45^\\circ$, with $AB=10$. The paper is folded such that the corresponding point $A'$ of point $A$ lies on the side $BC$, and the fold line $EF$ intersects $AB$, $AD$, and $AA'$ at points $E$, $F$, and $G$, respectively. Continue to fold the paper so that the corresponding point $C'$ of point $C$ falls on $A'F$, and connect $GC'$. What is the distance from point $G$ to $AD$? What is the minimum value of $GC'$?\n",
    "completion": "\\textbf{Solution:} Construct $BH \\perp AD$ at $H$, and $GP \\perp AD$ at $P$, with $GQ \\perp A'F$ at $Q$, and draw $A'R \\perp AD$ at $R$, as shown in the diagram: \\\\\nSince $\\angle BAD = 45^\\circ$ and $AB = 10$, \\\\\nit follows that $BH = \\frac{\\sqrt{2}}{2}AB = 5\\sqrt{2}$, \\\\\nSince quadrilateral $ABCD$ is a parallelogram, and $BH \\perp AD$, $A'R \\perp AD$, \\\\\nit follows that quadrilateral $BHR{A'}$ is a rectangle, \\\\\nthus, $A'R = BH = 5\\sqrt{2}$, \\\\\nSince $GP \\perp AD$ and $A'R \\perp AD$, \\\\\nit follows that $GP \\parallel A'R$, \\\\\nSince the paper is folded so that point $A$ maps to $A'$ on side $BC$, with fold line $EF$, \\\\\nit follows that $AG = A'G$ and $\\angle AFE = \\angle A'FE$, \\\\\ntherefore, $GP$ is the median of $\\triangle A'AR$, \\\\\nthus, $GP = \\frac{1}{2}A'R = \\frac{5\\sqrt{2}}{2}$, \\\\\nSince $GP \\perp AD$ and $GQ \\perp A'F$, \\\\\nit follows that $GP = GQ = \\frac{5\\sqrt{2}}{2}$, \\\\\nSince the paper is folded so that point $C$ maps to $C'$ on $A'F$, \\\\\nit follows that when $C'$ coincides with $Q$, $GC'$ is minimized, with the minimum value being the length of $GQ$, \\\\\nthus, the minimum value of $GC'$ is $\\boxed{\\frac{5\\sqrt{2}}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53490733.png"
  },
  {
    "solution": "\\boxed{\\frac{7}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, $AC=4$, and $\\tan A=\\frac{3}{4}$. A circle with point $C$ as the center and the length of $CB$ as the radius intersects $AB$ at point $D$. What is the length of $AD$?\n",
    "completion": "\\textbf{Solution:} Connect CD, and through point C draw $CE \\perp AB$ at point E,\\\\\nWithin $\\triangle ABC$, $\\tan \\angle A = \\frac{BC}{AC} = \\frac{3}{4}$, solve it to get: $BC=3$,\\\\\n$\\therefore AB = \\sqrt{AC^2 + BC^2} = \\sqrt{3^2 + 4^2} = 5$,\\\\\n$\\because S_{\\triangle ABC} = \\frac{1}{2}AC \\cdot BC = \\frac{1}{2}AB \\cdot CE$,\\\\\n$\\therefore 3 \\times 4 = 5CE$, solve it to get: $CE = \\frac{12}{5}$,\\\\\n$\\therefore BE = \\sqrt{BC^2 - CE^2} = \\sqrt{3^2 - \\left(\\frac{12}{5}\\right)^2} = \\frac{9}{5}$,\\\\\n$\\therefore BD = 2BE = \\frac{18}{5}$,\\\\\n$\\therefore AD = AB - BD = 5 - \\frac{18}{5} = \\boxed{\\frac{7}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51284005.png"
  },
  {
    "solution": "\\boxed{\\frac{11}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $l_{1}\\parallel l_{2}\\parallel l_{3}$. If $AB=2$, $BC=3$, $AD=1$, and $CF=4$, what is the length of $BE$?\n",
    "completion": "\\textbf{Solution:} Construct line $DN \\parallel AC$ through point D to intersect $BE$ at point $M$, and $CF$ at point $N$;\\\\\n$\\therefore$ Quadrilaterals $ABMD$ and $BCNM$ are both parallelograms\\\\\n$\\therefore$ $AB=DM=2$,\\\\\n$BC=MN=3$,\\\\\n$AD=BM=CN=1$\\\\\nFrom the problem, we know\\\\\n$\\frac{AB}{BC}=\\frac{DE}{EF}=\\frac{2}{3}$\\\\\n$\\because BE\\parallel CF$\\\\\n$\\therefore$ $\\triangle DME\\sim \\triangle DNF$\\\\\n$\\therefore$ $\\frac{ME}{NF}=\\frac{DE}{DF}=\\frac{DM}{DN}=\\frac{DM}{DM+MN}=\\frac{2}{5}$\\\\\n$\\because NF=CF-CN=4-1=3$\\\\\n$\\therefore$ $ME=\\frac{6}{5}$\\\\\n$\\therefore$ $BE=BM+ME=1+\\frac{6}{5}=\\boxed{\\frac{11}{5}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51433277.png"
  },
  {
    "solution": "\\boxed{130^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of the semicircle $O$, and $C$ is a point on the semicircle $O$. Connect $AC$ and $CB$, where $D$ is a point on the arc $\\widehat{BC}$, and connect $CD$ and $BD$. If $\\angle ABC=40^\\circ$, what is the degree measure of $\\angle D$?\n",
    "completion": "\\textbf{Solution:} Given that $AB$ is the diameter of semicircle O,\\\\\nit follows that $\\angle ACB=90^\\circ$,\\\\\nsince $\\angle ABC=40^\\circ$,\\\\\nit follows that $\\angle CAB=50^\\circ$,\\\\\nsince the quadrilateral ABCD is inscribed in semicircle O,\\\\\nit follows that $\\angle D=180^\\circ-50^\\circ=\\boxed{130^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51352039.png"
  },
  {
    "solution": "\\boxed{(-4, 8)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the Cartesian coordinate system $xOy$, given $A\\left(0, 4\\right)$ and $B\\left(4, 0\\right)$, the graph of the linear function $y=-2x$ intersects with line $AB$ at point $P$. Find the coordinates of point $P$.\n",
    "completion": "\\textbf{Solution:} According to the problem, let the equation of line $AB$ be: $y=kx+m$. Also, since $A\\left(0, 4\\right)$ and $B\\left(4, 0\\right)$ are on line $AB$, it follows that $m=4$, and $4k+4=0$. Solving this gives $k=-1$. Therefore, the equation of line $AB$ is $y=-x+4$. By solving the system of equations, we have $\\begin{cases}y=-x+4\\\\ y=-2x\\end{cases}$, which yields $\\begin{cases}x=-4\\\\ y=8\\end{cases}$. Therefore, the coordinates of point $P$ are $\\boxed{(-4, 8)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50560455_69.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ is an equilateral triangle. Point E is the midpoint of AC. A line through point E is perpendicular to AB at point F. The line BC is extended to intersect the extension of EF at point D. If EF = 1, what is the length of DF?\n",
    "completion": "\\textbf{Solution:} Connect BE, \\\\\nsince $\\triangle ABC$ is an equilateral triangle and point E is the midpoint of AC, \\\\\nit follows that $\\angle ABC=60^\\circ$, $\\angle ABE=\\angle CBE=30^\\circ$, \\\\\nsince EF$\\perp$AB, \\\\\nit follows that $\\angle D=90^\\circ-\\angle ABC=30^\\circ$, that is, $\\angle D=\\angle CBE=30^\\circ$, \\\\\nthus BE=DE, \\\\\nin $\\triangle BEF$, EF=1, \\\\\nhence BE=2EF=2, \\\\\nthus BE=DE=2, \\\\\ntherefore DF=EF+DE=\\boxed{3},",
    "image_path": "dataset/mm_math/images/MM_Math/51434958.png"
  },
  {
    "solution": "\\boxed{4\\,cm}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $BC = \\frac{1}{2}AB$, D is the midpoint of $AC$, and $DC = 3\\,cm$. What is the length of $AB$?\n",
    "completion": "\\textbf{Solution:} Since $D$ is the midpoint of $AC$ and $DC=3\\,cm$,\\\\\n$\\therefore AC=6\\,cm$,\\\\\nand since $BC=\\frac{1}{2}AB$,\\\\\n$\\therefore BC=\\frac{1}{3}AC=\\frac{1}{3}\\times 6=2\\,cm$,\\\\\n$\\therefore AB=AC-BC=6-2=\\boxed{4\\,cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51212587.png"
  },
  {
    "solution": "\\boxed{x \\geq 2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $l_{1}: y=kx-1$ intersects with the line $l_{2}: y=-x+5$ at point $P(2, m)$. What is the solution set of the inequality $kx-1 \\ge -x+5$ with respect to $x$?\n",
    "completion": "\\textbf{Solution:} Since the graph of the function shows that when $x \\geq 2$, the line $l_1\\colon y=kx-1$ is not below the line $l_2\\colon y=-x+5$, \\\\\ntherefore, the solution set for the inequality in terms of $x$, $kx-1 \\geq -x+5$, is $\\boxed{x \\geq 2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52386507.png"
  },
  {
    "solution": "\\boxed{152^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $O$ is a point on the line $MN$, $\\angle AOB$ is a right angle, and $OC$ bisects $\\angle MOB$. If $OD$ bisects $\\angle CON$, and $\\angle DON - \\angle AOM = 21^\\circ$, what is the measure of $\\angle BON$ in degrees?\n",
    "completion": "\\textbf{Solution:} Since $OC$ bisects $\\angle MOB$,\\\\\n$\\therefore \\angle BOC = \\angle MOC$,\\\\\nSince $OD$ bisects $\\angle CON$ and $\\angle AOB$ is a right angle,\\\\\n$\\therefore \\angle COD = \\angle NOD = \\frac{1}{2}\\left(180^\\circ - \\angle BOC\\right)$,\\\\\n$\\angle AOB = 90^\\circ$,\\\\\nSince $\\angle AOM = 90^\\circ - 2\\angle BOC$,\\\\\nSince $\\angle DON - \\angle AOM = 21^\\circ$,\\\\\n$\\therefore \\left\\{\\begin{array}{l}\n\\frac{1}{2}\\angle BOC + \\angle AOM = 69^\\circ \\\\\n\\angle AOM + 2\\angle BOC = 90^\\circ\n\\end{array}\\right.$,\\\\\nSolving yields $\\left\\{\\begin{array}{l}\n\\angle AOM = 62^\\circ \\\\\n\\angle BOC = 14^\\circ\n\\end{array}\\right.$,\\\\\n$\\therefore \\angle BON = 180^\\circ - 2\\angle BOC = 180^\\circ - 28^\\circ = \\boxed{152^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50083331.png"
  },
  {
    "solution": "\\boxed{y = -\\frac{3}{5}x + 3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, point B is in the first quadrant. $BA \\perp x$-axis at A, $BC \\perp y$-axis at C, $BA = 3$, and $BC = 5$. There is an inverse proportion function graph that exactly passes through point B. Find the expressions of the inverse proportion function that passes through point B and the linear function that passes through points A and C, respectively.\n",
    "completion": "\\textbf{Solution:} Given $\\because$ BA $\\perp$ x-axis at A, and BC $\\perp$ y-axis at C, with BA $= 3$ and BC $=5$, \\\\\n$\\therefore$ the coordinates of point B are $(5,3)$, point A $(5,0)$, and point C $(0,3)$.\\\\\nLet the inverse proportional function be $y = \\frac{k}{x}$ ($x \\neq 0$),\\\\\n$\\because$ point B lies on the graph of the inverse proportional function,\\\\\n$\\therefore k=3 \\times 5 = 15$\\\\\n$\\therefore$ the equation of the inverse proportional function is $y = \\frac{15}{x}$;\\\\\nLet the linear function be $y = ax + b$,\\\\\nSubstituting A$(5, 0)$ and C$(0, 3)$, we get\\\\\n$\\left\\{\\begin{array}{l}\nb=3\\\\\n5a+b=0\n\\end{array}\\right.$\\\\\nSolving gives:\\\\\n$\\left\\{\\begin{array}{l}\na=-\\frac{3}{5}\\\\\nb=3\n\\end{array}\\right.$\\\\\n$\\therefore$ the equation of the linear function is $\\boxed{y = -\\frac{3}{5}x + 3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52160945_160.png"
  },
  {
    "solution": "\\boxed{15}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In square $ABCD$, $E$ and $F$ are points on sides $AD$ and $CD$ respectively, where $AE=ED$ and $DF=\\frac{1}{4}DC$. Extend line $EF$ to intersect the extension of $BC$ at point $G$. $\\triangle ABE\\sim \\triangle DEF$; if the side length of the square is $6$, what is the length of $BG$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral ABCD is a square,\\\\\nit follows that $ED\\parallel BG$,\\\\\ntherefore $\\angle DEF=\\angle G$,\\\\\nsince $\\angle DFE=\\angle GFC$, $\\triangle EDF\\sim \\triangle GCF$,\\\\\nit follows that $\\frac{ED}{CG}=\\frac{DF}{CF}$,\\\\\nalso, given that $DF=\\frac{1}{4}DC$ and the side length of the square is 6,\\\\\nit follows that $ED=3$, $\\frac{ED}{CG}=\\frac{DF}{CF}=\\frac{1}{3}$,\\\\\ntherefore $CG=9$,\\\\\nthus $BG=BC+CG=\\boxed{15}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52793383_158.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the length of line segment $AD$ is 30 cm, and points $C$ and $B$ are both on line segment $AD$. Point $E$ is the midpoint of $AB$. The length of $AE$ is 12 cm. If $AC = \\frac{1}{3} AD$, and point $F$ is the midpoint of line segment $CD$, find the length of line segment $EF$.\n",
    "completion": "\\textbf{Solution:} As shown in the figure,\\\\\n$\\because AC= \\frac{1}{3} AD= \\frac{1}{3}\\times30=10\\,\\text{cm}$,\\\\\n$\\therefore CD=AD-AC=30-10=20\\,\\text{cm}$,\\\\\n$\\because$ Point F is the midpoint of line segment CD,\\\\\n$\\therefore DF=\\frac{1}{2}CD=10\\,\\text{cm}$,\\\\\n$\\therefore EF=AD-AE-DF=30-12-10=\\boxed{8}\\,\\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51445267_108.png"
  },
  {
    "solution": "\\boxed{60}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the cross section of the dam is a trapezoid with the top width of the dam $DC = 10m$, the height of the dam $15m$, the slope of the incline $AD$ having a gradient ${l}_{1} = 1 : 2$, and the slope of the incline $BC$ having a gradient ${l}_{2} = 3 : 4$. What is the width $AB$ at the bottom of the slope in meters?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, draw $DE \\perp AB$ at point E through point D, and draw $CF \\perp AB$ at point F through point C,\\\\\nthen $DE = CF = 15\\,m$, and quadrilateral $DEFC$ is a rectangle,\\\\\n$\\therefore EF = DC = 10\\,m$,\\\\\n$\\because$ the slope of ramp $AD$ has a gradient $l_{1} = 1\\colon 2$, and the slope of ramp $BC$ has a gradient $l_{2} = 3\\colon 4$,\\\\\n$\\therefore \\frac{DE}{AE} = \\frac{1}{2}$, $\\frac{CF}{BF} = \\frac{3}{4}$, that is, $\\frac{15}{AE} = \\frac{1}{2}$, $\\frac{15}{BF} = \\frac{3}{4}$,\\\\\nsolving these, we get $AE = 30\\,m$, $BF = 20\\,m$,\\\\\ntherefore, the width at the bottom of the slope $AB = AE + EF + BF = 30 + 10 + 20 = \\boxed{60}\\,m$.",
    "image_path": "dataset/mm_math/images/MM_Math/51318167.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\triangle ABC$, $E$ and $F$ are located on sides $AB$ and $AC$ respectively, with $EF\\parallel BC$. Given that $AB=3$, $AE=2$, and $EF=4$, what is the length of $BC$?\n",
    "completion": "\\textbf{Solution:} Since $EF\\parallel BC$,\\\\\nthus $\\frac{EF}{BC} = \\frac{AE}{AB}$, so $\\frac{4}{BC} = \\frac{2}{3}$\\\\\nSolving this yields: $BC=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51369877.png"
  },
  {
    "solution": "\\boxed{\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, P is a point on the diagonal AC of the rhombus ABCD, $PE\\perp AB$ at point E, and $PF\\perp AD$ at point F. Given that $\\angle BAD=60^\\circ$, $PE=1$, find the length of AE.\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rhombus and $\\angle BAD=60^\\circ$,\\\\\nit follows that $\\angle PAE=30^\\circ$\\\\\nSince $PE \\perp AB$,\\\\\nit follows that $AP=2PE=2$,\\\\\ntherefore $AE=\\sqrt{AP^{2}-PE^{2}}=\\boxed{\\sqrt{3}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53436646_123.png"
  },
  {
    "solution": "\\boxed{\\frac{7}{2}\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that in $\\triangle ABC$, $\\angle ACB=90^\\circ$, point D is on side $AC$, connecting $BD$, and constructing an isosceles right triangle $BDE$ (point E is on the right side of line $BD$) with $BD$ as the hypotenuse, and connecting $CE$. $\\triangle ABD \\sim \\triangle CBE$; If $BC=12$ and $CD=5$, find the length of segment $CE$.\n",
    "completion": "\\textbf{Solution:} Let us cut $CH=CB$ on the side $CA$, and connect $BH$.\\\\\nSimilarly, it can be proved: $\\triangle HBD \\sim \\triangle CBE$.\\\\\n$\\therefore \\frac{HD}{CE}=\\frac{BD}{BE}$.\\\\\n$\\because BC=12$, $\\therefore CH=12$.\\\\\nGiven that $CD=5$, $\\therefore HD=7$. Since $\\triangle BDE$ is an isosceles triangle, $\\therefore \\frac{BD}{BE}=\\sqrt{2}$.\\\\\n$\\therefore CE=\\boxed{\\frac{7}{2}\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52942184_159.png"
  },
  {
    "solution": "\\boxed{50^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the isosceles $\\triangle ABC$, where $AB=AC$ and $\\angle BAC=50^\\circ$, the angle bisector of $\\angle BAC$ intersects the perpendicular bisector of $AB$ at point $E$. By folding along $FG$ such that point $C$ coincides with point $E$, what is the measure of $\\angle CFG$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect BE,\n\n$\\because$ $\\angle BAC=50^\\circ$ and AE is the bisector of $\\angle BAC$,\n\n$\\therefore$ $\\angle BAE = \\frac{1}{2} \\times 50^\\circ = 25^\\circ$.\n\nMoreover, $\\because$ AB=AC,\n\n$\\therefore$ $\\angle ABC=\\angle ACB=65^\\circ$.\n\n$\\because$ DE is the perpendicular bisector of AB,\n\n$\\therefore$ EA=EB,\n\n$\\therefore$ $\\angle ABE=\\angle BAE=25^\\circ$,\n\n$\\therefore$ $\\angle EBC=\\angle ABC-\\angle ABE=65^\\circ-25^\\circ=40^\\circ$.\n\n$\\because$ AE is the bisector of $\\angle BAC$ and AB=AC,\n\n$\\therefore$ The line AE bisects BC perpendicularly,\n\n$\\therefore$ EB=EC,\n\n$\\therefore$ $\\angle ECB=\\angle EBC=40^\\circ$,\n\n$\\because$ $\\angle ACB$ is folded along FG,\n\n$\\therefore$ EF=CF,\n\n$\\therefore$ $\\angle CEF=\\angle ECF=40^\\circ$;\n\nIn $\\triangle ECF$, $\\angle EFC=180^\\circ-\\angle CEF-\\angle ECF=180^\\circ-40^\\circ-40^\\circ=100^\\circ$,\n\n$\\therefore$ $\\angle CFG = \\frac{1}{2} \\times \\angle CFE = \\boxed{50^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51545407.png"
  },
  {
    "solution": "\\boxed{\\frac{5}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $C$ is on line segment $AB$, $D$ is the midpoint of $AC$, and $E$ is the midpoint of $BC$. If $AB=9\\, \\text{cm}$ and $AC=5\\, \\text{cm}$. Find: What is the length of $AD$ in centimeters?\n",
    "completion": "\\textbf{Solution:} Given that AC$=5\\,\\text{cm}$ and D is the midpoint of AC,\\\\\nit follows that AD$=$DC$=\\frac{1}{2}$AC$=\\boxed{\\frac{5}{2}}\\,\\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52794293_223.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{3}a}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, BD is the diagonal of rhombus ABCD with side length \\(a\\), where \\( \\angle BAD=60^\\circ\\). Points M and N start simultaneously from points A and B, respectively, moving towards points B and D at the same speed along AB and BD. Lines DM and AN are drawn and intersect at point P. Line BP is then drawn. What is the minimum value of BP?\n",
    "completion": "\\textbf{Solution:} Extend $CD$ such that $CD=DG$,\\\\\n$\\because$ Quadrilateral $ABCD$ is a rhombus, $\\angle BAD=60^\\circ$\\\\\n$\\therefore$ $\\angle ADG=60^\\circ$, $AD=GD$\\\\\n$\\therefore$ $\\triangle ADG$ is an equilateral triangle\\\\\nIn $\\triangle ADM$ and $\\triangle BAN$\\\\\n$\\because$ $\\left\\{\\begin{array}{l}AD=BA\\\\ \\angle DAM=\\angle ABN=60^\\circ\\\\ AM=BN\\end{array}\\right.$\\\\\n$\\therefore$ $\\triangle ADM\\cong \\triangle BAN\\left(SAS\\right)$\\\\\n$\\therefore$ $\\angle ADM=\\angle BAN$\\\\\n$\\because$ $\\angle DAP+\\angle BAN=60^\\circ$\\\\\n$\\therefore$ $\\angle GAD+\\angle DAP+\\angle GDA+\\angle ADP=180^\\circ$\\\\\n$\\therefore$ Points A, P, D, G are concyclic, the radius of the circle is $\\frac{\\frac{AD}{2}}{\\cos30^\\circ}=\\frac{\\sqrt{3}a}{3}$\\\\\n$GB=2AB\\cos30^\\circ=\\sqrt{3}a$\\\\\n$\\therefore$ When points G, P, B are collinear, namely when GP is the diameter of the circle, $BP$ is the shortest,\\\\\n$\\therefore$ $BP=GB−GP=\\sqrt{3}a−2\\times \\frac{\\sqrt{3}a}{3}=\\boxed{\\frac{\\sqrt{3}a}{3}}$",
    "image_path": "dataset/mm_math/images/MM_Math/52950185.png"
  },
  {
    "solution": "\\boxed{143^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in parallelogram $ABCD$, $O$ is the midpoint of the diagonal $AC$, $AC\\perp AB$, point $E$ is the midpoint of $AD$, and $OF\\perp BC$, $\\angle D=53^\\circ$, then what is the degree measure of $\\angle FOE$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a parallelogram, \\\\\n$\\therefore$ AB $\\parallel$ CD, AD $\\parallel$ BC, \\\\\nSince AC $\\perp$ AB, \\\\\n$\\therefore$ $\\angle$BAC = $\\angle$DCA = 90$^\\circ$, \\\\\nSince point O is the midpoint of AC, and point E is the midpoint of AD, \\\\\n$\\therefore$ OE $\\parallel$ CD, \\\\\n$\\therefore$ $\\angle$COE + $\\angle$ACD = 180$^\\circ$, \\\\\n$\\therefore$ $\\angle$COE = 90$^\\circ$ \\\\\nSince $\\angle$D = $\\angle$B = 53$^\\circ$, and OF $\\perp$ BC, \\\\\n$\\therefore$ $\\angle$FOC = $\\angle$B = 53$^\\circ$, \\\\\n$\\therefore$ $\\angle$EOF = $\\angle$EOC + $\\angle$FOC = 90$^\\circ$ + 53$^\\circ$ = \\boxed{143^\\circ}, \\\\",
    "image_path": "dataset/mm_math/images/MM_Math/52386223.png"
  },
  {
    "solution": "\\boxed{50^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $PA$ and $PB$ are tangent to the circle $\\odot O$ at points $A$ and $B$, respectively. Point $C$ is a point on the circle $\\odot O$. Connect $AC$ and $BC$. If $\\angle P = 80^\\circ$, what is the degree measure of $\\angle ACB$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $OA$ and $OB$.\\\\\nSince $PA$ and $PB$ are tangent to the circle $\\odot O$ at points A and B, respectively,\\\\\nit follows that $OA\\perp PA$ and $OB\\perp PB$,\\\\\nwhich means $\\angle OAP=\\angle OBP=90^\\circ$,\\\\\nthus $\\angle AOB=180^\\circ-\\angle P=180^\\circ-80^\\circ=100^\\circ$,\\\\\ntherefore $\\angle ACB=\\frac{1}{2}\\angle AOB=\\frac{1}{2}\\times 100^\\circ=\\boxed{50^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53066089.png"
  },
  {
    "solution": "\\boxed{85}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of the circle $\\odot O$ with $AB=26$, and point $C$ is a point on the semicircle of $\\odot O$. $CE \\perp AB$ at point $E$, the angle bisector of $\\angle OCE$ intersects $\\odot O$ at point $D$, and the chord $AC=10$. What is the area of the triangle $\\triangle ACD$?\n",
    "completion": "\\textbf{Solution:} Connect OD and draw AF $\\perp$ CD with the foot at F, \\\\\n$\\because$ AB is the diameter of $\\odot$O, \\\\\n$\\therefore \\angle ACB=90^\\circ$, \\\\\n$\\therefore \\angle ACE+\\angle ECB=90^\\circ$, \\\\\n$\\because CE\\perp AB$, \\\\\n$\\therefore \\angle CEB=90^\\circ$, \\\\\n$\\therefore \\angle CBE+\\angle ECB=90^\\circ$, \\\\\n$\\therefore \\angle ACE=\\angle CBE$, \\\\\n$\\because OB=OC$, \\\\\n$\\therefore \\angle CBE=\\angle OCB$, \\\\\n$\\therefore \\angle ACE=\\angle OCB$, \\\\\n$\\because CD$ bisects $\\angle OCE$, \\\\\n$\\therefore \\angle OCD=\\angle ECD$, \\\\\n$\\therefore \\angle ACE+\\angle DCE=\\angle OCB+\\angle OCD= \\frac{1}{2} \\angle ACB=45^\\circ$, \\\\\n$\\therefore \\angle ACD=45^\\circ$, \\\\\n$\\therefore \\angle AOD=2\\angle ACD=90^\\circ$, \\\\\nIn $\\triangle ACF$, AC=10, \\\\\n$\\therefore AF=AC\\cdot\\sin45^\\circ=10\\times \\frac{\\sqrt{2}}{2} =5\\sqrt{2}$, \\\\\n$CF=AC\\cdot\\cos45^\\circ=10\\times \\frac{\\sqrt{2}}{2} =5\\sqrt{2}$, \\\\\nIn $\\triangle AOD$, AO=OD=$\\frac{1}{2}$AB=13, \\\\\n$\\therefore AD=\\sqrt{2}AO=13\\sqrt{2}$, \\\\\n$\\therefore DF=\\sqrt{AD^2-AF^2}=\\sqrt{(13\\sqrt{2})^2-(5\\sqrt{2})^2}=12\\sqrt{2}$, \\\\\n$\\therefore CD=CF+DF=17\\sqrt{2}$, \\\\\n$\\therefore$ the area of $\\triangle ACD$= $\\frac{1}{2}$CD$\\cdot$AF\\\\\n= $\\frac{1}{2}\\times17\\sqrt{2}\\times5\\sqrt{2}$\\\\\n= $\\boxed{85}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53124720.png"
  },
  {
    "solution": "\\boxed{46^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, a quadrilateral is inscribed in $\\odot O$. If one of its exterior angles $\\angle DCE=46^\\circ$, then what is the measure of $\\angle A$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral $ABCD$ is inscribed in circle $\\odot O$,\\\\\nit follows that $\\angle A+\\angle BCD=180^\\circ$,\\\\\nSince $\\angle DCE+\\angle BCD=180^\\circ$,\\\\\nit follows that $\\angle A=\\angle DCE=46^\\circ$,\\\\\nTherefore, the answer is: $\\boxed{46^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52837279.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the midpoint of line segment AB, which has a length of 18, is M. Point C divides line segment MB in the ratio MC:CB=1:2. Find the length of line segment AC.\n",
    "completion": "\\textbf{Solution:} Given that the midpoint of line segment AB, which has a length of 18, is M, \\\\\n$ \\therefore AM=BM=\\frac{1}{2}AB=9, $ \\\\\n$ \\because $ the ratio of MC to CB is 1:2, \\\\\n$ \\therefore MC=\\frac{1}{3}\\times 9=3, $ \\\\\n$ \\therefore AC=AM+MC=9+3=\\boxed{12}.$",
    "image_path": "dataset/mm_math/images/MM_Math/51355677.png"
  },
  {
    "solution": "\\boxed{y=\\frac{1}{4}x^{2}-1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the parabola $y=ax^2+bx-1\\ (a\\ne 0)$ intersects the x-axis at points $A(-2, 0)$ and $B(2, 0)$, and intersects the y-axis at point D. The line through point $C(0, 1)$ intersects the parabola at points A and F. What is the equation of the parabola?\n",
    "completion": "\\textbf{Solution:} Substituting the points $A\\left(-2, 0\\right)$, $B\\left(2, 0\\right)$ into the equation $y=ax^{2}+bx-1\\left(a\\ne 0\\right)$, we get\n\\[\n\\begin{cases}\n4a-2b-1=0 \\\\\n4a+2b-1=0\n\\end{cases}\n\\]\nSolving these, we find\n\\[\n\\begin{cases}\na=\\frac{1}{4} \\\\\nb=0\n\\end{cases}\n\\]\n$\\therefore$ The equation of the parabola is $\\boxed{y=\\frac{1}{4}x^{2}-1}$",
    "image_path": "dataset/mm_math/images/MM_Math/52620662_108.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the line $AB: y=-\\frac{1}{2}x+3$ and the line $CD: y=kx-2$ intersect at point $M(4, a)$, and intersect the coordinate axes at points A, B, C, and D, respectively. What are the values of $a$ and $k$?\n",
    "completion": "\\textbf{Solution:} Let's substitute the coordinates of point M into $y=-\\frac{1}{2}x+3$ and solve to get: $a=1$,\\\\\n$\\therefore$ for point $M(4, 1)$, substituting $M(4, 1)$ into $y=kx-2$, we get $4k-2=1$, solving this yields: $k=\\frac{3}{4}$,\\\\\n$\\therefore a=\\boxed{1}$, $k=\\boxed{\\frac{3}{4}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52381257_178.png"
  },
  {
    "solution": "\\boxed{P\\left(2, -2\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the line $y=-\\frac{1}{2}x-1$ intersects with the line $y=-2x+2$ at point $P$, and intersects with the $x$-axis at points $A$ and $B$, respectively. What are the coordinates of the intersection point $P$?\n",
    "completion": "\\textbf{Solution:} From the given, we have $\\begin{cases}y=-\\frac{1}{2}x-1 \\\\ y=-2x+2\\end{cases}$,\\\\\nSolving this, we get $\\begin{cases}x=2 \\\\ y=-2\\end{cases}$,\\\\\n$\\therefore \\boxed{P\\left(2, -2\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52250673_88.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $AB$ is the diameter of circle $\\odot O$, $C$ is a point on circle $\\odot O$, and the bisector of $\\angle OCB$ intersects circle $\\odot O$ at point $D$. A tangent to circle $\\odot O$ passing through point $D$ intersects the extension line of $CB$ at point $E$. $CE\\perp DE$; If $AB=10$, $\\tan A = \\frac{1}{3}$, find the length of $DE$.\n",
    "completion": "\\textbf{Solution:} Given that $AB$ is the diameter of $\\odot O$,\\\\\nit follows that $\\angle ADB=90^\\circ$,\\\\\nsince $\\tan A=\\frac{BD}{AD}=\\frac{1}{3}$,\\\\\nlet $BD=x$, then $AD=3x$,\\\\\nhence $A{D}^{2}+B{D}^{2}=A{B}^{2}$, which leads to ${\\left(3x\\right)}^{2}+{x}^{2}=10^{2}$,\\\\\nsolving for $x$ yields $\\sqrt{10}$,\\\\\nsince $\\angle ODE=\\angle ADB=90^\\circ$,\\\\\nit follows that $\\angle BDE=\\angle ADO$,\\\\\nsince $OA=OD$,\\\\\nit follows that $\\angle ADO=\\angle OAD$,\\\\\nhence $\\angle BDE=\\angle OAD$,\\\\\nsince $\\angle E=\\angle ADB=90^\\circ$,\\\\\nit follows that $\\triangle BDE\\sim \\triangle BAD$,\\\\\nhence $\\frac{DE}{BD}=\\frac{AD}{AB}$, which means $\\frac{DE}{\\sqrt{10}}=\\frac{3\\sqrt{10}}{10}$,\\\\\nyielding $DE=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53066284_139.png"
  },
  {
    "solution": "\\boxed{15}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the side length of square $ABCD$ is $12$, and the radius of $\\odot B$ is $6$. Point $P$ is a moving point on $\\odot B$. What is the minimum value of $PD + \\frac{1}{2}PC$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, take $BE=3$ on $BC$ and connect $BP$, $PE$,\n\n$\\because$ the side length of square $ABCD$ is 12 and the radius of $\\odot B$ is 6,\n\n$\\therefore$ $BC=12=CD$, $BP=6$, $EC=9$,\n\n$\\because$ $\\frac{BP}{BC}=\\frac{1}{2}=\\frac{BE}{BP}$, and $\\angle PBE=\\angle PBC$,\n\n$\\therefore$ $\\triangle PBE \\sim \\triangle CBP$,\n\n$\\therefore$ $\\frac{BE}{BP}=\\frac{PE}{PC}=\\frac{1}{2}$,\n\n$\\therefore$ $PE=\\frac{1}{2}PC$,\n\n$\\therefore$ $PD+\\frac{1}{2}PC=PD+PE$,\n\n$\\therefore$ when points $D$, $P$, and $E$ are collinear, $PD+PE$ has the minimum value, that is, $PD+\\frac{1}{2}PC$ has the minimum value,\n\n$\\therefore$ the minimum value of $PD+\\frac{1}{2}PC$ is $DE=\\sqrt{DC^2+CE^2}=\\boxed{15}$,",
    "image_path": "dataset/mm_math/images/MM_Math/51367426.png"
  },
  {
    "solution": "\\boxed{25\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, Xiao Ming wants to measure the height of the tower $CD$. He looks up at the top of the tower from point $A$ and measures the angle of elevation to be $30^\\circ$. Then, he moves $50$ meters towards the tower to point $B$, where he measures the angle of elevation to be $60^\\circ$. It is given that $AB=BD$; find the height of the tower $CD$. (Ignore Xiao Ming's height, and leave the result in square root form.)\n",
    "completion": "\\textbf{Solution:} Given: $BD=AB=50$ meters,\\\\\nIn $\\triangle BCD$, $\\angle C=90^\\circ$,\\\\\n$\\therefore \\sin \\angle DBC = \\frac{DC}{BD}$,\\\\\n$\\therefore DC=BD \\cdot \\sin 60^\\circ = 50 \\times \\frac{\\sqrt{3}}{2} = \\boxed{25\\sqrt{3}}$ (meters).",
    "image_path": "dataset/mm_math/images/MM_Math/51446320_101.png"
  },
  {
    "solution": "\\boxed{-1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, it is known that the graph of the linear function $y=\\frac{1}{2}x+1$ intersects the x-axis and y-axis at points A and B, respectively. A square ABCD is constructed in the second quadrant with AB as a side. Is there a point M on the x-axis such that the perimeter of $\\triangle MDB$ is minimized? Find the coordinates of point M.\n",
    "completion": "\\textbf{Solution:} As shown in the figure, find the point $B^{\\prime}$, which is the reflection of point B across the x-axis. Connect $B\\prime D$. If it intersects the x-axis at point M, then the perimeter of $\\triangle BMD$ is minimized at this moment.\\\\\n$\\because$B(0,1),\\\\\n$\\therefore B^{\\prime}$(0,−1)\\\\\nLet the equation of the line $B\\prime D$ be $y=kx+b(k\\ne 0)$,\\\\\nSubstituting the coordinates of $B^{\\prime}$ and D gives: $\\begin{cases}b=−1\\\\−3k+b=2\\end{cases}$,\\\\\nSolving these, we get: $\\begin{cases}k=−1\\\\b=−1\\end{cases}$,\\\\\n$\\therefore$The equation of the line $B\\prime D$ is $y=−x−1$.\\\\\nFor $y=−x−1$, let $y=0$, we find $x=−1$,\\\\\n$\\therefore M(\\boxed{-1},0)$.",
    "image_path": "dataset/mm_math/images/MM_Math/53706609_107.png"
  },
  {
    "solution": "\\boxed{x>-2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the abscissa of the intersection point of the linear functions $y_1=kx+b$ and $y_2=x+a$ is $-2$. What is the solution set of $kx+b<x+a$?\n",
    "completion": "\\textbf{Solution:} From the graph, we have $kx+b<x+a$ when $x>-2$, \\\\\nthus the answer is $\\boxed{x>-2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52424042.png"
  },
  {
    "solution": "\\boxed{2\\pi -\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points C and D are the trisection points of line segment AB. Construct an equilateral triangle $\\triangle OCD$ upwards with CD as its base. With O as the center and the length of OA as the radius, draw an arc that intersects the extensions of OC and OD at points E and F, respectively. If $AB = 6$, what is the area of the shaded region?\n",
    "completion": "\\textbf{Solution:} Connect OA, OB, and construct $OH\\perp AB$ at point H as shown in the diagram,\\\\\n$\\because AB=6$, C and D are the trisection points of the line segment AB, and $\\triangle COD$ is an equilateral triangle,\\\\\n$\\therefore \\angle OHA=90^\\circ$, $\\angle COH=30^\\circ$, $AC=CD=DB=2$, $AH=BH=3$, $\\angle OCH=60^\\circ$,\\\\\n$\\therefore OH=2\\times \\sin 60^\\circ=\\sqrt{3}$,\\\\\n$\\therefore OA=\\sqrt{3^2+(\\sqrt{3})^2}=2\\sqrt{3}$, $\\tan \\angle OAH=\\frac{OH}{AH}=\\frac{\\sqrt{3}}{3}$, $\\therefore \\angle OAH=30^\\circ$, $\\therefore \\angle AOH=60^\\circ$,\\\\\n$\\therefore \\angle AOC=30^\\circ$,\\\\\n$\\therefore$ the area of the shaded part is: $\\frac{2\\times \\sqrt{3}}{2}+\\left[\\frac{30\\times \\pi \\times (2\\sqrt{3})^{2}}{360}-\\frac{2\\times \\sqrt{3}}{2}\\right]\\times 2=\\boxed{2\\pi -\\sqrt{3}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51299527.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in diamond $ABCD$, $\\angle BAD=80^\\circ$, the perpendicular bisector of $AB$ intersects diagonal $AC$ at point $F$ and $AB$ at point $E$. Connecting $DF$, what is the value of $\\angle CDF$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect BF, \\\\\n$\\because$ quadrilateral ABCD is a rhombus, \\\\\n$\\therefore$ $\\angle BAC=\\frac{1}{2}\\angle BAD=\\frac{1}{2}\\times 80^\\circ=40^\\circ$, \\\\\n$BC=CD$, $\\angle DCF=\\angle BCF$, AD$\\parallel$BC, \\\\\n$\\therefore$ $\\angle ABC=180^\\circ-\\angle BAD=100^\\circ$, \\\\\n$\\because$EF is the perpendicular bisector of AB, \\\\\n$\\therefore$AF=BF, \\\\\n$\\therefore$ $\\angle ABF=\\angle BAC=40^\\circ$, \\\\\n$\\therefore$ $\\angle CBF=\\angle ABC-\\angle ABF=60^\\circ$, \\\\\nIn $\\triangle DCF$ and $\\triangle BCF$, \\\\\n$\\left\\{\\begin{array}{c}DC=BC\\\\ \\angle DCF=\\angle BCF\\\\ CF=CF\\end{array}\\right.$, \\\\\n$\\therefore$ $\\triangle DCF\\cong \\triangle BCF$ (SAS), \\\\\n$\\therefore$ $\\angle CDF=\\angle CBF= \\boxed{60^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/51510274.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in an equilateral triangle $ABC$, $D$ is the midpoint of $AC$, and $P$ is a moving point on side $AB$. Draw $PE\\perp AB$, intersecting $BC$ at point $E$, and connect $DP$, $DE$. If $AB=8$ and $\\triangle PDE$ is an isosceles triangle, what is the length of $BP$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, construct DM $\\perp$ AB at point M, and DN $\\perp$ BC at point N.\\\\\n$\\therefore \\angle AMD=\\angle DNC=90^\\circ$,\\\\\nthen $\\triangle AMD$ and $\\triangle DNC$ are both right-angled triangles.\\\\\n$\\because \\triangle ABC$ is an equilateral triangle, and AB=8,\\\\\n$\\therefore \\angle A=\\angle B=\\angle C=60^\\circ$.\\\\\n$\\because D$ is the midpoint of AC,\\\\\n$\\therefore AD=CD=\\frac{1}{2}AC=4$.\\\\\nIn $\\triangle AMD$,\\\\\nAM=$AD\\cdot\\cos\\angle A=4\\cdot\\cos60^\\circ=2$,\\\\\nDM=$AD\\cdot\\sin\\angle A=4\\cdot\\sin60^\\circ=2\\sqrt{3}$,\\\\\nSimilarly, we can derive CN=2, DN=$2\\sqrt{3}$.\\\\\n$\\therefore BM=AB-AM=6$,\\\\\nBN=BC-CN=6.\\\\\nLet BP=a,\\\\\n$\\because$ EP $\\perp$ AB\\\\\n$\\therefore \\angle EPB=90^\\circ$.\\\\\nIn $\\triangle EPB$,\\\\\n$PE=BP\\cdot\\tan\\angle B=a\\cdot\\tan60^\\circ=\\sqrt{3}a$,\\\\\nBE=$\\frac{BP}{\\cos\\angle B}=\\frac{a}{\\cos60^\\circ}=2a$.\\\\\n$\\therefore MP=BM-BP=6-a$,\\\\\nEN=BN-BE=6-2a.\\\\\nWhen $\\triangle PDE$ is an isosceles triangle,\\\\\n(1) When PE=DE,\\\\\nin $\\triangle DEN$, by the Pythagorean theorem,\\\\\n$EN^2+DN^2=DE^2$.\\\\\nWhich gives $(6-2a)^2+(2\\sqrt{3})^2=(\\sqrt{3}a)^2$.\\\\\nSolving this, we get $a_1=12-4\\sqrt{6}$, $a_2=12+4\\sqrt{6}$ (discard as it does not meet the requirements).\\\\\nHence, BP$=\\boxed{12-4\\sqrt{6}}$.\\\\\n(2) When PE=PD,\\\\\nin $\\triangle DMP$, by the Pythagorean theorem,\\\\\n$MP^2+DM^2=PD^2$.\\\\\nWhich gives $(6-a)^2+(2\\sqrt{3})^2=(\\sqrt{3}a)^2$.\\\\\nSolving this, we get $a_1=\\sqrt{33}-3$, $a_2=-\\sqrt{33}-3$ (discard as it does not meet the requirements).\\\\\nHence, BP$=\\boxed{\\sqrt{33}-3}$.\\\\\n(3) When PD=DE, for this to hold, point E must be to the right of point N, with EN=a-6. Since DM=DN, DP=DE, both being right-angled triangles, hence MP=EN, which gives 6-a=2a-6, solving this gets a=4.\\\\\nIn conclusion, the length of BP could be $12-4\\sqrt{6}$ or $\\sqrt{33}-3$ or $\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51005089.png"
  },
  {
    "solution": "\\boxed{\\frac{4}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $E$ is a point on the extension line of side $BA$ of rectangle $ABCD$. Connect $ED$ and $EC$, where $EC$ intersects $AD$ at point $G$. Construct $CF \\parallel ED$ intersecting $AB$ at point $F$, with $DC = DE$. Under the condition of (2), what is the length of $AG$?\n",
    "completion": "\\textbf{Solution:} Since $AD \\parallel BC$,\\\\\n$\\therefore \\frac{AG}{BC} = \\frac{AE}{EB}$,\\\\\n$\\therefore \\frac{AG}{3} = \\frac{4}{4+5}$,\\\\\n$\\therefore AG = \\boxed{\\frac{4}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52934013_179.png"
  },
  {
    "solution": "\\boxed{\\frac{24}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In parallelogram $ABCD$, $E$ is the trisection point of $AD$ close to $D$. Connect $BE$ and intersect $AC$ at point $F$. Given $AC = 12$, what is the length of $AF$?\n",
    "completion": "\\textbf{Solution:} Given in $\\square ABCD$, where AD = BC and AD$\\parallel$BC,\\\\\nsince E is the trisection point of AD,\\\\\nit follows that AE = $\\frac{2}{3}$ AD = $\\frac{2}{3}$ BC,\\\\\nsince AD$\\parallel$BC,\\\\\nit implies $\\triangle AEF\\sim \\triangle CBF$,\\\\\ntherefore, $\\frac{AF}{FC}=\\frac{AE}{BC}=\\frac{2}{3}$,\\\\\ngiven AC = 12,\\\\\nthus, AF = $\\frac{2}{5}\\times 12=\\boxed{\\frac{24}{5}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51368327.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given: Point $P$ is a point on segment $AC$, $BP=DP$, $AB=3$, $CD=7$. As shown in Figure 2, if $\\angle A = \\angle C = \\angle BPD \\neq 90^\\circ$, find the length of $AC$?\n",
    "completion": "\\textbf{Solution:}\\\\\nSince $\\angle A=\\angle C=\\angle BPD$, $\\angle A+\\angle B=\\angle BPD+\\angle CPD$\\\\\nThus, $\\angle B=\\angle CPD$\\\\\nSince $BP=DP$\\\\\nTherefore, $\\triangle APB \\cong \\triangle CDP$\\\\\nThus, $AB=PC=3$, $AP=CD=7$\\\\\nTherefore, $AC=\\boxed{10}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52827553_121.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ is an equilateral triangle with side length 2, $D$ is a point on the extension line of $CA$, an equilateral triangle $BDE$ is constructed with $BD$ as a side, and $AE$ is connected. Find the degree measure of $\\angle EAD$.\n",
    "completion": "\\textbf{Solution:} Given that $\\triangle ABC$ and $\\triangle BDE$ are equilateral triangles,\\\\\n$\\therefore AB=BC=AC=2$, $BD=BE$, $\\angle ABC=\\angle C=\\angle BAC=\\angle DBE=60^\\circ$,\\\\\n$\\therefore \\angle ABC+\\angle ABD=\\angle DBE+\\angle ABD$,\\\\\nwhich means $\\angle CBD=\\angle ABE$,\\\\\nIn $\\triangle CBD$ and $\\triangle ABE$,\\\\\n$ \\left\\{\\begin{array}{l}\nBC=AB \\\\\n\\angle CBD=\\angle ABE \\\\\nBD=BE\n\\end{array}\\right.$\\\\\n$\\therefore \\triangle CBD\\cong \\triangle ABE$ (SAS),\\\\\n$\\therefore \\angle BAE=\\angle BCD=60^\\circ$,\\\\\n$\\therefore \\angle EAD=180^\\circ-60^\\circ-60^\\circ=\\boxed{60^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52653743_94.png"
  },
  {
    "solution": "\\boxed{8cm}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, B is a point on line segment AC, and $AB=24\\,cm$, $BC=\\frac{1}{3}AB$. If O is the midpoint of line segment AC, then what is the length of line segment $OB$ in $cm$?\n",
    "completion": "\\textbf{Solution:} Given that $AB=24cm$, and $BC=\\frac{1}{3}AB$,\\\\\ntherefore, $BC=\\frac{1}{3} \\times 24=8cm$,\\\\\nthus, $AC=AB+BC=24+8=32cm$,\\\\\nsince $O$ is the midpoint of the line segment $AC$,\\\\\nthus, $AO=\\frac{1}{2}AB=\\frac{1}{2} \\times 32=16cm$,\\\\\ntherefore, $OB=AB-AO=24-16=\\boxed{8cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53365412.png"
  },
  {
    "solution": "\\boxed{\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, and $\\angle A=30^\\circ$, it then follows that $AB=2BC$. Based on this conclusion, further consider: if $AC=2$ and point $D$ is a moving point on side $AB$, what is the minimum value of $CD+ \\frac{1}{2} AD$?\n",
    "completion": "\\textbf{Solution:} Construct ray AG such that $\\angle BAG=30^\\circ$,\\\\\nDraw $DE \\perp AG$ at E from D, and draw $CF \\perp AG$ at F from C,\\\\\n$\\therefore DE= \\frac{1}{2} AD$,\\\\\n$\\therefore CD+ \\frac{1}{2} AD=CD+DE \\geq CF$,\\\\\n$\\because \\angle CAG=\\angle CAB+\\angle BAG=60^\\circ$, $AC=2$,\\\\\n$\\therefore \\angle ACF=30^\\circ$,\\\\\n$\\therefore AF=1$,\\\\\n$\\therefore CF= \\sqrt{AC^2-AF^2}=\\sqrt{3}$,\\\\\n$\\therefore$ The minimum value of $CD+ \\frac{1}{2} AD$ is $\\boxed{\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51435091.png"
  },
  {
    "solution": "\\boxed{18}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated in the figure, point $P$ is on the diagonal $AC$ of rectangle $ABCD$. Draw $EF\\parallel BC$ passing through point $P$ and intersecting $AB$ and $CD$ at points $E$ and $F$, respectively. Connect $PB$ and $PD$. If $AE=3$ and $PF=6$, what is the area of the shaded region in the figure?\n",
    "completion": "\\textbf{Solution:} Let line PM be perpendicular to AD at point M and intersect BC at point N.\\\\\nSince quadrilateral ABCD is a rectangle,\\\\\nit follows that $\\angle BAD = \\angle ADC = \\angle DCB = \\angle ABC = 90^\\circ$, AD$\\parallel$BC, and AB$\\parallel$CD,\\\\\ntherefore MN$\\perp$BC.\\\\\nSince EF$\\parallel$BC,\\\\\nit follows that EF$\\perp$AB and EF$\\perp$CD,\\\\\ntherefore $\\angle AEP = \\angle AMP = 90^\\circ$,\\\\\nhence quadrilateral AEPM is a rectangle.\\\\\nSimilarly, quadrilaterals DFPM, BEPN, and CFPN are rectangles,\\\\\nthus $ {S}_{\\triangle ADC}={S}_{\\triangle ABC}$, ${S}_{\\triangle AMP}={S}_{\\triangle AEP}$, ${S}_{\\triangle PBE}={S}_{\\triangle PBN}$, ${S}_{\\triangle PFD}={S}_{\\triangle PDM}$, and ${S}_{\\triangle PFC}={S}_{\\triangle PCN}$,\\\\\ntherefore $ 2{S}_{\\triangle PBE}={S}_{\\triangle ABC}-{S}_{\\triangle APE}-{S}_{\\triangle PNC}$, $2{S}_{\\triangle PDF}={S}_{\\triangle ADC}-{S}_{\\triangle APM}-{S}_{\\triangle PFC}$,\\\\\nthus $ 2{S}_{\\triangle PBE}=2{S}_{\\triangle PDF}$, i.e., ${S}_{\\triangle PBE}={S}_{\\triangle PDF}$,\\\\\nSince $ AE=3$ and $PF=6$,\\\\\nit follows that PM=AE=3,\\\\\nthus $ {S}_{\\triangle DFP}={S}_{\\triangle PBE}=\\frac{1}{2}\\times 3\\times 6=9$,\\\\\ntherefore, the area of the shaded part is $9\\times 2=\\boxed{18}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53359569.png"
  },
  {
    "solution": "\\boxed{24}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\triangle ABC$, where $ \\angle BAC=90^\\circ$, the medians $ AD$ and $ BE$ intersect at point F, and $ EG \\parallel BC$ intersects $ AD$ at point G. Given that $ GF=2$, what is the length of $ BC$?\n",
    "completion": "\\textbf{Solution:} Let $AD$ and $BE$ be the medians intersecting at point $F$, \\\\\n$\\therefore$ Point $F$ is the centroid of $\\triangle ABC$, and $BD=CD$, \\\\\n$\\therefore$ $AF=2DF$, \\\\\n$\\because$ $EG\\parallel BC$, \\\\\n$\\therefore$ $\\triangle AGE\\sim \\triangle ADC$ \\\\\n$\\therefore$ $\\frac{GE}{DC}=\\frac{AE}{AC}=\\frac{1}{2}$, \\\\\n$\\therefore$ $\\frac{GE}{BD}=\\frac{1}{2}$, \\\\\n$\\because$ $EG\\parallel BC$, \\\\\n$\\therefore$ $\\triangle GEF\\sim \\triangle DBF$, \\\\\n$\\therefore$ $\\frac{DF}{GF}=\\frac{BD}{EG}=2$, \\\\\n$\\therefore$ $DF=2GF=4$,\\\\\n$\\therefore$ $AF=2DF=8$, \\\\\n$\\therefore$ $AD=DF+AF=4+8=12$, \\\\\n$\\because$ $\\angle BAC=90^\\circ$, and $AD$ is the median of $\\triangle ABC$, \\\\\n$\\therefore$ $BC=2AD=\\boxed{24}$,",
    "image_path": "dataset/mm_math/images/MM_Math/52977397.png"
  },
  {
    "solution": "\\boxed{}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, rectangle $ABCD$ ($AD > AB$), equilateral triangles are constructed inwards using $AD$ and $BC$ as sides respectively (Figure 1); similarly, equilateral triangles are constructed inwards using $AB$ and $CD$ as sides respectively (Figure 2). The overlapping areas of the two equilateral triangles are shown as the shaded areas. Let the area of the shaded portion in Figure 1 be $S_{1}$, and the area of the shaded portion in Figure 2 be $S_{2}$. If $\\frac{S_{1}}{S_{2}}=8$, then what is the value of $\\frac{AD}{AB}$?\n",
    "completion": "\\textbf{Solution:} Let us set $AD=BC=a$, $AB=CD=b$,\\\\\nAccording to the problem: $\\angle ADN=\\angle BCH=60^\\circ$,\\\\\n$\\therefore \\angle NDC=\\angle HCD=30^\\circ$,\\\\\n$\\therefore FD=FC$,\\\\\n$\\because$ quadrilateral $ABCD$ is a rectangle,\\\\\n$\\therefore AD\\parallel BC$,\\\\\n$\\therefore \\angle FNC=\\angle ADN=60^\\circ$,\\\\\n$\\therefore \\triangle FNC$ is an equilateral triangle,\\\\\n$\\therefore FN=FC$,\\\\\n$\\therefore FN=FD$,\\\\\n$\\therefore S\\triangle FNC=S\\triangle FDC=\\frac{1}{2}S\\triangle DCN$,\\\\\nIn $\\triangle DNC$, $DN=2CN$,\\\\\nBy the Pythagorean theorem, we get $NC=\\frac{\\sqrt{3}}{3}b$,\\\\\n$\\therefore S\\triangle FNC=S\\triangle FDC=\\frac{1}{2}S\\triangle DCN=\\frac{1}{2}\\times\\frac{1}{2}NC\\cdot CD=\\frac{\\sqrt{3}}{12}b^2$,\\\\\nSimilarly: $S\\triangle DHF=S\\triangle AGE=S\\triangle ABE=S\\triangle BEM=\\frac{\\sqrt{3}}{12}b^2$,\\\\\n$\\therefore S_1=S_{\\text{rectangle}}ABCD-S\\triangle NFC-S\\triangle DFC-S\\triangle DHF-S\\triangle MBE-S\\triangle ABE-S\\triangle AGE=ab-\\frac{\\sqrt{3}}{2}b^2$,\\\\\nDraw $HM\\perp AD$ at $M$ through point $H$, and $GN\\perp AB$ at point $N$ through point $G$,\\\\\nAccording to the problem: $\\angle HEF=\\angle HGF=\\angle GAB=\\angle EDC=60^\\circ$,\\\\\n$GA=AB=CD=ED=EC=GB$,\\\\\n$\\therefore \\angle HAD=\\angle HDA=30^\\circ$,\\\\\n$\\therefore HA=HD$,\\\\\n$\\because HM\\perp AD$,\\\\\n$\\therefore AM=\\frac{1}{2}AD=\\frac{1}{2}a$,\\\\\nBy the Pythagorean theorem, we get $MH=\\frac{\\sqrt{3}}{6}a$,\\\\\n$\\therefore S\\triangle HAD=\\frac{1}{2}\\cdot AD\\cdot MH=\\frac{\\sqrt{3}}{12}a^2$,\\\\\nSimilarly: $S\\triangle FBC=\\frac{\\sqrt{3}}{12}a^2$,\\\\\n$\\because \\triangle GAB$ is an equilateral triangle, $GN\\perp AB$,\\\\\n$\\therefore AN=\\frac{1}{2}AB=\\frac{1}{2}b$,\\\\\n$\\because AG=AB=b$,\\\\\n$\\therefore GN=\\sqrt{AG^2-AN^2}=\\frac{\\sqrt{3}}{2}b$,\\\\\n$\\therefore S\\triangle ABG=\\frac{1}{2}AB\\cdot NG=\\frac{1}{2}b\\cdot\\frac{\\sqrt{3}}{2}b=\\frac{\\sqrt{3}}{4}b^2$,\\\\\nSimilarly: $S\\triangle CDE=\\frac{\\sqrt{3}}{4}b^2$,\\\\\n$\\therefore S_2=S\\triangle ABG+S\\triangle CDE+S\\triangle ADH+S\\triangle BFC-S_{\\text{rectangle}}ABCD=\\frac{\\sqrt{3}}{6}a^2+\\frac{\\sqrt{3}}{2}b^2-ab$,\\\\\n$\\because \\frac{S_1}{S_2}=8$,\\\\\n$\\therefore \\frac{ab-\\frac{\\sqrt{3}}{2}b^2}{\\frac{\\sqrt{3}}{6}a^2+\\frac{\\sqrt{3}}{2}b^2-ab}=\\frac{8}{1}$,\\\\\nSolving, we get: $a=\\frac{3\\sqrt{3}}{2}b$ or $a=\\frac{3\\sqrt{3}}{4}b$,\\\\\nAccording to the problem, we know: $a<\\sqrt{3}$",
    "image_path": "dataset/mm_math/images/MM_Math/53538212.png"
  },
  {
    "solution": "\\boxed{\\frac{9\\sqrt{5}}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within the rectangular paper $ABCD$, where $AD=9$, $E$ is a point on $CD$. Connect $AE$, and after folding $\\triangle ADE$ along the straight line $AE$, point $D$ falls on point $F$. Draw $FG\\perp AD$ at $G$. If $AG=2GD$, what is the value of $DE$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw line EH $\\perp$ FG at point H, then $\\angle$EHG = 90$^\\circ$. It is easy to deduce from the folding property that $\\triangle AED \\cong \\triangle AEF$, thus $AF = AD = 9$, $DE = EF$. Since $AG = 2GD$, it follows that $GD = 3$, $AG = 6$. Given $FG \\perp AD$, it follows that $\\angle FGD = \\angle AGF = 90^\\circ$. Therefore, in $\\triangle AGF$, $FG = \\sqrt{AF^2 - AG^2} = \\sqrt{9^2 - 6^2} = 3\\sqrt{5}$. Since ABCD is a rectangle, it follows that $\\angle D = 90^\\circ$, hence quadrilateral GHED is a rectangle. Therefore, $GH = DE$, $HE = GD = 3$. Let $DE = x$, then $GH = EF = x$, $HF = 3\\sqrt{5} - x$. Therefore, in $\\triangle HEF$, $HF^2 + HE^2 = EF^2$, which gives $(3\\sqrt{5} - x)^2 + 3^2 = x^2$. Solving for $x$ gives $x = \\frac{9\\sqrt{5}}{5}$. Therefore, $DE = \\boxed{\\frac{9\\sqrt{5}}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51638142.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{3}+2\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a square ABCD with a side length of 2 rolls clockwise on the outside of a regular hexagon with equal sides. It stops when it returns to the starting position after rolling around once. What is the maximum distance from point A to the starting point during the rolling process?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, the trajectory of point A is the arc in the diagram. Extend AE to intersect the arc at H, the length of segment AH, which is the maximum distance from point A to the starting point during the rolling process.\\\\\n$\\because$ EH＝EA$_{2}$＝$\\sqrt{2^2+2^2}=2\\sqrt{2}$,\\\\\nIn $\\triangle$AEF, $\\because$AF＝EF＝2, $\\angle$AFE＝$120^\\circ$,\\\\\n$\\therefore$AE＝$2\\times AF\\times \\cos30^\\circ=2\\sqrt{3}$,\\\\\n$\\therefore$AH＝AE+EH＝$2\\sqrt{3}+2\\sqrt{2}$,\\\\\n$\\therefore$The maximum distance from point A to the starting point during the rolling process is $\\boxed{2\\sqrt{3}+2\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51458140.png"
  },
  {
    "solution": "\\boxed{\\frac{4\\sqrt{5}}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points \\(M\\) and \\(N\\) are on side \\(AD\\) and diagonal \\(BD\\) of rectangle \\(ABCD\\), respectively. Connect \\(BM\\), \\(MN\\), where \\(AB=1\\), \\(BC=2\\), what is the minimum value of \\(BM+MN\\)?\n",
    "completion": "\\textbf{Solution:} The minimum value of $BM+MN$ is $\\boxed{\\frac{4\\sqrt{5}}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53151360_75.png"
  },
  {
    "solution": "\\boxed{(1, 0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the analytical expression of the line $l_1$ is: $y = -3x + 3$, and the line $l_1$ intersects the x-axis at point D. The line $l_2$ passes through points A$(4,0)$ and B$\\left(3, -\\frac{3}{2}\\right)$. The lines $l_1$ and $l_2$ intersect at point C. What are the coordinates of point D?\n",
    "completion": "\\textbf{Solution}: Since point D is on the x-axis, the equation of line $l_{1}$ is $y=-3x+3$. \\\\\nTherefore, setting $y=0$ yields $-3x+3=0$, solving for $x$ gives us $x=1$, \\\\\nTherefore, the coordinates of point D are $\\boxed{(1, 0)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52395164_93.png"
  },
  {
    "solution": "\\boxed{y=x^2-3x-4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the parabola $y = ax^2 + bx + c$ ($a \\neq 0$) passes through points $A(4, 0)$, $B(-1, 0)$, and $C(0, -4)$. Find the equation of this parabola.\n",
    "completion": "\\textbf{Solution:} Substitute the coordinates of A, B, and C into the equation of the parabola to get\n\\[\n\\begin{cases}\n16a+4b+c=0 \\\\\na-b+c=0 \\\\\nc=-4\n\\end{cases}\n\\]\nSolving this system gives\n\\[\n\\begin{cases}\na=1 \\\\\nb=-3 \\\\\nc=-4\n\\end{cases}\n\\]\nThus, the equation of the parabola is $y=x^2-3x-4$. Therefore, the answer is $\\boxed{y=x^2-3x-4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53019895_83.png"
  },
  {
    "solution": "\\boxed{16}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point C is on line segment AB, point D is the midpoint of line segment AC, and point C is the one-fourth division point of line segment BD. If $CB = 2$, what is the length of line segment AB?\n",
    "completion": "\\textbf{Solution:} Since point D is the midpoint of segment AC, and point C is the one-fourth point of segment BD\\\\\n$\\therefore AD=BD=\\frac{1}{2}AB$, $BC=\\frac{1}{4}BD$,\\\\\n$\\therefore BC=\\frac{1}{8}AB$,\\\\\nSince $CB=2$,\\\\\n$\\therefore AB=\\boxed{16}$",
    "image_path": "dataset/mm_math/images/MM_Math/52795983.png"
  },
  {
    "solution": "\\boxed{(0,1)\\cup(2,+\\infty)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y_1=k_1x+b$ intersects with the hyperbola $y_2=\\frac{k_2}{x}\\ (x>0)$ at points $A$ and $B$ within the first quadrant, where $A(1, m)$ and $B(2, 1)$. Based on the graph, directly write down the solution set for the inequality $y_1<y_2$.\n",
    "completion": "\\textbf{Solution:} The solution set for the inequality $y_1<y_2$ according to the graph is $0<x<1$ or $x>2$. Therefore, the answer is $x\\in\\boxed{(0,1)\\cup(2,+\\infty)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52601999_153.png"
  },
  {
    "solution": "\\boxed{14\\text{cm}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, point \\(C\\) lies on segment \\(AD\\), and \\(B\\) is the midpoint of \\(CD\\), with \\(AD=13 \\, \\text{cm}\\), and \\(BC=3 \\, \\text{cm}\\). If point \\(E\\) is on line \\(AD\\), and \\(EA=4 \\, \\text{cm}\\), what is the length of \\(BE\\) in cm?\n",
    "completion": "\\textbf{Solution:} As shown in Figure 1, when point E is on segment AC, \\\\\n$\\because AB=AC+BC=10\\text{cm}$, $EA=4\\text{cm}$, \\\\\n$\\therefore BE=AB-AE=10-4=\\boxed{6}\\text{cm}$. \\\\\nAs shown in Figure 2, when point E is on the extension line of segment CA, \\\\\n$\\because AB=10\\text{cm}$, $AE=4\\text{cm}$, \\\\\n$\\therefore BE=AE+AB=14\\text{cm}$. \\\\\nIn summary, the length of BE is either \\boxed{6\\text{cm}} or \\boxed{14\\text{cm}}.",
    "image_path": "dataset/mm_math/images/MM_Math/54962520_34.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $AB=12$, $AD=16$, $E$ and $F$ are the midpoints of $AD$ and $BC$, respectively. $G$ and $H$ are two moving points on diagonal $AC$, and they move towards each other from points $A$ and $C$ at a speed of $1$ unit length per second, for $t$ seconds, where $0\\le t<10$. If quadrilateral $EGFH$ is a rectangle, what is the value of $t$?\n",
    "completion": "\\textbf{Solution:} As shown in Figure 1, connect $EF$,\\\\\nsince $AE=BF$, $AE\\parallel BF$, and $\\angle B=90^\\circ$,\\\\\nthus quadrilateral $ABFE$ is a rectangle,\\\\\ntherefore $EF=AB=12$. When quadrilateral $EGFH$ is a rectangle,\\\\\nthus $GH=EF=12$,\\\\\nin $\\triangle ABC$, by the Pythagorean theorem, we get $AC=\\sqrt{AB^2+BC^2}=\\sqrt{12^2+16^2}=20$,\\\\\nsince $AG=CH=t$,\\\\\nthus $12+2t=20$,\\\\\ntherefore $t=\\boxed{4}$, i.e., when quadrilateral $EGFH$ is a rectangle, $t=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52416356_98.png"
  },
  {
    "solution": "\\boxed{74}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the four vertices of the square piece of paper \\(ABCD\\) lie on four parallel lines \\(l_1\\), \\(l_2\\), \\(l_3\\), and \\(l_4\\), respectively. The distances between each pair of adjacent lines among these four lines are \\(h_1\\), \\(h_2\\), and \\(h_3\\) sequentially (\\(h_1>0\\), \\(h_2>0\\), \\(h_3>0\\)). Given \\(h_1=5\\), \\(h_2=2\\), what is the area \\(S\\) of the square \\(ABCD\\)?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw AF $\\perp l_{3}$ through point A, intersecting $l_{2}$, $l_{3}$ at points E, F, respectively, and draw CG $\\perp l_{3}$ through point C, intersecting $l_{3}$, $l_{2}$ at points G, H, respectively,\n\n$\\because$ $l_{1}\\parallel l_{2}\\parallel l_{3}\\parallel l_{4}$,\n\n$\\therefore$ $\\angle AEB = \\angle AFD = \\angle DGC = \\angle BHC = 90^\\circ$,\n\n$\\because$ Quadrilateral ABCD is a square,\n\n$\\therefore$ AB=BC=CD=DA, $\\angle BAD = \\angle ADC = \\angle DCB = \\angle CBA = 90^\\circ$,\n\n$\\therefore$ $\\angle ABE = \\angle DAF = \\angle CDG = \\angle BCH$,\n\n$\\therefore$ $\\Delta ABE \\cong \\Delta DAF \\cong \\Delta CDG \\cong \\Delta BCH$,\n\n$\\therefore$ BE=AF=$h_{1}+h_{2}$,\n\n$\\therefore$ $S_{\\Delta ABE} = S_{\\Delta DAF} = S_{\\Delta CDG} = S_{\\Delta BCH} = \\frac{1}{2}h_{1}(h_{1}+h_{2})$,\n\n$\\therefore$ $S = 4S_{\\Delta ABE} + S_{\\text{square }EFGH} = 4 \\times \\frac{1}{2}h_{1}(h_{1}+h_{2}) + h_{2}^2 = \\boxed{74}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51419793.png"
  },
  {
    "solution": "\\boxed{12\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the diagonals AC and BD of rectangle ABCD intersect at point O. A line is drawn through point O such that $OE \\perp AC$, intersecting AD at point E. Another line is drawn through point E such that $EF \\perp BD$, with the foot of the perpendicular at F, where $EF=1$, $OE=2$, and $BD=4\\sqrt{3}$. What is the area of rectangle ABCD?\n",
    "completion": "\\textbf{Solution}: Since quadrilateral ABCD is a rectangle,\\\\\nit follows that $OD = OB = OA = OC = \\frac{1}{2}BD = 2\\sqrt{3}$,\\\\\nsince $EF \\perp BD$, and $OE \\perp AC$,\\\\\nthus, the area of $\\triangle AOE = \\frac{1}{2}AO \\times OE = \\frac{1}{2} \\times 2\\sqrt{3} \\times 2 = 2\\sqrt{3}$,\\\\\nthe area of $\\triangle DOE = \\frac{1}{2}OD \\times EF = \\frac{1}{2} \\times 2\\sqrt{3} \\times 1 = \\sqrt{3}$,\\\\\nhence, the area of $\\triangle AOD = {S}_{\\triangle AOE} + {S}_{\\triangle DOE} = 2\\sqrt{3} + \\sqrt{3} = 3\\sqrt{3}$,\\\\\nsince quadrilateral ABCD is a rectangle,\\\\\ntherefore, the area of rectangle ABCD = $4 \\times {S}_{\\triangle AOD} = 4 \\times 3\\sqrt{3} = \\boxed{12\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52416335.png"
  },
  {
    "solution": "\\boxed{36}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the square $ABCD$, $\\triangle ABC$ is rotated counterclockwise around point $A$ to obtain $\\triangle AB'C'$, with $AB'$ and $AC'$ intersecting the diagonal $BD$ at points $E$ and $F$, respectively. If $AE=6$, what is the value of $EF\\cdot ED$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a square,\\\\\nit follows that $\\angle BAC=\\angle ADB=45^\\circ$, \\\\\nsince triangle ABC is rotated counterclockwise around point A to obtain triangle $AB'C'$, \\\\\nit follows that $\\angle EAF=\\angle BAC=45^\\circ$, \\\\\nsince $\\angle AEF=\\angle DEA$, \\\\\nit follows that $\\triangle AEF\\sim \\triangle DEA$\\\\\ntherefore $\\frac{AE}{DE}=\\frac{EF}{AE}$, \\\\\ntherefore $EF\\cdot ED=AE^{2}$, \\\\\nsince $AE=6$, \\\\\ntherefore, the value of $EF\\cdot ED$ is $\\boxed{36}$;",
    "image_path": "dataset/mm_math/images/MM_Math/50322886.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC$, circle $\\odot O$ is the circumcircle of $\\triangle ABC$, and the extension of $BO$ intersects side $AC$ at point D. If $\\angle ACB=60^\\circ$ and $BC=8\\sqrt{3}$, find the radius of $\\odot O$.\n",
    "completion": "\\textbf{Solution:} Construct line $AO$ and extend $AO$ to meet $BC$ at point $E$. Draw a line from point $O$ such that $OH\\perp AB$, intersecting $AB$ at point $H$, as shown in the diagram.\\\\\nSince $AE$ passes through the center and $AB=AC$,\\\\\nit follows that $AE$ perpendicularly bisects $BC$,\\\\\nand thus $AE$ bisects $\\angle BAC$.\\\\\nGiven that in $\\triangle ABC$, $\\angle C=60^\\circ$,\\\\\nit implies that $\\triangle ABC$ is an equilateral triangle.\\\\\nTherefore, $\\angle BAO=\\frac{1}{2}\\angle BAC=30^\\circ$, and $AB=BC=8\\sqrt{3}$.\\\\\nIn right triangle $AOH$, $AH=\\frac{1}{2}AB=4\\sqrt{3}$, with $\\angle HAO=30^\\circ$,\\\\\nsince $\\cos\\angle HAO=\\frac{AH}{AO}$,\\\\\nit follows that $AO=\\frac{AH}{\\cos\\angle HAO}=8$.\\\\\nTherefore, the radius of circle $O$ is $\\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52745971_82.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, AD is the bisector of $\\angle BAC$ in $\\triangle ABC$, DE$\\perp$AB meeting AB at point E, and DF$\\perp$AC meeting AC at point F. If DE$=2$ and AC$=4$, what is the area of $\\triangle ADC$?\n",
    "completion": "Solution: Since $AD$ is the bisector of $\\angle BAC$, and $DE \\perp AB$, $DF \\perp AC$,\n\n$\\therefore DF = DE$,\n\nSince $DE = 2$,\n\n$\\therefore DF = 2$,\n\nTherefore, the area of $\\triangle ADC$ = $\\frac{1}{2} \\times AC \\times DF = \\frac{1}{2} \\times 4 \\times 2 = \\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52873809.png"
  },
  {
    "solution": "\\boxed{45^\\circ+\\theta}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $O$ is on line $AB$, with $OM$ and $ON$ on opposite sides of line $AB$ and $\\angle MON=90^\\circ$. $OE$ bisects $\\angle MOB$, and $OF$ bisects $\\angle AON$. Let $\\angle AOF = \\theta$, express $\\angle MOE$ in terms of $\\theta$.\n",
    "completion": "\\textbf{Solution:}\\\\\nSince $\\angle AOF=\\theta$,\\\\\nit follows that $\\angle AON=2\\angle AOF=2\\theta$,\\\\\nand since $\\angle MON=90^\\circ$,\\\\\nwe have $\\angle AOM=90^\\circ-2\\theta$,\\\\\nthus $\\angle BOM=180^\\circ-\\left(90^\\circ-2\\theta \\right)=90^\\circ+2\\theta$,\\\\\ntherefore $\\angle MOE=\\frac{1}{2}\\angle BOM=\\frac{1}{2}\\left(90^\\circ+2\\theta \\right)=45^\\circ+\\theta$.\\\\\nHence, $\\angle MOE$ can be expressed in a formula containing $\\theta$ as $\\angle MOE=\\boxed{45^\\circ+\\theta}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52948847_172.png"
  },
  {
    "solution": "\\boxed{12.5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, to measure the height of a flagpole, a comprehensive practice group designed the following plan: Use a 2.5m long bamboo pole as a measuring tool, move the bamboo pole to keep it parallel to the flagpole, so that the top ends of the bamboo pole and the flagpole cast shadows that exactly meet at the same point on the ground. At this moment, if the distance from the bamboo pole to this point is 5m, and the distance from the bamboo pole to the flagpole is 20m, what is the height of the flagpole in meters?\n",
    "completion": "\\textbf{Solution:} From the figure, we can see that,\\\\\nlet the height of the flagpole be $x$ meters,\\\\\n$\\because DE\\parallel BC$\\\\\n$\\therefore \\triangle ADE\\sim \\triangle ABC$\\\\\n$\\therefore \\frac{2.5}{5}=\\frac{x}{5+20}$\\\\\n$\\therefore x=\\boxed{12.5}$",
    "image_path": "dataset/mm_math/images/MM_Math/52797042.png"
  },
  {
    "solution": "\\boxed{70^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AD$ is the angle bisector of $\\triangle ABC$, and $AE$ is the altitude of $\\triangle ABC$. Given that $\\angle B=40^\\circ$ and $\\angle DAE=15^\\circ$, what is the degree measure of $\\angle C$?\n",
    "completion": "\\textbf{Solution:} Given the construction,\\\\\n$\\because AE$ is the altitude of $\\triangle ABC$,\\\\\n$\\therefore \\angle AEB=90^\\circ$,\\\\\n$\\because \\angle B=40^\\circ$,\\\\\n$\\therefore \\angle BAE=50^\\circ$,\\\\\nAlso, $\\because \\angle DAE=15^\\circ$,\\\\\n$\\therefore \\angle BAD=50^\\circ-15^\\circ=35^\\circ$,\\\\\n$\\because AD$ is the bisector of $\\triangle ABC$,\\\\\n$\\therefore \\angle BAD=\\angle CAD=35^\\circ$,\\\\\n$\\therefore \\angle BAC=70^\\circ$,\\\\\n$\\therefore \\angle C=180^\\circ-\\angle B-\\angle BAC=180^\\circ-40^\\circ-70^\\circ=\\boxed{70^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52653666.png"
  },
  {
    "solution": "\\boxed{6\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the cross-section of the dam, the height of the dam $BC$ is $6$ meters, and the slope of the inclined surface is $1\\colon 2$. What is the length of the slope $AB$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle ACB = 90^\\circ$,\\\\\nsince the slope of the inclined plane is $1:2$,\\\\\nit follows that $\\frac{BC}{AC} = \\frac{1}{2}$,\\\\\nthus $AC = 2BC = 12$ meters,\\\\\nhence $AB = \\sqrt{AC^2 + BC^2} = \\boxed{6\\sqrt{5}}$ meters.",
    "image_path": "dataset/mm_math/images/MM_Math/55604085.png"
  },
  {
    "solution": "\\boxed{1<x<4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y=kx+b$ and the graph of the inverse proportion function $y=\\frac{m}{x}$ (for $x>0$) intersect at points $A(1, 4)$ and $B(4, n)$. Based on the graph, directly write out the solution set for the inequality $kx+b-\\frac{m}{x}>0$ in terms of $x$.\n",
    "completion": "\\textbf{Solution:} According to the graph, we can directly write the solution set of the inequality $kx + b - \\frac{m}{x} > 0$ with respect to $x$ as $\\boxed{1<x<4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52205237_120.png"
  },
  {
    "solution": "\\boxed{24}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the large square $ABCD$ is composed of four congruent right-angled triangles and one small square. If $\\angle ADE = \\angle AED$ and $AD = 4\\sqrt{5}$, what is the area of $\\triangle ADE$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure:\\\\\n$\\because \\angle ADE = \\angle AED$,\\\\\n$\\therefore AD = AE = AB$,\\\\\n$\\therefore \\angle AEF = \\angle ABF$,\\\\\n$\\because AF \\perp BE$,\\\\\n$\\therefore EF = BF = \\frac{1}{2}BE$,\\\\\n$\\therefore GE = AH$,\\\\\n$\\because \\angle GEM = \\angle HAM, \\angle MGE = \\angle MHA$,\\\\\n$\\therefore \\triangle GEM \\cong \\triangle HAM$ (ASA),\\\\\n$\\therefore S_{\\triangle HAM} = S_{\\triangle GEM}$,\\\\\n$\\therefore S_{\\triangle ADE} = S_{\\triangle ADH} + S_{\\triangle DGE}$,\\\\\n$\\because AD = 4\\sqrt{5}$, DH = 2AH, $AD^2 = DH^2 + AH^2$,\\\\\n$\\therefore AH = 4, DH = 8$,\\\\\n$\\therefore DG = GE = 4$,\\\\\n$\\therefore S_{\\triangle ADE} = \\frac{1}{2} \\times 4 \\times 8 + \\frac{1}{2} \\times 4 \\times 4 = \\boxed{24}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52416276.png"
  },
  {
    "solution": "\\boxed{50^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $ \\angle AOB=40^\\circ$, $OB$ is the bisector of $ \\angle AOC$, and $OD$ is the bisector of $ \\angle COE$. If $ \\angle DOE=10^\\circ$, find the degree measure of $ \\angle BOD$.\n",
    "completion": "\\textbf{Solution:} Since $OB$ is the bisector of $\\angle AOC$, and $OD$ is the bisector of $\\angle COE$,\\\\\n$\\therefore \\angle BOC=\\angle AOB=40^\\circ$, $\\angle COD=\\angle DOE=10^\\circ$,\\\\\n$\\therefore \\angle BOD=\\angle BOC+\\angle COD=40^\\circ+10^\\circ=\\boxed{50^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51457701_187.png"
  },
  {
    "solution": "\\boxed{72^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the regular pentagon ABCDE is inscribed in $\\odot O$. Connecting AC, what is the degree measure of $\\angle ACD$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect OA, OE, OD, \\\\\nsince a regular pentagon ABCDE is inscribed in the circle $O$, \\\\\ntherefore, $\\angle AOE = \\angle DOE = \\frac{360^\\circ}{5} = 72^\\circ$, \\\\\nhence, $\\angle ACD = \\frac{1}{2} \\angle AOD = \\boxed{72^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51299784.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{5}}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within a $2\\times 4$ square grid, line segments $AB$ and $CD$ intersect at point $E$. If the side length of each small square is 1, what is the length of $DE$?\n",
    "completion": "\\textbf{Solution:} From the figure, we know: $BC \\parallel AD$, $\\therefore \\triangle CBE \\sim \\triangle DAE$, $\\therefore \\frac{CB}{AD} = \\frac{CE}{DE}$, $\\because CB = \\sqrt{2^2 + 2^2} = 2\\sqrt{2}$, $AD = \\sqrt{1^2 + 1^2} = \\sqrt{2}$, $\\therefore \\frac{CE}{DE} = \\frac{2}{1}$, $\\because CD = \\sqrt{1^2 + 2^2} = \\sqrt{5}$, $\\therefore DE = \\boxed{\\frac{\\sqrt{5}}{3}}$",
    "image_path": "dataset/mm_math/images/MM_Math/55497823.png"
  },
  {
    "solution": "\\boxed{B\\left(\\frac{9}{2}, \\frac{11}{2}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the Cartesian coordinate system $xOy$, the line $y=-2x+a$ intersects with the $y$-axis at point $A$ and intersects with the line $y=x+1$ at point $P(3,b)$. $B$ is a point on the line $y=x+1$. What are the coordinates of point $B$ when the segment $AB$ is at its shortest length?\n",
    "completion": "\\textbf{Solution:} When $AB\\perp$ line $y=x+1$, the segment $AB$ is shortest. Label the intersection point of line $y=x+1$ with the $y$-axis $(0,1)$ as $E$. Given the conditions, we have $A(0,10)$, and $\\angle AEB=45^\\circ$, $\\triangle AEB$ is an isosceles right triangle,\\\\\n$\\therefore AE=9$, $AB=BE=\\frac{9\\sqrt{2}}{2}$\\\\\n$\\therefore$ The x-coordinate of B is $\\frac{9\\sqrt{2}}{2}\\times \\frac{\\sqrt{2}}{2}=\\frac{9}{2}$, and the y-coordinate is $\\frac{9}{2}+1=\\frac{11}{2}$,\\\\\n$\\therefore \\boxed{B\\left(\\frac{9}{2}, \\frac{11}{2}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50677275_150.png"
  },
  {
    "solution": "\\boxed{(5,3)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, place $\\square ABCO$ in the Cartesian coordinate system $xOy$, where $O$ is the origin. If the coordinate of point $A$ is $(4,0)$ and the coordinate of point $C$ is $(1,3)$, then what are the coordinates of point $B$?\n",
    "completion": "\\textbf{Solution:} Extend BC to intersect the y-axis at point D,\\\\\n$\\because$ the coordinates of point A are (4,0),\\\\\n$\\therefore$ OA = 4,\\\\\n$\\because$ quadrilateral ABCD is a parallelogram,\\\\\n$\\therefore$ BC $\\parallel$ OA, BC = OA = 4,\\\\\n$\\because$ the coordinates of point C are (1,3),\\\\\n$\\therefore$ OD = 3, DC = 1,\\\\\n$\\therefore$ BD = DC + BC = 1 + 4 = 5,\\\\\n$\\therefore$ the coordinates of point B are \\boxed{(5,3)},",
    "image_path": "dataset/mm_math/images/MM_Math/52690931.png"
  },
  {
    "solution": "\\boxed{45\\text{cm}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, $AB$ is the diameter of circle $O$, $AC$ is a chord of circle $O$, $D$ is the midpoint of arc $BC$, and $DE$ is drawn perpendicular to $AC$ at point $E$, intersecting the extension of $AB$ at point $F$, and $DA$ is connected. If $AB=90\\text{cm}$, what is the distance from the center $O$ to the \"lever $EF$\"?\n",
    "completion": "\\textbf{Solution:} Connect $OD$,\\\\\nsince $D$ is the midpoint of the arc $BC$,\\\\\nit follows that $\\angle CAD=\\angle BAD$,\\\\\nsince $OA=OD$,\\\\\nit follows that $\\angle BAD=\\angle ADO$,\\\\\nhence $\\angle CAD=\\angle ADO$,\\\\\nsince $DE\\perp AC$,\\\\\nit follows that $\\angle E=90^\\circ$,\\\\\nhence $\\angle CAD+\\angle EDA=90^\\circ$, which means $\\angle ADO+\\angle EDA=90^\\circ$,\\\\\nit follows that $OD\\perp EF$,\\\\\nhence the length of $OD$ is the distance from the center $O$ to the \"lever $EF$\",\\\\\nsince $AB=90\\text{cm}$,\\\\\nit follows that $OD=OA=\\boxed{45\\text{cm}}$",
    "image_path": "dataset/mm_math/images/MM_Math/52457436_95.png"
  },
  {
    "solution": "\\boxed{100}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, two vehicles depart simultaneously from locations A and B, respectively, along the same road, traveling at constant speeds. It is known that the sedan travels $20km$ per hour faster than the truck. After meeting, the two vehicles take a break for a while and then continue traveling at the same time. The relationship between the distance $y(km)$ between the two vehicles and the truck's travel time $x(h)$ is represented by the piecewise linear graph $AB-BC-CD-DE$. What are the respective speeds of the two vehicles in $km/h$?\n",
    "completion": "\\textbf{Solution:} From the graph of the function, we learn that the distance between location A and B is 180 km. Let the speed of the truck be $x\\,km/h$, and the speed of the sedan is $(x+20)\\,km/h$. According to the problem, we have: $x + (x + 20) = 180$, solving this gives $x = \\boxed{80}$,\\\\\nthus, the speed of the truck is $80\\,km/h$, and the speed of the sedan is $\\boxed{100}\\,km/h$.",
    "image_path": "dataset/mm_math/images/MM_Math/52381910_138.png"
  },
  {
    "solution": "\\boxed{15}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB\\perp BC$, $CD\\perp BC$, with the perpendicular feet being $B$, $C$ respectively. Let $P$ be a point on the line segment $BC$. Connect $PA$ and $PD$. Given that $AB=5$, $DC=4$, and $BC=12$, what is the minimum value of $AP+DP$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, let's construct point $A'$ symmetrical to point $A$ with respect to $BC$. Draw $A'D$ intersecting $BC$ at point $P$. Through point $A'$, draw $A'M \\perp DC$ intersecting at point $M$,\n\\[\n\\therefore AP = A'P,\n\\]\n\\[\n\\therefore AP + PD = A'P + PD,\n\\]\nWhen points $A'$, $P$, and $D$ are collinear, $A'P + PD = A'D$. At this moment, the value of $A'P + PD$ is minimized,\n\\[\n\\because AB = 5, \\; DC = 4, \\; BC = 12,\n\\]\n\\[\n\\therefore AM = 12, \\; DM = 5 + 4 = 9,\n\\]\nIn $\\triangle A'DM$, $A'D = \\sqrt{A'M^{2} + DM^{2}} = \\sqrt{12^{2} + 9^{2}} = 15$,\n\\[\n\\therefore \\text{The minimum value of } AP + PD \\text{ is } \\boxed{15}.\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/52774844.png"
  },
  {
    "solution": "\\boxed{\\left(5,3\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As can be seen from the diagram, in the Cartesian coordinate system, an isosceles right-angled triangle board is placed as shown. If $A(3, 0), B(0, 2)$, then what are the coordinates of point $C$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $CH\\perp x$-axis at point H.\\\\\nSince $\\angle AHC=\\angle CAB=\\angle AOB=90^\\circ$,\\\\\nit follows that $\\angle BAO+\\angle CAH=90^\\circ$, $\\angle CAH+\\angle ACH=90^\\circ$,\\\\\nthus, $\\angle ACH=\\angle BAO$,\\\\\nin $\\triangle AHC$ and $\\triangle BOA$,\\\\\n$\\left\\{\\begin{array}{l}\n\\angle AHC=\\angle AOB \\\\\n\\angle ACH=\\angle OAB \\\\\nAC=AB\n\\end{array}\\right.$,\\\\\ntherefore, $\\triangle AHC\\cong \\triangle BOA\\left(AAS\\right)$,\\\\\nhence, $AH=OB$, $CH=OA$,\\\\\nsince $A(3,0)$, $B(0,2)$,\\\\\nit follows that $OA=CH=3$, $OB=AH=2$,\\\\\ntherefore, $OH=OA+AH=5$,\\\\\nthus, $C\\left(5,3\\right)$.\\\\\nTherefore, the answer is $C=\\boxed{\\left(5,3\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52826953.png"
  },
  {
    "solution": "\\boxed{\\frac{7}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in right-angled triangle $ABC$ with $\\angle ABC=90^\\circ$, $AB=4$, and $BC=3$, point $D$ is a moving point on circle $\\odot A$ with radius $2$, and point $M$ is the midpoint of $CD$. What is the maximum value of $BM$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, take N as the midpoint of AC, and connect MN, BN,\\\\\n$\\because$ in $\\triangle ABC$, $\\angle ABC=90^\\circ$, AB=4, BC=3,\\\\\n$\\therefore AC=\\sqrt{3^2+4^2}=5$\\\\\n$\\because$AN=NC=$\\frac{1}{2}AC=\\frac{5}{2}$,\\\\\n$\\therefore$BN=$\\frac{1}{2}AC=\\frac{5}{2}$\\\\\n$\\because$ point M is the midpoint of CD,\\\\\n$\\therefore$DM=MC,\\\\\n$\\therefore$MN=$\\frac{1}{2}AD=1$\\\\\n$\\therefore$BM$\\leq$BN+NM,\\\\\n$\\therefore$BM$\\leq\\frac{5}{2}+1=\\boxed{\\frac{7}{2}}$,\\\\\ni.e., the maximum value of BM is $\\frac{7}{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53147806.png"
  },
  {
    "solution": "\\boxed{x \\in (-\\infty, -3) \\cup (0, 4)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y_{1}=kx+b$ ($k\\neq 0$) intersects with the graph of the inverse proportion function $y_{2}=\\frac{m}{x}$ ($m\\neq 0$) at points $A(-3, 2)$ and $B(n, -\\frac{3}{2})$. Please directly write down the range of $x$ when $y_{1}>y_{2}$.\n",
    "completion": "\\textbf{Solution:} When $y_{1}>y_{2}$, the range of $x$ is $x<-3$ or $0<x<4$. Therefore, the answer is $\\boxed{x \\in (-\\infty, -3) \\cup (0, 4)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52076225_178.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{7}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that point $E$ is a moving point on the diagonal $AC$ of rectangle $ABCD$, and the vertices $G$ and $H$ of square $EFGH$ are both on side $BC$. If $AB = 3$ and $BC = 4$, then what is the value of $\\tan \\angle CFE$?\n",
    "completion": "Solution: In the square EFGH, we have EF$\\parallel$GH and EH$\\perp$BC,\\\\\nwhich means EF$\\parallel$BC,\\\\\n$\\therefore \\angle CFE=\\angle BCF$,\\\\\nIn the rectangle ABCD, AB$\\perp$BC,\\\\\n$\\therefore$EH$\\parallel$AB,\\\\\n$\\triangle ABC\\sim \\triangle EHC$,\\\\\n$\\therefore \\frac{EH}{AB}=\\frac{CH}{BC}$, i.e., $\\frac{AB}{BC}=\\frac{EH}{CH}=\\frac{3}{4}$,\\\\\nLet $EH=3a$, $CH=4a$, then $FG=3a$, $GH=3a$,\\\\\n$\\therefore CG=7a$,\\\\\n$\\therefore \\tan\\angle CFE=\\tan\\angle BCF=\\frac{FG}{CG}=\\frac{3a}{7a}=\\boxed{\\frac{3}{7}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51457863.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: \\textit{\"Does there exist a value of $t$ such that $\\frac{{S}_{\\triangle BQP}}{{S}_{\\text{quadrilateral } CDPQ}} = 1:4$? If so, determine the value of $t$.\"}\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $PF\\perp BC$ through point $P$,\\\\\nsince the ratio $\\frac{{S}_{\\triangle BQP}}{{S}_{\\text{quadrilateral} CDPQ}}=1:4$,\\\\\nthus the ratio $\\frac{{S}_{\\triangle BQP}}{{S}_{\\triangle BCD}}=1:5$.\\\\\nFrom (2), it is known that $\\triangle PBF\\sim \\triangle CBD$,\\\\\nthus $\\frac{BF}{BD}=\\frac{PB}{BC}=\\frac{PF}{CD}$, that is $\\frac{t}{10}=\\frac{PF}{8}$,\\\\\nhence $PF=\\frac{4t}{5}\\text{cm}$,\\\\\nthus ${S}_{\\triangle BPQ}=\\frac{1}{2}BQ\\cdot PF=\\frac{1}{2}\\times(10-2t)\\times \\frac{4}{5}t=\\left(-\\frac{4}{5}t^2+4t\\right)\\text{cm}^2$.\\\\\nSince ${S}_{\\triangle BCD}=\\frac{1}{2}BD\\cdot CD=\\frac{1}{2}\\times 6\\times 8=24\\text{cm}^2$,\\\\\ntherefore $\\left(-\\frac{4}{5}t^2+4t\\right)=\\frac{1}{5}\\times 24$,\\\\\nsolving gives: $t_{1}=\\boxed{2}$, $t_{2}=\\boxed{3}$,\\\\\nthus the value of $t$ is \\boxed{2} or \\boxed{3}.",
    "image_path": "dataset/mm_math/images/MM_Math/53043392_177.png"
  },
  {
    "solution": "\\boxed{\\frac{24}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rhombus $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$, with $BD=8$ and $BC=5$. If $AE \\perp BC$ at point $E$, what is the length of $AE$?\n",
    "completion": "\\textbf{Solution:} Given that $ABCD$ is a rhombus, with $BC=5$ and $BD=8$, \\\\\nthen $OC=\\sqrt{5^2-4^2}=3$, \\\\\nhence $AC=2OC=6$, \\\\\nsince $AE \\perp BC$, \\\\\nthen $BC \\cdot AE = \\frac{1}{2} AC \\cdot BD$, \\\\\nthus $5AE = \\frac{1}{2} \\times 6 \\times 8$, \\\\\ntherefore $AE = \\boxed{\\frac{24}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52205054.png"
  },
  {
    "solution": "\\boxed{6+2\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that in the parallelogram $OABC$, point $O$ is the origin of the coordinates, point $A(3, 0)$, and point $C(1, 2)$. The graph of the function $y=\\frac{k}{x} (k \\neq 0)$ passes through point $C$. What is the perimeter of the parallelogram $OABC$?\n",
    "completion": "\\textbf{Solution:} Construct $CD\\perp OA$ at point $D$, \\\\\n$\\because$ $C(1, 2)$, \\\\\n$\\therefore$ $OC=\\sqrt{1^2+2^2}=\\sqrt{5}$, \\\\\nIn parallelogram $OABC$, $CB=OA=3$, $AB=OC=\\sqrt{5}$, \\\\\n$\\therefore$ the perimeter of parallelogram $OABC$ is: $3+3+\\sqrt{5}+\\sqrt{5}=6+2\\sqrt{5}$, \\\\\nhence, the perimeter of parallelogram $OABC$ is $\\boxed{6+2\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50677341_144.png"
  },
  {
    "solution": "\\boxed{}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 3, in $\\triangle ABC$, $\\angle A=45^\\circ$, $\\angle ABC=120^\\circ$, $BC=2\\sqrt{3}$, and $D$ is a point in the plane containing $\\triangle ABC$. When the quadrilateral $ABCD$ is a semisymmetric quadrilateral and $AC$ is the diagonal that divides it, what is the area of the quadrilateral $ABCD$?\n",
    "completion": "\\textbf{Solution:} Construct $CE\\perp AB$, intersecting the extension of $AB$ at point $E$, as shown in the diagram,\\\\\nsince $CE\\perp AB$, $\\angle A=45^\\circ$, $\\angle ABC=120^\\circ$,\\\\\ntherefore $\\angle ACE=45^\\circ$, $\\angle EBC=60^\\circ$,\\\\\ntherefore $AE=EC$, $\\angle ECB=30^\\circ$,\\\\\ntherefore $BE=\\frac{1}{2}BC=\\sqrt{3}$,\\\\\ntherefore $EC=\\sqrt{BC^{2}-BE^{2}}=\\sqrt{(2\\sqrt{3})^{2}-(\\sqrt{3})^{2}}=3$,\\\\\ntherefore $AE=EC=3$,\\\\\ntherefore $AC=\\sqrt{2}EC=3\\sqrt{2}$.\\\\\ntherefore $AB=AE-BE=3-\\sqrt{3}$.\\\\\ntherefore $S_{\\triangle ABC}=\\frac{1}{2}AB\\cdot EC=\\frac{1}{2}(3-\\sqrt{3})\\times 3=\\frac{9}{2}-\\frac{3\\sqrt{3}}{2}$.\\\\\n(1) When $DA=DC$ and $AC$ bisects $\\angle BAD$, as shown in the diagram,\\\\\naccording to the problem: $\\angle DAC=\\angle BAC=45^\\circ$,\\\\\ntherefore $DA=DC$,\\\\\ntherefore $\\angle DCA=\\angle DAC=45^\\circ$,\\\\\ntherefore $\\angle ADC=90^\\circ$,\\\\\ntherefore $\\triangle ADC$ is an isosceles right triangle,\\\\\ntherefore $AD=CD=\\frac{\\sqrt{2}}{2}AC=3$,\\\\\ntherefore $S_{\\triangle DAC}=\\frac{1}{2}AD\\cdot CD=\\frac{9}{2}$,\\\\\ntherefore $S_{\\text{quadrilateral} ABCD}=S_{\\triangle ABC}+S_{\\triangle ADC}=\\frac{9}{2}-\\frac{3\\sqrt{3}}{2}+\\frac{9}{2}=9-\\frac{3\\sqrt{3}}{2}$;\\\\\n(2) When $DA=DC$ and $AC$ bisects $\\angle BCD$, as shown in the diagram,\\\\\naccording to the problem: $\\angle ACD=\\angle BCA=15^\\circ$,\\\\\ntherefore $DA=DC$,\\\\\ntherefore $\\angle DCA=\\angle DAC=15^\\circ$,\\\\\ntherefore $\\angle ADC=150^\\circ$,\\\\\nConstruct $CF\\perp AD$ through point $C$, intersecting the extension of $AD$ at point $F$,\\\\\nthen $\\angle CDF=30^\\circ$,\\\\\ntherefore $CF=\\frac{1}{2}CD$,\\\\\ntherefore $DF=\\frac{\\sqrt{3}}{2}CD$.\\\\\nLet $CD=x$, then $AD=x$, $CF=\\frac{1}{2}x$, $AF=AD+DF=(1+\\frac{\\sqrt{3}}{2})x$,\\\\\nIn $Rt\\triangle ACF$,\\\\\nsince $AC^{2}=AF^{2}+CF^{2}$,\\\\\ntherefore $(3\\sqrt{2})^{2}=[(1+\\frac{\\sqrt{3}}{2})x]^{2}+(\\frac{1}{2}x)^{2}$,\\\\\ntherefore $x=3+3\\sqrt{3}$ (contradicts the problem statement, thus discarded) or $x=3\\sqrt{3}-3$,\\\\\ntherefore $AD=3\\sqrt{3}-3$, $CF=\\frac{3\\sqrt{3}-3}{2}$.\\\\\ntherefore $S_{\\triangle ADC}=\\frac{1}{2}AD\\cdot CF=\\frac{18-9\\sqrt{3}}{2}$,\\\\\ntherefore $S_{\\text{quadrilateral} ABCD}=S_{\\triangle ABC}+S_{\\triangle ADC}=9-\\frac{3\\sqrt{3}}{2}+\\frac{18-9\\sqrt{3}}{2}=18-6\\sqrt{3}$.\\\\\nIn summary, the quadrilateral $ABCD$",
    "image_path": "dataset/mm_math/images/MM_Math/54654114_104.png"
  },
  {
    "solution": "\\boxed{(3,-6)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure: $\\triangle AOB$ and $\\triangle A_1OB_1$ are similar figures with the origin as the center of similarity, and the scale factor of similarity is $1\\colon 3$. If the coordinates of point B are ($-1$, $2$), what are the coordinates of point $B_1$?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle AOB$ and $\\triangle A_1OB_1$ are homothetic figures with the origin as the center of homothety, and the ratio of homothety is $1\\colon 3$, the coordinates of point $B$ are ($-1$, $2$),\\\\\ntherefore, the coordinates of point $B_1$ are ($-1\\times(-3)$, $2\\times(-3)$), which is $\\boxed{(3,-6)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51806898.png"
  },
  {
    "solution": "\\boxed{\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In rectangle $ABCD$, $AB=2BC$, and points $F$ and $G$ are on sides $AB$ and $DC$ respectively. The quadrilateral $AFGD$ is folded along $GF$ to obtain quadrilateral $EFGP$, where point $E$ lies on side $BC$, and $EP$ intersects $CD$ at point $H$. If $\\sin\\angle CGP=\\frac{3}{5}$ and $GF=2\\sqrt{5}$, what is the length of $CE$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect $AE$, and draw $GQ\\perp AB$ through point $G$,\\\\\n$\\because$ by folding,\\\\\n$\\therefore AE\\perp FG$,\\\\\n$\\therefore \\angle QGF+\\angle AFG=\\angle BAE+\\angle AFG$,\\\\\n$\\therefore \\angle QGF=\\angle BAE$,\\\\\n$\\because \\angle B=\\angle FQG$,\\\\\n$\\therefore \\triangle ABE\\sim \\triangle GQF$,\\\\\n$\\because AB=2BC$, $GF=2\\sqrt{5}$,\\\\\n$\\therefore \\frac{GF}{AE}=\\frac{QG}{AB}=\\frac{1}{2}$,\\\\\n$\\therefore AE=4\\sqrt{5}$,\\\\\n$\\because GP\\parallel FE$, $DC\\parallel AB$,\\\\\n$\\therefore \\angle PGH=\\angle EFB$,\\\\\n$\\because \\sin\\angle CGP=\\frac{3}{5}$,\\\\\n$\\therefore \\sin\\angle BFE=\\frac{3}{5}$,\\\\\nLet $BE=3k$, then $EF=5k$,\\\\\n$\\therefore FB=4k$,\\\\\n$\\therefore AB=AF+BF=EF+BF=5k+4k=9k$,\\\\\n$\\therefore AE=\\sqrt{AB^2+BE^2}=\\sqrt{(9k)^2+(3k)^2}=3\\sqrt{10}k$,\\\\\n$\\therefore 3\\sqrt{10}k=4\\sqrt{5}$,\\\\\nSolving, we find $k=\\frac{2}{3}\\sqrt{2}$,\\\\\n$\\because BC=\\frac{1}{2}AB=\\frac{9}{2}k$, $BE=3k$,\\\\\n$\\therefore EC=BC-BE=\\frac{9}{2}k-3k=\\frac{3}{2}k=\\boxed{\\sqrt{2}}$",
    "image_path": "dataset/mm_math/images/MM_Math/53080024.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown, it is known that in $\\triangle ABC$, $AD$ is the median to $BC$, and point $E$ is on $AD$ with $\\frac{DE}{AE} = \\frac{1}{3}$. The ray $CE$ intersects $AB$ at point $F$. Find the value of $\\frac{AF}{FB}$.\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $DM\\parallel AB$ through $D$, intersecting $CF$ at $M$, \\\\\n$\\therefore \\triangle DME\\sim \\triangle AFE, \\triangle CDM\\sim \\triangle CBF,$ \\\\\n$\\therefore \\frac{DE}{AE}=\\frac{DM}{AF}=\\frac{1}{3}, \\frac{CD}{CB}=\\frac{DM}{BF},$ \\\\\n$\\because D$ is the midpoint of $BC$, \\\\\n$\\therefore \\frac{CD}{CB}=\\frac{DM}{BF}=\\frac{1}{2},$ \\\\\n$\\therefore \\frac{AF}{BF}=\\boxed{\\frac{3}{2}}.$",
    "image_path": "dataset/mm_math/images/MM_Math/52977204.png"
  },
  {
    "solution": "\\boxed{32}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, points D, E, and F are the midpoints of $BC$, $AD$, and $CE$, respectively, and $S_{\\triangle AEF} = 4\\, \\text{cm}^2$, then what is the area of $\\triangle ABC$ in $\\text{cm}^2$?\n",
    "completion": "\\textbf{Solution:} Since point F is the midpoint of CE and the area of $\\triangle AEF = 4\\text{cm}^2$,\\\\\nit follows that AF is the median of $\\triangle AEC$, \\\\\nthus the area of $\\triangle AEC = 2$ times the area of $\\triangle AEF = 2 \\times 4 = 8$;\\\\\nSince point E is the midpoint of AD,\\\\\nit follows that CD is the median of $\\triangle ADC$,\\\\\nthus the area of $\\triangle ADC = 2$ times the area of $\\triangle AEC = 2 \\times 8 = 16$;\\\\\nSince point D is the midpoint of BC,\\\\\nit follows that AD is the median of $\\triangle ABC$,\\\\\nthus the area of $\\triangle ABC = 2$ times the area of $\\triangle ADC = 2 \\times 16 = \\boxed{32}\\text{cm}^2$;",
    "image_path": "dataset/mm_math/images/MM_Math/52733093.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC$, AD is the altitude, $AE\\parallel BC$, $BC=2AE$. The quadrilateral ADCE is a rectangle, point F is the midpoint of AB, connect DF, EF, if $\\angle DFE=90^\\circ$, $AB=4$, find the length of EF.\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect DE.\\\\\nSince quadrilateral ADCE is a rectangle,\\\\\nit follows that $DE=AC=AB=4$.\\\\\nSince point F is the midpoint of AB, and $\\triangle ABD$ is a right-angled triangle,\\\\\nit follows that $DF=\\frac{1}{2}AB=2$.\\\\\nIn $\\triangle DEF$, by the Pythagorean theorem, we have $EF=\\sqrt{DE^{2}-DF^{2}}=2\\sqrt{3}$.\\\\\nTherefore, the length of EF is $\\boxed{2\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51968626_117.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, with point $O$ as the center and the length of $AB$ as the diameter, a circle is drawn. A point $C$ is taken on $\\odot O$, and $AB$ is extended to point $D$. Line $DC$ is drawn, with $\\angle DCB = \\angle DAC$. A line is drawn through point $A$ such that $AE \\perp AD$ and it intersects the extension of $DC$ at point $E$. $CD$ is the tangent line of $\\odot O$. Given that $CD = 4$ and $DB = 2$, find the length of $AE$.\n",
    "completion": "\\textbf{Solution:} Let the radius of the circle $\\odot O$ be $r$, thus $OA = OB = OC = r$,\\\\\n$\\because BD = 2$,\\\\\n$\\therefore OD = OB + BD = r + 2, AD = OA + OD = 2r + 2$,\\\\\nIn right triangle $\\triangle OCD$, $CD^{2} + OC^{2} = OD^{2}$, that is $4^{2} + r^{2} = (r + 2)^{2}$,\\\\\nSolving for $r$ yields $r = 3$,\\\\\n$\\therefore OC = 3, AD = 8$,\\\\\nIn triangles $\\triangle ADE$ and $\\triangle CDO$, $\\left\\{\\begin{array}{l} \\angle DAE = \\angle DCO = 90^\\circ \\\\ \\angle D = \\angle D \\end{array}\\right.$,\\\\\n$\\therefore \\triangle ADE \\sim \\triangle CDO$,\\\\\n$\\therefore \\frac{AE}{OC} = \\frac{AD}{CD}$, that is $\\frac{AE}{3} = \\frac{8}{4}$,\\\\\nSolving yields $AE = \\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51352021_184.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{3}}{6}x^{2}-\\frac{\\sqrt{3}}{3}x-\\frac{\\sqrt{3}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that in $\\triangle ABC$, $\\angle BAC=30^\\circ$, $\\angle C=90^\\circ$, the coordinate of point A is $(-1, 0)$, and the coordinate of point B is $(3, 0)$. The vertex of the parabola $y_1$ is denoted as Q, and it passes through the three vertices A, B, and C of $\\triangle ABC$ (with point A to the left of point B, and point C below the x-axis). The parabola $y_2$ also intersects the x-axis at points A and B, and its vertex is P. When the value of $BP+CP$ is minimized, find the equation of the parabola $y_2$.\n",
    "completion": "Solution: Since the parabola $y_{1}$ and the parabola $y_{2}$ intersect the $x$-axis at the same points, the axis of symmetry of the parabola $y_{2}$ is $x=1$. Since point A is symmetrical to point B about the line $x=1$, the value of $BP+CP$ is minimized when $P$ is the intersection of line $AC$ and the axis of symmetry. Let the equation of line $AC$ be: $y=kx+b$. Then we have\n\\[\n\\left\\{\n\\begin{array}{l}\n-k+b=0\\\\\n2k+b=-\\sqrt{3}\n\\end{array}\n\\right.\n\\]\nSolving this, we get\n\\[\n\\left\\{\n\\begin{array}{l}\nk=-\\frac{\\sqrt{3}}{3}\\\\\nb=-\\frac{\\sqrt{3}}{3}\n\\end{array}\n\\right.\n\\]\nTherefore, the equation of line $AC$ is: $y=-\\frac{\\sqrt{3}}{3}x-\\frac{\\sqrt{3}}{3}$. Thus, the coordinates of point $P$ are $\\left(1，-\\frac{2\\sqrt{3}}{3}\\right)$. Let $y_{2}=m(x+1)(x-3)$, then $-4m=-\\frac{2\\sqrt{3}}{3}$, therefore $m=\\frac{\\sqrt{3}}{6}$, hence the equation of the parabola $y_{2}$ is: $y_{2}=\\frac{\\sqrt{3}}{6}(x+1)(x-3)=\\frac{\\sqrt{3}}{6}x^{2}-\\frac{\\sqrt{3}}{3}x-\\frac{\\sqrt{3}}{2}$. Therefore, the answer is $y_{2}=\\boxed{\\frac{\\sqrt{3}}{6}x^{2}-\\frac{\\sqrt{3}}{3}x-\\frac{\\sqrt{3}}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51317409_151.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C = 90^\\circ$. If $AC = 4$ and $BC = 3$, what is the value of $\\sin A$?\n",
    "completion": "\\textbf{Solution:} In right triangle $ABC$, $\\angle C=90^\\circ$, AC=4, BC=3,\\\\\nthen AB$=\\sqrt{AC^{2}+BC^{2}}=\\sqrt{4^{2}+3^{2}}=5$,\\\\\n$\\therefore \\sin A=\\frac{BC}{AB}=\\boxed{\\frac{3}{5}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51351168.png"
  },
  {
    "solution": "\\boxed{P_{4}(-3-4\\sqrt{2},0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $l_{1}: y = x + 3$ intersects with the line $l_{2}$ passing through point A$(3, 0)$ at point C$(1, m)$, and with the x-axis at point B. Is there a point P on the x-axis that makes the triangle BCP an isosceles triangle? If such a point exists, please write down the coordinates of point P; if not, please explain why.\n",
    "completion": "\\textbf{Solution:} As shown in Fig. 2, $\\because B(-3,0), C(1,4)$.\n$\\therefore BC= \\sqrt{(1+3)^{2}+(4-0)^{2}}=4\\sqrt{2}$.\nLet $P(x,0)$,\nWhen $PC=BC$, at this time point $P$ is symmetric to point $B$ about the line $x=1$, then $P_{1}(5,0)$;\nWhen $PC=PB$, $(x+3)^{2}=(x-1)^{2}+(0-4)^{2}$.\nSolving this yields $x=1$. At this time, $P_{2}(1,0)$;\nWhen $BP=BC$, $\\left|x+3\\right|=4\\sqrt{2}$,\nSolving this yields $x=-3+4\\sqrt{2}$ or $x=-3-4\\sqrt{2}$.\nAt this time, $P_{3}(-3+4\\sqrt{2},0)$, $P_{4}(-3-4\\sqrt{2},0)$.\nIn summary, the coordinates of point $P$ that meet the conditions are \\boxed{P_{1}(5,0)}, \\boxed{P_{2}(1,0)}, \\boxed{P_{3}(-3+4\\sqrt{2},0)}, \\boxed{P_{4}(-3-4\\sqrt{2},0)}.",
    "image_path": "dataset/mm_math/images/MM_Math/50979784_104.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the diagram, in $\\triangle ABC$, $\\angle BAC=90^\\circ$, $AB=5\\,cm$, $BC=13\\,cm$, point $D$ is on line segment $AC$, and $CD=7\\,cm$. A moving point $P$ starts from point $E$, which is $15\\,cm$ away from point $B$, and moves in the direction of the ray $EA$ at a speed of $2\\,cm$ per second, for a time of $t$ seconds. Find the length of $AD$.\n",
    "completion": "\\textbf{Solution:} According to the Pythagorean theorem, we have $AC=\\sqrt{BC^{2}-AB^{2}}=\\sqrt{13^{2}-5^{2}}=12\\,(\\text{cm})$,\\\\\n$\\therefore AD=AC-CD=5$,\\\\\n$\\therefore$ the length of $AD$ is $\\boxed{5}\\,\\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55588005_80.png"
  },
  {
    "solution": "\\boxed{2:1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point F is on the side AD of $ \\square ABCD$, and the extensions of BA and CF intersect at point E. If $AE:AB = 1:2$, what is the ratio of the area of the quadrilateral $ABCF$ to the area of $\\triangle CDF$?\n",
    "completion": "Solution: Since $AE: AB = 1: 2$,\\\\\ntherefore $AE: BE=1: 3$, $AE: DC=1: 2$.\\\\\nIn parallelogram $ABCD$ where $AB \\parallel CD$, $AD \\parallel BC$,\\\\\nhence $\\triangle AEF \\sim \\triangle DCF$, $\\triangle AEF \\sim \\triangle BEC$,\\\\\nthus ${S}_{\\triangle AEF}:{S}_{\\triangle DCF}=1:4$, ${S}_{\\triangle AEF}:{S}_{\\triangle BEC}=1:9$,\\\\\ntherefore ${S}_{\\triangle AEF}:{S}_{\\text{quadrilateral }ABCF}=1:8$.\\\\\nHence ${S}_{\\text{quadrilateral }ABCF}:{S}_{\\triangle CDF}=8:4=2:1$.\\\\\nThus the answer is: $\\boxed{2:1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53080811.png"
  },
  {
    "solution": "\\boxed{6-2\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, equilateral $\\triangle ABC$ has a side length of $4\\sqrt{3}$. The angle bisectors of $\\angle ABC$ and $\\angle ACB$ intersect at point O. $\\triangle OBC$ is rotated $30^\\circ$ counterclockwise around point O to obtain $\\triangle OB_{1}C_{1}$. $B_{1}C_{1}$ intersects BC at point D, and $B_{1}C_{1}$ intersects AC at point E. What is the length of DE?\n",
    "completion": "\\textbf{Solution:} Draw OH$\\perp$BC at H and OB$_{1}$ intersecting BC at M. Through M, draw MF$\\perp$BO at F, as shown in the diagram below:\\\\\nSince $\\triangle ABC$ is an equilateral triangle, and OB, OC are the angle bisectors of $\\angle ABC$, $\\angle ACB$ respectively,\\\\\ntherefore, $\\angle 1= \\frac{1}{2}\\angle ABC=30^\\circ$, $\\angle 3= \\frac{1}{2}\\angle ACB=30^\\circ$,\\\\\nthus, $\\triangle OBC$ is an isosceles triangle, and by \"the three lines coincide\", \\\\\nBH=CH=$ \\frac{1}{2}$BC=$ 2\\sqrt{3}$,\\\\\ntherefore, BO=$ \\frac{2\\sqrt{3}}{3}$BH=4,\\\\\nsince $\\triangle OBC$ is rotated $30^\\circ$ counterclockwise around point O to get $\\triangle OB_{1}C_{1}$,\\\\\ntherefore, $\\angle 2=30^\\circ=\\angle 1$,\\\\\nthus, $\\triangle OBM$ is an isosceles triangle, and by \"the three lines coincide\", \\\\\nBF=$ \\frac{1}{2}$BO=2,\\\\\nthus, MO=BM=$ \\frac{2\\sqrt{3}}{3}$BF=$ \\frac{4\\sqrt{3}}{3}$,\\\\\nthus, MB$_{1}$=OB$_{1}$−OM=OB−OM=$ 4−\\frac{4\\sqrt{3}}{3}$,\\\\\nmoreover, by the rotation $\\angle B=\\angle B_{1}=30^\\circ$, and the vertically opposite angle $\\angle BMO=\\angle DMB_{1}=120^\\circ$,\\\\\ntherefore, $\\angle MDB_{1}=180^\\circ-\\angle B_{1}-\\angle DMB_{1}=180^\\circ-30^\\circ-120^\\circ=30^\\circ$,\\\\\ntherefore, $\\triangle MB_{1}D$ is an isosceles triangle,\\\\\ntherefore, MD=MB$_{1}=$ $4-\\frac{4\\sqrt{3}}{3}$,\\\\\ntherefore, CD=BC−MD−BM=$ 4\\sqrt{3}-(4-\\frac{4\\sqrt{3}}{3})-\\frac{4\\sqrt{3}}{3}=4\\sqrt{3}-4$,\\\\\nsince the vertically opposite angle $\\angle EDC=\\angle MDB_{1}=30^\\circ$, and $\\angle ACB=60^\\circ$,\\\\\ntherefore, $\\angle DEC=180^\\circ-\\angle EDC-\\angle ACB=90^\\circ$,\\\\\ntherefore, $\\triangle CDE$ is a right-angled triangle with angles $30^\\circ$, $60^\\circ$, $90^\\circ$,\\\\\ntherefore, DE=$ \\frac{\\sqrt{3}}{2}$CD=$ \\frac{\\sqrt{3}}{2}(4\\sqrt{3}-4)=\\boxed{6-2\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51807221.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ is translated downward along the line containing the median $AD$ from side $BC$ to the position of $\\triangle ABC$. It is known that the area of $\\triangle ABC$ is $9$, and the area of the shaded triangle is $4$. If $AA'=1$, then what is the length of $A'D$?\n",
    "completion": "\\textbf{Solution:} Let $AB$, $AC$ intersect with $BE$ at points $E$, $F$, \\\\\nsince $\\triangle ABC$ is translated downwards along the line containing the median $AD$ from the side $BC$ to the position of $\\triangle ABC$, \\\\\ntherefore $\\triangle ABC \\sim \\triangle AEF$, \\\\\ntherefore $\\frac{S_{\\triangle ABC}}{S_{\\triangle AEF}}=\\left(\\frac{AD}{AD}\\right)^2$, \\\\\nsince the area of $\\triangle ABC$ is $9$, and the area of the shaded part of the triangle is $4$, $AA=1$, \\\\\ntherefore $\\frac{9}{4}=\\left(\\frac{AD+1}{AD}\\right)^2$, \\\\\nsolving this gives $AD=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51402947.png"
  },
  {
    "solution": "\\boxed{\\frac{13}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in rectangle $ABCD$, $AB=4$, $BC=6$. The perpendicular bisector of diagonal $AC$ intersects $AC$, $AD$, and $BC$ at points $O$, $E$, and $F$, respectively. Connecting $AF$ and $CE$, then what is the value of $\\frac{AE}{BF}$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral ABCD is a rectangle, it follows that AD$\\parallel$BC and $\\angle B=90^\\circ$. Since EF is the perpendicular bisector of AC, we have AF=FC. Moreover, since AD$\\parallel$BC, it follows that $\\angle$EAO=$\\angle$FCO. In $\\triangle$AOE and $\\triangle$COF, we have\n\\[\n\\left\\{\n\\begin{array}{l}\n\\angle EAO=\\angle FCO\\\\\nOA=OC\\\\\n\\angle AOE=\\angle COF\n\\end{array}\n\\right.\n\\]\nTherefore, $\\triangle$AOE$\\cong$$\\triangle$COF by ASA, which implies AE=FC. Consequently, AF=AE=FC. Letting AE=$x$, we find BF=BC−FC=6−$x$. In right triangle ABF, we have AB$^{2}$+BF$^{2}$=AF$^{2}$. Thus, $4^{2}$+$x^{2}$=(6−$x$)$^{2}$, which solves to $x=\\frac{13}{3}$. Therefore, BF=BC−FC=6−$\\frac{13}{3}$=$\\frac{5}{3}$. Hence, $\\frac{AE}{BF}=\\boxed{\\frac{13}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51510393.png"
  },
  {
    "solution": "\\boxed{4.8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rhombus ABCD, the diagonals AC and BD intersect at point O, BE $\\perp$ AD at point E, and $OA = 4$, $OB = 3$. What is the length of BE?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rhombus, \\\\\n$\\therefore$ $AC\\perp BD$, AC=2OA, BD=2OB, \\\\\nSince $OA=4$, $OB=3$, \\\\\n$\\therefore$ AD=$AB=\\sqrt{OA^2+OB^2}=5$, AC=8, BD=6, \\\\\n$\\therefore$ $S_{\\text{rhombus} ABCD}=\\frac{1}{2}AC\\cdot BD=AD\\cdot BE$, \\\\\nHence $24=5BE$, \\\\\n$\\therefore$ $BE=\\boxed{4.8}$",
    "image_path": "dataset/mm_math/images/MM_Math/52386539.png"
  },
  {
    "solution": "\\boxed{3:2:6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $AB \\parallel CD$, $AC$ intersects $BD$ at point $E$, and point $F$ is on line segment $BC$, with $\\frac{AB}{CD} = \\frac{1}{2}$, $\\frac{BF}{CF} = \\frac{1}{2}$. $AB \\parallel EF$; what is the ratio of $AB : EF : CD$?\n",
    "completion": "\\textbf{Solution:} Since $AB\\parallel EF$,\\\\\nwe have $\\frac{EF}{AB}=\\frac{CF}{BC}=\\frac{2}{3}$,\\\\\nSince $AB\\parallel CD$,\\\\\nwe have $\\frac{AB}{CD}=\\frac{AE}{EC}=\\frac{1}{2}=\\frac{3}{6}$,\\\\\nTherefore, $AB: EF: CD=\\boxed{3:2:6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52917680_94.png"
  },
  {
    "solution": "\\boxed{19.5+6\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the mathematics interest group measures the height of the flagpole $CD$ and the academic building $AB$. Firstly, at point E, they use an altimeter $EF$ of 1.5 meters high to measure the elevation angle of the top of the academic building A, $\\angle AFH$, as $45^\\circ$. At this moment, the top of the flagpole D is exactly in line with sight $FA$. Then, moving forward 12 meters to point G (G is on $CD$), the elevation angle of the top of the academic building A, $\\angle AGH$, is measured as $60^\\circ$. Points B, C, and E are on the same horizontal line. Find the height of the academic building $AB$ (express the result as an exact value).\n",
    "completion": "\\textbf{Solution:} Let GH = x meters. Given that AB$\\parallel$DC$\\parallel$EF, EF = CG = BH, and $\\angle$ABE = $90^\\circ$, \\\\\n$\\therefore$ the quadrilateral BCGH is a rectangle, $\\therefore$ $\\angle$AHF = DGF = $90^\\circ$. From (1), we have: DG = FG = 12 meters, BH = EF = 1.5 meters, \\\\\n$\\because$ $\\angle$AFH = $45^\\circ$, $\\therefore$ $\\triangle$AFH is an isosceles right triangle, \\\\\n$\\therefore$ AH = FH, $\\because$ $\\angle$AGH = $60^\\circ$, $\\therefore$ $\\tan$ $\\angle$AGH = $\\frac{AH}{GH} = \\tan60^\\circ = \\sqrt{3}$, \\\\\n$\\therefore$ AH = $\\sqrt{3}$GH = $\\sqrt{3}x$ (meters), \\\\\n$\\because$ FH = GH + FG = (x + 12) meters, $\\therefore$ $\\sqrt{3}x = x + 12$, \\\\\nSolving gives: x = $6\\sqrt{3}+6$, $\\therefore$ GH = ($6\\sqrt{3}+6$) meters, AH = (18 + $6\\sqrt{3}$) meters, \\\\\n$\\therefore$ AB = BH + AH = (19.5 + $6\\sqrt{3}$) meters, \\\\\nTherefore, the height of the teaching building AB is \\boxed{19.5+6\\sqrt{3}} meters.",
    "image_path": "dataset/mm_math/images/MM_Math/52602002_145.png"
  },
  {
    "solution": "\\boxed{2\\alpha}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the quadrilateral $ABCD$ is a rhombus, with diagonals $AC$ and $BD$ intersecting at point $O$. Let $DH \\perp AB$ at $H$ and connect $OH$. If $\\angle DHO = \\alpha$, what is the measure of $\\angle DAB$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rhombus,\\\\\nit follows that $OB=OD$, $AC\\perp BD$, $AB=AD$,\\\\\nhence $\\angle ABD=\\angle ADB$,\\\\\nsince $DH\\perp AB$,\\\\\nthen $OH=OB=\\frac{1}{2}BD$,\\\\\nsince $\\angle DHO=\\alpha $,\\\\\nthus $\\angle OHB=90^\\circ-\\angle DHO=90^\\circ-\\alpha $,\\\\\ntherefore $\\angle ABD=\\angle OHB=\\angle ADB=90^\\circ-\\alpha $,\\\\\nhence $\\angle DAB=180^\\circ-\\angle ABD-\\angle ADB=180^\\circ-90^\\circ+\\alpha -90^\\circ+\\alpha =\\boxed{2\\alpha}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52160356.png"
  },
  {
    "solution": "\\boxed{\\frac{25}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AD$ and $BE$ are the altitudes of $\\triangle ABC$, and $DE$ is connected. $\\triangle ACD \\sim \\triangle BCE$; If point $D$ is the midpoint of $BC$, $CE=6$, and $BE=8$, find the length of $AB$.\n",
    "completion": "\\textbf{Solution:} Since point D is the midpoint of BC, and AD is perpendicular to BC,\\\\\nit follows that AB=AC,\\\\\nIn right triangle $\\triangle BEC$,\\\\\nsince CE=6 and BE=8,\\\\\nwe have BC=$\\sqrt{CE^{2}+BE^{2}}=\\sqrt{6^{2}+8^{2}}=10$,\\\\\nthus CD=$\\frac{1}{2}$BC=5,\\\\\nsince $\\triangle ACD\\sim\\triangle BCE$,\\\\\nit follows that $\\frac{AD}{CD}=\\frac{BE}{EC}$,\\\\\nthus AD=$\\frac{8\\times 5}{6}=\\frac{20}{3}$,\\\\\ntherefore, AC=$\\sqrt{AD^{2}+CD^{2}}=\\sqrt{\\left(\\frac{20}{3}\\right)^{2}+5^{2}}=\\frac{25}{3}$,\\\\\nhence AB=AC=$\\boxed{\\frac{25}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52623814_119.png"
  },
  {
    "solution": "\\boxed{25^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of $\\odot O$, and C, D are two points on $\\odot O$. If $\\angle AOC=130^\\circ$, then what is the value of $\\angle D$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle AOC=130^\\circ$,\\\\\nit follows that $\\angle BOC=180^\\circ-130^\\circ=50^\\circ$,\\\\\nhence $\\angle D=\\frac{1}{2}\\angle BOC=25^\\circ$.\\\\\nTherefore, the answer is $\\boxed{25^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51261850.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{2}\\sqrt{6}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given, as shown in the diagram, in parallelogram $ABCD$, $\\angle B=60^\\circ$, $BC=6$, and point $E$ is the midpoint of side $BC$. Triangle $ABE$ is folded to the right along $AE$ so that point $B$ falls at point $B'$. Line $CB'$ is drawn and extended to intersect $AD$ at point $F$. When $AB' \\perp CD$, what is the length of $AE$?\n",
    "completion": "\\textbf{Solution:} Given $AB \\parallel CD$ and $AB' \\perp CD$, \\\\\nit follows that $AB' \\perp AB$, which means $\\angle BAB' = 90^\\circ$, \\\\\nby folding, we have $\\angle BAE = \\angle EAB' = 45^\\circ$, \\\\\nas shown in the diagram, draw $EH \\perp AB$ at point H, \\\\\nin $\\triangle BEH$, $\\angle B = 60^\\circ$ and $BE = 3$, \\\\\nthus $BH = \\frac{3}{2}$, $HE = \\sqrt{BE^2 - BH^2} = \\sqrt{3^2 - \\left(\\frac{3}{2}\\right)^2} = \\frac{3}{2}\\sqrt{3}$, \\\\\nin $\\triangle AEH$, $\\angle HAE = \\angle AEH = 45^\\circ$, $AH = EH = \\frac{3}{2}\\sqrt{3}$, \\\\\ntherefore $AE = \\sqrt{2}HE = \\boxed{\\frac{3}{2}\\sqrt{6}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51825231_77.png"
  },
  {
    "solution": "\\boxed{\\text{undetermined}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $C$ is a point on line $AB$, and point $M$ is the midpoint of segment $AC$. If $AB=8$, point $C$ is on segment $AB$, and $AC=3BC$, then what is the length of $AM$?\n",
    "completion": "\\textbf{Solution:} Since the information given in the problem is insufficient, it is not possible to provide a solution. Hence, the length of $AM$ cannot be determined. Therefore, the length of $AM$ is \\boxed{\\text{undetermined}}.",
    "image_path": "dataset/mm_math/images/MM_Math/53364701_167.png"
  },
  {
    "solution": "\\boxed{300\\,\\text{m}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AC=DB$, $AO=DO$, and $CD=300\\,m$. What is the distance in meters between points $A$ and $B$?\n",
    "completion": "\\textbf{Solution:} Since $AC=DB$ and $AO=DO$,\n\nit follows that $BO=CO$.\n\nIn triangles $AOB$ and $DOC$,\n\n\\[\n\\left\\{\\begin{array}{l}\nAO=DO\\\\\n\\angle AOB=\\angle DOC\\\\\nBO=CO\n\\end{array}\\right.,\n\\]\n\nthus $\\triangle AOB \\cong \\triangle DOC$ (SAS),\n\ntherefore $AB=DC$.\n\nSince $CD=300\\,\\text{m}$,\n\nit follows that $AB=\\boxed{300\\,\\text{m}}$,\n\nwhich means the distance between points $A$ and $B$ is $300\\,\\text{m}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52950949.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system, the graph of the linear function $y=kx+b$ intersects the x-axis and y-axis at points A and B respectively, and passes through the point $D\\left(-2,6\\right)$. It intersects the graph of the direct proportionality function $y=3x$ at point C, where the x-coordinate of point C is 1. What is the area of $\\triangle BOC$?\n",
    "completion": "\\textbf{Solution:} When $x=0$, then $y=4$, \\\\\n$\\therefore$ $S_{\\triangle BOC}=\\frac{1}{2}\\times 4\\times 1=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50520008_184.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, diagonals $AC$ and $BD$ intersect at point $E$. $BF \\parallel AC$ and $CF \\parallel BD$. If the area of quadrilateral $BECF$ is 2, what is the area of the rectangle $ABCD$?\n",
    "completion": "\\textbf{Solution:} Given that $BF\\parallel AC$ and $CF\\parallel BD$, \\\\\nit follows that quadrilateral BECF is a parallelogram, \\\\\nsince rectangle ABCD, \\\\\nwe have BE=CE, \\\\\nthus, quadrilateral BECF is a rhombus; \\\\\nhence, the area of rhombus BECF, $S_{\\text{rhombus} BECF}=2S_{\\triangle BEC}=2$, \\\\\nthus, the area of $\\triangle BEC=1$; \\\\\nsince rectangle ABCD, \\\\\nit follows that the area of rectangle ABCD, $S_{\\text{rectangle} ABCD}=4S_{\\triangle BEC}=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52188268.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated in the figure, AB is a chord of the circle $\\odot O$, $OP\\perp OA$ intersects AB at point P, and a line passing through point B intersects the extension of $OP$ at point C, with $BC$ being a tangent of $\\odot O$. Let the area of $\\triangle AOP$ be $S_1$ and the area of $\\triangle BCP$ be $S_2$, with $\\frac{S_1}{S_2}=\\frac{2}{5}$. Given that the radius of $\\odot O$ is $6$ and $BP=4\\sqrt{5}$, find $\\tan\\angle APO$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, construct $CD\\perp BP$ at $D$,\\\\\n$\\because PC=CB$,\\\\\n$\\therefore PD=BD=\\frac{1}{2}PB=2\\sqrt{5}$,\\\\\n$\\because \\angle PDC=\\angle AOP=90^\\circ$, $\\angle APO=\\angle CPD$,\\\\\n$\\therefore \\triangle AOP\\sim \\triangle PCD$,\\\\\n$\\because \\frac{{S}_{1}}{{S}_{2}}=\\frac{2}{5}$,\\\\\n$\\therefore \\frac{{S}_{\\triangle AOP}}{{S}_{\\triangle PCD}}=\\frac{4}{5}$,\\\\\n$\\therefore \\frac{OA^{2}}{CD^{2}}=\\frac{4}{5}$,\\\\\n$\\because OA=6$,\\\\\n$\\therefore CD=3\\sqrt{5}$,\\\\\n$\\therefore \\tan\\angle APO=\\tan\\angle CBP=\\frac{CD}{BD}=\\frac{3\\sqrt{5}}{2\\sqrt{5}}=\\boxed{\\frac{3}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51322517_140.png"
  },
  {
    "solution": "\\boxed{\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, rhombus OABC is in the first quadrant, with $\\angle AOC = 60^\\circ$, and the graph of the inverse proportion function $y = \\frac{k}{x}$ ($k > 0$) passes through point A and intersects side BC at point D. If the area of $\\triangle AOD$ is $\\sqrt{3}$, what is the value of $k$?\n",
    "completion": "\\textbf{Solution:} Join AC, through point A draw AE$\\perp$OC at E, \\\\\n$\\because$ Quadrilateral ABCO is a rhombus, \\\\\n$\\therefore$AO$\\parallel$CB, OA=OC, and $\\angle$AOC$=60^\\circ$, \\\\\n$\\therefore$$\\triangle$AOC is an equilateral triangle, and AE$\\perp$OC, \\\\\n$\\therefore$S$_{\\triangle AOE}=\\frac{1}{2}$S$_{\\triangle AOC}$, \\\\\n$\\because$OA$\\parallel$BC, \\\\\n$\\therefore$S$_{\\triangle OAD}$=S$_{\\triangle OAC}=\\sqrt{3}$, \\\\\n$\\therefore$S$_{\\triangle AOE}=\\frac{1}{2}$S$_{\\triangle AOC}=\\frac{\\sqrt{3}}{2}=\\frac{k}{2}$, \\\\\n$\\therefore$k=$\\boxed{\\sqrt{3}}$,",
    "image_path": "dataset/mm_math/images/MM_Math/52623629.png"
  },
  {
    "solution": "\\boxed{120}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, for the rhombus $ABCD$, if $DB = 10$ and $AB = 13$, then what is the area of the rhombus $ABCD$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect $AC$, intersecting $DB$ at point O, \\\\\nsince quadrilateral $ABCD$ is a rhombus and $DB=10$,\\\\\ntherefore $AC=2OA$, $OB=\\frac{1}{2}DB=5$, $AC\\perp BD$,\\\\\nsince $AB=13$,\\\\\ntherefore $OA=\\sqrt{AB^{2}-OB^{2}}=12$,\\\\\nthus $AC=24$,\\\\\nthen the area of the rhombus $ABCD$ is $\\frac{1}{2}AC\\cdot DB=\\frac{1}{2}\\times 24\\times 10=\\boxed{120}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52212584.png"
  },
  {
    "solution": "\\boxed{44^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $OC$ bisects $\\angle AOB$, and $\\angle BOC=22^\\circ$. What is the measure of $\\angle AOB$?\n",
    "completion": "\\textbf{Solution:} Since OC bisects $\\angle AOB$ and $\\angle BOC=22^\\circ$,\\\\\nit follows that $\\angle AOB =2\\angle BOC=44^\\circ$.\\\\\nTherefore, $\\angle AOB = \\boxed{44^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51407576.png"
  },
  {
    "solution": "\\boxed{4.8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, $AC=6$, $BC=8$. Point $D$ is on side $AB$, $DE \\perp AC$, and $DF \\perp BC$ with the feet of the perpendiculars being points $E$ and $F$, respectively. Join $EF$. What is the minimal value of segment $EF$?\n",
    "completion": "\\textbf{Solution:} Therefore, the minimum value of line segment $EF$ is $\\boxed{4.8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52211946.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the $8\\times 8$ grid shown in the figure, where the side length of each small square is $1$, and points $A$, $B$, $C$, and $D$ are all on the grid points. If $AB$ and $CD$ intersect at point $E$, what is the value of the tangent of $\\angle AED$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, select the lattice point K, and draw lines AK and BK.\\\\\nObserving the figure, it is clear that $AK \\perp BK$, $BK=2AK$, and $BK \\parallel CD$,\\\\\n$\\therefore \\angle AED=\\angle ABK$,\\\\\n$\\therefore \\tan \\angle AED=\\tan \\angle ABK= \\frac{AK}{BK}=\\boxed{\\frac{1}{2}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51322489.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, AB is the diameter of $\\odot O$, and points C and D are on $\\odot O$. If $\\angle D=120^\\circ$, what is the measure of $\\angle CAB$?\n",
    "completion": "\\textbf{Solution}: Since $\\angle D + \\angle B = 180^\\circ$ and $\\angle D = 120^\\circ$,\n\n$\\therefore \\angle B = 60^\\circ$,\n\nSince $AB$ is the diameter,\n\n$\\therefore \\angle ACB = 90^\\circ$,\n\nHence, $\\angle CAB = 90^\\circ - \\angle B = \\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51262127.png"
  },
  {
    "solution": "\\boxed{\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, square $ABCD$ has a side length of $2$. Points $E$ and $F$ are two moving points on side $AD$, satisfying $AE=DF$. Lines $BE$ and $CF$ are drawn, and $BE$ intersects the diagonal $AC$ at point $G$. Line $DG$ is drawn and intersects $CF$ at point $H$. When line $BH$ is drawn, what is the minimum value of $BH$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, \\\\\nsince square ABCD, it follows that $AB=CD$, $\\angle BAE=\\angle CDF=90^\\circ$ thus $\\triangle ABE \\cong \\triangle DCF$ (SAS), hence $\\angle ABE=\\angle DCF$. Since the median line where AC is located is the axis of symmetry of the square, it follows that $\\angle ABE=\\angle ADG=\\angle DCF$. Since $\\angle ADG + \\angle GDC = 90^\\circ$, it implies $\\angle DCF + \\angle GDC = 90^\\circ$, hence $\\angle DHC = 90^\\circ$ which means $DH \\perp CF$. Therefore, point H moves on the circle with CD as its diameter. When point H moves to the midpoint of AC, BH is the shortest. Connect BD and intercept AC at point O, since $BH + DH \\geq BD$, when points B, H, and D are on the same line, since the perpendicular segment is the shortest, hence when BH $\\perp$ AC, that is, when point H coincides with point O, the value of BH is minimum. In right $\\triangle BOC$, $BO=BC\\sin\\angle BCO=2\\times \\sin45^\\circ=2\\times \\frac{\\sqrt{2}}{2}=\\sqrt{2}$. Therefore, the minimum value of BH is $\\boxed{\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51407044.png"
  },
  {
    "solution": "\\boxed{\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, circle $O$ is the incircle of $\\triangle ABC$, where $AB=7, BC=5$, $AC=8$. Determine the radius of the incircle.\n",
    "completion": "\\textbf{Solution:} Construct BD $\\perp$ AC at D, with tangency points E, F, and G, respectively. Connect OE, OF, OG,\\\\\nLet AD = x, CD = 8−x, and the radius of the inscribed circle be r,\\\\\nAccording to the Pythagorean theorem, $AB^{2}-AD^{2}=BC^{2}-CD^{2}$, that is $7^{2}-x^{2}=5^{2}-(8-x)^{2}$,\\\\\nSolving the equation, we get $x=\\frac{11}{2}$,\\\\\nTherefore, BD = $\\sqrt{AB^{2}-AD^{2}}=\\sqrt{7^{2}-\\left(\\frac{11}{2}\\right)^{2}}=\\frac{5}{2}\\sqrt{3}$,\\\\\nBecause circle O is the inscribed circle of $\\triangle ABC$,\\\\\nTherefore, OE $\\perp$ AC, OF $\\perp$ AB, OG $\\perp$ BC, OE = OF = OG = r,\\\\\nTherefore, $S_{\\triangle ABC}=\\frac{1}{2}AC\\cdot BD=\\frac{1}{2}AB\\cdot OF+\\frac{1}{2}BC\\cdot OG+\\frac{1}{2}AC\\cdot OE=\\frac{1}{2}(AB+BC+AC)\\cdot r$,\\\\\nTherefore, AC$\\cdot$BD = $(AB+BC+AC)\\cdot r$,\\\\\nTherefore, $r=\\frac{AC\\cdot BD}{AB+BC+AC}=\\frac{8\\times \\frac{5}{2}\\sqrt{3}}{20}=\\boxed{\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53065827.png"
  },
  {
    "solution": "\\boxed{72^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\angle AOB$ and $\\angle BOC$ are adjacent supplementary angles. $OD$ is the angle bisector of $\\angle AOB$, and $OE$ is within $\\angle BOC$ such that $\\angle BOE=\\frac{1}{2}\\angle EOC$, and $\\angle DOE={72}^\\circ$. Find the degree measure of $\\angle EOC$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle BOE = \\frac{1}{2}\\angle EOC$, \\\\\nthen $\\angle BOC = \\angle BOE + \\angle EOC = 3\\angle BOE$, \\\\\nsince $\\angle DOE = 72^\\circ$, \\\\\nthus $\\angle BOD = \\angle DOE - \\angle BOE = 72^\\circ - \\angle BOE$, \\\\\nsince $OD$ bisects $\\angle AOB$, \\\\\nthus $\\angle AOB = 2\\angle BOD = 144^\\circ - 2\\angle BOE$, \\\\\nsince $\\angle AOB$ and $\\angle BOC$ are supplementary, \\\\\nthus $\\angle AOB + \\angle BOC = 180^\\circ$, \\\\\ntherefore $144^\\circ - 2\\angle BOE + 3\\angle BOE = 180^\\circ$, \\\\\ntherefore $\\angle BOE = 36^\\circ$, \\\\\ntherefore $\\angle EOC = 2\\angle BOE = \\boxed{72^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50815659.png"
  },
  {
    "solution": "\\boxed{(3, 4)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the rectangle $ABCD$ has side $AB$ on the $x$ axis, with $AB=3$ and $AD=2$. The line through point $C$, given by $y=x-2$, intersects the $x$ axis and the $y$ axis at points $E$ and $F$, respectively. Determine the coordinate of point $M$ in the Cartesian coordinate system such that the quadrilateral formed by points $M$, $D$, $C$, and $E$ is a parallelogram. Please directly provide the coordinates of point $M$.\n",
    "completion": "\\textbf{Solution:} When $y=0$, we have $x-2=0$, solving this gives $x=2$, thus $OE=2$. Since the quadrilateral with vertices $M$, $D$, $C$, $E$ is a parallelogram, if $DE$ is a diagonal, then $EM=CD=3$, therefore $OM=EM-OE=3-2=1$, in this case, the coordinates of point $M$ are $(-1, 0)$. If $CE$ is a diagonal, then $EM=CD=3$, and $OM=OE+EM=2+3=5$, in this case, the coordinates of point $M$ are $(5, 0)$. If $CD$ is a diagonal, then the center of the parallelogram has coordinates $\\left(\\frac{5}{2}, 2\\right)$, let the coordinates of point $M$ be $(x, y)$, then we have $\\frac{x+2}{2}=\\frac{5}{2}$ and $\\frac{y+0}{2}=2$, solving these gives $x=3$ and $y=4$, in this case, the coordinates of point $M$ are $(3, 4)$. In summary, the coordinates of point $M$ can be $(-1, 0)$, $(5, 0)$, or \\boxed{(3, 4)}.",
    "image_path": "dataset/mm_math/images/MM_Math/50561201_59.png"
  },
  {
    "solution": "\\boxed{35^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AD$ and $CE$ are respectively the median and angle bisector of $\\triangle ABC$, with $AB=AC$. If $\\angle CAD=20^\\circ$, find the measure of $\\angle ACE$.\n",
    "completion": "\\textbf{Solution:} Given that, since $AD$ is the median of $\\triangle ABC$, $AB=AC$, $\\angle CAD=20^\\circ$, \\\\\nthus $\\angle CAB=2\\angle CAD=40^\\circ$, $\\angle B=\\angle ACB=\\frac{1}{2}\\left(180^\\circ-\\angle CAB\\right)=70^\\circ$.\\\\\nSince $CE$ is the angle bisector of $\\triangle ABC$,\\\\\nthus $\\angle ACE=\\frac{1}{2}\\angle ACB=\\boxed{35^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52861537_106.png"
  },
  {
    "solution": "\\boxed{36^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the rhombus $ABCD$, the perpendicular bisector of $AD$ intersects $AD$ at point $E$, and intersects $AC$ at point $F$, $\\angle ABC=4\\angle DFC$, what is the measure of $\\angle BAD$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram,\\\\\n$\\because EF$ perpendicularly bisects $AD$.\\\\\n$\\therefore AF=DF$.\\\\\n$\\therefore \\angle FAD=\\angle FDA$.\\\\\n$\\because \\angle DFC=\\angle FAD+\\angle FDA$.\\\\\n$\\therefore \\angle FDC=2\\angle FAD$.\\\\\nIn rhombus $ABCD$, $AC$ is the diagonal,\\\\\n$\\therefore \\angle BAD=2\\angle FAD$.\\\\\n$\\therefore \\angle BAD=\\angle DFC$.\\\\\n$\\because \\angle ABC=4\\angle DFC$.\\\\\n$\\therefore \\angle ABC=4\\angle BAD$.\\\\\n$\\because \\angle BAD+\\angle ABC=180^\\circ$.\\\\\n$\\therefore \\angle BAD+4\\angle BAD=180^\\circ$.\\\\\n$\\therefore \\angle BAD=\\boxed{36^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/52204767.png"
  },
  {
    "solution": "\\boxed{-\\frac{1}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the function $y=kx+b$ (where $k$ and $b$ are constants) is illustrated. What is the value of $k$?\n",
    "completion": "\\textbf{Solution:} From the given problem, we see that the slope $k$ of the function $y=kx+b$ is $-\\frac{1}{2}$,\n\n$\\therefore k=\\boxed{-\\frac{1}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52422319_28.png"
  },
  {
    "solution": "\\boxed{\\text{Insufficient information}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within circle $O$, the diameter $AB$ intersects the chord $CD$ at point $P$, $\\angle CAB=40^\\circ$, $\\angle APD=65^\\circ$. If $AD=6$, what is the distance from the center of the circle $O$ to $BD$?\n",
    "completion": "\\textbf{Solution:} It is insufficient to deduce the distance from the center $O$ to $BD$ based on the given information, therefore a specific solution cannot be provided. $\\boxed{\\text{Insufficient information}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52814915_122.png"
  },
  {
    "solution": "\\boxed{65^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, it is known that $\\triangle ABC \\cong \\triangle DBE$, with point D on $AC$, and $BC$ intersects $DE$ at point P. If $\\angle ABE=160^\\circ$ and $\\angle DBC=30^\\circ$, what is the degree measure of $\\angle CBE$?\n",
    "completion": "\\textbf{Solution:} Given that $\\triangle ABC \\cong \\triangle DBE$,\\\\\nit follows that $\\angle ABC = \\angle DBE$, that is $\\angle ABD + \\angle DBC = \\angle DBC + \\angle CBE$,\\\\\ntherefore $\\angle ABD = \\angle CBE$,\\\\\nsince $\\angle ABE = 160^\\circ$ and $\\angle DBC = 30^\\circ$,\\\\\nit follows that $\\angle ABD + \\angle DBC + \\angle CBE = \\angle ABE = 160^\\circ$,\\\\\ntherefore $\\angle ABD = \\angle CBE = \\frac{1}{2}(\\angle ABE - \\angle DBC) = \\frac{1}{2}(160^\\circ - 30^\\circ) = \\boxed{65^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52978985.png"
  },
  {
    "solution": "\\boxed{15}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, it is known that points $D$ and $E$ are the midpoints of sides $AB$ and $CB$ of $\\triangle ABC$, respectively. If $AB = 8$, $CE = 6$, and $AC = 10$, what is the perimeter of $\\triangle BDE$?\n",
    "completion": "\\textbf{Solution:} Given that points D and E are the midpoints of sides AB and CB of $\\triangle ABC$ respectively,\n\n$\\therefore$ DE= $\\frac{1}{2}$ AC=5, BE=CE=6, BD= $\\frac{1}{2}$ AB=4,\n\n$\\therefore$ the perimeter of $\\triangle BDE$=BD+BE+DE=4+6+5=\\boxed{15}.",
    "image_path": "dataset/mm_math/images/MM_Math/52252059.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point O is a point on line AB, with $\\angle BOC : \\angle AOC = 1 : 2$. OD bisects $\\angle BOC$, and OE is perpendicular to OD at point O. What is the measure of $\\angle BOC$ in degrees?\n",
    "completion": "\\textbf{Solution:} Since $\\angle AOB = \\angle BOC + \\angle AOC = 180^\\circ$ and $\\angle BOC : \\angle AOC = 1 : 2$, it follows that $\\angle AOC = 2\\angle BOC$. Therefore, $\\angle BOC + 2\\angle BOC = 180^\\circ$, which implies $\\angle BOC = \\boxed{60^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51434664_64.png"
  },
  {
    "solution": "\\boxed{150^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, with point A of square ABCD as the center and AB as the radius, draw the arc $\\overset{\\frown}{BD}$. Choose a point F on arc $\\overset{\\frown}{BD}$ such that $DF=DC$, and let point E be another point on arc $\\overset{\\frown}{BD}$ (not coinciding with points D or F). What is the value of $\\angle DEF$?\n",
    "completion": "\\textbf{Solution:} Draw AE and AF.\\\\\n$\\because$ With A as the center of the circle and AB as the radius, construct arc $\\overset{\\frown}{BD}$, and choose a point F on arc $\\overset{\\frown}{BD}$ such that DF=DC.\\\\\n$\\therefore$AD=AE=AF=DF,\\\\\n$\\therefore$$\\triangle$ADF is an equilateral triangle,\\\\\n$\\therefore$$\\angle$DAF=$60^\\circ$, $\\angle$ADE=$\\angle$DEA, $\\angle$AEF=$\\angle$AFE,\\\\\n$\\because$$\\angle$DAE+$\\angle$ADE+$\\angle$DEA=$180^\\circ$, then 2$\\angle$ADE=$180^\\circ$−$\\angle$DAE,\\\\\n$\\angle$EAF+$\\angle$AEF+$\\angle$AFE=$180^\\circ$, then 2$\\angle$AEF=$180^\\circ$−$\\angle$EAF,\\\\\n$\\therefore$2$\\angle$ADE+2$\\angle$AEF=$360^\\circ$−($\\angle$DAE+$\\angle$EAF)=$360^\\circ$−$\\angle$DAF=$300^\\circ$,\\\\\n$\\therefore$$\\angle$ADE+$\\angle$AEF=$150^\\circ$, thus $\\angle$DEF=\\boxed{150^\\circ}.",
    "image_path": "dataset/mm_math/images/MM_Math/52733109.png"
  },
  {
    "solution": "\\boxed{145^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the quadrilateral $ABCD$ is an inscribed quadrilateral of the circle $\\odot O$. If $\\angle BOD = 70^\\circ$, what is the measure of $\\angle C$ in degrees?\n",
    "completion": "\\textbf{Solution:} Since $\\angle BOD=70^\\circ$,\\\\\nit follows that $\\angle A=\\frac{1}{2}\\angle BOD=35^\\circ$,\\\\\nsince quadrilateral $ABCD$ is inscribed in circle $\\odot O$,\\\\\nit follows that $\\angle C+\\angle A=180^\\circ$,\\\\\ntherefore, $\\angle C=\\boxed{145^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53147971.png"
  },
  {
    "solution": "\\boxed{3a}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\triangle ABC$, $ AC=2$, $ BC=4$, and D is a point on BC, $ \\angle CAD=\\angle B$, the area of $ \\triangle ADC$ is $a$, then what is the area of $ \\triangle ABD$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle CAD=\\angle B$ and $\\angle ACD=\\angle BCA$,\\\\\nit follows that $\\triangle ACD\\sim \\triangle BCA$,\\\\\nhence $\\frac{{S}_{\\triangle ACD}}{{S}_{\\triangle BCA}}=\\left(\\frac{AC}{BC}\\right)^{2}$,\\\\\ntherefore $\\frac{a}{{S}_{\\triangle BCA}}=\\frac{1}{4}$,\\\\\nsolving this, the area of $\\triangle BCA$ is $4a$,\\\\\nthus, the area of $\\triangle ABD$ is: $4a-a=\\boxed{3a}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52977726.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=2x$ intersects the line $y=kx+b$ at point $P(m, 2)$. What is the solution of the equation $kx+b=2$ with respect to $x$?\n",
    "completion": "\\textbf{Solution:} Since the line $y=2x$ intersects with $y=kx+b$ at point $P(1,2)$, \\\\\ntherefore, when $x=1$, $y=kx+b=2$, \\\\\nthus, the solution to the equation in terms of $x$, $kx+b=2$, is $x=\\boxed{1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52424459.png"
  },
  {
    "solution": "\\boxed{y=x^{2}+2x-3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system, A(-3, 0) and B(1, 0) are given points. Point A is rotated counterclockwise about the origin until it lies on point C on the negative half of the y-axis. The parabola $y=ax^2+bx+c$ passes through points A, B, and C. Find the equation of the parabola by connecting points A and C.\n",
    "completion": "\\textbf{Solution:} \\text{Given that, we know } C(0, -3) \\\\\n\\text{Substituting points } \\[A(-3, 0), B(1, 0), \\text{ and } C(0, -3) \\text{ into the parabola equation } y=ax^{2}+bx+c \\text{ yields}\\] \\\\\n\\[\n\\begin{cases}\n9a+3b+c=0 \\\\\na+b+c=0 \\\\\nc=-3\n\\end{cases}\n\\text{Solving this, we get: }\n\\begin{cases}\na=1 \\\\\nb=2 \\\\\nc=-3\n\\end{cases}\n\\]\n\\text{Therefore, the quadratic function is given by } \\boxed{y=x^{2}+2x-3}",
    "image_path": "dataset/mm_math/images/MM_Math/52602789_87.png"
  },
  {
    "solution": "\\boxed{\\left(\\frac{3}{5}t, \\frac{40-4t}{5}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the line $y=-\\frac{4}{3}x+8$ intersects the x-axis at point A and the y-axis at point B. A moving point C starts from point B and moves towards point A along line $BA$ at a constant speed of 1 unit per second. Simultaneously, another moving point D starts from point A and moves towards point O along line $AO$ at a constant speed of 2 units per second. The motion of one point stops when the other point stops. Let the time of motion be t seconds ($t>0$). Express the coordinates of point C using an algebraic equation involving t.\n",
    "completion": "\\textbf{Solution:} Construct $CM\\perp x$ axis and $CN\\perp y$ axis through point C,\\\\\n$\\therefore CN\\parallel x$ axis and $CM\\parallel y$ axis,\\\\\nIn $\\triangle AOB$, $AB=\\sqrt{OA^2+OB^2}=10$,\\\\\nAccording to the given information, $BC=t$, thus $AC=10-t$,\\\\\n$\\therefore \\frac{CN}{OA}=\\frac{BC}{AB}$, which means $\\frac{CN}{6}=\\frac{t}{10}$,\\\\\nSolving this gives: $CN=\\frac{3}{5}t$,\\\\\n$\\frac{CM}{OB}=\\frac{AC}{BC}$, which means $\\frac{CM}{8}=\\frac{10-t}{10}$,\\\\\nSolving this gives: $CM=\\frac{40-4t}{5}$,\\\\\n$\\therefore$The coordinates of point C are $\\boxed{\\left(\\frac{3}{5}t, \\frac{40-4t}{5}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52917352_79.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $BC = 1$, $AB = 2$. A perpendicular line is drawn from a point $P$ on the diagonal $BD$ to intersect $AB$ at point $F$ and $CD$ at point $E$. Line $EG$ is drawn parallel to $BD$ from point $E$ and intersects $BC$ at point $G$. Lines $FG$ and $DF$ are drawn to intersect $BD$ at point $H$. Calculate the value of $ \\frac{DE}{BG}$.\n",
    "completion": "\\textbf{Solution:} As shown in Figure 1, since quadrilateral ABCD is a rectangle,\\\\\nit follows that $\\angle ABC=90^\\circ$,\\\\\nsince EF$\\perp$AB,\\\\\nit follows that $\\angle AFE=90^\\circ$,\\\\\nthus $\\angle ABC=\\angle AFE$,\\\\\nthus EF$\\parallel$BC,\\\\\nsince EG$\\parallel$BD,\\\\\nit follows that quadrilateral EPBG is a parallelogram,\\\\\nthus EP=BG,\\\\\nthus $\\tan \\angle EDP= \\frac{EP}{DE} = \\frac{BC}{CD} = \\frac{1}{2}$,\\\\\nthus $\\frac{DE}{BG} = \\boxed{2}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51722675_124.png"
  },
  {
    "solution": "\\boxed{2.8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, points $D$ and $E$ are located on the sides $AB$ and $AC$ respectively, with $DE \\parallel BC$. Lines $BE$ and $CD$ intersect at point $O$. Given $AE=2$, $AD=EC=3$, $BD=4.5$, and $DE=4$. If the area of $\\triangle BOC$ is $5$, what is the area of $\\triangle CED$?\n",
    "completion": "\\textbf{Solution:} Since $DE \\parallel BC$,\n\\[\\therefore \\triangle ODE \\sim \\triangle OBC,\\]\n\\[\\therefore \\frac{S_{\\triangle DOE}}{S_{\\triangle COB}} = \\left( \\frac{DE}{BC} \\right)^2 = \\left( \\frac{4}{10} \\right)^2 = \\frac{4}{25},\\] \\[\\frac{OE}{OB} = \\frac{DE}{BC} = \\frac{2}{5},\\]\nSince $S_{\\triangle BOC} = 5$,\n\\[\\therefore S_{\\triangle DOE} = \\frac{4}{5},S_{\\triangle EOC} = 2,\\]\n\\[\\therefore S_{\\triangle CED} = S_{\\triangle EOD} + S_{\\triangle EOC} = \\frac{4}{5} + 2 = \\boxed{2.8}.\\]",
    "image_path": "dataset/mm_math/images/MM_Math/52917570_61.png"
  },
  {
    "solution": "\\boxed{P_{2}(2, -1)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, it is known that the parabola $y=ax^2+bx+c\\ (a\\ne 0)$ has its vertex at point Q$(2, -1)$, and it intersects with the y-axis at point C$(0, 3)$ and with the x-axis at points A and B (point A to the right of point B). Point P is a moving point on the parabola, moving from point C towards point A along the parabola (point P does not coincide with A). A line through point P is drawn, $PD \\parallel y$-axis, intersecting AC at point D. When $\\triangle ADP$ is a right-angled triangle, what are the coordinates of point P?\n",
    "completion": "\\textbf{Solution:} There are two cases to consider: When point $P_1$ is the right angle vertex, point $P_1$ coincides with point B (as shown in the diagram)\\\\\n    Let $y=0$, then $x^{2}-4x+3=0$\\\\\n    Solving this, we get $x_{1}=1$, $x_{2}=3$\\\\\n    Since point A is to the right of point B,\\\\\n    Therefore, B(1, 0), A(3, 0)\\\\\n    Hence, $P_1$(1, 0)\n    When point A is the right angle vertex of $\\triangle APD_{2}$ (as shown in the diagram)\\\\\n    Since OA=OC and $\\angle AOC=90^\\circ$,\\\\\n    Therefore, $\\angle OAD_{2}=45^\\circ$\\\\\n    When $\\angle D_{2}AP_{2}=90^\\circ$, $\\angle OAP_{2}=45^\\circ$,\\\\\n    Hence, AO bisects $\\angle D_{2}AP_{2}$\\\\\n    Also, since $P_{2}D_{2}\\parallel y$-axis,\\\\\n    Therefore, $P_{2}D_{2}\\perp$ AO,\\\\\n    Therefore, $P_{2}$ and $D_{2}$ are symmetric about the $x$-axis\\\\\n    Let the equation of line AC be $y=kx+b$\\\\\n    Substituting A(3, 0), C(0, 3) into the equation, we get\\\\\n    $\\left\\{\\begin{array}{l}\n    0=3k+b\\\\\n    3=b\n    \\end{array}\\right.$,\\\\\n    Therefore, $\\left\\{\\begin{array}{l}\n    k=-1\\\\\n    b=3\n    \\end{array}\\right.$\\\\\n    Hence, $y=-x+3$\\\\\n    Since $D_{2}$ is on $y=-x+3$ and $P_{2}$ is on $y=x^{2}-4x+3$,\\\\\n    Let $D_{2}$($x$, $-x+3$), $P_{2}$($x$, $x^{2}-4x+3$)\\\\\n    Hence, ($-x+3$) + ($x^{2}-4x+3$) = 0\\\\\n    That is, $x^{2}-5x+6=0$,\\\\\n    Therefore, $x_{1}=2$, $x_{2}=3$ (discard)\\\\\n    Therefore, when $x=2$, $y=x^{2}-4x+3=2^{2}-4\\times 2+3=-1$\\\\\n    Hence, the coordinates of $P_{2}$ are $P_{2}$(2, $-1$) (i.e., the vertex of the parabola)\\\\\n    Therefore, the coordinates of point P are \\boxed{P_{1}(1, 0)}, \\boxed{P_{2}(2, -1)}",
    "image_path": "dataset/mm_math/images/MM_Math/52622423_222.png"
  },
  {
    "solution": "\\boxed{90^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, we have already learned the concept of angle bisectors. So, can you use it to solve related problems? Assuming another angle is also folded over such that side $CD$ coincides with $CA'$, with the crease being $CE$, what are the measures of $\\angle 1$ and $\\angle BCE$?\n",
    "completion": "\\textbf{Solution}: From the conclusion of (1), we have $\\angle A'CD=110^\\circ$,\\\\\ndue to the property of folding, we obtain $\\angle 1= \\frac{1}{2} \\angle A'CD= \\frac{1}{2} \\times 110^\\circ=55^\\circ$,\\\\\nby the property of angle bisector, we get $\\angle 2=\\angle ACB=35^\\circ$;\\\\\n$\\therefore \\angle BCE=\\angle 1+\\angle 2=35^\\circ+55^\\circ=\\boxed{90^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53206857_139.png"
  },
  {
    "solution": "\\boxed{1.1\\le \\rho \\le 3.3\\,(kg/m^{3})}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, a sealed container holds a certain mass of carbon dioxide. When the volume $V$ (in cubic meters, $m^{3}$) of the container changes, the density $\\rho$ (in kilograms per cubic meter, $kg/m^{3}$) of the gas changes accordingly. It is known that the relationship between density $\\rho$ and volume $V$ is an inverse proportion, as depicted in the graph. When $V = 5m^{3}$, $\\rho = 1.98kg/m^{3}$. For $3 \\leq V \\leq 9$, determine the range of the carbon dioxide density $\\rho$.\n",
    "completion": "\\textbf{Solution:} By observing the function graph, it is known that $\\rho$ decreases with the increase of $V$,\\\\\nwhen $V=3\\,m^{3}$, $\\rho=\\frac{9.9}{3}=3.3\\,kg/m^{3}$,\\\\\nwhen $V=9\\,m^{3}$, $\\rho=\\frac{9.9}{9}=1.1\\,kg/m^{3}$,\\\\\n$\\therefore$ when $3\\le V\\le 9$, $1.1\\le \\rho \\le 3.3\\,(kg/m^{3})$\\\\\nThat is, the range of change in the density of carbon dioxide $\\rho$ is $\\boxed{1.1\\le \\rho \\le 3.3\\,(kg/m^{3})}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52977197_179.png"
  },
  {
    "solution": "\\boxed{65^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $AB$ is the diameter of the circumcircle of $\\triangle ABC$, and $\\angle A=25^\\circ$, then what is the degree measure of $\\angle B$?\n",
    "completion": "\\textbf{Solution:} Since $AB$ is the diameter of the circumcircle of $\\triangle ABC$, \\\\\nthen $\\angle ACB = 90^\\circ$, \\\\\nthus $\\angle A + \\angle B = 90^\\circ$, \\\\\nmoreover, since $\\angle A = 25^\\circ$, \\\\\ntherefore $\\angle B = 90^\\circ - \\angle A = 90^\\circ - 25^\\circ = \\boxed{65^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52874631.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, the diagonals $AC$ and $BD$ of square $\\square ABCD$ intersect at point $O$. A line $EF\\perp AC$ passing through point $O$ intersects $AB$ and $DC$ at points $E$ and $F$, respectively. Lines $AF$ and $CE$ are connected. If $OE= \\frac{3}{2}$, what is the length of $EF$?\n",
    "completion": "\\textbf{Solution:} From the given, since $\\square ABCD$, we have AO=OC, AB$\\parallel$DC, therefore $\\angle$FCO=$\\angle$EAO, $\\angle$CFO=$\\angle$AEO, hence $\\triangle FCO\\cong \\triangle EAO$ (AAS), thus FC=AE, therefore quadrilateral AECF is a parallelogram, thus OE=OF=$\\frac{1}{2}$EF, since OE=$\\frac{3}{2}$, therefore EF=\\boxed{3}.",
    "image_path": "dataset/mm_math/images/MM_Math/52205074_103.png"
  },
  {
    "solution": "\\boxed{24}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC$, $D$ is the midpoint of side $BC$, $\\angle B=25^\\circ$. If $AB=13$ and $AD=5$, what is the length of $BC$?\n",
    "completion": "\\textbf{Solution:} Given that $\\because AB=13$ and $AD=5$,\\\\\n$\\therefore BD= \\sqrt{AB^{2}-AD^{2}}=\\sqrt{13^{2}-5^{2}}=12$,\\\\\n$\\because AD\\perp BD$,\\\\\n$\\therefore BC=2BD=2\\times 12=\\boxed{24}.$",
    "image_path": "dataset/mm_math/images/MM_Math/52604307_110.png"
  },
  {
    "solution": "\\boxed{Q(5,0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, $AB\\parallel OC$, with $A(0, 12)$, $B(a, c)$, $C(b, 0)$, and $a, b$ satisfying $\\left|a-21\\right|+\\left(b-16\\right)^2=0$. A moving point $P$ starts from point $A$ and moves towards point $B$ along the line segment $AB$ at a speed of 2 units per second; a moving point $Q$ starts from point $O$ and moves towards point $C$ along the line segment $OC$ at a speed of 1 unit per second. Points $P$ and $Q$ start moving from points $A$ and $O$ simultaneously, and when point $P$ reaches point $B$, it stops moving, and point $Q$ stops moving as well. Let the movement time be $t$ seconds. At what value of $t$ will the quadrilateral $PQCB$ be a parallelogram? What are the coordinates of points $P$ and $Q$ at this time?\n",
    "completion": "\\textbf{Solution:} Since $BQ \\parallel CP$, when $BQ=CP$, the quadrilateral $PQCB$ is a parallelogram,\\\\\nthus, $21-2t=16-t$, solving this gives: $t=\\boxed{5}$,\\\\\ntherefore, when $t=5$, the quadrilateral $PQCB$ is a parallelogram,\\\\\nthus, $AP=2t=10$,$OQ=t=5$,\\\\\n$\\boxed{P(10,12)}$, $\\boxed{Q(5,0)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52345544_116.png"
  },
  {
    "solution": "\\boxed{10\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within rectangle $ABCD$, $E$ and $F$ are two points on $BC$, and $BE=CF$. Lines $AF$ and $DE$ intersect at point $O$. $\\triangle AOD$ is an isosceles triangle; if $AF\\perp DE$, $OF= \\frac{1}{3} OA=1$, calculate the perimeter of rectangle $ABCD$.\n",
    "completion": "Solution: Since $OF = \\frac{1}{3} OA = 1$,\\\\\nit follows that $OA = 3, AF = 4$,\\\\\nwhich means $OA = OD = 3$,\\\\\nGiven $AF \\perp DE$,\\\\\nit follows that $\\angle AOD = 90^\\circ$,\\\\\ntherefore, $\\angle OAD = \\angle ODA = 45^\\circ$,\\\\\n$AD = \\sqrt{3^2 + 3^2} = 3\\sqrt{2}$,\\\\\nIn rectangle ABCD, $\\angle BAD = \\angle ADC = 90^\\circ$,\\\\\nit follows that $\\angle BAF = \\angle CDE = 45^\\circ$,\\\\\nthus $\\angle AFB = 45^\\circ = \\angle BAF$,\\\\\nwhich leads to $AB = BF$,\\\\\nGiven $AB^2 + BF^2 = AF^2$,\\\\\nit follows that $AB = BF = 2\\sqrt{2}$,\\\\\nHence, the perimeter of rectangle ABCD is $2 \\times (2\\sqrt{2} + 3\\sqrt{2}) = \\boxed{10\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52076923_91.png"
  },
  {
    "solution": "\\boxed{\\frac{n}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, rectangle $OABC$ is in a Cartesian coordinate system, with the coordinate of vertex $B$ being $\\left(4, n\\right)$. Diagonals $OB$ and $AC$ intersect at $D$. The line $y=\\frac{3}{4}nx-\\frac{3}{2}n$ intersects $OA$, $AC$, and $OB$ at $P$, $M$, and $N$, respectively. Find the length of $DP$ (expressed in terms of $n$).\n",
    "completion": "\\textbf{Solution:} Since quadrilateral OABC is a rectangle,\\\\\nit follows that point D is the midpoint of OB,\\\\\nSince $O(0,0)$ and $B(4,n)$,\\\\\nit follows that $D(2,\\frac{n}{2})$,\\\\\nFrom $\\frac{3}{4}nx-\\frac{3}{2}n=0$ we get: $x=2$, therefore $P(2,0)$,\\\\\nthus $DP\\parallel$ y-axis,\\\\\nthus $DP=\\boxed{\\frac{n}{2}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52303875_134.png"
  },
  {
    "solution": "\\boxed{20^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AD \\perp BC$, $AE$ bisects $\\angle BAC$, $\\angle B = 70^\\circ$, $\\angle C = 30^\\circ$. What is the measure of $\\angle DAE$?\n",
    "completion": "\\textbf{Solution:} Since $AD \\perp BC$, \\\\\nit follows that $\\angle ADC=90^\\circ$, \\\\\nwhich implies $\\angle DAC=90^\\circ - \\angle C=60^\\circ$, \\\\\nthus $\\angle DAE=\\angle DAC - \\angle EAC=\\boxed{20^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52905934_142.png"
  },
  {
    "solution": "\\boxed{(-6, 0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=ax^2+bx+c\\,(a\\ne 0)$ passes through the points $(-1, 0)$ and $(0, -3)$, and its vertex is in the fourth quadrant. Let $P=a+b+c$. What is the range of values for $P$?\n",
    "completion": "\\textbf{Solution:} Given the parabola $y=ax^2+bx+c$ (where $a\\neq 0$) passes through the points $(-1, 0)$ and $(0, -3)$,\\\\\nthus $\\because$ $0=a(-1)^2-b+c$ and $-3=c$,\\\\\nit leads to $\\therefore$ $b=a-3$,\\\\\nConsidering $\\because$ when $x=1$, $y=ax^2+bx+c=a+b+c$,\\\\\nwe find that $\\therefore$ $P=a+b+c=a+a-3-3=2a-6$,\\\\\nGiven that the vertex is in the fourth quadrant, it implies that $a>0$,\\\\\ntherefore $\\therefore$ $b=a-3<0$,\\\\\nwhich implies $\\therefore$ $a<3$,\\\\\nHence $\\therefore$ $0<a<3$,\\\\\nresulting in $\\therefore$ $-6<2a-6<0$, i.e., $-6<P<0$. \\\\\nTherefore, the range of $P$ is $P \\in \\boxed{(-6, 0)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53080744.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $ \\triangle ABC$ is an isosceles right triangle, with $ \\angle C=90^\\circ$. Point D is on side $BC$, and segment $AD$ is drawn. Through point B, draw $ BE\\perp AD$, intersecting the extension of $AD$ at point E. If $ \\frac{CD}{BD}=\\frac{1}{2}$, what is the value of $ \\frac{BE}{AD}$?\n",
    "completion": "\\textbf{Solution:} Since $\\frac{CD}{BD}=\\frac{1}{2}$,\n\nwe can assume $CD=k$, $BD=2k$, hence $CB=CA=3k$,\n\nsince $\\angle C=90^\\circ$,\n\nthen $AD=\\sqrt{AC^2+CD^2}=\\sqrt{(3k)^2+k^2}=\\sqrt{10}k$,\n\nsince $BE\\perp AD$,\n\nthen $\\angle E=\\angle C=90^\\circ$,\n\nsince $\\angle CDA=\\angle BDE$,\n\nthen $\\triangle ACD\\sim \\triangle BED$,\n\nhence $\\frac{AC}{BE}=\\frac{AD}{BD}$,\n\nthus $\\frac{3k}{BE}=\\frac{\\sqrt{10}k}{2k}$,\n\ntherefore $BE=\\frac{3\\sqrt{10}}{5}k$,\n\nfinally, $\\frac{BE}{AD}=\\frac{\\frac{3\\sqrt{10}}{5}k}{\\sqrt{10}k}=\\boxed{\\frac{3}{5}}$",
    "image_path": "dataset/mm_math/images/MM_Math/52917331.png"
  },
  {
    "solution": "\\boxed{\\frac{2021}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: \\textit{When $n$ squares, each with a side length of 1, are arranged as shown in the figure, with $A_1$, $A_2$, ..., $A_n$ being the intersection points of the diagonals of the squares, what is the total area of the overlapping regions formed by overlapping 2022 such squares in this manner?}\n",
    "completion": "\\textbf{Solution:}\n\nAs shown in the figure, the center of square ABCD is $A_{1}$, and BC, CD intersect with the square where $A_{2}$ is located at points E, and F, respectively. Connect $A_{1}C$, $A_{1}D$,\n\nIn square ABCD, $\\angle A_{1}CB = \\angle A_{1}DA_{2} = 45^\\circ$, $A_{1}C = A_{1}D$, $\\angle CA_{1}D = 90^\\circ$,\n\nAlso, because $\\angle EA_{1}F = 90^\\circ$,\n\nTherefore, $\\angle EA_{1}C = \\angle FA_{1}D$,\n\nIn $\\triangle EA_{1}C$ and $\\triangle FA_{1}D$, $\\left\\{\\begin{array}{l}\\angle E{A}_{1}C=\\angle F{A}_{1}D \\\\ \\angle {A}_{1}CE=\\angle {A}_{1}DF \\\\ {A}_{1}C={A}_{1}D\\end{array}\\right.$,\n\nTherefore, $\\triangle EA_{1}C \\cong \\triangle FA_{1}D$ (AAS),\n\nTherefore, $S_{\\triangle EA_{1}C} = S_{\\triangle FA_{1}D}$,\n\nTherefore, $S_{\\text{Quadrilateral} EA_{1}FC} = S_{\\triangle A_{1}DC} = \\frac{1}{4}S_{\\text{Square} ABCD} = \\frac{1}{4} \\times 1 \\times 1 = \\frac{1}{4}$,\n\nSimilarly, the area of each shaded part is also $\\frac{1}{4}$,\n\nBecause 2022 squares overlap in this manner, and the overlap area of every two squares is $\\frac{1}{4}$, making a total of 2021 overlaps,\n\nTherefore, the sum of the overlap areas formed by the 2022 squares overlapped in this way is $\\boxed{\\frac{2021}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53360049.png"
  },
  {
    "solution": "\\boxed{18}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, the graph of the linear function $y=kx-3$ (where $k\\neq 0$) intersects the y-axis at point B and intersects the graph of the inverse proportion function $y=\\frac{m}{x}$ (where $x>0$) at point A (8, 1). Point C is located on the line segment AB (distinct from points A and B). A line parallel to the y-axis passing through point C intersects the graph of the inverse proportion function at point D. Lines OC, OD, and AD are connected. When CD equals 6, what are the coordinates of point C and the area of $\\triangle ACD$?\n",
    "completion": "\\textbf{Solution:} Let C$(a, \\frac{1}{2}a-3)$ (where $0<a<8$), then we have D$(a, \\frac{8}{a})$, \\\\\n$\\therefore$CD$=\\frac{8}{a}-\\left(\\frac{1}{2}a-3\\right)=\\frac{8}{a}-\\frac{1}{2}a+3$, \\\\\n$\\because$CD$=6$, \\\\\n$\\therefore \\frac{8}{a}-\\frac{1}{2}a+3=6$, \\\\\nSolving this, we get: $a=-8$ (discard) or $a=2$, \\\\\n$\\therefore \\frac{1}{2}a-3=1-3=-2$, \\\\\n$\\therefore$C$(2, -2)$, \\\\\nDraw AE$\\perp$CD at point E from A, then AE$=8-2=6$, \\\\\n$\\therefore$Area of $\\triangle ACD=\\frac{1}{2}\\times \\text{CD}\\times \\text{AE}=\\frac{1}{2}\\times 6\\times 6=\\boxed{18}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51445655_275.png"
  },
  {
    "solution": "\\boxed{120^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure is an axisymmetric figure. If $\\angle A=36^\\circ, \\angle C'=24^\\circ$, then what is the degree measure of $\\angle B$?\n",
    "completion": "\\textbf{Solution:} Since the figure is an axisymmetric figure, \\\\\ntherefore, $\\angle C = \\angle C' = 24^\\circ$,\\\\\nsince $\\angle A = 36^\\circ$,\\\\\ntherefore, $\\angle B = 180^\\circ - \\angle A - \\angle C = \\boxed{120^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52978487.png"
  },
  {
    "solution": "\\boxed{x>-1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y_1=x+b$ intersects with the graph of the linear function $y_2=kx-1$ at point P. Then, what is the solution set of the inequality concerning $x$, $x+b-kx+1>0$?\n",
    "completion": "\\textbf{Solution:} Given the graph of the function, when the graph of $y_1=x+b$ is above the graph of $y_2=kx-1$, we have $x+b>kx-1$, which leads to $x>-1$; \\\\\nHence, the answer is: $\\boxed{x>-1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51699602.png"
  },
  {
    "solution": "\\boxed{20^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point O is a point on line AB, $\\angle COD$ is a right angle, $OE$ bisects $\\angle BOC$. As shown in Figure (1), if $\\angle AOC = 40^\\circ$, what is the degree measure of $\\angle DOE$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle AOC + \\angle BOC = 180^\\circ$,\\\\\nwe have $\\angle BOC = 180^\\circ - \\angle AOC = 180^\\circ - 40^\\circ = 140^\\circ$,\\\\\nSince $OE$ bisects $\\angle BOC$,\\\\\nwe get $\\angle COE = \\frac{1}{2} \\angle BOC = \\frac{1}{2} \\times 140^\\circ = 70^\\circ$,\\\\\nGiven $\\angle COD$ is a right angle,\\\\\nthus $\\angle COE + \\angle DOE = 90^\\circ$,\\\\\nleading to $\\angle DOE = 90^\\circ - \\angle COE = 90^\\circ - 70^\\circ = \\boxed{20^\\circ};$",
    "image_path": "dataset/mm_math/images/MM_Math/51438807_182.png"
  },
  {
    "solution": "\\boxed{x<2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the graph of the linear function $y_1=kx+b$ (where $k\\ne 0$, and $k$, $b$ are constants) $l_1$ intersects the $x$ axis and $y$ axis at points $A$ and $B$ respectively. The coordinates of point $B$ are $(0, 5)$. The graph of the directly proportional function $y_2=2x$, $l_2$, intersects $l_1$ at point $C(m, 4)$. For what value of $x$ is $y_1 > y_2$?\n",
    "completion": "\\textbf{Solution:} Given the problem, the line $l_{1}: y=-\\frac{1}{2}x+5$ and the line $l_{2}: y_{2}=2x$ intersect at point $C(2, 4)$,\\\\\n$\\therefore$ when $x<2$, $y_{1}>y_{2}$.\\\\\nThus, the answer is $\\boxed{x<2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50677475_137.png"
  },
  {
    "solution": "\\boxed{115^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points A, D, B, E are on the same line, $AC=EF$, $AD=BE$, $\\angle C=\\angle F=90^\\circ$, $\\angle ABC=65^\\circ$, find the degree of $\\angle ADF$.\n",
    "completion": "\\textbf{Solution:} Given that $AD = BE$,\\\\\nit follows that $AD + BD = BE + BD$, which means $AB = DE$,\\\\\nfurthermore, since $AC = EF$,\\\\\nwe have Rt$\\triangle ABC \\cong$ Rt$\\triangle EDF$ (by HL criterion),\\\\\nwhich implies $\\angle FDE = \\angle ABC = 65^\\circ$,\\\\\ntherefore, $\\angle ADF = 180^\\circ - \\angle FDE = 180^\\circ - 65^\\circ = \\boxed{115^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52774849.png"
  },
  {
    "solution": "\\boxed{\\sqrt{33}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle AED$, $\\angle AED=90^\\circ, AB=AC=AD, EC=3, BE=11$, then what is the value of $ED$?\n",
    "completion": "\\textbf{Solution:} As the figure shows: draw $AF\\perp BC$ with foot F \\\\\n$\\because$ $EC=3$, $BE=11$ \\\\\n$\\therefore$ $BC=BE+EC=11+3=14$ \\\\\n$\\because$ $AB=AC$ \\\\\n$\\therefore$ $BF=CF=\\frac{1}{2}BC=7$ \\\\\n$\\therefore$ $EF=FC-EC=7-3=4$ \\\\\nIn $\\triangle ADE$, by Pythagoras' theorem, $DE^2=AD^2-AE^2$ \\\\\nIn $\\triangle AEF$, by Pythagoras' theorem, $AE^2=AF^2+EF^2$ \\\\\nAnd $\\because$ $AB=AD$ \\\\\n$\\therefore$ $DE^2=AB^2-(AF^2+EF^2)$ \\\\\nIn $\\triangle ABF$, by Pythagoras' theorem, $AB^2=AF^2+BF^2$ \\\\\n$\\therefore$ $DE^2=AF^2+BF^2-(AF^2+EF^2)=BF^2-EF^2=7^2-4^2=33$ \\\\\n$\\therefore$ $DE=\\sqrt{33}$ \\\\\nThus, the answer is: $DE=\\boxed{\\sqrt{33}}$",
    "image_path": "dataset/mm_math/images/MM_Math/52828219.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{15}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, a circle $\\odot O$ with radius 8 is folded along line $AB$, such that the arc $\\overset{\\frown}{AB}$ exactly passes through point $D$, which is the midpoint of the radius $OC$ that is perpendicular to $AB$. What is the length of the crease $AB$?\n",
    "completion": "\\textbf{Solution:} Extend $CO$ to meet $AB$ at point $E$, and connect $OB$,\\\\\nsince $CE\\perp AB$,\\\\\nthus, $E$ is the midpoint of $AB$,\\\\\naccording to the problem, we have $CD=4$, $OD=4$, and $OB=8$,\\\\\n$DE=\\frac{1}{2}\\left(8\\times 2-4\\right)=\\frac{1}{2}\\times 12=6$,\\\\\n$OE=6-4=2$,\\\\\nin $\\triangle OEB$, by the Pythagorean theorem, we have: $OE^{2}+BE^{2}=OB^{2}$,\\\\\nsubstituting the values, we find $BE=2\\sqrt{15}$,\\\\\nthus, $AB=4\\sqrt{15}$.\\\\\nSo, the answer is: $AB=\\boxed{4\\sqrt{15}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53009644.png"
  },
  {
    "solution": "\\boxed{y=x^2-2x-3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the axis of symmetry of the parabola is the line $x=1$, and it intersects the x-axis at points A and B, and intersects the y-axis at point $C\\left(0, -3\\right)$, with $OB=OC$. Find the equation of the parabola.\n",
    "completion": "\\textbf{Solution:} Given $C\\left(0, -3\\right)$ and $OB=OC$, it follows that $OB=OC=3$ and thus $B\\left(3, 0\\right)$. Suppose the equation of the parabola is $y=ax^2+bx+c$, then we have:\n\\[\n\\begin{cases}\n-\\frac{b}{2a}=1\\\\\n9a+3b+c=0\\\\\nc=-3\n\\end{cases}\n\\]\nSolving these, we obtain:\n\\[\n\\begin{cases}\na=1\\\\\nb=-2\\\\\nc=-3\n\\end{cases}\n\\]\nTherefore, the equation of the parabola is \\boxed{y=x^2-2x-3}.",
    "image_path": "dataset/mm_math/images/MM_Math/51433738_207.png"
  },
  {
    "solution": "\\boxed{112.5^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point E is a point on the extension line of side BC of square ABCD, and $CE=BD$. AE intersects DC at point F. What is the measure of $\\angle AFC$?\n",
    "completion": "\\textbf{Solution:} Construct line AC,\\\\\nsince quadrilateral ABCD is a square,\\\\\nthus AC=BD,\\\\\nsince CE=BD,\\\\\nthus CE=AC,\\\\\nthus $\\angle$E=$\\angle$CAE,\\\\\nsince AC is the diagonal of square ABCD,\\\\\nthus $\\angle$ACB=$45^\\circ$,\\\\\nthus $\\angle$E+$\\angle$CAE=$45^\\circ$,\\\\\nthus $\\angle$E=$ \\frac{1}{2} \\times 45^\\circ=22.5^\\circ$,\\\\\nIn $\\triangle$CEF, $\\angle$AFC=$\\angle$E+$\\angle$ECF=$22.5^\\circ+90^\\circ=\\boxed{112.5^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53728711.png"
  },
  {
    "solution": "\\boxed{(1, -\\frac{16}{3})}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=\\frac{4}{3}x^{2}+bx+c$ intersects the x-axis at points $A(-1,0)$ and $B$ (point $B$ is to the right of point $A$), and intersects the y-axis at point $C(0,-4)$. Find the analytical expression of the parabola and the coordinates of the vertex $E$.\n",
    "completion": "\\textbf{Solution:} Given that the parabola $y=\\frac{4}{3}x^{2}+bx+c$ intersects the $y$ axis at point $C(0, -4)$,\\\\\ntherefore, the equation of the parabola is $y=\\frac{4}{3}x^{2}+bx-4$.\\\\\nSince the parabola $y=\\frac{4}{3}x^{2}+bx-4$ intersects the $x$ axis at $A(-1, 0)$,\\\\\nthus, $0=\\frac{4}{3} \\times (-1)^{2}-b-4$.\\\\\nSolving this, we get: $b=-\\frac{8}{3}$,\\\\\ntherefore, the equation of the parabola is $y=\\frac{4}{3}x^{2}-\\frac{8}{3}x-4=\\frac{4}{3}(x-1)^{2}-\\frac{16}{3}$.\\\\\nHence, the vertex $E$ has the coordinates $\\boxed{(1, -\\frac{16}{3})}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52719833_111.png"
  },
  {
    "solution": "\\boxed{50^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $O$ is a point on the straight line $AB$, $\\angle DOB=90^\\circ$, $\\angle EOC=90^\\circ$. If $\\angle DOE=50^\\circ$, what is the degree measure of $\\angle BOC$?\n",
    "completion": "\\textbf{Solution:} From the given conditions, we have \\\\\n$\\because \\angle BOC + \\angle COD = 90^\\circ$, \\\\\n$\\angle COD + \\angle DOE = 90^\\circ$, \\\\\n$\\therefore \\angle BOC = \\angle DOE = \\boxed{50^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51457662_193.png"
  },
  {
    "solution": "\\boxed{35^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown, the $\\triangle ODC$ is obtained by rotating $\\triangle OAB$ clockwise by $30^\\circ$ around point $O$. If point $D$ falls exactly on $AB$, and the measure of $\\angle AOC$ is $100^\\circ$, then what is the measure of $\\angle B$?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ODC$ is rotated clockwise by $30^\\circ$ around point O to get $\\triangle OAB$,\\\\\nit follows that AO=DO, and $\\angle AOD=\\angle BOC=30^\\circ$,\\\\\nhence $\\angle A=\\angle ADO=(180^\\circ-30^\\circ)\\div 2=75^\\circ$,\\\\\nGiven that $\\angle AOC=100^\\circ$,\\\\\nit then follows that $\\angle DOB=\\angle AOC-\\angle AOD-\\angle BOC=100^\\circ-30^\\circ-30^\\circ=40^\\circ$,\\\\\nSince $\\angle ADO=\\angle B+\\angle DOB$,\\\\\nit follows that $\\angle B=\\angle ADO-\\angle DOB=75^\\circ-40^\\circ=\\boxed{35^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51445131.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB = 90^\\circ$, a square ABDE is constructed externally on the hypotenuse AB of the triangle, and the diagonals of the square meet at point $O$. Line OC is connected. Given that $AC=3$ and $OC=4\\sqrt{2}$, what is the length of BC?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw line OF $\\perp$ BC at F, and draw line AM $\\perp$ OF at M,\\\\\n$\\because$ $\\angle ACB=90^\\circ$,\\\\\n$\\therefore$ $\\angle AMO=\\angle OFB=90^\\circ$, $\\angle ACB=\\angle CFM=\\angle AMF=90^\\circ$,\\\\\n$\\therefore$ Quadrilateral ACFM is a rectangle,\\\\\n$\\therefore$ AM=CF, AC=MF=3,\\\\\n$\\because$ Quadrilateral ABDE is a square,\\\\\n$\\therefore$ $\\angle AOB=90^\\circ$, OA=OB,\\\\\n$\\therefore$ $\\angle AOM+\\angle BOF=90^\\circ$,\\\\\nAlso $\\because$ $\\angle AMO=90^\\circ$,\\\\\n$\\therefore$ $\\angle AOM+\\angle OAM=90^\\circ$,\\\\\n$\\therefore$ $\\angle BOF=\\angle OAM$,\\\\\n$\\therefore$ $\\triangle AOM \\cong \\triangle OBF$ (AAS),\\\\\n$\\therefore$ AM=OF, OM=FB,\\\\\n$\\therefore$ OF=CF,\\\\\n$\\because$ $\\angle CFO=90^\\circ$,\\\\\n$\\therefore$ $\\triangle CFO$ is an isosceles right-angle triangle,\\\\\n$\\because$ OC=$4\\sqrt{2}$,\\\\\nBy the Pythagorean theorem: CF=OF=4,\\\\\n$\\therefore$ BF=OF-MF=4-3=1,\\\\\n$\\therefore$ BC=CF+BF=4+1=\\boxed{5}.",
    "image_path": "dataset/mm_math/images/MM_Math/51510282.png"
  },
  {
    "solution": "\\boxed{\\frac{8}{3}\\pi +4\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC$, point $E$ is on $AC$, the circle $O$ passing through points $A$, $B$, and $E$ intersects $BC$ at point $D$, and point $D$ is the midpoint of arc $BE$, with $AB$ being the diameter of the circle; if $AB=8$ and $\\angle C=60^\\circ$, what is the area of the shaded region?\n",
    "completion": "\\textbf{Solution:}\n\nLet $OF \\perp BE$,\\\\\n$\\therefore \\angle OFB=90^\\circ$, $BE=2BF$,\\\\\n$\\because AB=AC$, $\\angle C=60^\\circ$,\\\\\n$\\therefore \\triangle ABC$ is an equilateral triangle,\\\\\n$\\therefore \\angle BAC=60^\\circ$,\\\\\n$\\because \\overset{\\frown}{BE}=\\overset{\\frown}{BE}$,\\\\\n$\\therefore \\angle BOE=2\\angle BAC=120^\\circ$, $\\angle AOE=180^\\circ-120^\\circ=60^\\circ$,\\\\\n$\\because AB=8$,\\\\\n$\\therefore$ the radius of the circle is 4,\\\\\n$\\because OB=OE$, $\\angle BOE=120^\\circ$, $OF\\perp BE$,\\\\\n$\\therefore \\angle BOF= \\frac{1}{2}\\angle BOE=60^\\circ$,\\\\\n$\\therefore \\angle OBF=90^\\circ-60^\\circ=30^\\circ$,\\\\\n$\\therefore OF= \\frac{1}{2}OB=2$,\\\\\n$\\therefore BF=\\sqrt{4^2-2^2}=2\\sqrt{3}$,\\\\\n$\\therefore BE=4\\sqrt{3}$,\\\\\n$\\therefore S_{\\text{shaded part}}= S_{\\text{sector} AOE}+S_{\\triangle BOE}=\\frac{60^\\circ \\pi \\times 4^2}{360}+\\frac{1}{2}\\times 2\\times 4\\sqrt{3}=\\boxed{\\frac{8}{3}\\pi +4\\sqrt{3}}$",
    "image_path": "dataset/mm_math/images/MM_Math/52861398_115.png"
  },
  {
    "solution": "\\boxed{\\left(2, 0\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the vertex $C$ of the parabola passing through points $A\\left(0, 4\\right)$ and $B\\left(5, 9\\right)$ is on the positive half of the $x$-axis. Find the coordinates of point $C$.\n",
    "completion": "\\textbf{Solution:} Substituting $y=0$ into $y=(x-2)^{2}$ yields:\\\\\n$(x-2)^{2}=0$,\\\\\n$\\therefore x=2$,\\\\\n$\\therefore$ the coordinates of point $C$ are $\\boxed{\\left(2, 0\\right)}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52551573_132.png"
  },
  {
    "solution": "\\boxed{\\frac{5}{6}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, we have $AB=6$ and $AD=4$. Points $E$ and $F$ are the midpoints of $BC$ and $CD$, respectively. Lines $BF$ and $DE$ intersect at point $G$. Through point $E$, draw $EH \\parallel CD$, intersecting $BF$ at point $H$. What is the length of segment $GH$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral $ABCD$ is a rectangle, $AB=6$, $AD=4$,\\\\\nthus $DC=AB=6$, $BC=AD=4$, $\\angle C=90^\\circ$,\\\\\nsince points E and F are the midpoints of $BC$ and $CD$, respectively,\\\\\nthus $DF=CF=\\frac{1}{2}DC=3$, $CE=BE=\\frac{1}{2}BC=2$,\\\\\nsince $EH\\parallel CD$,\\\\\nthus $FH=BH$,\\\\\nsince $BE=CE$,\\\\\nthus $EH=\\frac{1}{2}CF=\\frac{3}{2}$,\\\\\nBy the Pythagorean theorem, $BF=\\sqrt{BC^2+CF^2}=\\sqrt{4^2+3^2}=5$,\\\\\nthus $BH=FH=\\frac{1}{2}BF=\\frac{5}{2}$,\\\\\nsince $EH\\parallel CD$,\\\\\nthus $\\triangle EHG\\sim \\triangle DFG$,\\\\\nthus $\\frac{EH}{DF}=\\frac{GH}{FG}$,\\\\\nthus $\\frac{\\frac{3}{2}}{3}=\\frac{GH}{\\frac{5}{2}-GH}$,\\\\\nsolving for: $GH=\\boxed{\\frac{5}{6}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52866224.png"
  },
  {
    "solution": "\\boxed{\\frac{24}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, quadrilateral ABCD is a rhombus, with $BE \\perp AD$ and $BF \\perp CD$, where the feet of the perpendiculars are points E and F, respectively. $BE = BF$; When the diagonals of the rhombus ABCD satisfy $AC = 8$ and $BD = 6$, what is the length of $BE$?\n",
    "completion": "\\textbf{Solution:} Since the diagonals of rhombus ABCD are $AC=8$ and $BD=6$,\\\\\nit follows that $OA=OC=\\frac{1}{2}AC=4$, $OB=OD=\\frac{1}{2}BD=3$, and $\\angle AOD=90^\\circ$,\\\\\nthus $AD=\\sqrt{OA^{2}+OD^{2}}=5$,\\\\\nGiven that the area of rhombus ABCD is $S_{\\text{rhombus} ABCD}=\\frac{1}{2}AC\\cdot BD=\\frac{1}{2}AD\\cdot BE$,\\\\\nit follows that $BE=\\frac{AC\\cdot BD}{2AD}=\\boxed{\\frac{24}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53333382_89.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within a square $ABCD$ with a side length of 2, points $E$ and $F$ are located on sides $BC$ and $AD$, respectively. By connecting $EF$ and folding quadrilateral $ABEF$ along $EF$, the corresponding point of $B$ precisely falls on side $CD$, and the corresponding point of $A$ is $H$. After connecting $BH$, what is the minimum value of $BH+EF$?\n",
    "completion": "\\textbf{Solution:} Construct $FK \\perp BC$ at point $K$, extend $BC$ to point $M$ such that $CM = BC$, connect $AM$ and intersect $CD$ at point $N$, connect $MG$, $GA$, $BG$,\n\n$\\because$ Quadrilateral $ABCD$ is a square, \n\n$\\therefore$ $\\angle BAD = \\angle ABC = \\angle BCD = 90^\\circ$, $AB = BC$,\n\n$\\therefore$ $CD \\perp BM$,\n\n$\\therefore$ $CD$ is the perpendicular bisector of $BM$,\n\n$\\therefore$ $MG = BG$,\n\n$\\because$ Folding quadrilateral $ABEF$ along $EF$, the corresponding point of $B$ on side $CD$ is exactly $G$, the corresponding point of $A$ is $H$,\n\n$\\therefore$ $AB = HG$, $BE = EG$, $\\angle ABE = \\angle HGE$, $EF \\perp BG$,\n\n$\\therefore$ $\\angle GBE = \\angle BGE$,\n\n$\\therefore$ $\\angle ABG = \\angle HGB$,\n\nIn $\\triangle ABG$ and $\\triangle HGB$\n\n$\\left\\{\\begin{array}{l} BG = GB\\\\ \\angle ABG = \\angle HGB\\\\ AB = HG \\end{array}\\right.$\n\n$\\therefore$ $\\triangle ABG \\cong \\triangle HGB$ (SAS),\n\n$\\therefore$ $GA = BH$,\n\n$\\because$ $EF \\perp BG$,\n\n$\\because$ $FK \\perp BC$,\n\n$\\therefore$ $\\angle FKE = \\angle BCG = 90^\\circ$,\n\n$\\therefore$ $\\angle EFK + \\angle FEK = \\angle GBC + \\angle FEK = 90^\\circ$,\n\n$\\therefore$ $\\angle EFK = \\angle GBC$,\n\n$\\because$ $\\angle BAD = \\angle ABC = \\angle BKF = 90^\\circ$,\n\n$\\therefore$ Quadrilateral $ABKF$ is a rectangle,\n\n$\\therefore$ $AB = FK$,\n\n$\\therefore$ $FK = BC$,\n\nIn $\\triangle FEK$ and $\\triangle BGC$\n\n$\\left\\{\\begin{array}{l} \\angle FKE = \\angle BCG\\\\ FK = BC\\\\ \\angle EFK = \\angle GBC \\end{array}\\right.$\n\n$\\therefore$ $\\triangle FEK \\cong \\triangle BGC$ (ASA),\n\n$\\therefore$ $EF = BG$,\n\n$\\therefore$ $EF = MG$,\n\n$\\therefore$ $BH + EF = AG + MG$,\n\n$\\because$ $AG + MG \\geq AM$,\n\n$\\therefore$ $BH + EF \\geq AM$,\n\n$\\therefore$ When point $G$ coincides with point $N$, $AG + MA = AM$, at this time $AG + MA$ has the minimum value,\n\n$\\therefore$ The value of $BH + EF = AM$ is also the minimum,\n\n$\\because$ $\\angle ABM = 90^\\circ$, $AB = 2$, $BM = 2BC = 4$,\n\n$\\therefore$ $AM = \\sqrt{AB^{2} + BM^{2}} = \\sqrt{2^{2} + 4^{2}} = \\boxed{2\\sqrt{5}}$,\n\n$\\therefore$ The minimum value of $BH + EF$ is $\\boxed{2\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52205279.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{1}{2}x+3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y = -\\frac{4}{3}x + 8$ intersects the x-axis and y-axis at points A and B, respectively. Point M is on OB. If $\\triangle ABM$ is folded along AM, point B falls exactly on point B$^{\\prime}$ on the x-axis. Find the equation of line AM.\n",
    "completion": "Solution: Let the equation of line AM be \\(y=kx+b\\). Substituting \\((0,3)\\) and \\((6,0)\\) into it, we get\n\\[\n\\left\\{\n\\begin{array}{l}\nb=3\\\\\n6k+b=0\n\\end{array}\n\\right.\n\\]\nSolving this, we find\n\\[\n\\left\\{\n\\begin{array}{l}\nb=3\\\\\nk=-\\frac{1}{2}\n\\end{array}\n\\right.\n\\]\nTherefore, the equation of line AM is \\(\\boxed{y=-\\frac{1}{2}x+3}\\).",
    "image_path": "dataset/mm_math/images/MM_Math/50590744_147.png"
  },
  {
    "solution": "\\boxed{11}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In $\\triangle ABC$, $E$, $F$, $G$, and $H$ are the midpoints of $AC$, $CD$, $BD$, and $AB$, respectively. If $AD=3$ and $BC=8$, then what is the perimeter of the quadrilateral $EFGH$?\n",
    "completion": "\\textbf{Solution:} Given that E, F, G, H are the midpoints of AC, CD, BD, AB respectively,\\\\\nit follows that EF=$\\frac{1}{2}$AD=1.5, GH=$\\frac{1}{2}$AD=1.5, EH=$\\frac{1}{2}$BC=4, FG=$\\frac{1}{2}$BC=4,\\\\\ntherefore, the perimeter of the quadrilateral EFGH is EF+FG+GH+EH=1.5+4+1.5+4=\\boxed{11}.",
    "image_path": "dataset/mm_math/images/MM_Math/53183117.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB = 90^\\circ$, $AC = 4$, $BC = 8$, $ED$ bisects $AB$ perpendicularly at point $D$ and intersects $BC$ at $E$. What is the perimeter of $\\triangle ACE$?\n",
    "completion": "\\textbf{Solution:} Given that $ED$ is the perpendicular bisector of side $AB$,\\\\\nit follows that $EA=EB$,\\\\\nthus, the perimeter of $\\triangle ACE$, which is $AC+EC+EA=AC+EC+EB=AC+BC$,\\\\\ngiven $AC=4$ and $BC=8$,\\\\\ntherefore, the perimeter of $\\triangle ACE$ is $AC+BC=4+8=\\boxed{12}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52826732.png"
  },
  {
    "solution": "\\boxed{60}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the isosceles $\\triangle ABC$, with $AB=AC=5$ and $BC=8$, point B is on the y-axis, $BC\\parallel x$-axis. The graph of the inverse proportional function $y=\\frac{k}{x}$ ($k>0, x>0$) passes through point A and intersects $BC$ at point D. If $AB=BD$, what is the value of $k$?\n",
    "completion": "\\textbf{Solution:} Given that $AB=AC=5$, $BC=8$, \\\\\n$\\therefore BE=\\frac{1}{2}BC=4$, \\\\\n$\\therefore AE=\\sqrt{AB^{2}-BE^{2}}=\\sqrt{5^{2}-4^{2}}=3$, \\\\\nLet $OB=m$, \\\\\nAs $AB=AC=5$, $AB=BD$, \\\\\n$\\therefore$ Point A's coordinates are $(4, m+3)$, and Point D's coordinates are $(5, m)$, \\\\\nSince the inverse proportion function $y=\\frac{k}{x}$ $(k>0, x>0)$ passes through point A and intersects BC at point D, \\\\\n$\\therefore \\left\\{\\begin{array}{l}m+3=\\frac{k}{4}\\\\ m=\\frac{k}{5}\\end{array}\\right.$, solving these equations yields: $m=12$, \\\\\n$\\therefore k=\\boxed{60}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53080023.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of circle $\\odot O$, and points $C$ and $D$ are two points on circle $\\odot O$. Lines $BC$ and $DC$ are drawn, $BC=CD$, and $CE\\perp DA$, intersecting the extension of $DA$ at point $E$. $CE$ is the tangent to circle $\\odot O$; if $AE=2$ and $CE=4$, what is the length of $AD$?\n",
    "completion": "\\textbf{Solution:} Draw line $AC$, as shown in the diagram,\\\\\nfrom $\\left(1\\right)$ we know $\\angle ECA+\\angle ACO=90^\\circ$,\\\\\nsince $AB$ is the diameter of $\\odot O$,\\\\\ntherefore $\\angle ACB=90^\\circ$,\\\\\ntherefore $\\angle ACO+\\angle OCB=90^\\circ$,\\\\\nthus $\\angle ECA=\\angle OCB$.\\\\\nSince $OC=OB$,\\\\\nthus $\\angle B=\\angle OCB$,\\\\\ntherefore $\\angle ECA=\\angle B$,\\\\\nsince $\\angle B=\\angle ADC$,\\\\\nthus $\\angle ADC=\\angle ECA$.\\\\\nSince $\\angle E=\\angle E$,\\\\\nthus $\\triangle EAC\\sim \\triangle ECD$,\\\\\ntherefore $\\frac{EA}{EC}=\\frac{EC}{ED}$,\\\\\nsince $AE=2$, $CE=4$,\\\\\nthus $ED=8$,\\\\\ntherefore $AD=DE−AE=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52457443_124.png"
  },
  {
    "solution": "\\boxed{16.8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, $AD$ bisects $\\angle CAB$, $BD=2CD$, and the distance from point D to $AB$ is 5.6. What is the length of $BC$ in $cm$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, construct DE $\\perp$ AB at E, \\\\\n$\\because$ $\\angle C=90^\\circ$, \\\\\n$\\therefore$ CD $\\perp$ AC, \\\\\n$\\because$ AD bisects $\\angle$BAC, \\\\\n$\\therefore$ CD = DE, \\\\\n$\\because$ the distance from D to AB is equal to 5.6 cm, \\\\\n$\\therefore$ CD = DE = 5.6 cm, \\\\\nFurthermore, $\\because$ BD = 2CD, \\\\\n$\\therefore$ BD = 11.2 cm, \\\\\n$\\therefore$ BC = 5.6 + 11.2 = $\\boxed{16.8}$ cm,",
    "image_path": "dataset/mm_math/images/MM_Math/51368989.png"
  },
  {
    "solution": "\\boxed{\\left(3, 4\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the vertices $A$ and $C$ of the rectangle $OABC$ are located on the positive semi-axis of the $x$-axis and $y$-axis, respectively, and the coordinates of point $B$ are $\\left(8, 4\\right)$. The rectangle is folded along $OB$, with point $A$ corresponding to point $D$. $OD$ intersects $BC$ at point $E$. Given that $EO=EB$, find the coordinates of point $E$?\n",
    "completion": "\\textbf{Solution:} From equation (1), it is known that $EO=EB$, and since the vertices $A$, $C$ of the rectangle $OABC$ are located on the positive halves of the $x$-axis and $y$-axis respectively, with the coordinate of point $B$ being $\\left(8, 4\\right)$, it follows that $OC=AB=4$, and $OA=CB=8$. Let $OE=x$, then $DE=8-x$, and consequently $BE=OE=x$, $BD=4$. In $\\triangle BDE$, we have $BD^2+DE^2=BE^2$, hence $4^2+(8-x)^2=x^2$, leading to $x=5$, thus $BE=5$. Therefore, $CE=BC-BE=8-5=3$, which yields that the coordinate of point $E$ is $\\boxed{\\left(3, 4\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52302554_143.png"
  },
  {
    "solution": "\\boxed{-2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, on a foldable number line, points $A$ and $B$ represent the numbers $-8$ and $3$, respectively. If point $C$ is used as the folding point and the number line is folded to the right, such that point $A$ falls to the right of point $B$ and the distance between points $A$ and $B$ is $1$ unit, what is the number represented by point $C$?\n",
    "completion": "\\textbf{Solution:} Given that points A and B represent the numbers -8 and 3, respectively, \\\\\n$\\therefore$ $AB=3-\\left(-8\\right)=11$, \\\\\nGiven that the number line is folded to the right at point C, if point A falls on the right side of point B, and points A and B are 1 unit length apart, \\\\\n$\\therefore$ $BC=\\frac{11-1}{2}=5$, \\\\\n$\\therefore$ the number represented by point C is: $3-5=\\boxed{-2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55219238.png"
  },
  {
    "solution": "\\boxed{\\frac{4}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, $AB=5$, and $AC=4$, what is the value of $\\tan B$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle C=90^\\circ$ and $AB=5$, $AC=4$, \\\\\nthus $BC=\\sqrt{5^2-4^2}=3$,\\\\\ntherefore, $\\tan B=\\frac{AC}{BC}=\\frac{4}{3}=\\boxed{\\frac{4}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51299785.png"
  },
  {
    "solution": "\\boxed{\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that point $G$ is on the diagonal $AC$ of the square $ABCD$, $GE \\perp BC$ with the foot of the perpendicular as point $E$, and $GF \\perp CD$ with the foot of the perpendicular as point $F$. Quadrilateral $CEGF$ is a square. Find the value of $\\frac{AG}{BE}$.\n",
    "completion": "\\textbf{Solution:} Since quadrilateral $CEGF$ is a square, it follows that $\\angle CEG=\\angle B=90^\\circ$ and $\\angle ECG=45^\\circ$. Consequently, $GC=\\sqrt{2}EC$ and $AC=\\sqrt{2}BC$. Therefore, $AG=AC-GC=\\sqrt{2}(BC-EC)=\\sqrt{2}(m-n)$ and $BE=m-n$. Thus, $\\frac{AG}{BE}=\\boxed{\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51614839_82.png"
  },
  {
    "solution": "\\boxed{\\frac{6\\sqrt{10}}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the parabola $y=ax^2+bx-4$ passes through points $A(-8, 0)$ and $B(2, 0)$. The line $x=-4$ intersects the x-axis at point $C$ and intersects the parabola at point $D$. If the distances from points $B$, $D$, $C$ to the same line are $d_1$, $d_2$, $d_3$ respectively, is there a line $l$ such that $d_1=d_2=\\frac{d_3}{2}$? If such a line exists, please directly write down the value of $d_3$; if it does not exist, please explain the reason.\n",
    "completion": "\\textbf{Solution:} Exists. Connect BD, through point C draw $CH\\perp BD$ at point H,\n\n$\\therefore$ Point C ($-4,0$), Point B ($2,0$), Point D ($-4,-6$),\n\n$\\therefore OC=4$, $OB=2$, $CD=6$,\n\n$\\therefore \\triangle CDB$ is an isosceles right triangle,\n\n$\\therefore CH=CD\\cdot\\sin 45^\\circ=6\\times \\frac{\\sqrt{2}}{2}=3\\sqrt{2}$,\n\n$\\because BD=2CH$,\n\n$\\therefore BD=6\\sqrt{2}$,\n\n$\\because CO:OB=2:1$,\n\n$\\therefore$ The line $l_1$ passing through point O and parallel to BD satisfies the condition,\n\nDraw line $BE\\perp l_1$, $DF\\perp l_1$ at point F, $CH$ intersects line $l_1$ at point G, outside $\\triangle CDB$ draw line $l_2\\parallel DB$, extending CH to intersect line $l_2$ at point G', making $CH=HG'$,\n\n$\\because BE=DF$, i.e., $d_1=d_2$,\n\n$\\therefore \\frac{CG}{BE}=\\frac{CO}{BO}=\\frac{2}{1}$, $\\frac{CG}{GH}=\\frac{2}{1}$,\n\n$\\therefore \\frac{d_3}{d_1}=\\frac{2}{1}$,\n\n$\\therefore d_1=d_2=\\frac{1}{2}d_3$,\n\n$\\therefore CG=\\frac{2}{3}CH$,\n\n$\\therefore d_3=\\frac{2}{3}\\times 3\\sqrt{2}=2\\sqrt{2}$;\n\n$\\therefore d_3=CG' =2CH=6\\sqrt{2}$;\n\nAs shown in the diagram, through points H, O draw line $l_3$, draw $BE\\perp l_3$ at point E, $DF\\perp l_3$ at point F, $CG\\perp l_3$ at point G, draw $HI\\perp x$-axis at point I,\n\n$\\therefore DH=BH$, $BE=DF$, i.e., $d_1=d_2$,\n\n$\\because CO:OB=2:1$,\n\n$\\therefore d_1:d_2=\\frac{d_3}{2}$,\n\n$\\therefore HI=CI=\\frac{1}{2}CB=3$,\n\n$\\therefore OI=4−3=1$,\n\n$\\therefore OH=\\sqrt{HI^2+OI^2}=\\sqrt{3^2+1^2}=\\sqrt{10}$,\n\n$\\therefore S_{\\triangle COH}=\\frac{1}{2}\\times 4\\times 3=\\frac{1}{2}\\times \\sqrt{10}d_3$,\n\nSolving for this: $d_3=\\frac{6\\sqrt{10}}{5}$\n\n$\\therefore$ The value of $d_3$ is: $2\\sqrt{2}$ or $6\\sqrt{2}$ or $\\boxed{\\frac{6\\sqrt{10}}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52936367_232.png"
  },
  {
    "solution": "\\boxed{\\sqrt{2}+\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, $\\angle B=90^\\circ$, $\\angle BCA=60^\\circ$, $AC=2\\sqrt{2}$, $DA=1$, $CD=3$. Find the area of quadrilateral $ABCD$.\n",
    "completion": "\\textbf{Solution:}\n\nGiven $\\because$ $\\angle B=90^\\circ$, $\\angle BCA=60^\\circ$, $\\therefore$ $\\angle A=30^\\circ$, $\\therefore$ BC=$\\frac{1}{2}$AC=$\\sqrt{2}$, AB=$\\sqrt{6}$, $\\because$ AC=2$\\sqrt{2}$, DA=1, CD=3, $\\therefore$ AC$^{2}$+DA$^{2}$=9=CD$^{2}$, $\\therefore$ $\\angle CAD=90^\\circ$, $\\therefore$ $S_{\\text{quadrilateral} ABCD}=S_{\\triangle ABC}+S_{\\triangle ACD}$, =$\\frac{1}{2}\\cdot$AB$\\cdot$BC+$\\frac{1}{2}\\cdot$AC$\\cdot$DA, =$\\frac{1}{2}\\times\\sqrt{6}\\times\\sqrt{2}+\\frac{1}{2}\\times2\\sqrt{2}\\times1$, =$\\boxed{\\sqrt{2}+\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52733494.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In right triangle $ABC$, $\\angle B=90^\\circ$, $AB=4$, $BC=8$. Points $D$ and $E$ are located on sides $AC$ and $BC$, respectively. If $\\triangle ABC$ is folded along $DE$ such that points $A$ and $C$ coincide, what is the length of $EC$?\n",
    "completion": "\\textbf{Solution:} Let $EC = x$, then $BE = 8 - x$,\\\\\nsince folding $\\triangle ABC$ along $DE$ causes point $A$ to coincide with point $C$,\\\\\nit follows that $AE = EC = x$,\\\\\nIn $\\text{Rt}\\triangle ABE$, we have $AB^2 + BE^2 = AE^2$,\\\\\n$4^2 + (8 - x)^2 = x^2$,\\\\\nsolving for $x$ yields $x = \\boxed{5}$,\\\\\ntherefore, $EC = 5$.",
    "image_path": "dataset/mm_math/images/MM_Math/52991703.png"
  },
  {
    "solution": "\\boxed{20^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C = 90^\\circ$, $\\angle B = 35^\\circ$. The perpendicular bisector MN of segment AB intersects AB at point E and intersects BC at point D. If AD is connected, what is the measure of $\\angle DAC$ in degrees?\n",
    "completion": "\\textbf{Solution:} Since $\\angle B=35^\\circ$ and $\\angle C=90^\\circ$, \\\\\nit follows that $\\angle CAB=55^\\circ$, \\\\\nSince MN is the perpendicular bisector of line segment AB, \\\\\nit follows that AD=BD, \\\\\nThus, $\\angle B=\\angle BAD=35^\\circ$, \\\\\nHence, $\\angle DAC=\\boxed{20^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51368488.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the figure, for trapezoid $ABCD$, where $AD\\parallel BC$, point $E$ is on line segment $AD$, $CE$ and $BD$ intersect at point $H$, and $CE$ intersects the extension line of $BA$ at point $G$. Given $DE:AE=2:3$, $BC=4DE$, and $CE=10$ cm, find the lengths of $EH$ and $GE$.\n",
    "completion": "\\textbf{Solution:}\n\nSince $AD \\parallel BC$,\n\n$\\therefore \\frac{DE}{BC}=\\frac{EH}{HC}$.\n\nSince $BC=4DE$,\n\n$\\therefore \\frac{DE}{BC}=\\frac{1}{4}$.\n\nSince $CE=10$,\n\n$\\therefore HC=10-EH$\n\n$\\therefore \\frac{1}{4}=\\frac{EH}{10-EH}$,\n\n$\\therefore EH=\\boxed{2}$.\n\nSince $BC=4DE$, $DE:AE=2:3$,\n\n$\\therefore \\frac{AE}{BC}=\\frac{3}{8}$.\n\nSince $AD \\parallel BC$,\n\n$\\therefore \\triangle AGE \\sim \\triangle BGC$,\n\n$\\therefore \\frac{AE}{BC}=\\frac{GE}{GC}$.\n\nSince $CE=10$,\n\n$\\therefore GC=10+GE$,\n\n$\\therefore \\frac{3}{8}=\\frac{GE}{10+GE}$,\n\n$\\therefore GE=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51379447.png"
  },
  {
    "solution": "\\boxed{\\frac{8}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $B$ is on ray $AN$, an equilateral triangle $\\triangle ABC$ is constructed with $AB$ as a side, $M$ is the midpoint of $AN$, and $AN=4$. $P$ is the midpoint of $BC$. When $PM+PN$ is minimized, what is the length of $AB$?\n",
    "completion": "\\textbf{Solution:} Draw $AP$ and construct the point $D$ as the symmetric point of $M$ with respect to $AP$, and connect $DP$,\\\\\n$\\therefore PD=PM$, $AD=AM$,\\\\\n$\\therefore PM+PN=PD+PN\\ge DN$,\\\\\nthus, when $D$, $P$, $N$ are collinear and $DN\\perp AC$, $PM+PN$ is minimized;\\\\\n$\\because M$ is the midpoint of $AN$, and $AN=4$,\\\\\n$\\therefore AD=AM=\\frac{1}{2}AN=2$;\\\\\n$\\because P$ is the midpoint of $BC$, and $\\triangle ABC$ is an equilateral triangle,\\\\\n$\\therefore AP\\perp BC$, $\\angle CAP=30^\\circ$, $\\angle C=60^\\circ$,\\\\\nin $Rt\\triangle ADP$, $AP=2PD$, by the Pythagorean theorem $DP^{2}+2^{2}=(2DP)^{2}$,\\\\\n$\\therefore DP=\\frac{2\\sqrt{3}}{3}$;\\\\\n$\\because \\angle C=60^\\circ$, $DN\\perp AC$,\\\\\n$\\therefore \\angle DPC=30^\\circ$,\\\\\n$\\therefore PC=2DC$\\\\\nin $Rt\\triangle PDC$, by the Pythagorean theorem $DC^{2}+DP^{2}=(2DC)^{2}$,\\\\\nthat is $DC^{2}+\\left(\\frac{2\\sqrt{3}}{3}\\right)^{2}=(2DC)^{2}$, solving gives: $DC=\\frac{2}{3}$,\\\\\n$\\therefore AC=AD+DC=2+\\frac{2}{3}=\\frac{8}{3}$,\\\\\n$\\therefore AB=AC=\\boxed{\\frac{8}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53078593.png"
  },
  {
    "solution": "\\boxed{y=\\frac{1}{3}x+2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of a linear function intersects the x-axis at point A and the y-axis at point B. What is the analytical expression of the linear function?\n",
    "completion": "Solution: From the graph, we can find $A(-6, 0)$ and $B(0, 2)$. Assume the equation of the linear function is $y=kx+b$. Substituting the coordinates of $A$ and $B$ into the equation, we obtain\n\\[\n\\begin{cases}\n-6k+b=0 \\\\\nb=2\n\\end{cases}\n\\]\nFrom which we can solve\n\\[\n\\begin{cases}\nk=\\frac{1}{3} \\\\\nb=2\n\\end{cases}\n\\]\nTherefore, the equation of the linear function is $\\boxed{y=\\frac{1}{3}x+2}$",
    "image_path": "dataset/mm_math/images/MM_Math/50555422_88.png"
  },
  {
    "solution": "\\boxed{\\frac{5}{6}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated, both $\\triangle ADB$ and $\\triangle BCD$ are equilateral triangles. Extend $AD$ to point $E$ such that $\\angle AEC=90^\\circ$, with $AD=5$. A moving point $M$ starts from point $B$, travelling towards $D$ at a speed of 1 unit/second, while simultaneously, point $N$ moves from $D$ towards $C$ at a speed of 2 units/second. The movement of both points ceases the moment either reaches its destination. Connect $AM$, $CM$, $MN$, and $NE$. Given the duration of movement as $t$ ($0 \\leq t \\leq 2.5$), connect $BN$. At what value of $t$ will points $B$, $N$, and $E$ be collinear?\n",
    "completion": "\\textbf{Solution:} Since $\\angle ADE$ is a straight angle, \\\\\nit follows that $\\angle CDE=180^\\circ-\\angle ADB-\\angle BDC=180^\\circ-60^\\circ-60^\\circ=60^\\circ$,\\\\\nSince $\\angle CEA=90^\\circ$,\\\\\nit follows that $\\angle DCE=180^\\circ-\\angle CDE-\\angle CED=180^\\circ-60^\\circ-90^\\circ=30^\\circ$,\\\\\ntherefore $DE=\\frac{1}{2}CD=\\frac{1}{2}\\times 5=2.5$, and $CE=\\sqrt{CD^{2}-DE^{2}}=\\sqrt{5^{2}-2.5^{2}}=\\frac{5}{2}\\sqrt{3}$\\\\\nSince points B, N, and E are collinear;\\\\\nit follows that $\\angle BNC=\\angle END$,\\\\\nSince $\\angle BCD=\\angle CDE=60^\\circ$,\\\\\nit follows that BC$\\parallel$DE,\\\\\ntherefore $\\triangle BCN\\sim \\triangle EDN$,\\\\\nthus $\\frac{BC}{ED}=\\frac{CN}{DN}$ that is $\\frac{5}{2.5}=\\frac{5-DN}{DN}$,\\\\\nsolving gives $DN=\\frac{5}{3}$,\\\\\nhence $2t=\\frac{5}{3}$,\\\\\nsolving for $t$ yields $t=\\boxed{\\frac{5}{6}}$,\\\\\ntherefore, when $t=\\boxed{\\frac{5}{6}}$, points B, N, and E are collinear.",
    "image_path": "dataset/mm_math/images/MM_Math/51403181_163.png"
  },
  {
    "solution": "\\boxed{90^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, lines a and b are intersected by lines c and d. If $a\\parallel b$ and $\\angle 1=120^\\circ$, $\\angle 2=30^\\circ$, then what is the measure of $\\angle 3$ in degrees?\n",
    "completion": "\\textbf{Solution:} As shown in the figure,\\\\\n$\\because a \\parallel b$,\\\\\n$\\therefore \\angle 4 = \\angle 1 = 120^\\circ$,\\\\\n$\\because \\angle 4 = \\angle 2 + \\angle 3$,\\\\\n$\\therefore \\angle 3 = \\angle 4 - \\angle 2 = 120^\\circ - 30^\\circ = \\boxed{90^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52907830.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points $C$ and $E$ are two points on segment $AB$, point $D$ is the midpoint of segment $AB$, and $AB=6$, $CD=1$. What is the length of $BC$?\n",
    "completion": "\\textbf{Solution:} Since point D is the midpoint of line segment AB, \\\\\n$\\therefore$ AD=DB=$\\frac{1}{2}$AB=3, \\\\\n$\\therefore$ BC=DB−DC=3−1=\\boxed{2};",
    "image_path": "dataset/mm_math/images/MM_Math/51561259_160.png"
  },
  {
    "solution": "\\boxed{\\left(-\\frac{3}{2},-\\frac{19}{4}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, it is known that the parabola $y=x^{2}+bx+c$ intersects with line $AB$ at points $A$ and $B$, where $A(-3,-4)$ and $B(0,-1)$. The axis of symmetry of the parabola intersects with the x-axis at point $Q$. Point $P$ is any point on the parabola below line $AB$, and lines $PA$, $PB$, $QA$, $QB$ are connected. Find the maximum area of quadrilateral $PAQB$ and the coordinates of point $P$ at this time?\n",
    "completion": "\\textbf{Solution:} Let the equation of line AB be $y=kx+b$. Substituting A$(-3, -4)$ and B$(0, -1)$ into it, we get\n\\[\n\\left\\{\n\\begin{array}{l}\n-3k+b=-4 \\\\\nb=-1\n\\end{array}\n\\right.\n\\]\n$\\therefore k=1$, $b=-1$, thus $y=x-1$,\n\nDraw a perpendicular line from point P to the x-axis, intersecting the line AB at point F.\n\nLet point P be $(a,a^2+4a-1)$, then the coordinates of point F are $(a,a-1)$,\n\n$S_{\\triangle PAB}=\\frac{1}{2} |PF|\\cdot|x_B-x_A| =\\frac{3}{2}(a-1-a^2-4a+1)$\n\n$=\\frac{3}{2}(-a^2-3a)=-\\frac{3}{2}(a+\\frac{3}{2})^2+\\frac{27}{8}$,\n\nWhen $a=-\\frac{3}{2}$, $S_{\\triangle PAB}$ reaches its maximum value of $\\frac{27}{8}$. Since the area of $\\triangle QAB$ is $\\frac{9}{2}$,\n\n$\\therefore$ the maximum area of quadrilateral PAQB is $\\boxed{\\frac{63}{8}}$, at which point the coordinates of P are $\\boxed{\\left(-\\frac{3}{2},-\\frac{19}{4}\\right)}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52694121_91.png"
  },
  {
    "solution": "\\boxed{(2,0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the right-angled triangle $ABC$, where $\\angle ABC=90^\\circ$ and the side $AB$ is on the x-axis, taking $O$ as the center of dilation, the triangle $A_1B_1C_1$ is similar to the triangle $ABC$. If the corresponding point of $C(4,6)$ is $C_1(2,3)$, what are the coordinates of $B_1$?\n",
    "completion": "\\textbf{Solution:} Given that in the right triangle $ABC$, $\\angle ABC=90^\\circ$, and $C(4,6)$,\\\\\nthus $B(4,0)$,\\\\\ntherefore, taking $O$ as the center of similarity, construct $\\triangle A_1B_1C_1$ similar to $\\triangle ABC$, with the corresponding point of $C(4,6)$ being $C_1(2,3)$,\\\\\nthus, the coordinates of ${B}_1$ are: $(4\\times \\frac{1}{2},0)$, that is $\\boxed{(2,0)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51806942.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $OC\\perp AB$ at point $O$, $\\angle COD=\\frac{1}{4}\\angle BOD$, and $OE$ bisects $\\angle BOD$. A ray $OF$ is drawn through point $O$, and if $OF\\perp OE$, what is the measure of $\\angle BOF$?\n",
    "completion": "Solution: As shown in the figure, when OF is above line AB,\n\n$\\because$ OF$\\perp$OE,\n\n$\\therefore$ $\\angle EOF=90^\\circ$,\n\n$\\because$ $\\angle BOE=60^\\circ$,\n\n$\\therefore$ $\\angle BOF=\\angle BOE+\\angle EOF=60^\\circ+90^\\circ=150^\\circ$;\n\nWhen OF is below line AB,\n\n$\\because$ OF$\\perp$OE,\n\n$\\therefore$ $\\angle EOF=90^\\circ$,\n\n$\\because$ $\\angle BOE=60^\\circ$,\n\n$\\therefore$ $\\angle BOF=\\angle EOF-\\angle BOE=90^\\circ-60^\\circ=\\boxed{30^\\circ}$,\n\nTherefore, the measure of $\\angle BOF$ is $150^\\circ$ or $\\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52774109_159.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $OB$ is the bisector of $\\angle AOC$, and $OD$ is the bisector of $\\angle COE$. If $\\angle AOB=30^\\circ$ and $\\angle COE=60^\\circ$, then what is the measure of $\\angle BOD$?\n",
    "completion": "\\textbf{Solution:} Since OB is the bisector of $\\angle AOC$, and OD is the bisector of $\\angle COE$, \\\\\n$\\therefore \\angle COD= \\frac{1}{2} \\angle COE$ and $\\angle BOC=\\angle AOB$, \\\\\nAlso, since $\\angle AOB=30^\\circ$ and $\\angle COE=60^\\circ$, \\\\\n$\\therefore \\angle BOC=30^\\circ$ and $\\angle COD=30^\\circ$, \\\\\n$\\therefore \\angle BOD=\\angle BOC+\\angle COD=30^\\circ+30^\\circ=\\boxed{60^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52948978.png"
  },
  {
    "solution": "\\boxed{\\frac{49}{16}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given the parabola \\(y=ax^2+bx+c\\) intersects the x-axis at points \\(A(-1, 0)\\) and \\(B(3, 0)\\), and intersects the y-axis at point C. It also passes through point \\(D(2, -3)\\). Points P and Q are moving points on the parabola \\(y=ax^2+bx+c\\). When point P is located below the line OD, what is the maximum area of \\(\\triangle POD\\)?\n",
    "completion": "\\textbf{Solution:} Let the line PD intersect the y-axis at point G, assume point $P\\left(m, m^{2}-2m-3\\right)$,\\\\\nSubstituting the coordinates of points P, D into the linear function equation: $y=mx+t$ and solving, we get the expression for the line PD as: $y=mx-3-2m$, thus $OG=3+2m$,\\\\\n$S_{\\triangle POD}=\\frac{1}{2}\\times OG(x_{D}-x_{P})=\\frac{1}{2}(3+2m)(2-m)=-m^{2}+\\frac{1}{2}m+3$,\\\\\nSince $-1<0$, it follows that $S_{\\triangle POD}$ has a maximum value, when $m=\\frac{1}{4}$, its maximum value is $\\boxed{\\frac{49}{16}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51284522_193.png"
  },
  {
    "solution": "\\boxed{10\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, points D and E are the midpoints of sides BC and AC, respectively, with $BE = 2DE$. Line $AF$ is drawn parallel to $BE$ and intersects the extension of line DE at point F. Quadrilateral ABEF is a rhombus. If $\\angle ABE = 45^\\circ$ and $AB = 4$, what is the area of quadrilateral ABDF?\n",
    "completion": "\\textbf{Solution:} Draw $EM\\perp AB$ at point M, \\\\\n$AB=2DE$, $AB=4$, \\\\\n$DE=2$, \\\\\nBecause quadrilateral ABEF is a rhombus, \\\\\nTherefore $BE=EF=AB=4$, \\\\\nTherefore $DF=DE+EF=2+4=6$, \\\\\nIn $\\triangle EBM$, $\\angle ABE=45^\\circ$, \\\\\nTherefore $\\triangle EBM$ is an isosceles right triangle, \\\\\nThus $EM^2+BM^2=BE^2$, \\\\\nTherefore $EM=\\frac{\\sqrt{2}}{2}BE=\\frac{\\sqrt{2}}{2}\\times 4=2\\sqrt{2}$, \\\\\nTherefore, the area of quadrilateral ABDE is: $\\frac{1}{2}(AB+DF)\\cdot EM = \\frac{1}{2}\\times (4+6)\\times 2\\sqrt{2} = \\boxed{10\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52188880_126.png"
  },
  {
    "solution": "\\boxed{108^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: It is known that $O$ is a point on line $AB$, $\\angle COD$ is a right angle, and $OE$ bisects $\\angle BOC$. In the process where $\\angle DOC$ in Figure 1 is rotated clockwise around vertex $O$ to the position in Figure 2, with ray $OC$ remaining above line $AB$, at what degree of $\\angle AOC$ will $\\angle COE = 2\\angle DOB$?\n",
    "completion": "\\textbf{Solution:}\n\n(1) When OD is above the line AB, let $\\angle DOB=\\alpha$, then $\\angle COE=2\\alpha$.\\\\\nSince OE bisects $\\angle BOC$,\\\\\nit follows that $\\angle COE=\\angle BOE=2\\alpha$.\\\\\nSince $\\angle DOE=2\\alpha-\\alpha=\\alpha$,\\\\\nit follows that $\\angle COD=90^\\circ=\\angle COE+\\angle DOE=2\\alpha+\\alpha=3\\alpha$.\\\\\nThus, $\\alpha=30^\\circ$.\\\\\nTherefore, from (2) we have $\\angle AOC=2\\angle DOE=60^\\circ$.\\\\\n(2) When OD is below the line AB, let $\\angle DOB=\\alpha$, then $\\angle COE=2\\alpha$.\\\\\nSince OE bisects $\\angle BOC$,\\\\\nit follows that $\\angle COE=\\angle BOE=2\\alpha$.\\\\\nSince $\\angle COD=90^\\circ=\\angle COE+\\angle BOE+\\angle DOB=5\\alpha$,\\\\\nit follows that $\\alpha=18^\\circ$.\\\\\nThus, $\\angle COE+\\angle BOE=4\\alpha=72^\\circ$.\\\\\nTherefore, $\\angle AOC=180^\\circ-72^\\circ=\\boxed{108^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51317331_62.png"
  },
  {
    "solution": "\\boxed{\\left(\\frac{2\\sqrt{3}-3}{2}, \\frac{\\sqrt{3}}{2}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a right-angled triangle paper $ABO$ is placed in the Cartesian coordinate system, where point $A(\\sqrt{3}, 0)$, point $B(0, 1)$, and point $O(0, 0)$. $P$ is a moving point on side $AB$ (point $P$ does not coincide with points $A$ or $B$), and the paper is folded along $OP$ to obtain the corresponding point $A'$ of point $A$. When $\\angle BPA' = 30^\\circ$, directly write down the coordinates of point $P$.\n",
    "completion": "\\textbf{Solution:} When $\\angle BPA'=30^\\circ$, the coordinates of point $P$ are $\\boxed{\\left(\\frac{3-\\sqrt{3}}{2}, \\frac{3-\\sqrt{3}}{2}\\right)}$ or $\\boxed{\\left(\\frac{2\\sqrt{3}-3}{2}, \\frac{\\sqrt{3}}{2}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53572668_146.png"
  },
  {
    "solution": "\\boxed{60}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: The great ancient Chinese mathematician Liu Hui divided the right-angled triangle (referred to as the Gougu shape by ancients) into one square and two pairs of congruent right-angled triangles. As shown in the figure, if $AF = 5$ and $CE = 12$, then what is the area of this triangle?\n",
    "completion": "Solution: Let the side length of the small square be $x$,\\\\\n$\\because$AF=5, CE=12,\\\\\n$\\therefore$AB=5+x, BC=12+x,\\\\\n$\\because$$\\triangle$AFM$\\cong$$\\triangle$ADM, $\\triangle$CDM$\\cong$$\\triangle$CEM,\\\\\n$\\therefore$AD=AF=5, CD=CE=12,\\\\\n$\\therefore$AC=AD+CD=5+12=17,\\\\\nIn $\\triangle$ABC, $AC^{2}=AB^{2}+BC^{2}$,\\\\\n$\\therefore$17$^{2}=(5+x)^{2}+(12+x)^{2}$,\\\\\nSolving, we get $x=3$ (negative values are discarded),\\\\\n$\\therefore$AB=8, BC=15,\\\\\n$\\therefore$The area of $\\triangle$ABC is: $\\frac{1}{2}\\times 8\\times 15=\\boxed{60}$,",
    "image_path": "dataset/mm_math/images/MM_Math/52019441.png"
  },
  {
    "solution": "\\boxed{\\frac{5}{13}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that acute triangle $ABC$ is inscribed in circle $\\odot O$. When $AB>AC$, connect $BO$ and extend it to intersect side $AC$ at point $D$. If $\\angle A=45^\\circ$, $\\frac{OD}{OB}=\\frac{2}{3}$, what is $\\frac{AD}{DC}$?\n",
    "completion": "\\textbf{Solution:}\n\nExtend $BD$ to intersect $\\odot O$ at point $H$, and connect $CH$, $CO$, and $AH$. \\\\\n\nSince $\\frac{OD}{OB} = \\frac{2}{3}$, \\\\\n\nThus, set $BO = 3a = OC = OH$, and $OD = 2a$, \\\\\n\nHence, $DH = a$, \\\\\n\nSince $\\angle BAC = 45^\\circ$, \\\\\n\nThus $\\angle BOC = 2\\angle BAC = 90^\\circ$, \\\\\n\nHence, $CD = \\sqrt{OD^2 + OC^2} = \\sqrt{13}a$, $CH = \\sqrt{OH^2 + OC^2} = 3\\sqrt{2}a$, \\\\\n\nSince $BH$ is a diameter, hence $\\angle BAH = 90^\\circ$, \\\\\n\nSince $\\angle BAC = \\angle BHC = 45^\\circ$ \\\\\n\nThus $\\angle CAH = 90^\\circ - \\angle BAC = 45^\\circ = \\angle CHD$ \\\\\n\nAlso, since $\\angle ACH = \\angle DCH$, \\\\\n\nTherefore, $\\triangle ACH \\sim \\triangle HCD$, \\\\\n\nTherefore, $\\frac{AC}{CH} = \\frac{CH}{CD}$, \\\\\n\nThus, $\\frac{AC}{3\\sqrt{2}a} = \\frac{3\\sqrt{2}a}{\\sqrt{13}a}$, \\\\\n\nHence, $AC = \\frac{18\\sqrt{13}a}{\\sqrt{13}}$, \\\\\n\nThus, $AD = AC - CD = \\frac{18\\sqrt{13}a}{\\sqrt{13}} - \\sqrt{13}a = \\frac{5\\sqrt{13}a}{\\sqrt{13}}$, \\\\\n\nTherefore, $\\frac{AD}{CD} = \\frac{\\frac{5\\sqrt{13}a}{\\sqrt{13}}}{\\sqrt{13}a} = \\boxed{\\frac{5}{13}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52872084_84.png"
  },
  {
    "solution": "\\boxed{\\frac{3\\sqrt{7}}{7}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that after the line segment AD is moved downward by $m$ units and then moved to the right by $n$ units to the line segment BC, the corresponding points of points A and D are points B and C, respectively. Connect AB, CD, AC, and BD, with AC and BD intersecting at point O. It is given that $OB = OD$. Draw $DM \\perp BC$ at M, let N be the midpoint of CD, and connect MN. If $\\angle ADB = 45^\\circ$ and $OD = 3\\sqrt{2}$, with MN equal to 4, what is the value of $\\frac{BM}{MC}$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a parallelogram, \\\\\nit follows that $\\angle ADB=\\angle CBD$. $OB=OD=3\\sqrt{2}$, $BD=6\\sqrt{2}$.\\\\\nSince $\\angle ADB=45^\\circ$, and $DM\\perp BC$, \\\\\nit follows that $\\triangle BDM$ is an isosceles right triangle, hence $BM=MD=6$.\\\\\nSince $\\triangle CMD$ is a right triangle, with $N$ being the midpoint of $CD$, and $MN=4$, it follows that $CD=2MN=8$.\\\\\nIn $\\triangle DCM$, by the Pythagorean theorem, we find $MC=2\\sqrt{7}$, hence $\\frac{BM}{MC}=\\frac{6}{2\\sqrt{7}}=\\boxed{\\frac{3\\sqrt{7}}{7}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52386348_127.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AD$ is the altitude on the side $BC$, $CE$ is the median on the side $AB$, $DG \\perp CE$ at $G$, and $CD = AE$. Given that $BD = 6$, $CD = 5$, and $CG = EG$; find the area of $\\triangle CDG$.\n",
    "completion": "\\textbf{Solution:} Given $CE$ is the median of side $AB$,\\\\\nhence $AE=BE$,\\\\\nsince $BD=6, CD=5$,\\\\\nit follows that $AB=10$,\\\\\nthus $AD=\\sqrt{10^{2}-6^{2}}=8$,\\\\\ntherefore, the area of $\\triangle ABC=\\frac{1}{2}\\times 11\\times 8=44$, the area of $\\triangle ABD=\\frac{1}{2}\\times 6\\times 8=24$,\\\\\nsince $CE$ is the median of side $AB$,\\\\\nit follows that the area of $\\triangle BEC=\\frac{1}{2}$ area of $\\triangle ABC=22$,\\\\\nsince $DE$ is the median of side $AB$,\\\\\nit follows that the area of $\\triangle BDE=\\frac{1}{2}$ area of $\\triangle ABD=12$,\\\\\nthus the area of $\\triangle EDC$ is area of $\\triangle BEC$ - area of $\\triangle EDB=22-12=10$,\\\\\nadditionally, since $CG=EG$,\\\\\nthe area of $\\triangle CDG=\\frac{1}{2}$ area of $\\triangle EDC=\\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52828236_118.png"
  },
  {
    "solution": "\\boxed{16}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, if $A$ represents the area of the square, what is the value of $A$?\n",
    "completion": "\\textbf{Solution:} Let the side length of the square represented by A be $x$,\\\\\nby the Pythagorean theorem we have $x^{2}+3^{2}=5^{2}$,\\\\\n$\\therefore x^{2}=16,$ which means the area of the square represented by A is $\\boxed{16}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51737108.png"
  },
  {
    "solution": "\\boxed{y=x^{2}+2x-3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, the parabola $y=x^{2}+bx+c$ intersects the x-axis at points $A(-3,0)$ and $B(1,0)$, and intersects the y-axis at point $D$. The vertex is at point $C$, and the line $BC$ intersects the y-axis at point $E$. Find the analytical expression of the parabola?\n",
    "completion": "Solution: Since the parabola $y=x^{2}+bx+c$ intersects the x-axis at points $A(-3,0)$ and $B(1,0)$, \\\\\nit follows that the roots of $x^{2}+bx+c=0$ are $-3$ and $1$. \\\\\nHence, $c=-3\\times 1=-3$, and $b=-(-3+1)=2$. \\\\\nTherefore, the equation of the parabola is \\boxed{y=x^{2}+2x-3};",
    "image_path": "dataset/mm_math/images/MM_Math/51445907_258.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the Cartesian coordinate system, the line $l_{1}\\colon y=3x$ intersects with the line $l_{2}\\colon y=kx+b$ at point $A(a, 3)$, and point $B(2, 4)$ lies on the line $l_{2}$. Write directly the solution set of the inequality $kx+b<3x$ in terms of $x$.\n",
    "completion": "\\textbf{Solution:} The solution to the inequality $kx + b < 3x$ in terms of $x$ is $x > \\boxed{1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51983442_184.png"
  },
  {
    "solution": "\\boxed{72^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$, and $AE \\perp BD$ at point $E$. If $\\angle BAE = 2\\angle OAE$, what is the measure of $\\angle AOB$?\n",
    "completion": "\\textbf{Solution}: Given that quadrilateral ABCD is a rectangle, \\\\\nthus AO=BO, \\\\\nthus $\\angle$OAB=$\\angle$OBA.\\\\\nGiven that $\\angle$BAE=2$\\angle$OAE, \\\\\nthus let $\\angle$OAE=x, then $\\angle$BAE=2x, $\\angle$ABO=$\\angle$OAB=x+2x=3x, $\\angle$AOE=$90^\\circ-$.\\\\\nGiven that $\\angle$AOB+$\\angle$OAB+$\\angle$OBA=$180^\\circ$, \\\\\nthus 3x+3x+$90^\\circ-x=180^\\circ$, \\\\\nsolving for x gives x=$18^\\circ$, \\\\\nthus $\\angle$AOB=$90^\\circ-18^\\circ=\\boxed{72^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50475264.png"
  },
  {
    "solution": "\\boxed{}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, point E is a point on the side CD of the rhombus ABCD. Folding $\\triangle ADE$ along AE, the corresponding point of point D, which is point F, exactly falls on the side BC. Let $\\frac{DE}{CE} = k$. If point F coincides with point C, what is the value of k?\n",
    "completion": "The information you have provided seems to be incomplete. You have only given a number \"1\", without providing any LaTeX code or a specific description of the mathematical problem. In order to assist you with cleaning and correcting the LaTeX code, please provide the complete math problem and the corresponding LaTeX code.",
    "image_path": "dataset/mm_math/images/MM_Math/53565373_56.png"
  },
  {
    "solution": "\\boxed{1.5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that point $A(0, 2)$, $B(4, 0)$, point $C$ is on the $x$ axis, $CD\\perp x$ axis, intersecting line segment $AB$ at point $D$, and point $D$ does not coincide with points $A$, $B$. Fold $\\triangle ABO$ along $CD$, making point $B$ fall at point $E$ on the $x$ axis. Let the abscissa of point $C$ be $x$. Then, when $\\triangle ADE$ is a right-angled triangle, what is the value of $x$?\n",
    "completion": "\\textbf{Solution:} Given points A(0, 2), and B(4, 0),\\\\\n$\\therefore OA=2$, $OB=4$. Consider two cases:\\\\\n(1) When $\\angle DAE=90^\\circ$,\\\\\nthen $\\angle EAO+\\angle BAO=90^\\circ$,\\\\\n$\\because \\angle AOB=90^\\circ$,\\\\\n$\\therefore \\angle ABO+\\angle BAO=90^\\circ$,\\\\\n$\\therefore \\angle EAO=\\angle ABO$,\\\\\nalso, $\\because \\angle AOE=\\angle BOA=90^\\circ$,\\\\\n$\\therefore \\triangle AOE \\sim \\triangle BOA$,\\\\\n$\\therefore \\frac{OE}{OA}=\\frac{OA}{OB}=\\frac{1}{2}$,\\\\\n$\\therefore OE=\\frac{1}{2}OA=1$,\\\\\n$\\therefore BE=OE+OB=5$,\\\\\nBy the property of folding: $CE=CB=\\frac{1}{2}BE=2.5$,\\\\\n$\\therefore OC=CE-OE=1.5$,\\\\\n$\\therefore x=1.5$;\\\\\n(2) When $\\angle AED=90^\\circ$,\\\\\nSimilar to (1), $\\triangle AOE \\sim \\triangle BOA$,\\\\\n$\\therefore \\frac{OE}{OA}=\\frac{OA}{OB}=\\frac{1}{2}$,\\\\\n$\\therefore OE=\\frac{1}{2}OA=1$,\\\\\n$\\therefore BE=OB-OE=3$,\\\\\nBy the property of folding: $CE=CB=\\frac{1}{2}BE=1.5$,\\\\\n$\\therefore OC=CE+OE=2.5$,\\\\\n$\\therefore x=2.5$;\\\\\nIn summary, when $\\triangle ADE$ is a right triangle, the value of $x$ can be either $\\boxed{2.5}$ or $\\boxed{1.5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52513834.png"
  },
  {
    "solution": "\\boxed{72^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ is folded along line MN such that point A coincides with point E on side AB. If $\\angle B=54^\\circ$ and $\\angle C=90^\\circ$, what is the measure of $\\angle ENC$?\n",
    "completion": "\\textbf{Solution:} \n\\[\n\\because \\angle B=54^\\circ, \\angle C=90^\\circ,\n\\]\n\\[\n\\therefore \\angle A=90^\\circ-\\angle B=36^\\circ,\n\\]\nFrom the property of folding, we get:\n\\[\n\\angle AEN=\\angle A=36^\\circ,\n\\]\n\\[\n\\therefore \\angle ENC=\\angle AEN+\\angle A=\\boxed{72^\\circ}.\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/52891044.png"
  },
  {
    "solution": "\\boxed{24}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system, the line $y=\\frac{1}{2}x+b$ intersects the x-axis at point A, and the line $y=-x+2$ intersects the x-axis at point B. The two lines intersect at point $C(-2, a)$. What is the area of $\\triangle ABC$?\n",
    "completion": "\\textbf{Solution:} For the equation $y=\\frac{1}{2}x+5$, when $y=0$, we have $x=-10$, thus $A(-10, 0)$. In the equation $y=-x+2$, when $y=0$, we get $x=2$, hence $B(2, 0)$. Consequently, $AB=12$ and the area of $\\triangle ABC$ is given by ${S}_{\\triangle ABC}=\\frac{1}{2}\\times 12\\times 4=\\boxed{24}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52423819_217.png"
  },
  {
    "solution": "\\boxed{50}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a circular pipe in a residential area has ruptured, and the repair workers are preparing to replace a section of the pipe. They have measured the width of the sewage water surface to be $80\\, \\text{cm}$, and the distance from the water surface to the top of the pipe to be $20\\, \\text{cm}$. What should be the inner diameter of the new pipe that the repair workers need to prepare, in $\\text{cm}$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, let the center of the pipe be point O and the top of the pipe be point A, with BD representing the surface of the sewage. Connect AO, where AO intersects BD perpendicularly at point C.\\\\\nLet AO=OB=r, then OC=r−20, and BC=$\\frac{1}{2}BD=40$.\\\\\nWe have $OB^{2}=OC^{2}+BC^{2}$,\\\\\n$r^{2}=(r−20)^{2}+40^{2}$,\\\\\nSimplifying, we get $r=\\boxed{50}$.\\\\\nTherefore, the internal diameter of the new pipe is $100\\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51318341.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, behind a building \\textup{AB} there is an artificial hill with a slope \\textup{CD} of $i=1\\colon 2$. At point \\textup{E} on the slope, there is a pavilion. It is measured that the horizontal distance from the foot of the hill to the building $BC=24$ meters, and the distance to the pavilion $CE=8\\sqrt{5}$ meters. Xiao Li measures the angle of depression of point \\textup{E} from the top of the building to be $45^\\circ$. What is the distance from point \\textup{E} to the horizontal ground in meters?\n",
    "completion": "\\textbf{Solution:} Draw line $EF \\perp BC$ extending through point $E$ to $F$. In $\\triangle CEF$,\\\\\nsince the slope of $CD$ is $i=EF:CF=1:2$,\\\\\nit follows that $EF:CF:CE=1:2:\\sqrt{5}$,\\\\\nsince $CE=8\\sqrt{5}$,\\\\\nthus $EF=8$, and $CF=16$ meters,\\\\\ntherefore, the distance from point $E$ to the flat ground is $\\boxed{8}$ meters.",
    "image_path": "dataset/mm_math/images/MM_Math/51457882_87.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the parabola $y=ax^2+bx-1\\ (a\\neq 0)$ intersects the x-axis at points $A(-2, 0)$ and $B(2, 0)$, and intersects the y-axis at point D. A line passing through point $C(0, 1)$ intersects the parabola at points A and F. Under these conditions, shift the parabola $y=ax^2+bx-1\\ (a\\neq 0)$ to the right by 2 units and then downward by 1 unit to obtain a new parabola $y_1$. Let point M be a point on the axis of symmetry of $y_1$, and point N be another point on $y_1$. If the quadrilateral with vertices D, P, M, N is a parallelogram, write down the coordinates of all possible points M, and select one of these points M to explain the process of finding its coordinates.\n",
    "completion": "\\textbf{Solution:} \nSince $y=\\frac{1}{4}x^{2}-1$ intersects the y-axis at point D, \\\\\n$\\therefore$ D(0, $-1$),\\\\\nSince the parabola $y=ax^{2}+bx-1\\ (a\\ne 0)$ is first translated 2 units to the right and then 1 unit downward to get the new parabola $y_{1}=\\frac{1}{4}(x-2)^{2}-2$, \\\\\n$\\therefore$ the x-coordinate of point M is 2, \\\\\n(1) When DP is the diagonal, the x-coordinate of point N is 1, \\\\\n$\\therefore$ the y-coordinate of point N is $-\\frac{7}{4}$, \\\\\n$\\therefore$ translating point N 1 unit to the left and $\\frac{3}{4}$ units upward yields point D; then applying the same translation to point P yields point M, i.e., M(2, 2); \\\\\n(2) When DM is the diagonal, the x-coordinate of point N is $-1$, \\\\\n$\\therefore$ the y-coordinate of point N is $\\frac{1}{4}$, \\\\\n$\\therefore$ translating point D 1 unit to the left and $\\frac{5}{4}$ units upward yields point N; then applying the same translation to point P yields point M, i.e., M(2, $\\frac{5}{2}$); \\\\\n(3) When DN is the diagonal, the x-coordinate of point N is 5, \\\\\n$\\therefore$ the y-coordinate of point N is $\\frac{1}{4}$, \\\\\n$\\therefore$ translating point P 2 units to the right and 1 unit downward yields point N; then applying the same translation to point D yields point M, i.e., M(2, $-2$); \\\\\nOverall, the coordinates of M are (2, 2), (2, $\\frac{5}{2}$), or (2, $-2$). Choosing M(2, 2) for the solution, we find the coordinates of M to be M(2, $\\boxed{2}$).",
    "image_path": "dataset/mm_math/images/MM_Math/52620662_110.png"
  },
  {
    "solution": "\\boxed{\\frac{100}{21}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within $\\triangle ABC$, $\\angle ACB=90^\\circ$, $AB=25$, and $AC=15$. Point $P$ starts from point $B$ and moves towards $A$ at a speed of $2$ units per second, while point $Q$ starts from $C$ and moves towards $B$ at a speed of $1$ unit per second. The movement of one point ceases immediately when the other point arrives at its destination. Connect $PQ$, and on ray $PB$, take $PN=PQ$, then construct $\\square PQMN$ with $PN$ and $PQ$ as edges. Let the movement time be $t$ seconds ($t>0$). Construct the symmetric point $C'$ of point $C$ with respect to the line $PQ$. When points $C$, $Q$, and $C'$ are not collinear, and $\\angle CQC'$ equals twice the internal angle of $\\triangle ABC$, directly write out the value of $t$.\n",
    "completion": "\\textbf{Solution:} We have $t=\\boxed{\\frac{40}{7}}$ or $t=\\boxed{\\frac{100}{21}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51445941_164.png"
  },
  {
    "solution": "\\boxed{x>1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given the figure, the graphs of the linear functions $y_{1}=kx+4$ and $y_{2}=x+m$ intersect at point $P\\left(1, 3\\right)$. What is the solution set of the inequality $kx+4<x+m$ with respect to $x$?\n",
    "completion": "\\textbf{Solution:} Given that $x>1$, the graph of the function $y_1=kx+4$ lies entirely below the graph of the function $y_2=x+m$, \\\\\n$\\therefore$ the solution set of the inequality $kx+4<x+m$ with respect to $x$ is $x>1$, \\\\\nthus, the answer is: $\\boxed{x>1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53359921.png"
  },
  {
    "solution": "\\boxed{\\frac{24}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $E$ is the midpoint of $BC$, $DF \\perp AE$ with the foot of the perpendicular being $F$. If $AB = 4$ and $BC = 6$, find the length of $DF$.\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rectangle, \\\\\n$\\therefore AD=BC$, $AD\\parallel BC$, $\\angle B=90^\\circ$, \\\\\nSince $BC=AD=6$, $E$ is the midpoint of $BC$, \\\\\n$\\therefore BE=3$, \\\\\nSince $\\angle B=90^\\circ$, \\\\\n$\\therefore AE=\\sqrt{BE^2+AB^2}=\\sqrt{3^2+4^2}=5$, \\\\\nSince $AD\\parallel BC$, \\\\\n$\\therefore \\angle DAF=\\angle AEB$, \\\\\nSince $DF\\perp AE$, \\\\\n$\\therefore \\angle AFD=90^\\circ=\\angle B$, \\\\\n$\\therefore \\triangle ADF\\sim \\triangle EAB$. \\\\\n$\\therefore \\frac{DF}{AB}=\\frac{AD}{AE}$, \\\\\nThat is $\\frac{DF}{4}=\\frac{6}{5}$, \\\\\nWe find that: $DF=\\boxed{\\frac{24}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52765937.png"
  },
  {
    "solution": "\\boxed{(-2,-1)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the line $y_1 = x + m$ intersects the x-axis and y-axis at points A and B, respectively, and intersects the hyperbola $y_2 = \\frac{k}{x} \\, (x < 0)$ at points C and D, with the coordinates of point C being $(-1, 2)$. Using the graph, directly write out: when $x$ takes values in what range does $y_1 > y_2$?\n",
    "completion": "\\textbf{Solution}: It can be known from the graph that when $-2<x<-1$, $y_1>y_2$. Therefore, when $x$ takes values in the range $(-2,-1)$, $y_1>y_2$ holds. Hence, the answer is $x\\in\\boxed{(-2,-1)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53101933_154.png"
  },
  {
    "solution": "\\boxed{150^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given in the figure, it is known that point $P$ is a point inside the equilateral triangle $ABC$, and $PA=6$, $PB=8$, $PC=10$. What is the measure of $\\angle APB$ in degrees?\n",
    "completion": "\\textbf{Solution}: Draw EP, \\\\\n$\\therefore$BE=BP=8, AE=PC=10, $\\angle$PBE=60$^\\circ$\\\\\n$\\therefore$ $\\triangle BPE$ is an equilateral triangle,\\\\\n$\\therefore$PE=PB=8, $\\angle$BPE=60$^\\circ$,\\\\\nIn $\\triangle AEP$, AE=10, AP=6, PE=8,\\\\\n$\\therefore$AE$^{2}$=PE$^{2}$+PA$^{2}$,\\\\\n$\\therefore$ $\\triangle APE$ is a right-angled triangle, and $\\angle$APE=90$^\\circ$,\\\\\n$\\therefore$ $\\angle APB=90^\\circ+60^\\circ=\\boxed{150^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51212147_98.png"
  },
  {
    "solution": "\\boxed{16}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In Figure 1, the hypotenuse of the right-angled triangle is 4. Four of these right-angled triangles from Figure 1 are assembled into a square as shown in Figure 2. As illustrated, the four right-angled triangles are assembled into a square, where the areas of the shaded parts are denoted as $S_1$ and $S_2$, respectively. What is the value of $S_1+S_2$?\n",
    "completion": "\\textbf{Solution}: Let the length of the longer leg of the right triangle be $a$, and the length of the shorter leg be $b$,\\\\\n$\\therefore$ $a^{2}+b^{2}=4^{2}$\\\\\n$\\because$ $S_{1}=a^{2}$, $S_{2}=b^{2}$\\\\\n$\\therefore$ $S_{1}+S_{2}=16$\\\\\nTherefore, the answer is: $\\boxed{16}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52076836.png"
  },
  {
    "solution": "\\boxed{-16}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point A is on the graph of the inverse proportion function $y = \\frac{k}{x}$ ($x < 0$), AB is perpendicular to the x-axis at point B, and C is the midpoint of OB. Connect AO and AC. If the area of $\\triangle ABC$ is 4, what is the value of $k$?\n",
    "completion": "\\textbf{Solution:} Since $C$ is the midpoint of $OB$, and the area of $\\triangle ABC$ is 4, \\\\\nit follows that the area of $\\triangle AOB$ is 8, \\\\\nLet $A(a, b)$, \\\\\nSince $AB$ is perpendicular to the x-axis at point $B$, \\\\\nthen $-ab = 16$, \\\\\nSince point $A$ lies on the graph of the inverse proportion function $y = \\frac{k}{x}$ (x < 0), \\\\\nthus $k = \\boxed{-16}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52602099.png"
  },
  {
    "solution": "\\boxed{3\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in $\\triangle ABC$, $AD$ is the altitude on side $BC$, and $BE$ is an angle bisector, intersecting at point P. Given that $\\angle BAC=75^\\circ$ and $BD=\\sqrt{3}$, find the length of $AC$.\n",
    "completion": "\\textbf{Solution:} From (1) we have $ \\angle BAD=30^\\circ$, hence $ \\angle DAC=45^\\circ$,\\\\\n$\\therefore$ $ \\angle DAC=\\angle C=45^\\circ$,\\\\\n$\\therefore$ $ AD=CD$,\\\\\nIn $\\triangle ABD$, $ \\angle BAD=30^\\circ$, $ BD=\\sqrt{3}$,\\\\\n$\\therefore$ $ AB=2BD=2\\sqrt{3}$,\\\\\nBy the Pythagorean theorem, we have: $ AD=\\sqrt{AB^{2}-BD^{2}}=3$,\\\\\n$ AC=\\sqrt{AD^{2}+CD^{2}}=3\\sqrt{2}$,\\\\\ni.e., $ AC=\\boxed{3\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52906079_139.png"
  },
  {
    "solution": "\\boxed{\\frac{5}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In quadrilateral $ABCD$, $AD \\parallel BC$, $AB= \\sqrt{5}$, $AD=2$, $DC= 2\\sqrt{5}$, $\\tan \\angle ABC=2$ (as shown in the figure). Point $E$ is on ray $AD$, and point $F$ is on side $BC$. Lines $BE$ and $EF$ are drawn such that $\\angle BEF=\\angle DCB$. When $FB=FE$, what is the length of segment $BF$?\n",
    "completion": "\\textbf{Solution:} Construct $EM\\perp BC$ at point M. It follows that $EM=2$, \\\\\nFrom (1), we have $\\tan\\angle C= \\frac{DG}{GC}=\\frac{1}{2}$, \\\\\nSince $FB=FE$, \\\\\nit follows that $\\angle FEB=\\angle FBE$, \\\\\nSince $\\angle FEB=\\angle C$, \\\\\nit follows that $\\angle FBE=\\angle C$, \\\\\nthus, $\\tan\\angle FBE= \\frac{1}{2}$, \\\\\nwhich leads to $\\frac{EM}{BM}=\\frac{1}{2}$, \\\\\ntherefore, $BM=4$, \\\\\nGiven that $FM^2+EM^2=FE^2$, \\\\\nit follows that $(4−FB)^2+2^2=FB^2$, \\\\\nthus, $BF=\\boxed{\\frac{5}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51379498_104.png"
  },
  {
    "solution": "\\boxed{3\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that quadrilateral $ABCD$ is inscribed in circle $\\odot O$, with the diameter $AC=12$. The diagonals $AC$ and $BD$ intersect at point $E$, and $AB=BD$, $EC=2$. What is the length of $AD$?\n",
    "completion": "\\textbf{Solution:} Connect BO to intersect AD at point F, and draw OD, \\\\\n$\\because$ BA = BD, OA = OD, \\\\\n$\\therefore$ BF is the perpendicular bisector of line segment AD, \\\\\n$\\therefore$ BF $\\perp$ AD, \\\\\n$\\because$ AC is the diameter of $\\odot$O, \\\\\n$\\therefore$ $\\angle$ADC = $90^\\circ$, i.e., AD $\\perp$ DC, \\\\\n$\\therefore$ BF $\\parallel$ CD, \\\\\n$\\therefore$ $\\triangle$BOE $\\sim$ $\\triangle$DCE, \\\\\n$\\therefore$ OB : CD = EO : EC, \\\\\n$\\because$ AC = 12, EC = 2, \\\\\n$\\therefore$ OB = OC = 6, \\\\\n$\\therefore$ OE = 4, \\\\\n$\\therefore$ 4CD = 12, \\\\\n$\\therefore$ CD = 3, \\\\\n$\\therefore$ In $\\triangle$ADC, with AD = $\\sqrt{12^{2}-3^{2}}$ = $\\boxed{3\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53059666.png"
  },
  {
    "solution": "\\boxed{-\\frac{5}{3}n+\\frac{17}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $AB$ intersects the coordinate axes at points $A(0, -3)$ and $B(-5, 0)$, respectively. Point $C$ is the midpoint of segment $AB$, and point $P$ is a point on the $y$-axis. Connecting $CP$, a perpendicular line to $CP$ is drawn from point $C$ intersecting segment $BO$ at point $Q$. Let $AP=m$ and $BQ=n$. Find the expression of $m$ as a function of $n$.\n",
    "completion": "\\textbf{Solution:} Let line $CE\\perp OA$ at point E, and $CF\\perp OB$ at point F,\\\\\nthen $E(0,-\\frac{3}{2})$, $F(-\\frac{5}{2},0)$, $\\angle CEP=\\angle CFQ=90^\\circ$,\\\\\nbecause $AP=m$, $BQ=n$,\\\\\nthus $OQ=5-n$, $OP=3-m$,\\\\\ntherefore $QF=OQ-OF=5-n-\\frac{5}{2}=\\frac{5}{2}-n$, $PE=OE-OP=\\frac{3}{2}-(3-m)=m-\\frac{3}{2}$,\\\\\nbecause $\\angle CEP+\\angle AOB=180^\\circ$,\\\\\ntherefore $CE\\parallel OB$,\\\\\ntherefore $\\angle ECF=\\angle CFQ=90^\\circ$,\\\\\ntherefore $\\angle PCE+\\angle PCF=90^\\circ$,\\\\\nbecause $\\angle QCF+\\angle PCF=90^\\circ$,\\\\\ntherefore $\\angle PCE=\\angle QCF$,\\\\\ntherefore $\\triangle PCE\\sim \\triangle QCF$,\\\\\ntherefore $\\frac{PE}{CE}=\\frac{QF}{CF}$, i.e., $\\frac{m-\\frac{3}{2}}{\\frac{5}{2}}=\\frac{\\frac{5}{2}-n}{\\frac{3}{2}}$,\\\\\ntherefore $m=-\\frac{5}{3}n+\\frac{17}{3}$, hence the function expression of $m$ in terms of $n$ is $m=\\boxed{-\\frac{5}{3}n+\\frac{17}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51419593_125.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y = -x + 2$ passes through point $A(2, 0)$ on the $x$-axis and intersects the parabola $y = ax^2$ at points $B$ and $C$, with the coordinates of point $B$ being $(1, 1)$. Connect $OC$. What is the area of $\\triangle AOC$?\n",
    "completion": "\\textbf{Solution:} Given the problem, we know the analytical expression for line AB is $y=-x+2$.\\\\\nBy solving the system of equations formed by the two functions, $\\left\\{\\begin{array}{l}y=x^{2}\\\\ y=-x+2\\end{array}\\right.$,\\\\\nwe get: $\\left\\{\\begin{array}{l}x=1\\\\ y=1\\end{array}\\right.$ or $\\left\\{\\begin{array}{l}x=-2\\\\ y=4\\end{array}\\right.$,\\\\\n$\\therefore$ the coordinates of point C are (-2, 4).\\\\\n$\\therefore$ the area of $\\triangle AOC$ is $\\frac{1}{2}\\times 2\\times 4=\\boxed{4}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52550623_139.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, $\\angle A=30^\\circ$, $CD\\perp AB$ at point D, $DE\\parallel BC$ intersects AC at point E. If $BD=2$, what is the length of $DE$?\n",
    "completion": "\\textbf{Solution:}\n\nGiven $\\because \\angle ACB=90^\\circ$ and $\\angle A=30^\\circ$,\n\n$\\therefore \\angle B=60^\\circ$ and $AB=2BC$,\n\nSince $CD\\perp AB$ at point $D$,\n\n$\\therefore \\angle CDB=\\angle CDA=90^\\circ$,\n\n$\\therefore \\angle BCD=30^\\circ$,\n\nThus, $BC=2BD$,\n\nGiven $BD=2$,\n\n$\\therefore BC=4$,\n\nTherefore, $AB=2BC=8$,\n\nThus, $AD=AB-BD=6$,\n\nSince $DE\\parallel BC$,\n\n$\\therefore \\angle DEA=\\angle ACB=90^\\circ$,\n\nGiven in $\\triangle ADE$, $\\angle A=30^\\circ$,\n\n$\\therefore DE=\\frac{1}{2}AD=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51378784.png"
  },
  {
    "solution": "\\boxed{46^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the quadrilateral $ABDC$, the diagonal $AD$ bisects $\\angle BAC$, $\\angle ACD=136^\\circ$, $\\angle BCD=44^\\circ$, what is the measure of $\\angle ADB$?\n",
    "completion": "\\textbf{Solution:} Draw line DE$\\perp$AB through point D intersecting the extension of AB at E, and draw DF$\\perp$AC through point D intersecting the extension of AC at F, and draw DG$\\perp$BC through point D at point G, as shown in the figure: \\\\\n$\\because$ $AD$ bisects $\\angle BAC$, $DE\\perp AB$, $DF\\perp AC$\\\\\n$\\therefore$ $\\angle BAD=\\frac{1}{2}\\angle BAC$, $DE=DF$\\\\\n$\\because$ $\\angle ACD=136^\\circ$\\\\\n$\\therefore$ $\\angle DCF=180^\\circ-\\angle ACD=44^\\circ$\\\\\n$\\because$ $\\angle BCD=44^\\circ$, $\\angle ACB=\\angle ACD-\\angle BCD=92^\\circ$\\\\\n$\\therefore$ $CD$ bisects $\\angle BCF$\\\\\n$\\because$ $DF\\perp AC$, $DG\\perp BC$\\\\\n$\\therefore$ $DF=DG$\\\\\n$\\therefore$ $DE=DG$\\\\\n$\\because$ $DE\\perp AB$, $DG\\perp BC$\\\\\n$\\therefore$ $BD$ bisects $\\angle CBE$\\\\\n$\\therefore$ $\\angle DBE=\\frac{1}{2}\\angle CBE$\\\\\n$\\therefore$ $\\angle ADB=\\angle DBE-\\angle BAD$\\\\\n$=\\frac{1}{2}\\angle CBE-\\frac{1}{2}\\angle BAC$\\\\\n$=\\frac{1}{2}\\left(\\angle CBE-\\angle BAC\\right)$\\\\\n$=\\frac{1}{2}\\angle BCA$\\\\\n$=\\boxed{46^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52699727.png"
  },
  {
    "solution": "\\boxed{x>3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given the line $y=kx+5$ intersects the $x$ axis at point $A$ and the $y$ axis at point $B$, and the coordinates of point $A$ are $(5,0)$, the line $y=2x-4$ intersects the $x$ axis at point $D$, and intersects the line $AB$ at point $C$. Based on the graph, write the solution set of the inequality $2x-4>kx+5$ with respect to $x$.\n",
    "completion": "Observe the graph of the function, we know that for $x>3$, the line $y=2x-4$ is above the line $y=kx+5$, \\\\\n$\\therefore$ the solution set of the inequality $2x-4>kx+5$ is $\\boxed{x>3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52416375_90.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, there is a square $ABCD$ with a side length of $10$. $E$ is a moving point on side $BC$ (not coinciding with $B$ or $C$), and $AE$ is connected. $G$ is a point on the extension of $BC$. A perpendicular line through point $E$ to $AE$ intersects the angle bisector of $\\angle DCG$ at point $F$. If $FG \\perp BG$, then $\\triangle ABE \\sim \\triangle EGF$; if $EC = 2$, find the area of $\\triangle CEF$.\n",
    "completion": "\\textbf{Solution:} Since quadrilateral $ABCD$ is a square,\\\\\n$ \\therefore \\angle DCG=90^\\circ$,\\\\\nSince $CF$ bisects $\\angle DCG$,\\\\\n$ \\therefore \\angle FCG=\\frac{1}{2}\\angle DCG=45^\\circ$,\\\\\nSince $\\angle CGF=90^\\circ$,\\\\\n$ \\therefore \\angle GCF=\\angle CFG=45^\\circ$,\\\\\n$ \\therefore FG=CG$,\\\\\nSince $AB=BC=10$, $CE=2$,\\\\\n$ \\therefore BE=8$,\\\\\n$ \\therefore EG=CE+CG=2+FG$,\\\\\nFrom $(1)$, we know that $\\triangle BAE \\sim \\triangle GEF$,\\\\\n$ \\therefore \\frac{AB}{EG}=\\frac{BE}{FG}$,\\\\\n$ \\therefore \\frac{10}{2+FG}=\\frac{8}{FG}$,\\\\\n$ \\therefore FG=8$,\\\\\n$ \\therefore {S}_{\\triangle ECF}=\\frac{1}{2}CE\\cdot FG=\\frac{1}{2}\\times 2\\times 8=\\boxed{8}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52550935_92.png"
  },
  {
    "solution": "\\boxed{\\frac{4}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, $AC=4$, $BC=3$, then what is the value of $\\sin B$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle C=90^\\circ$, AC=4, BC=3, \\\\\ntherefore AB=5, \\\\\ntherefore $\\sin B= \\frac{AC}{AB}=\\boxed{\\frac{4}{5}}$",
    "image_path": "dataset/mm_math/images/MM_Math/52602893.png"
  },
  {
    "solution": "\\boxed{260^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle A=80^\\circ$. After cutting off $\\angle A$, we obtain the quadrilateral $BCDE$. Then, what is the sum of $\\angle 1 + \\angle 2$?\n",
    "completion": "\\textbf{Solution:} Given: $\\because$ $\\angle 1$ and $\\angle 2$ are exterior angles of $\\triangle ADE$,\\\\\n$\\therefore$ $\\angle 1 = \\angle ADE + \\angle A$, $\\angle 2 = \\angle AED + \\angle A$,\\\\\n$\\therefore$ $\\angle 1 + \\angle 2 = \\angle ADE + \\angle A + \\angle AED + \\angle A$,\\\\\nMoreover, $\\because$ $\\angle ADE + \\angle A + \\angle AED = 180^\\circ$,\\\\\n$\\therefore$ $\\angle 1 + \\angle 2 = 180^\\circ + \\angle A = \\boxed{260^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51177766.png"
  },
  {
    "solution": "\\boxed{9\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: A triangular piece of material is shown in the figure, where $ \\angle A=30^\\circ$, $ \\angle C=90^\\circ$, and $ AB=12$. A rectangle CDEF is cut from this material, with points D, E, and F located on AC, AB, and BC, respectively. Let the length of AE be $x$, and the area of rectangle CDEF be $S$. What is the length of AE when the area of rectangle CDEF is maximized? What is the maximum area?\n",
    "completion": "\\textbf{Solution:} Given: $S_{\\text{rectangle } CDEF} = -\\frac{\\sqrt{3}}{4}x^2 + 3\\sqrt{3}x\\ (0 < x < 12)$\\\\\n$= -\\frac{\\sqrt{3}}{4}(x^2 - 12x)$\\\\\n$= -\\frac{\\sqrt{3}}{4}(x^2 - 12x + 6^2 - 6^2)$\\\\\n$= -\\frac{\\sqrt{3}}{4}[(x-6)^2 - 36]$\\\\\n$= -\\frac{\\sqrt{3}}{4}(x-6)^2 + 9\\sqrt{3}$,\\\\\nTherefore, when $x=\\boxed{6}$, the area of rectangle CDEF is maximized,\\\\\nmeaning when point E is the midpoint of AB, the area of rectangle CDEF is maximized, with a maximum area of $\\boxed{9\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52602949_93.png"
  },
  {
    "solution": "\\boxed{116^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, it is known that $\\angle COD$ is inside $\\angle AOB$, and $\\angle AOB=160^\\circ$, $\\angle COD=\\frac{1}{4}\\angle AOB$, ray $OE$ bisects $\\angle AOD$, $\\angle COE=18^\\circ$. Find the measure of $\\angle AOD$ in degrees.\n",
    "completion": "\\textbf{Solution:} Given: $\\because$ $\\angle AOB=160^\\circ$ and $\\angle COE=18^\\circ$,\\\\\n$\\therefore$ $\\angle COD=\\frac{1}{4}\\angle AOB=\\frac{1}{4}\\times 160^\\circ=40^\\circ$,\\\\\n$\\therefore$ $\\angle EOD=\\angle EOC+\\angle COD=18^\\circ+40^\\circ=58^\\circ$,\\\\\n$\\because$ ray $OE$ bisects $\\angle AOD$,\\\\\n$\\therefore$ $\\angle AOE=\\angle EOD=\\frac{1}{2}\\angle AOD$,\\\\\n$\\therefore$ $\\angle AOD=2\\angle EOD=2\\times 58^\\circ=\\boxed{116^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53211137_114.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ is an equilateral triangle with side length 2. Point $D$ is on the extension of side $CA$. An equilateral triangle $BDE$ is constructed with $BD$ as a side. Connect $AE$. Find the value of $AE-AD$.\n",
    "completion": "\\textbf{Solution:} Since $\\triangle CBD \\cong \\triangle ABE$,\\\\\nit follows that $CD=AE$,\\\\\nthus $AE-AD=CD-AD=AC=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52653743_95.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, D and E are the midpoints of AB and AC, respectively. If $S_{\\triangle ADE}=2$, what is the area of quadrilateral DBCE?\n",
    "completion": "\\textbf{Solution:}\\\\\nSince points D and E are the midpoints of AB and AC, respectively,\\\\\ntherefore, DE is the midline of $\\triangle ABC$,\\\\\ntherefore, DE$\\parallel$BC, and DE=$\\frac{1}{2}$BC,\\\\\ntherefore, $\\triangle ADE\\sim \\triangle ABC$,\\\\\ntherefore, $\\frac{S_{\\triangle ADE}}{S_{\\triangle ABC}}=\\left(\\frac{DE}{BC}\\right)^2=\\frac{1}{4}$\\\\\ntherefore, $S_{\\triangle ABC}=4S_{\\triangle ADE}=8$\\\\\ntherefore, $S_{\\text{quadrilateral DBCE}}=S_{\\triangle ABC}-S_{\\triangle ADE}=8-2=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52991506.png"
  },
  {
    "solution": "\\boxed{y = -\\frac{15}{8}x + \\frac{15}{8}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: The graph of the quadratic function $y=ax^2+bx+4$ ($a\\neq 0$) passes through points $A(-4,0)$ and $B(1,0)$, and intersects the $y$-axis at point $C$. Point $P$ is a point on the parabola in the second quadrant. Lines $BP$ and $AC$ are drawn and intersect at point $Q$. A perpendicular line $PD$ is drawn from point $P$ to the $x$-axis at point $D$. Line $BC$ is drawn. When $\\angle DPB=2\\angle BCO$, find the equation of line $BP$.\n",
    "completion": "\\textbf{Solution}: Let $BP$ intersect the $y$-axis at point $E$, \\\\\nsince $PD \\parallel y$-axis, \\\\\n$\\therefore \\angle DPB = \\angle OEB$, \\\\\nsince $\\angle DPB = 2\\angle BCO$, \\\\\n$\\therefore \\angle OEB = 2\\angle BCO$, \\\\\n$\\therefore \\angle ECB = \\angle EBC$, \\\\\n$\\therefore BE = CE$, let $OE = a$, then $CE = 4-a$, \\\\\n$\\therefore BE = 4-a$, \\\\\nin $\\triangle BOE$, by Pythagoras theorem, we have $BE^2 = OE^2 + OB^2$, \\\\\n$\\therefore (4-a)^2 = a^2 + 1^2$, solving this gives $a = \\frac{15}{8}$, \\\\\n$\\therefore E\\left(0, \\frac{15}{8}\\right)$, \\\\\nlet the equation of the line containing $BE$ be $y = kx + e (k \\neq 0)$, \\\\\n$\\therefore \\left\\{\\begin{array}{l}k \\cdot 0 + e = \\frac{15}{8}, \\\\ k \\cdot 1 + e = 0.\\end{array}\\right.$\\\\\nSolving this gives $\\left\\{\\begin{array}{l}k = -\\frac{15}{8}, \\\\ e = \\frac{15}{8}.\\end{array}\\right.$\\\\\n$\\therefore$ The equation of line $BP$ is $\\boxed{y = -\\frac{15}{8}x + \\frac{15}{8}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53059889_152.png"
  },
  {
    "solution": "\\boxed{(-\\sqrt{3}, 1)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the Cartesian coordinate system, \\( \\triangle AOB\\) is an equilateral triangle, and the coordinates of point \\( B\\) are \\((2,0)\\). If \\( \\triangle AOB\\) is rotated \\(90^\\circ\\) counterclockwise around the origin, what are the coordinates of point \\( A'\\)?\n",
    "completion": "\\textbf{Solution:}\n\nGiven by rotation, $\\triangle A'B'O \\cong \\triangle ABO$,\\\\\nThrough point $A'$ draw $A'D \\perp$ y-axis at $D$,\\\\\nSince $\\triangle ABC$ is an equilateral triangle,\\\\\nTherefore, $OD=DB'= \\frac{1}{2}OB' =1$,\\\\\nHence, $A'{D}=\\sqrt{A'{O}^{2}-OD^{2}}=\\sqrt{2^{2}-1^{2}}=\\sqrt{3}$,\\\\\nTherefore, the coordinates of point $A'$ are $(-\\sqrt{3}, 1)$,\\\\\nSo, the answer is $\\boxed{(-\\sqrt{3}, 1)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51368830.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $ABC$ is an acute-angled triangle with $AB=4$, $\\angle BAC=60^\\circ$. The bisector of $\\angle BAC$ intersects $BC$ at point $D$. $M$ and $N$ are points on $AD$ and $AB$, respectively. When $BM+MN$ reaches its minimum value, what is the length of $AN$?\n",
    "completion": "\\textbf{Solution:} Let $E$ be the point symmetrical to $B$ with respect to $AD$. Draw $EN \\perp AB$ intersecting $AB$ at point $N$ and $AD$ at the extension of $CM$ at point $M$. Connect $BM$.\n\nSince $\\angle BAC=60^\\circ$ and $AD$ bisects $\\angle BAC$,\n\nit follows that point $E$ lies on $AC$,\n\nSince $BM + MN = EM + MN = EN$, the value of $BM + MN$ is minimized at this moment,\n\nBy symmetry, $AE = AB$,\n\nSince $AB=4$,\n\nTherefore, $AE=4$,\n\nIn $\\triangle ABE$, $\\angle EAN=60^\\circ$,\n\nTherefore, $\\angle AEN=30^\\circ$,\n\nTherefore, $AN=\\frac{1}{2}AE=\\boxed{2}$",
    "image_path": "dataset/mm_math/images/MM_Math/52856093.png"
  },
  {
    "solution": "\\boxed{y=\\frac{5}{3}x-\\frac{25}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, point $A$ has coordinates $\\left(5, 0\\right)$, and point $B$ lies on the positive half of the $y$-axis (with $OB<OA$). The line segment $AB$ is rotated $90^\\circ$ clockwise about point $A$ to obtain the line segment $AC$. Perpendicular lines are drawn from point $C$ to the $x$-axis and the $y$-axis, with the feet of the perpendiculars being $D$ and $E$, respectively. If $CE=4BE$, what is the equation of line $AC$?\n",
    "completion": "\\textbf{Solution:} Let $OB=a$. According to equation (1), we know that $OA=OE$. Since $OE=CD$, it follows that $OA=CD$. Since $\\triangle AOB$ and $\\triangle CDA$ are both right-angled triangles, and $AB=AC$, it therefore follows that $\\triangle AOB \\cong \\triangle CDA$. Thus, $AD=OB=a$, and hence $CE=OD=a+5$, $BE=5-a$. Since $CE=4BE$, we have $a+5=4(5-a)$. Solving this, we find $a=3$. Therefore, $C(8, 5)$. Also, since $A(5, 0)$, let the equation of line AC be $y=kx+b$. Then $\\begin{cases}5k+b=0\\\\ 8k+b=5\\end{cases}$, from which we find $\\begin{cases}k=\\frac{5}{3}\\\\ b=-\\frac{25}{3}\\end{cases}$. Therefore, the equation of line AC is $y=\\frac{5}{3}x-\\frac{25}{3}$. Thus, the expression for line AC is $\\boxed{y=\\frac{5}{3}x-\\frac{25}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50545780_115.png"
  },
  {
    "solution": "\\boxed{x<-2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: The graph of the linear function $y=kx+b$ (where $k$ and $b$ are constants and $k\\neq 0$) is shown in the figure, and it passes through the point $(-2, 0)$. What is the solution set of the inequality $kx+b>0$ with respect to $x$?\n",
    "completion": "\\textbf{Solution}: Since the linear function $y=kx+b$ passes through the point $(-2, 0)$, \\\\\nit follows that the solution set for $kx+b>0$ is $\\boxed{x<-2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52189048.png"
  },
  {
    "solution": "\\boxed{125^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle A=70^\\circ$, and $\\odot O$ is the inscribed circle of $\\triangle ABC$. What is the value of $\\angle BOC$?\n",
    "completion": "\\textbf{Solution:} Since $\\odot O$ is the inscribed circle of $\\triangle ABC$,\n\nit follows that OB bisects $\\angle ABC$, and OC bisects $\\angle ACB$,\n\ntherefore, $\\angle OBC = \\frac{1}{2}\\angle ABC$, and $\\angle OCB = \\frac{1}{2}\\angle ACB$,\n\nhence, $\\angle OBC + \\angle OCB = \\frac{1}{2}(\\angle ABC + \\angle ACB) = \\frac{1}{2}(180^\\circ - \\angle A)$,\n\nthus, $\\angle BOC = 180^\\circ - (\\angle OBC + \\angle OCB) = 180^\\circ - \\frac{1}{2}(180^\\circ - \\angle A) = 90^\\circ + \\frac{1}{2}\\angle A = 90^\\circ + \\frac{1}{2} \\times 70^\\circ = \\boxed{125^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/52765344.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{3}a}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, by connecting two non-adjacent vertices within a regular hexagon with side length $a$, a shape with a hexagonal interior is formed, as illustrated. What is the perimeter of the shaded area?\n",
    "completion": "\\textbf{Solution:} Let $PH \\perp AB$ at H,\\\\\nsince the outer hexagon is a regular hexagon with side length $a$,\\\\\ntherefore, the perimeter of the outer hexagon is $6a$,\\\\\nsince the hexagon is regular,\\\\\ntherefore $\\angle CAB=\\frac{(6-2)\\times 180^\\circ}{6}=120^\\circ$, CA=BA,\\\\\ntherefore $\\angle ABP=30^\\circ$,\\\\\nalso, since PA=PB,\\\\\ntherefore $\\angle PAB=30^\\circ$,\\\\\ntherefore $PA=\\frac{AH}{\\cos\\angle PAH}=\\frac{\\sqrt{3}}{3}a$,\\\\\nas the problem states, the shaded area is a regular hexagon,\\\\\ntherefore, the perimeter of the shaded area is $\\frac{\\sqrt{3}}{3}a\\times 6=2\\sqrt{3}a$.\\\\\nHence, the answer is: $2\\sqrt{3}a=\\boxed{2\\sqrt{3}a}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51299005.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the line segment $AB=12\\, \\text{cm}$, point $C$ is on the line segment $AB$, $AC=2BC$, then how many centimeters is $BC$?\n",
    "completion": "\\textbf{Solution:} Let $BC=x\\, \\text{cm}$, then $AC=2x\\, \\text{cm}$,\\\\\n$x+2x=12$,\\\\\nsolving for $x$, we get: $x=4$,\\\\\n$\\therefore BC=\\boxed{4}\\, \\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53078560.png"
  },
  {
    "solution": "\\boxed{72^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within $\\triangle ABC$, we have $AB=AC$ and $\\angle A=36^\\circ$. Point $D$ is located on $AC$, and $BD=BC$. What is the measure of $\\angle BDC$?\n",
    "completion": "\\textbf{Solution:} Since $AB=AC$ and $\\angle A=36^\\circ$,\n\nit follows that $ \\angle ABC=\\angle ACB=\\frac{1}{2}(180^\\circ−\\angle A)=72^\\circ$.\n\nSince $BD=BC$,\n\nit follows that $\\angle BDC=\\angle ACB=\\boxed{72^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51368829.png"
  },
  {
    "solution": "\\boxed{75^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, lines MD and CN intersect at point O. OA is a ray inside $\\angle MOC$, and OB is a ray inside $\\angle NOD$. $\\angle MON=70^\\circ$. If $\\angle BOD=\\frac{1}{2}\\angle COD$, find the degree measure of $\\angle BON$.\n",
    "completion": "\\textbf{Solution:}\n\nSince $\\angle MON=70^\\circ$,\\\\\nit follows that $\\angle COD= \\angle MON=70^\\circ$,\\\\\nthus $\\angle BOD= \\frac{1}{2}\\angle COD= \\frac{1}{2}\\times70^\\circ=35^\\circ$,\\\\\ntherefore $\\angle BON= 180^\\circ-\\angle MON-\\angle BOD= 180^\\circ-70^\\circ- 35^\\circ=\\boxed{75^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/54927588_139.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $BC=20$, $D$ and $E$ are the midpoints of $AB$ and $AC$ respectively, $F$ is a point on $DE$, $DF=4$, connecting $AF$ and $CF$. If $\\angle AFC=90^\\circ$, then what is the length of $AC$?\n",
    "completion": "\\textbf{Solution:} Since $D$ and $E$ are the midpoints of $AB$ and $AC$ respectively,\n\nit follows that $DE$ is the midline of $\\triangle ABC$,\n\ntherefore $DE= \\frac{1}{2}BC=10$,\n\nthus $EF=DE−DF=10−4=6$,\n\nin the right triangle $\\triangle AFC$, $AE=EC$,\n\ntherefore $AC=2EF=\\boxed{12}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51635741.png"
  },
  {
    "solution": "\\boxed{\\sqrt{6}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, $CD$ is the altitude on the hypotenuse $AB$. If $AD=3$ and $BD=2$, what is the length of $CD$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle ACB=90^\\circ$,\\\\\nit follows that $\\angle ACD+\\angle BCD=90^\\circ$.\\\\\nSince $CD$ is the altitude from the hypotenuse $AB$,\\\\\nit follows that $\\angle ADC=\\angle CDB=90^\\circ$, $\\angle ACD+\\angle CAD=90^\\circ$,\\\\\ntherefore $\\angle CAD=\\angle BCD$,\\\\\ntherefore $\\triangle CAD\\sim \\triangle BCD$,\\\\\ntherefore $\\frac{AD}{CD}=\\frac{CD}{BD}$,\\\\\ntherefore $CD^2=3\\times 2=6$,\\\\\ntherefore $CD=\\sqrt{6}$ (negative values discarded)\\\\\nThus, the answer is: $CD=\\boxed{\\sqrt{6}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52916837.png"
  },
  {
    "solution": "\\boxed{40^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in circle $O$, $\\overset{\\frown}{AB} = \\overset{\\frown}{AC}$. If $\\angle B = 70^\\circ$, then what is the measure of $\\angle A$?\n",
    "completion": "\\textbf{Solution:} Given that in circle $\\odot O$, $\\widehat{AB}=\\widehat{AC}$ and $\\angle B=70^\\circ$,\\\\\nit follows that $\\angle C=\\angle B=70^\\circ$,\\\\\nthus $\\angle A=180^\\circ-\\angle B-\\angle C=180^\\circ-70^\\circ-70^\\circ=\\boxed{40^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51367508.png"
  },
  {
    "solution": "\\boxed{x>1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=x^2+bx+c$ passes through point $A(-4,0)$ and point $C(0,-4)$, with another intersection point with the x-axis being $B$. When $y>0$, find the range of values for $x$.\n",
    "completion": "\\textbf{Solution:} Let $y=0$, then $x^{2}+3x-4=0$, \\\\\nsolving this yields $x_{1}=-4$ and $x_{2}=1$, \\\\\n$\\therefore$ the coordinates of point B are $\\left(1,0\\right)$, \\\\\n$\\therefore$ from the graph of the function, it is known that when $y>0$, \\boxed{x<-4} or \\boxed{x>1}.",
    "image_path": "dataset/mm_math/images/MM_Math/52871856_291.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $AB$ is the diameter of $\\odot O$, with $AB=6$, and $\\sin C= \\frac{1}{3}$. Find the length of the chord $AD$.\n",
    "completion": "\\textbf{Solution:} Draw line BD, \\\\\n$\\therefore \\angle C=\\angle B$, \\\\\n$\\therefore \\sin B= \\frac{1}{3}$, \\\\\n$\\because$ AB is the diameter, \\\\\n$\\therefore \\angle ADB=90^\\circ$, \\\\\n$\\therefore \\sin B =\\frac{AD}{AB}=\\frac{1}{3}$, \\\\\n$\\therefore AD=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51299813_123.png"
  },
  {
    "solution": "\\boxed{x>1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, the graphs of the functions $y=3x$ and $y=kx+4$ intersect at the point $P(m, 3)$. What is the solution set of the inequality $3x>kx+4$ with respect to $x$?\n",
    "completion": "\\textbf{Solution:} Substituting $P(m, 3)$ into $y=3x$, we have $3m=3$, from which we solve $m=1$. Thus, the coordinates of point P are $(1, 3)$, \\\\ \nhence for $x>1$, we have $3x>kx+4$, \\\\ \nmeaning the solution set for the inequality $3x>kx+4$ is $\\boxed{x>1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50519342.png"
  },
  {
    "solution": "\\boxed{92^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle PAB$, $PA=PB$, $\\angle A=\\angle B$, points $M$, $N$, and $K$ are located on $PA$, $PB$, and $AB$, respectively, and $AM=BK$, $BN=AK$. If $\\angle MKN=44^\\circ$, what is the degree measure of $\\angle P$?\n",
    "completion": "\\textbf{Solution:} Given $\\because$PA=PB,\\\\\n$\\therefore \\angle A=\\angle B$;\\\\\nIn $\\triangle AMK$ and $\\triangle BKN$\n\\[\n\\left\\{\n\\begin{array}{l}\nAM=BK\\\\\n\\angle A=\\angle B\\\\\nBN=AK\n\\end{array}\n\\right.\n\\]\n$\\therefore \\triangle AMK \\cong \\triangle BKN$ (SAS),\\\\\n$\\therefore \\angle AKM=\\angle BNK$;\\\\\n$\\because \\angle AKN=\\angle B+\\angle BNK=\\angle AKM+\\angle MKN$,\\\\\n$\\therefore \\angle MKN=\\angle B=\\angle A=44^\\circ$,\\\\\n$\\therefore \\angle P=180^\\circ-\\angle A-\\angle B=180^\\circ-2\\times 44^\\circ=\\boxed{92^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52661191.png"
  },
  {
    "solution": "\\boxed{\\frac{21\\sqrt{2}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, AB is a diameter of circle $O$, point P is on circle $O$, and $PA=PB$. Point M is outside circle $O$, and $MB$ is tangent to circle $O$ at point B. Connect $OM$, and draw $AC\\parallel OM$ through point A intersecting circle $O$ at point C, then connect $BC$ and intersect $OM$ at point D. $MC$ is a tangent to circle $O$; if $OB=\\frac{15}{2}$ and $BC=12$, connect $PC$. Find the length of $PC$.\n",
    "completion": "\\textbf{Solution:} Given that AB is the diameter of circle O,\\\\\nit follows that $\\angle ACB=\\angle APB=90^\\circ$,\\\\\ngiven that $OB=\\frac{15}{2}$,\\\\\nit follows that $AB=15$,\\\\\nthus $PA=PB=\\frac{15\\sqrt{2}}{2}$,\\\\\ngiven that $BC=12$,\\\\\nit follows that $AC=\\sqrt{AB^{2}-BC^{2}}=\\sqrt{15^{2}-12^{2}}=9$,\\\\\ndraw line AH perpendicular to PC at point H,\\\\\nthus $\\angle AHC=\\angle AHP=90^\\circ$,\\\\\ngiven that $\\angle ACH=\\angle ABP=45^\\circ$,\\\\\nit follows that $AH=CH=AC\\times \\sin 45^\\circ=9\\times \\frac{\\sqrt{2}}{2}=\\frac{9\\sqrt{2}}{2}$,\\\\\n$PH=\\sqrt{PA^{2}-AH^{2}}=\\sqrt{(\\frac{15\\sqrt{2}}{2})^{2}-(\\frac{9\\sqrt{2}}{2})^{2}}=6\\sqrt{2}$,\\\\\nthus $PC=PH+CH=\\boxed{\\frac{21\\sqrt{2}}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52746246_97.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $AB = 5$, $BC = 6$, point $E$ is on side $BC$, and $BE = 2$. $F$ is a moving point on side $AB$. Connect $EF$, and construct an equilateral $\\triangle EFG$ with $EF$ as a side, and point $G$ is inside rectangle $ABCD$. Connect $CG$. What is the minimum value of $CG$?\n",
    "completion": "\\textbf{Solution:} Given the problem, we know that point F is the driving point, and point G is the driven point. Point F moves on a line segment, and point G also definitely moves on a linear trajectory.\\\\\nRotate $\\triangle EFB$ about point E by $60^\\circ$, making EF coincide with EG, and thus obtaining $\\triangle EFB \\cong \\triangle EGH$,\\\\\ntherefore, it can be inferred that $\\triangle EBH$ is an equilateral triangle, and point G is on the line HN, which is perpendicular to HE,\\\\\nDraw CM$\\perp$HN, then CM is the minimum value of CG,\\\\\nDraw EP$\\perp$CM, it can be known that quadrilateral HEPM is a rectangle,\\\\\nthen $CM=MP+CP=HE+ \\frac{1}{2}EC=2+2=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53151323.png"
  },
  {
    "solution": "\\boxed{D_4(3+3\\sqrt{3}, 2-2\\sqrt{3})}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system, the graph of the linear function $y_{1}=kx+b$ $(k\\ne 0)$ intersects with the graph of the inverse proportion function $y_{2}=\\frac{m}{x}$ $(m\\ne 0)$ at points A and B within the second and fourth quadrants, respectively, and intersects with the x-axis at point C. Point B is located at $(n, -2)$. The segment OA has a length of 5, and point E is located on the negative half-axis of the x-axis, with $\\sin\\angle AOE=\\frac{4}{5}$. On the graph of the inverse proportion function, find a point D such that the area of $\\triangle ABD$ is equal to the area of $\\triangle AOB$. Please write down the coordinates of point D directly.\n",
    "completion": "\\textbf{Solution}: The coordinates of point D are $\\boxed{D_1(-3\\sqrt{2}, 2\\sqrt{2})}$, $\\boxed{D_2(3\\sqrt{2}, 2\\sqrt{2})}$, $\\boxed{D_3(3-3\\sqrt{3}, 2+2\\sqrt{3})}$, and $\\boxed{D_4(3+3\\sqrt{3}, 2-2\\sqrt{3})}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52765948_101.png"
  },
  {
    "solution": "\\boxed{53^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that lines $AB$ and $CD$ intersect at point $O$, and $\\angle COE = 90^\\circ$. If $\\angle AOC = 37^\\circ$, what is the measure of $\\angle BOE$?\n",
    "completion": "\\textbf{Solution:} Given that $\\because \\angle COE=90^\\circ$, $\\angle AOC=37^\\circ$, \\\\\n$\\therefore \\angle BOE=180^\\circ-\\angle AOC-\\angle COE$ \\\\\n$=180^\\circ-37^\\circ-90^\\circ$ \\\\\n$=\\boxed{53^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51420172_71.png"
  },
  {
    "solution": "\\boxed{\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the isosceles $\\triangle ABC$ is inscribed in $\\odot O$, with $AB=AC$, and $\\angle BAC=120^\\circ$. Point $D$ is on the arc $AC$, and line $BD$ is drawn. Point $E$ is on line $BD$, satisfying $\\angle ABE=\\angle ECB$. If $CD=2$, what is the area of $\\triangle AEC$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect $AD$,\\\\\nsince $AB=AC$ and $\\angle BAC=120^\\circ$,\\\\\nthus $\\angle ABC=\\angle ACB=\\frac{1}{2}\\left(180^\\circ-\\angle BAC\\right)=30^\\circ$,\\\\\ntherefore $\\angle ADB=\\angle ACB=30^\\circ$,\\\\\nsince $\\angle ABE=\\angle ECB$,\\\\\nthus $\\angle DEC=\\angle DBC+\\angle ECB=\\angle DBC+\\angle ABE=\\angle ABC=30^\\circ$,\\\\\ntherefore $\\angle ADB=\\angle DEC=30^\\circ$,\\\\\nhence $AD\\parallel EC$,\\\\\nthus the distance between $AD$ and $EC$ is equal everywhere,\\\\\ntherefore $\\text{Area of }\\triangle AEC=\\text{Area of }\\triangle DEC$,\\\\\nsince point D is a point on the arc $\\overset{\\frown}{AC}$,\\\\\nthus $\\angle EDC=\\angle BAC=120^\\circ$,\\\\\nconstruct $DH\\perp EC$ at point H,\\\\\nsince $\\angle ABD=\\angle ECB$ and $\\angle ACD=\\angle ABD$,\\\\\nthus $\\angle ACD=\\angle ECB$,\\\\\nsince $\\angle ACB=30^\\circ$,\\\\\nthus $\\angle ACE+\\angle ECB=30^\\circ$,\\\\\nthus $\\angle ACE+\\angle ACD=30^\\circ$,\\\\\nthat is, $\\angle DCE=30^\\circ=\\angle DEC$,\\\\\nthus $\\triangle DEC$ is an isosceles triangle,\\\\\nsince $CD=2$,\\\\\nthus $DH=\\frac{1}{2}CD=1$,\\\\\nthus $CH=\\sqrt{CD^{2}-DH^{2}}=\\sqrt{3}$,\\\\\nthus $EC=2\\sqrt{3}$,\\\\\nthus $\\text{Area of }\\triangle DEC=\\frac{1}{2}EC\\cdot DH=\\frac{1}{2}\\times 2\\sqrt{3}\\times 1=\\boxed{\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52991516.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\triangle ABC$, $AD$ is the altitude from $A$ to side $BC$, and $BE$ is the altitude from $B$ to side $AC$, with $AD$ and $BE$ intersecting at point $F$. If $BF=AC$, $CD=3$, and $BD=8$, what is the length of line segment $AF$?\n",
    "completion": "\\textbf{Solution:} Given that $AD$ is the altitude from $A$ on side $BC$ and $BE$ is the altitude from $B$ on side $AC$,\\\\\ntherefore $\\angle ADC=\\angle BDF=\\angle AEB=90^\\circ$,\\\\\nthus, $\\angle DAC+\\angle C=90^\\circ$, $\\angle C+\\angle DBF=90^\\circ$,\\\\\nwhich means $\\angle DAC=\\angle DBF$,\\\\\nin $\\triangle ADC$ and $\\triangle BDF$,\\\\\ngiven that $\\left\\{\\begin{array}{l}\\angle ADC=\\angle BDF \\\\ \\angle DAC=\\angle DBF \\\\ AC=BF \\end{array}\\right.$,\\\\\nwe can conclude that $\\triangle ADC\\cong \\triangle BDF$ (AAS),\\\\\nhence, $CD=FD=3$, $AD=BD=8$,\\\\\nthus, $AF=AD-FD=8-3=\\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53147702.png"
  },
  {
    "solution": "\\boxed{37^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle A=90^\\circ$, $AB=3$, $AC=4$. $D$, $E$, and $F$ are the midpoints of sides $AB$, $BC$, and $AC$ respectively. Connect $DE$, $DF$, and $EF$, with the measure of $\\angle ADF$ being $53^\\circ$. What is the measure of $\\angle C$?\n",
    "completion": "\\textbf{Solution:} Since $D$ and $F$ are the midpoints of sides $AB$ and $AC$ respectively, \\\\\n$\\therefore DF \\parallel BC$, \\\\\nSince $\\angle ADF = 53^\\circ$, \\\\\n$\\therefore \\angle B = \\angle ADF = 53^\\circ$, \\\\\nSince $\\angle A = 90^\\circ$, \\\\\n$\\therefore \\angle C = 180^\\circ - 90^\\circ - 53^\\circ = \\boxed{37^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53359538_147.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, rotate the rectangle $ABCD$ counterclockwise around point $A$ by an angle of $\\alpha$ $(0^\\circ<\\alpha<90^\\circ)$, and connect $EC$ and $ED$. At what value of $\\alpha$ will $EC=ED$?\n",
    "completion": "\\textbf{Solution:} Draw $BE$, and through point $E$, draw $EH \\perp CD$ at point $H$, intersecting $AB$ at point $M$, as shown in the diagram below,\\\\\nTo ensure $EC=ED$, then $EH\\perp CD$, and $CH=CD$,\\\\\n$\\because AB\\parallel CD$,\\\\\n$\\therefore EF\\perp AB$,\\\\\n$\\therefore \\angle BMH=\\angle AMH=\\angle MBC=\\angle BCH=\\angle ADH=\\angle MAD=90^\\circ$,\\\\\n$\\therefore$ Quadrilateral $ADHM$ and quadrilateral $BCHM$ are both rectangles,\\\\\n$\\therefore BM=CH=DH=AM$,\\\\\n$\\therefore EM$ bisects $AB$ perpendicularly,\\\\\n$\\therefore AE=BE$,\\\\\nBy the property of rotation, $AB=AE$,\\\\\n$\\therefore AB=AE=BE$,\\\\\n$\\therefore \\triangle ABE$ is an equilateral triangle,\\\\\n$\\therefore \\angle BAE=60^\\circ$,\\\\\nTherefore, when $\\alpha$ is $\\boxed{60^\\circ}$, then $EC=ED$.",
    "image_path": "dataset/mm_math/images/MM_Math/52091239.png"
  },
  {
    "solution": "\\boxed{\\frac{5}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points $D$ and $E$ are located on sides $AB$ and $AC$ of $\\triangle ABC$, respectively, with $AD=2$, $AC=6$, and $\\frac{AE}{AB}=\\frac{1}{3}$, $BC=\\frac{15}{4}$. Given that $\\triangle ADE\\sim \\triangle ACB$; what is the length of $DE$?\n",
    "completion": "\\textbf{Solution:} Given that $\\because \\triangle ADE \\sim \\triangle ACB$,\\\\\n$\\therefore \\frac{DE}{BC}=\\frac{AD}{AC}=\\frac{1}{3}$,\\\\\n$\\because BC = \\frac{15}{4}$,\\\\\n$\\therefore DE = \\frac{1}{3} \\times \\frac{15}{4} = \\boxed{\\frac{5}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52977407_126.png"
  },
  {
    "solution": "\\boxed{\\frac{12}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle BAC=90^\\circ$, $AB=3$, and $AC=4$. Let point $P$ be any point on $BC$. Connect $PA$, and construct parallelogram $PAQC$ with $PA$ and $PC$ as adjacent sides. Connect $PQ$. What is the minimum value of $PQ$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle BAC=90^\\circ$ and $AB=3$, $AC=4$,\\\\\n$\\therefore$ $BC=\\sqrt{AC^2+AB^2}=5$,\\\\\nSince quadrilateral $APCQ$ is a parallelogram,\\\\\n$\\therefore$ $PO=QO$, $CO=AO$,\\\\\nSince the shortest distance is $PQ$ which equates to the shortest distance $PO$,\\\\\n$\\therefore$ draw the perpendicular line from $O$ to $BC$, labelled as $OP^{\\prime }$,\\\\\nSince $\\angle ACB=\\angle P^{\\prime }CO$ and $\\angle CP^{\\prime }O=\\angle CAB=90^\\circ$,\\\\\n$\\therefore$ $\\triangle CAB\\sim \\triangle CP^{\\prime }O$,\\\\\n$\\therefore$ $\\frac{CO}{BC}=\\frac{OP^{\\prime }}{AB}$,\\\\\n$\\therefore$ $\\frac{2}{5}=\\frac{OP^{\\prime }}{3}$,\\\\\n$\\therefore$ $OP^{\\prime }=\\frac{6}{5}$,\\\\\n$\\therefore$ the minimum value of $PQ$ is $2OP^{\\prime }=\\boxed{\\frac{12}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53010226.png"
  },
  {
    "solution": "\\boxed{4:1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the pentagon ABCDE is obtained through a similarity transformation from the pentagon FGHMN, with point O as the center of similarity. Points F, G, H, M, and N are the midpoints of OA, OB, OC, OD, and OE, respectively. What is the ratio of the areas of the pentagons ABCDE and FGHMN?\n",
    "completion": "\\textbf{Solution:} Since $F$ is the midpoint of $AO$,\\\\\nit follows that $OF:OA=1:2$,\\\\\nsince pentagon $ABCDE$ is obtained from pentagon $FGHMN$ through a similarity transformation,\\\\\nit follows that $FN \\parallel AE$,\\\\\nthus $\\triangle OFN \\sim \\triangle OAE$,\\\\\nhence $OF:OA=FN:AE=1:2$,\\\\\ntherefore, the ratio of the areas of pentagons $ABCDE$ and $FGHMN$ is: $4:1$,\\\\\nso, the ratio of the areas of pentagons $ABCDE$ and $FGHMN$ is $\\boxed{4:1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51554733.png"
  },
  {
    "solution": "\\boxed{\\frac{3\\sqrt{2}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the rectangle $ABCD$, where $AB=4$. $M$ is a point on $AB$ with $AM=1$, and $N$ is a moving point on $BC$. When $\\triangle BMN$ is folded along $MN$ such that point $B$ coincides with point $B'$, if the distance from point $B'$ to $AD$ is 2, then what is the length of $BN$?\n",
    "completion": "\\textbf{Solution:} Construct $EF\\parallel AB$ through $B'$ intersecting $AD$ at $E$ and $BC$ at $F$. Draw $MG \\perp EF$ at $G$ as shown in the figure: \\\\\n$\\therefore$ Quadrilateral AMGE and quadrilateral MBFG are rectangles, \\\\\n$\\therefore$ $EG=AM=1$, $BM=GF=AB-AM=3$, \\\\\n$\\because$ The distance from point $B'$ to $AD$ is 2, \\\\\n$\\therefore$  $B'G=BE-EG=1$, \\\\\n$\\because$ Fold $\\triangle BMN$ along $MN$ making point $B$ coincide with point $B'$, \\\\\n$\\therefore$ $B'M=BM=3$, \\\\\nIn $\\triangle B'GM$ right-angled at $G$, $MG=\\sqrt{B'M^{2}-B'G^{2}}=\\sqrt{3^{2}-1^{2}}=2\\sqrt{2}=BF$,\\\\\nLet $BN=x$, then $B'N=x$,\\\\\n$NF=2\\sqrt{2}-x$ and $B'F=EF-B'E=2$,\\\\\nIn $\\triangle B'NF$ right-angled at $F$, $NF^{2}+B'F^{2}=B'N^{2}$,\\\\\n$\\therefore$ $(2\\sqrt{2}-x)^{2}+2^{2}=x^{2}$, solving for $x$ we get $x=\\boxed{\\frac{3\\sqrt{2}}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53728514.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the length of segment $AB$ is $10\\, \\text{cm}$. Point $C$ is a point on segment $AB$ with $BC=3\\, \\text{cm}$. Points $D$ and $E$ are the midpoints of segments $AC$ and $AB$, respectively. Find the length of segment $DE$.\n",
    "completion": "\\textbf{Solution:} Given that point $C$ lies on the segment $AB$, with segment $AB$ measuring $10cm$, and $BC=3cm$,\\\\\ntherefore, $AC=AB−BC=10−3=7cm$,\\\\\nsince point $D$ is the midpoint of $AC$,\\\\\nthus $AD=\\frac{1}{2}AC=\\frac{1}{2}\\times 7=\\frac{7}{2}$cm;\\\\\nsince point $E$ is the midpoint of $AB$,\\\\\nthus $AE=\\frac{1}{2}AB=\\frac{1}{2}\\times 10=5$cm,\\\\\nhence $DE=AE−AD=5−\\frac{7}{2}=\\boxed{\\frac{3}{2}}$cm.",
    "image_path": "dataset/mm_math/images/MM_Math/51438062.png"
  },
  {
    "solution": "\\boxed{x < -1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graphs of the linear function $y_{1}=mx\\ (m\\ne 0)$ and $y_{2}=x+3$ intersect at point $A(-1, 2)$. What is the solution set of the inequality $mx>x+3$ with respect to $x$?\n",
    "completion": "\\textbf{Solution:} \\\\\nSince $mx > x + 3$, \\\\\nit follows that $y_1 > y_2$, \\\\\nthus $x < -1$. \\\\\nHence, the answer is $\\boxed{x < -1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52284879.png"
  },
  {
    "solution": "\\boxed{6\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $A$ is inside $\\angle MON=45^\\circ$ with $OA=6\\text{cm}$. Points $B$ and $C$ are moving points on sides $OM$ and $ON$, respectively. What is the minimum perimeter of $\\triangle ABC$ in $\\text{cm}$?\n",
    "completion": "\\textbf{Solution:} Let $A'$ be the symmetric point of A with respect to $OM$, and let $A''$ be the symmetric point of A with respect to $ON$. Draw line $A'A''$, intersecting $OM$ at point B and $ON$ at point C. Draw lines $OA'$ and $OA''$. At this moment, the perimeter of $\\triangle ABC$ is minimal. \\\\ \n$\\because$ $AB = A'B$, $AC = A''C$, $OA = OA' = OA'' = 6$, $\\angle BOA' = \\angle BOA$, $\\angle COA'' = \\angle COA$, \\\\ \n$\\because$ $\\angle MON = 45^\\circ$, \\\\\n$\\therefore$ $\\angle A''OA' = 90^\\circ$, \\\\\n$\\therefore$ $A''A' = 6\\sqrt{2}$, \\\\\n$\\therefore$ The perimeter of $\\triangle ABC = AB + BC + AC = A'B + BC + A''C = A''A' = 6\\sqrt{2}$, \\\\\n$\\therefore$ The minimum perimeter of $\\triangle ABC$ is $\\boxed{6\\sqrt{2}}$ cm.",
    "image_path": "dataset/mm_math/images/MM_Math/52874444.png"
  },
  {
    "solution": "\\boxed{2\\sin 40^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given a right-angled $\\triangle ABC$, with $\\angle C=90^\\circ$, $\\angle A=50^\\circ$, and $AB=2$, what is the value of $AC$?\n",
    "completion": "\\textbf{Solution}: We have $\\angle B=90^\\circ-\\angle A=90^\\circ-50^\\circ=40^\\circ$,\\\\\nFurthermore, since $\\sin B= \\frac{AC}{AB}$,\\\\\nit follows that $AC=AB \\cdot \\sin 40^\\circ = \\boxed{2\\sin 40^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51262164.png"
  },
  {
    "solution": "\\boxed{24}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $D$ and $E$ are the midpoints of sides $AB$ and $AC$ of $\\triangle ABC$, respectively. If $DE = 12$, what is the length of $BC$?\n",
    "completion": "Solution: Since $D$ and $E$ are the midpoints of sides $AB$ and $AC$ of $\\triangle ABC$ respectively, \\\\\nit follows that $DE$ is the midline of $\\triangle ABC$, \\\\\nhence $BC = 2DE$, \\\\\nsince $DE = 12$, \\\\\nit follows that $BC = \\boxed{24}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53313048.png"
  },
  {
    "solution": "\\boxed{70^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that ship A is in the direction $30^\\circ$ north of east from lighthouse P, ship B is in the direction $70^\\circ$ south of east from lighthouse P, and ship C is on the bisector of $\\angle APB$. How many degrees north of east is ship C from lighthouse P?\n",
    "completion": "\\textbf{Solution:} Since $PC$ bisects $\\angle APB$ and $\\angle APB=80^\\circ$,\n\\[\\therefore \\angle APC= \\frac{1}{2}\\angle APB= \\frac{1}{2}\\times 80^\\circ=40^\\circ.\\]\n\\[\\therefore \\angle NPC=\\angle APN+\\angle APC=30^\\circ+40^\\circ=\\boxed{70^\\circ}.\\]\nTherefore, the ship C is in the direction of 70$^\\circ$ north by east from the lighthouse P.",
    "image_path": "dataset/mm_math/images/MM_Math/55431959_71.png"
  },
  {
    "solution": "\\boxed{12^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, two bamboo poles $AB$ and $DB$ are leaning against the wall $CE$ at angles such that $\\angle CAB = 33^\\circ$ and $\\angle CDB = 21^\\circ$. What is the measure of $\\angle ABD$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle CAB$ is an exterior angle of $\\triangle ABD$ and $\\angle CAB=33^\\circ$, $\\angle CDB=21^\\circ$,\\\\\nthen $\\angle CAB=\\angle ABD+\\angle CDB$,\\\\\nthus, $\\angle ABD=\\angle CAB-\\angle CDB=12^\\circ$,\\\\\ntherefore, the answer is: $\\boxed{12^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52836657.png"
  },
  {
    "solution": "\\boxed{\\frac{5}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $O$ is the centroid of $\\triangle ABC$. A straight line passing through $O$ intersects $AB$ and $AC$ at $G$ and $H$ (neither coinciding with the vertices of $\\triangle ABC$) respectively. Let ${S}_{\\mathrm{quadrilateral\\ BCHG}}$ and ${S}_{\\triangle AGH}$ represent the areas of the quadrilateral $BCHG$ and the triangle $AGH$, respectively. What is the maximum value of $\\frac{{S}_{\\mathrm{quadrilateral\\ BCHG}}}{{S}_{\\triangle AGH}}$?\n",
    "completion": "\\textbf{Solution}: Draw MN$\\parallel$BC through O intersecting AB at N and AC at M. Draw ME$\\parallel$AB through M intersecting GH at E. \\\\\nSince O is the centroid of $\\triangle ABC$, \\\\\ntherefore $\\frac{AO}{OD}=2$, where D is the midpoint of BC \\\\\nthus, BD=CD, $\\frac{AO}{AD}=\\frac{2}{3}$ \\\\\nSince MN$\\parallel$BC \\\\\ntherefore $\\triangle AMN\\sim \\triangle ACB$ \\\\\ntherefore $\\frac{NO}{BD}=\\frac{MO}{CD}=\\frac{AO}{AD}=\\frac{2}{3}$, $\\frac{{S}_{\\triangle AMN}}{{S}_{\\triangle ABC}}=\\left(\\frac{AO}{AD}\\right)^{2}=\\frac{4}{9}$ \\\\\nthus, MO=NO \\\\\nSince ME$\\parallel$AB \\\\\ntherefore $\\angle MOH=\\angle NOG$ \\\\\ntherefore $\\triangle MOE\\cong \\triangle NOG$ (AAS) \\\\\ntherefore ${S}_{\\triangle MOE}={S}_{\\triangle NOG}$ \\\\\nLet ${S}_{\\triangle MEH}=x$, ${S}_{\\triangle ABC}=9y$ \\\\\ntherefore ${S}_{\\triangle AMN}=4y$, ${S}_{\\text{quadrangle}BCMN}=5y$, \\\\\ntherefore ${S}_{\\text{quadrangle}BCHG}={S}_{\\text{quadrangle}BCMN}-{S}_{\\triangle HOM}+{S}_{\\triangle NOG}$ \\\\\n$={S}_{\\text{quadrangle}BCMN}-({S}_{\\triangle MOE}+{S}_{\\triangle MEH})+{S}_{\\triangle NOG}$ \\\\\n$=5y-x$ \\\\\n${S}_{\\triangle AHG}={S}_{\\triangle AMN}-{S}_{\\triangle NOG}+{S}_{\\triangle HOM}$ \\\\\n$={S}_{\\triangle AMN}-{S}_{\\triangle NOG}+({S}_{\\triangle MOE}+{S}_{\\triangle MEH})$ \\\\\n$=4y+x$ \\\\\ntherefore $\\frac{{S}_{\\text{quadrangle}BCHG}}{{S}_{\\triangle AGH}}=\\frac{5y-x}{4y+x}$ \\\\\nSince $x$ is a constant \\\\\ntherefore, as $y$ becomes smaller, the value of $\\frac{{S}_{\\text{quadrangle}BCHG}}{{S}_{\\triangle AGH}}=\\frac{5y-x}{4y+x}$ becomes larger \\\\\ntherefore, when $y=0$, $\\frac{{S}_{\\text{quadrangle}BCHG}}{{S}_{\\triangle AGH}}=\\boxed{\\frac{5}{4}}$ is maximized, at this point GH$\\parallel$BC",
    "image_path": "dataset/mm_math/images/MM_Math/51322733.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle ABCD, $AB=\\sqrt{3}$, $\\angle BAC=60^\\circ$, point E is on the extension of AB, and $BE=AC$. If DE is connected, what is the length of DE?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral ABCD is a rectangle,\\\\\nit follows that AD=BC, $\\angle$DAB=$\\angle$ABC=$90^\\circ$,\\\\\nsince $\\angle BAC=60^\\circ$,\\\\\nit follows that $\\angle$ACB=$30^\\circ$,\\\\\nsince AB=$\\sqrt{3}$,\\\\\nit follows that AC=2AB=$2\\sqrt{3}$,\\\\\nin $\\triangle ABC$, $BC=\\sqrt{AC^2-AB^2}=3$,\\\\\nthus AD=BC=3,\\\\\nsince AC=BE=$2\\sqrt{3}$ and AB=$\\sqrt{3}$,\\\\\nthus AE=AB+BE=$3\\sqrt{3}$,\\\\\nin $\\triangle DAE$, DE=$\\sqrt{AD^2+AE^2}=\\sqrt{3^2+(3\\sqrt{3})^2}=\\boxed{6}$",
    "image_path": "dataset/mm_math/images/MM_Math/52387647.png"
  },
  {
    "solution": "\\boxed{\\frac{2}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, given that $AB\\parallel CD\\parallel EF$, if $\\frac{AC}{AE}=\\frac{2}{3}$, what is the value of $\\frac{BD}{BF}$?\n",
    "completion": "\\textbf{Solution:} Since $AB \\parallel CD \\parallel EF$,\\\\\nit follows that $\\frac{AC}{AE}=\\frac{BD}{BF}=\\boxed{\\frac{2}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52991840.png"
  },
  {
    "solution": "\\boxed{y_1<y_2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graphs of the functions $y_1=2x+b$ and $y_2=ax-3$ intersect at point $P(-2, -5)$. The graphs of these two functions intersect the x-axis at points A and B, respectively. Based on the graph, directly write the range of values for $x$ when $y_1<y_2$.\n",
    "completion": "\\textbf{Solution:} It can be inferred from the graph of the function that, when $x<-2$, \\boxed{y_1<y_2}.",
    "image_path": "dataset/mm_math/images/MM_Math/53706598_204.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, DE is the perpendicular bisector of AC, AB=5, BC=7, then what is the perimeter of $\\triangle ABD$?\n",
    "completion": "Solution: Since $DE$ is the perpendicular bisector of $AC$, \\\\\n$\\therefore$ $AD=DC$, \\\\\n$\\therefore$ the perimeter of $\\triangle ABD$ is $AD+BD+AB=DC+BD+AB=BC+AB=5+7=\\boxed{12}$",
    "image_path": "dataset/mm_math/images/MM_Math/52861122.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is a chord of the circle $\\odot O$, $C$ is the midpoint of $AB$, and $OC$ intersects the arc $\\overset{\\frown}{AB}$ at point $D$. If $AB=8$ cm and $CD=2$ cm, what is the radius of the circle $\\odot O$ in centimeters?\n",
    "completion": "\\textbf{Solution:} Draw line OB. Since point C is the midpoint of arc $\\overset{\\frown}{AB}$, it follows that OC$\\perp$AB and BD=$\\frac{1}{2}$AB=4. Let the radius of the circle be r, then OD=r−2. In $\\triangle$OBD, $OD^{2}$+$BD^{2}$=$OB^{2}$. Therefore, $(r−2)^{2}$+$4^{2}$=$r^{2}$. Solving this equation, we find $r=\\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53081444.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{2}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Fold the rectangle $ABCD$ as shown in the diagram, with $BE$, $EG$, and $FG$ being the fold lines. If vertices $A$, $C$, $D$ all fall on point $O$, and points $B$, $O$, $G$ are on the same straight line, while points $E$, $O$, $F$ are on another straight line, what is the value of $\\frac{EG}{BE}$?\n",
    "completion": "\\textbf{Solution:} From the context, we know that the value of $\\frac{EG}{BE}$ is \\boxed{\\frac{\\sqrt{2}}{2}}.",
    "image_path": "dataset/mm_math/images/MM_Math/51212662_28.png"
  },
  {
    "solution": "\\boxed{y_{1}=-x+3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y_1=k_1x+b$ intersects with the hyperbola $y_2=\\frac{k_2}{x} \\left(x>0\\right)$ at points $A$ and $B$ within the first quadrant. Given $A(1, m)$ and $B(2, 1)$, find the equations of the line and the hyperbola.\n",
    "completion": "\\textbf{Solution:} Given point $B(2, 1)$ is on the hyperbola $y_{2}=\\frac{k_{2}}{x}$, \\\\\n$\\therefore k_{2}=2\\times 1=2$,\\\\\n$\\therefore$ the equation of the hyperbola is $y_{2}=\\frac{2}{x}$\\\\\nSince $A(1, m)$ is on the hyperbola $y_{2}=\\frac{2}{x}$, \\\\\n$\\therefore m=\\frac{2}{1}=2$,\\\\\n$\\therefore A(1, 2)$\\\\\nSince line $y_{1}=k_{1}x+b$ passes through points $A(1, 2)$ and $B(2, 1)$,\\\\\n$\\therefore \\left\\{\\begin{array}{l}k_{1}+b=2\\\\ 2k_{1}+b=1\\end{array}\\right.$, solving this gives $\\left\\{\\begin{array}{l}k_{1}=-1\\\\ b=3\\end{array}\\right.$,\\\\\n$\\therefore$ the equation for line $AB$ is $\\boxed{y_{1}=-x+3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52601999_152.png"
  },
  {
    "solution": "\\boxed{\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: \\[\nIn an isosceles \\triangle ABC with a right angle at C, a point D is on the side AC. Connect BD and fold \\triangle ABC along BD so that point A falls on side BC. Connect AE. Then, what is the value of \\frac{{S}_{\\triangle BDE}}{{S}_{\\triangle ACE}}?\n\\]\n",
    "completion": "Solution:Let's fold $\\triangle ABC$ along BD so that point A falls on side BC,\\\\\nhence, $BE=AB=AC$, $\\angle BED=\\angle BAC=90^\\circ$, and $AD=DE$,\\\\\nthus, $\\angle CED=90^\\circ$,\\\\\nsince $\\angle C=45^\\circ$,\\\\\nit follows that $\\triangle DEC$ is an isosceles right triangle,\\\\\nthus, $DE=CE$,\\\\\nlet $AD=DE=CE=x$,\\\\\nhence, $CD=\\sqrt{2}x$,\\\\\nthus, $BE=AC=x+\\sqrt{2}x$,\\\\\ndraw EH $\\perp$ CD at H from point E,\\\\\nhence, $EH=CH=\\frac{\\sqrt{2}}{2}CE=\\frac{\\sqrt{2}}{2}x$,\\\\\nsince $S_{\\triangle BDE}=\\frac{1}{2}BE\\cdot DE=\\frac{1}{2}x(x+\\sqrt{2}x)=\\frac{1+\\sqrt{2}}{2}x^2$, and $S_{\\triangle ACE}=\\frac{1}{2}AC\\cdot EH=\\frac{1}{2}\\times \\frac{\\sqrt{2}}{2}x(x+\\sqrt{2}x)=\\frac{\\sqrt{2}+2}{4}x^2$,\\\\\ntherefore, $\\frac{S_{\\triangle BDE}}{S_{\\triangle ACE}}=\\frac{\\frac{1+\\sqrt{2}}{2}x^2}{\\frac{\\sqrt{2}+2}{4}x^2}=\\boxed{\\sqrt{2}}$,",
    "image_path": "dataset/mm_math/images/MM_Math/52914198.png"
  },
  {
    "solution": "\\boxed{\\frac{25}{8}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that in triangle $ABC$, the side $AB$ is tangent to circle $O$ at point $B$. $AC$ passes through the center $O$ and intersects the circle at points $D$ and $C$. A line $CE$ is drawn through $C$ perpendicular to $AB$, intersecting the extension of $AB$ at point $E$, and $CB$ bisects $\\angle ACE$; if $BE=3$ and $CE=4$, what is the radius of circle $O$?\n",
    "completion": "\\textbf{Solution:} As shown in Figure 2, draw line BD, \\\\\nsince $CE \\perp AB$, \\\\\nhence $\\angle E=90^\\circ$, \\\\\nthus $BC=\\sqrt{BE^2+CE^2}=\\sqrt{3^2+4^2}=5$, \\\\\nsince CD is the diameter of the circle $O$, \\\\\nhence $\\angle DBC=90^\\circ$, \\\\\nthus $\\angle E=\\angle DBC$, \\\\\nhence $\\triangle DBC \\sim \\triangle CBE$, \\\\\nthus $\\frac{CD}{BC}=\\frac{BC}{CE}$, \\\\\nhence $BC^2=CD \\cdot CE$, \\\\\nthus $CD=\\frac{5^2}{4}=\\frac{25}{4}$, \\\\\nhence $OC=\\frac{1}{2}CD=\\frac{25}{8}$, \\\\\nthus the radius of circle $O$ is $\\boxed{\\frac{25}{8}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52766358_168.png"
  },
  {
    "solution": "\\boxed{15^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As the figure shows, it is known that: $O$ is a point on line $AB$, $\\angle COD$ is a right angle, $OE$ bisects $\\angle BOC$. If $\\angle AOC=30^\\circ$, what is the degree measure of $\\angle DOE$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle COD$ is a right angle, and $\\angle AOC=30^\\circ$,\\\\\nit follows that $\\angle BOD=180^\\circ-90^\\circ-30^\\circ=60^\\circ$,\\\\\nand $\\angle COB=180^\\circ-30^\\circ=150^\\circ$. Since OE bisects $\\angle BOC$,\\\\\nwe have $\\angle BOE=\\frac{1}{2}\\angle BOC=75^\\circ$,\\\\\nthus, $\\angle DOE=\\angle BOE-\\angle BOD=75^\\circ-60^\\circ=\\boxed{15^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/9153886_49.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC=4$, and $\\angle B=\\angle C=15^\\circ$. What is the area of $\\triangle ABC$?\n",
    "completion": "\\textbf{Solution:} Draw $CD \\perp AB$ through point $C$ intersecting the extended line of $BA$ at $D$,\\\\\n$\\because$ $\\angle B = \\angle ACB = 15^\\circ$,\\\\\n$\\therefore$ $\\angle CAD = \\angle B + \\angle ACB = 15^\\circ + 15^\\circ = 30^\\circ$,\\\\\n$\\because$ $AC = 4$ and $CD$ is the height from the base $AB$,\\\\\n$\\therefore$ $CD = \\frac{1}{2} AC = \\frac{1}{2} \\times 4 = 2$,\\\\\n$\\therefore$ the area of $\\triangle ABC = \\frac{1}{2} \\times 4 \\times 2 = \\boxed{4}$",
    "image_path": "dataset/mm_math/images/MM_Math/53010111.png"
  },
  {
    "solution": "\\boxed{120^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $OD$ bisects $\\angle AOC$, and $OE$ bisects $\\angle COB$. If $\\angle AOD = 20^\\circ$ and $\\angle EOB = 40^\\circ$, what is the measure of $\\angle AOB$ in degrees?\n",
    "completion": "\\textbf{Solution:} Since OD bisects $\\angle AOC$ and OE bisects $\\angle COB$, \\\\\n$\\therefore \\angle AOC = 2\\angle AOD$ and $\\angle COB = 2\\angle EOB$, \\\\\nSince $\\angle AOD = 20^\\circ$ and $\\angle EOB = 40^\\circ$. \\\\\n$\\therefore \\angle AOC = 2 \\times 20^\\circ = 40^\\circ$ and $\\angle BOC = 2 \\times 40^\\circ = 80^\\circ$, \\\\\n$\\therefore \\angle AOB = \\angle AOC + \\angle BOC = 40^\\circ + 80^\\circ = \\boxed{120^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/51284715.png"
  },
  {
    "solution": "\\boxed{\\frac{9}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $C$ is a point on the line segment $AB$, $D$ is the midpoint of $AC$, and $E$ is the midpoint of $BC$. If $AB=9\\,\\text{cm}$, $AC=5\\,\\text{cm}$. Find: the length of $DE$ in centimeters?\n",
    "completion": "\\textbf{Solution:} Given that $AB=9\\text{cm}$, $AC=5\\text{cm}$,\\\\\nthus $BC=AB-AC=9-5=4\\text{cm}$,\\\\\nsince $E$ is the midpoint of $BC$,\\\\\nthus $CE=\\frac{1}{2}BC=2\\text{cm}$,\\\\\ntherefore $DE=CD+CE=\\frac{5}{2}+2=\\boxed{\\frac{9}{2}}\\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52794293_224.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ is an equilateral triangle, with D and E located on $AB$ and $BC$ respectively, and $AD=BE$, $AE$ intersects $CD$ at point F. As shown in Figure 2, draw $CH\\perp AE$ at point H through point C. If $AE=5$ and $HF=2$, what is the length of $DF$?\n",
    "completion": "\\textbf{Solution:} As shown in Figure 2,\\\\\n$\\because CH\\perp AE$\\\\\n$\\therefore \\angle FHC=90^\\circ$\\\\\nIn $\\triangle FHC$, $\\angle FCH=90^\\circ-\\angle CFH=30^\\circ$\\\\\n$\\because HF=2$\\\\\n$\\therefore FC=2FH=4$\\\\\n$\\because \\triangle ABE\\cong \\triangle CAD$\\\\\n$\\therefore AE=CD=5$\\\\\n$\\therefore DF=CD-CF=5-4=\\boxed{1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52891027_105.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the rhombus $ABCD$ shown in the figure, we have $AB=4$ and $\\angle ABC=60^\\circ$. Points $P$ and $M$ are moving points on $BD$ and $BC$, respectively, with $M$ being distinct from points $B$ and $C$. What is the minimum value of $PM+PC$?\n",
    "completion": "Solution: As shown in Figure 1, connect PA,\n\n$\\because$ rhombus ABCD,\n\n$\\therefore$ AD=DC, $\\angle ADB=\\angle BDC$,\n\nIn $\\triangle ADP$ and $\\triangle CDP$,\n\n$\\because$ $\\left\\{\\begin{array}{l}AD=DC\\\\ \\angle ADP=\\angle PDC\\\\ PD=PD\\end{array}\\right.$,\n\n$\\therefore$ $\\triangle ADP\\cong \\triangle CDP\\left(SAS\\right)$,\n\n$\\therefore$ $PA=PC$, that is $PM+PC=PM+PA\\ge AM$.\n\n$\\because$ M is a moving point on BC,\n\n$\\therefore$ when AM$\\perp$BC, AM has the smallest value.\n\nAs shown in Figure 2, draw AM$\\perp$BC at point M through A,\n\n$\\because$ $\\angle ABC=60^\\circ$, $\\angle AMB=90^\\circ$, $AB=4$,\n\n$\\therefore$ $AM=2\\sqrt{3}$,\n\nSo the minimum value of $PM+PC$ is $\\boxed{2\\sqrt{3}}$,",
    "image_path": "dataset/mm_math/images/MM_Math/52204022.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB \\perp x$-axis, with $B$ being the foot of the perpendicular. The hyperbola $y = \\frac{k}{x}$ (where $x > 0$) intersects the two sides $OA$ and $AB$ of $\\triangle AOB$ at points $C$ and $D$, respectively, with $OC = CA$. If the area of $\\triangle ACD$ is $3$, what is the value of $k$?\n",
    "completion": "\\textbf{Solution:} Draw OD, and construct CE $\\perp$ x-axis through point C,\\\\\n$\\because$ OC = CA,\\\\\n$\\therefore$ OE : OB = 1 : 2;\\\\\nAssume the area of $\\triangle OBD$ as x, according to the meaning of the inverse ratio function k, the area of triangle OCE is x,\\\\\n$\\because$ $\\triangle COE \\sim \\triangle AOB$,\\\\\n$\\therefore$ the ratio of the areas of triangle COE to triangle BOA is 1 : 4,\\\\\n$\\because$ the area of $\\triangle ACD$ is 3,\\\\\n$\\therefore$ the area of $\\triangle OCD$ is 3,\\\\\n$\\therefore$ the area of triangle BOA is 6 + x,\\\\\nThat is, the area of triangle BOA is 6 + x = 4x,\\\\\nSolving this gives x = 2,\\\\\n$\\therefore$ $\\frac{1}{2}|k|=2$,\\\\\n$\\because$ k > 0,\\\\\n$\\therefore$ k = \\boxed{4}.",
    "image_path": "dataset/mm_math/images/MM_Math/51445482.png"
  },
  {
    "solution": "\\boxed{36^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the figure, within $\\triangle ABC$, we have $AB=AC$, and $D$ is a point on $BC$ such that $CD=AD$ and $AB=BD$. What is the measure of $\\angle B$?\n",
    "completion": "\\textbf{Solution:} We have AB=AC,\\\\\nthus $\\angle B=\\angle C$,\\\\\nand since CD=DA,\\\\\nit follows that $\\angle C=\\angle DAC$,\\\\\nalso given BA=BD,\\\\\nit implies $\\angle BDA=\\angle BAD=2\\angle C=2\\angle B$,\\\\\nfurthermore, since $\\angle B+\\angle BAD+\\angle BDA=180^\\circ$,\\\\\nwe conclude $5\\angle B=180^\\circ$,\\\\\nhence $\\angle B=\\boxed{36^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53080965.png"
  },
  {
    "solution": "\\boxed{\\sqrt{41} + 1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, within the Cartesian coordinate system, the vertices A and B of square ABCD are located on the x-axis, and the parabola $y=-x^2+bx+c$ passes through points A and $C(4, -5)$, and intersects with line DC at another point E. Let P be a point on the y-axis, and draw a perpendicular line from P to the axis of symmetry of the parabola, with Q being the foot of the perpendicular. Connect EQ and AP. Find the minimum value of $EQ+PQ+AD$.\n",
    "completion": "\\textbf{Solution:} Connect OC, intersecting the line of symmetry $x=1$ at point Q.\\\\\nSince $PQ \\perp y$-axis,\\\\\nit follows that $AO \\parallel PQ$,\\\\\nGiven that $AO = PQ = 1$,\\\\\nit follows that quadrilateral AOQP is a parallelogram,\\\\\nHence, $AP=OQ$,\\\\\nTherefore, $EQ + PQ + AP = EQ + 1 + OQ$\\\\\nTo minimize the value of $EQ + PQ + AP$, the sum $EQ + OQ$ must be minimized.\\\\\nSince E and C are symmetric about the axis $x=1$,\\\\\nit follows that $EQ = CQ$,\\\\\ntherefore, $EQ + OQ = CQ + OQ$,\\\\\nat this point, the value of $EQ + OQ$ is minimized, with the minimum value being the length of segment OC.\\\\\nGiven $C(4, -5)$,\\\\\nthus $OC = \\sqrt{4^{2}+5^{2}} = \\sqrt{41}$,\\\\\nSo, the minimum value of $EQ + PQ + AP$ is $\\sqrt{41} + 1$,\\\\\nmeaning the minimum value of $EQ + PQ + AP$ is $\\boxed{\\sqrt{41} + 1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52132998_102.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the figure, in $ \\triangle ABC$, $ \\angle C=90^\\circ$, $ \\angle ABC=60^\\circ$, $BD$ bisects $ \\angle ABC$, if $AD=8$cm, what is the length of $CD$ in cm?\n",
    "completion": "\\textbf{Solution}: Since $\\angle C=90^\\circ$ and $\\angle ABC=60^\\circ$\\\\\nTherefore, $\\angle A=30^\\circ$\\\\\nSince $BD$ bisects $\\angle ABC$\\\\\nTherefore, $\\angle ABD=\\angle CBD=30^\\circ$\\\\\nTherefore, $\\angle A=\\angle ABD$\\\\\nTherefore, $DB=AD=8$\\\\\nSince $\\angle C=90^\\circ$ and $\\angle CBD=30^\\circ$\\\\\nTherefore, $CD=\\frac{1}{2}DB=4$\\\\\nHence, the length of $CD$ is $\\boxed{4}$cm.",
    "image_path": "dataset/mm_math/images/MM_Math/53058560.png"
  },
  {
    "solution": "\\boxed{(4, 1)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated in the figure, rectangle $OCBA$ is positioned within the Cartesian coordinate system. Point $C$ is located on the positive half of the $x$-axis, and point $A$ is on the positive half of the $y$-axis. It is known that point $B$ has the coordinates $(4,2)$. The graph of the inverse proportion function $y=\\frac{k}{x}$ passes through the midpoint $D$ of $AB$ and intersects $BC$ at point $E$. Suppose the equation of line $DE$ is $y=mx+n$. Connect $OD$ and $OE$. What are the expression of the inverse proportion function $y=\\frac{k}{x}$ and the coordinates of point $E$?\n",
    "completion": "\\textbf{Solution:} Given $\\because$ quadrilateral OABC is a rectangle, and point B is (4, 2),\\\\\n$\\therefore$ AB = 4, BC = 2,\\\\\n$\\because$ the midpoint of AB is D,\\\\\n$\\therefore$ D is (2, 2),\\\\\n$\\because$ the graph of the inverse proportion function $y=\\frac{k}{x}$ passes through the midpoint D of AB,\\\\\n$\\therefore$ 2 = $\\frac{k}{2}$,\\\\\n$\\therefore$ k = 4,\\\\\n$\\therefore$ the equation of the inverse proportion function is: $y=\\frac{4}{x}$;\\\\\nwhen $x=4$, $y=\\frac{4}{4}=1$,\\\\\n$\\therefore$ the coordinates of point E are $\\boxed{(4, 1)}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52536619_156.png"
  },
  {
    "solution": "\\boxed{y=\\frac{60}{x}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As the diagram shows, a certain school in Xi'an city has carried out \"fumigation disinfection\" in all classrooms to further prevent the \"novel coronavirus\". It is known that during the burning and releasing process of the disinfectant, the concentration of the disinfectant in the classroom air, \\(y\\) (mg) per cubic meter, and the burning time \\(x\\) (min) have the following functional relationship: when \\(x < 6\\), \\(y\\) is directly proportional to \\(x\\); when \\(x \\geq 6\\), \\(y\\) is inversely proportional to \\(x\\). Based on the information provided by the graph, find the functional relation between \\(y\\) and \\(x\\) when \\(x \\geq 6\\).\n",
    "completion": "\\textbf{Solution}: From the graph, it is known that when $x \\geq 6$, $y$ is an inverse proportion function of $x$, let $y=\\frac{k}{x}$ ($k \\neq 0$).\\\\\n$\\because$ point $(15,4)$,\\\\\n$\\therefore k=15 \\times 4=60$,\\\\\nThe functional relationship between $y$ and $x$ is $\\boxed{y=\\frac{60}{x}}\\left(x\\geq 6\\right)$.",
    "image_path": "dataset/mm_math/images/MM_Math/52304593_211.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a point $E$ is taken outside the square $ABCD$, and lines $AE$, $BE$, $DE$ are drawn. A perpendicular line is drawn from point $A$ to line $AE$, intersecting $DE$ at point $P$. If $AE=AP=1$ and $PB= \\sqrt{10}$, what is the distance from point $B$ to line $AE$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle EAP=\\angle BAD=90^\\circ$, \\\\\nit follows that $\\angle EAB+\\angle BAP=90^\\circ$, and $\\angle PAD+\\angle BAP=90^\\circ$, \\\\\nthus, $\\angle EAB=\\angle PAD$, \\\\\nmoreover, since AE=AP and AB=AD, \\\\\nin triangles $\\triangle APD$ and $\\triangle AEB$, we have \\\\\n$\\begin{cases}\nAE=AP \\\\\n\\angle EAB=\\angle PAD \\\\\nAB=AD\n\\end{cases}$, \\\\\ntherefore, $\\triangle APD \\cong \\triangle AEB$ (SAS), \\\\\nextend a line from B to F such that BF$\\perp$AE and BF intersects the extended line of AE at F, \\\\\ngiven AE=AP and $\\angle EAP=90^\\circ$, \\\\\nit implies that $\\angle AEP=\\angle APE=45^\\circ$, \\\\\nthus, $\\angle AEB=\\angle APD=180^\\circ-45^\\circ=135^\\circ$, \\\\\ntherefore, $\\angle BEP=135^\\circ-45^\\circ=90^\\circ$, \\\\\nwhich implies EB$\\perp$ED, \\\\\nsince BF$\\perp$AF, \\\\\nit follows that $\\angle FEB=\\angle FBE=45^\\circ$, \\\\\ngiven PE=$\\sqrt{1^2+1^2}=\\sqrt{2}$, \\\\\nthus, BE=$\\sqrt{BP^2-PE^2}=2\\sqrt{2}$, \\\\\nhence, BF=EF=$\\frac{2\\sqrt{2}}{\\sqrt{2}}=\\boxed{2}$, \\\\\ntherefore, the distance from point B to line AE is $\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52387190.png"
  },
  {
    "solution": "\\boxed{y_1=\\frac{4}{x}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the inverse proportion function $y_{1}=\\frac{k}{x}$ (where $k$ is a constant and $k\\neq 0$) and the graph of the linear function $y_{2}=2x+2$ both pass through points $A(1, m)$ and $B(a, b)$. Find the coordinates of point A and the expression for the inverse proportion function.\n",
    "completion": "\\textbf{Solution:} Substituting point A into $y_2=2x+2$, we get $m=4$,\\\\\n$\\therefore$ the coordinates of A are $\\boxed{(1,4)}$;\\\\\n$\\because$ $y_1=\\frac{k}{x}$ passes through point A,\\\\\n$\\therefore$ $k=1\\times 4=4$,\\\\\n$\\therefore$ the expression for the inverse proportion function is $\\boxed{y_1=\\frac{4}{x}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52948375_255.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in a right-angled triangle paper with $\\angle C=90^\\circ$, $AB=10$, $BC=8$, and $AC=6$, the triangle is folded along a line passing through point $B$ such that point $C$ falls on point $E$ on side $AB$, with the fold line being $BD$. What is the perimeter of $\\triangle ADE$?\n",
    "completion": "\\textbf{Solution:} From the properties of folding, we know that $BE=BC=8$, $DE=CD$,\\\\\n$\\therefore AE=AB-BE=2$,\\\\\n$\\therefore$ the perimeter of $\\triangle ADE$ $=AD+AE+DE=AD+DE+AE=AC+AE=\\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51368158_113.png"
  },
  {
    "solution": "\\boxed{\\frac{4}{9}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, points D and E are located on sides $AB$ and $AC$ respectively, with $DE\\parallel BC$. If $DE:BC = 2:3$, then what is the ratio of ${S}_{\\triangle ADE}$ to ${S}_{\\triangle ABC}$?\n",
    "completion": "\\textbf{Solution:} Since $DE\\parallel BC$,\\\\\nit follows that $\\triangle ADE\\sim \\triangle ABC$,\\\\\ntherefore $\\frac{{S}_{\\triangle ADE}}{{S}_{\\triangle ABC}}=\\left(\\frac{DE}{BC}\\right)^{2}=\\left(\\frac{2}{3}\\right)^{2}=\\boxed{\\frac{4}{9}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51213309.png"
  },
  {
    "solution": "\\boxed{40}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given that segment $ AD=80$, points B and C are both on segment AD. As shown in Figure 1, $ BC: AB: CD=1:3:4$, point E is the midpoint of segment AC, and point F is the midpoint of segment BD. Find the length of EF.\n",
    "completion": "\\textbf{Solution:} Let $BC\\colon AB\\colon CD=1\\colon 3\\colon 4$,\\\\\nThus, let BC=x, then AB=3x, CD=4x,\\\\\nTherefore, AD=3x+x+4x=80, which means x=10,\\\\\nHence, AB=30, BC=10, CD=40,\\\\\nSince point E is the midpoint of segment AC, and point F is the midpoint of segment BD,\\\\\nThus, AE=20, DF=20,\\\\\nTherefore, EF=80−20−20=\\boxed{40};",
    "image_path": "dataset/mm_math/images/MM_Math/51439120_172.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, $OC \\perp AB$ at point O, $OD$ bisects $\\angle BOC$, $OE$ bisects $\\angle AOD$. Find the degree measure of $\\angle BOD$.\n",
    "completion": "\\textbf{Solution:} Since $OC\\perp AB$,\\\\\nit follows that $\\angle BOC=\\angle AOC=90^\\circ$,\\\\\nSince $OD$ bisects $\\angle BOC$,\\\\\nit follows that $\\angle BOD=\\angle COD=\\frac{1}{2}\\angle BOC=45^\\circ$,\\\\\nTherefore, the measure of $\\angle BOD$ is $\\boxed{45^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53364075_82.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, given the quadratic function $y = x^{2} - 4x + 3$, let the points where its graph intersects the x-axis be A and B (with point A to the left of point B), and let the point where it intersects the y-axis be C. Find: What is the area of $\\triangle ABC$?\n",
    "completion": "\\textbf{Solution}: From (1), we know $A(1,0)$, $B(3,0)$, $C(0,3)$,\\\\\nthus, the area of $\\triangle ABC = \\frac{1}{2} \\times 2 \\times 3 = \\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52992235_172.png"
  },
  {
    "solution": "\\boxed{26}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within $ \\square ABCD$, points E and F are both on side AD. BE bisects $\\angle ABC$ and CF bisects $\\angle BCD$. If $BE=8$, $CF=6$, and $EF=2$, what is the perimeter of $\\square ABCD$?\n",
    "completion": "\\textbf{Solution:} Extend BC to point P such that CP = EF = 2,\\\\\nsince $\\square ABCD$,\\\\\nthus EF $\\parallel$ CP, $\\angle ABC + \\angle DCB = 180^\\circ$,\\\\\nthus quadrilateral EFCP is a parallelogram,\\\\\nthus EP = CF = 6, EP $\\parallel$ CF,\\\\\nsince BE bisects $\\angle ABC$, CF bisects $\\angle BCD$,\\\\\nthus $\\angle EBC + \\angle FCB = \\frac{1}{2}\\angle ABC + \\frac{1}{2}\\angle DCB = 90^\\circ$,\\\\\nthus CF $\\perp$ BE,\\\\\nthus PE $\\perp$ BE,\\\\\nthus BP = $\\sqrt{BE^2 + PE^2}$ = 10,\\\\\nthus BC = 8,\\\\\nsince $\\angle ABE = \\angle EBC$, $\\angle EBC = \\angle AEB$,\\\\\nthus $\\angle ABE = \\angle AEB$,\\\\\nthus AB = AE,\\\\\nSimilarly, it is derived that: CD = DF,\\\\\nthus AE = DF,\\\\\nsince AD + EF = AE + DF,\\\\\nthus AE = DF = 5,\\\\\nthus AB = CD = 5,\\\\\nthus the perimeter of $\\square ABCD$ is equal to 2 $\\times$ (5 + 8) = \\boxed{26}.",
    "image_path": "dataset/mm_math/images/MM_Math/52344117.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the rectangle $ABCD$, $AB = 8$, $AD = 10$. Point $E$ is on the side $DC$. The triangle $ADE$ is folded along the straight line $AE$, and point $E$ precisely falls on point $F$ on the side $BC$. What is the length of $CE$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rectangle,\\\\\nit follows that CD = AB = 8, BC = AD = 10, and $\\angle B = \\angle D = \\angle C = 90^\\circ$\\\\\nSince $\\triangle ADE$ is folded along the straight line AE, with point D exactly landing on point F on side BC,\\\\\nit follows that AF = AD = 10, and EF = DE\\\\\nIn $\\triangle ABF$, BF = $\\sqrt{AF^2 - AB^2} = \\sqrt{10^2 - 8^2} = 6$\\\\\nTherefore, CF = BC - BF = 10 - 6 = 4,\\\\\nLet CE = x, then DE = EF = 8 - x\\\\\nIn $\\triangle CEF$,\\\\\nsince $CF^2 + CE^2 = EF^2$,\\\\\nit follows that $4^2 + x^2 = (8 - x)^2$, solving for x gives $x = \\boxed{3}$,\\\\\nTherefore, the length of CE is 3.",
    "image_path": "dataset/mm_math/images/MM_Math/53728940.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, for the line segments $AB=21, BC=15$, point $M$ is the midpoint of $AC$. A point $N$ is chosen on $CB$ such that $CN:NB=1:2$. What is the length of $MN$?\n",
    "completion": "\\textbf{Solution:} Since $BC=15$ and $CN:NB=1:2$,\\\\\nit follows that $CN=\\frac{1}{3}BC=\\frac{1}{3}\\times 15=5$.\\\\\nAdditionally, since point M is the midpoint of AC and $AC=6$,\\\\\nit follows that $MC=\\frac{1}{2}AC=3$,\\\\\ntherefore $MN=MC+NC=3+5=\\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53211052_87.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In square $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$. Point $E$ is on $DO$ with $DE = 2EO$. Connect $AE$ and fold $\\triangle ADE$ along $AD$ to get $\\triangle ADE'$. Point $F$ is the midpoint of $AE$, and connect $E'F$. If $DE = 2\\sqrt{2}$, what is the area of $\\triangle AFE'$?\n",
    "completion": "\\textbf{Solution:} Connect $EE'$,\\\\\nsince quadrilateral $ABCD$ is a square,\\\\\nit follows that $OA=OD$, $\\angle ADO=45^\\circ$, $\\angle AOD=90^\\circ$\\\\\nGiven $DE=2\\sqrt{2}$, $DE=2EO$\\\\\nit follows that $OD=OA=3\\sqrt{2}$\\\\\nthus, ${S}_{\\triangle ADE}=\\frac{1}{2}DE\\times OA=6$\\\\\nAccording to the problem, $\\angle ADE'=\\angle ADO=45^\\circ$, $DE=DE'=2\\sqrt{2}$\\\\\nthus, ${S}_{\\triangle DEE'}=\\frac{1}{2}DE\\times DE=4$, ${S}_{\\text{quadrilateral} AED{E}'}=2{S}_{\\triangle ADE}=12$\\\\\ntherefore, ${S}_{\\triangle AEE'}={S}_{\\text{quadrilateral}AED{E}'}-{S}_{\\triangle DEE'}=8$\\\\\nAlso, since $F$ is the midpoint of $AE$\\\\\nit follows that ${S}_{\\triangle AEE'F}=\\frac{1}{2}{S}_{\\triangle AEE'}=\\boxed{4}$",
    "image_path": "dataset/mm_math/images/MM_Math/50481647.png"
  },
  {
    "solution": "\\boxed{\\frac{6\\sqrt{3}}{7}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\triangle ABC$, $ BD \\perp AB$, with BD and AC intersecting at point D, $ AD = \\frac{4}{7}AC$, $ AB = 4$, and $ \\angle ABC = 150^\\circ$. What is the area of $ \\triangle DBC$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, draw a perpendicular from point C to BD, intersecting the extension of BD at point E, \\\\\nthen $ \\angle E=90^\\circ$, \\\\\nsince $BD\\perp AB$ and $CE\\perp BD$, \\\\\nit follows that $AB\\parallel CE$, $\\angle ABD=90^\\circ$, \\\\\nhence $\\triangle ABD\\sim \\triangle CED$, \\\\\ntherefore $\\frac{AD}{CD}=\\frac{AB}{CE}=\\frac{BD}{DE}$, \\\\\nsince $AD=\\frac{4}{7}AC$, \\\\\nit follows that $\\frac{AD}{CD}=\\frac{4}{3}$, \\\\\nfurthermore, since $AB=4$, \\\\\nit follows that $\\frac{4}{CE}=\\frac{4}{3}=\\frac{BD}{DE}$, \\\\\ntherefore $CE=3$, $BD=\\frac{4}{7}BE$, \\\\\nsince $\\angle ABC=150^\\circ$, $\\angle ABD=90^\\circ$, \\\\\nit follows that $\\angle CBE=60^\\circ$, \\\\\ntherefore, in $Rt\\triangle BCE$, $\\tan\\angle CBE=\\frac{CE}{BE}=\\sqrt{3}$, \\\\\nhence $BE=\\frac{\\sqrt{3}}{3}CE=\\sqrt{3}$, \\\\\nthus $BD=\\frac{4}{7}BE=\\frac{4\\sqrt{3}}{7}$, \\\\\ntherefore, the area of $\\triangle BCD=\\frac{1}{2}\\cdot BD\\cdot CE=\\frac{1}{2}\\times 3\\times \\frac{4\\sqrt{3}}{7}=\\boxed{\\frac{6\\sqrt{3}}{7}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51106894.png"
  },
  {
    "solution": "\\boxed{80^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $O$ is a point on line $AC$, $OB$ is a ray, $OD$ bisects $\\angle AOB$, $OE$ is within $\\angle BOC$, $2\\angle BOE = \\angle EOC$, $\\angle DOE = 70^\\circ$, find the degree measure of $\\angle EOC$.\n",
    "completion": "\\textbf{Solution:} Let $\\angle AOB = x^\\circ$, so $\\angle BOC = (180-x)^\\circ$.\\\\\nSince OD bisects $\\angle AOB$,\\\\\nit follows that $\\angle DOB= \\frac{1}{2}\\angle AOB$,\\\\\nthus $\\angle DOB= \\frac{1}{2}x^\\circ$\\\\\nSince $2\\angle BOE=\\angle EOC$,\\\\\nit follows that $\\angle BOE= \\frac{1}{3}\\angle BOC= \\frac{1}{3}(180-x)^\\circ$.\\\\\nSince $\\angle DOE=\\angle DOB+\\angle BOE=70^\\circ$,\\\\\nit follows that $\\frac{1}{2}x+ \\frac{1}{3} (180-x)=70$,\\\\\nSolving this, we find $x=60$,\\\\\nTherefore $\\angle BOC= 180^\\circ-60^\\circ= 120^\\circ$,\\\\\nTherefore $\\angle EOC= \\frac{2}{3}\\angle BOC= \\frac{2}{3}\\times120^\\circ= \\boxed{80^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55192431.png"
  },
  {
    "solution": "\\boxed{9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the similarity ratio between $\\triangle ADE$ and $\\triangle ABC$ is $1:2$, and the area of $\\triangle ADE$ is 3. What is the area of the quadrilateral $DBCE$?\n",
    "completion": "Since the similarity ratio between $\\triangle ADE$ and $\\triangle ABC$ is $1\\colon 2$,\\\\\nthe ratio of their areas is $1\\colon 4$.\\\\\nTherefore, the area ratio between $\\triangle ADE$ and quadrilateral $DBCE$ is $1\\colon 3$.\\\\\nSince the area of $\\triangle ADE$ is 3,\\\\\nthe area of quadrilateral $DBCE$ is $\\boxed{9}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53106512.png"
  },
  {
    "solution": "\\boxed{(2^{12}-1, 2^{11})}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, the squares $A_1B_1C_1O$, $A_2B_2C_2C_1$, $A_3B_3C_3C_2\\ldots$ are placed as illustrated. Points $A_1$, $A_2$, $A_3\\ldots$ and points $C_1$, $C_2$, $C_3\\ldots$ are respectively on the line $y=kx+b$ (where $k>0$) and the x-axis, given that points $B_1(1, 1)$, $B_2(3, 2)$, then what are the coordinates of $B_{12}$?\n",
    "completion": "\\textbf{Solution:} \\\\\nLet $B_{1}$ have coordinates (1, 1), and point $B_{2}$ have coordinates (3, 2), \\\\\ntherefore, the side length of square $A_{1}B_{1}C_{1}O_{1}$ = 1, and the side length of square $A_{2}B_{2}C_{2}C_{1}$ = 2, \\\\\nhence, the coordinates of $A_{1}$ are (0, 1), and the coordinates of $A_{2}$ are: (1, 2), \\\\\nplugging into $y=kx+b$, we get: $\\left\\{\\begin{array}{l}b=1 \\\\ k+b=2\\end{array}\\right.$, solving gives: $\\left\\{\\begin{array}{l}k=1 \\\\ b=1\\end{array}\\right.$, \\\\\ntherefore, the equation of the line is: $y=x+1$.\\\\\nSince $A_{1}B_{1}=1$, and the coordinates of point $B_{2}$ are (3, 2), \\\\\nthus, the coordinates of point $A_{3}$ are (3,4), \\\\\nhence, $A_{3}C_{2}=A_{3}B_{3}=B_{3}C_{3}=4$, \\\\\nthus, the coordinates of point $B_{3}$ are (7, 4), \\\\\ntherefore, the vertical coordinate of $B_{1}$ is: 1 = $2^{0}$, and the horizontal coordinate of $B_{1}$ is: 1 = $2^{1}$ - 1, \\\\\ntherefore, the vertical coordinate of $B_{2}$ is: 2 = $2^{1}$, and the horizontal coordinate of $B_{2}$ is: 3 = $2^{2}$ - 1, \\\\\ntherefore, the vertical coordinate of $B_{3}$ is: 4 = $2^{2}$, and the horizontal coordinate of $B_{3}$ is: 7 = $2^{3}$ - 1, \\\\\ntherefore, the vertical coordinate of $B_{n}$ is: $2^{n-1}$, and the horizontal coordinate is: $2^{n}$ - 1, \\\\\nthus $B_{n}(2^{n}-1, 2^{n-1})$. \\\\\nTherefore, the coordinates of $B_{12}$ are: $\\boxed{(2^{12}-1, 2^{11})}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52250632.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, $AC$ and $BD$ intersect at point $O$, $AD\\parallel BC$, and $BO=DO$. Quadrilateral $ABCD$ is a parallelogram. A line through point $O$ is drawn to be perpendicular to $BD$ and intersects $BC$ at point $E$, and $DE$ is connected. If $\\angle CDE=\\angle CBD=15^\\circ$, what is the degree measure of $\\angle ABC$?\n",
    "completion": "\\textbf{Solution:} Given that $OE\\perp BD$, that is $\\angle BOE=\\angle DOE=90^\\circ$,\\\\\ntherefore, in $\\triangle BOE$ and $\\triangle DOE$, we have $\\left\\{\\begin{array}{l}BO=DO \\\\ \\angle BOE=\\angle DOE=90^\\circ \\\\ OE=OE \\end{array}\\right.$,\\\\\nhence $\\triangle BOE\\cong \\triangle DOE$ (SAS),\\\\\nthus $\\angle EBD=\\angle EDB$, which means $\\angle CBD=\\angle EDB$.\\\\\nSince $\\angle CDE=\\angle CBD=15^\\circ$,\\\\\nthen $\\angle CDE=\\angle CBD=\\angle EDB=15^\\circ$,\\\\\ntherefore $\\angle BDC=\\angle EDB+\\angle CDE=30^\\circ$.\\\\\nSince quadrilateral $ABCD$ is a parallelogram,\\\\\nthen $\\angle ABD=\\angle BDC=30^\\circ$,\\\\\nhence $\\angle ABC=\\angle ABD+\\angle CBD=\\boxed{45^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52416409_105.png"
  },
  {
    "solution": "\\boxed{110^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the quadrilateral $ABCD$ is inscribed in $\\odot O$, with $AB$ being the diameter of $\\odot O$, and $BC=CD$. If $\\angle DAB=40^\\circ$, then what is the measure of $\\angle ADC$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect $AC$,\\\\\nSince $BC=CD$,\\\\\nTherefore, $\\angle CAB=\\frac{1}{2}\\angle DAB=20^\\circ$.\\\\\nTherefore, $AB$ is the diameter of $\\odot O$,\\\\\nTherefore, $\\angle ACB=90^\\circ$.\\\\\nSince $\\angle DAB=40^\\circ$,\\\\\nTherefore, $\\angle DCB=140^\\circ$.\\\\\nTherefore, $\\angle DCA=140^\\circ-90^\\circ=50^\\circ$.\\\\\nTherefore, $\\angle ADC=180^\\circ-20^\\circ-50^\\circ=\\boxed{110^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51262453.png"
  },
  {
    "solution": "\\boxed{25^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in circle $\\odot O$, C and D are two points on circle $\\odot O$, and AB is the diameter of $\\odot O$. Given that $\\angle AOC = 130^\\circ$, what is the measure of $\\angle BDC$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle AOC=130^\\circ$ and AB is the diameter of $\\odot O$,\\\\\nit follows that $\\angle BOC=180^\\circ-\\angle AOC=50^\\circ$,\\\\\nthus $\\angle BDC= \\frac{1}{2}\\angle BOC=\\boxed{25^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51379109.png"
  },
  {
    "solution": "\\boxed{40^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AD=DC$. If $\\angle BAD=20^\\circ$, what is the measure of $\\angle C$ in degrees?\n",
    "completion": "\\textbf{Solution:}\n\nGiven:\n\\[ \\because AB=AD \\]\n\\[ \\therefore \\angle ABD=\\angle ADB=\\frac{180^\\circ-\\angle BAD}{2}=\\frac{180^\\circ-20^\\circ}{2}=80^\\circ \\]\n\\[ \\because DA=DC \\]\n\\[ \\therefore \\angle C=\\angle DAC \\]\nSince $\\angle ADB$ is an exterior angle of $\\triangle ADC$,\n\\[ \\therefore \\angle ADB=\\angle C+\\angle DAC \\]\n\\[ \\therefore 80^\\circ=\\angle C+\\angle C \\]\nSolving, we find $\\angle C=\\boxed{40^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/52845731_111.png"
  },
  {
    "solution": "\\boxed{71^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, quadrilateral ABCD is inscribed in $\\odot O$, and point M is on the extension of AD. If $\\angle AOC = 142^\\circ$, what is the degree measure of $\\angle CDM$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle AOC=142^\\circ$,\\\\\nit follows that $\\angle B=71^\\circ$,\\\\\nsince quadrilateral ABCD is inscribed in $\\odot O$,\\\\\nit implies that $\\angle B+\\angle ADC=180^\\circ$,\\\\\nthus $\\angle ADC=142^\\circ$,\\\\\nwhich means $\\angle CDM=180^\\circ-142^\\circ=71^\\circ$,\\\\\nhence, the answer is: $\\boxed{71^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52808494.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points C and D lie on a circle with diameter AB, where AB = 2, and $\\angle ACD = 30^\\circ$. What is the length of AD?\n",
    "completion": "\\textbf{Solution:} Since AB is the diameter,\\\\\nit follows that $\\angle ADB = 90^\\circ$,\\\\\nand since $\\angle B = \\angle ACD = 30^\\circ$,\\\\\nit implies that AD = $\\frac{1}{2}$AB = $\\frac{1}{2} \\times 2 =$ \\boxed{1}.",
    "image_path": "dataset/mm_math/images/MM_Math/53009949.png"
  },
  {
    "solution": "\\boxed{0 < x < 1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the inverse proportion function $y_{1}=\\frac{k}{x}$ (where $k$ is a constant and $k \\neq 0$) and the graph of the linear function $y_{2}=2x+2$ both pass through points $A(1, m)$ and $B(a, b)$. Determine the coordinates of point B and, by referring to the graph, directly write out the range of values for $x$ when $y_{1} > y_{2}$.\n",
    "completion": "\\textbf{Solution:} Since $y_{1} = \\frac{k}{x}$ intersects with $y_{2} = 2x + 2$ at points A and B,\\\\\nit follows that $\\frac{4}{x} = 2x + 2$,\\\\\nthus $x = -2$ or $x = 1$,\\\\\ntherefore $y = -2$ or $y = 4$,\\\\\nhence, the coordinates of point B are \\boxed{(-2, -2)};\\\\\nWhen $y_{1} > y_{2}$, the range of $x$ is \\boxed{x < -2} or \\boxed{0 < x < 1}.",
    "image_path": "dataset/mm_math/images/MM_Math/52948375_256.png"
  },
  {
    "solution": "\\boxed{55^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that point $O$ is the circumcenter of $\\triangle ABC$, with $\\angle A=35^\\circ$. By connecting $BO$ and $CO$, what is the measure of $\\angle OBC$?\n",
    "completion": "\\textbf{Solution:} Since point O is the circumcenter of $\\triangle ABC$ and $\\angle A=35^\\circ$,\n\n$\\therefore \\angle BOC=2\\angle A=70^\\circ$, and BO=OC,\n\n$\\therefore \\angle OBC=(180^\\circ-\\angle BOC)\\div 2$,\n\n$\\therefore \\angle OBC=\\boxed{55^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53059657.png"
  },
  {
    "solution": "\\boxed{20}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point C is on segment $AB$, and point D is the midpoint of $BC$, with $AB=30\\,cm$ and $AC=4CD$. What is the length of $AC$ in $cm$?\n",
    "completion": "\\textbf{Solution:} Given that point $D$ is the midpoint of $BC$, we have \\\\\n$BC = 2CD = 2BD$, \\\\\nBy the segment addition and subtraction property, we have \\\\\n$AB = AC + BC$, which means $4CD + 2CD = 30$, \\\\\nSolving for $CD$ gives $CD = 5$, \\\\\nTherefore, $AC = 4CD = 4 \\times 5 = \\boxed{20}\\,\\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53042295.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated in the diagram, in $\\triangle ABC$, $\\angle ACB = 90^\\circ$, $AC = BC$, and point $O$ lies inside $\\triangle ABC$. Circle $O$ passes through points $B$ and $C$, and intersects line $AB$ at point $D$. Extend line $CO$ to intersect line segment $AB$ at point $G$, and construct parallelogram $GDEC$ with $GD$ and $GC$ as adjacent sides. Line $DE$ is a tangent to circle $O$; given $DE = 7$ and $CE = 5$, what is the radius of circle $O$?\n",
    "completion": "\\textbf{Solution:} Let the radius of $\\odot O$ be $r$,\\\\\nSince quadrilateral GDEC is a parallelogram,\\\\\nit follows that CG = DE = 7, and DG = CE = 5,\\\\\nSince $\\angle GOD=90^\\circ$,\\\\\nit follows that OD$^{2}$+OG$^{2}$=DG$^{2}$, that is, $r^{2}+(7-r)^{2}=5^{2}$,\\\\\nSolving this gives: $r_{1}=3$, $r_{2}=4$,\\\\\nWhen $r=3$, OG = 4 > 3, meaning point G is outside $\\odot O$, which is against the problem statement, thus discarded,\\\\\nTherefore, $r=\\boxed{4}$, meaning the radius of $\\odot O$ is \\boxed{4}.",
    "image_path": "dataset/mm_math/images/MM_Math/51402215_132.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, there is a lamp post $AB$ beside a straight road. Under the light, Xiao Li moves straight from the bottom of the lamp post at point $B$ for $4\\,\\text{m}$ to reach point $D$. At point $D$, the length of her shadow is measured to be $DE=1\\,\\text{m}$. Xiao Li's height is $CD=1.2\\,\\text{m}$. What is the height of the lamp post $AB$?\n",
    "completion": "\\textbf{Solution:} As shown in figure 1, according to the given problem: $AB\\parallel CD$, $BE=1+4=5$ (meters),\\\\\n$\\therefore \\triangle EAB \\sim \\triangle ECD$,\\\\\n$\\therefore \\frac{AB}{CD}=\\frac{BE}{DE}$,\\\\\nthus $\\frac{AB}{1.2}=\\frac{5}{1}$,\\\\\nSolving gives: $AB=\\boxed{6}$ (meters).",
    "image_path": "dataset/mm_math/images/MM_Math/52550926_99.png"
  },
  {
    "solution": "\\boxed{y=-x^2-2x+3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given that the graph of the parabola $y=ax^2+bx+3$ intersects the x-axis at point A and point $B(1,0)$, and intersects the y-axis at point C. Connect $AC$. Let a moving point D be on segment $AC$. Draw a perpendicular to the x-axis from point D, which intersects the parabola at point E and intersects the x-axis at point F. Given $AB=4$, let the x-coordinate of point D be m. Find the equation of the parabola?\n",
    "completion": "Solution: Since point $B(1,0)$ and $AB=4$, \\\\\ntherefore $A(-3,0)$, \\\\\nsubstituting $B(1,0)$ and $A(-3,0)$ into $y=ax^2+bx+3$, \\\\\ntherefore $\\begin{cases}\na+b+3=0\\\\\n9a-3b+3=0\n\\end{cases}$, \\\\\nthus $\\begin{cases}\na=-1\\\\\nb=-2\n\\end{cases}$, \\\\\ntherefore the equation of the parabola is $\\boxed{y=-x^2-2x+3}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53107133_271.png"
  },
  {
    "solution": "\\boxed{x>3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y=-2x+8$ intersects with the graph of the function $y=\\frac{k}{x}$ (where $x>0$) at points $A(m, 6)$ and $B(n, 2)$. Line segment $AC$ is perpendicular to the $y$-axis at point $C$, and line segment $BD$ is perpendicular to the $x$-axis at point $D$. Based on the graph, directly write down the range of values of $x$ for which $-2x+8-\\frac{k}{x}<0$.\n",
    "completion": "\\textbf{Solution:} The range of values for $x$ is \\boxed{0<x<1} or \\boxed{x>3}.",
    "image_path": "dataset/mm_math/images/MM_Math/53078921_54.png"
  },
  {
    "solution": "\\boxed{14}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points $B$ and $C$ are on the line segment $AD$, and $AB:BC:CD=2:3:4$. Point $M$ is the midpoint of the line segment $AC$, and point $N$ is a point on the line segment $CD$, with $MN=9$. If point $N$ is the midpoint of line segment $CD$, what is the length of $BD$?\n",
    "completion": "\\textbf{Solution}: As shown in the figure,\\\\\nsince point M is the midpoint of line segment AC, and point N is the midpoint of line segment CD,\\\\\nit follows that $CM= \\frac{1}{2} AC$, $CN= \\frac{1}{2} CD$,\\\\\nthus $MN=CM+CN= \\frac{1}{2} (AC+CD)= \\frac{1}{2} AD=9$,\\\\\ntherefore $AD=18$,\\\\\nsince $AB:BC:CD=2:3:4$,\\\\\nit follows that $AB= \\frac{2}{9} \\times AD=4$,\\\\\ntherefore $BD=AD-AB=18-4=\\boxed{14}$;",
    "image_path": "dataset/mm_math/images/MM_Math/47162039_157.png"
  },
  {
    "solution": "\\boxed{\\left(0, -6\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=2x^2-4x-6$ intersects the x-axis at points A and B, with point A to the left of point B, and intersects the y-axis at point C. Point D is a moving point on the parabola. Find the coordinates of points A, B, and C.\n",
    "completion": "\\textbf{Solution:} Substitute $y=0$ into $y=2x^{2}-4x-6$,\nwe have $2x^{2}-4x-6=0$.\\\\\nSolving this, we find $x_{1}=-1$ and $x_{2}=3$.\\\\\nTherefore, the coordinates of point A are $\\boxed{\\left(-1, 0\\right)}$, and the coordinates of point B are $\\boxed{\\left(3, 0\\right)}$.\\\\\nSubstituting $x=0$ into $y=2x^{2}-4x-6$, we get $y=-6$.\\\\\nTherefore, the coordinates of point C are $\\boxed{\\left(0, -6\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52977885_209.png"
  },
  {
    "solution": "\\boxed{y=\\frac{1}{2}x^2-x-4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: The parabola $y=ax^2+bx-4\\ (a\\neq 0)$ intersects the x-axis at points $A(-2, 0)$ and $B(4, 0)$, and intersects the y-axis at point $C$. Connect $BC$. Point $P$ is a moving point on the parabola below the line segment $BC$ (not coinciding with points $B$ or $C$). A line parallel to the y-axis passing through point $P$ intersects $BC$ at $M$ and the x-axis at $N$. Let the x-coordinate of point $P$ be $t$. What is the analytic expression of this parabola?\n",
    "completion": "\\textbf{Solution:} Substitute $A(-2, 0)$ and $B(4, 0)$ into $y=ax^2+bx-4$ to get\\\\\n\\[\n\\therefore \\left\\{\\begin{array}{l}\n4a-2b-4=0\\\\\n16a+4b-4=0\n\\end{array}\\right. \\text{ which is } \\left\\{\\begin{array}{l}\n2a-b=2\\\\\n4a+b=1\n\\end{array}\\right. \\therefore \\left\\{\\begin{array}{l}\na=\\frac{1}{2}\\\\\nb=-1\n\\end{array}\\right. \\therefore \\text{The equation of the parabola is: } \\boxed{y=\\frac{1}{2}x^2-x-4}\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/52774825_53.png"
  },
  {
    "solution": "\\boxed{48^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given that point O is a point on line AB, place the right-angled triangle MON as shown in the figure, with the right-angled vertex at O. Draw a ray OC inside $ \\angle MON$ such that OC exactly bisects $ \\angle BOM$. If $ \\angle CON=24^\\circ$, find the degree measure of $ \\angle AOM$.\n",
    "completion": "\\textbf{Solution:} Since $\\angle MON=90^\\circ$ and $\\angle CON=24^\\circ$, \\\\\nit follows that $\\angle MOC=90^\\circ-\\angle CON=66^\\circ$. \\\\\nSince OC bisects $\\angle MOB$, \\\\\nit follows that $\\angle BOM=2\\angle MOC=132^\\circ$, \\\\\nhence $\\angle AOM=180^\\circ-\\angle BOM=\\boxed{48^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51284834_166.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $E$ is a point on the extension of side $BA$ of rectangle $ABCD$. Lines $ED$ and $EC$ are drawn, and $EC$ intersects $AD$ at point $G$. Construct $CF\\parallel ED$ intersecting $AB$ at point $F$, and $DC = DE$. Quadrilateral $CDEF$ is a rhombus; if $BC = 3$ and $CD = 5$, what is the length of $AE$?\n",
    "completion": "\\textbf{Solution:} Given that $BC=3$, $CD=5$, it follows that $AD=3$, $ED=5$. Therefore, $AE= \\sqrt{DE^{2}-AD^{2}}=\\sqrt{5^{2}-3^{2}}=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52934013_178.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $OC$ is the bisector of $\\angle AOD$, and $OE$ is the bisector of $\\angle BOD$. If $OB$ and $OD$ are perpendicular to each other, and $\\angle AOD=30^\\circ$, then what is the degree measure of $\\angle COE$?\n",
    "completion": "Solution: Since $OB$ and $OD$ are perpendicular to each other, \\\\\n$\\therefore \\angle BOD=90^\\circ$, \\\\\nSince $OE$ is the bisector of $\\angle BOD$, \\\\\n$\\therefore \\angle DOE=\\frac{1}{2}\\angle BOD=\\frac{1}{2}\\times 90^\\circ=45^\\circ$, \\\\\nSince $OC$ is the bisector of $\\angle AOD$, and $\\angle AOD=30^\\circ$, \\\\\n$\\therefore \\angle COD=\\frac{1}{2}\\angle AOD=\\frac{1}{2}\\times 30^\\circ=15^\\circ$, \\\\\n$\\therefore \\angle COE=\\angle DOE+\\angle COD=45^\\circ+15^\\circ=\\boxed{60^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51508981_133.png"
  },
  {
    "solution": "\\boxed{169}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\angle ABC=\\angle FAC=90^\\circ$, the length of $BC$ is 3m, the length of $AB$ is 4cm, and the length of $AF$ is 12cm. Calculate the area of the square $CDEF$.\n",
    "completion": "Solution: In right triangle $ABC$,\\\\\n$\\because \\angle ABC=90^\\circ, BC=3\\text{cm}, AB=4\\text{cm}$\\\\\n$\\therefore AC=\\sqrt{AB^{2}+BC^{2}}=\\sqrt{3^{2}+4^{2}}=5\\text{cm}$\\\\\nIn right triangle $ACF$, $\\because \\angle FAC=90^\\circ, AF=12\\text{cm}$\\\\\n$\\therefore CF=\\sqrt{AF^{2}+AC^{2}}=\\sqrt{12^{2}+5^{2}}=13\\text{cm}$\\\\\n$\\therefore$ The area of the square $CDEF$ $=13\\times 13=\\boxed{169}\\text{cm}^{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53136572.png"
  },
  {
    "solution": "\\boxed{\\frac{25}{6}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the line $y=4x-6$ intersects the $x$-axis at point $A$, and the line $y=-2x-2$ intersects the $x$-axis at point $B$, and these two lines intersect at point $C$. What is the area of $\\triangle ABC$?\n",
    "completion": "Solution: Since $A\\left(\\frac{3}{2}, 0\\right)$ and $B\\left(-1, 0\\right)$,\n\n$\\therefore AB=\\frac{3}{2}-(-1)=\\frac{5}{2}$,\n\n$\\therefore {S}_{\\triangle ABC}=\\frac{1}{2}AB\\cdot\\left|{y}_{c}\\right|=\\frac{1}{2}\\times \\frac{5}{2}\\times \\frac{10}{3}=\\boxed{\\frac{25}{6}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52423437_62.png"
  },
  {
    "solution": "\\boxed{(1,8)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y=-2x+10$ intersects the graph of the inverse proportion function $y=\\frac{k}{x}$ at points $A$ and $B$ (with $A$ being to the right of $B$), and the x-coordinate of point $A$ is 4. Determine the equation of the inverse proportion function and the coordinates of point $B$.\n",
    "completion": "\\textbf{Solution:} Substitute $x=4$ into $y=-2x+10$ to obtain $y=2$, \\\\\n$\\therefore A(4,2)$, \\\\\nSubstitute $A(4,2)$ into $y=\\frac{k}{x}$, to get $k=4\\times 2=8$. \\\\\n$\\therefore$ The equation of the inverse proportion function is $y=\\frac{8}{x}$, \\\\\nSolve the system of equations $\\left\\{\\begin{array}{l}y=-2x+10 \\\\ y=\\frac{8}{x}\\end{array}\\right.$, to obtain $\\left\\{\\begin{array}{l}x=1 \\\\ y=8\\end{array}\\right.$, or $\\left\\{\\begin{array}{l}x=4 \\\\ y=2\\end{array}\\right.$\\\\\n$\\therefore$ The coordinates of point $B$ are $\\boxed{(1,8)}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53080947_108.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the diameter $AB$ of $\\odot O$ is perpendicular to the chord $CD$, with the foot of the perpendicular being $E$. If $\\angle B = 60^\\circ$ and $AC = 6$, what is the length of $CD$?\n",
    "completion": "\\textbf{Solution:} Draw $AD$,\\\\\nsince the diameter $AB$ of the circle $O$ is perpendicular to the chord $CD$ at the foot $E$,\\\\\nit follows that $AE=AE$, $CE=DE$, $\\angle AED=\\angle AEC=90^\\circ$,\\\\\nthus $\\triangle AED\\cong \\triangle AEC$ (by SAS),\\\\\nhence $AD=AC$,\\\\\nsince $\\angle B=60^\\circ$, $AC=6$,\\\\\nit follows that $\\angle BCE=30^\\circ$,\\\\\nsince $\\angle ACB=90^\\circ$,\\\\\nit follows that $\\angle ACE=\\angle ACB-\\angle BCE=60^\\circ$,\\\\\nsimilarly, it can be found that: $\\angle ADE=60^\\circ$,\\\\\nhence $\\triangle ACD$ is an equilateral triangle,\\\\\ntherefore $CD=AC=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52623298.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the figure, in the equilateral triangle $ABC$, the bisector of $\\angle ABC$ intersects with the bisector of $\\angle ACB$ at point $D$. Through point $D$, line segment $EF$ is drawn parallel to $BC$, intersecting $AB$ at $E$ and $AC$ at $F$, with $EF=4$. What is the length of $BC$?\n",
    "completion": "\\textbf{Solution:} Given that in equilateral triangle ABC, EF$\\parallel$BC,\\\\\nthus $\\angle$AEF=$\\angle$ABC=$\\angle$AFE=$\\angle$ACB=$60^\\circ$, and AB=AC=BC,\\\\\nwhich implies that $\\triangle$AEF is also an equilateral triangle,\\\\\ntherefore AE=EF=FA=4,\\\\\nresulting in EB=FC,\\\\\nsince the bisector of $\\angle$ABC intersects with the bisector of $\\angle$ACB at point D, and EF$\\parallel$BC,\\\\\nit follows that $\\angle$EBD=$\\angle$DBC=$\\angle$EDB=$\\angle$FCD=$\\angle$FDC=$\\angle$DCB=$30^\\circ$,\\\\\nhence EB=ED=DF=FC,\\\\\ngiven EF=4,\\\\\nit leads to EB=ED=DF=FC=2,\\\\\ntherefore AB=AC=BC=AE+EB=\\boxed{6},",
    "image_path": "dataset/mm_math/images/MM_Math/51434965.png"
  },
  {
    "solution": "\\boxed{115^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ABC=80^\\circ$, $\\angle ACB=50^\\circ$. $BP$ bisects $\\angle ABC$, $CP$ bisects $\\angle ACB$. Find the measure of $\\angle BPC$ in degrees?\n",
    "completion": "\\textbf{Solution:} Given that $BP$ bisects $\\angle ABC$ and $CP$ bisects $\\angle ACB$,\\\\\nit follows that $\\angle PBC= \\frac{1}{2}\\angle ABC$ and $\\angle PCB= \\frac{1}{2}\\angle ACB$,\\\\\nhence $\\angle PBC+\\angle PCB= \\frac{1}{2}(\\angle ABC+\\angle ACB)=65^\\circ$,\\\\\ntherefore, $\\angle BPC=180^\\circ-(\\angle PBC+\\angle PCB)=\\boxed{115^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/54368654_149.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given the diagram, we know that the length of segment $AB=26$ and $BC=18$. Point $M$ is the midpoint of $AC$. A point $N$ is chosen on $CB$ such that the ratio of $CN: NB=1:2$. What is the length of segment $MN$?\n",
    "completion": "\\textbf{Solution:} Since point $M$ is the midpoint of line segment $AC$,\\\\\nit follows that $MC= \\frac{1}{2}AC$,\\\\\nSince $AC=8$,\\\\\nit follows that $MC=4$. Additionally, since $BC=18$ and $CN:NB=1:2$,\\\\\nit follows that $CN= \\frac{1}{3}BC=6$,\\\\\ntherefore, $MN=MC+CN=6+4=\\boxed{10}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51251319_182.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the medians $AD$ and $BE$ of $\\triangle ABC$ intersect each other at point $F$. Through point $E$, draw $EG\\parallel AD$ where it intersects $BC$ at point $G$. What is the value of the ratio $EG\\colon AF$?\n",
    "completion": "\\textbf{Solution:} Connect DE.\\\\\nAD, BE are the medians of the triangle\\\\\n$\\therefore$DE$\\parallel$AB, DE=$\\frac{1}{2}$AB\\\\\n$\\therefore$$\\triangle$DEF$\\sim$$\\triangle$ABF\\\\\n$\\therefore$$\\frac{DF}{AF}=\\frac{DE}{AB}=\\frac{1}{2}$\\\\\n$\\therefore$AF=$\\frac{2}{3}AD$\\\\\n$\\because$ED$\\parallel$AD\\\\\n$\\therefore$$\\triangle$EGC$\\sim$$\\triangle$ADC\\\\\n$\\therefore$$\\frac{EG}{AD}=\\frac{CE}{AC}=\\frac{1}{2}$\\\\\n$\\therefore$EG=$\\frac{1}{2}AD$\\\\\n$\\therefore$EG$\\colon$AF=$\\boxed{\\frac{3}{4}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51445952.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{1}{2}x+2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the graph of the linear function $y=kx+b$ intersects the $x$ axis and the $y$ axis at points $A$ and $B$, respectively. It intersects with the graph of the inverse proportional function $y=\\frac{m}{x}$ at points $C$ and $D$, with $DE\\perp x$ axis at point $E$. Given that the coordinates of point $C$ are $(8, -2)$ and $DE=4$, find the expressions for the inverse proportional function and the linear function?\n",
    "completion": "Solution: Since point C (8, -2) lies on the graph of the inverse proportion function $y=\\frac{m}{x}$,\n\nit follows that $m=-16$,\n\nthus, the equation of the inverse proportion function is $y=-\\frac{16}{x}$.\n\nSince point D lies on the graph of $y=-\\frac{16}{x}$, and DE=4,\n\nit follows that $y=4$,\n\nthus, point D is (-4, 4).\n\nAlso, since C, D lie on the graph of $y=kx+b$,\n\nit follows that $\\left\\{\\begin{array}{l}8k+b=-2 \\\\ -4k+b=4\\end{array}\\right.$,\n\nsolving this gives $\\left\\{\\begin{array}{l}k=-\\frac{1}{2} \\\\ b=2\\end{array}\\right.$,\n\ntherefore, the equation of the linear function is \\boxed{y=-\\frac{1}{2}x+2}.",
    "image_path": "dataset/mm_math/images/MM_Math/52551855_127.png"
  },
  {
    "solution": "\\boxed{-8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $O$ is the origin of coordinates, point $A$ lies on the graph of the function $y=\\frac{k}{x}$ ($x<0$), line segment $AB$ is perpendicular to the $x$-axis at point $B$, and the area of $\\triangle AOB$ is 4. What is the value of $k$?\n",
    "completion": "\\textbf{Solution:} Since $AB \\perp x$-axis, \\\\\ntherefore $S_{\\triangle ABO} = \\frac{1}{2}|k|$, which means $\\frac{1}{2}|k| = 4$, \\\\\nSince $k < 0$, \\\\\ntherefore $k = \\boxed{-8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52977189.png"
  },
  {
    "solution": "\\boxed{x<-1 or x>4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the quadratic function $y_{1}=-x^{2}+bx+c$ intersects with the graph of the linear function $y_{2}=kx+2$ at points $A(-1, 3)$ and $B(4, m)$. In order for $y_{1}<y_{2}$, what is the range of values for $x$?\n",
    "completion": "Solution: Since $y_1 < y_2$, \\\\\nit follows that the range of $x$ for which the line is above the parabola is, \\\\\naccording to the graph, when $x < -1$ or $x > 4$, \\\\\ntherefore $x < -1$ or $x > 4$, \\\\\nhence, the answer is:$\\boxed{ x<-1 or x>4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53009879.png"
  },
  {
    "solution": "\\boxed{156^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a regular triangle, a square, and a regular pentagon are placed side by side with $AB$ as their common edge. What is the measure of $\\angle CDE$?\n",
    "completion": "\\textbf{Solution:}\n\nGiven that the sum of internal angles of a pentagon is $(5-2) \\times 180^\\circ = 540^\\circ$,\n\nit follows that $\\angle BAC = \\frac{1}{5} \\times 540^\\circ = 108^\\circ$,\n\nsince the internal angle of a triangle ABE ($\\angle BAE$) is $60^\\circ$ and the internal angle of a square ($\\angle BAD$) is $90^\\circ$,\n\nthus $\\angle DAC = \\angle BAC - \\angle BAD = 108^\\circ - 90^\\circ = 18^\\circ$, and $\\angle DAE = \\angle BAD - \\angle BAE = 90^\\circ - 60^\\circ = 30^\\circ$,\n\ngiven AD = AC,\n\nit implies $\\angle ADC = \\angle ACD = \\frac{180^\\circ - \\angle DAC}{2} = \\frac{180^\\circ - 18^\\circ}{2} = 81^\\circ$,\n\nalso, given EA = AD,\n\nit implies $\\angle ADE = \\angle AED = \\frac{180^\\circ - \\angle DAE}{2} = \\frac{180^\\circ - 30^\\circ}{2} = 75^\\circ$,\n\nthus $\\angle CDE = \\angle ADC + \\angle ADE = 81^\\circ + 75^\\circ = \\boxed{156^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52836684.png"
  },
  {
    "solution": "\\boxed{65^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, PA and PB are tangent to the circle $O$ at points A and B, respectively. C is a moving point on the major arc AB. If $\\angle P = 50^\\circ$, then what is the degree measure of $\\angle ACB$?\n",
    "completion": "\\textbf{Solution:} Draw $OA, OB$ as shown in the diagram,\\\\\n$\\because$ PA and PB are tangent to $\\odot O$ respectively\\\\\n$\\therefore \\angle OAP=\\angle OBP=90^\\circ$\\\\\n$\\therefore \\angle AOB=360^\\circ-\\angle OAP-\\angle OBP-\\angle P=130^\\circ$\\\\\n$\\because \\overset{\\frown}{AB}=\\overset{\\frown}{AB}$\\\\\n$\\therefore \\angle ACB=\\frac{1}{2}\\angle AOB=\\boxed{65^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/51350894.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AD\\perp BC$ at point $D$, $E$ is a point on $AD$, and $AE: ED = 7: 5$. $CE$ is extended to intersect side $AB$ at point $F$, $AC = 13$, $BC = 8$, $\\cos\\angle ACB = \\frac{5}{13}$. Find the value of $\\tan\\angle DCE$?\n",
    "completion": "\\textbf{Solution:} Since $AD \\perp BC$, it follows that $\\angle ADC=90^\\circ$. In $\\triangle ADC$, given that $AC=13$ and $\\cos\\angle ACB=\\frac{5}{13}$, it can be deduced that $CD=5$ and therefore $AD=\\sqrt{AC^{2}-CD^{2}}=\\sqrt{13^{2}-5^{2}}=12$. Given that the ratio $AE:ED=7:5$, it follows that $ED=12\\times \\frac{5}{7+5}=5$. Consequently, $\\tan\\angle DCE=\\frac{ED}{CD}=\\boxed{1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55595013_155.png"
  },
  {
    "solution": "\\boxed{\\frac{8}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $BD$ is the bisector of $\\angle ABC$, $DE \\perp AB$ at point $E$, $DF \\perp BC$ at point $F$, $AB=12$, $BC=15$, and the area of $\\triangle ABC$ is $36$. What is the length of $DE$?\n",
    "completion": "\\textbf{Solution:} Given that $BD$ is the angle bisector of $\\angle ABC$, and that $DE\\perp AB$, $DF\\perp BC$\\\\\nTherefore, $DE=DF$\\\\\nSince the area of $\\triangle ABC = S_{\\triangle ABD} + S_{\\triangle BCD} = \\frac{1}{2}BC\\cdot DF + \\frac{1}{2}AB\\cdot DE = 36$, and $AB=12$, $BC=15$\\\\\nTherefore, $\\frac{1}{2} \\times 12 \\cdot DE + \\frac{1}{2} \\times 15 \\cdot DF = 36$\\\\\nTherefore, $6DE + \\frac{15}{2}DF = 36$\\\\\nFurthermore, since $DE=DF$\\\\\nTherefore, $6DE + \\frac{15}{2}DE = 36$\\\\\nTherefore, $DE = \\boxed{\\frac{8}{3}}$",
    "image_path": "dataset/mm_math/images/MM_Math/52603600.png"
  },
  {
    "solution": "\\boxed{40^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Understanding Calculation: As shown in Figure (1), $ \\angle AOB=80^\\circ$, $ \\angle AOC=40^\\circ$. Ray $ OM$ bisects $ \\angle BOC$, $ ON$ bisects $ \\angle AOC$, find the degree measure of $ \\angle MON$?\n",
    "completion": "\\textbf{Solution:} Given $\\because \\angle BOC=\\angle AOB+\\angle AOC=80^\\circ+40^\\circ=120^\\circ$,\\\\\nray $OM$ bisects $\\angle BOC$,\\\\\n$\\therefore \\angle COM=\\frac{1}{2}\\angle BOC=\\frac{1}{2}\\times 120^\\circ=60^\\circ$,\\\\\n$\\because ON$ bisects $\\angle AOC$,\\\\\n$\\therefore \\angle CON=\\frac{1}{2}\\angle AOC=\\frac{1}{2}\\times 40^\\circ=20^\\circ$,\\\\\n$\\therefore \\angle MON=\\angle COM-\\angle CON=60^\\circ-20^\\circ=\\boxed{40^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51407373_74.png"
  },
  {
    "solution": "\\boxed{13.5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the figure, within $\\triangle ABC$, $CD\\perp AB$ at point $D$, and $E$, $F$ are the midpoints of $AC$, $BC$, respectively. Given $AB=10$, $BC=8$, and $DE=4.5$, what is the perimeter of $\\triangle DEF$?\n",
    "completion": "\\textbf{Solution:} Since $E$ and $F$ are the midpoints of $AC$ and $BC$ respectively,\\\\\ntherefore $EF=\\frac{1}{2}AB=5$,\\\\\nsince $CD\\perp AB$,\\\\\ntherefore $\\angle CDB=90^\\circ$,\\\\\ntherefore $\\triangle CDB$ is a right-angled triangle,\\\\\n$DF=\\frac{1}{2}BC=4$,\\\\\ntherefore the perimeter of $\\triangle DEF$ is $DF+EF+DE=4+5+4.5=\\boxed{13.5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53333661.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the linear function $y=x+b$ and the inverse proportion function $y=\\frac{k}{x}$ (where $k\\neq 0$) intersect at points $A$ and $B$. The coordinates of point $A$ are $(1,3)$. The linear function $y=x+b$ intersects the $x$-axis at point $C$. Connect $OA$ and $OB$. What are the coordinates of point $B$ and the area of $\\triangle AOB$?\n",
    "completion": "\\textbf{Solution:} From the system of equations\n\\[\n\\begin{cases}\ny=x+2\\\\\ny=\\frac{3}{x}\n\\end{cases}\n\\]\nwe find:\n\\[\n\\begin{cases}\nx=-3\\\\\ny=-1\n\\end{cases}\n\\]\nor\n\\[\n\\begin{cases}\nx=1\\\\\ny=3\n\\end{cases}\n\\]\n$\\therefore \\boxed{B(-3, -1)}$.\n\nLet $y=0$ and substitute into $y=x+2$ to get: $x=-2$,\n\n$\\therefore C(-2, 0)$,\n\n$\\therefore {S}_{\\triangle AOB}={S}_{\\triangle AOC}+{S}_{\\triangle BOC}=\\frac{1}{2}\\times 2\\times 3+\\frac{1}{2}\\times 2\\times 1=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51322344_128.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of the circle $O$. The chord $CD \\perp AB$ at point $E$, $OC = 5 \\text{cm}$, $CD = 6 \\text{cm}$, then what is the value of $BE$?\n",
    "completion": "\\textbf{Solution:} Given that AB is the diameter and CD$\\perp$AB,\\\\\nit follows that CE=$\\frac{1}{2}$CD=$\\frac{1}{2} \\times 6=3$, and $\\angle$OEC=$90^\\circ$,\\\\\nthus, OE=$\\sqrt{OC^{2}-CE^{2}}=\\sqrt{5^{2}-3^{2}}=4$,\\\\\nwhich means BE=OB−OE=5−4=\\boxed{1}cm.",
    "image_path": "dataset/mm_math/images/MM_Math/53059930.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ABCD$, $AE \\perp BD$ at $E$, $\\angle EAC = 30^\\circ$, and $AE = 3$. What is the length of $AC$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, in the right-angled $\\triangle AOE$,\\\\\n$\\cos\\angle EAO = \\frac{AE}{OA}$,\\\\\n$\\therefore OA = \\frac{AE}{\\cos\\angle EAO} = \\frac{3}{\\frac{\\sqrt{3}}{2}} = 2\\sqrt{3}$.\\\\\nFurthermore, $\\because$ quadrilateral ABCD is a parallelogram,\\\\\n$\\therefore AC = 2OA = 4\\sqrt{3}$.\\\\\nTherefore, the length of $AC$ is $\\boxed{4\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51174827.png"
  },
  {
    "solution": "\\boxed{(1,4)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=-x^{2}+2x+c$ intersects the $x$-axis at points $A(-1, 0)$ and $B(3, 0)$, and intersects the $y$-axis at point $C$. Point $D$ is the vertex of the parabola. What are the coordinates of point $D$?\n",
    "completion": "\\textbf{Solution:} The coordinates of point $D$ are $\\boxed{(1,4)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51445487_36.png"
  },
  {
    "solution": "\\boxed{(-3, -1)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the coordinates of points A and B are $(-2,0)$ and $(0,3)$, respectively. Segment AB is rotated counterclockwise by $90^\\circ$ about point B to obtain segment BC. Through point C, line CD is drawn perpendicular to OB, with D being the foot of the perpendicular. The graph of the inverse proportion function $y=\\frac{k}{x}$ passes through point C. Point P is on the graph of the inverse proportion function $y=\\frac{k}{x}$. When the area of $\\triangle PCD$ is 3, find the coordinates of point P.\n",
    "completion": "\\textbf{Solution:} Let $P\\left( \\frac{3}{m}, m \\right)$,\\\\\nsince $CD \\perp y$-axis, and $CD = 3$,\\\\\nfrom the area of $\\triangle PCD$ being 3, we have: $\\frac{1}{2}CD \\cdot |m-1|=3$,\\\\\nthus, $\\frac{1}{2} \\times 3 |m-1| = 3$,\\\\\nthus, $m-1 = \\pm 2$,\\\\\nthus, $m=3$ or $m=-1$,\\\\\nwhen $m=3$, $\\frac{3}{m}=1$; when $m=-1$, $\\frac{3}{m}=-3$,\\\\\nthus, the coordinates of point P are $\\boxed{(1, 3)}$ or $\\boxed{(-3, -1)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53147515_69.png"
  },
  {
    "solution": "\\boxed{\\text{pentagonal prism}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, what will be the shape after the planar figure is pulled into a three-dimensional form?\n",
    "completion": "\\textbf{Solution:} From the development diagram, we know: the two bases are regular pentagons, $\\therefore$ after the plane figure is extruded into a three-dimensional shape, it will form a \\boxed{\\text{pentagonal prism}}.",
    "image_path": "dataset/mm_math/images/MM_Math/55138467.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the grid is composed of small squares with a side length of 1, arranged in a $4\\times 4$ pattern. What is the value of $\\tan\\angle BAC$?\n",
    "completion": "Solution: Connect BC,\\\\\nBy Pythagoras' theorem, it is known that:\\\\\n$AC = \\sqrt{1^2 + 2^2} = \\sqrt{5}$,\\\\\n$BC = \\sqrt{2^2 + 4^2} = 2\\sqrt{5}$,\\\\\n$AB = \\sqrt{3^2 + 4^2} = 5$,\\\\\nSince $5^2 = (\\sqrt{5})^2 + (2\\sqrt{5})^2$,\\\\\nHence $AB^2 = AC^2 + BC^2$,\\\\\nTherefore, $\\triangle ABC$ is a right-angled triangle,\\\\\nTherefore, $\\tan\\angle BAC = \\frac{BC}{AC} = \\frac{2\\sqrt{5}}{\\sqrt{5}} = \\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52601992.png"
  },
  {
    "solution": "\\boxed{\\left(\\frac{40}{3}, -\\frac{10}{3}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system xOy, point A lies on the positive half of the y-axis, and point B lies on the positive half of the x-axis, with $OA=OB=10$. If point P is a moving point on line AB, and $S_{\\triangle OBP} = \\frac{1}{4} S_{\\triangle OAP}$, find the coordinates of point P.\n",
    "completion": "Solution: Given equation (1) and the conditions, we can assume point $P\\left(m, -m+10\\right)$. Thus, for $\\triangle OBP$ with OB as the base, the absolute value of the y-coordinate of point P is its height, and for $\\triangle OAP$ with OA as the base, the absolute value of the x-coordinate of point P is its height.\\\\\nSince $S_{\\triangle OBP} = \\frac{1}{4} S_{\\triangle OAP}$,\\\\\nit follows that $\\frac{1}{2}\\cdot OB\\cdot \\left|-m+10\\right|=\\frac{1}{4}\\cdot \\frac{1}{2}\\cdot OA\\cdot \\left|m\\right|$,\\\\\nGiven $OA=OB=10$,\\\\\nit implies $\\left|-m+10\\right|=\\frac{1}{4}\\left|m\\right|$,\\\\\nSolving this gives: $m=8$ or $m=\\frac{40}{3}$,\\\\\nTherefore, the coordinates of point P are $\\boxed{\\left(8, 2\\right)}$ or $\\boxed{\\left(\\frac{40}{3}, -\\frac{10}{3}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50493321_220.png"
  },
  {
    "solution": "\\boxed{BN}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Point C lies on segment AB, and point M is the midpoint of AC, where $AM=1$ and $BC=4$. If point N lies on ray AB with $AN=7$, please complete the figure and directly state the length of $BN$.\n",
    "completion": "\\textbf{Solution:} Given the information in the problem, it is not sufficient to complete the figure, thus we cannot directly determine the length of $\\boxed{BN}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51212253_188.png"
  },
  {
    "solution": "\\boxed{\\sqrt{19}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, square $ABCD$ has a side length of $3$, and points $E$ and $F$ are two moving points on the diagonal $AC$ (with point $E$ to the left of point $F$), and $EF=1$. What is the minimum value of $DE+BF$?\n",
    "completion": "\\textbf{Solution:} Let us solve for when $DE=BF$. The minimum value of $DE+BF$ is obtained when $DE=BF$. Connect $BD$ and $AC$, intersecting at point $O$, and connect $BE$,\\\\\n$\\because$ $ABCD$ is a square,\\\\\n$\\therefore$ $AD=AB$, $\\angle DAE=\\angle BAE=45^\\circ$, $AC\\perp BD$, $BD=\\sqrt{3^2+3^2}=3\\sqrt{2}$,\\\\\nIn $\\triangle ADE$ and $\\triangle ABE$ we have\\\\\n$\\left\\{\\begin{array}{l}\nAD=AB \\\\\n\\angle DAE=\\angle BAE=45^\\circ \\\\\nAE=AE,\n\\end{array}\\right.$\\\\\n$\\therefore$ $\\triangle ADE \\cong \\triangle ABE$,\\\\\n$\\therefore$ $DE=BE=BF$,\\\\\n$BO\\perp EF$,\\\\\n$OE=OF=\\frac{1}{2}EF=\\frac{1}{2}$,\\\\\n$OB=\\frac{1}{2}BD=\\frac{3\\sqrt{2}}{2}$,\\\\\n$\\therefore DE=BE=BF=\\sqrt{\\left(\\frac{3\\sqrt{2}}{2}\\right)^2+\\left(\\frac{1}{2}\\right)^2}=\\frac{\\sqrt{19}}{2}.$\\\\\n$DE+BF =\\frac{\\sqrt{19}}{2}+\\frac{\\sqrt{19}}{2}=\\boxed{\\sqrt{19}}.$",
    "image_path": "dataset/mm_math/images/MM_Math/52225773.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated, to calculate the height of the school's flagpole, Xiao Ming pulled the flag-raising rope to the bottom of the flagpole and tied a knot at the point where the rope was as long as the flagpole. Then, he pulled the bottom of the rope to a point on the ground 5 meters away from the bottom of the flagpole, and found that the distance from the bottom end of the rope to the knot was about 1 meter. What is the height of the flagpole?\n",
    "completion": "\\textbf{Solution:} Let the height of the flagpole be $x$ m, thus $AC = x$ m, $AB= (x+1)$ m, $BC=5$ m,\\\\\nIn $\\triangle ABC$, by the Pythagorean theorem we have $AC^2+BC^2=AB^2$\\\\\n$\\therefore x^2+5^2=(x+1)^2$\\\\\nSolving yields: $x=\\boxed{12}$",
    "image_path": "dataset/mm_math/images/MM_Math/47324133.png"
  },
  {
    "solution": "\\boxed{PH=2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ and $\\triangle ADE$ are equilateral triangles, with the segments BE, BD, CD, EC connected. As in figure 3, the side $BC=6$ of $\\triangle ABC$, and through point D, $EC=2\\sqrt{19}$. Let N be a moving point on line AB, connect DN, and fold $\\triangle DBN$ along DN to get $\\triangle DHN$. When AH is maximized, a perpendicular line is drawn from H through point AH. Let M be a moving point on the perpendicular line, connect MA, and rotate the segment MA $60^\\circ$ counterclockwise around point M to obtain the segment MP. Connect PH, and directly write out the minimum value of PH.\n",
    "completion": "textbf{Solution}: According to the problem, $AC=AB=6$, $\\triangle ABC$ and $\\triangle ADE$ are equilateral triangles,\\\\\n$\\therefore AC=AB, AE=AD$, $\\angle ADE=60^\\circ$, $\\angle EAD=\\angle CAB=60^\\circ$, $\\angle ACB=60^\\circ$\\\\\n$\\because \\angle EAD=\\angle CAB=60^\\circ$\\\\\n$\\therefore \\angle EAD+\\angle DAB=\\angle DAB+\\angle AC$\\\\\nThat is, $\\angle CAD=\\angle BAE$\\\\\n$\\therefore \\triangle EAB \\cong \\triangle DAC$\\\\\n$\\because \\angle ACB=60^\\circ$\\\\\n$\\therefore EB=DC, \\angle ABE=\\angle ACD$\\\\\nLet $BD=2x$, extend $BE$ to $K$, such that $EK=BD=2x$, connect $KA, CK$, and draw $ES\\perp KC$ at point $S$ with $AB$ intersecting $KC$ at point $O$,\\\\\n$\\therefore KB=BC=AB, \\angle ABK=\\angle ACB=60^\\circ$\\\\\n$\\therefore \\triangle ABK$ is an equilateral triangle\\\\\n$\\therefore AK=KB=AB=BC=AC$, $ \\angle AKB=60^\\circ$\\\\\n$\\therefore$ Quadrilateral $AKBC$ is a rhombus\\\\\n$\\therefore \\angle SKE=\\frac{1}{2}\\angle AKB=30^\\circ$\\\\\n$\\therefore SE=\\frac{1}{2}KE=\\frac{1}{2}BD=x$, $KS=\\sqrt{3}x$\\\\\n$\\because AC=AB=6$\\\\\n$\\therefore KC=2OC=2AC\\times \\sin\\angle BAC=6\\sqrt{3}$\\\\\n$\\therefore CS=6\\sqrt{3}-\\sqrt{3}x$\\\\\nIn $\\triangle CSE$, $CS=6\\sqrt{3}-\\sqrt{3}x$, $SE=x$, $CE=2\\sqrt{19}$\\\\\n$\\therefore SC^2+SE^2=EC^2$\\\\\nThat is, ${\\left(6\\sqrt{3}-\\sqrt{3}x\\right)}^2+{x}^2={\\left(2\\sqrt{19}\\right)}^2$\\\\\nSolve to get ${x}_1=1, {x}_2=-1$ (discard)\\\\\n$\\therefore BD=2$\\\\\n$\\because$ N is a moving point on line AB, connect DN, fold $\\triangle DBN$ along DN to obtain $\\triangle DHN$,\\\\\n$\\therefore DB=DH=2$\\\\\n$\\because AH\\leq AD+DH$\\\\\n$\\therefore$ When points $A, D, H$ are collinear, and $D$ is on segment $AH$, $AH$ is maximum, the maximum value is $AH$,\\\\\nDraw $AJ\\perp BC$ through point $A$,\\\\\n$\\because \\triangle ABC$ is an equilateral triangle, $AJ\\perp BC$\\\\\n$\\therefore BJ=\\frac{1}{2}BC=3$, $AJ=AB\\cdot \\sin60^\\circ=3\\sqrt{3}$,\\\\\n$\\because BD=2$,\\\\\n$\\therefore JD=1$,\\\\\nIn $\\triangle ADJ$, $AD=\\sqrt{AJ^2+JD^2}=\\sqrt{{(3\\sqrt{3})}^2+1}=2\\sqrt{7}$\\\\\n$\\therefore AH=AD+DH=2+2\\sqrt{7}$\\\\\nConnect $AP$, extend $AE$ to $T$, such that $ET=DH=2$, connect $TP$, let the perpendicular from $AH$ to line $TP$ at $HM$ intersect at point $Q$, connect $AQ$,\\\\\n$\\because MA=MP, \\angle AMP=60^\\circ$\\\\\n$\\therefore \\triangle AMP$ is an equilateral triangle\\\\\n$\\therefore AM=AP$, $ \\angle PAM=60^\\circ$\\\\\n$\\because \\angle TAP=\\angle TAH$\\\\\n$\\therefore \\boxed{PH=2}$",
    "image_path": "dataset/mm_math/images/MM_Math/51918275_107.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within the circle $\\odot O$, $OA=2$, and $\\angle ACB=30^\\circ$, what is the length of the chord $AB$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle AOB=2\\angle ACB$ and $\\angle ACB=30^\\circ$,\\\\\nit follows that $\\angle AOB=60^\\circ$,\\\\\nSince $OA=OB$,\\\\\nit follows that $\\triangle OAB$ is an equilateral triangle,\\\\\nTherefore, $AB=OA= \\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52755077.png"
  },
  {
    "solution": "\\boxed{\\frac{6\\sqrt{5}}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, rectangle $ABCD$ is folded along diagonal $BD$, and point $C$ lands on point $E$. Connect $CE$ and intersect side $AD$ at point $F$. If $DF=1$ and $BC=4$, what is the length of $AE$?\n",
    "completion": "\\textbf{Solution:} Based on the property of folding, we have $CE\\perp BD$,\\\\\n$\\therefore \\angle ADB + \\angle DFC = 90^\\circ$\\\\\n$\\because$ Quadrilateral $ABCD$ is a rectangle,\\\\\n$\\therefore \\angle ADC = \\angle BCD = 90^\\circ$,\\\\\n$\\therefore \\angle ADB + \\angle CDB = 90^\\circ$,\\\\\n$\\therefore \\angle CDB = \\angle DFC$,\\\\\n$\\because \\angle CDF = \\angle BCD = 90^\\circ$,\\\\\n$\\therefore \\triangle CDF \\sim \\triangle BCD$,\\\\\n$\\therefore \\frac{FD}{CD} = \\frac{CD}{BC}$, i.e., $\\frac{1}{CD} = \\frac{CD}{4}$\\\\\n$\\therefore CD = 2$ (discarding the negative value)\\\\\n$\\because$ Quadrilateral $ABCD$ is a rectangle, $\\triangle BED$ is obtained by folding $\\triangle BCD$,\\\\\n$\\therefore BC = BE = AD$, $\\angle CBD = \\angle DBE$, $AD \\parallel BC$,\\\\\n$\\therefore \\angle ADB = \\angle DBC$,\\\\\n$\\therefore \\angle ADB = \\angle DBE$,\\\\\n$\\therefore \\angle EAD = \\angle ADB = \\angle DBC$,\\\\\n$\\therefore AE \\parallel BD$,\\\\\n$\\therefore CE \\perp AE$,\\\\\nIn $\\triangle DCB$, $\\tan \\angle DBC = \\frac{CD}{BC} = \\frac{2}{4} = \\frac{1}{2}$,\\\\\n$\\because \\angle EAD = \\angle DBC$\\\\\n$\\therefore \\tan \\angle EAD = \\frac{1}{2}$\\\\\nIn $\\triangle AEF$, $EF = \\frac{1}{2}AE$\\\\\nBy the Pythagorean theorem, $AE^2 = AF^2 - EF^2$\\\\\ni.e., $AE^2 = (AD - FD)^2 - \\left(\\frac{1}{2}AE\\right)^2$\\\\\n$\\therefore AE^2 = (4 - 1)^2 - \\frac{1}{4}AE^2$\\\\\nSolving yields: $AE = \\boxed{\\frac{6\\sqrt{5}}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51578585.png"
  },
  {
    "solution": "\\boxed{22.5^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, extend the side $BA$ of the square $ABCD$ to point $E$ such that $AE=BD$, then what is the measure of $\\angle E$?\n",
    "completion": "\\textbf{Solution:} Draw line AC, \\\\\n$\\because$ Quadrilateral ABCD is a square, \\\\\n$\\therefore$ AC=BD, and $\\angle$CAB=$45^\\circ$, \\\\\nFurthermore, $\\because$BD=AE, \\\\\n$\\therefore$ AE=CA, \\\\\n$\\therefore$ $\\angle$E=$\\angle$ACE, \\\\\n$\\because$ $\\angle$CAB=$\\angle$ACE+$\\angle$E=2$\\angle$E=$45^\\circ$, \\\\\n$\\therefore$ $\\angle$E=$\\boxed{22.5^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52387414.png"
  },
  {
    "solution": "\\boxed{27^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, it is known that line AD $\\parallel$ BC, and BE bisects $\\angle ABC$ and intersects line DA at point E. If $\\angle DAB=54^\\circ$, what is the value of $\\angle E$?\n",
    "completion": "\\textbf{Solution}: Since AD$\\parallel$BC,\\\\\nit follows that $\\angle$ABC=$\\angle$DAB=54$^\\circ$, $\\angle$EBC=$\\angle$E,\\\\\nsince BE bisects $\\angle$ABC,\\\\\nit follows that $\\angle$EBC= $\\frac{1}{2}$ $\\angle$ABC=27$^\\circ$,\\\\\nhence, $\\angle$E=\\boxed{27^\\circ}.",
    "image_path": "dataset/mm_math/images/MM_Math/51434644.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle BAC=90^\\circ$, $AB=8$, $AC=6$, and point D is the midpoint of $BC$. What is the length of $AD$?\n",
    "completion": "\\textbf{Solution}: Given that in $\\triangle ABC$, $\\angle BAC=90^\\circ$, AB=8, AC=6, \\\\\ntherefore, $BC=\\sqrt{AB^2+AC^2}=\\sqrt{8^2+6^2}=10$, \\\\\nsince point D is the midpoint of BC, \\\\\nthus, AD is the median of $\\triangle ABC$, \\\\\ntherefore, AD=$\\frac{1}{2}BC=\\frac{1}{2}\\times 10=\\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52807790.png"
  },
  {
    "solution": "\\boxed{9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $A$, $B$, $C$, and $D$ are four points in sequence on a straight line, $M$ and $N$ are the midpoints of $AB$ and $CD$ respectively, and $MN=6\\,\\text{cm}$, $BC=3\\,\\text{cm}$. What is the length of $AD$?\n",
    "completion": "\\textbf{Solution:} Let $M$ and $N$ be the midpoints of $AB$ and $CD$ respectively.\\\\\nTherefore, $AB = 2MB$ and $CD = 2CN$.\\\\\nSince $MN = 6$ and $BC = 3$,\\\\\nit follows that $MB + CN = MN - BC = 6 - 3 = 3$,\\\\\nand therefore $AB + CD = 2MB + 2CN = 2(MB + CN) = 2 \\times 3 = 6$,\\\\\nhence $AD = AB + CD + BC = 6 + 3 = \\boxed{9}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55393326.png"
  },
  {
    "solution": "\\boxed{\\left(-3，2\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=-x+1$ intersects with the graph of the inverse proportion function $y=\\frac{k}{x}$. One of the intersection points has an abscissa of $-2$. After vertically translating the line $y=-x+1$ downwards by 2 units, find the coordinates of the intersection points between the translated line and the graph of the inverse proportion function $y=\\frac{k}{x}$.\n",
    "completion": "\\textbf{Solution:} The graph of the linear function $y=-x+1$ is translated downward by 2 units to obtain $y=-x-1$,\\\\\nby solving the system $\\left\\{\\begin{array}{l}y=-\\frac{6}{x}\\\\ y=-x-1\\end{array}\\right.$, we obtain: $\\left\\{\\begin{array}{l}x=2 \\\\ y=-3\\end{array}\\right.$ or $\\left\\{\\begin{array}{l}x=-3 \\\\ y=2\\end{array}\\right.$,\\\\\n$\\therefore$ the coordinates of the intersections between the translated line and the graph of the hyperbolic function $y=\\frac{k}{x}$ are $\\boxed{\\left(2，-3\\right)}$ and $\\boxed{\\left(-3，2\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52091246_98.png"
  },
  {
    "solution": "\\boxed{63^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABF$, $\\angle AFB=90^\\circ$. The vertex $E$ of $\\triangle CDE$ is on the side $BF$ of $\\triangle ABF$, point $C$ is on the extension of $BF$, $\\angle C=\\angle B$, and $CD=CE$. If $\\angle A=36^\\circ$, what is the degree measure of $\\angle D$?\n",
    "completion": "\\textbf{Solution:} In right triangle $ABF$, $\\angle B=90^\\circ-\\angle A=90^\\circ-36^\\circ=54^\\circ$,\\\\\n$\\therefore \\angle C=\\angle B=54^\\circ$.\\\\\nBecause $CD=CE$,\\\\\n$\\therefore \\angle D=\\angle DEC$,\\\\\nIn triangle $CDE$,\\\\\n$\\angle C+\\angle D+\\angle DEC=180^\\circ$,\\\\\n$\\therefore \\angle D=\\boxed{63^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52457273.png"
  },
  {
    "solution": "\\boxed{\\frac{32}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the parallelogram $ABCD$, $CE\\perp AD$ at point $E$. Extend $DA$ to point $F$ such that $AF=DE$. Connect $BF$ and $CF$. Quadrilateral $BCEF$ is a rectangle; If $AB=6$, $CF=8$, $DF=10$, what is the length of $EF$?\n",
    "completion": "\\textbf{Solution:} Since according to the properties of parallelograms we know $AB=CD$, \\\\\nthus $CD=6$, \\\\\nsince in $\\triangle FCD$, $DF^2=10^2=6^2+8^2=CF^2+DC^2$, \\\\\ntherefore $\\triangle FCD$ is a right triangle, $\\angle FCD=90^\\circ$, \\\\\nbased on the formula for the area of a triangle: $\\frac{1}{2}\\times FC\\times CD=\\frac{1}{2}\\times FD\\times EC$, \\\\\nthen $EC=\\frac{24}{5}$, \\\\\nsince in rectangle $FBCE$, $\\angle FBC=90^\\circ$, $EC=BF$, \\\\\nthus $BF=\\frac{24}{5}$, \\\\\ngiven $EF\\parallel BC$, it means $\\angle FCB=\\angle CFD$, \\\\\ntherefore $\\triangle FCB\\sim \\triangle DFC$, \\\\\nthus $\\frac{BC}{FC}=\\frac{FB}{DC}$, \\\\\nthus $BC=\\frac{FB}{DC}\\times FC=\\frac{\\frac{24}{5}\\times 8}{6}=\\frac{32}{5}$, \\\\\nthus $EF=BC=\\boxed{\\frac{32}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52225539_138.png"
  },
  {
    "solution": "\\boxed{x<1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graphs of the functions \\(y=3x\\) and \\(y=kx+5\\) intersect at point \\(A(a, 3)\\). Based on the graph, directly write the solution set of the inequality \\(3x<kx+5\\).\n",
    "completion": "\\textbf{Solution:} From the graph, it can be concluded that the solution set of the inequality $3x<kx+5$ is $\\boxed{x<1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52416343_87.png"
  },
  {
    "solution": "\\boxed{BG=4-\\sqrt{13}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $BC=1$, $AB=2$. A perpendicular line is drawn from a point $P$ on the diagonal $BD$ to intersect $AB$ at point $F$ and $CD$ at point $E$. A line is drawn through point $E$ such that $EG \\parallel BD$ and intersects $BC$ at point $G$. The line $FG$ intersects $BD$ at point $H$, and $DF$ is connected. When the quadrilateral $DFGE$ has one pair of adjacent sides equal, what is the length of $BG$?\n",
    "completion": "\\textbf{Solution:}\n\n(1) As shown in Figure 1, when $DE=EG$, let $BG=x$, then $DE=EG=2x$, $CE=2-2x$, $CG=1-x$. In $\\triangle CEG$ with a right angle, $CE^{2}+CG^{2}=EG^{2}$,\n\\[\n\\therefore (2-2x)^{2}+(1-x)^{2}=(2x)^{2},\n\\]\nsolving this yields $x=5-2\\sqrt{5}$,\n\\[\n\\therefore \\boxed{BG=5-2\\sqrt{5}};\n\\]\n\n(2) As shown in Figure 1, when $EG=GF$,\n\\[\n\\because CE=BF, \\angle C=\\angle ABC=90^\\circ,\n\\]\n\\[\n\\therefore \\text{Rt}\\triangle ECG\\cong \\text{Rt}\\triangle FBG \\text{ (by the HL congruence theorem)},\n\\]\n\\[\n\\therefore \\boxed{BG=CG=\\frac{1}{2}CB=\\frac{1}{2}},\n\\]\n\n(3) As shown in Figure 1, when $DF=GF$, let $BG=x$, then $AF=DE=2x$,\n\\[\n\\because DF^{2}=GF^{2},\n\\]\n\\[\n\\therefore DA^{2}+AF^{2}=BG^{2}+BF^{2},\n\\]\n\\[\n\\therefore 1^{2}+(2x)^{2}=(2-2x)^{2}+x^{2},\n\\]\nsolving this yields $x=4\\pm\\sqrt{13}$,\n\\[\n\\because BG<1,\n\\]\n\\[\n\\therefore \\boxed{BG=4-\\sqrt{13}};\n\\]\n\n(4) $\\because \\angle DEF=90^\\circ$,\n\\[\n\\therefore DF>DE, \\text{ i.e., } DF=DE \\text{ does not exist};\n\\]\n\nIn summary, the length of $BG$ can be \\boxed{5-2\\sqrt{5}}, \\boxed{\\frac{1}{2}}, or \\boxed{4-\\sqrt{13}}.",
    "image_path": "dataset/mm_math/images/MM_Math/51722675_125.png"
  },
  {
    "solution": "\\boxed{\\frac{10}{3}\\pi}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the radius of circle $O$ is 6, and $AB$, $BC$ are chords of circle $O$. If $\\angle ABC=50^\\circ$, then what is the length of arc $\\overset{\\frown}{AC}$?\n",
    "completion": "\\textbf{Solution:} Since $AB$ and $BC$ are chords of the circle $\\odot O$, and $\\angle ABC=50^\\circ$,\\\\\nit follows that $\\angle AOC=2\\angle ABC=100^\\circ$,\\\\\nGiven that the radius of $\\odot O$ is 6,\\\\\nit follows that $\\overset{\\frown}{AC}=\\frac{100}{360}\\times 2\\pi \\times 6=\\boxed{\\frac{10}{3}\\pi}$",
    "image_path": "dataset/mm_math/images/MM_Math/53042783.png"
  },
  {
    "solution": "\\boxed{x<2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graphs of the functions $y=x$ and $y=ax+4$ intersect at point $A(2, 2)$. What is the solution set of the inequality $x<ax+4$?\n",
    "completion": "\\textbf{Solution:} By observing the graph of the function, we conclude that when $x<2$,\\\\\nthe graph of $y=x$ is entirely below the line $y=ax+4$,\\\\\n$\\therefore$ the solution set for the inequality $x<ax+4$ is $\\boxed{x<2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52251028.png"
  },
  {
    "solution": "\\boxed{1\\colon 3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AD \\perp BC$ with the foot of the perpendicular being D, $AD=5$, $BC=10$. Quadrilaterals $EFGH$ and $HGNM$ are both squares, and points E, F, G, N, M are all on the sides of $\\triangle ABC$. What is the ratio of the area of $\\triangle AEM$ to the area of quadrilateral $BCME$?\n",
    "completion": "\\textbf{Solution:} Let quadrilaterals EFGH and HGNM both be squares, and $AD\\perp BC$, \\\\\ntherefore quadrilateral EFDP is a rectangle, and $AP\\perp EM$,\\\\ \nlet us set $PD=EF=EH=HM=x$,\\\\\ntherefore $EM=2x$, $AP=5-x$,\\\\\nsince $EM\\parallel BC$,\\\\\ntherefore $\\angle AEM=\\angle B$, $\\angle AME=\\angle C$,\\\\\ntherefore $\\triangle AEM\\sim \\triangle ABC$,\\\\\nsince AP and AD are altitudes of $\\triangle AEM$ and $\\triangle ABC$ respectively,\\\\\ntherefore $\\frac{AP}{AD}=\\frac{EM}{BC}$, substituting we get: $\\frac{5-x}{5}=\\frac{2x}{10}$,\\\\\nsolving yields: $x=\\frac{5}{2}$, which means $PD=\\frac{5}{2}$,\\\\\ntherefore $AP=5-\\frac{5}{2}=\\frac{5}{2}$,\\\\\ntherefore $\\frac{AP}{AD}=\\frac{\\frac{5}{2}}{5}=\\frac{1}{2}$,\\\\\ntherefore $\\frac{S_{\\triangle AEM}}{S_{\\triangle ABC}}=\\left(\\frac{AP}{AD}\\right)^2=\\left(\\frac{1}{2}\\right)^2=\\frac{1}{4}$,\\\\\nsince $S_{\\triangle ABC}=S_{\\triangle AEM}+S_{\\text{quadrilateral} BCME}$,\\\\\ntherefore $S_{\\triangle AEM} \\colon S_{\\text{quadrilateral} BCME}=1\\colon 3$.\\\\\nHence, the answer is: $\\boxed{1\\colon 3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51284390.png"
  },
  {
    "solution": "\\boxed{x<-1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, the line $l_{1}: y=x+3$ and the line $l_{2}: y=kx+b$ intersect at a point in the same Cartesian coordinate system. What is the solution set of the inequality $x+3<kx+b$ with respect to $x$?\n",
    "completion": "\\textbf{Solution:} Since the line $l_{1}: y=x+3$ intersects with the line $l_{2}: y=kx+b$ at the point $(-1,2)$, \\\\\nit follows that when $x<-1$, $x+3<kx+b$, \\\\\nthus, the solution set for the inequality $x+3<kx+b$ with respect to $x$ is $x<-1$. \\\\\nTherefore, the answer is $\\boxed{x<-1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50545714.png"
  },
  {
    "solution": "\\boxed{20}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $AB=60\\,\\text{cm}$, point C is the midpoint of line segment AB, and point D is a point on line segment AB, with the length ratio of $AD$ to $DB$ being $2\\colon 1$. What is the length of $BD$ in $\\text{cm}$?\n",
    "completion": "\\textbf{Solution:} Given that $AB=60\\text{cm}$ and the ratio of the lengths of AD to DB is $2\\colon 1$, \\\\\nit follows that $BD=60\\times \\frac{1}{3}=\\boxed{20}\\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/47112699_208.png"
  },
  {
    "solution": "\\boxed{4-\\frac{8\\pi }{9}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $BC=4$, a circle $\\odot A$ with center A and radius 2 is tangent to $BC$ at point D, intersects $AB$ at point E and $AC$ at point F. Point P is a point on $\\odot A$, and $\\angle EPF = 40^\\circ$. What is the area of the shaded region in the figure?\n",
    "completion": "\\[ \\text{\\bf Solution:} \\text{ As shown in the figure, connect AD.} \\]\n\\[ \\because \\angle EPF=40^\\circ, \\]\n\\[ \\therefore \\angle EAF=2\\angle EPF=80^\\circ, \\]\n\\[ \\therefore S_{\\text{sector} EAF}=\\frac{80\\pi \\times AE^{2}}{360}=\\frac{80\\pi \\times 2^{2}}{360}=\\frac{8\\pi }{9}. \\]\n\\[ \\because \\text{BC is the tangent of } \\odot O \\text{ with the tangent point at D}, \\]\n\\[ \\therefore AD\\perp BC, AD=2, \\]\n\\[ \\therefore S_{\\triangle ABC}=\\frac{1}{2}AD\\cdot BC=\\frac{1}{2}\\times 2\\times 4=4. \\]\n\\[ \\because S_{\\text{shaded}}=S_{\\triangle ABC}-S_{\\text{sector} EAF}, \\]\n\\[ \\therefore S_{\\text{shaded}}=4-\\frac{8\\pi }{9}. \\]\n\\[ \\text{Therefore, the answer is: } \\boxed{4-\\frac{8\\pi }{9}}. \\]",
    "image_path": "dataset/mm_math/images/MM_Math/51318343.png"
  },
  {
    "solution": "\\boxed{A\\left(-1, 0\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=-\\frac{1}{2}x^2+bx+c$ intersects the $x$-axis at points $A$ and $B(4, 0)$, and intersects the $y$-axis at point $C$. The line $y=-\\frac{1}{2}x+n$ passes through points $B$ and $C$. Point $D$ is a moving point on line $BC$. A perpendicular line $DQ$ is drawn from point $D$ to the $x$-axis, with the foot of the perpendicular being $Q$, intersecting the parabola at point $P$. Find the equation of the parabola and the coordinates of point $A$.\n",
    "completion": "\\textbf{Solution:} Substitute $B(4, 0)$ into $y=-\\frac{1}{2}x+2$,\\\\\n$\\therefore -2+2=0$,\\\\\nwe find $n=2$,\\\\\n$\\therefore y=-\\frac{1}{2}x+2$,\\\\\nlet $x=0$, then $y=2$,\\\\\n$\\therefore C(0, 2)$,\\\\\nSubstituting $B(4, 0)$ and $C(0, 2)$ into $y=-\\frac{1}{2}x^2+\\frac{3}{2}x+2$,\\\\\n$\\therefore \\left\\{\\begin{array}{l}c=2 \\\\ -8+4b+c=0\\end{array}\\right.$,\\\\\nwe find $\\left\\{\\begin{array}{l}c=2 \\\\ b=\\frac{3}{2}\\end{array}\\right.$,\\\\\n$\\therefore y=-\\frac{1}{2}x^2+\\frac{3}{2}x+2$,\\\\\nlet $y=0$, then $-\\frac{1}{2}x^2+\\frac{3}{2}x+2=0$,\\\\\nwe find $x=-1$ or $x=4$,\\\\\n$\\therefore \\boxed{A\\left(-1, 0\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52634644_173.png"
  },
  {
    "solution": "\\boxed{20}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\odot O$ is the circumcircle of the right-angled triangle $ABC$, with $OE \\perp AB$ intersecting $\\odot O$ at point $E$, and the foot of the perpendicular is point $D$. The extensions of $AE$ and $CB$ intersect at point $F$. If $ED=4$ and $AB=16$, what is the length of $FC$?\n",
    "completion": "\\textbf{Solution:} \\\\\nSince $OE\\perp AB$,\\\\\nit follows that $\\angle ADO=90^\\circ$,\\\\\nalso, since $\\angle ABC=90^\\circ$,\\\\\nit follows that $OD\\parallel BC$,\\\\\nsince the extensions of $AE$ and $CB$ meet at point F,\\\\\nit follows that $OE\\parallel CF$,\\\\\nsince $OA=OC$,\\\\\nit follows that $\\frac{AO}{AC}=\\frac{AD}{AB}=\\frac{AE}{EF}=1$,\\\\\ntherefore, $DE$ is the midline of $\\triangle ABF$,\\\\\nthus $AD=BD=\\frac{1}{2}AB=8$, $FB=2DE=8$,\\\\\nIn the right triangle $ADO$,\\\\\nsince $AD=8$, $OD=OE-DE=OA-DE=OA-4$,\\\\\nit follows that $OD^2+AD^2=OA^2$,\\\\\nhence $(OA-4)^2+8^2=OA^2$,\\\\\nwhich gives $OA=10$,\\\\\ntherefore, $OD=6$,\\\\\nhence $CB=2OD=12$,\\\\\nthus $CF=FB+BC=8+12=\\boxed{20}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53102148.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, $\\angle B=30^\\circ$, and $BC=4\\sqrt{3}$. Point P is a movable point on the right-angled side BC (P does not coincide with B or C). Connect AP and rotate line segment AP clockwise by $60^\\circ$ around point A to obtain line segment AD. Connect CD. What is the minimum value of line segment CD?\n",
    "completion": "\\textbf{Solution:} Rotate the line segment AC clockwise by $60^\\circ$ around point A to get the line segment AE, and connect DE. Then, as point D moves on DE, when CD$\\perp$DE, the value of the line segment CD is minimized. Draw CF$\\perp$AE at point F through point C. By the property of rotation: $\\triangle AED\\cong \\triangle ACP$, $\\therefore \\angle AED=\\angle ACP=90^\\circ$. Since CD$\\perp$DE and CF$\\perp$AE, $\\therefore$ quadrilateral EDCF is a rectangle, $\\therefore$ CD=EF. In $\\triangle ABC$, $\\angle ACB=90^\\circ$, $\\angle B=30^\\circ$, BC=$4\\sqrt{3}$, $\\therefore$ AC=$BC\\tan30^\\circ=4\\sqrt{3}\\times\\frac{\\sqrt{3}}{3}=4$, $\\therefore$ AE=AC=4. In $\\triangle ACF$, $\\angle AFC=90^\\circ$, $\\angle FAC=60^\\circ$, $\\therefore$ AF=$AC\\times\\cos60^\\circ=4\\times\\frac{1}{2}=2$, $\\therefore$ EF=AE−AF=4−2=2, $\\therefore$ CD=\\boxed{2}.",
    "image_path": "dataset/mm_math/images/MM_Math/55487875.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in parallelogram $ABCD$, $E$ is the midpoint of side $AD$. Line segments $AC$ and $BE$ intersect at point $F$. If the area of $\\triangle AEF$ is 2, then what is the area of $\\triangle ABC$?\n",
    "completion": "\\textbf{Solution:} Given $\\because$ parallelogram ABCD\\\\\n$\\therefore$ $AD\\parallel BC$ and $AD=BC$\\\\\n$\\because$ E is the midpoint of side AD\\\\\n$\\therefore$ BC=2AE\\\\\n$\\because$ $AE\\parallel BC$\\\\\n$\\therefore$ $\\angle EAC=\\angle BCA$\\\\\nAlso, $\\because$ $\\angle EFA=\\angle BFC$\\\\\n$\\therefore$ $\\triangle AEF\\sim \\triangle CBF$\\\\\nAs illustrated, draw $FH\\perp AD$ at point H and $FG\\perp BC$ at point G,\\\\\nthen $\\frac{EF}{BF}=\\frac{AF}{CF}=\\frac{AE}{BC}=\\frac{HF}{FG}=\\frac{1}{2}$,\\\\\n$\\therefore$ $\\frac{S_{\\triangle AEF}}{S_{\\triangle ABC}}=\\frac{\\frac{1}{2}\\cdot AE\\cdot FH}{\\frac{1}{2}\\cdot BC\\cdot HG}=\\frac{\\frac{1}{2}BC\\cdot FH}{BC\\cdot 3FH}=\\frac{1}{6}$,\\\\\n$\\because$ the area of $\\triangle AEF$ is 2\\\\\n$\\therefore$ $S_{\\triangle ABC}=6S_{\\triangle AEF}=6\\times 2=\\boxed{12}$",
    "image_path": "dataset/mm_math/images/MM_Math/52622068.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y=kx+2$ intersects with the graph of the inverse proportion function $y=\\frac{m}{x}$ at point P, which lies in the first quadrant. $PA\\perp x$ axis at point A, and the graph of the linear function intersects the x-axis and y-axis at points C and D, respectively, with $S_{\\triangle COD}=1$ and $\\frac{CO}{OA}=\\frac{1}{2}$. Based on the figure, directly write the range of $x$ for which $x>0$ and the value of the linear function is greater than or equal to the value of the inverse proportion function.\n",
    "completion": "\\textbf{Solution:} According to the graph, we directly write that for $x>0$, the range of x for which the value of the linear function is greater than or equal to the value of the inverse proportion function is $x\\ge \\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53043682_123.png"
  },
  {
    "solution": "\\boxed{\\frac{6\\sqrt{85}}{85}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $CD$ is the median to side $AB$, $\\angle B=45^\\circ$, $\\tan \\angle ACB=3$, and $AC= \\sqrt{10}$. Find the value of $\\sin \\angle ACD$?\n",
    "completion": "\\textbf{Solution:} Draw $DF \\perp BC$ at $F$, and $DE$ perpendicular to $AC$ at $E$. \\\\\nSince the area of $\\triangle ACD= \\frac{1}{2} \\times \\sqrt{10} \\times DE=3$, \\\\\nit follows that $DE= \\frac{3}{5}\\sqrt{10}$, \\\\\nBecause $AH \\perp BC$, $DF \\perp BC$, and $CD$ is the median to side $AB$, \\\\\nit follows that $DF= \\frac{1}{2}AH= \\frac{3}{2}$, \\\\\nTherefore, $BF=DF= \\frac{3}{2}$, \\\\\nIn $\\triangle CDF$, $CD= \\sqrt{DF^{2}+FC^{2}}=\\sqrt{\\left(\\frac{3}{2}\\right)^{2}+\\left(\\frac{5}{2}\\right)^{2}}=\\frac{\\sqrt{34}}{2}$, \\\\\nTherefore, in $\\triangle CDE$, $\\sin\\angle ACD= \\frac{DE}{CD}=\\boxed{\\frac{6\\sqrt{85}}{85}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51051567_107.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\triangle ABC$, $ \\angle BAC=60^\\circ$. If the radius $ OC$ of circle $ O$ is 2, what is the length of chord $ BC$?\n",
    "completion": "\\textbf{Solution:} Construct line OD $\\perp$ BC at point D, \\\\\n$\\therefore \\angle$ODC$=90^\\circ$, BC$=2$CD,\\\\\n$\\because$ arc BC$=$ arc BC,\\\\\n$\\therefore \\angle$BOC$=2\\angle$BAC$=2\\times 60^\\circ=120^\\circ$,\\\\\n$\\therefore \\angle C= \\frac{1}{2}(180^\\circ-120^\\circ)=30^\\circ$,\\\\\n$\\therefore$OD$=\\frac{1}{2}$CO$=1$ in $\\triangle$COD,\\\\\n$CD=\\sqrt{OC^{2}-OD^{2}}=\\sqrt{2^{2}-1^{2}}=\\sqrt{3}$,\\\\\n$\\therefore$BC$=2\\sqrt{3}$. Hence, the answer is: \\boxed{2\\sqrt{3}}",
    "image_path": "dataset/mm_math/images/MM_Math/52861149.png"
  },
  {
    "solution": "\\boxed{(0, 3)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, the parabola $y=\\frac{3}{4}x^{2}-\\frac{15}{4}x+3$ intersects the x-axis at points A and B, with point A to the left of point B, and intersects the y-axis at point C. What are the coordinates of points A, B, and C?\n",
    "completion": "Solution: Substituting \\(x=0\\) into \\(y=\\frac{3}{4}x^{2}-\\frac{15}{4}x+3\\), we get \\(y=3\\).\n\\\\\nTherefore, the coordinates of point \\(C\\) are \\((0, 3)\\).\n\\\\\nSubstituting \\(y=0\\) into \\(y=\\frac{3}{4}x^{2}-\\frac{15}{4}x+3\\), we get \\(\\frac{3}{4}x^{2}-\\frac{15}{4}x+3=0\\).\n\\\\\nSolving this, we find \\(x_{1}=1\\) and \\(x_{2}=4\\).\n\\\\\nTherefore, the coordinates of point \\(A\\) are \\((1, 0)\\), and the coordinates of point \\(B\\) are \\((4, 0)\\).\n\\\\\nTherefore, the coordinates of point \\(A\\) are \\(\\boxed{(1, 0)}\\), the coordinates of point \\(B\\) are \\(\\boxed{(4, 0)}\\), and the coordinates of point \\(C\\) are \\(\\boxed{(0, 3)}\\).",
    "image_path": "dataset/mm_math/images/MM_Math/52719985_71.png"
  },
  {
    "solution": "\\boxed{8 + 4\\sqrt{17}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, points $E$ and $F$ are the midpoints of $AD$ and $DC$, respectively, $BF\\perp CD$. Given that $BF=8$ and $EF=5$, what is the perimeter of quadrilateral $ABCD$?\n",
    "completion": "\\textbf{Solution:}\n\nAs shown in the figure, connect BE, extend BE to intersect the extension line of CD at point H, and draw EG $\\perp$ BF.\n\nSince parallelogram ABCD,\n\n$\\therefore$ AB $\\parallel$ DC, AB = CD,\n\n$\\therefore$ $\\angle$EAB = $\\angle$EDH, $\\angle$EBA = $\\angle$EHD,\n\nAlso, since E is the midpoint of AD,\n\n$\\therefore$ AE = ED,\n\n$\\therefore$ $\\triangle$AEB $\\cong$ $\\triangle$DEH,\n\n$\\therefore$ AB = DH, BE = EH,\n\nSince BF $\\perp$ CD, EG $\\perp$ BF,\n\n$\\therefore$ EG $\\parallel$ HF\n\n$\\therefore$ EG is the midline of $\\triangle$BHF,\n\n$\\therefore$ GF = $\\frac{1}{2}$BF = $\\frac{1}{2} \\times 8 = 4$,\n\n$\\therefore$ EG = $\\sqrt{EF^2 - GF^2} = \\sqrt{5^2 - 4^2} = 3$,\n\n$\\therefore$ HF = 6,\n\nSince F is the midpoint of CD,\n\n$\\therefore$ AB = DH = DC = 2DF = 2FC,\n\nLet FC = x, then HF = DH + DF = 2x + x = 3x = 6,\n\n$\\therefore$ x = 2,\n\n$\\therefore$ AB = 4, BC = $\\sqrt{BF^2 + FC^2} = \\sqrt{8^2 + 2^2} = 2\\sqrt{17}$,\n\n$\\therefore$ The perimeter of parallelogram ABCD = 2(AB + BC) = 2(4 + 2$\\sqrt{17}$) = $\\boxed{8 + 4\\sqrt{17}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51770458.png"
  },
  {
    "solution": "\\boxed{\\sqrt{13}-2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the quadrilateral $ABCD$ is a rectangle with $AB=3$ and $BC=4$. Point $P$ is a moving point on line segment $BC$, and point $M$ is a point on line segment $AP$ such that $\\angle ADM=\\angle BAP$. What is the minimum value of $BM$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, let O be the midpoint of AD and connect OB and OM.\\\\\n$\\because$ Quadrilateral ABCD is a rectangle,\\\\\n$\\therefore\\angle$BAD$=90^\\circ$, AD$=$BC$=4$,\\\\\n$\\therefore\\angle$BAP$+\\angle$DAM$=90^\\circ$,\\\\\n$\\because\\angle$ADM$=\\angle$BAP,\\\\\n$\\therefore\\angle$ADM$+\\angle$DAM$=90^\\circ$,\\\\\n$\\therefore\\angle$AMD$=90^\\circ$,\\\\\n$\\because$AO$=$OD$=2$,\\\\\n$\\therefore$OM$= \\frac{1}{2}$AD$=2$,\\\\\n$\\therefore$ the trajectory of point M is a circle $\\odot$O with O as the center and 2 as the radius.\\\\\n$\\because$OB$= \\sqrt{AB^{2}+AO^{2}} = \\sqrt{3^{2}+2^{2}} = \\sqrt{13}$,\\\\\n$\\therefore$BM$\\geq$OB$-$OM$=\\sqrt{13}-2$,\\\\\n$\\therefore$ the minimum value of BM is $\\boxed{\\sqrt{13}-2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52807129.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in square $ABCD$, side $BC$ is rotated counterclockwise around point $B$ to $BC'$, and lines $CC'$ and $DC'$ are drawn. If $\\angle CC'D = 90^\\circ$ and $BC' = 2\\sqrt{5}$, what is the length of segment $C'D$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $BE\\perp CC'$ at point E,\\\\\nsince quadrilateral ABCD is a square,\\\\\ntherefore $BC=CD$, $\\angle BCD=90^\\circ$,\\\\\ntherefore $\\angle BCE + \\angle C'CD = 90^\\circ$,\\\\\nsince $\\angle BCE + \\angle CBE = 90^\\circ$,\\\\\ntherefore $\\angle C'CD = \\angle CBE$,\\\\\nalso since $\\angle BEC = \\angle CC'D = 90^\\circ$,\\\\\ntherefore $\\triangle BCE \\cong \\triangle CDC'$ (AAS),\\\\\ntherefore $CE=C'D$, $BE=CC'$,\\\\\nsince side BC is rotated counterclockwise around point B to BC',\\\\\ntherefore $BC=BC'$,\\\\\nalso since $BE\\perp CC'$,\\\\\ntherefore $CE=C'E=C'D= \\frac{1}{2}CC'= \\frac{1}{2}BE$,\\\\\nsince $BC'= 2\\sqrt{5}$,\\\\\ntherefore $CE^2+BE^2=CE^2+(2CE)^2=BC^2=(2\\sqrt{5})^2$,\\\\\nsolving for: CE=2,\\\\\ntherefore, the length of segment C'D is $\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51737328.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC$, $\\angle A=120^\\circ$, and $BC=15$ cm. The perpendicular bisector of $AB$ intersects $AB$ at point D and intersects $BC$ at point E; the perpendicular bisector of $AC$ intersects $AC$ at point G, and intersects $BC$ at point F. What is the length of $EF$?\n",
    "completion": "\\textbf{Solution:} Let us connect $AE$ and $AF$,\\\\\nsince the perpendicular bisector of $AB$ intersects $AB$ at point D and $BC$ at point E; the perpendicular bisector of $AC$ intersects $AC$ at point G,\\\\\nit follows that $BE=AE$, $CF=AF$,\\\\\ntherefore $\\angle EAB=\\angle B$, $\\angle CAF=\\angle C$,\\\\\nsince $\\angle BAC=120^\\circ$ and $AB=AC$,\\\\\nit follows that $\\angle B=\\angle C=30^\\circ$,\\\\\ntherefore $\\angle BAE+\\angle CAF=60^\\circ$, $\\angle AEF=\\angle AFE=60^\\circ$,\\\\\nthus $\\triangle AEF$ is an equilateral triangle,\\\\\nhence $AE=AF=EF$,\\\\\nconsequently, $BE=EF=FC$,\\\\\nsince $BC=15=BE+EF+FC=3EF$,\\\\\nit follows that $EF=\\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52827537.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC$, the perpendicular bisector of $AB$, $DE$, intersects $AC$ at point $D$ and $AB$ at point $E$, $\\angle C=75^\\circ$. What is the degree measure of $\\angle A$?\n",
    "completion": "\\textbf{Solution:} Given that $AB=AC$ and $\\angle C=75^\\circ$, \\\\\nthus $\\angle ABC=\\angle C=75^\\circ$, \\\\\ntherefore $\\angle A=180^\\circ-75^\\circ-75^\\circ=30^\\circ$, \\\\\nthus the measure of $\\angle A$ is $\\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51545371_112.png"
  },
  {
    "solution": "\\boxed{26^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within rectangle $ABCD$, the diagonals $AC$ and $BD$ intersect at point $O$. If $\\angle COD=52^\\circ$, then what is the degree measure of $\\angle CAD$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral $ABCD$ is a rectangle,\\\\\nit follows that $OA=OB=OC=OD$,\\\\\nhence $\\angle CAD=\\angle ADO$,\\\\\nthus $\\angle COD=2\\angle CAD$,\\\\\ntherefore $\\angle CAD=\\boxed{26^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51435834.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $C$ is a point on line segment $AB$, and points $M$, $N$, $P$ are the midpoints of line segments $AC$, $BC$, $AB$, respectively. If $AB = 10\\, \\text{cm}$, what is the length of line segment $MN$?\n",
    "completion": "\\textbf{Solution:} Since $M$ and $N$ are the midpoints of $AC$ and $BC$ respectively,\n\n$\\therefore$ $MC = \\frac{1}{2}AC$, $CN = \\frac{1}{2}BC$,\n\n$\\therefore$ $MN = MC + CN = \\frac{1}{2}AC + \\frac{1}{2}BC = \\frac{1}{2}(AC + BC) = \\frac{1}{2}AB = \\frac{1}{2} \\times 10 = \\boxed{5}$ (cm).",
    "image_path": "dataset/mm_math/images/MM_Math/53848489_44.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$. Point $D$ is the midpoint of hypotenuse $AB$, $DE\\perp AC$ at foot $E$. If $DE=2$ and $CD=2\\sqrt{5}$, what is the length of $BE$?\n",
    "completion": "\\textbf{Solution:} Given that in the right-angled triangle $ABC$, angle $\\angle ACB=90^\\circ$ and $DE \\perp AC$, we have $DE \\parallel BC$. Since point $D$ is the midpoint of $AB$, $DE$ is the median of triangle $ABC$. Also, given $DE=2$, we infer $BC=4$. In the right-angled triangle $DCE$, with $DE=2$ and $CD=2\\sqrt{5}$, by Pythagoras' theorem, we have $CE=\\sqrt{CD^2-DE^2}=\\sqrt{(2\\sqrt{5})^2-2^2}=4$. Therefore, in the right-angled triangle $BCE$, applying Pythagoras' theorem again, we find $BE=\\sqrt{BC^2+CE^2}=\\sqrt{4^2+4^2}=4\\sqrt{2}$. Hence, the length of $BE$ is $\\boxed{4\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50482143.png"
  },
  {
    "solution": "\\boxed{B(2, 1)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=-x+3$ intersects the graph of the inverse proportion function $y=\\frac{2}{x}\\ (x>0)$ at points A and B. What are the coordinates of points A and B?\n",
    "completion": "\\textbf{Solution}: From the given problem, we have\n\\[\n\\left\\{\n\\begin{array}{l}\ny=-x+3 \\\\\ny=\\frac{2}{x}\n\\end{array}\n\\right.\n\\]\nTherefore, $-x+3=\\frac{2}{x}$, solving this gives $x_1=1$, $x_2=2$, $y_1=2$, $y_2=1$,\ntherefore, the points are \\boxed{A(1, 2)}, \\boxed{B(2, 1)}.",
    "image_path": "dataset/mm_math/images/MM_Math/51407111_115.png"
  },
  {
    "solution": "\\boxed{\\left(\\frac{1}{3}\\right)^{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the isosceles right-angled triangle $OAB$ with a hypotenuse length of 1, an inscribed square ${A}_{1}{B}_{1}{C}_{1}{D}_{1}$ is constructed (with the vertices of the square lying on the sides of $\\triangle OAB$). In the isosceles right-angled triangle $O{A}_{1}{B}_{1}$, another inscribed square ${A}_{2}{B}_{2}{C}_{2}{D}_{2}$ is constructed; in the isosceles right-angled triangle $O{A}_{2}{B}_{2}$, an inscribed square ${A}_{3}{B}_{3}{C}_{3}{D}_{3}$ is constructed; and so on. Continuing this process, what is the side length of the fifth square ${A}_{5}{B}_{5}{C}_{5}{D}_{5}$?\n",
    "completion": "\\textbf{Solution:} Given an isosceles right triangle OAB with a hypotenuse of 1, constructing a square $A_{1}B_{1}C_{1}D_{1}$,\n\n\\(\\therefore \\angle A = \\angle B = \\angle AA_{1}D_{1} = \\angle BB_{1}C_{1} = 45^\\circ\\), AO=BO, $A_{1}D_{1} = D_{1}C_{1} = B_{1}C_{1}$,\n\n\\(\\therefore AD_{1} = A_{1}D_{1} = D_{1}C_{1} = A_{1}B_{1} = C_{1}B\\),\n\n\\(\\therefore\\) The edge length of the square $A_{1}B_{1}C_{1}D_{1}$ is $A_{1}B_{1} = \\frac{1}{3}AB = \\frac{1}{3}\\);\n\nSimilarly, the edge length of the square $A_{2}B_{2}C_{2}D_{2}$ is $A_{2}B_{2} = \\left(\\frac{1}{3}\\right)^{2}$;\n\nThe edge length of the square $A_{3}B_{3}C_{3}D_{3}$ is $A_{3}B_{3} = \\left(\\frac{1}{3}\\right)^{3}$;\n\n\\(\\therefore\\) The edge length of the 5th square $A_{5}B_{5}C_{5}D_{5}$ is \\(\\boxed{\\left(\\frac{1}{3}\\right)^{5}}\\).",
    "image_path": "dataset/mm_math/images/MM_Math/51614696.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ABC=45^\\circ$, D is a point on BC, $\\angle ADC=60^\\circ$, $AE\\perp BC$ at point E, $CF\\perp AD$ at point F, $AE$ and $CF$ intersect at point G, $\\angle CAE=15^\\circ$. What is the degree measure of $\\angle ACF$?\n",
    "completion": "\\textbf{Solution:} Given that $\\because \\angle ADC=60^\\circ$ and $AE \\perp BC$,\\\\\n$\\therefore \\angle DAE=30^\\circ$,\\\\\n$\\because \\angle CAE=15^\\circ$,\\\\\n$\\therefore \\angle CAF=\\angle CAE+\\angle DAE=45^\\circ$,\\\\\n$\\because CF \\perp AD$,\\\\\n$\\therefore \\angle ACF=90^\\circ-\\angle CAF=45^\\circ$.\n\nThus, the answer is $\\angle ACF=\\boxed{45^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52978576_123.png"
  },
  {
    "solution": "\\boxed{22^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the straight lines AB and CD intersect at point O, $\\angle COE$ is a right angle, $OF$ bisects $\\angle AOE$, $\\angle COF=34^\\circ$, find the degree measure of $\\angle BOD$.\n",
    "completion": "\\textbf{Solution:} Given that $\\angle COE$ is a right angle, and $\\angle COF=34^\\circ$,\\\\\n$\\therefore \\angle EOF=90^\\circ-34^\\circ=56^\\circ$,\\\\\nSince $OF$ bisects $\\angle AOE$,\\\\\n$\\therefore \\angle AOF=\\angle EOF=56^\\circ$,\\\\\nGiven that $\\angle COF=34^\\circ$,\\\\\n$\\therefore \\angle AOC=56^\\circ-34^\\circ=22^\\circ$,\\\\\nSince $\\angle AOC$ and $\\angle BOD$ are vertically opposite angles,\\\\\n$\\therefore \\angle BOD=\\angle AOC=\\boxed{22^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/41691975.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure: It is known that the length of line segment $AB=16\\,\\text{cm}$, point $N$ is on the line segment $AB$, $NB=3\\,\\text{cm}$, $M$ is the midpoint of $AB$. If there is a point $C$ on the line segment $AB$ satisfying $BC=10\\,\\text{cm}$, find the length of line segment $MC$.\n",
    "completion": "\\textbf{Solution:} Since there exists a point $C$ on line segment $AB$, and $BC=10\\,\\text{cm}$, \\\\\nit follows that $MC=BC-MB=10-8=\\boxed{2}\\,\\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51407587_114.png"
  },
  {
    "solution": "\\boxed{\\frac{25}{9}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the right-angled triangle $ABC$, the perpendicular sides $AC=3\\text{cm}$, $BC=4\\text{cm}$. Let $P$ and $Q$ be points moving on $AB$ and $BC$ respectively. Point $P$ moves from point $A$ towards $B$ along $AB$ uniformly, and point $Q$ moves from point $B$ towards $C$ along $BC$ uniformly, with both their speeds being $1\\text{cm}$ per second. When point $Q$ reaches point $C$, point $P$ stops moving. Let the time for $P$ and $Q$ to move be $t$ seconds. $\\triangle PBQ$ is similar to the right-angled triangle $ABC$, what is the value of $t$?\n",
    "completion": "Solution: When $\\triangle PBQ \\sim \\triangle ABC$, we have $\\frac{BQ}{BC} = \\frac{BP}{AB}$, that is, $\\frac{t}{4} = \\frac{5-t}{5}$, solving this we get: $t = \\boxed{\\frac{20}{9}}$; \\\\\nWhen $\\triangle PBQ \\sim \\triangle CBA$, we have $\\frac{BQ}{AB} = \\frac{BP}{BC}$, that is, $\\frac{t}{5} = \\frac{5-t}{4}$, solving this we get: $t = \\frac{25}{9}$, \\\\\nTherefore, when $t = \\boxed{\\frac{20}{9}}$ or $\\boxed{\\frac{25}{9}}$, $\\triangle PBQ$ is similar to the right-angled triangle ABC.",
    "image_path": "dataset/mm_math/images/MM_Math/53043122_180.png"
  },
  {
    "solution": "\\boxed{72^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $O$ is a point on line $AB$. Three rays $OC$, $OD$, and $OE$ are drawn upward from point $O$ to line $AB$, and $OC$ bisects $\\angle AOD$, with $\\angle 2=3\\angle 1$. Suppose $\\angle 1=18^\\circ$, what is the degree measure of $\\angle COE$?\n",
    "completion": "\\textbf{Solution:} From the given conditions, we have \\\\\n$\\because \\angle 1=18^\\circ$ and $\\angle 2=3\\angle 1$, \\\\\n$\\therefore \\angle 2=54^\\circ$, \\\\\n$\\therefore \\angle AOD=180^\\circ-\\angle 1-\\angle 2=108^\\circ$. \\\\\n$\\because OC$ bisects $\\angle AOD$, \\\\\n$\\therefore \\angle 3=54^\\circ$, \\\\\n$\\therefore \\angle COE=\\angle 1+\\angle 3=\\boxed{72^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53206772_92.png"
  },
  {
    "solution": "\\boxed{200m}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, this is a schematic of a rifle being aimed. The distance from the eye to the sight, OE, is 80 cm, the width of the sight on the rifle, AB, is 0.2 cm, and the frontal width of the target, CD, is 50 cm. What is the distance from the eye to the target, OF, in meters?\n",
    "completion": "\\textbf{Solution:} Because $AB \\parallel CD$, \\\\\ntherefore $\\triangle OAB \\sim \\triangle OCD$, \\\\\ntherefore $\\frac{AB}{CD} = \\frac{OE}{OF}$, that is $\\frac{0.2}{50} = \\frac{80}{OF}$, \\\\\ntherefore $OF = \\frac{80 \\times 50}{0.2} = 20000\\,cm = 200\\,m$. \\\\\nHence, the answer is: $OF = \\boxed{200m}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51402774.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, where $E$ is the midpoint of $AD$, $\\triangle ABE$ is folded along line $BE$ to obtain $\\triangle GBE$. Extend $BG$ to intersect $CD$ at point $F$. If $AB = 3$ and $BC = 2\\sqrt{6}$, what is the length of $FD$?\n",
    "completion": "\\textbf{Solution:}\n\nGiven that $E$ is the midpoint of $AD$, it follows that $AE=DE$.\n\nSince $\\triangle ABE$ is folded along $BE$ to obtain $\\triangle GBE$, we have $AE=EG$ and $AB=BG$.\n\nThus, $ED=EG$.\n\nConsidering rectangle $ABCD$ implies that $\\angle A=\\angle D=90^\\circ$, leading to $\\angle EGF=90^\\circ$.\n\nBy connecting $EF$ and considering right triangles $EDF$ and $EGF$ with $ED=EG$ and $EF=EF$, it is concluded that right triangle $EDF$ is congruent to right triangle $EGF$ by HL congruence.\n\nHence, $DF=FG$.\n\nLet $DF=x$, then $BF=3+x$ and $CF=3-x$.\n\nIn right triangle $BCF$, using the Pythagorean theorem $BC^{2}+CF^{2}=BF^{2}$ translates to $(2\\sqrt{6})^{2}+(3-x)^{2}=(3+x)^{2}$.\n\nSolving this equation gives $x=\\boxed{2}$, which means $DF=2$.",
    "image_path": "dataset/mm_math/images/MM_Math/50482128.png"
  },
  {
    "solution": "\\boxed{40^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AD=DC$. If $\\angle BAD=20^\\circ$, find the degree measure of $\\angle C$?\n",
    "completion": "\\textbf{Solution}: Given the problem, \\\\\n$\\because DA=DC$, \\\\\n$\\therefore \\angle DAC=\\angle C$, \\\\\n$\\because \\angle ADB$ is an exterior angle of $\\triangle ADC$, \\\\\n$\\therefore \\angle ADB=\\angle DAC+\\angle C=2\\angle C$, \\\\\n$\\because AB=AD$, \\\\\n$\\therefore \\angle B=\\angle ADB=2\\angle C$, \\\\\n$\\because \\angle B+\\angle ADB+\\angle BAD=180^\\circ$, \\\\\n$\\therefore 2\\angle C+20^\\circ+2\\angle C=180^\\circ$, \\\\\n$\\therefore \\angle C=\\boxed{40^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53019706_105.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In right-angled triangle $ABC$, where $\\angle C=90^\\circ$, if $AC=6$ and $BC=8$, then what is the value of $\\cos A$?\n",
    "completion": "\\textbf{Solution:} In a right triangle $ABC$, with $\\angle C=90^\\circ$, $AC=6$, and $BC=8$,\\\\\nit follows that $AB=\\sqrt{AC^2+BC^2}=\\sqrt{6^2+8^2}=10$,\\\\\ntherefore, $\\cos A= \\frac{AC}{AB}=\\frac{6}{10}=\\boxed{\\frac{3}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51446126.png"
  },
  {
    "solution": "\\boxed{55^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, it is known that $ON$ and $OM$ respectively bisect $\\angle AOC$ and $\\angle BON$. If $\\angle MON=20^\\circ$ and $\\angle AOM=35^\\circ$, what is the measure of $\\angle AOB$?\n",
    "completion": "\\textbf{Solution:} Since $ON$ bisects $\\angle AOC$ and $OM$ bisects $\\angle BON$,\\\\\nit follows that $\\angle AON=\\angle NOC=\\frac{1}{2}\\angle AOC$ and $\\angle BOM=\\angle NOM=\\frac{1}{2}\\angle BON$.\\\\\nGiven that $\\angle MON=20^\\circ$ and $\\angle AOM=35^\\circ$,\\\\\nwe have $\\angle AON=\\angle AOM-\\angle MON=35^\\circ-20^\\circ=15^\\circ$.\\\\\nTherefore, $\\angle AOC=2\\times 15^\\circ=30^\\circ$ and $\\angle BOM=\\angle NOM=20^\\circ$.\\\\\nHence, $\\angle AOB=\\angle BOM+\\angle MON+\\angle AON=20^\\circ+20^\\circ+15^\\circ=\\boxed{55^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53211002.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given that in $\\triangle ABC$, $\\angle C=90^\\circ$, $AB=4$, $AC=\\sqrt{7}$. Find $\\sin A$.\n",
    "completion": "\\textbf{Solution:} We have $\\sin A= \\frac{BC}{AB}=\\boxed{\\frac{3}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51103676_84.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $AB \\perp AC$, $AB = AC$, and $DE$ passes through point A, with $CD \\perp DE$, $BE \\perp DE$, where the feet of the perpendiculars are points D and E, respectively. Given that $CD = 5$ and $BE = 3$, what is the length of $DE$?\n",
    "completion": "\\textbf{Solution:} Since $AB\\perp AC$, $CD\\perp DE$, and $BE\\perp DE$, \\\\\nit follows that $\\angle BAC=\\angle D=\\angle E=90^\\circ$, \\\\\ntherefore, $\\angle CAD+\\angle BAE=90^\\circ$ and $\\angle DCA+\\angle CAD=90^\\circ$, \\\\\nhence, $\\angle DCA=\\angle EAB$, \\\\\nIn $\\triangle ADC$ and $\\triangle BEA$, we have \\\\\n$\\left\\{\\begin{array}{l}\n\\angle D=\\angle E \\\\\n\\angle DCA=\\angle EAB \\\\\nAC=AB\n\\end{array}\\right.$, \\\\\nthus, $\\triangle ADC\\cong \\triangle BEA$ by AAS, \\\\\nwhich implies $DC=AE$, $AD=BE$, \\\\\nsince $DE=AD+AE$, \\\\\nit follows that $DE=DC+BE=5+3=\\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53044375.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated, it is known that the parabola $y=-\\frac{3}{16}(x+4)(x-4)$ intersects the $x$ axis at points $A$ and $B$, and intersects the $y$ axis at point $C$. The radius of $\\odot C$ is 2. Let $G$ be a moving point on $\\odot C$, and $P$ be the midpoint of $AG$. What is the minimum value of $AG$? What is the maximum value of $OP$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, when $y=0$, we have $0=-\\frac{3}{16}(x+4)(x-4)$. Solving this yields: $x_{1}=-4$, $x_{2}=4$. Therefore, we have point A ($-4$, $0$) and point B ($4$, $0$). Hence, $OB=OA=4$. When $x=0$, we have $y=3$. Therefore, point C ($0$, $3$) and hence $OC=3$; Since the radius of circle C is 2, therefore $CG=2$. Since point P is the midpoint of AG and point O is the midpoint of AB, we have $OP=\\frac{1}{2}BG$. Hence, when BG is at its maximum, the value of OP is maximum. Therefore, when points B, C, G lie on the same straight line, BG is at its maximum, consequently, OP reaches its maximum value. In $\\triangle$BOC, we have $BC=\\sqrt{3^2+4^2}=5$. Therefore, $BG=CG+BC=2+5=7$. Thus, the maximum value of OP is $\\boxed{\\frac{7}{2}}$; As illustrated, connecting AC and intersecting circle C at point $G_{1}$, then point G coincides with point $G_{1}$. At this moment, the length of AG is the shortest. Since $OC$ perpendicularly bisects $AB$, we have $AC=BC=5$. Thus, the minimum value of AG is $5-2=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51420128.png"
  },
  {
    "solution": "\\boxed{-2<x<-1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=kx+b$ passes through point $A(-1, -3)$ and point $B(-2, 0)$, and the line $y=3x$ passes through point $A$. What is the solution set of the inequality $3x<kx+b<0$?\n",
    "completion": "\\textbf{Solution:} According to the problem, we have $y=kx+b$ intersects with $y=3x$ at point $A(-1, -3)$,\\\\\nsolving the inequality $3x<kx+b<0$ for its solution set, which refers to the part of the graph between points $A$ and $B$,\\\\\nfurthermore, since $B(-2,0)$,\\\\\nthe range of the independent variable $x$ is $-2<x<-1$.\\\\\nTherefore, the solution set of the inequality $3x<kx+b<0$ is: $-2<x<-1$.\\\\\nHence, the answer is: $\\boxed{-2<x<-1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50482308.png"
  },
  {
    "solution": "\\boxed{34^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, AB is the diameter of $\\odot O$, point C is on the extension of AB, and CD is tangent to $\\odot O$ at point D. If $\\angle CDA=118^\\circ$, what is the degree measure of $\\angle C$?\n",
    "completion": "Solution:As shown in the diagram, connect OD, \\\\\n$\\because$CD is tangent to $\\odot$O at point D,\\\\\n$\\therefore \\angle$ODC$=90^\\circ$,\\\\\n$\\because \\angle$CDA$=118^\\circ$,\\\\\n$\\therefore \\angle$ODA$=\\angle$CDA$-\\angle$ODC$=118^\\circ-90^\\circ=28^\\circ$,\\\\\n$\\because$OD$=$OA,\\\\\n$\\therefore \\angle$OAD$=\\angle$ODA$=28^\\circ$,\\\\\n$\\therefore \\angle$DOC$=2\\angle$ODA$=56^\\circ$,\\\\\n$\\therefore \\angle$C$=90^\\circ-\\angle$DOC$=\\boxed{34^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52766462.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $AB=4$ and $AD=10$. When the right-angle vertex $P$ of the square slides on $AD$ (point $P$ does not coincide with $A$ or $D$), one right-angle side passes through point $C$, and the other intersects $AB$ at point $E$. Is there such a point $P$ that makes the perimeter of $\\triangle DPC$ equal to twice the perimeter of $\\triangle AEP$? Find the length of $DP$.\n",
    "completion": "\\textbf{Solution:} Let's assume that there is a point $P$ that satisfies the given conditions, let $DP=x$, then $AP=10-x$.\n\nSince $\\triangle AEP \\sim \\triangle DPC$, according to the perimeter of $\\triangle CDP$ being twice the perimeter of $\\triangle PAE$, we get the ratio of similarity between the two triangles is 2,\n\nthus $\\frac{CD}{AP}=\\frac{4}{10-x}=2$,\n\nsolving this equation gives $x=\\boxed{8}$,\n\ntherefore, there exists a point $P$ that satisfies the conditions, with the length of $DP$ being \\boxed{8}.",
    "image_path": "dataset/mm_math/images/MM_Math/52387502_135.png"
  },
  {
    "solution": "\\boxed{36^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $D$ is a point on side $BC$, and $AB=AD=DC$. If $AC=BC$, what is the degree measure of $\\angle C$?\n",
    "completion": "\\textbf{Solution:} Let $ \\angle C=x$\\\\\nSince $AD=DC$\\\\\nThen $\\angle DAC=\\angle C=x$\\\\\nThus $\\angle ADB=2x$\\\\\nSince $AB=AD$\\\\\nThen $\\angle B=\\angle ADB=2x$\\\\\nSince $AC=BC$\\\\\nThen $\\angle BAC=\\angle B=2x$\\\\\nIn $ \\triangle ABC$,\\\\\n$2x+2x+x=180^\\circ$\\\\\nTherefore $x=36^\\circ$\\\\\nThus, the measure of $\\angle C$ is $\\boxed{36^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/53044239_141.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given: As shown in the figure, in rectangle $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$. Lines parallel to $AC$ and $BD$ are drawn passing through points $B$ and $C$, respectively, and intersect at point $F$. Quadrilateral $BFCO$ is a rhombus. If $AB=3$ and $BC=4$, what is the perimeter of quadrilateral $BFCO$?\n",
    "completion": "\\textbf{Solution:} Because quadrilateral ABCD is a rectangle, it follows that $\\angle ABC=90^\\circ$. By the Pythagorean theorem, we have $AC=\\sqrt{AB^2+BC^2}=\\sqrt{3^2+4^2}=5$. Therefore, $OC=\\frac{1}{2}AC=\\frac{5}{2}$. From statement (1), we know quadrilateral BFCO is a rhombus. Therefore, $BF=FC=OC=OB$, hence the perimeter of quadrilateral BFCO is $4OC=4\\times \\frac{5}{2}=\\boxed{10}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52386182_124.png"
  },
  {
    "solution": "\\boxed{115^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC$, and the semicircle $O$ with diameter $AB$ intersects $BC$ and $AC$ at points $D$ and $E$, respectively. Connect $DE$ and $OD$. $\\overset{\\frown}{BD}=\\overset{\\frown}{ED}$. When the ratio of the degrees of $\\overset{\\frown}{AE}$ and $\\overset{\\frown}{BE}$ is $4\\colon 5$, find the degrees of the four interior angles of quadrilateral $ABDE$.\n",
    "completion": "\\textbf{Solution:} Given $\\because$ $\\overset{\\frown}{AE}$ + $\\overset{\\frown}{BE}$ = $180^\\circ$ and the ratio of the degrees of $\\overset{\\frown}{AE}$ to $\\overset{\\frown}{BE}$ is $4\\colon 5$, $\\therefore$ $\\overset{\\frown}{AE}$ = $80^\\circ$, $\\overset{\\frown}{BE}$ = $100^\\circ$. Consequently, $\\overset{\\frown}{BD}$ = $\\overset{\\frown}{ED}$ = $50^\\circ$, hence $\\overset{\\frown}{AD}$ = $\\overset{\\frown}{AE}$ + $\\overset{\\frown}{ED}$ = $130^\\circ$. Therefore, $\\angle BAE$ = $\\frac{1}{2} \\overset{\\frown}{BE}$ = $50^\\circ$, $\\angle B$ = $\\frac{1}{2} \\overset{\\frown}{AD}$ = $65^\\circ$, $\\because$ $\\angle AED$ + $\\angle B$ = $180^\\circ$ and $\\angle BDE$ + $\\angle A$ = $180^\\circ$, $\\therefore$ $\\angle AED$ = $115^\\circ$, $\\angle BDE$ = $130^\\circ$, thus $\\angle BAE$ = $\\boxed{50^\\circ}$, $\\angle B$ = $\\boxed{65^\\circ}$, $\\angle BDE$ = $\\boxed{130^\\circ}$, $\\angle AED$ = $\\boxed{115^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53081384_85.png"
  },
  {
    "solution": "\\boxed{120^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, the perpendicular bisectors of sides $AB$ and $AC$ intersect $BC$ at points $D$ and $E$, respectively. If $\\angle BAD + \\angle CAE = 60^\\circ$, what is the degree measure of $\\angle BAC$?\n",
    "completion": "\\textbf{Solution:} Since $DA=DB$ and $EA=EC$,\\\\\nit follows that $\\angle DAB = \\angle B$ and $\\angle EAC = \\angle C$,\\\\\ntherefore, $\\angle B + \\angle C = \\angle DAB + \\angle EAC = 60^\\circ$,\\\\\nhence, $\\angle BAC = \\boxed{120^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52733754_116.png"
  },
  {
    "solution": "\\boxed{6\\sqrt{7}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB = AC$, $AD$ is drawn through points $A$ and $C$ parallel to $BC$, and $CD$ is drawn parallel to $AB$ intersecting at point $D$. Extend $DC$ to point $E$ such that $DC = CE$, and connect $BE$. Quadrilateral $ACEB$ is a rhombus; If $AB = 4$, $BC = 6$, what is the area of the quadrilateral $ACEB$?\n",
    "completion": "\\textbf{Solution:} Construct line AE intersecting BC at point O, \\\\\nsince quadrilateral ACEB is a rhombus, \\\\\nthus AE$\\perp$BC, \\\\\nsince AB＝4, BC＝6, \\\\\nthus OB＝$\\frac{1}{2}$BC＝3, \\\\\nthus OA＝$\\sqrt{AB^{2}−OB^{2}}=\\sqrt{4^{2}−3^{2}}=\\sqrt{7}$, \\\\\nthus AE＝2OA＝2$\\sqrt{7}$, \\\\\nthus the area of quadrilateral $ACEB$, $S=\\frac{1}{2}AE\\cdot BC=\\frac{1}{2}\\times 6\\times 2\\sqrt{7}=\\boxed{6\\sqrt{7}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52188906_116.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{3}-2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the equilateral triangle $\\triangle ABC$, where $AB=6$ and $BD=4$, point $E$ is a moving point on side $AC$. Connect $DE$ and fold $\\triangle CDE$ along $DE$ to obtain $\\triangle FDE$. What is the minimum distance from point $F$ to line $AB$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, draw DT$\\perp$AB at T from point D.\\\\\n$\\because \\triangle ABC$ is an equilateral triangle,\\\\\n$\\therefore \\angle B=60^\\circ$, $BC=AB=6$,\\\\\n$\\because \\angle DTB=90^\\circ$, $BD=4$,\\\\\n$\\therefore CD=DF=2$, $DT=BD\\cdot\\sin60^\\circ=2\\sqrt{3}$,\\\\\nBy observing the diagram, it is known that when point F is on DT, the distance from point F to AB is the smallest, with the minimum value being $2\\sqrt{3}-2$,\\\\\nTherefore, the answer is: $2\\sqrt{3}-2=\\boxed{2\\sqrt{3}-2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51368331.png"
  },
  {
    "solution": "\\boxed{(-3, -2)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given the diagram, the graph of the linear function $y=k_1x-4$ intersects with the graph of the inverse proportion function $y=\\frac{k_2}{x}$ ($x>0$) at point $A(3, -6)$, and intersects with the x-axis at point $B$. Point $D$ lies on line segment $AB$. Connecting $OD$ and $OA$, and given that the area ratio $S_{\\triangle BOD}: S_{\\triangle BOA}=1:3$, what are the coordinates of point $D$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $DM\\perp x$-axis through point D, with M being the foot of the perpendicular, and draw $AN\\perp x$-axis through point A, with N being the foot of the perpendicular, \\\\\n$\\because \\frac{S_{\\triangle BOD}}{S_{\\triangle BOA}}=\\frac{\\frac{1}{2}OB\\cdot DM}{\\frac{1}{2}OB\\cdot AN}=\\frac{1}{3}$, \\\\\n$\\therefore \\frac{DM}{AN}=\\frac{1}{3}$, \\\\\n$\\because$ the coordinates of point A are (3, -6), \\\\\n$\\therefore AN=6$, \\\\\n$\\therefore DM=2$, i.e., the ordinate of point D is -2, \\\\\nSubstituting $y=-2$ into $y=-\\frac{2}{3}x-4$, \\\\\nwe get $x=-3$, \\\\\n$\\therefore$ the coordinates of point D are $\\boxed{(-3, -2)}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51445903_255.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $ \\triangle ABC \\cong \\triangle A'B'C'$, $ \\angle BCB' = 30^\\circ$, what is the degree measure of $ \\angle ACA'$?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABC \\cong \\triangle A'B'C$,\\\\\nit follows that $\\angle ACB = \\angle A'CB'$,\\\\\ntherefore $\\angle ACB - \\angle A'CB = \\angle A'CB' - \\angle A'CB$,\\\\\nwhich means $\\angle ACA' = \\angle BCB' = 30^\\circ$,\\\\\nthus A meets the conditions stated in the problem.\\\\\nHence, the answer is: $\\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53044170.png"
  },
  {
    "solution": "\\boxed{18}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the side length of rhombus ABCD is $6\\sqrt{3}$, and point M is a moving point on diagonal AC, with $\\angle ABC=120^\\circ$. What is the degree measure of $\\angle DAC$? What is the minimum value of $MA+MB+MD$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $ME\\perp AB$ at point E and connect BD.\\\\\nSince in rhombus ABCD, $\\angle ABC=120^\\circ$,\\\\\nit follows that $\\angle DAB=60^\\circ$, $AD=AB=DC=BC$, AC perpendicularly bisects BD,\\\\\nthus $\\triangle ADB$ is an equilateral triangle, DM=BM,\\\\\nhence $\\angle DAC=\\angle BAC=30^\\circ$,\\\\\nthus AM=2ME,\\\\\nsince MD=MB,\\\\\nit follows that MA+MB+MD=2ME+2DM,\\\\\nthus when points D, M, E are collinear, ME+DM is minimized, i.e., $MA+MB+MD$ is minimized,\\\\\nsince the side length of rhombus ABCD is $6\\sqrt{3}$, $\\angle DAE=60^\\circ$, $\\angle AED=90^\\circ$,\\\\\nit follows that AE=$\\frac{1}{2}AD=3\\sqrt{3}$,\\\\\nthus DE=$\\sqrt{AD^2-AE^2}=\\sqrt{(6\\sqrt{3})^2-(3\\sqrt{3})^2}=9$,\\\\\nhence $2DE=18$.\\\\\nTherefore, the minimum value of $MA+MB+MD$ is $\\boxed{18}$.\\\\\nTherefore, the answer is: $\\angle DAC=\\boxed{30^\\circ}$, the minimum value of $MA+MB+MD$ is $\\boxed{18}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51737151.png"
  },
  {
    "solution": "\\boxed{25}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, $AC=4$, and $BC=3$. A square $ABDE$ is constructed with $AB$ as its side. What is the area of square $ABDE$?\n",
    "completion": "\\textbf{Solution:} In $\\triangle ABC$, $\\angle C=90^\\circ$, $AC=4$, and $BC=3$,\\\\\n$\\therefore$ $AB=\\sqrt{AC^2+BC^2}=\\sqrt{4^2+3^2}=5$,\\\\\n$\\therefore$ the area of the square $ABDE$ is $AB^2=5^2=\\boxed{25}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53182793.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the length of the segment $AB$ is $20$, point $C$ is the midpoint of segment $AB$, point $D$ is the midpoint of segment $BC$, and point $E$ is on the segment $AC$ with $CE = \\frac{2}{5}AC$. Find the length of the segment $DB$.\n",
    "completion": "\\textbf{Solution:} Since point C is the midpoint of line segment AB,\\\\\nit follows that $AC=BC=\\frac{1}{2}AB=\\frac{1}{2}\\times 20=10$,\\\\\nFurthermore, since point D is the midpoint of line segment BC,\\\\\nit follows that $BD=CD=\\frac{1}{2}BC=\\frac{1}{2}\\times 10=\\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51251486_74.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $\\triangle ABC$ is an equilateral triangle with a side length of 3, $\\angle BDC = 120^\\circ$, $BD = CD$, points M and N are located on sides $AB$ and $AC$ respectively, and $\\angle MDN = 60^\\circ$. Connecting $MN$, what is the perimeter of $\\triangle AMN$?\n",
    "completion": "\\textbf{Solution:} Extend $AB$ to $F$ such that $BF=CN$, connect $DF$, \\\\\n$\\because$ $\\triangle BDC$ is an isosceles triangle, $\\angle BDC=120^\\circ$, \\\\\n$\\therefore$ $\\angle BCD=\\angle DBC=30^\\circ$, \\\\\n$\\because$ $\\triangle ABC$ is an equilateral triangle with side length of 3, \\\\\n$\\therefore$ $\\angle ABC=\\angle BAC=\\angle BCA=60^\\circ$, \\\\\n$\\therefore$ $\\angle DBA=\\angle DCA=90^\\circ$, \\\\\nIn $\\triangle DBF$ and $\\triangle CND$, $\\left\\{\\begin{array}{l}BF=CN \\\\ \\angle DBF=\\angle DCN \\\\ DB=DC\\end{array}\\right.$, \\\\\n$\\therefore$ $\\triangle BDF\\cong \\triangle CDN\\left(SAS\\right)$, \\\\\n$\\therefore$ $\\angle BDF=\\angle CDN$, $DF=DN$, \\\\\n$\\because$ $\\angle MDN=60^\\circ$, \\\\\n$\\therefore$ $\\angle BDM+\\angle CDN=60^\\circ$, \\\\\n$\\therefore$ $\\angle BDM+\\angle BDF=60^\\circ=\\angle FDM=\\angle MDN$, \\\\\nIn $\\triangle DMN$ and $\\triangle DMF$, $\\left\\{\\begin{array}{l}DN=DF \\\\ \\angle MDN=\\angle MDF \\\\ DM=MD\\end{array}\\right.$, \\\\\n$\\therefore$ $\\triangle DMN\\cong \\triangle DMF\\left(SAS\\right)$, \\\\\n$\\therefore$ $MN=MF$, \\\\\n$\\therefore$ $MF=BF+BM=BM+CN=MN$, \\\\\n$\\therefore$ The perimeter of $\\triangle AMN$ $AM+AN+MN=AM+MB+BF+AN=AB+AN+CN=AB+AC=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53080055.png"
  },
  {
    "solution": "\\boxed{3.5cm}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the length of segment $AB=6\\text{cm}$, $C$ is the midpoint of $AB$, point $D$ is on $AC$, and $CD=2AD$, $E$ is the midpoint of $BC$. Find the length of segment $DE$.\n",
    "completion": "\\textbf{Solution:} Given that $AB=6\\text{cm}$, and C is the midpoint of AB,\\\\\nthus $AC=BC=\\frac{1}{2}AB=3\\text{cm}$,\\\\\nalso, since $AB=6\\text{cm}$,\\\\\nthus $AC=BC=3\\text{cm}$,\\\\\nsince E is the midpoint of BC,\\\\\nthus $CE=\\frac{1}{2}BC=1.5\\text{cm}$,\\\\\nsince $CD=2AD$, and $AD+DC=AC$,\\\\\nthus $AD+2AD=AC=3AD$,\\\\\nthus $AD=1\\text{cm}$, $CD=2\\text{cm}$,\\\\\nthus $DE=CD+CE= 2+1.5=\\boxed{3.5cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51284798.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{2}\\alpha}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Extended exploration: As shown in the figure, $ \\angle AOB = \\alpha $, $ \\angle AOC = \\beta $ ($\\alpha$ and $\\beta$ are acute angles). Ray $OM$ bisects $\\angle BOC$, and ray $ON$ bisects $\\angle AOC$. What is the degree measure of $\\angle MON$?\n",
    "completion": "\\textbf{Solution:}\n\n\\[\n\\because \\angle BOC = \\angle AOB + \\angle AOC = \\alpha + \\beta,\n\\]\n\\[\n\\because \\text{ray } OM \\text{ bisects } \\angle BOC,\n\\]\n\\[\n\\therefore \\angle COM = \\frac{1}{2}\\angle BOC = \\frac{1}{2}(\\alpha + \\beta),\n\\]\n\\[\n\\because \\text{ray } ON \\text{ bisects } \\angle AOC,\n\\]\n\\[\n\\therefore \\angle CON = \\frac{1}{2}\\angle AOC = \\frac{1}{2}\\beta,\n\\]\n\\[\n\\therefore \\angle MON = \\angle COM - \\angle CON = \\frac{1}{2}(\\alpha + \\beta) - \\frac{1}{2}\\beta = \\boxed{\\frac{1}{2}\\alpha}.\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/51407373_75.png"
  },
  {
    "solution": "\\boxed{48}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the diagonals AC and BD of the rhombus ABCD intersect at point O. Through point D, draw DH $\\perp$ AB at point H, and connect OH. If OA = 6 and OH = 4, what is the area of the rhombus ABCD?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rhombus,\\\\\nit follows that OA=OC=6, OB=OD, and AC$\\perp$BD,\\\\\nthus AC=12,\\\\\nsince DH$\\perp$AB,\\\\\nit follows that $\\angle$BHD=90$^\\circ$,\\\\\nthus BD=2OH=$2\\times$ 4=8,\\\\\ntherefore, the area of rhombus ABCD = $\\frac{1}{2}AC\\cdot BD=\\frac{1}{2}\\times 12\\times 8=\\boxed{48}$",
    "image_path": "dataset/mm_math/images/MM_Math/52425012.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $OABC$, $OA=3$, $OC=4$. By respectively using the lines on which $OA$ and $OC$ lie as the $x$-axis and $y$-axis, we establish a Cartesian coordinate system. Let $D$ be a variable point on side $CB$ (not coinciding with $C$ or $B$). The graph of the inverse proportion function $y=\\frac{k}{x}$ (where $k>0$) passes through point $D$ and intersects side $BA$ at point $E$. Line $DE$ is drawn. When point $D$ moves to the midpoint of $BC$, find the value of $k$?\n",
    "completion": "\\textbf{Solution:} Since $OA=3$, $OC=4$, and quadrilateral $OABC$ is a rectangle,\\\\\nit follows that $BC=OA=3$, and the coordinates of point $B$ are $(3,4)$. Since point $D$ is the midpoint of side $BC$,\\\\\nit follows that $CD=\\frac{1}{2}BC=\\frac{3}{2}$\\\\\nTherefore, the coordinates of point $D$ are  $\\left(\\frac{3}{2}, 4\\right)$\\\\\nFurthermore, since point $D$ lies on the graph of the inverse proportion function $y=\\frac{k}{x}(k>0)$,\\\\\nit follows that $k=\\frac{3}{2}\\times 4=\\boxed{6}$",
    "image_path": "dataset/mm_math/images/MM_Math/52934017_238.png"
  },
  {
    "solution": "\\boxed{8cm}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB=24\\,cm$, $C$ is the midpoint of $AB$, point $D$ is on line segment $AC$, and $AD:CB=1:3$, what is the length of $CD$?\n",
    "completion": "\\textbf{Solution:} Given $AB=24cm$ and $C$ is the midpoint of $AB$,\\\\\nit follows that $AC=BC=\\frac{1}{2}AB=12cm$,\\\\\ngiven $AD\\colon CB=1\\colon 3$,\\\\\nit follows that $AD=\\frac{1}{3}CB=4cm$,\\\\\ntherefore, $CD=AC-AD=12-4=\\boxed{8cm}$",
    "image_path": "dataset/mm_math/images/MM_Math/53211321.png"
  },
  {
    "solution": "\\boxed{20^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, point D is on line AB. When $\\triangle ABC$ is folded along CD, point A falls on side BC at point A. If $\\angle B=35^\\circ$, what is the measure of $\\angle BDA$?\n",
    "completion": "\\textbf{Solution:} \\\\\nSince $\\angle ACB=90^\\circ$ and $\\angle B=35^\\circ$, \\\\\nit follows that $\\angle A=180^\\circ-\\angle ACB-\\angle B=180^\\circ-90^\\circ-35^\\circ=55^\\circ$, \\\\\nSince $\\triangle CDA$ is obtained by folding $\\triangle CDA$, \\\\\nthus $\\angle CAD=\\angle A=55^\\circ$, \\\\\nSince $\\angle CAD=\\angle B+\\angle BDA=\\angle B+20^\\circ$, \\\\\nit follows that $\\angle BDA=\\angle CAD-\\angle B=55^\\circ-35^\\circ=\\boxed{20^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51378814.png"
  },
  {
    "solution": "\\boxed{32^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $\\odot O$ is the circumcircle of $\\triangle ABD$, where AB is the diameter of $\\odot O$, and CD is a chord of $\\odot O$. If $\\angle ABD = 58^\\circ$, then what is the measure of $\\angle BCD$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle ABD=58^\\circ$, we can then find that $\\angle A=90^\\circ-\\angle ABD=32^\\circ$. Furthermore, in the same or congruent circles, angles subtended by the same or congruent arcs are equal, $\\therefore \\angle BCD=\\angle A=\\boxed{32^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51261962.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AD\\perp BC$, $BE\\perp AC$, with the feet of the perpendiculars being points $D$ and $E$ respectively, and $AD$ and $BE$ intersecting at point $F$. If $BF=AC$, then what is the measure of $\\angle ABC$ in degrees?\n",
    "completion": "\\textbf{Solution:} Since $AD \\perp BC$ at $D$, and $BE \\perp AC$ at $E$,\\\\\ntherefore, $\\angle EAF + \\angle AFE = 90^\\circ$, and $\\angle DBF + \\angle BFD = 90^\\circ$,\\\\\nAlso, since $\\angle BFD = \\angle AFE$ (vertical angles are equal),\\\\\ntherefore, $\\angle EAF = \\angle DBF$,\\\\\nIn right triangles $\\triangle ADC$ and $\\triangle BDF$,\\\\\n$\\left\\{\\begin{array}{l}\n\\angle CAD = \\angle FBD \\\\\n\\angle BDF = \\angle ADC \\\\\nBF = AC\n\\end{array}\\right.$,\\\\\ntherefore, $\\triangle ADC \\cong \\triangle BDF$ (by AAS),\\\\\ntherefore, $BD = AD$,\\\\\nsince $\\angle ADB = 90^\\circ$.\\\\\nTherefore, $\\angle ABC = \\angle BAD = \\boxed{45^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51213250.png"
  },
  {
    "solution": "\\boxed{55^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB\\parallel CD$, and point E is a point between the two lines. If $\\angle BAE=35^\\circ$ and $\\angle DCE=20^\\circ$, then what is the degree measure of $\\angle AEC$?\n",
    "completion": "\\textbf{Solution:} We have $\\angle AEC = \\boxed{55^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52949146_167.png"
  },
  {
    "solution": "\\boxed{1<x<2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the function $y=kx+b$ ($k\\neq 0$) passes through point $B(2,0)$ and intersects with the graph of the function $y=2x$ at point $A$. Then, what is the solution set of the inequality $0<kx+b<2x$?\n",
    "completion": "\\textbf{Solution:} When $y=2$, solving $2=2x$ gives $x=1$, \\\\\n$\\therefore$ point A is (1, 2), \\\\\n$\\therefore$ for $x>1$, $2x>kx+b$, \\\\\n$\\because$ the graph of the function $y=kx+b$ ($k \\ne 0$) passes through point B (2, 0), \\\\\n$\\therefore$ for $x<2$, $kx+b>0$, \\\\\n$\\therefore$ the solution set for the inequality $0<kx+b<2x$ is $1<x<2$.\\\\\nThus, the answer is: $\\boxed{1<x<2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52381181.png"
  },
  {
    "solution": "\\boxed{65^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is a chord of the circle $\\odot O$. Through point $O$, draw $OC\\perp OA$, where $OC$ intersects $AB$ at $P$, and $CP=BC$. Point $Q$ lies on the arc $\\overset{\\frown}{AmB}$. $BC$ is a tangent line to the circle $\\odot O$; Given that $\\angle BAO=25^\\circ$, what is the measure of $\\angle AQB$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle BAO=25^\\circ$,\\\\\nit follows that $\\angle ABO=25^\\circ$, $\\angle APO=65^\\circ$,\\\\\nthus $\\angle POB=\\angle APO-\\angle ABO=40^\\circ$,\\\\\ntherefore $\\angle AQB=\\frac{1}{2}(\\angle AOP+\\angle POB)=\\frac{1}{2}\\times 130^\\circ=\\boxed{65^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51445866_125.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the inverse proportion function $y=\\frac{1}{x}$ intersects with the graph of the direct proportion function $y=x$ at points A and C. $AB\\perp x$-axis and $CD\\perp x$-axis, with the perpendicular feet at points B and D, respectively. What is the area of the quadrilateral ABCD?\n",
    "completion": "\\textbf{Solution:} Solving the system of equations for the two functions gives \\(\\begin{cases}y=\\frac{1}{x}, \\\\ y=x, \\end{cases}\\) which results in \\(\\begin{cases}x=1, \\\\ y=1, \\end{cases}\\) or \\(\\begin{cases}x=-1, \\\\ y=-1. \\end{cases}\\)\n\n\\(\\therefore A(1, 1), C(-1, -1).\\)\n\n\\(\\because AB\\perp x\\) axis, \\(CD\\perp x\\) axis,\n\n\\(\\therefore B(1, 0), D(-1, 0),\\)\n\n\\(\\therefore BD=2, AB=CD=1.\\)\n\n\\(\\because AB\\perp BD, CD\\perp BD\\)\n\n\\(\\therefore AB\\parallel CD\\)\n\n\\(\\therefore\\) Quadrilateral ABCD is a parallelogram,\n\n\\(\\therefore\\) The area of quadrilateral ABCD \\(=AB\\cdot BD=1\\times 2=\\boxed{2}\\).",
    "image_path": "dataset/mm_math/images/MM_Math/51615099.png"
  },
  {
    "solution": "\\boxed{150^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points $A$, $B$, $C$ are on the circle $\\odot O$, $\\angle ACB=105^\\circ$, then what is the value of $\\angle \\alpha$?\n",
    "completion": "\\textbf{Solution:} Let's take any point $D$ on the major arc $AB$ and connect $AD$ and $BD$,\\\\\nsince quadrilateral $ACBD$ is inscribed in circle $O$, and $\\angle ACB=105^\\circ$,\\\\\nit follows that $\\angle ADB=180^\\circ-\\angle C=180^\\circ-105^\\circ=75^\\circ$,\\\\\nhence $\\angle AOB=2\\angle ADB=2\\times 75^\\circ=150^\\circ$.\\\\\nTherefore, the answer is: $\\boxed{150^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52536735.png"
  },
  {
    "solution": "\\boxed{B}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ABC=90^\\circ$, $\\angle A=60^\\circ$. Point $D$ is on side $AC$ with $\\angle DBC=30^\\circ$ and $AC=12\\,\\text{cm}$. What is the perimeter of $\\triangle ABD$?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABC\\cong \\triangle ADE$, and $\\angle D=30^\\circ$,\\\\\nit follows that $\\angle B=\\angle D=30^\\circ$,\\\\\nSince $\\angle C=70^\\circ$,\\\\\nthen $\\angle BAC=180^\\circ-\\angle B-\\angle C=180^\\circ-70^\\circ-30^\\circ=80^\\circ$,\\\\\nSince $\\triangle ABC\\cong \\triangle ADE$,\\\\\nit follows that $\\angle EAD=\\angle BAC=80^\\circ$,\\\\\nAlso, given that $\\angle CAD=35^\\circ$,\\\\\nthen $\\angle EAC=\\angle EAD-\\angle DAC=80^\\circ-35^\\circ=45^\\circ$,\\\\\nThus, the answer is: $\\boxed{B}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52905587.png"
  },
  {
    "solution": "\\boxed{y=2x+8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the graph of the linear function $y=kx+b$ passes through point $A(-2, 4)$, and intersects with the graph of the direct proportion function $y=-\\frac{2}{3}x$ at point $B(a, 2)$. What is the value of $a$ and what is the equation of the linear function $y=kx+b$?\n",
    "completion": "\\textbf{Solution:} Given the graph of the direct proportion function $y=-\\frac{2}{3}x$ passes through point $B(a, 2)$,\\\\\n$\\therefore 2=-\\frac{2}{3}a$, solving for $a$, we find $a=\\boxed{-3}$,\\\\\n$\\therefore B(-3, 2)$.\\\\\nGiven that the graph of the linear function $y=kx+b$ passes through points $A(-2, 4)$, $B(-3, 2)$,\\\\\n$\\therefore \\left\\{\\begin{array}{l}\n-2k+b=4 \\\\\n-3k+b=2\n\\end{array}\\right.$, solving for, $\\left\\{\\begin{array}{l}\nk=\\boxed{2} \\\\\nb=\\boxed{8}\n\\end{array}\\right.$,\\\\\n$\\therefore$ the equation of the linear function $y=kx+b$ is $\\boxed{y=2x+8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50553639_111.png"
  },
  {
    "solution": "\\boxed{3\\colon4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ABC = 90^\\circ$, $\\angle A = 60^\\circ$. Let D be a point on side $AC$, with $\\angle DBC = 30^\\circ$, and $AC = 12\\,cm$. What is the perimeter of $\\triangle ABD$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, since quadrilateral ABCD is a parallelogram, \\\\\ntherefore, AD $\\parallel$ BC and AD = BC, \\\\\ntherefore, $\\triangle$ADQ $\\sim$ $\\triangle$RCD, \\\\\ntherefore, AD $\\colon$ RC = DQ $\\colon$ CQ, that is AD $\\colon$ RC $=3\\colon1$, \\\\\ntherefore, AD = 3RC, \\\\\nSimilarly, $\\triangle$ADP $\\sim$ $\\triangle$RBP, \\\\\ntherefore, AD $\\colon$ BR = AP $\\colon$ PR, that is AD $\\colon$ (AD + RC) = AP $\\colon$ PR, \\\\\ntherefore, 3RC $\\colon$ 4RC = AP $\\colon$ PR, that is AP $\\colon$ PR $=\\boxed{3\\colon4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52874565.png"
  },
  {
    "solution": "\\boxed{\\frac{15}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In $\\triangle ABC$, $\\angle C = 90^\\circ$, $AC = 3$, and points $D$ and $E$ are located on sides $AC$ and $AB$ respectively. $\\triangle ADE$ is folded along line $DE$ such that point $A$ lands on side $BC$, denoted as point $F$. If $CF=1$ and $DF \\parallel AB$, what is the length of $AB$?\n",
    "completion": "\\textbf{Solution:} Given that $\\triangle ADE$ is folded along the line de, point A falls on side BC, denoted as point F, with CF=1,\\\\\nlet $DC=x$, then $DF=3-x$,\\\\\nIn $Rt\\triangle DCF$,\\\\\n$DC^2+FC^2=DF^2$,\\\\\nwe get $x^2+1^2=(3-x)^2$,\\\\\nSolving this gives: $x=\\frac{4}{3}$, so $DF=3-\\frac{4}{3}=\\frac{5}{3}$,\\\\\nBecause $DF\\parallel AB$,\\\\\n$\\triangle DCF \\sim \\triangle ACB$,\\\\\nthus $\\frac{DC}{AC}=\\frac{DF}{AB}$,\\\\\nwhich gives $\\frac{\\frac{4}{3}}{3}=\\frac{\\frac{5}{3}}{AB}$,\\\\\nSolved to get: $AB=\\boxed{\\frac{15}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51999252.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in parallelogram $ABCD$, $E$ is a point on side $DC$, and $\\angle EAB = \\angle EBC$. If $BE=2$ and $\\triangle ABE \\sim \\triangle BEC$; what is the value of $AB \\cdot CE$?\n",
    "completion": "\\textbf{Solution}: Since $\\triangle ABE \\sim \\triangle BEC$,\\\\\nit follows that $\\frac{AB}{EB}=\\frac{EB}{EC}$,\\\\\nthus $BE^2=AB \\cdot CE$,\\\\\nand since $BE=2$,\\\\\nwe have $AB \\cdot CE=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53110843_166.png"
  },
  {
    "solution": "\\boxed{65^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC \\cong \\triangle AED$, with point D on side BC. If $\\angle EAB = 50^\\circ$, what is the degree measure of $\\angle ADE$?\n",
    "completion": "\\textbf{Solution:} Given that $\\triangle ABC \\cong \\triangle AED$,\\\\\n$\\therefore \\angle BAC = \\angle EAD$,\\\\\n$\\therefore \\angle EAB + \\angle BAD = \\angle DAC + \\angle BAD$,\\\\\n$\\therefore \\angle DAC = \\angle EAB = 50^\\circ$,\\\\\n$\\because AD = AC$,\\\\\n$\\therefore \\angle ADC = \\angle C = \\angle ADE = \\frac{1}{2}(180^\\circ - \\angle DAC) = \\boxed{65^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52603090.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, extend the side $DC$ of the quadrilateral $ABCD$ to point $E$ such that $CE=DC$. Connect $AE$, which intersects $BC$ at point $F$, and connect $AC$, $BE$. Given that $BF=CF$; if $AB=2$, $AD=4$, and $\\angle AFC=2\\angle D$, what is the area of the quadrilateral $ABCD$?\n",
    "completion": "\\textbf{Solution:} From (1), we know that quadrilateral ABEC is a parallelogram, $\\therefore$ FA=FE, FB=FC. $\\because$ Quadrilateral ABCD is a parallelogram $\\therefore$ $\\angle ABC=\\angle D$. Also $\\because$ $\\angle AFC=2\\angle D$, $\\therefore$ $\\angle AFC=2\\angle ABC$. $\\because$ $\\angle AFC=\\angle ABC+\\angle BAF$, $\\therefore$ $\\angle ABC=\\angle BAF$, $\\therefore$ FA=FB, $\\therefore$ FA=FE=FB=FC, $\\therefore$ AE=BC, $\\therefore$ Quadrilateral ABEC is a rectangle, $\\therefore$ $\\angle BAC=90^\\circ$ $\\therefore$ BC=AD=4, $\\therefore$ $AC=\\sqrt{BC^2-AB^2}=\\sqrt{4^2-2^2}=2\\sqrt{3}$, $\\therefore$ ${S}_{\\square ABCD}=AB\\cdot AC=2\\times 2\\sqrt{3}=\\boxed{4\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51614441_119.png"
  },
  {
    "solution": "\\boxed{\\frac{12}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$ where $\\angle C=90^\\circ$, $BC=5$, and $AC=12$, what is the value of $\\tan B$?\n",
    "completion": "Solution: In the right triangle $\\triangle ABC$, $\\angle C=90^\\circ$, $BC=5$, and $AC=12$, \\\\\nthus $\\tan B=\\frac{AC}{BC}=\\boxed{\\frac{12}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51322348.png"
  },
  {
    "solution": "\\boxed{150^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of $\\odot O$, points C and D lie on $\\odot O$, $\\angle BCD=15^\\circ$, what is the degree measure of $\\angle AOD$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle BOD=2\\angle BCD$ and $\\angle BCD=15^\\circ$, \\\\\nit follows that $\\angle BOD=30^\\circ$, \\\\\nthus $\\angle AOD=180^\\circ-\\angle BOD=150^\\circ$, \\\\\ntherefore, $\\angle AOD=\\boxed{150^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52623149.png"
  },
  {
    "solution": "\\boxed{y=x^{2}-2x-3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the Cartesian coordinate system, the graph of the quadratic function $y = x^2 + bx + c$ passes through points $A(3, 0)$ and $B(2, -3)$. What is the expression of this quadratic function?\n",
    "completion": "Solution: By substituting the coordinates of points $A(3, 0)$ and $B(2, -3)$ into $y=x^{2}+bx+c$, we get:\n\\[\n\\begin{cases}\n9+3b+c=0\\\\\n4+2b+c=-3\n\\end{cases}\n\\]\nSolving this, we find:\n\\[\n\\begin{cases}\nb=-2\\\\\nc=-3\n\\end{cases}\n\\]\nTherefore, the expression for the quadratic function is \\boxed{y=x^{2}-2x-3};",
    "image_path": "dataset/mm_math/images/MM_Math/52731589_75.png"
  },
  {
    "solution": "\\boxed{140^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, rays OA, OB, OC, OD, and OE are drawn sequentially in a clockwise direction from point O as the endpoint. Moreover, OB is the bisector of $\\angle AOC$, and OD is the bisector of $\\angle COE$. If $\\angle AOD = 110^\\circ$ and $\\angle BOE = 100^\\circ$, what is the measure of $\\angle AOE$?\n",
    "completion": "\\textbf{Solution:} Given $OD$ bisects $\\angle COE$, let $\\angle EOD = \\angle DOC = x$, $OB$ bisects $\\angle AOC$, hence $\\angle AOB = \\angle COB$. Given $\\angle BOE = 100^\\circ$, we have: $\\angle AOB = \\angle COB = \\angle BOE - \\angle COE = 100^\\circ - 2x$. From $\\angle AOD = 110^\\circ$, we have: $\\angle AOD = \\angle COD + \\angle COB + \\angle AOB = 110^\\circ$, thus $x + (100^\\circ - 2x) + (100^\\circ - 2x) = 110^\\circ$. Solving this gives $x = 30^\\circ$, i.e., $\\angle EOD = \\angle DOC = 30^\\circ$. Therefore, $\\angle AOE = \\angle AOD + \\angle DOE = 110^\\circ + 30^\\circ = \\boxed{140^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52579788_200.png"
  },
  {
    "solution": "\\boxed{8\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the quadrilateral $ABCD$ is a square, with $E$ and $F$ being two points on the diagonal $AC$, and $AE=CF$. $\\triangle ADE\\cong \\triangle CBF$; the quadrilateral $BEDF$ forms a rhombus. If $AC=8, AE=2$, what is the perimeter of the quadrilateral $BEDF$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect $BD$, intersecting $AC$ at point O, \\\\\nsince quadrilateral $ABCD$ is a square and $AC=8$, \\\\\ntherefore $AC\\perp BD, OD=OA=\\frac{1}{2}AC=4$, \\\\\nsince $AE=2$, \\\\\ntherefore $OE=OA-AE=2$, \\\\\nIn right triangle $\\triangle DOE$, $DE=\\sqrt{OD^{2}+OE^{2}}=2\\sqrt{5}$, \\\\\nthus, the perimeter of quadrilateral $BEDF$ is $4DE=4\\times 2\\sqrt{5}=\\boxed{8\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52354771_100.png"
  },
  {
    "solution": "\\boxed{\\frac{56}{25}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the rectangle $ABCD$, point $E$ is on side $BC$, and $AE$ is connected. Folding $\\triangle ABE$ along line $AE$ results in $\\triangle AFE$, with $EF$ intersecting $AC$ at point $M$. Given $AB=8$, $BC=10$, and $BE= \\frac{3}{5}BC$, what is the distance from point $F$ to line $AD$?\n",
    "completion": "\\textbf{Solution:} Draw line $MN \\perp BC$ through point $F$,\\\\\nsince $BE= \\frac{3}{5}BC$ and $BC=10$,\\\\\ntherefore, $BE=6$,\\\\\nsince a fold is made,\\\\\ntherefore, $\\triangle ABE \\cong \\triangle AFE$,\\\\\ntherefore, $EF=BE=6$, $\\angle AFE=\\angle B= 90^\\circ$, $AF= AB=8$,\\\\\ntherefore, $\\angle AFN+\\angle EFM=90^\\circ$,\\\\\nsince $\\angle AFN+\\angle FAN=90^\\circ$,\\\\\ntherefore, $\\angle FAN=\\angle EFM$,\\\\\ntherefore, $\\triangle AFN \\sim \\triangle FEM$,\\\\\ntherefore, $\\frac{AF}{EF}=\\frac{AN}{FM}=\\frac{FN}{EM}=\\frac{4}{3}$,\\\\\nLet $AN=4x$, $FM=3x$, $FN=8−3x$, $EM=4x−6$,\\\\\ntherefore, $FN=8−3x$, $EM=4x−6$,\\\\\ntherefore, $\\frac{8−3x}{4x−6}=\\frac{4}{3}$,\\\\\ntherefore, $x=\\frac{48}{25}$,\\\\\nUpon verification: $x=\\frac{48}{25}$ is the root of the original equation,\\\\\ntherefore, $FN=\\boxed{\\frac{56}{25}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50497051.png"
  },
  {
    "solution": "\\boxed{38^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the diagonals AC and BD of rhombus ABCD intersect at point O. If $\\angle BAC = 52^\\circ$, then what is the measure of $\\angle CDO$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rhombus, \\\\\nit follows that AD=CD, and $\\angle CAD=\\angle BAC=52^\\circ$,\\\\\ntherefore, $\\angle ADC=180^\\circ - \\angle CAD - \\angle ACD=180^\\circ - 52^\\circ - 52^\\circ=76^\\circ$,\\\\\nhence, $\\angle CDO= \\frac{1}{2}\\angle ADC=\\boxed{38^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51614461.png"
  },
  {
    "solution": "\\boxed{(1,0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the line $y=\\frac{5}{2}x+5$ intersects the x-axis at point $A$ and the y-axis at point $B$. Another line passing through point $B$ intersects the positive half of the x-axis at point $C$, and the area of $\\triangle ABC$ is $15$. Under the condition of (2), point $E$ is a moving point on the line $AM$. Is there a point $D$ on the x-axis such that the quadrilateral with vertices $D$, $E$, $B$, and $C$ is a parallelogram? If it exists, write down the coordinates of point $D$ directly; if not, explain why.\n",
    "completion": "\\textbf{Solution:} There exist points satisfying the condition, and the coordinates of point $D$ are $\\boxed{(7,0)}$ or $\\boxed{(-11,0)}$ or $\\boxed{(1,0)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52019535_95.png"
  },
  {
    "solution": "\\boxed{42^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, fold $\\triangle ABC$ along the line $BD$, so that point $A$ falls exactly on point $E$ on side $BC$. If $CE=DE$ and $\\angle C=32^\\circ$, what is the degree measure of $\\angle DBE$?\n",
    "completion": "\\textbf{Solution}: Given folding $\\triangle ABC$ along the line $BD$ such that point A precisely falls on point E on side $BC$,\\\\\n$\\therefore \\angle DBE=\\angle ABD=\\frac{1}{2}\\angle ABC$, $\\angle A=\\angle BED$,\\\\\n$\\because CE=DE$, $\\angle C=32^\\circ$\\\\\n$\\therefore \\angle C=\\angle EDC=32^\\circ$,\\\\\n$\\therefore \\angle BED=\\angle C+\\angle EDC=64^\\circ$,\\\\\n$\\therefore \\angle A=64^\\circ$,\\\\\n$\\therefore \\angle ABC=180^\\circ-\\angle A-\\angle C=180^\\circ-64^\\circ-32^\\circ=84^\\circ$,\\\\\n$\\therefore \\angle DBE=\\frac{1}{2}\\angle ABC=\\boxed{42^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52827124.png"
  },
  {
    "solution": "\\boxed{90^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, quadrilateral $ABCD$ is inscribed in circle $\\odot O$, with diagonal $AC$ being the diameter of $\\odot O$. Through point $C$, draw $CE \\perp AC$, intersecting the extension of $AD$ at point $E$. Let $F$ be the midpoint of $CE$, and connect $DB$ and $DF$. What is the measure of $\\angle CDE$?\n",
    "completion": "\\textbf{Solution:} Since $AC$ is the diameter of $\\odot O$,\\\\\n\\(\\therefore \\angle ADC=90^\\circ\\),\\\\\n\\(\\therefore \\angle CDE=\\boxed{90^\\circ}\\);",
    "image_path": "dataset/mm_math/images/MM_Math/51010356_123.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Figure 1 shows a rectangular material $ABCD$, which is divided into three pieces, $\\angle AEB=30^\\circ$, and $GF\\perp AD$. These three pieces are seamlessly and non-overlappingly assembled into the shape of Figure 2, which is exactly an axisymmetric figure. Therefore, $\\frac{AB}{BC}=($     $)$\n",
    "completion": "\\textbf{Solution:} As shown in the figure, $\\because$ Figure 2 is an axisymmetric figure,\\\\\n$\\therefore BM=BE$, $MN=TE$,\\\\\nLet $AB=m$, then $BE=\\sqrt{3}m$, $AM=FG=DF=(\\sqrt{3}-1)m$,\\\\\n$\\because AF=\\sqrt{3}FG$,\\\\\n$\\therefore AF=(3-\\sqrt{3})m$,\\\\\n$\\therefore AD=BC=DF+AF=2m$,\\\\\n$\\therefore \\frac{AB}{BC}=\\frac{m}{2m}=\\boxed{\\frac{1}{2}}$,",
    "image_path": "dataset/mm_math/images/MM_Math/50849413.png"
  },
  {
    "solution": "\\boxed{\\frac{3+\\sqrt{3}}{6}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in a square $ABCD$ with side length 1, the diagonals $AC$ and $BD$ intersect at point $O$. Points $Q$ and $R$ are located on sides $AD$ and $DC$, respectively. Line $BR$ intersects line segment $OC$ at point $P$, where $QP \\perp BP$ and $QP$ intersects $BD$ at point $E$. $\\triangle APQ \\sim \\triangle DBR$; when $\\angle QED$ equals $60^\\circ$, find the value of $\\frac{AQ}{DR}$.\n",
    "completion": "\\textbf{Solution:} Let $OE=a$. In square $ABCD$, $\\angle POE=90^\\circ$, $OA=OB=OD$, \\\\\n$\\because \\angle QED=60^\\circ$, \\\\\n$\\therefore \\angle BEP=60^\\circ$, \\\\\nIn $\\triangle OEP$, \\\\\n$PE=\\frac{OE}{\\cos 60^\\circ}=2a$, $OP=OE\\cdot \\tan 60^\\circ=\\sqrt{3}a$, \\\\\n$\\because QP\\perp BP$, $\\angle BEP=60^\\circ$, \\\\\n$\\therefore \\angle PBE=30^\\circ$, \\\\\n$\\therefore BE=2PE=4a$, $BP=PE\\cdot \\tan 60^\\circ=2\\sqrt{3}a$, \\\\\n$\\therefore$ OA=OB=BE−OE=3a, \\\\\n$\\therefore$ BD=2OB=6a, \\\\\n$\\therefore AP=OA+OP=3a+\\sqrt{3}a=(3+\\sqrt{3})a$, \\\\\n$\\because \\triangle APQ\\sim \\triangle DBR$, \\\\\n$\\therefore \\frac{AQ}{DR}=\\frac{AP}{BD}=\\frac{(3+\\sqrt{3})a}{6a}=\\boxed{\\frac{3+\\sqrt{3}}{6}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51403342_140.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in circle $\\odot O$, the diameter $AB\\perp$ chord $CD$. An arc is drawn with $C$ as the center and $CD$ as the radius, intersecting the diameter $AB$ at point $E$. Connect $CE$ and extend it to meet $\\odot O$ at point $F$. Connect $DF$. If $AB=8$, what is the length of $DF$?\n",
    "completion": "\\textbf{Solution:} Construct diameter DG passing through point D, and connect DE and GF, \\\\\nsince a arc is drawn with center C and radius CD intersecting diameter AB at point E, \\\\\nit follows that CD=CE, \\\\\nsince diameter AB is perpendicular to chord CD, \\\\\nit follows that AB perpendicularly bisects CD, \\\\\nthus, we have CE=DE=CD, \\\\\ntherefore, $\\triangle CDE$ is an equilateral triangle, \\\\\nwhich implies $\\angle C=60^\\circ$, \\\\\nsince $\\overset{\\frown}{DF}=\\overset{\\frown}{DF}$, \\\\\nit follows that $\\angle G=\\angle C=60^\\circ$, \\\\\nsince DG is a diameter, \\\\\nit follows that $\\angle DFG=90^\\circ$, \\\\\ntherefore, $\\angle GDF=90^\\circ-60^\\circ=30^\\circ$, \\\\\nhence, FG=$\\frac{1}{2}$DG=$\\frac{1}{2}$×8=4, \\\\\ntherefore, DF=$\\sqrt{DG^{2}-FG^{2}}=\\sqrt{8^{2}-4^{2}}=\\boxed{4\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52874676.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: The graph of the linear function $y = ax + b$ (where $a$ and $b$ are constants, with $a \\neq 0$) is shown as in the diagram. What is the solution to the equation $ax + b = 0$?\n",
    "completion": "\\textbf{Solution:} Method 1: It can be seen from the graph that the coordinates of the intersection points of the line with the coordinate axes are $(0, -1)$ and $(2, 0)$,\\\\\nsubstituting $(0, -1)$ and $(2, 0)$ into the linear function $y=ax+b$ yields: $\\left\\{\\begin{array}{l}0=2a+b\\\\ -1=b\\end{array}\\right.$,\\\\\nsolving this gives $\\left\\{\\begin{array}{l}a=\\frac{1}{2}\\\\ b=-1\\end{array}\\right.$,\\\\\n$\\therefore \\frac{1}{2}x-1=0$, solving for $x$ gives $x=2$,\\\\\n$\\therefore$ the solution of the equation $ax+b=0$ is $x=\\boxed{2}$;\\\\\nMethod 2: According to the fact that the abscissa of the intersection point of the linear function $y=ax+b$ with the x-axis is the solution of the linear equation $ax+b=0$, it is known that the solution is $x=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52056146.png"
  },
  {
    "solution": "\\boxed{Q_2(-4,-2)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the line $AB$ passing through point $A(4,0)$ intersects the $y$-axis at point $B(0,4)$. The parabola $y=-x^2+bx+c$ that goes through the origin $O$ intersects line $AB$ at points $A$ and $C$, with the vertex of the parabola being $D$. Let $P$ be a moving point on the parabola and $Q$ a point in the Cartesian coordinate system. Is there a quadrilateral with vertices $A$, $C$, $P$, $Q$ that is a rectangle? If it exists, write down the coordinates of point $Q$ directly; if not, please explain why.\n",
    "completion": "\\textbf{Solution:}\n\nGiven: When AC is the side of a rectangle, \\\\\n$\\because -x^2+4x=-x+4$ Solve for: $x_1=4$, $x_2=1$, \\\\\nWhen $x=1$, we have $y=-1+4=3$, \\\\\n$\\therefore$ Point C is at $(1,3)$, \\\\\n$\\because y=-x^2+4x=-(x-2)^2+4$, \\\\\n$\\therefore$ Point D is at $(2,4)$ when $x=2$ and $y=-2+4=2$, \\\\\n$\\therefore$ Point R is at $(2,2)$. Lines perpendicular to AB through points C and A intersect the parabola at points $P_1$ and $P_2$, respectively, \\\\\n$\\because$C is at $(1,3)$ and D is at $(2,4)$, \\\\\n$\\therefore CD^2=(1-2)^2+(4-3)^2=2$, $CR^2=1$, and $RD=2$, \\\\\n$\\therefore CD^2+CR^2=DR^2$, \\\\\n$\\therefore \\angle RCD=90^\\circ$, \\\\\n$\\therefore$ Point $P_1$ coincides with point D, \\\\\nWhen $CP_1\\parallel AQ_1$ and $CP_1=AQ_1$, the quadrilateral $ACP_1Q_1$ is a rectangle, \\\\\n$\\because$ Moving C$(1,3)$ 1 unit to the right and 1 unit up yields $P_1(2,4)$, \\\\\n$\\therefore$ Moving A$(4,0)$ 1 unit to the right and 1 unit up yields $Q_1(5,1)$, \\\\\nNow, the equation of line $P_1C$ is: $y=x+2$, \\\\\n$\\because$ Line $P_2A$ is parallel to $P_1C$ and passes through A$(4,0)$, \\\\\n$\\therefore$ The equation of line $P_2A$ is: $y=x-4$, \\\\\n$\\because$ Point $P_2$ is at the intersection of the line $y=x-4$ and the parabola $y=-x^2+4x$, \\\\\n$\\therefore -x^2+4x=x-4$, solving for: $x_1=-1$, $x_2=4$ (discard), \\\\\n$\\therefore P_2$ lies at $(-1,-5)$, \\\\\nWhen $AC\\parallel P_2Q_2$, the quadrilateral $ACQ_2P_2$ is a rectangle, \\\\\n$\\because$ Moving A$(4,0)$ 3 units to the left and 3 units up yields C$(1,3)$, \\\\\n$\\therefore$ Moving $P_2(-1,-5)$ 3 units to the left and 3 units up yields $Q_2(-4,-2)$; \\\\\nWhen AC is the diagonal of the rectangle, \\\\\nLet $P_3(m, -m^2+4m)$ \\\\\nWhen $\\angle AP_3C=90^\\circ$, draw $P_3H\\perp x$-axis through $P_3$, and $CK\\perp P_3H$ through C, \\\\\n$\\therefore \\angle P_3KC=\\angle AHP_3=90^\\circ$, $\\angle P_3CK=\\angle AP_3H$, \\\\\n$\\therefore \\triangle P_3CK\\sim \\triangle AP_3H$, \\\\\n$\\therefore \\frac{P_3K}{CK}=\\frac{AH}{P_3H}$ \\\\\n$\\therefore \\frac{-m^2+4m-3}{m-1}=\\frac{4-m}{-m^2+4m}$, \\\\\n$\\because$ Point P does not coincide with points A or C, \\\\\n$\\therefore m\\neq 1$ and $m\\neq 4$, \\\\\n$\\therefore -m^2-3m+1=0$, \\\\\n$\\therefore m=\\frac{3\\pm \\sqrt{5}}{2}$, \\\\\n$\\therefore$ As illustrated, the conditions are satisfied. \\boxed{Q_1(5,1)}, \\boxed{Q_2(-4,-2)}.",
    "image_path": "dataset/mm_math/images/MM_Math/52936243_166.png"
  },
  {
    "solution": "\\boxed{x<2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the graph of the linear function $y=kx+b$, what is the solution set of the linear inequality $kx+b>0$?\n",
    "completion": "\\textbf{Solution:} From the graph, it is known that $y$ decreases as $x$ increases, and when $x<2$, the line is above the x-axis,\\\\\ntherefore, the solution set for $kx+b>0$ is: $\\boxed{x<2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51698893.png"
  },
  {
    "solution": "\\boxed{\\frac{32}{13}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Synthesis and Inquiry\\\\\nAs shown in the figure, the parabola $y=\\frac{3}{4}x^{2}-\\frac{15}{4}x+3$ intersects the x-axis at points $A$ and $B$, with point $A$ located to the left of point $B$, and intersects the y-axis at point $C$. Point $P$ starts from point $B$ and moves towards the endpoint $C$ along $BC$ at a speed of $1$ unit length per second. Simultaneously, point $Q$ starts from point $O$ and moves towards the endpoint $B$ at the same speed along the positive half of the x-axis. Once arrived, both points stop moving. Connect $PQ$. When $\\triangle BPQ$ is an isosceles triangle, please directly write down the time of movement.\n",
    "completion": "\\textbf{Solution:} Let the motion time be $t$. According to the problem statement, in order to form $\\triangle BPQ$, points P and Q must not coincide with B, hence the range of $t$ is $0<t<4$,\\\\\n$\\therefore PB=OQ=t$, $BQ=4-t$,\\\\\nAs shown in the figure, draw $PD\\perp x$ axis at point D, let the coordinates of P be $\\left(a, -\\frac{3}{4}a+3\\right)$, then $BD=4-a$, $PD=-\\frac{3}{4}a+3$,\\\\\nAccording to the Pythagorean theorem, in $\\triangle PDB$,\\\\\n$PD^2+DB^2=PB^2$,\\\\\n$\\left(-\\frac{3}{4}a+3\\right)^2+\\left(4-a\\right)^2=t^2$,\\\\\nWe find $a_1=4-\\frac{4}{5}t$, $a_2=4+\\frac{4}{5}t$ (this does not meet the conditions and is excluded),\\\\\n$\\therefore$The coordinates of point P are $\\left(4-\\frac{4t}{5}, \\frac{3t}{5}\\right)$,\\\\\n$\\because$The coordinates of point Q are $\\left(t, 0\\right)$,\\\\\n$\\therefore PQ^2=\\left(4-\\frac{4t}{5}-t\\right)^2+\\left(\\frac{3t}{5}\\right)^2=\\frac{90}{25}t^2-\\frac{72t}{5}+16$,\\\\\n(1) When $BP=BQ$,\\\\\nthat is, $t=4-t$,\\\\\nWe solve: $t=\\boxed{2}$;\\\\\n(2) When $BP=PQ$,\\\\\n$t^2=\\frac{90}{25}t^2-\\frac{72t}{5}+16$,\\\\\nWe solve: $t_1=\\boxed{\\frac{20}{13}}$, $t_2=4$ (this does not meet the conditions and is excluded),\\\\\n(3) When $BQ=PQ$,\\\\\n$\\left(4-t\\right)^2=\\frac{90}{25}t^2-\\frac{72t}{5}+16$,\\\\\nWe solve: $t_1=\\boxed{\\frac{32}{13}}$, $t_2=0$ (this does not meet the conditions and is excluded),\\\\\nIn summary: when $\\triangle BPQ$ is an isosceles triangle, the times are \\boxed{2} seconds, \\boxed{\\frac{20}{13}} seconds, \\boxed{\\frac{32}{13}} seconds.",
    "image_path": "dataset/mm_math/images/MM_Math/52719985_73.png"
  },
  {
    "solution": "\\boxed{40^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ and $CD$ are chords of the circle $\\odot O$, and $AB \\parallel CD$. If $\\angle AOC = 80^\\circ$, what is the measure of $\\angle BAD$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle AOC = 80^\\circ$,\\\\\nit follows that $\\angle ADC = \\frac{1}{2} \\angle AOC = 40^\\circ$,\\\\\nSince $AB \\parallel CD$,\\\\\nit follows that $\\angle BAD = \\angle ADC = \\boxed{40^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51433703.png"
  },
  {
    "solution": "\\boxed{\\frac{8}{3}\\pi -4\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, it is known that $AB$ is the diameter of $\\odot O$, and $AB=8C$, $D$ is a point on $\\odot O$, $OC\\parallel BD$, intersecting $AD$ at point $E$, and $BC$ is connected, $\\angle CBD=30^\\circ$. Calculate the area of the shaded region in the diagram (the result should retain $\\pi$).\n",
    "completion": "\\[ \\text{Solution: Connect } OD, \\]\n\\[ \\because \\angle CBD = \\angle OBC = 30^\\circ, \\]\n\\[ \\therefore \\angle BOD = 60^\\circ, \\]\n\\[ \\because OB = OD, \\]\n\\[ \\therefore \\triangle BOD \\text{ is an equilateral triangle}, \\]\n\\[ \\therefore \\text{Area of the shaded region} = \\text{Area of sector } BOD - \\text{Area of } \\triangle BOD = \\frac{60\\pi \\times 4^2}{360} - \\frac{1}{2} \\times 4 \\times \\frac{\\sqrt{3}}{2} \\times 4 = \\boxed{\\frac{8}{3}\\pi -4\\sqrt{3}}. \\]",
    "image_path": "dataset/mm_math/images/MM_Math/52872426_109.png"
  },
  {
    "solution": "\\boxed{48^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, AB is the diameter of $\\odot O$, D and C are the trisection points of arc BE, $\\angle COD=32^\\circ$. What is the degree measure of $\\angle E$?\n",
    "completion": "\\textbf{Solution:} Since D and C are the trisecting points of the arc BE,\\\\\nthus, arc DE = arc CD = arc BC,\\\\\nthus, $\\angle$BOC = $\\angle$COD = $\\angle$DOE = $32^\\circ$,\\\\\nthus, $\\angle$BOE = $3 \\times 32^\\circ = 96^\\circ$,\\\\\nthus, $\\angle$A = $\\frac{1}{2}\\angle$BOE = $48^\\circ$.\\\\\nTherefore, the answer is: $\\boxed{48^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52874569.png"
  },
  {
    "solution": "\\boxed{\\frac{64}{21}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, quadrilateral $OABC$ is a rectangle with points $A$ and $C$ on the coordinate axes. $\\triangle ODE$ is obtained by rotating $\\triangle OCB$ $90^\\circ$ clockwise about point $O$. Point $D$ is on the $x$ axis, and line $BD$ intersects the $y$ axis at point $F$ and $OE$ at point $H$. The lengths of segments $BC$ and $OC$ are the solutions of the system of equations $\\left\\{\\begin{array}{l}2x+3y=14 \\\\ 4x-5y=6\\end{array}\\right.$. Given that $OC > BC$, find the area of $\\triangle OFH$.\n",
    "completion": "\\textbf{Solution:} Given $y=-\\frac{2}{3}x+\\frac{8}{3}$, by setting $x=0$, we obtain $y=\\frac{8}{3}$\\\\\n$\\therefore F\\left(0, \\frac{8}{3}\\right)$\\\\\nLet the equation of line OE be $y=k_1x$,\\\\\nSubstituting point E yields $k_1=\\frac{1}{2}$,\\\\\n$\\therefore y=\\frac{1}{2}x$,\\\\\n$\\begin{cases} y=\\frac{1}{2}x \\\\ y=-\\frac{2}{3}x+\\frac{8}{3} \\end{cases}$,\\\\\nSolving this system gives $\\begin{cases} x=\\frac{16}{7} \\\\ y=\\frac{8}{7} \\end{cases}$,\\\\\n$\\therefore H\\left(\\frac{16}{7}, \\frac{8}{7}\\right)$,\\\\\nTherefore, the area of $\\triangle OFH$ $= \\frac{1}{2} \\times \\frac{8}{3} \\times \\frac{16}{7} = \\boxed{\\frac{64}{21}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52189515_148.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, within the same coordinate system, the line $l_{1}: y=-x+2$ intersects the x-axis at point P, and the line $l_{2}: y=ax-4$ passes through point P. What is the value of $a$?\n",
    "completion": "\\textbf{Solution:} Since $l_{1}\\colon y=-x+2$ intersects the x-axis at point P, \\\\\nit follows that $P(2,0)$, \\\\\nAlso, since $l_{2}\\colon y=ax-4$ passes through point P, \\\\\nthus $0=2a-4$ yields $a=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50431231_186.png"
  },
  {
    "solution": "\\boxed{\\frac{2}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that in $\\triangle ABC$, $\\angle C=90^\\circ$, and $\\tan A=\\frac{5}{12}$. After rotating $\\triangle ABC$ $90^\\circ$ counterclockwise around point $A$, we obtain $\\triangle ADE$, where point $B$ falls on point $D$, and point $C$ falls on point $E$. By connecting $BE$ and $CD$, and drawing the bisector $AN$ of $\\angle CAD$, which intersects line segment $BE$ at point $M$ and line segment $CD$ at point $N$, what is the value of $\\frac{AM}{AN}$?\n",
    "completion": "\\textbf{Solution:} Let's establish a Cartesian coordinate system with point C as the origin. Through point N, draw $NF\\perp AC$ extending to intersect BP at point P, and $NH\\perp AD$ intersecting at point H, and $NG\\perp x$-axis intersecting at point G. Through point D, draw $DQ\\perp x$-axis intersecting at point Q,\\\\\n$\\because$ $\\angle C=90^\\circ$, $\\tan A=\\frac{5}{12}$,\\\\\n$\\therefore$ let $BC=5k$, $AC=12k$, $AB=13k$,\\\\\nBy rotation, we can obtain: $AE=AC=12k$, $DE=BC=5k$, $AB=AD=13k$,\\\\\n$\\therefore$ $CQ=12k$, $DQ=7k$,\\\\\n$\\therefore$ $B(-5k,0)$, $E(12k,12k)$, $D(12k,7k)$,\\\\\n$\\because$ AN is the bisector of $\\angle CAD$,\\\\\n$\\therefore$ $NF=NH$,\\\\\n$\\therefore$ $\\frac{S_{\\triangle ANC}}{S_{\\triangle AND}}=\\frac{AC}{AD}=\\frac{12k}{13k}=\\frac{12}{13}$, hence $\\frac{CN}{DN}=\\frac{12}{13}$,\\\\\n$\\therefore$ $N(\\frac{144k}{25},\\frac{84k}{25})$,\\\\\nLet the equation of line BE be $y=mx+n$,\\\\\nSubstitute $B(-5k,0)$, $E(12k,12k)$ into it: $\\left\\{\\begin{array}{l}-5km+n=0 \\\\ 12km+n=12k\\end{array}\\right.$,\\\\\nSolving them, we get: $\\left\\{\\begin{array}{l}m=\\frac{12}{17} \\\\ n=\\frac{60k}{17}\\end{array}\\right.$,\\\\\n$\\therefore$ $y=\\frac{12}{17}x+\\frac{60k}{17}$,\\\\\nWhen $y=\\frac{84k}{25}$, $\\frac{12}{17}x+\\frac{60k}{17}=\\frac{84k}{25}$,\\\\\nSolving for $x$: $x=-\\frac{6k}{25}$,\\\\\n$\\therefore$ $P(-\\frac{6k}{25},\\frac{84k}{25})$,\\\\\n$\\therefore$ $NP=\\frac{144k}{25}-(-\\frac{6k}{25})=6k$,\\\\\n$\\because$ $NF\\perp AC$, $\\angle CAE=90^\\circ$,\\\\\n$\\therefore$ $\\angle CFN=\\angle CAE=90^\\circ$,\\\\\n$\\therefore$ $AE\\parallel PN$,\\\\\n$\\therefore$ $\\triangle MAE\\sim \\triangle MNP$,\\\\\n$\\therefore$ $\\frac{AM}{NM}=\\frac{AE}{NP}=\\frac{12k}{6k}=2$,\\\\\n$\\therefore$ $\\frac{AM}{AN}=\\boxed{\\frac{2}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51379554.png"
  },
  {
    "solution": "\\boxed{16}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point \\(C\\) is a point on segment \\(AB\\), point \\(M\\) is the midpoint of segment \\(AC\\), and point \\(N\\) is the midpoint of segment \\(BC\\). If \\(MN = 8 \\, \\text{cm}\\), what is the length of \\(AB\\) in centimeters?\n",
    "completion": "\\textbf{Solution:} Since point M is the midpoint of line segment AC, and point N is the midpoint of line segment BC, \\\\\n$\\therefore$ BC$=2$NC, AC$=2$MC, \\\\\nSince MN$=$NC$+$MC$=8$cm, \\\\\n$\\therefore$ AB$=$BC$+$AC$=2$MN$=2\\times 8=\\boxed{16}$cm.",
    "image_path": "dataset/mm_math/images/MM_Math/52916470_78.png"
  },
  {
    "solution": "\\boxed{9\\,\\text{cm}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, there are three points $C$, $D$, $E$ sequentially placed on the segment $AB$, which divide the segment $AB$ into four parts with ratios $2:3:4:5$. Given that $AB=28$, and if $M$, $N$ are the midpoints of segments $DE$ and $EB$ respectively, what is the length of segment $MN$?\n",
    "completion": "Solution: From the given problem,\\\\\nsince $M$ is the midpoint of $DE$,\\\\\ntherefore $ME = \\frac{1}{2}DE = 4\\,\\text{cm}$,\\\\\nsince $N$ is the midpoint of $EB$,\\\\\ntherefore $EN = \\frac{1}{2}EB = 5\\,\\text{cm}$,\\\\\ntherefore $MN = ME + EN = \\boxed{9\\,\\text{cm}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52794225_129.png"
  },
  {
    "solution": "\\boxed{4.8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, $CD\\perp AB$ with foot D, $AF$ bisects $\\angle CAB$ and intersects $CD$ at point E and $CB$ at point F. If $AC=6$ and $BC=8$, directly write out the length of $CD$.\n",
    "completion": "\\textbf{Solution:} We have $CD = \\boxed{4.8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52765791_141.png"
  },
  {
    "solution": "\\boxed{100^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $\\angle MON=40^\\circ$, P is a fixed point inside $\\angle MON$, point A is on OM, and point B is on ON. When the perimeter of $\\triangle PAB$ is at its minimum, what is the degree measure of $\\angle APB$?\n",
    "completion": "\\textbf{Solution:} Construct the symmetric points $P_{1}$ and $P_{2}$ of $P$ with respect to $OM$ and $ON$, respectively. Draw the line segment $P_{1}P_{2}$, intersecting $OM$ and $ON$ at points $A$ and $B$, respectively.\\\\\n$\\therefore$ $PA=P_{1}A$, $PB=P_{2}B$, at this moment the perimeter of $\\triangle PAB$ is minimal,\\\\\n$\\therefore$ $\\angle P_{1}PA=\\angle P_{1}$, $\\angle P_{2}PB=\\angle P_{2}$,\\\\\nAccording to the problem, $\\angle P_{1}PP_{2}=180^\\circ-\\angle MON=180^\\circ-40^\\circ=140^\\circ$,\\\\\n$\\therefore$ $\\angle P_{1}PA+\\angle P_{2}PB=\\angle P_{1}+\\angle P_{2}=180^\\circ-\\angle P_{1}PP_{2}=40^\\circ$,\\\\\n$\\therefore$ $\\angle APB=140^\\circ-40^\\circ=\\boxed{100^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52733748.png"
  },
  {
    "solution": "\\boxed{x<1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=x+2$ intersects with the line $y=ax+4$ at point $P(m, 3)$. What is the solution set of the inequality $x+2<ax+4$ with respect to $x$?\n",
    "completion": "\\textbf{Solution:} Given the problem statement, we know that point $P$ is also on $y=x+2$. Thus, when $y=3$, $x=1$\\\\\n$\\therefore$ when $x<1$, $x+2<ax+4$\\\\\nHence, the answer is: $\\boxed{x<1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52386831.png"
  },
  {
    "solution": "\\boxed{100^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC=4$, $\\angle B=\\angle C=40^\\circ$. Point D moves on line segment $BC$ (D does not coincide with B or C). Connect $AD$ and construct $\\angle ADE=40^\\circ$, with $DE$ intersecting line segment $AC$ at $E$. When $\\angle BDA=100^\\circ$, what is the value of $\\angle DEC$?\n",
    "completion": "\\textbf{Solution:} Given that in $\\triangle BAD$, $\\angle B=\\angle C=40^\\circ$, $\\angle BDA=100^\\circ$, \\\\\nthus, $\\angle BAD=180^\\circ-\\angle B-\\angle BDA=180^\\circ-40^\\circ-100^\\circ=40^\\circ$;\\\\\n$\\angle EDC=180^\\circ-\\angle ADB-\\angle ADE=180^\\circ-100^\\circ-40^\\circ=40^\\circ$.\\\\\n$\\angle DEC=180^\\circ-\\angle C-\\angle EDC=180^\\circ-40^\\circ-40^\\circ=\\boxed{100^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52913913_102.png"
  },
  {
    "solution": "\\boxed{40^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the lines $l_{1}$ and $l_{2}$ intersect two sides $AB$ and $BC$ of $\\triangle ABC$ respectively, and $l_{1} \\parallel l_{2}$. If $\\angle B=35^\\circ$ and $\\angle 1=105^\\circ$, what is the measure of $\\angle 2$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle B = 35^\\circ$ and $\\angle 1 = 105^\\circ$,\n\nit follows that $\\angle 3 = 180 - \\angle 1 - \\angle B = 40^\\circ$,\n\nGiven $l_{1} \\parallel l_{2}$,\n\ntherefore $\\angle 2 = \\angle 3 = \\boxed{40^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51435311.png"
  },
  {
    "solution": "\\boxed{x<1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As the figure shows, the line $y=kx-\\frac{3}{2}$ intersects with $y=mx-5$ at point $A\\left(1, -3\\right)$. What is the solution set of the inequality $kx-\\frac{3}{2}>mx-5$ with respect to $x$?\n",
    "completion": "\\textbf{Solution:} By observing the graph, we know that when $x=1$, $kx-\\frac{3}{2}=mx-5$.\\\\\nTherefore, when $x<1$, $kx-\\frac{3}{2}>mx-5$.\\\\\nThus, the answer is $\\boxed{x<1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52395569.png"
  },
  {
    "solution": "\\boxed{y=x^{2}-4x+4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the vertex $C$ of the parabola passing through points $A\\left(0, 4\\right)$ and $B\\left(5, 9\\right)$ is located on the positive half of the $x$-axis. Find the equation of the parabola.\n",
    "completion": "\\textbf{Solution:} Since the vertex of the parabola is on the x-axis, we can assume the equation of the parabola to be $y=a(x-h)^{2}$. Therefore,\n\\[\n\\left\\{\n\\begin{array}{l}\na(0-h)^{2}=4\\\\\na(5-h)^{2}=9\n\\end{array}\n\\right.\n\\]\nSolving this gives $h_{1}=2$, $h_{2}=-10$ (which is not consistent with the problem statement). When $h=2$, $a=1$, therefore the equation of the parabola is $y=(x-2)^{2}=x^{2}-4x+4$. Thus, the equation of the parabola is $\\boxed{y=x^{2}-4x+4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51369908_192.png"
  },
  {
    "solution": "\\boxed{-2<x<-1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y=-x+1$ intersects with the graph of $y=2x+m$ at point $P\\left(n, 2\\right)$. What is the solution set of the inequality $-x+1>2x+m>0$ with respect to $x$?\n",
    "completion": "Solution: Since the graph of the linear function $y=-x+1$ passes through point $P(-1,2)$,\nwe have $2=-(-1)+1$, solving this gives $n=-1$,\ntherefore $P(-1,2)$,\nsubstituting $P(-1,2)$ into $y=2x+m$, we get $2=-2+m$,\nsolving this gives $m=4$,\nthus $y=2x+4$,\nwhen $y=0$, $2x+4=0$, solving this gives $x=-2$,\ntherefore, the intersection of $y=2x+4$ and the x-axis is $(-2,0)$,\nthus, the solution set of the inequality in $x$, $-x+1>2x+m>0$, is $-2<x<-1$.\nTherefore, the answer is $\\boxed{-2<x<-1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50553935.png"
  },
  {
    "solution": "\\boxed{6\\sqrt{3}-2\\pi}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the diameter AB of the circle $\\odot O$ forms an angle $\\angle A=30^\\circ$ with the chord AC, where $AC=CP$. $CP$ is a tangent to $\\odot O$; if $PC=6$, what is the area of the shaded region in the figure?\n",
    "completion": "\\textbf{Solution:}\n\nSince $AC=PC$,\n\nit follows that $\\angle P=\\angle A=30^\\circ$,\n\nthus $\\angle COB=60^\\circ$ and $OP=2OC$.\n\nGiven $PC=6$,\n\nwe have $OP^2-OC^2=6^2$,\n\nwhich leads to $OC=2\\sqrt{3}$.\n\nTherefore, the area of the shaded region in the diagram $S_{\\triangle POC}-S_{\\text{sector} COB}$\n\n$=\\frac{1}{2}\\times 6\\times 2\\sqrt{3}-\\frac{60^\\circ \\cdot \\pi \\times (2\\sqrt{3})^2}{360^\\circ}=6\\sqrt{3}-2\\pi$.\n\nHence, the area of the shaded region in the diagram is $\\boxed{6\\sqrt{3}-2\\pi}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53019892_82.png"
  },
  {
    "solution": "\\boxed{\\sqrt{3} - 1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle A=90^\\circ$, $\\sin\\angle ABC=\\frac{1}{2}$. Circle $O$ is the inscribed circle of $\\triangle ABC$, tangent to $AC$, $AB$, $BC$ at points $F$, $P$, $E$, respectively. If $BC=4$, then what is the length of $AP$?\n",
    "completion": "\\textbf{Solution:} We have $AP = \\boxed{\\sqrt{3} - 1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55502222_59.png"
  },
  {
    "solution": "\\boxed{\\frac{20\\sqrt{3}}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within circle $O$, chord $AB$ perpendicularly bisects radius $OC$. If the length of chord $AB$ is $10$, what is the diameter of circle $O$?\n",
    "completion": "\\textbf{Solution:} Given that AB is a chord, OC is the radius, and $OC\\perp AB$ with $AB=10$,\\\\\nhence $BD= \\frac{1}{2}AB=5$,\\\\\nsince the chord AB bisects the radius OC vertically and $OB=OC$,\\\\\nthus $OD= \\frac{1}{2}OC= \\frac{1}{2}OB$,\\\\\nin $\\triangle ODB$, we have $OD^{2}+BD^{2}=OB^{2}$, that is $\\left( \\frac{1}{2}OB\\right)^{2}+5^{2}=OB^{2}$,\\\\\nsolving this, we get: $OB= \\frac{10\\sqrt{3}}{3}$ (neglecting the negative value),\\\\\ntherefore, the diameter of $\\odot O$ is $2\\times \\frac{10\\sqrt{3}}{3}=\\boxed{\\frac{20\\sqrt{3}}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52977640_148.png"
  },
  {
    "solution": "\\boxed{\\frac{8}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $DC=6$. There is a point $E$ on $DC$. By folding $\\triangle ADE$ along the line $AE$, such that point $D$ falls exactly on the side $BC$, and let this point be $F$. If the area of $\\triangle ABF$ is $24$, what is the length of $CE$?\n",
    "completion": "\\textbf{Solution:} We have $CE = \\boxed{\\frac{8}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53065890_52.png"
  },
  {
    "solution": "\\boxed{8\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the diameter $AB=20$ of the semicircle $O$, and the chord $AC=12$. The chord $AD$ bisects $\\angle BAC$. What is the length of $AD$?\n",
    "completion": "\\textbf{Solution:} Connect BD, BC, OD, intersecting BC at point F,\\\\\n$\\because$ AB is the diameter,\\\\\n$\\therefore \\angle ACB = \\angle ADB = 90^\\circ$,\\\\\n$\\therefore BC = \\sqrt{AB^2 - AC^2} = \\sqrt{20^2 - 12^2} = 16$,\\\\\n$\\because$ AD bisects $\\angle BAC$,\\\\\n$\\therefore \\angle CAD = \\angle DAB$,\\\\\n$\\therefore \\overset{\\frown}{CD} = \\overset{\\frown}{BD}$,\\\\\n$\\therefore$ OD$\\perp$CB,\\\\\n$\\therefore$ BF = $\\frac{1}{2}BC = 8$, $\\angle BFO = 90^\\circ$,\\\\\n$\\therefore OF = \\sqrt{OB^2 - BF^2} = \\sqrt{10^2 - 8^2} = 6$,\\\\\n$\\therefore DF = OD - OF = 10 - 6 = 4$,\\\\\nIn the right $\\triangle DFB$,\\\\\n$BD = \\sqrt{DF^2 + BF^2} = \\sqrt{4^2 + 8^2} = 4\\sqrt{5}$,\\\\\nIn the right $\\triangle ABD$,\\\\\n$AD = \\sqrt{AB^2 - BD^2} = \\sqrt{20^2 - (4\\sqrt{5})^2} = \\boxed{8\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53081442.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the figure, in the equilateral triangle $ABC$ with a side length of $2\\sqrt{3}$, the moving points $D$ and $E$ are on the sides $BC$ and $AC$, respectively, maintaining $AE=CD$. Lines $BE$ and $AD$ are drawn and intersect at point $P$. What is the minimum value of $CP$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram:\n\n$\\because$ CD = AE, BC = AC,\n\n$\\therefore$ BD = CE,\n\nIn $\\triangle ABD$ and $\\triangle BCE$,\n\n$\\left\\{\\begin{array}{l}\nAB=BC\\\\\n\\angle ABD=\\angle BCE\\\\\nBD=CE\n\\end{array}\\right.$,\n\n$\\therefore \\triangle ABD \\cong \\triangle BCE$ (SAS),\n\n$\\therefore \\angle BAD=\\angle CBE$,\n\n$\\because \\angle APE=\\angle ABE+\\angle BAD$, $\\angle APE=\\angle BPD$, $\\angle ABE+\\angle CBE=60^\\circ$,\n\n$\\therefore \\angle BPD=\\angle APE=\\angle ABC=60^\\circ$,\n\n$\\therefore \\angle APB=120^\\circ$,\n\n$\\therefore$ the path of point P is $\\overset{\\frown}{AB}$, with $\\angle AOB=120^\\circ$, connect CO,\n\n$\\because$ OA = OB, CA = CB, OC = OC,\n\n$\\therefore \\triangle AOC \\cong \\triangle BOC$ (SSS),\n\n$\\therefore \\angle OAC=\\angle OBC$, $\\angle ACO=\\angle BCO=30^\\circ$,\n\n$\\because \\angle AOB+\\angle ACB=180^\\circ$,\n\n$\\therefore \\angle OAC+\\angle OBC=180^\\circ$,\n\n$\\therefore \\angle OAC=\\angle OBC=90^\\circ$,\n\n$\\therefore OC=AC \\div \\cos 30^\\circ=4$, OA= $\\frac{1}{2}$ OC=2,\n\n$\\therefore OP=2$,\n\n$\\because PC \\geq OC-OP$,\n\n$\\therefore PC \\geq 2$,\n\n$\\therefore$ The minimum value of PC is $\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51384545.png"
  },
  {
    "solution": "\\boxed{55^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C = 90^\\circ$, $\\angle A = 20^\\circ$, $BD$ is the bisector of $\\angle ABC$, then what is the measure of $\\angle BDC$ in degrees?\n",
    "completion": "\\textbf{Solution:} Given that in $\\triangle ABC$, $\\angle C=90^\\circ$, $\\angle A=20^\\circ$,\\\\\nthus $\\angle ABC=180^\\circ-\\angle C-\\angle A=70^\\circ$,\\\\\nsince $BD$ is the bisector of $\\angle ABC$,\\\\\nthus $\\angle DBC=\\frac{1}{2}\\angle ABC=35^\\circ$,\\\\\nin $\\triangle DBC$,\\\\\nsince $\\angle C=90^\\circ$, $\\angle DBC=35^\\circ$,\\\\\nthus $\\angle BDC=180^\\circ-\\angle C-\\angle DBC=\\boxed{55^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53033285.png"
  },
  {
    "solution": "\\boxed{\\frac{4}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the square $ABCD$ with a side length of $4$, where point $E$ is a point on side $CD$, and point $F$ is the reflection of point $D$ across line $AE$. Connecting $AF$ and $BF$, if $\\tan \\angle ABF = 2$, what is the length of $DE$?\n",
    "completion": "\\textbf{Solution:} Draw a line FN $\\perp$ AB at point N, and extend NF to intersect CD at point M,\\\\\nsince AB$\\parallel$CD,\\\\\nthus MN$\\perp$CD,\\\\\nthus $\\angle$FME $= 90^\\circ$,\\\\\nsince $\\tan\\angle$ABF $= 2$,\\\\\nthus $\\frac{FN}{BN} = 2$,\\\\\nlet BN $= x$, then FN $= 2x$,\\\\\nthus AN $= 4 - x$,\\\\\nsince point F is symmetric to point D with respect to line AE,\\\\\nthus DE $= EF$, DA $= AF = 4$,\\\\\nsince AE $= AE$,\\\\\nthus $\\triangle$ADE $\\cong$ $\\triangle$AFE (SSS),\\\\\nthus $\\angle$D $= \\angle$AFE $= 90^\\circ$,\\\\\nsince AN$^2$ + NF$^2$ = AF$^2$,\\\\\nthus $(4 - x)^2 + (2x)^2 = 4^2$,\\\\\nthus $x_1 = 0$ (discard), $x_2 = \\frac{8}{5}$,\\\\\nthus AN $= 4 - x = 4 - \\frac{8}{5} = \\frac{12}{5}$, MF $= 4 - 2x = 4 - \\frac{16}{5} = \\frac{4}{5}$,\\\\\nsince $\\angle$EFM + $\\angle$AFN $= \\angle$AFN + $\\angle$FAN $= 90^\\circ$,\\\\\nthus $\\angle$EFM $= \\angle$FAN,\\\\\nthus $\\cos\\angle$EFM $= \\cos\\angle$FAN,\\\\\nthus $\\frac{FM}{EF} = \\frac{AN}{AF}$, thus $\\frac{\\frac{4}{5}}{EF} = \\frac{\\frac{12}{5}}{4}$,\\\\\nthus EF $= \\frac{4}{3}$,\\\\\nthus DE $= EF = \\boxed{\\frac{4}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51212656.png"
  },
  {
    "solution": "\\boxed{a \\ parallelogram}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, $DE \\perp AC$, $BF \\perp AC$, with the perpendicular feet being $E$ and $F$ respectively. Extend $DE$ and $BF$ to meet $AB$ at point $H$ and $BC$ at point $G$ respectively. If $AD \\parallel BC$ and $AE=CF$, then prove that quadrilateral $ABCD$ is a parallelogram.\n",
    "completion": "Proof:\n\nSince $DE \\perp AC$ and $BF \\perp AC$, \n\nit follows that $\\angle AED = \\angle CFB = 90^\\circ$ and $DE \\parallel BF$.\n\nGiven $AD \\parallel BC$,\n\nit implies $\\angle DAE = \\angle BCF$.\n\nIn $\\triangle DAE$ and $\\triangle BCF$,\n\n\\[\n\\left\\{\\begin{array}{l}\n\\angle DEA = \\angle BFC = 90^\\circ \\\\\nAE = CF \\\\\n\\angle DAE = \\angle BCF\n\\end{array}\\right.\n\\]\n\nTherefore, $\\triangle DAE \\cong \\triangle BCF$ (ASA),\n\nwhich means $AD = CB$.\n\nSince $AD \\parallel BC$,\n\nthe quadrilateral $ABCD$ is \\boxed{a \\ parallelogram}.",
    "image_path": "dataset/mm_math/images/MM_Math/52354589_102.png"
  },
  {
    "solution": "\\boxed{17280}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, there is an irregular quadrilateral plot of land, ABCD, at a certain middle school in Liupanshui City. The school plans to pave the space with a floating floor. After measurement, $\\angle ABC=90^\\circ$, $BC=6\\text{m}$, $AB=8\\text{m}$, $AD=26\\text{m}$, and $CD=24\\text{m}$. If it costs $120\\text{ yuan}$ to pave $1\\text{ square meter}$ of floating floor, how much in total will need to be invested?\n",
    "completion": "\\textbf{Solution:} Solve: $144\\times 120=\\boxed{17280}$ (yuan).",
    "image_path": "dataset/mm_math/images/MM_Math/51367124_121.png"
  },
  {
    "solution": "\\boxed{38^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, point D is on side AB. $\\triangle CBD$ is folded along CD such that point B falls exactly on point E on side AC. If $\\angle A=26^\\circ$, what is the degree measure of $\\angle ADE$?\n",
    "completion": "\\textbf{Solution:} From the folding, it can be seen that $\\angle ACD=\\angle BCD$, $\\angle BDC=\\angle CDE$,\\\\\n$\\because$ $\\angle ACB=90^\\circ$,\\\\\n$\\therefore$ $\\angle ACD=45^\\circ$,\\\\\n$\\because$ $\\angle A=26^\\circ$,\\\\\n$\\therefore$ $\\angle BDC=\\angle ACD+\\angle A=71^\\circ$,\\\\\n$\\therefore$ $\\angle CDE=71^\\circ$,\\\\\n$\\therefore$ $\\angle ADE=180^\\circ-71^\\circ-71^\\circ=\\boxed{38^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53078478.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the Cartesian coordinate system, the position of square $ABCD$ is shown in the diagram, with point $A$ having coordinates $(0, 2)$ and point $B$ having coordinates $(-3, 0)$. What is the distance from point $C$ to the $y$-axis?\n",
    "completion": "\\textbf{Solution:} Construct line CE perpendicular to the x-axis at point E, as shown in the diagram,\\\\\nhence, the distance from point C to the y-axis is OE.\\\\\n$\\because$ The coordinates of point A are $(0,2)$, and the coordinates of point B are $(-3,0)$,\\\\\n$\\therefore OA=2$, $OB=3$.\\\\\n$\\because CE \\perp x$-axis,\\\\\n$\\therefore \\angle CEB = 90^\\circ$.\\\\\n$\\therefore \\angle ECB + \\angle EBC = 90^\\circ$.\\\\\n$\\because$ Quadrilateral ABCD is a square,\\\\\n$\\therefore BC=AB$, $\\angle CBA=90^\\circ$.\\\\\n$\\therefore \\angle EBC + \\angle ABO = 90^\\circ$.\\\\\n$\\therefore \\angle ECB = \\angle ABO$.\\\\\nIn $\\triangle CBE$ and $\\triangle BAO$,\\\\\n$\\left\\{\\begin{array}{l}\n\\angle ECB=\\angle OBA\\\\\n\\angle CEB=\\angle BOA=90^\\circ\\\\\nCB=BA\n\\end{array}\\right.$,\\\\\n$\\therefore \\triangle CBE \\cong \\triangle BAO$ (AAS).\\\\\n$\\therefore EB=OA=2$.\\\\\n$\\therefore OE=OB+BE=3+2=\\boxed{5}$.\\\\\n$\\therefore$ The distance from point C to the y-axis is 5.",
    "image_path": "dataset/mm_math/images/MM_Math/52250936.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $O$ is the intersection point of the diagonals of rectangle $ABCD$, with $DE \\parallel AC$ and $CE \\parallel BD$. Quadrilateral $OCED$ is a rhombus. If $AB=3$ and $BC=4$, what is the area of the quadrilateral $OCED$?\n",
    "completion": "\\textbf{Solution:} Given that $AB=3$, $BC=4$, \\\\\nTherefore, the area of rectangle $ABCD$ is equal to $3\\times 4=12$, \\\\\nSince the area of $\\triangle ODC=\\frac{1}{4}$ of the area of rectangle $ABCD=3$, \\\\\nHence, the area of quadrilateral $OCED$ equals $2$ times the area of $\\triangle ODC=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53437314_121.png"
  },
  {
    "solution": "\\boxed{3\\sqrt{5}−2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In rectangle $ABCD$, $AB = 3$, $AD = 5$, the point $E$ is on side $AD$ with $AE = 4$. Point $P$ is inside the rectangle and $\\angle APE = 90^\\circ$. Point $M$ is on side $BC$. Connecting $PA$, $DM$, what is the minimum value of $PM + DM$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, \\\\\n$\\because$ $\\angle APE=90^\\circ$, \\\\\n$\\therefore$ point P moves on circle O with diameter $AE$, \\\\\nConstruct point G as the mirror image of D with respect to $BC$, connect $OG$ to intersect circle $O$ at point P and $BC$ at M, the minimum value of $PM+DM$ is the length of $PG$, \\\\\n$\\because$ quadrilateral $ABCD$ is a rectangle, \\\\\n$\\therefore$ $CG=CD=AB=3$, $\\angle ADC=90^\\circ$, \\\\\nIn $\\triangle ODG$, $DG=CD+CG=6$, $OD=AD−OA=5−2=3$, \\\\\n$\\therefore$ $OG=\\sqrt{OD^{2}+DG^{2}}=3\\sqrt{5}$, \\\\\n$\\therefore$ $PG=OG−OP=3\\sqrt{5}−2$, \\\\\n$\\therefore$ the minimum value of $PM+DM$ is: $\\boxed{3\\sqrt{5}−2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53147473.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in square $ABCD$, the diagonals $AC$ and $BD$ intersect at point $O$. Let $M$ be a point on side $AD$, and draw $OM$. Construct $ON \\perp OM$ intersecting $CD$ at point $N$. If the area of quadrilateral $MOND$ is $1$, what is the length of $AB$?\n",
    "completion": "\\textbf{Solution:} In square ABCD, diagonal BD $\\perp$ AC, \\\\\n$\\therefore \\angle AOD=90^\\circ$ \\\\\n$\\because ON\\perp OM$ \\\\\n$\\therefore \\angle MON=90^\\circ$ \\\\\n$\\therefore \\angle AOM=\\angle DON$ \\\\\nAlso, $\\because \\angle MAO=\\angle NDO=45^\\circ$, $AO=DO$ \\\\\n$\\therefore \\triangle MAO \\cong \\triangle NDO$ (ASA) \\\\\n$\\therefore S_{\\triangle MAO}=S_{\\triangle NDO}$ \\\\\n$\\because$ the area of quadrilateral MOND is 1, \\\\\n$\\therefore S_{\\triangle DAO}=1$ \\\\\n$\\therefore$ the area of square ABCD is 4, \\\\\n$\\therefore AB^2=4$ \\\\\n$\\therefore AB=\\boxed{2}$",
    "image_path": "dataset/mm_math/images/MM_Math/52423824.png"
  },
  {
    "solution": "\\boxed{13}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $A$, $B$, and $C$ are three consecutive points on line $L$, with $M$ being the midpoint of $AB$ and $N$ being the midpoint of $MC$. Given that $AB=6\\,\\text{cm}$ and $NC=8\\,\\text{cm}$, what is the length of $BC$ in centimeters?\n",
    "completion": "Solution: Since M is the midpoint of AB,\\\\\nit follows that AM = BM = $\\frac{1}{2}$AB = 3,\\\\\nSince N is the midpoint of MC,\\\\\nit follows that MN = NC = 8.\\\\\nTherefore, BN = MN - BM = 5,\\\\\nTherefore, BC = BN + NC = 5 + 8 = \\boxed{13} (cm)",
    "image_path": "dataset/mm_math/images/MM_Math/17130735.png"
  },
  {
    "solution": "\\boxed{75^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle COD$ is the shape obtained by rotating $\\triangle AOB$ clockwise around point $O$ by $30^\\circ$, and point $C$ is exactly on $AB$. What is the degree measure of $\\angle A$?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle COD$ is the shape obtained by rotating $\\triangle AOB$ $30^\\circ$ clockwise around point O,\\\\\nthus $\\angle AOC=30^\\circ$, $OA=OC$,\\\\\ntherefore $\\angle A=\\frac{180^\\circ-\\angle AOC}{2}=75^\\circ$,\\\\\nhence, the answer is: $\\boxed{75^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51212771.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, square $EFGH$ is inscribed in right-angled triangle $ABC$, with $\\angle A=90^\\circ$ and $BC=12$. If the area of $\\triangle ABC$ is 36, what is the length of $EH$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $AD\\perp BC$ at $D$, intersecting $EH$ at $M$, \\\\\n$\\because$ The area of $\\triangle ABC$ is $36$, $BC=12$, \\\\\n$\\therefore$ $\\frac{1}{2}BC\\times AD=36$, \\\\\n$\\therefore$ $\\frac{1}{2}\\times 12\\times AD=36$, \\\\\n$\\therefore$ $AD=6$, \\\\\n$\\because$ Square $EFGH$ is inscribed in $\\angle ABC$, \\\\\n$\\therefore$ $EH\\parallel FG$, let $EH=EF=FG=HG=x$, \\\\\n$\\therefore$ $\\angle AEH=\\angle B$, $\\angle AHE=\\angle C$, \\\\\n$\\therefore$ $\\triangle AEH\\sim \\triangle ABC$, \\\\\n$\\therefore$ $\\frac{AE}{AB}=\\frac{EH}{BC}$, \\\\\n$\\because$ $AD\\perp BC$, \\\\\n$\\therefore$ $\\angle ADG=90^\\circ$, \\\\\n$\\because$ $EH\\parallel FG$, \\\\\n$\\therefore$ $\\angle ADG=\\angle AMH=90^\\circ$, \\\\\n$\\therefore$ $AM\\perp EH$, \\\\\nAgain, $\\because$ $EH\\parallel FG$, $AD\\perp BC$, \\\\\n$\\therefore$ $DM=EF=EH=x$, $\\frac{AE}{AB}=\\frac{AM}{AD}$, \\\\\n$\\therefore$ $AM=6-x$, \\\\\n$\\because$ $\\frac{AE}{AB}=\\frac{EH}{BC}$, \\\\\n$\\therefore$ $\\frac{EH}{BC}=\\frac{AM}{AD}$, \\\\\n$\\therefore$ $\\frac{x}{12}=\\frac{6-x}{6}$, \\\\\n$\\therefore$ $x=\\boxed{4}$, \\\\\n$\\therefore$ $EH=4$.",
    "image_path": "dataset/mm_math/images/MM_Math/52917632.png"
  },
  {
    "solution": "\\boxed{(1,3)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y=-x+4$ intersects with the graph of the inverse proportion function $y=\\frac{k}{x}$ at points $A(3, a)$ and $B$, and intersects the y-axis at point $C$. Based on the graph, please directly write the solution set of the inequality $-x+4>\\frac{k}{x}$.\n",
    "completion": "\\textbf{Solution:} The solution set of the inequality $-x+4>\\frac{k}{x}$ is $1<x<3$. Hence, the solution set is $x \\in \\boxed{(1,3)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51353246_281.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{5}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a semicircle $O$ with diameter $AB=6$ is rotated counterclockwise about endpoint $A$. When the intersection point $H$ on the circular arc and the diameter satisfies $BH:AH=1:2$, what is the value of $\\tan\\angle B'AB$?\n",
    "completion": "\\textbf{Solution:} Connect $HB'$,\n\n$\\because$ When semi-circle $O$ with diameter $AB=6$ is rotated counterclockwise around endpoint $A$,\n\n$\\therefore$ $AB'=AB=6$, $\\angle AHB'=90^\\circ$,\n\n$\\because$ $AH+BH=6$, $BH\\colon AH=1\\colon 2$,\n\n$\\therefore$ $BH=2$, $AH=4$\n\n$\\therefore$ $B'H=\\sqrt{AB'^2-AH^2}=\\sqrt{6^2-4^2}=2\\sqrt{5}$;\n\nIn $\\triangle AHB'$,\n\n$\\tan\\angle B'AB=\\frac{B'H}{AH}=\\frac{2\\sqrt{5}}{4}=\\boxed{\\frac{\\sqrt{5}}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51369888.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, where $AB=AC=4$, $D$ and $E$ are the midpoints of $AB$ and $AC$, respectively. Connect $CD$, and draw $EF \\parallel DC$ through $E$ to intersect the extension of $BC$ at $F$; $DE=CF$; If $\\angle B=60^\\circ$, find the length of $EF$.\n",
    "completion": "\\textbf{Solution:} Given $\\because AB=AC=4$, $\\angle B=60^\\circ$,\\\\\n$\\therefore BC=AB=AC=4$,\\\\\nAlso, $\\because D$ is the midpoint of $AB$,\\\\\n$\\therefore CD\\perp AB$,\\\\\n$\\therefore$ in $\\triangle BCD$,\\\\\n$BD= \\frac{1}{2}AB=2$,\\\\\n$\\therefore CD= \\sqrt{BC^{2}-BD^{2}}=2\\sqrt{3}$,\\\\\n$\\because$ quadrilateral $CDEF$ is a parallelogram,\\\\\n$\\therefore EF=CD=2\\sqrt{3}$,\\\\\nThus, the length of $EF$ is $\\boxed{2\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53232760_141.png"
  },
  {
    "solution": "\\boxed{(0, -3)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system \\(xOy\\), the parabola \\(y=x^2+bx+c\\) intersects with the x-axis at points A \\((-1,0)\\) and B \\((3,0)\\), and intersects with the y-axis at point C, with the vertex being point D. Find the expression of the parabola and the coordinates of point C?\n",
    "completion": "\\textbf{Solution:} Substitute A ($-1$, $0$) and B ($3$, $0$) into $y=x^{2}+bx+c$, we have\n\\[\n\\left\\{\n\\begin{array}{l}\n1-b+c=0, \\\\\n9+3b+c=0.\n\\end{array}\n\\right.\n\\]\nSolving we get:\n\\[\n\\left\\{\n\\begin{array}{l}\nb=-2, \\\\\nc=-3.\n\\end{array}\n\\right.\n\\]\nTherefore, $y=x^{2}-2x-3$.\n\nWhen $x=0$, $y=-3$. $\\therefore$ The coordinates of point C are $\\boxed{(0, -3)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51379494_83.png"
  },
  {
    "solution": "\\boxed{7.5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: It is known that the length of segment $AB=20$. $M$ is the midpoint of segment $AB$, and $P$ is any point on segment $AB$. $N$ is the midpoint of segment $PB$. When $P$ is the midpoint of segment $AM$, what is the length of segment $NB$?\n",
    "completion": "\\textbf{Solution}: As shown in the figure, since $M$ is the midpoint of the segment $AB$, $AB=20$\\\\\ntherefore $MA=\\frac{1}{2}AB=10$\\\\\nsince $P$ is the midpoint of the segment $AM$,\\\\\ntherefore $AP=\\frac{1}{2}AM=5$\\\\\ntherefore $PB=AB-AP=20-5=15$\\\\\nsince $N$ is the midpoint of the segment $PB$\\\\\ntherefore $NB=\\frac{1}{2}PB=\\boxed{7.5}$",
    "image_path": "dataset/mm_math/images/MM_Math/50056753_141.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the Cartesian coordinate system, point O is the origin, and the vertices of the parallelogram OABC, with vertex A lying on the reciprocal function $y = \\frac{1}{x}$ and vertex B on the reciprocal function $y = \\frac{4}{x}$, and point C on the positive half of the x-axis, what is the area of the parallelogram OABC?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram,\\\\\ndraw AF $\\perp$ x-axis at point F passing through point A, draw BE $\\perp$ x-axis at point E passing through point B, extend BA to intersect the y-axis at point G, then\\\\\nquadrilateral OFAG and quadrilateral OEBG are rectangles,\\\\\n$\\because$ point A is on the inverse proportion function $y = \\frac{1}{x}$, point B is on the inverse proportion function $y = \\frac{4}{x}$,\\\\\n$\\therefore$ area of rectangle OFAG = 1, area of rectangle OEBG = 4,\\\\\n$\\therefore$ area of quadrilateral OABC = area of rectangle ABEF = area of rectangle OEBG - area of rectangle OFAG = 4 - 1 = \\boxed{3}.",
    "image_path": "dataset/mm_math/images/MM_Math/52622743.png"
  },
  {
    "solution": "\\boxed{-2x^{2}-4x+6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the line $y = 2x + 6$ intersects the x-axis at point A and intersects the y-axis at point C. The parabola $y = -2x^2 + bx + c$ passes through points A and C, and intersects the x-axis at another point B. Find the analytical equation of the parabola.\n",
    "completion": "\\textbf{Solution:} Let's solve for $y=2x+6$. When $x=0$, $y=6$, and when $y=0$, $x=-3$,\\\\\n$\\therefore C(0, 6)$ and $A(-3, 0)$,\\\\\n$\\because$ the parabola $y=-2x^{2}+bx+c$ passes through points A and C,\\\\\n$\\therefore \\left\\{\\begin{array}{l}\n-18-3b+c=0 \\\\\nc=6\n\\end{array}\\right.$,\\\\\nSolving this, we get $\\left\\{\\begin{array}{l}\nb=-4 \\\\\nc=6\n\\end{array}\\right.$,\\\\\n$\\therefore$ the equation of the parabola is $y=-2x^{2}-4x+6=\\boxed{-2x^{2}-4x+6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53066295_93.png"
  },
  {
    "solution": "\\boxed{\\left(\\frac{3}{2},0\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, line AB intersects the x-axis and y-axis at points A and B, respectively. Point C is located on the negative half-axis of the x-axis. The coordinates of these three points are \\(A(4,0)\\), \\(B(0,4)\\), and \\(C(-1,0)\\) respectively. Connect BC. Suppose point E is a movable point on segment AC (not coinciding with A or C). Draw \\(EF \\parallel BC\\) intersecting AB at point F. When the area of \\(\\triangle BEF\\) is \\(\\frac{5}{2}\\), find the coordinates of point E.\n",
    "completion": "\\textbf{Solution:} Let the coordinates of point E be $(m,0)$,\\\\\nthe equation of line BC be $y=4x+4$,\\\\\nsince EF$\\parallel$BC,\\\\\nthus, let the equation of line EF be: $y=4x+n$, substituting the coordinates of point E into the above equation yields: $0=4m+n$,\\\\\ntherefore $n=-4m$,\\\\\nthus, the equation of line EF is: $y=4x-4m$,\\\\\nhence $-x+4=4x-4m$,\\\\\nsolving this gives: $x=\\frac{4}{5}(m+1)$,\\\\\nsubstituting the value of $x$ into $y=-x+4$, we get $y=\\frac{16-4m}{5}$,\\\\\ntherefore, the coordinates of point F are $\\left(\\frac{4m+4}{5},\\frac{16-4m}{5}\\right)$,\\\\\n${S}_{\\triangle BEF}={S}_{\\triangle OAB}-{S}_{\\triangle OBE}-{S}_{\\triangle AEF}=\\frac{1}{2}\\times 4\\times 4-\\frac{1}{2}\\times 4m-\\frac{1}{2}\\left(4-m\\right)\\times \\frac{16-4m}{5}=\\frac{5}{2}$,\\\\\nsolving this yields: $m=\\frac{3}{2}$,\\\\\nthus, the coordinates of point E are $\\boxed{\\left(\\frac{3}{2},0\\right)}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51353152_180.png"
  },
  {
    "solution": "\\boxed{13}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in rectangle $OBAC$, the coordinates of point $A$ are $(5, 12)$. What is the length of $BC$?\n",
    "completion": "\\textbf{Solution:} Draw line OA, and draw AM perpendicular to the x-axis at M, \\\\\nsince the coordinates of point A are (5, 12), \\\\\ntherefore, OM = 5, AM = 12, \\\\\nby the Pythagorean theorem, we get: OA = $\\sqrt{5^{2} + 12^{2}} = 13$, \\\\\nsince quadrilateral OBAC is a rectangle, \\\\\ntherefore, BC = OA = \\boxed{13},",
    "image_path": "dataset/mm_math/images/MM_Math/53359741.png"
  },
  {
    "solution": "\\boxed{y=2x+4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y=kx+b$ intersects with the graph of the inverse proportion function $y=\\frac{m}{x}$ at points $A(1,n)$ and $B(-3,-2)$. Try to determine the analytical expressions of the above-mentioned inverse proportion function and linear function.\n",
    "completion": "\\textbf{Solution}: Let's substitute point $B(-3, -2)$ into the inverse proportion function $y=\\frac{m}{x}$, we get: $m=-3\\times (-2)=6$,\\\\\nthus, the analytic expression of the inverse proportion function is $y=\\frac{6}{x}$,\\\\\nsubstituting point $A(1, n)$ into the inverse proportion function $y=\\frac{6}{x}$, we find: $n=\\frac{6}{1}=6$, that is $A(1, 6)$,\\\\\nsubstituting points $A(1, 6)$ and $B(-3, -2)$ into the linear function $y=kx+b$, we get: $\\left\\{\\begin{array}{l}k+b=6 \\\\ -3k+b=-2\\end{array}\\right.$,\\\\\nfrom which we obtain $\\left\\{\\begin{array}{l}k=2 \\\\ b=4\\end{array}\\right.$,\\\\\nthus, the analytic expression of the linear function is $\\boxed{y=2x+4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53066101_83.png"
  },
  {
    "solution": "\\boxed{C(-7, 0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the graph of the inverse proportional function $y_1 = \\frac{k}{x}$ intersects with the graph of the linear function $y_2 = ax + b$ at points $A(1,4)$ and $B(m,-2)$. If a point $C$ is found on the $x$-axis such that the area of $\\triangle ABC$ equals $18$, determine the coordinates of point $C$.\n",
    "completion": "\\textbf{Solution:} Assume the line AB intersects the x-axis at point D, it is easy to find D (-1, 0).\\\\\nAssume C(x, 0), $\\because s_{\\triangle ABC}=s_{\\triangle ADC}+s_{\\triangle BCD}$,\\\\\n$\\therefore \\frac{1}{2}y_A|x+1|+\\frac{1}{2}|y_B||x+1|=18$,\\\\\n$\\frac{1}{2}\\times 4\\times |x+1|+\\frac{1}{2}\\times 2\\times |x+1|=18$\\\\\n$3|x+1|=18$,\\\\\nSolving, we get: $x=5$ or $x=-7$,\\\\\n$\\therefore \\boxed{C(5, 0)}$ or $\\boxed{C(-7, 0)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52586052_125.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in an equilateral $\\triangle ABC$, $D$ and $E$ are points on $BC$ and $AC$ respectively, and $BD = CE$, $AD$ intersects $B$ at point $P$. What is the measure of $\\angle APE$ in degrees?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABC$ is an equilateral triangle,\\\\\nwe have $AB=BC$ and $\\angle ABC=\\angle C=60^\\circ$,\\\\\nGiven $BD=CE$,\\\\\nit follows that $\\triangle ABD\\cong \\triangle BCE$ (SAS),\\\\\nhence $\\angle BAD=\\angle CBE$,\\\\\nSince $\\angle ABP+\\angle CBE=\\angle ABC=60^\\circ$,\\\\\nwe get $\\angle ABP+\\angle BAD=60^\\circ$,\\\\\nthus $\\angle APE = \\angle ABP+\\angle BAD=60^\\circ$;\\\\\nTherefore, the answer is: $\\boxed{60^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52856180.png"
  },
  {
    "solution": "\\boxed{65^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given that O is a point on line AB, and OC $\\perp$ OE. As shown in the diagram, when ray OC is below line AB, OF bisects $\\angle AOE$, $\\angle BOE = 130^\\circ$, find the degree measure of $\\angle COF$.\n",
    "completion": "\\textbf{Solution:} Given $\\angle BOE = 130^\\circ$,\\\\\nit follows that $\\angle AOE = 180^\\circ - \\angle BOE = 50^\\circ$\\\\\nSince OF bisects $\\angle AOE$,\\\\\nwe have $\\angle EOF = \\frac{1}{2} \\angle AOE = 25^\\circ = \\angle AOF$\\\\\nSince $OC \\perp OE$,\\\\\nit follows that $\\angle COF = 90^\\circ - \\angle EOF = \\boxed{65^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51438726_165.png"
  },
  {
    "solution": "\\boxed{(2, 0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the vertex \\(C\\) of the parabola that passes through points \\(A(0, 4)\\) and \\(B(5, 9)\\) is on the positive x-axis. Find the coordinates of point \\(C\\).\n",
    "completion": "\\textbf{Solution:} Substitute $y=0$ into $y=(x-2)^{2}$ to obtain $(x-2)^{2}=0$,\\\\\n$\\therefore x=2$,\\\\\n$\\therefore$ the coordinates of point $C$ are $\\boxed{(2, 0)}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52514156_166.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given that point $O$ is a point on line $AB$, a right-angled triangle ruler $MON$ is placed as shown in the figure, with the right-angle vertex at point $O$. A ray $OC$ is drawn inside $\\angle MON$, and $OC$ exactly bisects $\\angle MOB$. If $\\angle BON = 2 \\angle NOC$, what is the measure of $\\angle AOM$ in degrees?\n",
    "completion": "\\textbf{Solution:} Since $\\angle BON=2\\angle NOC$ and $OC$ bisects $\\angle MOB$,\\\\\nit follows that $\\angle MOC=\\angle BOC=3\\angle NOC$\\\\\nSince $\\angle MOC+\\angle NOC=\\angle MON=90^\\circ$,\\\\\nthen $3\\angle NOC+\\angle NOC=90^\\circ$\\\\\nThus, $4\\angle NOC=90^\\circ$,\\\\\nwhich implies $\\angle BON=2\\angle NOC=45^\\circ$,\\\\\nTherefore, $\\angle AOM=180^\\circ−\\angle MON−\\angle BON=180^\\circ−90^\\circ−45^\\circ=\\boxed{45^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52916218_157.png"
  },
  {
    "solution": "\\boxed{8+2\\sqrt{13}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $O$ is the midpoint of the diagonal $AC$ of rectangle $ABCD$, and $E$ is the midpoint of $AD$. If $AB=6$ and $BC=8$, then what is the perimeter of $\\triangle BOE$?\n",
    "completion": "\\textbf{Solution:} Since point $O$ is the midpoint of the diagonal $AC$ of rectangle $ABCD$, and point $E$ is the midpoint of $AD$, \\\\\n$\\therefore AB=CD=6$, $AD=BC=8$, $OE=\\frac{1}{2}CD=3$, $AE=\\frac{1}{2}AD=4$, \\\\\nIn $Rt\\triangle ABE$, $BE=\\sqrt{AB^{2}+AE^{2}}=\\sqrt{6^{2}+4^{2}}=2\\sqrt{13}$, \\\\\nIn $Rt\\triangle ABC$, $AC=\\sqrt{AB^{2}+BC^{2}}=\\sqrt{6^{2}+8^{2}}=10$, \\\\\n$\\therefore BO=\\frac{1}{2}AC=5$, \\\\\nThus, the perimeter of $\\triangle BOE$ is: $5+3+2\\sqrt{13}=8+2\\sqrt{13}$, \\\\\nTherefore, the answer is: $\\boxed{8+2\\sqrt{13}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52423423.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AD \\parallel BE \\parallel CF$, $AB=3$, $AC=9$, $DE=2$, then what is the value of $EF$?\n",
    "completion": "\\textbf{Solution:} Given that $AD\\parallel BE\\parallel CF$,\\\\\nit follows that $\\frac{AB}{BC}=\\frac{DE}{EF}$,\\\\\nGiven that $AB=3$, $AC=9$, $DE=2$,\\\\\nwe have $BC=6$,\\\\\nthus, $\\frac{3}{6}=\\frac{2}{EF}$,\\\\\ntherefore, $EF=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52719891.png"
  },
  {
    "solution": "\\boxed{2.4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the diagonals $AC$ and $BD$ of the rhombus $ABCD$ intersect at point $O$. Given that $AC=8$ and $BD=6$, what is the length of the side $AB$ of the rhombus $ABCD$? What is its area? If a perpendicular line $OH$ is drawn from point $O$ to side $AB$, with the foot of the perpendicular being point $H$, what is the distance $OH$ from point $O$ to side $AB$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral $ABCD$ is a rhombus, $\\therefore$ $AC\\perp BD$, $OB=OD=3$, $OA=OC=4$. Therefore, $AB= \\sqrt{OA^{2}+OB^{2}}=\\boxed{5}$, $S_{\\text{rhombus}ABCD}= \\frac{1}{2}AC\\cdot BD=\\boxed{24}$, $S_{\\triangle AOB}= \\frac{1}{2}OA\\cdot OB= \\frac{1}{2}AB\\cdot OH$. Therefore, $OH= \\frac{OA\\cdot OB}{AB}=\\boxed{2.4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51614463.png"
  },
  {
    "solution": "\\boxed{6.5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the quadrilateral $ABCD$ is a square, $BE \\perp EF$, $DF \\perp EF$, with $BE = 2.5 \\, \\text{dm}$, and $DF = 4 \\, \\text{dm}$. What is the length of $EF$?\n",
    "completion": "\\textbf{Solution:} Since ABCD is a square,\\\\\nit follows that BC=CD, and BC$\\perp$CD,\\\\\ntherefore, $\\angle$BCE+$\\angle$FCD=$90^\\circ$,\\\\\nalso, since BE$\\perp$EF and DF$\\perp$EF,\\\\\nit follows that $\\angle$BEC=$\\angle$CFD=$90^\\circ$,\\\\\ntherefore, $\\angle$BCE+$\\angle$EBC=$90^\\circ$,\\\\\nthus, $\\angle$EBC=$\\angle$FCD,\\\\\nhence, $\\triangle$BEC$\\cong$$\\triangle$CFD,\\\\\ntherefore, BE=CF=2.5dm, and DF=EC=4dm,\\\\\nthus, EF=EC+CF=$\\boxed{6.5}$dm.",
    "image_path": "dataset/mm_math/images/MM_Math/51435945.png"
  },
  {
    "solution": "\\boxed{\\frac{60}{13}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the rhombus $ABCD$, the diagonal $AC=5$, and $BD=12$. What is the height of the rhombus?\n",
    "completion": "\\textbf{Solution:} Let AC and BD intersect at point O. Through point A, construct AH $\\perp$ BC at point H,\\\\\nSince rhombus ABCD,\\\\\nTherefore, AC $\\perp$ BD, OB=$\\frac{1}{2}$BD=6, OC=$\\frac{1}{2}$AC=$\\frac{5}{2}$,\\\\\nTherefore, $BC=\\sqrt{OB^{2}+OC^{2}}=\\sqrt{6^{2}+\\left(\\frac{5}{2}\\right)^{2}}=\\frac{13}{2}$\\\\\nTherefore, the area of rhombus ABCD, ${S}_{\\mathrm{rhombus} ABCD}=\\frac{1}{2}AC\\cdot BD=BC\\cdot AH$\\\\\nTherefore, $\\frac{1}{2}\\times 5\\times 12=\\frac{13}{2}AH$\\\\\nSolving gives: $AH=\\boxed{\\frac{60}{13}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52251660.png"
  },
  {
    "solution": "\\boxed{50^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, AB is the diameter of $\\odot O$, CD is a chord of $\\odot O$, and $\\angle ACD = 40^\\circ$, what is the measure of $\\angle BAD$?\n",
    "completion": "\\textbf{Solution:} Connect BD,\\\\\nsince the angles subtended by the same arc in a circle are equal,\\\\\ntherefore, $\\angle B = \\angle C = 40^\\circ$,\\\\\nsince AB is a diameter,\\\\\ntherefore, $\\angle ADB = 90^\\circ$,\\\\\nthus, $\\angle DAB + \\angle B = 90^\\circ$,\\\\\ntherefore, $\\angle DAB = 90^\\circ - 40^\\circ = \\boxed{50^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53080558.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, points A, D, and C are on the same straight line, and $\\triangle ABC \\cong \\triangle DBE$. If $\\angle A = 60^\\circ, \\angle C = 30^\\circ$, then what is the measure of $\\angle DBC$?\n",
    "completion": "\\textbf{Solution:} Given $\\because$ $\\triangle ABC\\cong \\triangle DBE$ and $\\angle A=60^\\circ$,\\\\\n$\\therefore$ $AB=BD$,\\\\\n$\\therefore$ $\\angle ADB=\\angle A=60^\\circ$,\\\\\n$\\therefore$ $\\angle ABD=60^\\circ$,\\\\\n$\\because$ $\\angle A=60^\\circ$ and $\\angle C=30^\\circ$,\\\\\n$\\therefore$ $\\angle ABC=90^\\circ$,\\\\\n$\\therefore$ $\\angle DBC=\\angle ABC-\\angle ABD=\\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52978491.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, through the centroid \\(G\\) of \\(\\triangle ABC\\), construct \\(ED\\parallel AB\\) meeting sides \\(AC\\) and \\(BC\\) at points \\(E\\) and \\(D\\) respectively. Connect \\(AD\\). If \\(AD\\) bisects \\(\\angle BAC\\) and \\(AB=6\\), what is the length of \\(EC\\)?\n",
    "completion": "\\textbf{Solution}:\nDraw and extend CG to meet AB at H,\\\\\nsince G is the centroid of $\\triangle ABC$\\\\\ntherefore $\\frac{CG}{GH}=2$\\\\\ntherefore $\\frac{CG}{CH}=\\frac{2}{3}$\\\\\nsince $ED\\parallel AB$\\\\\ntherefore $\\frac{CG}{CH}=\\frac{2}{3}=\\frac{EC}{AC}$, $\\angle ADE=\\angle BAD$, $\\triangle ECD\\sim \\triangle ACB$\\\\\ntherefore $\\frac{EC}{AC}=\\frac{DE}{AB}=\\frac{2}{3}$\\\\\ntherefore $DE=\\frac{2}{3}AB=4$\\\\\nsince AD bisects $\\angle BAC$\\\\\ntherefore $\\angle EAD=\\angle BAD$\\\\\ntherefore $\\angle EAD=\\angle ADE$\\\\\ntherefore $DE=AE=4$\\\\\ntherefore $\\frac{2}{3}=\\frac{EC}{AC}=\\frac{EC}{EC+AE}$,\\\\\ntherefore $EC=\\boxed{8}$",
    "image_path": "dataset/mm_math/images/MM_Math/51379367.png"
  },
  {
    "solution": "\\boxed{1+\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, D is an arbitrary point on side AB, F is the midpoint of AC, and a line is drawn through point C such that $CE \\parallel AB$ intersects the extension of DF at point E. Line segments AE and CD are then drawn. If $\\angle B=30^\\circ$, $\\angle CAB=45^\\circ$, and $AC=\\sqrt{2}$, find the length of AB.\n",
    "completion": "\\textbf{Solution:} Construct $CG\\perp AB$ at point G,\\\\\nsince $\\angle CAB=45^\\circ$,\\\\\nit follows that $AG=CG$,\\\\\nin $\\triangle ACG$, $\\angle AGC=90^\\circ$,\\\\\nthus $AG^2+CG^2=AC^2$,\\\\\nsince $AC=\\sqrt{2}$,\\\\\nit follows that $CG=AG=1$,\\\\\nsince $\\angle B=30^\\circ$,\\\\\nit follows that $CG=\\frac{1}{2}BC$,\\\\\nthus $BC=2$,\\\\\nin $\\triangle BCG$, $BG=\\sqrt{BC^2-CG^2}=\\sqrt{4-1}=\\sqrt{3}$,\\\\\nthus $AB=AG+BG=1+\\sqrt{3}=\\boxed{1+\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51737267_152.png"
  },
  {
    "solution": "\\boxed{20^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the figure, within $\\triangle ABC$, $AD$ bisects $\\angle BAC$, $AE\\perp BC$. If $\\angle BAD=40^\\circ$ and $\\angle C=70^\\circ$, find the measure of $\\angle DAE$?\n",
    "completion": "\\textbf{Solution:} Given that $AD$ bisects $\\angle BAC$,\\\\\nthus $\\angle BAC = 2\\angle BAD = 80^\\circ$ (definition of angle bisector),\\\\\nGiven that $\\angle C = 70^\\circ$,\\\\\nin $\\triangle ABC$, thus $\\angle B = 180^\\circ - \\angle BAC - \\angle C\\\\\n= 180^\\circ - 70^\\circ - 80^\\circ = 30^\\circ$ (sum of interior angles in a triangle is $180^\\circ$),\\\\\nGiven that $AE \\perp BC$,\\\\\nthus $\\angle AEB = 90^\\circ$ (definition of perpendicular),\\\\\nin $\\triangle ABE$, $\\angle BAE = 180^\\circ - \\angle B - \\angle AEB\\\\\n= 180^\\circ - 30^\\circ - 90^\\circ = 60^\\circ$ (sum of interior angles in a triangle is $180^\\circ$),\\\\\nthus $\\angle DAE = \\angle BAE - \\angle BAD = 60^\\circ - 40^\\circ = \\boxed{20^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52891255.png"
  },
  {
    "solution": "\\boxed{\\frac{40}{17}\\text{cm}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Five rectangular books are placed in a square bookshelf as shown in the diagram, with four of them placed vertically, and the fifth one placed diagonally, where point $G$ precisely aligns with the edge of the bookshelf. Each book has a thickness of $5\\,cm$ and a height of $20\\,cm$, and the shelf has a width of $40\\,cm$. What is the length of $FI$ in $cm$?\n",
    "completion": "\\textbf{Solution:} From the given information, we have: $FG=5\\text{cm}$, $BC=5\\times 4=20\\text{cm}$, $BI=40\\text{cm}$, $EF=20\\text{cm}$,\\\\\n$CD\\perp BI$, $BI\\perp GI$, $EF\\perp GF$,\\\\\n$\\therefore \\angle I=\\angle ECF=90^\\circ$, $\\angle GFI+\\angle EFC=\\angle FEC+\\angle EFC=90^\\circ$,\\\\\n$\\therefore \\angle GFI=\\angle FEC$,\\\\\nIn $\\triangle GFI$ and $\\triangle FEC$, $\\left\\{\\begin{array}{l}\\angle I=\\angle ECF=90^\\circ \\\\ \\angle GFI=\\angle FEC\\end{array}\\right.$,\\\\\n$\\therefore \\triangle GFI\\sim \\triangle FEC$,\\\\\n$\\therefore \\frac{FI}{CE}=\\frac{FG}{EF}$,\\\\\nLet $FI=x\\text{cm}$, then $CF=BI-BC-FI=(20-x)\\text{cm}$,\\\\\n$\\therefore \\frac{x}{CE}=\\frac{5}{20}$, solving for $CE$ we get $CE=4x\\text{cm}$,\\\\\nIn right triangle $Rt\\triangle FEC$, $CF^2+CE^2=EF^2$, which gives $(20-x)^2+(4x)^2=20^2$,\\\\\nSolving, we get $x=\\frac{40}{17}$ or $x=0$ (not meeting the requirement, hence discarded),\\\\\nThus, $FI=\\boxed{\\frac{40}{17}\\text{cm}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51058862.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $E$ is a point on the extension of side $BA$ of $\\square ABCD$, $CE$ intersects $AD$ at point $F$, $AE=1$, $AB=2$, $BC=3$. What is the value of $AF$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a parallelogram, \\\\\nthen AD $\\parallel$ BC, \\\\\nthus $\\triangle EAF \\sim \\triangle EBC$, \\\\\nhence $\\frac{AE}{BE} = \\frac{AF}{BC}$, \\\\\nsince $AE=1$, $AB=2$, $BC=3$, \\\\\nthus $BE=3$, \\\\\nhence $\\frac{1}{3} = \\frac{AF}{3}$, \\\\\ntherefore $AF = \\boxed{1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51353482.png"
  },
  {
    "solution": "\\boxed{20^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the right-angled triangle paper ABC, where $\\angle ACB = 90^\\circ$ and $\\angle A = 50^\\circ$, it is folded along the midpoint CE of side AB, making point A fall on point $A'$, then what is the degree measure of $\\angle A'EB$?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABC$ is a right-angled triangle, and CE is the median, \\\\\nthus $AE=CE=BE$, \\\\\nby the property of folding, we obtain \\\\\n$AE=A'E$, $\\angle AEC=\\angle A'EC$, \\\\\ntherefore, $AE=CE=BE=A'E$, \\\\\nsince $\\angle A=50^\\circ$, \\\\\nhence $\\angle ACE=50^\\circ$, \\\\\ntherefore, $\\angle AEC=\\angle A'EC=180^\\circ-50^\\circ-50^\\circ=80^\\circ$, \\\\\nsince $\\angle BEC=50^\\circ+50^\\circ=100^\\circ$, \\\\\nhence $\\angle A'EB=100^\\circ-80^\\circ=\\boxed{20^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51368822.png"
  },
  {
    "solution": "\\boxed{(5, 0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $l_{1}\\colon y=x+1$ intersects the x-axis at point A and intersects the line $l_{2}\\colon y=\\frac{1}{2}x+2$ at point B. Point C is a point on the x-axis. If $\\triangle ABC$ is a right triangle, what is the x-coordinate of point C?\n",
    "completion": "\\textbf{Solution:} Given the line $l_{1}\\colon y=x+1$ intersects the x-axis at point A,\\\\\nthus, A is at (-1, 0),\\\\\nFrom $\\left\\{\\begin{array}{l}y=x+1 \\\\ y=\\frac{1}{2}x+2\\end{array}\\right.$, we get $\\left\\{\\begin{array}{l}x=2 \\\\ y=3\\end{array}\\right.$,\\\\\ntherefore, B is at (2, 3),\\\\\nWhen $\\angle ACB=90^\\circ$, the x-coordinate of point C is the same as that of B,\\\\\nthus, C is at (2, 0);\\\\\nWhen $\\angle ABC=90^\\circ$, then $AC^{2}=AB^{2}+BC^{2}$,\\\\\nLet C be at (x, 0), then $AC^{2}=(x+1)^{2}$, $AB^{2}=(2+1)^{2}+3^{2}$, $BC^{2}=(2-x)^{2}+3^{2}$,\\\\\nthus, $(x+1)^{2}=(2+1)^{2}+3^{2}+(2-x)^{2}+3^{2}$,\\\\\nSolving it, we get $x=\\boxed{5}$,\\\\\ntherefore, C is \\boxed{(5, 0)},\\\\\nIn summary, the coordinates of point C are either (2, 0) or (5, 0).",
    "image_path": "dataset/mm_math/images/MM_Math/51419577.png"
  },
  {
    "solution": "\\boxed{\\sqrt{11}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\triangle ABC$, D is the midpoint of AC, and CE $\\perp$ AB at point E. If DE = 3 and AE = 5, then what is the length of CE?\n",
    "completion": "\\textbf{Solution:} Since CE$\\perp$AB, it follows that $\\angle$AEC$=90^\\circ$. In the right-angled triangle AEC, since point D is the midpoint of the hypotenuse AC, and DE=3, it follows that AC=2DE=6. By the Pythagorean theorem, we have $CE=\\sqrt{AC^{2}-AE^{2}}=\\sqrt{6^{2}-5^{2}}=\\sqrt{11}=\\boxed{\\sqrt{11}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55498341.png"
  },
  {
    "solution": "\\boxed{x \\geq 2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the same Cartesian coordinate system, the graphs of the functions $y_1=kx+b$ and $y_2=mx+n$ are shown in the figure. What is the solution set of the inequality $kx+b\\geq mx+n$ with respect to $x$?\n",
    "completion": "\\textbf{Solution:} The solution set of the inequality $kx+b \\geq mx+n$ is: $\\boxed{x \\geq 2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50469234.png"
  },
  {
    "solution": "\\boxed{$0<x<8$}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the graph of the linear function $y=kx+b$ intersects the $x$-axis and $y$-axis at points $A$ and $B$, respectively, and intersects the graph of the inverse proportion function $y=\\frac{m}{x}$ at points $C$ and $D$. $DE\\perp x$-axis at point $E$. Given that the coordinate of point $C$ is $(8, -2)$ and $DE=4$, answer directly based on the graph: For which values of $x$ is the value of the linear function greater than that of the inverse proportion function?\n",
    "completion": "\\textbf{Solution:} Since a linear function and an inverse proportion function intersect at point D ($-4$, $4$) and point C ($8$, $-2$), \\\\\nit follows that for \\boxed{$x<-4$} or \\boxed{$0<x<8$}, the value of the linear function is greater than the value of the inverse proportion function.",
    "image_path": "dataset/mm_math/images/MM_Math/52551855_129.png"
  },
  {
    "solution": "\\boxed{40^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AD$ is the diameter of the circumcircle $\\odot O$ of $\\triangle ABC$. If $\\angle ACB = 50^\\circ$, what is the degree measure of $\\angle BAD$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect BD,\\\\\nsince AD is the diameter of the circumcircle $\\odot O$ of $\\triangle ABC$,\\\\\ntherefore $\\angle ABD=90^\\circ$,\\\\\nsince $\\angle D=\\angle C=50^\\circ$,\\\\\ntherefore $\\angle BAD=90^\\circ-\\angle D=90^\\circ-50^\\circ=\\boxed{40^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51261712.png"
  },
  {
    "solution": "\\boxed{\\text{indeterminate without further information}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that points $A(-4, 2)$ and $B(n, -4)$ are two intersection points between the graph of the linear function $y=kx+b$ and the graph of the inverse proportion function $y=\\frac{m}{x}$. By observing the graphs, directly write the solution set of the inequality $kx+b-\\frac{m}{x}>0$.\n",
    "completion": "\\textbf{Solution:} It follows from the problem statement that the solution set for $kx+b-\\frac{k}{x}>0$ is $\\boxed{\\text{indeterminate without further information}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52622329_235.png"
  },
  {
    "solution": "\\boxed{Q\\left(-\\frac{8}{3},6\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated, the line $y=-\\frac{3}{2}x-2$ intersects the $x$-axis and $y$-axis at points $A$ and $B$ respectively, and intersects the hyperbola $y=\\frac{m}{x}$ (where $m\\neq 0$) at point $C$ in the second quadrant. Line segment $CD$ is perpendicular to the $y$-axis at point $D$, and $CD=4$. Let point $Q$ be a point on the hyperbola, and the area of triangle $QOB$ is twice the area of triangle $AOB$. Find the coordinates of point $Q$.\n",
    "completion": "\\textbf{Solution:} Insert $x=0$ into $y=-\\frac{3}{2}x-2$, we get $y=-\\frac{3}{2}\\times0-2=-2$,\\\\\n$\\therefore B(0,-2)$,\\\\\nInsert $y=0$ into $y=-\\frac{3}{2}x-2$, we get $0=-\\frac{3}{2}x-2$,\\\\\n$\\therefore x=-\\frac{4}{3}$,\\\\\n$\\therefore A\\left(-\\frac{4}{3},0\\right)$,\\\\\n$\\because S_{\\triangle QOB}=2S_{\\triangle AOB}$,\\\\\n$\\therefore \\frac{1}{2}OB\\cdot |x_{Q}|=2\\times \\frac{1}{2}OA\\cdot OB$,\\\\\n$\\therefore \\frac{1}{2}\\times 2|x_{Q}|=2\\times \\frac{1}{2}\\times \\left|-\\frac{4}{3}\\right|\\times 2$,\\\\\nsolving this gives $x_Q=\\pm \\frac{8}{3}$,\\\\\nInsert $x=\\frac{8}{3}$ into $y=-\\frac{16}{x}$, we get $y=-6$,\\\\\nInsert $x=-\\frac{8}{3}$ into $y=-\\frac{16}{x}$, we get $y=6$,\\\\\n$\\therefore Q\\left(\\frac{8}{3},-6\\right)$ or $\\boxed{Q\\left(-\\frac{8}{3},6\\right)}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52189644_141.png"
  },
  {
    "solution": "\\boxed{(2,3)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, point O is the origin. The line $y=2x-1$ intersects the line $y=\\frac{3}{4}x+\\frac{3}{2}$ at point A. A perpendicular line to the x-axis is drawn through point A, with point B being the foot of the perpendicular. The x-coordinate of point C is $-1$, and point C lies on the line $y=2x-1$. Connect BC. What are the coordinates of point A?\n",
    "completion": "\\textbf{Solution:} Solve the system of equations:\n\\[\n\\left\\{\n\\begin{array}{l}\ny=2x-1 \\\\\ny=\\frac{3}{4}x+\\frac{3}{2}\n\\end{array}\n\\right.\n\\]\nto get\n\\[\n\\left\\{\n\\begin{array}{l}\nx=2\\\\\ny=3\n\\end{array}\n\\right.\n\\]\n$\\therefore$ The coordinates of point A are $\\boxed{(2,3)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50676594_164.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the vertex of the quadratic function $y=ax^2+bx+c$ is $M(2, 1)$, and it passes through the point $N(3, 2)$. What is the equation of this quadratic function?\n",
    "completion": "\\textbf{Solution:} Let the quadratic function be defined as $y=a(x-2)^2+1$.\\\\\nBy substituting $x=3$ and $y=2$, we get $a+1=2$, solving for $a$ gives $a=\\boxed{1}$.\\\\\nTherefore, the equation of the quadratic function is $y=(x-2)^2+1$ (or written as $y=x^2-4x+5$).",
    "image_path": "dataset/mm_math/images/MM_Math/52457074_126.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, it is known that $M$ and $N$ are two points on the side $BC$ of $\\triangle ABC$, and satisfy $BM=MN=NC$. A line parallel to $AC$ intersects the extensions of $AB$, $AM$, and $AN$ at points $D$, $E$, and $F$, respectively. What is the value of $\\frac{EF}{DE}$?\n",
    "completion": "\\textbf{Solution:} Draw line $MG \\parallel DF$ with point $G$ on $AB$, and draw line $NH \\parallel DF$ with point $H$ on $AB$, where $NH$ intersects $AM$ at $I$,\n\nthen we have $MG \\parallel DF \\parallel NH \\parallel AC$\n\n$\\because$ $GM \\parallel NH$,\n\n$\\therefore$ $\\triangle BMG \\sim \\triangle BNH$\n\n$\\therefore$ $\\frac{MG}{NH} = \\frac{BM}{BN}$\n\nFurthermore, $\\because$ $BM = \\frac{1}{2}BN$,\n\n$\\therefore$ $\\frac{MG}{NH} = \\frac{BM}{BN} = \\frac{1}{2}$\n\n$\\because$ $MG \\parallel NH \\parallel AC$, $MN = NC$\n\n$\\therefore$ $\\frac{GH}{AH} = \\frac{MN}{NC} = 1$\n\n$\\therefore$ $\\frac{AH}{AG} = \\frac{1}{2}$\n\n$\\because$ $MG \\parallel NH$\n\n$\\therefore$ $\\triangle AHI \\sim \\triangle AGM$\n\n$\\therefore$ $\\frac{IH}{MG} = \\frac{AH}{AG} = \\frac{1}{2}$\n\nAdditionally, $\\because$ $\\frac{MG}{NH} = \\frac{1}{2}$\n\n$\\therefore$ $\\frac{IH}{NH} = \\frac{IH}{MG} \\cdot \\frac{MG}{NH} = \\frac{1}{2} \\times \\frac{1}{2} = \\frac{1}{4}$\n\n$\\therefore$ $\\frac{IH}{IN} = \\frac{1}{3}$\n\nAlso, $\\because$ $DF \\parallel NH$\n\n$\\therefore$ $\\triangle AHI \\sim \\triangle ADE$, $\\triangle ANI \\sim \\triangle AFE$,\n\n$\\therefore$ $\\frac{IH}{ED} = \\frac{AI}{AE} = \\frac{IN}{EF}$\n\n$\\therefore$ $\\frac{ED}{EF} = \\frac{IH}{IN} = \\frac{1}{3}$\n\n$\\therefore$ $\\frac{EF}{DE} = \\boxed{3}$",
    "image_path": "dataset/mm_math/images/MM_Math/52586157.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in circle $\\odot O$, the diameter $CD$ is perpendicular to chord $AB$ at point $E$, and $OB$ and $CB$ are connected. Given that the radius of $\\odot O$ is $2$ and $AB=2\\sqrt{3}$, what is the measure of $\\angle BCD$?\n",
    "completion": "\\textbf{Solution:} Given that diameter CD is perpendicular to chord AB at point E, and AB = 2$\\sqrt{3}$, \\\\\nhence, EB = $\\frac{1}{2}$AB = $\\sqrt{3}$, \\\\\nsince the radius of circle O is 2, \\\\\ntherefore, $\\sin\\angle EOB = \\frac{EB}{OB} = \\frac{\\sqrt{3}}{2}$, \\\\\nthus, $\\angle EOB = 60^\\circ$, \\\\\ntherefore, $\\angle BCD = \\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52807528.png"
  },
  {
    "solution": "\\boxed{1.2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the right-angled triangle $ABC$ with $ \\angle ACB=90^\\circ$, $AC=3$, and $BC=4$, point $M$ is a point on side $AB$ (not coinciding with points $A$ or $B$). Draw $ME\\perp AC$ at point $E$ and $MF\\perp BC$ at point $F$. If point $P$ is the midpoint of $EF$, what is the minimum value of $CP$?\n",
    "completion": "\\textbf{Solution:} Connect $CM$, as shown in the diagram:\\\\\nSince $\\angle ACB=90^\\circ$, $AC=3$, $BC=4$,\\\\\nthen $AB=\\sqrt{AC^2+BC^2}=\\sqrt{3^2+4^2}=5$,\\\\\nSince $ME \\perp AC$ at point $E$, $MF \\perp BC$ at point $F$, $\\angle ACB=90^\\circ$,\\\\\nthen quadrilateral $CEMF$ is a rectangle,\\\\\nthus $EF=CM$,\\\\\nSince point $P$ is the midpoint of $EF$,\\\\\nthen $CP=\\frac{1}{2}EF$,\\\\\nWhen $CM \\perp AB$, $CM$ is the shortest,\\\\\nAt this time, $EF$ is also the smallest, thus $CP$ is the smallest,\\\\\nSince the area of $\\triangle ABC =\\frac{1}{2}AB\\cdot CM=\\frac{1}{2}AC\\cdot BC$,\\\\\nthus $CM=\\frac{AC\\cdot BC}{AB}=\\frac{3\\times 4}{5}=2.4$,\\\\\nthus $CP=\\frac{1}{2}EF=\\frac{1}{2}CM=\\boxed{1.2}$,",
    "image_path": "dataset/mm_math/images/MM_Math/53728800.png"
  },
  {
    "solution": "\\boxed{80^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, quadrilateral $ABCD$ is inscribed in circle $\\odot O$. If $\\angle ADC = 140^\\circ$, what is the degree measure of $\\angle AOC$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral $ABCD$ is inscribed in circle $O$, and $\\angle ADC=140^\\circ$\\\\\nTherefore, $\\angle ABC=40^\\circ$,\\\\\nTherefore, $\\angle AOC=2\\angle ABC=80^\\circ$,\\\\\nHence, $\\angle AOC=\\boxed{80^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/51262495.png"
  },
  {
    "solution": "\\boxed{2:3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, $DE \\parallel BC$, and $AD:DB = 2:1$, what is the ratio of similarity between $\\triangle ADE$ and $\\triangle ABC$?\n",
    "completion": "\\textbf{Solution:} Given that $DE\\parallel BC$ and $AD:DB=2:1$, \\\\\nit follows that $\\triangle ADE\\sim \\triangle ABC$ and $AD:AB=2:3$, \\\\\nhence, the ratio of similarity between $\\triangle ADE$ and $\\triangle ABC$ is $\\boxed{2:3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52871849.png"
  },
  {
    "solution": "\\boxed{20}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, there is a broken line OABC inside circle O, where OA = 8, AB = 12, and $\\angle A = \\angle B = 60^\\circ$. What is the length of BC?\n",
    "completion": "\\textbf{Solution:} Extend AO and intersect BC at D, and draw OE$\\perp$BC at E.\\\\\nSince $\\angle A = \\angle B = 60^\\circ$,\\\\\nthen $\\angle ADB = 60^\\circ$,\\\\\nthus $\\triangle ADB$ is an equilateral triangle,\\\\\nthus BD = AD = AB = 12,\\\\\nthus OD = 4,\\\\\nAdditionally, since $\\angle ADB = 60^\\circ$,\\\\\nthen DE = OD $\\cdot \\cos 60^\\circ = \\frac{1}{2}$ OD = 2,\\\\\nthus BE = 10,\\\\\nSince OE $\\perp$ BC,\\\\\nthus BC = 2BE = \\boxed{20}.",
    "image_path": "dataset/mm_math/images/MM_Math/52814831.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{2}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB = 90^\\circ$, $CD \\perp AB$ at D. If $AC=2\\sqrt{3}$ and $AB=3\\sqrt{2}$, then what is the value of $\\tan\\angle BCD$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle ACB=90^\\circ$, $AC=2\\sqrt{3}$, $AB=3\\sqrt{2}$, \\\\\nhence $BC=\\sqrt{(3\\sqrt{2})^{2}-(2\\sqrt{3})^{2}}=\\sqrt{6}$, and $\\angle B+\\angle A=90^\\circ$, \\\\\nsince $CD\\perp AB$ at $D$, \\\\\nthus $\\angle B+\\angle BCD =90^\\circ$, \\\\\ntherefore $\\angle BCD=\\angle A$, \\\\\nthus $\\tan\\angle BCD=\\tan\\angle A=\\frac{BC}{AC}=\\frac{\\sqrt{6}}{2\\sqrt{3}}=\\boxed{\\frac{\\sqrt{2}}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52511894.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{34}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, points $E$ and $F$ are the midpoints of sides $AB$ and $BC$, respectively. Lines $EC$ and $FD$ are connected, and points $G$ and $H$ are the midpoints of $EC$ and $FD$, respectively. Line $GH$ is then connected. If $AB=6$ and $BC=10$, what is the length of $GH$?\n",
    "completion": "\\textbf{Solution:} Draw line CH and extend it to intersect AD at P, and connect PE, as shown in the figure: \\\\\n$\\because$ Quadrilateral ABCD is a rectangle, \\\\\n$\\therefore$ $\\angle$A＝$90^\\circ$, $ AD\\parallel BC$, $ AD=BC=10$, \\\\\n$\\because$ E, F are the midpoints of side AB, BC respectively, AB＝6, BC＝10, \\\\\n$\\therefore$ AE＝$ \\frac{1}{2}$AB＝$ \\frac{1}{2}\\times $6＝3, CF＝$ \\frac{1}{2}$BC＝$ \\frac{1}{2}\\times $10＝5, \\\\\n$\\because$ $ AD\\parallel BC$, \\\\\n$\\therefore$ $\\angle$DPH＝$\\angle$FCH, \\\\\n$\\because$ H is the midpoint of DF, \\\\\n$\\therefore$ DH=FH \\\\\n$\\because$ In $\\triangle $PDH and $\\triangle $CFH $\\left\\{\\begin{array}{l}\\angle DPH=\\angle FCH\\hfill \\\\ \\angle DHP=\\angle FHC\\hfill \\\\ DH=FH\\hfill \\end{array} \\right.$, \\\\\n$\\therefore$ $\\triangle $PDH$\\cong $$\\triangle $CFH (AAS), \\\\\n$\\therefore$ PD＝CF＝5, CH＝PH, \\\\\n$\\therefore$ AP＝AD−PD＝5, \\\\\n$\\therefore$ PE＝$ \\sqrt{AP^{2}+AE^{2}}$＝$ \\sqrt{5^{2}+3^{2}}$＝$ \\sqrt{34}$, \\\\\n$\\because$ Point G is the midpoint of EC, CH＝PH, \\\\\n$\\therefore$ GH is the midline of $\\triangle $CEP, \\\\\n$\\therefore$ GH＝$ \\frac{1}{2}$EP＝$ \\frac{\\sqrt{34}}{2}$, \\\\\nHence, the answer is: $GH = \\boxed{\\frac{\\sqrt{34}}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52354792.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\triangle ABC$, $AB=AC$, the line EF is the perpendicular bisector of AB, D is the midpoint of BC, M is a moving point on EF, the area of $ \\triangle ABC$ is 12, and $BC=4$, then what is the minimum value of the perimeter of $ \\triangle BDM$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, we connect AD, and AM,\\\\\n\\(\\because\\) EF is the perpendicular bisector of line segment AB,\\\\\n\\(\\therefore\\) AM=BM,\\\\\n\\(\\therefore\\) the perimeter of \\(\\triangle BDM\\) = BD + BM + DM = AM + DM + BD,\\\\\n\\(\\therefore\\) to minimize the perimeter of \\(\\triangle BDM\\), we need to minimize the value of AM + DM,\\\\\n\\(\\therefore\\) when points A, M, and D are collinear, AM + DM is minimized, which equals to AD,\\\\\n\\(\\because\\) AB=AC, and D is the midpoint of BC,\\\\\n\\(\\therefore\\) AD\\(\\perp\\)BC, \\(BD=\\frac{1}{2}BC=2\\),\\\\\n\\(\\therefore\\) \\(S_{\\triangle ABC}=\\frac{1}{2}AD\\cdot BC=12\\),\\\\\n\\(\\therefore\\) AD=6,\\\\\n\\(\\therefore\\) The minimum perimeter of \\(\\triangle BDM\\) = AD + BD = \\boxed{8},",
    "image_path": "dataset/mm_math/images/MM_Math/51367866.png"
  },
  {
    "solution": "\\boxed{100^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure: It is known that $O$ is a point on line $CD$, $OA$ bisects $\\angle BOC$, $\\angle AOC=40^\\circ$, find the degree measure of $\\angle BOD$.\n",
    "completion": "\\textbf{Solution:} Given that $O$ is a point on the line $CD$, and $OA$ bisects $\\angle BOC$, $\\angle AOC=40^\\circ$,\\\\\nthen $\\angle BOC=2\\angle AOC=80^\\circ$,\\\\\nthus $\\angle BOD=180^\\circ-\\angle BOC=\\boxed{100^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51284638.png"
  },
  {
    "solution": "\\boxed{72^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC$, and $\\odot O$ is the circumcircle of $\\triangle ABC$, with the extension line of $BO$ intersecting side $AC$ at point $D$. When $\\triangle BCD$ is an isosceles triangle, find the measure of $\\angle BCD$.\n",
    "completion": "\\textbf{Solution:} Let $\\angle ABD=x$,\\\\\nfrom (1) we know that $\\angle BAC=2\\angle ABD=2x$,\\\\\n$\\therefore \\angle BDC=3x$,\\\\\n$\\triangle BCD$ is an isosceles triangle,\\\\\n(1) If $BD=BC$,\\\\\nthen $\\angle C=\\angle BDC=3x$,\\\\\n$\\because AB=AC$,\\\\\n$\\therefore \\angle ABC=\\angle C=3x$,\\\\\nIn $\\triangle ABC$, $\\angle ABC+\\angle C+\\angle BAC=180^\\circ$,\\\\\n$\\therefore 3x+3x+2x=180^\\circ$,\\\\\nsolving for $x$, we get $x=22.5^\\circ$,\\\\\n$\\therefore \\angle BCD=3x=67.5^\\circ$,\\\\\n(2) If $BC=CD$, then $\\angle BDC=\\angle CBD=3x$,\\\\\n$\\therefore \\angle ABC=\\angle ACB=4x$,\\\\\nIn $\\triangle ABC$, $\\angle ABC+\\angle C+\\angle BAC=180^\\circ$,\\\\\n$\\therefore 4x+4x+2x=180^\\circ$,\\\\\n$\\therefore x=18^\\circ$,\\\\\n$\\therefore \\angle BCD=4x=72^\\circ$,\\\\\nIn summary, $\\triangle BCD$ is an isosceles triangle, $\\angle BCD$ is either $\\boxed{67.5^\\circ}$ or $\\boxed{72^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53009799_130.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$, $AC$, $BD$ are the tangents to circle $\\odot O$ with the points of tangency at $P$, $C$, and $D$, respectively. If $AB=5$ and $AC=3$, what is the length of $BD$?\n",
    "completion": "\\textbf{Solution:} Since $AB$, $AC$, and $BD$ are tangents to the circle $\\odot O$,\\\\\nit follows that AP=AC and BP=BD,\\\\\nand since $AB=5$ and $AC=3$,\\\\\nit follows that AP=3,\\\\\nthus, BD=BP=AB-AP=\\boxed{2}.",
    "image_path": "dataset/mm_math/images/MM_Math/51213413.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{2}{3}x+\\frac{8}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the quadrilateral OABC is a rectangle, where points A and C are on the coordinate axes, and $\\triangle ODE$ is obtained by rotating $\\triangle OCB$ $90^\\circ$ clockwise around point O. Point D is on the x-axis, and line BD intersects the y-axis at point F and intersects OE at point H. The lengths of segments BC and OC are the solutions to the system of equations $\\begin{cases} 2x+3y=14 \\\\ 4x-5y=6 \\end{cases}$, with OC > BC. What is the analytical expression of line BD?\n",
    "completion": "\\textbf{Solution}: From the given problem, we have\n\\[\n\\begin{cases}\n2x+3y=14\\\\\n4x-5y=6\n\\end{cases}\n\\]\nSolving these equations, we obtain\n\\[\n\\begin{cases}\nx=4\\\\\ny=2\n\\end{cases},\n\\]\nSince $OC>BC$,\\\\\nHence, $CO=4$ and $BC=2$,\\\\\nThus, point B is $(-2,4)$,\\\\\nSince $\\triangle ODE$ is obtained by rotating $\\triangle OCB$ $90^\\circ$ clockwise around point O,\\\\\nTherefore, $\\triangle ODE \\cong \\triangle OCB$,\\\\\nHence, $OD=OC$ and $DE=BC$,\\\\\nThus, point D is $(4,0)$ and point E is $(4,2)$,\\\\\nLet the equation of line BD be $y=kx+b$,\\\\\nSubstituting points B and D, we get\n\\[\n\\begin{cases}\n-2k+b=4\\\\\n4k+b=0\n\\end{cases},\n\\]\nSolving these, we find\n\\[\n\\begin{cases}\nk=-\\frac{2}{3}\\\\\nb=\\frac{8}{3}\n\\end{cases},\n\\]\nTherefore, the equation of line BD is $y=-\\frac{2}{3}x+\\frac{8}{3}$; hence, the equation of line BD is $\\boxed{y=-\\frac{2}{3}x+\\frac{8}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52189515_147.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the diagonals of rhombus ABCD intersect at point O. Lines are drawn through points C and D such that $CE \\parallel BD$ and $DE \\parallel AC$, and these parallel lines intersect at point E. Quadrilateral OCED is a rectangle; if $CE=1$ and $DE=2$, calculate the area of rhombus ABCD.\n",
    "completion": "\\textbf{Solution:} From equation (1), we know that quadrilateral OCED is a rectangle, \\\\\n$\\therefore OD=CE=1, OC=DE=2$\\\\\n$\\because$ABCD is a rhombus,\\\\\n$\\therefore BD=2OD=2, AC=2OC=4$\\\\\n$\\therefore S_{ABCD}=\\frac{1}{2}\\times 2\\times 4=\\boxed{4}$",
    "image_path": "dataset/mm_math/images/MM_Math/53359977_145.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in square $ABCD$, $AB = 8\\,cm$. Extend $BC$ to point $E$ so that $CE = 2\\,cm$. Connect $DE$. A moving point $P$ starts from point $A$ and moves to the endpoint $A$ at a velocity of $2\\,cm/s$, along the path $AB \\rightarrow BC \\rightarrow CD \\rightarrow DA$. Let the movement time of point $P$ be $t$ seconds. When $\\triangle PBC$ and $\\triangle DCE$ are congruent, what is the value of $t$?\n",
    "completion": "\\textbf{Solution:} As shown in Figure A, when $\\triangle CBP_{1} \\cong \\triangle DCE$, we have $BP_{1} = CE = 2\\text{cm}$, \\\\\nthus $8-2t=2$. Solving for $t$ gives $t=\\boxed{3}$, \\\\\nAs shown in Figure B, when $\\triangle BCP_{2} \\cong \\triangle DCE$, we have $CP_{2} = CE=2\\text{cm}$\\\\\nthus $2t-2\\times 8=2$. Solving for $t$ gives $t=9$.",
    "image_path": "dataset/mm_math/images/MM_Math/52252065.png"
  },
  {
    "solution": "\\boxed{y=\\frac{2}{x}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the graph of the linear function $y=kx+b$ intersects the $x$ axis and the $y$ axis at points $A$ and $B$, respectively, and intersects the graph of the inverse proportion function $y= \\frac{m}{x}$ at point $C$ in the first quadrant. The line $CD$ is perpendicular to the $x$ axis, and the foot of the perpendicular is $D$. If $OA=OB=OD=1$, find the expression of this inverse proportion function.\n",
    "completion": "\\textbf{Solution:} Substitute $C(1,2)$ into $y=\\frac{m}{x}$,\n\n$\\therefore m=2$,\n\n$\\therefore$ the equation of the inverse proportion function is $\\boxed{y=\\frac{2}{x}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51546388_68.png"
  },
  {
    "solution": "\\boxed{-4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the Cartesian coordinate system, a part of the graph of the inverse proportion function $y=\\frac{k}{x}$ is shown in the figure. $AB\\perp y$-axis at point B, and point P is on the x-axis. If the area of $\\triangle ABP$ is 2, what is the value of $k$?\n",
    "completion": "\\textbf{Solution:} Let the equation of the inverse proportion function be: $y=\\frac{k}{x}$. Assume the coordinates of point A are (m, n).\\\\\nTherefore, AB = -m, OB = n, and mn = k.\\\\\nSince the area of $\\triangle$ABP is 2,\\\\\nit follows that $\\frac{1}{2}$AB$\\cdot$OB=2, i.e., -$\\frac{1}{2}$mn=2\\\\\nThus, mn=-4,\\\\\nresulting in $k=mn=\\boxed{-4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52602837.png"
  },
  {
    "solution": "\\boxed{\\frac{50}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that point $E$ is a point inside rectangle $ABCD$, satisfying $\\angle AEB=90^\\circ$. Extend $DE$ to intersect the semicircle with diameter $CD$ at point $F$. When $AE=20$, $BE=15$, and $DF=24$, what is the length of the side $AD$ of the rectangle?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, extend a line from point E to draw EH $\\perp$ CD at H, intersecting AB at point T, and connect CF.\\\\\nSince quadrilateral ABCD is a rectangle,\\\\\ntherefore AB=CD, AB$\\parallel$CD,\\\\\nsince $\\angle$AEB=90$^\\circ$, AE=20, BE=15,\\\\\ntherefore AB=$\\sqrt{AE^2+BE^2}=\\sqrt{20^2+15^2}=25$,\\\\\nsince ET$\\perp$AB,\\\\\ntherefore $\\frac{1}{2}AE \\cdot EB= \\frac{1}{2}AB \\cdot ET$,\\\\\ntherefore ET=$\\frac{20\\times 15}{25}=12$,\\\\\ntherefore AT=DH=$\\sqrt{AE^2-ET^2}=\\sqrt{20^2-12^2}=16$,\\\\\nsince CD is the diameter,\\\\\ntherefore $\\angle$CFD=90$^\\circ$,\\\\\ntherefore CF=$\\sqrt{CD^2-DF^2}=\\sqrt{25^2-24^2}=7$,\\\\\nsince $\\angle$EDH=$\\angle$CDF, $\\angle$EHD=$\\angle$CFD=90$^\\circ$,\\\\\ntherefore $\\triangle$DHE$\\sim$$\\triangle$DFC,\\\\\ntherefore $\\frac{DH}{DF}=\\frac{EH}{CF}$,\\\\\ntherefore $\\frac{16}{24}=\\frac{EH}{7}$,\\\\\ntherefore EH=$\\frac{14}{3}$,\\\\\ntherefore AD=BC=HT=ET+EH=12+$\\frac{14}{3}=\\boxed{\\frac{50}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52872143.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{6}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, circle $ \\odot O$ intersects with side $ OA$ and $ OE$ of the regular hexagon $OABCDE$ at points $F$ and $G$, respectively. Point $M$ is the midpoint of the minor arc $FG$. If $FM=4\\sqrt{2}$, what is the distance from point $O$ to line $FM$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect OM, and draw OH$\\perp$FM through point O,\\\\\n$\\because$ hexagon ABCDEO is a regular hexagon,\\\\\n$\\therefore$ $\\angle$AOE=120$^\\circ$,\\\\\n$\\because$ point M is the midpoint of the minor arc FG,\\\\\n$\\therefore$ $\\angle$AOM=60$^\\circ$,\\\\\n$\\because$ OM=OF,\\\\\n$\\therefore$ $\\triangle$OMF is an equilateral triangle,\\\\\n$\\therefore$ MH=FH=$\\frac{1}{2}$MF=2$\\sqrt{2}$, $\\angle$OFH=60$^\\circ$,\\\\\n$\\therefore$ OH=$\\tan60^\\circ\\cdot$ HF=$\\sqrt{3}\\times 2\\sqrt{2}=\\boxed{2\\sqrt{6}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51179568.png"
  },
  {
    "solution": "\\boxed{3\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the radius of $\\odot O$ is 5, $AB\\perp CD$, with the foot of the perpendicular as $P$, and $AB=CD=8$, then what is the length of $OP$?\n",
    "completion": "\\textbf{Solution:} Connect $OB$, draw $OE\\perp AB$ at $E$, and $OF\\perp CD$ at $F$,\\\\\nthen $BE=\\frac{1}{2}AB=4$, the quadrilateral $PEOF$ is a rectangle,\\\\\nsince $AB=CD$, $OE\\perp AB$, $OF\\perp CD$,\\\\\nthus $OE=OF$,\\\\\ntherefore, the rectangle $PEOF$ is a square,\\\\\nthus $OE=PE$,\\\\\nin $\\triangle OEB$, $OE=\\sqrt{OB^{2}-BE^{2}}=3$,\\\\\nthus $OP=\\sqrt{OE^{2}+PE^{2}}=\\boxed{3\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53009648.png"
  },
  {
    "solution": "\\boxed{(2+\\sqrt{3}, 1-\\sqrt{3})}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, given point $A(1,0)$, the $x$-axis is rotated clockwise by $60^\\circ$ about point $A$ to intersect the $y$-axis at point $B$, and then point $B$ is rotated clockwise by $90^\\circ$ about point $A$ to obtain point $C$. If point $Q$ is a point in the Cartesian coordinate system, and it satisfies that the quadrilateral $ABCQ$ is a parallelogram, what are the coordinates of point $Q$?\n",
    "completion": "\\textbf{Solution:} Let $Q(x, y)$, \\\\\nsince quadrilateral $ABCQ$ is a parallelogram, \\\\\ntherefore $AC$ and $BQ$ are the diagonals of the parallelogram, \\\\\nthe midpoint of $AC$ is $\\left(\\frac{2+\\sqrt{3}}{2}, \\frac{1}{2}\\right)$, the midpoint of $BQ$ is $\\left(\\frac{x}{2}, \\frac{y+\\sqrt{3}}{2}\\right)$, \\\\\ntherefore $\\frac{2+\\sqrt{3}}{2}=\\frac{x}{2}$, $\\frac{1}{2}=\\frac{y+\\sqrt{3}}{2}$, \\\\\ntherefore $x=2+\\sqrt{3}$, $y=1-\\sqrt{3}$, \\\\\ntherefore $Q(x, y)=\\boxed{(2+\\sqrt{3}, 1-\\sqrt{3})}$;",
    "image_path": "dataset/mm_math/images/MM_Math/50682648_122.png"
  },
  {
    "solution": "\\boxed{44^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, it is known that $\\angle AOB = 160^\\circ$, $\\angle COE$ is a right angle, and $OF$ bisects $\\angle AOE$. If $\\angle COF = 32^\\circ$, what is the measure of $\\angle BOE$ in degrees?\n",
    "completion": "\\textbf{Solution:} We have $\\angle BOE = \\boxed{44^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51482328_178.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, CD is the median of the base of the isosceles triangle $\\triangle ABC$, BE bisects $\\angle ABC$ and intersects CD at point E. Given that AC = 8 and DE = 2, what is the area of $\\triangle BCE$?\n",
    "completion": "\\textbf{Solution:} Construct EF $\\perp$ BC at F, \\\\\nsince AC = BC = 8, and CD is the median of the isosceles triangle $\\triangle ABC$ on the base, \\\\\ntherefore CD $\\perp$ AB, \\\\\nsince BE bisects $\\angle ABC$, ED $\\perp$ AB, and EF $\\perp$ BC, \\\\\ntherefore EF = DE = 2, \\\\\nthus, the area of $\\triangle BCE$ = $\\frac{1}{2} \\times BC \\times EF = \\frac{1}{2} \\times 8 \\times 2 = \\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51369696.png"
  },
  {
    "solution": "\\boxed{(-3, -2)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the graph of the linear function $y=2x+4$ intersects the x-axis at point A and the y-axis at point B. It intersects the graph of the inverse proportion function $y=\\frac{k}{x}$ ($k\\neq 0$) at points C and D, where the coordinate of point C is $(n, 6)$. Determine the coordinate of point D.\n",
    "completion": "\\textbf{Solution:} Solve by combining the two function expressions, \n\\[\n\\begin{cases}\ny=2x+4\\\\ \ny=\\frac{6}{x}\n\\end{cases}\n\\]\n, and find $x_{1}=-3$, $x_{2}=1$ (discard $x_{2}$).\\\\\nWhen $x=-3$, $y=2\\times(-3)+4=-2$,\\\\\n$\\therefore$ the coordinates of point D are $\\boxed{(-3, -2)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52720084_76.png"
  },
  {
    "solution": "\\boxed{16^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $AB \\parallel CD$, and the line $MN$ intersects $AB$ and $CD$ at points $E$ and $F$, respectively, with $\\angle NEB = 58^\\circ$. If $EG \\perp AB$, and $EH$ bisects $\\angle NEG$ and intersects $CD$ at point $H$, what is the measure of $\\angle BEH$ in degrees?\n",
    "completion": "\\textbf{Solution:} Since $EG\\perp AB$, it follows that $\\angle BEG=90^\\circ$, \\\\\nhence, $\\angle BEH=90^\\circ-\\angle GEH$, \\\\\nsince $EH$ bisects $\\angle NEG$, \\\\\nit follows that $\\angle GEH= \\frac{1}{2}\\angle NEG=\\frac{1}{2}(180^\\circ-\\angle MEG)$, \\\\\nfrom (1) we know $\\angle AEM=58^\\circ$, and since $EG\\perp AB$, \\\\\nthus, $\\angle MEG=90^\\circ-\\angle AEM=90^\\circ-58^\\circ=32^\\circ$, \\\\\nhence, $\\angle GEH=\\frac{1}{2}(180^\\circ-32^\\circ)=74^\\circ$, \\\\\nthus, $\\angle BEH=90^\\circ-74^\\circ=\\boxed{16^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52375765_143.png"
  },
  {
    "solution": "\\boxed{16}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the figure, within $\\triangle ABE$, the perpendicular bisector MN of AE intersects BE at point C, and AC is connected. If $AB=AC$, $CE=5$, and $BC=6$, then what is the perimeter of $\\triangle ABC$?\n",
    "completion": "\\textbf{Solution:} Since $MN$ is the perpendicular bisector of $AE$ and $CE=5$,\\\\\nit follows that $AC=CE=5$,\\\\\nSince $AB=AC$,\\\\\nand given that $AB=5$,\\\\\nand $BC=6$,\\\\\nthus, the perimeter of $\\triangle ABC$ is $AB+AC+BC=5+5+6=\\boxed{16}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51378731.png"
  },
  {
    "solution": "\\boxed{225^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, what is the sum of $\\angle A+\\angle B+\\angle C+\\angle D$?\n",
    "completion": "\\textbf{Solution}: Draw lines AD and BC.\\\\\nIn quadrilateral ABCD, we have $\\angle DAE + \\angle EAB + \\angle ABF + \\angle FBC + \\angle DCF + \\angle BCF + \\angle CDE + \\angle ADE = 360^\\circ$,\\\\\nSince $\\angle DEA = 105^\\circ$ and $\\angle BFC = 120^\\circ$,\\\\\nIt follows that $\\angle DAE + \\angle ADE = 180^\\circ - 105^\\circ = 75^\\circ$ and $\\angle FBC + \\angle BCF = 180^\\circ - 120^\\circ = 60^\\circ$,\\\\\nTherefore, $\\angle EAB + \\angle ABF + \\angle DCF + \\angle CDE = 360^\\circ - 75^\\circ - 60^\\circ = 225^\\circ$.\\\\\nHence, $\\angle A + \\angle B + \\angle C + \\angle D = \\boxed{225^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52861343.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $AB$ is the diameter of the circle $\\odot O$, and $\\odot O$ intersects side $AE$ of $\\triangle ABE$ at point $D$. Connect $OD$, and it satisfies $OD\\parallel BE$. Point $P$ is on the extension line of $BA$, and $PD$ intersects $BE$ at point $C$. $AB=BE$; If $PA=2$, $\\angle B=60^\\circ$, $PC\\perp BE$, find the length of diameter $AB$.\n",
    "completion": "\\textbf{Solution:} Since $\\angle PCB=90^\\circ$ and $\\angle B=60^\\circ$, it follows that $\\angle P=30^\\circ$. From this, we have $OD\\parallel BE$, hence $\\angle B=\\angle DOA=60^\\circ$. Also, since $OA=OD$, it follows that $\\triangle AOD$ is an equilateral triangle, thus $AD=OA$, and $\\angle DAO=60^\\circ$. Therefore, $\\angle ADP=\\angle P=30^\\circ$, which implies that $PA=AD=OA=2$, hence $AB=2OA=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52458135_90.png"
  },
  {
    "solution": "\\boxed{1.4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, Xiaofang and her father are taking a walk. Her father is 1.8m tall and his shadow on the ground is 2.1m long. If Xiaofang is only 1.2m tall, how long is her shadow?\n",
    "completion": "\\textbf{Solution:} Let the shadow length of Xiaofang be $x$ meters.\\\\\nAccording to the proportionality of height to shadow length at the same moment, we get: $\\frac{1.8}{2.1}=\\frac{1.2}{x}$,\\\\\nSolving this, we obtain: $x=\\boxed{1.4}$.\\\\\nUpon verification, this is consistent with the problem statement.",
    "image_path": "dataset/mm_math/images/MM_Math/51212652.png"
  },
  {
    "solution": "\\boxed{24^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the circle $\\odot O$, the chord $AC\\parallel$ to the radius $OB$, $\\angle BOC=48^\\circ$, what is the degree measure of $\\angle OAB$?\n",
    "completion": "\\textbf{Solution:} Given that $AC \\parallel OB$,\\\\\nit follows that $\\angle BOC = \\angle ACO = 48^\\circ$,\\\\\nsince $OA = OC$,\\\\\nthen $\\angle OAC = \\angle ACO = 48^\\circ$,\\\\\nsince $\\angle CAB = \\frac{1}{2} \\angle BOC = 24^\\circ$,\\\\\nhence $\\angle BAO = \\angle OAC - \\angle CAB = \\boxed{24^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51212770.png"
  },
  {
    "solution": "\\boxed{(6,8)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the Cartesian coordinate system, the position of the rhombus OABC is shown as in the figure, with $B(16, 8)$. What are the coordinates of point C?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $BD\\perp x$ axis at point $D$ through point $B$,\n\n$\\because B(16,8)$,\n\n$\\therefore D(16,0)$,\n\ni.e., $OD=16$, $BD=8$,\n\n$\\because$ for diamond $OABC$, let $OA=x$, then $AB=OA=x$, $AD=16-x$,\n\nIn $\\triangle ABD$ right, $AB^{2}=AD^{2}+BD^{2}$,\n\n$x^{2}=(16-x)^{2}+8^{2}$,\n\nSolving for $x$ yields $x=10$,\n\n$\\therefore BC=10$,\n\n$\\therefore C(6,8)$.\n\nThus, the answer is: $C(6,8)=\\boxed{(6,8)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52386401.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the parabola $y=ax^2-3ax-4a$ intersects the x-axis at points $A$ and $B$ (with $A$ to the left of $B$) and intersects the negative half of the y-axis at point $C$. Let $P$ be a point on the parabola in the fourth quadrant. The line $AP$ intersects the y-axis at point $M$, and the line $BP$ intersects the y-axis at point $N$. Find the value of $\\frac{4OM+ON}{OC}$.\n",
    "completion": "\\textbf{Solution:} Given from (1) that points $A(-1,0)$, $B(4,0)$, and $C(0,-4a)$. Let point $P(m,am^{2}-3am-4a)$, by substituting the coordinates of points P and A into the expression of a linear function and solving, we find the equation of line $PA$ to be: $y=a(m-4)x+a(m-4)$. Therefore, $OM=-a(m-4)$. Similarly, the equation of line $BP$ is $y=a(m+1)x-4a(m+1)$, $ON=4a(m+1)$. Hence, $4OM+ON=-4a(m-4)+4a(m+1)=20a$. Since $C(0,-4a)$, it follows that $OC=4a$. Thus, $\\frac{4OM+ON}{OC}=\\frac{20a}{4a}=\\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52916643_244.png"
  },
  {
    "solution": "\\boxed{4+\\sqrt{6}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a point E is taken outside the square ABCD, and AE, BE, and DE are connected respectively. A perpendicular line to AE is drawn from point A and intersects DE at point P. If AE = AP = 1, and PB = $ \\sqrt{5}$. What is the area of the square ABCD, $S_{\\text{square ABCD}}$=?\n",
    "completion": "\\textbf{Solution:}\n\nSince quadrilateral ABCD is a square,\n\\begin{align*}\n&\\therefore AB=AD, \\angle BAD=90^\\circ,\\\\\n&\\because AE\\perp AP, AE=AP=1,\\\\\n&\\therefore \\angle AEP=\\angle APE=45^\\circ, \\angle EAP=\\angle BAD=90^\\circ,\\\\\n&\\therefore \\angle EAB=\\angle PAD,\\\\\n&\\text{In } \\triangle EAB \\text{ and } \\triangle PAD,\\\\\n&\\because AB=AD, \\angle EAB=\\angle PAD, AE=AP,\\\\\n&\\therefore \\triangle EAB \\cong \\triangle PAD \\text{ (SAS)},\\\\\n&\\therefore \\angle EBA=\\angle ADP, BE=DP, \\angle APD=\\angle AEB=180^\\circ-45^\\circ=135^\\circ,\\\\\n&\\therefore \\angle PEB=135^\\circ-45^\\circ=90^\\circ, \\text{ i.e., } \\triangle BEP\\text{ is a right-angled triangle},\\\\\n&\\because AE=AP=1,\\\\\n&\\therefore EP=\\sqrt{1^2+1^2}=\\sqrt{2},\\\\\n&\\therefore BE=DP=\\sqrt{BP^2-EP^2}=\\sqrt{3},\\\\\n&\\text{Draw line BF }\\perp\\text{ AE extended to meet at point F, connect BD, then }\\angle FEB=180^\\circ-135^\\circ=45^\\circ,\\\\\n&\\therefore \\angle EBF=45^\\circ=\\angle FEB,\\\\\n&\\therefore EF=BF,\\\\\n&\\because BE=\\sqrt{3},\\\\\n&\\therefore BF=EF=\\frac{\\sqrt{6}}{2},\\\\\n&\\therefore {S}_{\\triangle APB}+{S}_{\\triangle APD}={S}_{\\triangle APB}+{S}_{\\triangle AEB}={S}_{\\text{quadrilateral }AEBP}={S}_{\\triangle AEP}+{S}_{\\triangle PEB}\\\\\n&=\\frac{1}{2}\\times 1\\times 1+\\frac{1}{2}\\times \\sqrt{2}\\times \\sqrt{3}\\\\\n&=\\frac{1}{2}+\\frac{1}{2}\\sqrt{6},\\\\\n&\\because {S}_{\\triangle DPB}=\\frac{1}{2}\\times DP\\times BE=\\frac{1}{2}\\times \\sqrt{3}\\times \\sqrt{3}=\\frac{3}{2},\\\\\n&\\therefore {S}_{\\text{square }ABCD}=2{S}_{\\triangle ABD}=2({S}_{\\triangle BPD}+{S}_{\\triangle APD}+{S}_{\\triangle APB})=2\\times (\\frac{1}{2}+\\frac{1}{2}\\sqrt{6}+\\frac{3}{2})=\\boxed{4+\\sqrt{6}}.\n\\end{align*}",
    "image_path": "dataset/mm_math/images/MM_Math/51968677.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $D$, $E$, and $F$ are the midpoints of the sides $AB$, $BC$, and $AC$ respectively. If the perimeter of $\\triangle ABC$ is 6, what is the perimeter of $\\triangle DEF$?\n",
    "completion": "\\textbf{Solution:} Given $\\because$ in $\\triangle ABC$, D, E, F are the midpoints of the sides $AB$, $BC$, and $AC$ respectively,\\\\\n$\\therefore$DE= $\\frac{1}{2}$ AC, EF= $\\frac{1}{2}$ AB, DF= $\\frac{1}{2}$ BC,\\\\\n$\\because$ the perimeter of $\\triangle ABC$ is AB+BC+AC=6,\\\\\n$\\therefore$ the perimeter of $\\triangle DEF$ is EF+DF+DE= $\\frac{1}{2}$ （AB+BC+AC）= $\\frac{1}{2} \\times 6=\\boxed{3}$",
    "image_path": "dataset/mm_math/images/MM_Math/50496952.png"
  },
  {
    "solution": "\\boxed{-2<x<\\frac{5}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the parabola \\(y_1=ax^2+bx+c\\ (a\\neq 0)\\) and the line \\(y_2=mx+n\\ (m\\neq 0)\\) intersect at points \\(A\\) and \\(B\\), with the \\(x\\)-coordinates of points \\(A\\) and \\(B\\) being \\(-2\\) and \\(\\frac{5}{4}\\), respectively. What is the solution to the inequality \\(ax^2+bx+c<mx+n\\)?\n",
    "completion": "Solution: From the graph, it is known that the solution to $ax^2+bx+c<mx+n$ occurs when the line is above the parabola, \\\\\n$\\therefore$ $-2<x<\\frac{5}{4}$, \\\\\nTherefore, the answer is $\\boxed{-2<x<\\frac{5}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53102185.png"
  },
  {
    "solution": "\\boxed{y=-x+4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the graph of the linear function $y=kx+b$ passes through point $A(-2, 6)$ and intersects with the $x$-axis at point $B$, and intersects with the graph of the direct proportion function $y=3x$ at point $C$, where the $x$-coordinate of point $C$ is 1. What is the analytic expression of the linear function?\n",
    "completion": "\\textbf{Solution:} Given that the x-coordinate of point C is 1,\\\\\n$\\therefore C(1, 3)$,\\\\\nsubstitute points $A(-2, 6)$ and $C(1, 3)$ into $y=kx+b$,\\\\\n$\\therefore \\left\\{\\begin{array}{l}k+b=3 \\\\ -2k+b=6\\end{array}\\right.$,\\\\\nsolving this yields $\\left\\{\\begin{array}{l}k=-1 \\\\ b=4\\end{array}\\right.$,\\\\\n$\\therefore \\boxed{y=-x+4}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53706908_198.png"
  },
  {
    "solution": "\\boxed{2.5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, point D is the midpoint of $AB$, and $CD$ is connected. If $AC=4$ and $BC=3$, then what is the length of $CD$?\n",
    "completion": "\\textbf{Solution:} Since in $\\triangle ABC$, $\\angle ACB=90^\\circ$, $AC=4$, and $BC=3$, by the Pythagorean theorem we have: $AB= \\sqrt{AC^{2}+BC^{2}}=\\sqrt{4^{2}+3^{2}}=5$. Since point D is the midpoint of AB, it follows that $CD= \\frac{1}{2}AB= \\frac{1}{2}\\times 5=\\boxed{2.5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51385683.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, with a circle having diameter $BC$ intersecting $AB$ at point $D$, and $\\angle ACD = \\angle ABC$. If point $E$ is on $BC$, given $BE = 6$, $\\cos\\angle ABC = \\frac{3\\sqrt{13}}{13}$, $\\tan\\angle AEC = \\frac{5}{3}$, and $CA$ is a tangent to the circle. What is the diameter of the circle?\n",
    "completion": "\\textbf{Solution:} In $\\triangle AEC$, $\\tan \\angle AEC = \\frac{5}{3}$, thus $\\frac{AC}{EC} = \\frac{5}{3}$, and $EC = \\frac{3}{5} AC$. In $\\triangle ABC$, $\\cos \\angle ABC = \\frac{3\\sqrt{13}}{13}$, thus $\\frac{BC}{AB} = \\frac{3\\sqrt{13}}{13}$, and $\\frac{AC}{BC} = \\frac{2}{3}$, thus $BC = \\frac{3}{2} AC$. Since $BC - EC = BE$ and $BE = 6$, it follows that $\\frac{3}{2} AC - \\frac{3}{5} AC = 6$. Solving this gives $AC = \\frac{20}{3}$, thus $BC = \\frac{3}{2} AC = \\frac{3}{2} \\times \\frac{20}{3} = 10$. The diameter of the circle is $\\boxed{10}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51265343_146.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the two diagonals of rectangle ABCD intersect at point O. If $\\angle AOD = 60^\\circ$ and AD = 2, then what is the length of CD?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rectangle,\\\\\nit follows that OA=OC, OB=OD, and AC=BD,\\\\\nwhich means OA=OB=OD,\\\\\nSince $\\angle AOD = 60^\\circ$,\\\\\nit follows that $\\triangle AOD$ is an equilateral triangle,\\\\\nthus, OD=AD=2,\\\\\nwhich implies BD=2OD=4,\\\\\ntherefore, $DC = \\sqrt{AC^{2} - AD^{2}} = \\sqrt{16 - 4} = 2\\sqrt{3}$,\\\\\nHence, the answer is: $DC = \\boxed{2\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53706883.png"
  },
  {
    "solution": "\\boxed{50^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Take point O on line AB as the endpoint to draw ray OC so that $\\angle BOC=40^\\circ$. Place the right-angled vertex of a right-angle set square at point O, i.e., $\\angle DOE=90^\\circ$. As shown in Figure 1, if one side OE of the right-angle set square DOE lies on the ray OA, what is the value of $\\angle COD$?\n",
    "completion": "\\textbf{Solution:} Hence, $\\angle COD = \\boxed{50^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52916141_149.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, points $D$ and $E$ are on sides $AB$ and $AC$, respectively, with $DE \\parallel BC$ and $BC = 6$. Given that the area of quadrilateral $BCED$ is $8$ times the area of $\\triangle ADE$, what is the length of $DE$?\n",
    "completion": "\\textbf{Solution:} Since ${S}_{\\text{quadrilateral} BCED}=8{S}_{\\triangle ADE}$,\nit follows that ${S}_{\\triangle ABC}={S}_{\\triangle ADE}+{S}_{\\text{quadrilateral} BCED}={S}_{\\triangle ADE}+8{S}_{\\triangle ADE}=9{S}_{\\triangle ADE}$,\nthus $\\frac{{S}_{\\triangle ADE}}{{S}_{\\triangle ABC}}=\\frac{1}{9}$,\nSince DE$\\parallel$BC, it follows that $\\angle$ADE=$\\angle$B, $\\angle$AED=$\\angle$C,\nwhich implies $\\triangle ADE\\sim \\triangle ABC$,\ntherefore $\\frac{{S}_{\\triangle ADE}}{{S}_{\\triangle ABC}}=\\left(\\frac{DE}{BC}\\right)^{2}=\\frac{1}{9}$,\nhence $\\frac{DE}{BC}=\\frac{1}{3}$,\nSince BC＝6, it follows that $DE=\\frac{1}{3}BC=\\frac{1}{3}\\times 6=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52601759.png"
  },
  {
    "solution": "\\boxed{\\frac{4}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, $E$ and $F$ are the midpoints of $AB$ and $AD$ respectively. If $EF=2$, $BC=5$, and $CD=3$, then what is the value of $\\sin C$?\n",
    "completion": "\\textbf{Solution:} Draw line $BD$, \\\\\nsince points E and F are the midpoints of $AB$ and $AD$ respectively, \\\\\ntherefore, $EF=\\frac{1}{2}BD$, \\\\\nsince $EF=2$, \\\\\ntherefore, $BD=2EF=4$, \\\\\nalso, since $BC=5$ and $CD=3$, \\\\\ntherefore, $BD^{2}+CD^{2}=BC^{2}$, \\\\\nthus, $\\triangle BCD$ is a right-angled triangle with $\\angle BDC=90^\\circ$, \\\\\ntherefore, $\\sin C=\\frac{BD}{BC}=\\boxed{\\frac{4}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50934151.png"
  },
  {
    "solution": "\\boxed{120^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AD$ is the angle bisector of the exterior angle $\\angle EAC$ of $\\triangle ABC$, intersecting the circumcircle $\\odot O$ of $\\triangle ABC$ at point D, with $\\angle EAC=120^\\circ$. What is the degree measure of arc $\\overset{\\frown}{BC}$?\n",
    "completion": "\\textbf{Solution:} Draw lines OB and OC,\n\nsince $\\angle EAC = 120^\\circ$,\n\ntherefore $\\angle BAC = 180^\\circ - \\angle EAC = 60^\\circ$,\n\ntherefore $\\angle BDC = \\angle BAC = 60^\\circ$,\n\ntherefore $\\angle BOC = 2\\angle BDC = 2 \\times 60^\\circ = 120^\\circ$,\n\nthus, the degree measure of arc $\\overset{\\frown}{BC}$ is $\\boxed{120^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52874761_88.png"
  },
  {
    "solution": "\\boxed{15^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of circle $O$. Points $C$ and $D$ are on circle $O$, and segments $AC$, $BC$, and $CD$ are connected respectively. Given that $\\angle DOB=150^\\circ$, what is the measure of $\\angle ACD$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle DOB=150^\\circ$ and $\\angle AOD+\\angle DOB=180^\\circ$,\\\\\nit follows that $\\angle AOD=30^\\circ$,\\\\\nthus $\\angle ACD=\\frac{1}{2}\\angle AOD=15^\\circ$.\\\\\nTherefore, the answer is: $\\boxed{15^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52793780.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AC$ is the diagonal of rhombus $ABCD$. If $\\angle ABC = 60^\\circ$, what is the degree measure of $\\angle CAD$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rhombus,\\\\\n$AB=BC=CD=AD$, and $AD\\parallel BC$,\\\\\nsince $\\angle ABC=60^\\circ$,\\\\\nthen $\\triangle ABC$ is an equilateral triangle, i.e., $\\angle ACB = \\angle CAB = 60^\\circ$,\\\\\nsince $AD\\parallel BC$,\\\\\nthen $\\angle CAD=\\angle ACB=60^\\circ$.\\\\\nTherefore, the answer is: $\\boxed{60^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52065343.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{5}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, we have $AB=AC=\\sqrt{5}$, and $BC=2$. A line is drawn from point $B$ such that $BD \\perp AC$ with $D$ being the foot of the perpendicular. Point $E$ is a point on the extension of $BD$. Connect $CE$. When $\\angle E = \\angle A$, find the length of the segment $CE$.\n",
    "completion": "\\textbf{Solution:} Since $S_{\\triangle ABC}=\\frac{1}{2}AG\\cdot BC=\\frac{1}{2}BD\\cdot AC$,\\\\\nit follows that $BD= \\frac{4\\sqrt{5}}{5}$,\\\\\nand since $\\cot\\angle ACB=\\frac{1}{2}$,\\\\\nit further follows that $\\frac{CD}{BD}=\\frac{1}{2}$,\\\\\nthus, $CD= \\frac{2\\sqrt{5}}{5}$,\\\\\nand since $\\angle BAC=\\angle E$,\\\\\nit can be concluded that $\\triangle ADB\\sim \\triangle EDC$,\\\\\ntherefore, $\\frac{EC}{AB}=\\frac{CD}{BD}=\\frac{1}{2}$,\\\\\nhence, $EC=\\frac{1}{2}AB=\\boxed{\\frac{\\sqrt{5}}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52977907_123.png"
  },
  {
    "solution": "\\boxed{-6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, on the graph of the inverse proportion function $y=\\frac{2}{3x}$, there is a moving point A. Connect AO and extend it to intersect another branch of the graph at point B. In the fourth quadrant, there is a point C, satisfying $AC=BC$. As point A moves, point C always moves on the graph of the function $y=\\frac{k}{x}$. If $\\tan\\angle CAB=3$, what is the value of $k$?\n",
    "completion": "\\textbf{Solution:} Connect OC, and draw AE $\\perp$ y-axis at point E, and draw CF $\\perp$ x-axis at point F, as shown in the diagram.\\\\\nDue to the symmetry of line AB and the inverse function $y=\\frac{2}{3x}$, points A and B are symmetrical about point O,\\\\\n$\\therefore$ AO = BO,\\\\\nAlso, since AC = BC,\\\\\n$\\therefore$ CO $\\perp$ AB.\\\\\nSince $\\angle$AOE + $\\angle$AOF = $90^\\circ$, and $\\angle$AOF + $\\angle$COF = $90^\\circ$,\\\\\n$\\therefore$ $\\angle$AOE = $\\angle$COF,\\\\\nAlso, since $\\angle$AEO = $\\angle$CFO = $90^\\circ$,\\\\\n$\\therefore$ $\\triangle AOE \\sim \\triangle COF$,\\\\\n$\\therefore$ $\\frac{AE}{CF} = \\frac{OE}{OF} = \\frac{OA}{OC}$,\\\\\nSince $\\tan \\angle$CAB = $\\frac{OC}{OA}$ = 3,\\\\\n$\\therefore$ CF = 3AE, OF = 3OE,\\\\\nAlso, since AE$\\cdot$OE = $\\left|\\frac{2}{3}\\right|$ = $\\frac{2}{3}$, CF$\\cdot$OF = $|k|$,\\\\\n$\\therefore$ |k| = 6,\\\\\n$\\therefore$ k = $\\pm$6,\\\\\nSince point C is in the fourth quadrant,\\\\\n$\\therefore$ k = \\boxed{-6}.",
    "image_path": "dataset/mm_math/images/MM_Math/52977770.png"
  },
  {
    "solution": "\\boxed{\\sqrt{2}-1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: To calculate $\\tan 15^\\circ$, as shown in the figure, in $\\text{Rt}\\triangle ABC$, $\\angle C=90^\\circ$, $\\angle ABC=30^\\circ$. Extend $CB$ to $D$, so that $BD=AB$, and connect $AD$, then $\\angle D=15^\\circ$. Thus, $\\tan 15^\\circ =\\frac{AC}{CD}=\\frac{1}{2+\\sqrt{3}}=\\frac{2-\\sqrt{3}}{(2+\\sqrt{3})(2-\\sqrt{3})}=2-\\sqrt{3}$. Following this method, what is the value of $\\tan 22.5^\\circ$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, in the isosceles right-angled $\\triangle ABC$, $\\angle C=90^\\circ$. Extend $CB$ to point $D$ such that $AB=BD$, then $\\angle BAD=\\angle D$.\\\\\n$\\because \\angle ABC=45^\\circ$,\\\\\n$\\therefore 45^\\circ=\\angle BAD+\\angle D=2\\angle D$,\\\\\n$\\therefore \\angle D=22.5^\\circ$,\\\\\nLet $AC=1$, then $BC=1$, $AB=\\sqrt{2}AC=\\sqrt{2}$,\\\\\n$\\therefore CD=CB+BD=CB+AB=1+\\sqrt{2}$,\\\\\n$\\therefore \\tan 22.5^\\circ=\\tan D=\\frac{AC}{CD}=\\frac{1}{1+\\sqrt{2}}=\\frac{1-\\sqrt{2}}{(1+\\sqrt{2})(1-\\sqrt{2})}=\\boxed{\\sqrt{2}-1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51320411.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the area of square $ABCD$ is $100 \\text{cm}^2$. Triangle $\\triangle ABP$ is a right-angled triangle with $\\angle P = 90^\\circ$, and $PB = 6\\text{cm}$. What is the length of $AP$?\n",
    "completion": "\\textbf{Solution:} Given that the area of square ABCD is $100\\,\\text{cm}^{2}$,\\\\\nit follows that AB=10cm,\\\\\nsince \\(\\triangle\\)ABP is a right triangle with \\(\\angle P=90^\\circ\\) and PB=6cm,\\\\\ntherefore AP=$\\sqrt{AB^{2}-BP^{2}}=\\sqrt{10^{2}-6^{2}}=\\boxed{8}\\,\\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52699329.png"
  },
  {
    "solution": "\\boxed{y=-x+5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y=kx+b$ intersects the graph of the inverse proportion function $y=\\frac{m}{x}$ where $x>0$, at points $A(1, 4)$ and $B(4, n)$. What are the expressions for the linear function and the inverse proportion function?\n",
    "completion": "\\textbf{Solution:} Substituting point A into the inverse proportion function $y=\\frac{m}{x}\\left(x>0\\right)$, we have $m=xy=4$. Therefore, the expression of the inverse proportion function is: $y=\\frac{4}{x}$. Hence, $4n=4$, solving this equation, we find $n=1$. Therefore, point $B(4, 1)$ is determined. This leads to the system $\\left\\{\\begin{array}{l}1=4k+b\\\\ 4=k+b\\end{array}\\right.$, from which we solve $\\left\\{\\begin{array}{l}k=-1\\\\ b=5\\end{array}\\right.$. Therefore, the equation of the linear function is: \\boxed{y=-x+5};",
    "image_path": "dataset/mm_math/images/MM_Math/52205237_119.png"
  },
  {
    "solution": "\\boxed{\\frac{9}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $D$ is the midpoint of $BC$, $DE\\perp AB$, and $DF\\perp AC$ with the perpendicular feet being $E$ and $F$ respectively, and $BE=CF$. $AD$ is the angle bisector of $\\triangle ABC$. If $AB=8$ and the area of $\\triangle ABC$ is $36$, what is the length of $DE$?\n",
    "completion": "\\textbf{Solution:} From (1), we have $\\triangle DEB \\cong \\triangle DFC$,\\\\\n$DE=DF$,\\\\\n$\\therefore \\angle B=\\angle C$,\\\\\n$\\therefore AB=AC=8$,\\\\\n$\\because S_{\\triangle ABC}=36$,\\\\\n$\\therefore \\frac{AB\\cdot DE}{2}+\\frac{AC\\cdot DF}{2}=36$,\\\\\n$\\therefore \\frac{8DE}{2}+\\frac{8DE}{2}=36$,\\\\\nFrom this, we find $DE=\\boxed{\\frac{9}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52845740_96.png"
  },
  {
    "solution": "\\boxed{\\frac{7}{8}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, it is known that the parabola $y=x^2-2x-3$ has its vertex at point $A$, intersects the $y$-axis at point $C$, and the segment $CB\\parallel x$-axis intersects the parabola at another point $B$. The parabola $y=x^2-2x-3$ is translated so that its vertex always moves along the line $AC$. When the translated parabola has only one common point with the ray $BA$, let the $x$-coordinate of the vertex of the parabola at this time be $n$. Please write down the range of values of $n$ directly.\n",
    "completion": "\\textbf{Solution:} The range of values for $n$ is $1<n\\le 4$ or $n=\\boxed{\\frac{7}{8}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52512802_171.png"
  },
  {
    "solution": "\\boxed{(3,0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given the graph of the parabola $y = x^2 - 2x - 3$ as shown in the figure. What are the coordinates of the points where the parabola intersects the x-axis and y-axis?\n",
    "completion": "\\textbf{Solution:} Let $x=0$, then $y=-3$. Therefore, the intersection point of the parabola with the $y$-axis is \\boxed{(0,-3)}. Let $y=0$, then $x^2-2x-3=0$. Solving this yields: $x_1=-1$, $x_2=3$. Therefore, the intersection points of the parabola with the $x$-axis are \\boxed{(-1,0)} and \\boxed{(3,0)}.",
    "image_path": "dataset/mm_math/images/MM_Math/52808108_36.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated, a triangular piece of material is given, with $\\angle A=30^\\circ$, $\\angle C=90^\\circ$, and $AB=12$. From this material, a rectangle CDEF is to be cut out, where points D, E, and F are located on AC, AB, and BC, respectively. Let the length of AE be $x$, and the area of rectangle CDEF be $S$. When the area of rectangle CDEF is $8\\sqrt{3}$, what is the length of AE?\n",
    "completion": "\\textbf{Solution:} According to the problem statement, we have $S_{\\text{rectangle }CDEF}=-\\frac{\\sqrt{3}}{4}x^2+3\\sqrt{3}x=8\\sqrt{3}$\\\\\n$\\therefore -\\frac{\\sqrt{3}}{4}\\left[(x-6)^2-36\\right]=8\\sqrt{3}$\\\\\n$-\\frac{\\sqrt{3}}{4}(x-6)^2+9\\sqrt{3}=8\\sqrt{3}$,\\\\\n$(x-6)^2=4$\\\\\n$\\therefore x-6=\\pm 2$\\\\\nSolving gives: $x_1=4$, $x_2=8$,\\\\\n$\\therefore$ The length of AE is either $\\boxed{4}$ or $\\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52602949_92.png"
  },
  {
    "solution": "\\boxed{120^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the straight lines AB and CD intersect at point O. OE is the bisector of $\\angle AOD$. If $\\angle AOC=60^\\circ$ and $OF\\perp OE$, what is the measure of $\\angle BOE$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle AOC=60^\\circ$,\\\\\nit follows that $\\angle BOD=\\angle AOC=60^\\circ$,\\\\\n$\\angle AOD=180^\\circ-60^\\circ=120^\\circ$,\\\\\nsince $OE$ is the bisector of $\\angle AOD$,\\\\\nit follows that $\\angle DOE= \\frac{1}{2} \\angle AOD=60^\\circ$,\\\\\nthus, $\\angle BOE=\\angle BOD+\\angle DOE=60^\\circ+60^\\circ=\\boxed{120^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/17148529_69.png"
  },
  {
    "solution": "\\boxed{67.5^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In $\\triangle ABC$ and $\\triangle ADE$, where point $E$ is on $BC$. It is known that $\\angle B = \\angle D$, $\\angle DAB = \\angle EAC$. Thus, $\\triangle ABC \\sim \\triangle ADE$. If $AC \\parallel DE$ and $\\angle AEC = 45^\\circ$, what is the degree measure of $\\angle C$?\n",
    "completion": "\\textbf{Solution:} Given that $AC \\parallel DE$,\\\\\n$\\therefore \\angle AED = \\angle EAC$,\\\\\nSince $\\triangle ABC \\sim \\triangle ADE$,\\\\\n$\\therefore \\angle AED = \\angle C$,\\\\\n$\\therefore \\angle EAC = \\angle C$,\\\\\nGiven that $\\angle AEC = 45^\\circ$,\\\\\n$\\therefore \\angle C = (180^\\circ - 45^\\circ) \\div 2 = 67.5^\\circ$,\\\\\nHence, the measure of $\\angle C$ is $\\boxed{67.5^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53065899_136.png"
  },
  {
    "solution": "\\boxed{\\frac{24}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rhombus $ABCD$, $AC=6$, $DB=8$, $AH \\perp BC$. What is the length of $AH$?\n",
    "completion": "\\textbf{Solution}: As shown in the figure, the diagonals $AC$ and $BD$ intersect at point O,\\\\\n$\\because$ Quadrilateral $ABCD$ is a rhombus, $AC=6$, $BD=8$\\\\\n$\\therefore AC\\perp BD$, $OA=OC=3$, $OB=OD=4$,\\\\\n$\\therefore BC=\\sqrt{OB^{2}+OC^{2}}=\\sqrt{4^{2}+3^{2}}=5$,\\\\\n$\\because AH\\perp BC$,\\\\\n$\\therefore$ The area of rhombus $ABCD$ $=\\frac{1}{2}AC\\cdot BD=BC\\cdot AH$,\\\\\nthat is, $\\frac{1}{2}\\times 6\\times 8=5AH$,\\\\\n$\\therefore AH=\\boxed{\\frac{24}{5}}$",
    "image_path": "dataset/mm_math/images/MM_Math/52424722.png"
  },
  {
    "solution": "\\boxed{64^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AE=AB$, $\\angle E=\\angle B$, $EF=BC$. If $\\angle EAB=52^\\circ$, then what is the measure of $\\angle EFA$?\n",
    "completion": "\\textbf{Solution:} In $\\triangle AEF$ and $\\triangle ABC$, since $AE=AB$, $\\angle E=\\angle B$, and $EF=BC$, then $\\triangle AEF \\cong \\triangle ABC$ by SAS, thus $\\angle EAF=\\angle BAC$, $AF=AC$, and $\\angle C=\\angle EFA$. Therefore, $\\angle EAF-\\angle BAF=\\angle BAC-\\angle BAF$, that is, $\\angle FAC=\\angle EAB=52^\\circ$. Consequently, $\\angle C=\\angle EFA= \\frac{1}{2}(180^\\circ-\\angle FAC)=\\boxed{64^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55498347.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, fold a rectangular strip of paper successively along the line segments $EF$ and $HI$, with $EF\\parallel HI$, to form a \"Z\" shaped pattern. Given that $\\angle DFE = 60^\\circ$ and $AE = 2\\text{cm}$, in order for point $H$ and point $K$ to be on the extensions of $AD$ and $EF$ respectively (not coincident with $D$ or $F$), what is the length of $AB$ in centimeters?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect $DH$ and $FK$, where points H and K are on the extensions of $AD$ and $EF$ (not coinciding with D and F), respectively, and point M is a point on the extension of $EH$.\\\\\nIn the rectangle strip $ABCD$, $\\angle A=\\angle ADC=90^\\circ$, $AB\\parallel CD$,\\\\\n$\\therefore$ $\\angle DFE=\\angle BEF=60^\\circ$, $EH\\parallel FI$,\\\\\nFrom the folding, it is known that: $\\angle GEF=\\angle BEF=60^\\circ$,\\\\\n$\\therefore$ $\\angle AEG=60^\\circ$,\\\\\n$\\therefore$ $\\angle AHE=30^\\circ$,\\\\\n$\\therefore$ $AE=\\frac{1}{2}EH$,\\\\\n$\\because$ $EH\\parallel FI$,\\\\\n$\\therefore$ $\\angle EHI+\\angle GEF=180^\\circ$,\\\\\n$\\therefore$ $\\angle EHI=120^\\circ$,\\\\\n$\\therefore$ $\\angle MHI=60^\\circ$,\\\\\nFrom the folding, it is known that: $\\angle JHI=\\angle MHI=60^\\circ$,\\\\\n$\\therefore$ $\\angle EHK=60^\\circ$,\\\\\n$\\because$ $\\angle GEF=60^\\circ$,\\\\\n$\\therefore$ $\\angle EKH=60^\\circ$,\\\\\n$\\therefore$ $\\triangle EHK$ is an equilateral triangle,\\\\\n$\\therefore$ $EH=HK=EK$,\\\\\n$\\because$ $AE=\\frac{1}{2}EH$, $AE=2\\text{cm}$\\\\\n$\\therefore$ $EH=HK=EK=4\\text{cm}$,\\\\\nFrom the folding, it is known: $AE+EH+HK=AB$,\\\\\n$\\therefore$ $AB=2+4+4=\\boxed{10}\\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52828917.png"
  },
  {
    "solution": "\\boxed{21}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Unfortunately, you haven't provided any specific text to translate. Please provide the text you'd like translated into English along with the required LaTeX formatting, and I'd be more than happy to assist you.\n",
    "completion": "\\textbf{Solution:} Since $AB = 18$, and point D is the midpoint of line segment AB, \\\\\nthus $BD = 18 \\div 2 = 9$; \\\\\nSince $BD = 3BC$, \\\\\nthus $BC = 9 \\div 3 = 3$, \\\\\ntherefore $AC = AB + BC = 18 + 3 = \\boxed{21}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53042337.png"
  },
  {
    "solution": "\\boxed{\\frac{65}{23}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In $\\triangle ABC$, $\\angle C=90^\\circ$. Point P is a moving point on side $AC$, and line $PQ$ is drawn parallel to $AB$ intersecting $BC$ at point Q. Point D is the midpoint of segment $PQ$, and $AD$ bisects $\\angle BAC$. Given $\\triangle ABC \\sim \\triangle PQC$ and $AB=13$, $BC=12$, find the length of $AP$.\n",
    "completion": "\\textbf{Solution:} Since $\\angle C=90^\\circ$ and $AB=13$, $BC=12$, it follows that $AC= \\sqrt{AB^{2}-BC^{2}} = \\sqrt{13^{2}-12^{2}} =5$. Given that $PQ\\parallel AB$, it implies that $\\angle BAD=\\angle ADP$. Furthermore, since $AD$ bisects $\\angle BAC$, we have $\\angle BAD=\\angle PAD$, which leads to $\\angle ADP=\\angle PAD$, thus $PA=PD$. Given that $D$ is the midpoint of segment $PQ$, we have $PD= \\frac{1}{2}PQ$, hence $QP=2PA. Since \\triangle ABC \\sim \\triangle PQC$, it follows that $\\frac{CP}{CA}=\\frac{PQ}{AB}$, which means $\\frac{5-AP}{5}=\\frac{2AP}{13}$. Solving this gives $AP=\\boxed{\\frac{65}{23}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50496482_107.png"
  },
  {
    "solution": "\\boxed{50^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle CAB=65^\\circ$. When $\\triangle ABC$ is rotated counterclockwise about point A in the plane to the position of $\\triangle AB'C'$, such that $CC'\\parallel AB$, what is the degree measure of the rotation angle?\n",
    "completion": "\\textbf{Solution:} Since $CC^{\\prime} \\parallel AB$,\\\\\nit follows that $\\angle ACC^{\\prime} = \\angle CAB = 65^\\circ$,\\\\\nSince $\\triangle ABC$ is rotated about point A to obtain $\\triangle AB^{\\prime}C^{\\prime}$,\\\\\nit follows that AC = AC$^{\\prime}$,\\\\\nTherefore, $\\angle CAC^{\\prime} = 180^\\circ - 2 \\times 65^\\circ = 50^\\circ$,\\\\\nHence, $\\angle CAC^{\\prime} = \\angle BAB^{\\prime} = \\boxed{50^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51212133.png"
  },
  {
    "solution": "\\boxed{40^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $PA$ and $PB$ are tangents to the circle $O$, with $A$ and $B$ being the points of tangency. Point $C$ is on the minor arc $\\overset{\\frown}{AB}$. If $\\angle ACB = 110^\\circ$, what is the measure of $\\angle P$?\n",
    "completion": "\\textbf{Solution:} \\textit{Draw OA, OB,\\\\\nsince PA, PB are tangents to the circle $\\odot O$,\\\\\nit follows that $\\angle OAP = \\angle OBP = 90^\\circ$,\\\\\nsince $\\angle ACB = 110^\\circ$,\\\\\nthe central angle corresponding to the major arc AB is $220^\\circ$,\\\\\nhence, $\\angle AOB = 360^\\circ - 220^\\circ = 140^\\circ$,\\\\\ntherefore, $\\angle P = 180^\\circ - \\angle AOB = \\boxed{40^\\circ}$,}",
    "image_path": "dataset/mm_math/images/MM_Math/52950176.png"
  },
  {
    "solution": "\\boxed{3\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Let point E be chosen on the side of rectangle ABCD. Fold $ \\triangle BCE$ along BE so that point C precisely lands on point F on side AD. If $ \\angle CBE=15^\\circ$, $ BC=2BA$, as shown in the figure, when $ AB=5$ and $ AF\\cdot FD=10$, find the length of BC.\n",
    "completion": "\\textbf{Solution:} Let's fold $\\triangle BCE$ along $BE$ such that point $C$ precisely lands on point $F$ on side $AD$, \\\\\nthen $\\angle BFE=\\angle C=90^\\circ$, $CE=EF$, \\\\\nalso, since in rectangle $ABCD$, $\\angle A=\\angle D=90^\\circ$, \\\\\nthen $\\angle AFB+\\angle DFE=90^\\circ$, $\\angle DEF+\\angle DFE=90^\\circ$, \\\\\ntherefore $\\angle AFB=\\angle DEF$, \\\\\nthus $\\triangle FAB\\sim \\triangle EDF$, \\\\\nthus $\\frac{AF}{DE}=\\frac{AB}{DF}$, \\\\\nthus $AF\\cdot DF=AB\\cdot DE$, \\\\\nsince $AF\\cdot DF=10$, $AB=5$, \\\\\nthen $DE=2$, \\\\\ntherefore $CE=DC-DE=5-2=3$, \\\\\nthus $EF=3$, \\\\\ntherefore $DF=\\sqrt{EF^2-DE^2}=\\sqrt{3^2-2^2}=\\sqrt{5}$, \\\\\nthus $AF=\\frac{10}{\\sqrt{5}}=2\\sqrt{5}$, \\\\\ntherefore $BC=AD=AF+DF=2\\sqrt{5}+\\sqrt{5}=\\boxed{3\\sqrt{5}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51402971_126.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, rectangle $ABCD$ is folded along $EF$, such that point $B$ exactly coincides with point ${B}^{\\prime}$ on side $AD$. If $AE=5$ and $\\angle EFB=60^\\circ$, then what is the length of $EF$?\n",
    "completion": "\\textbf{Solution:} Since it is folded,\\\\\n$\\therefore \\angle EFB = \\angle EFB' = 60^\\circ$,\\\\\n$\\therefore \\angle CFB' = 180^\\circ - \\angle EFB - \\angle EFB' = 60^\\circ$,\\\\\nSince quadrilateral ABCD is a rectangle,\\\\\n$\\therefore AD \\parallel BC$,\\\\\n$\\therefore \\angle EB'F = \\angle CFB' = 60^\\circ$,\\\\\n$\\therefore \\triangle EFB'$ is an equilateral triangle,\\\\\n$\\therefore EB' = EF$,\\\\\nSince $\\angle A'B'E = 90^\\circ - \\angle EB'F = 90^\\circ - 60^\\circ = 30^\\circ$,\\\\\nAnd $\\angle EA'B' = 90^\\circ$,\\\\\n$\\therefore EB' = 2EA' = 10$,\\\\\n$\\therefore EF = EB' = \\boxed{10}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52205218.png"
  },
  {
    "solution": "\\boxed{triangular prism}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the shapes of a certain geometric body viewed from three different directions are as shown. What is this geometric body?\n",
    "completion": "\\textbf{Solution:} From the top view, it can be concluded that the solid is a prism. Since both the front view and the left view are triangles, the solid is identified as a \\boxed{triangular prism}.",
    "image_path": "dataset/mm_math/images/MM_Math/55218712.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{3}}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, within square $ABCD$, $\\triangle BPC$ is an equilateral triangle. The extensions of $BP$ and $CP$ intersect $AD$ at points $E$ and $F$, respectively. Connect $BD$, $DP$, and $BD$ intersects $CF$ at point $H$. Given that $\\triangle BDE \\sim \\triangle DPE$, what is the value of $\\frac{FP}{PH}$?\n",
    "completion": "\\textbf{Solution:} Given: $PC=CD$, $\\angle PCD=30^\\circ$,\\\\\n$\\therefore \\angle PDC=75^\\circ$,\\\\\n$\\therefore \\angle FDP=15^\\circ$,\\\\\n$\\because \\angle DBA=45^\\circ$,\\\\\n$\\therefore \\angle PBD=15^\\circ$,\\\\\n$\\therefore \\angle FDP=\\angle PBD$,\\\\\n$\\because \\angle DFP=\\angle BPC=60^\\circ$,\\\\\n$\\therefore \\triangle DFP\\sim \\triangle BPH$,\\\\\n$\\therefore \\frac{PF}{PH}=\\frac{DF}{PB}=\\frac{DF}{CD}=\\boxed{\\frac{\\sqrt{3}}{3}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53019686_124.png"
  },
  {
    "solution": "\\boxed{20}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the diagonals $AC$ and $BD$ of the rhombus $ABCD$ intersect at point $O$. If $AC=8$ and $BD=6$, what is the perimeter of the rhombus?\n",
    "completion": "\\textbf{Solution:} It is known from the problem that $BD\\perp AC$, $AO=4$, and $BO=3$, \\\\\n$\\therefore AB= \\sqrt{3^2+4^2}=5$, \\\\\n$\\therefore$ the perimeter of the rhombus is $4\\times 5=\\boxed{20}$",
    "image_path": "dataset/mm_math/images/MM_Math/52076093.png"
  },
  {
    "solution": "\\boxed{110^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: We define two triangles as \"opposite triangles\" if their internal angles are vertically opposite angles. For example, in Figure 1, since the internal angle $\\angle AOB$ of $\\triangle AOB$ and the internal angle $\\angle COD$ of $\\triangle COD$ are vertically opposite angles, then $\\triangle AOB$ and $\\triangle COD$ are considered \"opposite triangles\". According to the triangle angle sum theorem, opposite triangles have the following property: $\\angle A+\\angle B=\\angle C+\\angle D$. Given two triangles defined as \"opposite triangles\", such as $\\triangle AOB$ and $\\triangle COD$ in Figure 1, with $\\angle AOB=70^\\circ$, what is the degree measure of $\\angle C+\\angle D$?\n",
    "completion": "\\textbf{Solution}: We have $\\angle C + \\angle D = \\boxed{110^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53413072_99.png"
  },
  {
    "solution": "\\boxed{0}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the radius of $ \\odot P$ is 1, and the center P moves along the parabola $ y=-\\frac{1}{2}x^{2}+1$. When $ \\odot P$ is tangent to the x-axis, what is the x-coordinate of the center P?\n",
    "completion": "\\textbf{Solution:} When $y=1$, we have $1=-\\frac{1}{2}x^2+1$, giving $x=0$.\\\\\nWhen $y=-1$, we have $-1=-\\frac{1}{2}x^2+1$, giving $x= \\pm 2$.\\\\\nTherefore, the answer is: $x=\\boxed{2}$ or $x=\\boxed{-2}$ or $x=\\boxed{0}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51323915.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that in the right $\\triangle ABC$, $\\angle ACB=90^\\circ$, $AC=6$, $BC=4$. Rotating $\\triangle ABC$ clockwise by $90^\\circ$ around the right angle vertex $C$ yields $\\triangle DEC$. If the point $F$ is the midpoint of $DE$, and $AF$ is connected, what is the length of $AF$?\n",
    "completion": "\\textbf{Solution:} Let FG$\\perp$AC, \\\\\naccording to the properties of rotation, EC=BC=4, DC=AC=6, $\\angle$ACD=$\\angle$ACB=$90^\\circ$, \\\\\nsince point F is the midpoint of DE, \\\\\ntherefore FG$\\parallel$CD \\\\\ntherefore $\\triangle EFG\\sim \\triangle EDC$ \\\\\ntherefore $\\frac{GF}{CD}=\\frac{EF}{ED}=\\frac{EG}{EC}=\\frac{1}{2}$ \\\\\ntherefore GF=$\\frac{1}{2}$CD=$\\frac{1}{2}$AC=3 \\\\\nEG=$\\frac{1}{2}$EC=$\\frac{1}{2}$BC=2 \\\\\nsince AC=6, EC=BC=4 \\\\\ntherefore AE=2 \\\\\ntherefore AG=4 \\\\\nAccording to the Pythagorean theorem, $AF=\\sqrt{AG^{2}+GF^{2}}=\\sqrt{4^{2}+3^{2}}=\\boxed{5}$",
    "image_path": "dataset/mm_math/images/MM_Math/52977800.png"
  },
  {
    "solution": "\\boxed{63^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB = 90^\\circ$, $\\angle A = 36^\\circ$. The bisector of the external angle $\\angle CBD$ of $\\triangle ABC$ intersects the extension of line $AC$ at point $E$. Point $F$ is located on the extension of line $AC$, and line $DF$ is drawn. Find the degree measure of $\\angle CBE$?\n",
    "completion": "\\textbf{Solution:} Given $\\because \\angle ACB=90^\\circ$ and $\\angle A=36^\\circ$. Since $\\angle CBD$ is an exterior angle of $\\triangle ABC$,\\\\\n$\\therefore \\angle CBD=\\angle ACB+\\angle A=126^\\circ$.\\\\\nSince $BE$ bisects $\\angle CBD$,\\\\\n$\\therefore \\angle CBE= \\frac{1}{2} \\angle CBD=\\boxed{63^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52837442_98.png"
  },
  {
    "solution": "\\boxed{\\frac{16}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\square ABCD$, $AB=8$, $BC=12$. $\\square ABCD$ is rotated counterclockwise about point A to obtain $\\square AB'C'D'$. Point $B'$ is on side BC, $BB'=4$, and $B'C'$ intersects side CD at point E. Side $C'D'$ passes through point D. What is the length of $CE$?\n",
    "completion": "\\textbf{Solution:} Given the properties of a parallelogram and rotation, it is known that $C'D'=CD=AB'=AB=8$, $AD'=AD=BC'=BC=12$, $\\angle BAD=\\angle B'AD'$, and $\\angle B'CE=\\angle DC'E$. \\\\\n$\\therefore \\angle B=\\angle AB'B=\\angle ADD'=\\angle D'$, hence $\\angle BAD-\\angle B'AD=\\angle B'AD'-\\angle B'AD$. \\\\\n$\\therefore \\angle BAB'=\\angle DAD'$. \\\\\n$\\therefore \\triangle BAB' \\sim \\triangle DAD'$, \\\\\nwhich yields $\\frac{AB}{AD}=\\frac{BB'}{DD'}$, i.e., $\\frac{8}{12}=\\frac{4}{DD'}$. \\\\\nSolving for $DD'=6$. \\\\\n$\\therefore DC'=D'C'-DD'=2$. \\\\\nSince $\\angle B'CE=\\angle DC'E$ and $\\angle B'EC=\\angle DEC'$, \\\\\n$\\therefore \\triangle B'EC \\sim \\triangle DEC'$, \\\\\nthus, $\\frac{B'E}{DE}=\\frac{CE}{C'E}=\\frac{BC}{DC'}=\\frac{12-4}{2}$. \\\\\n$\\therefore B'E=4DE$, and $CE=4C'E$. \\\\\nGiven $B'E=12-C'E=12-\\frac{1}{4}CE$, and $DE=8-CE$. \\\\\n$\\therefore 12-\\frac{1}{4}CE=4\\times (8-CE)$. \\\\\nSolving yields $CE=\\boxed{\\frac{16}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52719814.png"
  },
  {
    "solution": "\\boxed{140^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in $\\triangle ABC$ and $\\triangle ADC$, $AB=AD$, $BC=DC$, $\\angle B=140^\\circ$, what is the measure of $\\angle D$ in degrees?\n",
    "completion": "\\textbf{Solution:} In $\\triangle ABC$ and $\\triangle ADC$, AB = AD, BC = DC, AC = AC,\\\\\n$\\therefore \\triangle ABC \\cong \\triangle ADC$,\\\\\n$\\therefore \\angle D = \\angle B = \\boxed{140^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52808258.png"
  },
  {
    "solution": "\\boxed{40^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the tangent to circle $O$ at point $A$, $OB$ intersects circle $O$ at point $C$, point $D$ is on circle $O$, and $AD$, $CD$, $OA$ are connected. If $\\angle ADC = 25^\\circ$, what is the degree measure of $\\angle B$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle ADC=25^\\circ$,\\\\\nthus, $\\angle AOC=50^\\circ$,\\\\\nsince $AB$ is a tangent of $\\odot O$ at point A,\\\\\nthus, $\\angle OAB=90^\\circ$,\\\\\ntherefore, $\\angle B=90^\\circ-\\angle AOC=90^\\circ-50^\\circ=\\boxed{40^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51318914.png"
  },
  {
    "solution": "\\boxed{38cm}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, arcs are drawn with points A and C as the centers and with radii longer than $AC$, intersecting at points M and N; line $MN$ intersects $BC$ and $AC$ at points D and E, respectively. If $AE=6\\,\\text{cm}$ and the perimeter of $\\triangle ABD$ is $26\\,\\text{cm}$, what is the perimeter of $\\triangle ABC$?\n",
    "completion": "\\textbf{Solution:} Given that: $MN$ is the perpendicular bisector of $AC$,\\\\\n$\\therefore$ $AD=CD$, $AC=2AE=12cm$,\\\\\n$\\because$ the perimeter of $\\triangle ABD$ is 26cm,\\\\\nwhich means $AB+AD+BD=26cm$,\\\\\n$\\therefore$ $AB+CD+BD=AB+BC=26cm$,\\\\\n$\\therefore$ the perimeter of $\\triangle ABC$ is $AB+BC+AC=26+12=\\boxed{38cm}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52826835.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, it is known that $O$ is a point on line $AB$, and three rays $OC$, $OD$, and $OE$ are drawn upward from point $O$ to line $AB$, with $OC$ bisecting $\\angle AOD$ and $\\angle 2 = 3\\angle 1$. If $\\angle COE = 70^\\circ$, what is the degree measure of $\\angle 2$?\n",
    "completion": "\\textbf{Solution:} Let $ \\angle 1=x^\\circ$, $OC$ bisects $ \\angle AOD$, $\\angle COE=\\angle 1+\\angle 3=70^\\circ$, \\\\\n$\\therefore \\angle 3=\\angle 4=70^\\circ-x^\\circ$. \\\\\nMoreover, since $\\angle 1+\\angle 2+\\angle 3+\\angle 4=180^\\circ$, \\\\\n$\\therefore x^\\circ+\\angle 2+2(70^\\circ-x^\\circ)=180^\\circ$, \\\\\n$\\therefore \\angle 2=40^\\circ+x^\\circ$. \\\\\nGiven $\\angle 2=3\\angle 1$, that is $40^\\circ+x^\\circ=3x^\\circ$, \\\\\nsolving this gives $x=20$, \\\\\n$\\therefore \\angle 2=3\\angle 1=3\\times 20^\\circ=\\boxed{60^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53206772_93.png"
  },
  {
    "solution": "\\boxed{\\frac{13}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given that the vertex of the parabola is $P(1, 4)$, it intersects the y-axis at point $C(0, 3)$, and it intersects the x-axis at points $A$ and $B$. If the parabola has only one intersection point with the line $y = x + m$, what is the value of $m$?\n",
    "completion": "\\textbf{Solution:} Since the parabola and the line $y=x+m$ intersect at only one point,\\\\\nit follows that $-{x}^{2}+2x+3=x+m$, which simplifies to ${x}^{2}-x+(m-3)=0$,\\\\\nhence, $\\Delta=1-4(m-3)=0$,\\\\\nsolving for $m$, we get $m=\\boxed{\\frac{13}{4}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52049799_106.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, point D is located on side AB, and $\\angle ADC=\\angle ACB$, with $AD=2$ and $BD=6$. Given that $\\triangle ACD \\sim \\triangle ABC$, find the length of side AC.\n",
    "completion": "\\textbf{Solution:} Given that $AD=2, BD=6$,\n\n$\\therefore AB=AD+BD=8$,\n\nFrom (1) it has been proven that $\\triangle ACD\\sim \\triangle ABC$,\n\n$\\therefore \\frac{AC}{AB}=\\frac{AD}{AC}$, that is $\\frac{AC}{8}=\\frac{2}{AC}$,\n\nSolving this yields $AC=\\boxed{4}$ or $AC=-4<0$ (which is discarded as it does not meet the problem requirements),\n\nUpon verification, $AC=4$ is a solution of the listed fractional equation,\n\nTherefore, the length of side $AC$ is \\boxed{4}.",
    "image_path": "dataset/mm_math/images/MM_Math/53080096_99.png"
  },
  {
    "solution": "\\boxed{3\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the radius of sector $OAB$ is $6$, $C$ is any point on arc $AB$ (not coinciding with $A$ or $B$), $CM\\perp OA$ with the foot of the perpendicular being $M$, $CN\\perp OB$ with the foot of the perpendicular being $N$, connecting $MN$. If $\\angle AOB=45^\\circ$, what is the length of $MN$?\n",
    "completion": "\\textbf{Solution:} Connect OC, extend OA and NC to intersect at D, then OC = 6, \\\\\n$\\because$ CM $\\perp$ OA, CN $\\perp$ OB, \\\\\n$\\therefore$ $\\angle$DMC = $\\angle$DNO = $90^\\circ$, \\\\\n$\\because$ $\\angle$D = $\\angle$D, \\\\\n$\\therefore$ $\\triangle$DMC $\\sim$ $\\triangle$DNO, \\\\\n$\\therefore$ $\\frac{DM}{DN}=\\frac{DC}{DO}$, which implies $\\frac{DM}{DC}=\\frac{DN}{DO}$, \\\\\n$\\because$ $\\angle$D = $\\angle$D, \\\\\n$\\therefore$ $\\triangle$DMN $\\sim$ $\\triangle$DCO, \\\\\n$\\therefore$ $\\frac{MN}{CO}=\\frac{DN}{DO}$, \\\\\n$\\because$ CN $\\perp$ OB, $\\angle$AOB = $45^\\circ$, \\\\\n$\\therefore$ sin$\\angle$AOB = $\\frac{DN}{OD}=\\frac{\\sqrt{2}}{2}$, \\\\\n$\\therefore$ $\\frac{MN}{OC}=\\frac{\\sqrt{2}}{2}$, \\\\\n$\\because$ OC = 6, \\\\\n$\\therefore$ $\\frac{MN}{6}=\\frac{\\sqrt{2}}{2}$, \\\\\n$\\therefore$ MN = $\\boxed{3\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51265085.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=x+2$ and the hyperbolic function $y=\\frac{k}{x}$ intersect at point P in the first quadrant. If $OP=\\sqrt{20}$, what is the value of $k$?\n",
    "completion": "\\textbf{Solution:} Let $P(x, x+2)$,\\\\\nsince $OP=\\sqrt{20}$,\\\\\nit follows that $x^{2}+(x+2)^{2}=(\\sqrt{20})^{2}$,\\\\\nsimplifying, we get:\\\\\n$x^{2}+2x-8=0$,\\\\\nthus, $(x+4)(x-2)=0$,\\\\\nhence, $x_{1}=-4, x_{2}=2$,\\\\\nsince $P$ is in the first quadrant, then\\\\\n$P(2, 4)$,\\\\\ntherefore, $k=xy=2\\times 4=\\boxed{8}$",
    "image_path": "dataset/mm_math/images/MM_Math/51433085.png"
  },
  {
    "solution": "\\boxed{(0, -4)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the Cartesian coordinate system, line $l_1$ intersects the x-axis at point A and the y-axis at point B, with the coordinates of point B being $(0, 3)$. Line $l_2$: $y=2x$ intersects line $l_1$ at point C, with the x-coordinate of point C being 1. If point D is a point on the y-axis, and the area of $\\triangle OCD$ is $\\frac{2}{3}$ of the area of $\\triangle AOC$, find the coordinates of point D.\n",
    "completion": "\\textbf{Solution:} Given that point A is the intersection of line $l_{1}$ and the $x$-axis,\\\\\ntherefore, substituting $y=0$ into $y=-x+3$, we find $x=3$,\\\\\ntherefore, the coordinates of point A are $(3, 0)$.\\\\\nAs shown in the figure, draw $CM\\perp x$-axis through point C, with M being the foot of the perpendicular; draw $CN\\perp y$-axis through point C, with N being the foot of the perpendicular.\\\\\nGiven the coordinates of point C as $(1, 2)$, we obtain $CM=2$, $CN=1$,\\\\\nLet the coordinates of point D be $(0, y)$.\\\\\nAccording to the problem, we have $\\frac{1}{2}\\cdot OD\\cdot CN=\\frac{2}{3}\\times \\frac{1}{2}\\cdot OA\\cdot CM$,\\\\\nwhich is $\\frac{1}{2}\\times |y|\\times 1=\\frac{2}{3}\\times \\frac{1}{2}\\times 3\\times 2$.\\\\\nSolving this, we get $|y|=4$, that is $y=\\pm 4$.\\\\\nTherefore, the coordinates of point D are $\\boxed{(0, 4)}$ or $\\boxed{(0, -4)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51662414_211.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB = 90^\\circ$. Point D is on the extension line of AB, and CD is connected. If $AB = 2BD$ and $\\tan\\angle BCD = \\frac{2}{3}$, what is the value of $\\frac{AC}{BC}$?\n",
    "completion": "\\textbf{Solution:} Draw $DM\\perp BC$ from point D, intersecting the extension of BC at point M,\\\\\n$\\because$ $\\angle ACB=\\angle DMB=90^\\circ$, $\\angle ABC=\\angle DBM$,\\\\\n$\\therefore$ $\\triangle ABC\\sim \\triangle DBM$,\\\\\n$\\therefore$ $\\frac{BD}{AB}=\\frac{BM}{BC}=\\frac{DM}{AC}$,\\\\\n$\\because$ AB＝2BD,\\\\\n$\\therefore$ $\\frac{BD}{AB}=\\frac{BM}{BC}=\\frac{DM}{AC}=\\frac{1}{2}$,\\\\\nIn $\\triangle CDM$,\\\\\nsince $\\tan\\angle MCD=\\frac{2}{3}=\\frac{DM}{CM}$, let DM＝2k, then CM＝3k,\\\\\nalso $\\because$ $\\frac{BM}{BC}=\\frac{1}{2}=\\frac{DM}{AC}$,\\\\\n$\\therefore$ BC＝2k, AC＝4k,\\\\\n$\\therefore$ $\\frac{AC}{BC}=\\frac{4k}{2k}=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51103634.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, fold the rectangular paper ABCD in the manner shown, so that vertex C falls at the position of $C^{\\prime}$. If $AB = 4$ and $DE = 8$, what is the value of $\\sin\\angle C^{\\prime}ED$?\n",
    "completion": "\\textbf{Solution:} Given that ABCD is a rectangle,\\\\\nthus CD=AB=4, and $\\angle$C$=90^\\circ$,\\\\\nby the properties of paper folding, we have C$^{\\prime }$D=CD=4, $\\angle$C$^{\\prime }$=$\\angle$C$=90^\\circ$,\\\\\ntherefore, $\\sin\\angle {C}^{\\prime }ED=\\frac{{C}^{\\prime }D}{ED}=\\frac{4}{8}=\\boxed{\\frac{1}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51322890.png"
  },
  {
    "solution": "\\boxed{1:4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in trapezoid $ABCD$, where $AD \\parallel BC$ and $BD^2 = AD \\cdot BC$, if $AB : CD = 1 : 2$, what is the ratio of $AD : BC$?\n",
    "completion": "\\textbf{Solution:} Since $AD\\parallel BC$,\n\nit follows that $\\angle ADB=\\angle DBC$.\n\nSince $BD^{2}=AD\\cdot BC$,\n\nit follows that $\\frac{BD}{BC}=\\frac{AD}{BD}$,\n\ntherefore $\\triangle ADB\\sim \\triangle DBC$,\n\nhence $\\frac{BD}{BC}=\\frac{AD}{BD}=\\frac{AB}{CD}=\\frac{1}{2}$,\n\nthus $BD=2AD$, $BC=2BD$,\n\nhence $BC=4AD$,\n\nthus $AD:BC=\\boxed{1:4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52917562.png"
  },
  {
    "solution": "\\boxed{96^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC$, and the points $D$, $E$, $F$ are on the sides $BC$, $AB$, $AC$, respectively, with $CD=BE$ and $BD=CF$. If $\\angle EDF = 42^\\circ$, what is the measure of $\\angle BAC$?\n",
    "completion": "\\textbf{Solution:} Since $AB=AC$, \\\\\nit follows that $\\angle B=\\angle C$, \\\\\nIn $\\triangle BDE$ and $\\triangle CFD$, \\\\\n$\\begin{cases}\nBD=CF\\\\\n\\angle B=\\angle C\\\\\nBE=CD\n\\end{cases}$, \\\\\nhence, $\\triangle BDE \\cong \\triangle CFD$, \\\\\ntherefore, $\\angle EDB=\\angle DFC$, $\\angle FDC=\\angle BED$, \\\\\nsince $\\angle EDF+\\angle BDE+\\angle FDC=180^\\circ$, $\\angle B+\\angle BED+\\angle EDB=180^\\circ$, \\\\\nit follows that $\\angle B=\\angle EDF=42^\\circ$, \\\\\nthus, $\\angle BAC=180^\\circ-\\angle B-\\angle C=\\boxed{96^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53010086.png"
  },
  {
    "solution": "\\boxed{10^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AD$ is the altitude on side $BC$, and $AE$ bisects $\\angle BAC$. If $\\angle BAC : \\angle B : \\angle C = 4 : 3 : 2$, find the measure of $\\angle DAE$?\n",
    "completion": "\\textbf{Solution:} In $\\triangle ABC$,\n\nsince $\\angle BAC : \\angle B : \\angle C = 4 : 3 : 2$,\n\nthus $\\angle BAC = 180^\\circ \\times \\frac{4}{9} = 80^\\circ$, $\\angle B = 180^\\circ \\times \\frac{3}{9} = 60^\\circ$,\n\nsince $AD$ is the altitude to side $BC$,\n\nthus $\\angle ADB = 90^\\circ$,\n\nthus $\\angle BAD = 90^\\circ - \\angle B = 30^\\circ$,\n\nsince $AE$ bisects $\\angle BAC$,\n\nthus $\\angle BAE = \\frac{1}{2} \\angle BAC = 40^\\circ$,\n\nthus $\\angle DAE = \\angle BAE - \\angle BAD = \\boxed{10^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52693957.png"
  },
  {
    "solution": "\\boxed{[40s, 65s]}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given the figure, $\\angle AOD = 130^\\circ$, $\\angle BOC : \\angle COD = 1 : 2$, $\\angle AOB$ is $\\frac{1}{3}$ of the complement of $\\angle COD$. Fix $\\angle COD$ and rotate rays OA and OB clockwise at a speed of $2^\\circ/s$ until OA overlaps with OD. During the rotation, if ray OP bisects $\\angle AOB$ and OQ bisects $\\angle COD$, when $\\angle POQ + \\angle AOD = 50^\\circ$, find the range of the rotation time $t$ (in seconds).\n",
    "completion": "\\textbf{Solution:} From (1) we have: $\\angle AOB=40^\\circ$, $\\angle COD=60^\\circ$\\\\\nSince ray OP is the bisector of $\\angle AOB$, and OQ is the bisector of $\\angle COD$,\\\\\ntherefore $\\angle AOP=\\frac{1}{2}\\angle AOB=20^\\circ$, $\\angle DOQ=\\frac{1}{2}\\angle COD=30^\\circ$,\\\\\nIt can be observed that: $\\angle AOP+\\angle DOQ=50^\\circ$,\\\\\nTo find the range of time $t$, we need to find the critical condition,\\\\\nWhen $OP$ coincides with $OQ$, at this moment $\\angle AOD=\\angle AOP+\\angle DOQ=50^\\circ$, $\\angle POQ=0^\\circ$,\\\\\nIt is observed that, if $OP$ and $OQ$ do not coincide, there must be $\\angle AOD>\\angle AOP+\\angle DOQ=50^\\circ$ which does not satisfy $\\angle POQ+\\angle AOD=50^\\circ$, hence the time $t$ exactly reaches its minimum value at this moment,\\\\\nAccording to the problem statement: $OA$ rotates a total of $130^\\circ-\\angle AOD=80^\\circ$, thus time $t=80\\div 2=40s$,\\\\\n$\\therefore t\\ge 40s$,\\\\\nWhen $OA$ coincides with $OD$, at this moment $\\angle AOD=0^\\circ$, $\\angle POQ=\\angle AOP+\\angle DOQ=50^\\circ$,\\\\\nIf $OA$ continues to rotate downwards at this moment, there must be $\\angle POQ>\\angle AOP+\\angle DOQ=50^\\circ$, which does not satisfy $\\angle POQ+\\angle AOD=50^\\circ$, hence the time $t$ exactly reaches its maximum value at this moment,\\\\\nAccording to the problem statement: $OA$ rotates a total of $130^\\circ-\\angle AOD=130^\\circ$, thus time $t=130\\div 2=65s$,\\\\\n$\\therefore t\\le 65s$,\\\\\nIn conclusion: $40s\\le t\\le 65s$, hence the range of rotation time $t$ is $t\\in\\boxed{[40s, 65s]}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51324353_156.png"
  },
  {
    "solution": "\\boxed{16}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, the perpendicular bisector DM of AC intersects AC at D, and the perpendicular bisector EN of BC intersects BC at E, where DM and EN intersect at point F. If the perimeter of $\\triangle CMN$ is 16 cm, what is the length of AB in cm?\n",
    "completion": "Solution: Since $DM$ is the perpendicular bisector of side $AC$,\\\\\nit follows that $MA=MC$,\\\\\nand since $EN$ is the perpendicular bisector of side $BC$,\\\\\nit follows that $NB=NC$,\\\\\nThus, $AB=AM+MN+NB=MC+MN+NC=$ the perimeter of $\\triangle CMN = \\boxed{16}$cm;",
    "image_path": "dataset/mm_math/images/MM_Math/51367169_129.png"
  },
  {
    "solution": "\\boxed{\\frac{16}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the coordinate system, the line $l_{1}\\colon y=-x+2$ intersects the $x$-axis at point $P$, and the line $l_{2}\\colon y=ax-4$ passes through point $P$. Points $M$ and $N$ lie on lines $l_{1}$ and $l_{2}$, respectively, and are symmetric about the origin. What is the area of $\\triangle PMN$?\n",
    "completion": "\\textbf{Solution:} Given $a=2$, we have $l_2\\colon y=2x-4$. Let the x-coordinate of point $M$ be $x$, hence $M\\left(x, -x+2\\right)$, $N\\left(-x, x-2\\right)$. Furthermore, since $N\\left(-x, x-2\\right)$ lies on the line $l_2\\colon y=2x-4$, $\\therefore$ $x-2=-2x-4$ yields $x= -\\frac{2}{3}$. Thus, $M\\left(-\\frac{2}{3}, \\frac{8}{3}\\right)$, $N\\left(\\frac{2}{3}, -\\frac{8}{3}\\right)$. $\\therefore S_{\\triangle PMN}=\\frac{1}{2} \\cdot OP\\cdot |My|+ \\frac{1}{2} \\cdot OP\\cdot |Ny| = \\frac{1}{2} \\times 2\\times \\frac{8}{3} + \\frac{1}{2} \\times 2\\times \\frac{8}{3} = \\boxed{\\frac{16}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50431231_187.png"
  },
  {
    "solution": "\\boxed{14}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, the perpendicular bisector $MN$ of $AB$ intersects $AC$ at point $D$, and $BD$ is drawn. If $AC=9$ and $BC=5$, what is the perimeter of $\\triangle BDC$?\n",
    "completion": "\\textbf{Solution:} Since the perpendicular bisector $MN$ of $AB$ intersects $AC$ at point D,\\\\\nit follows that $AD=BD$,\\\\\nSince $AC=9$ and $BC=5$,\\\\\nthe perimeter of $\\triangle BDC$ is $BD+CD+BC=AD+CD+BC=AC+BC=9+5=\\boxed{14}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53066466.png"
  },
  {
    "solution": "\\boxed{\\frac{15}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the line $y_1=k_1x+b$ intersects with the hyperbola $y_2=\\frac{k_2}{x}$ at points $A(-2, 3)$ and $B(m, -2)$. A line through point B, $BP\\parallel x$-axis, intersects the y-axis at point P. What is the area of $\\triangle ABP$?\n",
    "completion": "\\textbf{Solution:} From (1) we have: $A(-2, 3)$, $B(3, -2)$,\\\\\nsince $BP\\parallel x$-axis,\\\\\ntherefore, $BP=3$,\\\\\nthus, the distance from point A to PB is $3-(-2)=5$,\\\\\nhence, the area of $\\triangle ABP$ is $\\frac{1}{2}\\times 3\\times 5=\\boxed{\\frac{15}{2}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53043125_137.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, the graph of the inverse proportion function $y = \\frac{k}{x}$ passes through points $A(1, m)$ and $B(2, 1)$. The coordinates of point $P$ are $(6, 1)$. Connect $PA$ and $PB$. What is the value of $\\tan \\angle P$?\n",
    "completion": "\\textbf{Solution:} Since the graph of the inverse proportion function $y = \\frac{2}{x}$ passes through point $A(1, 2)$,\\\\\nit follows that $m = 2$,\\\\\nDraw a perpendicular from point $A$ to line $BP$, extending to point $M$ on the extension of $PB$,\\\\\nthen $AM = 2 - 1 = 1$, and $MP = 6 - 1 = 5$,\\\\\nIn $\\triangle PAM$, $\\tan \\angle P = \\frac{AM}{MP} = \\frac{1}{5}$;\\\\\ntherefore, $\\tan \\angle P = \\boxed{\\frac{1}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51445690_117.png"
  },
  {
    "solution": "\\boxed{16}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the parallelogram $ABCD$, $AB=10$, $AD=15$, and the bisector of $\\angle BAD$ intersects $BC$ at point $E$ and the extension of $DC$ at point $F$. $BG\\perp AE$ at point $G$. If $BG=8$, then what is the perimeter of $\\triangle CEF$?\n",
    "completion": "\\textbf{Solution:} Let us consider quadrilateral $ABCD$, where $CD=AB=10$, $BC=AD=15$. The bisector of $\\angle BAD$ meets $BC$ at point $E$,\\\\\ntherefore, $AB\\parallel DC$, $\\angle BAF=\\angle DAF$,\\\\\nthus, $\\angle BAF=\\angle F$,\\\\\nand hence, $\\angle DAF=\\angle F$,\\\\\nleading to $DF=AD=15$, similarly $BE=AB=10$,\\\\\nhence, $CF=DF-CD=15-10=5$;\\\\\nsince $BG\\perp AE$,\\\\\nit follows that $AE=2AG$,\\\\\nwithin $\\triangle ABG$, where $BG\\perp AE$, and $AB=10$, $BG=8$,\\\\\nin right triangle $\\triangle ABG$, $AG=\\sqrt{AB^{2}-BG^{2}}=\\sqrt{100-64}=6$,\\\\\nthus, $AE=2AG=12$,\\\\\nhence, the perimeter of $\\triangle ABE$ equals $10+10+12=32$,\\\\\nsince quadrilateral ABCD is a parallelogram,\\\\\nit follows that $AB\\parallel CF$,\\\\\nhence, $\\triangle CEF\\sim \\triangle BEA$,\\\\\nthus, $\\frac{CF}{AB}=\\frac{5}{10}=\\frac{1}{2}$, i.e., the ratio of similarity is $1\\colon 2$,\\\\\nhence, the perimeter of $\\triangle CEF$ is $\\boxed{16}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52917519.png"
  },
  {
    "solution": "\\boxed{\\frac{2\\sqrt{10}}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, there is a triangular piece $ABC$, folded along the median $AD$ on the side $BC$, such that point $C$ falls on point $E$, and $AE$ intersects $BC$ at point $F$. If $AD$ is perpendicular to $AB$, and $AB=2$, $AD=3$, what is the length of $AF$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $CG \\perp AD$ at point $G$, and draw $DM \\perp AD$, $DN \\perp AD$ with the perpendicular foot at $D$. Since $AD$ is the median of $\\triangle ABC$, it is easy to prove $\\triangle ABD \\cong \\triangle GCD$. Therefore, $CG=AB=2$, $GD=AD=3$, hence $AC=\\sqrt{AG^{2}+CG^{2}}=\\sqrt{6^{2}+2^{2}}=2\\sqrt{10}$. Since $CG \\perp AD$ and $DM \\perp AD$, it follows that $DM \\parallel CG$, therefore $\\triangle AMD \\sim \\triangle ACG$. Since $GD=AD$, it follows that $AM=MC=\\frac{1}{2}AC=\\sqrt{10}$, $DM=\\frac{1}{2}CG=1$. Folding makes it easy to prove $\\triangle AMD \\cong \\triangle AND$, therefore $AN=AM=\\sqrt{10}$, $DN=DM=1$. Since $AD \\perp AB$, $DN \\perp AD$, it follows that $AB \\parallel DN$, therefore $\\frac{AF}{FN}=\\frac{AB}{DN}=\\frac{2}{1}$, hence $AF=\\frac{2}{3}AN=\\frac{2\\sqrt{10}}{3}$. Therefore, the answer is: $\\boxed{\\frac{2\\sqrt{10}}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55452505.png"
  },
  {
    "solution": "\\boxed{5\\colon 3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\triangle ABC$ and $\\triangle ADE$, $\\angle ACB=\\angle AED=90^\\circ$, $\\angle ABC=\\angle ADE$. Connect BD and CE. If $AC:BC=3:4$, then what is the ratio of $BD:CE$?\n",
    "completion": "\\textbf{Solution:} Given that $\\because$ $\\angle ACB=\\angle AED=90^\\circ$ and $\\angle ABC=\\angle ADE$,\\\\\n$\\therefore$ $\\triangle ABC\\sim \\triangle ADE$,\\\\\n$\\therefore$ $\\angle BAC=\\angle DAE$ and $\\frac{AC}{AB}=\\frac{AE}{AD}$,\\\\\n$\\because$ $\\angle BAC+\\angle BAE=\\angle DAE+\\angle BAE$,\\\\\n$\\therefore$ $\\angle CAE=\\angle BAD$,\\\\\n$\\because$ $\\frac{AC}{AB}=\\frac{AE}{AD}$,\\\\\n$\\therefore$ $\\triangle ACE\\sim \\triangle ABD$,\\\\\n$\\therefore$ $\\frac{BD}{CE}=\\frac{AB}{AC}$,\\\\\n$\\because$ $AC\\colon BC=3\\colon 4$ and $\\angle ACB=\\angle AED=90^\\circ$,\\\\\nlet $AC=3x$, $BC=4x$,\\\\\nIn $Rt\\triangle ABC$, $AB^2=AC^2+BC^2$,\\\\\n$\\therefore$ $AB=\\sqrt{AC^2+BC^2}=\\sqrt{(3x)^2+(4x)^2}=5x$,\\\\\n$\\therefore$ $AC\\colon BC\\colon AB=3\\colon 4\\colon 5$,\\\\\n$\\therefore$ $BD\\colon CE=\\boxed{5\\colon 3}$,",
    "image_path": "dataset/mm_math/images/MM_Math/53032950.png"
  },
  {
    "solution": "\\boxed{P_2(-\\sqrt{21}, 0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=x+7$ intersects with the line $y=-2x-2$ at point C, and they intersect the y-axis at points A and B, respectively. Point P is on the x-axis, such that $\\angle PBO=2\\angle PAO$. Directly write down the coordinates of point P.\n",
    "completion": "\\textbf{Solution:} Take point $B'$(0, 2) on the positive semi-axis of $y$, then $OB=OB'=2$, connect $PB'$,\\\\\n$\\therefore PB=PB'$, hence $\\angle PBO=\\angle PB'O=2\\angle PAO$,\\\\\n$\\therefore \\angle B'AP=\\angle B'PA$,\\\\\ntherefore $PB'=B'A=7-2=5$,\\\\\nAssume point $P(x, 0)$,\\\\\nthen $PB'^2=2^2+x^2=5^2$, solving for $x$ yields $x=\\pm \\sqrt{21}$,\\\\\nTherefore, the coordinates of point P are $\\boxed{P_1(\\sqrt{21}, 0)}$ or $\\boxed{P_2(-\\sqrt{21}, 0)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50561111_63.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{10}}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $E$ is a point on the side $CB$ of rectangle $ABCD$, and $AF \\perp DE$ at point $F$. $\\triangle EDC \\sim \\triangle DAF$. If $AB = 3$, $AD = 2$, and $CE = 1$, find the length of segment $DF$.\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rectangle,\\\\\nit follows that DC = AB = 3, and $\\angle ADC = \\angle C = 90^\\circ$.\\\\\nSince CE = 1,\\\\\nthen DE = $\\sqrt{DC^2 + CE^2} = \\sqrt{10}$.\\\\\nSince $\\triangle EDC \\sim \\triangle DAF$,\\\\\nit follows that $\\frac{DE}{AD} = \\frac{CE}{FD}$, that is $\\frac{\\sqrt{10}}{3} = \\frac{1}{FD}$,\\\\\ntherefore, FD = $\\boxed{\\frac{\\sqrt{10}}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52917314_67.png"
  },
  {
    "solution": "\\boxed{20^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, take a point $O$ on the line $AB$ and draw a ray $OC$ such that $\\angle BOC = 70^\\circ$. Place the right angle vertex of a right-angle triangle at point $O$. (Note: $\\angle DOE = 90^\\circ$) As shown in Figure 1, if one side $OD$ of the right-angle triangle $DOE$ falls on the ray $OB$, then what is the measure of $\\angle COE$?\n",
    "completion": "\\textbf{Solution:} Therefore, $\\angle COE = \\boxed{20^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55426249_4.png"
  },
  {
    "solution": "\\boxed{\\frac{17}{12}\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, it is known that the parabola $y= -\\frac{\\sqrt{3}}{3}x^{2}+\\frac{2\\sqrt{3}}{3}x+\\sqrt{3}$ intersects the $x$-axis at points $A$ and $B$ (with point $A$ to the left of point $B$), and intersects the $y$-axis at point $C$. Point $D$ is the symmetrical point of point $C$ with respect to the axis of symmetry of the parabola. Connect $CD$, and draw $DH \\perp x$-axis at point $H$ through point $D$. Draw $AE \\perp AC$ through point $A$, and let it intersect the extension of $DH$ at point $E$. As shown in Figure 2, try to find a point $F$ on segment $AE$ and a point $P$ on the parabola above the straight line $PF$. Calculate the maximum value of the area of $\\triangle MPF$ when the perimeter of $\\triangle CPF$ is at its minimum.\n",
    "completion": "\\textbf{Solution:} As shown in Fig. 2, find the symmetric point N (4, $\\sqrt{3}$) of point C with respect to DE, and the symmetric point G ($-$2, $-\\sqrt{3}$) of point C with respect to AE. Connect GN, intersect AE at point F and DE at point P. Connect CP and CF. When points G, F, P, and N are collinear, the perimeter of $\\triangle CPF = CF + PF + CP = GF + PF + PN$ is minimized,\\\\\nThe equation of line GN: $y = \\frac{\\sqrt{3}}{3}x - \\frac{\\sqrt{3}}{3}$, and the equation of line AE: $y = -\\frac{\\sqrt{3}}{3}x - \\frac{\\sqrt{3}}{3}$,\\\\\nBy solving these equations simultaneously, we get: F (0, $-\\frac{\\sqrt{3}}{3}$), P (2, $\\frac{\\sqrt{3}}{3}$). A line parallel to the y-axis through point M intersects FP at point Q. Let point M (m, $-\\frac{\\sqrt{3}}{3}m^{2}+\\frac{2\\sqrt{3}}{3}m+\\sqrt{3}$), thus Q (m, $\\frac{\\sqrt{3}}{3}m-\\frac{\\sqrt{3}}{3}$), (0 < m < 2),\\\\\n$\\therefore \\text{Area of }\\triangle MFP = \\text{Area of }\\triangle MQF + \\text{Area of }\\triangle MQP = \\frac{1}{2} \\times 2 \\cdot MQ = MQ = -\\frac{\\sqrt{3}}{3}m^{2}+\\frac{\\sqrt{3}}{3}m+\\frac{4\\sqrt{3}}{3}$,\\\\\n$\\therefore \\text{Area of }\\triangle MFP = -\\frac{\\sqrt{3}}{3}(m-\\frac{1}{2})^{2}+\\frac{17}{12}\\sqrt{3}$,\\\\\n$\\because$ The axis of symmetry is: $m=\\frac{1}{2}<2$, and the parabola opens downwards,\\\\\n$\\therefore$ When $m=\\frac{1}{2}$, $\\text{Area of }\\triangle MFP$ is maximized, with the maximum value being $\\boxed{\\frac{17}{12}\\sqrt{3}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51708944_113.png"
  },
  {
    "solution": "\\boxed{7}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in triangle $ABC$, $AD \\perp BC$, $CE \\perp AB$, with the perpendicular feet being D and E respectively. $AD$ and $CE$ intersect at point H. Given that $EH = EB = 5$ and $CH = 2$, then what is the length of $AE$?\n",
    "completion": "\\textbf{Solution:} Given: $EH=EB=5$, $CH=2$,\\\\\nTherefore $EC=EH+CH=7$,\\\\\nSince $AD\\perp BC$, $CE\\perp AB$,\\\\\nThus $\\angle ADB=\\angle AEH=90^\\circ$,\\\\\nSince $\\angle AHE=\\angle CHD$,\\\\\nThus $\\angle BAD=\\angle BCE$,\\\\\nIn $\\triangle HEA$ and $\\triangle BEC$, $\\left\\{\\begin{array}{l}\\angle EAH=\\angle BCE \\\\ \\angle AEH=\\angle BEC=90^\\circ \\\\ EH=EB\\end{array}\\right.$,\\\\\nTherefore $\\triangle HEA\\cong \\triangle BEC\\left(AAS\\right)$,\\\\\nHence $AE=EC=\\boxed{7}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52826808.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ is an equilateral triangle, with $D$ and $E$ located on $AB$ and $BC$ respectively, and $AD=BE$, $AE$ and $CD$ intersect at point $F$. What is the measure of $\\angle CFE$ in degrees?\n",
    "completion": "\\textbf{Solution:} As shown in Figure 1,\\\\\nFigure 1\\\\\n$\\because$ $\\triangle ABC$ is an equilateral triangle\\\\\n$\\therefore$ $AB=BC=CA$, and $\\angle ABC=\\angle BCA=\\angle CAB=60^\\circ$\\\\\nIn $\\triangle ABE$ and $\\triangle CAD$,\\\\\n$\\because$ $\\left\\{\\begin{array}{l}\nAB=CA\\\\\n\\angle ABE=\\angle CAD\\\\\nBE=AD\n\\end{array}\\right.$\\\\\n$\\therefore$ $\\triangle ABE \\cong \\triangle CAD$ (SAS)\\\\\n$\\therefore$ $\\angle ACD=\\angle BAE$\\\\\nIn $\\triangle AFC$, $\\angle CFE=\\angle ACD+\\angle CAF=\\angle BAE+\\angle CAF=\\angle BAC=60^\\circ$\\\\\n$\\therefore$ $\\angle CFE=\\boxed{60^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/52891027_104.png"
  },
  {
    "solution": "\\boxed{2\\,\\text{cm}^2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, it is known that points D, E, and F are the midpoints of sides BC, AD, and CE, respectively, and the area of $\\triangle ABC$ is $8\\,\\text{cm}^2$. What is the area of the shaded region?\n",
    "completion": "\\textbf{Solution:} Since $AD$ is the median of $\\triangle ABC$,\\\\\nthen $S_{\\triangle ABD}=S_{\\triangle ACD}=\\frac{1}{2}S_{\\triangle ABC}=4\\,\\text{cm}^2$,\\\\\nSince point $E$ is the midpoint of $AD$,\\\\\nthen $S_{\\triangle BDE}=\\frac{1}{2}S_{\\triangle ABD}=2\\,\\text{cm}^2$, $S_{\\triangle CDE}=\\frac{1}{2}S_{\\triangle CAD}=2\\,\\text{cm}^2$,\\\\\nthus $S_{\\triangle BCE}=S_{\\triangle BDE}+S_{\\triangle CDE}=2+2=4\\,\\text{cm}^2$,\\\\\nSince point $F$ is the midpoint of $CE$,\\\\\nthen $S_{\\triangle BFE}=\\frac{1}{2}S_{\\triangle BCE}=2\\,\\text{cm}^2$.\\\\\nTherefore, the area of the shaded part is $\\boxed{2\\,\\text{cm}^2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52808239.png"
  },
  {
    "solution": "\\boxed{80^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points $A,B,C$ are on circle $\\odot O$. If $\\angle BAC=40^\\circ$, what is the measure of $\\angle BOC$?\n",
    "completion": "\\textbf{Solution:} Given that $ \\because \\angle BOC=2\\angle BAC$ and $\\angle BAC=40^\\circ$,\\\\\n$\\therefore \\angle BOC=\\boxed{80^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52872096.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the diagonals $AC$ and $BD$ of rectangle $ABCD$ intersect at point $O$. $AE$ bisects $\\angle BAD$ and intersects $BC$ at point $E$. Connect $OE$. If $OE \\perp BC$ and $OE = 1$, what is the length of $AC$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral $ABCD$ is a rectangle, \\\\\n$\\therefore \\angle ABC=\\angle BAD=90^\\circ$, $OA=OC$, $OB=OD$, $AC=BD$, \\\\\n$\\therefore OB=OC$, \\\\\nSince $OE\\perp BC$, \\\\\n$\\therefore BE=CE$, \\\\\n$\\therefore OE$ is the median of $\\triangle ABC$, \\\\\n$\\therefore AB=2OE=2$, \\\\\nSince $AE$ bisects $\\angle BAD$, \\\\\n$\\therefore \\angle BAE=45^\\circ$, \\\\\n$\\therefore \\triangle ABE$ is an isosceles right triangle, \\\\\n$\\therefore BE=AB=2$, \\\\\n$\\therefore BC=2BE=4$, \\\\\n$\\therefore AC=\\sqrt{AB^{2}+BC^{2}}=\\sqrt{2^{2}+4^{2}}=\\boxed{2\\sqrt{5}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53729080.png"
  },
  {
    "solution": "\\boxed{50^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, it is known that $\\angle AOB=100^\\circ$. The rays $OE$ and $OF$ are the angle bisectors of $\\angle AOC$ and $\\angle COB$, respectively. If the ray $OC$ is inside the angle $\\angle AOB$ and $\\angle AOC=30^\\circ$, what is the degree measure of $\\angle EOF$?\n",
    "completion": "\\textbf{Solution:} By construction,\\\\\nsince $OE$ is the bisector of $\\angle AOC$, and $\\angle AOC=30^\\circ$,\\\\\nthen $\\angle COE=\\frac{1}{2}\\angle AOC=15^\\circ$,\\\\\nsince $\\angle AOB=100^\\circ$,\\\\\nthen $\\angle COB=\\angle AOB-\\angle AOC=70^\\circ$,\\\\\nsince $OF$ is the bisector of $\\angle COB$,\\\\\nthen $\\angle COF=\\frac{1}{2}\\angle COB=35^\\circ$,\\\\\nthus $\\angle EOF=\\angle COE+\\angle COF=15^\\circ+35^\\circ=\\boxed{50^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50518431_212.png"
  },
  {
    "solution": "\\boxed{\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=-\\frac{1}{2}x+2$ intersects the $x$-axis and the $y$-axis at points $B$ and $A$, respectively. Let $Q$ be a variable point (not coinciding with $A$ or $B$) on segment $AB$. When point $Q$ is rotated clockwise by $90^\\circ$ about point $P(1, 0)$, point $Q'$ is obtained. What is the minimum value of $OQ'$ when connecting $OQ'$?\n",
    "completion": "\\textbf{Solution}: Construct $QM \\perp x$-axis at point $M$, $Q'N \\perp x$-axis at $N$,\\\\\n$\\because$ $\\angle PMQ=\\angle PNQ'=\\angle QPQ'=90^\\circ$,\\\\\n$\\therefore$ $\\angle QPM+\\angle NPQ'=\\angle PQ'N+\\angle NPQ'$,\\\\\n$\\therefore$ $\\angle QPM=\\angle PQ'N$,\\\\\nIn $\\triangle PQM$ and $\\triangle Q'PN$,\\\\\n$\\left\\{\\begin{array}{l}\n\\angle PMQ=\\angle PNQ' \\\\\n\\angle QPM=\\angle PQ'N \\\\\nPQ=PQ'\n\\end{array}\\right.$,\\\\\n$\\therefore$ $\\triangle PQM\\cong \\triangle Q'PN$ (AAS),\\\\\n$\\therefore$ $PN=QM$, $Q'N=PM$,\\\\\nLet $Q\\left(m, -\\frac{1}{2}m+2\\right)$,\\\\\n$\\therefore$ $PM=\\left|m-1\\right|$, $QM=\\left|-\\frac{1}{2}m+2\\right|$,\\\\\n$\\therefore$ $ON=\\left|3-\\frac{1}{2}m\\right|$,\\\\\n$\\therefore$ $Q'\\left(3-\\frac{1}{2}m, 1-m\\right)$,\\\\\n$\\therefore$ $OQ'^2=\\left(3-\\frac{1}{2}m\\right)^2+\\left(1-m\\right)^2=\\frac{5}{4}m^2-5m+10=\\frac{5}{4}\\left(m-2\\right)^2+5$,\\\\\nWhen $m=2$, $OQ'^2$ reaches its minimum value of 5,\\\\\n$\\therefore$ The minimum value of $OQ'$ is $\\boxed{\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52536754.png"
  },
  {
    "solution": "\\boxed{\\frac{4}{9}\\pi + 4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\angle MON=40^\\circ$. With O as the center and 4 as the radius, an arc is drawn intersecting $OM$ at point A and $ON$ at point B. Arcs with centers at point A and point B, respectively, and with a radius greater than half the length of $AB$, intersect inside $\\angle MON$ at point C. A ray $OC$ is drawn intersecting $\\overset{\\frown}{AB}$ at point D. Let E be a moving point on $OA$, and connect $BE$ and $DE$. What is the minimum perimeter of the shaded area?\n",
    "completion": "\\textbf{Solution:} Given the problem statement: OC bisects $\\angle MON$,\\\\\n$\\therefore$ $\\angle BOD = \\frac{1}{2}\\angle MON = 20^\\circ$,\\\\\n$\\therefore$ the length of $\\overset{\\frown}{BD}$ $= \\frac{20\\pi \\times 4}{180} = \\frac{4}{9}\\pi$,\\\\\nConstruct the symmetric point D' of D about OM, draw BD' intersecting OM at point E', and connect OD', then BE'+ DE'= BE'+ D'E'=BD'. At this time, the minimum value of BE+DE= BD',\\\\\n$\\because$ $\\angle AOD' = \\angle AOD = \\angle BOD = 20^\\circ$,\\\\\n$\\therefore$ $\\angle BOD' = 60^\\circ$,\\\\\n$\\because$ OD'=OD=OB,\\\\\n$\\therefore$ $\\triangle BOD'$ is an equilateral triangle,\\\\\n$\\therefore$ BD'=OB=4,\\\\\n$\\therefore$ the minimum perimeter of the shaded area $= \\frac{4}{9}\\pi + 4 = \\boxed{\\frac{4}{9}\\pi + 4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51806751.png"
  },
  {
    "solution": "\\boxed{\\left(\\frac{3}{2}, \\frac{3}{2}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the parabola $y=ax^2+2x+c$ intersects the x-axis at points $A(-1,0)$ and $B$, and intersects the y-axis at point $C(0,3)$. Connect $NB$, $NC$. Is there a point $M$ that maximizes the area of $\\triangle BNC$? If it exists, find the coordinates of point $M$; if it does not exist, explain why.\n",
    "completion": "\\textbf{Solution:} Yes, let's connect $CN$ and $BN$, as shown in the figure:\\\\\n$S_{\\triangle BNC}=S_{\\triangle CMN}+S_{\\triangle MNB}=\\frac{1}{2}|MN|\\cdot |OB|$\\\\\n$\\therefore$ When $|MN|$ is at its maximum, the area of $\\triangle BNC$ is maximized,\\\\\n$MN=-m^2+3m=-(m-\\frac{3}{2})^2+\\frac{9}{4}$,\\\\\nWhen $m=\\frac{3}{2}$, $MN$ reaches its maximum value of $\\frac{9}{4}$,\\\\\nTherefore, when $m=\\frac{3}{2}$, the area of $\\triangle BNC$ is maximized, and the coordinates of point $M$ are $\\boxed{\\left(\\frac{3}{2}, \\frac{3}{2}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52766412_105.png"
  },
  {
    "solution": "\\boxed{\\left(0,\\frac{5}{2}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system, points $E$, $F$, $G$ are on the sides of rectangle $ABCO$. When $\\triangle EFO$ is folded along $EF$, point $O$ coincides exactly with point $G$, $GH\\perp x$-axis at point $H$, and point $M$ is the intersection of $GH$ and $EF$. If $CG=2$, and $B(6,4)$, find the coordinates of point $F$.\n",
    "completion": "\\textbf{Solution:} Let $OF=x$, then $CF=4-x$, and $GF=x$,\\\\\nIn $\\triangle CGF$, by the Pythagorean theorem, we have\\\\\n$CF^2+CG^2=GF^2$,\\\\\n$\\therefore (4-x)^2+2^2=x^2$, solving this gives $x=\\frac{5}{2}$,\\\\\n$\\therefore F\\boxed{\\left(0,\\frac{5}{2}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52019464_107.png"
  },
  {
    "solution": "\\boxed{\\frac{24}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the rhombus $ABCD$, the diagonals are $AC=6$ and $BD=8$. Point $E$ is the midpoint of side $AB$, and points $F$ and $P$ are moving points on $BC$ and $AC$, respectively. What is the minimum value of $PE+PF$?\n",
    "completion": "\\textbf{Solution:} Let AC and BD intersect at point H. Construct the symmetric point $E'$ of E with respect to AC. Draw a perpendicular from $E'$ to AD, intersecting BC at point $F'$. Connect $E'F'$ and $E'P$ as shown in the figure.\\\\\nBy the property of symmetry, we know that $E'P=PE$, thus PE+PF=$E'P+PF$.\\\\\nThen, when $E'$, P, and F are collinear and $E'P\\perp BC$, $E'P+PF$ is minimized,\\\\\nTherefore, the length of $E'F'$ is the minimum value of PE+PF,\\\\\n$\\because$ Quadrilateral ABCD is a rhombus with diagonals AC=6 and BD=8,\\\\\n$\\therefore$ AC$\\perp$BD, AH=HC=3, BH=HD=4,\\\\\n$\\therefore AD=\\sqrt{AH^2+DH^2}=\\sqrt{3^2+4^2}=5$,\\\\\n$\\because$ The area of the rhombus $S_{\\text{rhombus }ABCD}=\\frac{1}{2}\\times AC\\times BD=AD\\times E'F'$,\\\\\n$\\therefore E'F'=\\frac{AC\\times BD}{2AD}=\\frac{6\\times 8}{2\\times 5}=\\frac{24}{5}$,\\\\\nTherefore, the answer is: $\\boxed{\\frac{24}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53729246.png"
  },
  {
    "solution": "\\boxed{122^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $AB\\parallel CD$, and the line $MN$ intersects $AB$ and $CD$ at points $E$ and $F$ respectively, and $\\angle NEB=58^\\circ$. What is the degree measure of $\\angle CFE$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle NEB$ and $\\angle AEF$ are vertical angles, and $\\angle NEB=58^\\circ$,\\\\\nit follows that $\\angle AEF=\\angle NEB=58^\\circ$.\\\\\nSince $AB \\parallel CD$,\\\\\nwe have $\\angle AEF + \\angle CFE = 180^\\circ$,\\\\\ntherefore $\\angle CFE = 180^\\circ - 58^\\circ = \\boxed{122^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52375765_142.png"
  },
  {
    "solution": "\\boxed{\\frac{1+\\sqrt{5}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points A, B, C, and D are on the circle, with $ \\overset{\\frown}{AB}=\\overset{\\frown}{AD}=\\overset{\\frown}{CD}$. Line BD intersects AC at point E. As shown in figure 2, the bisector of $\\angle ACB$ intersects BD at point M and AB at point N. If point N is the midpoint of AB, and BE equals DM, what is the value of $\\frac{DM}{BM}$?\n",
    "completion": "\\textbf{Solution:} \\\\\nSince CN bisects $\\angle ACB$, it follows that NH=NF, \\\\\nSince AN=BN, it follows that Rt$\\triangle ANH \\cong$ Rt$\\triangle BNF$, \\\\\nHence, $\\angle BAC = \\angle ABC$, \\\\\nTherefore, $\\overset{\\frown}{BC} = \\overset{\\frown}{AC} = 2\\overset{\\frown}{CD} = 2\\overset{\\frown}{AD} = 2\\overset{\\frown}{AB}$, \\\\\nTherefore, $\\angle ABD = \\angle CBD = \\angle ACB = 36^\\circ$, \\\\\n$\\angle BAC = \\angle BDC = 72^\\circ$, \\\\\nTherefore, $\\angle DCM = 54^\\circ$, $\\angle DMC = 54^\\circ$, \\\\\nThus, CD=DM, \\\\\nSince $\\angle BDC = 72^\\circ$, $\\angle ACD = 36^\\circ$, it follows that $\\angle DEC = 72^\\circ$, \\\\\nHence, CD=CE, \\\\\nSince $\\angle CBD = \\angle ACB$, it follows that BE=CE, \\\\\nTherefore, BE=CE=CD=DM. \\\\\nLet DE=1, CD=x, then BE=CE=CD=x. \\\\\nIt is easy to see that $\\triangle DCE \\sim \\triangle DBC$, hence $\\frac{DE}{CD} = \\frac{CD}{BD}$, \\\\\nTherefore, $\\frac{1}{x} = \\frac{x}{1+x}$, \\\\\nSolving this obtains $x=\\frac{1+\\sqrt{5}}{2}$ ($x>0$). \\\\\nSince BE=DM, it follows that BM=DE. \\\\\nTherefore, $\\frac{DM}{BM}=\\frac{DM}{DE}=\\frac{CD}{DE}=\\boxed{\\frac{1+\\sqrt{5}}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55452515_66.png"
  },
  {
    "solution": "\\boxed{\\rho=\\frac{9.9}{V}\\left(V>0\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, there is a certain mass of carbon dioxide in a sealed container. When the volume $V$ (in $m^3$) of the container changes, the density $\\rho$ (in $kg/m^3$) of the gas changes accordingly. It is known that the density $\\rho$ and the volume $V$ have an inverse proportional function relationship. The graph of this relationship is shown in the figure. When $V=5m^3$, $\\rho=1.98kg/m^3$. Find the analytic formula of the density $\\rho$ as a function of the volume $V$?\n",
    "completion": "\\textbf{Solution:} Since density $\\rho$ is inversely proportional to volume $V$,\\\\\nwe set $\\rho=\\frac{k}{V}\\left(V>0\\right)$.\\\\\nGiven $V=5\\mathrm{m}^3$ and $\\rho=1.98\\mathrm{kg/m}^3$,\\\\\nhence $1.98=\\frac{k}{5}$,\\\\\nthus $k=1.98\\times 5=9.9$.\\\\\nTherefore, the formula relating density $\\rho$ to volume $V$ is: \\boxed{\\rho=\\frac{9.9}{V}\\left(V>0\\right)}",
    "image_path": "dataset/mm_math/images/MM_Math/52977197_178.png"
  },
  {
    "solution": "\\boxed{\\frac{4\\pi }{3}-\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of $\\odot O$, and chord $CD\\perp AB$ at point $E$. Connect $AC$, $BC$, $BD$, $OF\\perp AC$ at point $F$, and $OF=1$. When $\\angle D = 30^\\circ$, find the length of $\\overset{\\frown}{AC}$ and the area of the shaded region (keep the results in terms of square roots and $\\pi$).\n",
    "completion": "\\textbf{Solution:} Connect $OC$ as shown below:\\\\\n$\\because \\angle CAB=\\angle D=30^\\circ$, $OA=OC$,\\\\\n$\\therefore \\angle OAC=\\angle OCA=30^\\circ$,\\\\\n$\\therefore \\angle AOC=120^\\circ$,\\\\\nIn $Rt\\triangle ABC$, $\\because \\angle ACB=90^\\circ$, $BC=2$, $\\angle CAB=30^\\circ$,\\\\\n$\\therefore AB=2BC=4$, $AC=\\sqrt{3}BC=2\\sqrt{3}$,\\\\\n$\\therefore$ the length of $\\overset{\\frown}{AC}$ $=\\frac{120\\pi \\cdot 2}{180}=\\boxed{\\frac{4\\pi }{3}}$,\\\\\nThe area of the shaded region $=\\frac{120\\pi \\cdot{2}^{2}}{360}-\\frac{1}{2}\\times 2\\sqrt{3}\\times 1=\\boxed{\\frac{4\\pi }{3}-\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51262145_131.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, place the rectangle OABC into the Cartesian coordinate system xO, such that OA and OC lie on the positive semi-axes of the x and y axes respectively, where AB = 15. The equation of the line containing the diagonal AC is given by $y=-\\frac{5}{3}x+b$. Fold the rectangle OABC along BE such that point A falls on point D located on side OC. What is the length of EA?\n",
    "completion": "\\textbf{Solution:} In $\\triangle BCD$, we have $BC=9$, $BD=AB=15$,\\\\\n$\\therefore CD= \\sqrt{15^2-9^2} = 12$,\\\\\n$\\therefore OD=15-12=3$. Let $DE=AE=x$,\\\\\nIn $\\triangle DEO$, $\\because DE^2=OD^2+OE^2$,\\\\\n$\\therefore x^2=3^2+(9-x)^2$,\\\\\n$\\therefore x=5$,\\\\\n$\\therefore AE=\\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50675997_172.png"
  },
  {
    "solution": "\\boxed{y=2x+2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=ax^{2}+2ax+1$ has only one common point A with the x-axis. A line passing through point A intersects the parabola at point B and intersects the y-axis at point C, and point C is the midpoint of line segment AB. Find the analytical expression of the line AB.\n",
    "completion": "\\textbf{Solution:} \nGiven that $y=(x+1)^{2}$,\\\\\nhence, the coordinates of vertex A are (-1, 0),\\\\\nsince point C is the midpoint of segment AB,\\\\\nmeaning that points A and B are symmetric with respect to point C,\\\\\nthus, the x-coordinate of point B is 1,\\\\\nwhen $x=1$, $y=x^{2}+2x+1=1+2+1=4$, so point B is (1, 4),\\\\\nlet the equation of line AB be $y=kx+b$,\\\\\nsubstituting A (-1, 0) and B (1, 4) into it, we get\\\\\n$\\begin{cases}\n-k+b=0\\\\\nk+b=4\n\\end{cases}$\\\\\nSolving this, we find $\\begin{cases}\nk=2\\\\\nb=2\n\\end{cases}$\\\\\nthus, the equation of line AB is $\\boxed{y=2x+2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52978187_218.png"
  },
  {
    "solution": "\\boxed{125^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AD$ is the altitude to side $BC$, $AE$ and $BF$ are the angle bisectors of $\\angle BAC$ and $\\angle ABC$ respectively, $\\angle BAC=56^\\circ$, $\\angle C=70^\\circ$. What is the measure of $\\angle BOA$?\n",
    "completion": "\\textbf{Solution}: In $\\triangle ABC$, we have $\\angle ABC=180^\\circ-\\angle C-\\angle BAC=180^\\circ-70^\\circ-56^\\circ=54^\\circ$ \\\\\nSince $BF$ bisects $\\angle ABC$ \\\\\nTherefore, $\\angle 1=\\frac{1}{2}\\angle ABC=\\frac{1}{2}\\times 54^\\circ=27^\\circ$ \\\\\nTherefore, $AE$ bisects $\\angle BAC$ \\\\\nTherefore, $\\angle 2=\\frac{1}{2}\\angle BAC=\\frac{1}{2}\\times 56^\\circ=28^\\circ$ \\\\\nTherefore, in $\\triangle AOB$, $\\angle AOB=180^\\circ-(\\angle 1+\\angle 2)=180^\\circ-(27^\\circ+28^\\circ)=\\boxed{125^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/52845737_115.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the paper $\\triangle ABC$, $\\angle A=65^\\circ$, $\\angle B=75^\\circ$, fold one corner of the paper so that point C falls inside $\\triangle ABC$ (as shown in the figure). If $\\angle 1=20^\\circ$, what is the degree measure of $\\angle 2$?\n",
    "completion": "\\textbf{Solution:} Given in $\\triangle ABC$, $\\angle A=65^\\circ$, $\\angle B=75^\\circ$,\\\\\n$\\therefore$ $\\angle C=180^\\circ-\\angle A-\\angle B=180^\\circ-65^\\circ-75^\\circ=40^\\circ$,\\\\\nsince $\\angle 1=20^\\circ$,\\\\\n$\\therefore$ $\\angle CED=\\frac{180^\\circ-\\angle 1}{2}=80^\\circ$,\\\\\nin $\\triangle CDE$, $\\angle CDE=180^\\circ-\\angle C-\\angle CED=180^\\circ-40^\\circ-80^\\circ=60^\\circ$,\\\\\n$\\therefore$ $\\angle 2=180^\\circ-2\\angle CDE=180^\\circ-2\\times60^\\circ=\\boxed{60^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53078475.png"
  },
  {
    "solution": "\\boxed{8\\text{cm}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the line segment \\( AC=12\\,cm \\), and point \\( B \\) is on line segment \\( AC \\), satisfying \\( BC= \\frac{1}{2} AB \\). What is the length of \\( AB \\)?\n",
    "completion": "\\textbf{Solution:} Since $BC=\\frac{1}{2}AB$, \\\\\nit follows that $AB=\\frac{2}{3}AC=\\frac{2}{3}\\times 12=\\boxed{8\\text{cm}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51457698_190.png"
  },
  {
    "solution": "\\boxed{x>1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the graph, in the same Cartesian coordinate system, the graph of the linear function $y=-2x+4$ intersects with the graph of the direct proportion function $y=kx\\ (k>0)$ at point A, and the y-coordinate of point A is 2. Then, what is the solution set for the inequality $kx>-2x+4$?\n",
    "completion": "\\textbf{Solution:} Since the graph of the linear function $y=-2x+4$ intersects with the graph of the directly proportional function $y=kx\\ (k>0)$ at point $A$, and the ordinate of point $A$ is 2,\\\\\nwe substitute ${y}_{A}=2$ into $y=-2x+4$ to get: ${x}_{A}=1$,\\\\\nSince $kx>-2x+4$,\\\\\nit follows that $x>1$,\\\\\nTherefore, the answer is: $\\boxed{x>1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52395426.png"
  },
  {
    "solution": "\\boxed{\\left(-\\frac{12}{5}, \\frac{16}{5}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the quadrilateral $ABCO$ is a rectangle. The coordinates of point $A$ are $(4, 0)$, and the coordinates of point $C$ are $(0, 8)$. When the rectangle $ABCO$ is folded along $OB$, point $A$ falls on point $D$. What are the coordinates of point $D$?\n",
    "completion": "\\textbf{Solution:} Draw $DE \\perp x$-axis at $E$, as shown in the diagram:\\\\\n$\\because$ Quadrilateral $ABCO$ is a rectangle, where the coordinates of point $A$ are $(4, 0)$, and the coordinates of point $C$ are $(0, 8)$,\\\\\n$\\therefore OA = BC = 4, AB = OC = 8$,\\\\\n$\\because$ Folding rectangle $ABCO$ along $OB$,\\\\\n$\\therefore \\triangle ABO \\cong \\triangle DBO$,\\\\\n$\\therefore \\angle FDO = 90^\\circ, OD = OA = 4$,\\\\\nIn $\\triangle DFO$ and $\\triangle CFB$,\\\\\n$\\left\\{\\begin{array}{l}\n\\angle BFC = \\angle OFD\\\\\n\\angle FDO = \\angle BCF = 90^\\circ\\\\\nDO = 4 = BC\n\\end{array}\\right.$,\\\\\n$\\therefore \\triangle DFO \\cong \\triangle CFB (AAS)$,\\\\\n$\\therefore FD = FC$,\\\\\nLet $FD = FC = x$, then $FO = OC - FC = 8 - x$,\\\\\nIn $\\triangle DFO$, where $\\angle FDO = 90^\\circ, DF = x, OF = 8 - x, DO = 4$, according to the Pythagorean theorem $OF^{2} = DF^{2} + DO^{2}$, we get $(8-x)^{2} = x^{2} + 4^{2}$, solving for $x$ yields $x = 3$,\\\\\n$\\therefore S_{\\triangle DFO} = \\frac{1}{2}DF \\cdot DO = \\frac{1}{2}FO \\cdot OE$, substituting values we get $OE = \\frac{12}{5}$,\\\\\nIn $\\triangle DEO$, where $\\angle DEO = 90^\\circ, OE = \\frac{12}{5}, DO = 4$, according to the Pythagorean theorem, we get $DE = \\sqrt{DO^{2} - OE^{2}} = \\sqrt{4^{2} - \\left(\\frac{12}{5}\\right)^{2}} = \\frac{16}{5}$,\\\\\n$\\because$ Point $D$ is in the second quadrant,\\\\\n$\\therefore D\\left(-\\frac{12}{5}, \\frac{16}{5}\\right)$,\\\\\nHence, the answer is: $\\boxed{\\left(-\\frac{12}{5}, \\frac{16}{5}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52050224.png"
  },
  {
    "solution": "\\boxed{\\frac{15}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, in the Cartesian coordinate system, the parabola $y=ax^2-2ax-3a$ (where $a<0$) intersects the x-axis at points A, B (with A to the left of B), and intersects the y-axis at point C. Point D is the vertex of the parabola, and its axis of symmetry intersects the x-axis at point E. Also, OB=OC. As shown in Figure 2, point P is a moving point on the line $y=5$ (point P is not on the axis of symmetry of the parabola). There are two lines $l_1$ and $l_2$ passing through point P, each having only one common point with the parabola, and neither is parallel to the y-axis. Lines $l_1$ and $l_2$ intersect the axis of symmetry of the parabola at points M and N, respectively. Point G is a point on the axis of symmetry below points M and N, and it always satisfies $GP^2=GM\\cdot GN$. Find the coordinates of point G.\n",
    "completion": "\\textbf{Solution:}\n\nLet $P(p, 5)$, and let $y=5$ intersect with the axis of symmetry at $H$, \n\nthen the equation of the line passing through point $P$ is $y=k(x-p)+5$,\n\n\\[\n\\therefore \\left\\{\n\\begin{array}{l}\ny=-x^{2}+2x+3\\\\ \ny=k(x-p)+5\n\\end{array}\n\\right.,\n\\]\n\n\\[\n\\therefore -x^{2}+(2-k)x+kp-2=0,\n\\]\n\nlet $\\Delta =b^{2}-4ac=(2-k)^{2}+4(kp-2)=0$,\n\n\\[\n\\therefore k^{2}+(4p-4)k-4=0,\n\\]\n\n\\[\n\\therefore k_{1}+k_{2}=-\\frac{b}{a}=4-4p, \\quad k_{1}k_{2}=\\frac{c}{a}=-4,\n\\]\n\nwhen $x=1$, $y=k(1-p)+5$,\n\nthus $y_{M}=k_{1}(1-p)+5=y_{1}$, $y_{N}=k_{2}(1-p)+5=y_{2}$,\n\nlet $G(1, t)$,\n\n\\[\n\\because GP^{2}=GM\\cdot GN,\n\\]\n\n\\[\n\\therefore GH^{2}+PH^{2}=GM\\cdot GN, \\text{ i.e., } (5-t)^{2}+(p-1)^{2}=(y_{1}-t)(y_{2}-t),\n\\]\n\n\\[\n\\therefore (5-t)^{2}+(p-1)^{2}=(y_{1}-t)(y_{2}-t),\n\\]\n\n\\[\n\\therefore (5-t)^{2}+(p-1)^{2}=y_{1}y_{2}-t(y_{1}+y_{2})+t^{2},\n\\]\n\n\\[\n\\therefore (5-t)^{2}+(p-1)^{2}=[k_{1}(1-p)+5][k_{2}(1-p)+5]-t[k_{1}(1-p)+5+k_{2}(1-p)+5]+t^{2},\n\\]\n\n\\[\n\\therefore (5-t)^{2}+(p-1)^{2}=k_{1}k_{2}(1-p)^{2}+5(k_{1}+k_{2})(1-p)+25-t(k_{1}+k_{2})(1-p)-10t+t^{2},\n\\]\n\n\\[\n\\therefore (5-t)^{2}+(p-1)^{2}=-4(1-p)^{2}+20(1-p)^{2}+25-4t(1-p)^{2}-10t+t^{2},\n\\]\n\n\\[\n\\therefore 25-10t+t^{2}+(p-1)^{2}=-4(1-p)^{2}+20(1-p)^{2}+25-4t(1-p)^{2}-10t+t^{2},\n\\]\n\n\\[\n\\therefore (15-4t)(1-p)^{2}=0,\n\\]\n\n\\[\n\\because \\text{ The equation always holds, so it is independent of } p,\n\\]\n\n\\[\n\\therefore 15-4t=0,\n\\]\n\nSolving for $t$, we get $t=\\frac{15}{4}$,\n\nThe coordinates of point $G$ are $\\left(1, \\boxed{\\frac{15}{4}}\\right)$.",
    "image_path": "dataset/mm_math/images/MM_Math/51367443_125.png"
  },
  {
    "solution": "\\boxed{\\sqrt{7}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in an equilateral triangle $ABC$ with side length of 4, $D$ and $E$ are the midpoints of sides $BC$ and $AC$ respectively. $DF \\perp AB$ at point $F$. Connect $EF$. What is the length of $EF$?\n",
    "completion": "\\textbf{Solution:} Since $D$ and $E$ are the midpoints of sides $BC$ and $AC$ respectively, and the equilateral triangle $ABC$ has a side length of $4$,\\\\\n$\\therefore AB=BC=AC=4$, $\\angle B=60^\\circ=\\angle C$, $AE=CE=2$, $BD=CD=2$,\\\\\n$\\therefore \\triangle DEC$ is an equilateral triangle,\\\\\n$\\therefore DE=2$, $\\angle EDC=60^\\circ$,\\\\\nSince $DF\\perp AB$, then $\\angle BDF=30^\\circ$,\\\\\n$\\therefore BF=\\frac{1}{2}BD$, $BF\\colon DF\\colon BD=1\\colon \\sqrt{3}\\colon 2$,\\\\\n$\\therefore DF=\\sqrt{3}$,\\\\\nSince $\\angle BDF=30^\\circ$, $\\angle EDC=60^\\circ$,\\\\\n$\\therefore \\angle FDE=180^\\circ-30^\\circ-60^\\circ=90^\\circ$,\\\\\n$EF=\\sqrt{{\\left(\\sqrt{3}\\right)}^{2}+{2}^{2}}=\\sqrt{7}.$\\\\\nTherefore, the answer is $EF=\\boxed{\\sqrt{7}}.$",
    "image_path": "dataset/mm_math/images/MM_Math/51546408.png"
  },
  {
    "solution": "\\boxed{4.8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $P$ is a moving point on the hypotenuse $AC$ (not coinciding with $A$ or $C$) of $\\triangle ABC$. Draw $PM\\perp AB$ intersecting $AB$ at point $M$, and draw $PN\\perp BC$ intersecting $BC$ at point $N$. Connect $MN$. If $AB=6$ and $BC=8$, then what is the minimum value of $MN$ when point $P$ moves along $AC$?\n",
    "completion": "\\textbf{Solution:} Draw a line PB, \\\\\n$\\because$ $\\angle ABC=90^\\circ$, AB = 6, BC = 8, \\\\\n$\\therefore$ $AC=\\sqrt{AB^2+BC^2}=\\sqrt{6^2+8^2}=10$, \\\\\n$\\because$ PM $\\perp$ AB, PN $\\perp$ BC, $\\angle C=90^\\circ$, \\\\\n$\\therefore$ $\\angle PMB=\\angle PNB=\\angle B=90^\\circ$, \\\\\n$\\therefore$ Quadrilateral BNPM is a rectangle, \\\\\n$\\therefore$ MN = BP, \\\\\n$\\because$ The perpendicular segment is the shortest, \\\\\n$\\therefore$ When BP $\\perp$ AC, the length of MN is minimized, \\\\\n$\\therefore$ ${S}_{\\triangle ABC}=\\frac{1}{2}BC\\cdot AB=\\frac{1}{2}AC\\cdot BP$, \\\\\n$\\therefore$ $\\frac{1}{2}\\times 8\\times 6=\\frac{1}{2}\\times 10\\cdot BP$, \\\\\nSolve for it: BP = 4.8. \\\\\n$\\therefore$ The minimum value of MN is $\\boxed{4.8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52345445.png"
  },
  {
    "solution": "\\boxed{0<m\\leq 4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=-\\frac{1}{2}x^2+bx+c$ intersects the $x$-axis at points $A$ and $B(4, 0)$, and intersects the $y$-axis at point $C$. The line $y=-\\frac{1}{2}x+n$ passes through points $B$ and $C$. Point $D$ is a movable point on line $BC$. A perpendicular line $DQ$ from point $D$ to the $x$-axis is drawn, with $Q$ being the foot of the perpendicular and intersecting the parabola at point $P$. When point $D$ is moved 5 units to the right to arrive at point $E$, and segment $DE$ intersects with the parabola at only one point, please directly write down the range of values for the $x$-coordinate $m$ of point $D$.\n",
    "completion": "\\textbf{Solution:} The range of the x-coordinate $m$ of point $D$ is $m=-\\frac{9}{4}$ or $0<m\\leq 4$. Therefore, the answer is $m=\\boxed{-\\frac{9}{4}}$ or $\\boxed{0<m\\leq 4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52634644_175.png"
  },
  {
    "solution": "\\boxed{110^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the lines AB and CD intersect at point O, $\\angle 1 = 35^\\circ$, $\\angle 2 = 75^\\circ$. What is the measure of $\\angle EOB$ in degrees?\n",
    "completion": "\\textbf{Solution:} Given that $\\because$ $\\angle 1=35^\\circ$,\\\\\n$\\therefore$ $\\angle BOD=\\angle 1=35^\\circ$,\\\\\n$\\therefore$ $\\angle EOB=\\angle 2+\\angle BOD=75^\\circ+35^\\circ=\\boxed{110^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55190266.png"
  },
  {
    "solution": "\\boxed{\\frac{20}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, point $E$ is the midpoint of $BC$, and $EF\\perp AE$ intersects $CD$ at point $F$. $\\triangle ABE\\sim \\triangle ECF$; if $AB=3$ and $BC=8$, find the length of $EF$.\n",
    "completion": "\\textbf{Solution:} Since $E$ is the midpoint of $BC$, and $BC=8$\\\\\n$ \\therefore BE=EC=\\frac{1}{2}BC=4$\\\\\nSince $\\angle B=90^\\circ$, and $AB=3$\\\\\n$ \\therefore AE=5$\\\\\n$ \\therefore \\triangle ABE\\sim \\triangle ECF$\\\\\n$ \\therefore \\frac{AB}{EC}=\\frac{AE}{EF}$     i.e., $ \\frac{3}{4}=\\frac{5}{EF}$\\\\\n$ \\therefore EF=\\boxed{\\frac{20}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52536687_106.png"
  },
  {
    "solution": "\\boxed{x<0$ or $x>2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y_{1} = -x^{2} + 4x$ and the line $y_{2} = 2x$. What is the range of $x$ when $y_{1} < y_{2}$?\n",
    "completion": "\\textbf{Solution:} Given $-x^{2}+4x=2x$, we find $x_{1}=0$ and $x_{2}=2$. Thus, the $x$-coordinates of the intersection points of the parabola and the linear function are 0 and 2. Hence, when $x<0$ or $x>2$, $y_{1}<y_{2}$.\\\\\nTherefore, the answer is: $\\boxed{x<0$ or $x>2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52807530.png"
  },
  {
    "solution": "\\boxed{(2,-6)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given the parabola $y=ax^{2}-3ax-4a$ intersects the x-axis at points $A$ and $B$ (with $A$ to the left of $B$) and intersects the negative y-axis at point $C$. Let $P$ be a point on the parabola in the fourth quadrant. If $a=1$, a line through point $P$ is drawn perpendicular to the x-axis, intersecting line segment $BC$ at point $Q$. Find the maximum value of the line segment $PQ$ and the coordinates of point $P$ at this time.\n",
    "completion": "\\textbf{Solution:} When $a=1$, the parabola is $y=x^{2}-3x-4$. By substituting the coordinates of points $B(4,0)$ and $C(0,-4)$ into the expression of the linear function, we obtain the equation of line $BC$ as: $y=x-4$. Let point $P$ be $(t,t^{2}-3t-4)$, then point $Q$ is $(t,t-4)$, $\\therefore PQ=(t-4)-(t^{2}-3t-4)=-t^{2}+4t=-(t-2)^{2}+4$, $\\therefore$ when $t=2$, $PQ$ reaches its maximum value $\\boxed{4}$, at which point $P$ is $\\boxed{(2,-6)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52916643_243.png"
  },
  {
    "solution": "\\boxed{\\frac{11}{18}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the vertices of triangle $ \\triangle OAB$ are respectively $ O(0, 0)$, $ A(3, 4)$, and $ B(6, 0)$. Moving points P and Q start simultaneously from point O and move along the positive direction of the x-axis and the y-axis at speeds of 3 units per second and 2 units per second, respectively. The motion of points P and Q stops simultaneously when point P reaches point B. Through point Q, draw $ MN\\parallel OB$ to intersect $AO$ and $AB$ at points M and N, respectively, and connect PM and PN. Let the motion time be t (seconds). Connect AP. Determine whether there exists a point P such that $ \\angle OAP=\\angle BPN$. If it exists, find the value of t at that time; if it does not exist, explain the reason.\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect AP,\\\\\nsince $A(3, 4)$, $M\\left(\\frac{3}{2}t, 2t\\right)$,\\\\\nthus $OA=\\sqrt{3^{2}+4^{2}}=5$, $OM=\\sqrt{\\left(\\frac{3}{2}t\\right)^{2}+\\left(2t\\right)^{2}}=\\frac{5}{2}t$.\\\\\nAccording to the given conditions, $AO=AB$, $AM=AN$,\\\\\nthus $OM=BN=\\frac{5}{2}t$.\\\\\n(1) When $t=0$, points $M$, $P$ are both at the origin, at this time point $N$ coincides with point $B$,\\\\\nthus $\\angle BPN=\\angle OAP=0^\\circ$.\\\\\n(2) When $0<t<2$,\\\\\nsince $OA=OB$,\\\\\nthus $\\angle AOB=\\angle PBN$,\\\\\nand $\\angle OAP=\\angle BPN$,\\\\\nthus $\\triangle AOP\\sim \\triangle PBN$,\\\\\nthus $\\frac{OA}{BP}=\\frac{OP}{BN}$,\\\\\nnamely $\\frac{5}{6-3t}=\\frac{3t}{\\frac{5}{2}t}$,\\\\\nsolving this gives $t_{1}=\\frac{11}{18}$, $t_{2}=0$ (discard).\\\\\n(3) When $t=2$, it contradicts the given conditions, discard it.\\\\\nIn summary, $t=0$ or $t=\\boxed{\\frac{11}{18}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51434142_205.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\triangle ABC$, $AD$ is the median to side $BC$, $E$ is a point on side $AB$, and a line through point $C$ is drawn parallel to $AB$ intersecting the extension of $ED$ at point $F$. When $AD \\perp BC$, $AE=1$, and $CF=2$, what is the length of $AC$?\n",
    "completion": "\\textbf{Solution:}\n\nSince $AD$ is the median to side $BC$ and $AD\\perp BC$,\n\nit follows that $AB=AC$.\n\nSince $CF\\parallel AB$,\n\nit follows that $\\angle B=\\angle DCF$.\n\nIn $\\triangle BED$ and $\\triangle CFD$,\n\n\\[\n\\left\\{\\begin{array}{l}\n\\angle B=\\angle DCF \\\\\nBD=CD \\\\\n\\angle BDE=\\angle CDF\n\\end{array}\\right.\n\\]\n\nTherefore, $\\triangle BED \\cong \\triangle CFD$,\n\nwhich implies that $CF=BE$.\n\nGiven $AE=1$ and $CF=2$,\n\nthus $AB=AE+BE=AE+CF=1+2=3$,\n\nwhich leads to $AC=AB=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52720359.png"
  },
  {
    "solution": "\\boxed{y=x^{2}+x-\\frac{7}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Synthesis and Inquiry: As shown in the figure, in the Cartesian coordinate system, the graph of the quadratic function $y=x^2+bx+c$ passes through point $A\\left(0, -\\frac{7}{4}\\right)$ and point $B\\left(1, \\frac{1}{4}\\right)$. What is the analytical expression of this quadratic function?\n",
    "completion": "\\textbf{Solution:} Substitute $A\\left(0, -\\frac{7}{4}\\right)$ and point $B\\left(1, \\frac{1}{4}\\right)$ into $y=x^{2}+bx+c$ to get: \\\\\n\\[\n\\begin{cases}\n-\\frac{7}{4}=c\\\\\n\\frac{1}{4}=1+b+c\n\\end{cases}\n,\\] solving this system of equations gives\n\\[\n\\begin{cases}\nb=1\\\\\nc=-\\frac{7}{4}\n\\end{cases},\n\\]\n$\\therefore y=x^{2}+x-\\frac{7}{4}$. Therefore, the quadratic function is $\\boxed{y=x^{2}+x-\\frac{7}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52719950_148.png"
  },
  {
    "solution": "\\boxed{y=\\frac{3}{x}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the coordinates of points A and B are respectively $(-2,0)$ and $(0,3)$. Rotating line segment AB $90^\\circ$ counterclockwise around point B gives line segment BC. Draw line CD $\\perp$ OB with the foot of the perpendicular being D. The graph of the inverse proportion function $y=\\frac{k}{x}$ passes through point C. Write down the coordinates of point C directly, and find the equation of the inverse proportion function.\n",
    "completion": "\\textbf{Solution:} We have $C(3,1)$; $k=3\\times1=3$,\\\\\n$\\therefore$ the equation of the inverse proportion function is $\\boxed{y=\\frac{3}{x}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53147515_68.png"
  },
  {
    "solution": "\\boxed{3\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, within the rectangle $ABCD$, point $E$ lies on the extension of $AB$. If $BE=AC=6$ and $\\angle E=15^\\circ$, then what is the length of $AB$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect BD, \\\\\nsince quadrilateral ABCD is a rectangle, \\\\\ntherefore BD=AC, \\\\\nalso since BE=AC, \\\\\ntherefore BD=AC=BE=6, \\\\\ntherefore $\\angle E=\\angle BDE$, \\\\\ntherefore $\\angle ABD=\\angle E+\\angle BDE=2\\angle E=30^\\circ$, \\\\\ntherefore AB=$BD\\cos\\angle ABD$=6×$\\frac{\\sqrt{3}}{2}$=$\\boxed{3\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52050118.png"
  },
  {
    "solution": "\\boxed{6.5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $C$ is the midpoint of segment $AB$, point $E$ is a point on segment $AB$, and point $D$ is the midpoint of segment $AE$. If $AB=16$ and $CE=5$, find the length of segment $AD$.\n",
    "completion": "\\textbf{Solution:} Since point C is the midpoint of segment AB, and $AB=16$, \\\\\n$\\therefore BC=\\frac{1}{2}AB=8$, \\\\\nbecause $CE=5$, \\\\\n$\\therefore BE=BC-CE=8-5=3$, \\\\\nhence $AE=AB-BE=16-3=13$, \\\\\nbecause point D is the midpoint of segment AE, \\\\\n$\\therefore AD=\\frac{1}{2}AE=\\boxed{6.5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51323815.png"
  },
  {
    "solution": "\\boxed{(14,12)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given that, in the Cartesian coordinate system, for square $ABCD$, point $A(0,10)$, point $B(8,4)$, and point $C$ is in the first quadrant. Point $P$ starts from vertex $A$ and moves counterclockwise along the sides of the square for one round. Construct $\\angle OPQ = \\angle OAB$ on the right side of $OP$ intersecting the $x$-axis at point $Q$. Find the coordinates of point $C$.\n",
    "completion": "\\textbf{Solution:} As shown in Figure 1, since quadrilateral ABCD is a square, it follows that $\\angle ABC=90^\\circ$, and AB=BC. Therefore, $\\angle ABE+\\angle CBF=90^\\circ$. Draw $BE\\perp OA$ at E, and draw $CF\\perp BE$ at F. Thus, $\\angle AEB=\\angle ABC=\\angle F=90^\\circ$, which implies that $\\angle EAB+\\angle ABE=90^\\circ$. Therefore, $\\angle EAB=\\angle CBF$. Consequently, $\\triangle ABE\\cong \\triangle BCF$ (by AAS). Therefore, BF=AE, CF=BE. Given that point A is at (0,10) and point B is at (8,4), we deduce AE=6, BE=8, OE=4. Therefore, BF=6, CF=8. Hence, EF=BE+BF=14 and CF+OE=12. Thus, the coordinates of point C are \\boxed{(14,12)}.",
    "image_path": "dataset/mm_math/images/MM_Math/51317370_134.png"
  },
  {
    "solution": "\\boxed{25^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given that AP is a ray outside of square ABCD, and $B'$ is the reflection point of B about line AP, connect $B'D$. As shown in Figure 1. If $\\angle BAP=20^\\circ$, find the magnitude of the angle $\\angle ADB'$.\n",
    "completion": "\\textbf{Solution:} Draw $A{B}^{\\prime}$, as shown in figure 1,\\\\\n$\\because$ ${B}^{\\prime}$ is the symmetric point of B with respect to the line AP,\\\\\n$\\therefore$ $AB=A{B}^{\\prime}$,\\\\\n$\\therefore$ $\\angle BAP=\\angle {B}^{\\prime}AP=20^\\circ$\\\\\n$\\because$ quadrilateral ABCD is a square,\\\\\n$\\therefore$ $A{B}^{\\prime}=AB=AD$,\\\\\n$\\therefore$ $\\angle A{B}^{\\prime}D=\\angle AD{B}^{\\prime}$,\\\\\n$\\because$ $\\angle {B}^{\\prime}AD=\\angle {B}^{\\prime}AB+\\angle BAD=90^\\circ+40^\\circ=130^\\circ$,\\\\\n$\\therefore$ $\\angle AD{B}^{\\prime}=\\boxed{25^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53599320_97.png"
  },
  {
    "solution": "\\boxed{y=-4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the inverse proportion function $y=\\frac{k}{x}$ ($k \\neq 0$) passes through the intersection point P of the diagonals of rectangle $ABCD$. It is known that points A, C, and D are on the coordinate axes, $BD \\perp DC$, and the area of rectangle $ABCD$ is 8. What is the value of $k$?\n",
    "completion": "\\textbf{Solution:} Draw $PE\\perp y$-axis at point E.\\\\\nSince quadrilateral ABCD is a parallelogram,\\\\\nit follows that AB=CD.\\\\\nAlso, since BD$\\perp x$-axis,\\\\\nquadrilateral ABDO is a rectangle.\\\\\nHence, AB=DO.\\\\\nTherefore, the area of rectangle ABDO, which is also the area of parallelogram ABCD, is 8.\\\\\nSince P is the intersection point of the diagonals and $PE\\perp y$-axis,\\\\\nquadrilateral PDOE is a rectangle with area 4.\\\\\nThat is, $DO\\cdot EO=4$.\\\\\nLet the coordinates of point P be $(x,y)$.\\\\\nThen $k=x\\boxed{y=-4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51352010.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $OM$ is the bisector of $\\angle AOC$, and $ON$ is the bisector of $\\angle BOC$. As shown in Figure 1, when $\\angle AOB$ is a right angle and $\\angle BOC=60^\\circ$, what is the degree measure of $\\angle MON$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle AOB=90^\\circ$ and $\\angle BOC=60^\\circ$, \\\\\nthus $\\angle AOC=\\angle AOB+\\angle BOC=90^\\circ+60^\\circ=150^\\circ$. \\\\\nSince $OM$ bisects $\\angle AOC$ and $ON$ bisects $\\angle BOC$, \\\\\nthus $\\angle MOC=\\frac{1}{2}\\angle AOC=75^\\circ$ and $\\angle NOC=\\frac{1}{2}\\angle BOC=30^\\circ$. \\\\\nTherefore, $\\angle MON=\\angle MOC-\\angle NOC=75^\\circ-30^\\circ=\\boxed{45^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53363928_110.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{2}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $AB$ is the diameter of $\\odot O$, $BC$ is tangent to $\\odot O$ at point $B$. Connect $AC$ and $OC$. If $\\sin\\angle BAC=\\frac{1}{3}$, what is the value of $\\tan\\angle BOC$?\n",
    "completion": "\\textbf{Solution:} Given that $AB$ is the diameter of circle $O$, and $BC$ is tangent to circle $O$ at point $B$, \\\\\n$\\therefore AB \\perp BC$, \\\\\n$\\therefore \\angle ABC = 90^\\circ$, \\\\\n$\\because \\sin\\angle BAC = \\frac{1}{3}$, \\\\\n$\\therefore$ let $BC = x$, $AC = 3x$, \\\\\n$\\therefore AB = \\sqrt{AC^{2} - BC^{2}} = \\sqrt{(3x)^{2} - x^{2}} = 2\\sqrt{2}x$, \\\\\n$\\therefore OB = \\frac{1}{2}AB = \\sqrt{2}x$, \\\\\n$\\therefore \\tan\\angle BOC = \\frac{BC}{OB} = \\frac{x}{\\sqrt{2}x} = \\boxed{\\frac{\\sqrt{2}}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52537008.png"
  },
  {
    "solution": "\\boxed{30}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, in rectangle $ABCD$, point $P$ starts from point $C$ and moves in the direction of $C \\rightarrow D \\rightarrow A \\rightarrow B$ and stops at point $B$. Let the distance traveled by point $P$ be $x$, and the area of $\\triangle PBC$ be $y$. Given the functional relationship between $y$ and $x$ as shown in Figure 2, what is the area of rectangle $ABCD$?\n",
    "completion": "Solution: From the problem statement, we know that: when point P is on side CD, \\(y\\) increases as \\(x\\) increases;\\\\\nwhen point P is on side AD, \\(y\\) does not change as \\(x\\) changes;\\\\\nwhen point P is on side AB, \\(y\\) decreases as \\(x\\) increases.\\\\\nConsidering the graph of a linear function, we find that CD=5, AD=6,\\\\\n\\(\\therefore\\) the area of rectangle ABCD is: AD\\(\\times\\)CD=5\\(\\times\\)6=\\boxed{30}.",
    "image_path": "dataset/mm_math/images/MM_Math/53339388.png"
  },
  {
    "solution": "\\boxed{y=3x+6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $N(0,6)$ and point $M$ is on the negative half axis of the $x$-axis, with $ON=3OM$. $A$ is a point on the line segment $MN$. $AB \\perp x$-axis with the foot of the perpendicular as point $B$, and $AC \\perp y$-axis with the foot of the perpendicular as point $C$. What is the equation of line $MN$?\n",
    "completion": "\\textbf{Solution:} To find the equation of line $MN$, we utilize the equation $y=kx+b$. Substituting $M(-2,0)$ and $N(0,6)$ into it, we get\n\\[\n\\left\\{\n\\begin{array}{l}\n-2k+b=0 \\\\\nb=6\n\\end{array}\n\\right.\n\\]\nSolving this system yields\n\\[\n\\left\\{\n\\begin{array}{l}\nk=3 \\\\\nb=6\n\\end{array}\n\\right.\n\\]\n$\\therefore$ The equation of line $MN$ is $\\boxed{y=3x+6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52225566_110.png"
  },
  {
    "solution": "\\boxed{(3,1)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the line $AB$ passing through point $A(4,0)$ intersects the y-axis at point $B(0,4)$. The parabola $y=-x^2+bx+c$ passing through the origin $O$ intersects the line $AB$ at points $A$ and $C$, with the vertex of the parabola being $D$. $M$ is a point on segment $AB$, and $N$ is a point on the parabola. When $MN$ is parallel to the y-axis and $MN=2$, what are the coordinates of point $M$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram,\\\\\n$\\because$ point A (4, 0), point B (0, 4),\\\\\n$\\therefore$ let the equation of line AB be $y=kx+4$,\\\\\n$\\therefore$ $4k+4=0$, solving it: $k=-1$,\\\\\n$\\therefore$ the equation of line AB is $y=-x+4$;\\\\\nlet point N ($m$, $-m^{2}+4m$), point M ($m$, $-m+4$),\\\\\n$\\because$ point M is on the line segment AB,\\\\\n$\\therefore$ $0<m<4$,\\\\\n$MN=|-m^{2}+4m-(-m+4)|=|-m^{2}+5m-4|=2$,\\\\\n$\\therefore$ $-m^{2}+5m-4=\\pm2$,\\\\\nwhen $-m^{2}+5m-4=2$, solving it: $m_{1}=3$, $m_{2}=2$,\\\\\nwhen $m=3$ then $-m+4=1$;\\\\\nwhen $m=2$ then $-m+4=2$;\\\\\n$\\therefore$ the coordinates of point M are $\\boxed{(2,2)}$ or $\\boxed{(3,1)}$;\\\\\nwhen $-m^{2}+5m-4=-2$, solving it:\\\\\n$m_{3}=\\frac{5-\\sqrt{17}}{2}$, $m_{4}=\\frac{5+\\sqrt{17}}{2}>4$ (discard),\\\\\n$\\therefore$ $-m+4=-\\frac{5-\\sqrt{17}}{2}+4=\\frac{3+\\sqrt{17}}{2}$,\\\\\n$\\therefore$ the coordinates of point M are $\\boxed{\\left(\\frac{5-\\sqrt{17}}{2}, \\frac{3+\\sqrt{17}}{2}\\right)}$.\\\\\nThere are three possible coordinates for point M that satisfy the conditions: $\\boxed{\\left(\\frac{5-\\sqrt{17}}{2}, \\frac{3+\\sqrt{17}}{2}\\right)}$ or $\\boxed{(2,2)}$ or $\\boxed{(3,1)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52936243_165.png"
  },
  {
    "solution": "\\boxed{\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Let $A$ and $B$ be two fixed points on the side $OP$ of $\\angle POQ$, and let $E$ be a moving point on the side $OQ$. Construct squares $ABCD$ and $AEFG$ on the same side above $OP$, with $AB$ and $AE$ as their respective sides. As shown in the figure, $\\angle POQ=45^\\circ$, $OB=6$, and $OA=2$. Connect $BG$ and $DE$. When point $E$ moves along side $OQ$, the length of segment $BG$ has a minimum value. Determine this minimum value.\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, let $OQ$ intersect $AD$ at point $M$, and $CD$ at point $N$. Draw $DE' \\perp OQ$ at point $E'$,\\\\\n$\\because \\angle OAD=90^\\circ$, $\\angle QOA=45^\\circ$,\\\\\n$\\therefore \\angle QOA=\\angle AMO=45^\\circ$,\\\\\n$\\therefore OA=AM=2$,\\\\\n$\\because OB=6$,\\\\\n$\\therefore AD=AB=OB-OA=4$,\\\\\n$\\therefore DM=AD-AM=2$,\\\\\n$\\because \\angle DMN=\\angle AMO=45^\\circ$,\\\\\n$\\therefore DE'=DM\\sin\\angle DMN=2 \\times \\frac{\\sqrt{2}}{2}=\\sqrt{2}$,\\\\\n$\\therefore$ when point $E$ coincides with $E'$, the value of $DE$ is at its minimum,\\\\\n$\\because BG=DE$,\\\\\n$\\therefore$ the minimum value of $BG$ is $\\boxed{\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52050144_113.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{2}t+2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, the line $x=t$ intersects with the line $y=-x$ and the line $y=\\frac{1}{2}x+2$ at points E and D (where D is above E), respectively. What is the length of the line segment DE (expressed as an algebraic expression containing $t$)?\n",
    "completion": "\\textbf{Solution:} When $x=t$, $y=-t$; when $x=t$, $y=\\frac{1}{2}t+2$.\\\\\nTherefore, the coordinates of point D are $\\left(t, \\frac{1}{2}t+2\\right)$, and the coordinates of point E are $\\left(t, -t\\right)$.\\\\\nSince D is above E,\\\\\ntherefore the length of segment DE is $DE=\\frac{3}{2}t+2$, and $t>-\\frac{4}{3}$.\\\\\nHence, the length of the line segment DE is $DE=\\boxed{\\frac{3}{2}t+2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52395484_113.png"
  },
  {
    "solution": "\\boxed{\\frac{2\\sqrt{3}}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, it is known that $\\odot O$ is the circumscribed circle of quadrilateral $ABCD$ with $O$ as the center of the circle. If $\\angle BCD=120^\\circ$ and $AB=AD=2\\, \\text{cm}$, then what is the length of the radius of $\\odot O$?\n",
    "completion": "\\textbf{Solution:} Draw BD and construct OE $\\perp$ AD at E, and draw OD.\\\\\nSince $\\odot O$ is the circumcircle of quadrilateral ABCD, and $\\angle BCD=120^\\circ$,\\\\\nit follows that $\\angle BAD=60^\\circ$.\\\\\nGiven AD=AB=2cm,\\\\\ntriangle ABD is an equilateral triangle.\\\\\nHence, DE= $\\frac{1}{2}$ AD=1cm, and $\\angle ODE= \\frac{1}{2} \\angle ADB=30^\\circ$,\\\\\ntherefore, OD= $\\frac{DE}{\\cos 30^\\circ}$ = $\\boxed{\\frac{2\\sqrt{3}}{3}}$ cm.",
    "image_path": "dataset/mm_math/images/MM_Math/51130827.png"
  },
  {
    "solution": "\\boxed{\\left(4, 0\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=ax^2+bx+6$ passes through points $A(-2, 0)$ and $B(4, 0)$, and intersects the $y$-axis at point $C$. Point $D$ is a moving point on the parabola with the $x$-coordinate denoted as $m(1<m<4)$. Connect $AC$, $BC$, $DB$, and $DC$. When $m=2$, if point $M$ is a moving point on the $x$-axis, and point $N$ is a moving point on the parabola, determine whether there exists such a point $M$ that the quadrilateral with vertices $B$, $D$, $M$, and $N$ forms a parallelogram. If it exists, please write down the coordinates of point $M$ directly; if it does not exist, please explain the reason.\n",
    "completion": "\\textbf{Solution:} When $m=2$, the point $D(2, 6)$, assume point $M(x, 0)$, point $N(t, n)$, then $n=-\\frac{3}{4}t^2+\\frac{3}{2}t+6$. When $BD$ is a side, moving point $B$ 1 unit to the left and 6 units up gets point $D$, similarly moving point $M(N)$ 1 unit to the left and 6 units up gets point $N(M)$, hence\n\\[\n\\left\\{\n\\begin{array}{l}\nx-1=t \\\\\n0+6=n\n\\end{array}\n\\right.\n\\]\nor\n\\[\n\\left\\{\n\\begin{array}{l}\nx+1=t \\\\\n0-6=n\n\\end{array}\n\\right.\n\\]\nSolving this system gives:\n\\[\n\\left\\{\n\\begin{array}{l}\nx=3 \\\\\nt=2 \\\\\nn=6\n\\end{array}\n\\right.\n\\]\nor\n\\[\n\\left\\{\n\\begin{array}{l}\nx=\\frac{\\sqrt{33}-1}{2} \\\\\nt=\\frac{1+\\sqrt{33}}{2} \\\\\nn=-6\n\\end{array}\n\\right.\n\\]\nor\n\\[\n\\left\\{\n\\begin{array}{l}\nx=\\frac{-\\sqrt{33}-1}{2} \\\\\nt=\\frac{1-\\sqrt{33}}{2} \\\\\nn=-6\n\\end{array}\n\\right.\n\\]\n(Values that do not meet the premise have been omitted); hence the coordinates of point $M$ are $\\left(3, 0\\right)$ or $\\left(\\frac{\\sqrt{33}-1}{2}, 0\\right)$ or $\\left(\\frac{-\\sqrt{33}-1}{2}, 0\\right)$.\n\nWhen $BD$ is a diagonal, using the midpoint formula we get:\n\\[\n\\left\\{\n\\begin{array}{l}\n\\frac{1}{2}(3+4)=\\frac{1}{2}(x+m) \\\\\n\\frac{1}{2}(6+0)=\\frac{1}{2}(n+0)\n\\end{array}\n\\right.\n\\]\nSolving this system gives\n\\[\n\\left\\{\n\\begin{array}{l}\nm=3 \\\\\nn=6 \\\\\nx=4\n\\end{array}\n\\right.\n\\]\nTherefore, the coordinates of point $M$ are $\\left(4, 0\\right)$.\n\nIn conclusion, the coordinates of point $M$ are $\\boxed{\\left(3, 0\\right)}$ or $\\boxed{\\left(\\frac{\\sqrt{33}-1}{2}, 0\\right)}$ or $\\boxed{\\left(\\frac{-\\sqrt{33}-1}{2}, 0\\right)}$ or $\\boxed{\\left(4, 0\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52458461_179.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{7}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the quadrilateral $ABCD$ is a square. Point $M$ is on the side $BC$ (not coinciding with the endpoints $B$, $C$), and point $N$ is on the diagonal $AC$, with $MN \\perp AC$. Draw line $AM$, and let point $G$ be the midpoint of $AM$. Connect $NG$ and $DN$. If $AB = 10$ and $BM = 2\\sqrt{3}$, what is the length of $NG$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a square, $\\angle B=90^\\circ$. In $\\triangle ABM$, since AB＝10 and BM＝$2\\sqrt{3}$, we have $AM=\\sqrt{AB^2+BM^2}=4\\sqrt{7}$. Since MN$\\perp$AC and point G is the midpoint of AM, we have GN＝$\\frac{1}{2}AM=2\\sqrt{7}$. Therefore, the length of NG is $\\boxed{2\\sqrt{7}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50481659_134.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $\\angle AOB = 90^{\\circ}$, $\\angle EOF = 60^{\\circ}$, $OE$ bisects $\\angle AOB$, and $OF$ bisects $\\angle BOC$. What are the measures of $\\angle AOC$ and $\\angle COB$?\n",
    "completion": "\\textbf{Solution:} Since OE bisects $\\angle AOB$ and $\\angle AOB=90^\\circ$,\n\\[\n\\therefore \\angle BOE = \\angle AOB = 45^\\circ,\n\\]\nAlso, since $\\angle EOF=60^\\circ$,\n\\[\n\\therefore \\angle BOF = \\angle EOF - \\angle BOE= 15^\\circ,\n\\]\nAnd since OF bisects $\\angle BOC$,\n\\[\n\\therefore \\angle BOC=2\\angle BOF=30^\\circ,\n\\]\nThus, $\\angle AOC=\\angle AOB + \\angle BOC=120^\\circ$,\nHence, $\\angle AOC=\\boxed{120^\\circ}$, $\\angle COB=\\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53211177.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, it is known that $AB$ is the diameter of $\\odot O$ and $AB=8C$. Point $D$ is on $\\odot O$, $OC \\parallel BD$, intersecting $AD$ at point $E$. Connect $BC$, where $\\angle CBD=30^\\circ$. What is the degree measure of $\\angle COA$?\n",
    "completion": "\\textbf{Solution}: Since $OC \\parallel BD$,\\\\\nit follows that $\\angle OCB = \\angle CBD = 30^\\circ$,\\\\\nSince $OC = OB$,\\\\\nit follows that $\\angle OCB = \\angle OBC = 30^\\circ$,\\\\\nTherefore, $\\angle COA = \\angle OCB + \\angle OBC = \\boxed{60^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52872426_107.png"
  },
  {
    "solution": "\\boxed{12cm}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in parallelogram $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$, and point $E$ is the midpoint of $BC$. If $OE = 6\\,cm$, what is the length of $AB$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a parallelogram, \\\\\nit follows that $AO = CO$, \\\\\nsince point E is the midpoint of CB, \\\\\nthus OE is the median of $\\triangle ABC$, \\\\\ntherefore $AB = 2OE$, \\\\\nsince $OE = 6\\,cm$, \\\\\nit follows that $AB = \\boxed{12cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53639827.png"
  },
  {
    "solution": "\\boxed{40^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a ship sails in the direction indicated by the arrow. There is a lighthouse at point C. When the ship sails from point A to point B, $\\angle ACB = 90^\\circ$.\n",
    "completion": "\\textbf{Solution:} Since $\\angle CBD = \\angle A + \\angle C$,\\\\\nthus $\\angle C = \\angle CBD - \\angle A$\\\\\n$= 70^\\circ - 30^\\circ$\\\\\n$= \\boxed{40^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52815574.png"
  },
  {
    "solution": "\\boxed{y=-x+8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the figure, in the Cartesian coordinate plane, point $A(2, m)$ and point $B(6, 2)$ lie on the graph of an inverse proportionality function. What is the equation of the line $AB$?\n",
    "completion": "\\textbf{Solution:} Let the inverse proportion function be $y=\\frac{k}{x}$,\\\\\n$\\because$ point $A(2, m)$ and point $B(6, 2)$ lie on the graph of $y=\\frac{k}{x}$,\\\\\n$\\therefore k=2m=6\\times 2$,\\\\\nSolving this gives $m=6$,\\\\\n$\\therefore$ the coordinates of point $A$ are $(2, 6)$,\\\\\nLet the expression for line $AB$ be $y=ax+b$,\\\\\nSubstituting $A(2, 6)$ and $B(6, 2)$ into it gives $\\left\\{\\begin{array}{l}2a+b=6 \\\\ 6a+b=2\\end{array}\\right.$,\\\\\nSolving this yields $\\left\\{\\begin{array}{l}a=-1 \\\\ b=8\\end{array}\\right.$,\\\\\n$\\therefore$ the expression for line $AB$ is $y=-x+8$, hence the expression for line $AB$ is $\\boxed{y=-x+8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50447041_223.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the length of line segment $AB$ is $a$. Extend $BA$ to point $C$, making $AC=\\frac{1}{2}AB$. Let $D$ be the midpoint of line segment $BC$. If $AD=3\\, \\text{cm}$, what is the value of $a$?\n",
    "completion": "Solution: Since $AD=AB-BD=a-\\frac{3}{4}a=3\\text{cm}$, \\\\\nit follows that $a=\\boxed{12}\\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55327759_86.png"
  },
  {
    "solution": "\\boxed{\\frac{2}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, rectangle $ABCD$ is folded along $CE$ such that point B falls exactly on point F on $AD$. If $AB\\colon BC=2\\colon 3$, then what is the value of $\\cos\\angle DCF$?\n",
    "completion": "\\textbf{Solution:} Since rectangle $ABCD$ has $AB:BC=2:3$,\\\\\nlet $AB=2m$, then $BC=3m$,\\\\\nthus $AB=CD=2m$, and $AD=BC=CF=3m$, $\\angle D=90^\\circ$,\\\\\ntherefore $\\cos\\angle DCF=\\frac{DC}{FC}=\\frac{2m}{3m}=\\boxed{\\frac{2}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51213317.png"
  },
  {
    "solution": "\\boxed{17}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the known $\\triangle ABC$, proceed with the following steps: \n(1) With centers at B and C, respectively, draw arcs with radii longer than $\\frac{1}{2}BC$, intersecting at points M and N; \n(2) Draw line MN intersecting AB at D, and connect CD. If $AC = 5$ and $AB = 12$, what is the perimeter of $\\triangle ACD$?\n",
    "completion": "\\textbf{Solution:} From the construction, we have that $MN$ bisects $BC$ perpendicularly,\\\\\n$\\therefore DB=DC$,\\\\\n$\\therefore$ the perimeter of $\\triangle ACD = AC+AD+CD = AC+AD+BD = AC+AB = 5+12 = \\boxed{17}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52950677.png"
  },
  {
    "solution": "\\boxed{\\frac{16}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in $\\triangle ABC$, $\\angle BAC=90^\\circ$ and $\\angle B=45^\\circ$. A line is drawn through point A such that $AD \\parallel BC$, with point D to the right of point A. Points P and Q are located on ray $AD$ and ray $CB$, respectively. Point E is a point on line segment $CQ$, and $CQ=2AP$. Let $AP=x$ and $CE=y$, then $y=2x-2$. When point Q is the midpoint of $BC$, $y=3$. If $PE \\perp BC$, find the length of $BQ$.\n",
    "completion": "\\textbf{Solution}: As shown in the diagram, draw $AM\\perp BC$ at point M, PE intersects AC at point N,\\\\\n$\\because$ $\\angle BAC=90^\\circ$, $\\angle B=45^\\circ$,\\\\\n$\\therefore$ $\\angle C=180^\\circ-\\angle BAC-\\angle B=45^\\circ$,\\\\\n$\\therefore$ $\\angle B=\\angle C$, $BM=MC=\\frac{1}{2}BC=\\frac{1}{2}\\times 10=5$,\\\\\n$\\because$ $AM\\perp BC$, $PE\\perp BC$,\\\\\n$\\therefore$ $AM\\parallel PE$,\\\\\nAlso, $\\because$ $AD\\parallel BC$,\\\\\n$\\therefore$ Quadrilateral AMEP is a parallelogram,\\\\\n$\\therefore$ $ME=AP=x$,\\\\\n$\\because$ $CE=y=2x-2$,\\\\\nFrom $MC=CE+ME$, we get, $5=2x-2+x$,\\\\\nSolving yields $x=\\frac{7}{3}$, that is $AP=\\frac{7}{3}$,\\\\\n$\\therefore$ $CQ=2AP=2\\times \\frac{7}{3}=\\frac{14}{3}$,\\\\\n$\\therefore$ $BQ=BC-CQ=10-\\frac{14}{3}=\\boxed{\\frac{16}{3}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53498546_73.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$ and $\\triangle ADE$, $\\angle BAC = \\angle DAE = 90^\\circ$, $\\angle B = \\angle ADE = 30^\\circ$, AC and DE intersect at point F, connect CE, point D lies on side BC. $\\triangle ABD \\sim \\triangle ACE$; If $\\frac{AD}{BD} = \\sqrt{3}$, find the value of $\\frac{DF}{CF}$.\n",
    "completion": "\\textbf{Solution:} Given that $\\triangle ABD \\sim \\triangle ACE$,\\\\\n$\\therefore \\frac{AD}{AE}=\\frac{BD}{EC}$, which means $\\frac{AD}{BD}=\\frac{AE}{EC}=\\sqrt{3}$,\\\\\nIn $Rt\\triangle ADE$, $\\angle ADE=30^\\circ$,\\\\\n$\\therefore \\tan 30^\\circ =\\frac{AE}{AD}=\\frac{\\sqrt{3}}{3}$,\\\\\n$\\therefore \\frac{AD}{CE}=\\frac{AE}{EC} \\div \\frac{AE}{AD}=\\sqrt{3} \\times \\sqrt{3}=3$,\\\\\n$\\because \\triangle BAD \\sim \\triangle CAE$,\\\\\n$\\therefore \\angle B=\\angle ACE=\\angle ADE$,\\\\\n$\\because \\angle AFD=\\angle EFC$,\\\\\n$\\therefore \\triangle ADF \\sim \\triangle ECF$,\\\\\n$\\therefore \\frac{DF}{CF}=\\frac{AD}{EC}=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51353525_174.png"
  },
  {
    "solution": "\\boxed{\\frac{135}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in rectangle $ABCO$, point $C$ is on the x-axis, and point $A$ is on the y-axis, with the coordinates of point $B$ being $(-9, 12)$. The rectangle $ABCO$ is folded along line $BD$ such that point $A$ falls on point $E$ on the diagonal $OB$, and line $BD$ intersects $OA$ and the x-axis at points $D$ and $F$, respectively. What is the area of $\\triangle OBD$?\n",
    "completion": "\\textbf{Solution:} Let $OD=x$, \\\\\nsince quadrilateral $ABCO$ is a rectangle, \\\\\ntherefore $\\angle BAD=90^\\circ$. \\\\\nAccording to the problem, $\\triangle BAD\\cong \\triangle BED$, \\\\\ntherefore $BE=BA=9$, $AD=ED=12-x$, $\\angle BED=\\angle BAD=90^\\circ$, \\\\\nthus $\\angle OED=90^\\circ$, $EO=BO-BE=15-9=6$. \\\\\nIn $\\triangle DEO$, $OD^2=OE^2+DE^2$, $x^2=6^2+(12-x)^2$, \\\\\n$x=\\frac{15}{2}$, i.e., $OD=\\frac{15}{2}$, \\\\\ntherefore, the area of $\\triangle OBD=\\frac{1}{2}OD\\cdot AB=\\boxed{\\frac{135}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52056028_208.png"
  },
  {
    "solution": "\\boxed{\\frac{9}{16}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in parallelogram $ABCD$, point $E$ is on side $BC$. Connect $AE$ and extend it to intersect diagonal $BD$ at point $F$ and the extension of $DC$ at point $G$. If $\\frac{CE}{BE}=\\frac{2}{3}$, what is the value of $\\frac{FE}{EG}$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral ABCD is a parallelogram,\\\\\nthus AD=BC, and AD$\\parallel$BC,\\\\\ntherefore, $\\triangle ADF\\sim \\triangle EBF$, $\\triangle GEC\\sim \\triangle GAD$,\\\\\nthus, $\\frac{EF}{AF}=\\frac{BE}{AD}$, $\\frac{EG}{AG}=\\frac{EC}{AD}$,\\\\\ngiven that $\\frac{CE}{BE}=\\frac{2}{3}$,\\\\\nthus, $\\frac{BE}{AD}=\\frac{3}{5}$, $\\frac{CE}{AD}=\\frac{2}{5}$,\\\\\nthus, $\\frac{FE}{AF}=\\frac{3}{5}$, $\\frac{EG}{AG}=\\frac{2}{5}$,\\\\\nthus, $\\frac{FE}{AE}=\\frac{3}{8}$, $\\frac{EG}{AE}=\\frac{2}{3}$,\\\\\nthus, $\\frac{FE}{EG}=\\boxed{\\frac{9}{16}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52977604.png"
  },
  {
    "solution": "\\boxed{\\frac{6\\pi }{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC$, $\\angle BAC=54^\\circ$. The circle $\\odot O$ with diameter $AB$ intersects $AC$ and $BC$ at points $D$ and $E$, respectively. A tangent line to $\\odot O$ passing through point $B$ intersects the extension of $AC$ at point $F$. Given $AB=6$, what is the length of $\\overset{\\frown}{AD}$?\n",
    "completion": "\\textbf{Solution:} Draw line OD, \\\\\nsince $OA = OD$, $\\angle BAC = 54^\\circ$, \\\\\ntherefore $\\angle BOD = 72^\\circ$, $\\angle AOD = 72^\\circ$. \\\\\nAlso, since $AB = 6$, \\\\\ntherefore $OA = 3$. \\\\\nTherefore, $\\overset{\\frown}{AD} = \\frac{72\\pi \\times 3}{180} = \\boxed{\\frac{6\\pi }{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51318358_122.png"
  },
  {
    "solution": "\\boxed{\\left(-\\frac{3}{2}, 0\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the line $y=kx+b$ passes through point $A$ and point $B$. Find the coordinates of point $C$, the intersection of this line with the $x$-axis.\n",
    "completion": "\\textbf{Solution:} To solve this, we substitute points $A(-3, 3)$ and $B(0, -3)$ into $y=kx+b$ to get $\\begin{cases}-3k+b=3 \\\\ b=-3\\end{cases}$. Solving this system, we obtain $\\begin{cases}k=-2 \\\\ b=-3\\end{cases}$. Therefore, the equation of line $AB$ is $y=-2x-3$. When $y=0$, solving $-2x-3=0$ yields $x=-\\frac{3}{2}$. Therefore, the coordinates of point $C$ are $\\left(-\\frac{3}{2}, 0\\right)$. Hence, the intersection point $C$ of this line with the $x$-axis is $\\boxed{\\left(-\\frac{3}{2}, 0\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50676993.png"
  },
  {
    "solution": "\\boxed{(-10,0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle OABC, point A is on the x-axis, point C is on the y-axis, and the coordinates of point B are $(6, 8)$. The rectangle OABC is folded along the line BD, making point C exactly coincide with point E on the diagonal OB. The fold line BD intersects the y-axis and x-axis at points D and F, respectively. What are the coordinates of point F?\n",
    "completion": "\\textbf{Solution:} Let the coordinates of point D be $(0,a)$, so $OD=a$, $CD=8-a$,\\\\\nsince $BC=6$, $CD=DE=8-a$, $OB=10$,\\\\\nsince $\\frac{S_{\\triangle ODB}}{2}=\\frac{OD\\cdot BC}{2}=\\frac{OB\\cdot DE}{2}$,\\\\\nthus $\\frac{a\\times 6}{2}=\\frac{10(8-a)}{2}$,\\\\\nsolving, we get $a=5$,\\\\\nwhich means the coordinates of point D are $(0,5)$,\\\\\nLet the equation of the line containing the crease BD be $y=kx+b$,\\\\\nsince points D $(0,5)$ and B $(6,8)$ lie on line BD,\\\\\nthus $\\left\\{\\begin{array}{l}b=5 \\\\ 6k+b=8\\end{array}\\right.$, gives $\\left\\{\\begin{array}{l}k=0.5 \\\\ b=5\\end{array}\\right.$,\\\\\nhence, the equation of the line containing the crease BD is $y=0.5x+5$,\\\\\nwhen $y=0$, $0.5x+5=0$,\\\\\nsolving, we get $x=-10$,\\\\\nthus, the coordinates of point F are $\\boxed{(-10,0)}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52094893_171.png"
  },
  {
    "solution": "\\boxed{-3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the intersection of the line $y=-x+m$ with $y=nx+4n\\ (n\\ne 0)$ has an abscissa of $-2$. How many integer solutions are there for the inequality $-x+m>nx+4n>0$ with respect to $x$?\n",
    "completion": "\\textbf{Solution}: When $y=0$, for $y=nx+4n\\ (n\\ne 0)$, then $x=-4$. Therefore, the solution set for $nx+4n>0$ is $x>-4$. Since the $x$-coordinate of the intersection point between $y=-x+m$ and $y=nx+4n\\ (n\\ne 0)$ is $-2$, observing the graph shows that the solution set for $-x+m>nx+4n$ is $x<-2$. Hence, the solution set for $-x+m>nx+4n>0$ is $-4<x<-2$. Since $x$ is an integer, it follows that $x=\\boxed{-3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50560202.png"
  },
  {
    "solution": "\\boxed{48}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, behind the building \\(AB\\) there is a rockery \\(CD\\) with a slope of \\(i=1:2\\). At point \\(E\\) on the slope, there is a pavilion. It is measured that the horizontal distance from the bottom of the rockery to the building is \\(BC=24\\) meters, and the distance to the pavilion is \\(CE=8 \\sqrt{5}\\) meters. Xiao Li measures the angle of depression to \\(E\\) from the top of the building to be \\(45^\\circ\\). What is the height of the building \\(AB\\) in meters?\n",
    "completion": "\\textbf{Solution:} Construct $EH \\perp AB$ at point H. Since $AB \\perp BF$ and $EF \\perp BF$, it follows that quadrilateral BFEH is a rectangle; thus, $BH=EF=8$ and $HE=BF$. Given $BC=24$ and $CF=16$, we have $HE=BF=BC+CF=24+16=40$. In right triangle $AHE$, since $\\angle HAE=90^\\circ-45^\\circ=45^\\circ$, it follows that $AH=HE=40$. Therefore, $AB=AH+HB=40+8=\\boxed{48}$ meters.",
    "image_path": "dataset/mm_math/images/MM_Math/51457882_88.png"
  },
  {
    "solution": "\\boxed{x>2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: The graph of the function $y=kx+b$ is shown in the figure. What is the solution set of the inequality $kx+b<0$ with respect to $x$?\n",
    "completion": "\\textbf{Solution:} From the graph, we know that the function $y=kx+b$ intersects the x-axis at the point $(2,0)$. Therefore, the intersection point of the function $y=kx+b$ and the x-axis is $(2,0)$. Moreover, $y$ decreases as $x$ increases. \\\\\n$\\therefore$ When $x>2$, we have $kx+b<0$, \\\\\n$\\therefore$ The solution set of the inequality $kx+b<0$ with respect to $x$ is $x>2$. \\\\\nThus, the answer is: $\\boxed{x>2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52380922.png"
  },
  {
    "solution": "\\boxed{1 \\leq m \\leq 3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $l_{1}\\colon y=-x-b$ intersects the $x$-axis at $A(6, 0)$ and the $y$-axis at point $B$. Another line $l_{2}$ passing through point $B$ intersects the negative semi-axis of the $x$-axis at point $C$, and $OB:OC = 3:1$. If the point $P(m+1, m-1)$ is inside $\\triangle ABC$ (including the boundary), what is the range of $m$?\n",
    "completion": "\\textbf{Solution:} Therefore, the range of values for $m$ is $1 \\leq m \\leq 3$,\\\\\n$\\therefore$ the range of values for $m$ is $\\boxed{1 \\leq m \\leq 3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52354930_196.png"
  },
  {
    "solution": "\\boxed{36^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, lines AB and CD intersect at point O, with OA bisecting $\\angle EOC$. If $\\angle EOC = \\frac{2}{3}\\angle EOD$, find the degree measure of $\\angle BOD$.\n",
    "completion": "\\textbf{Solution}: Since $\\angle EOC = \\frac{2}{3}\\angle EOD$ and $\\angle EOC + \\angle EOD = 180^\\circ$,\\\\\nthus $\\frac{2}{3}\\angle EOD + \\angle EOD = 180^\\circ$,\\\\\nwhich implies $\\frac{5}{3}\\angle EOD = 180^\\circ$,\\\\\ntherefore $\\angle EOD = 108^\\circ$,\\\\\nthus $\\angle EOC = \\frac{2}{3} \\times 108^\\circ = 72^\\circ$,\\\\\nsince OA bisects $\\angle EOC$,\\\\\nit follows that $\\angle AOC = \\frac{1}{2}\\angle EOC = \\frac{1}{2} \\times 72^\\circ = 36^\\circ$,\\\\\nsince lines AB and CD intersect at point O,\\\\\nthus $\\angle BOD = \\angle AOC = \\boxed{36^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53223463_90.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system, the line $y=\\frac{4}{3}x+8$ intersects the x-axis and y-axis at points A and B, respectively. Point C is on the positive half of the x-axis, and $AB=AC$. Draw the line $BC$. What is the length of $AB$?\n",
    "completion": "\\textbf{Solution:} Given $y=\\frac{4}{3}x+8$,\\\\\nthus, when $x=0$, $y=8$,\\\\\ntherefore, $B(0, 8)$,\\\\\ntherefore, $OB=8$,\\\\\nwhen $y=0$, we have $\\frac{4}{3}x+8=0$, solving for $x$ gives $x=-6$,\\\\\ntherefore, $A(-6, 0)$,\\\\\ntherefore, $OA=6$,\\\\\ntherefore, $AB=\\sqrt{OA^2+OB^2}=\\boxed{10}$,",
    "image_path": "dataset/mm_math/images/MM_Math/53728859_91.png"
  },
  {
    "solution": "\\boxed{\\frac{8\\sqrt{10}}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, rectangle $ABCD$ is folded such that point $D$ falls on the trisection point $G$ of side $AB$, with $BG<AG$. The corresponding point of point $C$ after folding is $H$, and the fold line is $EF$. Connect $BH$. If $\\tan\\angle AEG = \\frac{4}{3}$ and $EF=3\\sqrt{5}$, what is the length of $BH$?\n",
    "completion": "\\textbf{Solution:}\n\nConstruct $MN\\perp AB$ extending to $N$ on line $AB$, and $FM\\perp AD$ intersecting $AD$ at $M$. Draw $DG$ intersecting $EF$ at $l$. Since quadrilateral $ABCD$ is a rectangle, it follows that $\\angle ADC=\\angle A=\\angle ABC=90^\\circ$, $AD\\parallel BC$, and $CD=AB$. Therefore, $\\angle AGE+\\angle AEG=90^\\circ$. By folding, we obtain $EG=DE$, $\\angle HGE=\\angle ADC=90^\\circ$. $EF$ bisects $DG$ perpendicularly, hence $\\angle HGN+\\angle AGE=90^\\circ$, $\\angle FED+\\angle ADG=90^\\circ$, $HG=CD$. Therefore, $\\angle AGE=\\angle HGN$. Given that $\\tan\\angle AEG=\\frac{4}{3}$, we can set $AG=4x$ and $AE=3x$, thus $DE=EG=5x$, $BG=2x$, $AD=8x$. $\\tan\\angle ADG=\\frac{1}{2}$. Since $FM\\perp AD$, it follows that $\\angle FED+\\angle EFM=90^\\circ$, $FM=AB=AG+BG=6x$. Therefore, $\\angle ADG=\\angle EFM$, that is, $\\tan\\angle EFM=\\tan\\angle ADG=\\frac{1}{2}$. Therefore, $ME=3k$, hence $ME^{2}+FM^{2}=EF^{2}$, which implies $(3x)^{2}+(6x)^{2}=(3\\sqrt{5})^{2}$. Solving gives $k=1$ or $-1$ (discard), therefore, $BG=2$, $HG=CD=AB=6$. Given that $\\tan\\angle HGN=\\frac{4}{3}$, we can let $HN=4a$ and $NG=3a$. Hence, $HN^{2}+NG^{2}=HG^{2}$, which implies $(4a)^{2}+(3a)^{2}=6^{2}$. Solving gives $a=\\frac{6}{5}$, hence $HN=\\frac{24}{5}$, $NG=\\frac{18}{5}$. Therefore, $NB=NG-BF=\\frac{8}{5}$, therefore $BH=\\sqrt{HN^{2}+BN^{2}}=\\boxed{\\frac{8\\sqrt{10}}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55533862.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C = 90^\\circ$, points D and E are located on sides AC and AB respectively, BD bisects $\\angle ABC$, $DE \\perp AB$, $AE = 6$, $\\cos A = \\frac{3}{5}$. Find the value of $\\tan\\angle DBC$?\n",
    "completion": "\\textbf{Solution:} Given that $AD=10$ and $DC=8$, we have $AC=AD+DC=18$.\\\\\nIn triangles $\\triangle ADE$ and $\\triangle ABC$, $\\angle A=\\angle A$, and $\\angle AED=\\angle ACB$,\\\\\ntherefore $\\triangle ADE \\sim \\triangle ABC$. It follows that $\\frac{DE}{BC}=\\frac{AE}{AC}$, that is $\\frac{8}{BC}=\\frac{6}{18}$, hence $BC=24$,\\\\\nTherefore, $\\tan\\angle DBC=\\frac{CD}{BC}=\\frac{8}{24}=\\boxed{\\frac{1}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52537023_112.png"
  },
  {
    "solution": "\\boxed{30}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, $ \\angle AOB=40^\\circ$, $ OB$ is the bisector of $ \\angle AOC$, and $ OD$ is the bisector of $ \\angle COE$. If $ \\angle AOD$ and $ \\angle BOD$ are complementary angles, find the degree measure of $ \\angle COE$?\n",
    "completion": "\\textbf{Solution:} \n\nGiven that $OB$ is the bisector of $\\angle AOC$, and $OD$ is the bisector of $\\angle COE$,\\\\\n$\\therefore \\angle BOC=\\angle AOB=40^\\circ$,\\\\\nLet $\\angle COD=\\angle DOE=x^\\circ$,\\\\\n$\\therefore \\angle BOD=\\angle BOC+\\angle COD=(40+x)^\\circ$, $\\angle AOD=\\angle AOB+\\angle BOC+\\angle COD=(80+x)^\\circ$,\\\\\nSince $\\angle AOD$ and $\\angle BOD$ are complementary,\\\\\n$\\therefore \\angle AOD+\\angle BOD=(40+x)^\\circ+(80+x)^\\circ=180^\\circ$,\\\\\n$\\therefore x=\\boxed{30}$,\\\\\n$\\therefore \\angle COD=\\angle DOE=30^\\circ$,\\\\\n$\\therefore \\angle COE=2\\angle COD=60^\\circ$.",
    "image_path": "dataset/mm_math/images/MM_Math/51457701_188.png"
  },
  {
    "solution": "\\boxed{9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given: As shown in the diagram, the graph of the directly proportional function $y=2x$ and the graph of the linear function $y=-\\frac{2}{3}x+4$ intersect at point $A\\left(\\frac{3}{2}, 3\\right)$, and the graph of the linear function $y=-\\frac{2}{3}x+4$ intersects the $x$-axis at point $B$. Find the area of $\\triangle AOB$.\n",
    "completion": "\\textbf{Solution:} As shown in Figure 1 \\\\\n$\\because A\\left(\\frac{3}{2}, 3\\right)$, $B(6, 0)$ \\\\\n$\\therefore OB = 6$, the height on side OB is 3, \\\\\n$\\therefore S_{\\triangle AOB} = \\frac{1}{2} \\times 6 \\times 3 = \\boxed{9}$",
    "image_path": "dataset/mm_math/images/MM_Math/51545374_103.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB \\parallel CD \\parallel EF$, $AF$ intersects $BE$ at point $G$, and $AG=2$, $GD=1$, $DF=5$. What is the value of $\\frac{BC}{CE}$?\n",
    "completion": "\\textbf{Solution}: We have $AD=AG+GD=3$, \\\\\nsince $AB\\parallel CD\\parallel EF$, \\\\\nthus $\\frac{BC}{CE}=\\frac{AD}{DF}=\\boxed{\\frac{3}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51285012.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is a diameter of the circle $O$ with $AB=4$ and $CD=2\\sqrt{2}$. The length of the minor arc $BC$ is twice that of the minor arc $BD$. What is the length of $AC$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect $OC$, $OD$, and $BC$,\n\n$\\because AB=4$\n\n$\\therefore OC=OD=2$\n\n$\\because OC^2+OD^2=8$, $CD^2=8$\n\n$\\therefore OC^2+OD^2=CD^2$\n\n$\\therefore \\triangle OCD$ is a right-angled triangle, and $\\angle COD=90^\\circ$\n\n$\\therefore \\overset{\\frown}{CB}=2\\overset{\\frown}{DB}$\n\n$\\therefore \\overset{\\frown}{BC}=\\frac{2}{3}\\overset{\\frown}{CD}$\n\n$\\therefore \\angle BOC=\\frac{2}{3}\\times \\angle COD=60^\\circ$\n\n$\\because OC=OB$\n\n$\\therefore \\triangle OBC$ is an equilateral triangle\n\n$\\therefore BC=OC=2$\n\n$\\because AB$ is the diameter, $AB=4$\n\n$\\therefore \\angle ACB=90^\\circ$\n\n$\\therefore AC=\\sqrt{3}BC=2\\sqrt{3}$\n\nThus, the answer is: $AC=\\boxed{2\\sqrt{3}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51350810.png"
  },
  {
    "solution": "\\boxed{(1,0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system, the line $y=2x+10$ intersects with the $x$ axis at point $A$ and with the $y$ axis at point $B$. Another line passing through point $B$ intersects the positive $x$ axis at point $C$, and the area of $\\triangle ABC$ is $60$. When the area of $\\triangle ABM$ is $20$, point $E$ is a moving point on line $AM$. Does there exist a point $D$ on the $x$ axis such that the quadrilateral formed by points $D$, $E$, $B$, and $C$ is a parallelogram? If it exists, write down the coordinates of point $D$ directly; if not, please explain the reason.\n",
    "completion": "\\textbf{Solution:} There exist coordinates for point $D$ as $(13,0)$, $(-23,0)$, or $(1,0)$, for the following reasons:\\\\\n$\\because S_{\\triangle ACM}=S_{\\triangle ABC}-S_{\\triangle ABM}=60-20=40$,\\\\\n$\\therefore \\frac{1}{2}AC \\cdot y_M=40$,\\\\\n$\\because AC=12$,\\\\\n$\\therefore y_M=\\frac{20}{3}$,\\\\\n$\\therefore \\frac{20}{3}=-\\frac{10}{7}m+10$,\\\\\nSolving this yields: $m=\\frac{7}{3}$,\\\\\n$\\therefore M\\left(\\frac{7}{3}, \\frac{10}{3}\\right)$,\\\\\nLet the equation of line AM be $y=kx+b$, then $\\left\\{\\begin{array}{l}\\frac{7}{3}m+n=\\frac{20}{3}\\\\ -5m+n=0\\end{array}\\right.$,\\\\\n$\\therefore \\left\\{\\begin{array}{l}m=\\frac{10}{11}\\\\ n=\\frac{50}{11}\\end{array}\\right.$,\\\\\n$\\therefore$ The equation of line AM is $y=\\frac{10}{11}x+\\frac{50}{11}$,\\\\\n(1) When BC is a side of a parallelogram, and quadrilateral BCDE forms a parallelogram,\\\\\n$\\because BE\\parallel CD$, $B(0,10)$,\\\\\n$\\therefore y_E=10$,\\\\\n$\\therefore 10=\\frac{10}{11}x+\\frac{50}{11}$,\\\\\nSolving for $x$ gives $x=6$,\\\\\n$\\therefore E(6,10)$,\\\\\n$\\because$ quadrilateral BCDE forms a parallelogram,\\\\\n$\\therefore CD=BE=6$,\\\\\n$\\therefore OD=OC+CD=7+6=13$,\\\\\n$\\therefore \\boxed{D(13,0)}$;\\\\\n(2) When BC is a side of a parallelogram, and quadrilateral BDEC forms a parallelogram, by drawing EF$\\perp$CD at point F,\\\\\n$\\because$ quadrilateral BDCE forms a parallelogram,\\\\\n$\\therefore BC=DE$, $BC\\parallel DE$,\\\\\n$\\therefore \\angle EDF=\\angle BCO$,\\\\\nAnd $\\because \\angle EFD=\\angle BOC$,\\\\\n$\\therefore \\triangle BDC\\cong \\triangle ECD$ (AAS),\\\\\n$\\therefore EF=OB=10$, $DF=OC=7$,\\\\\n$\\therefore y_E=-10$,\\\\\n$\\therefore -10=\\frac{10}{11}x+\\frac{50}{11}$,\\\\\nSolving for $x$ gives $x=-16$,\\\\\n$\\therefore E(-16,10)$,\\\\\n$\\therefore OF=16$,\\\\\n$\\therefore OD=OF+FD=16+7=23$,\\\\\n$\\therefore \\boxed{D(-23,0)}$;\\\\\n(3) When BC is a diagonal of a parallelogram, and quadrilateral BDEC forms a parallelogram,\\\\\n$\\because BE\\parallel CD$, $B(0,10)$,\\\\\n$\\therefore y_E=10$,\\\\\n$\\therefore 10=\\frac{10}{11}x+\\frac{50}{11}$,\\\\\nSolving for $x$ gives $x=6$,\\\\\n$\\therefore E(6,10)$,\\\\\n$\\because$ quadrilateral BCDE forms a parallelogram,\\\\\n$\\therefore CD=BE=6$,\\\\\n$\\because OC=7$,\\\\\n$\\therefore OD=OC-CD=1$,\\\\\n$\\therefore \\boxed{D(1,0)}$;\\\\\nIn summary, the coordinates of point $D$ are: \\boxed{(13,0)}, \\boxed{(-23,0)}, or \\boxed{(1,0)}.",
    "image_path": "dataset/mm_math/images/MM_Math/52205244_125.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, $AD$ is the altitude of $\\triangle ABC$, $CE$ is the angle bisector of $\\angle ACB$ in $\\triangle ACB$, and $F$ is the midpoint of $AC$. Given $\\angle ACB=50^\\circ$ and $\\angle BAD=65^\\circ$, if the difference in the perimeters of $\\triangle BCF$ and $\\triangle BAF$ is 3, with $AB=7$ and $AC=4$, then what is the length of $BC$?\n",
    "completion": "\\textbf{Solution:} From the given, we have $BC=\\boxed{10}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52815585_122.png"
  },
  {
    "solution": "\\boxed{\\frac{5}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, given rectangle ABCD, point E is on the side AB, and $BE=2$, $BC=3$. Fold $\\triangle CBE$ along the line CE so that point B falls on point G, and extend EG to intersect CD at point F. What is the length of the segment FG?\n",
    "completion": "\\textbf{Solution:} Given that when $\\triangle CBE$ is folded along line CE such that point B falls on point G, \\\\\nit follows that $\\angle BEC = \\angle GEC$, $GE = BE = 2$, and $CG = BC = 3$, \\\\\nsince quadrilateral ABCD is a rectangle, \\\\\nit implies $CD \\parallel AB$, \\\\\nthus $\\angle BEC = \\angle FCE$, \\\\\nleading to $\\angle GEC = \\angle FCE$, \\\\\ntherefore, $CF = EF$, \\\\\nlet $FG = x$, then $CF = EF = x + 2$, \\\\\nin $\\triangle CFG$, we have $FG^2 + CG^2 = CF^2$, \\\\\nwhich gives $x^2 + 3^2 = (x + 2)^2$, \\\\\nsolving for $x$ yields $x = \\boxed{\\frac{5}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52873810.png"
  },
  {
    "solution": "\\boxed{y = -\\frac{16}{x}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=-\\frac{3}{2}x-2$ intersects the $x$-axis and $y$-axis at points $A$ and $B$ respectively, and intersects with the hyperbola $y=\\frac{m}{x}$ (where $m \\neq 0$) at point $C$ in the second quadrant. $CD$ is perpendicular to the $y$-axis at point $D$, and $CD=4$. What is the equation of the hyperbola?\n",
    "completion": "\\textbf{Solution:} Given that the point of intersection in the second quadrant is C, and CD is perpendicular to the y-axis at point D, with CD equal to 4,\\\\\nthus the x-coordinate of point C is $-4$,\\\\\nsubstituting $x = -4$ into $y = -\\frac{3}{2}x - 2$, we get $y = -\\frac{3}{2} \\times (-4) - 2 = 4$,\\\\\ntherefore, C$(-4,4)$,\\\\\nsubstituting C$(-4,4)$ into $y = \\frac{m}{x}$, we get $4 = \\frac{m}{-4}$,\\\\\nhence, $m = -16$,\\\\\ntherefore, the equation of the hyperbola is: \\boxed{y = -\\frac{16}{x}}.",
    "image_path": "dataset/mm_math/images/MM_Math/52189644_140.png"
  },
  {
    "solution": "\\boxed{54^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the lines AB and CD intersect at O, and the ray OE divides $\\angle BOD$ into two angles. Given that $\\angle BOE = \\frac{1}{3} \\angle AOC$ and $\\angle EOD = 36^\\circ$, what is the measure of $\\angle AOC$ in degrees?\n",
    "completion": "\\textbf{Solution:} Because $\\angle AOC$ and $\\angle BOD$ are vertically opposite angles, \\\\\ntherefore $\\angle BOD = \\angle AOC$, \\\\\nbecause $\\angle BOE = \\frac{1}{3} \\angle AOC$, $\\angle EOD = 36^\\circ$, \\\\\ntherefore $\\angle EOD = 2\\angle BOE = 36^\\circ$, \\\\\ntherefore $\\angle EOD = 18^\\circ$, \\\\\ntherefore $\\angle AOC = \\angle BOE + \\angle EOD = 18^\\circ + 36^\\circ = \\boxed{54^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/8928624.png"
  },
  {
    "solution": "\\boxed{40^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that point $O$ is a point on line $AB$. A right-angled triangle ruler $MON$ is placed as shown, with the right-angled vertex at point $O$. Ray $OC$ is drawn inside $\\angle MON$, and $OC$ exactly bisects $\\angle MOB$. If $\\angle CON=20^\\circ$, what is the degree measure of $\\angle AOM$?\n",
    "completion": "\\textbf{Solution:} Given, \\\\\n$\\because \\angle MON=90^\\circ$ and $\\angle CON=20^\\circ$, \\\\\n$\\therefore \\angle MOC=90^\\circ-\\angle CON=70^\\circ$, \\\\\n$\\because OC$ bisects $\\angle MOB$, \\\\\n$\\therefore \\angle BOM=2\\angle MOC=140^\\circ$, \\\\\n$\\therefore \\angle AOM=180^\\circ-\\angle BOM=\\boxed{40^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52916218_156.png"
  },
  {
    "solution": "\\boxed{\\frac{15}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In right triangle $ABC$, where $AC=8$, $BC=6$, and $\\angle C=90^\\circ$, the triangle $ABC$ is folded such that point $A$ coincides with point $B$, with $DE$ being the crease. What is the length of $DE$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle C=90^\\circ$, AC=8, and BC=6,\\\\\nthen AB=$\\sqrt{AC^{2}+BC^{2}}$=10,\\\\\nSince folding $\\triangle ABC$ as shown in the figure, making point A coincide with point B, where the crease is DE,\\\\\nthen BE=AE, AD=BD=5, DE$\\perp$AB,\\\\\nIn $\\triangle BEC$, BE$^{2}$=BC$^{2}$+CE$^{2}$,\\\\\nthen BE$^{2}$=36+(8-BE)$^{2}$,\\\\\nthus BE=$\\frac{25}{4}$,\\\\\nIn $\\triangle BDE$, DE=$\\sqrt{BE^{2}-BD^{2}}$=$\\boxed{\\frac{15}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52807502.png"
  },
  {
    "solution": "\\boxed{(-2,-1)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $l_{1}\\colon y=-x-3$ intersects with the line $l_{2}$ passing through point $A(0,3)$ at point $C(m,1)$, and intersects with the y-axis at point $B$. Point $P$ is on the line $l_{1}$, $PQ\\parallel x$-axis, and intersects line $l_{2}$ at point $Q$. If $PQ=AB$, find the coordinates of point $P$.\n",
    "completion": "\\textbf{Solution:} Solve: In the line $l_{1}\\colon y=-x-3$, let $x=0$, we get $y=-3$,\\\\\n$\\therefore B(0,-3)$,\\\\\n$\\therefore AB=3-(-3)=6$,\\\\\nLet $P(-b-3,b)$, since $PQ\\parallel x$ axis, we get $Q(2b-6,b)$,\\\\\n$PQ=\\left|-b-3-\\left(2b-6\\right)\\right|=AB=6$,\\\\\nSolving, we get $b=3$ or $b=-1$,\\\\\n$\\therefore P=\\boxed{(-6,3)}$ or \\boxed{(-2,-1)}.",
    "image_path": "dataset/mm_math/images/MM_Math/50682805_119.png"
  },
  {
    "solution": "\\boxed{\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the quadrilateral $ABCD$ is inscribed in circle $\\odot O$, with $OC=2$, and $AC=2\\sqrt{2}$. The quadrilateral $AOCD$ is a rhombus. Find the distance from point $O$ to $AC$.\n",
    "completion": "\\textbf{Solution:} Connect OA and construct OH $\\perp$ AC at H,\\\\\nOA$^{2}$+OC$^{2}=8$, AC$^{2}=8$,\\\\\n$\\therefore$ OA$^{2}$+OC$^{2}$=AC$^{2}$,\\\\\n$\\therefore$ $\\triangle AOC$ is an isosceles right triangle, $\\angle AOC=90^\\circ$,\\\\\n$\\therefore$ OH=$\\frac{1}{2}$AC=$\\sqrt{2}$, that is, the distance from point O to AC is $\\boxed{\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53078807_105.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $AB=8$, point $P$ is a moving point on line segment $AB$, a square $APCD$ is constructed with $AP$ as a side, and an isosceles triangle $\\triangle PBE$ is constructed with $PB$ as the base. Connecting $CE$, what is the maximum area of $\\triangle PCE$?\n",
    "completion": "\\textbf{Solution:} Consider a line EH$\\perp$AB at point H,\\\\\nLet AP=x, then PB=8−x,\\\\\n$\\because$ $\\triangle PBE$ is an isosceles triangle with BP as its base,\\\\\n$\\therefore$ PH=$\\frac{1}{2}$(8−x)=4−$\\frac{1}{2}$x,\\\\\n$\\because$ square APCD,\\\\\n$\\therefore$AP=PC=x,\\\\\n$\\therefore$ ${S}_{\\triangle PCE}=\\frac{1}{2}PC \\cdot PH=\\frac{1}{2}x(4−\\frac{1}{2}x)=-\\frac{1}{4}(x−4)^2+4$,\\\\\n$\\because$ the parabola opens downward,\\\\\n$\\therefore$ when x=4, the maximum value of $s$ is $\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52755991.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: According to the figure, it is a schematic diagram of the pinhole imaging principle in physics. Given that the object $AB=30$, based on the dimensions in the figure $(AB\\parallel CD)$, what should be the length of $CD$?\n",
    "completion": "Solution: According to the problem, $\\triangle ODC \\sim \\triangle OAB$\\\\\n$\\therefore \\frac{AB}{CD} = \\frac{36}{12}$\\\\\n$\\because AB = 30$\\\\\n$\\therefore CD = \\frac{1}{3} \\times 30 = \\boxed{10}$",
    "image_path": "dataset/mm_math/images/MM_Math/53297427.png"
  },
  {
    "solution": "\\boxed{\\frac{8}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the edge OA of $\\triangle OABC$ is on the positive half of the x-axis, point D is on edge AB, and the extension of segment CD intersects the x-axis at point E. Lines OB and OD are connected. The inverse proportion function $y= \\frac{k}{x}$ ($k>0$) passes through points B and D. If the area of $\\triangle OBC$ is 4 and the area of $\\triangle BDE$ is 2, what is the value of $k$?\n",
    "completion": "\\textbf{Solution:} In parallelogram $ABCD$, $CB \\parallel OA$, therefore $S_{\\triangle OBC} = S_{\\triangle EBC} = 4$, because $S_{\\triangle BDE} = 2$, therefore $S_{\\triangle BDC} = S_{\\triangle BDE} = 2$ therefore $CD = DE$. Because $\\triangle CBD \\sim \\triangle EAD$, therefore $\\frac{CD}{DE} = \\frac{BD}{DA}$ therefore $BD = DA$, $D$ is the midpoint of $AB$. Draw $BM \\perp OA$, $DN \\perp OA$ then $MN = AN$, $DN = \\frac{1}{2}BM$ because the inverse proportion function $y = \\frac{k}{x}$ (k>0) passes through points $B$, $D$ therefore $S_{\\triangle BOM} = S_{\\triangle DON}$ therefore $OM \\cdot BM = ON \\cdot DN$ therefore $\\frac{OM}{ON} = \\frac{DN}{BM} = \\frac{1}{2}$. Let $OM = t$, then $ON = 2t$, $MN = t = AN$, $OA = 3t$ therefore $OM = \\frac{1}{3}OA$ therefore $S_{\\triangle BOM} = \\frac{1}{3}S_{\\triangle BOA} = \\frac{1}{3}S_{\\triangle BOC} = \\frac{4}{3}$. Because the inverse proportion function $y = \\frac{k}{x}$ (k>0) passes through point $D$ therefore $S_{\\triangle DON} = \\frac{1}{2}k = \\frac{4}{3}$ therefore $k = \\boxed{\\frac{8}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55452474.png"
  },
  {
    "solution": "\\boxed{-\\frac{1}{2}x^2+4x-5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the vertex of the parabola $y=ax^2+bx+c$ is $A(4, 3)$, and it intersects the y-axis at point $B(0, -5)$. The axis of symmetry is line $l$, and point $M$ is the midpoint of segment $AB$. What is the equation of the parabola?\n",
    "completion": "\\textbf{Solution:} Let the function be expressed as: $y=a(x-4)^2+3$, \\\\\nSubstituting the coordinates of point B into the equation and solving, we obtain: $a=-\\frac{1}{2}$, \\\\\nTherefore, the equation of the parabola is: $y=-\\frac{1}{2}x^2+4x-5=\\boxed{-\\frac{1}{2}x^2+4x-5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53042870_132.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the distance from the midpoint \\(M\\) of one side of the rhombus \\(ABCD\\) to the intersection point \\(O\\) of the diagonals is \\(1.5\\, \\text{cm}\\). What is the perimeter of the rhombus \\(ABCD\\) in centimeters?\n",
    "completion": "\\textbf{Solution:} Let point $M$ be the midpoint of side $AB$ of rhombus $ABCD$, and let point $O$ be the midpoint of diagonal $BD$, with $OM=1.5\\,cm$, \\\\\n$\\therefore$ in $\\triangle ABD$, we have $OM=\\frac{1}{2}AD=1.5$, \\\\\n$\\therefore$ $AB=BC=CD=DA=2OM=3$, \\\\\nthus, the perimeter of the rhombus $ABCD$ is $\\boxed{12}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53729102.png"
  },
  {
    "solution": "\\boxed{4.5cm}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $D$ is the midpoint of segment $AC$, and $E$ is the midpoint of segment $AB$. Given that $AD=6\\,cm$ and $BC=3\\,cm$, find the lengths of segment $AB$ and $EC$ in centimeters.\n",
    "completion": "\\textbf{Solution:} Since $D$ is the midpoint of segment $AC$, and $E$ is the midpoint of segment $AB$\\\\\nThus, $AD=CD=\\frac{1}{2}AC$, $AE=EB=\\frac{1}{2}AB$\\\\\nSince $AD=6cm$, $BC=3cm$\\\\\nThus, $AB = AC + CB = 2AD + BC = 12 + 3 = \\boxed{15cm}$\\\\\n$EC = EB - BC = \\frac{1}{2}AB - BC = \\frac{15}{2} - 3 = \\frac{9}{2} = \\boxed{4.5cm}$",
    "image_path": "dataset/mm_math/images/MM_Math/53404045.png"
  },
  {
    "solution": "\\boxed{P\\left(\\frac{2}{3}, \\frac{7}{3}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y=k_1x+b$ intersects with the graph of the hyperbolic function $y=\\frac{k_2}{x}$ at points $A$ and $B$, where the coordinates of point $A$ are $(-1, 4)$, and the coordinates of point $B$ are $(4, n)$. Point $P$ is on segment $AB$, and $\\frac{S_{\\triangle AOP}}{S_{\\triangle BOP}}=1\\colon 2$. Find the coordinates of point $P$.\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, \\\\\n$\\because S_{\\triangle AOP} : S_{\\triangle BOP} = 1 : 2$, \\\\\n$\\therefore S_{\\triangle AOP} = \\frac{1}{3} \\times 7.5 = 2.5$, $S_{\\triangle BOP} = 5$, \\\\\nMoreover, $S_{\\triangle AOC} = \\frac{1}{2} \\times 3 \\times 1 = 1.5$, \\\\\n$\\therefore$ point $P$ is in the first quadrant, \\\\\n$\\therefore S_{\\triangle COP} = 2.5 - 1.5 = 1$, \\\\\nMoreover, $OC = 3$, \\\\\n$\\therefore \\frac{1}{2} \\times 3 \\times x_{P} = 1$, thus solving gives $x_{P} = \\frac{2}{3}$, \\\\\nSubstituting $x_{P} = \\frac{2}{3}$ into $y = -x + 3$, we get $y_{P} = \\frac{7}{3}$, \\\\\n$\\therefore \\boxed{P\\left(\\frac{2}{3}, \\frac{7}{3}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51446477_241.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point A is an arbitrary point on the graph of the inverse proportion function $y= \\frac{2}{x}$ ($x>0$). Line segment $AB$ is parallel to the $x$-axis and intersects the graph of the inverse proportion function $y= -\\frac{3}{x}$ at point B. A square $\\square ABCD$ is formed with $AB$ as a side, where points $C$ and $D$ lie on the $x$-axis. What is the area $S_{\\square ABCD}$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $AF \\perp x$-axis at point $F$ and $BE \\perp x$-axis at point $E$ from points $A$ and $B$, respectively.\n\nSince $\\square ABCD$, \n\nit follows that $AB \\parallel x$-axis,\n\nhence quadrilateral $ABEF$ is a rectangle,\n\nthus $S_{\\square ABCD} = S_{\\text{rectangle } ABEF}$,\n\nsince $A$ and $B$ are located on the graphs of inverse proportion functions $y=\\frac{2}{x}$ and $y=-\\frac{3}{x}$, respectively,\n\nit follows that $S_{\\text{rectangle } ABEF} = 2 + 3 = 5$,\n\ntherefore, $S_{\\square ABCD} = \\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52205057.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the rectangle $ABCD$, where $AB=4$ and $AD=8$, points $E$ and $F$ are located on sides $AD$ and $BC$ respectively, with $AE=3$. Perform the following steps: First, fold along line $EF$ such that the corresponding point of $A$ falls exactly on the diagonal $AC$, and the corresponding point of $B$ is $B'$. What is the value of $\\tan\\angle AEF$? Second, take points $M$ on $EF$ and $N$ on $A'B'$, and continue folding along line $MN$ so that point $F$ coincides with point $E$. What is the length of segment $MN$?\n",
    "completion": "\\textbf{Solution:} Let point $O$ be the intersection of $AC$ and $EF$,\\\\\nsince quadrilateral $ABCD$ is a rectangle,\\\\\ntherefore $CD=AB=4$,\\\\\nsince along the line $EF$ when folded, the corresponding point of $A$ exactly falls on the diagonal $AC$, and the corresponding point of $B$ is $B$,\\\\\ntherefore $\\angle AOE=\\angle AOE=90^\\circ$,\\\\\ntherefore $\\angle AOE=\\angle ADC=90^\\circ$,\\\\\nsince $\\angle EAO=\\angle CAD$,\\\\\ntherefore $\\triangle AOE\\sim \\triangle ADC$,\\\\\ntherefore $\\angle AEO=\\angle ACD$,\\\\\ntherefore $\\tan\\angle AEF=\\tan\\angle ACD=\\frac{AD}{CD}=\\frac{8}{4}=\\boxed{2}$,\\\\\nConnect NE, NF, and draw $FT\\perp AD$ at point $T$ through point $F$,\\\\\nsince $\\triangle AOE\\sim \\triangle ADC$,\\\\\ntherefore $\\angle FET=\\angle ACD$,\\\\\nsince $\\angle FTE=\\angle ADC=90^\\circ$,\\\\\ntherefore $\\triangle FTE\\sim \\triangle ADC$,\\\\\ntherefore $\\frac{FT}{AD}=\\frac{ET}{CD}$, that is, $\\frac{4}{8}=\\frac{ET}{4}$,\\\\\ntherefore $ET=2$,\\\\\ntherefore $BF=AT=AE-ET=3-2=1$,\\\\\nBased on the folding property: $BF=BF=1$, $MN\\perp EF$, $NF=NE$,\\\\\nLet $BN=m$,\\\\\nthen $NF^2=1^2+m^2=NE^2=3^2+(4-m)^2$,\\\\\nSolving we get: $m=3$,\\\\\ntherefore $NF=\\sqrt{10}$,\\\\\nsince $EF=\\sqrt{2^2+4^2}=2\\sqrt{5}$,\\\\\ntherefore $MF=\\sqrt{5}$,\\\\\ntherefore $MN=\\sqrt{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51402157.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AD$ is the angle bisector of $\\triangle ABC$, $DE \\perp AB$ at point E, the area of $\\triangle ABC$ is 9, $DE=2$, and $AB=5$. What is the length of $AC$?\n",
    "completion": "Solution: As shown in the figure, construct $DF \\perp AC$,\\\\\nsince $DE \\perp AB$ and $AD$ is the angle bisector of $\\triangle ABC$,\\\\\ntherefore $DE = DF = 2$.\\\\\nSince $S_{\\triangle ABC} = S_{\\triangle ABD} + S_{\\triangle ACD}$, $AB = 5$, $S_{\\triangle ABC} = 9$,\\\\\ntherefore $S_{\\triangle ABC} = \\frac{1}{2}AB \\times DE + \\frac{1}{2}AB \\times DF$\\\\\nthat is $9 = \\frac{1}{2} \\times 2 \\times (5 + AC)$\\\\\nSolving gives $AC = \\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51367757.png"
  },
  {
    "solution": "\\boxed{90^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, there are two slides of equal length. The height \\(AC\\) of the left slide is equal to the horizontal length \\(DF\\) of the right slide. What is the sum of the degrees of the incline angles \\(\\angle ABC\\) and \\(\\angle DFE\\)?\n",
    "completion": "\\textbf{Solution:} Given $\\angle ABC + \\angle DFE = 90^\\circ$, the reasoning is as follows:\\\\\nFrom the problem, we know that both $\\triangle ABC$ and $\\triangle DEF$ are right-angled triangles, with $BC = EF$ and $AC = DF$.\\\\\nIn $\\triangle ABC$ and $\\triangle DEF$,\\\\\n$\\left\\{\\begin{array}{l}\nBC = EF\\\\\nAC = DF\n\\end{array}\\right.$,\\\\\n$\\therefore \\triangle ABC \\cong \\triangle DEF$ (by HL criterion),\\\\\n$\\therefore \\angle ABC = \\angle DEF$,\\\\\n$\\because \\angle DEF + \\angle DFE = 90^\\circ$,\\\\\n$\\therefore \\angle ABC + \\angle DFE = \\boxed{90^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52551631.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $ \\angle AOB=30^\\circ$, P is a point on the bisector of $\\angle AOB$, $PM \\perp OB$ at point M, $PN \\parallel OB$ intersects OA at point N, if PM=1, what is the length of PN?\n",
    "completion": "\\textbf{Solution:} Draw $PD\\perp AO$ at point D, \\\\\nsince P lies on the angle bisector of $\\angle AOB$ and $PM\\perp OB$ at point M, \\\\\ntherefore $\\angle AOP=\\angle BOP=15^\\circ$, and $PD=PM=1$, \\\\\nsince $PN\\parallel OB$, \\\\\ntherefore $\\angle NPO=\\angle BOP=15^\\circ$, \\\\\nthus $\\angle DNP=30^\\circ$, \\\\\nhence $PN=2PD=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52655235.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, it is known that $AD \\parallel BE \\parallel CF$, and they intersect lines $l_{1}$ and $l_{2}$ at points $A$, $B$, $C$ and $D$, $E$, $F$ respectively. If $AB = 6$, $BC = 8$, and $DF = 7$, what is the length of $EF$?\n",
    "completion": "\\textbf{Solution:} Since $AD \\parallel BE \\parallel CF$, \\\\\nit follows that $\\frac{AB}{BC} = \\frac{DE}{EF} = \\frac{3}{4}$, \\\\\ngiven that $DE=7$, \\\\\nthus $EF=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52861195_137.png"
  },
  {
    "solution": "\\boxed{5x+30}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the concentration level of students during class can be represented by an attention index. The variation of the attention index $y$ of a class of students during a math class with time $x$ (in minutes) is shown in the figure. At the beginning of the class, the attention index is $30$, and at the 2nd minute, the attention index is $40$. Within the first $10$ minutes, the attention index $y$ is a linear function of time $x$. After $10$ minutes, the attention index $y$ becomes an inverse proportional function of $x$. When $0\\leq x\\leq 10$, what is the functional relationship between $y$ and $x$?\n",
    "completion": "\\textbf{Solution:} Solve for when $0 \\leq x \\leq 10$, let $y=kx+b$.\\\\\nSubstituting the points $(0,30)$ and $(2,40)$ into the equation yields:\\\\\n$\\left\\{\\begin{array}{l}\nb=30\\\\\n2k+b=40\n\\end{array}\\right.$\\\\\nSolving this system, we find: $k=5$ and $b=30$,\\\\\nTherefore, $y=5x+30=\\boxed{5x+30}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52536612_148.png"
  },
  {
    "solution": "\\boxed{\\left(-\\frac{10}{3},\\frac{20}{3}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCO$, point $C$ is on the x-axis, point $A$ is on the y-axis, and the coordinates of point $B$ are ($-6$, $8$). Rectangle $ABCO$ is folded along line $BD$, such that point $A$ lands on point $E$ located on diagonal $OB$. Line $BD$ intersects $OA$ and the x-axis at points $D$ and $F$, respectively. Point $M$ is a moving point on line $BD$. Through point $M$, construct $MN\\perp x$ axis, with $N$ being the foot of the perpendicular. During the movement of point $M$, is there a scalene triangle with vertices $N$, $E$, $O$? Write down the coordinates of point $N$ and find the coordinates of point $M$.\n",
    "completion": "\\textbf{Solution:} Exist;\\\\\n(1) The coordinates of point N are $(4,0)$ or $(-4,0)$, corresponding to the coordinates of point M being $(4,3)$ or $(-4,7)$;\\\\\n(2) $\\therefore$ The coordinates of point N are $\\boxed{\\left(-\\frac{24}{5},0\\right)}$, corresponding to the coordinates of point M being $\\boxed{\\left(-\\frac{24}{5},\\frac{37}{5}\\right)}$;\\\\\n(3) The coordinates of point N are $\\boxed{\\left(-\\frac{10}{3},0\\right)}$, corresponding to the coordinates of point M being $\\boxed{\\left(-\\frac{10}{3},\\frac{20}{3}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51864146_102.png"
  },
  {
    "solution": "\\boxed{x\\leq 2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the direct proportion function $y=ax$ and the graph of the linear function $y=kx+2$ intersect at the point $(2, 1)$, what is the solution set of the inequality $kx+2\\ge ax$?\n",
    "completion": "\\textbf{Solution:} The intersection point of the two lines is at $(2,1)$,\\\\\nwhen $x=2$, the values of the two functions are equal,\\\\\nwhen $x<2$, the line $y=kx+2$ is above the line $y=ax$, thus $kx+2>ax$,\\\\\ntherefore, the solution set for the inequality $kx+2\\ge ax$ is: $\\boxed{x\\leq 2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52211879.png"
  },
  {
    "solution": "\\boxed{\\frac{13}{6}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, $AC=12$, and $BC=5$, what is the distance between the centroid $P$ and the circumcenter $Q$ of $\\triangle ABC$?\n",
    "completion": "\\textbf{Solution:} Given the centroid $P$ and circumcenter $Q$ of the right-angled triangle $ABC$, it's known that points $C$, $P$, and $Q$ are collinear.\\\\\nIn right-angled triangle $ABC$, $\\angle C=90^\\circ$, $AC=12$, $BC=5$,\\\\\n$\\therefore AB= \\sqrt{AC^{2}+BC^{2}}=\\sqrt{5^{2}+12^{2}}=13$\\\\\nSince the circumcenter of right-angled triangle $ABC$ is $Q$,\\\\\n$\\therefore Q$ is the midpoint of hypotenuse $AB$,\\\\\n$\\therefore CQ= \\frac{1}{2}AB= \\frac{13}{2}$,\\\\\nSince the centroid of right-angled triangle $ABC$ is $P$,\\\\\n$\\therefore PQ= \\frac{1}{3}CQ= \\boxed{\\frac{13}{6}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52872369.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, there is a grid composed of small squares with equal side lengths. Points $A$, $B$, $P$, and $Q$ are all located on the vertices of the square grid. The line segments $AB$ and $PQ$ intersect at point $E$. What is the value of $\\tan\\angle AEP$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, let the side length of the small square be 1,\\\\\naccording to the grid characteristics, $\\angle PQF=\\angle CBF=45^\\circ$,\\\\\n$\\therefore PQ\\parallel BC$,\\\\\n$\\therefore \\angle QEB=\\angle ABC$,\\\\\n$\\because \\angle AEP=\\angle QEB$,\\\\\n$\\therefore \\angle AEP=\\angle ABC$,\\\\\n$\\because AB=\\sqrt{1^2+3^2}=\\sqrt{10}$, $AC=\\sqrt{1^2+1^2}=\\sqrt{2}$, $BC=\\sqrt{2^2+2^2}=\\sqrt{8}=2\\sqrt{2}$,\\\\\n$\\therefore AC^2+BC^2=AB^2$,\\\\\n$\\therefore \\triangle ABC$ is a right triangle, and $\\angle ACB=90^\\circ$,\\\\\n$\\therefore \\tan\\angle ABC= \\frac{AC}{BC}=\\frac{\\sqrt{2}}{2\\sqrt{2}}=\\frac{1}{2}$,\\\\\n$\\therefore \\tan\\angle AEP=\\tan\\angle ABC= \\boxed{\\frac{1}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51265251.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, an arc is drawn with point B as the center and the length of BC as the radius, intersecting line segment AB at D; an arc is drawn with point A as the center and AD as the radius, intersecting line segment AC at point E, and CD is connected. If $BC=5$ and $AC=12$, what is the length of $AD$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle ACB=90^\\circ$ and $BC=5$, $AC=12$,\\\\\nby the Pythagorean theorem, we have $AB=\\sqrt{AC^2+BC^2}=\\sqrt{12^2+5^2}=13$,\\\\\nSince $AB=AD+BD$ and $BD=BC=5$,\\\\\nit follows that $AD=13-5=\\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52913907_116.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the lines AB and CD intersect at point O, with OM $\\perp$ AB, and ON $\\perp$ CD. If $\\angle 1 = \\frac{1}{3}\\angle BOC$, find the degree measure of $\\angle BOD$.\n",
    "completion": "\\textbf{Solution:} Given: $\\because \\angle 1 = \\frac{1}{3} \\angle BOC$,\\\\\n$\\therefore \\angle BOM = 2\\angle 1 = 90^\\circ$,\\\\\nWe find: $\\angle 1 = 45^\\circ$,\\\\\n$\\therefore \\angle BOD = 90^\\circ - 45^\\circ = \\boxed{45^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53206669_96.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{8}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the diagonals $AC$ and $BD$ of $\\square ABCD$ intersect at point $O$, and $AC \\perp BD$, $BD = 12\\,cm$, $AC = 6\\,cm$. Point $E$ moves from point $B$ towards point $O$ along the segment $BO$ at a speed of $1\\,cm/s$, and point $F$ moves from point $O$ towards point $D$ along the segment $OD$ at a speed of $2\\,cm/s$. If points $E$ and $F$ start moving at the same time, and the movement time is t seconds, is there a value of t such that $\\triangle AEF$ is an isosceles triangle with $AF$ as its base? Find the value of t.\n",
    "completion": "\\textbf{Solution:} Existence is shown as follows:\\\\\n$\\because \\triangle AEF$ is an isosceles triangle, with $AF$ as the base,\\\\\n$\\therefore AE=EF,$\\\\\n$\\because AC\\perp BD, OE=6-t, OA=3,$\\\\\n$\\therefore AE^{2}=3^{2}+(6-t)^{2},$\\\\\nwhile $EF^{2}=(6-t+2t)^{2}=(6+t)^{2},$\\\\\n$\\therefore 9+(6-t)^{2}=(6+t)^{2},$\\\\\nSolving gives: $t=\\boxed{\\frac{3}{8}}.$",
    "image_path": "dataset/mm_math/images/MM_Math/52250892_121.png"
  },
  {
    "solution": "\\boxed{9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure: Point D is on the side AB of $\\triangle ABC$, connect CD, $\\angle 1 = \\angle B$, $AD = 4$, $AC = 6$, find the length of AB.\n",
    "completion": "Solution: Since $\\angle 1 = \\angle B$, and $\\angle A = \\angle A$,\\\\\n$\\therefore$ $\\triangle ACD \\sim \\triangle ABC$,\\\\\n$\\therefore$ $\\frac{AD}{AC} = \\frac{AC}{AB}$,\\\\\n$\\therefore$ $\\frac{4}{6} = \\frac{6}{AB}$,\\\\\n$\\therefore$ AB $= \\boxed{9}$",
    "image_path": "dataset/mm_math/images/MM_Math/52866377.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{2}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the vertices of $\\triangle ABC$ are lattice points of a square grid. What is the value of $\\sin B$?\n",
    "completion": "\\textbf{Solution:} By the Pythagorean theorem, we have: $AB=\\sqrt{3^2+3^2}=3\\sqrt{2}$\\\\\n$\\therefore \\sin B=\\frac{3}{3\\sqrt{2}}=\\frac{\\sqrt{2}}{2}$\\\\\nTherefore, $\\sin B=\\boxed{\\frac{\\sqrt{2}}{2}}$",
    "image_path": "dataset/mm_math/images/MM_Math/52550162.png"
  },
  {
    "solution": "\\boxed{y = -x + 4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, point $O$ is the origin. The line $BC$ intersects the y-axis at point $B$ and the x-axis at point $C$. The coordinates of point $B$ are $(0,4)$, and $OB=OC$. What is the analytical expression of line $BC$?\n",
    "completion": "\\textbf{Solution:} Since $B(0, 4)$, \\\\\nit follows that $OB=4$, \\\\\nthus $OC=OB=4$, \\\\\nimplying $C(4, 0)$ \\\\\nLet the equation of line $BC$ be $y=kx+b\\ (k \\neq 0)$, \\\\\ntherefore $\\left\\{\\begin{array}{l}b = 4 \\\\ 4k + b = 0 \\end{array}\\right.$ \\\\\nhence $\\left\\{\\begin{array}{l}k = -1 \\\\ b = 4 \\end{array}\\right.$ \\\\\nThus, the equation of line $BC$ is $y = -x + 4$, so the answer is \\boxed{y = -x + 4}.",
    "image_path": "dataset/mm_math/images/MM_Math/52189291_144.png"
  },
  {
    "solution": "\\boxed{\\frac{2}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, inside the quadrilateral $ABCD$, point $E$ lies on the extension of side $BA$. Lines $EC$ and $BD$ are drawn and they intersect at point $G$. Line $EC$ intersects with line $AD$ at point $F$. It is known that $EA:AB=1:2$. What is the ratio of $FG:GC$?\n",
    "completion": "\\textbf{Solution:} Given that $AD\\parallel BC$,\\\\\n$\\therefore \\triangle FGD\\sim \\triangle CGB$,\\\\\n$\\therefore \\frac{FG}{GC}=\\frac{FD}{BC}$,\\\\\nFrom $\\triangle EAF\\sim \\triangle EBC$,\\\\\nBased on the conclusion from (1), $\\frac{AF}{BC}=\\frac{1}{3}$,\\\\\n$\\therefore AF=\\frac{BC}{3}$,\\\\\n$\\therefore FD=AD-AF=BC-\\frac{BC}{3}=\\frac{2BC}{3}$,\\\\\n$\\therefore \\frac{FG}{GC}=\\frac{\\frac{2BC}{3}}{BC}=\\boxed{\\frac{2}{3}}$,\\\\\n$\\therefore FG\\colon GC=2\\colon 3$.",
    "image_path": "dataset/mm_math/images/MM_Math/53102159_78.png"
  },
  {
    "solution": "\\boxed{120^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within $\\triangle ABC$, $BM$ bisects $\\angle ABC$, and $CM$ bisects $\\angle ACB$, with $BM$ and $CM$ intersecting at point $M$. If $\\triangle ABC$ is folded along $DE$ such that point $A$ coincides with point $M$, and given that $\\angle 1+\\angle 2=120^\\circ$, what is the measure of $\\angle BMC$?\n",
    "completion": "\\textbf{Solution:} From the property of folding, we have $\\angle AED=\\angle MED$, $\\angle ADE=\\angle MDE$,\n\\\\\n$\\because \\angle 1+\\angle ADM=180^\\circ$, $\\angle 2+\\angle AEM=180^\\circ$, $\\angle 1+\\angle 2=120^\\circ$,\n\\\\\n$\\therefore \\angle ADM+\\angle AEM=240^\\circ$,\n\\\\\n$\\therefore 2\\angle AED+2\\angle ADE=240^\\circ$,\n\\\\\n$\\therefore \\angle AED+\\angle ADE=120^\\circ$,\n\\\\\n$\\therefore \\angle A=180^\\circ-\\angle AED-\\angle ADE=60^\\circ$,\n\\\\\n$\\therefore$ By the same principle as in (1), we can conclude that $\\angle BMC=\\boxed{120^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/54661756_92.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AD$ is the altitude from $A$ to side $BC$, and $BE$ is an angle bisector, intersecting at point $P$. Given that $\\angle EPD=120^\\circ$, what is the degree measure of $\\angle BAD$?\n",
    "completion": "\\textbf{Solution:} Let $\\angle BAD=x^\\circ$\\\\\nGiven the problem, we have: $\\angle BDA=90^\\circ$, $\\angle ABE=\\angle CBE=\\frac{1}{2}\\angle ABD$,\\\\\n$\\therefore$ $\\angle ABD=90^\\circ-x^\\circ$, $\\angle ABE=\\angle CBE=45^\\circ-\\frac{x^\\circ}{2}$,\\\\\nBy the exterior angle property of triangles, we get $\\angle EBD+\\angle BDA=\\angle DPE$,\\\\\nthat is $45^\\circ-\\frac{x^\\circ}{2}+90^\\circ=120^\\circ$, solving this gives $x=\\boxed{30}$,\\\\\nhence $\\angle BAD=\\boxed{30^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52906079_138.png"
  },
  {
    "solution": "\\boxed{15^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, line $EF$ intersects line $CD$ at point $O$, $\\angle AOB=90^\\circ$, $OC$ bisects $\\angle AOF$. If $\\angle AOE=30^\\circ$, please write directly the measure of $\\angle BOD$ in degrees.\n",
    "completion": "\\textbf{Solution:} We have $\\angle BOD = \\boxed{15^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53223346_86.png"
  },
  {
    "solution": "\\boxed{15^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within the rhombus ABCD, point E is a point on side AB, and $DE=AD$. Connect EC. If $\\angle ADE = 40^\\circ$, what is the degree measure of $\\angle BCE$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rhombus,\\\\\nthus $AD=CD$, $\\angle A=\\angle BCD$, $CD\\parallel AB$,\\\\\nsince $DE=AD$, $\\angle ADE=40^\\circ$,\\\\\nthus $\\angle DAE=\\angle DEA=70^\\circ$,\\\\\nsince $CD\\parallel AB$,\\\\\nthus $\\angle CDE=\\angle DEA=70^\\circ$, and $DE=DC=DA$,\\\\\nthus $\\angle DCE=55^\\circ$,\\\\\nsince $\\angle DCB=\\angle DAE=70^\\circ$,\\\\\nthus $\\angle BCE=\\angle DCB-\\angle DCE=\\boxed{15^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52159927.png"
  },
  {
    "solution": "\\boxed{\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $OA_1=A_1A_2=A_2A_3=A_3A_4=A_4A_5=1$, $\\angle OA_1A_2=\\angle OA_2A_3=\\angle OA_3A_4=\\angle OA_4A_5=90^\\circ$, what is the length of $OA_5$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle O{A}_{1}{A}_{2}=\\angle O{A}_{2}{A}_{3}=\\angle O{A}_{3}{A}_{4}=\\angle O{A}_{4}{A}_{5}=90^\\circ$ and $O{A}_{1}={A}_{1}{A}_{2}={A}_{2}{A}_{3}={A}_{3}{A}_{4}={A}_{4}{A}_{5}=1$, \\\\\n$\\therefore OA{}_{2}=\\sqrt{O{{A}_{1}}^{2}+{A}_{1}{{A}_{2}}^{2}}=\\sqrt{{1}^{2}+{1}^{2}}=\\sqrt{2}$, \\\\\n$OA{}_{3}=\\sqrt{O{{A}_{2}}^{2}+{A}_{2}{{A}_{3}}^{2}}=\\sqrt{{\\left(\\sqrt{2}\\right)}^{2}+{1}^{2}}=\\sqrt{3}$, \\\\\n$OA{}_{4}=\\sqrt{O{{A}_{3}}^{2}+{A}_{3}{{A}_{4}}^{2}}=\\sqrt{{\\left(\\sqrt{3}\\right)}^{2}+{1}^{2}}=2$, \\\\\n$OA{}_{5}=\\sqrt{O{{A}_{4}}^{2}+{A}_{4}{{A}_{5}}^{2}}=\\sqrt{{2}^{2}+{1}^{2}}=\\boxed{\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51366535.png"
  },
  {
    "solution": "\\boxed{2\\text{cm}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AC$ bisects $\\angle BAD$, $\\angle B+\\angle D=180^\\circ$, $CE\\perp AD$ at point $E$, $AE=10\\,\\text{cm}$, $AB=8\\,\\text{cm}$, what is the length of $DE$ in $\\text{cm}$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, take $EF=DE$ on $AD$,\\\\\n$\\because CE\\perp AD$,\\\\\n$\\therefore$ the line $CE$ is the perpendicular bisector of the segment $FD$,\\\\\n$\\therefore CF=CD$,\\\\\n$\\therefore \\angle D=\\angle CFD$, $\\angle AFC+\\angle CFD=180^\\circ$,\\\\\n$\\because \\angle B+\\angle D=180^\\circ$,\\\\\n$\\therefore \\angle AFC=\\angle B$,\\\\\n$\\because \\left\\{\\begin{array}{l}\n\\angle B=\\angle AFC \\\\\n\\angle BAC=\\angle FAC \\\\\nAC=AC\n\\end{array}\\right.$,\\\\\n$\\therefore \\triangle BAC\\cong \\triangle FAC$,\\\\\n$\\therefore AF=AB=8\\text{cm}$, $EF=DE=AE-AF=AE-AB=10\\text{cm}-8\\text{cm}=2\\text{cm}$.\\\\\nThus, the answer is: $DE=\\boxed{2\\text{cm}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53107083.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\triangle ABC$, the perpendicular bisector of side $AB$ intersects $AB$ at point D and intersects $AC$ at point E. Line $BE$ is drawn. If $AC=5$ and $BC=3$, what is the perimeter of $ \\triangle BCE$?\n",
    "completion": "\\textbf{Solution:} Given $DE$ is the perpendicular bisector of $AB$,\\\\\nit follows that $AE=BE$,\\\\\nGiven $AC=5$,\\\\\nit then follows that $BE+CE=AE+CE=AC=5$,\\\\\nGiven $BC=3$,\\\\\nthe perimeter of $\\triangle BCE$ is $BE+CE+BC=5+3=\\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52720357.png"
  },
  {
    "solution": "\\boxed{(3,-3)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, the graph of the linear function \\(y=-\\frac{1}{2}x+4\\) intersects the \\(x\\)-axis and \\(y\\)-axis at point \\(A\\) and point \\(B\\), respectively, and intersects the graph of the direct proportion function \\(y=\\frac{3}{2}x\\) at point \\(C\\). Point \\(C\\) is translated 1 unit to the right and 6 units downward to obtain point \\(D\\). Find the perimeter of \\(\\triangle OAB\\) and the coordinates of point \\(D\\).\n",
    "completion": "\\textbf{Solution:} Solve: In the equation $y=-\\frac{1}{2}x+4$, when $x=0$, we get $y=4$, and when $y=0$, from $0=-\\frac{1}{2}x+4$ we have: $x=8$,\\\\\n$\\therefore$ Points A(8,0) and B(0,4).\\\\\n$\\therefore$ $OA=8$, $OB=4$.\\\\\n$\\therefore$ $AB=\\sqrt{OA^{2}+OB^{2}}=\\sqrt{8^{2}+4^{2}}=4\\sqrt{5}$.\\\\\n$\\therefore$ The perimeter of $\\triangle OAB$ $=OA+OB+AB=12+4\\sqrt{5}=\\boxed{12+4\\sqrt{5}}$.\\\\\nBy solving the system of equations $y=-\\frac{1}{2}x+4$ and $y=\\frac{3}{2}x$, we get:\\\\\n$\\left\\{\\begin{array}{l}\nx=2\\\\\ny=3\n\\end{array}\\right.$.\\\\\n$\\therefore$ Point D is (3,−3); The coordinates of point D are $\\boxed{(3,-3)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50481662_82.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ABC=90^\\circ$. A line parallel to $AC$ is drawn through point $B$, intersecting the bisector of $\\angle BAC$ at point $D$. Point $E$ is a point on $AC$, and $BE\\perp AD$ at point $F$. Connect $DE$. The quadrilateral $ABDE$ is a rhombus. Given that $AB=2$ and $\\angle ADC=90^\\circ$, what is the length of $BC$?\n",
    "completion": "\\textbf{Solution:}\n\nGiven that quadrilateral $ABDE$ is a rhombus, it follows that $DE=AE=AB=2$, hence $\\angle EAD=\\angle EDA$. Given that $\\angle ADC=90^\\circ$, it follows that $\\angle EDC+\\angle EDA=90^\\circ$ and $\\angle EAD+\\angle ECD=90^\\circ$. Therefore, $\\angle EDC=\\angle ECD$, which implies $DE=EC=2$. Consequently, $AC=AE+CE=4$. Considering $\\angle ABC=90^\\circ$, it can be derived that $BC=\\sqrt{AC^{2}-AB^{2}}=\\sqrt{4^{2}-2^{2}}=\\boxed{2\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52387465_131.png"
  },
  {
    "solution": "\\boxed{3\\sqrt{10}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: The cross-section of a certain flood control embankment is the trapezoid ABCD as shown in the figure, with the dam height $AH=3$ meters and the slope of the water-facing side AB being $i=1\\colon3$. What is the length of the slope AB in meters?\n",
    "completion": "\\textbf{Solution:} Given that the slope of embankment AB has a slope $i=1\\colon 3$,\\\\\n$AH=3$ meters,\\\\\n$\\therefore BH=9$ meters.\\\\\nBy the Pythagorean theorem, it follows that $AB=\\sqrt{BH^{2}+AH^{2}}=3\\sqrt{10}$ meters.\\\\\nTherefore, the answer is: $AB=\\boxed{3\\sqrt{10}}$ meters.",
    "image_path": "dataset/mm_math/images/MM_Math/51407002.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{10}{7}x+10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system, the line $y=2x+10$ intersects the $x$ axis at point $A$ and the $y$ axis at point $B$. Another line passing through point $B$ intersects the positive half of the $x$ axis at point $C$, and the area of $\\triangle ABC$ is $60$. What are the coordinates of point $C$ and the equation of line $BC$?\n",
    "completion": "\\textbf{Solution:} Since point B is on the line $y=2x+10$, we have B(0,10). When $y=0$, $x=-5$, therefore OA=5. The area of $\\triangle ABC$, $S_{\\triangle ABC}=\\frac{1}{2}AC\\cdot OB=60$, solving this gives $AC=12$. Therefore, OC=AC−OA=7, so C(7,0). Let the equation of line BC be $y=kx+10$, thus $0=7k+10$. Solving for $k$ yields $k=-\\frac{10}{7}$, hence $y=-\\frac{10}{7}x+10$. Therefore, the coordinates of point C are $\\boxed{C(7,0)}$, and the equation of line BC is $\\boxed{y=-\\frac{10}{7}x+10}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52205244_123.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, point D is on side BC, AD=BD, and DE bisects $\\angle ADB$ and intersects AB at point E. If AC=12 and BC=16, what is the length of AE?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, in $\\triangle ABC$, $\\angle C=90^\\circ$, $AC=12$, $BC=16$,\\\\\nby the Pythagorean theorem, we have $AB=\\sqrt{AC^2+BC^2}=\\sqrt{12^2+16^2}=20$,\\\\\nsince $AD=BD$ and $DE$ bisects $\\angle ADB$ and intersects $AB$ at point $E$.\\\\\nTherefore, $AE=BE=\\frac{1}{2}AB=\\boxed{10}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51991503.png"
  },
  {
    "solution": "\\boxed{2.5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given: As shown in the figure, in rhombus $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$, $OE\\parallel DC$ intersects $BC$ at point $E$, $AC=6$, and $BD=8$. What is the length of $OE$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rhombus, with AC=6, and BD=8,\\\\\nit follows that OA=OC=$\\frac{1}{2}$AC=3, and OB=$\\frac{1}{2}$BD=4, also AC$\\perp$BD,\\\\\ntherefore AB=$\\sqrt{OA^{2}+OB^{2}}$=5,\\\\\nsince OE$\\parallel$DC$\\parallel$AB,\\\\\nit follows that OE is the midline of $\\triangle ABC$,\\\\\nthus OE=$\\frac{1}{2}$AB=\\boxed{2.5}.",
    "image_path": "dataset/mm_math/images/MM_Math/52019411.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points A, B, and C are on the circle $\\odot O$. If $\\triangle OAB$ is an equilateral triangle, then what is the measure of $\\angle ACB$?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle OAB$ is an equilateral triangle,\\\\\nit follows that $\\angle AOB=60^\\circ$,\\\\\ntherefore, $\\angle ACB = \\frac{1}{2} \\angle AOB = \\frac{1}{2} \\times 60^\\circ = \\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51351375.png"
  },
  {
    "solution": "\\boxed{3\\sqrt{10}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $AD \\parallel BE \\parallel CF$, and they intersect lines $l_{1}$ and $l_{2}$ at points $A$, $B$, $C$, and $D$, $E$, $F$, respectively. If $AB:AC = 2:5$ and $EF=9$, and segment $x$ is the mean proportional of segments $DE$ and $DF$, what is the value of $x$?\n",
    "completion": "\\textbf{Solution:} Given that $AD\\parallel BE\\parallel CF$,\\\\\nthus, $\\frac{AB}{AC} = \\frac{DE}{DF} = \\frac{2}{5}$,\\\\\ntherefore, $\\frac{DE}{EF}=\\frac{2}{3}$,\\\\\ngiven that EF=9,\\\\\ntherefore, DE=6, DF=15,\\\\\nsince $x^{2} = DE \\times DF = 90$, and $x>0$,\\\\\nthus, $x= \\boxed{3\\sqrt{10}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52861195_138.png"
  },
  {
    "solution": "\\boxed{17^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the line $l_{1} \\parallel l_{2}$, $\\angle A=125^\\circ$, $\\angle B=85^\\circ$, and $\\angle 1$ is $4^\\circ$ larger than $\\angle 2$. What is the degree measure of $\\angle 1$?\n",
    "completion": "\\textbf{Solution:} Extend AB to intersect the two parallel lines at C and D, \\\\\nsince $l_1 \\parallel l_2$ and $\\angle A=125^\\circ$, $\\angle B=85^\\circ$, \\\\\nit follows that $\\angle 4+\\angle 2=85^\\circ$ and $\\angle 1+\\angle 3=125^\\circ$, with $\\angle 3+\\angle 4=180^\\circ$, \\\\\nthus, $85^\\circ-\\angle 2+125^\\circ-\\angle 1=180^\\circ$, \\\\\ntherefore $\\angle 1+\\angle 2=30^\\circ$, \\\\\nsince $\\angle 1$ is $4^\\circ$ larger than $\\angle 2$, \\\\\nit follows that $\\angle 2=\\angle 1-4^\\circ$, \\\\\nhence, $2\\angle 1=34^\\circ$, \\\\\nthus, $\\angle 1=\\boxed{17^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51434340.png"
  },
  {
    "solution": "\\boxed{40^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, a set of triangular plates are combined together, with sides $OA$, $OC$ coinciding with line $EF$, where $\\angle AOB=45^\\circ$, $\\angle COD=60^\\circ$. In Figure 2, the triangular plate $COD$ is fixed, and the triangular plate $AOB$ is rotated clockwise around point $O$ by an angle. During the rotation, triangular plate $AOB$ always stays inside $\\angle EOD$, let $\\angle EOA=\\alpha$. If $OB$ bisects $\\angle EOD$, what is the value of $\\alpha$?\n",
    "completion": "\\textbf{Solution:}\\\\\n(1) Since $\\angle COD=60^\\circ$,\\\\\nthen $\\angle EOD=180^\\circ-\\angle COD=180^\\circ-60^\\circ=120^\\circ$,\\\\\nSince $OB$ bisects $\\angle EOD$,\\\\\nthen $\\angle EOB=\\frac{1}{2}\\angle EOD=\\frac{1}{2}\\times 120^\\circ=60^\\circ$,\\\\\nSince $\\angle AOB=45^\\circ$,\\\\\nthen $\\alpha =\\angle EOB-\\angle AOB=60^\\circ-45^\\circ=\\boxed{15^\\circ}$;\\\\\n(2) Since $\\angle AOB=45^\\circ$ and $\\angle COD=60^\\circ$,\\\\\nthen $\\angle AOC=\\angle AOB+\\angle BOD+\\angle COD=45^\\circ+\\angle BOD+60^\\circ=105^\\circ+\\angle BOD$,\\\\\nSince $\\angle AOC=4\\angle BOD$,\\\\\nthen $105^\\circ+\\angle BOD=4\\angle BOD$,\\\\\nSolving yields: $\\angle BOD=35^\\circ$,\\\\\nthen $\\angle AOC=4\\angle BOD=4\\times 35^\\circ=140^\\circ$,\\\\\nthen $\\alpha =180^\\circ-\\angle AOC=180^\\circ-140^\\circ=\\boxed{40^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51407218_197.png"
  },
  {
    "solution": "\\boxed{M(0,-7)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of a linear function passes through points $P(3, 2)$ and $B(0, -2)$, and intersects with the x-axis at point $A$. Point $M$ is on the y-axis, and the area of $\\triangle ABM$ is $\\frac{15}{4}$. Find the coordinates of point $M$.\n",
    "completion": "\\textbf{Solution:} When $y=0$, $\\frac{4}{3}x-2=0$, solving for $x$ gives $x=\\frac{3}{2}$, so $A\\left(\\frac{3}{2},0\\right)$,\\\\\n$\\therefore OA=\\frac{3}{2}$.\\\\\n$\\because$ point M is on the y-axis, and the area of $\\triangle ABM$ is $\\frac{15}{4}$,\\\\\n$\\therefore S_{\\triangle ABM}=\\frac{1}{2}BM\\cdot OA=\\frac{15}{4}$,\\\\\n$\\therefore BM=5$,\\\\\n$\\because B(0,-2)$,\\\\\n$\\therefore \\boxed{M(0,3)} or \\boxed{M(0,-7)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50677300_147.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ABC=90^\\circ$, $\\angle A=60^\\circ$, point D is on side $AC$, $\\angle DBC=30^\\circ$, and $AC=12\\,cm$. What is the perimeter of $\\triangle ABD$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle ABC$ and $\\angle AOC$ are the inscribed angle and central angle subtended by the same arc AC, and $\\angle AOC=90^\\circ$,\n\n$\\therefore$ $\\angle ABC=\\frac{1}{2}\\angle AOC=\\boxed{45^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51320363.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, $O$ is a point on line segment $PB$. Taking $O$ as the center and the length of $OB$ as the radius, circle $\\odot O$ intersects $PB$ at point A. Point C is on circle $\\odot O$. Line $PC$ is drawn, satisfying $PC^{2}=PA\\cdot PB$. $\\triangle PAC\\sim \\triangle PCB$. If $AB=3PA$, find the value of $\\frac{AC}{BC}$.\n",
    "completion": "\\textbf{Solution:} As shown in the diagram below, connect $OC$,\n\n$\\because \\triangle PAC\\sim \\triangle PCB$,\n\n$\\therefore \\frac{PA}{PC}=\\frac{AC}{BC}$, $ \\angle PCA=\\angle ABC$\n\n$\\because OC=OA$, $OC=OB$\n\n$\\therefore \\angle OCA=\\angle CAO$, $\\angle OCB=\\angle OBC$\n\n$\\because \\angle PCA=\\angle ABC$, $\\angle OCB=\\angle OBC$\n\n$\\therefore \\angle PCA=\\angle OCB$\n\n$\\because AB$ is the diameter,\n\n$\\therefore \\angle ACB=90^\\circ$,\n\n$\\therefore \\angle ACO+\\angle OCB=90^\\circ$,\n\n$\\therefore \\angle ACO+\\angle PCA=90^\\circ$,\n\n$\\therefore \\angle PCO=90^\\circ$,\n\n$\\because AB=3PA$, $PB=PA+AB$,\n\n$\\therefore PB=4PA$,\n\n$\\because AB=2OC=2OA$,\n\n$\\therefore OA=OC=1.5PA$,\n\n$\\therefore PO=2.5PA$,\n\n$\\because PC^2+OC^2=PO^2$\n\n$\\therefore PC=\\sqrt{PO^2−OC^2}=2PA$,\n\n$\\therefore \\frac{AC}{BC}=\\frac{PA}{PC}=\\boxed{\\frac{1}{2}}$",
    "image_path": "dataset/mm_math/images/MM_Math/52991574_116.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABD \\cong \\triangle EBC$, $AB=4\\text{cm}$, $BC=7\\text{cm}$, then what is the length of $DE$ in cm?\n",
    "completion": "\\textbf{Solution:}\n\nSince $\\triangle ABD \\cong \\triangle EBC$, and given $AB=4\\, \\text{cm}$, $BC=7\\, \\text{cm}$,\n\nit follows that $BE=AB=4\\, \\text{cm}$, $BD=BC=7\\, \\text{cm}$,\n\nthus $DE=BD-BE=\\boxed{3}\\, \\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52715503.png"
  },
  {
    "solution": "\\boxed{(3,2)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, given that $\\square ABCD$ has three vertices at $A(-1, 0)$, $B(-2, -3)$, and $C(2, -1)$, what are the coordinates of the fourth vertex $D$?\n",
    "completion": "\\textbf{Solution:} Draw $BE\\perp x$-axis at $E$, $DM\\perp x$-axis at $M$, and $CF\\perp BE$ at $F$, where $DM$ and $CF$ intersect at $N$,\\\\\nthen quadrilateral $EFNM$ is a rectangle,\\\\\nthus $EF=MN$, $EM=FN$, $FN\\parallel EM$,\\\\\n$\\therefore \\angle EAB = \\angle AQC$,\\\\\n$\\because$ Quadrilateral $ABCD$ is a parallelogram,\\\\\n$\\therefore AB=DC$, $AB\\parallel DC$,\\\\\n$\\therefore \\angle AQC = \\angle DCN$,\\\\\n$\\therefore \\angle DCN = \\angle EAB$,\\\\\nIn $\\triangle DCN$ and $\\triangle BAE$\\\\\n$\\lbrace \\begin{array}{l} \\angle N=\\angle BEA=90^\\circ\\hfill \\\\ \\angle DCN=\\angle EAB\\hfill \\\\ CD=AB\\hfill \\end{array}$, \\\\\n$\\therefore \\triangle DCN \\cong \\triangle BAE$ (AAS),\\\\\n$\\therefore BE=DN$, $AE=CN$,\\\\\n$\\because A(-1,0)$, $B(-2,-3)$, $C(2,-1)$,\\\\\n$\\therefore CN=AE=2-1=1$, $DN=BE=3$,\\\\\n$\\therefore DM=3-1=2$, $OM=2+1=3$,\\\\\n$\\therefore$ The coordinates of $D$ are $\\boxed{(3,2)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53273997.png"
  },
  {
    "solution": "\\boxed{0}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the length of segment $AB=m$ (where $m$ is a constant), point $C$ is on the line segment $AB$ (not coinciding with $A$ or $B$), points $P$ and $Q$ are respectively on the segments $BC$ and $AC$, and it satisfies $CQ=2AQ$, $CP=2BP$. As shown in figure 2, if point $C$ is to the left of point $A$, while point $P$ is on the segment $AB$ (not coinciding with the endpoints), find the value of $2AP+CQ-2PQ$.\n",
    "completion": "\\textbf{Solution:} Let $AQ=x$, and $BP=y$,\n\nthen we have $CQ= 2x$, and $CP= 2y$,\n\nthus $AP=CP-CA=2y- 3x$, $PQ=CP-CQ=2y- 2x$,\n\nhence $2AP+CQ- 2PQ=2(2y- 3x) + 2x - 2(2y-2x)=4y - 6x+2x-4y+4x=0$.\n\n$\\therefore$ The value of $2AP+CQ- 2PQ$ is $\\boxed{0}$.",
    "image_path": "dataset/mm_math/images/MM_Math/54927374_15.png"
  },
  {
    "solution": "\\boxed{140^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, $\\angle BAD=110^\\circ$, $\\angle B=\\angle D=90^\\circ$. Find points $M$ and $N$ on $BC$ and $CD$ respectively, such that the perimeter of $\\triangle AMN$ is minimal. What is the measure of $\\angle AMN + \\angle ANM$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, construct the symmetric point of point A about line $BC$ as $A^{\\prime}$, and the symmetric point about line $CD$ as $A^{\\prime \\prime}$,\\\\\njoin $A^{\\prime}A^{\\prime \\prime}$, and let the intersection points with $BC$ and $CD$ be the required points M and N, respectively,\\\\\n$\\because \\angle BAD=110^\\circ$, $\\angle B=\\angle D=90^\\circ$,\\\\\n$\\therefore \\angle A^{\\prime} +\\angle A^{\\prime \\prime} =180^\\circ-110^\\circ=70^\\circ$,\\\\\nfrom the property of axial symmetry, we get: $\\angle A^{\\prime} =\\angle A^{\\prime}AM$, $\\angle A^{\\prime \\prime} =\\angle A^{\\prime \\prime}AN$,\\\\\n$\\therefore \\angle AMN+\\angle ANM=2(\\angle A^{\\prime} +\\angle A^{\\prime \\prime})=2\\times 70^\\circ=\\boxed{140^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53273961.png"
  },
  {
    "solution": "\\boxed{40^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points $A$, $B$, and $C$ are on circle $\\odot O$, with $OB\\parallel AC$. Line $BC$ intersects $OA$ at point $D$. If $\\angle ADB=60^\\circ$, what is the degree measure of $\\angle AOB$?\n",
    "completion": "\\textbf{Solution:} Let $ \\angle ACB=x^\\circ$,\\\\\nthen $ \\angle AOB=2\\angle ACB=2x^\\circ$,\\\\\nsince $OB\\parallel AC$,\\\\\nthus $ \\angle OBD=\\angle ACB=x^\\circ$,\\\\\nsince $ \\angle ADB=60^\\circ$,\\\\\nthus $ \\angle AOB+\\angle OBD=\\angle ADB=60^\\circ$, that is $ 2x+x=60$,\\\\\nsolving for $x$ gives $ x=20$,\\\\\nthen $ \\angle AOB=2x^\\circ=\\boxed{40^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53009487.png"
  },
  {
    "solution": "\\boxed{\\frac{2}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\triangle ABC$, $ DE \\parallel BC$, $ DF \\parallel AC$, $ AD=3$, $ BD=2$, what is the value of $ BF: DE$?\n",
    "completion": "\\textbf{Solution:} Since $DE \\parallel BC$ and $DF \\parallel AC$, \\\\\nit follows that $\\angle ADE=\\angle B$ and $\\angle BDF=\\angle BAC$, \\\\\nwhich means $\\triangle DBF \\sim \\triangle ADE$, \\\\\nthus $\\frac{BD}{DA}=\\frac{BF}{DE}=\\frac{2}{3}$, \\\\\nhence $\\frac{BF}{DE}=\\boxed{\\frac{2}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51379443.png"
  },
  {
    "solution": "\\boxed{115^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC$, point D is the midpoint of $BC$, point E is on $AC$, $AD=AE$, if $\\angle BAD=50^\\circ$, what is the degree measure of $\\angle CED$?\n",
    "completion": "\\textbf{Solution:} Since $AB=AC$ and point D is the midpoint of $BC$, \\\\\nit follows that triangle $ABC$ is an isosceles triangle, making $AD$ a median of $BC$ and the bisector of $\\angle BAC$, hence $\\angle CAD=\\angle BAD=50^\\circ$, \\\\\nSince $AD=AE$, \\\\\nit follows that $\\angle ADE=\\angle AED$, \\\\\nSince $\\angle ADE+\\angle AED+\\angle CAD=180^\\circ$, \\\\\nit follows that $\\angle ADE=\\angle AED=\\frac{1}{2}(180^\\circ-50^\\circ)=65^\\circ$, \\\\\nTherefore, $\\angle CED=180^\\circ-65^\\circ=\\boxed{115^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52978401.png"
  },
  {
    "solution": "\\boxed{15}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given points $C$ and $D$ are on segment $AB$, and $AB=22$, $CD=8$. If points $M$ and $N$ are respectively the midpoints of $AC$ and $BD$, find the length of $MN$.\n",
    "completion": "\\textbf{Solution:} Given $\\because AB=22$, $CD=8$.\\\\\n$\\therefore AC+BD=AB-CD=14$.\\\\\nAlso, $\\because M$, $N$ are the midpoints of $AC$, $BD$ respectively,\\\\\n$\\therefore CM+DN=7$.\\\\\n$\\therefore MN=CM+DN+CD=\\boxed{15}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52837995_155.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the mathematics activity class, Xiao Dong wants to measure the radius of a circular gear's inner circle. As shown in the figure, Xiao Dong first selects points $A$ and $B$ on the inner circle, then constructs the perpendicular bisector of chord $AB$ with the foot of the perpendicular being $C$ and intersecting $\\overset{\\frown}{AB}$ at point $D$, and connects $CD$. After measuring, $AB=8\\text{cm}$ and $CD=2\\text{cm}$. So, what is the radius of this gear's inner circle in centimeters?\n",
    "completion": "\\textbf{Solution:} Let the center of the circle be O, and connect OB.\\\\\nIn $\\triangle OBC$, $BC = \\frac{1}{2}AB = 4\\text{cm}$,\\\\\nAccording to the Pythagorean theorem, $OC^2 + BC^2 = OB^2$, which is: $(OB - 2)^2 + 4^2 = OB^2$,\\\\\nSolving this, we get: $OB = \\boxed{5}$;\\\\\nTherefore, the radius of the wheel is $5\\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51402230.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $O$ is the center of the semicircle, and $C$, $D$ are two points on the semicircle. Connect $CD$, $BD$, $AD$, where $CD=BD$. Extend $AC$ to intersect the extension of $BD$ at point $E$. Given that $CD=DE$; if $AC=6$ and the radius $OB=5$, find the length of $BD$.\n",
    "completion": "\\textbf{Solution:} In right triangle $\\triangle ACB$, we have $BC=\\sqrt{AB^{2}-AC^{2}}=\\sqrt{10^{2}-6^{2}}=8$.\\\\\nSince $CD=DE$ and $CD=BD$,\\\\\nit follows that $BD=ED$.\\\\\nIn $\\triangle ADB$ and $\\triangle ADE$, we have\\\\\n$\\left\\{\\begin{array}{l}\nAD=AD \\\\\n\\angle ADB=\\angle EDA \\\\\nBD=ED\n\\end{array}\\right.$,\\\\\ntherefore, $\\triangle ADB \\cong \\triangle ADE \\text{ (SAS)}$,\\\\\nthus, $AE=AB=10$,\\\\\nhence, $CE=AE-AC=10-6=4$,\\\\\nIn right triangle $\\triangle ECB$, $BE=\\sqrt{CE^{2}+CB^{2}}=\\sqrt{4^{2}+8^{2}}=4\\sqrt{5}$,\\\\\ntherefore, $BD=\\frac{1}{2}BE=2\\sqrt{5}=\\boxed{2\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51545272_137.png"
  },
  {
    "solution": "\\boxed{20}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $O$ is a point inside $\\triangle ABC$. Connect $OA$, $OB$, and $OC$, and connect the midpoints $D$, $E$, $F$, $G$ of $AB$, $OB$, $OC$, $AC$ respectively, to form quadrilateral $DEFG$. Quadrilateral $DEFG$ is a parallelogram; if $BO$ is on $CO$ and $M$ is the midpoint of $EF$, and $OA=8$, $OM=3$, what is the perimeter of the parallelogram $DEFG$?\n",
    "completion": "Solution: From the given, $BO\\perp CO$ and M is the midpoint of EF, $OM=3$,\\\\\n$\\therefore EF=2OM=6$. Given that quadrilateral DEFG is a parallelogram,\\\\\n$\\therefore DG=EF=6$,\\\\\n$\\because$ D is the midpoint of AB and E is the midpoint of BO,\\\\\n$\\therefore DE=\\frac{1}{2}AO=4$,\\\\\n$\\therefore$ the perimeter of the quadrilateral DEFG is: $4+4+6+6=\\boxed{20}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52344766_175.png"
  },
  {
    "solution": "\\boxed{120}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in parallelogram $ABCD$, the heights on the adjacent sides $AD$ and $CD$ are equal, that is, $BE=BF$. Parallelogram $ABCD$ is a rhombus; if $DB=10$ and $AB=13$, what is the area of parallelogram $ABCD$?\n",
    "completion": "\\textbf{Solution:} Draw line $AC$ intersecting $BD$ at point $O$, \\\\\nsince quadrilateral $ABCD$ is a rhombus, it follows that $AC \\perp BD$, and $BO = \\frac{1}{2}BD = 5$,\\\\\nin $\\triangle ABO$, by the Pythagorean theorem, we have: $AO = \\sqrt{AB^{2} - BO^{2}} = 12$, \\\\\nthus $AC = 2AO = 24$, \\\\\ntherefore, the area of parallelogram $ABCD$ $= \\frac{1}{2}AC \\times BD = \\boxed{120}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52423203_88.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that in $\\triangle ABC$ ($AB > AC$), $\\angle BAC = 60^\\circ$, $AC = 4$, $D$ is the midpoint of side $BC$, and the line $DF$ passing through point $D$ bisects the perimeter of $\\triangle ABC$ and intersects $AB$ at point $F$. What is the length of $DF$?\n",
    "completion": "\\textbf{Solution:} Solution as shown in the figure:\\\\\nExtend BA to E, such that AE=AC=4, take the midpoint F of BE, join DF, join CE, and draw a line AG$\\perp$CE at point G,\\\\\n$\\because$D is the midpoint of side BC,\\\\\n$\\therefore$BD=CD,\\\\\n$\\because$EF=BF,\\\\\n$\\therefore$BD+BF=CD+AE+FA=CD+EF,\\\\\n$\\therefore$The line DF bisects the perimeter of $\\triangle ABC$,\\\\\n$\\because$AE=AC=4, $\\angle BAC=60^\\circ$,\\\\\n$\\therefore$$\\angle ACE=\\angle E=30^\\circ$,\\\\\n$\\therefore$AG=$\\frac{1}{2}$AE=2,\\\\\n$\\therefore$EG=$\\sqrt{AE^2-AG^2}=\\sqrt{4^2-2^2}=2\\sqrt{3}$,\\\\\n$\\because$AE=AC, AG$\\perp$CE,\\\\\n$\\therefore$GE=$\\frac{1}{2}$CE,\\\\\n$\\because$D is the midpoint of BC, F is the midpoint of BE,\\\\\n$\\therefore$FD is the mid-segment of $\\triangle BCE$,\\\\\n$\\therefore$DF=$\\frac{1}{2}$CE=EG=$\\boxed{2\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52304107.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the line segment $AB=4\\,m$. The line segment $AB$ is extended to point $C$, such that $BC=2AB$. If point $D$ is the midpoint of line segment $AC$, how many $cm$ is the line segment $BD$?\n",
    "completion": "\\textbf{Solution:} Given $\\because$AB=4, BC=2AB, \\\\\n$\\therefore$BC=$2\\times 4$=8, \\\\\n$\\therefore$AC=AB+BC=4+8=12, \\\\\n$\\because$ point D is the midpoint of segment AC, \\\\\n$\\therefore$AD=$\\frac{1}{2}$AC=6, \\\\\n$\\therefore$BD=AD−AB=6−4=\\boxed{2}",
    "image_path": "dataset/mm_math/images/MM_Math/55308242.png"
  },
  {
    "solution": "\\boxed{80^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, rays $BA$ and $CA$ intersect at point $A$, with $BC$ connected. Given that $AB=AC$ and $\\angle B=40^\\circ$, what is the value of $x$?\n",
    "completion": "\\textbf{Solution:} Since $AB = AC$ and $\\angle B = 40^\\circ$, it follows that $\\angle C = \\angle B = 40^\\circ$, hence $x^\\circ = 40^\\circ + 40^\\circ = \\boxed{80^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55553377.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in a square grid, it is known that the three vertices of $\\triangle ABC$ are all on the grid points. What is the value of the tangent of $\\angle ACB$?\n",
    "completion": "\\textbf{Solution:} Extend CB to intersect the grid at D and connect AD as shown in the figure:\\\\\nThen $\\angle ADC = 45^\\circ + 45^\\circ = 90^\\circ$,\\\\\n$\\because AD = \\sqrt{1^2 + 1^2} = \\sqrt{2}$, $CD = \\sqrt{2^2 + 2^2} = 2\\sqrt{2}$,\\\\\n$\\therefore$ the tangent of $\\angle ACB = \\frac{AD}{CD} = \\frac{\\sqrt{2}}{2\\sqrt{2}} = \\boxed{\\frac{1}{2}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53080562.png"
  },
  {
    "solution": "\\boxed{\\frac{9}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AC$ is the diameter of $\\odot O$, $BC$ is the chord of $\\odot O$, point $P$ is a point outside $\\odot O$, and $PB$ and $AB$ are connected, with $\\angle PBA=\\angle C$. $PB$ is the tangent of $\\odot O$; connect $OP$, if $OP\\parallel BC$ and $OP=8$, the radius of $\\odot O$ is $3$, find the length of $BC$.\n",
    "completion": "Solution:\\\\\nSince the radius of $\\odot O$ is $3$,\\\\\nhence $OB=3$, $AC=6$,\\\\\nSince $OP \\parallel BC$,\\\\\nthus $\\angle CBO = \\angle BOP$,\\\\\nSince $OC=OB$,\\\\\nthus $\\angle C = \\angle CBO$,\\\\\nthus $\\angle C = \\angle BOP$,\\\\\nAlso, since $\\angle ABC = \\angle PBO = 90^\\circ$,\\\\\nthus $\\triangle ABC \\sim \\triangle PBO$,\\\\\nthus $\\frac{BC}{OB} = \\frac{AC}{OP}$,\\\\\nthat is $\\frac{BC}{3} = \\frac{6}{8}$,\\\\\nhence $BC = \\boxed{\\frac{9}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51433255_140.png"
  },
  {
    "solution": "\\boxed{132^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within circle $O$, $AB$ is a side of the inscribed regular hexagon in circle $O$, and $BC$ is a side of the inscribed regular decagon in circle $O$. What is the measure of $\\angle ABC$ in degrees?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, connect AO, BO, CO,\n\n$\\because$ AB is a side of the inscribed hexagon in $\\odot O$,\n\n$\\therefore$ $\\angle AOB = \\frac{360^\\circ}{6} = 60^\\circ$,\n\n$AO = BO$,\n\n$\\therefore$ $\\angle ABO = \\frac{1}{2}(180^\\circ-60^\\circ) = 60^\\circ$,\n\n$\\because$ BC is a side of the inscribed decagon in $\\odot O$,\n\n$\\therefore$ $\\angle BOC = \\frac{360^\\circ}{10} = 36^\\circ$, BO=CO,\n\n$\\therefore$ $\\angle CBO = \\frac{1}{2}(180^\\circ-36^\\circ) = 72^\\circ$,\n\n$\\therefore$ $\\angle ABC = \\angle ABO + \\angle CBO = 60^\\circ + 72^\\circ = \\boxed{132^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51433278.png"
  },
  {
    "solution": "\\boxed{x>-2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the straight line $y=x+b$ intersects with the coordinate axes at two points. What is the range of $x$ when $y>0$?\n",
    "completion": "\\textbf{Solution:} From the graph, it is known that the intersection point of the line with the x-axis is ($-2, 0$),\\\\\nwhen $y>0$, the graph is above the x-axis, on the right side of the intersection point with the x-axis,\\\\\ntherefore, when $x>-2$, $y>0$\\\\\nHence, the answer is: $\\boxed{x>-2}$",
    "image_path": "dataset/mm_math/images/MM_Math/52212539.png"
  },
  {
    "solution": "\\boxed{}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the side length of square $ABCD$ is $6\\,cm$, and $E$ is a point on side $AB$ with $AE$ being $1\\,cm$ long. A moving point $P$ starts from point $B$ and moves along the direction of ray $BC$ at a speed of $1\\,cm$ per second. $\\triangle EBP$ is folded along $EP$, with point $B$ landing at point ${B}^{\\prime}$, and the movement time is $t$ seconds. When $t$ equals what value, is $\\angle {B}^{\\prime}PC$ a right angle?\n",
    "completion": "\\textbf{Solution:} Given that the problem information is insufficient, it is impossible to provide specific solution steps. Please provide a complete problem description for an accurate mathematical solution.",
    "image_path": "dataset/mm_math/images/MM_Math/53359333_48.png"
  },
  {
    "solution": "\\boxed{\\frac{5}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, given a rectangle $ABCD$ with two sides $AB=6$ and $AD=8$, point $E$ is the intersection of diagonals $AC$ and $BD$, point $P$ is a moving point on side $AD$. Construct point $D'$, which is the symmetric point of $D$ with respect to line $PE$. When $ED'$ is perpendicular to a side of the rectangle, what is the length of $PD$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, when $ED^{\\prime} \\perp AD$,\n\n$\\because$ rectangle ABCD,\n\n$\\therefore$ AE=CE, CD$\\perp$AD, AB=CD=6,\n\n$\\therefore$ EF$\\parallel$CD,\n\n$\\therefore$ EF is the median of $\\triangle ACD$,\n\n$\\therefore$ EF=$\\frac{1}{2}CD=\\frac{1}{2}\\times 6=3$, DF=$\\frac{1}{2}AD=4$, AC=BD=2DE, in $\\triangle ADC$\n\n$AC=\\sqrt{AD^{2}+CD^{2}}=\\sqrt{8^{2}+6^{2}}=10$,\n\n$\\therefore$ DE=5\n\n$\\because$ point D is symmetrical to point D$^{\\prime}$ regarding line PE,\n\n$\\therefore$ DE=ED$^{\\prime}=5$, DP=D$^{\\prime}P$,\n\n$\\therefore$ FD$^{\\prime}=5-3=2$, let DP=D$^{\\prime}P=x$, then PF=4-x, in $\\triangle PFD^{\\prime}$\n\n$D^{\\prime}F^{2}+PF^{2}=D^{\\prime}P^{2}$, which implies $x^{2}=(4-x)^{2}+4$\n\nSolving this: $x=\\boxed{\\frac{5}{2}}$;\n\nWhen $D^{\\prime}E\\perp AB$\n\n$\\because$ DA$\\perp$AB,\n\n$\\therefore$ AD$\\parallel D^{\\prime}E$,\n\n$\\therefore \\angle DPE=\\angle PED^{\\prime}$,\n\n$\\because$ point D is symmetrical to point D$^{\\prime}$ regarding line PE,\n\n$\\therefore \\triangle DPE$ and $\\triangle D^{\\prime}PE$ are symmetrical regarding line PE,\n\n$\\therefore \\angle DPE=\\angle D^{\\prime}PE$,\n\n$\\therefore \\angle PED^{\\prime}=\\angle D^{\\prime}PE$,\n\n$\\therefore DP^{\\prime}=DP=D^{\\prime}E=5$.\n\nHence, the length of PD is $\\boxed{\\frac{5}{2}}$ or 5.",
    "image_path": "dataset/mm_math/images/MM_Math/52204870.png"
  },
  {
    "solution": "\\boxed{(0,1)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle DEF$ is obtained by rotating $\\triangle ABC$ around a certain point. What are the coordinates of this point?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, \\\\ \nwe connect AD, BE and construct the perpendicular bisectors of line segments AD, BE, \\\\\nthe intersection of these lines is the center of rotation $O^{\\prime}$. Its coordinates are $\\boxed{(0,1)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51212130.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $\\angle A=\\angle B=90^\\circ$, $AB=6$, and E, F are moving points on segment $AB$ and ray $BD$ respectively, with $BF=2BE$. Point G is on ray $AC$, and $EG$ is connected. If $\\triangle AEG$ is congruent to $\\triangle BEF$, what is the length of segment $AG$?\n",
    "completion": "\\textbf{Solution:} \n(1) As shown in the figure:\\\\\nWhen $\\triangle GAE \\cong \\triangle EBF$, we have: $AG=BE$, $AE=BF$\\\\\nSince $BF=2BE$,\\\\\nit follows that $AE=2BE$,\\\\\nSince $AB=AE+BE=3BE=6$,\\\\\nit follows that $BE=2$,\\\\\nHence $AG=BE=\\boxed{2}$;\\\\\n(2) When $\\triangle GAE \\cong \\triangle FBE$, we have: $AE=BE$, $AG=BF$\\\\\nSince $AB=AE+BE=2BE=6$,\\\\\nit follows that $BE=3$,\\\\\nSince $BF=2BE$,\\\\\nit follows that $AG=2BE=\\boxed{6}$;\\\\",
    "image_path": "dataset/mm_math/images/MM_Math/52827570.png"
  },
  {
    "solution": "\\boxed{13}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, if quadrilateral $ABCD$ is a rhombus with $AC=24$ and $BD=10$, what is the side length of the rhombus $ABCD$?\n",
    "completion": "\\textbf{Solution:} Let AC and BD intersect at point O, \\\\\nsince quadrilateral ABCD is a rhombus, \\\\\ntherefore AC$\\perp$BD, $OC=\\frac{1}{2}AC=12$, $OD=\\frac{1}{2}BD=5$, \\\\\ntherefore $\\angle$COD$=90^\\circ$, \\\\\ntherefore $CD=\\sqrt{OC^{2}+OD^{2}}=\\boxed{13}$",
    "image_path": "dataset/mm_math/images/MM_Math/52225703.png"
  },
  {
    "solution": "\\boxed{-4<x<3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y_1=mx+n$ ($m\\neq 0$) and the graph of the quadratic function $y_2=ax^2+bx+c$ ($a\\neq 0$) are illustrated. What is the solution set of the inequality $ax^2+bx+c>mx+n$?\n",
    "completion": "\\textbf{Solution:} Given the linear function $y_1=mx+n$ ($m\\neq 0$) and the quadratic function $y_2=ax^2+bx+c$ ($a\\neq 0$), by examining their graphs, it is known that:\\\\\nThe solution set of the inequality $ax^2+bx+c>mx+n$ is: $-4<x<3$,\\\\\nHence, the solution set is $\\boxed{-4<x<3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52976992.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that: $O$ is a point on line $AB$, $\\angle COD$ is a right angle, and $OE$ bisects the obtuse angle $\\angle BOC$. As shown in Figure 2, $OF$ bisects $\\angle BOD$. What is the degree measure of $\\angle EOF$?\n",
    "completion": "\\textbf{Solution:} From the construction, we have $OE$ bisect $\\angle BOC$, $OF$ bisect $\\angle BOD$,\\\\\n$\\therefore \\angle BOE=\\frac{1}{2}\\angle BOC$, $\\angle BOF=\\frac{1}{2}\\angle BOD$,\\\\\n$\\therefore \\angle EOF=\\angle BOE-\\angle BOF=\\frac{1}{2}\\left(\\angle BOC-\\angle BOD\\right)=\\frac{1}{2}\\angle COD$,\\\\\n$\\because \\angle COD=90^\\circ$,\\\\\n$\\therefore \\angle EOF=\\boxed{45^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53119812_121.png"
  },
  {
    "solution": "\\boxed{\\frac{5}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, the quadrilateral ABCD and the quadrilateral CEFG are both rectangles, with points E and G located on the sides CD and CB respectively, and point $F$ on AC, with $AB=3$, $BC=4$. Calculate the value of $\\frac{AF}{BG}$.\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD and quadrilateral CEFG are both rectangles,\\\\\nit follows that $\\angle ABC=\\angle FGC=90^\\circ$\\\\\nThus, $AB\\parallel FG, \\ AC=\\sqrt{AB^{2}+BC^{2}}=5$\\\\\nTherefore, $\\frac{AF}{BG}=\\frac{CF}{CG}=\\frac{AC}{BC}=\\boxed{\\frac{5}{4}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51284029_180.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Overlay a right-angled triangle board DOE with AOC as shown in figure (1), where $\\angle O=90^\\circ$, $\\angle A=30^\\circ$, and $\\angle E=45^\\circ$, with $OD>OC$. Within the plane containing the two triangle boards, rotate the triangle board DOE clockwise around point O by $\\alpha$ degrees $(0^\\circ<\\alpha<90^\\circ)$ to the position $D_{1}OE_{1}$, so that $OD_{1}\\parallel AC$, as shown in figure (2). What is the value of $\\alpha$?\n",
    "completion": "\\textbf{Solution:} By the given information, we have: $\\alpha =\\angle AOD_{1}$\\\\\nSince $OD_{1}\\parallel AC$ and $\\angle A=30^\\circ$\\\\\nTherefore, $\\angle AOD_{1}=\\angle A=30^\\circ$\\\\\nHence, $\\alpha =\\boxed{30^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/55093220_106.png"
  },
  {
    "solution": "\\boxed{26}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in right-angled triangle $ABC$, with $\\angle ACB = 90^\\circ$, squares $ADEB$, $CBFG$, and $ACHI$ are constructed externally on sides $AB$, $BC$, and $AC$ of $\\triangle ABC$, respectively. The square $ABED$ is folded along the line $AB$ to obtain square $ABE'D'$, where $AD'$ intersects $CH$ at point $N$, point $E'$ lies on side $FG$, and $D'E'$ intersects $CG$ at point $M$. Let the area of $\\triangle ANC$ be $S_1$ and the area of quadrilateral $BCME'$ be $S_2$. Given that $CN = 2NH$ and $S_1 + S_2 = 14$, what is the area of square $ABED$?\n",
    "completion": "\\textbf{Solution:} Let $NH=x$, then $CN=2x$,\\\\\nAccording to the problem, we have $CA=CH=3x$,\\\\\nIn $\\triangle ACN$ and $\\triangle BCA$ right triangles,\\\\\n$\\angle ACN=\\angle BCA=90^\\circ$,\\\\\nSince $\\angle CAN+\\angle CNA=\\angle CAN+\\angle CAB=90^\\circ$,\\\\\nTherefore, $\\angle CNA=\\angle CAB$,\\\\\nThus, $\\triangle ACN \\sim \\triangle BCA$ right triangles,\\\\\nHence, $\\frac{CN}{AC}=\\frac{AC}{BC}=\\frac{2x}{3x}=\\frac{2}{3}$,\\\\\nSo, $BC=\\frac{9}{2}x$,\\\\\nIn the $\\triangle ABC$ right triangle, by the Pythagorean theorem, we find:\\\\\n$AB^{2}=AC^{2}+BC^{2}=9x^{2}+\\frac{81}{4}x^{2}=\\frac{117}{4}x^{2}$,\\\\\nSince $S_{ABED}=AB^{2}$,\\\\\nTherefore, $S_{ABED}=\\frac{117}{4}x^{2}$,\\\\\nIn $\\triangle ABN$ and $\\triangle D'AM$ right triangles, $\\angle ABN=\\angle D'AM$, AB=AD', $\\angle BAN=\\angle D'$,\\\\\nTherefore, $\\triangle ABN \\cong \\triangle D'AM$ (ASA),\\\\\nHence, $S_{CND'M}=S_{\\triangle ABC}$,\\\\\nSo, $S_{1}+S_{2}=S_{ABE'D'}-2S_{\\triangle ABC}=14$,\\\\\nThus, $\\frac{117}{4}x^{2}-2\\times \\frac{1}{2}\\cdot 3x\\cdot \\frac{9}{2}x=14$,\\\\\nSolving yields: $x^{2}=\\frac{56}{63}$,\\\\\nTherefore, $S_{ABED}=\\frac{117}{4}x^{2}=\\frac{117}{4}\\times \\frac{56}{63}=\\boxed{26}$,\\\\",
    "image_path": "dataset/mm_math/images/MM_Math/51322970.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{569}}{10}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\odot O$, $CP=2$, $PD=6$, $AP=5$, and chord $CD\\perp AB$ with the foot of the perpendicular at point $P$. Find the length of $OP$.\n",
    "completion": "\\textbf{Solution:} Draw $OE\\perp CD$ at point E, and $OF\\perp AB$ at point F, connect $OA$ and $OD$,\\\\\nsince $CD\\perp AB$,\\\\\ntherefore, quadrilateral $OEPF$ is a rectangle,\\\\\nthus $PF=OE$, $OF=PE$,\\\\\ngiven $CP=2$, $PD=6$,\\\\\nthus $CD=CP+DP=8$,\\\\\nsince $CD\\perp OE$,\\\\\nthus $DE=\\frac{1}{2}CD=4$,\\\\\nthus $OF=PE=PD-DE=2$,\\\\\nlet $OE=x$, then $PF=x$, $AF=5-x$,\\\\\nin $\\triangle AOF$, $AF^2+OF^2=OA^2$,\\\\\nin $\\triangle DOE$, $OE^2+DE^2=OD^2$,\\\\\nsince $OA=OD$,\\\\\nthus $AF^2+OF^2=OE^2+DE^2$, i.e., $(5-x)^2+2^2=x^2+4^2$,\\\\\nsolving for $x$ gives $x=\\frac{13}{10}$,\\\\\nin $\\triangle POE$, $OP=\\sqrt{OE^2+PE^2}=\\sqrt{\\left(\\frac{13}{10}\\right)^2+2^2}=\\boxed{\\frac{\\sqrt{569}}{10}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52866055.png"
  },
  {
    "solution": "\\boxed{21^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points $B$ and $D$ are on ray $AM$, and points $C$ and $E$ are on ray $AN$, with $AB=BC=CD=DE$. Given that $\\angle EDM=84^\\circ$, what is the degree measure of $\\angle A$?\n",
    "completion": "\\textbf{Solution:} Given that $AB = BC = CD = DE$,\\\\\nit follows that $\\angle A = \\angle BCA$, $\\angle CBD = \\angle BDC$, and $\\angle ECD = \\angle CED$,\\\\\nBased on the exterior angle property of triangles, $\\angle A + \\angle BCA = \\angle CBD$, $\\angle A + \\angle CDB = \\angle ECD$, and $\\angle A + \\angle CED = \\angle EDM$,\\\\\nAlso, since $\\angle EDM = 84^\\circ$,\\\\\nit follows that $\\angle A + 3\\angle A = 84^\\circ$,\\\\\nSolving for $\\angle A$, we find that $\\angle A = \\boxed{21^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53147939.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a \"clover\" pattern is composed of two congruent rhombuses (rhombus ABCD and rhombus EFGH), where the overlapping part is a regular octagon (shaded area). Points B and D are on EG, and points F and H are on AC. If CF equals 2, what is the length of BD?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram: connect DF, BD, CF, and extend lines BD and CF to meet at point O,\\\\\nsince it's a regular octagon,\\\\\ntherefore $\\angle CDA=(8-2)\\times 180^\\circ \\div 8=135^\\circ$,\\\\\ntherefore $\\angle DCB=180^\\circ - \\angle CDA=45^\\circ$,\\\\\ntherefore $\\angle DCO= \\frac{1}{2} \\angle DCB=22.5^\\circ$.\\\\\nSince $MF=MD$ and $\\angle DMF=135^\\circ$,\\\\\ntherefore $\\angle MDF= \\frac{1}{2}(180^\\circ - \\angle DMF)=22.5^\\circ$,\\\\\ntherefore $\\angle MDF=\\angle DCO$,\\\\\ntherefore $DF=FC=2$,\\\\\nsince $\\angle DOF=360^\\circ \\div 8 \\times 2=90^\\circ$,\\\\\ntherefore $DO^2+OF^2=DO^2+DO^2=DF^2=4$,\\\\\ntherefore $DO= \\sqrt{2}$,\\\\\ntherefore $BD=2DO=2\\sqrt{2}=\\boxed{2\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50475268.png"
  },
  {
    "solution": "\\boxed{\\frac{17}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, point $M$ is on diagonal $BD$. The circle passing through points $A$, $B$, and $M$ intersects $BC$ at point $E$. Given that $AB = 6$, $BC = 8$, and the ratio of $AM$ to $ME$. If $BM = 7$, what is the length of $BE$?\n",
    "completion": "\\textbf{Solution:} In $(1)$ we have already proved that chord $AE$ is the diameter of $\\odot O$,\\\\\nwhich gives $\\angle AME=90^\\circ$,\\\\\ntherefore, in $\\triangle AEM$ right angled at $M$, $AM: ME=\\tan\\angle AEM$,\\\\\nsince $\\angle AEM=\\angle ABM$,\\\\\nin $\\triangle ABD$ right angled at $B$, $\\tan\\angle ABM=\\frac{AD}{AB}$,\\\\\nsince in the rectangle $ABCD$, $AB=CD$, $AD=BC$, $AB=6$, $BC=8$\\\\\ntherefore, $\\tan\\angle ABM=\\frac{AD}{AB}=\\frac{8}{6}=\\frac{4}{3}$,\\\\\ntherefore, $AM: ME=\\tan\\angle AEM=\\tan\\angle ABD=\\frac{4}{3}$;\\\\\n$(2)$ Let $AD$ intersect $\\odot O$ at point $H$, connect $HM$ and $HE$, as shown in the diagram,\\\\\nAccording to the property that opposite angles supplement each other in a cyclic quadrilateral,\\\\\nwe have: $\\angle AHE=\\angle ABE=90^\\circ=\\angle BEH=\\angle BAH$, $\\angle BMH=\\angle BAH=90^\\circ$,\\\\\ntherefore, quadrilateral $ABEH$ is a rectangle, hence $AH=BE$,\\\\\nsince $\\angle BMH=\\angle BAH=90^\\circ$,\\\\\ntherefore, $\\angle DMH=\\angle BAD=90^\\circ$,\\\\\nsince $\\angle HDM=\\angle BDA$,\\\\\ntherefore, $\\triangle DMH\\sim \\triangle DAB$,\\\\\ntherefore, $\\frac{DH}{DM}=\\frac{DB}{DA}$, $DH=\\frac{DB\\times DM}{DA}$,\\\\\nsince $AB=6$, $AD=BC=8$,\\\\\ntherefore, $BD=\\sqrt{AD^2+AB^2}=\\sqrt{8^2+6^2}=10$,\\\\\nsince $BM=7$,\\\\\ntherefore, $DM=BD−BM=3$,\\\\\ntherefore, $DH=\\frac{DB\\times DM}{DA}=\\frac{10\\times 3}{8}=\\frac{15}{4}$,\\\\\ntherefore, $AH=AD−DH=8−\\frac{15}{4}=\\frac{17}{4}$,\\\\\ntherefore, $BE=AH=\\boxed{\\frac{17}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53102171_24.png"
  },
  {
    "solution": "\\boxed{x^{2}-4x-5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the line $y=x-5$ intersects the x-axis at point A and the y-axis at point B. The parabola $y=ax^2+bx+c$ passes through points A and B, and has another intersection point C with the x-axis, where $\\tan\\angle CBO=\\frac{1}{5}$. What is the equation of the parabola?\n",
    "completion": "\\textbf{Solution:} To solve this, in the line $y=x-5$, when $x=0$, $y=-5$, \\\\\nwhen $y=0$, $x=5$, \\\\\n$\\therefore B\\left(0, -5\\right)$, $A\\left(5, 0\\right)$, \\\\\n$\\therefore OB=5$, \\\\\n$\\because \\tan\\angle CBO=\\frac{CO}{OB}=\\frac{CO}{5}=\\frac{1}{5}$, \\\\\n$\\therefore CO=1$, \\\\\n$\\therefore C\\left(-1, 0\\right)$, \\\\\nLet the equation of the parabola be $y=a\\left(x+1\\right)\\left(x-5\\right)$, \\\\\nSubstituting $B\\left(0, -5\\right)$ into the equation yields: $-5=-5a$, \\\\\nSolving this gives: $a=1$, \\\\\n$\\therefore$ the equation of the parabola is $y=\\left(x+1\\right)\\left(x-5\\right)=x^{2}-4x-5=\\boxed{x^{2}-4x-5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52797118_49.png"
  },
  {
    "solution": "\\boxed{6cm}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point C is the midpoint of segment AB, and CD is $\\frac{1}{3}$ of AC. If AD equals 2 cm, what is the length of AB?\n",
    "completion": "\\textbf{Solution:} Let $CD=x$,\\\\\nsince $CD=\\frac{1}{3}AC$,\\\\\ntherefore $AC=3x$\\\\\nsince $AD=2cm$,\\\\\ntherefore $AD=AC-CD=3x-x=2x=2$\\\\\ntherefore $x=1$\\\\\ntherefore $AC=3cm$\\\\\nsince point C is the midpoint of the line segment AB,\\\\\ntherefore $AB=2AC= \\boxed{6cm}$",
    "image_path": "dataset/mm_math/images/MM_Math/52916059.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in $\\triangle ABC$, BD bisects $\\angle ABC$, and $AD \\perp BD$. Let E be the midpoint of AC, with $AD=6\\,\\text{cm}$, $BD=8\\,\\text{cm}$, and $BC=16\\,\\text{cm}$. What is the length of $DE$ in $\\,\\text{cm}$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, extend AD to meet BC at point F,\\\\\n$\\because$AD$\\perp$BD,\\\\\n$\\therefore$AB=$\\sqrt{AD^{2}+BD^{2}}$=$\\sqrt{6^{2}+8^{2}}=10$,\\\\\n$\\because$BD bisects $\\angle$BC, and AD$\\perp$BD, BD=BD,\\\\\n$\\therefore$$\\triangle$ABD$\\cong$$\\triangle$FBD (ASA),\\\\\n$\\therefore$AD=DF, BF=AB=10,\\\\\nAlso $\\because$AE=EC, FC=BC−BF=16−10=6,\\\\\n$\\therefore$DE is the median of $\\triangle$ACF,\\\\\n$\\therefore$DE=$\\frac{1}{2}$FC=\\boxed{3}.",
    "image_path": "dataset/mm_math/images/MM_Math/51770529.png"
  },
  {
    "solution": "\\boxed{\\frac{54}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that D and E are points on sides AC and AB of $\\triangle ABC$ respectively, with $\\angle AED = \\angle C$, $AE=5$, $AC=9$, and $DE=6$. $\\triangle ABC \\sim \\triangle ADE$. Find the length of BC.\n",
    "completion": "\\textbf{Solution:}\n\nSince $ \\triangle ABC\\sim \\triangle ADE$,\\\\\n$\\therefore \\frac{BC}{DE}=\\frac{AC}{AE}$,\\\\\n$\\therefore \\frac{BC}{6}=\\frac{9}{5}$,\\\\\n$\\therefore BC=\\boxed{\\frac{54}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52872384_93.png"
  },
  {
    "solution": "\\boxed{24^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in circle $O$, if $\\angle BAC=24^\\circ$ and $\\angle ACB=42^\\circ$, then what is the degree measure of $\\angle ACO$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle BAC = 24^\\circ$ and $\\angle ACB = 42^\\circ$,\\\\\nit follows that $\\angle BOC = 2\\angle BAC = 48^\\circ$ and $\\angle AOB = 2\\angle ACB = 84^\\circ$,\\\\\nhence $\\angle AOC = \\angle BOC + \\angle AOB = 132^\\circ$.\\\\\nSince OA = OC,\\\\\nit implies that $\\angle ACO = \\angle OAC = \\frac{1}{2}(180^\\circ - 132^\\circ) = \\boxed{24^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52872065.png"
  },
  {
    "solution": "\\boxed{54^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, within a regular pentagon, $BE$ is a diagonal, $AF \\perp BE$ intersects $CD$ at point $F$, then what is the degree measure of $\\angle BAF$?\n",
    "completion": "\\textbf{Solution:} Given that the polygon ABCDE is a regular pentagon,\\\\\n$\\therefore \\angle BAE=\\frac{1}{5}\\times \\left(5-2\\right)\\times 180^\\circ=108^\\circ$,\\\\\n$\\because AF\\perp BE$,\\\\\n$\\therefore \\angle BAF=\\frac{1}{2}\\angle BAE=\\frac{1}{2}\\times 108^\\circ=54^\\circ$.\\\\\nHence, the answer is: $\\boxed{54^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52991505.png"
  },
  {
    "solution": "\\boxed{x \\geq 3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y=kx+b$ intersects the x-axis and y-axis at points $A(3, 0)$ and $B(0, 5)$, respectively. What is the solution set of the inequality $kx+b\\le 0$?\n",
    "completion": "\\textbf{Solution:} From the graph, we can observe that when $y \\leq 0$, then $x \\geq 3$,\\\\\nwhich means the solution set of the inequality $kx+b \\leq 0$ is $\\boxed{x \\geq 3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50590993.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{3}{2}x+3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $l_1$ intersects the x-axis and y-axis at points A and B, respectively, with the coordinates of A and B being $(2, 0)$ and $(0, 3)$, respectively. The line $l_2$ passing through point B is given by $y=\\frac{1}{2}x+3$ and intersects the x-axis at point C. Point $D(n, 6)$ lies on the line $l_1$, and line $CD$ is drawn. What is the analytical expression for $l_1$?\n",
    "completion": "\\textbf{Solution:} Let the equation of line $l_{1}$ be $y=kx+b$, \\\\\nSubstituting points A(2, 0) and B(0,3) into the equation gives\\\\\n$\\begin{cases}2k+b=0\\\\ b=3\\end{cases}$,\\\\\nSolving this system of equations yields $\\begin{cases}k=-\\frac{3}{2}\\\\ b=3\\end{cases}$,\\\\\n$\\therefore$ the equation of line $l_{1}$ is $\\boxed{y=-\\frac{3}{2}x+3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51699679_175.png"
  },
  {
    "solution": "\\boxed{\\frac{5}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, within rectangle $ABCD$, $AC$ and $BD$ intersect at point $O$, and point $E$ is on $AD$, with $CE$ and $BD$ intersecting at point $F$. Let $\\frac{DE}{AD}=k (0<k<1)$, denote the area of $\\triangle CBF$ as $S_{1}$, and the area of quadrilateral $ABFE$ as $S_{2}$. Find the maximum value of $\\frac{S_{2}}{S_{1}}$.\n",
    "completion": ":\\textbf{Solution:}\n\nLet $S_{\\triangle DEF}=x$,\n\nsince AD$\\parallel$BC,\n\nhence $\\triangle DEF\\sim \\triangle CFB$,\n\nthus $\\frac{S_{\\triangle DEF}}{S_{\\triangle CFB}}=\\left(\\frac{DE}{BC}\\right)^2=k^2$,\n\ntherefore $S_1=\\frac{x}{k^2}$,\n\nsince $\\frac{DF}{BF}=\\frac{DE}{BC}=\\frac{DE}{AD}=k$,\n\nhence $\\frac{DF}{BD}=\\frac{k}{k+1}$,\n\nsince FH$\\perp$AD,\n\ntherefore $\\angle FHD=90^\\circ$,\n\nthus $\\angle FHD=\\angle BAD$,\n\nmoreover,\n\nsince $\\angle FDH=\\angle BDA$,\n\nhence $\\triangle FDH\\sim \\triangle BDA$,\n\nthus $\\frac{FH}{AB}=\\frac{DF}{BD}=\\frac{k}{k+1}$,\n\ntherefore $\\frac{S_{\\triangle DEF}}{S_{\\triangle DAB}}=\\frac{\\frac{1}{2}DE\\cdot FH}{\\frac{1}{2}AD\\cdot AB}=\\frac{DE}{AD}\\cdot \\frac{FH}{AB}=\\frac{k^2}{k+1}$,\n\ntherefore $S_{\\triangle DAB}=\\frac{(k+1)x}{k^2}$,\n\nthus $S_2=S_{\\triangle DAB}-S_{\\triangle DEF}=\\frac{(k+1-k^2)x}{k^2}$,\n\nhence $\\frac{S_2}{S_1}=k+1-k^2=-\\left(k-\\frac{1}{2}\\right)^2+\\frac{5}{4}$,\n\ntherefore, when $k=\\frac{1}{2}$, the maximum value of $\\frac{S_2}{S_1}$ is $\\boxed{\\frac{5}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53009704_165.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{10}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system, the line $l_1: y=2x-1$ intersects the $x$-axis and the $y$-axis at points $A$ and $B$, respectively, and the line $l_2: y=-\\frac{1}{2}x+1$ intersects the $x$-axis and the $y$-axis at points $P$ and $C$, respectively. Connecting $AC$, the lines $l_1$ and $l_2$ intersect at point $D$. What is the area of $\\triangle ACD$?\n",
    "completion": "\\textbf{Solution:} Given $y=2x-1$, let $y=0$, we find $x=\\frac{1}{2}$, thus $A\\left(\\frac{1}{2}, 0\\right)$,\\\\\nlet $x=0$, we find $y=-1$, thus $B\\left(0, -1\\right)$,\\\\\nGiven $y=-\\frac{1}{2}x+1$, let $x=0$, we find $y=1$, thus $C\\left(0, 1\\right)$,\\\\\nlet $y=0$, we find $x=2$, thus $P\\left(2, 0\\right)$,\\\\\n$\\therefore BC=2, AO=\\frac{1}{2}$,\\\\\n$\\therefore Area_{\\triangle ACD}=Area_{\\triangle BCD}-Area_{\\triangle ABC}=\\frac{1}{2}BC\\times \\left|x_{D}\\right|-\\frac{1}{2}BC\\times OA=\\frac{1}{2}\\times 2\\times \\left(\\frac{4}{5}-\\frac{1}{2}\\right)=\\boxed{\\frac{3}{10}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52386314_131.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $AB$ is the diameter of the semicircle $O$, $C$ is a point on the semicircle arc, $P$ is the midpoint of $BC$, and $PD\\parallel BC$ intersects the extension of $AB$ at point $D$. If $AB=10$ and $\\cos\\angle D=\\frac{4}{5}$, what is the value of $PC$?\n",
    "completion": "\\textbf{Solution:} Given that $AB$ is the diameter, then $\\angle ACB=90^\\circ$. Since $PD\\parallel BC$, it follows that $\\angle D=\\angle ABC$. Therefore, $\\cos\\angle D = \\cos\\angle ABC = \\frac{BC}{AB} = \\frac{4}{5}$. Solving this gives $BC=8$. Since $OP\\perp CB$, it follows that $CE=BE=\\frac{1}{2}BC=4$. In $\\triangle BOE$, we have $OE=\\sqrt{OB^2-BE^2}=\\sqrt{5^2-4^2}=3$. Thus, $PE=OP-OE=5-3=2$. In $\\triangle CEP$, we find that $PC=\\sqrt{CE^2+PE^2}=\\sqrt{4^2+2^2}=\\boxed{2\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51369905_95.png"
  },
  {
    "solution": "\\boxed{(2, 0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=ax^2+bx+3$ intersects the x-axis at points $A(-1, 0)$ and $B(3, 0)$, and intersects the y-axis at point C. Line $BC$ is drawn. When the parabola $y=ax^2+bx+3$ is translated 1 unit to the right, a new parabola is obtained. Point M is on the axis of symmetry of the new parabola and point N is a moving point on the new parabola. If the quadrilateral formed by points M, N, B, and C is a parallelogram, directly write down the coordinates of point M.\n",
    "completion": "\\textbf{Solution:} The coordinates of point $M$ are $\\boxed{(2, -8)}$, $\\boxed{(2, -2)}$, or $\\boxed{(2, 0)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53107063_280.png"
  },
  {
    "solution": "\\boxed{105^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $OA \\perp OB$, $\\angle COD = 60^\\circ$. If $\\angle BOC = \\frac{3}{7}\\angle AOD$, what is the degree measure of $\\angle AOD$?\n",
    "completion": "\\textbf{Solution:}\nGiven: $\\because$ $\\angle AOB=90^\\circ$,\\\\\n$\\therefore$ $\\angle BOD=\\angle AOD-\\angle AOB=\\angle AOD-90^\\circ$,\\\\\n$\\because$ $\\angle COD=60^\\circ$,\\\\\n$\\therefore$ $\\angle BOD=\\angle COD-\\angle BOC=60^\\circ-\\angle BOC$,\\\\\n$\\therefore$ $60^\\circ-\\angle BOC=\\angle AOD-90^\\circ$,\\\\\n$\\because$ $\\angle BOC=\\frac{3}{7}\\angle AOD$,\\\\\n$\\therefore$ $60^\\circ-\\frac{3}{7}\\angle AOD=\\angle AOD-90^\\circ$,\\\\\nSolving, we get: $\\angle AOD=\\boxed{105^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51482433_175.png"
  },
  {
    "solution": "\\boxed{40^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of $\\odot O$, $C$ and $D$ are points on $\\odot O$, $\\angle CDB=25^\\circ$. A tangent to $\\odot O$ passing through point $C$ intersects the extension of $AB$ at point $E$. What is the measure of $\\angle E$?\n",
    "completion": "\\textbf{Solution:} Draw line OC,\\\\\n$\\because$ CE is the tangent to $\\odot$O,\\\\\n$\\therefore$ OC$\\perp$CE,\\\\\nwhich means $\\angle$OCE$=90^\\circ$,\\\\\n$\\because$ $\\angle$COB$=2\\angle$CDB$=50^\\circ$,\\\\\n$\\therefore$ $\\angle$E$=90^\\circ-\\angle$COB$=\\boxed{40^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52623549.png"
  },
  {
    "solution": "\\boxed{9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Four congruent right-angled triangles are assembled as shown in the figure to form two differently sized squares, $ABCD$ and $EFGH$, where the ratio of the lengths of the two perpendicular sides of each right-angled triangle is $1:2$. If the area of square $EFGH$ is $5$, what is the area of square $ABCD$?\n",
    "completion": "\\textbf{Solution:} From the given problem, we have $Rt\\triangle AEH \\cong Rt\\triangle BFE$,\\\\\ntherefore $AH=BE$,\\\\\nsuppose $AH<AE$, and let $AH=BE=x$, then $AE=2x$, $AB=AE+BE=3x$,\\\\\nin $Rt\\triangle AEH$, by the Pythagorean theorem, we get: $EH^2=AE^2+AH^2=5x^2$,\\\\\nsince the area of square $EFGH$ is 5,\\\\\ntherefore $EH^2=5x^2=5$,\\\\\nsolving this yields $x^2=1$,\\\\\nthus, the area of square $ABCD$ is $AB^2=9x^2=\\boxed{9}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52188832.png"
  },
  {
    "solution": "\\boxed{130^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, AB is the diameter of the semicircle O, and $\\angle BAC=40^\\circ$. What is the measure of $\\angle D$ in degrees?\n",
    "completion": "\\textbf{Solution:} Let $AB$ be the diameter of the semicircle O,\\\\\n$\\therefore$ $\\angle ACB = 90^\\circ$,\\\\\n$\\therefore$ $\\angle BAC + \\angle B = 90^\\circ$,\\\\\nSince $\\angle BAC = 40^\\circ$,\\\\\n$\\therefore$ $\\angle B = 90^\\circ - \\angle BAC = 90^\\circ - 40^\\circ = 50^\\circ$,\\\\\nSince $\\angle B + \\angle D = 180^\\circ$,\\\\\n$\\therefore$ $\\angle D = 180^\\circ - 50^\\circ = \\boxed{130^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52874742.png"
  },
  {
    "solution": "\\boxed{(-4, -1)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, with O as the center and the length of OP as the radius, an arc is drawn intersecting the negative x-axis at point A. Thus, the coordinates of point A are ($- \\sqrt{17}$, 0). Given that the y-coordinate of point P is $-1$, what are the coordinates of point P?\n",
    "completion": "\\textbf{Solution:} Construct line $PB\\perp OA$ at point B, \\\\\n$\\because$ the coordinates of point A are ($-\\sqrt{17}$, 0), \\\\\n$\\therefore$ OP=OA=$\\sqrt{17}$, \\\\\n$\\because$ the ordinate of point P is $-1$, \\\\\n$\\therefore$ PB=1, \\\\\n$\\therefore$ OB=$\\sqrt{OP^{2}-PB^{2}}=4$, \\\\\n$\\therefore$ the coordinates of point P are $\\boxed{(-4, -1)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53359331.png"
  },
  {
    "solution": "\\boxed{}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, point $E$ is on side $AC$, and point $F$ is on side $AD$, with $AF=CE$. $EF$ intersects diagonal $BD$ at point $O$. $O$ is the midpoint of $BD$. If $EF\\perp BD$ and the perimeter of quadrilateral $ABCD$ is 24, when connecting $BF$, what is the perimeter of $\\triangle ABF$?\n",
    "completion": "\\textbf{Solution:} Given the insufficient information from the problem statement, it is not possible to determine the perimeter of $\\triangle ABF$ directly. Additional information is required to solve this problem.",
    "image_path": "dataset/mm_math/images/MM_Math/51747371_161.png"
  },
  {
    "solution": "\\boxed{9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, the bisector of $\\angle BAC$ intersects the circumcircle of $\\triangle ABC$ at point D and intersects BC at point E. The bisector of $\\angle ABC$ intersects AD at point F. If BD$=6$ and DF$=4$, what is the length of AD?\n",
    "completion": "\\textbf{Solution:} Given $\\angle 5=\\angle 2=\\angle 1$ and $\\angle FDB=\\angle BDA$,\\\\\n$\\therefore \\triangle DBF\\sim \\triangle ADB$,\\\\\n$\\therefore \\frac{BD}{DA}=\\frac{DF}{DB}$, thus $\\frac{6}{AD}=\\frac{4}{6}$,\\\\\n$\\therefore AD=\\boxed{9}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52745974_85.png"
  },
  {
    "solution": "\\boxed{7}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the line $l\\colon y=x+4$ intersects with the graph of the inverse proportion function $y=\\frac{k}{x}$ ($x<0$) at point $A(-1,n)$, and the line $l'$ passes through point $A$ and is symmetrical to $l$ with respect to the line $x=-1$. What is the area of the shaded region in the figure?\n",
    "completion": "\\textbf{Solution:} Let the line $l: y=x+4$ intersect the x-axis and y-axis at points B and C, respectively,\\\\\nthen B$(-4,0)$, C$(0,4)$.\\\\\nSince line $l'$ passes through point A and is symmetric to $l$ with respect to the line $x=-1$,\\\\\nlet the intersection of line $l'$ with the x-axis be point E, then E$(2,0)$.\\\\\nLet line $l': y=ax+b$, then $\\left\\{\\begin{array}{l}\n3=-a+b\\\\\n0=2a+b\n\\end{array}\\right.$ solves to $\\left\\{\\begin{array}{l}\na=-1,\\\\\nb=2,\n\\end{array}\\right.$\\\\\ntherefore $y=-x+2$. Let the intersection of line $l'$ with the y-axis be point D, then D$(0,2)$,\\\\\ntherefore, the area of the shaded part = area of $\\triangle BOC$ - area of $\\triangle ACD$ = $\\frac{1}{2}\\times4\\times4-\\frac{1}{2}\\times2\\times1=\\boxed{7}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55608665_58.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB\\parallel CD\\parallel EF$, $AF$ intersects with $BE$ at point $G$, and $AG=4$, $GD=2$, $DF=8$. What is the value of $\\frac{BC}{CE}$?\n",
    "completion": "\\textbf{Solution:} Since $AB\\parallel CD\\parallel EF$, \\\\\nit follows that $\\frac{BC}{CE}=\\frac{AD}{DF}=\\frac{AG+GD}{DF}$, \\\\\ngiven that $AG=4, GD=2, DF=8$, \\\\\nwe have $\\frac{BC}{CE}=\\frac{AD}{DF}=\\frac{AG+GD}{DF}=\\frac{4+2}{8}=\\boxed{\\frac{3}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52623770.png"
  },
  {
    "solution": "\\boxed{16\\colon 21}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, D is a point on side AB, with $AD:DB = 3:4$ and $DE \\parallel AC$, intersecting BC at point E. What is the value of $\\frac{S_{\\triangle BDE}}{S_{\\triangle AEC}}$?\n",
    "completion": "\\textbf{Solution:} Since $DE\\parallel AC$,\\\\\nit follows that $\\triangle BDE\\sim \\triangle BAC$, and $AD\\colon DB=3\\colon 4$,\\\\\nthus $BD\\colon AB=DE\\colon AC=4\\colon 7$, $S_{\\triangle BDE}\\colon S_{\\triangle BAC}=16\\colon 49$,\\\\\ntherefore $S_{\\triangle BDE}\\colon S_{\\text{quadrilateral }DECA}=16\\colon 33$,\\\\\nsince $DE\\colon AC=4\\colon 7$, and the heights of $\\triangle ADE$ and $\\triangle ACE$ are equal,\\\\\nit follows that $S_{\\triangle ADE}\\colon S_{\\triangle ACE}=4\\colon 7=12\\colon 21$,\\\\\ntherefore $S_{\\triangle BDE}\\colon S_{\\triangle AEC}=\\boxed{16\\colon 21}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53106640.png"
  },
  {
    "solution": "\\boxed{\\frac{49}{25}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, points $A_1$, $A_2$, $A_3$ lie on the line $y=\\frac{1}{6}x+b$, and points $B_1$, $B_2$, $B_3$ lie on the x-axis; $\\triangle O A_1 B_1$, $\\triangle B_1 A_2 B_2$, $\\triangle B_2 A_3 B_3$ are all isosceles right triangles. Given that point $A_1(1,1)$, what is the y-coordinate of point $A_3$?\n",
    "completion": "\\textbf{Solution:} Let's set the ordinate of points $A_{2}$ and $A_{3}$ to be $m$ and $n$ respectively, \\\\\nsince $A_{1}(1,1)$ lies on the line $y= \\frac{1}{6}x+b$, \\\\\nwe have $b= \\frac{5}{6}$, \\\\\nthus $y= \\frac{1}{6}x+ \\frac{5}{6}$. \\\\\nLet $A_{2}(m+2, m)$, substituting into $y=\\frac{1}{6}x+\\frac{5}{6}$, we get: $m=\\frac{1}{6}(m+2)+\\frac{5}{6}$, solving for $m$ gives: $m=\\frac{7}{5}$; \\\\\nLet $A_{3}\\left(2+2\\times \\frac{7}{5}+n, n\\right)$, substituting into $y=\\frac{1}{6}x+\\frac{5}{6}$, we get: $n=\\frac{1}{6}\\left(\\frac{24}{5}+n\\right)+\\frac{5}{6}$, solving for $n$ gives: $n=\\boxed{\\frac{49}{25}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50595458.png"
  },
  {
    "solution": "\\boxed{38^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, in $\\triangle ABC$, $\\angle B=80^\\circ$, $\\angle C=30^\\circ$. If $\\triangle ABC \\cong \\triangle ADE$, $\\angle DAC=32^\\circ$, then what is the degree measure of $\\angle EAC$?\n",
    "completion": "\\textbf{Solution:} Given: $\\because$ $\\angle B=80^\\circ$ and $\\angle C=30^\\circ$,\\\\\n$\\therefore$ $\\angle BAC=180^\\circ-80^\\circ-30^\\circ=70^\\circ$,\\\\\n$\\because$ $\\triangle ABC \\cong \\triangle ADE$,\\\\\n$\\therefore$ $\\angle BAC=\\angle DAE=70^\\circ$,\\\\\n$\\therefore$ $\\angle EAC=\\angle DAE-\\angle DAC=70^\\circ-32^\\circ=\\boxed{38^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/53043996.png"
  },
  {
    "solution": "\\boxed{75^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is assembled from a pair of set squares. What is the degree measure of $\\angle ABC$ in the figure?\n",
    "completion": "\\textbf{Solution:} Given $\\because$ $\\angle F=30^\\circ$ and $\\angle EAC=45^\\circ$, \\\\\n$\\therefore$ $\\angle ABF=\\angle EAC-\\angle F=45^\\circ-30^\\circ=15^\\circ$, \\\\\n$\\because$ $\\angle FBC=90^\\circ$, \\\\\n$\\therefore$ $\\angle ABC=\\angle FBC-\\angle ABF=90^\\circ-15^\\circ=\\boxed{75^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53106965.png"
  },
  {
    "solution": "\\boxed{160^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the circle $\\odot O$, points C and D lie on $\\odot O$, and are distributed on opposite sides of diameter AB. Extending CO intersects chord BD at point E. If $\\angle DEC=120^\\circ$, and point A is the midpoint of $\\overset{\\frown}{DC}$, then what is the degree measure of $\\overset{\\frown}{DC}$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, join OD, \\\\\n$\\because$ point A is the midpoint of $ \\overset{\\frown}{DC}$, and AB is the diameter, \\\\\n$\\therefore$ AB$\\perp$CD, and $\\angle$AOC = $\\angle$AOD, \\\\\n$\\therefore$ $\\angle$AOD = 2$\\angle$B, \\\\\nLet $\\angle$B = x, then $\\angle$AOC = $\\angle$AOD = 2x, and $\\angle$ODB = $\\angle$B = x, \\\\\n$\\because$ $\\angle$DEC = $120^\\circ$, and $\\angle$COD = $\\angle$DEC + $\\angle$ODB, \\\\\n$\\therefore$ 4x = x + $120^\\circ$, \\\\\nSolving for x, we get x = $40^\\circ$, \\\\\n$\\therefore$ 4x = $160^\\circ$, \\\\\nThat is, $\\angle$COD = $160^\\circ$, \\\\\n$\\therefore$ the degree of $ \\overset{\\frown}{CD}$ is $\\boxed{160^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52755183.png"
  },
  {
    "solution": "\\boxed{28\\text{cm}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, an isosceles right-angled triangle object $ABC$ is leaning against a corner of the wall ($\\angle O = 90^\\circ$). If $OA = 50cm$ and $OB = 28cm$, what is the distance in $cm$ of point $C$ from the ground?\n",
    "completion": "\\textbf{Solution:} Draw $CD \\perp OB$ at point D, as shown in the diagram:\\\\\n$\\therefore$ $\\angle CDB=\\angle AOB=90^\\circ$\\\\\n$\\because$ $\\triangle ABC$ is an isosceles right triangle\\\\\n$\\therefore$ AB=CB, $\\angle ABC=90^\\circ$\\\\\n$\\therefore$ $\\angle ABO+\\angle CBD=90^\\circ$\\\\\n$\\therefore$ $\\angle ABO=\\angle BCD$,\\\\\n$\\therefore$ $\\triangle ABO\\cong \\triangle BCD(AAS)$\\\\\n$\\therefore$ CD=BO=\\boxed{28\\text{cm}}",
    "image_path": "dataset/mm_math/images/MM_Math/55502385.png"
  },
  {
    "solution": "\\boxed{solution}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of $\\odot O$, and lines $BD$ and $CD$ are the tangents to $\\odot O$ at points $B$ and $C$, respectively. If $BD=2$, what is the length of $CD$?\n",
    "completion": "\\textbf{Solution:} Due to insufficient information provided in the question, it is impossible to offer a \\boxed{solution}.",
    "image_path": "dataset/mm_math/images/MM_Math/52551457_90.png"
  },
  {
    "solution": "\\boxed{y=2x+6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the line $y=\\frac{3}{4}x+6$ intersects the x-axis at point A and the y-axis at point B. Triangle $\\triangle AOB$ is folded along BC, where point O precisely lands on point D on side AB, with BC being the crease. Find the analytical expression of line BC.\n",
    "completion": "\\textbf{Solution:} According to the problem statement, we have: $\\angle ADC=\\angle BDC=\\angle BOC=90^\\circ$, $BD=OB=6$,\\\\\n$\\therefore AD=AB-BD=10-6=4$,\\\\\nlet $OC=x$, then $AC=OA-OC=8-x$,\\\\\nin $\\triangle ACD$, $AD^2+CD^2=AC^2$,\\\\\n$\\therefore 4^2+x^2=(8-x)^2$, solving for $x$ yields: $x=3$, that is $OC=3$,\\\\\n$\\therefore$ the coordinates of point C are $(-3,0)$,\\\\\nlet the equation of line BC be $y=kx+b\\ (k\\neq 0)$,\\\\\nsubstituting points B$(0,6)$, C$(-3,0)$ into the equation gives:\\\\\n$\\begin{cases} b=6 \\\\ -3k+b=0 \\end{cases}$, solving for $k$ and $b$ gives: $\\begin{cases} k=2 \\\\ b=6 \\end{cases}$,\\\\\n$\\therefore$ the equation of line BC is $y=2x+6$. Hence, the answer is $\\boxed{y=2x+6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52212500_114.png"
  },
  {
    "solution": "\\boxed{\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, there is a clock face. If we connect points $B$ and $D$ that represent 2 o'clock and 10 o'clock respectively, and extend this line to intersect the tangent line through the 8 o'clock point at point $P$, and if the length of the tangent line $PC=2$, what is the radius of the clock face?\n",
    "completion": "\\textbf{Solution:} Let the center of the clock be point $O$, and connect $BC$ and $OD$,\\\\\nFrom the given information,\\\\\nPoint $O$ is on $BC$, $\\angle DOC=2\\times 30^\\circ=60^\\circ$,\\\\\n$\\therefore \\angle DBC=\\frac{1}{2}\\angle DOC=30^\\circ$,\\\\\n$\\because$ Line $PC$ is tangent to circle $O$ at point $C$,\\\\\n$\\therefore \\angle BCP=90^\\circ$,\\\\\n$\\therefore BC=\\frac{CP}{\\tan 30^\\circ}=\\frac{2}{\\frac{\\sqrt{3}}{3}}=2\\sqrt{3}$,\\\\\n$\\therefore$ The radius of the clock face is $\\boxed{\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52457420.png"
  },
  {
    "solution": "\\boxed{\\frac{107}{33}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, the parabola $y=ax^2+bx+c$ passes through points $A(-1,0)$, $B(4,0)$, and $C(0,3)$. Point $D$ lies on the parabola above line $BC$, and $DE$ is perpendicular to $BC$ at point $E$. In diagram 2, the midpoint of $AB$ is marked as $F$, and lines $CD$ and $CF$ are drawn. There exists a point $D$ such that one of the angles in $\\triangle CDE$ is equal to $\\angle CFO$. Find the x-coordinate of point $D$.\n",
    "completion": "\\textbf{Solution:} Existence exists. Assume that there is such a point $D$, $\\triangle CDE$ where there is an angle equal to $\\angle CFO$,\\\\\n$\\because$ point $F$ is the midpoint of $AB$,\\\\\n$\\therefore OF=\\frac{3}{2}$, $\\tan\\angle CFO=\\frac{OC}{OF}=2$,\\\\\nDraw $BG\\perp BC$ through point $B$, intersecting the extension of $CD$ at point $G$, and draw $GH\\perp x$ axis through point $G$, with the foot of the perpendicular at $H$,\\\\\n(1) If $\\angle DCE = \\angle CFO$,\\\\\n$\\therefore\\tan\\angle DCE=\\frac{GB}{BC}=2$,\\\\\n$\\therefore BG=10$,\\\\\n$\\because\\triangle GBH\\sim \\triangle BCO$,\\\\\n$\\therefore\\frac{GH}{BO}=\\frac{HB}{OC}=\\frac{GB}{BC}$,\\\\\n$\\therefore GH=8$, $BH=6$,\\\\\n$\\therefore G(10,8)$,\\\\\nLet the analytical equation of line $CG$ be $y=kx+b'$,\\\\\n$\\therefore\\left\\{\\begin{array}{l}b'=3\\\\ 10k+b'=8\\end{array}\\right.$,\\\\\nSolving we get $\\left\\{\\begin{array}{l}k=\\frac{1}{2}\\\\ b'=3\\end{array}\\right.$,\\\\\n$\\therefore$ the analytical equation of line $CG$ is $y=\\frac{1}{2}x+3$,\\\\\n$\\therefore\\left\\{\\begin{array}{l}y=\\frac{1}{2}x+3\\\\ y=\\frac{3}{4}x^{2}+\\frac{9}{4}x+3\\end{array}\\right.$,\\\\\nSolving we get $x=\\boxed{\\frac{7}{3}}$, or $x=0$ (discard).\\\\\n(2) If $\\angle CDE = \\angle CFO$,\\\\\nSimilarly, we get $BG=\\frac{5}{2}$, $GH=2$, $BH=\\frac{3}{2}$,\\\\\n$\\therefore G\\left(\\frac{11}{2},2\\right)$,\\\\\nSimilarly, we can obtain that the analytical equation of line $CG$ is $y=-\\frac{2}{11}x+3$,\\\\\n$\\therefore\\left\\{\\begin{array}{l}y=-\\frac{2}{11}x+3\\\\ y=\\frac{3}{4}x^{2}+\\frac{9}{4}x+3\\end{array}\\right.$,\\\\\nSolving we get $x=\\boxed{\\frac{107}{33}}$ or $x=0$ (discard),\\\\\nIn conclusion, there exists a point $D$, such that there is an angle in $\\triangle CDE$ equal to $\\angle CFO$, and the $x$-coordinate of point $D$ is \\boxed{\\frac{7}{3}} or \\boxed{\\frac{107}{33}}.",
    "image_path": "dataset/mm_math/images/MM_Math/52551774_97.png"
  },
  {
    "solution": "\\boxed{}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ABC=90^\\circ$, $\\angle A=60^\\circ$. Point D lies on side $AC$, $\\angle DBC=30^\\circ$, $AC=12\\text{cm}$. What is the perimeter of $\\triangle ABD$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a parallelogram, \\\\\nit follows that OA=OC, \\\\\nand since E is the midpoint of AB, \\\\\nit follows that OE is the median of $\\triangle ABC$, \\\\\nthus OE$=\\frac{1}{2}$BC=5.",
    "image_path": "dataset/mm_math/images/MM_Math/52345515.png"
  },
  {
    "solution": "\\boxed{25\\sqrt{3}\\,mm}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in circle $\\odot O$ with a radius of $50mm$, the chord $AB$ has a length of $50mm$. Question: What is the distance in $mm$ from point $O$ to $AB$?\n",
    "completion": "\\textbf{Solution:} Draw $OC\\perp AB$ at point $O$ with the foot at point $C$, as shown in the figure,\\\\\nby the perpendicular diameter theorem, we have $AC=CB=\\frac{1}{2}AB=25\\,mm$,\\\\\nin $\\triangle OAC$, $OC^2=OA^2-AC^2=50^2-25^2=25^2\\times 3$,\\\\\n$\\therefore OC=\\sqrt{25^2\\times 3}=25\\sqrt{3}\\,mm$,\\\\\nthat is, the distance from point $O$ to $AB$ is \\boxed{25\\sqrt{3}\\,mm}.",
    "image_path": "dataset/mm_math/images/MM_Math/52978029_151.png"
  },
  {
    "solution": "\\boxed{C_{4}\\left(1, -1\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=kx$ intersects the parabola $y=x^2+c$ at points A and B, where the coordinates of point B are $(2, 2)$. Point C is a point on the parabola, and the area of $\\triangle ABC$ is 3. Find the coordinates of point C.\n",
    "completion": "\\textbf{Solution:} Solve the system of equations $\\begin{cases}y=kx\\\\ y=x^{2}+c\\end{cases}$, yielding $\\begin{cases}x=2\\\\ y=2\\end{cases}$ or $\\begin{cases}x=-1\\\\ y=-1\\end{cases}$.\\\\\nWhen C is above line AB, construct $CD\\perp x$ axis intersecting AB at D, with the coordinates of point C as $\\left(t, t^{2}+c\\right)$, hence the coordinates of point D are $\\left(t, kt\\right)$,\\\\\n$\\therefore CD=t^{2}+c-kt$,\\\\\n$\\therefore S_{\\triangle ABC}=S_{\\triangle DBC}-S_{\\triangle ADC}=\\frac{1}{2}CD\\times (2-t)-\\frac{1}{2}CD\\times (-1-t)=\\frac{3}{2}CD=3$,\\\\\n$\\therefore CD=2$,\\\\\n$\\therefore t^{2}+c-kt=2$,\\\\\nyielding $t_{1}=\\frac{1-\\sqrt{17}}{2}$, $t_{2}=\\frac{1+\\sqrt{17}}{2}$,\\\\\n$\\therefore t^{2}+c=\\frac{5-\\sqrt{17}}{2}$ or $\\frac{5+\\sqrt{17}}{2}$,\\\\\n$\\therefore C_{1}\\left(\\frac{1-\\sqrt{17}}{2}, \\frac{5-\\sqrt{17}}{2}\\right)$, $C_{2}\\left(\\frac{1+\\sqrt{17}}{2}, \\frac{5+\\sqrt{17}}{2}\\right)$,\\\\\nWhen C is below line AB, construct $CD\\perp x$ axis intersecting AB at D, with the coordinates of point C as $\\left(t, t^{2}+c\\right)$, hence the coordinates of point D are $\\left(t, kt\\right)$,\\\\\n$\\therefore CD=-t^{2}-c+kt$,\\\\\n$\\therefore S_{\\triangle ABC}=S_{\\triangle BBC}+S_{\\triangle ADC}=\\frac{1}{2}CD(2-t)+\\frac{1}{2}CD(t+1)=\\frac{3}{2}CD=3$,\\\\\n$\\therefore CD=2$,\\\\\n$\\therefore t-t^{2}+c=2$,\\\\\nyielding $t_{1}=0$, $t_{2}=1$,\\\\\n$\\therefore t^{2}+c=-2$ or $-1$,\\\\\n$\\therefore C_{3}\\left(0, -2\\right)$, $C_{4}\\left(1, -1\\right)$,\\\\\n$\\therefore$ the coordinates of point C are $\\boxed{C_{1}\\left(\\frac{1-\\sqrt{17}}{2}, \\frac{5-\\sqrt{17}}{2}\\right)}$, $\\boxed{C_{2}\\left(\\frac{1+\\sqrt{17}}{2}, \\frac{5+\\sqrt{17}}{2}\\right)}$, $\\boxed{C_{3}\\left(0, -2\\right)}$, $\\boxed{C_{4}\\left(1, -1\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53066184_90.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $AB = 8$, and $BC = 4$. The rectangle is folded along $AC$, and point $D$ falls on point $D'$. What is the area of the overlapping part, $\\triangle AFC$?\n",
    "completion": "\\textbf{Solution:} Given that rectangle $ABCD$ is folded,\\\\\nit follows that $AD=BC=AD^{\\prime}, \\angle D=\\angle B=90^\\circ=\\angle D^{\\prime},$ \\\\\nsince $\\angle AFD^{\\prime}=\\angle CFB,$ \\\\\nit can be concluded that $\\triangle AFD^{\\prime} \\cong \\triangle CFB,$ \\\\\nhence, $D^{\\prime}F=BF,$ \\\\\nlet $D^{\\prime}F=x$, then $AF=8-x$,\\\\\nin $\\triangle AFD^{\\prime}$, $(8-x)^{2}=x^{2}+4^{2}$,\\\\\nsolving this yields: $x=3$,\\\\\nthus, $AF=AB-FB=8-3=5$,\\\\\ntherefore, the area of $\\triangle AFC$ is $\\frac{1}{2} \\cdot AF \\cdot BC = \\boxed{10}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51918836.png"
  },
  {
    "solution": "\\boxed{\\left(3, 1\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y = -x + 4$ intersects with the graph of the inverse proportion function $y = \\frac{k}{x}$ at points $A(3, a)$ and $B$, and intersects the y-axis at point $C$. What is the expression of the inverse proportion function?\n",
    "completion": "\\textbf{Solution:} Substituting $A\\left(3, a\\right)$ into the linear equation $y=-x+4$, we have,\\\\\n$a=-3+4=1$,\\\\\n$\\therefore$ the coordinates of point A are $\\boxed{\\left(3, 1\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51353246_279.png"
  },
  {
    "solution": "\\boxed{(2,0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=-\\frac{1}{4}x^{2}+x+3$ intersects the $x$-axis at points $A$ and $B$, and intersects the $y$-axis at point $C$. Point $F$ is the vertex of the parabola. There is a point $G$ on the axis of symmetry of the parabola. When is the coordinate of point $G$ such that $\\triangle AFG$ is an isosceles triangle?\n",
    "completion": "\\textbf{Solution:} When $y=0$, we have $-\\frac{1}{4}x^{2}+x+3=0$,\\\\\nwhich gives us $x_{1}=-2$ and $x_{2}=6$,\\\\\n$\\therefore A(-2,0)$ and $B(6,0)$,\\\\\n$\\because y=-\\frac{1}{4}x^{2}+x+3=-\\frac{1}{4}(x-2)^{2}+4$,\\\\\n$\\therefore$ the axis of symmetry is the line $x=2$, and the coordinates of vertex $F$ are $(2,4)$,\\\\\n$\\therefore AF=\\sqrt{(2+2)^{2}+(4-0)^{2}}=4\\sqrt{2}$,\\\\\nLet point $G(2,t)$,\\\\\nIf $\\triangle AFG$ is an isosceles triangle,\\\\\n(1) For $GA=GF$,\\\\\n$(2+2)^{2}+t^{2}=(4-t)^{2}$,\\\\\nsolving this gives $t=0$,\\\\\n$\\therefore G(2,0)$;\\\\\n(2) For $FG=FA=4\\sqrt{2}$,\\\\\n$\\therefore G(2,4+4\\sqrt{2})$ or $(2,4-4\\sqrt{2})$;\\\\\n(3) For $AG=AF$, point $G$ is symmetric to point $F$ with respect to the $x$-axis,\\\\\n$\\therefore$ the coordinates of point $G$ are $(2,-4)$,\\\\\nIn summary, the coordinates of point $G$ are $\\boxed{(2,0)}$ or $(2,-4)$ or $(2,4+4\\sqrt{2})$ or $(2,4-4\\sqrt{2})$.",
    "image_path": "dataset/mm_math/images/MM_Math/52699827.png"
  },
  {
    "solution": "\\boxed{M\\left(-\\frac{12}{17}, -\\frac{20}{17}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the graph, it is known that the parabola $y=ax^2+bx+c$ ($a\\neq 0$) passes through points $A(4, 0)$, $B(-1, 0)$, and $C(0, -4)$. If the circle centered at point $A$ is tangent to line $BC$ at point $M$, find the coordinates of the tangent point $M$.\n",
    "completion": "\\textbf{Solution:} Draw $AM\\perp BC$ through point A, with M as the foot,\n\nthus, $\\angle AMB = \\angle COB = 90^\\circ$,\n\nsince $\\angle ABM = \\angle CBO$,\n\ntherefore, $\\triangle BOC \\sim \\triangle BMA$,\n\ntherefore, $\\frac{OB}{BM}=\\frac{BC}{AB}$,\n\nsince $BC = \\sqrt{OC^2 + OB^2} = \\sqrt{17}$,\n\ntherefore, $\\frac{1}{BM} = \\frac{\\sqrt{17}}{5}$,\n\ntherefore, $BM = \\frac{5}{\\sqrt{17}}$,\n\nDraw $MN \\perp AB$ through M, with N as the foot,\n\nthus, $MN \\parallel OC$,\n\ntherefore, $\\triangle BOC \\sim \\triangle BNM$,\n\ntherefore, $\\frac{OC}{MN} = \\frac{OB}{NB} = \\frac{\\sqrt{17}}{\\frac{5}{\\sqrt{17}}}$,\n\ntherefore, $MN = \\frac{4}{MN} = \\frac{1}{NB} = \\frac{20}{17}$, $NB = \\frac{5}{17}$,\n\ntherefore, $ON = \\frac{12}{17}$,\n\ntherefore, $\\boxed{M\\left(-\\frac{12}{17}, -\\frac{20}{17}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53019895_84.png"
  },
  {
    "solution": "\\boxed{28}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point B is a point on line segment AC, and $AB=21$, $BC= \\frac{1}{3}AB$. What is the length of line segment AC?\n",
    "completion": "\\textbf{Solution:} Given that $AB=21$ and $BC= \\frac{1}{3}AB$, \\\\\nit follows that $BC=7$ and $AC=AB+BC=21+7=\\boxed{28}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55307630_3.png"
  },
  {
    "solution": "\\boxed{(3, -3)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the graph of the linear function $y=kx+b$ passes through point $A(-2,6)$, and intersects with the x-axis at point $B$, and intersects with the graph of the direct proportion function $y=3x$ at point $C$, where the x-coordinate of point $C$ is $1$. In the coordinate plane, is there a point $P$ such that the quadrilateral with vertices $O$, $B$, $C$, and $P$ is a parallelogram? If it exists, directly write down the coordinates of point $P$.\n",
    "completion": "\\textbf{Solution:} Yes, there exists such a point $P$, and the coordinates of point $P$ are $\\boxed{(-3, 3)}$ or $\\boxed{(3, -3)}$ or $(5, 3)$.",
    "image_path": "dataset/mm_math/images/MM_Math/53706908_200.png"
  },
  {
    "solution": "\\boxed{7\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AC$ bisects $\\angle BAD$, and $\\angle BAD = \\angle BCD$. If $AD=6$, $AB=8$, then what is the length of $AC$?\n",
    "completion": "\\textbf{Solution:} We have $AC=\\boxed{7\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53102367_62.png"
  },
  {
    "solution": "\\boxed{P(-8, 0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the Cartesian coordinate system $xOy$, it is known that the line $y=-\\frac{1}{2}x+2$ intersects the $x$ axis and the $y$ axis at points $A$ and $B$, respectively. The graph of the line $y=kx-1$ intersects the $y$ axis at point $C$, and intersects the known line at point $D$, where the $x$-coordinate of point $D$ is $2$. Is there a point $P$ on the $x$ axis such that the quadrilateral formed by points $P$, $A$, $B$, and $C$ is a trapezoid? If so, find the coordinates of point $P$.\n",
    "completion": "\\textbf{Solution:}\\\\\nGiven the graph of the line $y=kx-1$ intersects the $y$ axis at point $C$,\\\\\n$\\therefore$ when $x=0$, $y=-1$,\\\\\n$\\therefore$ the coordinates of $C$ are $(0, -1)$.\\\\\n(1) When $CP\\parallel AB$, the quadrilateral $ABPC$ forms a trapezoid,\\\\\n$\\therefore$ the equation of line $CP$ is $y=-\\frac{1}{2}x-1$,\\\\\nLet $y=0$, then $x=-2$,\\\\\n$\\therefore \\boxed{P(-2, 0)}$.\\\\\n(2) When $BP\\parallel AC$, the quadrilateral $ABPC$ forms a trapezoid,\\\\\nLet the equation of line $AC$ be $y=mx-1$, substituting point $A(4, 0)$ gives $m=\\frac{1}{4}$,\\\\\n$\\therefore$ the equation of line $AC$ is $y=\\frac{1}{4}x-1$,\\\\\n$\\because BP\\parallel AC$,\\\\\n$\\therefore$ the equation of line $BP$ is $y=\\frac{1}{4}x+2$,\\\\\nLet $y=0$, then $x=-8$,\\\\\n$\\therefore \\boxed{P(-8, 0)}$.\\\\\nThus, such a point $P$ exists, making the quadrilateral formed by points $P$, $A$, $B$, and $C$ a trapezoid, with coordinates $(-2, 0)$ or $(-8, 0)$.",
    "image_path": "dataset/mm_math/images/MM_Math/50446789_235.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the rectangle ABCD, AB = 4 cm, and BC = 8 cm. Now it is folded along EF, making point C coincide with point A. What is the length of AF?\n",
    "completion": "\\textbf{Solution:} Let $AF=x$ cm, then $DF=(8-x)$ cm, \\\\\nsince in the rectangle ABCD, $AB=4$ cm, $BC=8$ cm, and it is folded along EF such that point C coincides with point A, \\\\\ntherefore $DF=D^\\prime F$, $AD^\\prime=CD=AB=4$, \\\\\nin $\\triangle AD^\\prime F$, since $AF^2=A{D^\\prime}^2+D^\\prime F^2$, \\\\\ntherefore $x^2=4^2+(8-x)^2$, \\\\\nSolving yields: $x=\\boxed{5}$ (cm).",
    "image_path": "dataset/mm_math/images/MM_Math/52699554.png"
  },
  {
    "solution": "\\boxed{15^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within square $ABCD$, an equilateral triangle $\\triangle ADE$ is constructed. By connecting $BE$ and $CE$, what is the measure of $\\angle CBE$ in degrees?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ADE$ is an equilateral triangle,\\\\\nit follows that $AE=DE=AD$, and $\\angle ADE=\\angle EAD=\\angle AED=60^\\circ$,\\\\\nSince quadrilateral ABCD is a square,\\\\\nit follows that $AB=AD=DC$, and $\\angle BAD=\\angle ADC=90^\\circ$,\\\\\nthus $\\angle BAE=\\angle CDE=90^\\circ-60^\\circ=30^\\circ$, and $AB=AE=DE=CD$,\\\\\nhence $\\angle ABE=\\frac{1}{2}\\left(180^\\circ-30^\\circ\\right)=75^\\circ=\\angle DCE$,\\\\\nthus $\\angle CBE=90^\\circ-75^\\circ=\\boxed{15^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52387024.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the segment $AB=24\\,cm$, extending $AB$ to $C$ such that $BC=\\frac{1}{2}AB$. If $D$ is the midpoint of $AB$, and $E$ is the midpoint of $AC$, what is the length of $DE$ in $cm$?\n",
    "completion": "\\textbf{Solution:} Since $D$ is the midpoint of $AB$, and $E$ is the midpoint of $AC$,\\\\\n$\\therefore$ $AD = \\frac{1}{2}AB = 12$cm, $AE = \\frac{1}{2}AC = 18$cm,\\\\\n$\\therefore$ $DE = 18 - 12 = \\boxed{6}$cm.",
    "image_path": "dataset/mm_math/images/MM_Math/51355086_60.png"
  },
  {
    "solution": "\\boxed{\\frac{18}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ABC=90^\\circ$, $BD\\perp AC$ at point $D$, $AC=10$, $\\cos C=\\frac{3}{5}$, then how long is $CD$?\n",
    "completion": "Solution: Since $BD\\perp AC$ and $\\angle ABC=90^\\circ$, \\\\\nit follows that $\\cos C=\\frac{CD}{BC}=\\frac{BC}{AC}=\\frac{3}{5}$, \\\\\ngiven that $AC=10$, \\\\\nit follows that $BC=6$, \\\\\nhence, $CD=\\frac{3}{5}BC=\\boxed{\\frac{18}{5}}$",
    "image_path": "dataset/mm_math/images/MM_Math/53079120.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, $AC=3$, and $AB=5$, what is the value of $\\sin B$?\n",
    "completion": "\\textbf{Solution:} In the right-angled triangle $ABC$, $\\angle C=90^\\circ$, $AC=3$, $AB=5$, \\\\\n$\\therefore$ $\\sin B=\\frac{AC}{AB}=\\boxed{\\frac{3}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50913431.png"
  },
  {
    "solution": "\\boxed{x>-2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the Cartesian coordinate system xOy, the graph of the line $l_1: y_1 = k_1x + 5$ and the graph of the line $l_2: y_2 = k_2x$ are shown as in the figure. What is the solution set of the inequality $k_2x < k_1x + 5$ with respect to $x$?\n",
    "completion": "\\textbf{Solution:} According to the graph, the coordinates of the intersection point of the two lines are $(-2, 3)$,\n\n$\\therefore$ the solution set of the inequality ${k}_{2}x<{k}_{1}x+5$ is $\\boxed{x>-2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52386623.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that rectangle $ABCD$ is folded along line $BD$, such that point $C$ falls on point $C'$, $BC$ intersects $AD$ at $E$, $AD=8$, and $AB=4$. What is the length of $DE$?\n",
    "completion": "\\textbf{Solution:} By the property of folding, we have $\\angle EBD = \\angle CBD$,\\\\\n$\\because$ in rectangle ABCD, AD$\\parallel$BC,\\\\\n$\\therefore \\angle ADB = \\angle CBD$,\\\\\n$\\therefore \\angle EBD = \\angle EDB$,\\\\\n$\\therefore DE = BE$,\\\\\nLet DE = x, then BE = x, AE = 8 - x,\\\\\n$\\because$ in $\\triangle ABE$, by the Pythagorean theorem we have $BE^2 = AB^2 + AE^2$,\\\\\n$\\therefore x^2 = 4^2 + (8 - x)^2$, solving for $x$ gives: $x = \\boxed{5}$,\\\\\n$\\therefore DE = \\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53296816.png"
  },
  {
    "solution": "\\boxed{\\left(-\\frac{1}{2}, \\frac{21}{8}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated, the line $y=\\frac{1}{2}x+2$ intersects the $x$-axis and the $y$-axis at points $A$ and $C$, respectively. The parabola $y=-\\frac{1}{2}x^2+bx+c$ passes through points $A$ and $C$, and intersects the $x$-axis at another point $B$. $D$ is a moving point on the parabola. $F$ is a moving point on the axis of symmetry of the parabola. There exists a parallelogram with vertices $B$, $C$, $D$, and $F$. Please find the coordinates of point $D$.\n",
    "completion": "\\textbf{Solution:} The coordinates of point $D$ are $\\boxed{\\left(-\\frac{1}{2}, \\frac{21}{8}\\right)}$ or $\\left(-\\frac{5}{2}, \\frac{21}{8}\\right)$ or $\\left(\\frac{5}{2}, -\\frac{39}{8}\\right)$.",
    "image_path": "dataset/mm_math/images/MM_Math/52622506_112.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in parallelogram $ABCD$, point $E$ is on $AD$. Connect $CE$ and extend it to intersect the extension of $BA$ at point $F$. If the ratio of $AE$ to $AD$ is $2:3$, and $CD=2$, what is the length of $AF$?\n",
    "completion": "Solution: Since quadrilateral ABCD is a parallelogram, \\\\\nit follows that AB $\\parallel$ CD, \\\\\ntherefore, $\\triangle AFE\\sim \\triangle DCE$,\\\\\nthus $\\frac{AE}{DE}=\\frac{AF}{CD}$, \\\\\nsince AE:AD=2:3 and CD=2, \\\\\nit follows that $\\frac{2}{1}=\\frac{AF}{2}$, \\\\\ntherefore, AF=\\boxed{4}.",
    "image_path": "dataset/mm_math/images/MM_Math/51433790.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\triangle ABC$, points D and E are located on sides $ AB$ and $ AC$ respectively, and $ DE\\parallel BC$. If $ AD=2\\,cm$, $ AB=6\\,cm$, and $ AE=1.5\\,cm$, what is the length of $ EC$ in $cm$?\n",
    "completion": "\\textbf{Solution:} Since $DE\\parallel BC$,\\\\\nit follows that $\\frac{AD}{AB}=\\frac{AE}{AC}=\\frac{2}{6}=\\frac{1}{3}$,\\\\\nthus $\\frac{1.5}{AC}=\\frac{1}{3}$,\\\\\nsolving this gives: $AC=4.5\\text{cm}$,\\\\\ntherefore $EC=AC-AE=4.5-1.5=\\boxed{3}\\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51032479.png"
  },
  {
    "solution": "\\boxed{x>-1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=kx+b$ passes through points $A(m, -2)$ and point $B(-2, 0)$. If the line $y=2x$ passes through point A, what is the solution set of the inequality $2x>kx+b$?\n",
    "completion": "\\textbf{Solution}: Since the line $y=2x$ passes through point A, \\\\\nhence $-2=2m$, solving yields: $m=-1$. \\\\\nAs known from the graph, the solution set of $2x>kx+b$ is to the right side of point A, that is, $\\boxed{x>-1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50612891.png"
  },
  {
    "solution": "\\boxed{88^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $O$ is a point inside $\\triangle ABC$ such that the distances from point $O$ to the three sides of $\\triangle ABC$ are equal. Given that $\\angle BOC = 134^\\circ$, what is the measure of $\\angle A$?\n",
    "completion": "\\textbf{Solution:} Since the point O is equidistant from the sides of $\\triangle ABC$,\\\\\nit follows that BO bisects $\\angle ABC$ and CO bisects $\\angle ACB$,\\\\\ntherefore $\\angle A = 180^\\circ - (\\angle ABC + \\angle ACB) = 180^\\circ - 2(\\angle OBC + \\angle OCB) = 180^\\circ - 2 \\times (180^\\circ - \\angle BOC) = 180^\\circ - 2 \\times (180^\\circ - 134^\\circ) = \\boxed{88^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/53105799.png"
  },
  {
    "solution": "\\boxed{55^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in $\\triangle ABC$, $D$ is a point on side $BC$, $AB=DB$, $BE$ bisects $\\angle ABC$, and intersects side $AC$ at point $E$, with $DE$ connected. $\\triangle ABE \\cong \\triangle DBE$; if $\\angle A=105^\\circ$, $\\angle C=50^\\circ$, what is the degree measure of $\\angle DEC$?\n",
    "completion": "\\textbf{Solution:} Given: $\\because \\angle A=105^\\circ$, $\\angle C=50^\\circ$, $\\triangle ABE \\cong \\triangle DBE$,\\\\\n$\\therefore \\angle BDE=\\angle A=105^\\circ$,\\\\\n$\\therefore \\angle EDC=180^\\circ-\\angle BDE=180^\\circ-105^\\circ=75^\\circ$,\\\\\n$\\therefore \\angle DEC=180^\\circ-\\angle EDC-\\angle C=180^\\circ-75^\\circ-50^\\circ=\\boxed{55^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52457281_93.png"
  },
  {
    "solution": "\\boxed{C\\left(-2, 4\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the graph of the linear function $y=x+b$ intersects the x-axis and y-axis at points A and B, respectively, and intersects the graph of the direct proportionality function $y=-2x$ at point C. The ordinate of point C is 4. Find the coordinates of points A, B, and C?\n",
    "completion": "\\textbf{Solution:} \\\\\nSince point C lies on the graph of $y=-2x$, when $y=4$, $-2x=4$, thus $x=-2$, \\\\\nhence the coordinates of point C are $\\left(-2, 4\\right)$, \\\\\nfurthermore, since $y=x+b$ passes through point $\\left(-2, 4\\right)$, \\\\\nit follows that $-2+b=4$, solving this gives $b=6$, \\\\\nhence $y=x+6$, when $x=0$, $y=6$, thus the coordinates of point A are $\\left(0, 6\\right)$, \\\\\nwhen $y=0$, $x=-6$, thus the coordinates of point B are $\\left(-6, 0\\right)$. \\\\\n$\\therefore \\boxed{A\\left(0, 6\\right)}, \\boxed{B\\left(-6, 0\\right)}, \\boxed{C\\left(-2, 4\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52190033_150.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB=AC$, $AD=AE$, $\\angle BAC=\\angle DAE$, and points $B$, $D$, $E$ are on the same line. If $CE=3$, $DE=5$, then what is the length of $BE$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle BAC = \\angle BAD + \\angle DAC$ and $\\angle DAE = \\angle CAE + \\angle DAC$, \\\\\nSince $\\angle BAC = \\angle DAE$, \\\\\n$\\therefore \\angle BAD = \\angle CAE$, \\\\\nIn $\\triangle BAD$ and $\\triangle CAE$, \\\\\n$\\left\\{\\begin{array}{l}\nAB = AC \\\\\n\\angle BAD = \\angle CAE \\\\\nAD = AE\n\\end{array}\\right.$, \\\\\n$\\therefore \\triangle BAD \\cong \\triangle CAE \\left( SAS \\right)$, \\\\\n$\\therefore BD = CE = 3$, \\\\\n$\\therefore BE = BD + DE = 3 + 5 = \\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53066421.png"
  },
  {
    "solution": "\\boxed{\\frac{9\\sqrt{5}}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, where $AD=9$, $E$ is a point on $CD$. Connect $AE$. Triangle $ADE$ is folded along the straight line $AE$ such that point $D$ falls onto point $F$. Draw $FG\\perp AD$ at $F$, with the foot being $G$. If $AG=2GD$, what is the value of $DE$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, \\\\\nfrom the properties of the folding, we have $AF=AD=9$, $DE=EF$, and $\\angle AED=\\angle AEF$, \\\\\n$\\because$ FG$\\perp$AD, \\\\\n$\\therefore$ FG$\\parallel$CD, \\\\\n$\\therefore$ $\\angle AED=\\angle AMG=\\angle EMF$, \\\\\n$\\therefore$ $\\angle EMF=\\angle AEF$, \\\\\n$\\therefore$ MF=EF=DE, \\\\\n$\\because$ AD=9, AG=2GD, \\\\\n$\\therefore$ AG=6, \\\\\n$\\therefore$ FG=$3\\sqrt{5}$, \\\\\n$\\because$ $\\frac{GM}{DE}=\\frac{AG}{AD}=\\frac{6}{9}=\\frac{2}{3}$, \\\\\n$\\therefore$ GM=$\\frac{2}{3}DE$, \\\\\n$\\therefore$ $\\frac{2}{3}DE+DE=3\\sqrt{5}$, \\\\\n$\\therefore$ DE=$\\boxed{\\frac{9\\sqrt{5}}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52936345.png"
  },
  {
    "solution": "\\boxed{\\frac{7}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within rectangle $ABCD$, $E$ is the midpoint of $BC$. Connect $AE$ and draw $EF \\perp AE$ intersecting $DC$ at point $F$. If $AB=4$ and $BC=6$, what is the length of $DF$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, since quadrilateral ABCD is a rectangle,\\\\\nhence AB = CD = 4, and $\\angle$B = $\\angle$C = $90^\\circ$,\\\\\nsince BC = 6, and E is the midpoint of BC,\\\\\nhence BE = EC = 3,\\\\\nsince EF $\\perp$ AE,\\\\\nhence $\\angle$AEF = $90^\\circ$,\\\\\nhence $\\angle$BAE = $90^\\circ$ - $\\angle$AEB = $\\angle$CEF,\\\\\nsince $\\triangle$ABE $\\sim$ $\\triangle$ECF,\\\\\nhence $\\frac{AB}{EC} = \\frac{BE}{CF}$,\\\\\nhence CF = $\\frac{BE \\cdot EC}{AB}$ = $\\frac{3 \\times 3}{4}$ = $\\frac{9}{4}$,\\\\\nhence DF = 4 - $\\frac{9}{4}$ = $\\boxed{\\frac{7}{4}}$,",
    "image_path": "dataset/mm_math/images/MM_Math/52536606.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{15}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, quadrilateral $ABCD$ is a parallelogram, with $AD=BD$. Through point $C$, construct $CE\\parallel BD$, intersecting the extension line of $AD$ at point $E$. Quadrilateral $BDEC$ is a rhombus; connect $BE$. If $AB=2$ and $AD=4$, what is the length of $BE$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect $BE$ intersecting $CD$ at $O$,\\\\\nsince quadrilateral $BDEC$ is a rhombus,\\\\\nthus $DO=CO=\\frac{1}{2}CD=1$, $BO=\\frac{1}{2}BE$, and $CD\\perp BE$,\\\\\nin $\\triangle BDO$, $AD=BD=4$, $DO=1$,\\\\\nthus $BO=\\sqrt{BD^2-DO^2}=\\sqrt{16-1}=\\sqrt{15}$,\\\\\ntherefore, $BE=2BO=2\\sqrt{15}$.\\\\\nHence, the length of $BE$ is $\\boxed{2\\sqrt{15}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50481338_142.png"
  },
  {
    "solution": "\\boxed{\\frac{75}{8}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that in the right-angled $\\triangle ABC$, $\\angle ACB=90^\\circ$, $AC=9$, $BC=12$, the perpendicular bisector of $AB$ intersects $AB$ at point $D$ and intersects $BC$ at point $E$. Connect $AE$. Find the length of $BE$.\n",
    "completion": "\\textbf{Solution:} In right-angled $\\triangle ABC$, by the Pythagorean theorem, we have \n\n\\[AB = \\sqrt{AC^{2} + BC^{2}} = \\sqrt{9^{2} + 12^{2}} = 15,\\]\n\nsince $DE$ is the perpendicular bisector of $AB$, \n\nthus $AE = BE$.\n\nLet $BE = AE = x$, then $CE = 12 - x$.\n\nIn right-angled $\\triangle ACE$, by the Pythagorean theorem, we get \n\n\\[AE^{2} = AC^{2} + CE^{2},\\]\n\nwhich implies\n\n\\[x^{2} = 9^{2} + (12-x)^{2},\\]\n\nsolving this, we find $x = \\boxed{\\frac{75}{8}}$,\n\nthus, the length of $BE$ is $\\frac{75}{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53009863.png"
  },
  {
    "solution": "\\boxed{112^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given: As shown in the figure, the line $BO \\perp AO$ at point $O$, $OB$ bisects $\\angle COD$, $\\angle AOC=68^\\circ$, what is the degree measure of $\\angle DOA$?\n",
    "completion": "\\textbf{Solution:} Since $AO \\perp BO$,\\\\\nit follows that $\\angle AOB = 90^\\circ$,\\\\\nSince $\\angle AOC = 68^\\circ$,\\\\\nit follows that $\\angle BOC = 90^\\circ - 68^\\circ = 22^\\circ$,\\\\\nSince $OB$ bisects $\\angle COD$,\\\\\nit follows that $\\angle COD = 2\\angle BOC = 44^\\circ$,\\\\\nthus $\\angle DOA = \\angle DOC + \\angle AOC = 44^\\circ + 68^\\circ = \\boxed{112^\\circ}$, therefore C meets the requirements of the problem.",
    "image_path": "dataset/mm_math/images/MM_Math/52796215.png"
  },
  {
    "solution": "\\boxed{135^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $O$ is a point on line $AB$, $\\angle DOB = 90^\\circ$, $\\angle EOC = 90^\\circ$. If $OE$ bisects $\\angle AOD$, what is the degree measure of $\\angle BOE$?\n",
    "completion": "Solution:\n\\[\n\\because OE \\text{ bisects } \\angle AOD, \n\\]\n\\[\n\\therefore \\angle DOE=\\frac{1}{2}\\angle AOD=45^\\circ,\n\\]\n\\[\n\\text{Also, } \\because \\angle BOE = \\angle BOD + \\angle DOE,\n\\]\n\\[\n\\therefore \\angle BOE = \\boxed{135^\\circ}.\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/51457662_194.png"
  },
  {
    "solution": "\\boxed{x>4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the proportional function $y_{1}=x$ and the quadratic function $y_{2}=x^{2}-bx$ intersect at points $O(0,0)$ and $A(4,4)$. When $y_{1} < y_{2}$, directly write the range of $x$.\n",
    "completion": "\\textbf{Solution:} Solve: When $y_{1}< y_{2}$, directly write the range of $x$ as $x<0$ or $x>4$, \\\\\nhence, the range of $x$ is $\\boxed{x<0}$ or $\\boxed{x>4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51458362_259.png"
  },
  {
    "solution": "\\boxed{50^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In quadrilateral $ABCD$, $\\triangle OAB$ and $\\triangle OCD$ share a common vertex $O$, and $\\triangle OAB \\cong \\triangle OCD$. As shown in Figure 1, $O$ is a point on side $BC$. Given that $AO=DO$, and if $\\angle AOD=80^\\circ$, connect $AC$ and $BD$, with the intersection point being $E$. Find the measure of $\\angle DEC$.\n",
    "completion": "\\textbf{Solution:} As shown, \\\\\n$\\because$ $\\triangle OAB \\cong \\triangle OCD$, \\\\\n$\\therefore$ $\\angle BOA = \\angle COD = \\frac{180^\\circ - 80^\\circ}{2} = 50^\\circ, OA = OC, OB = OD$, \\\\\n$\\therefore$ $\\angle AOC = \\angle DOB = 130^\\circ$, \\\\\n$\\therefore$ $\\angle ACO = \\angle DBO = \\frac{180^\\circ - 130^\\circ}{2} = 25^\\circ$, \\\\\n$\\therefore$ $\\angle DEC = \\boxed{50^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53033486_102.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Quadrilateral ABCD is a rectangular piece of paper, with point E on side AD. Fold $\\triangle ABE$ along BE such that point A falls onto the diagonal BD of the rectangle. Draw line CF. Given that $DE=1$, investigate the following question: As shown in Figure 1, when F is exactly at the midpoint of BD, what is the length of $AE$?\n",
    "completion": "\\textbf{Solution:} We have $AE = \\boxed{\\frac{1}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52601962_64.png"
  },
  {
    "solution": "\\boxed{110^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points D and E are located on line segments BC and AC, respectively, and lines AD and BE are connected. If $\\angle A=35^\\circ$, $\\angle B=30^\\circ$, and $\\angle C=45^\\circ$, what is the measure of $\\angle AFB$?\n",
    "completion": "\\textbf{Solution:} Given $\\because$ $\\angle A=35^\\circ$, $\\angle B=30^\\circ$, $\\angle C=45^\\circ$\\\\\n$\\therefore$ $\\angle ADC=180^\\circ-\\angle A-\\angle C=180^\\circ-35^\\circ-45^\\circ=100^\\circ$,\\\\\n$\\angle BEC=180^\\circ-\\angle B-\\angle C=180^\\circ-30^\\circ-45^\\circ=105^\\circ$,\\\\\nIn quadrilateral CDFE, $\\angle DFE=360^\\circ-\\angle FEC-\\angle FDC-\\angle C=360^\\circ-105^\\circ-100^\\circ-45^\\circ=110^\\circ$,\\\\\n$\\therefore$ $\\angle AFB=\\angle DFE=\\boxed{110^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52096296.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, diagonals AC and BD of rhombus ABCD intersect at point O. A perpendicular line is drawn from point C to the extended line of AB, intersecting it at point E. Line OE is then drawn. If AB = $2\\sqrt{5}$ and BD = 4, what is the length of OE?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral $ABCD$ is a rhombus,\\\\\nit follows that $OA=OC$, $BD\\perp AC$,\\\\\nSince $CE\\perp AB$,\\\\\nit follows that $OE=OA=OC$,\\\\\nSince $BD=4$,\\\\\nit follows that $OB=\\frac{1}{2}BD=2$,\\\\\nIn $\\triangle AOB$, $AB=2\\sqrt{5}$, $OB=2$,\\\\\nit follows that $OA=\\sqrt{AB^{2}-OB^{2}}=4$,\\\\\ntherefore, $OE=OA=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52387446.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in right $\\triangle ABC$, $\\angle ABC=90^\\circ$, point $E$ is the midpoint of $BC$, and $DE$ is drawn. $DE$ is a tangent to the semicircle $\\odot O$. If $\\angle BAC=30^\\circ$ and $DE=2$, find the length of $AD$.\n",
    "completion": "\\textbf{Solution:}\n\nGiven that \\\\\n$DE=2$, \\\\\nit follows that $BC=2ED=4$, \\\\\nGiven $\\angle BAC=30^\\circ$, \\\\\nit follows that $AC=2BC=8$, \\\\\nThus, $AB=\\sqrt{8^2-4^2}=4\\sqrt{3}$, \\\\\nSo, $BD=\\frac{1}{2}AB=2\\sqrt{3}$, \\\\\nTherefore, $AD=\\sqrt{(4\\sqrt{3})^2-(2\\sqrt{3})^2}=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52949265_104.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $AB\\colon y=2x-k$ passes through the point $M\\left(k, 2\\right)$, and intersects the $x$-axis at point $A$ and the $y$-axis at point $B$. Find the value of $k$.\n",
    "completion": "\\textbf{Solution:} Since the line $AB: y = 2x - k$ passes through the point $M(k, 2)$,\n\ntherefore, $2k - k = 2$. Solving this gives $k = \\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52345458_210.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, CE is the altitude to the hypotenuse AB, and the angle bisector BD intersects CE at point M. $\\triangle CDM$ is an isosceles triangle. If AB = 10 and AC = 8, find the length of CM.\n",
    "completion": "\\textbf{Solution:} Construct $DF\\perp AB$ at point $F$, as shown in the figure,\\\\\n$\\because \\angle DCB=90^\\circ$ and $BD$ bisects $\\angle ABC$,\\\\\n$\\therefore DC=DF$,\\\\\n$\\because \\angle ACB=90^\\circ$, $AB=10$, $AC=8$,\\\\\n$\\therefore BC= \\sqrt{AB^{2}-AC^{2}}=\\sqrt{10^{2}-8^{2}}=6$,\\\\\n$\\because S_{\\triangle ABC}=S_{\\triangle BCD}+S_{\\triangle ADB}$,\\\\\n$\\therefore \\frac{AC\\cdot BC}{2}=\\frac{BC\\cdot CD}{2}+\\frac{AB\\cdot DF}{2}$,\\\\\nthus $\\frac{8\\times 6}{2}=\\frac{6\\cdot CD}{2}+\\frac{10\\cdot DF}{2}$,\\\\\nsolving gives $CD=DF=3$,\\\\\nFrom (1) we know: $CM=CD$,\\\\\n$\\therefore CM=\\boxed{3}$,\\\\\nwhich means the length of $CM$ is $3$.",
    "image_path": "dataset/mm_math/images/MM_Math/55498361_78.png"
  },
  {
    "solution": "\\boxed{25^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is tangent to $\\odot O$ at point $B$, extending $AO$ intersects $\\odot O$ at point $C$, connecting $BC$. If $\\angle A=40^\\circ$, what is the value of $\\angle C$?\n",
    "completion": "\\textbf{Solution:} Given that AB is a tangent to circle O at point B,\\\\\nit follows that OB$\\perp$AB, meaning $\\angle$ABO=$90^\\circ$,\\\\\nhence $\\angle$AOB=$50^\\circ$ (the two acute angles of a right triangle are complementary),\\\\\nAlso, given that point C is on the extension of AO and lies on circle O,\\\\\nthus $\\angle$C=$\\frac{1}{2}\\angle$AOB=$\\boxed{25^\\circ}$ (the angle subtended by an arc at the circumference is half that at the center).",
    "image_path": "dataset/mm_math/images/MM_Math/51379153.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within circle $O$, the diameter $AB \\perp$ chord $CD$ at point $M$, and point $E$ is the midpoint of arc $AD$. Connect $AD$ and $CE$, and extend $AE$ and $CD$ to meet at point $F$. It is given that $\\angle AEC = 2\\angle C$. When $AM=4$ and $CD=6$, what is the length of $AF$?\n",
    "completion": "\\textbf{Solution:} Given the problem statement,\\\\\n$\\because \\angle AEC = 2\\angle C$ and $\\angle AEC = \\angle C + \\angle F$,\\\\\n$\\therefore \\angle C = \\angle F$,\\\\\nAlso, $\\because \\angle DAE = \\angle C$,\\\\\n$\\therefore \\angle DAE = \\angle F$,\\\\\n$\\therefore AD=DF$,\\\\\n$\\because AM=4$, $CD=6$, and diameter $AB\\perp$ chord CD,\\\\\n$\\therefore AD=DF=5$,\\\\\n$\\therefore AF=\\sqrt{4^2+8^2}=\\boxed{4\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52874685_99.png"
  },
  {
    "solution": "\\boxed{\\frac{12}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the rectangle $ABCD$, where $AB = 6$, $AD = 10$, point $E$ is moving on side $AD$. If triangle $DEC$ is folded along $EC$ such that point $D$ is placed at point $D'$, and if there exists a relationship between two sides of triangle $DEC$ where one side is twice the length of the other, what is the distance from point $D'$ to side $AD$?\n",
    "completion": "Solution:Let's solve the problem step by step. \n(1) When $DE=\\frac{1}{2}CE$, construct $D'H\\perp AD$ at H, as shown in the diagram: \\\\\nSince rectangle ABCD, AB=6, \\\\\n$\\therefore$$\\angle$D=$90^\\circ$, CD=6, \\\\\nIn $\\triangle DCE$, $DE^{2}+CD^{2}=CE^{2}$, CD=6, $DE=\\frac{1}{2}CE$, \\\\\n$\\therefore$$DE^{2}+{6}^{2}=(2DE)^{2}$, \\\\\n$\\therefore$$DE=2\\sqrt{3}$, \\\\\nSince $DE=\\frac{1}{2}CE$, $\\angle$D=$90^\\circ$, \\\\\n$\\therefore$$\\angle$DCE=$30^\\circ$, $\\angle$DEC=$60^\\circ$, \\\\\nSince $\\triangle DEC$ is folded along EC causing D to fall on D', \\\\\n$\\therefore$$\\angle$D'EC=$\\angle$DEC=$60^\\circ$, $D'E=2\\sqrt{3}$, \\\\\n$\\therefore$$\\angle$D'EH=$180^\\circ$-$\\angle D'EC-\\angle$DEC=$60^\\circ$, in right $\\triangle D'HE$, $D'H=D'E\\cdot\\sin\\angle D'EH=2\\sqrt{3}\\times \\sin60^\\circ=3$; \\\\\n(2) When $DE=\\frac{1}{2}DC$, construct D'H$\\perp$AD at H, extend HD' to intersect BC at F, as shown in the diagram: \\\\\nIn right $\\triangle DCE$, $DE^{2}+CD^{2}=CE^{2}$, CD=6, $DE=\\frac{1}{2}CD$, \\\\\n$\\therefore$$DE=3$, $CE=3\\sqrt{5}$, \\\\\nSince $\\triangle DEC$ is folded along EC causing D to fall on D', \\\\\n$\\therefore$$\\angle$ED'C=$\\angle$D=$90^\\circ$, DE=D'E, CD=CD', \\\\\n$\\therefore$$\\angle$HD'E=$90^\\circ$-$\\angle$FD'C=$\\angle$FCD', \\\\\nand $\\angle$D'HE=$\\angle$CFD'=$90^\\circ$, \\\\\n$\\therefore$$\\triangle D'HE\\sim \\triangle CFD'$, \\\\\n$\\therefore$$\\frac{HE}{D'F}=\\frac{D'H}{CF}=\\frac{D'E}{CD'}$, \\\\\nSince $DE=\\frac{1}{2}DC$, \\\\\n$\\therefore$$\\frac{DE}{DC}=\\frac{1}{2}$, \\\\\n$\\therefore$$\\frac{HE}{D'F}=\\frac{D'H}{CF}=\\frac{D'E}{CD'}=\\frac{1}{2}$, \\\\\nLet HE=x, D'H=y, then D'F=2x, CF=2y, \\\\\nSince D'H$\\perp$AD, and ABCD is a rectangle, \\\\\n$\\therefore$Quadrilateral HFCD is a rectangle, \\\\\n$\\therefore$$\\begin{cases} 2x+y=6 \\\\ 2y=3+x \\end{cases}$ \\\\\nSolving gives $y=\\frac{12}{5}$, \\\\\n$\\therefore$D'H=$\\boxed{\\frac{12}{5}}$; \\\\\n(3) When $DC=\\frac{1}{2}CE$, construct $D'H\\perp AD$ at H, as shown in the diagram: \\\\\nIn right $\\triangle DCE$, $DE^{2}+CD^{2}=CE^{2}$",
    "image_path": "dataset/mm_math/images/MM_Math/51817979.png"
  },
  {
    "solution": "\\boxed{145^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\angle 1$ and $\\angle 2$ are vertically opposite angles, $\\angle 1 = 180^\\circ - a$, $\\angle 2 = 35^\\circ$, then $a=$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle 1$ and $\\angle 2$ are vertical angles,\\\\\nit follows that $\\angle 1 = \\angle 2$,\\\\\nSince $\\angle 1=180^\\circ-a$  and  $\\angle 2=35^\\circ$,\\\\\nit follows that $180^\\circ-a=35^\\circ$,\\\\\nhence, $a=\\boxed{145^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/47362683.png"
  },
  {
    "solution": "\\boxed{22.5^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown, it is known that $OE$ is the angle bisector of $\\angle AOC$, and $OD$ is the angle bisector of $\\angle BOC$. If $\\angle DOE=45^\\circ$, and $\\angle AOC + \\angle BOC = 180^\\circ$, what is the degree measure of $\\angle COD$?\n",
    "completion": "\\textbf{Solution:} Since $OD$ is the angle bisector of $\\angle BOC$,\\\\\nlet $\\angle COD=\\angle BOD=x$.\\\\\nSince $\\angle DOE=45^\\circ$,\\\\\nthen $\\angle BOE=45^\\circ-x$, $\\angle COE=45^\\circ+x$.\\\\\nSince $OE$ is the angle bisector of $\\angle AOC$,\\\\\nthen $\\angle AOE=\\angle COE=45^\\circ+x$.\\\\\nSince $\\angle AOC+\\angle BOC=180^\\circ$,\\\\\nthen $2(45^\\circ+x)+2x=180^\\circ$,\\\\\nthus $x=\\boxed{22.5^\\circ}$, that is $\\angle COD=22.5^\\circ$.",
    "image_path": "dataset/mm_math/images/MM_Math/47111555_200.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, locations A and B are separated by a pond. In order to measure the distance between A and B, an appropriate point C outside of AB is chosen. The lines AC and BC are connected, and the midpoints E and F of the line segments AC and BC, respectively, are identified. It is measured that $EF = 6m$. What is the distance between A and B in meters?\n",
    "completion": "\\textbf{Solution:} Since $E$ and $F$ are the midpoints of $AC$ and $BC$ respectively,\\\\\ntherefore, $EF$ is a midsegment of $\\triangle ABC$,\\\\\ntherefore, $AB = 2EF = \\boxed{12}$m.",
    "image_path": "dataset/mm_math/images/MM_Math/52050259.png"
  },
  {
    "solution": "\\boxed{\\angle AEF=90^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the square $ABCD$, $E$ is the midpoint of $BC$, and $F$ is a point on $CD$ such that $CF = \\frac{1}{4}CD$. Prove what the degree measure of $\\angle AEF$ is.\n",
    "completion": "\\textbf{Solution:} Let the length of $CF$ be $a$ \\\\\nThen $EF=\\sqrt{5}a$, $AE=2\\sqrt{5}a$ \\\\\nConnect $AF$, thus $DF=3a$, and $AD=4a$, \\\\\nIn $Rt\\triangle ADF$, $AF=\\sqrt{AD^{2}+DF^{2}}=5a$ \\\\\nIn $\\triangle AEF$, \\\\\n$\\because AE^{2}+EF^{2}=AF^{2}$ \\\\\n$\\therefore \\boxed{\\angle AEF=90^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51798762_142.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, we have $AB=4$, $AC=6$, and $AD$ is the bisector of $\\angle BAC$. $M$ is the midpoint of $BC$, and $ME\\parallel AD$ intersects $AC$ at $F$ and intersects the extension of $BA$ at $E$. What is the value of $BE$?\n",
    "completion": "\\textbf{Solution:} Proof:\n\nSince $AD$ bisects $\\angle BAC$,\n\nthen $\\angle BAD=\\angle DAC$,\n\nSince $MF \\parallel AD$,\n\nthen $\\angle DAC=\\angle AFE$, $\\angle BAD=\\angle E$,\n\nthus $\\angle E=\\angle AFE$,\n\nhence $AE=AF$;\n\nExtend $FM$ to point $N$ such that $MN=FM$, $\\angle BMN=\\angle CMF$, $MB=CM$,\n\ntherefore $\\triangle BMN \\cong \\triangle CMF$ (SAS),\n\ntherefore $CF=BN$, $\\angle N=\\angle MFC$,\n\nSince $\\angle EFA=\\angle MFC$,\n\nthus $\\angle N=\\angle EFA$,\n\nthus $\\angle N=\\angle E$,\n\nthus $BN=BE$,\n\nthus $AB+AC=AB+AF+FC=AB+AE+FC=BE+FC$,\n\nSince $BE=FC$,\n\ntherefore $2BE=10$,\n\ntherefore $BE=\\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51368144.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, in $\\triangle ABC$, the endpoints $D$ and $E$ of the line segment $DE$ are located on the sides $AB$ and $AC$, respectively. If the product of the two line segments $AD$ and $AE$ above $DE$ is equal to the product of the two line segments $BD$ and $CE$ below $DE$, that is, $AD \\times AE = BD \\times CE$, then $DE$ is called a \"friendly partition\" segment of $\\triangle ABC$. As shown in Figure 1, if $DE$ is a \"friendly partition\" segment of $\\triangle ABC$, what is the length of $AC$?\n",
    "completion": "\\textbf{Solution:} Let $AE = x$, \\\\\nsince $DE$ is a \"friendly split\" line segment of $\\triangle ABC$, \\\\\nit follows that $AD \\cdot AE = BD \\cdot EC$. \\\\\nSince $AD = 2CE$ and $AB = 8$, \\\\\nit follows that $2EC \\cdot AE = (8 - AD) \\cdot EC$. \\\\\nTherefore, $2x = 8 - 2EC$. \\\\\nTherefore, $x = 4 - EC$, \\\\\nwhich means $AE = 4 - EC$. \\\\\nThus, $AC = AE + EC = \\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51458341_142.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the figure, in equilateral triangle $ABC$, we have $AB=2$. Let $BD$ be the altitude to side $AC$. Extend $BC$ to point $E$ such that $CE=CD$. What is the length of $BE$?\n",
    "completion": "\\textbf{Solution:} Since triangle $ABC$ is an equilateral triangle,\\\\\nwe have $BC=AC=2$,\\\\\nand since $BD$ is the altitude of side $AC$,\\\\\nwe have $DC=\\frac{1}{2}AC=\\frac{1}{2}\\times 2=1$,\\\\\nthus $CE=CD=1$,\\\\\ntherefore $BE=BC+CE=2+1=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51378776.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rhombus $ABCD$, point $E$ is on side $AD$, and ray $CE$ intersects the extension of $BA$ at point $F$. If $\\frac{AE}{ED}=\\frac{1}{2}$ and $AB=3$, then what is the length of $AF$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral $ABCD$ is a rhombus, \\\\\n$\\therefore$ $AE\\parallel BC$, $AD=AB=BC=3$, \\\\\n$\\therefore$ $\\triangle FAE\\sim \\triangle FBC$, \\\\\n$\\therefore$ $\\frac{AF}{BF}=\\frac{AE}{BC}$, \\\\\nSince $\\frac{AE}{ED}=\\frac{1}{2}$, \\\\\n$\\therefore$ $\\frac{AE}{BC}=\\frac{AE}{AD}=\\frac{1}{3}$, \\\\\n$\\therefore$ $\\frac{AF}{BF}=\\frac{AE}{BC}=\\frac{1}{3}$, that is: $\\frac{AF}{AF+AB}=\\frac{AF}{AF+3}=\\frac{1}{3}$, \\\\\nSolving gives: $AF=\\boxed{\\frac{3}{2}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53110752.png"
  },
  {
    "solution": "\\boxed{\\frac{8\\sqrt{3}}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, two tangents $PA$ and $PB$ are drawn from a point $P$ outside the circle $O$, and the points of tangency are $A$ and $B$, respectively. If $\\angle APB=60^\\circ$ and $PA=8$, then what is the radius of circle $O$?\n",
    "completion": "\\textbf{Solution:} Draw OA, OP.\\\\\nSince PA, PB are tangents to the circle $O$ and $\\angle APB=60^\\circ$,\\\\\nit follows that $\\angle OPA= \\frac{1}{2}\\angle APB=30^\\circ$ and OA$\\perp$OP,\\\\\nthus OA=AP$\\cdot \\tan \\angle OPA=8\\times \\frac{\\sqrt{3}}{3}=\\frac{8\\sqrt{3}}{3}$,\\\\\nTherefore, the radius of circle $O$ is $\\boxed{\\frac{8\\sqrt{3}}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52746228.png"
  },
  {
    "solution": "\\boxed{55^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the circumcircle of $\\triangle ABC$ has its center at $O$, and $\\angle BOC=110^\\circ$. What is the measure of $\\angle A$?\n",
    "completion": "\\textbf{Solution:} Since $\\odot O$ is the circumcircle of $\\triangle ABC$ and $\\angle BOC=110^\\circ$,\n\\[ \\therefore \\angle A=\\frac{1}{2}\\angle BOC=\\frac{1}{2}\\times 110^\\circ=\\boxed{55^\\circ}. \\]",
    "image_path": "dataset/mm_math/images/MM_Math/51445736.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in $\\triangle ABC$, $\\angle ABC=90^\\circ$, $\\angle A=60^\\circ$, and D is a point on side $AC$ such that $\\angle DBC=30^\\circ$, $AC=12\\text{cm}$. What is the perimeter of $\\triangle ABD$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral $ABCD$ is a parallelogram,\\\\\nit follows that $DC\\parallel AB$, $AD\\parallel BC$, $DC=AB$, $AD=BC$,\\\\\ntherefore, $\\triangle CDF\\sim \\triangle BEF$,\\\\\nthus, $BE\\colon DC=BF\\colon CF$,\\\\\nsince $BE\\colon AB=3\\colon 2$, $DC=AB$,\\\\\nit follows that $BE\\colon DC=BF\\colon CF=3\\colon 2$,\\\\\ntherefore, $CF\\colon BF=2\\colon 3$,\\\\\nthus, $CF\\colon BC=2\\colon 5$,\\\\\nsince $AD=BC=10$,\\\\\nit follows that $CF\\colon 10=2\\colon 5$,\\\\\ntherefore, $CF=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53043250.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{5}+3}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In an isosceles triangle with a vertex angle of $36^\\circ$, we refer to such a triangle as a \"golden triangle\", where the ratio of the base to the leg is the golden ratio. As shown in the figure, in $\\triangle ABC$, $\\angle A=36^\\circ$, $AB=AC$, $BD$ bisects $\\angle ABC$ and intersects $AC$ at point $D$. If $CD=1$, what is the length of $AC$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle A=36^\\circ$ and AB＝AC, with BD bisecting $\\angle ABC$ and intersecting AC at point D,\\\\\nit follows that $\\angle ABC=\\angle ACB=72^\\circ$, $\\angle A=\\angle DBC=36^\\circ$, and $\\angle BCD=\\angle BDC=72^\\circ$,\\\\\nimplying that $\\angle ACB=\\angle BCD$,\\\\\nwhich means $\\triangle BCD\\sim \\triangle ABC$,\\\\\nleading to the ratio BC$\\colon$ AB=DC$\\colon$ BC,\\\\\nand given BC$\\colon$ AB=$\\frac{\\sqrt{5}-1}{2}$ and DC=1,\\\\\nwe find BC=$\\frac{\\sqrt{5}+1}{2}$,\\\\\nthus $\\frac{\\sqrt{5}+1}{2}\\colon$ AB=$\\frac{\\sqrt{5}-1}{2}$,\\\\\nresulting in AB=$\\frac{\\sqrt{5}+3}{2}$,\\\\\nso AC=$\\boxed{\\frac{\\sqrt{5}+3}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53080323.png"
  },
  {
    "solution": "\\boxed{\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, $BC=1$, and $AC=\\sqrt{3}$. Find the values of $\\tan A$ and $\\tan B$.\n",
    "completion": "\\textbf{Solution:} Given in right-angled triangle $ABC$, $\\angle C=90^\\circ$, $BC=1$, and $AC=\\sqrt{3}$,\\\\\nthus, $\\tan A=\\frac{BC}{AC}=\\frac{1}{\\sqrt{3}}=\\frac{\\sqrt{3}}{3}$,\\\\\n$\\tan B=\\frac{AC}{BC}=\\frac{\\sqrt{3}}{1}=\\boxed{\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51322360.png"
  },
  {
    "solution": "\\boxed{3- \\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Fig. 1, there is a circular paper with point O as the center and AB as the diameter. Point C is on the circle $\\odot$O. The circular paper is folded along the diameter CF, causing point B to fall on point D on $\\odot$O (not coinciding with point A). After restoring the paper, connect CB, CD, AD. Assume CD intersects diameter AB at point E. As shown in Fig. 3, when $AD=DE$, if $BC=2$, what is the length of $AD$?\n",
    "completion": "\\textbf{Solution:} Given that $AD=DE$, \\\\\nit follows that $\\angle DAE=\\angle DEA$, \\\\\ngiven that $\\angle DEA=\\angle BEC$ and $\\angle DAE=\\angle BCE$, \\\\\nit follows that $\\angle BEC=\\angle BCE$, \\\\\nconsidering the circle paper folded along the line CF, \\\\\nit follows that $\\angle ECO=\\angle BCO$, \\\\\nfurthermore, given that $OB=OC$, \\\\\nit follows that $\\angle OCB=\\angle B$, \\\\\ngiven that $\\angle ECO=\\angle B$, and $\\angle CEO=\\angle CEB$, \\\\\nit follows that $\\triangle CEO\\sim \\triangle BEC$, \\\\\nresulting in $\\frac{CE}{ED} = \\frac{BE}{CE}$, \\\\\nand therefore $CE^2=EO\\cdot BE$, \\\\\nlet EO=x, EC=OC=OB=a, \\\\\nhence $a^2=x(x+a)$, \\\\\nsolving yields, x= $\\frac{\\sqrt{5}-1}{2}a$ (negative value discarded), \\\\\ntherefore, $OE= \\frac{\\sqrt{5}-1}{2}a$, \\\\\ntherefore, $AE=OA-OE=a- \\frac{\\sqrt{5}-1}{2}a= \\frac{3-\\sqrt{5}}{2}a$, \\\\\ngiven that $\\angle AED=\\angle BEC$, $\\angle DAE=\\angle BCE$, \\\\\nit follows that $\\triangle BCE\\sim \\triangle DAE$, \\\\\nresulting in $\\frac{BC}{AD} = \\frac{EC}{AE}$, \\\\\ngiven that $BC=2$, \\\\\n$\\frac{2}{AD} = \\frac{a}{\\frac{3-\\sqrt{5}}{2}a}$, \\\\\nthus, $AD=\\boxed{3- \\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53124745_90.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $CD$ is the diameter of the circle $\\odot O$, and $AB$ is a chord. $CD\\perp AB$ at point $E$. If $OA=5$ and $AB=8$, what is the length of $AD$?\n",
    "completion": "\\textbf{Solution:} Given that CD is the diameter of circle O and AB is a chord with CD $\\perp$ AB, and AB = 8, \\\\\n$\\therefore$ AE=BE=$\\frac{1}{2}AB=\\frac{1}{2}\\times 8=4$, \\\\\nIn $\\triangle OEB$, according to the Pythagorean theorem, OE=$\\sqrt{OA^{2}-EA^{2}}=\\sqrt{5^{2}-4^{2}}=3$, \\\\\n$\\therefore$ CD=OD+OE=5+3=8, \\\\\nIn $\\triangle AED$, AD=$\\sqrt{AE^{2}+ED^{2}}=\\sqrt{4^{2}+8^{2}}=\\boxed{4\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51446093.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{6}}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, A, B, and C are three points on circle \\(O\\) with a radius of 2, where AB is the diameter. The bisector of \\(\\angle BAC\\) intersects the circle at point D. A perpendicular line is drawn from point D to the extended line of AC, intersecting it at point E. The extended line ED intersects the extended line of AB at point F. The line EF is tangent to circle \\(O\\); if \\(DF = 4\\sqrt{2}\\), find the value of \\(\\cos\\angle EAD\\).\n",
    "completion": "\\textbf{Solution:} In $\\triangle \\text{ODF}$, we have $OD=2$, $DF=4\\sqrt{2}$,\\\\\n$\\therefore OF= \\sqrt{OD^2+DF^2}=6$,\\\\\n$\\because OD\\parallel AE$,\\\\\n$\\therefore \\frac{OD}{AE}=\\frac{OF}{AF}$,\\\\\n$\\therefore \\frac{2}{AE}=\\frac{6}{8}$,\\\\\n$\\therefore AE=\\frac{8}{3}$,\\\\\n$\\therefore ED=\\sqrt{AF^2-AE^2}=\\sqrt{8^2-\\left(\\frac{8}{3}\\right)^2}=\\frac{16}{3}\\sqrt{2}$,\\\\\n$\\therefore AD=\\sqrt{AE^2+ED^2}=\\sqrt{\\frac{64}{9}+\\frac{256}{9}}=\\frac{4\\sqrt{6}}{3}$,\\\\\n$\\therefore \\cos\\angle EAD=\\frac{AE}{AD}=\\boxed{\\frac{\\sqrt{6}}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51403036_137.png"
  },
  {
    "solution": "\\boxed{20^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the lines $EF$ and $CD$ intersect at the point $O$, with $\\angle AOB = 90^\\circ$, and $OC$ bisects $\\angle AOF$. If $\\angle AOE = 40^\\circ$, find the degree measure of $\\angle BOD$.\n",
    "completion": "\\textbf{Solution:} Given $\\because \\angle AOE+\\angle AOE=180^\\circ$, $\\angle AOE=40^\\circ$,\\\\\n$\\therefore \\angle AOF=180^\\circ-\\angle AOE=140^\\circ$\\\\\nSince $OC$ bisects $\\angle AOF$,\\\\\n$\\therefore \\angle AOC=\\frac{1}{2}\\angle AOF=\\frac{1}{2}\\times 140^\\circ=70^\\circ$\\\\\nGiven $\\angle AOB=90^\\circ$\\\\\n$\\therefore \\angle BOD=180^\\circ-\\angle AOC-\\angle AOB$\\\\\n$=180^\\circ-70^\\circ-90^\\circ$\\\\\n$=\\boxed{20^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/53223346_85.png"
  },
  {
    "solution": "\\boxed{\\triangle AHF \\text{ is an isosceles triangle}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: It is known that in the parallelogram $ABCD$, points $E$ and $F$ are respectively on sides $AB$ and $BC$, with $DE=AF$, $DE\\perp AF$ at point $G$. If $\\angle BAD=90^\\circ$, parallelogram $ABCD$ is a square; extending $CB$ to point $H$, making $BH=AE$, determine the shape of $\\triangle AHF$ and explain the reason.\n",
    "completion": "\\textbf{Solution:}\nTo prove: $\\triangle AHF$ is an isosceles triangle, for the following reasons:\\\\\nFrom (1) we have: $\\triangle ADE \\cong \\triangle BAF$,\\\\\n$\\therefore AE=BF$,\\\\\n$\\because BH=AE$,\\\\\n$\\therefore BF=BH$,\\\\\n$\\because$ quadrilateral ABCD is a square,\\\\\n$\\therefore \\angle ABC=90^\\circ$,\\\\\n$\\therefore AB \\perp BC$ i.e., AB bisects FH perpendicularly,\\\\\n$\\therefore AH=AF$,\\\\\n$\\therefore \\boxed{\\triangle AHF \\text{ is an isosceles triangle}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52344940_165.png"
  },
  {
    "solution": "\\boxed{48^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in $ \\triangle ABC$, extend $ BC$ to point D such that $ CD=AB$, draw $ DE\\parallel AB$ with $ DE=BC$, and connect $ CE, AE$. $ \\angle BAC=\\angle DCE$; If $ \\angle B=32^\\circ$ and $ \\angle ACD=58^\\circ$, what is the measure of $ \\angle CEA$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle B=32^\\circ$ and $\\angle ACD=58^\\circ$,\\\\\nit follows that $\\angle DCE=\\angle BAC=\\angle ACD-\\angle B=58^\\circ-32^\\circ=26^\\circ$,\\\\\ntherefore, $\\angle ACE=\\angle ACD+\\angle DCE=58^\\circ+26^\\circ=84^\\circ$,\\\\\nsince $\\triangle ABC\\cong \\triangle CDE$,\\\\\nit implies $AC=CE$,\\\\\nthus, $\\angle CEA=\\angle CAE=\\frac{1}{2}\\left(180^\\circ-\\angle ACE\\right)=\\frac{1}{2}\\left(180^\\circ-84^\\circ\\right)=\\boxed{48^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53066403_122.png"
  },
  {
    "solution": "\\boxed{(4, 0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system $xOy$, line $AB$ intersects the x-axis at point B and intersects the y-axis at point $C(0, 2)$. It also intersects the graph of the inverse proportional function $y=\\frac{6}{x}$ within the first quadrant at point A. Draw $AD\\perp x$ axis at point $D$, with $OD=2$. Suppose point E is on the x-axis, and $\\triangle EBC$ is an isosceles triangle. Directly write down the coordinates of point E.\n",
    "completion": "\\textbf{Solution:} The coordinates of point $E$ are $(2\\sqrt{5}-4, 0)$ or $(-2\\sqrt{5}-4, 0)$ or $(4, 0)$ or $(-1.5, 0)$. Therefore, the answer is $(2\\sqrt{5}-4, 0)$ or $(-2\\sqrt{5}-4, 0)$ or $\\boxed{(4, 0)}$ or $(-1.5, 0)$.",
    "image_path": "dataset/mm_math/images/MM_Math/51458203_266.png"
  },
  {
    "solution": "\\boxed{66^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, there are two congruent triangles. The letters in the figure represent the lengths of the sides of the triangles. What is the degree measure of $\\angle 1$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure,\\\\\nby the triangle interior angle sum theorem, $ \\angle 2=180^\\circ-60^\\circ-54^\\circ=66^\\circ$,\\\\\nsince the two triangles are congruent,\\\\\nhence $ \\angle 1=\\angle 2=\\boxed{66^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52837310.png"
  },
  {
    "solution": "\\boxed{8cm}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, square DEFG has a side length of 4 cm, and all four vertices are on the sides of $\\triangle ABC$. In right $\\triangle ABC$, with $\\angle A=90^\\circ$, where $BD > EC$, and $BC=14$ cm, $\\triangle BDG \\sim \\triangle FEC$; find the length of $BD$ in cm?\n",
    "completion": "\\textbf{Solution:} Given that the side length of the square DEFG is 4cm,\\\\\nthus DE=DG=FE=4cm. Let BD=xcm.\\\\\nGiven that BC=14cm,\\\\\nthus $EC=BC−BD−DE=(10−x)cm$.\\\\\nGiven that $\\triangle BDG\\sim \\triangle FEC$,\\\\\nthus $\\frac{BD}{FE}=\\frac{DG}{EC}$.\\\\\nTherefore, $\\frac{x}{4}=\\frac{4}{10−x}$. Solving gives $x_{1}=8$, $x_{2}=2$. Upon verification, $x=8$ and $x=2$ are solutions to the aforementioned fractional equation,\\\\\nthus BD=8cm or BD=2cm. When BD=8cm, EC=2cm, which satisfies BD>EC. When BD=2cm, EC=8cm, does not satisfy BD>EC.\\\\\nTherefore, BD=\\boxed{8cm}.",
    "image_path": "dataset/mm_math/images/MM_Math/51368188_110.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, both quadrilateral $OABC$ and quadrilateral $ODEF$ are squares, with points F, O, A lying on a straight line, point D on the side $OC$. A Cartesian coordinate system $xOy$ is established with $FA$ as the x-axis and $OC$ as the y-axis. The line $y = \\frac{1}{2}x + 3$ passes through points B and E. What are the side lengths of square $OABC$ and square $ODEF$?\n",
    "completion": "\\textbf{Solution:} Let the side lengths of square $OABC$ and square $ODEF$ be: $a$, $b$,\\\\\n$\\therefore B(a,a)$, $A(-b,b)$,\\\\\n$\\because$ the line $y=\\frac{1}{2}x+3$ passes through points B and E,\\\\\n$\\therefore \\left\\{\\begin{array}{l}\n\\frac{1}{2}a+3=a \\\\\n-\\frac{1}{2}b+3=b\n\\end{array}\\right.$, solving this we get:\\\\\n$\\left\\{\\begin{array}{l}\na=\\boxed{6} \\\\\nb=\\boxed{2}\n\\end{array}\\right.$,\\\\\n$\\therefore$ the side lengths of square $OABC$ and square $ODEF$ are: \\boxed{6} and \\boxed{2};",
    "image_path": "dataset/mm_math/images/MM_Math/50594162_140.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within a $4\\times4$ square grid, points A, B, and C are grid intersections, with $AD\\perp BC$ and the foot of the perpendicular being D. What is the value of $\\sin\\angle BAD$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect AC. \\\\\nIn $\\triangle BEC$, $BC=\\sqrt{BE^{2} + CE^{2}} = \\sqrt{4^{2} + 3^{2}} = 5$, \\\\\nsince $AD \\perp BC$, \\\\\nthus $\\frac{1}{2} \\times BC \\times AD = 8$, \\\\\nwhich means $\\frac{1}{2} \\times 5 \\times AD = 8$, \\\\\nsolving this we get $AD = \\frac{16}{5}$, \\\\\nin $\\triangle ADB$, $BD = \\sqrt{AB^{2} - AD^{2}} = \\frac{12}{5}$, \\\\\ntherefore $\\sin\\angle BAD = \\frac{BD}{AB} = \\frac{\\frac{12}{5}}{4} = \\boxed{\\frac{3}{5}}$",
    "image_path": "dataset/mm_math/images/MM_Math/53043774.png"
  },
  {
    "solution": "\\boxed{5\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AD$ is the median of $\\triangle ABC$, and $\\angle DAC=\\angle B$. Point $E$ is on $AD$, with $CD=CE$. $\\triangle ACE\\sim \\triangle BAD$: If $AB=10$, and $BC=6$, determine the length of the line segment $AD$.\n",
    "completion": "\\textbf{Solution:} Given that $AD$ is the median of $\\triangle ABC$,\\\\\nthus $CD = BD = CE = \\frac{1}{2} BC = 3$,\\\\\nsince $\\angle DAC = \\angle B$, it follows that $\\angle ACD = \\angle BCA$, and thus $\\triangle ACD \\sim \\triangle BCA$,\\\\\nso $\\frac{AC}{BC} = \\frac{CD}{AC}$, that is, $\\frac{AC}{6} = \\frac{3}{AC}$,\\\\\nwhich leads to $AC = 3\\sqrt{2}$. Given that $\\triangle ACE \\sim \\triangle BAD$,\\\\\nwe have $\\frac{AC}{BA} = \\frac{CE}{AD}$, that is, $\\frac{3\\sqrt{2}}{10} = \\frac{3}{AD}$,\\\\\ntherefore $AD = \\boxed{5\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52694111_72.png"
  },
  {
    "solution": "\\boxed{20}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in rectangle $ABCD$, $AC$ and $BD$ intersect at point $O$, with $M$ and $N$ being the midpoints of $BC$ and $OC$, respectively. If $MN = 4$, what is the length of $AC$? If the area of $\\triangle BOC$ is 5, what is the area of rectangle $ABCD$?\n",
    "completion": "\\textbf{Solution:} Since M and N are the midpoints of BC and OC, respectively,\n\nit follows that $BO = 2MN = 8$.\n\nSince quadrilateral ABCD is a rectangle,\n\nit follows that $AC = BD = 2BO = \\boxed{16}$.\n\nSince the area of $\\triangle BOC$ is 5,\n\nthe area of rectangle ABCD is $5 \\times 4 = \\boxed{20}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52354816.png"
  },
  {
    "solution": "\\boxed{\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point A is a trisection point on the semicircle, point B is the midpoint of arc AN, and point P is a moving point on the diameter MN. Given that the radius $O$ is 1, what is the minimum value of $AP+BP$?\n",
    "completion": "\\textbf{Solution:} Let $A'$ be the symmetrical point of A with respect to line $MN$, and draw line $A'B$, which intersects $MN$ at point $P$. Then, $PA+PB=PA'+PB=A'B$ is minimized.\\\\\nDraw lines $OA'$ and $AA'$.\\\\\nSince point $A$ and $A'$ are symmetrical with respect to $MN$, and point $A$ is a trisection point on the semicircle,\\\\\nhence, $\\angle A'ON=\\angle AON=60^\\circ$, and $PA=PA'$,\\\\\nSince point $B$ is the midpoint of arc $AN$,\\\\\ntherefore, $\\angle BON=30^\\circ$,\\\\\nthus, $\\angle A'OB=\\angle A'ON + \\angle BON=90^\\circ$,\\\\\nAlso, since $OA=OA'=1$,\\\\\ntherefore, $A'B= \\sqrt{2}$\\\\\nhence, $PA+PB=PA'+PB=A'B= \\boxed{\\sqrt{2}}$",
    "image_path": "dataset/mm_math/images/MM_Math/52991365.png"
  },
  {
    "solution": "\\boxed{50^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, fold a rectangular strip $ABCD$ along $EF$. If $\\angle 1=65^\\circ$, what is the measure of $\\angle AEG$ in degrees?\n",
    "completion": "\\textbf{Solution:} Since $AD \\parallel BC$,\n\n$\\therefore \\angle DEF = \\angle 1 = 65^\\circ$,\n\nBy the property of folding, $\\angle DEF = \\angle DEF = 65^\\circ$,\n\n$\\therefore \\angle AEG = 180^\\circ - 2\\times65^\\circ = \\boxed{50^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52574157.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $C$ is the midpoint of $\\overset{\\frown}{ACB}$, $\\angle AOC = 4\\angle B$, and $OC = 4$. What is the length of line segment $AB$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, extend $CO$ to intersect $AB$ at point D,\\\\\n$\\because C$ is the midpoint of $\\overset{\\frown}{ACB}$,\\\\\n$\\therefore \\overset{\\frown}{AC}=\\overset{\\frown}{BC}$, $CD\\perp AB$, $AD=DB$,\\\\\n$\\because OC=4$, $\\angle A=30^\\circ$,\\\\\n$\\therefore AO=4, OD=2$,\\\\\n$\\therefore AD=\\sqrt{AO^{2}-OD^{2}}=\\sqrt{4^{2}-2^{2}}=2\\sqrt{3}$,\\\\\n$\\therefore AB=2AD=4\\sqrt{3}=\\boxed{4\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52906015_80.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC$, $\\odot O$ is the circumcircle of $\\triangle ABC$, and the extension of $BO$ intersects side $AC$ at point $D$. If $\\angle ACB=60^\\circ$ and $BC=8\\sqrt{3}$, what is the radius of $\\odot O$?\n",
    "completion": "\\textbf{Solution:} Draw line OA and extend it to intersect BC at point E,\\\\\n$\\because$ AB = AC,\\\\\n$\\therefore$ $\\overset{\\frown}{AB}$ = $\\overset{\\frown}{AC}$,\\\\\n$\\because$ AE passes through the center O,\\\\\n$\\therefore$ AE is the perpendicular bisector of BC,\\\\\n$\\therefore$ AE bisects $\\angle$BAC, and BE = $\\frac{1}{2}$BC = $4\\sqrt{3}$,\\\\\n$\\therefore$ $\\angle$BAC = $2\\angle$BAE,\\\\\n$\\because$ OA = OB,\\\\\n$\\therefore$ $\\angle$ABD = $\\angle$BAE,\\\\\n$\\because$ AB = AC and $\\angle$ACB = $60^\\circ$,\\\\\n$\\therefore$ $\\angle$ABC = $\\angle$ACB = $\\angle$BAC = $60^\\circ$,\\\\\n$\\therefore$ $\\angle$ABD = $\\frac{1}{2}\\angle$BAC = $30^\\circ$,\\\\\n$\\therefore$ $\\angle$CBD = $30^\\circ$,\\\\\n$\\therefore$ OB = \\boxed{8},\\\\\nThus, the radius of $\\odot$O is 8.",
    "image_path": "dataset/mm_math/images/MM_Math/53009799_129.png"
  },
  {
    "solution": "\\boxed{\\frac{7}{6}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\cos B= \\frac{\\sqrt{2}}{2}$, $\\sin C= \\frac{3}{5}$, and $AC= \\frac{5}{3}$. What is the area of $\\triangle ABC$?\n",
    "completion": "\\textbf{Solution}: Draw AD$\\perp$BC at point D through point A,\\\\\n$\\because$ $\\sin C=\\frac{3}{5}$,\\\\\n$\\therefore$ $\\frac{AD}{AC}=\\frac{3}{5}$,\\\\\n$\\therefore$ AD=$\\frac{3}{5}AC=\\frac{3}{5}\\times\\frac{5}{3}=1$,\\\\\n$\\therefore$ CD=$\\sqrt{AC^{2}-AD^{2}}=\\sqrt{\\left(\\frac{5}{3}\\right)^{2}-1^{2}}=\\frac{4}{3}$,\\\\\n$\\because$ $\\cos B=\\frac{\\sqrt{2}}{2}$,\\\\\n$\\therefore$ $\\angle B=45^\\circ$,\\\\\n$\\therefore$ BD=AD=1,\\\\\n$\\therefore$ BC=BD+CD=1+$\\frac{4}{3}=\\frac{7}{3}$,\\\\\n$\\therefore$ The area of $\\triangle ABC$=$\\frac{1}{2}\\times\\frac{7}{3}\\times1=\\boxed{\\frac{7}{6}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51034140.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, there are three points $D$, $B$, and $E$ sequentially on the line segment $AC$, among which point $B$ is the midpoint of line segment $AC$, $AD=BE$, if $AC=8$, then what is the length of $DE$?\n",
    "completion": "\\textbf{Solution:} \nGiven that points D, B, and E lie sequentially on segment $AC$,\n\nit follows that $DE = DB + BE$,\n\nsince $AD = BE$,\n\nthus $DE = DB + AD = AB$,\n\ngiven $AC = 8$, and B is the midpoint of segment $AC$,\n\nit follows that $AB = \\frac{1}{2}AC = \\frac{1}{2} \\times 8 = 4$,\n\ntherefore, $DE = AB = \\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53404553.png"
  },
  {
    "solution": "\\boxed{\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, $O$ is the midpoint of $AB$, $OD$ bisects $\\angle AOC$ and intersects $AC$ at point $G$, $OD = OA$, $BD$ intersects $AC$ and $OC$ at points $E$ and $F$ respectively, connect $AD$ and $CD$. What is the value of $OG\\colon BC$? If $CE = CF$, what is the value of $CF\\colon OF$?\n",
    "completion": "\\textbf{Solution:}\n\n(1) Since $O$ is the midpoint of $AB$ and $\\angle ACB=90^\\circ$, \\\\\n$\\therefore$ $OB=OA=OC$, \\\\\nSince $OD$ bisects $\\angle AOC$ and intersects $AC$ at point $G$, \\\\\n$\\therefore$ point $G$ is the midpoint of $AC$, \\\\\n$\\therefore$ $OG$ is the median of $\\triangle ABC$, \\\\\n$\\therefore$ $OG\\parallel BC$ and $OG=\\frac{1}{2}BC$, that is, $\\frac{OG}{BC}=\\boxed{\\frac{1}{2}}$; \n\n(2) \\\\\nSince $OD=OA$, \\\\\n$\\therefore$ $OD=OB$, \\\\\n$\\therefore$ $\\angle ODB=\\angle OBD$, \\\\\nSince $OG\\perp AC$, \\\\\n$\\therefore$ $\\angle DGE=90^\\circ$, \\\\\n$\\therefore$ $\\angle GDE+\\angle DEG=90^\\circ$, \\\\\nSince $CE=CF$, \\\\\n$\\therefore$ $\\angle CEF=\\angle CFE$, \\\\\nSince $\\angle CEF=\\angle DEG$, $\\angle CFE=\\angle OFB$, and $\\angle ODB=\\angle OBD$, \\\\\n$\\therefore$ $\\angle OFB+\\angle OBD=90^\\circ$, \\\\\n$\\therefore$ $\\angle FOB=90^\\circ$, that is, $CO\\perp AB$, \\\\\n$\\therefore$ $\\triangle OBC$ is an isosceles right triangle, \\\\\n$\\therefore$ $BC: OB=\\sqrt{2}: 1$; \n\nKnowing from (1) that $OG\\parallel BC$, \\\\\n$\\therefore$ $\\triangle BCF\\sim \\triangle DOF$, \\\\\n$\\therefore$ $\\frac{CF}{OF}=\\frac{BC}{OD}=\\frac{BC}{OB}=\\boxed{\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52602144.png"
  },
  {
    "solution": "\\boxed{50^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ADB \\cong \\triangle ECB$, and the corresponding point of A is E, the corresponding point of D is C. If $\\angle CBD = 40^\\circ$ and $BD \\perp EC$, then what is the degree measure of $\\angle D$?\n",
    "completion": "\\textbf{Solution:} Let BD and EC intersect at point G,\\\\\nsince $\\triangle ADB \\cong \\triangle ECB$,\\\\\nit follows that $\\angle C = \\angle D$,\\\\\nsince BD$\\perp$EC,\\\\\nit follows that $\\angle BGC = 90^\\circ$,\\\\\ntherefore, $\\angle C = \\angle D = 90^\\circ - \\angle CBD = 90^\\circ - 40^\\circ = \\boxed{50^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52694135.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, fold the square paper $ABCD$ along the line that passes through the midpoint of the opposite sides and then unfold it; the crease is marked as $MN$. Fold the paper again through point $B$ making point $A$ land on point $F$ on $MN$, with the crease marked as $BE$. If $FN=3$, what is the side length of the square paper?\n",
    "completion": "\\textbf{Solution:} Let the side length of the square be $a$. From the folding, it is known that $BN=\\frac{1}{2}a$ and $BF=AB=a$. \\\\\nIn $\\triangle BFN$, according to the Pythagorean theorem, we have $BF^2=FN^2+BN^2$, \\\\\nwhich implies: $a^2=3^2+\\left(\\frac{a}{2}\\right)^2$, \\\\\nSolving this gives: $a=2\\sqrt{3}$ or $a=-2\\sqrt{3}$ (discard this solution). \\\\\nTherefore, the answer is: $a=\\boxed{2\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52690432.png"
  },
  {
    "solution": "\\boxed{2-\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within rectangle $ABCD$, the bisector of $\\angle BCD$ intersects side $AD$ at point $E$, the bisector of $\\angle AEC$ intersects the extended line of side $CB$ at point $G$ and intersects side $AB$ at point $F$. If $AB = 3\\sqrt{2}$ and $AF = 2BF$, then what is the length of $GB$?\n",
    "completion": "\\textbf{Solution:} From rectangle $ABCD$, the angle bisector of $\\angle BCD$, CE, intersects AD at E; \\\\\n$\\therefore CD=AB=3\\sqrt{2}$, $\\angle DCE=\\angle BCE=45^\\circ$, \\\\\n$\\therefore CD=DE=3\\sqrt{2}$, \\\\\nFrom right triangle CDE, \\\\\n$\\therefore CE=\\sqrt{CD^2+DE^2}=6$, \\\\\nAlso, the angle bisector of $\\angle AEC$, EG, intersects AB at point F, \\\\\n$\\therefore \\angle AEG=\\angle CEG$, \\\\\nSince AD$\\parallel$BC, \\\\\n$\\therefore \\angle G=\\angle AEG$, \\\\\n$\\therefore \\angle CEG=\\angle G$, \\\\\n$\\therefore CG=CE=6$, \\\\\nSince $\\angle G=\\angle AEF$, $\\angle AFE=\\angle BFG$, \\\\\n$\\therefore \\triangle AEF\\sim \\triangle BGF$, \\\\\n$\\therefore \\frac{GB}{AE}=\\frac{FG}{EF}=\\frac{FB}{AF}=\\frac{BF}{2BF}=\\frac{1}{2}$, \\\\\nLet $BG=x$, then $AE=2x$, thus $BC=AD=3\\sqrt{2}+2x$, \\\\\nBecause $CG=BC+BG$, \\\\\n$\\therefore 6=3\\sqrt{2}+2x+x$, solving for $x$ gives $x=\\boxed{2-\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51379481.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{5}-2}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated, in $\\triangle ABC$, $AD$ is the altitude to side $BC$, $AE$ is the median to side $BC$, $\\angle C=45^\\circ$, $\\sin B=\\frac{2}{3}$, and $AD=4$. What is the value of $\\tan \\angle DAE$?\n",
    "completion": "\\textbf{Solution:} Since AE is the median to side BC,\\\\\nhence CE = $\\frac{1}{2}$BC = $\\sqrt{5}+2$,\\\\\ntherefore DE = CE - CD = $\\sqrt{5} - 2$,\\\\\nthus $\\tan \\angle DAE = \\frac{DE}{AD} = \\boxed{\\frac{\\sqrt{5}-2}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51318176_97.png"
  },
  {
    "solution": "\\boxed{5x}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the Cartesian coordinate system $xOy$, the graph of the linear function $y=-x+6$, denoted as ${l}_{1}$, intersects the $x$ and $y$ axes at points A and B, respectively. The graph of the direct proportionality function, ${l}_{2}$, intersects ${l}_{1}$ at point C$(m,5)$. Find the value of $m$ and the equation of ${l}_{2}$.\n",
    "completion": "\\[ \\text{{Solution:}} \\] \\[ \\because \\] the graph of the linear function \\(y = -x + 6\\), denoted as \\(l_1\\), passes through the point \\(C(m, 5)\\), \\\\\n\\[ \\therefore \\] \\(-m + 6 = 5\\), \\\\\n\\[ \\therefore \\] \\(m = \\boxed{1}\\), \\\\\nLet the equation of \\(l_2\\) be \\(y = kx\\), \\\\\n\\[ \\therefore \\] \\(k = 5\\), \\\\\n\\[ \\therefore \\] the equation of \\(l_2\\) is \\(y = \\boxed{5x}\\).",
    "image_path": "dataset/mm_math/images/MM_Math/50555762_80.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that in $ \\triangle ABC$, $ \\angle BAC=45^\\circ$, $ AD\\perp BC$ with the foot of the perpendicular at D, $ BD=2$, and $ DC=3$. Determine the length of AD. Following the approach of folding the figure as Xiaoming did, explore and solve the following question: What is the length of AD?\n",
    "completion": "\\textbf{Solution:} Let $AD=x$, then $EG=GF=AE=x$,\\\\\nsince $BD=2$, $DC=3$,\\\\\nthus $BE=2$, $CF=3$,\\\\\ntherefore $BG=x-2$, $CG=x-3$,\\\\\nsince quadrilateral AEGF is a square,\\\\\nthus $\\angle EGC=90^\\circ$,\\\\\nin $\\triangle ABC$, $BG^2+CG^2=BC^2$,\\\\\nthus $(x-2)^2+(x-3)^2=5^2$,\\\\\nafter simplifying, $(x-6)(x+1)=0$, solving gives $x=6$, or $x=-1$ (discard),\\\\\ntherefore $AD=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53359676_135.png"
  },
  {
    "solution": "\\boxed{41\\pi}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, ABCD is a quadrilateral inscribed in circle O, and $AC \\perp BD$. If $AB=10$ and $CD=8$, what is the area of circle O?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, connect $AO$ and extend it to meet circle $O$ at point $E$, and connect $EB$, $EC$.\\\\\nThen $AB\\perp BE$, $AC\\perp CE$.\\\\\nSince $AC\\perp BD$,\\\\\nTherefore $BD$ $\\parallel$ $EC$,\\\\\nTherefore $\\overset{\\frown}{CD}=\\overset{\\frown}{BE}$\\\\\nTherefore $BE=CD$,\\\\\nSince $CD=8$\\\\\nTherefore $EB=CD=8$.\\\\\nIn $\\triangle ABE$, $AB=10$, $EB=8$\\\\\nSo, by the Pythagorean theorem, $AE=\\sqrt{AB^{2}+BE^{2}}=\\sqrt{10^{2}+8^{2}}=2\\sqrt{41}$\\\\\nTherefore $OA=\\frac{1}{2}AE=\\sqrt{41}$.\\\\\nThus, the area of circle $O$ is $ \\pi \\times OA^{2}=\\boxed{41\\pi}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51722329.png"
  },
  {
    "solution": "\\boxed{y=x^{2}+2x+1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=ax^{2}+2ax+1$ has only one common point A with the x-axis. The line passing through point A intersects the parabola at point B and the y-axis at point C, and point C is the midpoint of line segment AB. What is the analytical expression of the function corresponding to this parabola?\n",
    "completion": "\\textbf{Solution:} Since the parabola $y=ax^{2}+2ax+1$ has only one common point $A$ with the x-axis,\\\\\ntherefore, $\\Delta=4a^{2}-4a=0$, solving this yields $a_{1}=0$ (discard), $a_{2}=1$,\\\\\ntherefore, the equation of the parabola is $\\boxed{y=x^{2}+2x+1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52978187_217.png"
  },
  {
    "solution": "\\boxed{0 < x < 1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y_1=kx+b$ (where $k$ is a constant and $k\\ne 0$) intersects the hyperbola $y_2=\\frac{2}{x}$ at points $A$ and $D$, and intersects the $x$-axis and $y$-axis at points $B$ and $C$ respectively. The coordinates of point $A$ are $(m, 2)$, and the coordinates of point $D$ are $(-2, n)$. Based on the graph, directly write the range of values for $x$ when $y_1 < y_2$.\n",
    "completion": "\\textbf{Solution:} From the graph, it is known that when $y_1 < y_2$, the range of $x$ is $\\boxed{x < -2}$ or $\\boxed{0 < x < 1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50594263_138.png"
  },
  {
    "solution": "\\boxed{22.5^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, AB is a chord of the semicircle O, DE is the diameter. The tangent BC passing through point B is tangent to $\\odot$O at point B, and intersects the extension line of DE at point C. Connect BD. If the quadrilateral OABC is a parallelogram, then what is the degree measure of $\\angle BDC$?\n",
    "completion": "Solution: Since $BC$ is the tangent of circle $O$,\\\\\ntherefore $\\angle OBC=90^\\circ$,\\\\\nSince quadrilateral $ABCO$ is a parallelogram,\\\\\ntherefore $AO=BC$,\\\\\nAlso since $AO=BO$,\\\\\ntherefore $BO=BC$,\\\\\ntherefore $\\angle BOC=\\angle BCO=45^\\circ$,\\\\\ntherefore $\\angle BDC=\\boxed{22.5^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51407041.png"
  },
  {
    "solution": "\\boxed{y=\\frac{-12}{x}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=k_1x+b$ intersects with the hyperbola $y=\\frac{k_2}{x}$ at points $A$ and $B$. The line $AB$ intersects the $x$-axis at point $C$ and the $y$-axis at point $D(0, -5)$. The length of $OA$ is $5$, and $\\tan\\angle AOC=\\frac{3}{4}$. The $y$-coordinate of point $B$ is $-8$. What is the analytical expression of the inverse proportion function?\n",
    "completion": "Solution: As shown in the diagram, construct line $AE$ perpendicular to the $x$-axis at point $E$, \\\\\n$\\because \\tan\\angle AOC=\\frac{3}{4}$, let $AE=3a$, then $EO=4a$, \\\\\n$\\therefore OA=5a$ \\\\\n$\\because OA=5$, \\\\\n$\\therefore a=1$ \\\\\n$\\therefore AE=3$, $EO=4$, \\\\\n$\\therefore$ the coordinates of point $A$ are $\\left(-4, 3\\right)$, \\\\\n$\\therefore$ the equation of the hyperbola is $\\boxed{y=\\frac{-12}{x}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51446358_247.png"
  },
  {
    "solution": "\\boxed{2.8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, it is known that $AB$ is the diameter of $\\odot O$, $\\triangle ABC$ is inscribed in $\\odot O$, $AB=10$, $BC=6$, and point $D$ is a moving point on $\\odot O$ (point $D$ does not coincide with points $A$ or $B$). $OC \\parallel AD$, what is the length of $AD$?\n",
    "completion": "\\textbf{Solution:} Given that $AB=10$ and $BC=6$,\\\\\nthus $OC=AO=OB=5$,\\\\\nsince $\\angle BHO=\\angle BHC=90^\\circ$,\\\\\nit follows that $OB^2-OH^2=CB^2-CH^2=BH^2$,\\\\\nthus $5^2-OH^2=6^2-(5-OH)^2$,\\\\\ntherefore $OH=1.4$,\\\\\nsince $\\overset{\\frown}{BC}=\\overset{\\frown}{CD}$,\\\\\nit follows that $DH=BH$,\\\\\nhence $AD=2OH=\\boxed{2.8}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52793804_148.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\text{Rt}\\triangle ABC$, $\\angle ACB=90^\\circ$, points $D$ and $E$ are the midpoints of sides $AB$ and $AC$, respectively. Extend $BC$ to point $F$ such that $CF=\\frac{1}{2}BC$. If $AB=10$, what is the length of $EF$?\n",
    "completion": "\\textbf{Solution:} Given that points D and E are the midpoints of AB and AC, respectively,\n\n$\\therefore$ $AD=BD=\\frac{1}{2}AB=5$, DE is the midline of $\\triangle ACB$, AE=CE\n\n$\\therefore$ $DE\\parallel BC$, $DE=\\frac{1}{2}BC$,\n\n$\\because$ $\\angle ACB=90^\\circ$, $CF=\\frac{1}{2}BC$,\n\n$\\therefore$ $\\angle DEA=\\angle FCE=90^\\circ$, CF=ED,\n\n$\\therefore$ $\\triangle AED\\cong \\triangle ECF$ (SAS),\n\n$\\therefore$ EF=AD=\\boxed{5}.",
    "image_path": "dataset/mm_math/images/MM_Math/51968676.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, it is known that $\\angle AOB= 40^\\circ$. If $\\angle AOC= \\frac{1}{3}\\angle BOC$, what is the degree measure of $\\angle BOC$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, when OC is outside of $\\angle AOB$,\\\\\n$\\because \\angle AOC= \\frac{1}{3}\\angle BOC= \\frac{1}{3}(\\angle AOB+\\angle AOC)$,\\\\\n$\\therefore \\angle AOC= \\frac{1}{2}\\angle AOB= \\frac{1}{2}\\times 40^\\circ=20^\\circ$,\\\\\n$\\therefore \\angle BOC=\\angle AOB+\\angle AOC=40^\\circ+ 20^\\circ= 60^\\circ$;\\\\\nAs shown in the diagram, when OC is inside of $\\angle AOB$,\\\\\n$\\because \\angle AOC= \\frac{1}{3}\\angle BOC= \\frac{1}{3} (\\angle AOB-\\angle AOC)$,\\\\\n$\\therefore \\angle AOC= \\frac{1}{4}\\angle AOB= \\frac{1}{4}\\times 40^\\circ=10^\\circ$,\\\\\n$\\therefore \\angle BOC=\\angle AOB-\\angle AOC=40^\\circ- 10^\\circ= 30^\\circ$.\\\\\nHence, the degree of $\\angle BOC$ is $\\boxed{60^\\circ}$ or $\\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55192442_107.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, lines AB and CD intersect at point O. It is known that $\\angle BOD=75^\\circ$, and OE divides $\\angle AOC$ into two angles, with $\\angle AOE: \\angle EOC = 2: 3$. What is the degree measure of $\\angle AOE$?\n",
    "completion": "\\textbf{Solution:} From the given information, we have $\\angle AOC=\\angle BOD=75^\\circ$,\\\\\n$\\therefore \\angle AOE=75^\\circ \\times \\frac{2}{5}=\\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51244984_206.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, D and E are points on AB and AC respectively, $\\angle AED = \\angle B$, $AD = 2$, $AC = 3$, and the angle bisector of $\\triangle ABC$ intersects DE at point G and intersects BC at point F. $\\triangle ADE \\sim \\triangle ACB$; find the value of $\\frac{AG}{GF}$.\n",
    "completion": "\\textbf{Solution:} Given that $\\triangle ADE \\sim \\triangle ACB$,\\\\\n$\\therefore \\angle ADE = \\angle C$,\\\\\nas $AF$ bisects $\\angle BAC$,\\\\\n$\\therefore \\angle DAG = \\angle CAF$,\\\\\n$\\therefore \\triangle ADG \\sim \\triangle ACF$,\\\\\n$\\therefore \\frac{AG}{AF} = \\frac{AD}{AC}$,\\\\\ngiven that $AD = 2$ and $AC = 3$,\\\\\n$\\therefore \\frac{AG}{AF} = \\frac{2}{3}$,\\\\\n$\\therefore \\frac{AG}{GF} = \\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51545199_171.png"
  },
  {
    "solution": "\\boxed{140^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point O lies on line AB, and ray OD bisects $\\angle AOC$. If $\\angle AOD = 20^\\circ$, what is the measure of $\\angle COB$ in degrees?\n",
    "completion": "\\textbf{Solution:} Since OD bisects $\\angle$AOC,\\\\\nit follows that $\\angle$AOC = 2$\\angle$AOD = 40$^\\circ$,\\\\\ntherefore, $\\angle$COB = 180$^\\circ$ - $\\angle$COA = \\boxed{140^\\circ}",
    "image_path": "dataset/mm_math/images/MM_Math/53253563.png"
  },
  {
    "solution": "\\boxed{150^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $a\\parallel b$. Place a right-angled triangle, which includes a $30^\\circ$ angle, in the position as shown in the figure. If $\\angle 2 = 2\\angle 1$, then what is the degree measure of $\\angle 3$?\n",
    "completion": "\\textbf{Solution}: As shown in the diagram,\\\\\nfrom the given condition we have $\\angle 1+\\angle 2=90^\\circ$,\\\\\nsince $\\angle 2=2\\angle 1$,\\\\\nthus $3\\angle 1=90^\\circ$,\\\\\ntherefore $\\angle 1=30^\\circ$,\\\\\nhence $\\angle 2=2\\angle 1=60^\\circ$,\\\\\nsince $a\\parallel b$,\\\\\ntherefore $\\angle 2=\\angle ABD=60^\\circ$,\\\\\nsince $\\angle ABD=\\angle C+\\angle BDC$ and $\\angle C=30^\\circ$,\\\\\nthus $\\angle BDC=30^\\circ$,\\\\\ntherefore $\\angle 3=180^\\circ−30^\\circ=\\boxed{150^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52513499.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ and $\\triangle CDE$, with the right-angled vertices coinciding at point $C$, point $D$ on $AB$, $\\angle BAC=\\angle DEC$ and $\\sin\\angle BAC=\\frac{1}{3}$, connect $AE$. If $BD=2$, $AD=7$, then what is the length of $AE$?\n",
    "completion": "\\textbf{Solution:} Given that $BD=2$, $AD=7$,\n\ntherefore $AB=AD+BD=7+2=9$, in $\\triangle ABC$, $\\sin\\angle BAC=\\frac{BC}{AB}=\\frac{1}{3}$,\n\nhence $BC=\\frac{1}{3}AB=3$,\n\nthus $AC=\\sqrt{9^2-3^2}=6\\sqrt{2}$, and in $\\triangle BCA$ and $\\triangle DCE$, $\\angle BAC=\\angle DEC$,\n\nthus $\\tan\\angle BAC=\\tan\\angle DEC$,\n\ntherefore $\\frac{BC}{AC}=\\frac{DC}{EC}$,\n\ngiven $\\angle BCA=\\angle DCE=90^\\circ$,\n\nhence $\\angle BCA-\\angle DCA=\\angle DCE-\\angle DCA$, that is, $\\angle BCD=\\angle ACE$,\n\ntherefore $\\triangle BCD \\sim \\triangle ACE$,\n\nthus $\\angle CAE=\\angle B$, $\\frac{BC}{AC}=\\frac{BD}{AE}$,\n\nhence $\\angle DAE=\\angle DAC+\\angle CAE=90^\\circ$,\n\nthus $\\frac{3}{6\\sqrt{2}}=\\frac{2}{AE}$, solving gives: $AE=\\boxed{4\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55589668.png"
  },
  {
    "solution": "\\boxed{y=x-3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given the line AB: $y = \\frac{1}{2}x + 1$ intersects the x-axis and y-axis at points A and B, respectively. The line CD: $y = x + b$ intersects the x-axis and y-axis at points C and D, respectively. The line AB intersects line CD at point P, and $S_{\\triangle ABD} = 4$. Find the equation of line CD.\n",
    "completion": "\\textbf{Solution:} When $y=0$, that is, when $\\frac{1}{2}x+1=0$, we find $x=-2$,\\\\ \n$\\therefore \\boxed{A(-2,0)}$,\\\\\nwhen $x=0$, $y=\\frac{1}{2}\\times 0+1=1$,\\\\\n$\\therefore \\boxed{B(0,1)}$,\\\\\n$\\because$ ${S}_{\\triangle ABD} = \\frac{1}{2} \\text{BD} \\times \\text{OA} = \\frac{1}{2}\\times \\text{BD}\\times 2 = 4$,\\\\\n$\\therefore \\text{BD}=4$,\\\\\nAlso, $\\because$ when $x=0$ $y=0+b=b$,\\\\\n$\\therefore \\boxed{D(0,b)}$,\\\\\n$\\because \\text{BD}=1-b=4$,\\\\\n$\\therefore b=-3$,\\\\\n$\\therefore$ the equation of line CD is: $\\boxed{y=x-3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52251756_78.png"
  },
  {
    "solution": "\\boxed{24}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rhombus $ABCD$, $AB=13$ and $AC=10$, what is the length of $BD$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral ABCD is a rhombus,\\\\\nhence BD = 2OB, AC = 2OA, and AC $\\perp$ BD,\\\\\ntherefore OA = 5, and $\\angle$AOB = 90$^\\circ$\\\\\nIn right $\\triangle$AOB, $BO = \\sqrt{AB^{2}-OA^{2}} = \\sqrt{13^{2}-5^{2}} = 12$\\\\\ntherefore BD = $2 \\times 12 = \\boxed{24}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50475012.png"
  },
  {
    "solution": "\\boxed{0<b<80}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system $xOy$, given $A\\left(0, 4\\right)$ and $B\\left(4, 0\\right)$, the graph of the linear function $y=-2x$ intersects with line $AB$ at point $P$. If the line $y=-2x+b$ has exactly two intersection points with the three sides of $\\triangle AOB$, what is the range of values for $b$?\n",
    "completion": "\\textbf{Solution:} Let the range of $b$ be $0<b<80$, \\\\\n$\\therefore$ the range of $b$ is $\\boxed{0<b<80}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50560455_71.png"
  },
  {
    "solution": "\\boxed{336}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, three squares form a right-angled triangle, with $64$ and $400$ as the areas of the respective squares. What is the area of the square represented by the letter $M$ in the figure?\n",
    "completion": "\\textbf{Solution:} According to the relationship between the area and the side length of a square,\\\\\nthe square of the short leg of the right-angled triangle in the figure is 64, and the square of the hypotenuse is 400,\\\\\nbased on the Pythagorean theorem, the square of the long leg of the right-angled triangle is $400-64=336$,\\\\\nthus the area of the square represented by the letter $M$ is $\\boxed{336}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53033458.png"
  },
  {
    "solution": "\\boxed{65^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ABCD$, $E$ is a point on side $BC$. Taking $AE$ as a side, square $AEFG$ is constructed. If $\\angle BAE = 40^\\circ$ and $\\angle CEF = 15^\\circ$, then what is the measure of $\\angle D$?\n",
    "completion": "\\textbf{Solution:} Given that square $AEFG$,\\\\\nit follows that $\\angle AEF=90^\\circ$,\\\\\ngiven $\\angle BAE=40^\\circ$, $\\angle CEF=15^\\circ$,\\\\\nthus, $\\angle AEC=\\angle AEF+\\angle CEF=\\angle BAE+\\angle B$, which means $90^\\circ+15^\\circ=40^\\circ+\\angle B$,\\\\\nso, $\\angle B=65^\\circ$,\\\\\ngiven parallelogram $ABCD$,\\\\\nit follows that $\\angle D=\\angle B=\\boxed{65^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51435947.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{5}}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, point D is the midpoint of side AB, and CD is connected. If $BC=4$ and $CD=3$, what is the value of $\\sin\\angle DCB$?\n",
    "completion": "\\textbf{Solution:} Given that in $\\triangle ABC$, $\\angle ACB=90^\\circ$, and point D is the midpoint of side AB,\\\\\n$\\therefore$AD=BD=CD=$\\frac{1}{2}$AB,\\\\\n$\\therefore$ $\\angle DCB=\\angle B$,\\\\\nAlso, since CD=3,\\\\\n$\\therefore$AB=6,\\\\\n$AC=\\sqrt{AB^{2}-BC^{2}}=\\sqrt{6^{2}-4^{2}}=2\\sqrt{5}$,\\\\\n$\\therefore$ $\\sin\\angle DCB=\\sin B=\\frac{AC}{AB}=\\frac{2\\sqrt{5}}{6}=\\boxed{\\frac{\\sqrt{5}}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51318127.png"
  },
  {
    "solution": "\\boxed{1-\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=x^{2}+bx+c$ (where $b$ and $c$ are constants) intersects the x-axis at points $A(-1,0)$ and $B(3,0)$, and intersects the y-axis at point $C$. Let $P$ be a point on the parabola with an abscissa of $m$. When the difference between the maximum and minimum y-coordinates of the section of the parabola between points $C$ and $P$ (including points $C$ and $P$) is $2$, find the value of $m$?\n",
    "completion": "\\textbf{Solution:} From the given, we have $y=x^{2}-2x-3=(x-1)^{2}-4$,\\\\\n$\\therefore$ the vertex of the parabola is $(1,-4)$.\\\\\nLet $x=0$, then $y=-3$,\\\\\n$\\therefore$ point $C(0,-3)$ is established.\\\\\n$\\because$ the difference between the highest and lowest ordinate (y-coordinate) of the segment between point $C$ and point $P$ (including points $C$ and $P$) is $2$,\\\\\n$\\therefore$ point $P$ cannot be below point $C$.\\\\\n$\\therefore$ point $P$ is above point $C$.\\\\\n$\\therefore$ the ordinate of point $P$ is $-1$,\\\\\nLet $y=-1$, then $m^{2}-2m-3=-1$.\\\\\nSolving yields: $m=1\\pm\\sqrt{3}$.\\\\\n$\\therefore$ the value of $m$ is: $1+\\sqrt{3}$ or $1-\\sqrt{3}$.\\\\\nThus, the value of $m$ is $\\boxed{1+\\sqrt{3}}$ or $\\boxed{1-\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52623662_243.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, AB is the diameter of the circle $\\odot O$. Chord CD is perpendicular to AB at point E. Lines AC and BC are connected. If $\\angle BAC=30^\\circ$ and $CD=6\\text{cm}$, what is the diameter of $\\odot O$ in cm?\n",
    "completion": "\\textbf{Solution:} Since diameter AB$\\perp$CD and CD = 6cm,\\\\\nit follows that CE = 3cm,\\\\\nIn $\\triangle ACE$, with $\\angle A=30^\\circ$,\\\\\nit follows that AC = 6cm,\\\\\nSince AB is a diameter,\\\\\nit follows that $\\angle ACB=90^\\circ$,\\\\\nIn $\\triangle ACB$, AB = $\\frac{AC}{\\cos\\angle A}=\\frac{6}{\\cos 30^\\circ}=4\\sqrt{3}$ (cm).\\\\\nTherefore, the diameter of $\\odot O$ is $\\boxed{4\\sqrt{3}}$ cm.",
    "image_path": "dataset/mm_math/images/MM_Math/53009955_44.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, it is known that the graph of the linear function $y_1=kx+b$ and the graph of the inverse proportion function $y_2=\\frac{m}{x}$ intersect at points A(6,1) and B(a,-3) in the first and third quadrants, respectively. Lines OA and OB are drawn. What is the area of $\\triangle AOB$?\n",
    "completion": "\\textbf{Solution:}  Let the linear function $y=\\frac{1}{2}x-2$ intersect the y-axis at point C. Draw $AD\\perp y$ axis at point D through point A, and draw $BE\\perp y$ axis at point E through point B. In $y=\\frac{1}{2}x-2$, let $x=0$, then $y=-2$, thus $OC=2$. Given $A(6, 1)$, $B(-2, -3)$, and $AD\\perp y$ axis, $BE\\perp y$ axis. Therefore, $AD=6$, $BE=2$ \\\\\nTherefore, the area of $\\triangle AOB$ is $\\frac{1}{2}OC\\cdot AD+\\frac{1}{2}OC\\cdot BE= \\frac{1}{2}OC(AD+BE)=\\frac{1}{2}\\times 2\\times 8 = \\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51457885_276.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle AOB$ is an isosceles right triangle, with the hypotenuse $OB$ on the x-axis. The graph of the linear function $y=3x-4$ passes through point A and intersects the y-axis at point C. The graph of the inverse proportion function $y=\\frac{k}{x}$ (where $x>0$) also passes through point A. Draw $OD \\perp AC$ at point $D$ from point O. Find the value of $CD^2-AD^2$.\n",
    "completion": "\\textbf{Solution:} Given $A\\left(2, 2\\right)$,\\\\\ntherefore $AO^{2}=2^{2}+2^{2}=8$,\\\\\nsubstitute $x=0$ into $y=3x-4$ to obtain $y=-4$,\\\\\ntherefore $C\\left(0, -4\\right)$,\\\\\ntherefore $OC=4$,\\\\\nin $\\triangle COD$, we have $CD^{2}=OC^{2}-OD^{2}$ (1),\\\\\nin $\\triangle AOD$, we have $AD^{2}=OA^{2}-OD^{2}$ (2),\\\\\nSubtracting (2) from (1) yields: $CD^{2}-AD^{2}=OC^{2}-OA^{2}=16-8=\\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52602253_187.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{10}}{10}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the side length of the square $ABCD$ is $1$. Extend $BA$ to $E$ so that $AE=1$. Connect $EC$ and $ED$. Then, what is the value of $\\sin\\angle CED$?\n",
    "completion": "\\textbf{Solution:} Connect AC, draw EF $\\perp$ CA, meeting the extension of CA at point F.\\\\ \n$\\because$ Quadrilateral ABCD is a square, $\\therefore$ $\\angle$BAC = 45$^\\circ$, $\\angle$DAE = $\\angle$DAB = 90$^\\circ$.\\\\\n$\\because$ AD = AE = 1, $\\therefore$ $\\angle$AED = $\\angle$ADE = 45$^\\circ$, that is $\\angle$DEA = $\\angle$CAB = 45$^\\circ$, $\\therefore$ AC $\\parallel$ ED, $\\therefore$ $\\angle$CED = $\\angle$ECA.\\\\\n$\\because$ AE = 1, $\\therefore$ by the Pythagorean theorem: EF = AF = $\\frac{\\sqrt{2}}{2}$.\\\\\n$\\because$ in $\\triangle$EBC, by the Pythagorean theorem: CE$^{2}$ = $1^{2} + 2^{2} = 5$, $\\therefore$ CE = $\\sqrt{5}$, $\\therefore$ $\\sin\\angle$CED = $\\sin\\angle$ECF = $\\frac{EF}{CE} = \\frac{\\frac{\\sqrt{2}}{2}}{\\sqrt{5}} = \\boxed{\\frac{\\sqrt{10}}{10}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51213099.png"
  },
  {
    "solution": "\\boxed{y=\\frac{8}{x}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the linear function $y=kx+2$ intersects the $y$-axis at point A and intersects the graph of the inverse proportion function $y=\\frac{m}{x}$ at points B and C. Line $BD\\perp y$-axis intersects the $y$-axis at point D, $OA=OD$, and the area of triangle $\\triangle ABD$ is $8$. Find the expressions of the linear function and the inverse proportion function.\n",
    "completion": "Solution: From the given information, point $A$ is the intersection point of the linear function $y=kx+2$ with the $y$-axis,\\\\\n$\\therefore$ setting $x=0$, we get $y=k\\times 0+2=2$,\\\\\nwhich means $A(0,2)$,\\\\\n$\\therefore OA=2$,\\\\\nand since $OD=OA$,\\\\\n$\\therefore OD=2$,\\\\\n$\\therefore D(0,-2)$,\\\\\n$\\therefore AD=2OD=4$,\\\\\nsince $BD \\perp y$-axis,\\\\\n$\\therefore$ the ordinate of point $B$ is $-2$,\\\\\nsince $S_{\\triangle ABD}=8$,\\\\\n$\\therefore \\frac{1}{2}AD\\times BD=8$,\\\\\n$\\therefore \\frac{1}{2}\\times 4\\times BD=8$,\\\\\n$\\therefore BD=4$,\\\\\n$\\therefore$ the coordinates of point $B$ are $(-4,-2)$,\\\\\nSubstituting point $B(-4,-2)$ into the linear function $y=kx+2$ and the inverse proportion function $y=\\frac{m}{x}$ respectively,\\\\\nwe find: $-4=-2k+2$, $-2=\\frac{m}{-4}$,\\\\\n$\\therefore k=1$, $m=8$,\\\\\n$\\therefore$ the equation of the linear function is: \\boxed{y=x+2}, and the equation of the inverse proportion function is: \\boxed{y=\\frac{8}{x}};",
    "image_path": "dataset/mm_math/images/MM_Math/51445968_243.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in square ABCD, point F is on CD, and BF intersects AC at point E. What is the measure of $\\angle ACB$ in degrees?\n",
    "completion": "\\textbf{Solution:} We have $\\angle ACB = \\boxed{45^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53313466_134.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y_{1}=kx+2$ intersects with the graph of the inverse proportion function $y_{2}=-\\frac{8}{x}$ at points $A(a，-2a)$ and $B(4，-2)$. Connect $OA$ and $OB$. What is the area of $\\triangle AOB$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, AB intersects the x-axis at point C. According to (1), we have $y_1 = -x + 2$, $\\therefore$ the coordinates of point C are $(2, 0)$, $\\therefore OC = 2$, $\\therefore$ the area of $\\triangle AOB$, $S_{\\triangle AOB} = \\frac{1}{2} \\times OC \\times (y_A - y_B) = \\frac{1}{2} \\times 2 \\times (4 + 2) = \\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52050093_182.png"
  },
  {
    "solution": "\\boxed{\\frac{24}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in rectangle \\(ABCD\\), \\(AB=6\\), \\(BC=8\\), \\(M\\) is any point on \\(AD\\), and \\(ME \\perp AC\\) at point \\(E\\), \\(MF \\perp BD\\) at point \\(F\\), then what is the value of \\(ME+MF\\)?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, take point O as the intersection of AC and BD, and connect OM,\\\\\n$S_{\\triangle AOD}=S_{\\triangle AOM}+S_{\\triangle MOD}= \\frac{1}{2}$OA$\\cdot$ME+ $\\frac{1}{2}$OD$\\cdot$MF= $\\frac{1}{2}+$OA(ME+MF),\\\\\n$\\because$ Quadrilateral ABCD is a rectangle,\\\\\n$\\therefore$ $\\angle$ABC$=90^\\circ$,\\\\\n$\\therefore$ AC= $\\sqrt{AB^{2}+BC^{2}}=10$,\\\\\n$\\therefore$ OA= $\\frac{1}{2}$AC=5,\\\\\n$S_{\\triangle AOD}= \\frac{1}{4}S_{\\text{rectangle}ABCD}=12$,\\\\\n$\\therefore$ ME+MF=12÷ $\\frac{1}{2}$OA= $\\boxed{\\frac{24}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51510388.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AC=4$, $BC=3$, and $\\angle C=90^\\circ$, then what is the value of $\\sin A$?\n",
    "completion": "\\textbf{Solution:} Since $AC=4$, $BC=3$, and $\\angle C=90^\\circ$, \\\\\ntherefore $AB=\\sqrt{AC^{2}+BC^{2}}=\\sqrt{3^{2}+4^{2}}=5$, \\\\\nhence $\\sin A=\\frac{BC}{AB}=\\boxed{\\frac{3}{5}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53065838.png"
  },
  {
    "solution": "\\boxed{(-3, \\frac{1}{2})}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system (where $O$ is the origin), it is known that the line $y=kx+b$ intersects the $x$ axis and the $y$ axis at points $A(-2,0)$ and $B(0,-1)$, respectively. The coordinates of point $C$ are $(0,2)$. Suppose point $D$ is a point on line $AB$ and $CD=BD$. Find the coordinates of point $D$.\n",
    "completion": "\\textbf{Solution:} Let us draw $DH \\perp BC$ from point $D$, with the perpendicular foot at $H$, as shown in the figure:\\\\\n$\\because CD=BD$,\\\\\n$\\therefore HC=HB=\\frac{1}{2}BC$,\\\\\n$\\because B(0, -1), C(0, 2) \\therefore BC=3, OC=2$\\\\\n$\\therefore CH=\\frac{3}{2}$,\\\\\n$\\therefore OH=\\frac{1}{2}$,\\\\\n$\\therefore$ substitute $y=\\frac{1}{2}$ into the line equation $y=-\\frac{1}{2}x-1$,\\\\\nwe get $\\frac{1}{2}=-\\frac{1}{2}x-1$, solving for $x$, we find $x=-3$\\\\\n$\\therefore$ The coordinates of point $D$ are $\\boxed{(-3, \\frac{1}{2})}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55498635_92.png"
  },
  {
    "solution": "\\boxed{rectangle}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the perimeter of square $ABCD$ is 40. Point $P$ is a moving point on the diagonal $AC$ of square $ABCD$. Perpendiculars are drawn from point $P$ to $AB$ and $BC$, with the feet of the perpendiculars being $E$ and $F$, respectively. Prove that the quadrilateral $PEBF$ is a rectangle.\n",
    "completion": "\\textbf{Solution:} Proof:\\\\\nSince $PE\\perp AB$ and $PF\\perp BC$,\\\\\nit follows that $\\angle PEB=\\angle PFB=90^\\circ$.\\\\\nFurthermore, since $ABCD$ is a square,\\\\\nit follows that $\\angle ABC=90^\\circ$.\\\\\nTherefore, the quadrilateral PEBF is a \\boxed{rectangle}.",
    "image_path": "dataset/mm_math/images/MM_Math/52422339_93.png"
  },
  {
    "solution": "\\boxed{155^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $P$ is a point on the bisector $OC$ of $\\angle MON$, with $P$ as the vertex, the sides of $\\angle APB$ intersect rays $OM$ and $ON$ at points $A$ and $B$, respectively. If $\\angle APB$ always satisfies $OA \\cdot OB = OP^2$ while rotating around point $P$, we call $\\angle APB$ the associated angle of $\\angle MON$. If $\\angle MON = 50^\\circ$ and $\\angle APB$ is the associated angle of $\\angle MON$, what is the degree measure of $\\angle APB$?\n",
    "completion": "\\textbf{Solution:} \\\\\nSince $OA \\cdot OB = OP^{2}$, \\\\\nit follows that $\\frac{OB}{OP} = \\frac{OP}{OA}$, \\\\\nSince $OP$ bisects $\\angle MON$, \\\\\nit follows that $\\angle AOP = \\angle BOP$, \\\\\nthus, $\\triangle AOP \\sim \\triangle POB$, \\\\\ntherefore, $\\angle OPA = \\angle OBP$, \\\\\nIn $\\triangle OBP$, $\\angle BOP = \\frac{1}{2}\\angle MON = 25^\\circ$, \\\\\nthus, $\\angle OBP + \\angle OPB = 180^\\circ - 25^\\circ = 155^\\circ$, \\\\\nhence, $\\angle OPA + \\angle OPB = 155^\\circ$, \\\\\nwhich means $\\angle APB = \\boxed{155^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51367394.png"
  },
  {
    "solution": "\\boxed{16\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rhombus $ABCD$, $\\angle A=60^\\circ$, points $E$ and $F$ are respectively on sides $BC$ and $CD$, and $CE=DF$, $BF$ intersects $DE$ at point $G$. If $BG=3$ and $DG=5$, what is the area of quadrilateral $ABGD$?\n",
    "completion": "Solution:Since quadrilateral $ABCD$ is a rhombus,\\\\\n$\\therefore AB=AD$,\\\\\nSince $\\angle BAD=60^\\circ$,\\\\\n$\\therefore \\triangle ABD$ is an equilateral triangle,\\\\\n$\\therefore \\angle BAD=\\angle ABD=60^\\circ$,\\\\\n$\\therefore \\angle DBC=\\angle BDF=\\angle C=60^\\circ$,\\\\\nIn $\\triangle CDE$ and $\\triangle DBF$,\\\\\n$\\left\\{\\begin{array}{l}CD=DB\\\\ \\angle C=\\angle BDF\\\\ CE=DF\\end{array}\\right.$,\\\\\n$\\therefore \\triangle CDE \\cong \\triangle DBF\\left(SAS\\right)$,\\\\\n$\\therefore \\angle CDE=\\angle DBF$,\\\\\n$\\therefore \\angle GBE=\\angle BDE$,\\\\\n$\\therefore \\angle DBF+\\angle GBE=\\angle DBF+\\angle BDE=\\angle BGE=\\angle DGF=60^\\circ=\\angle BAD$,\\\\\n$\\therefore$ Quadrilateral $ABGD$ is a cyclic quadrilateral,\\\\\n$\\therefore \\angle BGD=120^\\circ$,\\\\\nAs shown in the figure, draw $AM\\perp DE$ and $AN\\perp BF$ from point $A$, with the feet $M$ and $N$ respectively,\\\\\nSince $AG$ bisects $\\angle BGD$,\\\\\n$\\therefore AN=AM$,\\\\\nIn $Rt\\triangle ABN$ and $Rt\\triangle ADM$,\\\\\n$\\left\\{\\begin{array}{l}AB=AD\\\\ AN=AM\\end{array}\\right.$,\\\\\n$\\therefore Rt\\triangle ABN \\cong Rt\\triangle ADM\\left(HL\\right)$,\\\\\n$\\therefore BN=DM$,\\\\\n$\\therefore GN+GM=BG+DG=3+5=8$,\\\\\nConnect $AG$,\\\\\nIn $Rt\\triangle AGN$ and $Rt\\triangle AGM$,\\\\\n$\\left\\{\\begin{array}{l}AG=AG\\\\ AN=AM\\end{array}\\right.$,\\\\\n$\\therefore Rt\\triangle AGN \\cong Rt\\triangle AGM\\left(HL\\right)$,\\\\\n$\\therefore NG=MG=\\frac{1}{2}\\left(BG+DG\\right)=4$, $\\angle AGN=\\frac{1}{2}\\angle BGD=60^\\circ$,\\\\\n$\\therefore AN=NG\\cdot \\tan\\angle AGN=4\\sqrt{3}$,\\\\\n$\\therefore S_{\\text{quadrilateral }ABGD}=S_{\\text{quadrilateral }ANGM}$,\\\\\n$S_{\\text{quadrilateral }ABGD}=2S_{\\triangle AGN}=2\\times \\frac{1}{2}\\times NG\\times AN=4\\times 4\\sqrt{3}=\\boxed{16\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52513362.png"
  },
  {
    "solution": "\\boxed{70^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $OB$ is the angle bisector of $\\angle AOC$, and $OD$ is the angle bisector of $\\angle COE$. If $\\angle AOB=40^\\circ$ and $\\angle COE=60^\\circ$, what is the degree measure of $\\angle BOD$?\n",
    "completion": "\\textbf{Solution:} Given that OB is the angle bisector of $\\angle AOC$, and OD is the angle bisector of $\\angle COE$, with $\\angle AOB=40^\\circ$ and $\\angle COE=60^\\circ$,\\\\\nthus, $\\angle BOC=\\angle AOB=40^\\circ$ and $\\angle COD= \\frac{1}{2}\\angle COE=\\frac{1}{2}\\times 60^\\circ=30^\\circ$,\\\\\nhence, $\\angle BOD=\\angle BOC+\\angle COD=40^\\circ+30^\\circ=\\boxed{70^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52576677.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the side length of the regular hexagon inscribed in a circle is 6. What is the radius of the circle?\n",
    "completion": "\\textbf{Solution:} Since $\\angle AOB$ is the central angle of a regular hexagon,\\\\\nit follows that $\\angle AOB=\\frac{1}{6}\\times 360^\\circ=60^\\circ$,\\\\\nand since $OA=OB$,\\\\\nit follows that $\\triangle AOB$ is an equilateral triangle,\\\\\nhence $OA=AB=\\boxed{6}$,\\\\\nwhich means the radius of this regular hexagon is 6.",
    "image_path": "dataset/mm_math/images/MM_Math/53147460.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in square $ABCD$, where $AB=6$, $E$ is a point on side $BC$ (point $E$ does not coincide with points $B$, $C$), and point $F$ is on the extension of $CD$, and $BE=DF$. Connect $EF$, intersecting $AD$ and $AC$ at points $M$ and $N$, respectively. Given that $MD=1$, what is the length of $BE$?\n",
    "completion": "\\textbf{Solution:} Given that the quadrilateral $ABCD$ is a square, \\\\\n$\\therefore CD=BC=AB=6$, $AD\\parallel BC$, \\\\\n$\\therefore \\triangle FDM\\sim \\triangle FCE$, \\\\\n$\\therefore \\frac{DM}{CE}=\\frac{DF}{CF}$, \\\\\n$\\because BE=DF$, \\\\\n$\\therefore CE=BC-BE=6-BE$, $CF=CD+DF=6+BE$, \\\\\n$\\therefore \\frac{1}{6-BE}=\\frac{BE}{6+BE}$, \\\\\n$\\therefore BE=\\boxed{2}$ or $BE=\\boxed{3}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52977916_147.png"
  },
  {
    "solution": "\\boxed{11\\,cm}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $C$ is the midpoint of segment $BD$, $AD=3\\,\\text{cm}$, $AC=7\\,\\text{cm}$. What is the length of $AB$ in centimeters?\n",
    "completion": "\\textbf{Solution:} Given that $AD=3\\,cm$ and $AC=7\\,cm$,\n\nthus, $CD=AC-AD=4\\,cm$,\n\nsince point $C$ is the midpoint of line segment $BD$,\n\nit follows that $BC=CD=4\\,cm$,\n\ntherefore, $AB=AC+BC=7\\,cm+4\\,cm=\\boxed{11\\,cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53404099.png"
  },
  {
    "solution": "\\boxed{-3<x<-1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: The graph of the line $l_{1}\\colon y=k_{1}x+b$ and the line $l_{2}\\colon y=k_{2}x$ in the same Cartesian coordinate system are shown in the figure. The line $l_{1}\\colon y=k_{1}x+b$ intersects the x-axis at the point $(-3, 0)$. What is the solution set of the inequality in terms of x, $k_{2}x<k_{1}x+b<0$?\n",
    "completion": "\\textbf{Solution:} Given the graph, it is apparent that the intersection point of line $l_{1}$ and line $l_{2}$ is at ($-1$, $-2$), and within line $l_{1}$, $y$ decreases as $x$ increases.\\\\\nBecause $y=k_{1}x+b$ intersects the $x$-axis at point ($-3$, $0$), and the solution set of the inequality $k_{2}x<k_{1}x+b$ with respect to $x$ is $x<-1$,\\\\\nTherefore, the solution set for the inequality $k_{2}x<k_{1}x+b<0$ with respect to $x$ is $-3<x<-1$,\\\\\nThus, the answer is: $\\boxed{-3<x<-1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52055941.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points A, B, and C are all on the lattice points of a $4 \\times 4$ square grid. What is the value of $\\tan\\angle BAC$?\n",
    "completion": "Solution: As shown in the figure, take lattice point D and connect BD.\\\\\nFrom the lattice graph, it can be inferred that $BD\\perp AC$,\\\\\nUsing the properties of lattice triangles, we find: $AD=\\sqrt{3^2+3^2}=3\\sqrt{2}$, $BD=\\sqrt{1^2+1^2}=\\sqrt{2}$,\\\\\n$\\therefore$ $\\tan\\angle BAC=\\frac{BD}{AD}=\\frac{\\sqrt{2}}{3\\sqrt{2}}=\\boxed{\\frac{1}{3}}$",
    "image_path": "dataset/mm_math/images/MM_Math/52765236.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AD$ bisects $\\angle BAC$, $AB=4$, $AC=2$. If the area of $\\triangle ACD$ is equal to $3$, what is the area of $\\triangle ABD$?\n",
    "completion": "\\textbf{Solution:} Draw $DE \\perp AB$ at $E$ and $DF \\perp AC$ at $F$, as shown in the figure,\\\\\nsince $AD$ bisects $\\angle BAC$,\\\\\nthus $DE = DF$,\\\\\nsince the area of $\\triangle ACD$ = $\\frac{1}{2} \\cdot DF \\cdot AC = 3$,\\\\\nthus $DF = \\frac{2 \\times 3}{AC} = 3$,\\\\\nthus $DE = 3$.\\\\\nTherefore, the area of $\\triangle ABD$ = $\\frac{1}{2} \\cdot DE \\cdot AB = \\frac{1}{2} \\times 3 \\times 4 = \\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53102443.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{3}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $AD$, $BE$, $CF$ are the three altitudes of $\\triangle ABC$ (with $D$, $E$, $F$ being the feet of the perpendiculars), $\\angle ABC=45^\\circ$, $\\angle C=60^\\circ$, then what is the value of $\\frac{DE}{DF}$?\n",
    "completion": "\\textbf{Solution:} Given $\\because \\angle ABC=45^\\circ$,\\\\\n$\\therefore \\angle BAD=45^\\circ$, $\\angle BCF=45^\\circ$,\\\\\n$\\therefore \\triangle ABD$ and $\\triangle BCF$ are both isosceles right triangles,\\\\\n$\\because \\frac{BF}{BC}=\\frac{BD}{AB}=\\frac{\\sqrt{2}}{2}$, $\\angle ABC=\\angle ABC$,\\\\\n$\\therefore \\triangle BFD\\sim \\triangle BCA$,\\\\\n$\\therefore \\frac{DF}{AC}=\\frac{\\sqrt{2}}{2}$,\\\\\nSimilarly, it can be derived that $\\triangle CDE\\sim \\triangle CBA$,\\\\\n$\\therefore \\frac{DE}{AB}=\\frac{CD}{AC}=\\frac{1}{2}$,\\\\\nHence, $DF= \\frac{\\sqrt{2}}{2} AC$, $DE= \\frac{1}{2} AB$,\\\\\n$\\therefore \\frac{DE}{DF}=\\frac{\\frac{1}{2}AB}{\\frac{\\sqrt{2}}{2}AC}=\\frac{\\sqrt{2}AB}{2AC}$,\\\\\nLet $AD=a$, then $AB= \\sqrt{2} a$, $AC= \\frac{2a}{\\sqrt{3}}$,\\\\\n$\\therefore \\frac{DE}{DF}=\\frac{\\sqrt{2}AB}{2AC}=\\frac{\\sqrt{2}\\times \\sqrt{2}a}{2\\times \\frac{2a}{\\sqrt{3}}}=\\boxed{\\frac{\\sqrt{3}}{2}}.$",
    "image_path": "dataset/mm_math/images/MM_Math/51317346.png"
  },
  {
    "solution": "\\boxed{1+\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in parallelogram $ABCD$, point $O$ is on side $AD$, and a circle $\\odot O$ with $O$ as the center and $OA$ as the radius passes exactly through point $B$ and intersects side $AD$ at point $E$. The line containing side $CD$ is tangent to $\\odot O$ at point $H$. Lines $AH$ and $EH$ are connected. If $2\\angle HAB + \\angle C = 180^\\circ$, $CB$ is tangent to $\\odot O$; if $DE = 1$, what is the radius of $\\odot O$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral $ABCD$ is a parallelogram, \\\\\nit follows that $AD\\parallel BC$ and $AB\\parallel CH$. \\\\\nSince $\\angle OBC=90^\\circ$, \\\\\nit follows that $\\angle BOA=90^\\circ$, \\\\\nthus, $\\triangle AOB$ is an isosceles right triangle, \\\\\ntherefore, $\\angle BAD=\\angle ABO=45^\\circ$, \\\\\nthus, $\\angle ODH=\\angle BAD=45^\\circ$. \\\\\nSince $\\angle CHO=90^\\circ$, which means $\\angle DHO=90^\\circ$, \\\\\ntherefore, $\\triangle DHO$ is an isosceles right triangle, \\\\\ntherefore, $OH=DH=OE$. \\\\\nLet $OH=x$, then $OE=DH=x$, $OD=OE+DE=x+1$. \\\\\nSince in $\\triangle DHO$, $OH^2+DH^2=OD^2$, which means $x^2+x^2=(x+1)^2$ \\\\\nSolving gives: $x_1=1+\\sqrt{2}$, $x_2=1-\\sqrt{2}$ (discard) \\\\\nTherefore, the radius of $\\odot O$ is $\\boxed{1+\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51318777_125.png"
  },
  {
    "solution": "\\boxed{\\frac{9\\sqrt{10}}{50}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the vertices of $\\triangle ABC$ are all on the grid points of the graph paper. What is the value of $\\sin C$?\n",
    "completion": "Solution: Let the length of each small square in the grid be 1,\\\\\nthus CD=3, and AD=4,\\\\\n$\\therefore$ in $\\mathrm{Rt}\\triangle ACD$, we have $AC=\\sqrt{AD^{2}+CD^{2}}=\\sqrt{4^{2}+3^{2}}=5$\\\\\nIn $\\mathrm{Rt}\\triangle BCD$, $CB=\\sqrt{BD^{2}+CD^{2}}=\\sqrt{1^{2}+3^{2}}=\\sqrt{10}$\\\\\nAs shown in the figure below, draw a perpendicular from point C to AB, and let it intersect the extension of AB at point D. Let AF=4, connect BF, which intersects AC at E,\\\\\nIn $\\triangle FAB$ and $\\triangle ADC$,\\\\\n$\\because$ AF=DA=4, $\\angle FAB=\\angle ADC=90^\\circ$, AB=DC=3,\\\\\n$\\therefore \\triangle FAB \\cong \\triangle ADC$ (SAS),\\\\\n$\\therefore \\angle F=\\angle CAD$, FB=AC=5\\\\\n$\\because \\angle FBA=\\angle ABE$,\\\\\n$\\therefore \\triangle ABE \\sim \\triangle FBA$,\\\\\n$\\therefore \\angle AEB=\\angle FAB=90^\\circ$,\\\\\n$\\therefore \\frac{AB}{EB}=\\frac{FB}{AB}$, that is $\\frac{3}{EB}=\\frac{5}{3}$\\\\\n$\\therefore EB=\\frac{9}{5}$\\\\\n$\\therefore \\sin C= \\frac{BE}{BC}=\\frac{\\frac{9}{5}}{\\sqrt{10}}=\\boxed{\\frac{9\\sqrt{10}}{50}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51367390.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, points $D$, $E$, $F$ are on sides $AB$, $AC$, $BC$ respectively, and $DE \\parallel BC$, $EF \\parallel AB$. If $AD = 2BD$, then what is the value of $\\frac{CF}{BF}$?\n",
    "completion": "\\textbf{Solution:} Since $DE\\parallel BC$, $EF\\parallel AB$, and $AD=2BD$,\n\\[\n\\therefore \\frac{AD}{BD}=\\frac{AE}{EC}=2, \\quad \\frac{AE}{EC}=\\frac{BF}{CF}=2,\n\\]\n\\[\n\\therefore \\frac{CF}{BF}=\\boxed{\\frac{1}{2}}.\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/51445440.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\angle AOD = 130^\\circ$, $\\angle BOC : \\angle COD = 1 : 2$, $\\angle AOB$ is $\\frac{1}{3}$ of the complement of $\\angle COD$. How many degrees is $\\angle COD$?\n",
    "completion": "\\textbf{Solution:} We have $\\angle COD = \\boxed{60^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51324353_154.png"
  },
  {
    "solution": "\\boxed{N(\\frac{20}{9}, -\\frac{8}{3})}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, the quadrilateral OABC is a rectangle, with points A and C located on the x-axis and y-axis, respectively. The triangle $ \\triangle ODE$ is obtained by rotating $ \\triangle OCB$ clockwise around point O by $90^\\circ$. Point D is on the x-axis, and the line BD intersects the y-axis at point F and intersects OE at point H. The coordinate of point B is $(-2, 4)$. Point M is on the x-axis. Is there a point N in the plane such that the quadrilateral formed by points D, F, M, N is a rectangle? If yes, please directly write down the coordinates of point N; if not, please explain why.\n",
    "completion": "\\textbf{Solution:} The coordinates of point N are $\\boxed{N(4, \\frac{8}{3})}$ or $\\boxed{N(\\frac{20}{9}, -\\frac{8}{3})}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52251770_76.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $C$ is the midpoint of segment $AB$, and $D$ is on segment $CB$, with $DA=6, DB=4$. Find the length of $CD$.\n",
    "completion": "\\textbf{Solution:} Since point C is the midpoint of the line segment $AB$,\\\\\n$\\therefore AC=\\frac{1}{2}AB=5$.\\\\\n$\\therefore CD=AD-AC=6-5=\\boxed{1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55000668_19.png"
  },
  {
    "solution": "\\boxed{2\\colon 5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the quadrilateral $ABCD$ is a parallelogram, point $E$ is on side $CD$, with $DE=2CE$, connect $AE$ and intersect $BD$ at point $F$. What is the ratio of $DF:BD$?\n",
    "completion": "\\textbf{Solution}: Since $DE=2CE$,\\\\\nit follows that $\\frac{DE}{DC}=\\frac{2}{3}$,\\\\\nsince quadrilateral $ABCD$ is a parallelogram,\\\\\nit follows that $AB=CD$, $AB\\parallel CD$,\\\\\ntherefore, $\\triangle DFE\\sim \\triangle BFA$,\\\\\nthus, $\\frac{DF}{BF}=\\frac{DE}{AB}=\\frac{DE}{CD}=\\frac{2}{3}$,\\\\\nhence, $DF\\colon BD=\\boxed{2\\colon 5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53066091.png"
  },
  {
    "solution": "\\boxed{\\frac{5}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, points $D$ and $E$ are located on sides $AB$ and $AC$ respectively, and $DE \\parallel BC$. If $AD = 2$, $DB = 3$, and $DE = 1$, what is the length of $BC$?\n",
    "completion": "\\textbf{Solution:} Given $\\because$ DE $\\parallel$ BC, \\\\\n$\\therefore$ $\\triangle ADE \\sim \\triangle ABC$, \\\\\n$\\therefore$ $\\frac{AD}{AB} = \\frac{DE}{BC}$, \\\\\n$\\because$ AD = 2, DB = 3, \\\\\n$\\therefore$ AB = 5, \\\\\n$\\therefore$ $\\frac{2}{5} = \\frac{1}{BC}$, solving this we get: $BC = \\boxed{\\frac{5}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53078795.png"
  },
  {
    "solution": "\\boxed{40^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AD\\perp BC$, $AE$ bisects $\\angle BAC$, $\\angle B=70^\\circ$, and $\\angle C=30^\\circ$. What is the measure of $\\angle BAE$?\n",
    "completion": "\\textbf{Solution:} Therefore, $\\angle BAE = \\boxed{40^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52905934_141.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, D and E are the midpoints on the sides AB and AC of $\\triangle ABC$, respectively, and F is a point on the segment DE. If $AB=10$, $BC=12$, and $\\angle AFB=90^\\circ$, then what is the length of segment EF?\n",
    "completion": "\\textbf{Solution:} Since $\\angle AFB=90^\\circ$ and D is the midpoint of AB,\\\\\nit follows that DF=$\\frac{1}{2}$AB=5,\\\\\nSince D and E are the midpoints of AB and AC respectively,\\\\\nDE is the midline of $\\triangle ABC$,\\\\\nthus DE=$\\frac{1}{2}$BC=6,\\\\\ntherefore EF=DE−DF=6−5=\\boxed{1}.",
    "image_path": "dataset/mm_math/images/MM_Math/52076714.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in rhombus $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$, with $AC=16$ and $BD=12$. Let $E$ be a point on side $AD$, and line $OE$ intersects $BC$ at point $F$. Fold the rhombus along line $EF$, making the corresponding point of $B$ as $B'$ and the corresponding point of $A$ as $A'$. If $AE=4$, what is the length of $B'F$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral $ABCD$ is a rhombus, $AC=16$, and $BD=12$,\\\\\nit follows that $OA=OC=\\frac{1}{2}AC=8$, $OB=\\frac{1}{2}BD=6$, $AC\\perp BD$, $AD\\parallel BC$,\\\\\nhence $BC=\\sqrt{OB^2+OC^2}=10$, $\\angle OCF=\\angle OAE$, $\\angle OFC=\\angle OEA$,\\\\\nin $\\triangle COF$ and $\\triangle AOE$, we have $\\left\\{\\begin{array}{l}\\angle OFC=\\angle OEA \\\\ \\angle OCF=\\angle OAE \\\\ OC=OA\\end{array}\\right.$,\\\\\ntherefore, $\\triangle COF\\cong \\triangle AOE$ (AAS),\\\\\nthus, $CF=AE=4$,\\\\\nhence $BF=BC-CF=10-4=6$,\\\\\nby the property of folding, we get: ${B'}^F=BF=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50482846.png"
  },
  {
    "solution": "\\boxed{y=\\frac{1}{4}x^{2}+x-2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the Cartesian coordinate system $xOy$, points $A$ and $B$ lie on the line $y=\\frac{1}{2}x$, as shown in the figure. The graph of the quadratic function $y=ax^2+bx-2$ also passes through points $A$ and $B$, and intersects the $y$-axis at point $C$. If $BC$ is parallel to the $x$-axis, and the $x$-coordinate of point $A$ is $2$, what is the analytical expression of this quadratic function?\n",
    "completion": "\\textbf{Solution:} Given that the quadratic function $y=\\frac{1}{4}x^{2}+bx-2$ intersects the $y$-axis at point $C$,\n\nthus the coordinates of point $C$ are $(0, -2)$,\n\nsince $BC\\parallel x$-axis,\n\nthus the ordinate of point $B$ is $-2$,\n\nsince points $A$ and $B$ lie on the line $y=\\frac{1}{2}x$ and the abscissa of point $A$ is $2$,\n\nthus the coordinates of point $A$ are $(2, 1)$, and those of point $B$ are $(-4, -2)$,\n\nsince the graph of the quadratic function also passes through points $A(2, 1)$ and $B(-4, -2)$, we get\n\n\\[\n\\left\\{\n\\begin{array}{l}\n4a+2b-2=1\\\\\n16a-4b-2=-2\n\\end{array}\n\\right.,\n\\]\n\nsolving this system of equations, we find $a=\\frac{1}{4}$, $b=1$,\n\ntherefore, the explicit formula of the quadratic function is $\\boxed{y=\\frac{1}{4}x^{2}+x-2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51379457_207.png"
  },
  {
    "solution": "\\boxed{\\frac{10\\sqrt{3}}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the vertices A and C of $\\triangle ABC$ are on circle O, and circle O intersects side AB at point D. Through point D, draw $DF \\parallel BC$, intersecting AC at point E and circle O at point F. Join DC, with point C being the midpoint of arc DF. Given these conditions, connect AF. If $BD = AF$, find the length of AD.\n",
    "completion": "\\textbf{Solution:} Let's extend CO and intersect $\\odot O$ at G, and let DF and CG intersect at point M, then connect DO, AF, CF,\\\\\naccording to the conclusion of (2), we have $OM\\perp DF$, $MC=2$, $DM=FM=2\\sqrt{2}$\\\\\n$\\therefore CD=CF=\\sqrt{MC^{2}+MF^{2}}=2\\sqrt{3}$\\\\\n$\\because$DF$\\parallel$BC,\\\\\n$\\therefore \\angle ADF=\\angle B$, $\\angle BCD=\\angle CDF$\\\\\n$\\because \\angle ADF=\\angle ACF$, $\\angle CAF=\\angle CDF$\\\\\n$\\therefore \\angle B=\\angle ACF$, $\\angle BCD=\\angle CAF$\\\\\n$\\therefore \\triangle BCD\\sim \\triangle CAF$\\\\\n$\\therefore \\frac{BD}{CF}=\\frac{CD}{AF}$\\\\\n$\\therefore \\frac{BD}{2\\sqrt{3}}=\\frac{2\\sqrt{3}}{AF}$\\\\\n$\\because BD=AF$\\\\\n$\\therefore BD=AF=2\\sqrt{3}$ or $BD=AF=-2\\sqrt{3}$ (discard)\\\\\n$\\therefore BD=AF=CF=CD=2\\sqrt{3}$\\\\\n$\\because$ point C is the midpoint of the arc DF.\\\\\n$\\therefore \\angle CFD=\\angle CAF$\\\\\n$\\because \\angle ECF=\\angle FCA$\\\\\n$\\therefore \\triangle ECF\\sim \\triangle FCA$\\\\\n$\\therefore \\frac{EF}{AF}=\\frac{CE}{CF}$\\\\\n$\\because AF=CF=2\\sqrt{3}$\\\\\n$\\therefore EF=CE$\\\\\nLet $CE=x$\\\\\n$\\therefore ME=FM-EF=FM-CE=2\\sqrt{2}-x$\\\\\n$\\because CE^{2}=MC^{2}+ME^{2}$\\\\\n$\\therefore x^{2}=4+{(2\\sqrt{2}-x)}^{2}$\\\\\n$\\therefore x=\\frac{3\\sqrt{2}}{2}$\\\\\n$\\therefore CA=\\frac{12}{CE}=4\\sqrt{2}$\\\\\n$\\therefore AE=CA-CE=\\frac{5}{2}\\sqrt{2}$\\\\\n$\\because \\angle ADF=\\angle B$, $\\angle DAE=\\angle BAC$\\\\\n$\\therefore \\triangle ADE\\sim \\triangle ABC$\\\\\n$\\therefore \\frac{AD}{AB}=\\frac{AE}{AC}=\\frac{5}{8}$\\\\\n$\\therefore \\frac{AD}{AD+BD}=\\frac{AD}{AD+2\\sqrt{3}}=\\frac{5}{8}$\\\\\n$\\therefore AD=\\boxed{\\frac{10\\sqrt{3}}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51367439_118.png"
  },
  {
    "solution": "\\boxed{x<-3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the intersection of the lines $y = -x + m$ and $y = nx + 4n$ ($n \\neq 0$) has an x-coordinate of $-3$. What is the solution set for the inequality $-x + m > nx + 4n$ with respect to $x$?\n",
    "completion": "\\textbf{Solution}: When $x<-3$, we have $-x+m>nx+4n$, \\\\\n$\\therefore$ the solution set for the inequality $-x+m>nx+4n$ with respect to $x$ is $\\boxed{x<-3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52386400.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, there is a lamppost $AB$ beside a straight road. Under the light, Xiao Li progresses in a straight line from the bottom of the lamppost $B$ for $4m$ to reach point $D$. At point $D$, her shadow $DE$ is measured to be $1m$. The height of Xiao Li, $CD$, is $1.2m$. If Xiao Li continues to move in a straight line from point $D$ for another $4m$ to reach point $G$ (as shown in Figure 2), what is the length of her shadow $GH$ at that point?\n",
    "completion": "\\textbf{Solution:} As shown in Figure 2, according to the given conditions: \\(AB\\parallel FG\\), \\(BE=5\\) (meters), \\\\\n\\(\\therefore \\triangle HGF \\sim \\triangle HBA\\), \\\\\n\\(\\therefore \\frac{AB}{FG}=\\frac{BH}{GH}\\), \\\\\nthat is \\(\\frac{6}{1.2}=\\frac{8+GH}{GH}\\), \\\\\nsolving this we get: \\(GH=\\boxed{2}\\) (meters).",
    "image_path": "dataset/mm_math/images/MM_Math/52550926_100.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $C$ is the midpoint of the semicircle, $AB$ is the diameter, and point $D$ is a point on the semicircle. $AC$ and $BD$ intersect at point $E$. If $AD=2$ and $BD=6$, what is the length of $AC$? What is the length of $CD$?\n",
    "completion": "\\textbf{Solution:} Since $AB$ is the diameter, it follows that $\\angle ADB = \\angle ACB = 90^\\circ$. Within $\\triangle ADB$, given AD = 2 and BD = 6, it follows $AB = \\sqrt{2^2 + 6^2} = 2\\sqrt{10}$. Since point $C$ is the midpoint of the semicircle, it follows that $\\angle CAB = 45^\\circ$. Therefore, $AB = \\sqrt{2}AC = \\sqrt{2}BC$, implying $AC = BC = 2\\sqrt{5}$. Also, since $\\angle DEA = \\angle CEB$ and $\\angle DAE = \\angle CBE$, it follows that $\\triangle BEC \\sim \\triangle AED$. Hence, $CE:DE = BE:AE = BC:AD = 2\\sqrt{5}:2$, which leads to $CE = \\sqrt{5}DE$ and $BE = \\sqrt{5}AE$. Since $BE = BD - DE$ and $AE = AC - CE$, we have $6 - DE = \\sqrt{5}(2\\sqrt{5} - CE) = \\sqrt{5}(2\\sqrt{5} - \\sqrt{5}DE)$. Solving this, we find $DE = 1$, thus $CE = \\sqrt{5}$ and $AE = 2\\sqrt{5} - \\sqrt{5} = \\sqrt{5}$. Since $\\angle DCA = \\angle DBA$ and $\\angle CED = \\angle BEA$, it follows that $\\triangle DEC \\sim \\triangle AEB$. Consequently, $DE:EA = CD:AB$, leading to $CD = \\frac{DE \\cdot AB}{EA} = \\frac{1 \\times 2\\sqrt{10}}{\\sqrt{5}} = 2\\sqrt{2}$. Therefore, the solutions are: $AC = \\boxed{2\\sqrt{5}}$ and $CD = \\boxed{2\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52808104.png"
  },
  {
    "solution": "\\boxed{25^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $O$ is a point on the line $AB$, $CO \\perp DO$, $OE$ bisects $\\angle BOC$. If $\\angle AOC=50^\\circ$, what is the degree measure of $\\angle DOE$?\n",
    "completion": "\\textbf{Solution:} Because $\\angle AOB$ is a straight angle,\\\\\ntherefore $\\angle BOC=180^\\circ-\\angle AOC=180^\\circ-50^\\circ=130^\\circ$,\\\\\ntherefore $\\angle EOC=\\frac{1}{2}\\angle BOC=\\frac{1}{2}\\times 130^\\circ=65^\\circ$,\\\\\nbecause $CO\\perp DO$,\\\\\ntherefore $\\angle DCC=90^\\circ$,\\\\\ntherefore $\\angle DOE=\\angle DOC=\\angle EOC=90^\\circ-65^\\circ=\\boxed{25^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51438609_161.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points D, B, and E are three points on line segment AC, with D being the midpoint of line segment AB. If $AC = 2DE = 20$ and $AD: EC = 3:2$, find the length of line segment EC.\n",
    "completion": "Solution: Since point E is the midpoint of BC, and D is the midpoint of segment AB,\\\\\nit follows that $BD=AD=\\frac{1}{2}AB$, and $BE=EC=\\frac{1}{2}BC$,\\\\\nSince $AC=AD+DC=AD+BD+BE+EC=2(AD+EC)=20$,\\\\\nit follows that $AD+EC=10$,\\\\\nSince $AD\\colon EC=3\\colon 2$,\\\\\nit follows that $EC=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51179538_123.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, the perpendicular bisectors of sides $AB$ and $AC$ respectively intersect $BC$ at $D$ and $E$. If $BC=5$, what is the perimeter of $\\triangle ADE$?\n",
    "completion": "\\textbf{Solution:}\n\nGiven that the perpendicular bisectors of sides AB and AC intersect BC at D and E, respectively,\n\n$\\therefore DA=DB$, $EA=EC$,\n\n$\\therefore$ the perimeter of $\\triangle ADE = AD + DE + AE = DB + DE + EC = BC = \\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52733754_115.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, within $\\triangle ABC$, $\\triangle BDE$ and $\\triangle FGH$ are two congruent equilateral triangles, and $BC=5$. What is the perimeter of the pentagon $DECHF$?\n",
    "completion": "\\textbf{Solution:} Given that $\\triangle GFH$ is an equilateral triangle,\\\\\nthus $FH=GH$, $\\angle FHG=60^\\circ$,\\\\\ntherefore, $\\angle AHF+\\angle GHC=120^\\circ$,\\\\\ngiven that $\\triangle ABC$ is an equilateral triangle,\\\\\nthus $AB=BC=AC=5$, $\\angle ACB=\\angle A=60^\\circ$,\\\\\ngiven that $\\angle AHF=180^\\circ-\\angle FHG-\\angle GHC =120^\\circ-\\angle GHC$,\\\\\n$\\angle HGC=180^\\circ-\\angle C-\\angle GHC =120^\\circ-\\angle GHC$,\\\\\ntherefore, $\\angle AHF=\\angle HGC$,\\\\\nin $\\triangle AFH$ and $\\triangle CHG$ we have\\\\\n$\\left\\{\\begin{array}{l}\n\\angle A=\\angle C \\\\\n\\angle AHF=\\angle HGC \\\\\nFH=GH\n\\end{array}\\right.$,\\\\\ntherefore, $\\triangle AFH\\cong \\triangle CHG$ by AAS,\\\\\nthus, AF=CH.\\\\\nGiven that $\\triangle BDE$ and $\\triangle FGH$ are two congruent equilateral triangles,\\\\\nthus, BE=FH,\\\\\ntherefore, the perimeter of pentagon DECHF=DE+CE+CH+FH+DF=BD+CE+AF+BE+DF,\\\\\n=(BD+DF+AF)+(CE+BE),\\\\\n=AB+BC=\\boxed{10}.",
    "image_path": "dataset/mm_math/images/MM_Math/51366350.png"
  },
  {
    "solution": "\\boxed{\\frac{12}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given: As shown in the figure, in $\\triangle ABC$, $AD$ is the altitude on the side $BC$, $E$ is the midpoint of side $AC$, $BC=14$, $AD=12$, $\\sin B=\\frac{4}{5}$. Find the value of $\\tan\\angle EDC$.\n",
    "completion": "\\textbf{Solution:} In $\\triangle ADC$, E is the midpoint of AC, \\\\\n$\\therefore DE=EC$, \\\\\n$\\therefore \\angle EDC=\\angle C$, \\\\\n$\\therefore \\tan\\angle EDC=\\tan\\angle C=\\frac{AD}{CD}=\\boxed{\\frac{12}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53043832_89.png"
  },
  {
    "solution": "\\boxed{90^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $O$ is a point on line $AC$, $OB$ is a ray, $OD$ bisects $\\angle AOB$, $OE$ is within $\\angle BOC$, $\\angle BOE = \\frac{1}{3}\\angle EOC$, $\\angle DOE = 60^\\circ$, then what is the degree measure of $\\angle EOC$?\n",
    "completion": "\\textbf{Solution:} Let $\\angle BOE$ be $x^\\circ$, then $\\angle DOB = 60^\\circ - x^\\circ$,\\\\\nsince $\\angle BOE = \\frac{1}{3}\\angle EOC$,\\\\\nit follows that $\\angle EOC = 3\\angle BOE = 3x^\\circ$,\\\\\ngiven that OD bisects $\\angle AOB$,\\\\\nwe get $\\angle AOB = 2\\angle DOB$,\\\\\nthus, we have $3x + x + 2(60 - x) = 180$,\\\\\nsolving the equation, we find $x = 30^\\circ$,\\\\\ntherefore, $\\angle EOC = \\boxed{90^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53207045.png"
  },
  {
    "solution": "\\boxed{y=\\frac{25x-5x^2}{x^2-5x+16}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the rectangle $ABCD$, where $AB=4$ and $BC=5$. Let $P$ be a moving point on ray $BC$. Construct $PE\\perp AP$ where $PE$ intersects ray $DC$ at point E, and ray $AE$ intersects ray $BC$ at point F. Let $BP=x$ and $CF=y$. As illustrated, when point P is on side $BC$ (excluding points B and C), find the function relationship between $y$ and $x$, and state its domain.\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral $ABCD$ is a rectangle, \\\\\n$\\therefore EC\\parallel AB$, \\\\\n$\\therefore \\triangle ECF \\sim \\triangle ABF$, \\\\\n$\\therefore \\frac{CF}{BF}=\\frac{CE}{AB}$, \\\\\nthat is, $\\frac{y}{5+y}=\\frac{CE}{4}$, \\\\\nsolving this, we get: $CE=\\frac{4y}{5+y}$, \\\\\n$\\because \\triangle PCE \\sim \\triangle ABP$, \\\\\n$\\therefore \\frac{CE}{x}=\\frac{5-x}{4}$, \\\\\nsolving this, we obtain: $CE=\\frac{x(5-x)}{4}$, \\\\\n$\\therefore \\frac{4y}{5+y}=\\frac{x(5-x)}{4}$, \\\\\nwhich simplifies to: $y=\\frac{25x-5x^2}{x^2-5x+16}$, for $0<x<5$. \\\\\nThus, the functional relationship of $y$ with respect to $x$ is $\\boxed{y=\\frac{25x-5x^2}{x^2-5x+16}}$, with the domain being $(0,5)$.",
    "image_path": "dataset/mm_math/images/MM_Math/52917616_87.png"
  },
  {
    "solution": "\\boxed{\\frac{2}{3}a}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: The given line segment $AB=a$ (as shown in the figure), where $C$ is a point on the extension of $AB$ in the opposite direction, and $AC=\\frac{1}{3}AB$, $D$ is the midpoint of line segment $BC$. What is the length of $CD$ expressed as an algebraic expression involving $a$?\n",
    "completion": "\\textbf{Solution:} We have $CD=\\boxed{\\frac{2}{3}a}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51495344_110.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, C is a point on line segment AB, and M, N are the midpoints of AC and BC, respectively. If AC = 8 cm and BC = 6 cm, what is the length of line segment MN?\n",
    "completion": "\\textbf{Solution}: Since $M$ and $N$ are the midpoints of $AC$ and $BC$ respectively,\\\\\n$\\therefore CM=\\frac{1}{2}AC=4\\text{cm}$, $CN=\\frac{1}{2}BC=6\\text{cm}$,\\\\\n$\\therefore MN=CM+CN=4+6=\\boxed{10}\\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53274798_98.png"
  },
  {
    "solution": "\\boxed{x<-1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=-x$ intersects with the line $y=x+b$ at point A. If the ordinate of point A is 1, what is the solution set of the inequality $x+b<-x$?\n",
    "completion": "\\textbf{Solution:} Given that the line $y=-x$ intersects with the line $y=x+b$ at point A, and the ordinate of point A is 1,\\\\\n$\\therefore$ $-x=1$,\\\\\nSolving this, we get $x=-1$,\\\\\nwhich means the abscissa of point A is $-1$,\\\\\n$\\therefore$ the solution set of the inequality $x+b<-x$ is $\\boxed{x<-1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52424160.png"
  },
  {
    "solution": "\\boxed{50^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within circle $\\odot O$, $OC \\perp AB$. If $\\angle BOC=40^\\circ$, what is the value of $\\angle OAB$?\n",
    "completion": "\\textbf{Solution:} In circle $\\odot O$, $OA=OB$, \\\\\n$\\therefore \\triangle AOB$ is an isosceles triangle,\\\\\n$\\because OC\\perp AB$,\\\\\n$\\therefore \\angle AOC=\\angle BOC=40^\\circ$,\\\\\n$\\therefore \\angle AOB=80^\\circ$,\\\\\n$\\therefore \\angle OAB=(180^\\circ-\\angle AOB)\\div 2=50^\\circ$.\\\\\nThus, $\\angle OAB=\\boxed{50^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51402310.png"
  },
  {
    "solution": "\\boxed{\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ is an equilateral triangle with side length 2, and the area of $\\triangle ABC$ is $\\sqrt{3}$. $D$ and $E$ are the midpoints of $BC$ and $AC$, respectively; $P$ is a moving point on $AD$. What is the minimum value of $PE+PC$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $BE$ intersecting $AD$ at point $P'$,\\\\\nsince $\\triangle ABC$ is an equilateral triangle, with $AB=2$, and $AD$ being the altitude on side $BC$, and E being the midpoint of $AC$,\\\\\ntherefore, $AD$ and $BE$ are the perpendicular bisectors of sides $BC$ and $AC$ of the equilateral triangle $ABC$ respectively,\\\\\ntherefore, $P'B=P'C$,\\\\\ntherefore, $P'E+P'C=P'E+P'B=BE$,\\\\\nbased on the principle that the line segment between two points is the shortest,\\\\\nwhen point P is at $P'$, $PE+PC$ has its minimum value, which is exactly the length of $BE$.\\\\\nsince the area of $\\triangle ABC$ is equal to $\\sqrt{3}$, and the side length is 2,\\\\\ntherefore, $\\frac{1}{2}\\times 2\\times BE=\\sqrt{3}$,\\\\\ntherefore, $BE=\\sqrt{3}$,\\\\\nthus, the minimum value of $PE+PC$ is $\\boxed{\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52905597.png"
  },
  {
    "solution": "\\boxed{\\frac{4}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated in the diagram, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, $CD\\perp AB$ with the foot of the perpendicular being D, $AC=4$, and $BC=3$. Find the tangent value of $\\angle ACD$.\n",
    "completion": "Solution: Since $\\angle ACD + \\angle BCD = 90^\\circ$ and $\\angle BCD + \\angle B = 90^\\circ$,\\\\\nit follows that $\\angle ACD = \\angle B$,\\\\\nthus $\\tan \\angle ACD = \\tan \\angle B = \\frac{AC}{BC} = \\boxed{\\frac{4}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53148902_120.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\triangle ABC$, $ \\angle ACB=90^\\circ$, $BC=4$, and $CD$ is the median. Extend $CB$ to point $F$ so that $BF=BC$, and connect $DF$. Let $E$ be the midpoint of $DF$, and connect $BE$. If $BE= 2$, what is the length of $AC$?\n",
    "completion": "\\textbf{Solution:} Given $\\because BF=BC$, and E is the midpoint of DF,\\\\\n$\\therefore BE=\\frac{1}{2}CD$,\\\\\n$\\because BE=2$,\\\\\n$\\therefore CD=4$,\\\\\n$\\because$ In $\\triangle ABC$ with $\\angle ACB=90^\\circ$, BC=4, and CD being the median,\\\\\n$\\therefore AB=2CD=8$, $AC=\\sqrt{AB^{2}-BC^{2}}=\\sqrt{8^{2}-4^{2}}=\\boxed{4\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53359597.png"
  },
  {
    "solution": "\\boxed{y={x}^{2}+3x-4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=x^2+bx+c$ passes through point $A(-4,0)$ and point $C(0,-4)$, with another intersection point with the x-axis being B. What is the analytical expression of this parabola?\n",
    "completion": "\\textbf{Solution}: Since the parabola $y={x}^{2}+bx+c$ passes through points $A\\left(-4,0\\right)$ and $C\\left(0,-4\\right)$,\\\\\nthus $\\begin{cases}16-4b+c=0\\\\ c=-4\\end{cases}$,\\\\\ntherefore $\\begin{cases}b=3\\\\ c=-4\\end{cases}$,\\\\\nhence the equation of the parabola is $\\boxed{y={x}^{2}+3x-4}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52871856_289.png"
  },
  {
    "solution": "\\boxed{x>5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: The graph of the linear functions $y_1=4x+5$ and $y_2=3x+10$ is shown in the figure. What is the solution set of $4x+5>3x+10$?\n",
    "completion": "\\textbf{Solution:} Solve: The graphs of the linear functions $y_1=4x+5$ and $y_2=3x+10$ intersect at point P(5, 25).\\\\\nTherefore, the solution set for $4x+5>3x+10$ is: $\\boxed{x>5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50677214.png"
  },
  {
    "solution": "\\boxed{3\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in rhombus $ABCD$, $ \\angle A=60^\\circ$. Points $E$ and $F$ are located on sides $AB$ and $AD$ respectively, and satisfy $ \\angle BCE=\\angle DCF$, $EF$ is connected. $ \\angle CFE=\\angle CEF$. If $AF=2\\sqrt{3}$ cm, what is the area of $\\triangle AEF$?\n",
    "completion": "\\textbf{Solution:} Construct $FM \\perp AB$ at $M$, \\\\\nsince $\\triangle CDF \\cong \\triangle CBE$, \\\\\nhence $FD=BE$, \\\\\nsince quadrilateral $ABCD$ is a rhombus, \\\\\nhence $AB=AD$, \\\\\nthus $AB-BE=AD-FD$, \\\\\ntherefore $AF=AE$, \\\\\nalso, since $\\angle A=60^\\circ$, \\\\\nhence $\\triangle AEF$ is an equilateral triangle, \\\\\nthus $AE=AF=2\\sqrt{3}$, \\\\\nhence $AM=\\sqrt{3}$, \\\\\ntherefore $FM=\\sqrt{AF^2-AM^2}=3$, \\\\\nhence the area of $\\triangle AEF$ = $\\frac{1}{2} \\times AE \\times FM=\\frac{1}{2} \\times 3 \\times 2\\sqrt{3}=\\boxed{3\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52205241_101.png"
  },
  {
    "solution": "\\boxed{y=-2x+8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y=kx+b$ ($k\\neq 0$) intersects with the graph of the inverse proportion function $y=\\frac{6}{x}$ ($x>0$) at points $A(m, 6)$ and $B(3, n)$. Find the explicit expression of the linear function.\n",
    "completion": "\\textbf{Solution:} Given points A(1, 6) and B(3, 2) are on the graph of the inverse proportional function $y=\\frac{6}{x}$ ($x>0$),\\\\\n$\\therefore$ 6m=3n=6,\\\\\n$\\therefore$ m=1, n=2,\\\\\nAlso, since points A(1, 6) and B(3, 2) are on the graph of the linear function $y=kx+b$ ($k\\neq 0$),\\\\\n$\\therefore$ $\\left\\{\\begin{array}{l}6=k+b \\\\ 2=3k+b \\end{array}\\right.$,\\\\\nSolving these, we get $\\left\\{\\begin{array}{l}k=-2 \\\\ b=8 \\end{array}\\right.$,\\\\\nThus, the equation of the linear function is: $\\boxed{y=-2x+8}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53043725_110.png"
  },
  {
    "solution": "\\boxed{\\left( \\frac{8}{3}, 3 \\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, the vertices A and C of the rectangle $OABC$ are located on the positive semi-axes of the x-axis and y-axis, respectively, and point B (4, 3). The graph of the inverse proportion function $y=\\frac{k}{x}$ ($x>0$) intersects with $AB$ and $BC$ at points $D$ and $E$, respectively, with $BD=1$. Let point $P$ be a moving point on the line segment $OA$. Determine the equation of the inverse proportion function and the coordinates of point $E$?\n",
    "completion": "\\textbf{Solution:} Because the coordinates of point B are (4, 3),\\\\\nthus OC=AB=3, OA=BC=4.\\\\\nBecause BD=1,\\\\\nthus AD=2,\\\\\nthus the coordinates of point D are (4, 2).\\\\\nBecause the graph of the inverse proportion function $y= \\frac{k}{x}$ (where $x>0$) passes through point D,\\\\\nthus $k=4\\times 2=8$,\\\\\nthus the equation of the inverse proportion function is $y= \\frac{8}{x}$.\\\\\nWhen $y=3$, $3= \\frac{8}{x}$, solving this gives: $x= \\frac{8}{3}$,\\\\\nthus the coordinates of point E are $\\boxed{\\left( \\frac{8}{3}, 3 \\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51446330_267.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated in the diagram, within the rectangle $ABCD$, the ratio of $AB$ to $BC$ is $3:4$, and $BP=3$. When $OABP$ is folded along $AP$ such that point $B$ precisely aligns with point $E$ on the diagonal $AC$, what is the length of $BC$?\n",
    "completion": "Solution: Let $AB=3x$ and $BC=4x$,\\\\\nsince quadrilateral $ABCD$ is a rectangle,\\\\\nthus $AC= \\sqrt{AB^{2}+BC^{2}}=5x$,\\\\\nsince it is folded,\\\\\nthus $BP=PE=3$, $AE=AB=3x$, $\\angle AEP=\\angle B=90^\\circ$,\\\\\nthus $PC=4x-3$, $EC=AC-AE=2x$,\\\\\nthus $PC^{2}=PE^{2}+EC^{2}$, i.e., $(4x-3)^{2}=3^{2}+(2x)^{2}$,\\\\\nSolving gives: $x=2$ or $0$ (discard $0$),\\\\\nthus $BC=4x=\\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52082448.png"
  },
  {
    "solution": "\\boxed{27^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in $\\triangle ABC$, $AB=AC$, and $\\angle BAC=54^\\circ$. The circle $\\odot O$ with diameter $AB$ intersects $AC$ and $BC$ at points $D$ and $E$, respectively. A tangent to $\\odot O$ passing through point $B$ intersects the extension of $AC$ at point $F$. Given $BE=CE$; find the measure of $\\angle CBF$ in degrees.\n",
    "completion": "\\textbf{Solution:} Since $\\angle BAC=54^\\circ$ and $AB=AC$, it follows that $\\angle ABC=63^\\circ$.\\\\\nAlso, since $BF$ is the tangent to the circle $O$, it follows that $\\angle ABF=90^\\circ$.\\\\\nTherefore, $\\angle CBF=\\angle ABF-\\angle ABC=\\boxed{27^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51318358_121.png"
  },
  {
    "solution": "\\boxed{7}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $C$ is a point on line segment $AB$, and $M$, $N$ are the midpoints of $AC$, $BC$, respectively. If $AC=8\\,cm$ and $BC=6\\,cm$, what is the length of line segment $MN$ in $cm$?\n",
    "completion": "\\textbf{Solution:} Given that $M$ and $N$ are the midpoints of $AC$ and $BC$ respectively, \\\\\n$\\therefore CM=\\frac{1}{2}AC=4\\,cm$, $CN=\\frac{1}{2}BC=3\\,cm$, \\\\\n$\\therefore MN=CM+CN=4+3=\\boxed{7}\\,cm$.",
    "image_path": "dataset/mm_math/images/MM_Math/53319958_80.png"
  },
  {
    "solution": "\\boxed{\\frac{9}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $E$ is on the side $BC$ of the square $ABCD$, where $BE=2CE$, and point $G$ is the foot of the perpendicular. If $FG=2$, what is the value of $DG$?\n",
    "completion": "In the square $ABCD$, we have: $AD=AB=BC=CD$\\\\\n$\\angle DAB=\\angle B=\\angle C=\\angle ADC=90^\\circ$\\\\\n$\\therefore \\angle 1+\\angle 3=90^\\circ$\\\\\n$\\because DF\\perp AE$\\\\\n$\\therefore \\angle DGE=\\angle DAF=90^\\circ$\\\\\n$\\therefore \\angle 2+\\angle 3=90^\\circ$\\\\\nAlso, $\\angle 1+\\angle 3=90^\\circ$\\\\\n$\\therefore \\angle 2=\\angle 1$\\\\\nIn $\\triangle ADF$ and $\\triangle BAE$, we have\\\\\n$\\left\\{\\begin{array}{l}\n\\angle 2=\\angle 1\\\\\nAD=AB\\\\\n\\angle DAB=\\angle B=90^\\circ\n\\end{array}\\right.$\\\\\n$\\therefore \\triangle ADF\\cong \\triangle BAE$ (ASA)\\\\\n$\\therefore AF=BE$\\\\\n$\\because BE=2CE$\\\\\nLet $CE=x$, then $BE=2x$, $BC=3x$, $AF=2x$, $BF=x$\\\\\n$\\because \\angle 1=\\angle 2$, $\\angle AGF=\\angle B=90^\\circ$\\\\\n$\\therefore \\triangle AGF\\sim \\triangle ABE$\\\\\n$\\therefore \\frac{FG}{BE}=\\frac{AF}{AE}$\\\\\n$\\therefore \\frac{2}{2x}=\\frac{2x}{AE}$\\\\\n$\\therefore AE=2x^2$\\\\\nIn $\\triangle ABE$, we have $AB^2+BE^2=AE^2$\\\\\n$\\therefore (3x)^2+(2x)^2=(2x^2)^2$\\\\\n$\\therefore 9x^2+4x^2=4x^4$\\\\\n$\\therefore 13x^2=4x^4$\\\\\n$\\because x\\neq 0$\\\\\n$\\therefore 13=4x^2$\\\\\n$\\therefore x=\\pm \\frac{\\sqrt{13}}{2}$ (discard the negative solution)\\\\\n$\\therefore x=\\frac{\\sqrt{13}}{2}$\\\\\n$\\therefore AE=\\frac{13}{2}$\\\\\n$AF=\\sqrt{13}$\\\\\nIn $\\triangle AGF\\sim \\triangle ABE$,\\\\\n$\\frac{AF}{AE}=\\frac{FG}{BE}$\\\\\n$\\therefore \\frac{\\sqrt{13}}{\\frac{13}{2}}=\\frac{FG}{\\sqrt{13}}$\\\\\n$\\therefore FG=2$\\\\\n$\\therefore DG=DF-GF=\\boxed{\\frac{9}{2}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51402614.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $D$ and $E$ are points on $AB$ and $AC$, respectively, in $\\triangle ABC$, with $\\angle AED = \\angle B$, $AD = 2$, $AC = 3$. The bisector of $\\angle A$ in $\\triangle ABC$, $AF$, intersects $DE$ at point $G$ and $BC$ at point $F$. Given that $\\triangle ADE \\sim \\triangle ACB$, find the value of $\\frac{AG}{GF}$.\n",
    "completion": "Solution: Since $\\triangle ADE \\sim \\triangle ACB$, it follows that $\\angle ADE = \\angle C$. Since AF bisects $\\angle BAC$, we have $\\angle DAG = \\angle CAF$. Consequently, $\\triangle ADG \\sim \\triangle ACF$, which means $\\frac{AG}{AF} = \\frac{AD}{AC}$. Given that $AD = 2$ and $AC = 3$, we find $\\frac{AG}{AF} = \\frac{2}{3}$. Letting $AG = 2a$ and $AF = 3a$, we get $FG = a$.\n\nTherefore, $\\frac{AG}{GF} = \\boxed{2}",
    "image_path": "dataset/mm_math/images/MM_Math/51458292_147.png"
  },
  {
    "solution": "\\boxed{130^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, given: In $\\triangle ABC$, $\\angle ABC=60^\\circ$, $\\angle ACB=40^\\circ$, $BD$ bisects $\\angle ABC$, and $CD$ bisects $\\angle ACB$. What is the measure of $\\angle BDC$?\n",
    "completion": "\\textbf{Solution:}\n\nSince BD bisects $\\angle ABC$, \\\\\nit follows that $\\angle DBC = \\frac{1}{2}\\angle ABC = \\frac{1}{2} \\times 60^\\circ = 30^\\circ$, \\\\\nSince CD bisects $\\angle ACB$, \\\\\nit follows that $\\angle DCB = \\frac{1}{2}\\angle ACB = \\frac{1}{2} \\times 40^\\circ = 20^\\circ$, \\\\\nTherefore, $\\angle BDC = 180^\\circ - \\angle DBC - \\angle DCB = 180^\\circ - 30^\\circ - 20^\\circ = \\boxed{130^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52978993_120.png"
  },
  {
    "solution": "\\boxed{-1<x<0}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line ${l}_{1}: y={k}_{1}x+b$ and the line ${l}_{2}: y={k}_{2}x$, with their graphs depicted in the same Cartesian coordinate system as illustrated. What is the solution set for the inequality regarding $x$, ${k}_{1}x+b>{k}_{2}x>0$?\n",
    "completion": "\\textbf{Solution:} Given the graph, when $-1<x<0$, it follows that $k_1x+b>k_2x>0$,\\\\\nhence the solution set for $k_1x+b>k_2x>0$ is $-1<x<0$.\\\\\nTherefore, the answer is: \\boxed{-1<x<0}.",
    "image_path": "dataset/mm_math/images/MM_Math/52395496.png"
  },
  {
    "solution": "\\boxed{\\frac{9}{16}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In $\\triangle ABC$, points D and E are located on sides AB and AC respectively, such that $DE\\parallel BC$ and $AD: BD = 3: 2$. What is the ratio of the area of $\\triangle ADE$ to that of quadrilateral $BCED$?\n",
    "completion": "\\textbf{Solution:} Given that $\\because \\frac{AD}{BD}=\\frac{3}{2}$,\\\\\n$\\therefore \\frac{AD}{AB}=\\frac{3}{5}$,\\\\\n$\\because DE \\parallel BC$,\\\\\n$\\therefore \\triangle ADE \\sim \\triangle ABC$,\\\\\n$\\therefore \\frac{S_{\\triangle ADE}}{S_{\\triangle ABC}}=\\left(\\frac{AD}{AB}\\right)^2=\\left(\\frac{3}{5}\\right)^2=\\frac{9}{25}$,\\\\\n$\\therefore \\frac{S_{\\triangle ADE}}{S_{\\text{quadrilateral} BCED}}=\\boxed{\\frac{9}{16}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51209210.png"
  },
  {
    "solution": "\\boxed{88^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $CA$ bisects $\\angle DCB$, $CB = CD$, and the extension line of $DA$ intersects $BC$ at point $E$. If $\\angle EAC = 46^\\circ$, what is the degree measure of $\\angle BAE$?\n",
    "completion": "\\textbf{Solution}: Since $CA$ bisects $\\angle DCB$,\\\\\nit follows that $\\angle BCA=\\angle DCA$,\\\\\nand since $CB=CD$, $AC=AC$,\\\\\nit follows that $\\triangle ABC\\cong \\triangle ADC\\ (SAS)$,\\\\\ntherefore $\\angle B=\\angle D$,\\\\\nhence $\\angle D+\\angle ACD=\\angle B+\\angle ACB=\\angle CAE=46^\\circ$,\\\\\ntherefore $\\angle BAE=180^\\circ-\\angle B-\\angle ACB-\\angle CAE=\\boxed{88^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/53106609.png"
  },
  {
    "solution": "\\boxed{}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ is rotated about point $A$ to $\\triangle ADE$. What is the angle of rotation?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABC$ is rotated about point A to $\\triangle ADE$,\\\\\nthe angle of rotation is either $\\angle BAD$ or $\\angle CAE$.",
    "image_path": "dataset/mm_math/images/MM_Math/51212803.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AD$ is the angle bisector of $\\triangle ABC$, $DE \\perp AB$ with the foot of the perpendicular at $E$, $DF \\perp AC$ with the foot of the perpendicular at $F$, if $AB=5$, $AC=3$, and $DF=2$, what is the area of $\\triangle ABC$?\n",
    "completion": "\\textbf{Solution:} Since AD is the angle bisector of $\\triangle ABC$, and DE$\\perp$AB, DF$\\perp$AC,\\\\\nthus DE=DF=2,\\\\\ntherefore, the area of $\\triangle ABC$ = $\\frac{1}{2} \\times 5 \\times 2 + \\frac{1}{2} \\times 3 \\times 2 = \\boxed{8}$",
    "image_path": "dataset/mm_math/images/MM_Math/51043532.png"
  },
  {
    "solution": "\\boxed{x>-1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the line $y_1=x+a$ and $y_2=kx+b$ intersect at point $P(-1,2)$. What is the correct solution set of the inequality $x+a>kx+b$ with respect to $x$?\n",
    "completion": "\\textbf{Solution:} Given the lines $y_1=x+a$ and $y_2=kx+b$ intersect at point P ($-1$, $2$), \\\\\nfor the inequality $x+a>kx+b$ to hold, it is necessary that $y_1=x+a$ is above $y_2=kx+b$,\\\\\nAccording to the graph, when $x>-1$, $y_1=x+a$ is above $y_2=kx+b$,\\\\\nTherefore, the solution set for the inequality $x+a>kx+b$ is $x>-1$,\\\\\nThus, the answer is: $\\boxed{x>-1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52250870.png"
  },
  {
    "solution": "\\boxed{95^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the quadrilateral $ABCD$ is inscribed in the circle $\\odot O$. If $\\angle ADC=85^\\circ$, then what is the measure of $\\angle B$ in degrees?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral $ABCD$ is inscribed in circle $O$,\\\\\nit follows that $\\angle ADC + \\angle B = 180^\\circ$,\\\\\nSince $\\angle ADC = 85^\\circ$,\\\\\nthen $\\angle B = 180^\\circ - \\angle ADC$\\\\\n$= 180^\\circ - 85^\\circ$\\\\\n$= \\boxed{95^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53078985.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{5}}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within a grid of small squares with side length $1$, points $A, B, C, D$ are all at the vertices of these small squares. Lines $AB$ and $CD$ intersect at point $O$. What is the value of $\\cos\\angle BOD$?\n",
    "completion": "\\textbf{Solution:} Draw $CE$ and $DE$ as shown in the figure:\\\\\nSince it is known from the figure that $\\angle 1=\\angle 2=\\angle 3=\\angle 4=\\angle ABE=45^\\circ$\\\\\nThus, $\\angle CED=\\angle 2+\\angle 3=90^\\circ$, $AB\\parallel CE$\\\\\nTherefore, $\\angle BOD=\\angle DCE$\\\\\nSince the side length of the small square is $1$\\\\\nTherefore, in $Rt\\triangle CDE$, $CE=\\sqrt{1^2+1^2}=\\sqrt{2}$, $CD=\\sqrt{1^2+3^2}=\\sqrt{10}$\\\\\nTherefore, $\\cos\\angle DCE=\\frac{CE}{CD}=\\frac{\\sqrt{2}}{\\sqrt{10}}=\\frac{\\sqrt{5}}{5}$\\\\\nTherefore, $\\cos\\angle BOD=\\cos\\angle DCE=\\boxed{\\frac{\\sqrt{5}}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51403017.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, line $AD\\parallel BE\\parallel CF$, if $\\frac{AB}{BC}=\\frac{1}{3}$, $AD=2$, and $CF=6$, then what is the length of the line segment $BE$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, draw DG $\\parallel$ AC through point D, intersecting CF at point G, and BE at point H, \\\\\n$\\because$ $AD\\parallel BE\\parallel CF$, \\\\\n$\\therefore$ $\\frac{DE}{EF}=\\frac{AB}{BC}=\\frac{1}{3}$, quadrilaterals ABHD and ACGD are parallelograms, \\\\\n$\\therefore$ BH=AD=CG=2, $\\frac{DE}{DF}=\\frac{1}{4}$, \\\\\n$\\because$ $CF=6$, \\\\\n$\\therefore$ FG=4, \\\\\n$\\because$ BE$\\parallel$CF, \\\\\n$\\therefore$ $\\triangle DEH\\sim \\triangle DFG$, \\\\\n$\\therefore$ $\\frac{HE}{FG}=\\frac{DE}{DF}=\\frac{1}{4}$, \\\\\n$\\therefore$ HE=1, \\\\\n$\\therefore$ BE=BH+HE=$\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51403270.png"
  },
  {
    "solution": "\\boxed{x \\geq 2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graphs of the functions $y=ax$ and $y=bx+5$ intersect at point A. What is the solution set of the inequality $ax\\geq bx+5$?\n",
    "completion": "\\textbf{Solution:} According to the graph of the function, when $x \\geq 2$, $ax \\geq bx + 5$.\\\\\nTherefore, the answer is: $\\boxed{x \\geq 2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52056392.png"
  },
  {
    "solution": "\\boxed{x>1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the hyperbola represented by the inverse proportion function $y=\\frac{2m}{x}$ and the straight line represented by the linear function $y=kx-1$ intersect at points $A(m, 2m)$ and $B$. Find the coordinates of point $B$ and, based on the graph, directly write out the range of $x$ values that satisfy the inequality $\\frac{2m}{x}<kx-1$.\n",
    "completion": "\\textbf{Solution:} From\n\\[\n\\left\\{\n\\begin{array}{l}\ny=\\frac{2}{x}\\\\\ny=3x-1\n\\end{array}\n\\right.\n\\]\nwe get \\boxed{B\\left(-\\frac{2}{3},-3\\right)}.\n\nTherefore, by the graphical representation, the range of $x$ that satisfies $\\frac{2m}{x}<kx-1$ is \\boxed{-\\frac{2}{3}<x<0} or \\boxed{x>1}.",
    "image_path": "dataset/mm_math/images/MM_Math/52622749_216.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, $\\angle B=22.5^\\circ$, $AC=2$. Taking points $A$ and $B$ as centers, arcs are drawn with radii longer than $\\frac{1}{2}AB$, intersecting at points $M$ and $N$. The line $MN$ intersects $BC$ at point $D$. Connecting $AD$, what is the length of $BD$?\n",
    "completion": "\\textbf{Solution:} Given the construction, $MN$ bisects $AB$ perpendicularly, hence $DA=DB$, \\\\\n$\\therefore \\angle DAB=\\angle B=22.5^\\circ$, \\\\\n$\\therefore \\angle CDA=22.5^\\circ+22.5^\\circ=45^\\circ$, \\\\\n$\\because \\angle C=90^\\circ$, \\\\\n$\\therefore \\triangle ACD$ is an isosceles right triangle, \\\\\n$\\therefore AD=\\sqrt{2}AC=2\\sqrt{2}$, \\\\\n$\\therefore BD=\\boxed{2\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55587990.png"
  },
  {
    "solution": "\\boxed{M(2, 5)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: It is known that, in the Cartesian coordinate system, points $A(-3, 0)$ and $B(1, 0)$ are given. Rotate point $A$ counterclockwise around the origin to let it fall on point $C$ located on the negative half of the y-axis. A parabola, $y=ax^2+bx+c$, passes through points $A$, $B$, and $C$. The line segment $AC$ is connected. The line $AC$ is translated upward, the translated line $DM$ intersects the y-axis at point $D$ and intersects the parabola to the right of the y-axis at point $M$. The segments $AM$ and $CM$ are connected. Given that the area of $\\triangle ACM$ is $15$, what are the coordinates of point $M$?\n",
    "completion": "\\textbf{Solution:} Let line $DM$ be obtained by translating line $AC$,\\\\\n$\\therefore DM\\parallel AC$,\\\\\n$\\therefore S_{\\triangle ACD}=S_{\\triangle ACM}=15$,\\\\\nSince $S_{\\triangle ACM}=\\frac{1}{2}CD\\cdot AO=15, AO=3$,\\\\\n$\\therefore CD=10$,\\\\\n$\\therefore D(0, 7)$,\\\\\n$\\therefore A(-3, 0), C(0, -3)$,\\\\\nLet $AC: y=kx+b(k\\ne 0)$,\\\\\nSolving gives: $\\left\\{\\begin{array}{l}k=-1 \\\\ b=-3\\end{array}\\right.$,\\\\\n$\\therefore AC: y=-x-3$,\\\\\nSince $DM\\parallel AC$,\\\\\nLet $DM: y=-x+b$,\\\\\nSubstituting $D(0, 7)$ gives,\\\\\n$DM: y=-x+7$,\\\\\nCombining it with the parabola, we solve to get,\\\\\n$\\left\\{\\begin{array}{l}x=2 \\\\ y=5\\end{array}\\right.$ or $\\left\\{\\begin{array}{l}x=-5 \\\\ y=12\\end{array}\\right.$ (which does not meet the conditions, hence disregarded),\\\\\n$\\therefore \\boxed{M(2, 5)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52602789_88.png"
  },
  {
    "solution": "\\boxed{6\\text{cm}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, $BC=4\\text{cm}$. A point $E$ is chosen on $AC$ such that $EC=BC$. A line $EF$ is drawn perpendicular to $AC$, and $CF$ is connected so that $CF=AB$. If $EF=10\\text{cm}$, what is the length of $AE$?\n",
    "completion": "\\textbf{Solution:} Given that $EF\\perp AC$,\\\\\nit follows that $\\angle CEF = \\angle ACB = 90^\\circ$,\\\\\nIn right $\\triangle ACB$ and right $\\triangle FEC$,\\\\\n$\\left\\{\\begin{array}{l}\nAB = FC \\\\\nBC = CE\n\\end{array}\\right.$,\\\\\nhence, right $\\triangle ACB \\cong$ right $\\triangle FEC$,\\\\\nthus, AC = $EF = 10\\text{cm}$, EC = $BC = 4\\text{cm}$,\\\\\nhence, AE = AC - EC = $\\boxed{6\\text{cm}}$,",
    "image_path": "dataset/mm_math/images/MM_Math/53033114.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, there is a circular paper with point O as the center and AB as the diameter. Point C is on the circle $\\odot$O. Folding the circular paper along the diameter CF, point B falls on point D located on $\\odot$O (not coinciding with point A). After unfolding the paper, connect CB, CD, and AD. Let CD intersect with diameter AB at point E. AD$\\parallel$OC; as shown in Figure 2, when CD$\\perp$AB, if OC＝2, what is the length of BC?\n",
    "completion": "\\textbf{Solution:} From the fold, it is known that: $\\angle ECO = \\angle BCO$,\\\\\n$\\because OB = OC$,\\\\\n$\\therefore \\angle OCB = \\angle B$,\\\\\n$\\therefore \\angle ECO = \\angle OCB = \\angle B$,\\\\\n$\\because CD \\perp AB$,\\\\\n$\\therefore \\angle ECO + \\angle OCB + \\angle B = 90^\\circ$,\\\\\n$\\therefore \\angle ECO = \\angle OCB = \\angle B = 30^\\circ$,\\\\\n$\\because OC = OB = 2$,\\\\\n$\\therefore OE = 1$,\\\\\n$\\therefore CE = \\sqrt{3}$,\\\\\n$\\therefore BC = 2CE = 2\\sqrt{3}$;\\\\\nHence, the length of BC is $BC = \\boxed{2\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53124745_89.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, EF passes through the intersection point O of the diagonals of rectangle ABCD, and intersects AD and BC at points E and F, respectively. Given that AB = 3 and BC = 4, what is the area of the shaded region in the figure?\n",
    "completion": "\\textbf{Solution:} Given rectangle ABCD, \\\\\nit follows that BO=DO=AO=CO, and AD$\\parallel$BC, \\\\\nhence, $\\angle$EDO=$\\angle$FBO, \\\\\nsince also $\\angle$EOD=$\\angle$FOB, \\\\\nit follows that $\\triangle$EDO$\\cong$$\\triangle$FBO (by ASA), \\\\\nthus, the area of $\\triangle$FBO = the area of $\\triangle$EDO, \\\\\nhence, the area of the shaded region = S$_{\\triangle BOC}$= $\\frac{1}{4}$S$_{\\text{rectangle }ABCD}$= $\\frac{1}{4}\\times4\\times3=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51614394.png"
  },
  {
    "solution": "\\boxed{15}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB=20$, point $C$ is the midpoint of segment $AB$, and point $D$ is the midpoint of segment $CB$. Find the length of $AD$.\n",
    "completion": "\\textbf{Solution:} Since point $C$ is the midpoint of segment $AB$, and $AB=20$, \\\\\ntherefore $AC=BC=\\frac{1}{2}AB=10$. \\\\\nSince point $D$ is the midpoint of segment $BC$, \\\\\ntherefore $CD=\\frac{1}{2}BC=5$. \\\\\nTherefore, $AD=AC+CD=10+5=\\boxed{15}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55253057.png"
  },
  {
    "solution": "\\boxed{\\frac{32}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In rectangle $ABCD$, $AB=6$, $AD=8$, point $P$ is a moving point on line segment $BD$ (not coinciding with points $B$, $D$). A line is drawn through point $P$ such that $PE \\perp BD$, intersecting the ray $DC$ at point $E$, and $BE$ is connected. As shown in Figure 1, when point $E$ coincides with point $C$, what is the length of $BP$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral $ABCD$ is a rectangle, with $AB=6$ and $AD=8$,\\\\\nthen $CD=6$, $BC=8$, $BD=10$, $\\angle A=90^\\circ$, and $AD\\parallel CB$,\\\\\nhence $\\angle ADB=\\angle CBP$,\\\\\nsince $CP\\perp BD$,\\\\\nthen $\\angle CPB=\\angle A=90^\\circ$,\\\\\nthus $\\triangle CPB\\sim \\triangle BAD$,\\\\\ntherefore $\\frac{BP}{AD}=\\frac{BC}{DB}$,\\\\\nthus $\\frac{BP}{8}=\\frac{8}{10}$,\\\\\ntherefore $BP=\\boxed{\\frac{32}{5}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52866596_115.png"
  },
  {
    "solution": "\\boxed{40^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC \\cong \\triangle ADE$, $\\angle C = 40^\\circ$, then what is the degree measure of $\\angle E$?\n",
    "completion": "\\textbf{Solution:} \nSince $\\triangle ABC \\cong \\triangle ADE$,\\\\\ntherefore $\\angle E = \\angle C = 40^\\circ$.\\\\\nHence, $\\angle E = \\boxed{40^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51419663.png"
  },
  {
    "solution": "\\boxed{100^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated in the diagram, line $m\\parallel n$, $\\triangle ABC$ is an equilateral triangle with vertex $B$ on line $n$. Line $m$ intersects $AB$ at point $E$ and $AC$ at point $F$. Given that $\\angle 1=140^\\circ$, what is the measure of $\\angle 2$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure: \\\\\nSince $\\triangle ABC$ is an equilateral triangle,\\\\\nit follows that $\\angle A=\\angle ABC=\\angle C=60^\\circ$.\\\\\nIn $\\triangle AFE$, since $\\angle 1=\\angle A+\\angle AEF=140^\\circ$,\\\\\nit follows that $\\angle AEF=140^\\circ-60^\\circ=80^\\circ$,\\\\\nhence, $\\angle DEB=\\angle AEF=80^\\circ$,\\\\\nSince $m\\parallel n$,\\\\\nit follows that $\\angle 2+\\angle DEB=180^\\circ$,\\\\\nthus, $\\angle 2=180^\\circ-80^\\circ=\\boxed{100^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52634468.png"
  },
  {
    "solution": "\\boxed{120^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $\\triangle ABC$ is an equilateral triangle, $BD$ is the median of $\\triangle ABC$. Extend $BC$ to $E$ so that $CE=CD$, and connect $DE$. What is the measure of $\\angle BDE$?\n",
    "completion": "\\textbf{Solution:} Given that $\\triangle ABC$ is an equilateral triangle and BD is the median,\\\\\ntherefore, $\\angle BDC=90^\\circ$, $\\angle ACB=60^\\circ$\\\\\nthus, $\\angle ACE=180^\\circ-\\angle ACB=180^\\circ-60^\\circ=120^\\circ$,\\\\\nsince CE=CD,\\\\\ntherefore, $\\angle CDE=\\angle CED=30^\\circ$,\\\\\nthus, $\\angle BDE=\\angle BDC+\\angle CDE=90^\\circ+30^\\circ=\\boxed{120^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/51434804.png"
  },
  {
    "solution": "\\boxed{x>-1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is the graph of the linear function $y=mx+n$. What is the solution set of the inequality $mx+n>0$ with respect to $x$?\n",
    "completion": "\\textbf{Solution:} From the graph, we know: \\\\\nThe solution set of the inequality $mx+n>0$ with respect to $x$ is: $x>-1$. \\\\\nThus, the answer is: $\\boxed{x>-1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52065351.png"
  },
  {
    "solution": "\\boxed{\\sin\\alpha}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Figure 1 is the emblem of the 7th International Congress on Mathematical Education (ICME), featuring two adjacent right-angled triangles in its main pattern, which together form the quadrilateral $OABC$ as shown in Figure 2. In the figure, if $AB=BC=1$ and $\\angle AOB=\\alpha$, then what is the value of $\\tan\\angle BOC$?\n",
    "completion": "\\textbf{Solution}: Since $AB=BC=1$,\\\\\nin the right triangle $OAB$, $\\sin\\alpha =\\frac{AB}{OB}$,\\\\\ntherefore $OB=\\frac{1}{\\sin\\alpha}$,\\\\\nin the right triangle $OBC$, $\\tan\\angle BOC=\\frac{BC}{OB}=\\frac{1}{\\frac{1}{\\sin\\alpha}}=\\sin\\alpha$.\\\\\nHence, the answer is: $\\boxed{\\sin\\alpha}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51318708.png"
  },
  {
    "solution": "\\boxed{10\\sqrt{3}-12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $AB=4$ and $AD=10$. When the vertex $P$ of the right angle of a right-angled ruler slides along side $AD$ (point $P$ does not coincide with $A$ or $D$), one of the right-angle sides passes through point $C$ and the other right-angle side intersects $AB$ at point $E$. $Rt\\triangle AEP \\sim Rt\\triangle DPC$; when $\\angle CPD=30^\\circ$, what is the length of $AE$?\n",
    "completion": "\\textbf{Solution:} In right triangle $\\triangle PCD$, $\\angle CPD=30^\\circ$ and $CD=4$, \\\\\ntherefore $PC=8$, \\\\\ntherefore $PD=\\sqrt{PC^{2}-CD^{2}}=4\\sqrt{3}$, \\\\\ntherefore $AP=AD-PD=10-4\\sqrt{3}$, \\\\\nAs right triangles $\\triangle AEP \\sim \\triangle DPC$ indicate: $\\frac{PD}{AE}=\\frac{CD}{AP}$, thus $\\frac{4\\sqrt{3}}{AE}=\\frac{4}{10-4\\sqrt{3}}$, \\\\\ntherefore $AE=\\boxed{10\\sqrt{3}-12}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52387502_134.png"
  },
  {
    "solution": "\\boxed{35^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, with $O$ in the center, point $C$ is on the arc $\\overset{\\frown}{AB}$, and $\\angle ADC$, $\\angle BEC$ are the angles subtended by arcs $\\overset{\\frown}{AC}$, $\\overset{\\frown}{BC}$ on the circumference, respectively. If $\\angle AOB=110^\\circ$ and $\\angle ADC=20^\\circ$, what is the measure of $\\angle BEC$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle AOB=110^\\circ$,\\\\\ntherefore $\\angle ADC+\\angle BEC= \\frac{1}{2} \\angle AOB=55^\\circ$,\\\\\nsince $\\angle ADC=20^\\circ$,\\\\\ntherefore $\\angle BEC=\\boxed{35^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52755175.png"
  },
  {
    "solution": "\\boxed{\\frac{5\\sqrt{2}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y = -x^2 - 4x + 5$ intersects the $x$-axis at points $A$ and $B$, and intersects the $y$-axis at point $C$. Point $P$ is any point on the axis of symmetry of the parabola. If points $D$, $E$, and $F$ are the midpoints of $BC$, $BP$, and $PC$, respectively, and $DE$ and $DF$ are connected, what is the minimum value of $DE + DF$?\n",
    "completion": "\\textbf{Solution:} Let points D, E, F be the midpoints of BC, BP, and PC, respectively.\\\\\nTherefore, DE and DF are the midlines of $\\triangle PBC$,\\\\\nHence, $DE=\\frac{1}{2}PC$, $DF=\\frac{1}{2}PB$,\\\\\nThus, $DE+DF=\\frac{1}{2}(PB+PC)$,\\\\\nWhen the value of PB+PC is minimum, the value of DE+DF is also at its minimum,\\\\\nWhen $y=0$, solving $-x^2-4x+5=0$, we find $x_1=-5$ and $x_2=1$, thus $A(-5,0)$, $B(1,0)$,\\\\\nWhen $x=0$, $y=-x^2-4x+5=5$, hence $C(0,5)$,\\\\\nConnect AC and the symmetry axis of the parabola at point P',\\\\\nSince $P'A=P'B$,\\\\\nIt follows that $P'B+P'C=P'A+P'C=AC$,\\\\\nTherefore, at this point, the value of $P'B+P'C$ is at its minimum, which is $\\sqrt{5^2+5^2}=5\\sqrt{2}$,\\\\\nHence, the minimum value of DE+DF is $\\boxed{\\frac{5\\sqrt{2}}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52515604.png"
  },
  {
    "solution": "\\boxed{\\frac{2\\sqrt{5}}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, let $ABCD$ be a square with side length $2$. Let points $E$ and $F$ be the midpoints of $AB$ and $CD$, respectively. Connect $CE$ and $AF$, and extend a line from $D$ perpendicular to $AF$ at point $G$. Extend $DG$ to intersect with $CE$ at point $H$. What is the length of $DG$?\n",
    "completion": "\\textbf{Solution:} Given a square ABCD, we have AD=CD=AB=2 and $\\angle ADC=90^\\circ$. Since point F is the midpoint of CD, it follows that DF=$\\frac{1}{2}$CD=1. In $\\triangle ADF$, we find $AF=\\sqrt{AD^2+DF^2}=\\sqrt{2^2+1^2}=\\sqrt{5}$. Considering $S_{\\triangle ADF}=\\frac{1}{2}AD\\cdot DF=\\frac{1}{2}AF\\cdot DG$ and DG$\\perp$AF, we deduce that $2\\times 1=\\sqrt{5}DG$. Solving this, we get $DG=\\boxed{\\frac{2\\sqrt{5}}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52160938_144.png"
  },
  {
    "solution": "\\boxed{-2\\sqrt{3}x^2 + 4\\sqrt{3}x}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the four vertices of rectangle DEFG are respectively on the sides of equilateral triangle ABC. Given that the side length of $\\triangle ABC$ is 4, and denoting the area of rectangle DEFG as $S$, and the line segment BE as $x$. Find the expression of $S$ as a function of $x$.\n",
    "completion": "\\textbf{Solution:}\n\nIn the equilateral triangle $ABC$, the length of line segment $BE$ is denoted as $x$,\n\n$\\therefore \\angle B = \\angle C = 60^\\circ$.\n\n$\\therefore DE = BE \\cdot \\tan 60^\\circ = \\sqrt{3}x$.\n\nSince the four vertices of rectangle $DEFG$ are located on the sides of equilateral triangle $ABC$,\n\n$\\therefore DE = GF$, and $\\angle BED = \\angle CFG = 90^\\circ$,\n\n$\\therefore \\triangle BED \\cong \\triangle CFG \\left(AAS\\right)$.\n\n$\\therefore BE = CF = x$.\n\nSince the side length of $\\triangle ABC$ is 4,\n\n$\\therefore EF = 4 - 2x$.\n\n$\\therefore S = DE \\cdot EF = \\sqrt{3}x(4 - 2x) = -2\\sqrt{3}x^2 + 4\\sqrt{3}x \\left(0 < x < 2\\right)$.\n\nThus, the function expression for $S$ with respect to $x$ is $S = \\boxed{-2\\sqrt{3}x^2 + 4\\sqrt{3}x}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51545233_180.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$. Extend $AB$ to point $D$ such that $BD=AB$. Connect $CD$. If $\\tan\\angle BCD=\\frac{1}{3}$, what is the value of $\\frac{BC}{AC}$?\n",
    "completion": "\\textbf{Solution:} Draw $BE\\parallel AC$ through point B, intersecting $CD$ at E, as shown in the diagram:\\\\\n$\\therefore$ $\\angle ACB=\\angle CBE=90^\\circ$,\\\\\n$\\because$ $\\tan\\angle BCD=\\frac{1}{3}$,\\\\\n$\\therefore$ in $\\triangle BCE$, $\\frac{BE}{BC}=\\frac{1}{3}$,\\\\\n$\\therefore$ $BC=3BE$,\\\\\n$\\because$ $BE\\parallel AC$,\\\\\n$\\therefore$ $\\triangle DBE\\sim \\triangle DAC$,\\\\\n$\\therefore$ $\\frac{BE}{AC}=\\frac{BD}{AD}$,\\\\\n$\\because$ $BD=AB$,\\\\\n$\\therefore$ $\\frac{BE}{AC}=\\frac{BD}{AD}=\\frac{1}{2}$,\\\\\n$\\therefore$ $AC=2BE$,\\\\\n$\\therefore$ $\\frac{BC}{AC}=\\frac{3BE}{2BE}=\\boxed{\\frac{3}{2}}$,",
    "image_path": "dataset/mm_math/images/MM_Math/55594993.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $OP$ bisects $\\angle AOB$, where point $E$ is on $OA$ with $OE = 4$, and the distance from point $P$ to $OB$ is $2$. What is the area of $\\triangle POE$?\n",
    "completion": "\\textbf{Solution:} Since OP bisects $\\angle$AOB, and the distance from point P to OB is 2, \\\\\n$\\therefore$ the distance from point P to OA is also 2, \\\\\n$\\therefore$ $ {S}_{\\triangle OEP}=\\frac{1}{2}\\times 4\\times 2=\\boxed{4}$",
    "image_path": "dataset/mm_math/images/MM_Math/52950518.png"
  },
  {
    "solution": "\\boxed{124^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $E$ is the incenter of $\\triangle ABC$. The extension line of $AE$ intersects the circumcircle of $\\triangle ABC$ at point $D$. Connect $BE$. If $\\angle CBD = 34^\\circ$, what is the degree measure of $\\angle BEC$?\n",
    "completion": "\\textbf{Solution}:\\\\\nSince $\\angle CBD=34^\\circ$,\\\\\nit follows that $\\angle CAD=34^\\circ$,\\\\\nsince point E is the incenter of $\\triangle ABC$,\\\\\nit follows that $\\angle BAC=2\\angle CAD=68^\\circ$,\\\\\ntherefore $\\angle EBC+\\angle ECB=(180^\\circ-68^\\circ)\\div 2=56^\\circ$,\\\\\nhence, $\\angle BEC=180^\\circ-56^\\circ=\\boxed{124^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52766514_133.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the right-angled triangle $ABC$, where $\\angle C = 90^\\circ$, $AD$ bisects $\\angle BAC$ and intersects $BC$ at point $D$. Line $DE \\parallel AB$ intersects $AC$ at point $E$. Given that $CE = 3$ and $CD = 4$, what is the length of $AD$?\n",
    "completion": "\\textbf{Solution:} In right $\\triangle ABC$, $\\angle C=90^\\circ$, $CE=3$, $CD=4$, \\\\\n$\\therefore DE^{2}=CD^{2}+CE^{2}=9+16=25$, \\\\\n$\\therefore DE=5$, \\\\\n$\\because DE\\parallel AB$, \\\\\n$\\therefore \\angle BAD=\\angle ADE$, \\\\\n$\\because AD$ bisects $\\angle BAC$, \\\\\n$\\therefore \\angle BAD=\\angle CAD$, \\\\\n$\\therefore \\angle CAD=\\angle ADE$, \\\\\n$\\therefore AE=DE=5$, \\\\\n$\\therefore AC=AE+EC=5+3=8$, \\\\\n$\\therefore$ in right $\\triangle ACD$, \\\\\n$\\therefore AD^{2}=AC^{2}+CD^{2}=64+16=80$, \\\\\n$\\therefore AD=\\boxed{4\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52774836.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $\\triangle OAC$ and $\\triangle ABC$ are both isosceles right triangles, with $\\angle ACO=\\angle ABC=90^\\circ$. The graph of the inverse proportion function $y=\\frac{k}{x}$ in the first quadrant passes through point $B$. Connect $OB$, and the area of $\\triangle OAB$ is $4$. What is the value of $k$?\n",
    "completion": "\\textbf{Solution:} Let $OC=AC=a$\\\\\nSince $\\triangle OAC$ is an isosceles right triangle,\\\\\nwe have $\\sin 45^\\circ=\\frac{OC}{AO}=\\frac{\\sqrt{2}}{2}$,\\\\\nthus $OA=\\sqrt{2}a$.\\\\\nSince $\\triangle ABC$ is also an isosceles right triangle,\\\\\nwe have $\\sin 45^\\circ=\\frac{AB}{AC}=\\frac{\\sqrt{2}}{2}$,\\\\\nthus $AB=BC=\\frac{\\sqrt{2}}{2}a$.\\\\\nSince both $\\triangle OAC$ and $\\triangle ABC$ are isosceles right triangles,\\\\\nwe have $\\angle OAC=\\angle BAC=45^\\circ$,\\\\\nhence $\\angle OAB=90^\\circ$,\\\\\ntherefore, ${S}_{\\triangle OAB}=\\frac{1}{2}OA\\cdot AB=\\frac{1}{2}\\cdot \\sqrt{2}a \\cdot \\frac{\\sqrt{2}}{2}a=4$.\\\\\nSolving, we find $a=2\\sqrt{2}$,\\\\\nthus $OC=2\\sqrt{2}$, $BC=2$.\\\\\nDraw $BE\\perp x$ axis through point $B$,\\\\\nSince $\\angle ACB=45^\\circ$,\\\\\nwe have $\\angle BCE=45^\\circ$,\\\\\nhence $\\triangle BCE$ is an isosceles right triangle,\\\\\ntherefore $\\sin 45^\\circ=\\frac{CE}{CB}=\\frac{\\sqrt{2}}{2}$,\\\\\nthus $CE=BE=\\sqrt{2}$,\\\\\nhence $OE=3\\sqrt{2}$,\\\\\ntherefore $B(3\\sqrt{2},\\sqrt{2})$.\\\\\nSince the inverse proportional function $y=\\frac{k}{x}$ passes through point $B$ in the first quadrant,\\\\\nwe have $k=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52585579.png"
  },
  {
    "solution": "\\boxed{15^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, a pair of set squares is overlaid. Given that $\\angle OAB=\\angle OCD=90^\\circ$, $\\angle AOB=45^\\circ$, and $\\angle COD=60^\\circ$, with $OB$ bisecting $\\angle COD$, what is the measurement of $\\angle AOC$ in degrees?\n",
    "completion": "\\textbf{Solution:} Given that OB bisects $\\angle$COD and $\\angle$COD = 60°,\\\\\nit follows that $\\angle$BOC = 30°,\\\\\nand since $\\angle$AOB = 45°,\\\\\nit implies $\\angle$AOC = $\\angle$AOB - $\\angle$BOC = 45° - 30° = \\boxed{15^\\circ}.",
    "image_path": "dataset/mm_math/images/MM_Math/52793014.png"
  },
  {
    "solution": "\\boxed{\\left(\\frac{6\\sqrt{13}+13}{13}, \\frac{4\\sqrt{13}+52}{13}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=ax^2-2ax-3a$ (where \\(a\\) is a constant, $a<0$) intersects the x-axis at points A and B (with point A to the left of point B), and intersects the y-axis at point C, with OB=OC. Point K is a point in the coordinate plane, and DK=2, point M is the midpoint of segment BK. Connect AM. When AM is maximized, find the coordinates of point K.\n",
    "completion": "\\textbf{Solution:} Connect $BD$ and take the midpoint $Q$ of $BD$. Connect $QM$. Therefore, $QM$ is the median of $\\triangle BDK$, thus $QM=\\frac{1}{2}DK=1$. According to the problem, point $K$ lies on the circle with center $D$ and radius 2, so $M$ moves on the circle with center $Q$ and radius 1. When points $A$, $Q$, $M$ are collinear, and $M$ lies on the extension of $AQ$, $AM$ is maximized. Since $B(3, 0)$ and $D(1, 4)$, thus $Q\\left(\\frac{1+3}{2}, \\frac{4+0}{2}\\right)$ i.e., $Q(2, 2)$. Since $A(-1, 0)$ and $Q(2, 2)$, let the equation of line $AM$ be $y=kx+d$. Substitute points $A(-1, 0)$ and $Q(2, 2)$ to get\n\\[\n\\begin{cases}\n0=-k+d\\\\\n2=2k+d\n\\end{cases}\n\\]\nSolving these equations gives\n\\[\n\\begin{cases}\nk=\\frac{2}{3}\\\\\nb=\\frac{2}{3}\n\\end{cases}\n\\]\nTherefore, the equation of line $AM$ is $y=\\frac{2}{3}x+\\frac{2}{3}$. Since $DK\\parallel QM$, let the equation of line $DK$ be $y=\\frac{2}{3}x+b$. Given $D(1, 4)$, thus $4=\\frac{2}{3}+b$, solving this gives $b=\\frac{10}{3}$, so the equation of $DK$ is $y=\\frac{2}{3}x+\\frac{10}{3}$. Let point $K(m, \\frac{2}{3}m+\\frac{10}{3})$ $(m>0)$, given $D(1, 4)$ and $DK=2$, thus $(m-1)^2+\\left(\\frac{2}{3}m+\\frac{10}{3}-4\\right)^2=2^2$, solving this gives $m_1=\\frac{6\\sqrt{13}+13}{13}$, $m_2=\\frac{-6\\sqrt{13}+13}{13}$ (discard), thus $m=\\frac{6\\sqrt{13}+13}{13}$, thus $\\frac{2}{3}m+\\frac{10}{3}=\\frac{2}{3}\\times \\frac{6\\sqrt{13}+13}{13}+\\frac{10}{3}=\\frac{4\\sqrt{13}+52}{13}$, thus $K\\boxed{\\left(\\frac{6\\sqrt{13}+13}{13}, \\frac{4\\sqrt{13}+52}{13}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51433217_217.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{5}m}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a supermarket's storage center has a slope $AB$, with a gradient of $i=1\\colon 2$, the height at its top $AC=4m$, and points B, C are on the same horizontal ground. What is the length of the slope $AB$?\n",
    "completion": "\\textbf{Solution:} Given that the slope ratio is $i=1\\colon 2$ and $AC=4m$, \\\\\ntherefore $i=\\frac{AC}{BC}=\\frac{1}{2}$, \\\\\nthus $BC=2AC=4\\times 2=8m$, \\\\\nIn the right triangle $\\triangle ABC$, according to the Pythagorean theorem, the slope is: \\\\\n$AB=\\sqrt{AC^{2}+BC^{2}}=\\sqrt{4^{2}+8^{2}}=\\boxed{4\\sqrt{5}m}$, \\\\\nhence, the length of the slope $AB$ is $4\\sqrt{5}m$.",
    "image_path": "dataset/mm_math/images/MM_Math/53065573_33.png"
  },
  {
    "solution": "\\boxed{x \\leq 1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given the figure, the graph of the linear function $y_{1}=x+b$ intersects with the graph of the linear function $y_{2}=kx+4$ at point $P(1, 3)$. What is the solution set of the inequality $x+b\\le kx+4$ with respect to $x$?\n",
    "completion": "\\textbf{Solution:} From the graph, we have, when $x \\leq 1$, $x + b \\leq kx + 4$,\n\nwhich means the solution set of the inequality $x + b \\leq kx + 4$ with respect to $x$ is $\\boxed{x \\leq 1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52050253.png"
  },
  {
    "solution": "\\boxed{y=3x+6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $l_1: y = -x - b$ intersects the $x$ and $y$ axes at points $A(6, 0)$ and $B$ respectively. The line $l_2$ passing through point $B$ intersects the negative $x$ axis at point $C$, and $OB:OC = 3:1$. Find the coordinates of points $B$ and $C$, and the function expression of line $BC$.\n",
    "completion": "\\textbf{Solution:} Substitute point $A(6,0)$ into the equation of line $AB$ to get: $0=-6-b$, solving this gives: $b=-6$, $\\therefore$ the equation of line $AB$ is $y=-x+6$, $\\therefore$ the coordinates of point $B$ are: $\\boxed{(0,6)}$. Since $OB:OC=3:1$, $\\therefore$ $OC=2$, $\\therefore$ the coordinates of point $C$ are $\\boxed{(-2,0)}$. Let the equation of line $BC$ be $y=kx+6$, $0=-2k+6$, solving this gives: $k=3$, $\\therefore$ the equation of line $BC$ is: $\\boxed{y=3x+6}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52354930_194.png"
  },
  {
    "solution": "\\boxed{\\left(\\frac{-1-\\sqrt{13}}{2}, \\frac{3+3\\sqrt{13}}{2}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=ax^2+bx+c$ intersects the x-axis at points $A(-1, 0)$ and $B(3, 0)$, and intersects the y-axis at point C, and passes through the point $D(2, -3)$. Points P and Q are moving points on the parabola $y=ax^2+bx+c$. The line OQ intersects the segment BC at point E. When $\\triangle OBE$ is similar to $\\triangle ABC$, what are the coordinates of point Q?\n",
    "completion": "\\textbf{Solution:} Given that $OB=OC=3$, it follows that $\\angle OCB=\\angle OBC=45^\\circ$, \\\\\nsince $\\angle ABC=\\angle OBE$, the triangle $\\triangle OBE$ and $\\triangle ABC$ are similar. This leads to two cases: \\\\\n(1) When $\\angle ACB=\\angle BOQ$, we have $AB=4$, $BC=3\\sqrt{2}$, $AC=\\sqrt{10}$, \\\\\ndraw $AH$ perpendicular to $BC$ at point H, \\\\\nthe area of $\\triangle ABC$ is $\\frac{1}{2}\\times AH\\times BC=\\frac{1}{2}AB\\times OC$, solving for $AH$ gives: $AH=2\\sqrt{2}$, \\\\\nthus $CH=\\sqrt{2}$ \\\\\ntherefore $\\tan\\angle ACB=2$, \\\\\nthe equation of line OQ is: $y=-2+x$... (2), \\\\\nsolving (1) and (2) gives: $x=\\pm \\sqrt{3}$, \\\\\nthus point $\\boxed{Q(\\sqrt{3}, -2\\sqrt{3})}$ or $\\boxed{(−\\sqrt{3}, 2\\sqrt{3})}$; \\\\\n(2) When $\\angle BAC=\\angle BOQ$, \\\\\n$\\tan\\angle BAC=\\frac{OC}{OA}=3=\\tan\\angle BOQ$, \\\\\nthe equation of line OQ is: $y=-3+x$... (3), \\\\\nsolving (1) and (3) gives: $x=\\frac{-1\\pm \\sqrt{13}}{2}$, \\\\\nthus point $\\boxed{Q\\left(\\frac{-1+\\sqrt{13}}{2}, \\frac{-1-\\sqrt{13}}{2}\\right)}$ or $\\boxed{\\left(\\frac{-1-\\sqrt{13}}{2}, \\frac{3+3\\sqrt{13}}{2}\\right)}$; \\\\\nIn summary, the point $\\boxed{Q(\\sqrt{3}, -2\\sqrt{3})}$ or $\\boxed{(−\\sqrt{3}, 2\\sqrt{3})}$ or $\\boxed{\\left(\\frac{-1+\\sqrt{13}}{2}, \\frac{-1-\\sqrt{13}}{2}\\right)}$ or $\\boxed{\\left(\\frac{-1-\\sqrt{13}}{2}, \\frac{3+3\\sqrt{13}}{2}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51284522_194.png"
  },
  {
    "solution": "\\boxed{y=x^{2}-2x+3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=x^{2}+bx+c$ (where $b$ and $c$ are constants) intersects the $x$-axis at points $A(-1,0)$ and $B(3,0)$, and intersects the $y$-axis at point $C$. Let $P$ be a point on the parabola with an abscissa of $m$. Find the equation of this parabola.\n",
    "completion": "\\textbf{Solution}: Since the parabola $y=x^{2}+bx+c$ (where $b$ and $c$ are constants) intersects the x-axis at points $A(-1, 0)$ and $B(3, 0)$,\\\\\nwe have $\\begin{cases} 1-b+c=0 \\\\ 9+3b+c=0 \\end{cases}$,\\\\\nwhich yields $\\begin{cases} b=-2 \\\\ c=3 \\end{cases}$.\\\\\nThus, the equation of the parabola is $\\boxed{y=x^{2}-2x+3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52623662_241.png"
  },
  {
    "solution": "\\boxed{27\\sqrt{2}-36}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in square $ABCD$, where $AB=6$, $E$ is a point on side $BC$ (point $E$ does not coincide with points $B$, $C$), and point $F$ lies on the extension of $CD$, with $BE=DF$. Connect $EF$, intersecting $AD$ and $AC$ at points $M$ and $N$, respectively. It's given that $EF^2=2EM\\cdot FN$; when $\\triangle AMN$ is an isosceles triangle, find the value of $S_{\\triangle AMN}$.\n",
    "completion": "\\textbf{Solution:} As shown in Figure 2, connect $BD$, then $BD\\perp AC$. When $AN=MN$, $\\angle AMN=\\angle CAD=45^\\circ$, at this time $\\angle ANNM=90^\\circ$, i.e., $MN\\perp AC$, $\\therefore BD\\parallel MN$, this situation does not exist. When $AM=MN$, $EF\\perp AD$, this situation also does not exist. As shown in Figure 3, when $AM=AN$, construct $EG\\perp AD$ at $G$, $\\therefore \\angle ANM=\\angle AMN=\\frac{180^\\circ-\\angle MAN}{2}=\\frac{180^\\circ-45^\\circ}{2}=67.5^\\circ$. $\\because AD\\parallel BC$, $\\therefore \\angle CEF=\\angle AMN=67.5^\\circ$. $\\because \\angle AEF=45^\\circ$, $\\therefore \\angle AEB=180^\\circ-45^\\circ-67.5^\\circ=67.5^\\circ$, $\\therefore \\angle AEB=\\angle CEF$. $\\because \\angle ABC=\\angle ADF=90^\\circ$, $\\therefore \\triangle ABE\\sim \\triangle FCE$, $\\therefore \\frac{AB}{CF}=\\frac{BE}{CE}$. Let $BE=CF=a$, $\\therefore CE=6-a$, $CF=6+a$, $\\therefore \\frac{6}{6+a}=\\frac{a}{6-a}$, $\\therefore a_1=6\\sqrt{2}-6$, $a_2=-6\\sqrt{2}-6$ (discard). $\\because \\angle AMN=\\angle AME$, $\\angle MAN=\\angle AEM=45^\\circ$, $\\therefore \\triangle AMN\\sim \\triangle AMA$, $\\therefore \\frac{AM}{AN}=\\frac{EM}{AE}$, $\\frac{S_{\\triangle AMN}}{S_{\\triangle AEM}}=\\left(\\frac{AM}{AE}\\right)^2$. $\\because AM=AN$, $\\therefore AE=EM$, $\\therefore GM=AG=BE$. $\\because \\left(\\frac{BE}{AE}\\right)^2=\\frac{BE^2}{BE^2+AB^2}=\\frac{3-2\\sqrt{2}}{4-2\\sqrt{2}}$, $\\therefore S_{\\triangle EAM}=\\frac{1}{2}AM\\cdot AB=\\frac{1}{2}\\times 6\\times (12\\sqrt{2}-12)=36\\sqrt{2}-36$, $\\therefore \\frac{S_{\\triangle AMN}}{36\\sqrt{2}-36}=\\frac{3-2\\sqrt{2}}{4-2\\sqrt{2}}$, $\\therefore S_{\\triangle AMN}=\\boxed{27\\sqrt{2}-36}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52977916_149.png"
  },
  {
    "solution": "\\boxed{y=\\frac{8}{x}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y=kx+2$ intersects the x-axis and the y-axis at points $A$ and $B$, respectively, and intersects the graph of the inverse proportion function $y=\\frac{n}{x}$ $(n \\neq 0)$ at point $C$ in the first quadrant. If $OA=OB$, and $B$ is the midpoint of segment $AC$, find the analytical expressions of the linear function and the inverse proportion function.\n",
    "completion": "\\textbf{Solution:} Construct $CD\\perp x$-axis at $D$,\\\\\nsince the graph of the linear function $y=kx+2$ intersects the $y$-axis at point $B$,\\\\\nthus $B(0, 2)$,\\\\\ntherefore $OB=2$,\\\\\ntherefore $OA=OB=2$,\\\\\nthus $A\\left(-2, 0\\right)$,\\\\\nsubstituting $A\\left(-2, 0\\right)$ into $y=kx+2$ gives $k=1$,\\\\\ntherefore, the equation of the linear function is $y=x+2$;\\\\\nFurthermore, since $B$ is the midpoint of the segment $AC$, and $OB\\parallel CD$,\\\\\nthus $OB$ is the median of $\\triangle ACD$,\\\\\nthus $OA=OD=2$, $CD=2OB=4$,\\\\\nthus $C(2, 4)$,\\\\\nsince the graph of the inverse proportionality function $y=\\frac{n}{x}(n\\ne 0)$ intersects in the first quadrant at point $C$,\\\\\nthus $n=2\\times 4=8$,\\\\\nthus, the equation of the inverse proportionality function is $y=\\frac{8}{x}$\\\\\nTherefore, the equations of the linear function and the inverse proportionality function are $\\boxed{y=x+2}$ and $\\boxed{y=\\frac{8}{x}}$, respectively;",
    "image_path": "dataset/mm_math/images/MM_Math/53065570_74.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB = 90^\\circ$, and D, E, and F are the midpoints of sides AB, AC, and BC, respectively. If the length of CD is 3, what is the length of EF?\n",
    "completion": "Solution: Since $\\angle ACB=90^\\circ$ and point D is the midpoint of side AB, with CD=3, it follows that AB=2CD=6.\nSince E and F are the midpoints of sides AC and BC, respectively, it follows that EF is the median of $\\triangle ABC$,\nwhich implies EF=$\\frac{1}{2}$AB=$\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50475086.png"
  },
  {
    "solution": "\\boxed{y=-2x+3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=-\\frac{3}{4}x+3$ intersects the x-axis at point A and the y-axis at point B. Triangle $AOB$ is folded along a certain line passing through point B such that point A falls on point D located on the negative y-axis. The fold intersects the x-axis at point C. Find the equation of line BC.\n",
    "completion": "\\textbf{Solution:} Let the expression for line BC be $y=kx+b$,\\\\\n$\\therefore \\begin{cases}3=b\\\\ 0=\\frac{3}{2}k+b\\end{cases}$,\\\\\nsolving yields $\\begin{cases}k=-2\\\\ b=3\\end{cases}$,\\\\\n$\\therefore$ the expression for line BC is $\\boxed{y=-2x+3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51894652_192.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In $\\triangle ABC$, $AD$ is the angle bisector, and points $E$ and $F$ are the midpoints of line segments $AC$ and $CD$, respectively. If the areas of $\\triangle ABD$ and $\\triangle EFC$ are 21 and 7, respectively, what is the value of $\\frac{AB}{AC}$?\n",
    "completion": "\\textbf{Solution:} Construct $DM \\perp AB$ at $M$, and $DN \\perp AC$ at $N$.\\\\\nSince the point F is the midpoint of DC,\\\\\nit follows that $S_{\\triangle DEF} = S_{\\triangle FCE} = 7$,\\\\\nSince E is the midpoint of AC,\\\\\nit follows that $S_{\\triangle ADE} = S_{\\triangle CDE} = 14 + 7 = 21$,\\\\\nTherefore, $S_{\\triangle ADC} = 28$,\\\\\nSince AD is the angle bisector, $DM \\perp AB$, and $DN \\perp AC$,\\\\\nit follows that $DM = DN$,\\\\\nSince $S_{\\triangle ABD} : S_{\\triangle ADC} = \\frac{1}{2} AB \\cdot DM : (\\frac{1}{2} \\cdot AC \\cdot DN) = 21:28$\\\\\nTherefore, $\\frac{AB}{AC} = \\boxed{\\frac{3}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51367320.png"
  },
  {
    "solution": "\\boxed{\\frac{16}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, quadrilateral $ABCD$ is a rhombus, and point $E$ is on the extension line of $AC$, $\\angle ACD = \\angle ABE$. $\\triangle ABC \\sim \\triangle AEB$; when $AD = 4$ and $AC = 3$, what is the length of $AE$?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABC\\sim \\triangle AEB$, \\\\\nit follows that $\\frac{AB}{AE}=\\frac{AC}{AB}$, \\\\\nSince $ABCD$ is a rhombus, we have $AB=AD=4$, $AC=3$, \\\\\nthus $\\frac{4}{AE}=\\frac{3}{4}$, \\\\\nhence $AE=\\boxed{\\frac{16}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52866279_128.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $ \\triangle ABC$ is an inscribed triangle of $ \\odot O$ with $ \\angle C=30^\\circ$ and the radius of $ \\odot O$ is $ 2\\,cm$. If point P is a point on $ \\odot O$ and $ PB=AB$, what is the length of $ PA$?\n",
    "completion": "\\textbf{Solution:} Connect $OA$ and $OP$, extend $OB$ to intersect $AP$ at $H$,\\\\\nby the Inscribed Angle Theorem, $\\angle AOB=2\\angle C=60^\\circ$,\\\\\nsince $PB=AB$,\\\\\nthus $\\angle POB=\\angle AOB=60^\\circ$, and $OB\\perp AP$, hence $\\angle OAP=30^\\circ$, therefore $OH=\\frac{1}{2}AO=1$,\\\\\nthen $AH=PH=\\sqrt{AO^2-AH^2}=\\sqrt{3}\\ (\\text{cm})$,\\\\\nthus $PA=2AH=2\\sqrt{3}\\ (\\text{cm})$,\\\\\ntherefore, the answer is: $PA=\\boxed{2\\sqrt{3}}\\ (\\text{cm})$.",
    "image_path": "dataset/mm_math/images/MM_Math/53102104.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the line $y=-\\frac{4}{3}x+8$ intersects the x-axis at point A and the y-axis at point B. A moving point C starts from point B, moving towards point A at a uniform speed of 1 unit per second along direction $BA$. Simultaneously, a moving point D starts from point A, moving towards point O at a uniform speed of 2 units per second along direction $AO$. When one point stops moving, the other point stops as well. Let the time of movement be t seconds ($t>0$). Directly state the value of t when the area of $\\triangle ACD$ is $\\frac{84}{5}$.\n",
    "completion": "\\textbf{Solution:} We have $t=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52917352_80.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that point $C$ is a point on $AB$, $AC=30\\,cm$, $BC=\\frac{2}{5}AC$, and $D$, $E$ are the midpoints of $AC$, $AB$ respectively. What is the length of $DE$ in $cm$?\n",
    "completion": "\\textbf{Solution:} Given that $AC=30\\,cm$, $BC=\\frac{2}{5}AC$,\\\\\nhence $BC=\\frac{2}{5}\\times 30=12\\,cm$, $AB=AC+BC=42\\,cm$,\\\\\nalso, since D and E are the midpoints of $AC$ and $AB$, respectively,\\\\\nthus $AD=DC=\\frac{1}{2}AC=15\\,cm$, $AE=BE=\\frac{1}{2}AB=21\\,cm$\\\\\ntherefore $DE=AE-AD=21-15=\\boxed{6}\\,cm$",
    "image_path": "dataset/mm_math/images/MM_Math/55532993_53.png"
  },
  {
    "solution": "\\boxed{55^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, points E and F are on sides BC and AD respectively, and $BE=DF$. Quadrilateral $AECF$ is a parallelogram; If $AE$ bisects $\\angle BAD$ and $\\angle B=70^\\circ$, what is the measure of $\\angle DFC$?\n",
    "completion": "\\textbf{Solution:} Since AD $\\parallel$ BC and $\\angle B=70^\\circ$, \\\\\nit follows that $\\angle B+\\angle BAD=180^\\circ$, \\\\\nhence $\\angle BAD=110^\\circ$, \\\\\nmoreover, since AE bisects $\\angle BAD$, \\\\\nit follows that $\\angle BAE=\\angle EAD=55^\\circ$, \\\\\nfrom (1) we have: quadrilateral AECF is a parallelogram, \\\\\nthus AE $\\parallel$ FC, \\\\\ntherefore, $\\angle DFC=\\angle DAE=55^\\circ$, \\\\\nhence, $\\angle DFC=\\boxed{55^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50497059_79.png"
  },
  {
    "solution": "\\boxed{6.7}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, a billboard CD is erected at the top of a building. Xiaoming and his classmates measure the angle of elevation to the bottom of the billboard, D, to be $53^\\circ$ from point A at the base of the slope. They then walk up the slope along AB to point B and measure the angle of elevation to the top of the billboard, C, to be $45^\\circ$. It is known that the slope gradient of AB is $i=1\\colon \\sqrt{3}$, $AB=10$ meters, and $AE=21$ meters. (The height of the protractor is negligible, and the result is accurate to 0.1 meters. Reference data: $\\sqrt{2}\\approx1.41$, $\\sqrt{3}\\approx1.73$, $\\sin53^\\circ\\approx\\frac{4}{5}$, $\\cos53^\\circ\\approx\\frac{3}{5}$, $\\tan53^\\circ\\approx\\frac{4}{3}$) What is the height of the billboard CD in meters?\n",
    "completion": "\\textbf{Solution:}\\\\\nSince $BM \\perp AE$, $BN \\perp CE$, and $CE \\perp AE$,\\\\\nit follows that quadrilateral $BMEN$ is a rectangle,\\\\\ntherefore, $NE = BM = 5$ meters, and $BN = ME$,\\\\\nIn $\\triangle ABM$, $\\angle BAM = 30^\\circ$,\\\\\ntherefore, $AM = AB \\cdot \\cos30^\\circ = \\frac{\\sqrt{3}}{2}AB = 5\\sqrt{3}$ (meters),\\\\\nthus, $ME = AM + AE = 5\\sqrt{3} + 21$ (meters) = $BN$,\\\\\nSince $\\angle CBN = 45^\\circ$,\\\\\nit follows that $CN = BN = 5\\sqrt{3} + 21$ meters,\\\\\nthus, $CE = CN + NE = 5\\sqrt{3} + 26$ meters,\\\\\nIn $\\triangle ADE$, $\\angle DAE = 53^\\circ$, and $AE = 21$ meters,\\\\\ntherefore, $DE = AE \\cdot \\tan53^\\circ \\approx 21 \\times \\frac{4}{3} = 28$ (meters),\\\\\nthus, $CD = CE - DE = 5\\sqrt{3} + 26 - 28 = 5\\sqrt{3} - 2 \\approx \\boxed{6.7}$ (meters),\\\\\ni.e., the height of the billboard $CD$ is approximately $6.7$ meters.",
    "image_path": "dataset/mm_math/images/MM_Math/52601742_133.png"
  },
  {
    "solution": "\\boxed{y=-\\sqrt{3}x+\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $l\\colon y = \\sqrt{3}x + \\sqrt{3}$ intersects the two coordinate axes at points $A$ and $B$, respectively, and point $M$ is the midpoint of line segment $AB$. The line $l'$ is symmetrical to line $l$ with respect to the $y$-axis. Write the equation of line $l'$.\n",
    "completion": "\\textbf{Solution:} Since line $l'$ is symmetric to line $l$ with respect to the $y$-axis, let line $l'$ intersect the $x$-axis at point $C$. Thus, point $C$ is symmetric to point $B$ with respect to the $y$-axis, that is $C(1, 0)$,\\\\\nSince line $l'$ passes through point $A(0, \\sqrt{3})$, we can set the equation of line $l'$ to be $y=kx+\\sqrt{3}$,\\\\\nGiven that line $l'$ passes through point $C(1, 0)$, $\\therefore 0=k\\times 1+\\sqrt{3}$, $\\therefore k=-\\sqrt{3}$,\\\\\nthus, the equation of the function is: \\boxed{y=-\\sqrt{3}x+\\sqrt{3}}.",
    "image_path": "dataset/mm_math/images/MM_Math/52690944_61.png"
  },
  {
    "solution": "\\boxed{3\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the parabola $y=\\frac{k}{8}(x-2)(x+4)$ (where $k$ is a constant and $k>0$) intersects the $x$-axis at points $A$ and $B$ from left to right, respectively, and intersects the $y$-axis at point $C$. Point $D$ lies on the graph of the parabola in the first quadrant, and $\\angle BAD=30^\\circ$. Under these conditions, let $F$ be a point on segment $AD$ (excluding the endpoints), and connect $BF$. A moving point $M$ starts from point $B$, moves along segment $BF$ at a speed of 1 unit per second to point $F$, then moves along segment $FD$ at a speed of 2 units per second to point $D$ and stops. When the coordinates of point $F$ are what values, will point $M$ spend the least time in the entire movement process, and what is the minimum time spent?\n",
    "completion": "\\textbf{Solution:} From (1), we know $D(5,3\\sqrt{3})$. As shown in the diagram, draw $DN\\perp x$-axis at point N, then $DN=3\\sqrt{3}$, $ON=5$, $AN=4+5=9$. Draw $DK\\parallel x$-axis through point D, then $\\angle KDF=\\angle DAB=30^\\circ$. Draw $FG\\perp DK$ at point G through point F, then $FG=\\frac{1}{2}DF$. According to the problem statement, the path of the moving point M is the polyline $BF+DF$, with the movement time: $t=BF+\\frac{1}{2}DF$, $\\therefore t=BF+FG$, that is, the time value of the movement is equal to the length value of the polyline $BF+FG$. It is known from the shortest vertical segment that the minimum length of the polyline $BF+FG$ is the vertical segment between $DK$ and the $x$-axis. Draw $BH\\perp DK$ at point B through point H, then $t_{\\text{minimum}}=BH$, and the intersection of line segment $BH$ with line $AD$ is the desired point F. Suppose the equation of line $AD$ is $y=kx+b$, substituting $(-4,0)$ and $(5,3\\sqrt{3})$, we get\n\\[\n\\begin{cases}\n-4k+b=0\\\\\n5k+b=3\\sqrt{3}\n\\end{cases}\n\\]\nSolving, we find\n\\[\n\\begin{cases}\nk=\\frac{\\sqrt{3}}{3}\\\\\nb=\\frac{4\\sqrt{3}}{3}\n\\end{cases}\n\\]\n$\\therefore$ The equation of line $AD$ is: $y=\\frac{\\sqrt{3}}{3}x+\\frac{4\\sqrt{3}}{3}$, $\\because$ the $x$-coordinate of point B is 2, $\\therefore y=\\frac{2\\sqrt{3}}{3}+\\frac{4\\sqrt{3}}{3}=2\\sqrt{3}$, $\\therefore F(2,2\\sqrt{3})$. $\\therefore HF=BH-BF=3\\sqrt{3}-2\\sqrt{3}=\\sqrt{3}$, $\\therefore FD=2\\sqrt{3}$.\n\nIn summary, when the coordinates of point F are $(2,2\\sqrt{3})$, point M takes the least amount of time in the entire motion, which is $2\\sqrt{3}\\div 1+2\\sqrt{3}\\div 2=2\\sqrt{3}+\\sqrt{3}=\\boxed{3\\sqrt{3}}$ (seconds).",
    "image_path": "dataset/mm_math/images/MM_Math/50890989_140.png"
  },
  {
    "solution": "\\boxed{y=2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the line $y_{1}=2x-2$ intersects the y-axis at point A, and the graph of the line $y_{2}=-2x+6$ intersects the y-axis at point B. The two lines intersect at point C. What is the solution to the system of equations $\\begin{cases}2x-y=2, \\\\ 2x+y=6?\\end{cases}$\n",
    "completion": "\\textbf{Solution:} Given the system of equations $\\begin{cases}2x-y=2, \\\\ 2x+y=6\\end{cases}$, we find the solution $x=2$, $y=2$.\n\n$\\therefore$ The solution to the system is $x=\\boxed{2}$, $\\boxed{y=2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52423206_69.png"
  },
  {
    "solution": "\\boxed{36^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the regular pentagon $ABCDE$, connect $AD$. What is the measure of $\\angle 1$?\n",
    "completion": "\\textbf{Solution:} From the polygon interior angle sum theorem, we can deduce that the sum of the interior angles of a pentagon is $(5-2)\\times 180^\\circ=540^\\circ$,\\\\\n$\\therefore$ $\\angle AED=540^\\circ\\div 5=108^\\circ$,\\\\\n$\\because$ in a regular pentagon $AE=DE$,\\\\\n$\\therefore$ $\\angle 1=\\angle EDA=\\left(180^\\circ-108^\\circ\\right)\\div 2=\\boxed{36^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/52837422.png"
  },
  {
    "solution": "\\boxed{\\frac{49}{13}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, we have $AB=AC$. Point $D$ is inside $\\triangle ABC$, and $AD$ bisects $\\angle BAC$. Connect $CD$, and fold $\\triangle ADC$ along $CD$ such that $AC$ lands on $CE$ and intersects $AB$ at $F$, where $CE\\perp AB$. If $BC=10$ and $AD=7$, what is the length of $EF$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, extend AD to intersect BC at point G, \\\\\n\\[\n\\because AB = AC, \\text{ and } AD \\text{ bisects } \\angle BAC,\n\\]\n\\[\n\\therefore \\angle B = \\angle ACB, \\text{ and } AG \\perp BC, \\text{ thus } BG = CG,\n\\]\n\\[\n\\because BC = 10,\n\\]\n\\[\n\\therefore BG = CG = \\frac{1}{2}BC = 5,\n\\]\n\\[\n\\because CE \\perp AB,\n\\]\n\\[\n\\therefore \\angle BFC = 90^\\circ,\n\\]\n\\[\n\\therefore \\angle B + \\angle BCF = 90^\\circ,\n\\]\n\\[\n\\because \\angle B = \\angle C = \\angle BCF + \\angle ACD + \\angle ECD,\n\\]\n\\[\n\\therefore 2\\angle BCF + \\angle ACD + \\angle ECD = 90^\\circ,\n\\]\nAccording to the property of folding, we know that \\(\\angle ACD = \\angle ECD\\), and \\(AC = EC\\),\n\\[\n\\therefore \\angle BCF + \\angle ECD = 45^\\circ,\n\\]\n\\[\n\\therefore \\triangle CDG is an isosceles right triangle,\n\\]\n\\[\n\\therefore AG = AD + DG = 7 + 5 = 12,\n\\]\nIn the right triangle \\(\\triangle ACG\\), \\(AC = \\sqrt{CG^{2}+AG^{2}} = \\sqrt{5^{2}+12^{2}} = 13\\)\n\\[\n\\therefore CE = AB = AC = 13,\n\\]\n\\[\n\\because S_{\\triangle ABC} = \\frac{1}{2}BC \\cdot AG = \\frac{1}{2}AB \\cdot CF,\n\\]\n\\[\n\\therefore \\frac{1}{2} \\times 10 \\times 12 = \\frac{1}{2} \\times 13 \\times CF,\n\\]\n\\[\n\\therefore CF = \\frac{120}{13},\n\\]\n\\[\n\\therefore EF = CE - CF = 13 - \\frac{120}{13} = \\boxed{\\frac{49}{13}}.\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/52874486.png"
  },
  {
    "solution": "\\boxed{\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given in the Cartesian coordinate system $xOy$ (as shown in the figure), the line $y=2x+2$, intersects the x-axis and y-axis at points A and B respectively, and the coordinates of point C are $(3, 2)$. Connecting AC and intersecting the y-axis at point D. Find the length of segment AB.\n",
    "completion": "\\textbf{Solution:} Let $x=0$, then $y=2$,\\\\\n$\\therefore B(0,2)$,\\\\\n$\\therefore OB=2$,\\\\\nLet $y=0$, then $x=-1$,\\\\\n$\\therefore A(-1,0)$,\\\\\n$\\therefore OA=1$,\\\\\n$\\therefore AB=\\boxed{\\sqrt{5}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52977945_46.png"
  },
  {
    "solution": "\\boxed{\\frac{25\\sqrt{13}}{13}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within square $ABCD$, point $E$ is on side $AD$. Triangle $ABE$ is folded along line $BE$ to obtain triangle $FBE$. Connect $CF$ and extend it to intersect the extension of $BE$ at point $P$. If $AB=5$ and $AE=1$, what is the measure of $\\angle P$? What is the length of $PC$?\n",
    "completion": "\\textbf{Solution:} Let a perpendicular be drawn from B to CP at foot N, and from C to BP at foot M, thus CM$\\perp$BP.\\\\\n$\\because$ $\\triangle ABE$ is folded along line BE to obtain $\\triangle FBE$,\\\\\n$\\therefore$ $Rt\\triangle ABE\\cong Rt\\triangle FEB$,\\\\\n$\\therefore$ $BF=AB=5$, $\\angle ABE=\\angle PBF$,\\\\\n$\\because$ Quadrilateral ABCD is a square,\\\\\n$\\therefore$ $BC=5$,\\\\\n$\\therefore$ $BC=BF$,\\\\\n$\\therefore$ $\\triangle BFC$ is an isosceles triangle,\\\\\nAlso, $\\because$ $BN\\perp FC$,\\\\\n$\\therefore$ $BN$ bisects $\\angle FBC$,\\\\\n$\\therefore$ $\\angle PBN=\\angle PBF+\\angle FBN=\\frac{1}{2}\\left(\\angle ABF+\\angle FBC\\right)=45^\\circ$,\\\\\nIn $Rt\\triangle BPN$, $\\angle PBN=45^\\circ$, $\\angle BNP=90^\\circ$,\\\\\n$\\therefore$ $\\angle P=45^\\circ$,\\\\\n$\\therefore$ $\\triangle CMP$ is an isosceles right triangle,\\\\\n$\\therefore$ $CM=MP$,\\\\\n$\\because$ $\\angle ABP+\\angle PBC=90^\\circ$, $\\angle PBC+\\angle MCB=90^\\circ$, $\\therefore$ $\\angle ABP=\\angle MCB$,\\\\\n$\\therefore$ $\\triangle EAB\\sim \\triangle BMC$,\\\\\n$\\therefore$ $\\frac{CM}{AB}=\\frac{CB}{BE}$,\\\\\nIn $Rt\\triangle ABE$, $BE=\\sqrt{AB^2+AE^2}=\\sqrt{26}$,\\\\\n$\\therefore$ $\\frac{CM}{5}=\\frac{5}{\\sqrt{26}}$, $CM=\\frac{25}{\\sqrt{26}}$,\\\\\n$\\therefore$ In $Rt\\triangle CMP$, $PM=CM=\\frac{25}{\\sqrt{26}}$,\\\\\n$\\therefore$ $CP=\\sqrt{CM^2+PM^2}=\\boxed{\\frac{25\\sqrt{13}}{13}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52991441.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, seven squares are arranged such that each pair of adjacent squares shares a common vertex, and the area of each square is represented by numeric or alphabetic characters. What is the value of $S_{1} + S_{2} + S_{3} + S_{4}$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, \\\\\nsince AB=BE, $\\angle$ACB=$\\angle$BDE=$90^\\circ$, \\\\\ntherefore, $\\angle$ABC+$\\angle$BAC=$90^\\circ$, $\\angle$ABC+$\\angle$EBD=$90^\\circ$, \\\\\nthus, $\\angle$BAC=$\\angle$EBD, \\\\\nhence, $\\triangle ABC\\cong \\triangle BDE$ (AAS), \\\\\ntherefore, BC=ED, \\\\\nsince $ AB^{2}=AC^{2}+BC^{2}$, \\\\\nthus $ AB^{2}=AC^{2}+ED^{2}=S_{1}+S_{2}$, \\\\\nnamely $ S_{1}+S_{2}=1$, \\\\\nsimilarly, $ S_{3}+S_{4}=3$.\\\\\nThen $ S_{1}+S_{2}+S_{3}+S_{4}=1+3=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53538155.png"
  },
  {
    "solution": "\\boxed{A(2,0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, two straight lines $l_{1}$ and $l_{2}$ passing through point A intersect the y-axis at points B and C, respectively, where point B is above the origin, and point C is below the origin. Given that AB = $\\sqrt{13}$ and B(0, 3), what are the coordinates of point A?\n",
    "completion": "\\textbf{Solution}: Since $B(0,3)$,\\\\\nit follows that $OB=3$,\\\\\nIn $\\triangle AOB$, $OA= \\sqrt{AB^{2}-OB^{2}}=\\sqrt{13-3^{2}}=2$,\\\\\nthus \\boxed{A(2,0)}.",
    "image_path": "dataset/mm_math/images/MM_Math/52096283_100.png"
  },
  {
    "solution": "\\boxed{7\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC=13$, and $BC=24$. Point $D$ is on $BC$ ($BD > CD$). $\\triangle AED$ and $\\triangle ACD$ are symmetric about line $AD$, with point $C$ being symmetric to point $E$. $AE$ intersects $BC$ at point $F$. $BE$ and $CE$ are connected. When $DE \\perp BC$, what is the degree measure of $\\angle ADE$? What is the length of $CE$?\n",
    "completion": "\\textbf{Solution:} Construct $AH \\perp BC$ at $H$, \\\\\nsince $AB=AC=13$, $BC=24$, \\\\\nthus $BH=CH=12$, \\\\\nthus $AH= \\sqrt{AB^{2}-BH^{2}} =5$, \\\\\nsince $\\triangle AED$ and $\\triangle ACD$ are symmetrical about the line $AD$, \\\\\nthus $\\angle ADC=\\angle ADE$, $CD=DE$, \\\\\nsince $DE\\perp BC$, \\\\\nthus $\\angle BDE=90^\\circ$, \\\\\nthus $\\angle ADE=90^\\circ+\\angle ADB=\\angle ADC$, \\\\\nthus $90^\\circ+\\angle ADB=180^\\circ-\\angle ADB$, \\\\\nthus $\\angle ADB=45^\\circ$, \\\\\nsince $\\angle AHC=90^\\circ$, \\\\\nthus $\\angle ADB=\\angle HAD=45^\\circ$, \\\\\nthus $AH=HD=5$, $\\angle ADE=\\angle ADB+\\angle BDE=\\boxed{135^\\circ}$, \\\\\nthus $BD=12+5=17$, \\\\\nthus $CD=DE=24-17=7$, \\\\\nthus $CE= \\sqrt{CD^{2}+DE^{2}} =7\\sqrt{2}$, \\\\\nHence, the answer is: $\\boxed{135^\\circ}$, $\\boxed{7\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51419578.png"
  },
  {
    "solution": "\\boxed{(0,4)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y = \\frac{4}{3}x + 4$ intersects the x-axis and the y-axis at points A and B, respectively. What are the coordinates of points A and B?\n",
    "completion": "\\textbf{Solution:} For $y=\\frac{4}{3}x+4$, setting $y=0$ gives $y=\\frac{4}{3}x+4=0$. Solving this, we find $x=-3$. Setting $x=0$, we get $y=4$. Therefore, the coordinates of points A and B are \\boxed{(-3,0)} and \\boxed{(0,4)}, respectively.",
    "image_path": "dataset/mm_math/images/MM_Math/52978850_129.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, all vertices of the shape are on the grid points of a $3 \\times 3$ square grid. Then, what is the sum of $\\angle 1+\\angle 2$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, \\\\\nfrom the given conditions, in right triangles $\\triangle ABC$ and $\\triangle EFC$, \\\\\n$\\because \\left\\{\\begin{array}{l}AB=EF\\\\ \\angle B=\\angle EFC=90^\\circ\\\\ BC=FC\\end{array}\\right.$  \\\\\n$\\therefore$ right triangles $\\triangle ABC\\cong \\triangle EFC$ (SAS) \\\\\n$\\therefore \\angle 3=\\angle 1$ \\\\\n$\\because \\angle 2+\\angle 3=45^\\circ$ \\\\\n$\\therefore \\angle 1+\\angle 2=\\angle 3+\\angle 2=\\boxed{45^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/51534188.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $CD$ is the bisector of $\\angle ACB$. Extend $CD$ to point $E$ such that $DE = \\frac{1}{2}CD$. Connect $BE$. If $AC = 2BC$ and the area of $\\triangle BDE$ is $1$, what is the area of $\\triangle ABC$?\n",
    "completion": "\\[\n\\textbf{Solution:} \\text{As shown in the figure, draw } DG \\perp AC \\text{ at } G, \\text{ and } DH \\perp CB \\text{ at } H, \\\\\n\\because DE = \\frac{1}{2}CD, \\text{ and the area of } \\triangle BDE \\text{ is 1,} \\\\\n\\therefore S\\triangle BCD = 2S\\triangle BDE = 2, \\\\\n\\because CD \\text{ is the angle bisector of } \\angle ACB, DH \\perp CB, \\text{ and } DG \\perp AC, \\\\\n\\therefore DG = DH, \\\\\n\\because AC = 2BC, S_{\\triangle ACD} = \\frac{1}{2}AC \\cdot GD, S_{\\triangle BCD} = \\frac{1}{2}BC \\cdot DH, \\\\\n\\therefore S\\triangle ACD = 2S\\triangle BCD, \\\\\n\\therefore S\\triangle ACD = 4, \\\\\n\\therefore S\\triangle ABC = S\\triangle ACD + S\\triangle BCD = 4 + 2 = \\boxed{6}.\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/52836748.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, $\\angle B=30^\\circ$, point $D$ is on the extension of $BA$, $\\angle CDB=45^\\circ$, point $M$ is on side $BC$, and quadrilateral $ACMN$ is a square. What is the value of $\\frac{BM}{AD}$?\n",
    "completion": "\\textbf{Solution:} As shown in Figure 1, draw $CQ\\perp AB$ at point $Q$ passing through point $C$,\\\\\nthen $\\angle AQC=\\angle BQC=90^\\circ$,\\\\\nlet $AQ=x$,\\\\\nsince $\\angle B=30^\\circ$,\\\\\nit follows that $BC=2CQ$,\\\\\nin $\\triangle ACQ$, $\\angle CAQ=60^\\circ$,\\\\\nthus $\\angle ACQ=90^\\circ-\\angle CAQ=30^\\circ$,\\\\\ntherefore $AC=2AQ=2x$,\\\\\ntherefore $CQ=\\sqrt{AC^{2}-AQ^{2}}=\\sqrt{(2x)^{2}-x^{2}}=\\sqrt{3}x$,\\\\\ntherefore $BC=2\\sqrt{3}x$,\\\\\nsince quadrilateral $ACMN$ is a square,\\\\\ntherefore $CM=AC=2x$,\\\\\ntherefore $BM=BC-CM=(2\\sqrt{3}-2)x$,\\\\\nsince $\\angle DQC=90^\\circ$, $\\angle BDC=45^\\circ$,\\\\\ntherefore $\\triangle CDQ$ is an isosceles right triangle,\\\\\ntherefore $DQ=CQ=\\sqrt{3}x$,\\\\\ntherefore $AD=DQ-AQ=(\\sqrt{3}-1)x$,\\\\\ntherefore $\\frac{BM}{AD}=\\frac{(2\\sqrt{3}-2)x}{(\\sqrt{3}-1)x}=\\boxed{2}$",
    "image_path": "dataset/mm_math/images/MM_Math/53182827_161.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $AD=8$, $AB=4$, and $E$, $F$ are points on sides $AD$ and $BC$, respectively (points $E$, $F$ do not coincide with the vertices). If the rectangle is folded along line $EF$, such that point $B$ coincides with point $D$, and the corresponding point of point $A$ is point $G$, then what is the length of line segment $EF$?\n",
    "completion": "\\textbf{Solution:} Construct line EM $\\perp$ BC at point M,\\\\\nsince quadrilateral ABCD is a rectangle,\\\\\ntherefore $\\angle$A=$\\angle$B=$\\angle$C=$\\angle$ADC=$90^\\circ$, CD=AB=4, AD$\\parallel$BC, BC=AD=8,\\\\\nsince EM$\\perp$BC,\\\\\ntherefore $\\angle$BME=$\\angle$A=$\\angle$B=$\\angle$C=$\\angle$ADC=$90^\\circ$,\\\\\ntherefore quadrilaterals ABME and CDEM are both rectangles,\\\\\ntherefore EM=CD=4, CM=DE,\\\\\nsince the rectangle is folded along line $EF$, point $B$ coincides exactly with point $D$, and the corresponding point of $A$ is point $G$,\\\\\ntherefore DF=BF, $\\angle$BFE=$\\angle$DFE,\\\\\ntherefore CF=BC−BF=8−BF=8−DF,\\\\\nIn right triangle $\\triangle$CDF, $CF^{2}+CD^{2}=DF^{2}$,\\\\\ntherefore $(8−DF)^{2}$+$4^{2}$=$DF^{2}$, solving gives: DF=5,\\\\\ntherefore CF=3,\\\\\nsince AD$\\parallel$BC,\\\\\ntherefore $\\angle$DEF=$\\angle$BFE,\\\\\ntherefore $\\angle$DEF=$\\angle$DFE,\\\\\ntherefore DE=DF=5,\\\\\ntherefore CM=DE=5,\\\\\ntherefore MF=CM−CF=5−3=2,\\\\\nIn right triangle $\\triangle$EFM, by the Pythagorean theorem,\\\\\n$EF=\\sqrt{EM^{2}+MF^{2}}=\\sqrt{4^{2}+2^{2}}=\\boxed{2\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52386290.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $PA$ and $PB$ are tangent to the circle $\\odot O$ at points $A$ and $B$ respectively. Point $C$ lies on circle $\\odot O$, and lines $AC$ and $BC$ are connected. If $\\angle ACB=60^\\circ$, what is the degree measure of $\\angle P$?\n",
    "completion": "\\textbf{Solution:} Connect OA, OB,\\\\\nsince $\\angle ACB=60^\\circ$,\\\\\nthus $\\angle AOB=2\\angle ACB =120^\\circ$,\\\\\nalso since PA and PB are tangent to the circle O at points A and B respectively,\\\\\nthus $\\angle PAO=\\angle PBO=90^\\circ$,\\\\\ntherefore $\\angle P=360^\\circ-\\angle PAO-\\angle PBO-\\angle AOB=\\boxed{60^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51261683.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the parabola $y=x^2+x-2$ intersects the x-axis at points A and B, and intersects the y-axis at point C. Find the area of $\\triangle ABC$.\n",
    "completion": "\\textbf{Solution:} Let $y=0$, which gives ${x}^{2}+x-2=0$. Solving this, ${x}_{1}=-2$, ${x}_{2}=1$,\\\\\n$\\therefore$ points $A(-2, 0)$, $B(1, 0)$,\\\\\n$\\therefore$ $AB=3$,\\\\\nLet $x=0$, which means $y=-2$,\\\\\n$\\therefore$ point $C(0, -2)$,\\\\\n$\\therefore$ $OC=2$,\\\\\n$\\therefore$ the area of $\\triangle ABC=\\frac{1}{2}\\times AB\\times OC=\\frac{1}{2}\\times 3\\times 2=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52747348_69.png"
  },
  {
    "solution": "\\boxed{70^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Point O is a point on line AB. A ray OC is drawn through point O such that $\\angle BOC = 65^\\circ$. Place the right-angled vertex of a right-angled triangle ruler at point O. When the triangle ruler MON is rotated counterclockwise around point O to the position shown in figure (3), $\\angle NOC = \\frac{1}{4}\\angle AOM$, what is the degree measure of $\\angle NOB$?\n",
    "completion": "\\textbf{Solution:} According to the given problem, \\\\\n$\\because \\angle NOC = \\frac{1}{4}\\angle AOM$, \\\\\n$\\therefore \\angle AOM = 4\\angle NOC$, \\\\\n$\\because \\angle BOC = 65^\\circ$, \\\\\n$\\therefore \\angle AOC = \\angle AOB - \\angle BOC = 180^\\circ - 65^\\circ = 115^\\circ$, \\\\\n$\\because \\angle MON = 90^\\circ$, \\\\\n$\\therefore \\angle AOM + \\angle NOC = \\angle AOC - \\angle MON = 115^\\circ - 90^\\circ = 25^\\circ$, \\\\\n$\\therefore 4\\angle NOC + \\angle NOC = 25^\\circ$, \\\\\n$\\therefore \\angle NOC = 5^\\circ$, \\\\\n$\\therefore \\angle NOB = \\angle NOC + \\angle BOC = \\boxed{70^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52169374_194.png"
  },
  {
    "solution": "\\boxed{x\\ge -2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the position of the line $y=kx+b$ in the Cartesian coordinate system is illustrated. What is the solution set for the inequality $kx+b\\le 2$?\n",
    "completion": "\\textbf{Solution:} According to the graph, the line $y=kx+b$ passes through the points (0,1) and (2,0),\\\\\nSubstituting these two points into $y=kx+b$ yields $\\left\\{\\begin{array}{l}b=1\\\\ 2k+b=0\\end{array}\\right.$, \\\\\nSolving this system gives $\\left\\{\\begin{array}{l}b=1\\\\ k=-\\frac{1}{2}\\end{array}\\right.$, \\\\\n$\\therefore$ the equation of the line is: $y=-\\frac{1}{2}x+1$, \\\\\nSubstituting y=2 we get $2=-\\frac{1}{2}x+1$, \\\\\nThis gives x=$-2$, \\\\\n$\\therefore$ the solution set of the inequality $kx+b\\le 2$ is $\\boxed{x\\ge -2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50482131.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, within circle $O$, $OC \\perp AB$ at point $C$. If the radius of circle $O$ is $10$ and $AB=16$, then what is the length of $OC$?\n",
    "completion": "\\textbf{Solution:} Connect OB, \\\\\n$\\because$OC$\\perp$AB, \\\\\n$\\therefore \\angle$OCB$=90^\\circ$, BC$= \\frac{1}{2}$AB$= \\frac{1}{2} \\times 16=8$, \\\\\n$\\therefore OC=\\sqrt{OB^{2}-BC^{2}}=\\sqrt{10^{2}-8^{2}}=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53081373.png"
  },
  {
    "solution": "\\boxed{\\frac{2}{3} \\pi - \\frac{\\sqrt{3}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within circle $\\odot O$, through the midpoint $C$ of radius $OD$, construct $AB\\perp OD$ intersecting $\\odot O$ at points $A$ and $B$, and $AB=2\\sqrt{3}$. What is the area of the shaded region?\n",
    "completion": "\\[\n\\textbf{Solution:} \\text{The area of the shaded part} = S_{\\text{sector } BOD} - S_{\\triangle COB} = \\frac{60^\\circ \\times \\pi \\times 2^2}{360^\\circ} - \\frac{1}{2} \\times \\sqrt{3} \\times 1 = \\boxed{\\frac{2}{3} \\pi - \\frac{\\sqrt{3}}{2}}.\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/52754810_77.png"
  },
  {
    "solution": "\\boxed{20^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $O$ is a point on line $AB$, $\\angle COD$ is a right angle, and $OE$ bisects the obtuse angle $\\angle BOC$. If $\\angle AOC=40^\\circ$, what is the degree measure of $\\angle DOE$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle AOC = 40^\\circ$, \\\\\nthus $\\angle BOC = 180^\\circ - \\angle AOC = 140^\\circ$, \\\\\nsince $\\angle COD$ is a right angle, \\\\\nthus $\\angle COD = 90^\\circ$, \\\\\ntherefore $\\angle BOD = \\angle BOC - \\angle COD = 140^\\circ - 90^\\circ = 50^\\circ$, \\\\\nsince $OE$ bisects $\\angle BOC$, \\\\\nthus $\\angle BOE = \\frac{1}{2} \\angle BOC = 70^\\circ$, \\\\\ntherefore $\\angle DOE = \\angle BOE - \\angle BOD = 70^\\circ - 50^\\circ = \\boxed{20^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53119812_120.png"
  },
  {
    "solution": "\\boxed{32^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point \\(P\\) is a point on the side \\(AB\\) of the square \\(ABCD\\) (not coinciding with points \\(A\\) or \\(B\\)), connect \\(PD\\) and rotate the line segment \\(PD\\) \\(90^\\circ\\) clockwise around point \\(P\\) to get the line segment \\(PE\\), which intersects side \\(BC\\) at point \\(F\\), connect \\(BE\\), \\(DF\\). If \\(\\angle ADP = 32^\\circ\\), what is the measure of \\(\\angle FPB\\)?\n",
    "completion": "\\textbf{Solution:} Since the quadrilateral ABCD is a square,\\\\\nit follows that $\\angle A=\\angle PBC=90^\\circ$ and AB=AD,\\\\\ntherefore, $\\angle ADP+\\angle APD=90^\\circ$,\\\\\nsince $\\angle DPE=90^\\circ$,\\\\\nit follows that $\\angle APD+\\angle EPB=90^\\circ$,\\\\\ntherefore, $\\angle ADP=\\angle FPB=\\boxed{32^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51322751_127.png"
  },
  {
    "solution": "\\boxed{\\frac{2}{7}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $O$ is the center of symmetry of $\\square ABCD$, point $E$ is on the side $BC$, $AD=7$, and $BE=3$. Rotate $\\triangle ABE$ around point $O$ by $180^\\circ$. Let the corresponding point of $E$ be $E'$. What is the value of $\\frac{S_{\\triangle AEE'}}{S_{\\square ABCD}}$?\n",
    "completion": "\\textbf{Solution:} Construct $\\triangle CD{E}^{\\prime }$ to be symmetrical to $\\triangle ABE$ with respect to point O, and connect $EE^{\\prime}$.\n\nSince $\\triangle CD{E}^{\\prime }$ is symmetrical to $\\triangle ABE$ with respect to point O,\n\nit follows that $BE=DE^{\\prime }=3$,\n\nGiven that $AD=7$,\n\nit follows that $AE^{\\prime }=4$,\n\nLet the height of quadrilateral $ABCD$ be $h$,\n\nthen the height of $\\triangle AE{E}^{\\prime }$ is also equal to $h$,\n\nThus, $\\frac{S_{\\triangle AE{E}^{\\prime }}}{S_{\\square ABCD}}=\\frac{\\frac{1}{2}\\times 4h}{7h}=\\boxed{\\frac{2}{7}}.$",
    "image_path": "dataset/mm_math/images/MM_Math/50322995.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ is an equilateral triangle, and $BD=CE$, $\\angle 1=15^\\circ$. What is the degree measure of $\\angle 2$?\n",
    "completion": "\\textbf{Solution:} Given that $\\triangle ABC$ is an equilateral triangle,\\\\\nhence, $AB=BC$ and $\\angle ABC=\\angle ACB$,\\\\\nin $\\triangle ABD$ and $\\triangle BCE$,\\\\\nwe have\n\\[\n\\left\\{\\begin{array}{l}\nAB=BC\\\\\n\\angle ABC=\\angle ACB\\\\\nBD=CE\n\\end{array}\\right.,\n\\]\ntherefore, $\\triangle ABD \\cong \\triangle BCE$ (SAS),\\\\\nthus, $\\angle 1 = \\angle CBE$,\\\\\nsince $\\angle 2 = \\angle 1 + \\angle ABE$,\\\\\nhence $\\angle 2 = \\angle CBE + \\angle ABE = \\angle ABC = 60^\\circ$.\\\\\nTherefore, the answer is: $\\boxed{60^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53106967.png"
  },
  {
    "solution": "\\boxed{-2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the line $y=mx$ intersects with the hyperbola $y=\\frac{n}{x}$ at points $A(-1, a)$ and $B$. $BC$ is perpendicular to the $x$-axis, with the foot of the perpendicular being $C$, and the area of $\\triangle AOC$ is $1$. What are the values of $m$ and $n$?\n",
    "completion": "\\textbf{Solution:} Since the line $y=mx$ intersects the hyperbola $y=\\frac{n}{x}$ at points A($-1$, $a$) and B,\\\\\nit follows that the x-coordinate of point B is $1$, that is, C($1$, $0$),\\\\\nsince the area of $\\triangle AOC$ is $1$,\\\\\nit follows that A($-1$, $2$),\\\\\nSubstituting A($-1$, $2$) into $y=mx$ and $y=\\frac{n}{x}$, we obtain $m=\\boxed{-2}$, $n=\\boxed{-2}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52551767_142.png"
  },
  {
    "solution": "\\boxed{20^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in rhombus ABCD, AC intersects BD at O, and DE $\\perp$ BC at E. Connect OE. If $\\angle BAD=40^\\circ$, then what is the degree measure of $\\angle OED$?\n",
    "completion": "\\textbf{Solution:}\nGiven that quadrilateral ABCD is a rhombus, and $\\angle BAD=40^\\circ$,\\\\\nit follows that AD$\\parallel$BC\\\\\nHence, $\\angle BAD+\\angle ABC=180^\\circ$\\\\\nTherefore, $\\angle ABC=180^\\circ-40^\\circ=140^\\circ$\\\\\nThus, $\\angle ABD=\\angle CBD=\\frac{1}{2}\\angle ABC=70^\\circ$, and BO=DO,\\\\\nSince DE$\\perp$BC,\\\\\nit follows that OE=OD=OB, and $\\angle BDE=20^\\circ$,\\\\\nTherefore, $\\angle ODE=\\angle OED=\\boxed{20^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/52189500.png"
  },
  {
    "solution": "\\boxed{15}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the Cartesian coordinate system xOy, the graph of the linear function $y=-x+6$, denoted as ${l}_{1}$, intersects the $x$ and $y$ axes at points A and B, respectively. The graph of the direct variation function, denoted as ${l}_{2}$, intersects ${l}_{1}$ at point C ($m$, 5). What is the area of $\\triangle AOC$?\n",
    "completion": "\\textbf{Solution:} The solution for the linear function $y=-x+6$ and its intersection point A with the x-axis is found by setting $y=0$, and solving $-x+6=0$ to get $x=6$,\\\\\n$\\therefore$OA=6,\\\\\nThe area of $\\triangle AOC$ is $S_{\\triangle AOC}=\\frac{1}{2}OA\\cdot y_c=\\frac{1}{2}\\times 6\\times 5=\\boxed{15}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50555762_81.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, three squares with the same side length overlap each other. $O_1$ and $O_2$ are the centers of two of the squares. If the sum of the areas of the shaded parts is 4, what is the side length of the squares?\n",
    "completion": "\\textbf{Solution:} Connect $O_1B$ and $O_1C$, as shown in the figure: \\\\\n$\\because \\angle BO_1F + \\angle FO_1C = 90^\\circ$ and $\\angle FO_1C + \\angle CO_1G = 90^\\circ$, \\\\\n$\\therefore \\angle BO_1F = \\angle CO_1G$, \\\\\n$\\because$ quadrilateral $ABCD$ is a square, \\\\\n$\\therefore \\angle O_1BF = \\angle O_1CG = 45^\\circ$, \\\\\nIn $\\triangle O_1BF$ and $\\triangle O_1CG$, \\\\\n$\\left\\{\\begin{array}{l}\n\\angle FO_1B = \\angle CO_1G \\\\\nBO_1 = CO_1 \\\\\n\\angle FBO_1 = \\angle GCO_1\n\\end{array}\\right.$, \\\\\n$\\therefore \\triangle O_1BF \\cong \\triangle O_1CG$ (by ASA), \\\\\n$\\therefore S_{\\triangle O_1BF} = S_{\\triangle O_1CG}$, \\\\\n$\\therefore$ The area of the shaded parts in squares $O_1$ and $O_2$ is $\\frac{1}{4}S_{\\text{square}} ABCD$, \\\\\nSimilarly, the area of the shaded parts in the other two squares is also $\\frac{1}{4}S_{\\text{square}} ABCD$, \\\\\n$\\therefore$ The total area of the shaded parts $= 4 = \\frac{1}{2}S_{\\text{square}} ABCD$, \\\\\n$\\therefore S_{\\text{square}} ABCD = 8 = AD^2$, \\\\\n$\\therefore AD = \\boxed{2\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52423083.png"
  },
  {
    "solution": "\\boxed{\\frac{4\\pi }{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the radius OA of the arc $\\overset{\\frown}{AB}$ is 2, OC is perpendicular to AB at point C, and $\\angle AOC=60^\\circ$. What is the length of the arc $\\overset{\\frown}{AB}$?\n",
    "completion": "\\textbf{Solution:} Since $OC\\perp AB$ and $\\angle AOC=60^\\circ$, \\\\\nit follows that $\\angle AOB=120^\\circ$, \\\\\nSince $OA=2$,\\\\\nthe length of the arc $\\overset{\\frown}{AB}$ is: $\\frac{120^\\circ\\pi \\times 2}{180^\\circ}=\\boxed{\\frac{4\\pi }{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52765293_136.png"
  },
  {
    "solution": "\\boxed{\\frac{16}{25}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in parallelogram $ABCD$, $DE: EC = 4: 1$, and $AE$ intersects $BD$ at point $F$, then what is the ratio of $S_{\\triangle DEF}$ to $S_{\\triangle BAF}$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a parallelogram, \\\\\n$\\therefore$DC$\\parallel$AB, DC=AB, \\\\\nSince DE: EC=4: 1, \\\\\n$\\therefore$ $\\frac{DE}{DC}=\\frac{4}{5}$, \\\\\n$\\therefore$ $\\frac{DE}{AB}=\\frac{4}{5}$, \\\\\nSince DE$\\parallel$AB, \\\\\n$\\therefore$ $\\triangle DEF\\sim \\triangle BAF$, \\\\\n$\\therefore$ $\\frac{S_{\\triangle DEF}}{S_{\\triangle BAF}}=\\left(\\frac{DE}{AB}\\right)^2=\\boxed{\\frac{16}{25}}$",
    "image_path": "dataset/mm_math/images/MM_Math/52602477.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, if $\\triangle ABC \\cong \\triangle DEF$, $AC = 4$, $AB = 3$, $EF = 5$, then what is the perimeter of $\\triangle ABC$?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABC \\cong \\triangle DEF$, \\\\\n\\[\n\\therefore AB = DE = 3, \\quad AC = DF = 4, \\quad BC = EF = 5,\n\\]\n\\[\n\\therefore \\text{the perimeter of} \\ \\triangle ABC = AB + AC + BC = 3 + 4 + 5 = \\boxed{12}\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/53044024.png"
  },
  {
    "solution": "\\boxed{80^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, quadrilateral $ABCD$ is inscribed in circle $\\odot O$, where $\\angle A=100^\\circ$. What is the degree measure of $\\angle C$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral ABCD is inscribed in a circle,\\\\\nit follows that $\\angle A+\\angle C=180^\\circ$,\\\\\nsince $\\angle A=100^\\circ$,\\\\\nthus $\\angle C=180^\\circ-\\angle A=180^\\circ-100^\\circ=\\boxed{80^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53102391.png"
  },
  {
    "solution": "\\boxed{N_5\\left(-2+2\\sqrt{2}, 3-\\sqrt{2}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, the line $y = -\\frac{1}{2}x + 2$ intersects the x-axis at point B and the y-axis at point C. A parabola with the axis of symmetry $x = \\frac{3}{2}$ passes through points B and C and intersects the negative half of the x-axis at point A. Let M be a moving point on the parabola, and N be on the line $BC$. When the quadrilateral formed by points O, C, M, and N is a parallelogram, directly write down the coordinates of point N.\n",
    "completion": "\\textbf{Solution:} Let $N_1(2, 1)$, $N_2\\left(2+2\\sqrt{2}, 1-\\sqrt{2}\\right)$, $N_3\\left(2-2\\sqrt{2}, 1+\\sqrt{2}\\right)$, $N_4\\left(-2\\sqrt{2}-2, 3+\\sqrt{2}\\right)$, and $N_5\\left(-2+2\\sqrt{2}, 3-\\sqrt{2}\\right)$. \n\n\\boxed{N_1(2, 1)}, \\boxed{N_2\\left(2+2\\sqrt{2}, 1-\\sqrt{2}\\right)}, \\boxed{N_3\\left(2-2\\sqrt{2}, 1+\\sqrt{2}\\right)}, \\boxed{N_4\\left(-2\\sqrt{2}-2, 3+\\sqrt{2}\\right)}, and \\boxed{N_5\\left(-2+2\\sqrt{2}, 3-\\sqrt{2}\\right)}.",
    "image_path": "dataset/mm_math/images/MM_Math/52977381_196.png"
  },
  {
    "solution": "\\boxed{y=x^2-3x-4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system $xOy$, the parabola $y=ax^{2}+bx-4$ intersects the $x$-axis at points $A$ and $B$ (with $B$ to the right of $A$) and intersects the $y$-axis at point $C$. It is known that $OA=1$ and $OB=4OA$. Connect $BC$. What is the equation of the parabola?\n",
    "completion": "\\[ \\text{Solution: From } OA=1 \\text{ we get } A(-1, 0). \\]\n\\[ \\because OB=4OA=4, \\]\n\\[ \\therefore B(4, 0). \\]\nSubstituting $A(-1, 0)$ and $B(4, 0)$ into $y=ax^2+bx-4$ we get:\n\\[ \\begin{cases} a-b-4=0 \\\\ 16a+4b-4=0 \\end{cases}, \\]\nSolving this yields\n\\[ \\begin{cases} a=1 \\\\ b=-3 \\end{cases}, \\]\n\\[ \\therefore y=x^2-3x-4; \\]\nThus, the equation of the parabola is $\\boxed{y=x^2-3x-4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51446444_219.png"
  },
  {
    "solution": "\\boxed{P\\left(\\frac{2}{3}, \\frac{7}{3}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y=k_1x+b$ intersects with the graph of the inverse proportion function $y=\\frac{k_2}{x}$ at points A and B, where the coordinates of point A are $(-1, 4)$, and the coordinates of point B are $(4, n)$. Point P is on line segment AB, and $\\frac{S_{\\triangle AOP}}{S_{\\triangle BOP}}=1:2$. Find the coordinates of point P.\n",
    "completion": "\\textbf{Solution:} Let the intersection of line $AB$ and the $y$-axis be point $C$, \\\\\nsince the analytical expression of the linear function is $y=-x+3$, \\\\\nthus $C(0, 3)$, \\\\\nsince the area of $\\triangle AOC = \\frac{1}{2} \\times 3 \\times 1 = \\frac{3}{2}$, \\\\\nthus the area of $\\triangle AOB = \\text{area of} \\ \\triangle AOC + \\text{area of} \\ \\triangle BOC = \\frac{1}{2} \\times 3 \\times 1 + \\frac{1}{2} \\times 3 \\times 4 = \\frac{15}{2}$, \\\\\nsince the ratio of the areas $S_{\\triangle AOP} : S_{\\triangle BOP} = 1:2$, \\\\\nthus $S_{\\triangle AOP} = \\frac{15}{2} \\times \\frac{1}{3} = \\frac{5}{2}$, \\\\\ntherefore, $S_{\\triangle AOC} < S_{\\triangle AOP}$, the area of $\\triangle COP = \\frac{5}{2} - \\frac{3}{2} = 1$, \\\\\ntherefore, $\\frac{1}{2} \\times 3 \\times x_{P} = 1$, \\\\\nthus $x_{P} = \\frac{2}{3}$, \\\\\nsince point $P$ is on the segment $AB$, \\\\\nthus $y = -\\frac{2}{3} + 3 = \\frac{7}{3}$, \\\\\ntherefore, \\boxed{P\\left(\\frac{2}{3}, \\frac{7}{3}\\right)}.",
    "image_path": "dataset/mm_math/images/MM_Math/51446676_224.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $OB$, $OC$, and $OD$ are three rays within $\\angle AOE$, with $OB$ bisecting $\\angle AOE$ and $OD$ bisecting $\\angle COE$. If $\\angle AOB=70^\\circ$ and $\\angle DOE=20^\\circ$, what is the degree measure of $\\angle BOC$?\n",
    "completion": "\\textbf{Solution:} Because $OB$ bisects $\\angle AOE$ and $OD$ bisects $\\angle COE$, \\\\\ntherefore $\\angle BOE = \\angle AOB = 70^\\circ$, $\\angle COE = 2\\angle DOE = 40^\\circ$, \\\\\nthus $\\angle BOC = \\angle BOE - \\angle COE = 70^\\circ - 40^\\circ = \\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/54927250_122.png"
  },
  {
    "solution": "\\boxed{\\frac{81}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the graph of the quadratic function $y=-x^2+bx+c$ intersects the coordinate axes at points $A$, $B$, and $C$, where point $B$ has the coordinates $(1,0)$, point $C$ has the coordinates $(0,4)$, and point $D$ has the coordinates $(0,2)$. Let point $P$ be a moving point on the graph of the quadratic function. When point $P$ is located in the second quadrant on the graph of the quadratic function, connect points $AD$ and $AP$, and construct the parallelogram $APED$ with $AD$ and $AP$ as adjacent sides. Let the area of parallelogram $APED$ be $S$. Find the maximum value of $S$.\n",
    "completion": "\\textbf{Solution:} Draw PD and construct PG $\\parallel$ y-axis intersecting AD at point G, as shown in the figure.\\\\\nLet P$(t，-t^{2}-3t+4)$ (where $-4<t<0$), then G$(t，\\frac{1}{2}t+2)$,\\\\\n$\\therefore PG=-t^{2}-3t+4-\\frac{1}{2}t-2=-t^{2}-\\frac{7}{2}t+2$,\\\\\n$\\therefore S=2S_{\\triangle APD}=2\\times \\frac{1}{2}PG\\cdot |x_{D}-x_{A}|=-4t^{2}-14t+8$,\\\\\n$=-4\\left(t+\\frac{7}{4}\\right)^{2}+\\frac{81}{4}$.\\\\\nSince $-4<0$ and $-4<t<0$,\\\\\n$\\therefore$ when $t=-\\frac{7}{4}$, $S$ attains its maximum value $\\boxed{\\frac{81}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51321592_201.png"
  },
  {
    "solution": "\\boxed{(2,1)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the points A, B, and C have integer coordinates for both their x- and y-values. If an arc is drawn through these three points, what are the coordinates of the center of the arc?\n",
    "completion": "\\textbf{Solution:} According to the corollary of the Perpendicular Diameter Theorem: the perpendicular bisector of a chord always passes through the center of the circle,\\\\\nthe perpendicular bisectors of chords AB and BC can be drawn, and their intersection point is the center of the circle.\\\\\nAs shown in the figure, the center of the circle is $\\boxed{(2,1)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51351385.png"
  },
  {
    "solution": "\\boxed{17}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, with $AB=AC=7$, the perpendicular bisector $DE$ of $AB$ intersects $BC$ and $AB$ at points $D$ and $E$, respectively. Line $AD$ is drawn. If $BC=10$, then what is the perimeter of $\\triangle ACD$?\n",
    "completion": "\\textbf{Solution:} Given $DE$ is the perpendicular bisector of $AB$,\\\\\n$\\therefore$ $AD=DB$,\\\\\n$\\therefore$ the perimeter of $\\triangle ADC$ is $AD+DC+AC=DB+DC+AC=BC+AC=10+7=\\boxed{17}$",
    "image_path": "dataset/mm_math/images/MM_Math/53033974.png"
  },
  {
    "solution": "\\boxed{130^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the central angle $\\angle AOB$ is $100^\\circ$. What is the measure of the inscribed angle $\\angle ACB$?\n",
    "completion": "\\textbf{Solution:} Let point E be a point on the major arc AB, and connect EA and EB. \\\\\n$\\because$ $\\angle$AOB=100$^\\circ$, \\\\\n$\\therefore$ $\\angle$E= $ \\frac{1}{2}\\angle$AOB=50$^\\circ$, \\\\\n$\\therefore$ $\\angle$ACB=180$^\\circ$−$\\angle$E=\\boxed{130^\\circ}.",
    "image_path": "dataset/mm_math/images/MM_Math/52622477.png"
  },
  {
    "solution": "\\boxed{\\frac{125}{16}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=ax^2+bx+\\frac{5}{2}$ intersects with line segment $AB$ at points $A(-1,0)$ and $B(4,\\frac{5}{2})$. Point $D$ is a moving point on the parabola between points $A$ and $B$ (not coinciding with $A$ or $B$), the line $CD$ is parallel to the y-axis and intersects line $AB$ at point $C$, and lines $AD$ and $BD$ are connected. Let the x-coordinate of point $D$ be $m$, and the area of $\\triangle ADB$ be $S$. Determine the function of $S$ in terms of $m$, and find the coordinates of point $C$ when $S$ reaches its maximum value.\n",
    "completion": "\\textbf{Solution:} As shown in the figure, construct $BF\\perp DE$ at point F passing through point B.\\\\\nLet the equation of line $AB$ be: $y=kx+m$,\\\\\nthen: $\\left\\{\\begin{array}{l}\n0=-k+m \\\\\n\\frac{5}{2}=4k+m\n\\end{array}\\right.$,\\\\\nsolving this gives: $\\left\\{\\begin{array}{l}\nk=\\frac{1}{2} \\\\\nm=\\frac{1}{2}\n\\end{array}\\right.$,\\\\\n$\\therefore y=\\frac{1}{2}x+\\frac{1}{2}$,\\\\\n$\\because$ the abscissa of point D is m and $CD\\parallel y$-axis,\\\\\n$\\therefore$ the coordinate of point C is $\\left(m, \\frac{1}{2}m+\\frac{1}{2}\\right)$, and the ordinate of point D is $\\left(-\\frac{1}{2}m^{2}+2m+\\frac{5}{2}\\right)$,\\\\\n$\\therefore AE=m+1, BF=4-m$, $CD=-\\frac{1}{2}m^{2}+\\frac{3}{2}m+2$,\\\\\n$\\therefore S=\\frac{1}{2}CD\\cdot (AE+BF)=\\frac{1}{2}\\times \\left(-\\frac{1}{2}m^{2}+\\frac{3}{2}m+2\\right)\\times (m+1+4-m)$\\\\\n$=-\\frac{5}{4}\\left(m-\\frac{3}{2}\\right)^{2}+\\frac{125}{16}$,\\\\\n$\\therefore$ when $m=\\frac{3}{2}$, $S$ reaches its maximum value $\\boxed{\\frac{125}{16}}$, at this point $C\\left(\\frac{3}{2}, \\frac{5}{4}\\right)$.",
    "image_path": "dataset/mm_math/images/MM_Math/53043850_141.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $AD=2$, $AB=4$, and $EF\\perp AC$, intersecting $AB$ and $CD$ at $E$ and $F$ respectively, what is the minimum value of $AF+CE$?\n",
    "completion": "\\textbf{Solution:} Let $DF=x$, then $FC=4-x$,\\\\\nDraw $CG\\parallel EF$ through point C, and let $CG=EF$, connect $FG$,\\\\\nWhen points A, F, G are collinear, find the minimum value of $AF+FG$;\\\\\n$\\because$ $CG\\parallel EF$, and $CG=EF$,\\\\\n$\\therefore$ quadrilateral $CEFG$ is a parallelogram;\\\\\n$\\therefore$ $EC\\parallel FG$, $EC=FG$,\\\\\nFurthermore, since points A, F, G are collinear,\\\\\n$\\therefore$ $AF\\parallel EC$,\\\\\nAlso, since quadrilateral $ABCD$ is a rectangle,\\\\\n$\\therefore$ $AE\\parallel DC$, $\\angle D=90^\\circ$,\\\\\n$\\therefore$ quadrilateral $AECF$ is a parallelogram,\\\\\n$\\therefore$ $OA=OC$, $OE=OF$,\\\\\nMoreover, since $EF\\perp AC$,\\\\\n$AF=CF=4-x$,\\\\\nIn $\\triangle ADF$, by the Pythagorean theorem, we have:\\\\\n$AD^2+DF^2=AF^2$,\\\\\nAlso, since $AD=2$, $DF=x$, then $FC=4-x$,\\\\\n$\\therefore$ $2^2+x^2=(4-x)^2$,\\\\\nSolving this, we get: $x=\\frac{3}{2}$,\\\\\n$\\therefore$ $AF=\\frac{5}{2}$,\\\\\nIn $\\triangle ADC$, by the Pythagorean theorem, we have:\\\\\n$AD^2+DC^2=AC^2$,\\\\\n$\\therefore$ $AC=2\\sqrt{5}$,\\\\\n$\\therefore$ $AO=\\frac{1}{2}AC=\\sqrt{5}$,\\\\\nMoreover, since $OF\\parallel CG$,\\\\\n$\\therefore$ $\\triangle AOF\\sim \\triangle ACG$,\\\\\n$\\therefore$ $\\frac{AO}{AC}=\\frac{AF}{AG}$,\\\\\n$\\therefore$ $AG=5$,\\\\\nFurthermore, since $AG=AF+FG$, $FG=EC$,\\\\\n$\\therefore$ $AF+EC=\\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52866230.png"
  },
  {
    "solution": "\\boxed{120^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rhombus $ABCD$, $AB=4$, $E$ is the midpoint of $BC$, $AE\\perp BC$ at point $E$, $AF\\perp CD$ at point $F$, $CG\\parallel AE$, $CG$ intersects $AF$ at point $H$, and intersects $AD$ at point $G$. What is the degree measure of $\\angle CHA$?\n",
    "completion": "\\textbf{Solution:} In equilateral triangle ABC,\\\\\n$\\because AE\\perp BC$,\\\\\n$\\therefore \\angle CAE=\\frac{1}{2}\\angle BAC=\\frac{1}{2}\\times 60^\\circ=30^\\circ.$\\\\\nSimilarly, $\\angle CAF=30^\\circ$.\\\\\n$\\therefore \\angle EAF=\\angle CAE+\\angle CAF=30^\\circ+30^\\circ=60^\\circ$\\\\\n$\\because AE\\parallel CG$\\\\\n$\\therefore \\angle CHA=180^\\circ-\\angle EAF=180^\\circ-60^\\circ=\\boxed{120^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/51614476_116.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{21}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in $\\triangle ABC$, $CD$ is the angle bisector, and $DE$ bisects $\\angle CDB$ and intersects $BC$ at point $E$, and $DE \\parallel AC$. It is given that $CD^2 = CA \\cdot CE$. If $\\frac{AD}{BD} = \\frac{4}{3}$ and $AC = 14$, what is the length of $AD$?\n",
    "completion": "\\textbf{Solution:} Since $CD$ bisects $\\angle ACB$, then $\\angle BCD = \\angle ACD$. Since $DE\\parallel BC$, then $\\angle EDC = \\angle BCD$, which leads to $\\angle ECD = \\angle EDC$; hence $DE=CE$. Given $DE\\parallel BC$, we have $\\frac{AD}{BD}=\\frac{4}{3}$, which implies $\\triangle BDE \\sim \\triangle BAC$. Therefore, $\\frac{DE}{AC}=\\frac{BD}{AB}=\\frac{3}{7}$. This simplifies to $\\frac{CE}{14}=\\frac{3}{7}$. Solving this gives $CE=DE=6$. Since $DE\\parallel BC$, it follows that $\\angle ACD = \\angle CDE$ and $\\angle A = \\angle EDB$. Given $\\angle CDE = \\angle EDB$, then $\\angle ACD = \\angle A$, which implies $AD=CD$. Knowing $CD^2 = CA \\cdot CE$, we find $AD^2 = 14 \\times 6$. Solving this gives $AD= \\boxed{2\\sqrt{21}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52661234_78.png"
  },
  {
    "solution": "\\boxed{\\frac{25}{13}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given that in $\\triangle ABC$, $AB=AC=5, BC=8$, point $E$ is a moving point on ray $CA$, and point $O$ is a moving point on side $BC$, with $OC=OE$, ray $OE$ intersects ray $BA$ at point $D$. Connect $AO$. If $\\triangle AEO$ is an isosceles triangle with $AE$ as the leg, what is the length of the line segment $OC$?\n",
    "completion": "\\textbf{Solution:}\\\\\nSince $\\triangle AEO$ is an isosceles triangle with $AE$ as its leg,\\\\\nthus $AE=OE$,\\\\\nsince $OC=OE$,\\\\\nthus, let $AE=OE=OC=x$,\\\\\nfrom (1), we get $\\triangle ABC \\sim \\triangle OEC$,\\\\\nthus, $\\frac{AC}{OC}=\\frac{BC}{CE}$,\\\\\nthus, $\\frac{5}{x}=\\frac{8}{5-x}$,\\\\\nsolving this, we obtain $x=\\boxed{\\frac{25}{13}}$, upon verification, $x=\\frac{25}{13}$ is a solution to the original equation;\\\\\nhence, the length of $OC$ is \\boxed{\\frac{25}{13}}.",
    "image_path": "dataset/mm_math/images/MM_Math/51319009_113.png"
  },
  {
    "solution": "\\boxed{6 - 2\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within the square paper ABCD, E is the midpoint of CD. The square paper is folded so that point B falls on point G on segment AE, with the crease being AF. If AD = 4 cm, what is the length of CF in centimeters?\n",
    "completion": "\\textbf{Solution:}\n\nGiven square ABCD, where point E is the midpoint of CD,\\\\\nthus $\\angle B = \\angle D = 90^\\circ$, AD = CD = 4, DE = CE = $\\frac{1}{2}$CD = 2,\\\\\nin $\\triangle ADE$,\\\\\n$AE = \\sqrt{AD^2 + DE^2} = \\sqrt{4^2 + 2^2} = 2\\sqrt{5}$,\\\\\nbecause folding the square paper such that point B falls on point G on segment AE, with the fold line being AF,\\\\\nthus DE = EF = 2, AG = AB = 4, BF = FG,\\\\\nthus GE = $2\\sqrt{5} - 4$, let BF = x, then CF = 4 - x,\\\\\nin $\\triangle GEF$, $EF^2 = GE^2 + GF^2 = (2\\sqrt{5} - 4)^2 + x^2$,\\\\\nthus in $\\triangle CEF$, $EF^2 = CE^2 + CF^2$,\\\\\nthus $(2\\sqrt{5} - 4)^2 + x^2 = (4 - x)^2 + 2^2$, solving it:\\\\\n$x = 2\\sqrt{5} - 2$,\\\\\nthus $FC = 4 - (2\\sqrt{5} - 2) = \\boxed{6 - 2\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51614714.png"
  },
  {
    "solution": "\\boxed{70^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $BE\\perp AC$ at point D, and $AD = CD$, $BD = ED$. If $\\angle E = 35^\\circ$, what is the value of $\\angle ABC$?\n",
    "completion": "\\textbf{Solution:} Given that $BE\\perp AC$ and $AD=DC$,\\\\\nit follows that $BA=BC$,\\\\\nwhich implies $\\angle ABD=\\angle CBD=\\frac{1}{2}\\angle ABC$,\\\\\nConsidering $\\triangle ADB$ and $\\triangle CDE$, we have\\\\\n\\[\n\\begin{cases}\nAD=DC\\\\\n\\angle ADB=\\angle EDC\\\\\nBD=DE\n\\end{cases},\n\\]\nhence, $\\triangle ADB\\cong \\triangle CDE$,\\\\\nwhich means $\\angle E=\\angle ABD=35^\\circ$,\\\\\nthus, $\\angle ABC=2\\angle ABD=\\boxed{70^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52906060.png"
  },
  {
    "solution": "\\boxed{35^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, two triangles OAB and OCD are placed with $\\angle ABO=30^\\circ$ and $\\angle ODC=45^\\circ$. If OM and ON bisect $\\angle AOD$ and $\\angle BOC$ respectively. Rotate the triangle OCD clockwise around point O from figure 1. In the process of rotation, is there a scenario where $\\angle BOD = \\angle COB + \\angle AOD$? If it exists, calculate the rotation angle $\\alpha$; if not, explain the reason.\n",
    "completion": "\\textbf{Solution:}\\\\\nSince $\\angle BOD = \\angle COB + \\angle AOD$,\\\\\nit follows that $\\angle DOC + \\angle AOB = 45^\\circ + 60^\\circ = 105^\\circ$, thus $\\angle COB + \\angle BOD + \\angle AOD + \\angle BOD = 105^\\circ$,\\\\\ntherefore $3\\angle BOD = 105^\\circ$,\\\\\nThe angle of rotation $\\angle BOD = \\alpha = \\boxed{35^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/8928654_74.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the circumcircle $O$ of $\\triangle ABC$ has a radius of $2\\text{cm}$. If $BC=2\\text{cm}$, what is the measure of $\\angle A$ in degrees?\n",
    "completion": "\\textbf{Solution:} Connect OB and OC,\\\\\nsince the radius of circle O is 2cm, and BC=2cm,\\\\\ntherefore OB=OC=BC,\\\\\nthus, $\\triangle$OBC is an equilateral triangle,\\\\\ntherefore, $\\angle$BOC=60$^\\circ$,\\\\\ntherefore, $\\angle$A=$\\frac{1}{2}\\angle$BOC=30$^\\circ$,\\\\\nhence, the answer is: $\\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51318557.png"
  },
  {
    "solution": "\\boxed{\\sqrt{3}+2\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\odot P$ intersects the x-axis at points $A(-5,0)$ and $B(1,0)$, and intersects the positive y-axis at point $C$. If $\\angle ACB=60^\\circ$, what is the ordinate of point $C$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $PH\\perp AB$ at point $H$, and $PD\\perp OC$ at point $D$. Connect $PA$, $PB$, and $PC$.\\\\\nSince $A(-5,0)$ and $B(1,0)$,\\\\\nthus $OA=5$ and $OB=1$,\\\\\ntherefore, $AB=6$,\\\\\nsince $PH\\perp AB$,\\\\\ntherefore, $AH=BH=\\frac{1}{2}AB=3$,\\\\\nthus $OH=2$,\\\\\nbecause $\\angle ACB=60^\\circ$,\\\\\nthus $\\angle APB=2\\angle ACB=2\\times60^\\circ=120^\\circ$,\\\\\nthus $\\angle APH=60^\\circ$, $\\angle PAH=30^\\circ$,\\\\\nsince in $\\triangle PAH$ right-angled, $PH=\\frac{\\sqrt{3}}{3}AH=\\sqrt{3}$,\\\\\nthus $PA=2PH=2\\sqrt{3}$,\\\\\nbecause $\\angle PHO=\\angle PDO=\\angle HOD=90^\\circ$,\\\\\nthus quadrilateral $PHOD$ is a rectangle,\\\\\nthus $OD=PH=\\sqrt{3}$, $PD=OH=2$,\\\\\nsince in right-angled $\\triangle PCD$, $PC=PA=2\\sqrt{3}$, $PD=2$,\\\\\nthus $CD=\\sqrt{PC^{2}-PD^{2}}=\\sqrt{(2\\sqrt{3})^{2}-2^{2}}=2\\sqrt{2}$,\\\\\nthus $OC=OD+CD=\\sqrt{3}+2\\sqrt{2}$,\\\\\nsince point $C$ is on the positive semi-axis of $y$,\\\\\nthus the $y$-coordinate of point $C$ is $\\boxed{\\sqrt{3}+2\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52874641.png"
  },
  {
    "solution": "\\boxed{7.2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $E$ is the midpoint of side $BC$, and $DF\\perp AE$ at point $F$. It is given that $\\frac{AF}{BE} = \\frac{AD}{AE}$. Knowing that $AB=8, BC=12$, what is the length of $AF$?\n",
    "completion": "\\textbf{Solution:} Given, $\\because E$ is the midpoint of $BC$,\\\\\n$\\therefore BE = \\frac{1}{2}BC = 6$,\\\\\n$\\therefore AE = \\sqrt{AB^{2} + BE^{2}} = 10$.\\\\\n$\\because \\frac{AF}{BE} = \\frac{AD}{AE}$,\\\\\n$\\therefore \\frac{AF}{6} = \\frac{12}{10}$,\\\\\n$\\therefore AF = \\boxed{7.2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53065905_146.png"
  },
  {
    "solution": "\\boxed{25}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, which consists of two right-angled triangles and three squares, what is the area of the shaded part?\n",
    "completion": "\\textbf{Solution:} Since $\\angle ACB = 90^\\circ$, \\\\\nit follows that $BC^{2} = AB^{2} - AC^{2} = 13^{2} - 12^{2} = 25$, \\\\\ntherefore, the area of the shaded region $S_{\\text{shaded}} = BC^{2} = \\boxed{25}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52225946.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, $AB=10$, $AC=8$, D is the midpoint of AB, M is a moving point on the segment AC, connect DM, construct a right triangle DEM with DM as the right angle side such that $\\angle DEM=30^\\circ$, the hypotenuse DE intersects ray MC at point F. If the area of $\\triangle MDF$ is $\\sqrt{3}$ times the area of $\\triangle MEF$, what is the length of CM?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw DG $\\perp$ AC at G, and draw EH $\\perp$ AC at H, \\\\\nthen $\\angle$DGM=$\\angle$MHE=$90^\\circ$, \\\\\nin $\\triangle ABC$, $BC=\\sqrt{AB^{2}-AC^{2}}=\\sqrt{10^{2}-8^{2}}=6$, \\\\\nsince D is the midpoint of AB, \\\\\nthus AD=$\\frac{1}{2}$AB=5, \\\\\nsince $\\angle$AGD=$\\angle$ACB=$90^\\circ$, $\\angle$DAG=$\\angle$BAC, \\\\\nthus $\\triangle ADG\\sim \\triangle ABC$, \\\\\nthus $\\frac{DG}{BC}=\\frac{AG}{AC}=\\frac{AD}{AB}$, that is $\\frac{DG}{6}=\\frac{AG}{8}=\\frac{5}{10}$, \\\\\nthus DG=3, AG=4, \\\\\nthus CG=AC−AG=8−4=4, \\\\\nsince the area of $\\triangle MDF$ is $\\sqrt{3}$ times the area of $\\triangle MEF$, \\\\\nthus $\\frac{1}{2}$FM•DG=$\\sqrt{3} \\times \\frac{1}{2}$FM•EH, \\\\\nthus DG=$\\sqrt{3}$EH, that is EH=$\\frac{\\sqrt{3}}{3}$DG=$\\sqrt{3}$, \\\\\nin $\\triangle DEM$, $\\angle$DME=$90^\\circ$, $\\angle$DEM=$30^\\circ$, \\\\\nthus $\\frac{DM}{ME}=\\tan\\angle DEM=\\tan30^\\circ=\\frac{\\sqrt{3}}{3}$, \\\\\nsince $\\angle$DMG+$\\angle$MDG=$90^\\circ$, $\\angle$DMG+$\\angle$EMH=$\\angle$DME=$90^\\circ$, \\\\\nthus $\\angle$MDG=$\\angle$EMH, \\\\\nthus $\\triangle DMG\\sim \\triangle MEH$, \\\\\nthus $\\frac{MG}{EH}=\\frac{DM}{ME}$, \\\\\nthus $\\frac{MG}{\\sqrt{3}}=\\frac{\\sqrt{3}}{3}$, \\\\\nthus MG=1, \\\\\nthus CM=CG+MG=4+1=\\boxed{5},",
    "image_path": "dataset/mm_math/images/MM_Math/52930758.png"
  },
  {
    "solution": "\\boxed{40^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AC=BC$, point $D$ is on side $AC$, and point $E$ is on the extension line of $CB$. $DE$ intersects $AB$ at point $F$. If $\\angle C=50^\\circ$ and $\\angle E=25^\\circ$, then what is the degree measure of $\\angle BFE$?\n",
    "completion": "\\textbf{Solution:} Given $AC=BC$, \\\\\n$\\therefore \\angle A=\\angle ABC$, \\\\\nSince $\\angle C=50^\\circ$, and $\\angle A+\\angle ABC+\\angle C=180^\\circ$, \\\\\n$\\therefore \\angle ABC= \\frac{1}{2}(180^\\circ-\\angle C)=65^\\circ$, \\\\\nGiven $\\angle E=25^\\circ$, and $\\angle ABC=\\angle E+\\angle BFE$, \\\\\n$\\therefore \\angle BFE=\\angle ABC-\\angle E=65^\\circ-25^\\circ=\\boxed{40^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53059015.png"
  },
  {
    "solution": "\\boxed{(2, 4)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a linear function $y=kx+2$ intersects the $y$-axis at point A and intersects the graph of the hyperbolic function $y=\\frac{m}{x}$ at points B and C. Line BD is perpendicular to the $y$-axis and intersects the $y$-axis at point D. It is given that OA＝OD and the area of $\\triangle ABD$ is $8$. Determine the coordinates of point C, and directly write the solution set of the inequality $kx+2>\\frac{m}{x}$.\n",
    "completion": "\\textbf{Solution:} From equation (1), we can set up the system of equations\n\\[\n\\begin{cases}\ny=x+2\\\\\ny=\\frac{8}{x}\n\\end{cases}\n\\]\nSolving this system, we get:\n\\[\n\\begin{cases}\nx=-4\\\\\ny=-2\n\\end{cases}\n\\]\nor\n\\[\n\\begin{cases}\nx=2\\\\\ny=4\n\\end{cases}\n\\]\n$\\because$ Point $C$ is in the first quadrant,\\\\\nthus the coordinates of point $C$ are $\\boxed{(2, 4)}$,\\\\\nFrom the graph, it is obtained that when $-4<x<0$ or $x>2$, $kx+2>\\frac{m}{x}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51445968_244.png"
  },
  {
    "solution": "\\boxed{15^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, on the exterior of the square $ABCD$, an equilateral triangle $ADE$ is constructed. What is the measure of $\\angle AEB$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral ABCD is a square,\\\\\nit follows that $\\angle$BAD=90$^\\circ$, and AB=AD.\\\\\nSince $\\triangle$ABD is an equilateral triangle,\\\\\nit follows that $\\angle$DAE=60$^\\circ$, and AD=AE.\\\\\nTherefore, $\\angle$BAE=$\\angle$DAE+$\\angle$BAD=90$^\\circ$+60$^\\circ$=150$^\\circ$, and AB=AE.\\\\\nHence, $\\triangle$BAE is an isosceles triangle,\\\\\nwhich means $\\angle AEB=\\frac{180^\\circ−\\angle BAE}{2}=\\frac{180^\\circ−150^\\circ}{2}=15^\\circ$.\\\\\nThus, the answer is: $\\boxed{15^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51770527.png"
  },
  {
    "solution": "\\boxed{84}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the parallelogram $ABCD$, the diagonals $AC$ and $BD$ intersect at point $O$, $AC=AB$, $E$ is the midpoint of side $AB$, $G$ and $F$ are points on side $BC$, and $OG$ and $EF$ are connected. If $AB=25$, $BC=14$, and $GF=7$, what is the area of the shaded region in the figure?\n",
    "completion": "\\textbf{Solution}: Connect OE, let OG intersect EF at point H, and draw AP$\\perp$BC at point P, intersecting OE at point Q, as shown in the figure.\\\\\n$\\because$AB=AC, AP$\\perp$BC,\\\\\n$\\therefore$ $BP=\\frac{1}{2}BC=7.$\\\\\n$\\therefore$ $AP=\\sqrt{AB^{2}-BP^{2}}=\\sqrt{25^{2}-7^{2}}=24.$\\\\\n$\\because$Quadrilateral ABCD is a parallelogram,\\\\\n$\\therefore$OA=OC.\\\\\n$\\because$EA=EB,\\\\\n$\\therefore$OE is the median line of $\\triangle ABC$.\\\\\n$\\therefore$OE$\\parallel$BC and $OE=\\frac{1}{2}BC=7.$\\\\\n$\\therefore \\angle AEO=\\angle ABC,$ $\\angle AOE=\\angle ACB.$\\\\\n$\\because AB=AC,$\\\\\n$\\therefore \\angle ABC=\\angle ACB.$\\\\\n$\\therefore \\angle AEO=\\angle AOE.$\\\\\n$\\therefore AO=AE.$ That is, $\\triangle AEO$ is an isosceles triangle.\\\\\n$\\because$OE$\\parallel$BC,\\\\\n$\\therefore$$\\angle$AQE=$\\angle$APB=90$^\\circ$.\\\\\n$\\therefore$AP$\\perp$OE at point Q.\\\\\n$\\therefore EQ=OQ=\\frac{1}{2}OE=\\frac{7}{2}.$\\\\\n$\\therefore$ $AQ=\\sqrt{AE^{2}-EQ^{2}}=\\sqrt{\\left(\\frac{25}{2}\\right)^{2}-\\left(\\frac{7}{2}\\right)^{2}}=12.$\\\\\n$\\therefore$ ${S}_{\\triangle AEO}=\\frac{1}{2}OE\\cdot AQ=\\frac{1}{2}\\times 7\\times 12=42.$\\\\\nConnect EG, OF,\\\\\n$\\because$OE$\\parallel$GF, OE=GF=7,\\\\\n$\\therefore$Quadrilateral OEGF is a parallelogram.\\\\\n$\\therefore$HE=HF, HO=HG.\\\\\n$\\because$PQ=AP−AQ=24−12=12,\\\\\n$\\therefore$ ${S}_{\\text{parallelogram} OEGF}=GF\\cdot PQ=7\\times 12=84.$\\\\\n$\\therefore$ ${S}_{\\triangle OEH}={S}_{\\triangle GFH}=\\frac{1}{4}{S}_{\\text{parallelogram} OEGF}=\\frac{1}{4}\\times 84=21.$\\\\\n$\\therefore$ ${S}_{\\text{shaded}}={S}_{\\triangle AEO}+{S}_{\\triangle OEH}+{S}_{\\triangle GFH}=42+21+21=\\boxed{84}.$",
    "image_path": "dataset/mm_math/images/MM_Math/50496534.png"
  },
  {
    "solution": "\\boxed{(-17,0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, the straight line AB intersects the x-axis and y-axis at points A and B respectively, with point C on the negative half of the x-axis. The coordinates of these three points are A(4,0), B(0,4), and C(-1,0). As shown in Figure 2, point B is translated 1 unit to the right to obtain point D. There exists a moving point P on the x-axis. If $\\angle DCO+\\angle DPO=\\alpha$ and $\\tan\\alpha=4$, please directly write down the coordinate of point P.\n",
    "completion": "\\textbf{Solution:} The coordinates of point $P$ are $(19,0)$ or $(-17,0)$, therefore the answer is \\boxed{(19,0)} or \\boxed{(-17,0)}.",
    "image_path": "dataset/mm_math/images/MM_Math/51353152_181.png"
  },
  {
    "solution": "\\boxed{\\frac{6}{\\cos\\alpha}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a ladder $AB$ rests against a wall that is perpendicular to the ground level. The distance $AC$, which is the gap between the base of the ladder $A$ and the wall, is 6 meters. If the angle between the ladder and the ground is $\\alpha$, what is the length of the ladder $AB$?\n",
    "completion": "\\textbf{Solution:} Given the problem, we have $\\cos\\alpha =\\frac{AC}{AB}$, where $AC=6$ meters.\\\\\nTherefore, $AB=\\frac{6}{\\cos\\alpha}$ meters,\\\\\nSo, the answer is: $AB=\\boxed{\\frac{6}{\\cos\\alpha}}$ meters.",
    "image_path": "dataset/mm_math/images/MM_Math/53066051.png"
  },
  {
    "solution": "\\boxed{5\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of circle $O$, points $C$ and $D$ are on circle $O$, $\\angle D=60^\\circ$, and $AB=10$. What is the length of $AC$?\n",
    "completion": "\\textbf{Solution:} Connect $BC$, as shown in the figure,\\\\\n$\\because$ $AB$ is the diameter of $\\odot O$, and $AB=10$,\\\\\n$\\therefore$ $\\angle ACB=90^\\circ$,\\\\\n$\\because$ $\\angle D=60^\\circ$,\\\\\n$\\therefore$ $\\angle ABC=60^\\circ$,\\\\\n$\\therefore$ $\\angle CAB=180^\\circ-\\angle ABC-\\angle ACB=30^\\circ$,\\\\\n$\\therefore$ in $\\triangle ABC$, we have: $BC=\\frac{1}{2}AB=5$,\\\\\n$\\therefore$ $AC=\\sqrt{AB^{2}-BC^{2}}=\\sqrt{10^{2}-5^{2}}=\\boxed{5\\sqrt{3}}$",
    "image_path": "dataset/mm_math/images/MM_Math/53102153.png"
  },
  {
    "solution": "\\boxed{y=-2x-3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line \\(l_{1}\\) intersects the y-axis at point A in the Cartesian coordinate system, and point B \\((-3, 3)\\) is also on line \\(l_{1}\\). By moving point B 1 unit to the right and then 2 units downward, we obtain point C, which is also on line \\(l_{1}\\). Determine the coordinates of point C and the equation of line \\(l_{1}\\).\n",
    "completion": "\\textbf{Solution:} By the translation rule, the coordinates of point C are $(-3+1, 3-2)$, i.e., $(-2, 1)$.\\\\\nLet the equation of line $l_{1}$ be $y=kx+c$,\\\\\nthen $\\begin{cases}3=-3k+c\\\\ 1=-2k+c\\end{cases}$, solving this gives: $\\begin{cases}k=-2\\\\ c=-3\\end{cases}$,\\\\\n$\\therefore$ the equation of line $l_{1}$ is $y=-2x-3$.\\\\\nTherefore, the coordinates of point C are $\\boxed{(-2, 1)}$ and the equation of line $l_{1}$ is $\\boxed{y=-2x-3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52019425_104.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the rectangle $OABC$ has its vertex $O$ at the origin. Side $OC$ is on the positive half of the $x$-axis, and side $OA$ is on the positive half of the $y$-axis, with $OA=3$ and $AB=4$. The graph of the inverse proportion function $y=\\frac{k}{x}$ ($k>0$) intersects sides $AB$ and $BC$ at points $D$ and $E$, respectively, and $BD=2AD$. Find the coordinates of point $D$ and the value of $k$?\n",
    "completion": "\\textbf{Solution:} Given that $AB=4$ and $BD=2AD$,\\\\\nit follows that $AB=AD+BD=AD+2AD=3AD=4$,\\\\\nhence $AD=\\frac{4}{3}$.\\\\\nAlso, given that $OA=3$,\\\\\nit follows that $D\\left(\\frac{4}{3}, 3\\right)$.\\\\\nSince point D lies on the hyperbola $y=\\frac{k}{x}$,\\\\\nit follows that $k=\\frac{4}{3}\\times 3=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51402967_34.png"
  },
  {
    "solution": "\\boxed{6-3\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $M$ and $N$ are points on sides $AB$ and $BC$, respectively. After folding the rectangle $ABCD$ along line $MN$, point $B$ falls on point $E$ on side $AD$. If $AB=4$, $AD=6$, and $AE=2\\sqrt{2}AM$, then what is the length of $CN$?\n",
    "completion": "\\textbf{Solution:} Given the property of folding, we know that $BM=EM$.\\\\\nLet $AM=x$, then $AE=2\\sqrt{2}x$, $BM=EM=AB-AM=4-x$,\\\\\nsince in $\\triangle AEM$, $AM^2+AE^2=EM^2$,\\\\\ntherefore $x^2+(2\\sqrt{2}x)^2=(4-x)^2$,\\\\\nsolving this, we get $x_1=1$, $x_2=-2$ (discard),\\\\\nhence $AM=1$, $AE=2\\sqrt{2}$.\\\\\nAs illustrated, draw $NF\\perp AD$ at point F through point N.\\\\\nTherefore, $\\angle MAE=\\angle EFN=90^\\circ$, $NF=AB=4$, $BN=AF$,\\\\\ntherefore, $\\angle FEN+\\angle FNE=90^\\circ$.\\\\\nSince $\\angle FEN+\\angle AEM=90^\\circ$,\\\\\ntherefore, $\\angle AEM=\\angle FNE$,\\\\\ntherefore, $\\triangle MAE\\sim \\triangle EFN$,\\\\\ntherefore, $\\frac{AM}{EF}=\\frac{AE}{NF}$, that is $\\frac{1}{EF}=\\frac{2\\sqrt{2}}{4}$,\\\\\ntherefore, $EF=\\sqrt{2}$,\\\\\ntherefore, $BN=AF=AE+EF=2\\sqrt{2}+\\sqrt{2}=3\\sqrt{2}$,\\\\\ntherefore, $CN=BC-BN=6-3\\sqrt{2}$.\\\\\nThus, the answer is: $CN=\\boxed{6-3\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52977901.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point \\(C\\) is on line segment \\(AB\\), \\(D\\) is the midpoint of line segment \\(AC\\), and \\(E\\) is the midpoint of line segment \\(BC\\).\n",
    "completion": "\\textbf{Solution:}\n\n(1) Since $D$ is the midpoint of segment $AC$, and $AC=8$,\\\\\nit follows that $DC= \\frac{1}{2}AC=\\frac{1}{2}\\times 8=4$\\\\\nSince $E$ is the midpoint of segment $BC$, and $BC=3$,\\\\\nit follows that $CE= \\frac{1}{2}BC=\\frac{1}{2}\\times 3=\\frac{3}{2}$,\\\\\ntherefore, $DE=DC+CE= 4+\\frac{3}{2}=\\boxed{5\\frac{1}{2}}$;\\\\\n(2) Since point $C$ is on segment $AB$, $D$ is the midpoint of segment $AC$, and $E$ is the midpoint of segment $BC$.\\\\\nit follows that $AC=2DC$, $BC=2CE$,\\\\\ntherefore, $AB=AC+BC=2DC+2CE=2(DC+CE)=2DE=2\\times 5=\\boxed{10}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51508769.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $C$ is a point on line segment $AB$, and $D$ is the midpoint of line segment $BC$. If $AB=10$ and $AC=6$, find the length of $AD$.\n",
    "completion": "\\textbf{Solution:} Since $C$ is a point on line segment $AB$, $AB=10$, $AC=6$,\\\\\nthus $BC=AB-AC=10-6=4$,\\\\\nsince $D$ is the midpoint of line segment $BC$,\\\\\nthus $BD=\\frac{1}{2}BC=2$,\\\\\ntherefore $AD=AB-BD=10-2=\\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55191292.png"
  },
  {
    "solution": "\\boxed{\\frac{3\\sqrt{2}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: The large square $ABCD$, as shown in the figure, is composed of four congruent right-angled triangles and a small square. The diagonal $EF$ of the small square is extended bidirectionally to intersect the side $AB$ and the extension of side $BC$ at points $G$ and $H$, respectively. If the ratio of the area of the large square to that of the small square is $5$, and $GH = 2\\sqrt{5}$, then what is the side length of the large square?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram:\\\\\n$\\because$ The ratio of the area of the large square to the small square is 5,\\\\\n$\\therefore$ $\\frac{AD}{EM} = \\sqrt{5}$,\\\\\n$\\therefore$ AD = $\\sqrt{5}$EM,\\\\\nLet EM = a, AE = b, then AD = $\\sqrt{5}$a,\\\\\nBy Pythagoras' theorem: AE$^2$ + DE$^2$ = AD$^2$,\\\\\n$\\therefore$ b$^2$ + (a + b)$^2$ = ($\\sqrt{5}$a)$^2$,\\\\\n$\\therefore$ 2b$^2$ + 2ab - 4a$^2$ = 0,\\\\\n(b - a)(b + 2a) = 0,\\\\\n$\\because$ b + 2a $\\neq$ 0,\\\\\n$\\therefore$ b - a = 0,\\\\\n$\\therefore$ b = a,\\\\\n$\\therefore$ AE = DM = a,\\\\\nAs shown in the diagram, extend BF to meet CD at N,\\\\\n$\\because$ BN $\\parallel$ DE, CF = FM,\\\\\n$\\therefore$ DN = CN,\\\\\n$\\therefore$ EN = $\\frac{1}{2}$DM = $\\frac{1}{2}$a,\\\\\n$\\because$ PN $\\parallel$ BG,\\\\\n$\\therefore$ $\\frac{FN}{BF} = \\frac{PN}{BG} = \\frac{FP}{GF} = \\frac{\\frac{1}{2}a}{2a} = \\frac{1}{4}$,\\\\\nLet PN = x, then BG = 4x,\\\\\n$\\because$ DE = BF, $\\angle$BFG = $\\angle$DEF, $\\angle$BGF = $\\angle$DPE,\\\\\n$\\therefore$ $\\triangle$BFG $\\cong$ $\\triangle$DEP (AAS),\\\\\n$\\therefore$ PD = BG = 4x,\\\\\nSimilarly, EG = FP,\\\\\n$\\therefore$ DN = 3x = CN,\\\\\n$\\therefore$ PC = 2x,\\\\\n$\\because$ CP $\\parallel$ BG,\\\\\n$\\therefore$ $\\frac{CP}{BG} = \\frac{PH}{GH}$, that is $\\frac{2x}{4x} = \\frac{PH}{2\\sqrt{5}}$,\\\\\n$\\therefore$ PH = PG = $\\sqrt{5}$,\\\\\n$\\because$ $\\frac{FP}{FG} = \\frac{1}{4}$,\\\\\n$\\therefore$ EF = $\\sqrt{2}$a = $\\frac{3}{5}$GP = $\\frac{3}{5}\\sqrt{5}$,\\\\\n$\\therefore$ a = $\\frac{3\\sqrt{10}}{10}$,\\\\\n$\\therefore$ AD = $\\sqrt{5}$a = $\\boxed{\\frac{3\\sqrt{2}}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52514055.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{2}{3}x+2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line \\(l_1\\) passes through points \\(A(0, 2)\\) and \\(C(6, -2)\\). The coordinates of point B are \\( (4, 2) \\). Point P is a moving point on segment AB (point P does not coincide with point A). The line \\(l_2: y = kx + 2k\\) passes through point P and intersects \\(l_1\\) at point M. Through point P, draw \\(PN \\perp l_2\\), intersecting \\(l_1\\) at point N. Find the function expression of \\(l_1\\)?\n",
    "completion": "\\textbf{Solution:} Let the equation of ${l}_{1}$ be: $y={k}_{1}x+b$,\\\\\nSubstituting points $A(0, 2)$ and $C(6, -2)$ into the equation, we get: $\\begin{cases}b=2 \\\\ 6{k}_{1}+b=-2\\end{cases}$,\\\\\nSolving these equations, we find: $\\begin{cases}{k}_{1}=-\\frac{2}{3} \\\\ b=2\\end{cases}$,\\\\\nTherefore, the equation of ${l}_{1}$ is: \\boxed{y=-\\frac{2}{3}x+2}.",
    "image_path": "dataset/mm_math/images/MM_Math/51662536_199.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the inverse proportion function $y_{1}=\\frac{k}{x}$ intersects with the graph of the linear function $y_{2}=ax+b$ at points $A(1, 4)$ and $B(m, -2)$. If point $C$ is symmetrical to point $A$ with respect to the $x$-axis, what is the area of $\\triangle ABC$?\n",
    "completion": "\\textbf{Solution:} Since point C is symmetric to point A with respect to the x-axis, and given A(1, 4), it follows that point C is at (1, -4). By plotting and connecting points AC and BC in the Cartesian coordinate system, we find that AC = 8 and BC = 3. Therefore, the area of triangle $ABC$ is given by $S_{\\triangle ABC} = \\frac{1}{2} \\cdot AC \\cdot BC = \\frac{1}{2} \\times 8 \\times 3 = \\boxed{12}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51445145_140.png"
  },
  {
    "solution": "\\boxed{35^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the circumcircle of quadrilateral $ABCD$ is $\\odot O$, $BC=CD$, $\\angle DAC=35^\\circ$. What is the measure of $\\angle BDC$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle DAC=35^\\circ$, \\\\\nthen $\\angle DBC=\\angle DAC=35^\\circ$, \\\\\nbecause $BC=CD$, \\\\\nthus $\\angle BDC=\\angle DBC=35^\\circ$, \\\\\nTherefore, the answer is: $\\boxed{35^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51262088.png"
  },
  {
    "solution": "\\boxed{144^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $O$ is a point on line $AB$, with $CO \\perp DO$, and $OE$ bisects $\\angle BOC$. If $\\angle COE = \\frac{1}{3}\\angle DOB$, determine the measure of $\\angle AOC$.\n",
    "completion": "\\textbf{Solution:} Let $\\angle COE=x^\\circ$, then $\\angle DOB=3x^\\circ$ \\\\\n$\\because OE$ bisects $\\angle BOC$ \\\\\n$\\therefore \\angle BOE=\\angle COE=x^\\circ$ \\\\\n$\\because CO\\perp DO$ \\\\\n$\\therefore \\angle DOC=90^\\circ$ \\\\\n$\\therefore x^\\circ+x^\\circ+3x^\\circ=90^\\circ$ \\\\\n$\\therefore x=18$ \\\\\n$\\therefore \\angle BOE=\\angle COE=18^\\circ$ \\\\\n$\\therefore \\angle AOC=\\angle AOB-\\angle BOE-\\angle COE=180^\\circ-18^\\circ-18^\\circ=\\boxed{144^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51438609_162.png"
  },
  {
    "solution": "\\boxed{\\frac{9}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, $CD$ is the median to the hypotenuse $AB$. A line through point A perpendicular to $CD$ intersects $CD$ and $CB$ at points E and F, respectively. If $\\tan B = \\frac{2}{3}$, what is the value of $\\frac{AE}{EF}$?\n",
    "completion": "\\textbf{Solution:} Given $\\because$ $\\angle ACB=90^\\circ$ and CD is the median to the hypotenuse AB,\\\\\n$\\therefore$ CD=BD, $\\angle ACD+\\angle BCD=90^\\circ$,\\\\\n$\\therefore$ $\\angle BCD=\\angle B$,\\\\\n$\\therefore$ $\\tan\\angle B=\\tan\\angle BCD=\\frac{EF}{CE}=\\frac{2}{3}$,\\\\\n$\\therefore$ let EF=2x, CE=3x,\\\\\n$\\because$ AE$\\perp$CD,\\\\\n$\\therefore$ $\\angle AEC=\\angle CEF=90^\\circ$,\\\\\n$\\therefore$ $\\angle CAF+\\angle ACD=90^\\circ$,\\\\\n$\\therefore$ $\\angle CAF=\\angle BCD$,\\\\\n$\\therefore$ $\\triangle AEC\\sim \\triangle CEF$,\\\\\n$\\therefore$ $\\frac{AE}{CE}=\\frac{CE}{EF}$,\\\\\n$\\therefore$ AE=$\\frac{CE^2}{EF}=\\frac{9}{2}x$,\\\\\n$\\therefore$ $\\frac{AE}{EF}=\\frac{\\frac{9}{2}x}{2x}=\\boxed{\\frac{9}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52942173.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the regular hexagon $ABCDEF$, the diagonals $AC$, $BD$ intersect at point $M$. What is the value of $\\frac{AM}{CM}$?\n",
    "completion": "\\textbf{Solution:} Since hexagon ABCDEF is a regular hexagon, \\\\\n$\\therefore \\angle BCD=\\angle ABC=120^\\circ$, AB=BC=CD, \\\\\n$\\therefore \\angle BAC=\\angle ACB=\\angle CBD=\\angle CDB=30^\\circ$, \\\\\n$\\therefore CM=BM$, $\\angle ABM=90^\\circ$, \\\\\nLet $BM=a$, then $CM=a$, \\\\\n$\\therefore AM=2BM=2a$, \\\\\n$\\therefore \\frac{AM}{CM}=\\frac{2a}{a}=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53080530.png"
  },
  {
    "solution": "\\boxed{H\\left(5, \\frac{5}{2}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line passing through the origin $y=2x$ intersects the inverse proportion function $y_{1}=\\frac{2}{x}$ at point $B$, and intersects another inverse proportion function $y_{2}=\\frac{k}{x}$ at point $C$, with $OB=BC$. Point $A$ has an abscissa of $4$ and is a point on $y_{1}=\\frac{2}{x}$. A perpendicular line from point $B$ to the x-axis is drawn, and the foot of the perpendicular is point $D$. Suppose a moving point $Q$ is on the graph of $y_{2}=\\frac{k}{x}$. Is there a point $H$ in the plane such that points $A$, $B$, $Q$, $H$ can form a rectangle with $AB$ as one side? If it exists, please write down the coordinates of point $H$ directly; if not, please explain the reason.\n",
    "completion": "\\textbf{Solution:} There exists a point $\\boxed{H\\left(5, \\frac{5}{2}\\right)}$ such that points $A$, $B$, $Q$, and $H$ can form a rectangle with $AB$ as a side.",
    "image_path": "dataset/mm_math/images/MM_Math/51322681_144.png"
  },
  {
    "solution": "\\boxed{quadrilateral \\, ADCE \\, is \\, a \\, parallelogram}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $D$ is any point on side $AB$, $F$ is the midpoint of $AC$, and line $CE \\parallel AB$ is drawn through point $C$ intersecting the extension of $DF$ at point $E$. $AE$ and $CD$ are connected. Prove that the quadrilateral $ADCE$ is a parallelogram.\n",
    "completion": "Proof: Since $AB \\parallel CE$, \\\\\nit follows that $\\angle CAD = \\angle ACE$ and $\\angle ADE = \\angle CED$. \\\\\nSince $F$ is the midpoint of $AC$, \\\\\nthen $AF = CF$. \\\\\nIn $\\triangle AFD$ and $\\triangle CFE$, \\\\\nwe have $\\left\\{\\begin{array}{l}\n\\angle CAD = \\angle ACE \\\\\n\\angle ADE = \\angle CED \\\\\nAF = CF\n\\end{array}\\right.$, \\\\\nthus $\\triangle AFD \\cong \\triangle CFE$ (AAS), \\\\\ntherefore $DF = EF$, \\\\\nhence \\boxed{quadrilateral \\, ADCE \\, is \\, a \\, parallelogram}.",
    "image_path": "dataset/mm_math/images/MM_Math/51737267_151.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC$, $\\angle BAC=110^\\circ$, $D$ is a point on $BC$, connect $AD$, make $\\angle ADE=35^\\circ$, and $DE$ intersects line segment $AC$ at point $E$. Write down the sizes of $\\angle B$ and $\\angle C$ directly.\n",
    "completion": "\\textbf{Solution}: Let $ \\angle B = \\angle C = \\boxed{30^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/53009760_108.png"
  },
  {
    "solution": "\\boxed{80^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $ OM $, $ OB $, and $ ON $ are three rays within $ \\angle AOC $, where $ OM $ and $ ON $ are the angle bisectors of $ \\angle AOB $ and $ \\angle BOC $, respectively. Furthermore, $ \\angle NOC $ is three times that of $ \\angle AOM $, and $ \\angle BON $ is $ 20^\\circ $ larger than $ \\angle MOB $. Determine the degree measure of $ \\angle AOC $.\n",
    "completion": "Let the measure of $\\angle AOM$ be $x$, then the measure of $\\angle NOC$ is $3x$,\\\\\nsince OM and ON are the bisectors of $\\angle AOB$ and $\\angle BOC$ respectively,\\\\\ntherefore $\\angle MOB=\\angle AOM=x$ and $\\angle BON=\\angle NOC=3x$,\\\\\nsince $\\angle BON$ is $20^\\circ$ greater than $\\angle MOB$,\\\\\ntherefore $3x-x=20^\\circ$,\\\\\ntherefore $x=10^\\circ$,\\\\\ntherefore $\\angle AOC=\\angle AOM+\\angle MOB+\\angle BON+\\angle NOC=8x=\\boxed{80^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52794817_105.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the line segment $AB=10\\,\\text{cm}$, and point C is a point on the line segment AB. If M is the midpoint of AC and $AM=2\\,\\text{cm}$, find the length of the line segment BC.\n",
    "completion": "\\textbf{Solution:} Given that $M$ is the midpoint of $AC$, and $AM=2\\,\\text{cm}$, \\\\\nthus $AC=2AM=4\\,\\text{cm}$, \\\\\nsince $AB=10\\,\\text{cm}$, \\\\\ntherefore $BC=AB-AC=\\boxed{6}\\,\\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55232695.png"
  },
  {
    "solution": "\\boxed{\\frac{40}{17}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the right-angled $\\triangle ABC$ with $\\angle C=90^\\circ$, $AC=8$, and $BC=6$, point $M$ moves from point $C$ along the segment $CA$ towards point $A$, and $BM$ is connected, with $MN\\perp BM$ intersecting side $AB$ at point $N$. If $CM=2$, then how long is the segment $AN$? When point $M$ moves from $C$ to the midpoint of $AC$, what is the length of the path traversed by point $N$?\n",
    "completion": "\\textbf{Solution:}\n\n(1) As shown in the diagram, construct $ND\\perp AC$, it is easy to determine $AB=10$, and it is found that $\\triangle BCM \\sim \\triangle MDN$, \\\\\n$\\therefore \\frac{MD}{ND}=\\frac{BC}{MC}=\\frac{6}{2}=3$, let $ND=x$, then $MD=3x$, \\\\\nthus $AD=AC-CM-MD=6-3x$, it is easy to determine $\\triangle DNA \\sim \\triangle CBA$, \\\\\n$\\therefore \\frac{AD}{AC}=\\frac{ND}{BC}=\\frac{AN}{AB}$, \\\\\n$\\therefore \\frac{6-3x}{8}=\\frac{x}{6}=\\frac{AN}{10}$, \\\\\n$\\therefore AN=\\boxed{\\frac{30}{13}}$. \\\\\n(2) Following the same reasoning as in (1), we obtain $AN=\\boxed{\\frac{40}{17}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51369939.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, rectangle $OABC$ is in the Cartesian coordinate system, where the coordinate of vertex $B$ is $(4, n)$, and the diagonals $OB$ and $AC$ intersect at $D$. The line $y=\\frac{3}{4}nx-\\frac{3}{2}n$ intersects $OA$, $AC$, and $OB$ at points $P$, $M$, and $N$, respectively. When $CN=2MN$, what is the value of $n$?\n",
    "completion": "\\textbf{Solution:} Since point M is the midpoint of the line segment PN,\\\\\n$PN=2MN=CN$,\\\\\nSince $CN=\\sqrt{3^2+\\left(n-\\frac{3}{4}n\\right)^2}=\\sqrt{9+\\frac{1}{16}n^2}$,\\\\\n$PN=\\sqrt{(2-3)^2+\\left(0-\\frac{3}{4}n\\right)^2}=\\sqrt{1+\\frac{9}{16}n^2}$,\\\\\nTherefore, $\\sqrt{9+\\frac{1}{16}n^2} = \\sqrt{1+\\frac{9}{16}n^2}$,\\\\\nSolving this, we get: $n=4$ or $n=-4$,\\\\\nSince $n>0$,\\\\\nTherefore, $n=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52303875_136.png"
  },
  {
    "solution": "\\boxed{(4,3)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, $OB=5$, $\\sin\\angle AOB= \\frac{3}{5}$, and the coordinate of point $A$ is $(10,0)$. What are the coordinates of point $B$?\n",
    "completion": "\\textbf{Solution:} Draw $BC\\perp OA$ at $C$ through point $B$. In $\\triangle BOC$, $\\sin\\angle AOB= \\frac{BC}{OB}=\\frac{3}{5}$,\\\\\ntherefore $BC=3$. According to the problem, $OB^2=BC^2+OC^2$,\\\\\ntherefore $OC=\\sqrt{OB^2-BC^2}=\\sqrt{5^2-3^2}=4$,\\\\\ntherefore the coordinates of point $B$ are $\\boxed{(4,3)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51083577_99.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{2}a}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the length of segment AB is $a$, and segment AB is extended to point C so that $BC = \\frac{1}{2}AB$. Find the length of segment AC (expressed as an algebraic expression involving $a$).\n",
    "completion": "\\textbf{Solution:} Given that $\\because AB=a$,\\\\\n$\\therefore BC=\\frac{1}{2}a$,\\\\\n$\\therefore AC=AB+BC=a+\\frac{1}{2}a=\\boxed{\\frac{3}{2}a}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53462104_53.png"
  },
  {
    "solution": "\\boxed{\\frac{4}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, composed of small squares with a side length of 1, points A, B, and C are all located at the intersections of the grid lines. Circle $O$ with AB as its diameter passes through point C. If point D is on circle $O$, then what is the value of $\\tan \\angle ADC$?\n",
    "completion": "\\textbf{Solution:} Construct $AC$ and $BC$, \\\\\nsince $AB$ is the diameter of the circle $O$, \\\\\nit follows that $\\angle ACB = 90^\\circ$, \\\\\nsince $\\angle ABC = \\angle ADC$, \\\\\nit follows that $\\tan\\angle ADC = \\tan\\angle ABC = \\frac{AC}{BC} = \\boxed{\\frac{4}{3}}$",
    "image_path": "dataset/mm_math/images/MM_Math/52601657.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, in quadrilateral $ABCD$, $AB \\parallel DC$, $AB=DC$, and the diagonals $AC$ and $BD$ intersect at point $O$, with $AC$ bisecting $\\angle BAD$. $\\angle DAC = \\angle DCA$; quadrilateral $ABCD$ is a rhombus. As shown in Figure 2, draw $CE \\perp AB$ extended to meet at point $E$, and connect $OE$. If $AB= \\sqrt{5}$ and $BD=2$, what is the length of $OE$?\n",
    "completion": "\\textbf{Solution:} \n\nSince quadrilateral ABCD is a rhombus and $OA=OC$, $OB=OD$, $BD\\perp AC$, and $CE\\perp AB$, it follows that $OE=OA=OC$. Given that $BD=2$, thus $OB=\\frac{1}{2}BD=1$. In $\\triangle AOB$, by the Pythagorean theorem, we have: $OA=\\sqrt{AB^{2}-OB^{2}}=\\sqrt{(\\sqrt{5})^{2}-1^{2}}=\\boxed{2}$, therefore $OE=OA=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52422726_85.png"
  },
  {
    "solution": "\\boxed{36}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the right-angled triangle $ABC$, $\\angle ACB=90^\\circ$, $\\angle ABC=30^\\circ$. Three squares are constructed externally on the sides AC, BC, and AB of $\\triangle ABC$, and the areas of these three squares are denoted as $S_1$, $S_2$, and $S_3$, respectively. Given that $S_1=9$, what is the value of $S_3$?\n",
    "completion": "\\textbf{Solution:} Since $S_{1}=AC^{2}=9$, it results that $AC=3$. In $\\triangle ABC$, $\\angle ACB=90^\\circ$ and $\\angle ABC=30^\\circ$, hence $AB=2AC=6$. Therefore, $S_{3}=AB^{2}=\\boxed{36}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52774835.png"
  },
  {
    "solution": "\\boxed{10^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $ \\angle AOB=30^\\circ$, $ \\angle AOE=140^\\circ$, $OB$ bisects $ \\angle AOC$, and $OD$ bisects $ \\angle AOE$. Find the degree measure of $ \\angle COD$.\n",
    "completion": "\\textbf{Solution:}\n\nSince $\\angle AOE = 140^\\circ$ and $\\angle AOC = 60^\\circ$,\n\nit follows that $\\angle COE = \\angle AOE - \\angle AOC = 140^\\circ - 60^\\circ = 80^\\circ$.\n\nAlso, since OD bisects $\\angle AOE$,\n\nwe have $\\angle DOE = \\frac{1}{2} \\angle AOE = \\frac{1}{2} \\times 140^\\circ = 70^\\circ$.\n\nTherefore, $\\angle COD = \\angle COE - \\angle DOE = 80^\\circ - 70^\\circ = \\boxed{10^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51495222_146.png"
  },
  {
    "solution": "\\boxed{50^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\angle AOB$ is a right angle, $OP$ bisects $\\angle AOB$, $OQ$ bisects $\\angle AOC$, $\\angle POQ=70^\\circ$, find the degree of $\\angle AOC$.\n",
    "completion": "\\textbf{Solution:}\\\\\nSince $\\angle AOB=90^\\circ$ and OP bisects $\\angle AOB$,\\\\\nit follows that $\\angle POA=45^\\circ$,\\\\\nSince $\\angle POQ=70^\\circ$,\\\\\nit results in $\\angle AOQ=\\angle POQ-\\angle POA=25^\\circ$,\\\\\nSince OQ bisects $\\angle AOC$,\\\\\nit concludes that $\\angle AOC=2\\angle AOQ=\\boxed{50^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/41968767.png"
  },
  {
    "solution": "\\boxed{40^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, the perpendicular bisector $OM$ of side $AB$ intersects with the perpendicular bisector $ON$ of side $AC$ at point $O$. These two perpendicular bisectors respectively intersect $BC$ at points $D$ and $E$. Given $\\angle ABC=30^\\circ$ and $\\angle ACB=40^\\circ$, what is the measure of $\\angle DAE$ in degrees?\n",
    "completion": "\\textbf{Solution:} Since $\\angle ABC=30^\\circ$ and $\\angle ACB=40^\\circ$, \\\\\nit follows that $\\angle BAC=180^\\circ-\\angle ABC-\\angle ACB=180^\\circ-30^\\circ-40^\\circ=110^\\circ$, \\\\\nSince $DM$ is the perpendicular bisector of segment $AB$, \\\\\nit follows that $DA=DB$, \\\\\nTherefore, $\\angle DAB=\\angle ABC=30^\\circ$, \\\\\nSimilarly, $EA=EC$, \\\\\nTherefore, $\\angle EAC=\\angle ACB=40^\\circ$, \\\\\nTherefore, $\\angle DAE=\\angle BAC-\\angle BAD-\\angle EAC=110^\\circ-30^\\circ-40^\\circ=\\boxed{40^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53147720_118.png"
  },
  {
    "solution": "\\boxed{115^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the circle $\\odot O$ is the incircle of $\\triangle ABC$, and $\\angle ABC=50^\\circ$, $\\angle ACB=80^\\circ$. What is the measure of $\\angle BOC$?\n",
    "completion": "\\textbf{Solution:} Since $\\odot O$ is the inscribed circle of $\\triangle ABC$,\\\\\nit follows that $OB$ and $OC$ are the angle bisectors of $\\triangle ABC$,\\\\\ntherefore $\\angle OBC=\\frac{1}{2}\\angle ABC=25^\\circ$, $\\angle OCB=\\frac{1}{2}\\angle ACB=40^\\circ$,\\\\\nhence $\\angle BOC=180^\\circ-\\angle OBC-\\angle OCB=\\boxed{115^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52551436.png"
  },
  {
    "solution": "\\boxed{(-7,0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=\\frac{4}{3}x+4$ intersects the x-axis and the y-axis at points $A$ and $B$ respectively. A line is drawn through point $B$ intersecting the x-axis at point $P$. If the area of $\\triangle ABP$ is $8$, find the coordinates of point $P$.\n",
    "completion": "\\textbf{Solution:} Let the point $P(x,0)$, \\\\\nthen the area of $\\triangle ABP = \\frac{1}{2} \\times AP \\times OB = \\frac{1}{2} \\times 4 \\times |x+3| = 2|x+3|$, solving for $x$ gives $x=1$ or $x=-7$, \\\\\n$\\therefore$ the coordinates of point $P$ are $\\boxed{(1,0)}$ or $\\boxed{(-7,0)}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52978850_130.png"
  },
  {
    "solution": "\\boxed{y=x^{2}+2x-3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the parabola $y=x^2+bx+c$ intersects the x-axis at points $A$ and $B$, where the coordinates of point $A$ are $(-3,0)$, and it intersects the y-axis at point $C$. Point $D(-2,-3)$ is on the parabola. What is the equation of the parabola?\n",
    "completion": "\\textbf{Solution:} Solve from the problem's statement, we obtain\n\\[\n\\left\\{\\begin{array}{l}\n9-3b+c=0\\\\\n4a-2b+c=-3\n\\end{array}\\right.\n\\]\nSolving yields:\n\\[\n\\left\\{\\begin{array}{l}\nb=2\\\\\nc=-3\n\\end{array}\\right.\n\\]\n$\\therefore$ The equation of the parabola is $\\boxed{y=x^{2}+2x-3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52586455_120.png"
  },
  {
    "solution": "\\boxed{P_{3}(8,9)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $OABC$, point $A$ is on the $x$-axis, point $C$ is on the $y$-axis, and the coordinates of point $B$ are $\\left(6, 8\\right)$. The rectangle $OABC$ is folded along the line $BD$ so that point $C$ exactly falls on point $E$ on the diagonal $OB$. The fold line $BD$ intersects with the $y$-axis and the $x$-axis at points $D$ and $F$, respectively. If point $M$ is on the line $y=-\\frac{1}{2}x$, then is there a point $P$ on the line $BD$ such that the quadrilateral with vertices $C$, $D$, $M$, and $P$ is a parallelogram? Find the coordinates of the point $P$ that satisfy the condition.\n",
    "completion": "\\textbf{Solution}: Let there exist a point $P$ on line $BD$ such that the quadrilateral constituted by vertices $C$, $D$, $M$, and $P$ forms a parallelogram,\\\\\n$\\because$ from (2) we know the equation of line $BD$ is $y=0.5x+5$,\\\\\n$\\therefore D(0,5)$,\\\\\nand $\\because C(0,8)$,\\\\\n$\\therefore CD=3$,\\\\\nPoint $M$ lies on the line $y=-0.5x$, and point $P$ lies on line $BD$,\\\\\nTo make the quadrilateral with vertices $C$, $D$, $M$, and $P$ a parallelogram,\\\\\nit requires either $CD$ to be parallel and equal to $MP$ or $CP$ to be parallel and equal to $MD$,\\\\\nWhen $CD$ is parallel and equal to $MP$, let the coordinate of point $P$ be $(m,0.5m+5)$, then $M(m,-0.5m)$,\\\\\n$\\therefore MP=|\\left(0.5m+5\\right)-\\left(-0.5m\\right)|=3$,\\\\\nFrom which, $m_{1}=-2$, $m_{2}=-8$,\\\\\n$\\therefore P_{1}(-2,4)$, $P_{2}(-8,1)$\\\\\nWhen $CP$ and $MD$ are parallel and equal, let the coordinate of point $P$ be $(m,0.5m+5)$, then $M(-m,0.5m)$,\\\\\n$\\therefore |8-(0.5m+5)|=|0.5m-5|$,\\\\\nFrom which, $m=8$,\\\\\n$\\therefore P_{3}(8,9)$\\\\\nThus, the coordinates of point $P$ satisfying the requirements are \\boxed{P_{1}(-2,4)}, \\boxed{P_{2}(-8,1)}, and \\boxed{P_{3}(8,9)}.",
    "image_path": "dataset/mm_math/images/MM_Math/52094893_172.png"
  },
  {
    "solution": "\\boxed{\\frac{9}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the analytic expression of line $ l_{1}$ is: $ y=-3x+3$, and line $ l_{1}$ intersects the x-axis at point D, line $ l_{2}$ passes through point A$(4,0)$ and point $B\\left(3, -\\frac{3}{2}\\right)$, lines $ l_{1}$ and $ l_{2}$ intersect at point C. What is the area of $\\triangle ADC$?\n",
    "completion": "Solution: By solving the system of equations of lines $l_1$ and $l_2$\n\\[\n\\begin{cases}\ny=-3x+3\\\\\ny=\\frac{3}{2}x-6\n\\end{cases}\n\\]\nwe obtain\n\\[\n\\begin{cases}\nx=2\\\\\ny=-3\n\\end{cases}\n\\]\n$\\therefore C(2, -3)$, $\\because$ in $\\triangle ADC$, $AD=3$, and the height from $AD$ is $|-3|=3$, $\\therefore$ the area of $\\triangle ADC$ is $\\frac{1}{2} \\times 3 \\times 3 = \\boxed{\\frac{9}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52395164_95.png"
  },
  {
    "solution": "\\boxed{9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y_{1}=x+2$ is the line $l_{1}$, and the graph of the linear function $y_{2}=kx+b$ is the line $l_{2}$. The two lines intersect at point $A(1,a)$. It is known that lines $l_{1}$ and $l_{2}$ intersect the $x$-axis at points $B$ and $C$ respectively, and line $l_{2}$ intersects the $y$-axis at point $E(0,4)$. Find the area of $\\triangle ABC$.\n",
    "completion": "\\textbf{Solution:} Let $y_2=0$, that is, $-x+4=0$ \\\\\nSolving this, we get $x=4$ \\\\\nHence, $C(4, 0)$ \\\\\nTherefore, the area of $\\triangle ABC = \\frac{1}{2} \\times BC \\times y_A = \\frac{1}{2}[4-(-2)]\\times 3 = \\boxed{9}$",
    "image_path": "dataset/mm_math/images/MM_Math/52395370_110.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{2}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $A$, $B$, $C$ are the vertices of small squares, and the side length of each small square is $1$. What is the value of $\\sin\\angle ABC$?\n",
    "completion": "\\textbf{Solution:} Connect AC, \\\\\nAC $= \\sqrt{5}$, BC $= \\sqrt{5}$, AB $= \\sqrt{10}$, \\\\\n$\\therefore \\triangle ABC$ is a right-angled triangle, \\\\\n$\\therefore \\sin\\angle ABC = \\frac{AC}{AB} = \\frac{\\sqrt{5}}{\\sqrt{10}} = \\boxed{\\frac{\\sqrt{2}}{2}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51298723.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the vertex $P$ of a $30^\\circ$ angle of a right-angled triangle board falls on the circle $\\odot O$, and the two sides intersect the circle $\\odot O$ at points $A$ and $B$, respectively. If $AO$ and $BO$ are connected, what is the degree measure of $\\angle AOB$?\n",
    "completion": "\\textbf{Solution}: Since $\\angle P=30^\\circ$, $\\angle AOB=2\\angle P$\\\\\nTherefore, $\\angle AOB=\\boxed{60^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/53080520.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the sector $ABC$ with a radius of $6$, where $\\angle AOB = 120^\\circ$, $C$ is a moving point on the arc $\\overset{\\frown}{AB}$ (not coinciding with $A$ or $B$), $OD \\perp AC$, $OE \\perp BC$, with the feet of the perpendiculars being points $D$ and $E$, respectively. Find the measures of each interior angle of the quadrilateral $ODCE$.\n",
    "completion": "\\textbf{Solution:} As the figure shows, take a point $H$ on the major arc $AB$, and connect $AH$, $BH$, \\\\\n$\\therefore \\angle AHB=\\frac{1}{2}\\angle AOB=60^\\circ$, \\\\\n$\\because$ quadrilateral $ACBH$ is a cyclic quadrilateral, \\\\\n$\\therefore \\angle DCE=180^\\circ-\\angle AHB=\\boxed{120^\\circ}$, \\\\\n$\\because OD\\perp AC$, $OE\\perp BC$, \\\\\n$\\therefore \\angle ODC=\\angle OEC=90^\\circ$, \\\\\n$\\therefore \\angle BOD=360^\\circ-\\angle ODC-\\angle OEC-\\angle DCE=\\boxed{60^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52871720_105.png"
  },
  {
    "solution": "\\boxed{y=\\frac{1}{3}x+2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $A(0,2)$, $B(3,3)$, and $C(7, -1)$, the quadrilateral $OBCD$ is a rectangle, and $BD$ intersects the x-axis at point $E$. Find the equation of line $AB$.\n",
    "completion": "\\textbf{Solution:} Let the equation of line AB be $y=kx+b$. Substituting point A$(0,2)$ and B$(3,3)$ into $y=kx+b$ gives\n\\[\n\\begin{cases}\nb=2\\\\\n3k+b=3\n\\end{cases}\n\\]\nSolving this, we get\n\\[\n\\begin{cases}\nk=\\frac{1}{3}\\\\\nb=2\n\\end{cases}\n\\]\n$\\therefore \\boxed{y=\\frac{1}{3}x+2}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52691041_53.png"
  },
  {
    "solution": "\\boxed{65^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, AB is a chord of $\\odot O$. Draw OC $\\perp$ OA, where OC intersects AB at P, and CP = BC. Point Q is a point on the arc $\\overset{\\frown}{AmB}$. BC is a tangent of $\\odot O$; given that $\\angle BAO = 25^\\circ$, what is the degree measure of $\\angle AQB$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle BAO=25^\\circ$,\\\\\nit follows that $\\angle ABO=25^\\circ$, $\\angle APO=65^\\circ$,\\\\\nthus $\\angle POB=\\angle APO-\\angle ABO=40^\\circ$,\\\\\ntherefore $\\angle AQB=\\frac{1}{2}(\\angle AOP+\\angle POB)=\\frac{1}{2}\\times 130^\\circ=\\boxed{65^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52622752_93.png"
  },
  {
    "solution": "\\boxed{\\frac{10\\sqrt{2}}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $AB$ is the diameter of circle $O$, and chord $CD$ intersects with diameter $AB$ at point $E$, with $AC=CD$. Extend line $CO$ to intersect chord $AD$ at point $F$. Connect $AC$, $BC$, $BD$, with $AB=10$ and $BE=2$. What is the length of $AF$?\n",
    "completion": "\\textbf{Solution:} Given that $AO=OB$ and $AF=FD$, \\\\\nhence $OF=\\frac{1}{2}BD=\\frac{5}{3}$, \\\\\nin $\\triangle AOF$, $AF=\\sqrt{OA^{2}-OF^{2}}=\\sqrt{5^{2}-\\left(\\frac{5}{3}\\right)^{2}}=\\boxed{\\frac{10\\sqrt{2}}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51317366_155.png"
  },
  {
    "solution": "\\boxed{4.8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given: As shown in the figure, $\\triangle ABC$ is inscribed in $\\odot O$, with $AB$ being the diameter. The bisector of $\\angle CBA$ intersects $AC$ at point $F$ and intersects $\\odot O$ at point $D$. $DE \\perp AB$ at point $E$, and it intersects $AC$ at point $P$. Connect $AD$. $\\angle DAC = \\angle DBA$; Connect $CD$. If $CD=6$ and $BD=8$, find the radius of $\\odot O$ and the length of $DE$.\n",
    "completion": "Solution: As shown in the diagram:\\\\\n$\\because \\angle CBD=\\angle DBA$,\\\\\n$\\therefore AD=CD=6$,\\\\\n$\\therefore AB=\\sqrt{AD^{2}+BD^{2}}=10$,\\\\\n$\\therefore$ the radius of $\\odot O$ is $\\boxed{5}$,\\\\\n$\\because \\frac{1}{2}AD\\cdot BD=\\frac{1}{2}AB\\cdot DE$,\\\\\n$\\therefore \\frac{1}{2}\\times 6\\times 8=\\frac{1}{2}\\times 10\\times DE$,\\\\\nSolving gives: $DE=\\boxed{4.8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53080543_100.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated, in $\\triangle ABC$, $\\angle ACB = 90^\\circ$, $CD$ and $CH$ are the median and the altitude on side $AB$, respectively, $BC = \\sqrt{14}$, $\\cos\\angle ACD = \\frac{3}{4}$, calculate the lengths of $AB$ and $CH$.\n",
    "completion": "\\textbf{Solution:} Draw DE $\\perp$ AC at E, then $\\angle AED = \\angle CED = 90^\\circ$,\\\\\nsince $\\angle ACB = 90^\\circ$,\\\\\nit follows that $\\angle AED = \\angle ACB$,\\\\\nhence, DE $\\parallel$ BC,\\\\\nsince CD is the median of $\\triangle ABC$,\\\\\nit follows that AD = BD,\\\\\nhence, CE = AE, i.e., AC = 2CE\\\\\nsince BC = $\\sqrt{14}$,\\\\\nit follows that DE = $\\frac{1}{2}$BC = $\\frac{\\sqrt{14}}{2}$,\\\\\nsince $\\cos\\angle ACD = \\frac{CE}{CD} = \\frac{3}{4}$\\\\\nlet CE = 3x, CD = 4x,\\\\\nby the Pythagorean theorem, we get: $DE = \\sqrt{CD^2 - CE^2} = \\sqrt{(4x)^2 - (3x)^2} = \\sqrt{7}x$\\\\\nhence, $\\sqrt{7}x = \\frac{\\sqrt{14}}{2}$, i.e., x = $\\frac{\\sqrt{2}}{2}$\\\\\nhence, AE = CE = 3x = $\\frac{3\\sqrt{2}}{2}$\\\\\nhence, AC = AE + CE = $3\\sqrt{2}$\\\\\nsince $\\cos\\angle ACD = \\frac{AC}{AB} = \\frac{3}{4}$, i.e., $\\frac{3\\sqrt{2}}{AB} = \\frac{3}{4}$\\\\\nhence, AB = $4\\sqrt{2}$\\\\\nsince $S_{\\triangle ABC} = \\frac{1}{2} \\times AC \\times BC = \\frac{1}{2} \\times AB \\times CH$\\\\\nhence, $\\frac{1}{2} \\times 3\\sqrt{2} \\times \\sqrt{14} = \\frac{1}{2} \\times 4\\sqrt{2} \\times CH$, solving this gives: CH = $\\frac{3\\sqrt{14}}{4}$.\\\\\nhence, the length of CH is $\\boxed{\\frac{3\\sqrt{14}}{4}}$, and the length of AB is $\\boxed{4\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51403336.png"
  },
  {
    "solution": "\\boxed{6-2\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Place a right triangular plate as shown in the figure, with point $C$ on the extension of $FD$, $AB\\parallel CF$, $\\angle F = \\angle ACB = 90^\\circ$, $\\angle E = 30^\\circ$, $\\angle A = 45^\\circ$, $AC = 6\\sqrt{2}$. What is the length of $CD$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, construct BM$\\perp$FD at M,\\\\\nIn $\\triangle ACB$, $\\angle ACB=90^\\circ$, $\\angle A=45^\\circ$, $AC=6\\sqrt{2}$,\\\\\n$\\therefore BC=AC=6\\sqrt{2}$,\\\\\n$\\because AB\\parallel CF$,\\\\\n$\\therefore BM=BC\\times\\sin45^\\circ=6\\sqrt{2}\\times \\frac{\\sqrt{2}}{2}=6$,\\\\\n$\\therefore CM=BM=6$,\\\\\nIn $\\triangle EFD$, $\\angle F=90^\\circ$, $\\angle E=30^\\circ$,\\\\\n$\\therefore \\angle EDF=60^\\circ$,\\\\\n$\\therefore MD=\\frac{BM}{\\tan60^\\circ}=\\frac{6}{\\sqrt{3}}=2\\sqrt{3}$,\\\\\n$\\therefore CD=CM-MD=6-2\\sqrt{3}$.\\\\\nThus, the answer is: $CD=\\boxed{6-2\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51231562.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in a square grid with a side length of 1, points A, O, and B are all on the grid points. What is the value of $\\tan\\angle AOB$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect AB.\\\\\nIn the right-angled $\\triangle AOB$, since $\\angle OBA=90^\\circ$, AB=2, OB=4,\\\\\ntherefore $\\tan\\angle AOB= \\frac{AO}{BO}=\\frac{2}{4}=\\boxed{\\frac{1}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52601836.png"
  },
  {
    "solution": "\\boxed{13}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, in the Cartesian coordinate system, point O is the origin, and the quadrilateral $ABCO$ is a rhombus. The coordinates of point A are $(-5, 12)$, and point C is on the positive half-axis of the x-axis. The line AC intersects the y-axis at point M, and the side AB intersects the y-axis at point H. What is the side length of the rhombus $ABCO$?\n",
    "completion": "Solution: Since the coordinates of point A are $(-5, 12)$, \\\\\nthus $AH = 5$, and $OH = 12$, \\\\\nBy the Pythagorean theorem, we have: $OA = \\sqrt{AH^{2}+OH^{2}} = \\sqrt{5^{2}+12^{2}} = \\boxed{13}$, \\\\\nTherefore, the side length of the rhombus $ABCO$ is \\boxed{13}.",
    "image_path": "dataset/mm_math/images/MM_Math/52395385_98.png"
  },
  {
    "solution": "\\boxed{2.5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral ABCD, $\\angle A=90^\\circ$, AB = 4, and AD = 3. Points M and N are moving points on line segments BC and AB, respectively (point M does not coincide with point B). Points E and F are the midpoints of DM and MN, respectively. What is the maximum length of EF?\n",
    "completion": "\\textbf{Solution:} Connect DN and DB as shown in the diagram: \\\\\nIn $\\triangle DAB$, where $\\angle A=90^\\circ$, $AB=4$, and $AD=3$, \\\\\n$\\therefore BD=\\sqrt{AD^2+AB^2}=\\sqrt{3^2+4^2}=5$, \\\\\n$\\because$ Points E and F are the midpoints of DM and MN respectively, \\\\\n$\\therefore EF=\\frac{1}{2}DN$, \\\\\nAccording to the problem, when point N coincides with point B, DN is at its maximum, which is 5, \\\\\n$\\therefore$ The maximum length of $EF$ is $\\boxed{2.5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53706958.png"
  },
  {
    "solution": "\\boxed{(2,2)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, let $A$ be a point on the negative half-axis of the x-axis. A line $AB$ perpendicular to the x-axis is drawn through point $A$ and intersects the line $y=x$ at point $B$. The triangle $\\triangle ABO$ is translated upwards along the line $y=x$ by $5\\sqrt{2}$ units to obtain $\\triangle A'B'O'$. If the coordinates of point $A$ are $(-3, 0)$, what are the coordinates of point $B'$?\n",
    "completion": "\\textbf{Solution:} Because the coordinates of point A are $(-3, 0)$, and AB is perpendicular to the x-axis and intersects the line $y = x$ at point B,\n$\\therefore$ the coordinates of B are $(-3, -3)$,\nMoving $\\triangle ABO$ along the line $y = x$ upwards by $5\\sqrt{2}$ units essentially translates $\\triangle ABO$ 5 units to the right and 5 units upwards,\n$\\therefore$ the coordinates of $B'$ are $(-3+5, -3+5)$, that is $B'(2,2)$,\nThus, the answer is: the coordinates of $B'$ are $\\boxed{(2,2)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50545589.png"
  },
  {
    "solution": "\\boxed{\\frac{2}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, $AB=5$, $BC=\\frac{10}{3}$, point $O$ is on $AB$, $OB=2$, and a circle $\\odot O$ with $OB$ as the radius intersects $BC$ at point $D$. $AC$ is a tangent to $\\odot O$; what is the length of $CD$?\n",
    "completion": "\\textbf{Solution:} Draw OF$\\perp$BC at point O, with F as the foot,\\\\\nsince $\\angle$OEA=$\\angle$C=$\\angle$OFC=90$^\\circ$,\\\\\nthus quadrilateral OFCE is a rectangle,\\\\\nthus OE=CF=2,\\\\\nsince BC=$\\frac{10}{3}$,\\\\\nthus BF=BC−CF=$\\frac{4}{3}$,\\\\\nsince OF$\\perp$BD,\\\\\nthus BD=2BF=$\\frac{8}{3}$,\\\\\nthus CD=BC−BD=$\\boxed{\\frac{2}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52601679_174.png"
  },
  {
    "solution": "\\boxed{(-5,0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given that in the Cartesian coordinate system, point $A(1,2)$ lies on the graph of the inverse proportion function $y=\\frac{k_1}{x}$, the line passing through point $A$ intersects the other branch of the hyperbola at point $B(-2,m)$. If point $C$ is a moving point on the $x$-axis, what are the coordinates of point $C$ when the area of $\\triangle ABC$ is 6?\n",
    "completion": "\\textbf{Solution:} Let the line AB intersect the x-axis at point D. Therefore, the coordinates of point D are (-1,0). Therefore, the area $S_{\\triangle ABC}=S_{\\triangle ADC}+S_{\\triangle BDC}=6$, that is, $\\frac{1}{2} \\times DC \\times 2 + \\frac{1}{2} \\times DC \\times 1 = 6$ $\\therefore CD = 4$. Therefore, the coordinates of point C are $\\boxed{(3,0)}$ or $\\boxed{(-5,0)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50555566_85.png"
  },
  {
    "solution": "\\boxed{(0, -1)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the vertices A and C of the parallelogram OABC are located at (2,0) and (1,2), respectively. Is there a point P on the y-axis such that $\\triangle PAC$ forms a right triangle with P as the right angle vertex? What are the coordinates of P?\n",
    "completion": "\\textbf{Solution:} Let the coordinates of point $P$ be $(0, m)$. Based on the problem statement, we have: $PC^{2}=1^{2}+(2-m)^{2}$, $AC^{2}=1^{2}+2^{2}=5$, $PA^{2}=2^{2}+m^{2}$,\n\n(1) When point $C$ is the right angle vertex, $PC^{2}+AC^{2}=PA^{2}$,\n\n$\\therefore 1^{2}+(2-m)^{2}+1^{2}+2^{2}=2^{2}+m^{2}$,\n\nSolving this, we get: $m=\\frac{3}{2}$,\n\nAt this point, the coordinates of $P$ are $(0, \\frac{3}{2})$;\n\n(2) When point $A$ is the right angle vertex, $PA^{2}+AC^{2}=PC^{2}$,\n\n$\\therefore 2^{2}+m^{2}=1^{2}+(2-m)^{2}+1^{2}+2^{2}$,\n\nSolving this, we get: $m=-1$,\n\nAt this point, the coordinates of $P$ are $(0, -1)$;\n\nIn summary, the coordinates of point $P$ are $\\boxed{(0, \\frac{3}{2})}$ or $\\boxed{(0, -1)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50682377_130.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AD$ is the median of $\\triangle ABC$. If $AB=AC=13$ and $BC=10$, then what is the length of $AD$?\n",
    "completion": "\\textbf{Solution:} Since $AD$ is the median of $\\triangle ABC$, if $AB=AC=13$, $BC=10$, \\\\\nthen $AD\\perp BC$, $BD=\\frac{1}{2}BC=5$, \\\\\nIn $Rt\\triangle ABD$, by the Pythagorean theorem, we have: $AD=\\sqrt{AB^{2}-BD^{2}}=\\boxed{12}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52826905.png"
  },
  {
    "solution": "\\boxed{36^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $OC$ is the angle bisector of $\\angle AOB$, $\\angle AOD = 2\\angle BOD$, and $\\angle COD = 18^\\circ$. Please find the measure of $\\angle BOD$ in degrees.\n",
    "completion": "\\textbf{Solution:} From the construction we get \\\\\nSince $OC$ is the bisector of $\\angle AOB$, \\\\\nTherefore $\\angle BOC=\\frac{1}{2}\\angle AOB$; \\\\\nSince $\\angle AOD=2\\angle BOD$, \\\\\nTherefore $\\angle AOB=3\\angle BOD$, i.e., $\\angle BOD=\\frac{1}{3}\\angle AOB$; \\\\\nTherefore $\\angle COD=\\frac{1}{2}\\angle AOB-\\frac{1}{3}\\angle AOB=\\frac{1}{6}\\angle AOB$, \\\\\nTherefore $\\angle BOD=2\\angle COD$, \\\\\nSince $\\angle COD=18^\\circ$, \\\\\nTherefore $\\angle BOD=\\boxed{36^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52746394.png"
  },
  {
    "solution": "\\boxed{70^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AD$ bisects $\\angle BAC$, $AE \\perp BC$ at point E, $\\angle B = 40^\\circ$, $\\angle DAE = 15^\\circ$, what is the measure of $\\angle C$?\n",
    "completion": "\\textbf{Solution:} Given that in $\\triangle ABC$, $AE\\perp BC$, \\\\\nit follows that $\\angle AEB=90^\\circ$, \\\\\nsince $\\angle B=40^\\circ$, \\\\\nit follows that $\\angle BAE = 90^\\circ - 40^\\circ = 50^\\circ$, \\\\\nsince $\\angle DAE = 15^\\circ$, \\\\\nit follows that $\\angle BAD = 50^\\circ - 15^\\circ = 35^\\circ$, \\\\\nsince $AD$ bisects $\\angle BAC$, \\\\\nit follows that $\\angle BAC = 2\\angle BAD = 70^\\circ$, \\\\\nthus, $\\angle C = 180^\\circ - \\angle B - \\angle BAC = \\boxed{70^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52978751.png"
  },
  {
    "solution": "\\boxed{36}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ is similar to $\\triangle DEF$ with point O as the center of similarity. If $OD=3OA$ and the area of $\\triangle ABC$ is $4$, what is the area of $\\triangle DEF$?\n",
    "completion": "\\textbf{Solution:}\\\\\nSince $\\triangle ABC$ and $\\triangle DEF$ are similar,\\\\\nthus $AC\\parallel DF$,\\\\\nthus $\\triangle OAC\\sim \\triangle ODF$,\\\\\nthus $\\frac{AC}{DF}=\\frac{OA}{OD}=\\frac{1}{3}$,\\\\\nthus $\\frac{{S}_{\\triangle ABC}}{{S}_{\\triangle DEF}}={(\\frac{AC}{DF})}^{2}=\\frac{1}{9}$,\\\\\nsince ${S}_{\\triangle ABC}=4$,\\\\\ntherefore ${S}_{\\triangle DEF}=\\boxed{36}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51407073.png"
  },
  {
    "solution": "\\boxed{48}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, Master Zhao uses a piece of triangular tinplate to cut out a piece of square tinplate for reserve. In $\\triangle ABC$, $BC=120$, the height $AD=80$, the square $EFGH$ has its side $GH$ on side $BC$, and $E, F$ are respectively on sides $AB, AC$. Then, what is the length of the side of the square $EFGH$?\n",
    "completion": "\\textbf{Solution:} Set the side length of the square part as $x$.\\\\\nIn square $EFGH$, $EF\\parallel BC$, $EH\\parallel AD$\\\\\n$\\therefore \\angle AEF=\\angle ABC$, $\\angle EAF=\\angle BAC$; $\\angle BHE=\\angle BDA$, $\\angle B=\\angle B$\\\\\n$\\therefore \\triangle AEF\\sim \\triangle ABC$, $\\triangle BEH\\sim \\triangle BAD$\\\\\n$\\therefore \\frac{EF}{BC}=\\frac{AE}{AB}$, $\\frac{EH}{AD}=\\frac{BE}{BA}$\\\\\n$\\therefore \\frac{EF}{BC}+\\frac{EH}{AD}=\\frac{AE}{AB}+\\frac{BE}{BA}=1$\\\\\n$\\therefore \\frac{x}{120}+\\frac{x}{80}=1$ \\\\\nSolving for $x$ gives: $x=\\boxed{48}$",
    "image_path": "dataset/mm_math/images/MM_Math/52601760.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a set of triangle rulers is placed as shown in figure a, where $\\angle ACB=\\angle DEC=90^\\circ$, $\\angle A=45^\\circ$, $\\angle D=30^\\circ$, the hypotenuse $AB=6\\text{cm}$, $DC=7\\text{cm}$. The triangle ruler DCE is rotated clockwise by $15^\\circ$ around point C to get $\\triangle D_{1}CE_{1}$ (as shown in figure b). At this time, $AB$ intersects $CD_{1}$ at point $O$, and intersects $D_{1}E_{1}$ at point $F$. What is the degree of $\\angle OFE_{1}$, and what is the length of the line segment $AD_{1}$ in cm?\n",
    "completion": "\\textbf{Solution:}\n\nSince $\\angle ACB = \\angle DEC = 90^\\circ$, $\\angle A = 45^\\circ$, $\\angle D = 30^\\circ$,\\\\\nthus $\\angle DCE = 60^\\circ$, $\\angle B = 45^\\circ$\\\\\nSince rotating triangle DCE around point C clockwise by $15^\\circ$ results in $\\triangle D_1CE_1$,\\\\\nthus $\\angle D_1CE_1 = 60^\\circ$, $\\angle E_1 = \\angle DEC = 90^\\circ$, $\\angle BCE_1 = 15^\\circ$\\\\\ntherefore $\\angle OCB = 45^\\circ$, and $\\angle B = 45^\\circ$,\\\\\nthus $\\angle COB = 90^\\circ$,\\\\\nsince $\\angle COB + \\angle D_1CE_1 + \\angle E_1 + \\angle OFE_1 = 360^\\circ$\\\\\ntherefore $\\angle OFE_1 = 120^\\circ$;\\\\\nsince $\\triangle ACB$ is an isosceles right triangle and $\\angle COB = 90^\\circ$,\\\\\nthus $AO = CO = BO = 3\\text{cm}$,\\\\\ntherefore $D_1O = 4\\text{cm}$,\\\\\nthus $AD_1 = \\sqrt{AO^2 + D_1O^2} = \\sqrt{3^2 + 4^2} = \\boxed{5}$",
    "image_path": "dataset/mm_math/images/MM_Math/52755391.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $AB=9$, $AD=3$, and $M$, $N$ are points on $AB$, $CD$ respectively. When quadrilateral $MBCN$ is folded along $MN$, point $B$ coincides exactly with point $D$ and point $C$ with point $E$. Connect $BN$. Quadrilateral $DMBN$ is a rhombus; determine the length of the line segment $AM$.\n",
    "completion": "\\textbf{Solution:} Let $AM=x$, then $DM=BM=AB-AM=9-x$,\\\\\nsince quadrilateral $ABCD$ is a rectangle,\\\\\ntherefore $\\angle A=90^\\circ$,\\\\\nin $\\triangle ADM$, $AD^2+AM^2=DM^2$,\\\\\ntherefore $3^2+x^2=(9-x)^2$, solving gives: $x=\\boxed{4}$,\\\\\nthat is, $AM=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52302802_139.png"
  },
  {
    "solution": "\\boxed{40^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $BD$ bisects $\\angle ABC$, $E$ is a point on $BD$, $EA \\perp AB$, and $EB=EC$, $\\angle EBC=\\angle ECB$. If $\\angle ABC=40^\\circ$, what is the measure of $\\angle DEC$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle ABC=40^\\circ$ and BD bisects $\\angle ABC$,\\\\\nit follows that $\\angle EBC= \\frac{1}{2}\\angle ABC=20^\\circ$,\\\\\nSince $EB=EC$,\\\\\nwe have $\\angle ECB=\\angle EBC=20^\\circ$,\\\\\nBecause $\\angle DEC is an exterior angle of \\triangle EBC$,\\\\\nit follows that $\\angle DEC=\\angle ECB+\\angle EBC=\\boxed{40^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52890230_142.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, AB is the diameter of $\\odot O$, point C is on $\\odot O$, extend BC to point D such that DC=CB, extend DA to another intersection point E with $\\odot O$, and connect AC, CE. If AB=13 and BC−AC=7, what is the length of CE?\n",
    "completion": "\\textbf{Solution:} Let $AC=x$,\\\\\nsince $BC-AC=7$,\\\\\ntherefore $BC=x+7$,\\\\\nin $\\triangle ABC$, where $AB=13$,\\\\\nthus $AB^2=AC^2+BC^2=13^2$, that is $x^2+(x+7)^2=13^2$,\\\\\nhence $x_1=5$, $x_2=-12$ (discard),\\\\\ntherefore $AC=5$,\\\\\ntherefore $BC=CD=12$,\\\\\nsince $\\angle D=\\angle E$,\\\\\nthus $CD=CE$,\\\\\ntherefore $CE=\\boxed{12}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52874653_102.png"
  },
  {
    "solution": "\\boxed{20^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Since there was no text provided in the initial query, I am unable to provide a translated version in LaTeX format. If you have specific text in mind, please share it, and I'll be happy to assist with the translation.\n",
    "completion": "\\textbf{Solution:} Given the problem,\\\\\n\\[\\because \\angle AOD=100^\\circ,\\text{ and }OB \\text{ bisects } \\angle AOD,\\]\n\\[\\therefore \\angle AOB=\\frac{1}{2}\\angle AOD=50^\\circ,\\]\n\\[\\because \\angle AOC : \\angle COB = 3 : 2,\\]\n\\[\\therefore \\angle COB=\\frac{2}{5}\\angle AOC=\\frac{2}{5}\\times 50^\\circ=\\boxed{20^\\circ}.\\]",
    "image_path": "dataset/mm_math/images/MM_Math/51385662_125.png"
  },
  {
    "solution": "\\boxed{137^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\angle AOB$ is a straight angle, $\\angle AOC = 28^\\circ$, $\\angle BOD = 58^\\circ$, $OM$ and $ON$ are the angle bisectors of $\\angle AOC$ and $\\angle BOD$, respectively. Find the measure of $\\angle MON$.\n",
    "completion": "\\textbf{Solution:} From the given, $OM$ and $ON$ are the angle bisectors of $\\angle AOC$ and $\\angle BOD$, respectively, with $\\angle AOC=28^\\circ$ and $\\angle BOD=58^\\circ$.\\\\\nTherefore, $\\angle AOM=\\frac{1}{2}\\angle AOC=14^\\circ$ and $\\angle BON=\\frac{1}{2}\\angle BOD=29^\\circ$.\\\\\nTherefore, $\\angle MON=\\angle AOB-\\angle AOM-\\angle BON=180^\\circ-14^\\circ-29^\\circ=\\boxed{137^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/46874295.png"
  },
  {
    "solution": "\\boxed{\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rhombus ABCD, where AB=2, and $\\angle A=120^\\circ$, points P, Q, and K are arbitrarily located on segments BC, CD, and BD, respectively. What is the minimum value of $PK+QK$?\n",
    "completion": "\\textbf{Solution}: As shown in the figure, construct point $P^{\\prime}$ symmetric to point P with respect to BD, and draw the intersection K between line $P^{\\prime}Q$ and BD,\\\\\n$\\because$ Quadrilateral ABCD is a rhombus, AB=2, $\\angle A=120^\\circ$,\\\\\n$\\therefore$ AD=2, $\\angle ADC=60^\\circ$,\\\\\nDraw AE$\\perp$CD at E, then AE=$P^{\\prime}Q$,\\\\\n$\\because$ AE=AD$\\cdot\\sin60^\\circ=2\\times\\frac{\\sqrt{3}}{2}=\\sqrt{3}$,\\\\\n$\\therefore$ the distance from point $P^{\\prime}$ to CD is $\\sqrt{3}$,\\\\\n$\\therefore$ the minimum value of PK+QK is $\\boxed{\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52688304.png"
  },
  {
    "solution": "\\boxed{\\frac{8}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, line AB intersects the coordinate axes at points A $(0, -3)$ and B $(-5, 0)$, respectively. Point C is the midpoint of line segment AB, and point P is a point on the y-axis. Connect CP, and draw a perpendicular from C to line segment BO intersecting at point Q. As shown in Figure 2, when point Q coincides with point B, connect PQ. Find the length of PO.\n",
    "completion": "\\[ \\text{Solution: From the given, point } C \\text{ is the midpoint of segment } AB, \\]\n\\[ \\therefore C\\left( -\\frac{5}{2}, -\\frac{3}{2} \\right), \\]\n\\[ \\text{Let } P(0,t), \\]\n\\[ \\because CP\\perp CQ \\text{ and point } Q \\text{ coincides with point } B, \\]\n\\[ \\therefore CP \\text{ is the perpendicular bisector of segment } AB, \\]\n\\[ \\therefore PA=PB, \\]\n\\[ \\text{In } \\triangle PBO, PB^2=OB^2+OP^2, OB=5, OP=t, \\]\n\\[ \\therefore PB^2=5^2+t^2=25+t^2, \\]\n\\[ \\because PA=OP+OA=t+3, \\]\n\\[ \\therefore (t+3)^2=25+t^2, \\]\n\\[ \\text{Solving, we get: } t= \\frac{8}{3}, \\]\n\\[ \\therefore P\\left(0, \\frac{8}{3}\\right), \\]\n\\[ \\therefore \\text{The length of } PO \\text{ is } \\boxed{\\frac{8}{3}}. \\]",
    "image_path": "dataset/mm_math/images/MM_Math/51419593_124.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{1}{2}x+5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCO$, point $C$ is on the x-axis, point $A$ is on the y-axis, and point $B$ has coordinates $(-6, 8)$. The rectangle $ABCO$ is folded along line $BD$ such that point $A$ falls on point $E$ on the diagonal $OB$, and the line containing $BD$ intersects $OA$ and the x-axis at points $D$ and $F$, respectively. What is the equation of line $BD$?\n",
    "completion": "\\textbf{Solution:} Let $D(0, b)$. Then, according to the problem, we have $\\angle DEO=90^\\circ$, $DE=DA=8-b$, $BE=BA=6$, and $EO=4$. Applying the Pythagorean theorem to the right triangle $DEO$: $OE^2+DE^2=DO^2$, that is, $4^2+(8-b)^2=b^2$, solving this, we get $b=5$, hence $D(0, 5)$. The equation of line $BD$ is given by $y=kx+b$. Given $B(-6, 8)$ and $D(0, 5)$, we can establish: $\\begin{cases}-6k+b=8\\\\ b=5\\end{cases}$, from which we find $\\begin{cases}k=-\\frac{1}{2}\\\\ b=5\\end{cases}$. Therefore, the equation of line $BD$ is: $y=-\\frac{1}{2}x+5$; hence, the equation of line $BD$ is $\\boxed{y=-\\frac{1}{2}x+5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51864146_101.png"
  },
  {
    "solution": "\\boxed{9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, it is known that lines $a \\parallel b \\parallel c$, and lines $m$, $n$ intersect $a$, $b$, $c$ at points $A$, $C$, $E$, $B$, $D$, $F$, respectively. If $AC=8$, $CE=12$, and $BD=6$, what is the value of $DF$?\n",
    "completion": "\\textbf{Solution:} Since line $a \\parallel b \\parallel c$,\\\\\nthus, $\\frac{AC}{CE}=\\frac{BD}{DF}$,\\\\\nthus, $\\frac{8}{12}=\\frac{6}{DF}$,\\\\\nwe get $DF=\\boxed{9}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51458031.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the Cartesian coordinate system, the graph of the linear function $y=kx+b$ intersects the x-axis at point $A(-3, 0)$ and the y-axis at point $B$, and its intersection point with the direct proportion function $y=\\frac{4}{3}x$ is $C(a, 4)$. What is the area of $\\triangle BOC$?\n",
    "completion": "\\textbf{Solution:} In $y=\\frac{2}{3}x+2$, let $x=0$, we get $y=2$, hence $B(0,2)$ and $OB=2$. Since point $C(3,4)$, we have $x_C=3$. Therefore, the area of $\\triangle BOC$ is $\\frac{1}{2}OB\\times x_C=\\frac{1}{2}\\times 2\\times 3=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52423287_75.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, $AC=8$, $BC=6$. Points D and E are respectively the midpoints of $AC$ and $AB$. When connecting $DE$, what is the length of $DE$?\n",
    "completion": "\\textbf{Solution:} Since $D$ and $E$ are the midpoints of $AC$ and $AB$ respectively, \\\\\nit follows that $DE$ is the midline of $\\triangle ABC$, \\\\\ntherefore $DE = \\frac{1}{2}BC = \\frac{1}{2} \\times 6 = \\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52386808.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\square ABCD$, the diagonals AC, BD intersect at point O, with OA=OB. If AD=4 and $\\angle AOD=60^\\circ$, what is the length of AB?\n",
    "completion": "\\textbf{Solution:} Since $ABCD$ is a square and $OA=OB$,\\\\\nit follows that $AO=BO=DO=OC$,\\\\\nwhich implies $AC=BD$,\\\\\ntherefore, quadrilateral ABCD is a rectangle,\\\\\nhence $\\angle DAB=90^\\circ$,\\\\\nsince $\\angle AOD=60^\\circ$,\\\\\nit means $\\triangle ADO$ is an equilateral triangle,\\\\\nthus $DO=AD=4$,\\\\\ntherefore $BD=OD+OB=8$,\\\\\nin $\\triangle ABD$, $AB=\\sqrt{BD^{2}-AD^{2}}=\\sqrt{64-16}=4\\sqrt{3}$.\\\\\nHence, the answer is: $AB=\\boxed{4\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51817954.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that point C is on the line segment AB, and M and N are the midpoints of AC and BC, respectively. Given the line segment DE, extend DE to F such that EF= $ \\frac{1}{2}$DE; extend ED to G such that DG=2DE, with P and Q being the midpoints of EF and DG, respectively. If PQ=18 cm, what is the length of DE in cm?\n",
    "completion": "\\textbf{Solution:} Let $DE=x$ cm, then $EF= \\frac{1}{2}x$ cm, $EP= \\frac{1}{2}EF= \\frac{1}{4}x$ cm, $DG=2x$ cm, and $DQ= \\frac{1}{2}DG=x$ cm.\\\\\nSince $PQ=18$ cm, we have $x+x+\\frac{1}{4}x=18$,\\\\\nsolving this yields $x=\\boxed{8}$,\\\\\nthus $DE=\\boxed{8}$ cm.",
    "image_path": "dataset/mm_math/images/MM_Math/54927380_131.png"
  },
  {
    "solution": "\\boxed{D (-1, 4)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the Cartesian coordinate system, the parabola $y = -x^2 - 2x + 3$ intersects the x-axis at points A and B (A is to the left of B) and intersects the y-axis at point C, with the vertex at D. Please directly write the coordinates of points A, C, and D.\n",
    "completion": "Solution: When $y=-x^2-2x+3$ and $y=0$, we have $-x^2-2x+3=0$,\\\\\nwe get: $x_1=-3$, $x_2=1$,\\\\\nSince A is on the left side of B,\\\\\nTherefore, \\boxed{A (-3, 0)}, \\boxed{B (1, 0)},\\\\\nWhen $y=-x^2-2x+3$ and $x=0$, then $y=3$,\\\\\nTherefore, \\boxed{C (0, 3)},\\\\\nSince $y=-x^2-2x+3=-(x+1)^2+4$,\\\\\nTherefore, the vertex \\boxed{D (-1, 4)};",
    "image_path": "dataset/mm_math/images/MM_Math/52720132_79.png"
  },
  {
    "solution": "\\boxed{P(0,-2)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the Cartesian coordinate system $xOy$, the line $AB$ intersects the x-axis at point $B$ and the y-axis at point $C(0, 2)$. It also intersects the graph of the inverse proportion function $y=\\frac{6}{x}$ within the first quadrant at point $A$. Line $AD$ is perpendicular to the x-axis at point $D$, with $OD=2$. Let point $P$ be a point on the y-axis. If the area of triangle $ACP$ is equal to 4, what are the coordinates of point $P$?\n",
    "completion": "\\textbf{Solution:} Since point P is on the y-axis and the area of $\\triangle ACP$ is equal to 4, with A(2,3),\\\\\nit follows that $S_{\\triangle ACP} = \\frac{1}{2} \\times \\text{CP} \\times |x_A| = \\frac{1}{2} \\times \\text{CP} \\times 2 = 4$,\\\\\nthus CP = 4.\\\\\nSince point C is (0,2) and point P is on the y-axis,\\\\\nit follows that \\boxed{P(0,6)} or \\boxed{P(0,-2)}.",
    "image_path": "dataset/mm_math/images/MM_Math/51458203_265.png"
  },
  {
    "solution": "\\boxed{\\frac{2\\sqrt{5}}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in a $3\\times 3$ grid, A and B are both lattice points. With point A as the center and the length of AB as the radius, an arc is drawn. Point C in the figure is the intersection point of the arc and the grid lines. Then, what is the value of $\\tan\\angle BAC$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, since $CD\\parallel AB$,\\\\\nit follows that $\\angle BAC = \\angle DCA$.\\\\\nSince the radii of the same circle are equal, it follows that $AC = AB = 3$, and $AD = 2$,\\\\\ntherefore, $CD = \\sqrt{AC^2 - AD^2} = \\sqrt{5}$,\\\\\nin $\\triangle ACD$, $\\tan \\angle ACD = \\frac{AD}{CD} = \\frac{2}{\\sqrt{5}} = \\frac{2\\sqrt{5}}{5}$.\\\\\nThus, $\\tan \\angle BAC = \\tan \\angle ACD = \\boxed{\\frac{2\\sqrt{5}}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51318944.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC \\cong \\triangle DEF$, $BC=7$, and $EC=4$, then what is the length of $CF$?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABC \\cong \\triangle DEF$,\\\\\nit follows that $BC = EF = 7$,\\\\\nhence $CF = EF - EC = 7 - 4 = \\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52733744.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{14}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In circle $ \\odot O$, where $AB$ is the diameter and point C is a point on the circle. The minor arc is folded along chord $AC$ to intersect $AB$ at point D. Connect $CD$. As shown in Figure 2, if $AD=6$ and $DB=2$, find the length of $AC$.\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect $OC$ and $CB$, and draw $CG\\perp AB$ at point G,\\\\\naccording to (2) we have $CB=CD$,\\\\\n$\\therefore BG=DG$,\\\\\n$\\because AD=6$, $DB=2$,\\\\\n$\\therefore BG=DG=\\frac{1}{2}DB=1$, $AB=AD+DB=6+2=8$,\\\\\n$\\therefore r=OC=\\frac{1}{2}AB=4$,\\\\\n$\\therefore OD=AD-OB=6-4=2$, $OG=OD+DG=1+2=3$,\\\\\n$\\therefore CG=\\sqrt{OC^{2}-DG^{2}}=\\sqrt{4^{2}-3^{2}}=\\sqrt{7}$, $AG=AD+DG=6+1=7$,\\\\\n$\\therefore AC=\\sqrt{AG^{2}+CG^{2}}=\\sqrt{7^{2}+(\\sqrt{7})^{2}}=\\boxed{2\\sqrt{14}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52871917_109.png"
  },
  {
    "solution": "\\boxed{(-2,4)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated in the figure, the linear function \\(y=kx+2\\) intersects the \\(y\\)-axis at point A and intersects the graph of the inverse proportion function \\(y=\\frac{m}{x}\\) at points B and C. Line BD is perpendicular to the \\(y\\)-axis and intersects it at point D. OA is equal to OD, and the area of triangle \\(ABD\\) is 8. In the given plane, there exists a point E such that the quadrilateral formed by points B, C, D, and E is a parallelogram. Please directly provide the coordinates of all possible points E that satisfy these conditions.\n",
    "completion": "\\textbf{Solution:} The coordinates of the point $E$ that meet the conditions are: $\\boxed{(6,4)}$, $\\boxed{(-6,-8)}$, $\\boxed{(-2,4)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51445968_245.png"
  },
  {
    "solution": "\\boxed{\\frac{5\\sqrt{5}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in square $ABCD$, point $G$ is located on $BC$ and $\\frac{GC}{BG}=\\frac{1}{2}$. Line $DG$ intersects the diagonal $AC$ at point $F$. A perpendicular line from $D$ to $DG$ intersects the extension of line $CA$ at point $E$. If $AE=5$, what is the length of $DF$?\n",
    "completion": "\\textbf{Solution:} Draw line $EH \\perp AD$ through point $E$, extending it to intersect the extended line of $DA$ at $H$,\\\\\n$\\therefore \\angle H=90^\\circ$,\\\\\nIn square $ABCD$, $AB=BC=CD=AD$, $\\angle BAD=\\angle B=\\angle BCD=\\angle ADC=90^\\circ$,\\\\\n$\\therefore \\angle 2+\\angle 3=90^\\circ$, $\\angle H=\\angle BCD$,\\\\\nBecause $DE\\perp DG$,\\\\\n$\\therefore \\angle EDG=90^\\circ$,\\\\\n$\\therefore \\angle 2+\\angle 1=90^\\circ$,\\\\\n$\\therefore \\angle 1=\\angle 3$,\\\\\n$\\therefore \\triangle DEH\\sim \\triangle DGC$,\\\\\n$\\therefore \\frac{EH}{GC}=\\frac{DH}{DC}$,\\\\\nSince $\\frac{GC}{BG}=\\frac{1}{2}$,\\\\\n$\\therefore$ let $GC=x$, then $BG=2x$, $DC=BC=3x$,\\\\\n$\\therefore \\frac{EH}{GC}=\\frac{DH}{3x}$,\\\\\n$\\therefore DH=3EH$,\\\\\nBecause $AC$ is the diagonal of square $ABCD$,\\\\\n$\\therefore \\angle DAC=45^\\circ$,\\\\\nSince $\\angle EAH=\\angle DAC=45^\\circ$,\\\\\n$\\therefore \\angle HEA=45^\\circ$,\\\\\n$\\therefore EH=HA$,\\\\\n$\\therefore EH^2+HA^2=25$,\\\\\n$\\therefore EH=HA=\\frac{5\\sqrt{2}}{2}$,\\\\\n$\\therefore DH=3EH=\\frac{15\\sqrt{2}}{2}$,\\\\\n$\\therefore AD=DH−AH=5\\sqrt{2}$,\\\\\n$\\therefore GC=\\frac{1}{3}BC=\\frac{1}{3}AD=\\frac{5\\sqrt{2}}{3}$,\\\\\n$\\therefore DG=\\sqrt{CD^2+CG^2}=\\frac{10\\sqrt{5}}{3}$,\\\\\nSince in square $ABCD$, $AD\\parallel BC$,\\\\\n$\\therefore \\frac{CG}{AD}=\\frac{GF}{DF}=\\frac{1}{3}$,\\\\\n$\\therefore DF=3GF$,\\\\\n$\\therefore DF=\\frac{3}{4}DG=\\frac{3}{4}\\times \\frac{10\\sqrt{5}}{3}=\\boxed{\\frac{5\\sqrt{5}}{2}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52793251.png"
  },
  {
    "solution": "\\boxed{30+10\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: The lookout tower $AB$ has a height of $20m$. From the base of the lookout tower $B$, the angle of elevation to the top of the opposite tower $C$ is measured to be $60^\\circ$, and from the top of the lookout tower $A$, the angle of elevation to the tower top $C$ is measured to be $45^\\circ$. Given that the terrain height of the lookout tower and tower $CD$ is the same, what is the height of tower $CD$ in meters?\n",
    "completion": "\\textbf{Solution:} Let the height of the tower $CD$ be $x$, thus $BD=\\frac{\\sqrt{3}}{3}x$,\\\\\nfrom $BD \\cdot \\tan 60^\\circ - BD \\cdot \\tan 45^\\circ = AB$, substituting $BD=\\frac{\\sqrt{3}}{3}x$,\\\\\nwe get: $x-\\frac{\\sqrt{3}}{3}x=20$,\\\\\nsolving this gives: $x=\\boxed{30+10\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51403135.png"
  },
  {
    "solution": "\\boxed{130^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, P is a point outside the circle $\\odot O$. The tangents PA and PB are drawn from P to $\\odot O$. If $\\angle APB=50^\\circ$, what is the degree measure of $\\overset{\\frown}{AB}$?\n",
    "completion": "\\textbf{Solution:} Since PA and PB are tangents to the circle O,\\\\\nit follows that $\\angle PAO=\\angle PBO=90^\\circ$,\\\\\nand since $\\angle APB=50^\\circ$,\\\\\nthen $\\angle AOB=360^\\circ-\\angle APB-\\angle PAO-\\angle PBO=130^\\circ$,\\\\\nthus, the degree measure of $\\overset{\\frown}{AB}$ is $\\boxed{130^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52536676.png"
  },
  {
    "solution": "\\boxed{(-6,-6)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the graph, in the Cartesian coordinate system, the parabola $y=\\frac{1}{2}x^{2}+bx+c$ passes through point $A(-4,0)$. Point $M$ is the vertex of the parabola, and point $B$ is on the $y$-axis, with $OA=OB$. The line $AB$ intersects the parabola in the first quadrant at point $C(2,6)$. Is there a point $N$ in the coordinate plane such that the quadrilateral with vertices $A$, $O$, $C$, and $N$ is a parallelogram? If it exists, please write the coordinates of point $N$ directly; if not, please explain why.\n",
    "completion": "\\textbf{Solution:} There exists, the coordinates of point $N$ are: $\\boxed{(6,6)}$, $\\boxed{(-2,6)}$, or $\\boxed{(-6,-6)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53043807_116.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=4$, $BC=5$, $CA=6$, and the three angle bisectors intersect at point $O$. The area of $\\triangle CAO$ equals $9$. What is the area of $\\triangle ABO$?\n",
    "completion": "\\textbf{Solution:} Draw $OM\\perp AB$ at point M, and $OD\\perp AC$ at point D, \\\\\nsince the three angle bisectors intersect at point O, \\\\\nhence $OM=OD$, \\\\\nsince the area of $\\triangle CAO$ is 9, \\\\\nthus ${S}_{\\triangle CAO}=\\frac{1}{2}AC\\cdot OD=\\frac{1}{2}\\times 6\\cdot OD=9$, \\\\\nSolving this, we get $OD=3$, \\\\\nthus $OM=3$, \\\\\ntherefore ${S}_{\\triangle ABO}=\\frac{1}{2}AB\\cdot OM=\\frac{1}{2}\\times 4\\times 3=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52815746.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $C$ is a point on segment $AB$ with $AC>BC$. Point $M$ is the midpoint of segment $AB$, and point $N$ is the midpoint of segment $BC$. Given $AM=8$ and $CN=3$, find the length of segment $MN$.\n",
    "completion": "\\textbf{Solution:} Since point M is the midpoint of line segment AB, $AM=8$,\\\\\nhence $BM=AM=8$,\\\\\nand since point N is the midpoint of BC, $CN=3$,\\\\\nhence $BN=CN=3$,\\\\\ntherefore $MC=BM-BN-CN=8-3-3=2$,\\\\\ntherefore $MN=MC+CN=2+3=\\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52948485.png"
  },
  {
    "solution": "\\boxed{110^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $ABCD$ is a quadrilateral inscribed in circle $O$, and $\\angle ABC = 125^\\circ$. What is the measure of $\\angle AOC$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral $ABCD$ is inscribed in circle $O$,\\\\\nit follows that $\\angle D+\\angle ABC=180^\\circ$ \\\\\nSince $\\angle ABC=125^\\circ$,\\\\\nwe have $\\angle D=180^\\circ−\\angle A=180^\\circ−125^\\circ=55^\\circ$,\\\\\nBy the Inscribed Angle Theorem, $\\angle AOC=2\\angle D= \\boxed{110^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/51446648.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, AB is the diameter of $\\odot O$, C is a point on $\\odot O$, and a tangent to $\\odot O$ through point C intersects the extension of AB at point E. If $\\angle A=30^\\circ$, what is the value of $\\sin E$?\n",
    "completion": "\\textbf{Solution:} Connect OC, \\\\\n$\\because$ CE is a tangent of $\\odot$O, \\\\\n$\\therefore$ OC$\\perp$CE, \\\\\n$\\because$ $\\angle$A=30$^\\circ$, and OC=OA, \\\\\n$\\therefore$ $\\angle$BOC=2$\\angle$A=60$^\\circ$, \\\\\n$\\therefore$ $\\angle$E=90$^\\circ$−$\\angle$BOC=30$^\\circ$, \\\\\n$\\therefore$ $\\sin\\angle$E=$\\sin30^\\circ$= \\boxed{\\frac{1}{2}}.",
    "image_path": "dataset/mm_math/images/MM_Math/51265193.png"
  },
  {
    "solution": "\\boxed{40^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $CD$ is the angle bisector of $\\angle ACB$, $DE \\parallel BC$, $\\angle A=65^\\circ$, $\\angle B=35^\\circ$, what is the measure of $\\angle EDC$?\n",
    "completion": "\\textbf{Solution:} In $\\triangle ABC$, since $\\angle A=65^\\circ$ and $\\angle B=35^\\circ$, it follows that $\\angle ACB=180^\\circ-\\angle A-\\angle B=180^\\circ-65^\\circ-35^\\circ=80^\\circ$. Since $CD$ is the angle bisector of $\\angle ACB$, it follows that $\\angle BCD=\\frac{1}{2}\\angle ACB=\\frac{1}{2}\\times 80^\\circ=40^\\circ$. Since $DE\\parallel BC$, it follows that $\\angle EDC=\\angle BCD=\\boxed{40^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52661295.png"
  },
  {
    "solution": "\\boxed{42}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, $AC=5$, $BC=12$. Rotating $\\triangle ABC$ $60^\\circ$ clockwise around point $B$ results in $\\triangle BDE$. When $DC$ is connected and intersects $AB$ at point $F$, what is the sum of the perimeters of $\\triangle ACF$ and $\\triangle BDF$?\n",
    "completion": "\\textbf{Solution:} Given that $\\triangle BDE$ is obtained by rotating $\\triangle BCA$, \\\\\nthus BD = BC = 12. \\\\\nBecause $\\angle CBD = 60^\\circ$, \\\\\nthus $\\triangle BCD$ is an equilateral triangle, \\\\\ntherefore CD = BC = 12. \\\\\nIn $\\text{Rt}\\triangle ABC$, with $\\angle ACB = 90^\\circ$, AC = 5, and BC = 12, \\\\\nthus AB = $\\sqrt{AC^{2}+BC^{2}}$ = 13, \\\\\ntherefore the perimeter of $\\triangle ACF$ + the perimeter of $\\triangle BDF$ = AC + CF + AF + BF + DF + BD = AC + AB + CD + BD = 5 + 13 + 12 + 12 = \\boxed{42}.",
    "image_path": "dataset/mm_math/images/MM_Math/51419666.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, the perimeter of parallelogram $ABCD$ is $20$. The diagonals $AC$ and $BD$ intersect at point $O$. Point $E$ is the midpoint of $CD$, and $BD = 6$. What is the perimeter of $\\triangle DOE$?\n",
    "completion": "\\textbf{Solution:} Given the perimeter of quadrilateral ABCD is 20,\\\\\nthen $2(BC+CD)=20$,\\\\\nthus $BC+CD=10$.\\\\\nGiven ABCD is a parallelogram, and the diagonals AC, BD intersect at point O, with BD = 6,\\\\\nthus $OD=OB=\\frac{1}{2}BD=3$.\\\\\nGiven point E is the midpoint of CD, and point O is the midpoint of BD,\\\\\nthus OE is the midsegment of $\\triangle BCD$ and $DE=\\frac{1}{2}CD$,\\\\\nhence $OE=\\frac{1}{2}BC$,\\\\\ntherefore, the perimeter of $\\triangle DOE = OD + OE + DE = \\frac{1}{2}BD + \\frac{1}{2}(BC + CD) = 5 + 3 = \\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52250772.png"
  },
  {
    "solution": "\\boxed{9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, $AC=BC$, and point E is a point inside $\\angle ACB$. Connect $CE$, draw $AD\\perp CE$ and $BE\\perp CE$, with the perpendicular feet being points D and E respectively. $\\triangle CAD\\cong \\triangle BCE$, if $AD=15\\,cm$ and $BE=6\\,cm$, find the length of $DE$.\n",
    "completion": "\\textbf{Solution:} Since $\\triangle CAD\\cong \\triangle BCE$,\\\\\nit follows that $AD=CE=15\\text{cm}$, and $BE=CD=6\\text{cm}$,\\\\\nthus $DE=CE-CD=\\boxed{9}\\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53106979_145.png"
  },
  {
    "solution": "\\boxed{135^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the graph, in the Cartesian coordinate system $xOy$, we have $OA=OB$ and $\\angle AOB=90^\\circ$. A line parallel to the $x$-axis is drawn through point B, and it intersects with the line passing through point A, which is $y=-x+b$, at point C. Given that the coordinates of point A are (2, 1), what is the degree measure of $\\angle ACB$?\n",
    "completion": "\\textbf{Solution:} Let the line intersect the y-axis at point E, as shown in the figure:\\\\\n$\\therefore OD=OE$,\\\\\n$\\therefore \\angle OEC=\\angle ODC=45^\\circ$,\\\\\nAlso, since $BC\\parallel OD$,\\\\\n$\\therefore \\angle BCA+\\angle CDO=180^\\circ$,\\\\\n$\\therefore \\angle BCA=135^\\circ$,\\\\\nTherefore, the measure of $\\angle ACB$ is $\\boxed{135^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50677077_158.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, within quadrilateral $ABCD$, $AC$ and $BD$ intersect at point $O$. Point $E$ is the midpoint of side $BC$. If $AB = 4$, what is the length of $OE$?\n",
    "completion": "\\textbf{Solution:} In quadrilateral $ABCD$, $AC$ and $BD$ intersect at point $O$,\\\\\n$\\therefore$ $BO=DO$,\\\\\n$\\because$ point $E$ is the midpoint of side $BC$,\\\\\n$\\therefore$ $OE$ is the median of $\\triangle ABC$,\\\\\n$\\therefore$ $OE = \\frac{1}{2}AB = \\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53706951.png"
  },
  {
    "solution": "\\boxed{90^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, \\(AD\\) is the angle bisector of \\( \\triangle ABC\\), \\(CE\\) is the height of \\( \\triangle ABC\\), \\(\\angle BAC=50^\\circ\\), \\(\\angle BCE=25^\\circ\\), what is the degree measure of \\( \\angle ADB\\)?\n",
    "completion": "\\textbf{Solution:} Given that $AD$ is the angle bisector of $\\triangle ABC$, and $\\angle BAC=50^\\circ$,\\\\\ntherefore $\\angle BAD=25^\\circ$,\\\\\nalso given that $CE$ is the altitude of $\\triangle ABC$, and $\\angle BCE=25^\\circ$,\\\\\ntherefore $\\angle BEC=90^\\circ$,\\\\\nthus we get $\\angle B=65^\\circ$,\\\\\nhence, $\\angle ADB=180^\\circ-\\angle B-\\angle BAD=180^\\circ-65^\\circ-25^\\circ=\\boxed{90^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52653437.png"
  },
  {
    "solution": "\\boxed{(1,-2)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given the parabola $y=x^2+bx+c$ as shown in the figure, it intersects the x-axis at point $A(-1,0)$ and the y-axis at point $C(0,3)$. On the axis of symmetry of the parabola, there is a moving point $P$. Find the coordinates of point $P$ when the value of $PA+PC$ is minimized.\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, construct the axis of symmetry and draw line BC. The intersection point P is the point we are seeking. Suppose line BC is $y=kx+b$. Substituting point $(3,0)$ and $C(0,-3)$ into it, we obtain\n\\[\n\\left\\{\n\\begin{array}{l}\n3k+b=0\\\\\nb=-3\n\\end{array}\n\\right.\n\\]\nSolving this, we get\n\\[\n\\left\\{\n\\begin{array}{l}\nk=1\\\\\nb=-3\n\\end{array}\n\\right.\n\\]\n$\\therefore$ The equation of line BC is: $y=x-3$, and because $y=x^2-2x-3=(x-1)^2-4$, thus the axis of symmetry of the parabola is the line $x=1$. Substituting $x=1$ into $y=x-3$ yields $y=-2$. $\\therefore$ The coordinates of point $P$ are $\\boxed{(1,-2)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52874723_284.png"
  },
  {
    "solution": "\\boxed{(0, 2.4)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, quadrilateral $OABC$ is a rectangle, with points $A$ and $C$ located on the $x$-axis and $y$-axis respectively. Connect $AC$. The coordinates of point $B$ are $(12, 5)$. The bisector of $\\angle CAO$ intersects the $y$-axis at point $D$. What are the coordinates of point $D$?\n",
    "completion": "\\textbf{Solution:} Let line DE be perpendicular to AC at E, \\\\\nbecause quadrilateral ABCO is a rectangle, with B(12,5), \\\\\ntherefore OC = AB = 5, OA = BC = 12, $\\angle COA = 90^\\circ$, \\\\\nbecause AD bisects $\\angle OAC$, \\\\\ntherefore OD = DE, \\\\\nby the Pythagorean theorem: OA$^{2}$ = AD$^{2}$ - OD$^{2}$, AE$^{2}$ = AD$^{2}$ - DE$^{2}$, \\\\\ntherefore OA = AE = 12, \\\\\nby the Pythagorean theorem: AC = $\\sqrt{5^{2} + 12^{2}} = 13$, \\\\\nin right triangle $\\triangle DEC$, DE$^{2}$ + EC$^{2}$ = CD$^{2}$, \\\\\nthat is, OD$^{2}$ + (13 - 12)$^{2}$ = (5 - OD)$^{2}$, \\\\\nsolving this gives: OD = 2.4, \\\\\ntherefore, the coordinates of D are \\boxed{(0, 2.4)}.",
    "image_path": "dataset/mm_math/images/MM_Math/52160043.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{5}}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the three vertices of $\\triangle ABC$ are all on the coordinate axes of the Cartesian coordinate system, $BC=6$, and the expression for the line containing side $AB$ is $y=x+2$. What is $\\sin\\angle ACB$?\n",
    "completion": "\\textbf{Solution:} Given that the equation of line AB is $y=x+2$,\\\\\nit follows that when $y=0$, $x=-2$, and when $x=0$, $y=2$,\\\\\ntherefore, point A is $(0,2)$ and point B is $(-2,0)$,\\\\\nthus, OA$=2$ and OB$=2$,\\\\\nsince BC$=6$,\\\\\nit follows that OC$=BC-OB=6-2=4$,\\\\\ntherefore, AC$=\\sqrt{OA^2+OC^2}=\\sqrt{2^2+4^2}=2\\sqrt{5}$,\\\\\nhence, $\\sin\\angle ACB=\\frac{OA}{AC}=\\frac{2}{2\\sqrt{5}}=\\boxed{\\frac{\\sqrt{5}}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52536543.png"
  },
  {
    "solution": "\\boxed{(-4, -1)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the graph of the linear function $y=x+b$ intersects with the $x$ and $y$ axes at points A and B, respectively, and intersects with the graph of the direct proportion function $y=-2x$ at point C, where the ordinate of point C is 4. If the point $Q(m, 2)$ is inside $\\triangle OBC$ (excluding the boundary), please write down the range of the values of $m$ directly.\n",
    "completion": "\\textbf{Solution:} The range of $-4<m<-1$ is $m \\in \\boxed{(-4, -1)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52190033_152.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, $CD\\perp AB$ at point $D$, it is known that $AD=5$, $BD=4$, what is the length of $BC$?\n",
    "completion": "\\textbf{Solution:} Since $CD \\perp AB$,\\\\\nthen $\\angle BDC = 90^\\circ$,\\\\\nthus $\\angle BDC = \\angle BCA$,\\\\\nSince $\\angle B = \\angle B$,\\\\\nthus $\\triangle BDC \\sim \\triangle BCA$,\\\\\ntherefore $\\frac{BD}{BC}=\\frac{BC}{BA}$, that is $\\frac{4}{BC}=\\frac{BC}{4+5}$,\\\\\nSolving gives: $BC=\\boxed{6}$ (negative values have been discarded).",
    "image_path": "dataset/mm_math/images/MM_Math/51446578.png"
  },
  {
    "solution": "\\boxed{108}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $D$ lies on side $BC$ of $\\triangle ABC$, with $AC=AB=BD$, $AD=CD$. What is the measure of $\\angle BAC$ in degrees?\n",
    "completion": "\\textbf{Solution:} \\boxed{108}",
    "image_path": "dataset/mm_math/images/MM_Math/55588141.png"
  },
  {
    "solution": "\\boxed{\\sqrt{7}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system $xOy$, a circle with its center at the origin $O$ passes through point $A(2, 0)$. The line $y=\\frac{\\sqrt{3}}{3}x+\\sqrt{3}$ intersects the circle $O$ at points $B$ and $C$. What is the length of chord $BC$?\n",
    "completion": "\\textbf{Solution:} Let us consider the line $y=\\frac{\\sqrt{3}}{3}x+\\sqrt{3}$ intersecting the two coordinate axes at points D and E. Draw OM$\\perp$BC at point M and connect OB,\\\\\nWhen $x=0$, $y=\\sqrt{3}$,\\\\\n$\\therefore$D$\\left(0, \\sqrt{3}\\right)$,\\\\\nWhen $y=0$, $x=-3$,\\\\\n$\\therefore$ point E$(-3,0)$,\\\\\n$\\therefore$ $DE=\\sqrt{\\left(\\sqrt{3}\\right)^{2}+3^{2}}=2\\sqrt{3}$,\\\\\n$\\therefore$DE=2OD,\\\\\n$\\therefore$$\\angle$DEO$=30^\\circ$,\\\\\n$\\therefore$OM$=\\frac{1}{2}$OE$=\\frac{3}{2}$,\\\\\n$\\because$ point A$(2,0)$,\\\\\n$\\therefore$OA=OB=2,\\\\\n$\\therefore$ $BM=\\sqrt{OB^{2}-OM^{2}}=\\sqrt{4-\\frac{9}{4}}=\\frac{\\sqrt{7}}{2}$,\\\\\n$\\because$OM$\\perp$BC,\\\\\n$\\therefore$BC=2BM$=\\boxed{\\sqrt{7}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52861387.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, in the Cartesian coordinate system, the line ${l}_{1}$: $y=kx+b\\ (k\\ne 0)$ intersects the x-axis at point $A(a, 0)$, and it intersects the y-axis at point $B(0, b)$. The line ${l}_{2}$ passing through the origin intersects line ${l}_{1}$ at point C. Given condition (1), when the x-coordinate of point C is 1, what is the area of $\\triangle AOC$?\n",
    "completion": "\\textbf{Solution:} Substitute $x=1$ into $y=-2x+4$, we get $y=-2\\times 1+4=2$,\\\\\n$\\therefore C(1, 2)$,\\\\\n$\\because A(2, 0)$,\\\\\n$\\therefore OA=2$,\\\\\n$\\therefore {S}_{\\triangle AOC}=\\frac{1}{2}OA\\times \\left|{y}_{C}\\right|=\\frac{1}{2}\\times 2\\times 2=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52386704_120.png"
  },
  {
    "solution": "\\boxed{24}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that in the rhombus $ABCD$, the diagonals $AC$ and $BD$ intersect at point $O$. Point $E$ is on the extension of side $AB$, and $AB=BE$. Connect $CE$. Given that $BD \\parallel EC$, if $AD=5$ and $CE=6$, what is the area of the rhombus $ABCD$?\n",
    "completion": "\\textbf{Solution:} Given that rhombus $ABCD$ with $AD=5$, hence $AB=5$. The diagonals $AC$ and $BD$ are perpendicular and bisect each other, hence $\\angle AOB=90^\\circ$, and $AO=OC$. Also, since $AB=BE$ and $CE=6$, we have $BO= \\frac{1}{2}CE=3$ and $BD=2BO=6$. Thus, $AO= \\sqrt{5^2-3^2}=4$ and $AC=2AO=8$. Therefore, the area of rhombus $ABCD$ is $S_{\\text{rhombus }ABCD}=\\frac{1}{2}AC\\cdot BD=\\frac{1}{2}\\times 8\\times 6=\\boxed{24}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51942351_127.png"
  },
  {
    "solution": "\\boxed{y=x-4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function \\(y=ax+b\\) intersects with the graph of the inverse proportion function \\(y=\\dfrac{k}{x}\\) (\\(k \\neq 0\\)) at points \\(A(1, -3)\\) and \\(B(m, -1)\\). Connect \\(OA\\) and \\(OB\\). What is the expression of the linear function?\n",
    "completion": "\\textbf{Solution:} Since the inverse proportion function is $y=\\frac{k}{x}$ (where $k \\neq 0$) and its graph passes through the point A(1, −3), \\\\\nwe have $k=1\\times(-3)=-3$, \\\\\nthus, the expression for the inverse proportion function is $y=-\\frac{3}{x}$. \\\\\nGiven that the graph of the inverse proportion function $y=-\\frac{3}{x}$ passes through point B(m, −1), \\\\\nit follows that $m=3$, \\\\\ntherefore, point B is (3, −1). \\\\\nSince points A(1, −3) and B(3, −1) lie on the graph of the linear function $y=ax+b$, \\\\\nwe obtain $\\left\\{\\begin{array}{l}a+b=-3 \\\\ 3a+b=-1\\end{array}\\right.$, from which we solve $\\left\\{\\begin{array}{l}a=1 \\\\ b=-4\\end{array}\\right.$, \\\\\nthus, the expression for the linear function is $\\boxed{y=x-4}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52602121_133.png"
  },
  {
    "solution": "\\boxed{13.5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the isosceles right triangle $ABC$, $\\angle BAC=90^\\circ$, $D$ is the midpoint of $BC$, $E$ is a point on the side $AC$ (not coinciding with the endpoints), a line is drawn through point $E$ such that $EG \\perp BC$ at point $G$, and $EH \\perp AD$ at point $H$, a line is drawn through point $B$ such that $BF \\parallel AC$ intersecting the extension of $EG$ at point $F$. If $AG=3$, what is the area of the shaded region?\n",
    "completion": "\\textbf{Solution:} Let $DG=a$, $CG=b$, then $CD=a+b$, \\\\\nsince $\\triangle ABC$ is an isosceles right triangle, $\\angle BAC=90^\\circ$, \\\\\nthus $\\angle ABC=\\angle ACB=45^\\circ$, $AB=AC$, \\\\\nalso since $D$ is the midpoint of $BC$, \\\\\nthus $BD=AD=CD=a+b$, $BC=2BD=2(a+b)$, \\\\\nsince $EG\\perp BC$, $EH\\perp AD$, \\\\\nthus quadrilateral DGEH is a rectangle, $\\angle GEC=45^\\circ$, \\\\\nthus $DH=EG=CG=b$, \\\\\nsince $BF\\parallel AC$, \\\\\nthus $\\angle FBG=\\angle ACB=45^\\circ$, \\\\\nsince $EF\\perp BC$, \\\\\nthus $\\angle F=45^\\circ$, \\\\\nthus $GF=BG=BD+DG=a+b+a=2a+b$, \\\\\nby the Pythagorean theorem, $AD^2+DG^2=AG^2$, \\\\\nthus $(a+b)^2+a^2=9$, \\\\\nwhich simplifies to $2a^2+2ab+b^2=9$, \\\\\nAccording to the problem, ${S}_{\\text{shaded}}={S}_{\\triangle ABC}+{S}_{\\triangle BGF}-{S}_{\\text{rectangle} DGEH}$\\\\\n$=\\frac{1}{2}BC\\cdot AD+\\frac{1}{2}BG\\cdot GF-DG\\cdot DH$\\\\\n$=BD\\cdot AD+\\frac{1}{2}BG^2-DG\\cdot DH$\\\\\n$=(a+b)^2+\\frac{1}{2}(2a+b)^2-ab$\\\\\n$=a^2+2ab+b^2+2a^2+ab+\\frac{1}{2}b^2-ab$\\\\\n$=\\frac{3}{2}(2a^2+2ab+b^2)$\\\\\n$=\\frac{3}{2}\\times 9$\\\\\n$=\\boxed{13.5}$",
    "image_path": "dataset/mm_math/images/MM_Math/52914191.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the diagonals of rectangle $ABCD$ intersect at point $O$. If $\\angle ACB = 30^\\circ$ and $BO = 2$, then what is the length of $BC$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rectangle,\\\\\n$\\therefore \\angle ABC=90^\\circ$, AC=BD, AO=CO, BO=DO,\\\\\n$\\therefore$ AO=OC=OB=OD,\\\\\nSince BO=2,\\\\\n$\\therefore$ AO=OC=BO=2,\\\\\n$\\therefore$ AC=2+2=4,\\\\\nSince $\\angle BCA=30^\\circ$,\\\\\n$\\therefore$ AB= $\\frac{1}{2}$ AC=2,\\\\\n$\\therefore$ BC= $\\sqrt{AC^2-AB^2}=2\\sqrt{3}$,\\\\\nTherefore, the answer is: $BC=\\boxed{2\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50474756.png"
  },
  {
    "solution": "\\boxed{44^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ is inscribed in circle $\\odot O$, with $\\angle C=46^\\circ$. Connecting $OA$, what is the measure of $\\angle OAB$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure:\\\\\nConnect OB,\\\\\n$\\because \\angle C=46^\\circ$,\\\\\n$\\therefore \\angle AOB=92^\\circ$,\\\\\nAlso, $\\because OA=OB$,\\\\\n$\\therefore \\angle OAB=\\angle OBA=(180-92^\\circ)÷2=\\boxed{44^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52774538.png"
  },
  {
    "solution": "\\boxed{4.8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a painting with a rectangular frame is to be drawn on the gable roof of a building for beautification. It is known that $BC=12m$, and the height $AD=8m$. If the frame is a square, and one side of the square lies on $BC$, with the other two vertices on $AB$ and $AC$ respectively, what is the side length of the square?\n",
    "completion": "\\textbf{Solution:} Let the side length of the square $PN=x$, then $AE=(8-x)$,\\\\\nsince quadrilateral $PQMN$ is a square,\\\\\ntherefore $PN\\parallel QM$,\\\\\nsince $QM$ lies on $BC$,\\\\\ntherefore $PM\\parallel BC$,\\\\\ntherefore $\\triangle APN\\sim \\triangle ABC$,\\\\\ntherefore $\\frac{PN}{BC}=\\frac{AE}{AD}$,\\\\\nthat is $\\frac{x}{12}=\\frac{8-x}{8}$,\\\\\nsolving for $x$ yields $x=\\boxed{4.8}$, which means: the side length of the square is \\boxed{4.8}m.",
    "image_path": "dataset/mm_math/images/MM_Math/53080410_112.png"
  },
  {
    "solution": "\\boxed{-5+2\\sqrt{13}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in parallelogram $ABCD$, point $E$ is on side $AB$. $\\triangle ADE$ is folded along $DE$ to obtain $\\triangle A'DE$. Given that $AB=14$, $AD=13$, $BD=15$, and let $BE=x$. When point $A'$ lies inside (including on the boundary) of $\\triangle CDE$, what is the range of values for $x$?\n",
    "completion": "\\textbf{Solution:} Construct $DH\\perp AB$ through point D, with the foot being H, and construct $CI\\perp AB$ through point C, with the foot being I.\\\\\nSince quadrilateral ABCD is a parallelogram,\\\\\ntherefore $AB\\parallel CD$, $AD=BC$, $AB=CD$,\\\\\nSince $DH\\perp AB$, $CI\\perp AB$,\\\\\ntherefore $DH\\parallel CI$, $\\angle DHA=\\angle DHB=\\angle CIB=90^\\circ$,\\\\\ntherefore quadrilateral DHIC is a rectangle, $\\triangle ADH$, $\\triangle BDH$, $\\triangle BCI$ are right-angled triangles,\\\\\ntherefore $DH=CI$,\\\\\nIn $\\triangle ADH$ and $\\triangle BCI$,\\\\\n$\\left\\{\\begin{array}{l}DH=CI\\\\ AD=BC\\end{array}\\right.$\\\\\ntherefore $\\triangle ADH \\cong \\triangle BCI\\ (HL)$,\\\\\ntherefore $AH=BI$,\\\\\nLet $AH=m$,\\\\\nSince $AB=14$,\\\\\ntherefore $HB=14-m$,\\\\\nIn $\\triangle ADH$, using Pythagoras' theorem, we get: $DH^2=AD^2-AH^2$,\\\\\nIn $\\triangle BDH$, using Pythagoras' theorem, we get: $DH^2=BD^2-HB^2$,\\\\\ntherefore $AD^2-AH^2=BD^2-HB^2$,\\\\\nSince $AD=13$, $BD=15$,\\\\\ntherefore $13^2-m^2=15^2-(14-m)^2$, simplifying gives: $28m=140$, solving gives: $m=5$,\\\\\ntherefore $AH=5$, hence $BI=5$,\\\\\ntherefore $DH=\\sqrt{AD^2-AH^2}=\\sqrt{13^2-5^2}=12$, hence $CI=12$,\\\\\n(1) When $A'$ is on CE, as shown in the figure,\\\\\nSince $\\triangle ADE$ folds along DE to form $\\triangle A'DE$\\\\\ntherefore $\\angle A=\\angle DA'E$, $AD=A'D$,\\\\\nSince $\\angle DA'E + \\angle DA'C=180^\\circ$,\\\\\ntherefore $\\angle A + \\angle DA'C =180^\\circ$,\\\\\nSince quadrilateral ABCD is a parallelogram,\\\\\ntherefore $AD\\parallel BC$, $AB\\parallel CD$, $AB=CD$,\\\\\ntherefore $\\angle CBE + \\angle A=180^\\circ$, $\\angle CEB=\\angle DCA'$, $AD=BC$,\\\\\ntherefore $\\angle DA'C=\\angle CBE$, $DA'=BC$,\\\\\nIn $\\triangle BCE$ and $\\triangle A'DC$,\\\\\n$\\left\\{\\begin{array}{l}\\angle CEB=\\angle DCA'\\\\ \\angle CBE=\\angle DA'C\\\\ BC=A'D\\end{array}\\right.$\\\\\n$\\triangle BCE \\cong \\triangle A'DC\\ (AAS)$,\\\\\ntherefore $EC=CD$,\\\\\nSince $CD=AB=14$,\\\\\ntherefore $EC=14$\\\\\nSince $BE=x$,\\\\\ntherefore $EI=BE+BI=x+5$,\\\\\nIn $\\triangle ECI$, using Pythagoras' theorem, we get: $EC^2=EI^2+CI^2$,\\\\\ntherefore $14^2=(x+5)^2+12^2$, simplifying gives: $x^2+10x-27=0$, solving gives: $x_1=-5+2\\sqrt{13}$, $x_2=-5-2\\sqrt{13}$ (discard),\\\\\ntherefore $x=\\boxed{-5+2\\sqrt{13}}$,\\\\",
    "image_path": "dataset/mm_math/images/MM_Math/52159971.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, the medians $AD$ and $BE$ intersect at point $O$. If the area of $\\triangle AOE$ is 4, what is the area of the quadrilateral $OECD$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect DE,\\\\\n$\\because$ AD, BE are medians on sides BC, AC respectively,\\\\\n$\\therefore$ D, E are the midpoints of BC, AC respectively,\\\\\n$\\therefore$ DE is the midline of $\\triangle ABC$,\\\\\n$\\therefore$ DE $=\\frac{1}{2}AB$, DE $\\parallel$ AB,\\\\\n$\\therefore$ $\\triangle ABO\\sim \\triangle DEO$, $\\triangle CDE\\sim \\triangle CBA$,\\\\\n$\\therefore$ $\\frac{OE}{OB}=\\frac{DE}{AB}=\\frac{1}{2}$,\\\\\n$\\therefore$ $\\frac{S_{\\triangle ABO}}{S_{\\triangle AOE}}=\\frac{BO}{EO}=2$,\\\\\n$\\therefore$ $S_{\\triangle ABO}=8$,\\\\\n$\\therefore$ $S_{\\triangle ABE}=S_{\\triangle ABO}+S_{\\triangle AOE}=12$,\\\\\n$\\therefore$ $S_{\\triangle ABC}=2S_{\\triangle ABE}=24$,\\\\\n$\\because$ $\\frac{S_{\\triangle DEO}}{S_{\\triangle ABO}}=\\left(\\frac{DE}{AB}\\right)^2=\\frac{1}{4}$,\\\\\n$\\therefore$ $S_{\\triangle DEO}=2$,\\\\\n$\\because$ $\\frac{S_{\\triangle CDE}}{S_{\\triangle ABC}}=\\left(\\frac{DE}{AB}\\right)^2=\\frac{1}{4}$,\\\\\n$\\therefore$ $S_{\\triangle CDE}=6$,\\\\\n$\\therefore$ $S_{\\text{quadrilateral} OECD}=S_{\\triangle DEO}+S_{\\triangle CDE}=\\boxed{8}$,",
    "image_path": "dataset/mm_math/images/MM_Math/51379404.png"
  },
  {
    "solution": "\\boxed{\\frac{49}{13}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the side length of the square paper $ABCD$ is $12$. Point $E$ is on side $CD$, and line $AE$ is connected. The paper is folded such that point $A$ exactly coincides with point $G$ on $AE$, resulting in the crease $BF$ intersecting $AD$ at point $F$. If $DE=5$, what is the length of $GE$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a square,\\\\\nwe have AB=AD=12 and $\\angle$BAD=$\\angle$D=90$^\\circ$,\\\\\nby the property of folding and axial symmetry, $\\triangle$ABF$\\cong$$\\triangle$GBF, BF bisects AG perpendicularly,\\\\\nthus BF$\\perp$AE, and AH=GH,\\\\\nwhich means $\\angle$BAH+$\\angle$ABH=90$^\\circ$,\\\\\nalso, since $\\angle$FAH+$\\angle$BAH=90$^\\circ$,\\\\\nwe get $\\angle$ABH=$\\angle$FAH,\\\\\nthus, $\\triangle$ABF$\\cong$$\\triangle$DAE (ASA),\\\\\ntherefore, AF=DE=5,\\\\\nin right triangle $\\triangle$ABF,\\\\\nBF=$\\sqrt{AB^{2}+AF^{2}}=\\sqrt{12^{2}+5^{2}}=13$,\\\\\n$Area_{\\triangle ABF}=\\frac{1}{2}$AB$\\cdot$AF=$\\frac{1}{2}$BF$\\cdot$AH,\\\\\ntherefore, 12$\\times$5=13AH,\\\\\nthus AH=$\\frac{60}{13}$,\\\\\ntherefore, AG=2AH=$\\frac{120}{13}$,\\\\\nsince AE=BF=13,\\\\\nthus, GE=AE−AG=13−$\\frac{120}{13}=\\boxed{\\frac{49}{13}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52211582_75.png"
  },
  {
    "solution": "\\boxed{\\frac{7}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABD$ and $\\triangle CBD$, $\\angle ADB=90^\\circ$, $\\angle ABD=\\angle DBC$, $AD=DC=1$. If $AB=4$, what is the length of $BC$?\n",
    "completion": "\\textbf{Solution:} Draw DE$\\perp$AB at point E, and DF$\\perp$BC, intersecting the extension of BC at point F, \\\\\n$\\therefore$ $\\angle$CFD$=90^\\circ$, \\\\\n$\\because$ $\\angle$ABD$=\\angle$DBC, \\\\\n$\\therefore$ DE=DF; \\\\\nIn right triangles $\\triangle$DEB and $\\triangle$DFB, \\\\\n$\\left\\{\\begin{array}{l}BD=BD\\\\ DE=DF\\end{array}\\right.$\\\\\n$\\therefore$ right $\\triangle$DEB$\\cong$ right $\\triangle$DFB (HL)\\\\\n$\\therefore$ BE=BF; \\\\\nIn right $\\triangle$ABD, \\\\\n$BD=\\sqrt{AB^{2}-AD^{2}}=\\sqrt{4^{2}-1^{2}}=\\sqrt{15}$; \\\\\n$\\because$ $S_{\\triangle ABC}=\\frac{1}{2}AD\\cdot BD=\\frac{1}{2}AB\\cdot DE$ \\\\\n$\\therefore$ DE=DF$=\\frac{1\\times \\sqrt{15}}{4}=\\frac{\\sqrt{15}}{4}$, \\\\\nIn right $\\triangle$BDE, \\\\\n$BE=BF=\\sqrt{BD^{2}-DE^{2}}=\\sqrt{\\left(\\sqrt{15}\\right)^{2}-\\left(\\frac{\\sqrt{15}}{4}\\right)^{2}}=\\frac{15}{4}$, \\\\\n$CF=\\sqrt{DC^{2}-DF^{2}}=\\sqrt{1^{2}-\\left(\\frac{\\sqrt{15}}{4}\\right)^{2}}=\\frac{1}{4}$ \\\\\n$\\therefore$ $BC=BF-CF=\\frac{15}{4}-\\frac{1}{4}=\\boxed{\\frac{7}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53033969.png"
  },
  {
    "solution": "\\boxed{\\frac{15}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C = 90^\\circ$, point D is on $AC$, and $\\angle DBC = \\angle A$. If $AC=4$ and $\\cos A = \\frac{4}{5}$, what is the length of $BD$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle C=90^\\circ$,\\\\\nhence $\\cos A=\\frac{AC}{AB}$,\\\\\nsince $AC=4$, $\\cos A=\\frac{4}{5}$,\\\\\nthus $AB=5$,\\\\\nby the Pythagorean theorem, we get $BC= \\sqrt{AB^{2}-AC^{2}}=3$,\\\\\nsince $\\angle DBC=\\angle A$,\\\\\nthus $\\cos \\angle DBC=\\cos A= \\frac{4}{5}$,\\\\\ntherefore $\\cos \\angle DBC= \\frac{BC}{BD}=\\frac{4}{5}$, which means $\\frac{3}{BD}=\\frac{4}{5}$\\\\\nhence $BD=\\boxed{\\frac{15}{4}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51212924.png"
  },
  {
    "solution": "\\boxed{x\\le -1$ or $x\\ge 2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, within the same Cartesian coordinate system, the graphs of the functions $y_1=|x|$ and $y_2=kx+b$ intersect at points $A(-1, 1)$ and $B(2, 2)$. Based on the graph, what is the solution set of the inequality $|x|-kx-b\\ge 0$?\n",
    "completion": "\\textbf{Solution:} Observing the graph, we find: when $x\\le -1$ or $x\\ge 2$, the graph of the function $y_1=|x|$ is above or intersects with the graph of $y_2=kx+b$, \\\\\n$\\because |x|-kx-b\\ge 0$, \\\\\n$\\therefore |x| \\ge kx+b$, \\\\\n$\\therefore$ the solution set of the inequality $|x|-kx-b\\ge 0$ is $x\\le -1$ or $x\\ge 2$. \\\\\nThus, the answer is: $\\boxed{x\\le -1$ or $x\\ge 2}$",
    "image_path": "dataset/mm_math/images/MM_Math/52211318.png"
  },
  {
    "solution": "\\boxed{70^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, rectangle $ABCD$ is folded along $EF$, causing point $D$ to coincide with point $B$. If $\\angle AEB = 40^\\circ$, what is the degree measure of $\\angle BFE$?\n",
    "completion": "\\textbf{Solution:} Given the method,\\\\\n$\\because \\angle AEB=40^\\circ$,\\\\\n$\\therefore \\angle BED=140^\\circ$,\\\\\n$\\therefore$ From the property of folding: $\\angle BEF=\\angle DEF=70^\\circ$,\\\\\n$BG=CD$, $GF=CF$, $\\angle G=\\angle D=90^\\circ$,\\\\\n$\\because$ Quadrilateral ABCD is a rectangle,\\\\\n$\\therefore AD\\parallel BC$, $CD=AB$, $BC=AD$,\\\\\n$\\therefore \\angle BFE=\\angle DEF=\\boxed{70^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51420070_124.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the parallelogram $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$. Point $E$ is taken on the extension line of $DC$ such that $CE=\\frac{1}{2}CD$. Line $OE$ intersects $BC$ at point $F$. If $BC=12$, what is the length of $CF$?\n",
    "completion": "\\textbf{Solution:} Construct $OM\\parallel BC$ intersecting $CD$ at $M$,\\\\\nsince in parallelogram $ABCD$, $BC=12$,\\\\\ntherefore, $BO=DO$,\\\\\ntherefore, $CM=DM=\\frac{1}{2}CD$, $OM=\\frac{1}{2}BC=6$\\\\\nsince $CE=\\frac{1}{2}CD$,\\\\\ntherefore, $CE=CM$,\\\\\nsince $OM\\parallel BC$,\\\\\ntherefore, $CF$ is the midline of $\\triangle EMO$, that is $CF=\\frac{1}{2}OM=\\boxed{3}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52423384.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\triangle MPN $, H is the intersection of the altitudes $ MQ $ and $ NR $, and $ MQ = NQ $. Given $ PQ = 3 $ and $ NQ = 7 $, what is the length of $ MH $?\n",
    "completion": "\\textbf{Solution:} Let $H$ be the intersection of altitudes $MQ$ and $NR$,\\\\\ntherefore $NR\\perp PM$, and $MQ\\perp PN$,\\\\\ntherefore $\\angle PMQ + \\angle MHR = 90^\\circ$, and $\\angle HNQ + \\angle QHN = 90^\\circ$.\\\\\nSince $\\angle MHR = \\angle QHN$,\\\\\ntherefore $\\angle PMQ = \\angle HNQ$.\\\\\nAlso, since $MQ = NQ = 7$ and $\\angle PQM = \\angle HQN = 90^\\circ$,\\\\\ntherefore $\\triangle PMQ \\cong \\triangle HNQ$ (by ASA),\\\\\nthus $PQ = HQ = 3$,\\\\\nhence $MH = MQ - HQ = 7 - 3 = \\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52978601.png"
  },
  {
    "solution": "\\boxed{\\frac{4}{15}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, point $E$ is a point on side $BC$, and $BE: EC=2:1$, $AE$ intersects $BD$ at point $F$. What is the ratio of the area of $\\triangle BEF$ to the area of $\\triangle ABD$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral ABCD is a parallelogram,\\\\\n$\\therefore$ $AD\\parallel BC$ and $AD=BC$,\\\\\n$\\therefore$ $\\angle ADB=\\angle CBD$,\\\\\nGiven $\\angle AFD=\\angle BFE$,\\\\\n$\\therefore$ $\\triangle AFD\\sim \\triangle EFB$,\\\\\nGiven $BE:EC=2:1$,\\\\\n$\\therefore$ $\\frac{BE}{AD}=\\frac{2}{3}$,\\\\\n$\\therefore$ $\\frac{{h}_{\\triangle BEF}}{{h}_{\\triangle DAF}}=\\frac{2}{3}$,\\\\\n$\\therefore$ $\\frac{{h}_{\\triangle BEF}}{{h}_{\\triangle ABD}}=\\frac{2}{5}$,\\\\\nLet $BE=2x$, then $AD=3x$,\\\\\nLet ${h}_{\\triangle BEF}=2a$, then ${h}_{\\triangle ABD}=5a$,\\\\\n$\\therefore$ $\\frac{{S}_{\\triangle BEF}}{{S}_{\\triangle ABD}}=\\frac{2x\\cdot 2a}{3x\\cdot 5a}=\\boxed{\\frac{4}{15}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51353232.png"
  },
  {
    "solution": "\\boxed{24}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, P is a point on segment AB, and Q is a point on the extension of AB. If $AB=10$ and $\\frac{AP}{PB} = \\frac{AQ}{QB} = \\frac{3}{2}$. Find the length of segment PQ.\n",
    "completion": "\\textbf{Solution:} Given $\\because AP:PB=3:2$,\\\\\n$\\therefore$ let AP=3a, and PB=2a,\\\\\n$\\because P$ is a point on segment AB, and AB=10,\\\\\n$\\therefore 3a+2a=10$, solving for $a$ we find $a=2$,\\\\\n$\\therefore AP=6$, PB=4. Let the length of BQ be $x$, then $AQ=AB+BQ=10+x$,\\\\\n$\\because \\frac{AQ}{BQ}=\\frac{3}{2}$,\\\\\n$\\therefore \\frac{10+x}{x}=\\frac{3}{2}$, solving for $x$ yields $x=20$, i.e., $BQ=20$,\\\\\n$\\therefore PQ=PB+BQ=\\boxed{24}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55034387.png"
  },
  {
    "solution": "\\boxed{36^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the figure, points A, B, and C are all on the circle $\\odot O$, and point C is on the major arc corresponding to chord AB. If $\\angle AOB=72^\\circ$, what is the degree measure of $\\angle ACB$?\n",
    "completion": "\\textbf{Solution:} \\\\\nSince the central angle $\\angle AOB$ and the inscribed angle $\\angle ACB$ both subtend the arc $\\overset{\\frown}{AB}$, \\\\\ntherefore $\\angle ACB = \\frac{1}{2}\\angle AOB = \\frac{1}{2} \\times 72^\\circ = \\boxed{36^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/51446532.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in trapezoid $ABCD$, where $AD\\parallel BC$ and $AC$ intersects $BD$ at point $O$, if $BC = 2AD$, what is the value of the ratio $S_{\\triangle AOD} : S_{\\triangle BOC}$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure below:\n\n$\\because$AD $\\parallel$ BC,\n\n$\\therefore\\triangle AOD\\sim \\triangle COB$,\n\n$\\therefore \\frac{S_{\\triangle AOD}}{S_{\\triangle COB}}=\\left(\\frac{AD}{BC}\\right)^2$\n\n$\\because$BC＝2AD,\n\n$\\therefore \\frac{S_{\\triangle AOD}}{S_{\\triangle COB}}=\\left(\\frac{AD}{2AD}\\right)^2=\\left(\\frac{1}{2}\\right)^2=\\boxed{\\frac{1}{4}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51433232.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AD$ and $AE$ are the median and altitude to side $BC$, respectively. If $AE = 5\\,\\text{cm}$ and the area of $\\triangle ABC = 30\\,\\text{cm}^2$, what is the length of $DC$ in cm?\n",
    "completion": "\\textbf{Solution:} Let $AD$ and $AE$ be the median and the height to the side $BC$ respectively, with $AE=5\\,\\text{cm}$ and $S_{\\triangle ABC}=30\\,\\text{cm}^{2}$,\\\\\ntherefore $S_{\\triangle ADC}=15\\,\\text{cm}^{2}$,\\\\\nthus $\\frac{1}{2} \\times AE \\times CD = 15$,\\\\\nhence $\\frac{1}{2} \\times 5 \\times CD = 15$,\\\\\nsolving this gives: $CD=\\boxed{6}\\,(\\text{cm})$;",
    "image_path": "dataset/mm_math/images/MM_Math/52861353_108.png"
  },
  {
    "solution": "\\boxed{AB + AC}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $D$, $E$, and $F$ are the midpoints of sides $BC$, $CA$, and $AB$, respectively. What is the perimeter of the quadrilateral $AFDE$?\n",
    "completion": "Solution: Since D, E, and F are the midpoints of sides $BC$, $CA$, and $AB$ respectively, \\\\\nit follows that $DF \\parallel AC$, $DE \\parallel AB$, and $DF = \\frac{1}{2} AC$, $DE = \\frac{1}{2} AB$, \\\\\ntherefore, quadrilateral AFDE is a parallelogram, \\\\\nhence, $DF = AE = \\frac{1}{2} AC$, $DE = AF = \\frac{1}{2} AB$, \\\\\ntherefore, the perimeter of parallelogram AFDE is equal to $\\boxed{AB + AC}$",
    "image_path": "dataset/mm_math/images/MM_Math/52211683.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{7}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, square $ABCD$ and square $CEFG$ are given, where point $D$ is on $CG$, $BC=2$, $H$ is the midpoint of $AF$, and $CH=4$. What is the length of $CE$?\n",
    "completion": "\\textbf{Solution:} Connect $AC, CF$,\n\nSince both quadrilateral $ABCD$ and quadrilateral $GCEF$ are squares,\n\n$AC = \\sqrt{AB^{2}+BC^{2}} = \\sqrt{2^{2}+2^{2}} = 2\\sqrt{2}$,\n\n$\\angle ACD=\\angle FCD=45^\\circ$,\n\ntherefore $\\angle ACF=90^\\circ$,\n\nAlso, since point $H$ is the midpoint of $AF$,\n\n$AF=2CH=8$．\n\nTherefore, in $\\triangle ACF$, $CF =\\sqrt{AF^{2}-AC^{2}}=\\sqrt{8^{2}-\\left(2\\sqrt{2}\\right)^{2}}=2\\sqrt{14}$,\n\nTherefore, in $\\triangle CEF$, $CE=EF$,\n\n$CE^{2}+EF^{2}=CF^{2}$,\n\n$2CE^{2}=\\left(2\\sqrt{14}\\right)^{2}$,\n\ntherefore, $CE=\\boxed{2\\sqrt{7}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50496996.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in a Cartesian coordinate system, the line $y=-\\frac{1}{2}x-1$ intersects with the line $y=-2x+2$ at point $P$, and they intersect the $x$-axis at point A and B, respectively. What is the area of $\\triangle PAB$?\n",
    "completion": "\\textbf{Solution:} Let $y=-\\frac{1}{2}x-1=0$,\\\\\nthen $-\\frac{1}{2}x-1=0$,\\\\\nsolving, we find: $x=-2$,\\\\\n$\\therefore \\boxed{A(-2, 0)}$,\\\\\nLet $y=-2x+2=0$,\\\\\nthen $-2x+2=0$,\\\\\nsolving, we find: $x=1$,\\\\\n$\\therefore \\boxed{B(1, 0)}$,\\\\\n$\\therefore AB=3$,\\\\\n$\\therefore {S}_{\\triangle PAB}=\\frac{1}{2}AB\\cdot \\left|{y}_{P}\\right|=\\frac{1}{2}\\times 3\\times 2=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52250673_89.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=6$, $BC=5$, points $D$ and $E$ are on $BC$ and $AC$ respectively, $CD=2BD$, $CE=2AE$, $BE$ intersects $AD$ at point $F$, what is the maximum area of $\\triangle AFE$?\n",
    "completion": "\\textbf{Solution:} Connect DE, and draw DH $\\perp$ AB at H, as shown in the figure:\\\\\n$\\because$ CD = 2BD, CE = 2AE,\\\\\n$\\therefore \\frac{CD}{BD}=\\frac{CE}{AE}=2$,\\\\\n$\\therefore$ DE $\\parallel$ AB,\\\\\n$\\therefore \\triangle CDE \\sim \\triangle CBA, \\triangle DEF \\sim \\triangle ABF$,\\\\\n$\\therefore \\frac{DE}{BA}=\\frac{CD}{CB}=\\frac{2}{3}$, $\\frac{DF}{AF}=\\frac{DE}{AB}$,\\\\\n$\\therefore \\frac{DF}{AF}=\\frac{DE}{AB}=\\frac{2}{3}$,\\\\\n$\\because$ DE $\\parallel$ AB,\\\\\n$\\therefore S_{\\triangle ABE}=S_{\\triangle ABD}$,\\\\\n$\\therefore S_{\\triangle ABE} - S_{\\triangle ABF}=S_{\\triangle ABD} - S_{\\triangle ABF}$,\\\\\n$\\therefore S_{\\triangle AEF}=S_{\\triangle BDF}$,\\\\\n$\\therefore S_{\\triangle AEF}=\\frac{2}{5}S_{\\triangle ABD}$,\\\\\n$\\because BD = \\frac{1}{3}BC = \\frac{5}{3}$, and $AB = 6$,\\\\\n$\\therefore$ when DH is maximal, the area of $\\triangle ABD$ is maximal,\\\\\n$\\because$ DH $\\leq$ BD,\\\\\n$\\therefore$ when DH = BD = $\\frac{5}{3}$, the area of $\\triangle ABD$ is maximal, and the maximum value is $\\frac{1}{2} \\times \\frac{5}{3} \\times 6 = 5$,\\\\\n$\\therefore$ the maximum area of $\\triangle AFE$ is $\\frac{2}{5} \\times 5 = \\boxed{2}$,",
    "image_path": "dataset/mm_math/images/MM_Math/52536566.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, it is known that point $B$ is on line segment $AC$, and point $D$ is on line segment $AB$. If $AB=6\\,\\text{cm}$, $BC=4\\,\\text{cm}$, and $D$ is the midpoint of line segment $AC$, what is the length of line segment $DB$ in centimeters?\n",
    "completion": "\\textbf{Solution:}\\\\\nSince $AC=AB+BC$, $AB=6\\text{cm}$, $BC=4\\text{cm}$\\\\\nTherefore $AC=6+4=10\\text{cm}$\\\\\nAlso, since $D$ is the midpoint of line segment $AC$\\\\\nTherefore $DC= \\frac{1}{2}AC= \\frac{1}{2}\\times10=5\\text{cm}$\\\\\nTherefore $DB=DC-BC=6-5=\\boxed{1}\\text{cm}$",
    "image_path": "dataset/mm_math/images/MM_Math/51438092_89.png"
  },
  {
    "solution": "\\boxed{10\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the rhombus $ABCD$, $\\angle ABC=120^\\circ$ and $AB=10$, with points $M$ and $N$ located on sides $AD$ and $AB$, respectively. What is the length of the diagonal $AC$?\n",
    "completion": "\\textbf{Solution:} The length of the diagonal $AC$ is $10\\sqrt{3}=\\boxed{10\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53151061_71.png"
  },
  {
    "solution": "\\boxed{2x^{2}-2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=ax^{2}-a$ ($a>0$) intersects the x-axis at points $A$ and $B$ ($A$ is to the left of $B$), and intersects the y-axis at point $C$, with $OC=2OA$. What is the analytical expression of the parabola?\n",
    "completion": "\\textbf{Solution:} \nGiven $A(-1, 0)$, \\\\\n$\\therefore OA=1$, \\\\\n$\\therefore OC=2OA=2$, \\\\\n$\\therefore C(0, -2)$, \\\\\nSubstituting into the parabola $y=ax^{2}-a(a>0)$, we get: $-a=-2$, \\\\\n$\\therefore a=2$, \\\\\n$\\therefore$ The equation of the parabola is $y=2x^{2}-2=\\boxed{2x^{2}-2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52917460_254.png"
  },
  {
    "solution": "\\boxed{-36}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the line $AB$ intersects the $x$-axis and $y$-axis at point $A$ and $B$, respectively, and the line $CD$ intersects the $x$-axis and $y$-axis at point $C$ and $D$, respectively. The lines $AB$ and $CD$ intersect at point $E$. The lengths of segments $OA$ and $OC$ are the two roots of the quadratic equation ${x}^{2}-18x+72=0$ (with $OA>OC$), and $BE=5$, $\\tan\\angle ABO=\\frac{3}{4}$. If the graph of the inverse proportion function $y=\\frac{k}{x}$ passes through point $E$, what is the value of $k$?\n",
    "completion": "\\textbf{Solution:} Draw a line $EH \\perp AC$ at point H through point E,\\\\\nbecause $\\tan\\angle ABO=\\frac{3}{4}$,\\\\\nthus $\\frac{OA}{OB}=\\frac{3}{4}$, solving gives: $OB=16$,\\\\\nhence $AB=\\sqrt{16^2+12^2}=20$,\\\\\nsince $BE=5$,\\\\\nthus $AE=15$,\\\\\nbecause $EH\\perp AC$, $OB\\perp AC$,\\\\\nthus $EH\\parallel OB$,\\\\\nhence $\\triangle AEH\\sim \\triangle AOB$,\\\\\nthus $\\frac{EH}{BO}=\\frac{AH}{AO}=\\frac{AE}{AB}$, that is $\\frac{EH}{16}=\\frac{AH}{12}=\\frac{15}{20}$,\\\\\nsolving gives: $EH=12$, $AH=9$,\\\\\nthus $OH=3$,\\\\\nhence point E is (−3,12),\\\\\nsince the graph of the inverse proportion function $y=\\frac{k}{x}$ passes through point E,\\\\\nthus $12=\\frac{k}{-3}$, solving gives: $k=\\boxed{-36}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52623363_231.png"
  },
  {
    "solution": "\\boxed{70^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, six rays OA, OB, OC, OD, OE, and OF are drawn from point O, and $\\angle AOB=90^\\circ$, OF bisects $\\angle BOC$, OE bisects $\\angle AOD$. If $\\angle EOF=170^\\circ$, what is the degree measure of $\\angle COD$?\n",
    "completion": "\\textbf{Solution:} Given the problem, we have \\\\\n\\(\\because \\angle AOB=90^\\circ\\), \\(\\angle EOF=170^\\circ\\), \\\\\n\\(\\therefore \\angle AOE + \\angle FOB = 360^\\circ - 90^\\circ - 170^\\circ = 100^\\circ\\). \\\\\n\\(\\because\\) OF bisects \\(\\angle COB\\), OE bisects \\(\\angle AOD\\), \\\\\n\\(\\therefore \\angle COF = \\angle FOB\\), \\(\\angle AOE = \\angle EOD\\). \\\\\n\\(\\therefore \\angle EOD + \\angle COF = 100^\\circ\\). \\\\\n\\(\\therefore \\angle COD = 170^\\circ - 100^\\circ = \\boxed{70^\\circ}\\).",
    "image_path": "dataset/mm_math/images/MM_Math/52890008.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the inverse proportion function $y=\\frac{k}{x}\\ (k\\ne 0)$ intersects with the graph of the direct proportion function $y=2x$ at points $A(1, a)$ and $B$. Point $C$ is located in the fourth quadrant, with $CA\\parallel y$-axis, and $AB\\perp BC$. What is the area of $\\triangle ABC$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $BH\\perp AC$ at $H$, and let $AC$ intersect the x-axis at point $D$, \\\\\n$\\because AB\\perp BC$, \\\\\n$\\therefore \\angle ABC=90^\\circ$, $\\angle BHC=90^\\circ$, \\\\\n$\\therefore \\angle C=\\angle ABH$, \\\\\n$\\because BH\\parallel x$-axis, \\\\\n$\\therefore \\angle AOD=\\angle ABH$, \\\\\n$\\therefore \\angle AOD=\\angle C$, \\\\\n$\\therefore \\tan C=\\tan\\angle AOD=\\frac{AD}{OD}=2$, \\\\\n$\\because A(1,2), B(-1,-2)$, \\\\\n$\\therefore AH=4$, $BH=2$, $OD=1$, $AD=2$, \\\\\n$\\therefore AB=\\sqrt{AH^2+BH^2}=\\sqrt{4^2+2^2}=2\\sqrt{5}$, the area of $\\triangle AOD=\\frac{1}{2}OD\\cdot AD=1$, \\\\\n$\\because \\angle AOD=\\angle C$, $\\angle ADO=\\angle ABC=90^\\circ$, \\\\\n$\\therefore \\triangle ADO\\sim \\triangle ABC$, \\\\\n$\\therefore$ the ratio $\\frac{S_{\\triangle ADO}}{S_{\\triangle ABC}}=\\left(\\frac{AD}{AB}\\right)^2$, hence $\\frac{1}{S_{\\triangle ABC}}=\\left(\\frac{2}{2\\sqrt{5}}\\right)^2$, \\\\\nSolving gives $S_{\\triangle ABC}=\\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52765898_184.png"
  },
  {
    "solution": "\\boxed{-1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=ax^2-2ax-3a$ (where $a$ is a constant and $a<0$) intersects the x-axis at points A and B (point A is to the left of point B), and intersects the y-axis at point C, and $OB=OC$. Find the value of $a$.\n",
    "completion": "\\textbf{Solution.} From the given information, we have\n\\[y = a{x}^{2}-2ax-3a \\\\\n= a({x}^{2}-2x-3) \\\\\n= a(x-3)(x+1)\\]\nSetting $y=0$, we find ${x}_{1}=-1$, ${x}_{2}=3$. Setting $x=0$, we have $y=-3a$.\n\nThe parabola $y=a{x}^{2}-2ax-3a$ (where $a$ is a constant, $a<0$) intersects the $x$-axis at points $A$ and $B$ (with $A$ to the left of $B$) and the $y$-axis at point $C$, making the intersection points with the $x$-axis $A(-1,0)$, $B(3,0)$, and with the $y$-axis $C(0,-3a)$.\n\nSince $OB=OC$, we have $OC=3$, hence $C(0,3)$, which implies $-3a=3$.\n\nSolving for $a$ gives $a=\\boxed{-1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51433217_215.png"
  },
  {
    "solution": "\\boxed{\\frac{80}{11}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given, in the Cartesian coordinate system, square $ABCD$, with point $A(0,10)$, point $B(8,4)$, and point $C$ in the first quadrant. Point $P$ starts from vertex $A$ and moves counterclockwise along the edges of the square for one round. On the right side of $OP$, make $\\angle OPQ = \\angle OAB$ intersecting the x-axis at point $Q$. When point $P$ coincides with point $B$, find the length of $OQ$.\n",
    "completion": "\\textbf{Solution:} As shown in Figure 2, draw $BE\\perp OQ$ at $E$, and $QF\\perp OB$ at $F$, \\\\\nsince $B(8,4)$, \\\\\ntherefore $OB=\\sqrt{8^2+4^2}=4\\sqrt{5}$, $\\tan\\angle BOE=\\frac{BE}{OE}=\\frac{1}{2}$, \\\\\n$\\tan\\angle OBQ=\\tan\\angle OAB=\\frac{BG}{AG}=\\frac{8}{6}=\\frac{4}{3}$, \\\\\nthus $\\frac{QF}{BF}=\\frac{4}{3}$, $\\frac{QF}{OF}=\\frac{1}{2}$. Let $FQ=4k$, $BF=3k$, $OF=8k$, since $OF+BF=OB$, \\\\\nthus $8k+3k=4\\sqrt{5}$, \\\\\nthus $k=\\frac{4\\sqrt{5}}{11}$, \\\\\nthus $OF=\\frac{32\\sqrt{5}}{11}$, $FQ=\\frac{16\\sqrt{5}}{11}$, \\\\\nthus $OQ=\\sqrt{OF^2+FQ^2}=\\boxed{\\frac{80}{11}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51317370_135.png"
  },
  {
    "solution": "\\boxed{x<-2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, the graph $l_1$ of the linear function $y=k_1x+b_1$ intersects with the graph $l_2$ of $y=k_2x+b_2$ at point P. What is the solution set for the inequality in terms of $x$, $\\left(k_1-k_2\\right)x+\\left(b_1-b_2\\right)>0$?\n",
    "completion": "\\textbf{Solution:} Given that $y_1-y_2=(k_1x+b_1)-(k_2x+b_2)=(k_1-k_2)x+(b_1-b_2)$,\\\\\nas can be seen from the graph: to the left of their intersection point, $y_1>y_2$,\\\\\nmeaning that when $x<-2$, $y_1-y_2>0$,\\\\\ntherefore, the solution set of $(k_1-k_2)x+(b_1-b_2)>0$ is $\\boxed{x<-2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52386068.png"
  },
  {
    "solution": "\\boxed{75^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of a semicircle, and $C$, $D$ are two points on the semicircle. Given that $ \\angle ADC=105^\\circ$, what is the measure of $ \\angle ABC$?\n",
    "completion": "\\textbf{Solution:} Since $A,B,C,D$ are points on the circle $\\odot O$ and $\\angle ADC=105^\\circ$,\n\nit follows that $\\angle ABC=180^\\circ-105^\\circ=\\boxed{75^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/51446341.png"
  },
  {
    "solution": "\\boxed{7}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point B is a point on line segment AC, and $AB=21$, $BC=\\frac{1}{3}AB$. If point O is the midpoint of line segment AC, calculate the length of line segment OB.\n",
    "completion": "\\textbf{Solution:} Since $O$ is the midpoint of $AC$,\\\\\ntherefore, $OC=\\frac{1}{2}AC=14$,\\\\\ntherefore, $OB=OC-BC=14-7=\\boxed{7}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52251794_179.png"
  },
  {
    "solution": "\\boxed{130^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, straight lines AB and CD intersect at point O. If $\\angle 1 + \\angle 2 = 100^\\circ$, what is the measure of $\\angle BOC$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle 1+\\angle 2=100^\\circ$ and $\\angle 1=\\angle 2$, \\\\\nit follows that $\\angle 1=\\angle 2=50^\\circ$. \\\\\nSince $\\angle 1+\\angle BOC=180^\\circ$, \\\\\nit follows that $\\angle BOC=\\boxed{130^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/54927575.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the graph of the linear function $y=kx+b$ intersects the $x$ axis and the $y$ axis at points $A$ and $B$ respectively, and intersects the graph of the hyperbolic function $y=\\frac{m}{x}$ at points $C$ and $D$. $DE \\perp x$ axis at point $E$. Given that the coordinates of point $C$ are $(8, -2)$, and $DE=4$. Find the area of $\\triangle ABO$.\n",
    "completion": "\\textbf{Solution:} Substituting $y=0$ into $y=-\\frac{1}{2}x+2$, we find $x=4$,\\\\\n$\\therefore A(4,0)$, $\\therefore OA=4$. Substituting $x=0$ into $y=-\\frac{1}{2}x+2$, we find $y=2$,\\\\\n$\\therefore B(0,2)$, $\\therefore OB=2$,\\\\\n$\\therefore$ The area of $\\triangle ABO$ is $\\frac{1}{2}\\cdot OA\\cdot OB=\\frac{1}{2}\\times 4\\times 2=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52551855_128.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In $\\triangle ABC$, $AB = AC$, $\\angle ABC = 30^\\circ$, $\\triangle BDE$ is an equilateral triangle, and $CD$ and $AE$ are connected. As shown in Figure 1, when points $A$, $B$, and $D$ are on the same line, $AE$ and $BC$ intersect at point $P$, and $AE \\perp AC$. If $PC = 4$, what is the length of $PE$?\n",
    "completion": "\\textbf{Solution:}\\\\\nSince $AB=AC$, and $\\angle ABC=30^\\circ$\\\\\nThen, $\\angle ACP=30^\\circ$\\\\\nSince $AE\\perp AC$\\\\\nThen, $\\angle PAC=90^\\circ$\\\\\nHence, $AP= \\frac{1}{2}PC=2$\\\\\nSince $\\angle APC=\\angle ABC+\\angle BAE=60^\\circ$\\\\\nThen, $\\angle ABC=\\angle BAE=30^\\circ$\\\\\nHence, $BP=AP=2$\\\\\nSince $\\angle ABE=60^\\circ$\\\\\nThen, $\\angle PBE=30^\\circ$, $\\angle PEB=90^\\circ$\\\\\nHence, $PE= \\boxed{1}$",
    "image_path": "dataset/mm_math/images/MM_Math/51737359_116.png"
  },
  {
    "solution": "\\boxed{25^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of $\\odot O$. If $\\angle AOC=130^\\circ$, what is the measure of $\\angle D$ in degrees?\n",
    "completion": "\\textbf{Solution:} Draw AD, \\\\\nsince AB is the diameter of circle O, and $\\angle$AOC $= 130^\\circ$, \\\\\ntherefore, $\\angle$BDA $= 90^\\circ$, and $\\angle$CDA $= 65^\\circ$, \\\\\nhence, $\\angle$BDC $=\\boxed{25^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51458271.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in $\\triangle ABC$, $AB=AC$, $\\angle ABC=45^\\circ$, $AD\\perp BC$ at point D. If $BC=8$, what is the length of $AD$?\n",
    "completion": "\\textbf{Solution:} Given that $AB=AC$ and $AD\\perp BC$,\\\\\nit follows that $BD=DC=4$,\\\\\nsince $\\angle ABC=45^\\circ$ and $AD\\perp BC$,\\\\\nit results in $\\angle ABD=\\angle BAD=45^\\circ$,\\\\\nthus $AD=BD=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52891246.png"
  },
  {
    "solution": "\\boxed{9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle A=90^\\circ$, $\\angle C=30^\\circ$, and $AD \\perp BC$ at $D$. $BE$ is the bisector of $\\angle ABC$, and it intersects $AD$ at point $P$. If $AP=3$, what is the length of $AC$?\n",
    "completion": "\\textbf{Solution:} Given $\\because \\angle A=90^\\circ$, $\\angle C=30^\\circ$, \\\\\n$\\therefore \\angle ABC=60^\\circ$, \\\\\n$\\because BE$ is the bisector of $\\angle ABC$, \\\\\n$\\therefore \\angle ABP=\\angle DBP=30^\\circ$, \\\\\n$\\because AD\\perp BC$, \\\\\n$\\therefore \\angle BAD=30^\\circ$, \\\\\n$\\therefore \\angle PAB=\\angle PBA$, \\\\\n$\\therefore BP=AP=3$, \\\\\nIn $Rt\\triangle PBD$, $PD=\\frac{1}{2}PB=\\frac{3}{2}$, \\\\\n$\\therefore AD=AP+PD=\\frac{9}{2}$, \\\\\nIn $Rt\\triangle ADC$, $AC=2AD=\\boxed{9}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52653195.png"
  },
  {
    "solution": "\\boxed{y=-x-1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the graph of the linear function $y=kx+b$ intersects with the graph of the inverse proportion function $y=\\frac{m}{x}$ at points $A$ and $B$. The coordinates of point $A$ are $\\left(-2, 1\\right)$, and the coordinates of point $B$ are $\\left(1, n\\right)$. Determine the analytical expressions of the two functions.\n",
    "completion": "\\textbf{Solution:} Since the graph of the inverse proportion function $y=\\frac{m}{x}$ passes through point $A(-2, 1)$,\n\n\\[m=-2\\times 1=-2,\\]\n\nthus, the analytical expression of the inverse proportion function is $y=-\\frac{2}{x}$.\n\nSince the graph of the inverse proportion function $y=-\\frac{2}{x}$ passes through point $B(1, n)$,\n\n\\[-2=1\\times n,\\]\n\nthus, $n=-2$,\n\nimplying $B(1, -2)$.\n\nSince the graph of the linear function $y=kx+b$ passes through points $A$, $B$,\n\nwe have the system of equations\n\\[\n\\left\\{\\begin{array}{l}\nk+b=-2 \\\\\n-2k+b=1\n\\end{array}\\right.,\n\\]\n\nsolving this system yields\n\\[\n\\left\\{\\begin{array}{l}\nk=-1 \\\\\nb=-1\n\\end{array}\\right.,\n\\]\n\nhence, the analytical expression of the linear function is $y=-x-1$;\n\nTherefore, the answers are \\boxed{y=-\\frac{2}{x}} and \\boxed{y=-x-1}.",
    "image_path": "dataset/mm_math/images/MM_Math/52623325_245.png"
  },
  {
    "solution": "\\boxed{9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, points D, E, and F are located on sides AB, BC, and AC, respectively, with $DE \\parallel AC$ and $EF \\parallel AB$. $\\triangle BDE \\sim \\triangle EFC$. If $\\frac{AF}{FC} = \\frac{1}{2}$ and $AD = 6$, find the length of AB.\n",
    "completion": "\\textbf{Solution:} Given $\\because$ EF $\\parallel$ AB, $\\therefore$ $\\frac{BE}{EC}=\\frac{AF}{FC}=\\frac{1}{2}$, and $\\because$ DE $\\parallel$ AC, $\\therefore$ $\\frac{BD}{AD}=\\frac{BE}{EC}=\\frac{1}{2}$, thus $\\frac{BD}{AD}=\\frac{AF}{FC}=\\frac{1}{2}$. Given AD＝6, $\\therefore$ $\\frac{BD}{6}=\\frac{1}{2}$, thus BD=3, hence AB＝\\boxed{9}.",
    "image_path": "dataset/mm_math/images/MM_Math/51433795_137.png"
  },
  {
    "solution": "\\boxed{24}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the diagonals $AC$ and $BD$ of the rhombus $ABCD$ intersect at point $O$, and $E$ is the midpoint of $AD$. If $OE=3$, then what is the perimeter of the rhombus $ABCD$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral $ABCD$ is a rhombus,\\\\\nit follows that $AC \\perp BD$ and $AB = BC = CD = DA$,\\\\\nwhich means $\\triangle AOD$ is a right-angled triangle.\\\\\nGiven that $OE = 3$, and point $E$ is the midpoint of segment $AD$,\\\\\nit follows that $AD = 2OE = 6$.\\\\\nTherefore, the perimeter of the rhombus is: $6 \\times 4 = \\boxed{24}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51918841.png"
  },
  {
    "solution": "\\boxed{(-3,0) \\text{ or } (7,0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=-\\frac{1}{2}x$ intersects the x-axis and y-axis at points B and A, respectively. Point C is a moving point on the x-axis. A circle $\\odot C$ with center C and radius $\\sqrt{5}$ is drawn. When $\\odot C$ is tangent to line AB, what are the coordinates of point C?\n",
    "completion": "\\textbf{Solution:}\n(1) When point C is on the negative half of the x-axis, let C be $(t,0)$. Construct CM $\\perp$ AB at M, as shown,\\\\\nFor\\\\\n$y=-\\frac{1}{2}x+1$, when $x=0$, $y=1$; when $y=0$, $x=2$\\\\\n$\\therefore$ A$(0,1)$, B$(2,0)$\\\\\n$\\therefore$ OA$=1$, OB$=2$,\\\\\n$\\therefore$ BC$=2-t$\\\\\nBy the Pythagorean theorem,\\\\\n$AB=\\sqrt{OA^2+OB^2}=\\sqrt{1^2+2^2}=\\sqrt{5}$\\\\\nSince line AB is tangent to circle C,\\\\\n$\\therefore$ $\\angle$CMB$=90^\\circ$\\\\\nAlso $\\angle$ABO$=\\angle$CBM,\\\\\n$\\triangle$BMC$\\sim$$\\triangle$BOA,\\\\\n$\\therefore$ $\\frac{AB}{BC}=\\frac{AO}{CM}$, i.e.,\\\\\n$\\frac{\\sqrt{5}}{2-t}=\\frac{1}{\\sqrt{5}}$\\\\\nSolving this,\\\\\n$t=-3$\\\\\n$\\therefore$ The coordinates of point C are ($-3$,0)\\\\\n(2) When point C is on the positive half of the x-axis, let C be $(t,0)$. Construct CN $\\perp$ AB at N, as shown,\\\\\n$\\therefore$ BC$=t-2$\\\\\nSince $\\angle$ABO$=\\angle$CBN,\\\\\n$\\angle$CNB$=\\angle$AOB\\\\\n$\\therefore$ $\\triangle$BNC$\\sim$$\\triangle$BOA,\\\\\n$\\therefore$ $\\frac{AB}{BC}=\\frac{AO}{CN}$, i.e.,\\\\\n$\\frac{\\sqrt{5}}{t-2}=\\frac{1}{\\sqrt{5}}$\\\\\nSolving this,\\\\\n$t=7$\\\\\n$\\therefore$ The coordinates of point C are $(7,0)$\\\\\nIn summary, the coordinates of point C are $(-3,0)$ or $(7,0)$\\\\\nTherefore, the answer is: $\\boxed{(-3,0) \\text{ or } (7,0)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51322578.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within circle $O$ with a radius of $10\\,\\text{cm}$, $AB=16\\,\\text{cm}$, and $OC\\perp$ chord $AB$ at point $C$, what is the length of $OC$ in $\\text{cm}$?\n",
    "completion": "\\textbf{Solution:} Join OA, as shown in the figure,\\\\\n$\\because OC\\perp AB$,\\\\\n$\\therefore AC=BC= \\frac{1}{2}AB=8$,\\\\\nIn Rt$\\triangle$ OAC, $OC=\\sqrt{OA^{2}-AC^{2}}=\\sqrt{10^{2}-8^{2}}=\\boxed{6}\\ (\\text{cm})$.",
    "image_path": "dataset/mm_math/images/MM_Math/53080595.png"
  },
  {
    "solution": "\\boxed{D(6,8)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, rectangle $ABCD$ is positioned in the Cartesian coordinate system as depicted, with $F$ being a point on $AB$. Fold $\\triangle BCF$ along $CF$ making point $B$ precisely fall on the intersection point $E$ of $AD$ and the y-axis. Connect $CE$. If the lengths of $AE$ and $AB$ satisfy $\\sqrt{AE-4}+(AB-8)^2=0$, what are the coordinates of point $D$?\n",
    "completion": "\\textbf{Solution:} Let $AE$ be $x$. According to the Pythagorean theorem, we have:\\\\\n$(8-x)^{2}-x^{2}=4^{2}$\\\\\nSolving this, we get: $x=3$\\\\\nLet $ED$ be $y$. According to the Pythagorean theorem, we have:\\\\\n$y^{2}+8^{2}=(y+4)^{2}$\\\\\nSolving this, we get: $y=6$\\\\\n$\\therefore \\boxed{D(6,8)}$",
    "image_path": "dataset/mm_math/images/MM_Math/53359586_201.png"
  },
  {
    "solution": "\\boxed{y=-2x+5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, the linear function $y=x+2$ intersects the x-axis and the y-axis at points A and B, respectively. Point $M\\left(1, m\\right)$ is a point on line AB, and line MC intersects the x-axis at point $C\\left(\\frac{5}{2}, 0\\right)$. Find the analytical expression of line MC.\n",
    "completion": "\\textbf{Solution:} Substituting the point $M\\left(1, m\\right)$ into $y=x+2$ yields: $m=1+2=3$, $\\therefore$ the point $M(1, 3)$. Let the equation of line $MC$ be $y=kx+b\\left(k\\ne 0\\right)$. Substituting the points $M(1, 3)$ and $C\\left(\\frac{5}{2}, 0\\right)$ into it gives: $\\begin{cases}\\frac{5}{2}k+b=0 \\\\ k+b=3\\end{cases}$, solving this system yields: $\\begin{cases}k=-2 \\\\ b=5\\end{cases}$, $\\therefore$ the equation of the line $MC$ is $\\boxed{y=-2x+5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52381160_167.png"
  },
  {
    "solution": "\\boxed{(-3, -14)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola passes through points $A\\left(4, 0\\right)$, $B\\left(1, 0\\right)$, and $C\\left(0, -2\\right)$. Let $P$ be a moving point on the parabola, and a perpendicular line is drawn from $P$ to the $x$-axis, with the foot of the perpendicular being $M$. Is there a point $P$ on the parabola such that the triangle with vertices $A$, $P$, and $M$ is similar to $\\triangle OAC$? If such a point exists, find the coordinates of point $P$; if not, please explain why.\n",
    "completion": "\\textbf{Solution:} A solution exists. Let the x-coordinate of point $P$ be $m$, then the y-coordinate of $P$ is $-\\frac{1}{2}m^{2}+\\frac{5}{2}m-2$. When $1<m<4$, $AM=4-m$, $PM=-\\frac{1}{2}m^{2}+\\frac{5}{2}m-2$.\\\\\nSince $\\angle COA=\\angle PMA=90^\\circ$,\\\\\nTherefore, (1) when $\\frac{AM}{PM}=\\frac{AO}{OC}$, since $C$ is on the parabola, $OC=2$, $OA=4$, thus $\\frac{AM}{PM}=\\frac{AO}{OC}=2$, $\\triangle APM \\sim \\triangle ACO$, that is, $4-m=2\\left(-\\frac{1}{2}m^{2}+\\frac{5}{2}m-2\\right)$.\\\\\nSolving yields $m_{1}=2$, $m_{2}=4$ (discard), thus $P(2, 1)$.\\\\\n(2) When $\\frac{AM}{PM}=\\frac{OC}{OA}=\\frac{1}{2}$, $\\triangle APM \\sim \\triangle CAO$, that is, $2(4-m)=-\\frac{1}{2}m^{2}+\\frac{5}{2}m-2$.\\\\\nSolving yields $m_{1}=4$, $m_{2}=5$ (neither fits the conditions, discard)\\\\\nTherefore, when $1<m<4$, $P(2, 1)$, when $m>4$, $AM=m-4$, $PM=\\frac{1}{2}m^{2}-\\frac{5}{2}m+2$,\\\\\n(1) $\\frac{PM}{AM}=\\frac{OC}{OA}=\\frac{1}{2}$ or (2) $\\frac{PM}{AM}=\\frac{OA}{OC}=2$,\\\\\nSubstituting $P\\left(m, -\\frac{1}{2}m^{2}+\\frac{5}{2}m-2\\right)$ yields: $2\\left(\\frac{1}{2}m^{2}-\\frac{5}{2}m+2\\right)=m-4$, $2\\left(m-4\\right)=\\frac{1}{2}m^{2}-\\frac{5}{2}m+2$,\\\\\nThe solutions are: the first equation yields $m=-2-2\\sqrt{3}<4$ (discard) $m=-2+2\\sqrt{3}<4$ (discard), the second equation yields $m=5$, $m=4$ (discard)\\\\\nSolving for $m=5$, $-\\frac{1}{2}m^{2}+\\frac{5}{2}m-2=-2$, then $P(5, -2)$, when $m<1$, $AM=4-m$, $PM=\\frac{1}{2}m^{2}-\\frac{5}{2}m+2$.\\\\\n(1) $\\frac{PM}{AM}=\\frac{OC}{OA}=\\frac{1}{2}$ or $\\frac{PM}{AM}=\\frac{OA}{OC}=2$,\\\\\nYields: $2\\left(\\frac{1}{2}m^{2}-\\frac{5}{2}m+2\\right)=4-m$, $2\\left(4-m\\right)=\\frac{1}{2}m^{2}-\\frac{5}{2}m+2$,\\\\\nThe solutions are: the first equation yields $m=0$ (discard), $m=4$ (discard), the second equation yields $m=4$ (discard), $m=-3$,\\\\\nFor $m=-3$, $-\\frac{1}{2}m^{2}+\\frac{5}{2}m-2=-14$, then $P(-3, -14)$,\\\\\nIn summary, the points $P$ that meet the conditions are $\\boxed{(2, 1)}$, $\\boxed{(5, -2)}$, or $\\boxed{(-3, -14)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52513841_162.png"
  },
  {
    "solution": "\\boxed{100^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, line $AB$ and line $CD$ intersect at point $O$, $\\angle DOE = 90^\\circ$, $OD$ bisects $\\angle BOF$, $\\angle BOE = 50^\\circ$. What is the measure of $\\angle AOF$ in degrees?\n",
    "completion": "\\textbf{Solution:} From the given conditions, we have\n\\[\n\\angle AOF = 180^\\circ - \\angle AOC - \\angle DOF = 180^\\circ - 40^\\circ - 40^\\circ = \\boxed{100^\\circ}.\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/52795107_102.png"
  },
  {
    "solution": "\\boxed{90^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, point D is the midpoint of side AB, and $CD = \\frac{1}{2}AB$. Prove that: $\\angle ACB = 90^\\circ$.\n",
    "completion": "\\textbf{Solution:} Proof: Since point D is the midpoint of side $AB$,\\\\\n$ \\therefore AD=BD=\\frac{1}{2}AB$,\\\\\nAlso since $CD=\\frac{1}{2}AB$,\\\\\n$ \\therefore AD=BD=CD$,\\\\\n$ \\therefore \\angle ACD=\\angle A$, $\\angle BCD=\\angle B$.\\\\\nSince $\\angle A+\\angle B+(\\angle ACD+\\angle BCD)=180^\\circ$,\\\\\n$ \\therefore \\angle ACD+\\angle BCD+(\\angle ACD+\\angle BCD)=180^\\circ$,\\\\\n$ \\therefore 2(\\angle ACD+\\angle BCD)=180^\\circ$,\\\\\n$ \\therefore \\angle ACD+\\angle BCD=90^\\circ$,\\\\\nwhich means $\\angle ACB=\\boxed{90^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52891228.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, place $ \\angle BAC$ on a $5\\times 5$ square grid. If the vertices A, B, and C are all on the grid points, then what is the value of the tangent of $ \\angle BAC$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect BC,\\\\\nthen $AB=BC=\\sqrt{1^2+3^2}=\\sqrt{10}$, $AC=\\sqrt{2^2+4^2}=2\\sqrt{5}$,\\\\\n$\\therefore AB^2+BC^2=10+10=20=AC^2$,\\\\\n$\\therefore \\triangle ABC$ is an isosceles right triangle, and $\\angle ABC=90^\\circ$,\\\\\n$\\therefore \\angle BAC=45^\\circ$,\\\\\nthen $\\tan\\angle BAC=\\boxed{1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51402363.png"
  },
  {
    "solution": "\\boxed{65^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AD$ is the bisector of $\\angle BAC$, and $E$ is a point on $AD$, with $EF \\perp BC$ at point $F$. If $\\angle C = 35^\\circ$ and $\\angle DEF = 15^\\circ$, what is the measure of $\\angle B$?\n",
    "completion": "\\textbf{Solution:} Since $EF\\perp BC$ and $\\angle DEF=15^\\circ$, we have $\\angle ADB=90^\\circ-15^\\circ=75^\\circ$. \\\\\nSince $\\angle C=35^\\circ$, it follows that $\\angle CAD=75^\\circ-35^\\circ=40^\\circ$. \\\\\nSince $AD$ is the bisector of $\\angle BAC$, we have $\\angle BAC=2\\angle CAD=80^\\circ$, hence $\\angle B=180^\\circ-\\angle BAC-\\angle C=180^\\circ-80^\\circ-35^\\circ=\\boxed{65^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52978928.png"
  },
  {
    "solution": "\\boxed{15}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $BC=10$, and $\\angle BAC=30^\\circ$. If points $M$ and $N$ are chosen on $AC$ and $AB$, respectively, such that the value of $BM+MN$ is minimized, what is this minimum value?\n",
    "completion": "\\textbf{Solution:} Construct the symmetric point $B^{\\prime}$ of point $B$ with respect to $AC$. Draw $B^{\\prime}N \\perp AB$, intersecting $AC$ at point $M$.\n\nSince rectangle $ABCD$,\n\nTherefore, $\\angle ABC = 90^\\circ$,\n\nSince in $\\triangle ABC$, $\\angle BAC = 30^\\circ$,\n\nTherefore, $\\angle ACB = 90^\\circ - 30^\\circ = 60^\\circ$,\n\nTherefore, $AB = BC \\tan \\angle ACB = 10 \\tan 60^\\circ = 10\\sqrt{3}$;\n\nSince points $B$ and $B^{\\prime}$ are symmetric about $AC$,\n\nTherefore, $AB = AB^{\\prime}$, $\\angle BAC = \\angle B^{\\prime}AC = 30^\\circ$,\n\nTherefore, $\\angle BAB^{\\prime} = 60^\\circ$,\n\nTherefore, $\\triangle BAB^{\\prime}$ is an equilateral triangle;\n\nSince $B^{\\prime}N \\perp AB$,\n\nTherefore, $B^{\\prime}N = BB^{\\prime} \\sin \\angle B^{\\prime}BN = 10\\sqrt{3} \\sin 60^\\circ = 10\\sqrt{3} \\times \\frac{\\sqrt{3}}{2} = 15$\n\nSince points $B$ and $B^{\\prime}$ are symmetric about $AC$,\n\nTherefore, $BM = B^{\\prime}M$,\n\nTherefore, $BM + MN = B^{\\prime}M + MN = B^{\\prime}N = \\boxed{15}$",
    "image_path": "dataset/mm_math/images/MM_Math/52331837.png"
  },
  {
    "solution": "\\boxed{6-3\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of the semicircle $O$, and points $E$, $C$, $D$ are three points on the semicircular arc, with $AC \\parallel OD$ and $AB \\parallel CD$. Given that $AB=12$ and $\\angle EAC = 15^\\circ$, and connecting $OE$ that intersects $AC$ at point $F$, what is the length of $EF$?\n",
    "completion": "\\textbf{Solution:} Let's connect OC. \\\\\nSince $AC\\parallel OD$, $AB\\parallel CD$, and $OA=OD$, \\\\\nit follows that quadrilateral AODC is a rhombus. \\\\\nThus, $AC=OD$. \\\\\nGiven $AB=12$, \\\\\nit follows that $OE=OC=AC=OA=\\frac{1}{2}AB=6$. \\\\\nHence, $\\triangle AOC$ is an equilateral triangle. \\\\\nThus, $\\angle AOC=60^\\circ$. \\\\\nSince $\\angle EAC=15^\\circ$, \\\\\nit follows that $\\angle EOC=2\\angle EAC=30^\\circ$. \\\\\nThus, $OE$ bisects $\\angle AOC$. \\\\\nTherefore, $OE\\perp AC$. \\\\\nIn $\\triangle OFC$, $\\cos\\angle EOC=\\frac{OF}{OC}=\\frac{\\sqrt{3}}{2}$. \\\\\nThus, $OF=\\frac{\\sqrt{3}}{2}OC=3\\sqrt{3}$. \\\\\nTherefore, $EF=OE-OF=6-3\\sqrt{3}$. \\\\\nHence, the answer is: $EF=\\boxed{6-3\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53080565.png"
  },
  {
    "solution": "\\boxed{\\frac{4}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, $AD\\parallel BC$, $\\angle ABC=90^\\circ$, $O$ is the midpoint of the diagonal $BD$, $OA=2$, $BC=5$, and $CD=3$. What is the value of $\\tan\\angle DCB$?\n",
    "completion": "\\textbf{Solution:} Since AD $\\parallel$ BC and $\\angle$ABC$=90^\\circ$, \\\\\nit follows that $\\angle$BAD$=90^\\circ$, \\\\\nGiven that O is the midpoint of the diagonal BD and OA=2, \\\\\nit follows that BD=2OA=4, \\\\\nGiven BC=5 and CD=3, \\\\\nit follows that BD$^{2}$+CD$^{2}$=BC$^{2}$, \\\\\nthus $\\angle$BDC$=90^\\circ$, \\\\\nhence $\\tan$($\\angle$DCB) = $\\frac{BD}{CD}$ = $\\boxed{\\frac{4}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51351896.png"
  },
  {
    "solution": "\\boxed{126^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, given two lines $AB$ and $CD$ intersect at point $O$, $OE$ bisects $\\angle BOD$, and $OF$ bisects $\\angle COB$. If $\\angle BOE = 36^\\circ$, what is the degree measure of $\\angle AOF$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle DOE + \\angle EOB + \\angle BOF + \\angle FOC = 180^\\circ$, and that $OE$ bisects $\\angle BOD$ and $OF$ bisects $\\angle COB$, \\\\\nthus $2\\angle EOB + 2\\angle BOF = 180^\\circ$, \\\\\nhence, $\\angle BOE + \\angle BOF = 90^\\circ$, \\\\\nsince $\\angle BOE = 36^\\circ$, \\\\\ntherefore $\\angle BOF = 90^\\circ - 36^\\circ = 54^\\circ$, \\\\\ngiven that $\\angle AOF + \\angle BOF = 180^\\circ$, \\\\\nthus $\\angle AOF = 180^\\circ - 54^\\circ = \\boxed{126^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53253495.png"
  },
  {
    "solution": "\\boxed{110^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, quadrilateral $ABCD$ is inscribed in circle $\\odot O$, with $E$ being a point on the extension line of diameter $AB$, and $AB \\parallel DC$. If $\\angle A=70^\\circ$, what is the degree measure of $\\angle CBE$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral $ABCD$ is inscribed in circle $O$,\\\\\nit follows that $\\angle A+\\angle C=180^\\circ$,\\\\\nSince $\\angle A=70^\\circ$,\\\\\nit follows that $\\angle C=110^\\circ$,\\\\\nSince $AB\\parallel DC$,\\\\\nit follows that $\\angle CBE=\\angle C=110^\\circ$;\\\\\nTherefore, the answer is: $\\boxed{110^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51402228.png"
  },
  {
    "solution": "\\boxed{12^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC$, point $D$ is the midpoint of $BC$, $\\angle BAD=24^\\circ$, $AD=AE$, what is the measure of $\\angle EDC$ in degrees?\n",
    "completion": "\\textbf{Solution:} Given $AB=AC$ and point $D$ is the midpoint of $BC$, with $\\angle BAD=24^\\circ$, \\\\\nit follows that $\\angle CAD=\\angle BAD=24^\\circ$, and $AD\\perp BC$,\\\\\nGiven $AD=AE$,\\\\\nit implies $\\angle ADE=\\angle AED=\\frac{1}{2}\\times (180^\\circ-24^\\circ)=78^\\circ$,\\\\\nthus $\\angle EDC=90^\\circ-\\angle ADE=\\boxed{12^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52604387.png"
  },
  {
    "solution": "\\boxed{C(0,4)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, given the quadratic function $y=-\\frac{4}{3}(x+1)^{2}+\\frac{16}{3}$ intersects the x-axis at points $A$ and $B$ (with point $A$ to the left of point $B$), and intersects the y-axis at point $C$. Point $D$ is the vertex of the parabola. Find the coordinates of points $A$ and $C$.\n",
    "completion": "\\textbf{Solution:} Let $y=0$, we get $-\\frac{4}{3}(x+1)^2+\\frac{16}{3}=0$,\\\\\nSolving for $x$, we find $x_1=-3$ and $x_2=1$,\\\\\n$\\therefore \\boxed{A(-3, 0)}, B(1, 0)$,\\\\\nLet $x=0$, we get $y=-\\frac{4}{3}(0+1)^2+\\frac{16}{3}=4$,\\\\\n$\\therefore \\boxed{C(0,4)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52872397_95.png"
  },
  {
    "solution": "\\boxed{25^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $O$ is a point on the line $AB$. Rays $OC$, $OD$, and $OM$ are drawn such that $OM$ bisects $\\angle AOC$. When $OC$, $OD$, and $OM$ are all above $AB$, given that $\\angle COD=60^\\circ$ and $\\angle BOC=10^\\circ$, what is the degree measure of $\\angle DOM$?\n",
    "completion": "\\textbf{Solution}: From the given information, we have $\\angle AOC=180^\\circ-10^\\circ=170^\\circ$, \\\\\nsince $OM$ bisects $\\angle AOC$, \\\\\nthus $\\angle MOC=\\frac{170^\\circ}{2}=85^\\circ$, \\\\\ntherefore $\\angle MOD=85^\\circ-60^\\circ=\\boxed{25^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55393357_80.png"
  },
  {
    "solution": "\\boxed{16}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in quadrilateral $ABCD$, with $AB=10$, $BC=18$, the angle bisectors of $\\angle ABC$ and $\\angle BCD$ intersect $AD$ at points E and F, respectively. If $BE=12$, then what is the length of $CF$?\n",
    "completion": "\\textbf{Solution:} In parallelogram $ABCD$, $BE$ bisects $\\angle ABC$, and $CF$ bisects $\\angle BCD$,\n\n$\\therefore$ $\\angle ABE=\\angle CBE$, $\\angle BCF=\\angle DCF$,\n\n$\\because$ $AB\\parallel CD$,\n\n$\\therefore$ $\\angle ABC+\\angle DCB=180^\\circ$, $\\angle CBE=\\angle AEB$, $\\angle BCF=\\angle DFC$,\n\n$\\therefore$ $\\angle CBE+\\angle BCF=90^\\circ$, $\\angle ABE=\\angle AEB$, $\\angle DCF=\\angle DFC$,\n\n$\\therefore$ $AE=AB=10$, $DF=DC=10$,\n\n$\\because$ $AD=BC=18$,\n\n$\\therefore$ $AF=AD-DF=8$,\n\n$\\therefore$ $EF=AE-AF=2$,\n\nExtend $CG$ such that $CG=EF$,\n\n$\\therefore$ $EFCG$ is a parallelogram,\n\n$\\therefore$ $CG=EF=2$, $BG=20$, $EG=CF$,\n\n$\\therefore$ $\\angle BCF=\\angle G$,\n\n$\\therefore$ $\\angle BEG=90^\\circ$,\n\n$\\because$ $BE=12$, $BG=20$,\n\n$\\therefore$ $EG=\\sqrt{BG^2-BE^2}=\\sqrt{20^2-12^2}=16$,\n\n$\\therefore$ $CF=EG=\\boxed{16}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53538277.png"
  },
  {
    "solution": "\\boxed{y=\\frac{800}{x}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the level of concentration of students during a class can be represented by the attention index. The variation of the attention index $y$ of a class of students during a math class over the class time $x$ (in minutes) is depicted in the figure. At the beginning of the class, the attention index is $30$, and at the $2$nd minute, the attention index is $40$. Within the first $10$ minutes, the attention index $y$ is a linear function of time $x$. After $10$ minutes, the attention index $y$ becomes inversely proportional to $x$. When $10\\leq x\\leq 40$, what is the functional relationship between $y$ and $x$?\n",
    "completion": "\\textbf{Solution:} When $10\\leq x\\leq 40$, let $y=\\frac{m}{x}$. Substituting $(10,80)$ into it, we get: $m=800$.\\\\\nThus $y=\\frac{800}{x}$.\\\\\nTherefore, the function relationship of $y$ with respect to $x$ is $\\boxed{y=\\frac{800}{x}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52536612_149.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{6}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in parallelogram $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$. $M$ is the midpoint of $AD$, and line $CM$ intersects $BD$ at point $N$. What is the value of $\\frac{S_{\\triangle MDN}}{S_{\\triangle BCD}}$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral ABCD is a parallelogram, \\\\\n$\\therefore AD\\parallel BC$ and $AD=BC$, \\\\\nGiven that point M is the midpoint of AD, \\\\\n$\\therefore MD = \\frac{1}{2}AD = \\frac{1}{2}BC$, \\\\\nSince $AD\\parallel BC$, \\\\\n$\\therefore \\triangle MDN \\sim \\triangle CBN$, \\\\\n$\\therefore \\frac{MD}{BC} = \\frac{DN}{BN} = \\frac{MN}{NC} = \\frac{1}{2}$, \\\\\n$\\therefore BN = 2DN$, $CN = 2MN$, \\\\\n$\\therefore S_{\\triangle CDN} = 2S_{\\triangle MDN}$, $S_{\\triangle BNC} = 2S_{\\triangle CDN}$, \\\\\n$\\therefore S_{\\triangle BCD} = 6S_{\\triangle MDN}$, \\\\\n$\\therefore \\frac{S_{\\triangle MDN}}{S_{\\triangle BCD}} = \\boxed{\\frac{1}{6}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52457419.png"
  },
  {
    "solution": "\\boxed{\\frac{3\\sqrt{3}}{8}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC$, $\\angle BAC=120^\\circ$, $BC=2\\sqrt{3}$, with $D$ being the midpoint of $BC$, and $DE\\perp AB$, what is the area of $\\triangle EBD$?\n",
    "completion": "\\textbf{Solution:} Since AB=AC and $\\angle$BAC=120$^\\circ$, \\\\\nit follows that $\\angle$B=$\\angle$C=30$^\\circ$, \\\\\nSince D is the midpoint of BC and BC=$2\\sqrt{3}$, \\\\\nthen in right $\\triangle$BED, BD=$\\frac{1}{2}$BC=$\\sqrt{3}$, thus DE=$\\frac{\\sqrt{3}}{2}$, BE=$\\frac{3}{2}$. \\\\\nTherefore, the area of $\\triangle$EBD is $\\frac{1}{2}\\times \\frac{\\sqrt{3}}{2}\\times \\frac{3}{2}=\\boxed{\\frac{3\\sqrt{3}}{8}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53105790.png"
  },
  {
    "solution": "\\boxed{\\frac{9}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $D$ is a point on $AB$, and $\\angle ABC = \\angle ACD$. $\\triangle ABC \\sim \\triangle ACD$; if $AC=3$, $AD=2$, find the length of $AB$.\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABC \\sim \\triangle ACD$, \\\\\n$\\therefore \\frac{AB}{AC}=\\frac{AC}{AD}$, \\\\\nwhich means $\\frac{AB}{3}=\\frac{3}{2}$, \\\\\n$\\therefore AB=\\boxed{\\frac{9}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52513755_118.png"
  },
  {
    "solution": "\\boxed{71^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\angle A = \\angle B$, $AE = BE$, point $D$ is on side $AC$, $\\angle 1 = \\angle 2$, $AE$ and $BD$ intersect at point $O$, if $\\angle 1 = 38^\\circ$, then what is the degree measure of $\\angle BDE$?\n",
    "completion": "\\textbf{Solution:} Given that $AE$ and $BD$ intersect at point $O$,\\\\\n$\\therefore \\angle AOD=\\angle BOE$,\\\\\nIn $\\triangle AOD$ and $\\triangle BOE$, $\\angle A=\\angle B$,\\\\\n$\\therefore \\angle BEO=\\angle 2$,\\\\\n$\\because \\angle 1=\\angle 2$,\\\\\n$\\therefore \\angle 1=\\angle BEO$,\\\\\n$\\therefore \\angle 1+\\angle DEO=\\angle BEO+\\angle DEO$,\\\\\nthat is, $\\angle AEC=\\angle BED$,\\\\\nIn $\\triangle AEC$ and $\\triangle BED$,\\\\\n$\\left\\{\\begin{array}{l}\n\\angle A=\\angle B \\\\\n\\angle AEC=\\angle BED \\\\\nAE=BE\n\\end{array}\\right.$\\\\\n$\\therefore \\triangle AEC\\cong\\triangle BED\\left(AAS\\right)$,\\\\\n$\\therefore CE=DE$,\\\\\n$\\because \\angle 1=38^\\circ$,\\\\\n$\\therefore \\angle C=\\angle EDC=\\frac{1}{2}\\times \\left(180^\\circ-\\angle 1\\right)=\\frac{1}{2}\\times \\left(180^\\circ-38^\\circ\\right)=71^\\circ$,\\\\\n$\\therefore \\angle BDE=\\boxed{71^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53078472.png"
  },
  {
    "solution": "\\boxed{127^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, place the right-angle vertex of a triangular ruler, which includes a $30^\\circ$ angle, against one side of a straightedge. If $\\angle 1 = 37^\\circ$, then what is the degree measure of $\\angle 2$?\n",
    "completion": "\\textbf{Solution:}\\\\\nSince $\\angle 1=37^\\circ$, the triangle is a right-angled triangle.\\\\\nTherefore, $\\angle 3=90^\\circ-\\angle 1=53^\\circ$.\\\\\nSince the two lines on the ruler are parallel,\\\\\nTherefore, $\\angle 4=\\angle 3=53^\\circ$.\\\\\nTherefore, $\\angle 2=180^\\circ-\\angle 4=\\boxed{127^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52949122.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, $AC<BC$, $CD$ bisects $\\angle ACB$ and intersects $AB$ at point $D$. A circle $\\odot O$ is drawn with $DB$ as its diameter, intersecting $CD$ and $BC$ at points $E$ and $F$, respectively. Lines $BE$ and $EF$ are drawn. What is the measure of $\\angle EBF$? If $DE=DC$ and $BC=8$, what is the length of $EF$?\n",
    "completion": "\\textbf{Solution:} Connect DF, and draw EG$\\perp$BC at point G,\\\\\n$\\because$BD is a diameter,\\\\\n$\\therefore \\angle CEB=90^\\circ$,\\\\\n$\\because \\angle ACB=90^\\circ$, CD bisects $\\angle ACD$,\\\\\n$\\therefore \\angle DCF= \\frac{1}{2}\\angle ACB=45^\\circ$,\\\\\n$\\therefore \\angle EBF=90^\\circ-\\angle DCF=90^\\circ-45^\\circ=\\boxed{45^\\circ}$;\\\\\n$\\because$BD is a diameter,\\\\\n$\\therefore \\angle DFG=90^\\circ$,\\\\\n$\\therefore DF\\perp BC$,\\\\\n$\\therefore DF\\parallel FG$,\\\\\n$\\because DE=DC$,\\\\\n$\\therefore CF=FG$,\\\\\n$\\because \\angle FCG=\\angle EBC=45^\\circ$,\\\\\n$\\therefore EC=BE$,\\\\\nIn $\\triangle CEB$, $\\angle EBC=45^\\circ$, BC=8,\\\\\n$\\therefore BE=CB\\sin\\angle EBC=8\\sin45^\\circ= 8\\times \\frac{\\sqrt{2}}{2}=4\\sqrt{2}$;\\\\\nIn $\\triangle EBG$, EG=CG=BE$\\sin\\angle EBC= 4\\sqrt{2}\\sin45^\\circ= 4\\sqrt{2}\\times \\frac{\\sqrt{2}}{2}=4$,\\\\\n$\\therefore FG=CG-4$,\\\\\n$\\therefore FG=2$\\\\\nIn $\\triangle EFG$, $EF=\\sqrt{FG^{2}+EG^{2}}=\\sqrt{2^{2}+4^{2}}=\\boxed{2\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52733116.png"
  },
  {
    "solution": "\\boxed{inside}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the Cartesian coordinate system, let $A(0,4)$, $B(2,4)$, $C(4,2)$. What is the position of point $D(3,-1)$ relative to the circle $\\odot M$? (Fill in: inside, outside, or on).\n",
    "completion": "\\textbf{Solution:} Point $D$ is \\boxed{inside} circle $\\odot M$.",
    "image_path": "dataset/mm_math/images/MM_Math/52623631_57.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points $C$ and $D$ are two points on line segment $AB$, with $AC:CD:BD=3:6:4$. If $AB=13$, what is the length of $CD$?\n",
    "completion": "\\textbf{Solution:} Given $\\because$ $AC: CD: BD = 3: 6: 4$, and $AB = 13$,\n\n$\\therefore$ $CD = \\frac{6}{3+6+4}AB = \\boxed{6}$",
    "image_path": "dataset/mm_math/images/MM_Math/53119446.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $C$, $D$, $E$ are three points on the line segment $AB$, where $E$ is the midpoint of $BD$, $AC = \\frac{1}{2} CD$, $AB = 5$, and $EB = 1$. What is the length of $CD$?\n",
    "completion": "\\textbf{Solution:} Since $E$ is the midpoint of $BD$, $EB=1$\\\\\nTherefore, $BD=2BE=2$\\\\\nSince $AB=5$\\\\\nTherefore, $AD=AB-BD=5-2=3$\\\\\nSince $AC=\\frac{1}{2}CD$,\\\\\nTherefore, $AD=AC+CD=\\frac{1}{2}CD+CD=3$, solving gives: $CD=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53253409.png"
  },
  {
    "solution": "\\boxed{32^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, a right-angled triangle ruler containing a $30^\\circ$ angle is positioned as shown together with a straight ruler. If $\\angle 1 = 62^\\circ$, what is the degree measure of $\\angle 2$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram:\\\\\n$\\because \\angle 1=62^\\circ$,\\\\\n$\\therefore$ by the property that parallel lines have equal alternate interior angles, we have $\\angle 3=62^\\circ$,\\\\\n$\\therefore$ by the exterior angle theorem of a triangle, we have $\\angle 4=\\angle 3-30^\\circ=32^\\circ$,\\\\\n$\\therefore \\angle 2=\\angle 4=\\boxed{32^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51213210.png"
  },
  {
    "solution": "\\boxed{M\\left(\\frac{1}{2}, \\frac{15}{4}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that line $AB$ intersects the parabola $C\\colon y=ax^2+2x+c$ at points $A(-1, 0)$ and $B(2, 3)$. If point $M$ is a moving point on the parabola above line $AB$, and a parallelogram $MANB$ is formed with $MA$ and $MB$ as two adjacent sides, when the area of the parallelogram $MANB$ is maximized, find the area $S$ of the parallelogram $MANB$ at this time and the coordinates of point $M$?\n",
    "completion": "\\textbf{Solution:} As shown in Figure 1, draw line $MH\\perp x$-axis at point $H$ through point $M$, and intersect line $AB$ at point $K$.\\\\\nSubstituting the points $\\left(-1,0\\right)$ and $\\left(2,3\\right)$ into $y=kx+b$, we get\n\\[\n\\begin{cases}\n-k+b=0\\\\\n2k+b=3\n\\end{cases},\n\\]\nfrom which we find\n\\[\n\\begin{cases}\nk=1\\\\\nb=1\n\\end{cases},\n\\]\nthus $\\therefore y_{AB}=x+1$.\\\\\nLet $M$ be the point $\\left(a,-a^{2}+2a+3\\right)$, then $K$ is $\\left(a,a+1\\right)$,\\\\\nthen $MK=-a^{2}+2a+3-(a+1)=-\\left(a-\\frac{1}{2}\\right)^{2}+\\frac{9}{4}$.\\\\\nAccording to the properties of quadratic functions, when $a=\\frac{1}{2}$, $MK$ has the maximum length $\\frac{9}{4}$.\\\\\n$\\therefore \\text{The maximum area of } \\triangle AMB=S_{\\triangle AMK}+S_{\\triangle BMK}$\\\\\n$=\\frac{1}{2}MK\\cdot AH+\\frac{1}{2}MK\\cdot \\left(x_{B}-x_{H}\\right)$\\\\\n$=\\frac{1}{2}MK\\cdot \\left(x_{B}-x_{A}\\right)$\\\\\n$=\\frac{1}{2}\\times \\frac{9}{4}\\times 3$\\\\\n$=\\frac{27}{8}$,\\\\\n$\\therefore$ For parallelogram $MANB$ constructed with sides $MA$ and $MB$ being adjacent sides, when the area of parallelogram $MANB$ is maximized,\\\\\n$\\text{The maximum area } S_{\\text{max}}=2S_{\\text{max } \\triangle AMB}=2\\times \\frac{27}{8}=\\boxed{\\frac{27}{4}}$, with $\\boxed{M\\left(\\frac{1}{2}, \\frac{15}{4}\\right)}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53101984_149.png"
  },
  {
    "solution": "\\boxed{\\frac{10\\sqrt{5}}{13}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in $\\triangle ABC$, $AB=AC$, $AD \\perp BC$ at $D$, draw $DE \\perp AC$ at $E$, and let $F$ be the midpoint of $AB$. Connect $EF$ and intersect $AD$ at point $G$. It is given that $AD^{2}=AB\\cdot AE$; if $AB=5$ and $AE=4$, find the value of $DG$.\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect DF.\\\\\n$\\because$AB $= 5$, $\\angle$ADB $= 90^\\circ$, BF $=$ AF,\\\\\n$\\therefore$DF $= \\frac{1}{2}$AB $= \\frac{5}{2}$,\\\\\n$\\because$AB $=$ AC, AD $\\perp$ BC,\\\\\n$\\therefore$BD $=$ DC,\\\\\n$\\therefore$DF $\\parallel$ AC,\\\\\n$\\therefore$ $ \\frac{DF}{AE} = \\frac{DG}{AG} = \\frac{\\frac{5}{2}}{4} = \\frac{5}{8}$,\\\\\n$\\therefore$ $ \\frac{DG}{AD} = \\frac{5}{13}$\\\\\n$\\because$AD$^{2}$ $=$ AB $\\cdot$ AE.\\\\\n$\\therefore$ $ AD = \\sqrt{AB \\cdot AE} = \\sqrt{5 \\times 4} = 2\\sqrt{5}$ \\\\ \n$\\therefore$ $ DG = \\frac{5}{13} \\times 2\\sqrt{5} = \\boxed{\\frac{10\\sqrt{5}}{13}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52977417_129.png"
  },
  {
    "solution": "\\boxed{y=3x+12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the equation of the straight line $l$ is $y=kx+b$. The line $l$ intersects the $x$ axis at point $A$ and intersects the $y$ axis at point $B(0,12)$. Additionally, the area of $\\triangle AOB$ is $24$. What is the equation of the straight line $l$?\n",
    "completion": "\\textbf{Solution:} Given $B(0,12)$ and the area of $\\triangle AOB = 24$,\\\\\nwe have $\\frac{1}{2} \\cdot OA \\cdot 12 = 24$,\\\\\ntherefore $OA=4$,\\\\\nthus $A(-4,0)$,\\\\\nSubstituting $A(-4,0)$ and $B(0,12)$ into $y=kx+b$ yields: $\\left\\{\\begin{array}{l} -4k+b=0 \\\\ b=12 \\end{array}\\right.$,\\\\\nsolving this we get $\\left\\{\\begin{array}{l} k=3 \\\\ b=12 \\end{array}\\right.$,\\\\\nthus, the equation of line $l$ is: \\boxed{y=3x+12};",
    "image_path": "dataset/mm_math/images/MM_Math/52424049_220.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the line segment $AB = 26$, $BC = 18$, and the point $M$ is the midpoint of $AC$. What is the length of the line segment $AC$?\n",
    "completion": "\\textbf{Solution:} Because $AB=26$ and $BC=18$, \\\\\ntherefore $AC=AB-BC=\\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51251319_181.png"
  },
  {
    "solution": "\\boxed{\\frac{125}{39}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle BCA=90^\\circ$, $BC=8$, $AC=6$. Point $D$ is the midpoint of side $AB$, and point $E$ is a moving point on side $BC$. Connect $DE$, and fold $\\triangle BDE$ along $DE$ to get $\\triangle FDE$. As shown in figure (2), when point $E$ coincides with point $C$, segment $EF$ intersects segment $AB$ at point $P$. Find the length of $DP$.\n",
    "completion": "\\[\n\\textbf{Solution:} \\text{Given that,}\n\\]\n\\[\n\\because \\angle PCD = \\angle BCD = \\angle B, \\\\\n\\therefore \\angle PCD = \\angle B, \\\\\n\\because \\angle CPD = \\angle BPC, \\\\\n\\therefore \\triangle CPD \\sim \\triangle BPC, \\\\\n\\therefore \\frac{DP}{CP} = \\frac{CP}{BP} = \\frac{CD}{BC} = \\frac{5}{8}, \\\\\n\\text{Let } DP = 5k, CP = 8k, \\\\\n\\because CP^2 = PD \\cdot PB, \\\\\n\\therefore 64k^2 = 5k(5k + 8k), \\\\\n\\therefore k = \\frac{25}{39}, \\\\\n\\therefore PD = 5k = \\boxed{\\frac{125}{39}}.\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/53151227_94.png"
  },
  {
    "solution": "\\boxed{50^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, it is known that $AB \\parallel CD$, $\\angle 1 = \\angle 2$, $\\angle E = 50^\\circ$. What is the measure of $\\angle F$?\n",
    "completion": "\\textbf{Solution:} Draw line BC, \\\\\n$\\because$AB$\\parallel$CD, \\\\\n$\\therefore$$\\angle$ABC=$\\angle$BCD, \\\\\n$\\because$$\\angle$1=$\\angle$2, \\\\\n$\\therefore$$\\angle$EBC=$\\angle$BCF, \\\\\n$\\therefore$EB$\\parallel$CF, \\\\\n$\\therefore$$\\angle$F=$\\angle$E=$50^\\circ$. \\\\\nTherefore, the answer is: $\\boxed{50^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/54769534.png"
  },
  {
    "solution": "\\boxed{D(-5, 4)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system \\(xOy\\), the side AB of the rhombus ABCD lies on the x-axis, and the vertex C is on the y-axis. It is known that the coordinates of points A and C are \\((-2, 0)\\) and \\((0, 4)\\), respectively. What are the coordinates of points B and D?\n",
    "completion": "\\textbf{Solution:} According to the problem, we know: $OA=2$, $OC=4$, $AB=BC$,\\\\\nLet $AB=BC=x$, $OB=AB-OA=x-2$,\\\\\n$\\therefore$ in right-angled $\\triangle OBC$, $BC^2=OB^2+OC^2$,\\\\\n$\\therefore$ $x^2=(x-2)^2+4^2$, solving this gives: $x=5$,\\\\\n$\\because$ rhombus ABCD, $AB=BC=CD$,\\\\\n$\\therefore$ $AB=CD=5$,\\\\\n$\\therefore$ $OB=3$,\\\\\n$\\therefore$ $B(3, 0)$,\\\\\n$\\because$ $CD \\parallel x$ axis, $CD=5$, $OC=4$,\\\\\n$\\therefore$ point $D(-5, 4)$,\\\\\nThus, the coordinates of points B and D are respectively \\boxed{B(3, 0)} and \\boxed{D(-5, 4)}.",
    "image_path": "dataset/mm_math/images/MM_Math/53182915_204.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated, quadrilateral $ABCD$ is inscribed in circle $\\odot O$, and diagonal $BD$ is the diameter. A tangent line $AE$ from point $A$ to circle $\\odot O$ intersects the extension of $CD$ at point $E$. It is given that $DA$ bisects $\\angle BDE$ and $AE \\perp DE$; if the radius of $\\odot O$ is $5$ and $CD=6$, what is the length of $AD$?\n",
    "completion": "\\textbf{Solution:} Construct $OF \\perp DC$ through point $O$, with the perpendicular foot at $F$,\\\\\n$\\therefore \\angle OFD=90^\\circ$,\\\\\n$\\because \\angle OAE=\\angle E=90^\\circ$,\\\\\n$\\therefore$ Quadrilateral $OAEF$ is a rectangle,\\\\\n$\\therefore OA=EF=5$, $AE=OF$,\\\\\n$\\because OF\\perp CD$,\\\\\n$\\therefore DF=\\frac{1}{2}CD=3$,\\\\\n$\\therefore DE=EF-DF=5-3=2$,\\\\\n$\\therefore OF=\\sqrt{OD^{2}-DF^{2}}=\\sqrt{5^{2}-3^{2}}=4$,\\\\\n$\\because AE=OF=4$,\\\\\n$\\therefore AD=\\sqrt{AE^{2}+DE^{2}}=\\sqrt{4^{2}+2^{2}}=\\boxed{2\\sqrt{5}}$,\\\\\n$\\therefore$ The length of $AD$ is $2\\sqrt{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52457114_93.png"
  },
  {
    "solution": "\\boxed{\\angle EAF=45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the square $ABCD$, the vertices $E$ and $F$ of $\\triangle AEF$ lie on $BC$ and $CD$ respectively. $AG \\perp EF$ with the foot being $G$ and $AG = AB$. Find the measure of $\\angle EAF$.\n",
    "completion": "\\textbf{Solution:} In right-angled triangles $Rt\\triangle ABE$ and $Rt\\triangle AGE$, we have $\\left\\{\\begin{array}{l}AB=AG\\\\ AE=AE\\end{array}\\right.$, \\\\\n$\\therefore \\triangle ABE\\cong \\triangle AGE\\left(HL\\right)$, \\\\\n$\\therefore \\angle BAE=\\angle GAE$, similarly $\\angle GAF=\\angle DAF$, \\\\\n$\\therefore \\boxed{\\angle EAF=45^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50496360_112.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $DE \\parallel BC$, and $EC:BD = 2:3$, $AD=6$. What is the length of $AE$?\n",
    "completion": "\\textbf{Solution:} Since DE$\\parallel$BC, and EC$\\colon$ BD=2$\\colon$3, AD=6, \\\\\n$\\therefore \\frac{AE}{EC}=\\frac{AD}{BD}$,\\\\\n$\\therefore \\frac{AE}{2}=\\frac{6}{3}$,\\\\\n$\\therefore$ AE=\\boxed{4}",
    "image_path": "dataset/mm_math/images/MM_Math/52976991.png"
  },
  {
    "solution": "\\boxed{25^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the rectangle \\(ABCD\\) is rotated counterclockwise around point \\(A\\) to obtain rectangle \\(AEFG\\), such that point \\(E\\) falls on the diagonal \\(BD\\). Connect \\(DG\\) and \\(DF\\). If \\(\\angle BAE = 50^\\circ\\), what is the degree measure of \\(\\angle DGF\\)?\n",
    "completion": "\\textbf{Solution:} Given by rotation, $AB=AE$, $AD=AG$, $\\angle BAD=\\angle EAG=\\angle AGF=90^\\circ$,\\\\\n$\\therefore \\angle BAE=\\angle DAG=50^\\circ$,\\\\\n$\\therefore \\angle AGD=\\angle ADG= \\frac{180^\\circ-50^\\circ}{2}=65^\\circ$,\\\\\n$\\therefore \\angle DGF=90^\\circ-65^\\circ=\\boxed{25^\\circ};$",
    "image_path": "dataset/mm_math/images/MM_Math/51737207_157.png"
  },
  {
    "solution": "\\boxed{64^\\circ25'}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\angle AOC = 90^\\circ$, OC bisects $\\angle DOB$, and $\\angle DOC = 25^\\circ 35'$. Find the degree measure of $\\angle BOA$?\n",
    "completion": "Solution: Since $OC$ bisects $\\angle DOB$ and $\\angle DOC=25^\\circ35'$, \\\\\nit follows that $\\angle DOC=\\angle BOC=25^\\circ35'$. \\\\\nSince $\\angle AOC=90^\\circ$, \\\\\nit follows that $\\angle BOA=90^\\circ-25^\\circ35'=64^\\circ25'$. \\\\\nHence, the answer is $\\boxed{64^\\circ25'}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55301112.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ is inscribed in $\\odot O$, where $AB=AC$, and $AO$ and $CO$ are connected. If $BC=6$ and $\\sin\\angle BAC=\\frac{3}{5}$, what is the radius of $\\odot O$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw and extend BO to intersect circle O at D, then BD is the diameter of circle O,\\\\\n$\\therefore$ $\\angle BCD=90^\\circ$, $\\angle BAC=\\angle BDC$,\\\\\n$\\therefore$ $\\sin\\angle BAC=\\sin\\angle BDC=\\frac{BC}{BD}=\\frac{3}{5}$,\\\\\n$\\therefore$ $BD=\\frac{5}{3}BC=10$,\\\\\n$\\therefore$ the radius of circle O is $\\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52930757.png"
  },
  {
    "solution": "\\boxed{\\frac{6}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In $\\triangle ABC$, $\\angle C=90^\\circ$, $AC=3$, $BC=4$. Point $N$ is on side $BC$, and point $M$ is a moving point on side $AB$. Points $D$ and $E$ are the midpoints of $CN$ and $MN$, respectively. What is the minimum value of $DE$?\n",
    "completion": "\\textbf{Solution:} Let's join CM,\\\\\nsince points D and E are the midpoints of CN and MN respectively,\\\\\ntherefore DE = $\\frac{1}{2}$ CM,\\\\\nwhen CM $\\perp$ AB, the value of CM is minimal, and at this point, the value of DE is also minimal.\\\\\nAccording to the Pythagorean theorem: $AB = \\sqrt{AC^{2} + BC^{2}} = \\sqrt{3^{2} + 4^{2}} = 5$,\\\\\nsince $S_{\\triangle ABC} = \\frac{1}{2} \\times AB \\times CM = \\frac{1}{2} \\times AC \\times BC$,\\\\\ntherefore $CM = \\frac{12}{5}$,\\\\\ntherefore $DE = \\frac{1}{2}CM = \\boxed{\\frac{6}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50496379.png"
  },
  {
    "solution": "\\boxed{\\frac{13}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in right-angled triangle $ABC$ with $\\angle ACB=90^\\circ$, $CD\\perp AB$, $AC=5$, and $BC=12$, point $P$ is a moving point on line segment $CD$. When a circle $\\odot P$ with radius $4$ is tangent to one side of $\\triangle ABC$, what is the length of $CP$?\n",
    "completion": "\\textbf{Solution:}\n\nGiven that in right triangle $ABC$, $AC=5$, $BC=12$,\\\\\ntherefore $AB=\\sqrt{AC^2+BC^2}=\\sqrt{5^2+12^2}=13$,\\\\\nsince $CD\\perp AB$,\\\\\nthe area of $\\triangle ABC = \\frac{1}{2}AB\\cdot CD=\\frac{1}{2}AC\\cdot BC$,\\\\\nthus, $13CD=5\\times 12$,\\\\\ntherefore, $CD= \\frac{60}{13}$,\\\\\nconsidering three cases:\\\\\n(1) When circle $P$ is tangent to side $BC$, as shown in the figure:\\\\\nDraw $PE\\perp BC$ at point E,\\\\\nsince $PE \\perp BC$,\\\\\nthus $\\angle PEC=90^\\circ$,\\\\\nhence $\\angle PCE+\\angle CPE=90^\\circ$, and $\\angle PCE+\\angle B=90^\\circ$,\\\\\ntherefore, $\\angle B=\\angle CPE$,\\\\\nsince $\\angle CEP=\\angle ACB=90^\\circ$,\\\\\nthus $\\triangle BCA\\sim \\triangle PEC$,\\\\\nhence $\\frac{BA}{PC}=\\frac{BC}{PE}$,\\\\\ntherefore, $\\frac{13}{PC}=\\frac{12}{4}$,\\\\\nthus, $PC=\\frac{13}{3}$,\\\\\n(2) When circle $P$ is tangent to side $AB$, as shown in the figure:\\\\\nsince $PD\\perp AB$,\\\\\ntherefore, $CP=CD-PD=\\frac{60}{13}-4=\\frac{8}{13}$,\\\\\n(3) When circle $P$ is tangent to side $AC$, as shown in the figure:\\\\\nDraw $PF\\perp AC$ at point F,\\\\\nsince $PF\\perp AC$,\\\\\nthus $\\angle PFC=90^\\circ$,\\\\\nhence $\\angle CPF+\\angle PCF=90^\\circ$, and $\\angle A+\\angle PCF=90^\\circ$\\\\\ntherefore, $\\angle A=\\angle CPF$,\\\\\nsince $\\angle CFP=\\angle ACB=90^\\circ$,\\\\\nthus $\\triangle BCA\\sim \\triangle CFP$,\\\\\nhence $\\frac{BA}{PC}=\\frac{CA}{FP}$,\\\\\ntherefore, $\\frac{13}{PC}=\\frac{5}{4}$,\\\\\nthus, $PC=\\frac{52}{5}$,\\\\\nsince $\\frac{52}{5}>\\frac{60}{13}$\\\\\ntherefore, $PC=\\frac{52}{5}$ (discarded)\\\\\nIn conclusion, when circle $P$ with radius 4 is tangent to one side of $\\triangle ABC$, the length of $CP$ could be: $\\boxed{\\frac{13}{3}}$ or $\\frac{8}{13}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51545296.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $l_{1}$ intersects the x-axis and y-axis at points A and B, respectively, with coordinates $(2, 0)$ and $(0, 3)$. The line $l_{2}: y=\\frac{1}{2}x+3$ passing through point B intersects the x-axis at point C. Point $D(n, 6)$ is a point on the line $l_{1}$, and line segment $CD$ is drawn.\n",
    "completion": "\\textbf{Solution:}\n\n(I) Let the equation of line $l_1$ be $y=kx+b$. Substituting A(2,0) and B(0,3) into it, we get\n\\[\n\\left\\{\n\\begin{array}{l}\n2k+b=0\\\\\nb=3\n\\end{array}\n\\right.\n\\]\nSolving this, we find\n\\[\n\\left\\{\n\\begin{array}{l}\nk=-\\frac{3}{2}\\\\\nb=3\n\\end{array}\n\\right.\n\\]\nTherefore, the equation of line $l_1$ is $y=-\\frac{3}{2}x+3$;\n\n(II) When $y=0$, $\\frac{1}{2}x+3=0$, solving this gives $x=-6$, therefore the coordinates of point C are (-6,0). Substituting D(n,6) into $y=-\\frac{3}{2}x+3$ gives $-\\frac{3}{2}n+3=6$, solving this gives $n=-2$, therefore the coordinates of point D are (-2,6);\n\n(III) $S_{\\triangle BCD}=S_{\\triangle DAC}-S_{\\triangle BAC}=\\frac{1}{2}\\times(2+6)\\times6-\\frac{1}{2}\\times(2+6)\\times3=\\boxed{12}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50469243.png"
  },
  {
    "solution": "\\boxed{225}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$. If $AB=15$, what is the sum of the areas of square $ADEC$ and square $BCFG$?\n",
    "completion": "\\textbf{Solution:} Let the area of square $ADEC$ be: $AC^{2}$, and the area of square $BCFG$ be: $BC^{2}$; \\\\\n$\\because$ $\\angle ACB=90^\\circ$, \\\\\n$\\therefore$ $\\triangle ABC$ is a right triangle, \\\\\nThat is, in $\\text{Rt}\\triangle ABC$, $AB^{2}=AC^{2}+BC^{2}$, with $AB=15$, \\\\\nThen $15^{2}=AC^{2}+BC^{2}=225$. \\\\\nTherefore, the combined area of square $ADEC$ and square $BCFG$ is $\\boxed{225}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53729004.png"
  },
  {
    "solution": "\\boxed{y=\\frac{3}{2}x-6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the analytical expression of line $l_1$ is $y=-3x+3$, and line $l_1$ intersects the x-axis at point D. Line $l_2$ passes through points A$(4,0)$ and $B\\left(3, -\\frac{3}{2}\\right)$. Lines $l_1$ and $l_2$ intersect at point C. What is the analytical expression of line $l_2$?\n",
    "completion": "\\textbf{Solution:}\nLet the analytical expression of line $l_2$ be: $y=kx+b (k\\ne 0)$. Since line $l_2$ passes through points $A(4, 0)$ and $B\\left(3, -\\frac{3}{2}\\right)$, substituting into the expression, we get:\n\\[\n\\begin{cases}\n4k+b=0\\\\ \n3k+b=-\\frac{3}{2}\n\\end{cases}\n\\]\nSolving this system of equations yields:\n\\[\n\\begin{cases}\nk=\\frac{3}{2}\\\\ \nb=-6\n\\end{cases}\n\\]\nTherefore, the analytical expression of line $l_2$ is: $\\boxed{y=\\frac{3}{2}x-6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52395164_94.png"
  },
  {
    "solution": "\\boxed{\\frac{5}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within rectangle $ABCD$, points $E$ and $F$ are the midpoints of sides $AB$ and $BC$ respectively. Lines $EC$ and $FD$ are connected, and points $H$ and $G$ are the midpoints of $EC$ and $FD$ respectively. Line $GH$ is connected. If $AB = 6$ and $BC = 8$, then what is the length of $GH$?\n",
    "completion": "\\textbf{Solution:} Join CH and extend it to intersect AD at P, connect PE,\n\n$\\because$ Quadrilateral ABCD is a rectangle,\n\n$\\therefore \\angle A=90^\\circ$, AD$\\parallel$BC,\n\n$\\because$ E and F are the midpoints of AB and BC, respectively, AB$=6$, BC$=8$,\n\n$\\therefore$ AE$=\\frac{1}{2}$AB$=3$, CF$=\\frac{1}{2}$BC$=4$,\n\n$\\because$ AD$\\parallel$BC,\n\n$\\therefore \\angle DPH=\\angle FCH$,\n\nIn $\\triangle PDH$ and $\\triangle CFH$, $\\left\\{\\begin{array}{l}\\angle DPH=\\angle FCH \\\\ \\angle DHP=\\angle FHC \\\\ DH=FH \\end{array}\\right.$,\n\n$\\therefore \\triangle PDH \\cong \\triangle CFH$ (AAS),\n\n$\\therefore$ PD$=$CF$=4$, CH$=$PH,\n\n$\\therefore$ AP$=$AD$-$PD$=4$,\n\n$\\therefore$ PE$=\\sqrt{AP^{2}+AE^{2}}=\\sqrt{4^{2}+3^{2}}=5$,\n\n$\\because$ Point G is the midpoint of EC,\n\n$\\therefore$ GH$=\\frac{1}{2}$EP$=\\boxed{\\frac{5}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52691056.png"
  },
  {
    "solution": "\\boxed{115^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the straight lines $AB$ and $CD$ intersect at point $O$, and $OE$ bisects $\\angle AOD$. If $\\angle AOC = 50^\\circ$, find the measure of $\\angle BOE$.\n",
    "completion": "\\textbf{Solution:} Given $\\because \\angle AOC=50^\\circ$, \\\\\n$\\therefore \\angle BOD=\\angle AOC=50^\\circ$ \\\\\n$\\therefore \\angle AOD=180^\\circ-\\angle AOC=180^\\circ-50^\\circ=130^\\circ$\\\\\nSince $OE$ bisects $\\angle AOD$\\\\\n$\\therefore \\angle DOE=\\frac{1}{2}\\angle AOD=\\frac{1}{2}\\times 130^\\circ=65^\\circ$\\\\\nTherefore, $\\angle BOE=\\angle BOD+\\angle DOE=50^\\circ+65^\\circ=\\boxed{115^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/53364698_173.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{11}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a circle with center at the origin $O$ passes through point $A(4, 0)$. Inside the circle, there is a fixed point $B(-1, 2)$. A line through point $B$ intersects the circle at points $M$ and $N$. What is the minimum value of $MN$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, connect $OB$ and $OM$. It is known that when $MN\\perp OB$, $MN$ is minimized,\\\\\n$\\because B(-1,2)$,\\\\\n$\\therefore OB^2=1^2+2^2=5$,\\\\\n$\\because OM=OA=4$,\\\\\n$\\therefore BM=\\sqrt{OM^2-OB^2}=\\sqrt{16-5}=\\sqrt{11}$,\\\\\n$\\because MN\\perp OB$,\\\\\n$\\therefore MN=2BM=2\\sqrt{11}$,\\\\\n$\\therefore$ The minimum value of $MN$ is $\\boxed{2\\sqrt{11}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52991639.png"
  },
  {
    "solution": "\\boxed{3\\sqrt{7}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ is an equilateral triangle, CE is an external angle bisector, and point D is on AC. Extend BD to intersect with CE at point E. $\\triangle ABD\\sim \\triangle CED$, if $AB=6$ and $AD=2CD$, find the length of $BE$.\n",
    "completion": "\\textbf{Solution:} Draw BM$\\perp$AC at point M, where AC=AB=6.\\\\\n$\\therefore$AM=CM=3, BM=AB$\\cdot\\sin 60^\\circ=3\\sqrt{3}$.\\\\\n$\\because$AD=2CD,\\\\\n$\\therefore$CD=2, AD=4, MD=1.\\\\\nIn $\\triangle BDM$, BD=$\\sqrt{BM^2+MD^2}=2\\sqrt{7}$.\\\\\nSince $\\triangle ABD\\sim\\triangle CED$,\\\\\n$\\frac{BD}{ED}=\\frac{AD}{CD}$, $\\frac{2\\sqrt{7}}{ED}=2$,\\\\\n$\\therefore$ED=$\\sqrt{7}$\\\\\n$\\therefore$BE=BD+ED=$3\\sqrt{7}$.\\\\\nHence, the length of $BE$ is $\\boxed{3\\sqrt{7}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52536896_115.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{3}}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given $\\angle AOB=60^\\circ$ and a circle $\\odot P$ with a radius of $3\\text{cm}$ moves parallel from the right to the left along side $OA$, the tangent point on side $OA$ is marked as point $C$. $\\odot P$ moves to intersect side $OB$ at points $E$ and $F$, given that $EF=4\\sqrt{2}\\text{cm}$, what is the length of $OC$ in $\\text{cm}$?\n",
    "completion": "\\textbf{Solution:} There are two cases to consider:\n\n(1) As shown in Figure 1, when P is inside $\\angle$AOB, draw PE and PC. Through point P, draw PM$\\perp$EF at point M and extend CP to intersect OB at point N,\n\n$\\because$EF $=$ 4$\\sqrt{2}$ cm,\n\n$\\therefore$EM $=$ 2$\\sqrt{2}$ cm,\n\nIn the right triangle $\\triangle$EPM, PM $=$ $\\sqrt{3^{2}-(2\\sqrt{2})^{2}}=1$ cm,\n\n$\\because$$\\angle$AOB $=$ $60^\\circ$,\n\n$\\therefore$$\\angle$PNM $=$ $30^\\circ$,\n\n$\\therefore$PN $=$ 2PM $=$ 2cm,\n\n$\\therefore$NC $=$ PN + PC $=$ 5cm,\n\nIn the right triangle $\\triangle$OCN, OC $=$ NC$\\times\\tan30^\\circ$ $=$ 5$\\times\\frac{\\sqrt{3}}{3}$ $=$ $\\boxed{\\frac{5\\sqrt{3}}{3}}$ cm.\n\n(2) As shown in Figure 2, when P is outside $\\angle$AOB, draw PF and PC, where PC intersects EF at point N. Through point P, draw PM$\\perp$EF at point M,\n\nFrom (1) we know, PN $=$ 2cm,\n\n$\\therefore$NC $=$ PC - PN $=$ 1cm,\n\nIn the right triangle $\\triangle$OCN, OC $=$ NC$\\times\\tan30^\\circ$ $=$ 1$\\times\\frac{\\sqrt{3}}{3}$ $=$ $\\boxed{\\frac{\\sqrt{3}}{3}}$ cm.\n\nIn conclusion, the length of OC is either $\\frac{\\sqrt{3}}{3}$ cm or $\\frac{5\\sqrt{3}}{3}$ cm.",
    "image_path": "dataset/mm_math/images/MM_Math/51106528.png"
  },
  {
    "solution": "\\boxed{9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line segment $AD=21\\,cm$, point $B$ is on line segment $AD$, $C$ is the midpoint of $BD$, and $AB=\\frac{1}{3}CD$, what is the length of $BC$?\n",
    "completion": "\\textbf{Solution:} Let $AB=x$, then $CD=3x$,\\\\\nsince C is the midpoint of BD,\\\\\nthus $BC=CD=3x$,\\\\\ntherefore $3x+3x+x=21$,\\\\\nsolving this equation yields $x=3$,\\\\\n$BC=3\\times 3=\\boxed{9}$ cm.",
    "image_path": "dataset/mm_math/images/MM_Math/51495019.png"
  },
  {
    "solution": "\\boxed{\\frac{27}{8}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=ax^2+bx+3$ ($a\\neq 0$) intersects the x-axis at points $A(1,0)$ and $B(-3,0)$, and intersects the y-axis at point $C$. The line segment $BC$ is drawn and intersects the axis of symmetry of the parabola at point $E$, with the vertex being point $D$. If $M$ is a moving point on the parabola located above the line segment $BC$, what is the maximum area of $\\triangle BCM$?\n",
    "completion": "\\textbf{Solution:} Draw $MF\\perp x$ axis from point M, intersecting BC at point G and the x-axis at point F, \\\\\nwhen $x=0$, $y=3$, thus the coordinates of C are $(0,3)$, \\\\\nit is easy to derive the equation of BC as $y=x+3$, \\\\\nlet $G(x, x+3)$, $M(x, -x^{2}-2x+3)$, \\\\\nthen $MG= -x^{2}-2x+3-(x+3)=-x^{2}-3x$, \\\\\nTherefore, ${S}_{\\triangle BCM}={S}_{\\triangle BMG}+{S}_{\\triangle CMG}=\\frac{1}{2}MG\\cdot BO=\\frac{1}{2}(-x^{2}-3x)\\times 3=-\\frac{3}{2}x^{2}-\\frac{9}{2}x$, \\\\\nsince $a=-\\frac{3}{2}<0$, \\\\\nTherefore, when $x=-\\frac{b}{2a}=-\\frac{3}{2}$, ${S}_{\\triangle BCM}$ reaches its maximum value, which is $\\boxed{\\frac{27}{8}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52602919_159.png"
  },
  {
    "solution": "\\boxed{6\\sqrt{2}-2\\sqrt{6}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, line $AB$ intersects the x-axis at point A and the y-axis at point B, where $ \\angle OAB=30^\\circ$, and the coordinates of point B are $(0, 2\\sqrt{3})$. The line $y=kx+b$ passes through point A and intersects the y-axis at point C, with $OC=OA$. Let P be a point on line $AC$, and Q be a point on line $l$. When $\\triangle APQ$ is an equilateral triangle, write down the side length of $\\triangle APQ$.\n",
    "completion": "\\textbf{Solution:}\n\n(1) As shown in Figure 2, when P is on the extension line of segment CA, connect $A'$P, and $A'$ with BQ,\\\\\n$\\because \\triangle APQ$ is an equilateral triangle,\\\\\n$\\therefore AP = AQ = PQ$,\\\\\n$\\angle BAA' = \\angle PAQ = 60^\\circ$,\\\\\n$\\therefore \\angle BAQ = \\angle A'AP$,\\\\\nIn $\\triangle ABQ$ and $\\triangle AA'P$,\\\\\n$\\left\\{\\begin{array}{l}\nAB = AA' \\\\\n\\angle BAQ = \\angle A'AP \\\\\nAQ = AP\n\\end{array}\\right.$,\\\\\n$\\therefore \\triangle ABQ \\cong \\triangle AA'P$ (SAS),\\\\\n$\\therefore BQ = A'P$,\\\\\nAlso, $\\because Q$ is a point on the perpendicular bisector of AB,\\\\\n$\\therefore BQ = AQ$,\\\\\n$\\therefore AQ = A'P$,\\\\\nAlso, $\\because AP = AQ$,\\\\\n$\\therefore AP = A'P$,\\\\\n$\\therefore P$ is on the perpendicular bisector of $AA'$,\\\\\n$\\because AA' \\perp OA$, and $AA' = 4\\sqrt{3}$,\\\\\n$\\therefore$ The ordinate of P is $2\\sqrt{3}$,\\\\\nLet $y = 2\\sqrt{3}$, then $x-6 = 2\\sqrt{3}$,\\\\\n$\\therefore x = 6 + 2\\sqrt{3}$,\\\\\n$\\therefore P(6+2\\sqrt{3}, 2\\sqrt{3})$,\\\\\n$\\therefore AP = \\sqrt{(6+2\\sqrt{3}-6)^2+(2\\sqrt{3})^2} = 2\\sqrt{6}$,\\\\\n(2) As shown in Figure 3, when P is on the ray AC, connect BP,\\\\\nSimilarly, we can get $\\triangle ABP \\cong AA'Q$,\\\\\n$\\therefore PB = QA'$, $\\angle PBA = \\angle QA'A$,\\\\\n$\\because \\angle BAA' = 60^\\circ$, and $A'Q$ vertically bisects AB,\\\\\n$\\therefore \\angle QA'A = 30^\\circ$,\\\\\n$\\therefore \\angle PBA = 30^\\circ$,\\\\\nAlso, $\\angle ABO = 90^\\circ - \\angle BAO = 60^\\circ$,\\\\\n$\\therefore \\angle PBO = \\angle ABO - \\angle PBA = 30^\\circ$,\\\\\nDraw $PG \\perp y$-axis at point G from P, let P$(m, m-6)$,\\\\\nIn $\\triangle PBG$, $\\angle PBO = 30^\\circ$,\\\\\n$\\therefore PB = 2PG = 2m$,\\\\\n$\\therefore BG = \\sqrt{3}m$,\\\\\n$\\therefore \\sqrt{3}m = 2\\sqrt{3}-(m-6)$,\\\\\n$\\therefore m = 2\\sqrt{3}$,\\\\\n$\\therefore P(2\\sqrt{3}, 2\\sqrt{3}-6)$,\\\\\n$\\therefore AP = \\sqrt{(2\\sqrt{3}-6)^2+(2\\sqrt{3}-6)^2} = 6\\sqrt{2}-2\\sqrt{6}$.\\\\\nIn summary: The side length of $\\triangle APQ$ is either \\boxed{2\\sqrt{6}} or \\boxed{6\\sqrt{2}-2\\sqrt{6}}.",
    "image_path": "dataset/mm_math/images/MM_Math/50482334_74.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $l_{1}\\parallel l_{2}\\parallel l_{3}$, lines $a$ and $b$ intersect $l_{1}$, $l_{2}$, and $l_{3}$ at points $A$, $B$, $C$ and $D$, $E$, $F$ respectively. Given that $\\frac{AB}{BC} = \\frac{2}{3}$ and $DE = 4$, what is the length of $EF$?\n",
    "completion": "\\textbf{Solution.} Given: $\\because$ ${l}_{1}\\parallel {l}_{2}\\parallel {l}_{3}$\\\\\n$\\therefore$ $\\frac{AB}{BC}=\\frac{DE}{EF}$, that is $\\frac{2}{3}=\\frac{4}{EF}$, solving gives $EF=\\boxed{6}$",
    "image_path": "dataset/mm_math/images/MM_Math/52866359.png"
  },
  {
    "solution": "\\boxed{y_1<y_2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given that point $M$ is the vertex of the quadratic function $y=-(x-b)^2+4b+1$, and the line $y=mx+5$ intersects the positive half of the $x$-axis and the $y$-axis at points $A$ and $B$ respectively. As shown in Figure 2, the coordinates of point $A$ are $(5,0)$, and point $M$ is inside $\\triangle AOB$. If points $C\\left(\\frac{1}{4},y_1\\right)$ and $D\\left(\\frac{3}{4},y_2\\right)$ are both on the graph of the quadratic function, compare the sizes of $y_1$ and $y_2$.\n",
    "completion": "\\textbf{Solution:}\n\nSubstituting $A(5,0)$ into $y=mx+5$ yields, \\\\\n$0=5m+5$, \\\\\nsolving for $m$ gives $m=-1$, \\\\\n$\\therefore y=-x+5$, \\\\\n$\\because M(b,4b+1)$ is inside $\\triangle AOB$, \\\\\n$\\therefore \\left\\{\\begin{array}{l}0<b<5 \\\\ 0<4b+1<-b+5\\end{array}\\right.$, \\\\\nsolving yields $0<b<\\frac{4}{5}$, \\\\\nWhen points C and D are symmetrical about the axis of symmetry, $b=\\frac{\\frac{1}{4}+\\frac{3}{4}}{2}=\\frac{1}{2}$, \\\\\n$\\therefore$ for $0<b<\\frac{1}{2}$, \\boxed{y_1>y_2}, \\\\\nfor $b=\\frac{1}{2}$, \\boxed{y_1=y_2}, \\\\\nand for $\\frac{1}{2}<b<\\frac{4}{5}$, \\boxed{y_1<y_2}.",
    "image_path": "dataset/mm_math/images/MM_Math/53009807_169.png"
  },
  {
    "solution": "\\boxed{40^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a right-angled triangular board $ABC$ (where $\\angle ACB=90^\\circ$, $\\angle A=30^\\circ$) is placed inside the acute angle $\\angle POQ$, making the right-angled side $BC$ lie on the side $OQ$, and let $\\angle POQ=\\alpha$. Now the triangular board $ABC$ is rotated counterclockwise around point $B$ at a speed of $m^\\circ$ per second for $t$ seconds (the right-angled side $BC$ rotates to the position shown in the figure, and point $A$ always remains within $\\angle POQ$). Through point $A$, draw $MN \\parallel OQ$ intersecting the ray $OP$ at point $M$, and $AD$ bisects $\\angle MAB$ intersecting the ray $OQ$ at point $D$, where the value of $m$ is such that the algebraic expression $\\left|10−m\\right|+3$ is minimized. When $t=5$ seconds, determine the degree measure of $\\angle NAC$.\n",
    "completion": "\\textbf{Solution:} Given $t=5$, then $\\angle CBQ=50^\\circ$,\\\\\nsince $\\angle ACB=90^\\circ, \\angle BAC=30^\\circ$,\\\\\nthus $\\angle ABC=90^\\circ-\\angle BAC=60^\\circ$,\\\\\nwhich means $\\angle ABQ=\\angle ABC+\\angle CBQ=110^\\circ$,\\\\\ngiven that $MN\\parallel OQ$,\\\\\nit follows that $\\angle NAB=180^\\circ-\\angle ABQ=70^\\circ$,\\\\\nthus $\\angle NAC=\\angle NAB-\\angle BAC=40^\\circ$,\\\\\ntherefore the measure of $\\angle NAC$ is $\\boxed{40^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/54883358_110.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, AC is the diameter of $\\odot O$, PA is the tangent to $\\odot O$, with A being the point of tangency. Point B lies on $\\odot O$, and PA = PB. Chord AB intersects PC at point M. PB is the tangent to $\\odot O$, and segment BC is connected. If $\\angle APB = 4\\angle BPC$ and AP = 6, what is the length of BC?\n",
    "completion": "\\textbf{Solution:} Connect PO.\\\\\nSince PA and PB are tangents to the circle $O$,\\\\\nit follows that PA=PB=6, and $\\angle APO=\\angle BPO$, that is, $\\angle 1=\\angle 2+\\angle 3$,\\\\\nSince $\\angle APB=4\\angle BPC$,\\\\\nit follows that $\\angle APM=3\\angle BPM$,\\\\\ntherefore $\\angle 1+\\angle 2=3\\angle 3$,\\\\\nthus $2\\angle 2+\\angle 3=3\\angle 3$, which means $\\angle 2=\\angle 3=\\frac{1}{2}\\angle 1$.\\\\\nLet $\\angle 2=\\angle 3=\\alpha$, then $\\angle 1=2\\alpha$, $\\angle APB=4\\alpha$,\\\\\nthus $\\angle PBA=\\frac{1}{2}(180^\\circ-4\\alpha)=90^\\circ-2\\alpha$,\\\\\nConsidering that AC is a diameter,\\\\\nthus $\\angle ABC=90^\\circ$,\\\\\ntherefore $\\angle PBC=180^\\circ-2\\alpha$,\\\\\nthus $\\angle 4=180^\\circ-\\angle 3-\\angle PBC=180^\\circ-\\alpha-(180^\\circ-2\\alpha)=\\alpha$,\\\\\ntherefore $\\angle 3=\\angle 4$,\\\\\nthus BC=PB=\\boxed{6}.",
    "image_path": "dataset/mm_math/images/MM_Math/51299739_169.png"
  },
  {
    "solution": "\\boxed{(-2, 1)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the line $y_1=x+m$ intersects the x-axis and y-axis at points A and B, respectively, and intersects the hyperbola $y_2=\\frac{k}{x}$ ($x<0$) at points C and D, where the coordinates of point C are $(-1, 2)$. Find the coordinates of point D.\n",
    "completion": "\\textbf{Solution:} Solving the systems of equations simultaneously, we get\n\\[\n\\begin{cases}\ny=x+3\\\\\ny=\\frac{-2}{x}\n\\end{cases}\n\\]\nwhich yields\n\\[\n\\begin{cases}\nx=-1\\\\\ny=2\n\\end{cases}\n\\]\nor\n\\[\n\\begin{cases}\nx=-2\\\\\ny=1\n\\end{cases}\n\\]\nSince the coordinates of point C are $(-1, 2)$,\\\\\ntherefore, the coordinates of point D are $\\boxed{(-2, 1)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53101933_153.png"
  },
  {
    "solution": "\\boxed{\\sqrt{5}\\colon 1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is a \"Gougu Circle and Square Diagram\" (also known as \"Zhao Shuang's Chord Diagram\") created by the ancient Chinese mathematician Zhao Shuang. It consists of four congruent right-angled triangles and a smaller square EFGH seamlessly pieced together to form a larger square ABCD. Point M is on FG, with $AB\\parallel EM$ and $AB=2EM$. Find the ratio of the perimeter of square ABCD to that of square EFGH.\n",
    "completion": "\\textbf{Solution:} Given that $AB\\parallel EM$, it follows that $\\angle ABE = \\angle FEM$ and $\\angle AEB = \\angle EFM = 90^\\circ$. Therefore, $\\triangle ABE \\sim \\triangle EMF$, which implies $\\frac{EM}{AB}=\\frac{EF}{BE}=\\frac{1}{2}$, so $BF=EF$. Thus, $AE=BF=EF$. Let EF=1, then $AB=\\sqrt{AE^2+BE^2}=\\sqrt{5}$. Therefore, the perimeter ratio of square ABCD to square EFGH is $AB\\colon EF = \\boxed{\\sqrt{5}\\colon 1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51385818_98.png"
  },
  {
    "solution": "\\boxed{1.4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, it is an ancient trebuchet capable of attacking from a distance. Figure 2 represents a schematic of the trebuchet during a moment in the stone-throwing process, where $GP$ is the lever, the sling hangs at point $G$, the counterweight is hung at point $P$, point $A$ is the fulcrum, and point $D$ is a point on the horizontal baseboard $BC$, with $AD=AC=3$ meters, and $CD=3.6$ meters. When the trebuchet is prepared, point $G$ exactly coincides with point $B$, at this time $AG$ and $AC$ are perpendicular. Then, how many meters is $BD$?\n",
    "completion": "\\textbf{Solution:} We have $BD=\\boxed{1.4}$ meters.",
    "image_path": "dataset/mm_math/images/MM_Math/52808005_36.png"
  },
  {
    "solution": "\\boxed{rectangle}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rhombus $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$. Draw a perpendicular line from point $A$ to $BC$ with the foot of the perpendicular being point $E$. Extend $BC$ to point $F$ such that $CF=BE$. Connect $DF$. Prove that the quadrilateral $AEFD$ is a rectangle.\n",
    "completion": "\\textbf{Solution:} \\\\\nSince $CF=BE$, \\\\\nit follows that $EF=BC=AD$, \\\\\nand since $EF\\parallel AD$, \\\\\nit implies that quadrilateral $AEFD$ is a parallelogram, \\\\\nsince $AE\\perp EF$, \\\\\nit follows that $\\angle AEF=90^\\circ$, \\\\\ntherefore, the quadrilateral AEFD is a \\boxed{rectangle}.",
    "image_path": "dataset/mm_math/images/MM_Math/52344804_171.png"
  },
  {
    "solution": "\\boxed{3\\sqrt{10}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, rectangle $ABCD$ is folded along diagonal $AC$, making point $B$ coincide with point $E$, and $CE$ intersects $AD$ at point $F$. If $AF=5$ and $DF=4$, then what is the length of $AC$?\n",
    "completion": "\\textbf{Solution:} From the property of folding, we have $\\angle ACF = \\angle ACB$. Since $AD \\parallel BC$, it follows that $\\angle ACB = \\angle FAC$, which implies $\\angle FAC = \\angle ACF$, hence $AF = CF = 5$. Given that $CF = 5$, $DF = 4$, and $\\angle D = 90^\\circ$, it follows $CD = \\sqrt{CF^2 - DF^2} = 3$. Since $AD = AD + DF = 5 + 4 = 9$, it follows that $AC = \\sqrt{AD^2 + CD^2} = \\sqrt{9^2 + 3^2} = \\boxed{3\\sqrt{10}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50475274.png"
  },
  {
    "solution": "\\boxed{y=-x+1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, it's known that the line $l_1\\colon y=kx+b$ ($k\\ne 0$), which passes through the point $B(1, 0)$, intersects with the line $l_2\\colon y=2x+4$ at point $P(-1, a)$. Find the equation of the line $l_1$.\n",
    "completion": "\\textbf{Solution}: Since point P ($-1$, $2$) lies on the line $l_2: y=2x+4$, \\\\\nit follows that $2\\times(-1)+4=2$, which means $a=2$, so the coordinates of P are ($-1$, $2$).\\\\\nAccording to the problem, we have\n\\[\n\\left\\{\n\\begin{array}{l}\nk+b=0\\\\\n-k+b=2\n\\end{array}\n\\right.,\n\\]\nfrom which we obtain:\n\\[\n\\left\\{\n\\begin{array}{l}\nk=-1\\\\\nb=1\n\\end{array}\n\\right.,\n\\]\nTherefore, the equation for $l_1$ is: \\boxed{y=-x+1};",
    "image_path": "dataset/mm_math/images/MM_Math/52386847_116.png"
  },
  {
    "solution": "\\boxed{x<1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given the graph, the functions $y=bx$ and $y=ax+4$ intersect at point $A(1,3)$. What is the solution set of the inequality $bx<ax+4$?\n",
    "completion": "\\textbf{Solution:} \\\\\nFrom the given condition, the solution set of the inequality $bx<ax+4$ is $x<1$,\\\\\nTherefore, the answer is: $\\boxed{x<1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50431063.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the length of line segment $AB$ is $8$, point $D$ is on $AB$, $\\triangle ACD$ is an equilateral triangle with a side length of $3$. A ray $DP$ is drawn perpendicular to $CD$ through point $D$, and a rectangle $CDGH$ is constructed with a moving point $G$ (not coinciding with $D$) on $DP$. Let the intersection point of the diagonals of rectangle $CDGH$ be $O$. Connect $OB$. What is the minimum value of the length of segment $BO$?\n",
    "completion": "\\textbf{Solution:} Draw line AO. Since quadrilateral CDGH is a rectangle, we have CG=DH, OC= $\\frac{1}{2}$ CG, and OD= $\\frac{1}{2}$ DH, which means OC=OD. Since $\\triangle$ACD is an equilateral triangle, we have AC=AD and $\\angle$CAD=$60^\\circ$. In triangles $\\triangle$ACO and $\\triangle$ADO,\n\\[\\left\\{\\begin{array}{l}\nAC=AD\\\\\nAO=AO\\\\\nCO=DO\n\\end{array}\\right.\\]\n, therefore, $\\triangle$ACO$\\cong$$\\triangle$ADO (SSS), thus $\\angle$OAB=$\\angle$CAO=$30^\\circ$. This means point O must move along the bisector of $\\angle$CAB. Therefore, when OB$\\perp$AO, the length of OB is minimized. Since $\\angle$OAB=$30^\\circ$ and $\\angle$AOB=$90^\\circ$, we get OB= $\\frac{1}{2}$ AB= $\\frac{1}{2}$ ×8=4, thus the minimum value of OB is $\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50496295.png"
  },
  {
    "solution": "\\boxed{80^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\angle A$ is the inscribed angle of $\\odot O$ and $\\angle A = 40^\\circ$, then what is the measure of $\\angle BOC$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle A$ is the peripheral angle of $\\odot O$, $\\angle A=40^\\circ$,\\\\\nhence, $\\angle BOC=2\\angle A=\\boxed{80^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52623225.png"
  },
  {
    "solution": "\\boxed{110^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in circle $O$, $\\angle BAC=25^\\circ$, $\\angle CED=30^\\circ$, then what is the degree measure of $\\angle BOD$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect OC,\\\\\nGiven that $\\angle BAC=25^\\circ$ and $\\angle CED=30^\\circ$,\\\\\nby the Inscribed Angle Theorem, we have $\\angle BOC=50^\\circ$ and $\\angle DOC=60^\\circ$,\\\\\ntherefore, $\\angle BOD=\\angle BOC+\\angle DOC=50^\\circ+60^\\circ=\\boxed{110^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52623522.png"
  },
  {
    "solution": "\\boxed{70^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $\\triangle OCA \\cong \\triangle OBD$ and $\\angle A=30^\\circ$, $\\angle AOC=80^\\circ$. What is the measure of $\\angle B$?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle OCA \\cong \\triangle OBD$, $\\angle A=30^\\circ$, $\\angle AOC=80^\\circ$,\n$\\therefore \\angle B=\\angle C=180^\\circ-\\angle A-\\angle AOC=180^\\circ-30^\\circ-80^\\circ=\\boxed{70^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52914187.png"
  },
  {
    "solution": "\\boxed{y=-x^2+2x+3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=ax^2+bx+c$ passes through the points $A(-1,0)$, $B(3,0)$, and $C(0,3)$. The vertex is denoted by $D$, and the axis of symmetry is the line $DE$. Determine the equation of the parabola.\n",
    "completion": "\\textbf{Solution:} Substitute $A(-1,0)$, $B(3,0)$, $C(0,3)$ into $y=ax^2+bx+c$,\\\\\n$\\therefore \\begin{cases}\na-b+c=0\\\\\n9a+3b+c=0\\\\\nc=3\n\\end{cases}$,\\\\\nSolving this, we get $\\begin{cases}\na=-1\\\\\nb=2\\\\\nc=3\n\\end{cases}$,\\\\\n$\\therefore y=-x^2+2x+3$; hence, the equation of the parabola is $\\boxed{y=-x^2+2x+3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52699702_86.png"
  },
  {
    "solution": "\\boxed{22.5^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within the square $ABCD$, $P$ is a point on the diagonal $AC$, and point $E$ lies on the extension of $BC$, with $PE=PB$. When $PC=CE$, what is the degree measure of $\\angle CDP$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a square,\\\\\nwe have AB=BC=CD=AD, $\\angle$BCP=$\\angle$DCP=$45^\\circ$, $\\angle$BCD=$\\angle$DCE=$90^\\circ$,\\\\\nthus, $\\angle$PCE=$45^\\circ+90^\\circ=135^\\circ$,\\\\\nIn $\\triangle$BCP and $\\triangle$DCP,\\\\\n$\\left\\{\\begin{array}{l}\nBC=DC\\\\\n\\angle BCP=\\angle DCP\\\\\nCP=CP\n\\end{array}\\right.$,\\\\\nthus, $\\triangle$BCP $\\cong$ $\\triangle$DCP (SAS),\\\\\ntherefore, BP=DP, $\\angle$CBP=$\\angle$CDP,\\\\\nsince PE=PB, PC=CE,\\\\\nthus, PD=PE, $\\angle$CBP=$\\angle$PEB=$\\angle$CPE= $\\frac{1}{2}(180^\\circ - 135^\\circ)=22.5^\\circ$,\\\\\ntherefore, $\\angle$CDP=$\\boxed{22.5^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53599241_138.png"
  },
  {
    "solution": "\\boxed{\\frac{65}{16}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, $AB=5$, $BC=3$. Point P starts from point A and moves at a speed of 2 units per second along the path $AC-CB-BA$ until it stops at point A. When point P does not coincide with the vertices of $\\triangle ABC$, a perpendicular line is drawn from point P to the edge it is on, intersecting another side of $\\triangle ABC$ at point Q. Let the movement time of point P be $t$ seconds. When point Q is on the right-angled sides of $\\triangle ABC$ and $PQ=\\frac{3}{2}$, find the value of $t$.\n",
    "completion": "\\textbf{Solution}: \n\nFrom the problem statement, we get:\n\n(1) When point $Q$ is on side $AC$, $AP=12-2t$,\n\nthen $PQ=\\frac{3}{2}=AP\\cdot \\tan A$,\n\nwhich means $\\frac{3}{2}=(12-2t)\\times \\frac{3}{4}$,\n\nsolving this yields: $t=\\boxed{5}$.\n\n(2) When point $Q$ is on side $BC$, $BP=2t-7$,\n\nthen $PQ=\\frac{3}{2}=BP\\cdot \\tan B$,\n\nwhich means $\\frac{3}{2}=(2t-7)\\times \\frac{4}{3}$,\n\nsolving this yields: $t=\\boxed{\\frac{65}{16}}$;\n\nIn summary, if $PQ=\\frac{3}{2}$, the value of $t$ is either \\boxed{5} or \\boxed{\\frac{65}{16}}.",
    "image_path": "dataset/mm_math/images/MM_Math/51318739_101.png"
  },
  {
    "solution": "\\boxed{120^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, the quadrilateral $ABCD$ is an inscribed quadrilateral of the circle $\\odot O$, and $\\angle BCD=120^\\circ$. What is the measure of $\\angle BOD$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle BCD = 120^\\circ$,\\\\\nit follows that $\\angle A = 180^\\circ - \\angle BCD = 60^\\circ$,\\\\\ntherefore $\\angle BOD = 2\\angle A = \\boxed{120^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52602899.png"
  },
  {
    "solution": "\\boxed{70^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within $\\angle AOB$, $OD$ is the bisector of $\\angle BOC$, and $OE$ is the bisector of $\\angle AOC$. If $\\angle AOB=140^\\circ$, then what is the measure of $\\angle EOD$ in degrees?\n",
    "completion": "\\textbf{Solution:} Since OD is the bisector of $\\angle$BOC, and OE is the bisector of $\\angle$AOC, \\\\\ntherefore, $\\angle$DOC=$\\frac{1}{2}$$\\angle$BOC, $\\angle$EOC=$\\frac{1}{2}$$\\angle$AOC, \\\\\nthus, $\\angle$DOE=$\\angle$DOC+$\\angle$EOC=$\\frac{1}{2}$$\\angle$AOB, \\\\\nsince $\\angle$AOB=$140^\\circ$, \\\\\ntherefore, $\\angle$EOD=\\boxed{70^\\circ}.",
    "image_path": "dataset/mm_math/images/MM_Math/51284661.png"
  },
  {
    "solution": "\\boxed{\\frac{9}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC=8$, $BC=4\\sqrt{7}$. A circle with center O on side BC is drawn to be tangent to AB and AC at D and E, respectively. What is the length of AD?\n",
    "completion": "\\textbf{Solution:} Join $AO$, $OD$, and $OB$.\\\\\nSince $AB$ and $AC$ are tangent to point $D$ and $E$, respectively,\\\\\nit follows that $OD\\perp AB$ and $OE\\perp AC$,\\\\\nand since $AO=AO$,\\\\\nit results in $\\triangle ADO \\cong \\triangle AEO$ (by Hypotenuse-Leg),\\\\\nthus $\\angle DAO = \\angle EAO$,\\\\\nand since $AB=AC$,\\\\\nit follows that $AO \\perp BC$ and $BO=OC=2\\sqrt{7}$,\\\\\nby applying the Pythagorean theorem we get\\\\\n$AO=\\sqrt{AB^{2}-BO^{2}}=6$,\\\\\nsince $\\angle DAO = \\angle BAO$ and $\\angle ADO = \\angle AOB$,\\\\\nit results in $\\triangle ADO \\sim \\triangle AOB$,\\\\\nthus $\\frac{AD}{AO} = \\frac{AO}{AB}$,\\\\\ntherefore, $AD=\\boxed{\\frac{9}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52977730.png"
  },
  {
    "solution": "\\boxed{x=1\\pm \\frac{\\sqrt{2}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the four vertices of rectangle DEFG each lie on the sides of equilateral triangle ABC. It is known that the side length of $\\triangle ABC$ is 4. Denote the area of rectangle DEFG as $S$, and the length of segment BE as $x$. When $S = \\sqrt{3}$, find the value of $x$.\n",
    "completion": "\\textbf{Solution:} When $S=\\sqrt{3}$, we get $-2\\sqrt{3}x^{2}+4\\sqrt{3}x=\\sqrt{3}$. \\\\\nSolving gives $x=1\\pm \\frac{\\sqrt{2}}{2}$. \\\\\nSince $0<x<2$, \\\\\nit follows that \\boxed{x=1\\pm \\frac{\\sqrt{2}}{2}}.",
    "image_path": "dataset/mm_math/images/MM_Math/51545233_181.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{1}{2}x+2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the graph of the linear function $y_1=kx+b$ intersects with the x-axis at point $A$ and with the y-axis at point $B$. It intersects with the graph of the inverse proportion function $y_2=\\frac{m}{x}$ at points $C$ and $D$, where point $C(-2,3)$, and the ordinate of point $D$ is $-1$. Find the analytic expressions of the inverse proportion function and the linear function?\n",
    "completion": "\\textbf{Solution:} Given the inverse proportional function $y_2=\\frac{m}{x}$ passes through point $C(-2, 3)$, substituting gives:\\\\\n$\\therefore m=-2\\times 3=-6$,\\\\\n$\\therefore$ the equation of the inverse proportional function is $y=-\\frac{6}{x}$;\\\\\nGiven the y-coordinate of point $D$ is $-1$,\\\\\n$\\therefore D(6, -1)$,\\\\\nSubstituting points $C$ and $D$ into $y_1=kx+b$,\\\\\nwe get $\\begin{cases}-2k+b=3\\\\ 6k+b=-1\\end{cases}$, solving this gives $\\begin{cases}k=-\\frac{1}{2}\\\\ b=2\\end{cases}$,\\\\\n$\\therefore$ the equation of the inverse proportional function is $\\boxed{y=-\\frac{6}{x}}$,\\\\\n$\\therefore$ the equation of the linear function is $\\boxed{y=-\\frac{1}{2}x+2}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52765598_142.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{3}}{4}-\\frac{\\pi }{12}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, there is a circle with a point $O$ on the side $BC$ of $\\triangle ABC$ as the center. This circle goes through points $A$ and $B$ and intersects with side $BC$ at point $E$. $DO \\perp BE$ at point $O$, and $AD$ is connected intersecting $BC$ at $F$, given $AC=FC$. If $\\angle ADB= 60^\\circ$ and $BD=1$, calculate the area of the shaded part (leave the result in radical form).\n",
    "completion": "\\textbf{Solution:} Since $BO=DO$, and $BD=1$, $\\angle DOB=90^\\circ$\\\\\nTherefore, in $\\triangle BOD$ right at $O$, by the Pythagorean theorem $BO^2+OD^2=BD^2$\\\\\nThus, we find $BO=DO=\\frac{\\sqrt{2}}{2}$\\\\\nThat is, the radius of the circle $O$ is $\\frac{\\sqrt{2}}{2}$\\\\\nSince $\\angle ADB=60^\\circ$\\\\\nTherefore $\\angle AOB=2\\angle ADB=120^\\circ$\\\\\nTherefore $\\angle AOC=180^\\circ-\\angle AOB=60^\\circ$\\\\\nSince $OA\\perp AC$\\\\\nTherefore $\\angle OAC=90^\\circ$\\\\\nTherefore, in $\\triangle OAC$ right at $O$, $\\tan AOC=\\tan 60^\\circ=\\frac{AC}{OA}=\\sqrt{3}$．\\\\\nSince $OA=\\frac{\\sqrt{2}}{2}$\\\\\nTherefore $AC=\\sqrt{3}OA=\\frac{\\sqrt{6}}{2}$\\\\\nTherefore, the area of $\\triangle OAC=\\frac{1}{2}OA\\cdot AC=\\frac{1}{2}\\times \\frac{\\sqrt{2}}{2}\\times \\frac{\\sqrt{6}}{2}=\\frac{\\sqrt{3}}{4}$, and the area of the sector $OAE=\\frac{60^\\circ\\pi (\\frac{\\sqrt{2}}{2})^2}{360^\\circ}=\\frac{\\pi }{12}$\\\\\nTherefore, the area of the shaded region $=S_{\\triangle OAC}-S_{\\text{sector }OAE}=\\boxed{\\frac{\\sqrt{3}}{4}-\\frac{\\pi }{12}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52623066_101.png"
  },
  {
    "solution": "\\boxed{\\frac{9\\sqrt{3}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle BCA=90^\\circ$, CD is the median to side AB, and lines parallel to BA and BC are drawn through points C and D, respectively, intersecting at point E, where DE intersects AC at point O, and AE is connected. It is given that AD=CD; if $\\angle B=60^\\circ$ and BC=3, what is the area of quadrilateral ADCE?\n",
    "completion": "\\textbf{Solution:} In the right-angled $\\triangle ABC$, where $CD$ is the median to side $AB$, and $\\angle B=60^\\circ$, $BC=3$,\\\\\n$\\therefore AD=DB=CD=3$.\\\\\n$\\therefore AB=6$, by the Pythagorean theorem, $AC=3\\sqrt{3}$.\\\\\nSince quadrilateral $DBCE$ is a parallelogram, $\\therefore DE=BC=3$.\\\\\nSince $EC=BD=AD$ and $CE\\parallel DB$, $\\therefore$ quadrilateral $ADCE$ is a parallelogram.\\\\\n$\\therefore ED\\parallel BC$.\\\\\n$\\therefore \\angle AOD=\\angle ACB$.\\\\\nSince $\\angle ACB=90^\\circ$, $\\therefore \\angle AOD=\\angle ACB=90^\\circ$.\\\\\n$\\therefore$ parallelogram $ADCE$ is a rhombus,\\\\\n$\\therefore$ Area of the rhombus $ADCE$ is $\\frac{AC\\cdot ED}{2}=\\frac{3\\sqrt{3}\\times 3}{2}=\\boxed{\\frac{9\\sqrt{3}}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52424024_129.png"
  },
  {
    "solution": "\\boxed{\\frac{5\\sqrt{2}-20}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that quadrilateral $ABCD$ is inscribed in circle $\\odot O$, $AB\\parallel CD$, $BC\\parallel AD$, $AB=6$, and $BC=8$. Point $E$ is a point on circle $\\odot O$ (point $E$ is not coincident with $A$ or $C$). Connect $EA$, $EC$, and $OE$. Point $I$ is the incenter of $\\triangle AEC$, and point $M$ is on line segment $OE$ such that $ME=2MO$. What is the minimum value of the length of segment $MI$?\n",
    "completion": "\\textbf{Solution:} The minimum value of segment $MI$ is $\\boxed{\\frac{5\\sqrt{2}-20}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51458371_140.png"
  },
  {
    "solution": "\\boxed{y=\\frac{12}{x}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, $AC=BC$, point $C(2,0)$, and point $B(0,4)$. The graph of the inverse proportion function $y=\\frac{k}{x} (k\\ne 0)$ passes through point A. What is the expression of the inverse proportion function?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $AD\\perp x$-axis at D, then $\\angle ADC=90^\\circ$, \\\\\n$\\because \\angle ACB=90^\\circ, AC=BC$, \\\\\n$\\therefore \\angle BCO+\\angle ACD=90^\\circ$, \\\\\n$\\because \\angle BCO+\\angle CBO=90^\\circ$, \\\\\n$\\therefore \\angle ACD=\\angle CBO$, \\\\\nIn $\\triangle BOC$ and $\\triangle CDA$, \\\\\n$\\left\\{\\begin{array}{l}\n\\angle BOC=\\angle CDA=90^\\circ \\\\\n\\angle ACD=\\angle CBO \\\\\nAC=BC\n\\end{array}\\right.$, \\\\\n$\\therefore \\triangle BOC\\cong \\triangle CDA\\left(AAS\\right)$, \\\\\n$\\because$ the coordinates of point C are $(2,0)$, and the coordinates of point B are $(0,4)$, \\\\\n$\\therefore OC=2, OB=4$, \\\\\n$\\therefore CD=OB=4, AD=OC=2$, \\\\\n$\\therefore OD=OC+CD=6$, \\\\\n$\\therefore$ the coordinates of point A are $(6,2)$, \\\\\nSubstituting the coordinates of point A into the inverse proportional function $y=\\frac{k}{x}$, we get $k=2\\times 6=12$, \\\\\n$\\therefore$ the analytic expression of the inverse proportional function is $\\boxed{y=\\frac{12}{x}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52977778_227.png"
  },
  {
    "solution": "\\boxed{\\frac{5\\sqrt{3}}{3}−\\frac{1}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated in the diagram, a rectangular wooden box slides down an inclined plane. When the box slides to the position shown in the diagram, we have $AB=2\\text{m}$. Given that the height of the box $BE=1\\text{m}$, and the angle of the inclined plane $\\angle BAC$ is $30^\\circ$, calculate the height of the endpoint $E$ above the ground $AC$.\n",
    "completion": "\\textbf{Solution:} Draw $EH\\perp AC$ intersecting $AC$ at $H$ and $AB$ at $G$, as shown in the figure, \\\\\nFrom the given, \\\\\n$\\angle BAC+\\angle AGH=90^\\circ$, $\\angle BEG+\\angle BGE=90^\\circ$, $\\angle AGH=\\angle BGE$, and $\\angle BAC$ is $30^\\circ$, \\\\\n$\\therefore \\angle BAC=\\angle BEG=30^\\circ$, \\\\\n$\\because \\cos\\angle BEG=\\cos30^\\circ=\\frac{BE}{EG}$, $\\tan\\angle BEG=\\tan30^\\circ=\\frac{BG}{BE}$, $BE=1m$, \\\\\n$\\therefore GE=\\frac{BE}{\\cos30^\\circ}=\\frac{2\\sqrt{3}}{3}$, $BG=BE\\times \\tan30^\\circ=\\frac{\\sqrt{3}}{3}$, \\\\\n$\\because AB=2m$, \\\\\n$\\therefore AG=AB−BG=2−\\frac{\\sqrt{3}}{3}$, \\\\\n$\\because \\cos\\angle BAC=\\cos30^\\circ=\\frac{GH}{GA}$, \\\\\n$\\therefore GH=AG\\times \\cos30^\\circ=(2−\\frac{\\sqrt{3}}{3})\\times \\frac{\\sqrt{3}}{2}=\\sqrt{3}−\\frac{1}{2}$, \\\\\n$EH=EG+GH=\\frac{2\\sqrt{3}}{3}+\\sqrt{3}−\\frac{1}{2}=\\boxed{\\frac{5\\sqrt{3}}{3}−\\frac{1}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53102380.png"
  },
  {
    "solution": "\\boxed{\\left(-\\frac{9}{2}, 0\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a rectangular paper ABCO is placed in the Cartesian coordinate system, with $BC=10$. After folding the paper, point B falls exactly on the x-axis, denoted as $B'$, and the crease is CE. Given that the ratio $OC:OB'=4:3$. If point P is a moving point on the y-axis, when $\\triangle CPE$ is an isosceles triangle, find the coordinates of point P.\n",
    "completion": "\\textbf{Solution:} Let the coordinates of point E be (10, 3), and those of C be (0, 8), $\\therefore$ the coordinates of B are (10, 8), $\\therefore$ CE= $\\sqrt{5^2+10^2}=\\sqrt{125}=5\\sqrt{5}$,\\\\\n(1) When PC=CE, the coordinates of P are ($8+5\\sqrt{5}$, 0) and ($8-5\\sqrt{5}$, 0);\\\\\n(2) When PE=CE, the coordinates of P are ($-2$, 0);\\\\\n(3) When PC=PE, let the coordinates of P be (0, y), then PC=$8-y$, PE= $\\sqrt{10^2+(3-y)^2}$, hence $(8-y)^2=100+(3-y)^2$, solving this gives: $64-16y+y^2=100+9-6y+y^2$, $-10y=45$, $y=-\\frac{9}{2}$, the coordinates of P are $\\left(-\\frac{9}{2}, 0\\right)$;\\\\\nOverall, the coordinates of P are ($8+5\\sqrt{5}$, 0), ($8-5\\sqrt{5}$, 0), ($-2$, 0), $\\boxed{\\left(-\\frac{9}{2}, 0\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50677242_155.png"
  },
  {
    "solution": "\\boxed{7}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: A cylindrical oil tank is partially filled with oil, with its cross-section shown in the figure. It is known that the radius of the cross-section $\\odot O$ is $5\\,\\text{cm}$, and the width of the oil surface $AB$ is $6\\,\\text{cm}$. If more oil is added into the tank, and the width of the oil surface becomes $8\\,\\text{cm}$, by how many centimeters does the oil surface $AB$ rise?\n",
    "completion": "\\textbf{Solution:} Solve in two cases:\\\\\n(1) As shown in Figure 1, for the oil surface CD with a width of 8cm, draw $ON\\perp AB$ intersecting CD and AB at points M and N respectively.\\\\\nFrom the problem statement, we know\\\\\n$OM\\perp CD$,\\\\\n$CM=MD=\\frac{1}{2}CD=4\\,\\text{cm}$,\\\\\n$AN=BN=\\frac{1}{2}AB=3\\,\\text{cm}$\\\\\nIn $\\triangle BON$, by the Pythagorean theorem, we have\\\\\n$ON=\\sqrt{OB^{2}-BN^{2}}=4\\,\\text{cm}$\\\\\nIn $\\triangle DOM$, by the Pythagorean theorem, we have\\\\\n$OM=\\sqrt{OD^{2}-DM^{2}}=3\\,\\text{cm}$\\\\\n$\\therefore MN=ON-OM=1\\,\\text{cm}$\\\\\n(2) As shown in Figure 2, for the oil surface EF with a width of 8cm, draw $PN\\perp EF$ intersecting AB and EF at points N and P respectively, and connect OB.\\\\\nFrom the problem statement, we know\\\\\n$PN\\perp AB$,\\\\\n$EP=PF=\\frac{1}{2}EF=4\\,\\text{cm}$,\\\\\n$AN=BN=\\frac{1}{2}AB=3\\,\\text{cm}$\\\\\nIn $\\triangle BON$, by the Pythagorean theorem, we have\\\\\n$ON=\\sqrt{OB^{2}-BN^{2}}=4\\,\\text{cm}$\\\\\nIn $\\triangle EPO$, by the Pythagorean theorem, we have\\\\\n$OP=\\sqrt{OE^{2}-EP^{2}}=3\\,\\text{cm}$\\\\\n$\\therefore NP=ON+OP=7\\,\\text{cm}$\\\\\n$\\therefore$ The oil surface AB rises to CD by 1cm, and the oil surface AB rises to EF by \\boxed{7}cm;",
    "image_path": "dataset/mm_math/images/MM_Math/51545289.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $E$ is a point inside the square $ABCD$. Connect $AE$, $BE$, $DE$, where $\\angle AEB=90^\\circ$, and the area of $\\triangle ADE$ is 4. What is the length of $AE$?\n",
    "completion": "\\textbf{Solution:} Extend AE, and draw DF perpendicular to AE from point D, where F is the foot of the perpendicular,\\\\\n$\\therefore$ $\\angle$DFA=$90^\\circ$,\\\\\n$\\therefore$ $\\angle$DAF+$\\angle$ADF=$90^\\circ$,\\\\\n$\\because$ In the square ABCD, $\\angle$DAB=$90^\\circ$, and AD=AB,\\\\\n$\\therefore$ $\\angle$DAF+$\\angle$BAE=$90^\\circ$,\\\\\n$\\therefore$ $\\angle$ADF=$\\angle$BAE,\\\\\n$\\because$ $\\angle$AEB=$90^\\circ$,\\\\\n$\\therefore$ $\\angle$AEB=$\\angle$DFA,\\\\\n$\\therefore$ $\\triangle$ABE$\\cong$ $\\triangle$DAF (by AAS),\\\\\n$\\therefore$ AE=DF,\\\\\nGiven $S_{\\triangle ADE}=4$, we have\\\\\n$\\frac{1}{2}AE\\cdot DF=\\frac{1}{2}AE^2=4$,\\\\\nSolving this gives AE=$\\boxed{2\\sqrt{2}}$,\\\\\n$\\therefore$ the length of AE is $2\\sqrt{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52189539.png"
  },
  {
    "solution": "\\boxed{12\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, the diagonals $AC$ and $BD$ of square $ABCD$ intersect at point $O$. Extend $BD$ in both directions to points $E$ and $F$ respectively, such that $BE=DF$. If $AB=3\\sqrt{2}$ and $BE=3$, what is the perimeter of quadrilateral $AECF$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral ABCD is a square, \\\\\ntherefore $AO=\\frac{1}{2}AC$, $BO=\\frac{1}{2}BD$, $AC=BD$, and $AC \\perp BD$, \\\\\nthus $AO=BO$, $\\angle AOB=90^\\circ$. \\\\\nIn the right triangle $\\triangle AOB$, by the Pythagorean theorem, $AB=\\sqrt{AO^{2}+BO^{2}}=3\\sqrt{2}$, \\\\\ntherefore $AO=BO=3$. \\\\\nHence $EO=OB+BE=6$. \\\\\nIn $\\triangle AOE$, $\\angle AOE = 90^\\circ$, $AE=\\sqrt{AO^{2}+EO^{2}}=\\sqrt{3^{2}+6^{2}}=3\\sqrt{5}$. \\\\\nGiven that quadrilateral AECF is a rhombus, \\\\\ntherefore $AE=EC=CF=AF$. \\\\\nThus, the perimeter of quadrilateral AECF is $4AE=12\\sqrt{5}$. \\\\\nTherefore, the perimeter of quadrilateral AECF is $\\boxed{12\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53359506_126.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the figure, within $\\triangle ABC$, point D is a point on side $AB$. If $\\angle ACD = \\angle B$, $AD = 1$, $AC = 2$, and the area of $\\triangle ADC$ is $1$, what is the area of $\\triangle BCD$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle ACD = \\angle B$, \\\\\nand $\\angle A = \\angle A$, \\\\\nthus $\\triangle ACD \\sim \\triangle ABC$, \\\\\nhence $\\frac{S_{\\triangle ACD}}{S_{\\triangle ABC}}=\\left(\\frac{AD}{AC}\\right)^2=\\frac{1}{4}$, \\\\\nsince $S_{\\triangle ACD} = 1$, \\\\\nit follows that $S_{\\triangle ABC} = 4$, \\\\\ntherefore, the area of $\\triangle BCD = S_{\\triangle ABC} - S_{\\triangle ACD} = \\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51545324.png"
  },
  {
    "solution": "\\boxed{18}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $B$ and $C$ are two points on the line segment $AD$, and $AB:BC:CD=2:3:4$, $M$ is the midpoint of $AD$, if $MC=1$. What is the length of the line segment $AD$?\n",
    "completion": "\\textbf{Solution:} Let $AB=2x$, hence $BC=3x, CD=4x$.\\\\\n$AD=2x+3x+4x=9x.$\\\\\nSince $M$ is the midpoint of $AD$,\\\\\nit follows that $MD=\\frac{1}{2}AD=\\frac{9}{2}x.$\\\\\n$MC=MD-CD=\\frac{9}{2}x-4x=\\frac{1}{2}x$,\\\\\nAccording to the problem, $\\frac{1}{2}x=1$,\\\\\nSolving this gives $x=2$,\\\\\nTherefore, $AD=9x=\\boxed{18}$.",
    "image_path": "dataset/mm_math/images/MM_Math/54661145_68.png"
  },
  {
    "solution": "\\boxed{y=x^{2}-2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=kx$ intersects the parabola $y=x^2+c$ at points A and B, where the coordinates of point B are $(2, 2)$. Find the equations of the line $AB$ and the parabola.\n",
    "completion": "Solution: Since the line $y=kx$ intersects the parabola $y=x^{2}+c$ at point $B(2, 2)$,\\\\\ntherefore $2k=2$ and $4+c=2$,\\\\\nthus $k=1$ and $c=-2$,\\\\\ntherefore the equation of line $AB$ is \\boxed{y=x}, and the equation of the parabola is \\boxed{y=x^{2}-2}.",
    "image_path": "dataset/mm_math/images/MM_Math/53066184_89.png"
  },
  {
    "solution": "\\boxed{169}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\angle ABC=\\angle FAC=90^\\circ$, the length of $BC$ is $3\\text{cm}$, the length of $AB$ is $4\\text{cm}$, and the length of $AF$ is $12\\text{cm}$. What is the area of the square $CDEF$ in $\\text{cm}^2$?\n",
    "completion": "\\textbf{Solution:} Given in right triangle $\\triangle ABC$, $AC= \\sqrt{AB^{2}+BC^{2}}=\\sqrt{3^{2}+4^{2}}=5\\,\\text{cm}$,\nin right triangle $\\triangle ACF$, $CF= \\sqrt{AF^{2}+AC^{2}}=\\sqrt{12^{2}+5^{2}}=13\\,\\text{cm}$,\n$\\therefore$ the area of square $CDEF$ is $13\\times13=\\boxed{169}\\,\\text{cm}^{2}$",
    "image_path": "dataset/mm_math/images/MM_Math/51554788.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the rectangle $ABCD$, the lengths of sides $AB$ and $BC$ ($AB < BC$) are the two roots of the equation ${x}^{2}-7x+12=0$. Point $P$ starts from point $A$ and moves along the edge $ABC$ in the direction $A\\to B\\to C\\to A$ at a speed of $1$ unit per second, with the movement time being $t$ seconds. What are the lengths of $AB$ and $BC$?\n",
    "completion": "\\textbf{Solution:} Since $x^{2}-7x+12=0$,\\\\\nthen $(x-3)(x-4)=0$,\\\\\ntherefore $x_{1}=3$, $x_{2}=4$.\\\\\nHence $AB=\\boxed{3}$, $BC=\\boxed{4}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51722482_163.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure: Given the line segment $AB=16\\,\\text{cm}$, point $N$ is on the line segment $AB$ with $NB=3\\,\\text{cm}$, and $M$ is the midpoint of $AB$. What is the length of the line segment $MN$ in centimeters?\n",
    "completion": "\\textbf{Solution:} \n\nGiven that $AB=16\\,\\text{cm}$, and $M$ is the midpoint of $AB$,\n\n$\\therefore AM=MB=\\frac{1}{2}AB=8\\,\\text{cm}$,\n\nGiven that $NB=3\\,\\text{cm}$,\n\n$\\therefore MN=MB-NB=\\boxed{5}\\,\\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51407587_113.png"
  },
  {
    "solution": "\\boxed{\\frac{9}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=x^2+bx+c$ intersects the x-axis at points $A$ and $B(3,0)$, and the y-axis at point $C(0,3)$. If point $M$ is a moving point on the parabola below the x-axis, and through point $M$ a line $MN$ is drawn parallel to the y-axis intersecting the line $BC$ at point $N$, what is the maximum value of the line segment $MN$?\n",
    "completion": "\\textbf{Solution:} Let the coordinates of point M be $(m, m^2-4m+3)$. Assume the equation of line BC is $y=-x+3$, \\\\\nSubstituting point B $(3,0)$ into $y=-x+3$, \\\\\nWe get: $0=3k+3$, solving for $k$ we find $k=-1$, \\\\\n$\\therefore$ The equation of line BC is $y=-x+3$.\\\\\n$\\because MN\\parallel y$-axis, \\\\\n$\\therefore$ The coordinates of point N are $(m, -m+3)$.\\\\\n$\\because$ The equation of the parabola is $y=x^2-4x+3=(x-2)^2-1$, \\\\\n$\\therefore$ The axis of symmetry for the parabola is $x=2$,\\\\\n$\\therefore$ The point $(1,0)$ lies on the graph of the parabola,\\\\\n$\\therefore$ $1<m<3$.\\\\\n$\\because$ The length of segment $MN=-m+3-(m^2-4m+3)=-m^2+3m=-\\left(m-\\frac{3}{2}\\right)^2+\\frac{9}{4}$,\\\\\n$\\therefore$ When $m=\\frac{3}{2}$, segment $MN$ reaches its maximum value, which is $\\boxed{\\frac{9}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53124516_62.png"
  },
  {
    "solution": "\\boxed{48^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\angle AOD = 84^\\circ$, $\\angle AOB = 18^\\circ$, and OB bisects $\\angle AOC$. What is the measure of $\\angle COD$?\n",
    "completion": "\\textbf{Solution:} Given that $OB$ bisects $\\angle AOC$, and $\\angle AOB=18^\\circ$,\\\\\nit follows that $\\angle AOC=2\\angle AOB=36^\\circ$,\\\\\nAlso, given that $\\angle AOD=84^\\circ$,\\\\\nit results in $\\angle COD=\\angle AOD-\\angle AOC=84^\\circ-36^\\circ=\\boxed{48^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51284590.png"
  },
  {
    "solution": "\\boxed{65^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, E is a point on the diagonal BD of the square ABCD. Connect AE and CE, and extend CE to meet AD at point F. If $\\angle AEC = 140^\\circ$, what is the measure of $\\angle DFE$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a square,\\\\\nit follows that AB = CB, and $\\angle ABE = \\angle CBE = \\angle ADB = 45^\\circ$.\\\\\nIn $\\triangle ABE$ and $\\triangle CBE$, we have\\\\\n\\[\n\\left\\{\n\\begin{array}{l}\nAB = CB\\\\\n\\angle ABE = \\angle CBE\\\\\nBE = BE\n\\end{array}\n\\right.,\n\\]\nthus $\\triangle ABE \\cong \\triangle CBE$,\\\\\nwhich implies $\\angle AEB = \\angle CEB$.\\\\\nAlso, since $\\angle AEC = 140^\\circ$,\\\\\nit follows that $\\angle CEB = 70^\\circ$.\\\\\nBecause $\\angle DEC + \\angle CEB = 180^\\circ$,\\\\\nthus $\\angle DEC = 180^\\circ - \\angle CEB = 110^\\circ$.\\\\\nSince $\\angle DFE + \\angle ADB = \\angle DEC$,\\\\\nit follows that $\\angle DFE = \\angle DEC - \\angle ADB = 110^\\circ - 45^\\circ = \\boxed{65^\\circ}.$",
    "image_path": "dataset/mm_math/images/MM_Math/53333892.png"
  },
  {
    "solution": "\\boxed{\\frac{32}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB \\parallel CD \\parallel EF$, and $AF$ intersects with $BE$ at point $G$. If $AG=2$, $GD=1$, $DF=5$, $BC=4$, then what is the length of $BE$?\n",
    "completion": "\\textbf{Solution:} Given that $AB\\parallel CD\\parallel EF$,\\\\\nit follows that $\\frac{AD}{DF}=\\frac{BC}{CE}$,\\\\\nsince $AD=AG+GD$, where $AG=2$, $GD=1$, $DF=5$, and $BC=4$,\\\\\nwe have $\\frac{2+1}{5}=\\frac{4}{CE}$,\\\\\nwhich leads to $CE=\\frac{20}{3}$,\\\\\nthus, $BE=BC+CE=4+\\frac{20}{3}=\\boxed{\\frac{32}{3}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52917325.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $I$ is the incenter of $\\triangle ABC$. Line segment $AI$ is extended to intersect the circumcircle of $\\triangle ABC$ at point $D$. Point $E$ is the midpoint of chord $AC$. Lines $CD$, $EI$, and $IC$ are connected. When $AI=2CD$, $IC=6$, and $ID=5$, what is the length of $IE$?\n",
    "completion": "\\textbf{Solution:} Extend $ID$ to $M$, such that $DM=ID$, and connect $CM$.\\\\\nSince $I$ is the incenter of $\\triangle ABC$,\\\\\nit follows that $\\angle IAC=\\angle IAB$, $\\angle ICA=\\angle ICB$,\\\\\nSince $\\angle DIC=\\angle IAC+\\angle ICA$, $\\angle DCI=\\angle BCD+\\angle ICB$,\\\\\nit follows that $\\angle DIC=\\angle DCI$,\\\\\nthus $DI=DC=DM$,\\\\\ntherefore $\\angle ICM=90^\\circ$,\\\\\nhence $CM=\\sqrt{IM^{2}-IC^{2}}=8$,\\\\\nSince $AI=2CD=10$,\\\\\nit follows that $AI=IM$,\\\\\nSince $AE=EC$,\\\\\nit follows that $IE$ is the midline of $\\triangle ACM$,\\\\\nthus $IE=\\frac{1}{2}CM=\\boxed{4}$",
    "image_path": "dataset/mm_math/images/MM_Math/52457422.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: On a line, seven squares are placed in sequence (as shown). It is known that the areas of the three diagonally placed squares are 1, 2, and 3, respectively, and the areas of the four upright squares are $S_1$, $S_2$, $S_3$, and $S_4$. What is the value of $S_3 + S_4 - S_1 - S_2$?\n",
    "completion": "\\textbf{Solution:} Observe that, \\\\\n$\\because AB=BE$, and $\\angle ACB=\\angle BDE=90^\\circ$, \\\\\n$\\therefore \\angle ABC+\\angle BAC=90^\\circ$, and $\\angle ABC+\\angle EBD=90^\\circ$, \\\\\n$\\therefore \\angle BAC=\\angle EBD$, \\\\\n$\\because$ in triangles $\\triangle ABC$ and $\\triangle BDE$, \\\\\n$\\left\\{\\begin{array}{l}\n\\angle ACB=\\angle BDE=90^\\circ \\\\\n\\angle BAC=\\angle EBD \\\\\nAB=BE\n\\end{array}\\right.$, \\\\\n$\\therefore \\triangle ABC\\cong \\triangle BDE(\\text{AAS})$, \\\\\n$\\therefore BC=ED$, \\\\\n$\\because AB^2=AC^2+BC^2$, \\\\\n$\\therefore AB^2=AC^2+ED^2={S}_{1}+{S}_{2}$, \\\\\ni.e., ${S}_{1}+{S}_{2}=1$, \\\\\nSimilarly, ${S}_{3}+{S}_{4}=3$. \\\\\nHence, ${S}_{3}+{S}_{4}-{S}_{1}-{S}_{2}={S}_{3}+{S}_{4}-({S}_{1}+{S}_{2})=3-1=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52731257.png"
  },
  {
    "solution": "\\boxed{(6,0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the direct proportion function $y_1=kx$ intersects with the graph of the inverse proportion function $y_2=\\frac{m}{x}$ at points $A(2,4)$ and point $B$. The coordinate of point $C$ is $(4,0)$, and point $D$ lies on the graph of $y_2=\\frac{m}{x}$. Let point $M$ be on the $x$-axis, and point $N$ be within the Cartesian coordinate system. When the quadrilateral $CDNM$ is a square, directly write down the coordinates of point $M$.\n",
    "completion": "\\textbf{Solution:} When the quadrilateral $CDNM$ is a square, the coordinates of the point $M$ are $\\boxed{(2,0)}$ or $\\boxed{(6,0)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51403218_44.png"
  },
  {
    "solution": "\\boxed{-\\frac{1}{10}\\le k\\le \\frac{3}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the graph of the linear function $y = -\\frac{1}{2}x + 5$, denoted as $l_1$, intersects the x and y axes at points $A$ and $B$, respectively. The graph of the direct proportion function, denoted as $l_2$, intersects $l_1$ at point $C(m, 4)$. The graph of the linear function $y = kx+1$ intersects line segment $AC$. Directly write down the value of $k$.\n",
    "completion": "\\textbf{Solution:} Solve: When $y=kx+1$ exactly passes through point C, we have $4=2k+1$, and get $k=\\frac{3}{2}$. When $y=kx+1$ exactly passes through point A, we have $0=10k+1$, and get $k=-\\frac{1}{10}$.\\\\\n$\\therefore$ When $-\\frac{1}{10}\\le k\\le \\frac{3}{2}$, the graph of the linear function $y=kx+1$ intersects with the line segment AC.\\\\\nThus, the value of $k$ is $\\boxed{-\\frac{1}{10}\\le k\\le \\frac{3}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50676403_169.png"
  },
  {
    "solution": "\\boxed{1\\colon 3\\colon 2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $A$ and $B$ are located on the horizontal lines of the circular grid. Points $C$ and $D$ are the intersections of the diameter $AB$ with the grid lines. What is the ratio of $BC:CD:DA$?\n",
    "completion": "\\textbf{Solution:} Given the figure, draw CE$\\perp$BE at point E, BF$\\perp$CF at point F, and AG$\\perp$DG at point G,\\\\\nthus $\\angle$BEC=$\\angle$CFD=$\\angle$AGD=$90^\\circ$,\\\\\nthus BE$\\parallel$CF$\\parallel$DG,\\\\\nthus $\\angle$B=$\\angle$FCD=$\\angle$ADG,\\\\\nthus $\\triangle$BEC$\\sim$ $\\triangle$CFD$\\sim$ $\\triangle$DGA,\\\\\nthus BC$\\colon$ CD$\\colon$ AD=EC$\\colon$ FD$\\colon$ GA=1$\\colon$ 3$\\colon$ 2.\\\\\nTherefore, the answer is: BC$\\colon$ CD$\\colon$ AD=\\boxed{1\\colon 3\\colon 2}.",
    "image_path": "dataset/mm_math/images/MM_Math/51369881.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the rhombus $ABCD$, the diagonals $AC$ and $BD$ intersect at point $O$, and $H$ is the midpoint of side $AD$. If the perimeter of the rhombus $ABCD$ is $24$, what is the length of $OH$?\n",
    "completion": "\\textbf{Solution:} Since in the rhombus $ABCD$, the perimeter is 24, \\\\\nit follows that $OA=OC$, $CD=24\\div 4=6$, \\\\\nsince $H$ is the midpoint of side $AD$, \\\\\nit follows that $OH=\\frac{1}{2}CD=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53360042.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, given points A, B, and C are three points on circle $\\odot O$, with radius $OC=2$, $\\angle ABC=30^\\circ$, the tangent line $AP$ intersects the extension line of $OC$ at point $P$, then how long is $AP$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, connect OA.\\\\\n$\\because$ $\\angle ABC=30^\\circ$,\\\\\n$\\therefore$ $\\angle AOC=2\\angle ABC=60^\\circ$.\\\\\n$\\because$ PA is the tangent of circle $O$,\\\\\n$\\therefore$ $\\angle OAP=90^\\circ$,\\\\\n$\\therefore$ $\\angle OPA=30^\\circ$.\\\\\n$\\because$ OA=OC=2,\\\\\n$\\therefore$ OP=2OA=4.\\\\\n$\\therefore$ $AP=\\sqrt{OP^{2}-OA^{2}}=2\\sqrt{3}$,\\\\\nThus, $AP=\\boxed{2\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52720285.png"
  },
  {
    "solution": "\\boxed{246}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, $BD$ bisects $\\angle ABC$, $\\angle BDC = \\angle A = 90^\\circ$, $\\cos\\angle ABD = \\frac{4}{5}$. If $BC=25$, find the area of the quadrilateral $ABCD$.\n",
    "completion": "\\textbf{Solution:} Since $\\angle ABD=\\angle CBD$,\\\\\nwe have $\\cos\\angle CBD=\\frac{BD}{BC}=\\frac{4}{5}$,\\\\\nSince $BC=25$,\\\\\nwe get $BD=20$.\\\\\nThus, $CD=\\sqrt{BC^{2}-BD^{2}}=15$.\\\\\nSince $\\cos\\angle ABD=\\frac{AB}{BD}=\\frac{4}{5}$,\\\\\nwe get $AB=16$,\\\\\nand thus $AD=\\sqrt{BD^{2}-AB^{2}}=12$.\\\\\nTherefore, the area of quadrilateral $ABCD$, $S_{\\text{quadrilateral}ABCD}=S_{\\triangle ABD}+S_{\\triangle BCD}$\\\\\n$=\\frac{1}{2}AB\\cdot AD+\\frac{1}{2}BD\\cdot CD$\\\\\n$=\\frac{1}{2}\\times 16\\times 12+\\frac{1}{2}\\times 20\\times 15$\\\\\n$=96+150$\\\\\n$=\\boxed{246}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52977904_142.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the lines $l_1 \\parallel l_2 \\parallel l_3$, and the lines AC and DF are intersected by $l_1$, $l_2$, $l_3$. Given that AB = 8, BC = 12, and EF = 9, what is the length of DE?\n",
    "completion": "\\textbf{Solution:} Since $l_1\\parallel l_2\\parallel l_3$,\\\\\nit follows that $\\frac{AB}{BC}=\\frac{DE}{EF}$,\\\\\ngiven $AB=8, BC=12, EF=9$,\\\\\nthus $\\frac{8}{12}=\\frac{DE}{9}$, solving gives $DE=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53043135.png"
  },
  {
    "solution": "\\boxed{21}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In $\\triangle ABC$, $D$ is a point on $BC$, $AC=10$, $CD=6$, $AD=8$, $AB=17$. Find the length of $BC$.\n",
    "completion": "\\textbf{Solution:} Given $\\because CD=6, AD=8$,\\\\\n$\\therefore CD^{2}+AD^{2}=6^{2}+8^{2}=100$,\\\\\n$\\because AC^{2}=10^{2}=100$,\\\\\n$\\therefore CD^{2}+AD^{2}=AC^{2}$,\\\\\n$\\therefore \\angle ADC=90^\\circ$,\\\\\n$\\therefore \\angle ADB=90^\\circ$,\\\\\n$\\therefore BD=\\sqrt{AB^{2}-AD^{2}}=\\sqrt{17^{2}-8^{2}}=15$,\\\\\n$\\therefore BC=BD+CD=15+6=\\boxed{21}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53043951.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle B = \\angle C$. Point $D$ is on side $AB$. A line is drawn through $D$, with $DE \\perp BC$ intersecting $BC$ at point $E$ and extending to intersect the extended line of $CA$ at point $F$. Given $AC = 5$ and $BD = 3$, what is the length of $AF$?\n",
    "completion": "\\textbf{Solution:} Given that $DE\\perp BC$,\\\\\nit follows that $\\angle DEB=\\angle FEC=90^\\circ$,\\\\\nthus $\\angle B+\\angle BDE=90^\\circ$, $\\angle C+\\angle F=90^\\circ$,\\\\\nsince $\\angle B=\\angle C$,\\\\\nit results in $\\angle F=\\angle BDE$,\\\\\nand since $\\angle FDA=\\angle BDE$,\\\\\nwe have $\\angle FDA=\\angle F$,\\\\\nhence $AF=AD$,\\\\\nas $\\angle B=\\angle C$ and $AC=5$,\\\\\nit implies $AB=AC=5$,\\\\\ngiven $BD=3$,\\\\\nit follows $AD=AB−BD=5−3=2$,\\\\\nthus $AF=AD=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52978719.png"
  },
  {
    "solution": "\\boxed{\\frac{2}{\\sqrt{5}}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within square $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$, with $E$ and $F$ located on $OB$ and $OC$ respectively. The extension of line $AE$ intersects $BF$ at point $M$, and $OE=OF$. $\\triangle AOE \\cong \\triangle BOF$; $AM \\perp BF$; Given $AE=\\sqrt{5}$ and $OE=1$, calculate the length of $EM$.\n",
    "completion": "Solution: Since $AE=\\sqrt{5}$ and $OE=1$, $\\angle AOE=90^\\circ$, \\\\\nit follows that $OA=\\sqrt{AE^2-OE^2}=2$, \\\\\nSince $\\triangle AOE \\cong \\triangle BOF$, \\\\\nit follows that $S_{\\triangle AOE}=S_{\\triangle BOF}=S_{\\triangle EOF}+S_{\\triangle BEF}$, with $BF=AE=\\sqrt{5}$, \\\\\nTherefore, $\\frac{1}{2}OE \\cdot AO=\\frac{1}{2}OE \\cdot OF+\\frac{1}{2}BF \\cdot EM$, \\\\\nThus, $\\frac{1}{2} \\times 1 \\times 2=\\frac{1}{2} \\times 1 \\times 1+\\frac{\\sqrt{5}}{2}EM$, \\\\\nHence, $EM=\\boxed{\\frac{2}{\\sqrt{5}}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51968629_124.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the graph of the linear function $y=-\\frac{1}{2}x+b$ passes through point $A(0,3)$. Point $P$ is a moving point on this line. Perpendiculars $PM$ and $PN$ are drawn from point $P$ to the $x$-axis at point $M$ and the $y$-axis at point $N$, respectively. On the quadrilateral $PMON$, the following segments are taken: $PC=\\frac{1}{3}MP$, $MB=\\frac{1}{3}OM$, $OE=\\frac{1}{3}ON$, $ND=\\frac{1}{3}NP$. What is the value of $b$?\n",
    "completion": "\\textbf{Solution:} We have $b=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52395032_110.png"
  },
  {
    "solution": "\\boxed{100^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $P$ is a point inside $\\angle AOB$. Construct the symmetrical points $P_{1}$ and $P_{2}$ of point $P$ with respect to $OB$ and $OA$ respectively. Connect $P_{1}P_{2}$, let it intersect $OB$ at $M$, and intersect $OA$ at $N$. If $\\angle AOB=40^\\circ$, what is the measure of $\\angle MPN$?\n",
    "completion": "\\textbf{Solution:} Since the symmetric point of $P$ about $OB$ is ${P}_{1}$ and the symmetric point of $P$ about $OA$ is ${P}_{2}$, \\\\\ntherefore, $PM={P}_{1}M$, $PN={P}_{2}N$, $\\angle {P}_{2}=\\angle {P}_{2}PN$, $\\angle {P}_{1}=\\angle {P}_{1}PM$, $PP_{1}\\perp OB$, $PP_{2}\\perp OA$, \\\\\nsince $\\angle AOB=40^\\circ$, \\\\\ntherefore, $\\angle {P}_{1}PP_{2}=140^\\circ$, \\\\\ntherefore, $\\angle {P}_{1}+\\angle {P}_{2}=180^\\circ-\\angle {P}_{1}PP_{2}=40^\\circ$, \\\\\nsince $\\angle PMN=\\angle {P}_{1}+\\angle MP{P}_{1}=2\\angle {P}_{1}$, $\\angle PNM=\\angle {P}_{2}+\\angle NP{P}_{2}=2\\angle {P}_{2}$, \\\\\ntherefore, $\\angle PMN+\\angle PNM=2(\\angle {P}_{1}+\\angle {P}_{2})=2\\times 40^\\circ=80^\\circ$, \\\\\ntherefore, $\\angle MPN=180^\\circ-(\\angle PMN+\\angle PNM)=180^\\circ-80^\\circ=\\boxed{100^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53044357.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{7}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, within rectangle $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$. $DF$ is perpendicular to and bisects $OC$, intersecting $AC$ at point $E$ and $BC$ at point $F$. $AF$ is connected. If $CD = 2\\sqrt{3}$, then what is the length of $AF$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rectangle, \\\\\nit follows that AB=$CD=2\\sqrt{3}$, and AO=CO=BO=DO, \\\\\nsince DF is perpendicular bisector of OC, \\\\\nit follows that OD=DC, \\\\\nthus OD=DC=OC, \\\\\ntherefore, $\\triangle$ODC is an equilateral triangle, \\\\\nthus OD=OC=$CD=2\\sqrt{3}$, \\\\\ntherefore, AC=$4\\sqrt{3}$, \\\\\nthus, BC=$\\sqrt{AC^{2}-AB^{2}}=6$, \\\\\nsince $\\triangle$ODC is an equilateral triangle and DE$\\perp$AC, \\\\\nit follows that $\\angle$CDE=$\\angle$ODE=$30^\\circ$, \\\\\nthus, DC=$\\sqrt{3}$CF=$2\\sqrt{3}$, \\\\\ntherefore, CF=2, \\\\\nthus, BF=4, \\\\\ntherefore, AF=$\\sqrt{AB^{2}+BF^{2}}=\\sqrt{12+16}=\\boxed{2\\sqrt{7}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53359605.png"
  },
  {
    "solution": "\\boxed{y=\\frac{4}{3}x-4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y=(m-2)x-4$ intersects the negative half-axis of the x-axis at point A, and intersects the y-axis at point B, and the area of $\\triangle OAB$ is 8. A line $CD \\parallel AB$ is drawn through point C (0, 3), intersecting the x-axis at point D. Find the equation of line BD.\n",
    "completion": "\\textbf{Solution:} Given that $C(0,3)$ and $AB\\parallel CD$, thus the equation of line CD is: $y=-x+3$. Hence, the coordinate of point D is $(3,0)$. Let the equation of line BD be $y=kx+b$. Substituting points B$(0,-4)$ and D$(3,0)$ into the equation, we get:\n\\[b=-4\\\\\n3k+b=0\\],\nsolving this, we obtain:\n\\[k=\\frac{4}{3}\\\\\nb=-4\\].\\\\\nTherefore, the equation of line BD is $\\boxed{y=\\frac{4}{3}x-4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52095492_168.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{18}{x}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y=k_1x-4$ intersects with the graph of the hyperbolic function $y=\\frac{k_2}{x}$ (where $x>0$) at point $A(3, -6)$, and intersects with the $x$-axis at point $B$. Point $D$ is a point on segment $AB$, and lines $OD$ and $OA$ are connected, with $\\frac{S_{\\triangle BOD}}{S_{\\triangle BOA}}=1\\colon 3$. Find the analytic expressions of the linear and hyperbolic functions?\n",
    "completion": "\\textbf{Solution:} Substituting point A (3, −6) into $y=k_1x-4$,\\\\\nwe get $-6=3k_1-4$,\\\\\nfrom which we find $k_1=-\\frac{2}{3}$,\\\\\nSubstituting point A (3, −6) into $y=\\frac{k_2}{x}$ (where $x>0$), we get $-6=\\frac{k_2}{3}$,\\\\\n$\\therefore k_2=-18$,\\\\\n$\\therefore$ the equation of the linear function is $\\boxed{y=-\\frac{2}{3}x-4}$, and the equation of the inverse proportion function is $\\boxed{y=-\\frac{18}{x}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51445903_254.png"
  },
  {
    "solution": "\\boxed{145^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, lines AB and CD intersect at point E, with EF$\\perp$AB at point E. If $\\angle FEC - \\angle AEC = 20^\\circ$, what is the measure of $\\angle AED$?\n",
    "completion": "\\textbf{Solution:} Let $\\angle AEC$ be $x$, then $\\angle FEC=x+20^\\circ$. \\\\\nSince EF$\\perp$AB, \\\\\nit follows that $\\angle AEF=90^\\circ$, \\\\\nwhich means $\\angle AEC+\\angle FEC= 90^\\circ$, \\\\\nhence $x+x+20^\\circ=90^\\circ$, \\\\\nSolving this gives: $x=35^\\circ$, \\\\\nthat is $\\angle AEC=35^\\circ$, \\\\\nTherefore, $\\angle AED=180^\\circ−35^\\circ=\\boxed{145^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/54962528.png"
  },
  {
    "solution": "\\boxed{48}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, two congruent right triangles overlap each other. One of the triangles is translated along the direction from point B to C to the position of $\\triangle DEF$, where $AB=10$, $DO=4$, and the translation distance is $6$. What is the area of the shaded region?\n",
    "completion": "\\textbf{Solution:} Given the property of translation, we have $BE=6$, $DE=AB=10$,\\\\\nthus $OE=DE-DO=10-4=6$,\\\\\ntherefore, the area of quadrilateral $ODFC$ is equal to the area of trapezoid $ABEO$, which is $\\frac{1}{2}\\times(AB+OE)\\times BE =\\frac{1}{2}\\times(10+6)\\times 6=\\boxed{48}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51212186.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the perimeter of $\\triangle ACD$ is $14$, $AB-AC=2$, and the perpendicular bisector of $BC$ intersects $AB$ at point $D$ and intersects $BC$ at point $E$. Find the lengths of $AB$ and $AC$.\n",
    "completion": "\\textbf{Solution}: Since the perimeter of $\\triangle ACD$ is 14,\\\\\n$\\therefore AC+AD+CD=14$,\\\\\nbecause $DE$ is the perpendicular bisector of $BC$,\\\\\n$\\therefore BD=CD$,\\\\\n$\\therefore AC+AD+BD=14$,\\\\\n$\\therefore AC+AB=14$,\\\\\nsince $AB-AC=2$,\\\\\n$\\therefore AB=\\boxed{8}$, $AC=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51369007.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, ABCD is a square with a side length of 4. Point H is on the extension of CD, and CEFH is also a square. What is the area of $\\triangle DBF$?\n",
    "completion": "\\textbf{Solution:} Let the side length of the square $CEFH$ be $a$,\n\naccording to the problem, we have: $S_{\\triangle BDF}=S_{\\text{square }ABCD}+S_{\\text{square }CEFH}-S_{\\triangle ABD}-S_{\\triangle DHF}-S_{\\triangle BEF}$,\n\n$=4^2+a^2 -\\frac{1}{2}\\times4\\times4-\\frac{1}{2}a(a-4)-\\frac{1}{2}a(a+4) =\\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51614626.png"
  },
  {
    "solution": "\\boxed{6\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the cross-section of a certain dam, the slope ratio $i$ of the slope AB is $1\\colon 2$, and the slope ratio $i$ of the water-facing slope CD is $1\\colon 1$. If the length of AB is $6\\sqrt{5}$ meters, what is the length of the slope CD?\n",
    "completion": "\\textbf{Solution:} \n\nLet lines BE and CF be perpendicular to AD, passing through B and C respectively,\n\n$\\therefore$ Quadrilateral BEFC is a rectangle,\n\nThe slope ratio of ramp AB is $i=1\\colon 2$, that is $\\frac{BE}{AE}=\\frac{1}{2}$. Assuming $BE=x$, then $AE=2x$,\n\nIn $\\triangle ABE$, according to the Pythagorean theorem, we have: $BE^2+AE^2=AB^2$,\n\n$x^2+(2x)^2=(6\\sqrt{5})^2$, solving this gives $x=6$ or $x=-6$ (which is inconsistent with the problem, thus discarded),\n\nMoreover, since the slope ratio of the uphill slope CD is $i=1\\colon 1$,\n\n$\\therefore$ $CF=DF=6$,\n\n$\\therefore$ In $\\triangle CFD$, according to the Pythagorean theorem, we have:\n\n$CD=\\sqrt{CF^2+DF^2}=\\sqrt{6^2+6^2}=\\boxed{6\\sqrt{2}}$",
    "image_path": "dataset/mm_math/images/MM_Math/53148883.png"
  },
  {
    "solution": "\\boxed{\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the diagonals $AC$ and $BD$ of the quadrilateral $ABCD$ intersect at point $O$, with $AC \\perp BD$. Let $E$ and $F$ be the midpoints of $AB$ and $CD$, respectively. If $AC = BD = 2$, what is the length of $EF$?\n",
    "completion": "\\textbf{Solution}: Let G be the midpoint of BC, and H be the midpoint of AD. Connect EG, GF, FH, and HE. \\\\\n$\\because$ E and G are the midpoints of AB and BC respectively, $AC=2$ \\\\\n$\\therefore EG=\\frac{1}{2}AC=1$, $EG\\parallel AC$,\\\\\nSimilarly: $FH=\\frac{1}{2}AC$, $FH\\parallel AC$, $GF\\parallel BD$, $GF=\\frac{1}{2}BD=1$,\\\\\n$\\therefore$ GE$\\parallel$FH, GF$\\parallel$EH\\\\\n$\\therefore$ Quadrilateral EGFH is a parallelogram,\\\\\n$\\because AC=BD$,\\\\\n$\\therefore GE=GF$,\\\\\n$\\therefore$ parallelogram EGFH is a rhombus,\\\\\n$\\because AC\\perp BD$, $EG\\parallel AC$, $GF\\parallel BD$,\\\\\n$\\therefore EG\\perp GF$,\\\\\n$\\therefore$ rhombus EGFH is a square,\\\\\n$\\therefore EF=\\sqrt{2}EG=\\boxed{\\sqrt{2}}$",
    "image_path": "dataset/mm_math/images/MM_Math/52284278.png"
  },
  {
    "solution": "\\boxed{540^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, what is the sum of $\\angle 1 + \\angle 2 + \\angle 3 + \\angle 4 + \\angle 5 + \\angle 6 + \\angle 7$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, \\\\\nby the internal angle sum of a quadrilateral, we have $\\angle 2+\\angle 3+\\angle 5+\\angle 8=360^\\circ$, $\\angle 6+\\angle 7+\\angle 9+\\angle 10=360^\\circ$, \\\\\n$\\therefore \\angle 2+\\angle 3+\\angle 5+\\angle 8+\\angle 6+\\angle 7+\\angle 9+\\angle 10=720^\\circ$, \\\\\n$\\because \\angle 8+\\angle 9=180^\\circ$, $\\angle 10=\\angle 1+\\angle 4$, \\\\\n$\\therefore \\angle 1+\\angle 2+\\angle 3+\\angle 5+\\angle 8+\\angle 6+\\angle 7=720^\\circ-180^\\circ=\\boxed{540^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52550870.png"
  },
  {
    "solution": "\\boxed{9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points $A\\left(-6, 2\\right)$ and $B\\left(n, -4\\right)$ are two intersections of the linear function $y=kx+b$ and the inverse proportion function $y=\\frac{m}{x}$. What is the area of $\\triangle AOB$?\n",
    "completion": "\\textbf{Solution:} Solving by substituting $y=0$ into $y=-\\frac{2}{3}x-2$, we get $0=-\\frac{2}{3}x-2$,\\\\\nwhich yields: $x=-3$,\\\\\nhence, the coordinates of point C are ($-3$, $0$),\\\\\nthus, $S_{\\triangle AOB}=S_{\\triangle AOC}+S_{\\triangle BOC}=\\frac{1}{2} \\times OC \\times y_A + \\frac{1}{2} \\times OC \\times (-y_B) = \\frac{1}{2} \\times OC \\times (y_A-y_B) = \\frac{1}{2} \\times 3 \\times (2+4) = \\boxed{9}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52602496_140.png"
  },
  {
    "solution": "\\boxed{2.5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In rhombus $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$. $E$ is the midpoint of $AB$, $AC=8\\,\\text{cm}$, and $BD=6\\,\\text{cm}$. What is the length of $OE$?\n",
    "completion": "\\textbf{Solution:} Given that in the rhombus ABCD, AC = 6cm, BD = 8cm,\\\\\nhence AC $\\perp$ BD, OB = $\\frac{1}{2}$BD = 4cm, OC = $\\frac{1}{2}$AC = 3cm,\\\\\ntherefore BC = $\\sqrt{OC^{2}+OB^{2}}$ = 5cm,\\\\\nsince E is the midpoint of BC,\\\\\nthus OE = $\\frac{1}{2}$BC = $\\boxed{2.5}$cm.",
    "image_path": "dataset/mm_math/images/MM_Math/52387015.png"
  },
  {
    "solution": "\\boxed{\\frac{5}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC$, the angle bisectors of $\\angle BAC$ and $\\angle ABC$ intersect at point $E$, and the extended line of $AE$ intersects the circumcircle of $\\triangle ABC$ at point $D$, connect $BD$. If $AB=5$, $BC=8$, and $\\angle BAD=\\angle DBC$; points $B$, $E$, $C$ are on the same circle centered at point $D$; what is the distance between the incenter and the circumcenter of $\\triangle ABC$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram:\\\\\n$\\because BD=DC, \\angle ABD=\\angle ACD=90^\\circ, AD=AD$,\\\\\n$\\therefore Rt\\triangle ABD\\cong Rt\\triangle ACD(HL)$,\\\\\n$\\therefore AB=AC$,\\\\\n$\\because AH=AH, \\angle BAH=\\angle CAH$,\\\\\n$\\therefore \\triangle ABH\\cong \\triangle ACH(SAS)$,\\\\\n$\\therefore BH=CH$,\\\\\n$\\therefore BH=\\frac{1}{2}BC=4$,\\\\\n$\\therefore \\angle AHB=\\angle AHC=90^\\circ$,\\\\\n$\\therefore AD\\perp BC$,\\\\\nIn $Rt\\triangle ABH$, $AH=3$,\\\\\nIn $Rt\\triangle BHO$, let $BO=x$, $OH=x-3$,\\\\\nThen $BO^2=BH^2+OH^2$,\\\\\ni.e., $x^2=16+(x-3)^2$,\\\\\nSolving yields: $x=\\frac{25}{6}$,\\\\\ni.e., $BO=\\frac{25}{6}$,\\\\\n$\\because AD$ is the diameter,\\\\\n$\\therefore \\angle ABD=90^\\circ$,\\\\\nIn $Rt\\triangle ABD$,\\\\\n$BD=\\sqrt{AD^2-AB^2}=\\frac{20}{3}$,\\\\\n$\\therefore DE=\\frac{20}{3}$,\\\\\n$\\therefore OE=\\frac{20}{3}-\\frac{25}{6}=\\frac{5}{2}$,\\\\\n$\\because E$ is the intersection of the angle bisectors of $\\triangle ABC$,\\\\\n$\\therefore E$ is the incenter,\\\\\n$\\therefore OE$ is the distance between the incenter and circumcenter of $\\triangle ABC$,\\\\\n$\\therefore The distance between the incenter and circumcenter of \\triangle ABC$ is $\\boxed{\\frac{5}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51445755_113.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, inside the circle $ \\odot O$, $ \\overset{\\frown}{AB}=\\overset{\\frown}{AC}$, and $ \\angle C=75^\\circ$, what is the degree measure of $ \\angle A$?\n",
    "completion": "\\textbf{Solution:} In $\\odot O$, $\\overset{\\frown}{AB}=\\overset{\\frown}{AC}$, \\\\\n$\\therefore \\angle B=\\angle C=75^\\circ$, \\\\\n$\\because \\angle A+\\angle B+\\angle C=180^\\circ$, \\\\\n$\\therefore \\angle A=180^\\circ-\\angle B-\\angle C=\\boxed{30^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/52720184.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{2}}{2}a}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the figure, in $\\triangle ABC$, $\\angle BAC=90^\\circ$, $AB=AC=a$, $AD$ is the altitude of $\\triangle ABC$, what is the length of $AD$?\n",
    "completion": "\\textbf{Solution:} Given that in $\\triangle ABC$, $\\angle BAC=90^\\circ$ and $AB=AC=a$, \\\\\n$\\therefore$ BC=$\\sqrt{AB^2+AC^2}=\\sqrt{a^2+a^2}=\\sqrt{2}a$,\\\\\nsince $AD$ is the height of $\\triangle ABC$,\\\\\n$\\therefore$ the area $S_{\\triangle ABC}=\\frac{1}{2}\\times AB\\times AC=\\frac{1}{2}\\times BC\\times AD$,\\\\\nthus $\\frac{1}{2}\\times a\\times a=\\frac{1}{2}\\times \\sqrt{2}a\\times AD$,\\\\\nwe find $AD=\\boxed{\\frac{\\sqrt{2}}{2}a}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53010582.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point C is the midpoint of segment AB, point D is the midpoint of segment AC, and DB equals 9. What is the length of AB?\n",
    "completion": "\\textbf{Solution:} Since point C is the midpoint of AB,\\\\\nit follows that AC=BC,\\\\\nSince point D is the midpoint of the segment AC,\\\\\nit follows that AC=BC=2AD=2DC\\\\\nSince DB=9,\\\\\nit follows that DC+BC=9,\\\\\nit follows that DC+2DC=9,\\\\\nSolving this gives: DC=AD=3,\\\\\nTherefore, AB=AD+BD=3+9=\\boxed{12}.",
    "image_path": "dataset/mm_math/images/MM_Math/51317300.png"
  },
  {
    "solution": "\\boxed{35^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that A, B, and C are three points on the circle $\\odot$O. If $\\angle AOB=70^\\circ$, what is the degree measure of $\\angle ACB$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle AOB=70^\\circ$,\\\\\nthen $\\angle ACB=\\frac{1}{2}\\angle AOB=35^\\circ$,\\\\\ntherefore, the answer is: $\\boxed{35^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52905954.png"
  },
  {
    "solution": "\\boxed{16}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, $E$ is the midpoint of line segment $AB$, point $C$ is on line segment $AB$, and $F$ is the midpoint of $AC$. If $EF = 5\\,cm$ and $AC = 6\\,cm$, find the lengths of line segments $CE$ and $AB$.\n",
    "completion": "Solution: Since F is the midpoint of AC and $AC=6\\,cm$, \\\\\nthus $CF=AF=\\frac{1}{2}AC=3\\,cm$, \\\\\nSince $EF=5\\,cm$, \\\\\nthus $CE=EF-CF=5-3=\\boxed{2}\\,(cm)$, \\\\\nthus $AE=AC+CE=6+2=8\\,(cm)$, \\\\\nSince E is the midpoint of the line segment AB, \\\\\nthus $AB=2AE=\\boxed{16}\\,cm$.",
    "image_path": "dataset/mm_math/images/MM_Math/52796148_175.png"
  },
  {
    "solution": "\\boxed{(2^{2021}-1,2^{2020})}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Squares $A_{1}B_{1}C_{1}O$, $A_{2}B_{2}C_{2}C_{1}$, $A_{3}B_{3}C_{3}C_{2}$, ..., are placed as shown in the diagram. Points $A_{1}$, $A_{2}$, $A_{3}$, ..., and points $C_{1}$, $C_{2}$, $C_{3}$, ..., are located on the line $y=kx+b$ $(k>0)$ and the $x$-axis, respectively. Given that point $B_{1}$ is at $(1, 1)$ and $B_{2}$ is at $(3, 2)$, what are the coordinates of point $B_{2021}$?\n",
    "completion": "\\textbf{Solution:} Since the coordinates of point B$_1$ are (1,1), and the coordinates of point B$_2$ are (3,2),\\\\\ntherefore, the side length of square A$_1$B$_1$C$_1$O$_1$ is 1, and the side length of square A$_2$B$_2$C$_2$C$_1$ is 2,\\\\\nthus, the coordinates of A$_1$ are (0,1), and the coordinates of A$_2$ are: (1,2),\\\\\nsubstituting into y=kx+b yields: $\\left\\{\\begin{array}{l}b=1 \\\\ k+b=2\\end{array}\\right.$,\\\\\nsolving gives: $\\left\\{\\begin{array}{l}k=1 \\\\ b=1\\end{array}\\right.$,\\\\\nhence, the equation of the line is: y=x+1.\\\\\nSince A$_1$B$_1$=1, and the coordinates of point B$_2$ are (3,2),\\\\\nthus, the coordinates of point A$_3$ are (3,4),\\\\\ntherefore, A$_3$C$_2$=A$_3$B$_3$=B$_3$C$_3$=4,\\\\\ntherefore, the coordinates of point B$_3$ are (7,4),\\\\\nthus, the y-coordinate of B$_1$ is: 1=$2^0$, the x-coordinate of B$_1$ is: 1=$2^1$−1,\\\\\nthus, the y-coordinate of B$_2$ is: 2=$2^1$, the x-coordinate of B$_2$ is: 3=$2^2$−1,\\\\\nthus, the y-coordinate of B$_3$ is: 4=$2^2$, the x-coordinate of B$_3$ is: 7=$2^3$−1,\\\\\nthus, the y-coordinate of B$_n$ is: $2^{n-1}$, the x-coordinate is: $2^{n}$−1,\\\\\ntherefore, the coordinates of B$_n$ are: ($2^{n}$−1,$2^{n-1}$).\\\\\ntherefore, the coordinates of B$_{2021}$ are: $\\boxed{(2^{2021}-1,2^{2020})}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50482068.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{1}{2}x+3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given the line $y=kx+b$ passes through points $A(6, 0)$ and $B(2, 2)$. Find the analytical expression of line $AB$.\n",
    "completion": "\\textbf{Solution}: Since the line $y=kx+b$ passes through points $A(6, 0)$ and $B(2, 2)$, \\\\\nthen $\\left\\{\\begin{array}{l}\n6k+b=0 \\\\\n2k+b=2\n\\end{array}\\right.$. Solving this system of equations, we find $\\left\\{\\begin{array}{l}\nk=-\\frac{1}{2} \\\\\nb=3\n\\end{array}\\right.$. \\\\\nTherefore, the equation of line $AB$ is: \\boxed{y=-\\frac{1}{2}x+3}.",
    "image_path": "dataset/mm_math/images/MM_Math/52056013_215.png"
  },
  {
    "solution": "\\boxed{\\frac{16}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the rectangle OABC has its vertex O at the origin. Side OC is on the positive x-axis, and side OA is on the positive y-axis, with OA = 3 and AB = 4. The graph of the inverse proportion function $y=\\frac{k}{x}$ (where $k>0$) intersects the sides AB and BC of the rectangle at points D and E, respectively, and BD = 2AD. Connect OD, OE, and DE. Find the area of $\\triangle DOE$.\n",
    "completion": "\\textbf{Solution:} From the given problem, we have \\\\\n$\\because AD=\\frac{4}{3}$, $OA=3$, \\\\\n$\\therefore S_{\\triangle AOD}=\\frac{1}{2}AD\\cdot AO=\\frac{1}{2}\\times \\frac{4}{3}\\times 3=2$. \\\\\n$\\because k=4$, \\\\\n$\\therefore$ the inverse proportion function can be expressed as $y=\\frac{4}{x}$. \\\\\n$\\because$ in rectangle ABCD, $BC=OA=3$, $OC=AB=4$, \\\\\nalso, $\\because$ point E lies on the graph of the inverse proportion function $y=\\frac{4}{x}$, \\\\\n$\\therefore E(4, 1)$. \\\\\n$\\therefore CE=1$, \\\\\n$\\therefore BE=2$, \\\\\n$\\therefore S_{\\triangle OCE}=\\frac{1}{2}OC\\cdot CE=2$. \\\\\n$\\because AB=4$, $AD=\\frac{4}{3}$. \\\\\n$\\therefore BD=\\frac{8}{3}$, \\\\\n$\\therefore S_{\\triangle BDE}=\\frac{1}{2}BD\\cdot BE=\\frac{8}{3}$, \\\\\n$\\because S_{\\text{rectangle} AOCB}=AO\\cdot OC=3\\times 4=12$, \\\\\n$\\therefore S_{\\triangle DOE}=S_{\\text{rectangle} AOCB}-S_{\\triangle AOD}-S_{\\triangle OCE}-S_{\\triangle BDE}=12-2-2-\\frac{8}{3}=\\boxed{\\frac{16}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51402967_35.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, $AB\\parallel CD$, the diagonals $AC$ and $BD$ intersect at point $O$, and $AO=CO$. It is known that $\\triangle ABO$ is an equilateral triangle. If quadrilateral $ABCD$ is a rectangle and $AB=2$, what is the length of $BC$?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABO$ is an equilateral triangle and $AB=2$, it follows that $OA=AB=2$. Given that quadrilateral $ABCD$ is a rectangle, we have $\\angle ABC=90^\\circ$, and thus $AC=2OA=4$. Therefore, in right triangle $\\triangle ABC$, we have $BC=\\sqrt{AC^{2}-AB^{2}}=\\sqrt{4^{2}-2^{2}}=2\\sqrt{3}$. Hence, the length of $BC$ is $\\boxed{2\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53360003_123.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $AB=4$, $BC=8$. The rectangle $ABCD$ is folded along $EF$ so that point $C$ coincides with point $A$. What is the length of the crease $EF$?\n",
    "completion": "\\textbf{Solution:}\\\\\nLet $BE = x$, then $CE = BC - BE = 8 - x$, \\\\\n$\\because$ After folding along $EF$, point $C$ coincides with point $A$, \\\\\n$\\therefore$ $AE = CE = 8 - x$, \\\\\nIn $\\triangle ABE$, $AB^{2} + BE^{2} = AE^{2}$, that is $4^{2} + x^{2} = (8 - x)^{2}$, \\\\\nSolving it, we get: $x = 3$, \\\\\n$\\therefore$ $AE = 8 - 3 = 5$, \\\\\nDue to the property of folding, $\\angle AEF = \\angle CEF$, \\\\\n$\\because$ $AD \\parallel BC$, \\\\\n$\\therefore$ $\\angle AFE = \\angle CEF$, \\\\\n$\\therefore$ $\\angle AEF = \\angle AFE$, \\\\\n$\\therefore$ $AE = AF = 5$, \\\\\nDraw a line from point $E$ to $H$ perpendicular to $AD$ at $H$, then quadrilateral $ABEH$ is a rectangle, \\\\\n$\\therefore$ $EH = AB = 4$, $AH = BE = 3$, \\\\\n$\\therefore$ $FH = AF - AH = 5 - 3 = 2$, \\\\\nIn $\\triangle EFH$, $EF = \\sqrt{4^{2} + 2^{2}} = \\boxed{2\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52387157.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, the slope ratio of the cross-section of the river embankment $BC: AC$ is $1: \\sqrt{3}$, with $AC = 6\\,m$. What is the length of the slope surface $AB$?\n",
    "completion": "\\textbf{Solution:} Given the slope ratio $BC:AC$ is $1:\\sqrt{3}$, and $AC=6$m, \\\\\nit follows that $\\frac{BC}{6} = \\frac{1}{\\sqrt{3}}$, \\\\\nsolving this gives $BC=2\\sqrt{3}$ (m), \\\\\nby the Pythagorean theorem, $AB=\\sqrt{AC^{2}+BC^{2}} =\\boxed{4\\sqrt{3}}$ (m).",
    "image_path": "dataset/mm_math/images/MM_Math/50934146.png"
  },
  {
    "solution": "\\boxed{1\\colon 3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, points D and E are the midpoints of AB and AC, respectively. What is the ratio of $S_{\\triangle ADE}$ to $S_{\\text{quadrilateral} BCED}$?\n",
    "completion": "\\textbf{Solution:} Since points D and E are the midpoints of AB and AC respectively,\\\\\nit follows that DE$\\parallel$BC,\\\\\nwhich implies $\\triangle ADE \\sim \\triangle ABC$,\\\\\nthus, DE$\\colon$ BC=1$\\colon$ 2,\\\\\nleading to $S_{\\triangle ADE}\\colon S_{\\triangle ABC}=1\\colon 4$,\\\\\nhence, $S_{\\triangle ADE}\\colon S_{\\text{quadrilateral} BCED}=1\\colon 3$.\\\\\nTherefore, the answer is: $S_{\\triangle ADE} \\colon S_{\\text{quadrilateral} BCED}=\\boxed{1\\colon 3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52963742.png"
  },
  {
    "solution": "\\boxed{1.5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, line $y_{1} = -x + 3$ and line $y_{2} = kx - 2$ intersect the $y$-axis at points $A$ and $B$, respectively, and intersect at point $C(2, m)$. What are the values of $m$ and $k$?\n",
    "completion": "\\textbf{Solution:} Substitute $C(2, m)$ into $y_1 = -x + 3$, \\\\\nwe get: $m = -2 + 3 = 1$, \\\\\n$\\therefore$ the coordinates of point $C$ are $(2, 1)$, \\\\\nsubstitute $C(2, 1)$ into $y_2 = kx - 2$, \\\\\nwe get: $2k - 2 = 1$, \\\\\nsolving for $k$, we obtain $k = \\boxed{1.5}$. \\\\\nIn conclusion, $m = \\boxed{1}$ and $k = \\boxed{1.5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51968557_184.png"
  },
  {
    "solution": "\\boxed{18}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, $DA\\perp AB$, $DA=6\\,\\text{cm}$, $\\angle B+\\angle C=150^\\circ$, point $A$ is exactly the midpoint of $EB$, and $P$, $Q$ are moving points on the line segments $CE$, $BE$ respectively. What is the minimum value of $BP+PQ$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, construct point $F$ as the symmetric point of $B$ with respect to $CE$. Draw $BF$ and $EF$. Then, $EB=EF$,\n\nsince $\\angle B + \\angle C = 150^\\circ$, it follows that $\\angle BEC = 180^\\circ - (\\angle B + \\angle C) = 30^\\circ$. Since point $B$ is symmetric to point $F$ with respect to $EC$, it follows that $\\angle BEC = \\angle FEC = 30^\\circ$, hence $\\angle BEF = 60^\\circ$, therefore $\\triangle BEF$ is an equilateral triangle. Draw $BP$, $PF$, $PQ$, then $BP=FP$, therefore $BP + QP = FP + PQ$. Thus, when $F$, $P$, and $Q$ are on the same line and $FQ \\perp EB$, then the minimum value of $BP+PQ$ is the length of $FQ$. At this moment, $Q$ coincides with the midpoint of $EB$, hence coincides with point $A$. Since $DA \\perp AB$ and $DA=6$cm, it follows that $AE= 6\\sqrt{3}$cm. In $\\triangle QEF$, $FQ = \\sqrt{3}AE = \\sqrt{3} \\times 6\\sqrt{3} = \\boxed{18}$cm. Therefore, the minimum value of $BP+PQ$ is $18$cm.",
    "image_path": "dataset/mm_math/images/MM_Math/55498626.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $E$ is a point on $AD$, connecting $AC$ and $BE$ intersecting at point $F$, and $FG \\perp CD$ at $G$. If $AE=DE=3$, find the length of $FG$.\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral ABCD is a rectangle, with AE=DE=3, it follows that AD$\\parallel$BC, and BC=AD=6. Hence $\\triangle$AEF$\\sim$$\\triangle$CBF, implying that $\\frac{CF}{AF}=\\frac{BC}{AE}=2$. Since FG$\\parallel$AD, $\\triangle$CFG$\\sim$$\\triangle$CAD, leading to $\\frac{FG}{AD}=\\frac{CF}{AC}=\\frac{2}{3}$, thus FG=\\boxed{4}.",
    "image_path": "dataset/mm_math/images/MM_Math/52935315_169.png"
  },
  {
    "solution": "\\boxed{\\frac{2\\sqrt{3}\\pi }{9}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the hexagon $ABCDEF$ is a regular hexagon inscribed in circle $\\odot O$. Let the area of the circle $\\odot O$ be $S_1$ and the area of the hexagon $ABCDEF$ be $S_2$. Find the value of $\\frac{S_1}{S_2}$.\n",
    "completion": "\\textbf{Solution:} Draw OG $\\perp$ DE at G, and connect OE,\\\\\nLet the radius of circle O be r,\\\\\n$\\therefore$ EF=BC=ED=r, AD=2r,\\\\\nIn the regular hexagon ABCDEF,\\\\\n$\\angle$OED=$\\angle$ODE=60$^\\circ$,\\\\\n$\\therefore$ $\\angle$EOG=30$^\\circ$,\\\\\n$\\therefore$ EG=$ \\frac{1}{2}$r,\\\\\n$\\therefore$ OG=$ \\sqrt{OE^{2}-EG^{2}}$=$ \\frac{\\sqrt{3}}{2}$r,\\\\\n$\\therefore$ The area of regular hexagon ABCDEF=$ 6\\times \\frac{1}{2}\\times r\\times \\frac{\\sqrt{3}}{2}r$=$ \\frac{3\\sqrt{3}}{2}r^{2}$,\\\\\nThe area of circle O=$ \\pi r^{2}$,\\\\\n$\\therefore$ $ \\frac{{S}_{1}}{{S}_{2}}$=$ \\frac{\\pi r^{2}}{\\frac{3\\sqrt{3}}{2}r^{2}}$=$ \\boxed{\\frac{2\\sqrt{3}\\pi }{9}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51366887_127.png"
  },
  {
    "solution": "\\boxed{\\frac{45}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $AB = 3\\,cm$, $BC = 6\\,cm$. Point $P$ starts moving from point $D$ towards point $A$ and stops upon reaching $A$; simultaneously, point $Q$ starts moving from point $B$ towards point $C$ and stops upon reaching $C$. The speeds of points $P$ and $Q$ are $1\\,cm/s$ each. Connect $PQ$, $AQ$, $CP$. Let the time taken for points $P$ and $Q$ to move be $t$ seconds. When $t=\\frac{9}{4}$ seconds, the quadrilateral $AQCP$ forms a rhombus; calculate the perimeter and area of the rhombus $AQCP$?\n",
    "completion": "\\textbf{Solution:} When $t=\\frac{9}{4}$, $AQ=\\frac{15}{4},\\ CQ=\\frac{15}{4}$, \\\\\nthen the perimeter is: $4AQ=4\\times \\frac{15}{4}=\\boxed{15}\\ (\\text{cm})$, \\\\\nand the area is: $CQ\\cdot AB=\\frac{15}{4}\\times 3=\\boxed{\\frac{45}{4}}\\ (\\text{cm}^2)$.",
    "image_path": "dataset/mm_math/images/MM_Math/53313476_112.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the rhombus $ABCD$, diagonal $AC=8$, $BD=10$, then what is the area of $\\triangle ADO$?\n",
    "completion": "\\textbf{Solution:} In diamond $ABCD$, the diagonals have lengths $AC=8$ and $BD=10$,\\\\\n$\\therefore$OA=4, OD=5, and $AC\\perp BD$,\\\\\n$\\therefore$the area of $\\triangle ADO$ is $\\frac{1}{2}OA\\cdot OD=\\frac{1}{2}\\times 4\\times 5=\\boxed{10}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52250704.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the figure, in $\\triangle OAB$, $\\angle OAB=90^\\circ$, point A is on the x-axis, the graph of the inverse proportion function $y=\\frac{k}{x}$ ($x>0$) passes through the midpoint D of hypotenuse OB and intersects AB at point C. If the area of $\\triangle OBC$ is 3, what is the value of $k$?\n",
    "completion": "\\textbf{Solution:} Draw $DE \\perp OA$ at point $E$, then $S_{\\triangle ODE}=S_{\\triangle OAC}=\\frac{1}{2}|k|$,\n\n$\\because$ D is the midpoint of OB,\n\n$\\therefore$ OD=BD=$\\frac{1}{2}$OB,\n\n$\\because$ DE$\\perp$OA and $\\angle$OAB$=90^\\circ$,\n\n$\\therefore$ DE$\\parallel$AB,\n\n$\\therefore$ $\\triangle ODE \\sim \\triangle OBA$,\n\n$\\therefore$ $\\frac{S_{\\triangle ODE}}{S_{\\triangle OBA}}=\\left(\\frac{OD}{OB}\\right)^2=\\frac{1}{4}$,\n\n$\\therefore$ $S_{\\triangle OAB}=4S_{\\triangle ODE}=2|k|$,\n\n$\\therefore$ $S_{\\triangle OBC}=3=S_{\\triangle OAB}-S_{\\triangle OAC}=\\frac{3}{2}|k|$,\n\nAgain, $\\because k>0$,\n\n$\\therefore k=\\boxed{2}$,",
    "image_path": "dataset/mm_math/images/MM_Math/52601930.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $O$ is on line $AB$, rays $OC$ and $OD$ are on the same side of line $AB$ (where $0^\\circ<\\angle AOC<90^\\circ$, $0^\\circ<\\angle BOD<90^\\circ$), ray $OE$ bisects $\\angle AOC$, and ray $OF$ bisects $\\angle BOD$. If $\\angle EOD$ and $\\angle COF$ are complementary, what is the angle of $\\angle COD$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle EOD$ and $\\angle COF$ are complementary,\\\\\nit follows that $\\angle EOD+\\angle COF=180^\\circ$,\\\\\nthus $\\angle EOF+\\angle COD=180^\\circ$,\\\\\nand since $\\angle EOF+\\angle AOE+\\angle BOF=180^\\circ$,\\\\\nit results in $\\angle COD=\\angle AOE+\\angle BOF$,\\\\\ngiven that ray OE bisects $\\angle AOC$ and ray OF bisects $\\angle BOD$,\\\\\nwe have $\\angle AOE=\\angle COE$, $\\angle BOF=\\angle DOF$,\\\\\nthus $\\angle COE+\\angle DOF=\\angle COD$,\\\\\nwhich implies $\\angle COD=180^\\circ\\div 3=\\boxed{60^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52773539.png"
  },
  {
    "solution": "\\boxed{\\frac{4}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the side length of square $ABCD$ is $4$, $E$ is the midpoint of side $AD$, connect $BE$, fold $\\triangle ABE$ along line $BE$ to $\\triangle FBE$, extend $EF$ to intersect $CD$ at point $G$, what is the length of $CG$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, construct line $BG$. Because quadrilateral $ABCD$ is a square, therefore $AB=BC=4$, $\\angle A=\\angle ABC=\\angle C=90^\\circ$. From folding, $AB=BF=BC=4$, $AE=EF=\\frac{1}{2}AD=2=DE$, $\\angle A=\\angle BFE=90^\\circ=\\angle C$. Because $\\angle BFE+\\angle BFG=180^\\circ$, therefore $\\angle C=\\angle BFG=90^\\circ$. In $\\triangle BFG$ and $\\triangle BCG$, we have $\\left\\{\\begin{array}{c}BF=BC\\\\ BG=BG\\end{array}\\right.$, therefore $\\triangle BFG\\cong \\triangle BCG$ (HL), therefore $FG=CG$. Let $CG=x$, then $DG=4-x$, $EG=2+x$. In $\\triangle DEG$, by the Pythagorean theorem, $EG^{2}=DE^{2}+DG^{2}$, solving $(2+x)^{2}=2^{2}+(4-x)^{2}$ yields $x=\\boxed{\\frac{4}{3}}$, which means the length of $CG$ is $\\frac{4}{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52204028.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the square grid paper shown in the figure, each small quadrilateral is an identical square. Points $A$, $B$, $C$, and $D$ are all located at the grid points, and $AB$ intersects with $CD$ at point $O$. What is the value of $\\tan\\angle AOC$?\n",
    "completion": "\\textbf{Solution:} Let $BE$ be connected. We have $OE= \\sqrt{1^2+1^2}=\\sqrt{2}$ and $BE= \\sqrt{3^2+3^2}=3\\sqrt{2}$, and $BO= \\sqrt{4^2+2^2}=2\\sqrt{5}$.\\\\\nTherefore, $BO^2=20=EO^2+EB^2=2+18=20$.\\\\\nHence, $\\triangle OBE$ is a right-angled triangle.\\\\\nTherefore, $\\tan\\angle AOC=\\tan\\angle BOE= \\frac{BE}{EO}=\\frac{3\\sqrt{2}}{\\sqrt{2}}=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51545328.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $AB\\parallel CD\\parallel EF$. If $AC=6$, $CE=3$, and $DF=2$, what is the length of $BD$?\n",
    "completion": "\\textbf{Solution:} Since $AB\\parallel CD\\parallel EF$, \\\\\ntherefore $\\frac{AC}{CE}=\\frac{BD}{DF}$, \\\\\nsince $AC=6$, $CE=3$, $DF=2$, \\\\\ntherefore $\\frac{6}{3}=\\frac{BD}{2}$, \\\\\nsolving, we get: $BD=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52513828.png"
  },
  {
    "solution": "\\boxed{\\frac{24}{7}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $A(0, 2)$, $B(3, 3)$, $C(7, -1)$, and quadrilateral $OBCD$ is a rectangle, with $BD$ intersecting the $x$ axis at point $E$. Find the length of the line segment $OE$.\n",
    "completion": "\\textbf{Solution:} Connect OC to intersect BD at point F.\\\\\nSince quadrilateral OBCD is a rectangle, with point C's coordinates at (7, -1) and point O's coordinates at (0, 0), and point F being the midpoint of OC, therefore the coordinates of point F are $\\left(\\frac{7}{2}, -\\frac{1}{2}\\right)$. Let the equation of line BD be $y=kx+b$. Substituting (3, 3) and $\\left(\\frac{7}{2}, -\\frac{1}{2}\\right)$ into $y=kx+b$ yields $\\left\\{\\begin{array}{l}3k+b=3 \\\\ \\frac{7}{2}k+b=-\\frac{1}{2}\\end{array}\\right.$. Solving this system of equations gives $\\left\\{\\begin{array}{l}k=-7 \\\\ b=24\\end{array}\\right.$, therefore $y=-7x+24$. Substituting $y=0$ into $y=-7x+24$ gives $0=-7x+24$, solving for $x$ yields $x=\\frac{24}{7}$. Therefore, the coordinates of point E are $\\left(\\frac{24}{7}, 0\\right)$, thus $OE=\\boxed{\\frac{24}{7}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52691041_54.png"
  },
  {
    "solution": "\\boxed{92^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $\\angle C=\\angle D$, $OC=OD$. If $\\angle O=72^\\circ$, $\\angle C=20^\\circ$, and $\\triangle OAD \\cong \\triangle OBC$; what is the degree measure of $\\angle DAC$?\n",
    "completion": "\\textbf{Solution:} Given that $ \\triangle OAD \\cong \\triangle OBC$,\\\\\n$\\therefore \\angle C=\\angle D=20^\\circ$,\\\\\n$\\therefore \\angle DAC=\\angle O+\\angle D=20^\\circ+72^\\circ=\\boxed{92^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52551303_90.png"
  },
  {
    "solution": "\\boxed{1.2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AC=3$, $AB=4$, and $BC=5$. $P$ is a moving point on $BC$, $PG\\perp AC$ at point $G$, $PH\\perp AB$ at point $H$, and $M$ is the midpoint of $GH$. What is the minimum value of $PM$ during the motion of $P$?\n",
    "completion": "\\textbf{Solution}: Given that $AC=3$, $AB=4$, and $BC=5$,\\\\\nthus we have $AC^{2}=9$, $AB^{2}=16$, $BC^{2}=25$,\\\\\nwhich leads us to $AC^{2}+AB^{2}=BC^{2}$,\\\\\ntherefore $\\angle A=90^\\circ$.\\\\\nSince $PG\\perp AC$ and $PH\\perp AB$,\\\\\nwe get $\\angle AGP=\\angle AHP=90^\\circ$,\\\\\nhence quadrilateral AGPH is a rectangle.\\\\\nBy connecting AP,\\\\\nthus $GH=AP$\\\\\nGiven that when $AP\\perp BC$, AP is at its minimum,\\\\\nthus $3 \\times 4 = 5AP$,\\\\\ntherefore $AP= \\frac{12}{5}$, \\\\\nthus the minimum value of PM is $\\boxed{1.2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52302532.png"
  },
  {
    "solution": "\\boxed{x>-2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Based on the graph, it is known that the functions $y=3x+b$ and $y=ax-3$ intersect at point $P(-2,-5)$. Then, according to the graph, what is the solution set for the inequality $3x+b>ax-3$?\n",
    "completion": "\\textbf{Solution:} Since the graphs of the functions $y=3x+b$ and $y=ax-3$ intersect at point $P(-2,-5)$,\\\\\nit follows from the graphical representation that the solution set of the inequality $3x+b>ax-3$ is $x>-2$,\\\\\ntherefore, the answer is: $\\boxed{x>-2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51983467.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in circle $O$, $AB$ is the diameter, and the length of chord $AC$ is 5 cm. Point $D$ is on the circle, and $\\angle ADC=30^\\circ$. What is the radius of circle $O$ in cm?\n",
    "completion": "\\textbf{Solution:} Connect \\textit{BC} as shown in the figure:\\\\\n$\\because$ $\\angle ADC=30^\\circ$,\\\\\n$\\therefore$ $\\angle ABC=\\angle ADC=30^\\circ$,\\\\\n$\\because$ \\textit{AB} is the diameter,\\\\\n$\\therefore$ $\\angle ACB=90^\\circ$,\\\\\n$\\because$ \\textit{AC} $=5\\text{cm}$,\\\\\n$\\therefore$ \\textit{AB} $=2AC=10\\text{cm}$,\\\\\n$\\therefore$ the radius of $\\odot O$ is $\\boxed{5}\\text{cm}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51261993.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, it is known that points D, E, and F are the midpoints of sides BC, AD, and CE, respectively, and the area of $\\triangle ABC$ is 4. What is the area of $\\triangle BEF$?\n",
    "completion": "\\textbf{Solution:} Given that point D is the midpoint of BC, and the area of $\\triangle ABC$ is 4,\\\\\nit follows that the area of $\\triangle ABD = \\triangle ADC = \\frac{1}{2}\\times$ the area of $\\triangle ABC = \\frac{1}{2}\\times4=2$,\\\\\nGiven that point E is the midpoint of AD,\\\\\nit follows that the area of $\\triangle BDE = \\frac{1}{2}\\times$ the area of $\\triangle ABD = \\frac{1}{2}\\times2=1$, and the area of $\\triangle CDE = \\frac{1}{2}\\times$ the area of $\\triangle ADC = \\frac{1}{2}\\times2=1$,\\\\\ntherefore, the area of $\\triangle BEC = $ the area of $\\triangle BED + \\triangle CDE = 2$,\\\\\nGiven that point F is the midpoint of CE,\\\\\nit follows that the area of $\\triangle BEF = \\frac{1}{2}\\times$the area of $\\triangle BEC = \\frac{1}{2}\\times2 = \\boxed{1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53009848.png"
  },
  {
    "solution": "\\boxed{-x^2+2x+3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the parabola $y = -x^2 + bx + c$ passes through points A$(-1, 0)$ and B$(3, 0)$. What is the analytic equation of the parabola?\n",
    "completion": "\\textbf{Solution:} Let's substitute points A ($-1$, $0$) and B ($3$, $0$) into $y=-x^2+bx+c$,\\\\\n$\\therefore \\begin{cases}0=-1^2-b+c\\\\ 0=-9+3b+c\\end{cases}$,\\\\\nSolving, we get $\\begin{cases}b=2\\\\ c=3\\end{cases}$,\\\\\n$\\therefore$ the equation of the parabola is: $y=-x^2+2x+3=\\boxed{-x^2+2x+3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52977350_187.png"
  },
  {
    "solution": "\\boxed{y=\\frac{1}{4}x^2-\\frac{3}{2}x-4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given the parabola: $y=ax^{2}-6ax-16a$ ($a>0$) intersects the x-axis at points $A$ and $B$ (with $A$ to the left of $B$) and intersects the y-axis at point $C$. Point $G$ is the midpoint of segment $AC$. The line $y=-\\frac{3}{2}x$ intersects the parabola at points $M$ and $N$, and $MO=NO$. Find the equation of the parabola.\n",
    "completion": "\\textbf{Solution:} Solve the system of equations\n\\[\n\\begin{cases}\ny=-\\frac{3}{2}x\\\\\ny=ax^2-6ax-16a\n\\end{cases}\n\\]\nUpon simplification, we have $ax^2-6ax+\\frac{3}{2}x-16a=0$,\\\\\nThus, $x_M+x_N=6-\\frac{3}{2a}$,\\\\\nSince $MO=NO$,\\\\\nIt follows that points M and N are symmetric about the origin,\\\\\nThus, $x_M+x_N=0$,\\\\\nTherefore, $6-\\frac{3}{2a}=0$,\\\\\nThus, $a=\\frac{1}{4}$,\\\\\nTherefore, $\\boxed{y=\\frac{1}{4}x^2-\\frac{3}{2}x-4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51545315_151.png"
  },
  {
    "solution": "\\boxed{\\left(\\frac{3}{2}, -\\frac{15}{4}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, the parabola $y=ax^2+bx-3\\ (a\\neq 0)$ intersects the x-axis at points $A(-1, 0)$ and $B(3, 0)$, and intersects the y-axis at point C. Let P be a moving point on the parabola in the fourth quadrant. Find the maximum value of the area $S$ of the triangle $\\triangle BPC$ and the coordinates of the point P at this time.\n",
    "completion": "\\textbf{Solution:} Draw $PD \\perp x$-axis at point D through point P. Assume the coordinates of point P as $(x, y)$, \\\\\nthen $S = S_{\\text{trapezoid }OCPD} + S_{\\triangle DPB} - S_{\\triangle OCB}$\\\\\n$= \\frac{1}{2}x\\left[3-(x^{2}-2x-3)\\right]+\\frac{1}{2} \\times (3-x) \\times [- (x^{2}-2x-3)]-\\frac{1}{2} \\times 3 \\times 3$\\\\\n$= -\\frac{3}{2}(x-\\frac{3}{2})^{2}+\\frac{27}{8}$\\\\\n$\\therefore$ when $x=\\frac{3}{2}$, $S_{\\text{max}}=\\boxed{\\frac{27}{8}}$.\\\\\nAt this point $y=(\\frac{3}{2})^{2}-2 \\times \\frac{3}{2} -3 = -\\frac{15}{4}$\\\\\nTherefore, the maximum value of the area $S$ of $\\triangle BPC$ is $\\boxed{\\frac{27}{8}}$, with the coordinates of point P being $\\boxed{\\left(\\frac{3}{2}, -\\frac{15}{4}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52796987_47.png"
  },
  {
    "solution": "\\boxed{y=\\frac{3}{4}x+\\frac{3}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, rectangle $ABCD$ has side $AB$ on the x-axis, with points $A$ and $B$ being symmetrically positioned about the origin. The coordinates of point $D$ are $(-2, 3)$. Find the analytical expression of line $AC$.\n",
    "completion": "\\textbf{Solution:} Since the side $AB$ of rectangle $ABCD$ lies on the x-axis, and points A, B are symmetrical about the origin, with point D's coordinates being $\\left(-2, 3\\right)$, \\\\\n$\\therefore A\\left(-2, 0\\right), B\\left(2, 0\\right), AB=CD, AB\\parallel CD, \\angle ABC=\\angle BCD=\\angle DAB=90^\\circ,$ \\\\\n$\\therefore C\\left(2, 3\\right),$ \\\\\nLet $AC$ be defined by $y=kx+b,$ \\\\\n$\\therefore \\begin{cases}-2k+b=0\\\\ 2k+b=3\\end{cases},$ \\\\\nSolving, we get: $\\begin{cases}k=\\frac{3}{4}\\\\ b=\\frac{3}{2}\\end{cases},$ \\\\\n$\\therefore$The equation of AC is: $\\boxed{y=\\frac{3}{4}x+\\frac{3}{2}}.$",
    "image_path": "dataset/mm_math/images/MM_Math/52691272.png"
  },
  {
    "solution": "\\boxed{19^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, extend the side BC of rectangle ABCD to point E so that $CE=BD$. Connect AE, intersecting the diagonal BD at point F, and intersecting side CD at point G. If $\\angle ADB = 38^\\circ$, what is the measure of $\\angle E$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure: Connect AC. Let the intersection point of AC and BD be O. \\\\\nSince ABCD is a rectangle, \\\\\nit follows that AC=BD, \\\\\nwhich means OA=OC=OB=OD, \\\\\nSince $\\angle ADB=38^\\circ$, \\\\\nit implies $\\angle BCA=\\angle DBC=38^\\circ$, \\\\\nSince CE=BD, \\\\\nit leads to AC=CE, \\\\\nTherefore, $\\angle E=\\boxed{19^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52211484.png"
  },
  {
    "solution": "\\boxed{x<6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system, the graph of the line $L_1\\colon y = kx + b$ passes through point $C(-2, 7)$, intersects with the $x$-axis at point $B$, and intersects with the line $L_2\\colon y = \\frac{1}{2}x$ at point $A$, where the $x$-coordinate of point $A$ is $6$. Write directly the solution set of the inequality $kx + b > \\frac{1}{2}x$ in terms of $x$.\n",
    "completion": "\\textbf{Solution:} The solution set of the inequality $kx+b>\\frac{1}{2}x$ with respect to $x$ is: $\\boxed{x<6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50520352_180.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, point A has coordinates $ \\left(3, 4\\right)$. What is the value of $ \\cos\\alpha $?\n",
    "completion": "\\textbf{Solution:} Construct $AB \\perp x$-axis at $B$, as shown in the figure,\\\\\nsince the coordinates of point A are (3,4),\\\\\nhence $OB = 3$, $AB = 4$,\\\\\ntherefore $OA = \\sqrt{3^2 + 4^2} = 5$,\\\\\nin $\\triangle AOB$ right, $\\cos\\alpha = \\frac{OB}{OA} = \\frac{3}{5}$ .\\\\\nThus $\\cos\\alpha = \\boxed{\\frac{3}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51457858.png"
  },
  {
    "solution": "\\boxed{70^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ is inscribed in $\\odot O$, and $AD$ is the diameter of $\\odot O$. If $\\angle B = 20^\\circ$, then what is the degree measure of $\\angle CAD$?\n",
    "completion": "\\textbf{Solution:} Draw line BD. Since AD is the diameter of circle O, it follows that $\\angle ABD = 90^\\circ$. Since $\\angle ABC = 20^\\circ$, it follows that $\\angle CBD = \\angle ABD - \\angle ABC = 70^\\circ$. Therefore, $\\angle CAD = \\angle CBD = \\boxed{70^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52807127.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $\\angle AOB=30^\\circ$, $\\angle AOE=140^\\circ$, $OB$ bisects $\\angle AOC$, and $OD$ bisects $\\angle AOE$. What is the measure of $\\angle AOC$?\n",
    "completion": "\\textbf{Solution:} From the given, \\\\\nsince $\\angle AOB = 30^\\circ$ and OB bisects $\\angle AOC$, \\\\\nit follows that $\\angle AOC = 2\\angle AOB = 2 \\times 30^\\circ = \\boxed{60^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51495222_145.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\triangle ABC$, $\\angle ACB=90^\\circ$, $AM=AC$, $BN=BC$. When $\\angle A=30^\\circ$, what is the measure of $\\angle MCN$ in degrees?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle A=30^\\circ$ and $AC=AM$,\\\\\nit follows that $\\angle AMC=\\frac{1}{2}\\times \\left(180^\\circ-30^\\circ\\right)=75^\\circ$,\\\\\nSince $\\angle B=60^\\circ$ and $BC=BN$,\\\\\nit implies $\\angle CNB=\\frac{1}{2}\\times \\left(180^\\circ-60^\\circ\\right)=60^\\circ$,\\\\\nThus, $\\angle MCN=180^\\circ-60^\\circ-75^\\circ=\\boxed{45^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52867075_157.png"
  },
  {
    "solution": "\\boxed{\\left(0, 6\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system, the vertices of the rectangle $OABC$ are such that vertex $O$ coincides with the origin, and vertices $A$ and $C$ are respectively on the x-axis and y-axis. The coordinates of point $B$ are $(2, 3)$. The graph of the inverse proportion function $y=\\frac{k}{x}$ passes through the midpoint $D$ of $BC$, and intersects with $AB$ at point $E$. Connect $DE$. If point $P$ is on the y-axis, and the area of $\\triangle OPD$ is equal to the area of quadrilateral $BDOE$, find the coordinates of point $P$.\n",
    "completion": "\\textbf{Solution:} Given that $S_{\\text{quadrilateral} BDOE}=S_{\\text{rectangle} OABC}-{S}_{\\triangle AOE}-{S}_{\\triangle COD}=2\\times 3-\\frac{1}{2}\\times 2\\times \\frac{3}{2}-\\frac{1}{2}\\times 3\\times 1=3$,\\\\\naccording to the problem, we have $\\frac{1}{2}OP\\cdot DC=3, DC=1$,\\\\\n$\\therefore OP=6$,\\\\\n$\\therefore$ the coordinates of point P are $\\boxed{\\left(0, 6\\right)}$ or $\\left(0, -6\\right)$.",
    "image_path": "dataset/mm_math/images/MM_Math/52827689_142.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, D is a point on $BC$, $DE\\perp AB$ at point E, $AE=AC$, connect $AD$, if $BC=8$, what is the value of $BD+DE$?\n",
    "completion": "\\textbf{Solution:}\\\\\nSince $DE \\perp AB$,\\\\\nit follows that $\\angle AED = \\angle C = 90^\\circ$,\\\\\nSince $AE = AC$, $AD = AD$,\\\\\nit follows that $\\triangle ADE \\cong \\triangle ADC\\ (HL)$,\\\\\ntherefore $DE = DC$,\\\\\nthus $BD + DE = BC + DC = BC = \\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52661286.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in quadrilateral $ABCD$, $E$ and $F$ are two points on the diagonal $AC$, and $AE=CF$. Quadrilateral $BEDF$ is a parallelogram; if $AB \\perp BF$, $AB=8$, $BF=6$, and $AC=16$. Find the length of the line segment $EF$.\n",
    "completion": "\\textbf{Solution:} In right triangle $Rt\\triangle ABF$, we have $AF=\\sqrt{AB^{2}+BF^{2}}=\\sqrt{8^{2}+6^{2}}=10$,\\\\\nsince $AC=16$,\\\\\nthen $CF=AC-AF=16-10=6$,\\\\\nsince $AE=CF$, thus $AE=6$,\\\\\ntherefore $EF=AF-AE=10-6=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52395580_115.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, AD is the median of $\\triangle ABC$, E is a point on AD, and BE intersects AC at F. If $BE=AC$, $BF=9$, and $CF=6$, then what is the length of $AF$?\n",
    "completion": "\\textbf{Solution:} Extend AD to G such that DG = AD, and connect BG,\\\\\n$\\because$ AD is the median of $\\triangle ABC$,\\\\\n$\\therefore$ CD = BD,\\\\\nIn $\\triangle ACD$ and $\\triangle GBD$,\\\\\n$ \\left\\{\\begin{array}{l}CD=BD \\\\ \\angle ADC= \\angle BDG \\\\ AD=DG \\end{array}\\right.$\\\\\n$\\therefore$ $\\triangle ACD \\cong \\triangle GBD$ (SAS),\\\\\n$\\therefore$ $\\angle CAD = \\angle G$, AC = BG,\\\\\n$\\because$ BE = AC,\\\\\n$\\therefore$ BE = BG,\\\\\n$\\therefore$ $\\angle G = \\angle BEG$,\\\\\n$\\because$ $\\angle BEG = \\angle AEF$,\\\\\n$\\therefore$ $\\angle AEF = \\angle EAF$.\\\\\n$\\therefore$ EF = AF,\\\\\n$\\therefore$ AF + CF = BF - AF,\\\\\ni.e., AF + 6 = 9 - AF,\\\\\n$\\therefore$ AF = $\\boxed{\\frac{3}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52693950.png"
  },
  {
    "solution": "\\boxed{y=\\frac{3}{x}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the linear function $y=x+b$ intersects with the inverse proportion function $y=\\frac{k}{x}$ ($k\\neq 0$) at points $A$ and $B$, and the coordinates of point $A$ are (1, 3). The linear function $y=x+b$ intersects with the x-axis at point $C$. Connect $OA$ and $OB$. What are the expressions for the linear function and the inverse proportion function?\n",
    "completion": "\\textbf{Solution:} Given point $A(1, 3)$, substituting into the linear function $y=x+b$ yields: $3=1+b$,\\\\\nSolving, we find: $b=2$,\\\\\nThus, the expression for the linear function is $y=x+2$,\\\\\nSubstituting $A(1, 3)$ into the inverse proportion function $y=\\frac{k}{x}$ yields: $3=\\frac{k}{1}$, implying $k=3$,\\\\\nTherefore, the expression for the inverse proportion function is $\\boxed{y=\\frac{3}{x}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51322344_127.png"
  },
  {
    "solution": "\\boxed{\\frac{4}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, where $AB = 3$ and $BC = 4$, connect $BD$ and draw the bisector of $\\angle CBD$ intersecting $CD$ at point $E$. What is the length of $CE$?\n",
    "completion": "\\textbf{Solution:} Construct EH$\\perp$BD at H, as shown in the figure,\\\\\nsince quadrilateral ABCD is a rectangle,\\\\\ntherefore AB＝CD＝3, BC＝AD＝4, $\\angle$C＝$90^\\circ$,\\\\\ntherefore BD＝$\\sqrt{BC^{2}+CD^{2}}$＝5,\\\\\nsince BE bisects $\\angle$CBD,\\\\\ntherefore $\\angle$EBC＝$\\angle$EBH,\\\\\nIn $\\triangle$EBH and $\\triangle$EBC,\\\\\n$\\left\\{\\begin{array}{l}\n\\angle EHB=\\angle C=90^\\circ\\\\\n\\angle EBH=\\angle EBC\\\\\nBE=BE\n\\end{array}\\right.$,\\\\\ntherefore $\\triangle$EBH$\\cong$$\\triangle$EBC,\\\\\ntherefore BC＝BH＝4, EC＝EH, let EC＝EH＝x,\\\\\nIn $\\triangle$DEH,\\\\\nsince DE$^{2}$＝DH$^{2}$+EH$^{2}$,\\\\\ntherefore $(3−x)^{2}$＝$1^{2}$+$x^{2}$,\\\\\ntherefore x＝$\\boxed{\\frac{4}{3}}$,\\\\\ntherefore CE＝$\\frac{4}{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52344112.png"
  },
  {
    "solution": "\\boxed{(1,3)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y_{1}=2x-2$ intersects the y-axis at point $A$, and the line $y_{2}=-2x+6$ intersects the y-axis at point $B$. The two lines intersect at point $C$. What is the range of values for $x$ when both $2x-2>0$ and $-2x+6>0$ are true?\n",
    "completion": "\\textbf{Solution:} When both $y_1>0$ and $y_2>0$ hold, \\\\\n$\\therefore$ the range of values for $x$ is: $x\\in\\boxed{(1,3)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52422271_66.png"
  },
  {
    "solution": "\\boxed{\\left(3, -1\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the vertex A of the right triangle \\(Rt\\triangle ABO\\) is the intersection point in the second quadrant of the hyperbola \\(y=\\frac{k}{x}\\) and the line \\(y=-x-(k+1)\\). \\(AB \\perp x\\)-axis at B, and the area of the triangle \\(S_{\\triangle ABO}=\\frac{3}{2}\\). Determine the coordinates of the two intersection points A and C of the line and the hyperbola.\n",
    "completion": "Solution: Combine the analytic expressions of the two functions into a system of equations, $\\begin{cases}y=-\\frac{3}{x}\\\\ y=-x+2\\end{cases}$. Solving this, we find: $\\begin{cases}x_1=-1\\\\ y_1=3\\end{cases}$ and $\\begin{cases}x_2=3\\\\ y_2=-1\\end{cases}$,\\\\\n$\\therefore$ the coordinates of point A are $\\boxed{\\left(-1, 3\\right)}$, and the coordinates of point C are $\\boxed{\\left(3, -1\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51458226_262.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system, given points $A(8, 1)$ and $B(0, -3)$, the graph of the inverse proportion function $y=\\frac{k}{x}$ $(x>0)$ passes through point A. The moving line $x=t$ $(0<t<8)$ intersects the graph of the inverse proportion function at point M and intersects line $AB$ at point N. What is the value of $k$?\n",
    "completion": "\\textbf{Solution:} Since the inverse proportion function $y=\\frac{k}{x}(x>0)$ passes through point $A\\left(8, 1\\right)$,\\\\\nthus, substituting the coordinates of point A into $y=\\frac{k}{x}$ yields: $1=\\frac{k}{8}$,\\\\\nsolving for $k$ gives $k=\\boxed{8}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53490748_197.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $\\triangle ABC$ is an equilateral triangle with a side length of 10. $D$ is the midpoint of $AC$, $\\angle EDF=120^\\circ$, $DE$ intersects line segment $AB$ at $E$, and $DF$ intersects the extension of $BC$ at $F$. If $AE=4BE$, then what is the length of $CF$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, construct $DM\\parallel BC$ intersecting $AB$ at $M$, \\\\\n$\\because$ $AE=4BE$, \\\\\n$\\therefore$ $BE=\\frac{1}{5}AB=2$, \\\\\n$\\because$ $D$ is the midpoint of $AC$, and $\\triangle ABC$ is an equilateral triangle, \\\\\n$\\therefore$ $MD$ is the median of $\\triangle ABC$, \\\\\n$\\therefore$ $MD=DC=BM=\\frac{1}{2}BC=5$, $\\angle DME=120^\\circ=\\angle DCF$, $\\angle ADM=60^\\circ$, \\\\\n$\\therefore$ $ME=BM-BE=3$, $\\angle MDC=120^\\circ$, \\\\\n$\\because$ $\\angle EDF=120^\\circ$, $\\angle MDC-\\angle EDC=\\angle EDF-\\angle EDC$, \\\\\n$\\therefore$ $\\angle MDE=\\angle CDF$, \\\\\nIn $\\triangle DEM$ and $\\triangle DFC$, \\\\\n$\\because$ $\\left\\{\\begin{array}{l}\\angle MDE=\\angle CDF \\\\ \\angle DME=\\angle DCF \\\\ DM=DC \\end{array}\\right.$, \\\\\n$\\therefore$ $\\triangle DEM\\cong \\triangle DFC\\left(AAS\\right)$, \\\\\n$\\therefore$ $CF=ME=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53010135.png"
  },
  {
    "solution": "\\boxed{60\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a ship is located at point $A$, which is at the northeast direction $45^\\circ$ from the lighthouse $C$ and at a distance of $60$ nautical miles from it. After sailing southward for a while, it arrives at point $B$, which is at the southeast direction $30^\\circ$ from the lighthouse $C$. What is the distance between the ship and the lighthouse $C$ at this point? (Keep the square root in the result.)\n",
    "completion": "\\textbf{Solution:} Draw CD $\\perp$ AB. In the right triangle $ACD$, $\\angle A=45^\\circ$ and AC=60 nautical miles,\\\\\n$\\therefore \\triangle ACD$ is an isosceles right triangle,\\\\\n$\\therefore$ AD=CD=$\\frac{\\sqrt{2}}{2}$AC=30$\\sqrt{2}$ nautical miles,\\\\\nIn right triangle $BCD$, $\\angle B=30^\\circ$,\\\\\n$\\therefore$ BC=2CD=2×30$\\sqrt{2}$=60$\\sqrt{2}$ nautical miles\\\\\nThus, at this time, the distance from the ship to the lighthouse is \\boxed{60\\sqrt{2}} nautical miles.",
    "image_path": "dataset/mm_math/images/MM_Math/52537080.png"
  },
  {
    "solution": "\\boxed{\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, $AB \\parallel CD$, $AB \\perp BD$, $AB = 5$, $BD = 4$, $CD = 3$. If point $E$ is the midpoint of $AC$, what is the length of $BE$?\n",
    "completion": "\\textbf{Solution:}\\\\\nSince $AB \\parallel CD$ and $AB \\perp BD$,\\\\\nit follows that $CD \\perp BD$,\\\\\nthus $\\angle ABD = \\angle CDB = 90^\\circ$.\\\\\nExtend $CD$ to $F$, and construct $AF \\perp CF$ at point $F$.\\\\\nThen, $\\angle BDF= 90^\\circ$, $\\angle F= 90^\\circ$.\\\\\nTherefore, quadrilateral $ABDF$ is a rectangle.\\\\\nTherefore, $AF=BD=4$, and $DF=AB=5$.\\\\\nSince $CD=3$,\\\\\nthus $CF=5+3=8$.\\\\\nSince $AC= \\sqrt{8^2+4^2}=\\sqrt{80}=4\\sqrt{5}$,\\\\\nin right triangle $BCD$ where $CD=3$ and $BD=4$,\\\\\nhence $BC=5$,\\\\\nthus $AB=BC$.\\\\\nSince $\\triangle ABC$ is an isosceles triangle,\\\\\nand point $E$ is the midpoint of $AC$,\\\\\nhence $AE=\\frac{1}{2}AC=2\\sqrt{5}$, and $BE \\perp AC$.\\\\\nSince $BE^2=AB^2−AE^2 = 5^2−(2\\sqrt{5})^2 =25-20=5$,\\\\\nthus $BE=\\sqrt{5}$.\\\\\nTherefore, the answer is: $BE=\\boxed{\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52252071.png"
  },
  {
    "solution": "\\boxed{6cm}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in parallelogram $ABCD$, point $E$ is on $AD$. Connect $CE$ and extend it to intersect the extension of $BA$ at point $F$. If $AE: AD = 2:3$ and $CD = 3$ cm, what is the length of $AF$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral ABCD is a parallelogram,\\\\\nit follows that AB $\\parallel$ CD,\\\\\nwhich implies $\\triangle AFE \\sim \\triangle DEC$,\\\\\nhence AE : DE = AF : CD,\\\\\ngiven AE : AD = 2 : 3,\\\\\nit follows that AE : DE = 2 : 1,\\\\\nsince CD = 3 cm,\\\\\nwe have AF = 2CD = \\boxed{6cm}.",
    "image_path": "dataset/mm_math/images/MM_Math/52963737.png"
  },
  {
    "solution": "\\boxed{\\frac{5}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in right-angled triangle $ABC$, where $\\angle C=90^\\circ$, $AC=3$, and $BC=4$, the line $AD$ bisects $\\angle CAB$ and intersects $BC$ at $D$, what is the length of $BD$?\n",
    "completion": "\\textbf{Solution:} Construct $DE \\perp AB$ at $E$, \\\\\nsince $AD$ is the bisector of $\\angle CAB$, $\\angle C=90^\\circ$, and $DE \\perp AB$ at $E$, \\\\\nit follows that $CD=DE$, \\\\\nIn the right triangles $\\triangle ACD$ and $\\triangle AED$, \\\\\n$\\left\\{\\begin{array}{l} AD=AD\\\\ DC=DE \\end{array}\\right.$, \\\\\nhence, right $\\triangle ACD \\cong$ right $\\triangle AED$ (HL), \\\\\nthus $AC=AE=3$, \\\\\nby the Pythagorean theorem, $AB=\\sqrt{AC^2+BC^2}=\\sqrt{3^2+4^2}=5$, \\\\\nthus $BE=AB-AE=5-3=2$, \\\\\nLet $BD=x$, then $CD=DE=4-x$ \\\\\nIn the right $\\triangle BDE$, $DE^2+BE^2=BD^2$ \\\\\nthus $(4-x)^2+2^2=x^2$ \\\\\nSolving gives $x=\\frac{5}{2}$ \\\\\nThat is, $BD=\\boxed{\\frac{5}{2}}$",
    "image_path": "dataset/mm_math/images/MM_Math/52827539.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the vertices of $\\triangle OAB$ are $O(0, 0)$, $A(3, 4)$, and $B(6, 0)$. Moving points P and Q start from point O simultaneously, heading along the positive directions of the x-axis and y-axis at speeds of 3 units per second and 2 units per second, respectively. Point P and Q stop moving when P reaches point B. A line through point Q, $MN \\parallel OB$, intersects $AO$ and $AB$ at points M and N, respectively. Lines $PM$ and $PN$ are then drawn. Let the movement time be t (seconds). What value of t will maximize the area of quadrilateral $MNPB$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $AD\\perp OB$ through point $A$, intersecting $MN$ and $OB$ at points $C$ and $D$, respectively.\\\\\n$\\because MN\\parallel OB$, $A(3, 4)$, $M\\left(\\frac{3}{2}t, 2t\\right)$,\\\\\n$\\therefore D(3, 0)$, $C(3, 2t)$.\\\\\n$\\therefore OD=BD$.\\\\\n$\\because MN\\parallel OB$,\\\\\n$\\therefore \\angle ANC=\\angle AMC$, $AC\\perp MN$,\\\\\n$\\therefore AM=AN$,\\\\\n$\\therefore MC=CN=\\frac{1}{2}MN$,\\\\\n$\\therefore MN=2MC=2\\left(3-\\frac{3}{2}t\\right)=6-3t$.\\\\\n$\\because$ Point P moves at a speed of $3$,\\\\\n$\\therefore OP=3t$.\\\\\n$\\because B(6, 0)$,\\\\\n$\\therefore PB=6-3t$.\\\\\n$\\therefore MN=PB$,\\\\\n$\\therefore$ Quadrilateral $MNPB$ is a parallelogram.\\\\\n$\\therefore S_{\\text{quadrilateral }MNBP}=\\frac{1}{2}\\left[(6-3t)+(6-3t)\\right]\\times 2t=-6(t-1)^2+6$.\\\\\n$\\because$ Point P stops moving when it reaches point B along with point Q,\\\\\n$\\therefore 0<3t<6$, which means $0<t<2$.\\\\\n$\\therefore$ When $t=\\boxed{1}$, the area of quadrilateral $MNBP$ is maximal, with the maximum value being $6$.",
    "image_path": "dataset/mm_math/images/MM_Math/51434142_204.png"
  },
  {
    "solution": "\\boxed{25^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Let point O be a point on line AB. A ray OC is drawn through point O such that $\\angle BOC = 65^\\circ$. Place the right-angled vertex of a right-angled triangle ruler at point O. As shown in Figure 2, rotate the triangle ruler MON counterclockwise around point O by a certain angle, such that OC is the bisector of $\\angle MOB$. What are the measures of $\\angle BON$ and $\\angle CON$?\n",
    "completion": "\\textbf{Solution}: Since $\\angle BOC=65^\\circ$ and OC is the bisector of $\\angle MOB$,\\\\\nit follows that $\\angle MOB=2\\angle BOC=130^\\circ$,\\\\\ntherefore $\\angle BON=\\angle MOB-\\angle MON=130^\\circ-90^\\circ=40^\\circ$,\\\\\n$\\angle CON=\\angle COB-\\angle BON=65^\\circ-40^\\circ=25^\\circ$,\\\\\nwhich means $\\angle BON=\\boxed{40^\\circ}$ and $\\angle CON=\\boxed{25^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51291351_163.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the line $l\\colon y=kx-1$ ($k\\neq 0$) intersects with the graph of the function $y=\\frac{m}{x}$ ($x>0$) at point $A(3,2)$. What are the values of $k$ and $m$?\n",
    "completion": "\\textbf{Solution:} Since the line $l\\colon y=kx-1$ ($k\\neq 0$) intersects with the graph of the function $y=\\frac{m}{x}$ ($x>0$) at point $A(3,2)$,\\\\\nwe have $2=3k-1$ and $2=\\frac{m}{3}$,\\\\\nsolving these equations, we find $k=\\boxed{1}, m=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53078816_45.png"
  },
  {
    "solution": "\\boxed{64^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $OB$ bisects $\\angle AOC$, and the complement of $\\angle AOC$ is $30^\\circ$ smaller than $\\angle BOC$. A ray $OD$ is drawn through point O such that $\\angle AOD = \\frac{1}{5}\\angle AOC$. Can you find the measure of $\\angle COD$ in degrees?\n",
    "completion": "\\textbf{Solution:} From (1) we have $\\angle AOC = 2x = 2\\times 40^\\circ = 80^\\circ$\\\\\nThen, $\\angle AOD = \\frac{1}{5}\\angle AOC = \\frac{1}{5}\\times 80^\\circ = 16^\\circ$.\\\\\nWhen ray $OD$ is below $OA$,\\\\\n$\\angle COD = \\angle AOC + \\angle AOD = 80^\\circ + 16^\\circ = \\boxed{96^\\circ}$\\\\\nWhen ray $OD$ is above $OA$,\\\\\n$\\angle COD = \\angle AOC - \\angle AOD = 80^\\circ - 16^\\circ = \\boxed{64^\\circ}$\\\\\nIn summary: The measure of $\\angle COD$ is either $96^\\circ$ or $64^\\circ$.",
    "image_path": "dataset/mm_math/images/MM_Math/53206051_133.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{3}-1}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $ABCD$ is a square with side length $1$, and $\\triangle BPC$ is an equilateral triangle. Then,what is the area of $\\triangle BPD$.\n",
    "completion": "\\textbf{Solution}: As shown in the figure, \\\\\ndraw PE $\\perp$ CD and PF $\\perp$ BC, \\\\\nsince the side length of square ABCD is 1, and $\\triangle BPC$ is an equilateral triangle, \\\\\ntherefore $\\angle PBC=\\angle PCB=60^\\circ$, PB=PC=BC=CD=1, \\\\\ntherefore $\\angle PCE=30^\\circ$, \\\\\nthus PF=PB$\\cdot \\sin60^\\circ=1\\times \\frac{\\sqrt{3}}{2}=\\frac{\\sqrt{3}}{2}$, PE=FC=$\\frac{1}{2}$, \\\\\n$S_{\\triangle BPD}=S_{\\text{quadrilateral} PBCD}-S_{\\triangle BCD}=S_{\\triangle PBC}+S_{\\triangle PDC}-S_{\\triangle BCD}$ \\\\\n=$\\frac{1}{2}\\times\\frac{\\sqrt{3}}{2}\\times1+\\frac{1}{2}\\times\\frac{1}{2}\\times1-\\frac{1}{2}\\times1\\times1=\\boxed{\\frac{\\sqrt{3}-1}{4}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53043322.png"
  },
  {
    "solution": "\\boxed{y=-x+3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y=k_1x+b$ intersects with the graph of the inverse proportion function $y=\\frac{k_2}{x}$ at points A and B. The coordinates of point A are $(-1, 4)$, and the coordinates of point B are $(4, n)$. What are the expressions for these two functions?\n",
    "completion": "\\textbf{Solution:} To solve, substitute $A(-1, 4)$ into $y=\\frac{k_2}{x}$, obtaining $k_2=-4$, \\\\\n$\\therefore y=-\\frac{4}{x}$,\\\\\n$\\because$ point $B(4, n)$ is on $y=-\\frac{4}{x}$,\\\\\n$\\therefore n=-1$,\\\\\n$\\therefore \\boxed{B(4, -1)}$,\\\\\nSubstituting $A(-1, 4)$, $B(4, -1)$ into $y=k_1x+b_1$ yields\\\\\n$\\begin{cases}\n-k_1+b=4\\\\\n4k_1+b=-1\n\\end{cases}$, solving gives $\\begin{cases}\nk_1=-1\\\\\nb=3\n\\end{cases}$,\\\\\n$\\therefore \\boxed{y=-x+3}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51446477_238.png"
  },
  {
    "solution": "\\boxed{70^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $\\angle AOB = 50^\\circ$, $\\angle BOC = 90^\\circ$, and $OM$ and $ON$ are the angle bisectors of $\\angle AOB$ and $\\angle BOC$ respectively. What is the degree measure of $\\angle MON$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle AOB=50^\\circ$, and $OM$ is the angle bisector of $\\angle AOB$,\\\\\n$\\therefore \\angle BOM=25^\\circ$.\\\\\nAlso, since $\\angle BOC=90^\\circ$, and $ON$ is the angle bisector of $\\angle BOC$,\\\\\n$\\therefore \\angle BON=45^\\circ$.\\\\\n$\\therefore \\angle MON=25^\\circ+45^\\circ=\\boxed{70^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/42811603.png"
  },
  {
    "solution": "\\boxed{(2,1)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown, it is known that the line $y=-\\frac{1}{2}x+2$ intersects the x-axis at point B and the y-axis at point A. Point P is a moving point on line AB (not coinciding with A or B). Line $PQ\\perp x$ axis at point Q. If the triangle with vertices P, O, Q is similar to $\\triangle AOB$, then what are the coordinates of point P?\n",
    "completion": "\\textbf{Solution:}\n\nSince point $P$ lies on the line $AB$ defined by $y=-\\frac{1}{2}x+2$,\n\nlet $P\\left(m, -\\frac{1}{2}m+2\\right)$. Therefore, $PQ=\\left|-\\frac{1}{2}m+2\\right|$ and $OQ=\\left|m\\right|$\n\nGiven $\\angle PQO=90^\\circ$ and $\\angle AOB=90^\\circ$,\n\nif the triangle formed by points $P$, $O$, and $Q$ is similar to $\\triangle AOB$,\n\nthen $\\triangle POQ\\sim \\triangle AOB$ or $\\triangle PQO\\sim \\triangle BOA$\n\nThus, $\\frac{OQ}{OB}=\\frac{QP}{OA}$ or $\\frac{OQ}{OA}=\\frac{QP}{OB}$\n\nGiven $y=-\\frac{1}{2}x+2$,\n\nby setting $x=0$, we get $y=2$, and by setting $y=0$, we get $x=4$\n\nHence, $OA=2$ and $BO=4$\n\nTherefore, $\\frac{\\left|m\\right|}{4}=\\frac{\\left|-\\frac{1}{2}m+2\\right|}{2}$ or $\\frac{\\left|m\\right|}{2}=\\frac{\\left|-\\frac{1}{2}m+2\\right|}{4}$\n\nSolving gives $m=2$, or $m=\\frac{4}{5}$, or $m=-\\frac{4}{3}$\n\nThus, the coordinates of $P$ are $\\boxed{(2,1)}$, $\\left(\\frac{4}{5}, \\frac{8}{5}\\right)$, or $\\left(-\\frac{4}{3}, \\frac{8}{3}\\right)$",
    "image_path": "dataset/mm_math/images/MM_Math/52765245.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, it is known that $\\angle BOC = 40^\\circ$, $OE$ bisects $\\angle AOC$, and $OF$ bisects $\\angle BOC$. If $AO \\perp BO$, then what is the measure of $\\angle EOF$ in degrees?\n",
    "completion": "\\textbf{Solution:} Given $\\because$ $\\angle BOC=40^\\circ$ and $AO\\perp BO$,\\\\\n$\\therefore$ $\\angle AOB=90^\\circ, \\angle AOC=\\angle AOB+\\angle BOC=130^\\circ$,\\\\\n$\\because$ $OE$ bisects $\\angle AOC$ and $OF$ bisects $\\angle BOC$,\\\\\n$\\therefore$ $\\angle EOC=\\frac{1}{2}\\angle AOC=65^\\circ, \\angle COF=\\frac{1}{2}\\angle BOC=20^\\circ$,\\\\\n$\\therefore$ $\\angle EOF=\\angle EOC−\\angle COF=65^\\circ−20^\\circ=\\boxed{45^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53364112_170.png"
  },
  {
    "solution": "\\boxed{96}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that quadrilateral $ABCD$ is a rhombus, with points $E$ and $F$ being the midpoints of sides $AB$ and $BC$, respectively. Lines $DE$, $EF$, and $DF$ are drawn. $\\triangle DEF$ is an isosceles triangle; if $AD=10$ and $EF=8$, what is the area of rhombus $ABCD$?\n",
    "completion": "\\textbf{Solution:} Connect $AC$ and $BD$ intersecting at point $O$. \\\\\nSince $ABCD$ is a rhombus, \\\\\nthen $AC \\perp BD$, $OA=OC$, $OB=OD$. \\\\\nSince points $E$ and $F$ are the midpoints of sides $AB$ and $BC$, respectively, \\\\\nthen $AC=2EF=16$, thus $OA=8$. \\\\\nSince $AD=10$, \\\\\nthen $OD=\\sqrt{10^2-8^2}=6$, \\\\\nthus $BD=12$. \\\\\nThe area of rhombus $ABCD$ is $\\frac{1}{2}AC \\cdot BD=\\boxed{96}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52204885_107.png"
  },
  {
    "solution": "\\boxed{x<2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=kx+3$ passes through the point $(2,0)$. What is the solution set of the inequality $kx+3>0$ with respect to $x$?\n",
    "completion": "\\textbf{Solution:} Since the line $y=kx+3$ passes through the point $(2,0)$,\\\\\nit follows that the solution set of the inequality $kx+3>0$ with respect to $x$ is $\\boxed{x<2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52226187.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the diagonals AC and BD of the rhombus ABCD intersect at point O, and E is the midpoint of AB. Connect OE. If the perimeter of the rhombus ABCD is 32, what is the length of OE?\n",
    "completion": "\\textbf{Solution:} Given that the perimeter of rhombus ABCD is 32,\\\\\nit follows that BA=8, and AC$\\perp$BD,\\\\\nwhich implies $\\angle$AOB=90$^\\circ$,\\\\\nsince point E is the midpoint of AB,\\\\\nit follows that OE=$ \\frac{1}{2}$AB=$\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52188263.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In quadrilateral $ABCD$, $AC$ and $BD$ intersect at point $O$, with $AD \\parallel BC$ and $AO=CO$. Quadrilateral $ABCD$ is a parallelogram. Through point $O$, draw $OE \\perp BD$ intersecting $BC$ at point $E$, and connect $DE$. If $\\angle CDE = \\angle CBD = 15^\\circ$, what is the measure of $\\angle ABC$ in degrees?\n",
    "completion": "\\textbf{Solution:} Given that OB = OD, and OE $\\perp$ BD, it follows that BE = ED. Consequently, $\\angle CBD = \\angle BDE = 15^\\circ$. Since $\\angle CDE = 15^\\circ$, we find that $\\angle BDC = 30^\\circ$. Given that quadrilateral ABCD is a parallelogram, it follows that AB $\\parallel$ CD, implying $\\angle ABD = \\angle BDC = 30^\\circ$. Therefore, $\\angle ABC = \\angle ABD + \\angle CBD = 30^\\circ + 15^\\circ = \\boxed{45^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51827682_71.png"
  },
  {
    "solution": "\\boxed{6.5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, D is the midpoint of AB, $AC=5$, and $BC=12$. What is the length of $CD$?\n",
    "completion": "\\textbf{Solution:} In right triangle $\\triangle ACB$, $\\angle ACB=90^\\circ$, $AC=5$, $BC=12$,\\\\\n$\\therefore AB= \\sqrt{AC^{2}+BC^{2}}=\\sqrt{5^{2}+12^{2}}=13$,\\\\\n$\\because D$ is the midpoint of $AB$,\\\\\n$\\therefore AD=BD$,\\\\\n$\\therefore CD= \\frac{1}{2} AB=\\boxed{6.5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52861956.png"
  },
  {
    "solution": "\\boxed{B(1, 0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=ax^2-a$ (where $a>0$) intersects the x-axis at points $A$ and $B$ (with point $A$ to the left of point $B$), and intersects the y-axis at point $C$. Given that $OC = 2OA$, what are the coordinates of points $A$ and $B$?\n",
    "completion": "\\textbf{Solution:} According to the given information, we have $y=0$, which implies $ax^2-a=0$, \\\\\nthus $a(x+1)(x-1)=0$, \\\\\nsolving this we get: $x=-1$ or $x=1$, \\\\\nthus, we have the points \\boxed{A(-1, 0)}, \\boxed{B(1, 0)}.",
    "image_path": "dataset/mm_math/images/MM_Math/52917460_253.png"
  },
  {
    "solution": "\\boxed{20}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in parallelogram \\(ABCD\\), diagonal \\(AC \\perp BC\\), \\(M\\) is on the bisector of \\(\\angle CAD\\), and \\(AM \\perp DM\\), point \\(N\\) is the midpoint of \\(CD\\), connect \\(MN\\), if \\(AD = 12\\), \\(MN = 2\\). What is the length of \\(AB\\)?\n",
    "completion": "\\textbf{Solution:} Extend $DM$ to intersect $AC$ at $E$, \\\\ \n$\\because AM$ bisects $\\angle CAD$ and $AM \\perp DM$, \\\\\n$\\angle DAM = \\angle EAM$, $\\angle AMD = \\angle AME = 90^\\circ$, \\\\\nIn $\\triangle ADM$ and $\\triangle AEM$, \\\\\n$\\left\\{\\begin{array}{l} \\angle DAM = \\angle EAM \\hfill \\\\ AM = AM \\hfill \\\\ \\angle AMD = \\angle AME \\hfill \\end{array}\\right.$, \\\\\n$\\therefore \\triangle ADM \\cong \\triangle AEM$ (ASA),\\\\\n$\\therefore DM = EM$, $AE = AD = 12$,\\\\\n$\\therefore$ Point $M$ is the midpoint of $DE$,\\\\\n$\\because N$ is the midpoint of $CD$,\\\\\n$\\therefore MN$ is the midline of $\\triangle CDE$,\\\\\n$\\because MN = 2$,\\\\\n$\\therefore CE = 2MN = 4$,\\\\\n$\\therefore AC = AE + CE = 12 + 4 = 16$,\\\\\nIn the parallelogram $ABCD$, $AB = CD$, $AD \\parallel BC$, $AC \\perp BC$,\\\\\n$\\therefore AC \\perp AD$,\\\\\n$\\therefore \\angle CAD = 90^\\circ$,\\\\\n$\\therefore AB = CD = \\sqrt{AD^2 + AC^2} = \\sqrt{12^2 + 16^2} = \\boxed{20}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52354486.png"
  },
  {
    "solution": "\\boxed{x < \\frac{3}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graphs of the functions $y=2x$ and $y=ax+4$ intersect at point $A(m, 3)$. What is the solution set of the inequality $2x < ax+4$?\n",
    "completion": "\\textbf{Solution:} Since the graphs of the functions $y=2x$ and $y=ax+4$ intersect at point A$(m,3)$,\\\\\nit follows that $3=2m$,\\\\\nyielding $m= \\frac{3}{2}$,\\\\\nthus, the coordinates of point A are $\\left( \\frac{3}{2}, 3\\right)$,\\\\\ntherefore, the solution set for the inequality $2x < ax+4$ is $\\boxed{x < \\frac{3}{2}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/5056405.png"
  },
  {
    "solution": "\\boxed{\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the side length of the square $ABCD$ is 1. Point $E$ is a moving point on side $AB$. Connect $ED$, and rotate $ED$ clockwise $90^\\circ$ around point $E$ to $EF$. Connect $DF$ and $CF$. What is the minimum value of $DF+CF$?\n",
    "completion": "\\textbf{Solution:} Connect BF, through point F draw $FG\\perp AB$ intersecting the extension of AB at point G, as shown in the figure,\\\\\n$\\because$ Rotating $ED$ $90^\\circ$ clockwise around point $E$ to EF,\\\\\n$\\therefore EF\\perp DE$, $EF=DE$,\\\\\n$\\therefore \\angle EDA=\\angle FEG$,\\\\\nIn $\\triangle AED$ and $\\triangle GFE$,\\\\\n$\\left\\{\\begin{array}{l}\n\\angle A=\\angle FGE\\\\\n\\angle EDA=\\angle FEG\\\\\nDE=EF\n\\end{array}\\right.$,\\\\\n$\\therefore \\triangle AED\\cong \\triangle GFE (AAS)$,\\\\\n$\\therefore FG=AE$,\\\\\n$\\therefore$ Point F moves along the ray BF,\\\\\nConstruct the symmetric point ${C}^{\\prime }$ of point C with respect to BF,\\\\\n$\\because EG=DA$, $FG=AE$,\\\\\n$\\therefore AE=BG$,\\\\\n$\\therefore BG=FG$,\\\\\n$\\therefore \\angle FBG=45^\\circ$,\\\\\n$\\therefore \\angle CBF=45^\\circ$,\\\\\n$\\therefore BF$ is the angle bisector of $\\angle CBC^{\\prime}$, hence point $F$ moves on the angle bisector of $\\angle CBC^{\\prime}$,\\\\\n$\\therefore$ Point ${C}^{\\prime }$ is on the extension line of AB,\\\\\nWhen points D, F, ${C}^{\\prime }$ are collinear, DF + CF= D${C}^{\\prime }$ is minimal,\\\\\nIn right $\\triangle$AD${C}^{\\prime }$, where AD= 1, A${C}^{\\prime }$= 2,\\\\\n$\\therefore D{C}^{\\prime }=\\sqrt{AD^{2}+AC^{2}}=\\sqrt{1^{2}+2^{2}}=\\boxed{\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51721939.png"
  },
  {
    "solution": "\\boxed{(2, -2)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the Cartesian coordinate system, line $l_{1}$ passes through points $A(1,0)$ and $B(0,2)$. By translating line $l_{1}$ downward by 4 units, we obtain line $l_{2}$. Line $l_{2}$ intersects the $x$ axis and the $y$ axis at points $C$ and $D$, respectively. Connect $AD$, $BC$, and $CP$. If segment $CP$ divides the quadrilateral $ABCD$ into two parts with an area ratio of $1:3$, find the coordinates of point $P$.\n",
    "completion": "\\textbf{Solution:} Since the line $l_{1}$ is translated 4 units downward to obtain line $l_{2}$,\\\\\nthe equation of line $l_{2}$ is $y=-2x-2$.\\\\\nWhen $y=-2x-2=0$, we solve to find $x=-1$; when $x=0$, $y=-2x-2=-2$.\\\\\nThus, points C$(-1,0)$ and D$(0,-2)$,\\\\\nhence OA$=$OC$=1$, OB$=$OD$=2$.\\\\\nTherefore, quadrilateral ABCD is symmetric about the $x$ axis and the $y$ axis.\\\\\nTherefore, when line CP passes through the midpoint of AB or AD, it divides quadrilateral ABCD into two parts with an area ratio of $1:3$.\\\\\nWhen point P is the midpoint of AB (denoted as $P_{1}$), take the midpoint of OA, denoted as E, and draw line $P_{1}E$ such that $P_{1}E \\parallel y$ axis, and $P_{1}E= \\frac{1}{2} OB=1$,\\\\\nthus $P_{1}\\left( \\frac{1}{2}, 1\\right)$.\\\\\nLet M be the midpoint of AD, by symmetry, M$\\left( \\frac{1}{2}, -1\\right)$,\\\\\nthe equation of line CM can be obtained from the coordinates of C and M as $y=-\\frac{2}{3}x-\\frac{2}{3}$.\\\\\nBy solving $\\begin{cases}y=-\\frac{2}{3}x-\\frac{2}{3}\\\\ y=-2x+2\\end{cases}$, we find $\\begin{cases}x=2\\\\ y=-2\\end{cases}$\\\\\nTherefore, $P_{2}(2, -2)$\\\\\nIn conclusion, the coordinates of point P are $\\boxed{\\left( \\frac{1}{2}, 1\\right)}$ or $\\boxed{(2, -2)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50561277_51.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AD$ is the median of $\\triangle ABC$, with points $E$ and $F$ located on $AD$, and $AE=EF=DF$. Connect $BE$ and $CF$. If the area of $\\triangle ABC$ is 18, what is the area of the shaded region?\n",
    "completion": "\\textbf{Solution:} Since $AD$ is the median of $\\triangle ABC$, and the area of $\\triangle ABC$, $S_{\\triangle ABC}=18$,\\\\\nthen $S_{\\triangle ABD}=S_{\\triangle ADC}=\\frac{1}{2}S_{\\triangle ABC}$,\\\\\nsince $S_{\\triangle ABC}=18$,\\\\\nthen $S_{\\triangle ABD}=S_{\\triangle ADC}=9$,\\\\\nsince $AE=EF=DF$,\\\\\nthen $S_{\\triangle ABE}=\\frac{1}{3}S_{\\triangle ABD}=3$, $S_{\\triangle DCF}=\\frac{1}{3}S_{\\triangle ADC}=3$,\\\\\ntherefore, the area of the shaded region $S_{\\text{shaded}}=S_{\\triangle ABE}+S_{\\triangle DCF}=3+3=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52634473.png"
  },
  {
    "solution": "\\boxed{-18}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the equation of line $l$ is $y=kx+b$. Line $l$ intersects the $x$-axis at point $A$ and intersects the $y$-axis at point $B(0,12)$, and the area ${S}_{\\triangle AOB}=24$. If point $P$ is a point on line $l$ and ${S}_{\\triangle AOP}=36$, determine the coordinates of point $P$.\n",
    "completion": "\\textbf{Solution:} Let $P(m, 3m+12)$, \\\\\nsince $S_{\\triangle AOP} = 36$, \\\\\nthus $\\frac{1}{2} \\cdot OA \\cdot |y_P| = 36$, that is $\\frac{1}{2} \\times 4 \\times |3m + 12| = 36$, \\\\\ntherefore $|3m + 12| = 18$, \\\\\nsolving for $m$, we get $m = 2$ or $m = -10$, \\\\\nthus $P(2, 18)$ or $P(-10, -18)$. \\\\\nHence, the coordinates of point $P$ are $P(2, \\boxed{18})$ or $P(-10, \\boxed{-18})$.",
    "image_path": "dataset/mm_math/images/MM_Math/52424049_221.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $C$ is the midpoint of line segment $AB$, and point $D$ is on $AB$ with $AD=\\frac{1}{3}AB$. If $CD=2$, what is the length of line segment $AB$?\n",
    "completion": "\\textbf{Solution:} Given that $C$ is the midpoint of $AB$, \\\\\nwe have $AC= \\frac{1}{2}AB$, \\\\\nsince $AD= \\frac{1}{3}AB$, \\\\\nit follows that $CD=AC-AD= \\frac{1}{6}AB$, \\\\\ngiven that $CD=2$, \\\\\nthus $AB=6CD=\\boxed{12}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51419997_164.png"
  },
  {
    "solution": "\\boxed{\\frac{25}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the graphs of the functions $y_{1}=2x+b$ and $y_{2}=ax-3$ intersect at point $P(-2, -5)$, while the graphs of these two functions intersect the x-axis at points A and B, respectively. What is the area of $\\triangle ABP$?\n",
    "completion": "\\textbf{Solution:} Since in $y_{1}=2x-1$, by setting $y_{1}=0$, we get $x=\\frac{1}{2}$.\\\\\nHence $A\\left(\\frac{1}{2}, 0\\right)$.\\\\\nSince in $y_{2}=x-3$, by setting $y_{2}=0$, we get $x=3$,\\\\\nHence $B\\left(3, 0\\right)$.\\\\\nTherefore, the area $S_{\\triangle ABP}=\\frac{1}{2}AB\\times 5=\\frac{1}{2}\\times \\frac{5}{2}\\times 5=\\boxed{\\frac{25}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53706598_203.png"
  },
  {
    "solution": "\\boxed{9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, quadrilateral ABCD is a square, with diagonals AC and BD intersecting at point O. Point E is any point on AD other than the endpoints. A line is drawn from point O such that $OF \\perp OE$ and intersects CD at point F. If $AB=6$, what is the area of quadrilateral EOFD?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a square, \\\\\nit follows that OA=OB=OC=OD, AC$\\perp$BD, $\\angle$ODA=$\\angle$OCD=45$^\\circ$,\\\\\nSince OE$\\perp$OF,\\\\\nit follows that $\\angle$EOF=$\\angle$COD=90$^\\circ$,\\\\\ntherefore $\\angle$EOD=$\\angle$FOC,\\\\\nthus $\\triangle$EOD$\\cong$$\\triangle$FOC,\\\\\ntherefore the area of $\\triangle$EOD = the area of $\\triangle$FOC,\\\\\nSince OA=OB and $\\angle$AOB=90$^\\circ$,\\\\\nit follows that 2OA$^{2}$=AB$^{2}$=6$^{2}$,\\\\\nSolving for OA$^{2}$ gives 18,\\\\\nSince the area of quadrilateral EOFD = area of $\\triangle$EOD + area of $\\triangle$FOD = area of $\\triangle$FOC + area of $\\triangle$FOD = area of $\\triangle$COD,\\\\\nTherefore, the area of quadrilateral EOFD = $\\frac{1}{2}OD^{2}=\\frac{1}{2}OA^{2}=\\boxed{9}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52386504.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, AB is the diameter of $\\odot O$, and the quadrilateral ABCD is inscribed in $\\odot O$. OD intersects AC at point E, where OD $\\parallel$ BC, and AD = CD; if AC = 8 and DE = 2, find the length of BC.\n",
    "completion": "\\textbf{Solution:} Given that $\\because$ OD$\\perp$AC, and AC$=8$,\n\n$\\therefore$ AE$= \\frac{1}{2}$ AC$=4$,\n\nLet the radius of $\\odot$O be r,\n\n$\\because$ DE$=2$,\n\n$\\therefore$ OE$=$OD$-$DE$=r-2$,\n\nIn the $\\triangle AEO$, $AE^{2}$ + $OE^{2}$ = $AO^{2}$,\n\n$\\therefore$ $16+(r-2)^{2}=r^{2}$,\n\nSolving for r we get r$=5$,\n\n$\\therefore$ AB$=2r=10$,\n\nIn $\\triangle ACB$, $BC= \\sqrt{AB^{2}-AC^{2}} = \\sqrt{10^{2}-8^{2}} = \\boxed{6}$,\n\n$\\therefore$ the length of BC is $\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53124735_79.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AD$ and $CE$ are respectively the median and the angle bisector of $\\triangle ABC$, with $AB = AC$. If the area of $\\triangle ABC$ is $20$ and $BC = 4$, what is the length of $AD$?\n",
    "completion": "\\textbf{Solution:} Given that $AD$ is the median of $\\triangle ABC$, and $AB=AC$. \\\\\n$\\therefore AD \\perp BC$,\\\\\nSince the area of $\\triangle ABC$ is 20, and $BC=4$,\\\\\n$\\therefore \\frac{1}{2}BC\\cdot AD=20$,\\\\\n$\\therefore \\frac{1}{2}\\times 4\\times AD=20$,\\\\\n$\\therefore AD=\\boxed{10}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52861537_105.png"
  },
  {
    "solution": "\\boxed{\\left(\\frac{5}{2}, \\frac{5}{4}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=-\\frac{1}{2}x+b$ intersects the $x$-axis at point $A$ and the $y$-axis at point $B$, and intersects the graph of the function $y=kx$ at point $M(1, 2)$. Point $C$ is on the line segment $MA$, and a perpendicular line to the $x$-axis is drawn through point $C$, intersecting the graph of the function $y=kx$ at point $D$. If $2CD=OB$, find the coordinates of point $C$.\n",
    "completion": "\\textbf{Solution:} According to the problem,\\\\\n$\\because b=\\frac{5}{2}$,\\\\\n$\\therefore OB= b=\\frac{5}{2}$,\\\\\n$\\because 2CD=OB$,\\\\\n$\\therefore CD=\\frac{1}{2}OB=\\frac{1}{2}\\times \\frac{5}{2}=\\frac{5}{4}$,\\\\\n$\\because$ point C is on the line $y=-\\frac{1}{2}x+\\frac{5}{2}$,\\\\\n$-\\frac{1}{2}x+\\frac{5}{2}=\\frac{5}{4}$,\\\\\nSolving this, we get $x=\\frac{5}{2}$,\\\\\nThe coordinates of point C are $\\boxed{\\left(\\frac{5}{2}, \\frac{5}{4}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50560215_78.png"
  },
  {
    "solution": "\\boxed{\\frac{8\\sqrt{3}}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, both $\\triangle ABC$ and $\\triangle PQR$ are equilateral triangles, and $AD=BE=CF=\\frac{1}{4}AB$. When $\\angle AFR=\\angle BDP=\\angle CEQ=30^\\circ$, the area of $\\triangle PQR$ is $\\sqrt{3}$. What is the side length of $\\triangle ABC$?\n",
    "completion": "\\textbf{Solution:} As illustrated in the figure, extend FP to intersect AB at point O,\\\\\nsince $\\triangle PQR$ is an equilateral triangle,\\\\\nit follows that $\\angle QPR=60^\\circ$\\\\\nsince $\\angle BDP=30^\\circ$\\\\\nit follows that $\\angle DOP=90^\\circ$\\\\\nLet $AD=BE=CF=a(a>0)$\\\\\ntherefore, $AB=4a$, $AF=3a$\\\\\nsince $\\angle AFP=30^\\circ$\\\\\nit follows that $AO=\\frac{1}{2}AF=\\frac{3}{2}a$, $OF=\\sqrt{AF^{2}-AO^{2}}=\\frac{3\\sqrt{3}}{2}a$,\\\\\nthus $DO=AO-AD=\\frac{3}{2}a-a=\\frac{1}{2}a$\\\\\nsince $\\angle BDP=30^\\circ$\\\\\nit follows that $DP=\\frac{OD}{\\cos 30^\\circ}=\\frac{\\frac{1}{2}a}{\\frac{\\sqrt{3}}{2}}=\\frac{\\sqrt{3}}{3}a$\\\\\ntherefore $OP=\\frac{1}{2}DP=\\frac{1}{2}\\times \\frac{\\sqrt{3}}{3}a=\\frac{\\sqrt{3}}{6}a$\\\\\ntherefore $PR=OF-OP-FR=\\frac{3\\sqrt{3}}{2}a-\\frac{\\sqrt{3}}{6}a-\\frac{\\sqrt{3}}{3}a=\\sqrt{3}a$\\\\\nsince the area of the equilateral triangle $\\triangle PQR$ is $\\sqrt{3}$,\\\\\nit follows that $\\frac{1}{2}\\times PR\\times \\frac{\\sqrt{3}}{2}PR=\\frac{1}{2}\\times \\sqrt{3}a\\times \\frac{\\sqrt{3}}{2}\\times \\sqrt{3}a=\\frac{3\\sqrt{3}}{4}a^{2}=\\sqrt{3}$\\\\\nthus $a^{2}=\\frac{4}{3}$\\\\\ntherefore $a=\\frac{2\\sqrt{3}}{3}$\\\\\ntherefore $AB=4a=4\\times \\frac{2\\sqrt{3}}{3}=\\boxed{\\frac{8\\sqrt{3}}{3}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51322798.png"
  },
  {
    "solution": "\\boxed{2\\pi}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of $\\odot O$, and point $C$ is a point on $\\odot O$. Given $OA=3$, and $\\angle OCA=40^\\circ$, what is the area of the shaded region?\n",
    "completion": "Solution: Since $OA=OC$,\\\\\nit follows that $\\angle OAC=\\angle OCA=40^\\circ$,\\\\\ntherefore, $\\angle COB=80^\\circ$,\\\\\nthus, the area of the shaded region is: $ \\frac{\\pi \\times {3}^{2}\\times 80^\\circ}{360^\\circ}=\\boxed{2\\pi}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52872256.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Point $C$ is on line segment $AB$, point $M$ is the midpoint of $AC$, $AM=1$, $BC=4$. If point $N$ is the midpoint of $BC$, what is the length of $MN$?\n",
    "completion": "\\textbf{Solution:} Since point $M$ is the midpoint of $AC$, $AM=1$,\\\\\n$\\therefore CM=AM=1$,\\\\\nSince point $N$ is the midpoint of $BC$, $BC=4$,\\\\\n$\\therefore CN=\\frac{1}{2}BC=2$,\\\\\n$\\therefore MN=CM+CN=1+2=\\boxed{3}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51212253_187.png"
  },
  {
    "solution": "\\boxed{10\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a square frame with a side length of 10 cm (ignoring the thickness of the frame) is pulled into a quadrilateral $ABCD$. If $\\angle BAD=60^\\circ$, what is the length of $AC$ in cm?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, construct $DE\\perp AB$ intersecting $AB$ at point E, and $CF\\perp AB$ at its extension at point F, and connect $AC$,\\\\\nsince $AB=BC=CD=AD=10$cm, and $\\angle BAD=60^\\circ$,\\\\\nthus $DE=CF=AD\\times \\sin 60^\\circ=10\\times \\frac{\\sqrt{3}}{2}=5\\sqrt{3}$,\\\\\n$BF=BC\\times \\cos 60^\\circ=10\\times \\frac{1}{2}=5$,\\\\\nin right triangle $AFC$, $AC=\\sqrt{AF^{2}+CF^{2}}$,\\\\\nthus $AC=\\sqrt{15^{2}+{(5\\sqrt{3})}^{2}}=\\boxed{10\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53043407.png"
  },
  {
    "solution": "\\boxed{15}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, rectangle ABCD is located in the first quadrant, and AB is parallel to the y-axis. The line M: $y=-x$ is translated in the positive direction of the x-axis, and the segment EF intercepted by rectangle ABCD has its length $l$ as a function of the translation distance $a$ as shown in the figure. What is the area of rectangle ABCD?\n",
    "completion": "\\textbf{Solution}: From the graph, it is known that when $a=1$, the line $M$ passes through point $A$.\\\\\nWhen $a=4$, line $M$ goes through point $B$.\\\\\nWhen $a=6$, line $M$ goes through point $D$.\\\\\nWhen $a=9$, line $M$ goes through point $C$.\\\\\nTherefore, when $F$ moves on segment $BC$, $4\\leq a \\leq 9$, and $BC=9-4=5$,\\\\\nWhen $F$ moves on segment $AB$, $1\\leq a \\leq 4$,\\\\\nAlso, at this time $AF=AE$,\\\\\n$\\therefore AB=4-1=3$.\\\\\nHence, the area of rectangle $ABCD$ is $5\\times 3=\\boxed{15}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52076831.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that isosceles $\\triangle ABC$, with $AB=BC$, and the circle with diameter $AB$ intersects $AC$ at point D. The tangent of $\\odot O$ through point D intersects $BC$ at point E. If $CD=4\\sqrt{5}, CE=8$, then what is the radius of $\\odot O$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, connect OD and BD,\\\\\n$\\because$ DE is a tangent,\\\\\n$\\therefore$ OD $\\perp$ DE,\\\\\n$\\because$ AB is a diameter,\\\\\n$\\therefore$ $\\angle ADB = 90^\\circ$, and AB=BC,\\\\\n$\\therefore$ AD=CD=$4\\sqrt{5}$, and AO=OB,\\\\\n$\\therefore$ DO $\\parallel$ BC, and DE $\\perp$ OD,\\\\\n$\\therefore$ DE $\\perp$ EC,\\\\\n$\\therefore$ DE$=\\sqrt{CD^{2}-CE^{2}}=\\sqrt{80-64}=4$,\\\\\n$\\because$ $\\tan C = \\frac{BD}{CD} = \\frac{DE}{EC} = \\frac{4}{8} = \\frac{1}{2}$,\\\\\n$\\therefore$ BD=$2\\sqrt{5}$,\\\\\n$\\therefore$ AB$=\\sqrt{AD^{2}+DB^{2}}=10$,\\\\\n$\\therefore$ OA=$\\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52917439.png"
  },
  {
    "solution": "\\boxed{90^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $AE$ is a straight line, and $O$ is a point on the straight line $AE$. $OB$ is the bisector of $\\angle AOC$, and $OD$ is the bisector of $\\angle COE$. Given that $\\angle AOB=30^\\circ$, what is the degree measure of $\\angle BOD$?\n",
    "completion": "\\textbf{Solution:} Since $OD$ is the bisector of $\\angle COE$,\\\\\nit follows that $\\angle COD= \\frac{1}{2}\\angle COE=60^\\circ$,\\\\\ntherefore, $\\angle BOD=\\angle COD+\\angle BOC=60^\\circ+30^\\circ=\\boxed{90^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52579729_197.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the grid is a square grid, and A, B, C, D, E, F are the intersections of the grid lines. What is the ratio of the area of $\\triangle ABC$ to the area of $\\triangle DEF$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, let the side length of the small squares in the square grid be 1,\\\\\nthen we have $AB=1$, $BC=\\sqrt{1^2+2^2}=\\sqrt{5}$, $AC=\\sqrt{1^2+1^2}=\\sqrt{2}$, $DE=2$, $EF=\\sqrt{2^2+2^2}=2\\sqrt{2}$, $DF=\\sqrt{2^2+4^2}=2\\sqrt{5}$,\\\\\n$\\therefore \\frac{AB}{DE}=\\frac{BC}{DF}=\\frac{AC}{EF}=\\frac{1}{2}$,\\\\\n$\\therefore \\triangle ABC\\sim \\triangle DEF$,\\\\\n$\\therefore S_{\\triangle ABC} : S_{\\triangle DEF}=\\left(\\frac{1}{2}\\right)^2=\\boxed{\\frac{1}{4}}$,",
    "image_path": "dataset/mm_math/images/MM_Math/51379110.png"
  },
  {
    "solution": "\\boxed{3\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given: As shown in the figure, in trapezoid $ABCD$, $AD\\parallel BC$, $AB=DC=6$, $E$ is a point on the diagonal $BD$, $DE=4$, and $\\angle BCE=\\angle ABD$. $\\triangle ABD\\sim \\triangle ECB$; if the ratio $AD:BC=3:5$, what is the length of $AD$?\n",
    "completion": "\\textbf{Solution:} In trapezoid ABCD, we have $AD\\parallel BC$ and $AB=DC=6$, \\\\\n$\\therefore \\angle ABC=\\angle BCD$, \\\\\nAlso, since $\\angle BCE=\\angle ABD$, \\\\\n$\\therefore \\angle DBC=\\angle DCE$, \\\\\nSince $\\angle BDC=\\angle CDE$, \\\\\n$\\therefore \\triangle BDC\\sim \\triangle CDE$, \\\\\n$\\therefore \\frac{CD}{DE}=\\frac{BD}{CD}$, \\\\\nSince $DC=6$ and $DE=4$, \\\\\n$\\therefore BD=9$, \\\\\n$\\therefore BE=BD-DE=5$, \\\\\nSince $\\triangle ABD\\sim \\triangle ECB$, \\\\\n$\\therefore \\frac{AD}{BE}=\\frac{BD}{BC}$, \\\\\nGiven $AD\\colon BC=3\\colon 5$, \\\\\nLet $AD=3x$ and $BC=5x$, \\\\\n$\\therefore \\frac{3x}{5}=\\frac{9}{5x}$, \\\\\nSolving, we get $x=\\sqrt{3}$ (neglecting the negative value), \\\\\n$\\therefore x=\\sqrt{3}$, hence $AD=3\\sqrt{3}$. Therefore, the length of $AD$ is $\\boxed{3\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51353491_177.png"
  },
  {
    "solution": "\\boxed{55^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $D$ is a point on $BC$, $CA=CD$, $CE$ bisects $\\angle ACB$, intersects $AB$ at point $E$, and $DE$ is connected. If $\\angle A=100^\\circ$ and $\\angle B=45^\\circ$, what is the measure of $\\angle BED$ in degrees?\n",
    "completion": "\\textbf{Solution:} Given that $CE$ bisects $\\angle ACB$, \\\\\nthus $\\angle ACE = \\angle DCE$, \\\\\nin $\\triangle ACE$ and $\\triangle DCE$, \\\\\n\\[\n\\left\\{\n\\begin{array}{l}\nCA = CD \\\\\n\\angle ACE = \\angle DCE \\\\\nCE = CE\n\\end{array}\n\\right.,\n\\]\ntherefore $\\triangle ACE \\cong \\triangle DCE$ (SAS), \\\\\nthus $\\angle CDE = \\angle A = 100^\\circ$, \\\\\nsince $\\angle B = 45^\\circ$, \\\\\ntherefore $\\angle BED = \\angle CDE - \\angle B = 100^\\circ - 45^\\circ = \\boxed{55^\\circ}$,",
    "image_path": "dataset/mm_math/images/MM_Math/52838016.png"
  },
  {
    "solution": "\\boxed{x \\geq 1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $I_1 \\colon y=x+1$ intersects with the line $I_2 \\colon y=mx+n$ at point $P(a, 2)$. What is the solution set of the inequality $x+1\\ge mx+n$ in terms of $x$?\n",
    "completion": "\\textbf{Solution:} Since $y=x+1$ intersects with the line $I_2 \\colon y=mx+n$ at point $P(1, 2)$,\\\\\nsubstituting $y=2$ into $y=x+1$ gives $x=1$,\\\\\ntherefore, the coordinates of point $P$ are $(1, 2)$;\\\\\nFrom the graph, it is known that when $x \\geq 1$, $x+1 \\geq mx+n$.\\\\\nThus, the answer is: $\\boxed{x \\geq 1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50482454.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, there are three points $D$, $B$, and $E$ successively located on the line segment $AC$, with $AD = \\frac{1}{2}DB$, $E$ being the midpoint of $BC$, and $BE = \\frac{1}{5}AC = 2$. Calculate the length of line segment $DE$.\n",
    "completion": "\\textbf{Solution:} Since $AD=\\frac{1}{2}DB$,\\\\\nit follows that $AB=AD+DB=\\frac{3}{2}DB$,\\\\\nSince $AB=6$,\\\\\nit follows that $DB=4$,\\\\\nAlso, since $BE=2$,\\\\\nit follows that $DE=DB+BE$\\\\\n$=4+2=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53119836_135.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{3}}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the slope of the water-facing side of the river embankment, with a height of $AC=1$ and a horizontal distance $BC=\\sqrt{3}$. What is the slope of the slope $AB$?\n",
    "completion": "\\textbf{Solution}: Let $\\because$ the height of the slope $AC=1$, the horizontal distance $BC=\\sqrt{3}$,\\\\\n$\\therefore$ the gradient of the slope $AB$ is $\\tan B=\\frac{AC}{BC}=\\frac{1}{\\sqrt{3}}=\\boxed{\\frac{\\sqrt{3}}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52602230.png"
  },
  {
    "solution": "\\boxed{130^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the quadrilateral $ABCD$ is an axisymmetric figure with line $AC$ as its axis of symmetry. If $\\angle BAC = 65^\\circ$ and $\\angle B = 50^\\circ$, what is the measure of $\\angle BCD$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle BAC=65^\\circ$ and $\\angle B=50^\\circ$,\\\\\nthus $\\angle BCA=180^\\circ-\\angle B-\\angle BAC=180^\\circ-50^\\circ-65^\\circ=65^\\circ$,\\\\\nsince quadrilateral ABCD is an axisymmetric figure,\\\\\ntherefore $\\angle BCA=\\angle DCA=65^\\circ$,\\\\\nhence $\\angle BCD=\\angle BCA+\\angle DCA=\\boxed{130^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53044213.png"
  },
  {
    "solution": "\\boxed{\\frac{12\\sqrt{5}}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, point O is on side AC. Circle $\\odot O$ passes through point C and is tangent to hypotenuse AB at point D, intersecting side AC at point E. $\\angle ACD=\\frac{1}{2}\\angle B$. Given that $BC=6$ and $AC=8$, what are the lengths of $AD$ and $CD$?\n",
    "completion": "\\textbf{Solution:} In $\\triangle ABC$, we have $AB=\\sqrt{6^2+8^2}=10$,\\\\\nsince $OC \\perp CB$ and $OC$ is the radius,\\\\\nthus $BC$ is the tangent,\\\\\ntherefore $BD=BC=6$,\\\\\nthus $AD=4$,\\\\\nlet the radius of $\\odot O$ be $r$, then $OD=OC=r$, $OA=8-r$,\\\\\nin $\\triangle AOD$, we have $r^2+4^2=(8-r)^2$,\\\\\nsolving this, we get $r=3$,\\\\\ntherefore, $OC=3$,\\\\\ndraw $OB$ intersecting $CD$ at $H$,\\\\\nsince $OC=OD$ and $BC=BD$,\\\\\nthus $OB$ perpendicularly bisects $CD$,\\\\\nin $\\triangle OCB$, we have $OB=\\sqrt{3^2+6^2}=3\\sqrt{5}$,\\\\\nbecause $\\frac{1}{2}OB \\cdot CH=\\frac{1}{2}OC \\cdot BC$,\\\\\nthus $CH=\\frac{3 \\times 6}{3\\sqrt{5}}=\\frac{6\\sqrt{5}}{5}$,\\\\\ntherefore, $CD=2CH=\\boxed{\\frac{12\\sqrt{5}}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51212830_91.png"
  },
  {
    "solution": "\\boxed{96}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, the perimeter of rhombus $ABCD$ is 40, and the diagonal $BD=12$. What is the area of the rhombus $ABCD$?\n",
    "completion": "\\textbf{Solution:} Given that the perimeter of rhombus $ABCD$ is 40,\\\\\nthus, the length of side $BC=10$,\\\\\nsince $BD=12$,\\\\\nthen $OB=\\frac{1}{2}BD=6$,\\\\\nthus $OC=\\sqrt{BC^{2}-OB^{2}}=\\sqrt{10^{2}-6^{2}}=8$,\\\\\ntherefore $AC=2OC=16$,\\\\\nhence the area $S_{\\text{rhombus}ABCD}=\\frac{1}{2}AC\\cdot BD=\\frac{1}{2}\\times 12\\times 16=\\boxed{96}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52076050.png"
  },
  {
    "solution": "\\boxed{24}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $BD$ bisects $\\angle ABC$. If $AB=8$, $BC=12$, and the area of $\\triangle ABD$ is $16$, then what is the area of $\\triangle CBD$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, draw $DE\\perp AB$ at E, and $DF\\perp BC$ at F, \\\\\n$\\because BD$ bisects $\\angle ABC$, $DE\\perp AB$, $DF\\perp BC$, \\\\\n$\\therefore DE=DF$, \\\\\n$\\because {S}_{\\triangle ABD}=\\frac{1}{2}AB\\times DE=16$, \\\\\n$\\therefore DE=4$, \\\\\n$\\therefore DF=4$, \\\\\n$\\because BC=12$, \\\\\n$\\therefore {S}_{\\triangle BCD}=\\frac{1}{2}\\times BC\\times DF=\\frac{1}{2}\\times 12\\times 4=\\boxed{24}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52604233.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the lines $l_{1} \\parallel l_{2} \\parallel l_{3}$, and the lines $a$ and $b$ intersect $l_{1}$, $l_{2}$, and $l_{3}$ at points A, B, C and D, E, F, respectively. If $AB: BC=1: 2$ and $DE=2$, then what is the length of $EF$?\n",
    "completion": "\\textbf{Solution:} Given $\\because$ lines $l_{1}\\parallel l_{2}\\parallel l_{3}$,\\\\\n$\\therefore$ $\\frac{AB}{BC}=\\frac{DE}{EF}=\\frac{1}{2}$,\\\\\n$\\therefore$ EF=2DE=2$\\times$2=$\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52661212.png"
  },
  {
    "solution": "\\boxed{-5\\leq x<0}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the rhombus $ABCD$, where $AD\\parallel$ the $x$-axis, the coordinates of point $A$ are $(0,3)$, and the coordinates of point $B$ are $(4,0)$. The line containing side $CD$, denoted by $y_1=mx+n$, intersects the $x$-axis at point $C$ and intersects the hyperbola $y_2= \\frac{k}{x}$ ($x<0$) at point $D$. What is the range of values for the independent variable $x$ such that $y_1 - y_2 \\leq 0$ when $x<0$?\n",
    "completion": "\\textbf{Solution:} When $x<0$, the range of the independent variable $x$ that makes $y_1-y_2\\leq 0$ is $\\boxed{-5\\leq x<0}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50591287_144.png"
  },
  {
    "solution": "\\boxed{\\frac{22}{9}\\pi -\\frac{3\\sqrt{7}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, $\\triangle ABC$ is an inscribed triangle of circle $\\odot O$ with $AB$ as the diameter. $BD$ is tangent to $\\odot O$ at point $B$ and intersects the extension line of $AC$ at point $D$. $E$ is the midpoint of $BD$, and $CE$ intersects the extension line of $BA$ at point $F$. $BD=8$, and $BE = \\frac{1}{3}EF$. If $\\angle F= 20^\\circ$ and $BC=3\\sqrt{2}$, what is the area of the shaded region in the diagram?\n",
    "completion": "\\textbf{Solution:} Let OM $\\perp$ BC at point M,\\\\\n$\\therefore$ BM=$\\frac{1}{2}$BC=$\\frac{3\\sqrt{2}}{2}$,\\\\\nIn $\\triangle BMO$, OM=$\\sqrt{OB^{2}-BM^{2}}=\\sqrt{(2\\sqrt{2})^{2}-\\left(\\frac{3\\sqrt{2}}{2}\\right)^{2}}=\\frac{\\sqrt{14}}{2}$,\\\\\n$\\therefore$ Area of $\\triangle BOC$=$\\frac{1}{2}$BC$\\cdot$OM=$\\frac{1}{2}\\times 3\\sqrt{2}\\times\\frac{\\sqrt{14}}{2}=\\frac{3\\sqrt{7}}{2}$,\\\\\n$\\because \\angle F=20^\\circ$,\\\\\n$\\therefore \\angle BOC=\\angle F+\\angle OCF=110^\\circ$,\\\\\n$\\therefore$ Area of the sector BOC=$\\frac{110}{360}\\pi \\times (2\\sqrt{2})^{2}=\\frac{22}{9}\\pi$,\\\\\n$\\therefore$ Area of the shaded region=Area of the sector BOC-Area of $\\triangle BOC$=$\\frac{22}{9}\\pi -\\frac{3\\sqrt{7}}{2}$,\\\\\n$\\therefore$ The area of the shaded part in the figure is $\\boxed{\\frac{22}{9}\\pi -\\frac{3\\sqrt{7}}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51446035_119.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, $AC=3$, $BC=4$. On segment $AB$, a moving point $M$ starts at point $A$ and moves towards point $B$ at a constant speed, while another moving point $N$ starts at point $B$ and moves towards point $A$ at a constant speed. When one of the points $M$ or $N$ stops moving, the other point stops moving as well. Perpendicular lines are drawn from points $M$ and $N$ to $AB$, intersecting the lines through the right-angle sides $AC$ and $BC$ at points $D$ and $E$, respectively. Line $DE$ is drawn. If the movement time is $t$ seconds, throughout the movement, the quadrilateral $DENM$ always forms a rectangle (except when points $M$ and $N$ coincide). As shown in the figure, let the length of $DM$ be $x$, and the area of rectangle $DENM$ be $S$. Find the functional relationship between $S$ and $x$; what value of $x$ maximizes the area of rectangle $DENM$? What is the maximum area?\n",
    "completion": "\\textbf{Solution:} Since in $\\triangle ABC$, $\\angle C=90^\\circ$, $AC=3$, $BC=4$, we have $AB=\\sqrt{AC^2+BC^2}=5$. Because $\\triangle ADM\\sim \\triangle ABC$, we get $\\frac{AM}{AC}=\\frac{MD}{BC}$. Considering $DM=x$, we find $\\frac{AM}{3}=\\frac{x}{4}$, which means $AM=\\frac{3}{4}x$. Therefore, $AD=\\sqrt{AM^2+DM^2}=\\frac{5}{4}x$. Thus, $CD=AC-AD=3-\\frac{5}{4}x$. Given $\\triangle ADM\\sim \\triangle ABC$ and $\\triangle DEC\\sim \\triangle ABC$, it follows that $\\triangle ADM\\sim \\triangle DEC$. Therefore, $\\frac{DE}{AD}=\\frac{CD}{AM}$, that is $\\frac{DE}{\\frac{5}{4}x}=\\frac{3-\\frac{5}{4}x}{\\frac{3}{4}x}$, which leads to $DE=5-\\frac{25}{12}x$. Thus, the area of quadrilateral $DENM$ is $DM\\cdot DE=5x-\\frac{25}{12}x^2=-\\frac{25}{12}\\left(x^2-\\frac{12}{5}x+\\frac{36}{25}\\right)+3=-\\frac{25}{12}\\left(x-\\frac{6}{5}\\right)^2+3$. Since $-\\frac{25}{12}<0$, the area of rectangle $DENM$ reaches its maximum when $x=\\frac{6}{5}$, with the maximal area being \\boxed{3}.",
    "image_path": "dataset/mm_math/images/MM_Math/51433258_212.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the direct proportion function $y = x$ intersects with the graph of the inverse proportion function $y = \\frac{1}{x}$ at points A and C. $AB \\perp x$-axis at B, and $CD \\perp x$-axis at D. What is the area of the quadrilateral $ABCD$?\n",
    "completion": "\\textbf{Solution:} Solve the system of equations:\n\\[\\begin{cases}\ny=x\\\\ \ny=\\frac{1}{x}\n\\end{cases}\\]\n\nWe obtain points A(1,1) and C(-1,-1).\n\nTherefore, the area of $\\triangle ADB = \\frac{1}{2} \\times (1+|-1|) \\times 1 = 1$, the area of $\\triangle BCD = \\frac{1}{2} \\times (1+|-1|) \\times |-1| = 1$,\n\nThus, the area of the quadrilateral $ABCD = S_{\\triangle ADB} + S_{\\triangle BCD} = 1 + 1 = \\boxed{2}$",
    "image_path": "dataset/mm_math/images/MM_Math/53019881.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, with $M(2, 3)$ as the center of the circle and $AB$ as its diameter, the circle is tangent to the x-axis and intersects the y-axis at points A and C. What is the length of $AC$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect $BC$, and assume the circle is tangent to the x-axis at point $D$, connect $MD$ and it intersects $BC$ at point $E$, so $MD\\perp x$ axis,\\\\\n$\\because AB$ is the diameter,\\\\\n$\\therefore \\angle ACB=\\angle AOD=90^\\circ$,\\\\\n$\\therefore BC\\parallel x$ axis,\\\\\n$\\therefore BC\\perp MD$,\\\\\n$\\because M\\left(2,3\\right)$,\\\\\n$\\therefore MB=MD=3$, $CE=EB=2$,\\\\\n$\\therefore CB=4$, $AB=2BM=6$,\\\\\n$\\therefore AC=\\sqrt{AB^{2}-BC^{2}}=\\sqrt{6^{2}-4^{2}}=\\boxed{2\\sqrt{5}}$,",
    "image_path": "dataset/mm_math/images/MM_Math/53019730.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $AB$ is the diameter of $\\odot O$, and $AB=8C$. Point $D$ is on $\\odot O$, $OC\\parallel BD$, intersecting $AD$ at point $E$. Connect $BC$, with $\\angle CBD=30^\\circ$. What is the length of $CE$?\n",
    "completion": "\\textbf{Solution:} Since AB is the diameter of $\\odot O$,\\\\\n$\\therefore \\angle ADB=90^\\circ$,\\\\\nSince OC $\\parallel$ BD,\\\\\n$\\therefore \\angle AEO=\\angle ADB=90^\\circ$,\\\\\nSince $\\angle AOC=60^\\circ$,\\\\\n$\\therefore \\angle OAE=30^\\circ$,\\\\\n$\\therefore$ OE=$\\frac{1}{2}$ OA,\\\\\n$\\therefore$ CE=$\\frac{1}{2}$ OC=$\\frac{1}{2} \\times 4=\\boxed{2}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52872426_108.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the area of $\\triangle ABC$ is 24. Point $D$ is a point on side $AC$, and extending $BD$ intersects the line parallel to $BC$ through point $A$ at point $E$. Connect $EC$, and construct parallelogram $DECF$ with $DE$ and $EC$ as adjacent sides. $DF$ intersects side $BC$ at point $H$. Connect $AH$. When $AD = \\frac{1}{2}CD$, what is the area of $\\triangle AHC$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, extend $HD$ to intersect $AG$ at point $Q$, \\\\\n$\\because$ quadrilateral $DECF$, \\\\\n$\\therefore$ $DF \\parallel CE$, $ED \\parallel CF$, \\\\\n$\\therefore$ $\\angle CFH = \\angle EDQ$, \\\\\nAlso, $\\because$ $BC \\parallel AQ$, \\\\\n$\\therefore$ quadrilateral $HQEC$ is a parallelogram, \\\\\n$\\therefore$ $EQ = CH$, \\\\\nAlso, $\\angle EQD = \\angle CHF$, \\\\\n$\\therefore$ $\\triangle EDQ \\cong \\triangle CHF$ (AAS), \\\\\n$\\therefore$ $S_{\\triangle EDQ} = S_{\\triangle CHF}$, \\\\\n$\\therefore$ $S_{\\square HQEC} = S_{\\square DECF}$, \\\\\n$\\because$ the area of $\\triangle ABC$ is 24, \\\\\n$\\therefore$ $S_{\\triangle BEC} = 24$, \\\\\nAlso, $\\because$ $AD = \\frac{1}{2}CD$, \\\\\n$\\therefore$ $S_{\\triangle BDC} = \\frac{2}{3}S_{\\triangle ABC} = 16$, \\\\\n$\\therefore$ $S_{\\triangle DEC} = S_{\\triangle BEC} - S_{\\triangle BDC} = 24 - 16 = 8$, \\\\\n$\\therefore$ $S_{\\square HQEC} = S_{\\square DECF} = 2S_{\\triangle DEC} = 16$, \\\\\n$\\therefore$ $S_{\\triangle AHC} = \\frac{1}{2}S_{\\square HQEC} = \\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51825215.png"
  },
  {
    "solution": "\\boxed{24}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 2, it is known that point $B$ is on line segment $AC$, and point $D$ is on line segment $AB$. If $BD = \\frac{1}{4}AB = \\frac{1}{3}CD$, with $E$ being the midpoint of line segment $AB$ and $EC=16\\, \\text{cm}$, find the length of line segment $AC$.\n",
    "completion": "\\textbf{Solution:} Let $BD=x$,\\\\\nsince $BD=\\frac{1}{4}AB=\\frac{1}{3}CD$,\\\\\nthus $AB=4BD=4x$ and $CD=3BD=3x$,\\\\\ntherefore $BC=DC-DB=3x-x=2x$,\\\\\nthus $AC=AB+BC=4x+2x=6x$.\\\\\nSince $E$ is the midpoint of the segment $AB$,\\\\\nthus $BE=\\frac{1}{2}AB=\\frac{1}{2}\\times 4x=2x$,\\\\\nthus $EC=BE+BC=2x+2x=4x$.\\\\\nFurthermore, since $EC=16$,\\\\\nthus $4x=16$,\\\\\nsolving this gives: $x=4$,\\\\\ntherefore $AC=6x=6\\times 4=\\boxed{24}$.",
    "image_path": "dataset/mm_math/images/MM_Math/54927244_9.png"
  },
  {
    "solution": "\\boxed{-1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the directly proportional function $y=-x$ intersects with the graph of the linear function $y=kx+ \\frac{5}{2}$ (where $k\\neq 0$) at point $P$. What is the solution to the equation $-x=kx+ \\frac{5}{2}$ in terms of $x$?\n",
    "completion": "\\textbf{Solution:} Let the coordinates of point P be $(x, 1)$,\\\\\nSubstituting $(x, 1)$ into $y=-x$,\\\\\n$\\therefore$ $-x=1$,\\\\\n$\\therefore$ $x=-1$,\\\\\nThus, the coordinates of point P are $(-1, 1)$,\\\\\n$\\therefore$ the solution to the equation concerning $x$, $-x=kx+ \\frac{5}{2}$, is $x=\\boxed{-1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50594473.png"
  },
  {
    "solution": "\\boxed{39^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, OD is the bisector of $\\angle BOC$, OE is the bisector of $\\angle AOC$, and $\\angle AOB : \\angle BOC = 3 : 2$. If $\\angle BOE = 13^\\circ$, what is the degree measure of $\\angle DOE$?\n",
    "completion": "\\textbf{Solution:} Let $\\angle AOB=3x$ and $\\angle BOC=2x$.\\\\\nThen, $\\angle AOC=\\angle AOB+\\angle BOC=5x$.\\\\\nSince OE is the bisector of $\\angle AOC$,\\\\\n$\\therefore \\angle AOE= \\frac{1}{2}\\angle AOC= \\frac{5}{2}x$,\\\\\n$\\therefore \\angle BOE=\\angle AOB-\\angle AOE=3x- \\frac{5}{2}x= \\frac{1}{2}x$,\\\\\nSince $\\angle BOE=13^\\circ$,\\\\\n$\\therefore \\frac{1}{2}x=13^\\circ$,\\\\\nSolving, we find: $x=26^\\circ$,\\\\\nSince OD is the bisector of $\\angle BOC$,\\\\\n$\\therefore \\angle BOD= \\frac{1}{2}\\angle BOC=x=26^\\circ$,\\\\\n$\\therefore \\angle DOE=\\angle DOB+\\angle BOE=26^\\circ+13^\\circ=\\boxed{39^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53404083.png"
  },
  {
    "solution": "\\boxed{y=\\frac{12}{x}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, the side $OB$ of the rhombus $OBAC$ lies on the x-axis. The point $A$ is located at $(8,4)$, and $\\tan\\angle COB=\\frac{4}{3}$. If the graph of the inverse proportion function $y=\\frac{k}{x}\\left(k\\neq 0\\right)$ passes through point $C$, what is the equation of the inverse proportion function?\n",
    "completion": "\\textbf{Solution:} Draw a perpendicular from point C to $OB$, intersecting at $M$, as shown in the figure.\\\\\nSince quadrilateral $OBAC$ is a rhombus, with $A(8,4)$,\\\\\nit follows that $CM=4$,\\\\\nAlso, since $\\tan\\angle COB=\\frac{CM}{OM}=\\frac{4}{3}$,\\\\\nit follows that $OM=3$,\\\\\nTherefore, the coordinates of point C are $C(3,4)$,\\\\\nSince point $C(3,4)$ lies on the graph of the inverse proportion function $y=\\frac{k}{x}$,\\\\\nit follows that $k=3\\times 4=12$,\\\\\nTherefore, the equation of the inverse proportion function is $\\boxed{y=\\frac{12}{x}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52602711.png"
  },
  {
    "solution": "\\boxed{9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=5$, $AC<BC$, and the perimeter of $\\triangle ABC$ is $14$. Construct the perpendicular bisector $l$ of $AB$, which intersects $BC$ at point $E$. Connect $AE$. What is the perimeter of $\\triangle ACE$?\n",
    "completion": "\\textbf{Solution:} Given that $AB=5$ and the perimeter of $\\triangle ABC$ is 14, \\\\\nthus $AC+BC=14-5=9$, \\\\\nsince $DE$ is the perpendicular bisector of the segment $AB$, \\\\\nit follows that $EA=EB$, \\\\\nhence the perimeter of $\\triangle ACE$ is $EA+EC+AC=EB+EC+AC=BC+AC=9$, \\\\\ntherefore, the answer is: $\\boxed{9}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52914188.png"
  },
  {
    "solution": "\\boxed{\\frac{15}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the parabola $C_1\\colon y=-\\frac{1}{4}x^2+bx+c$ passes through the points $(-3,0)$ and $(3,3)$, and intersects the y-axis at point C. The parabola $C_1$ is first translated 3 units to the left and then 2 units downward to obtain the parabola $C_2$. The vertex of parabola $C_2$ is D. What is the area of $\\triangle ODC$?\n",
    "completion": "Solution: Since $y=-\\frac{1}{4}x^2+\\frac{1}{2}x+\\frac{15}{4}=-\\frac{1}{4}(x-1)^2+4$,\\\\\nthe vertex of parabola $C_1$ is located at $(1,4)$, and it intersects the y-axis at point $C(0,\\frac{15}{4})$,\\\\\nsince moving parabola $C_1$ 3 units to the left and 2 units down yields parabola $C_2$,\\\\\nthe vertex $D$ of parabola $C_2$ has coordinates $(-2,2)$,\\\\\ntherefore, the area of $\\triangle ODC$ is $\\frac{1}{2}\\times 2\\times \\frac{15}{4}=\\boxed{\\frac{15}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52916776_258.png"
  },
  {
    "solution": "\\boxed{y=x^{2}+2x-3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the Cartesian coordinate system, the graph of the quadratic function \\(y=ax^2+bx+c (a \\ne 0)\\) intersects the x-axis at points \\(A(-3,0)\\) and \\(B(1,0)\\), and intersects the y-axis at point \\(C(0, -3)\\). The vertex is \\(D\\), and its axis of symmetry intersects the x-axis at point \\(E\\). What is the analytic expression of the quadratic function?\n",
    "completion": "\\textbf{Solution:} Let $y=ax^{2}+bx+c\\ (a\\ne 0)$, \\\\\nsince $A(-3,0)$, $B(1,0)$, $C(0, -3)$, \\\\\ntherefore $\\begin{cases}\n(-3)^{2}a+(-3)b+c=0\\\\\na+b+c=0\\\\\nc=-3\n\\end{cases}$, \\\\\nhence $\\begin{cases}\na=1\\\\\nb=2\\\\\nc=-3\n\\end{cases}$, \\\\\ntherefore the quadratic function is: \\boxed{y=x^{2}+2x-3}.",
    "image_path": "dataset/mm_math/images/MM_Math/53147831_63.png"
  },
  {
    "solution": "\\boxed{\\sqrt{5}\\pi}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within the rectangular coordinate system, an arc passes through grid points $A\\left(0, 4\\right)$, $B\\left(4, 4\\right)$, and $C\\left(6, 2\\right)$. Find the length of arc ABC.\n",
    "completion": "\\textbf{Solution:} Connect AC,\\\\\nBased on the grid, we can determine that OP=CQ=2, OA=PQ=4,\\\\\n$\\angle AOP=\\angle PQC=90^\\circ$,\\\\\nBy Pythagorean theorem,\\\\\nAP= $\\sqrt{OP^2+OA^2}=\\sqrt{2^2+4^2}=2\\sqrt{5}=PC$,\\\\\n$\\because AP^2=2^2+4^2=20$, CP$^2=2^2+4^2=20$, AC$^2=2^2+6^2=40$,\\\\\n$\\therefore AP^2+CP^2=AC^2$,\\\\\n$\\therefore \\angle APC=90^\\circ$,\\\\\n$\\therefore$ The length of arc ABC is $\\frac{90^\\circ \\pi \\times 2\\sqrt{5}}{180}=\\boxed{\\sqrt{5}\\pi}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51402739_128.png"
  },
  {
    "solution": "\\boxed{8\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rhombus $ABCD$, $\\angle A=60^\\circ$, $E$ is the midpoint of $AD$, and $F$ is the midpoint of $AB$. If $EF=\\sqrt{5}$, what is the perimeter of rhombus $ABCD$?\n",
    "completion": "\\textbf{Solution:} Since $E$ and $F$ are the midpoints of $AD$ and $AB$ respectively, \\\\\n$\\therefore EF=\\frac{1}{2}BD$, \\\\\nAlso, since $EF=\\sqrt{5}$, \\\\\n$\\therefore BD=2\\sqrt{5}$, \\\\\nSince quadrilateral $ABCD$ is a rhombus with $\\angle A=60^\\circ$, \\\\\n$\\therefore \\triangle ABD$ and $\\triangle BDC$ are both equilateral triangles, \\\\\n$\\therefore AB=2\\sqrt{5}$, \\\\\n$\\therefore$ the side length of the rhombus $ABCD$ is $2\\sqrt{5}\\times 4=\\boxed{8\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52380887.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, on two sides of $\\triangle MON$, segments $OA$ and $OB$ are intercepted such that $OA = OB$; arcs are drawn with points A and B as centers and the length of $OA$ as the radius, intersecting at point C; segments $AC$, $BC$, $AB$, and $OC$ are connected. If $AB = 2\\,\\text{cm}$, and the area of quadrilateral $OACB$ is $4\\,\\text{cm}^{2}$, what is the length of $OC$?\n",
    "completion": "Solution: According to the drawing method, we have $AC=BC=OA$,\\\\\n$\\because$ $OA=OB$,\\\\\n$\\therefore$ $OA=OB=BC=AC$,\\\\\n$\\therefore$ Quadrilateral $OACB$ is a rhombus.\\\\\n$\\because$ $AB=2\\text{cm}$, and the area of quadrilateral $OACB$ is $4\\text{cm}^2$,\\\\\n$\\therefore$ $\\frac{1}{2}AB\\times OC=\\frac{1}{2}\\times 2\\times OC=4$,\\\\\nSolving yields $OC=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53706435.png"
  },
  {
    "solution": "\\boxed{6\\pi -9\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $AB$ is the diameter of $\\odot O$, and the chord $AC$ is parallel to the radius $OD$. Point $D$ is the midpoint of the arc $\\overset{\\frown}{BC}$. If $AC=OD=6$, find the area of the shaded part (the segment $AC$).\n",
    "completion": "\\textbf{Solution:} Let $CE\\perp AB$ at E,\\\\\nas shown in Figure 2, $\\because AC=OD=OC=OA$,\\\\\n$\\therefore \\triangle AOC$ is an equilateral triangle,\\\\\n$\\therefore \\angle AOC=60^\\circ$,\\\\\n$\\therefore {S}_{\\text{sector} AOC}=\\frac{60\\pi \\times {6}^{2}}{360}=6\\pi$,\\\\\n$\\therefore \\angle OCE=30^\\circ$,\\\\\n$\\therefore OE=\\frac{1}{2}OC=\\frac{1}{2}\\times 6=3$,\\\\\n$\\therefore CE=\\sqrt{3}OE=3\\sqrt{3}$,\\\\\n$\\therefore {S}_{\\triangle AOC}=\\frac{1}{2}AO\\cdot CE=\\frac{1}{2}\\times 6\\times 3\\sqrt{3}=9\\sqrt{3}$,\\\\\n$\\therefore {S}_{\\text{shaded}}={S}_{\\text{sector} AOC}-{S}_{\\triangle AOC}=6\\pi -9\\sqrt{3}=\\boxed{6\\pi -9\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52872394_122.png"
  },
  {
    "solution": "\\boxed{\\frac{24}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle BAC=90^\\circ$, $AB=6$, $AC=8$. Point $P$ is any point on $BC$, connect $PA$, with $PA$ and $PC$ as adjacent sides, construct $\\square PAQC$, connect $PQ$, then what is the minimum value of $PQ$?\n",
    "completion": "\\textbf{Solution:} Let $PQ$ intersect $AC$ at point $O$, and draw $O{P}^{\\prime} \\perp BC$ at ${P}^{\\prime}$.\\\\\nIn $\\triangle ABC$, $BC=\\sqrt{AC^{2}+AB^{2}}=\\sqrt{6^{2}+8^{2}}=10$,\\\\\n$\\because \\angle OC{P}^{\\prime} = \\angle ACB$, $\\angle O{P}^{\\prime}C = \\angle CAB$,\\\\\n$\\therefore \\triangle CO{P}^{\\prime} \\sim \\triangle CBA$,\\\\\n$\\therefore \\frac{CO}{CB}=\\frac{O{P}^{\\prime}}{AB}$,\\\\\n$\\therefore \\frac{4}{10}=\\frac{O{P}^{\\prime}}{6}$,\\\\\n$\\therefore O{P}^{\\prime}=\\frac{12}{5}$,\\\\\nWhen $P$ coincides with ${P}^{\\prime}$, the value of $PQ$ is minimized, and the minimum value of $PQ=2O{P}^{\\prime}=\\boxed{\\frac{24}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52623724.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point A is on the graph of the inverse proportion function $y = \\frac{k}{x}$ (where $k > 0$), and AB is perpendicular to the x-axis at point B. If the area of $\\triangle AOB$ is 3, what is the value of $k$?\n",
    "completion": "\\textbf{Solution:} Since point $A$ lies on the graph of the inverse proportion function $y=\\frac{k}{x}$, we have $S_{\\triangle AOB}=\\frac{1}{2}|k|=3$;\\\\\nSince $k>0$, it follows that $k=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52514040.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{97}+5}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In $\\triangle ABC$, $AC=BC=10$, $\\tan A=\\frac{3}{4}$. Points $D$ and $E$ are moving points on sides $AB$ and $AC$ respectively, connecting $DE$. Construct the symmetric figure of $\\triangle ADE$ with respect to $DE$, denoted as $\\triangle A'DE$. As shown in Figure 3, $E$ is the midpoint of side $AC$. Connect $A'B$, let $F$ be the midpoint of $A'B$, and then connect $CF$. During the movement of point $D$, find the maximum length of segment $CF$.\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect $BE$, draw $CH\\perp AB$ at H, take the midpoint O of $BE$, connect $OF, OH$, and draw $OG\\perp CH$ at G,\\\\\n$\\because$ $\\tan A = \\frac{3}{4}$, $AC=10$,\\\\\n$\\therefore$ $AH=8$, $HC=6$,\\\\\n$\\because$ $E$ is the midpoint of side $AC$,\\\\\n$\\therefore$ $AE=CE=5$,\\\\\n$\\therefore$ $A'E=5$,\\\\\n$\\because$ point F is the midpoint of $A'B$, point O is the midpoint of $BE$,\\\\\n$\\therefore$ $FO=\\frac{1}{2}A'E=\\frac{5}{2}$,\\\\\n$\\therefore$ point F moves on a circle with center O and radius $OF$,\\\\\n$\\therefore$ when point F is on the extension line of $CO$, $CF$ has its maximum value,\\\\\n$\\because$ $AH=BH$, point O is the midpoint of $BE$,\\\\\n$\\therefore$ $OH\\parallel AE$, $OH=\\frac{1}{2}AE=\\frac{5}{2}$,\\\\\n$\\therefore$ $\\angle ACH = \\angle CHO$,\\\\\nAnd $\\because$ $\\angle AHC = \\angle OGH = 90^\\circ$,\\\\\n$\\therefore$ $\\triangle ACH \\sim \\triangle OHG$,\\\\\n$\\therefore$ $\\frac{OH}{AC}=\\frac{GH}{CH}=\\frac{OG}{AH}=\\frac{1}{4}$,\\\\\n$\\therefore$ $HG=\\frac{3}{2}, GO=2$,\\\\\n$\\therefore$ $CG=\\frac{9}{2}$,\\\\\nIn $\\triangle GCO$ right-angled at $G$, by Pythagoras theorem, we get: $CO=\\sqrt{\\frac{81}{4}+4}=\\frac{\\sqrt{97}}{2}$,\\\\\n$\\therefore$ the maximum value of $CF$ is $\\boxed{\\frac{\\sqrt{97}+5}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53080186_106.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, there is a rectangular piece of paper ABCD, with point E on side AB. Triangle $BCE$ is folded along line CE, such that point B falls on the diagonal AC, marked as point F. Connect DF. If points D, F, and E are on the same straight line, and $DF=2$, find the length of AE.\n",
    "completion": "\\textbf{Solution:} As shown in Figure 2, in the rectangle ABCD, since $DC\\parallel AB$, it follows that $\\angle DCE = \\angle BEC$. From the folding, we know that $\\angle BEC = \\angle FEC$, therefore $\\angle DCE = \\angle FEC$, which implies $DC = DE$. Moreover, since points D, F, and E lie on the same straight line and $\\angle EFC = \\angle B$, it follows that $\\angle DFC = 90^\\circ$, hence $\\angle DFC = \\angle DAE = 90^\\circ$. Given $CF = CB = DA$, it follows that $\\triangle CDF \\cong \\triangle DEA$, \\\\\ntherefore $AE = DF = \\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52122243_161.png"
  },
  {
    "solution": "\\boxed{-3<x<-1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of line $l_{1}: y=k_{1}x+b$ and the graph of line $l_{2}: y=k_{2}x$ are shown in the same Cartesian coordinate system. The line $l_{1}: y=k_{1}x+b$ intersects the x-axis at point $(-3,0)$. What is the solution set of the inequality $k_{2}x<k_{1}x+b<0$ with respect to $x$?\n",
    "completion": "\\textbf{Solution:} From the graph, it is known that the intersection point of line $l_{1}$ and line $l_{2}$ is ($-1$, $-2$), and on line $l_{1}$, $y$ decreases as $x$ increases.\\\\\nSince $y=k_{1}x+b$ intersects the $x$-axis at the point ($-3$, $0$), the solution set of the inequality $k_{2}x < k_{1}x+b$ with respect to $x$ is $x<-1$,\\\\\ntherefore, the solution set of the inequality $k_{2}x < k_{1}x+b < 0$ with respect to $x$ is $-3<x<-1$,\\\\\nhence, the answer is: $\\boxed{-3<x<-1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50676552.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle A=30^\\circ$, $AC=2\\sqrt{3}$, $\\tan B=\\frac{\\sqrt{3}}{2}$, find the length of $AB$.\n",
    "completion": "\\textbf{Solution}: As shown in the diagram, draw $CD \\perp AB$ at point D,\\\\\n$\\therefore \\angle ADC=\\angle BDC=90^\\circ$\\\\\n$\\because \\angle A=30^\\circ$, $AC=2\\sqrt{3}$,\\\\\n$\\therefore CD=\\frac{1}{2}AC=\\sqrt{3}$,\\\\\n$\\therefore AD=\\sqrt{AC^{2}-CD^{2}}=3$\\\\\n$\\because \\tan B=\\frac{CD}{BD}=\\frac{\\sqrt{3}}{2}$,\\\\\n$\\therefore BD=2$,\\\\\n$\\therefore AB=AD+BD=\\boxed{5}$",
    "image_path": "dataset/mm_math/images/MM_Math/51213326.png"
  },
  {
    "solution": "\\boxed{M(-1, 2)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $l_{1}\\colon y=x+3$ intersects with the line $l_{2}$ that passes through the point $A(3,0)$ at point $C(1,m)$, and intersects with the $x$-axis at point $B$. Point $M$ is on the line $l_{1}$, $MN$ is parallel to the $y$-axis, and intersects line $l_{2}$ at point $N$. If $MN=AB$, find the coordinates of point $M$.\n",
    "completion": "\\textbf{Solution:} In the equation $y = x + 3$, let $y = 0$ to find $x = -3$,\n\n$\\therefore B(-3,0)$,\n\n$\\therefore AB = 3 - (-3) = 6$,\n\nLet $M(a, a + 3)$. Since $MN$ is parallel to the y-axis, we get $N(a, -2a + 6)$,\n\n$MN = |a + 3 - (-2a + 6)| = |3a - 3|$,\n\nSince $MN = AB$,\n\n$\\therefore |3a - 3|= 6$,\n\nSolving this gives $a = 3$ or $a = -1$,\n\n$\\therefore \\boxed{M(3, 6)} or \\boxed{M(-1, 2)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50979784_103.png"
  },
  {
    "solution": "\\boxed{\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that diamond $ABCD$ has a side length of 2, and $\\angle BAD=60^\\circ$. If $DE\\perp AB$ with the foot of the perpendicular at point $E$, then what is the length of $DE$?\n",
    "completion": "\\textbf{Solution:} Since $DE \\perp AB$,\\\\\nhence $\\angle AED = 90^\\circ$,\\\\\nin $\\triangle ADE$, given $\\angle BAD = 60^\\circ$, and $AD = 2$,\\\\\nhence $\\sin60^\\circ = \\frac{DE}{AD}$,\\\\\nthus $DE = AD \\cdot \\sin60^\\circ = 2 \\times \\frac{\\sqrt{3}}{2} = \\boxed{\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51320294.png"
  },
  {
    "solution": "\\boxed{(0,9)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system, the line $y=-\\frac{3}{4}x+m$ intersects the x-axis and y-axis at points B and A, respectively, with the coordinates of point B being $(12,0)$. The line $y=\\frac{3}{8}x$ intersects with line AB at point C. Find the coordinates of point A.\n",
    "completion": "\\textbf{Solution:} Given the line $y=-\\frac{3}{4}x+m$, intersects the y-axis at point B$(12,0)$, \\\\\n$\\therefore$ $0=-\\frac{3}{4}\\times 12+m$, \\\\\n$\\therefore$ $m=9$, \\\\\n$\\therefore$ The equation of line AB is: $y=-\\frac{3}{4}x+9$, \\\\\nWhen $x=0$, $y=9$, \\\\\n$\\therefore$ The coordinates of point A are $\\boxed{(0,9)}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52381062_170.png"
  },
  {
    "solution": "\\boxed{\\frac{9\\sqrt{17}-9}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $E$ is the midpoint of $AD$. After folding $\\triangle ABE$ along $BE$, $\\triangle GBE$ is obtained, and point $G$ is inside the rectangle $ABCD$. Extending $BG$ intersects $DC$ at point $F$ such that $GF=DF$; If $DC=9$ and $DE=2CF$, what is the length of $AD$?\n",
    "completion": "\\textbf{Solution:} \\\\\nSince quadrilateral ABCD is a rectangle, \\\\\ntherefore AB=CD=BG=9, AD=BC, \\\\\nlet CF=x, then DF=9-x, DE=2CF=2x, BC=AD=2DE=4x, \\\\\nsince GF=DF, \\\\\ntherefore GF=9-x, \\\\\ntherefore BF=BG+GF=9+9-x=18-x, \\\\\nin $\\triangle BCF$, $BC^2+CF^2=BF^2$, \\\\\ntherefore $(4x)^2+x^2=(18-x)^2$, \\\\\nupon solving, we get $x=\\frac{9\\sqrt{17}-9}{8}$ or $x=\\frac{9-9\\sqrt{17}}{8}$ (discard), \\\\\ntherefore AD=4x=$\\boxed{\\frac{9\\sqrt{17}-9}{2}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53313106_104.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, AC is the diameter of $\\odot O$, PA is the tangent to $\\odot O$ with A being the point of tangency, point B is on $\\odot O$, PA = PB, and the chord AB intersects PC at point M. As shown in Figure 2, if AB = 4BM, find the value of $ \\frac{MC}{MB}$.\n",
    "completion": "\\textbf{Solution:} Connect PO and intersect AB at point Q, connect BO and BC.\\\\\n$\\because$ PA and PB are the tangents to $\\odot$O,\\\\\n$\\therefore$ PA＝PB, $\\angle$APO＝$\\angle$BPO,\\\\\n$\\therefore$ AQ＝BQ＝ $ \\frac{1}{2}$ AB, $\\angle$PQM＝$90^\\circ$,\\\\\n$\\because$ AB＝4BM,\\\\\n$\\therefore$2BQ＝4BM, that is BQ＝2BM,\\\\\n$\\therefore$ M is the midpoint of BQ.\\\\\nLet BM＝QM＝x, then AQ＝BQ＝2x, AM＝3x,\\\\\n$\\because$ $\\angle$PQM＝$\\angle$CBM＝$90^\\circ$, BM＝QM, $\\angle$PMQ＝CMB,\\\\\n$\\therefore$ $\\triangle$PQM$\\cong$$\\triangle$CBM (ASA),\\\\\n$\\therefore$ PM＝CM,\\\\\n$\\because$ $\\angle$PAC＝$90^\\circ$,\\\\\n$\\therefore$AM＝CM＝3x,\\\\\n$\\therefore$ $ \\frac{MC}{MB}=\\frac{3x}{x}=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51299739_170.png"
  },
  {
    "solution": "\\boxed{\\frac{10}{3}<x<10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $AB=8$, $BC=12$, point E is on $AB$ with $AE=5$, and P is a point on $AD$. The rectangle is folded along $PE$ such that point A falls on point ${A}^{\\prime}$, and $AC$ is connected, intersecting with $PE$ at point F. Let $AP=x$. If point ${A}^{\\prime}$ is inside $\\triangle ABC$, directly write down the range of $x$.\n",
    "completion": "\\textbf{Solution:} As shown in Figure 3-1, when point $A'$ falls on $AC$, \\\\\n$\\because$ $\\angle AEP+\\angle EAC=90^\\circ$, $\\angle ACB+\\angle EAC=90^\\circ$, \\\\\n$\\therefore$ $\\angle AEP=\\angle ACB$, \\\\\nAlso, $\\because$ in rectangle $ABCD$, we have $\\angle EAP=\\angle ABC=90^\\circ$, \\\\\n$\\therefore$ $\\triangle AEP\\sim \\triangle BCA$, \\\\\n$\\therefore$ $\\frac{AP}{AE}=\\frac{AB}{BC}$, \\\\\ni.e., $AP=\\frac{AB}{BC}\\times AE=\\frac{8}{12}\\times 5=\\frac{10}{3}$; \\\\\nIn Figure 3-2, when point $A'$ falls on $CB$, draw $PH\\perp BC$ at H, \\\\\nIt is easy to prove that quadrilateral $APHB$ is a rectangle, \\\\\nThen $PH=AB=8$, $PA=BH$. \\\\\nIn $\\triangle BEA'$, $BE=3$, $EA'=EA=5$, \\\\\n$\\therefore$ $BA'=4$, \\\\\n$\\because$ $\\angle B=\\angle EA'P=\\angle PHA'=90^\\circ$, \\\\\n$\\therefore$ $\\angle BA'E+\\angle PA'H=90^\\circ$, $\\angle PA'H+\\angle A'PH=90^\\circ$, \\\\\n$\\therefore$ $\\angle BA'E=\\angle A'PH$, \\\\\n$\\therefore$ $\\triangle BA'E\\sim \\triangle HPA'$, \\\\\n$\\therefore$ $\\frac{A'H}{HP}=\\frac{BE}{BA'}$, \\\\\n$\\because$ $BE=3$, $PH=AB=8$, $BA'=4$, \\\\\n$\\therefore$ $A'H=\\frac{BE}{BA'}\\times HP=\\frac{3}{4}\\times 8=6$, \\\\\n$\\therefore$ $AP=BH=BA'+A'H=10$. \\\\\nThus, when $\\frac{10}{3}<x<10$, point $A'$ is inside $\\triangle ABC$. Hence, the range of $x$ is $\\boxed{\\frac{10}{3}<x<10}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52793389_155.png"
  },
  {
    "solution": "\\boxed{\\frac{3\\sqrt{10}}{10}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the vertices of $\\triangle ABC$ are all at the vertices of a small square with equal sides. What is the value of $\\cos\\angle BAC$?\n",
    "completion": "\\textbf{Solution:} Let the side length of the small square be 1,\\\\\nDraw CD $\\perp$ AB at D,\\\\\n$S_{\\triangle ABC} = \\frac{1}{2} \\times 2 \\times 2 = 2$,\\\\\nBy Pythagoras theorem, we have: $AB = \\sqrt{2^2 + 4^2} = 2\\sqrt{5}$, $AC = \\sqrt{2^2 + 2^2} = 2\\sqrt{2}$,\\\\\n$\\therefore 2 = \\frac{1}{2} \\times 2\\sqrt{5} \\times CD$,\\\\\nSolving for this, we get: $CD = \\frac{2\\sqrt{5}}{5}$,\\\\\nUsing Pythagoras theorem again, we find: $AD = \\sqrt{AC^2 - CD^2} = \\sqrt{(2\\sqrt{2})^2 - \\left(\\frac{2\\sqrt{5}}{5}\\right)^2} = \\frac{6\\sqrt{5}}{5}$,\\\\\n$\\therefore \\cos\\angle BAC = \\frac{AD}{AC} = \\frac{\\frac{6\\sqrt{5}}{5}}{2\\sqrt{2}} = \\boxed{\\frac{3\\sqrt{10}}{10}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51083538.png"
  },
  {
    "solution": "\\boxed{\\frac{m^2}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $P$ is inside the square $ABCD$, and lines $AP$, $BP$, and $DP$ are connected, with $\\angle 1 = \\angle 2$. If $AB = m$ and $\\tan \\angle 2 = \\frac{1}{2}$, what is the area of $\\triangle ABP$? (Express in terms of $m$).\n",
    "completion": "\\textbf{Solution:} Therefore, the area of $\\triangle ABP$ is $\\boxed{\\frac{m^2}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53065849_137.png"
  },
  {
    "solution": "\\boxed{36}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the length of line segment AB is 24 cm. AB is extended to C such that BC is equal to $\\frac{1}{2}$ of AB. What is the length of AC?\n",
    "completion": "\\textbf{Solution:} Given that $BC = \\frac{1}{2}AB$ and $AB = 24$ cm, \\\\\nthen $BC = \\frac{1}{2} \\times 24 = 12\\, \\text{cm}$, \\\\\ntherefore $AC = AB + BC = \\boxed{36}\\, \\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51355086_59.png"
  },
  {
    "solution": "\\boxed{-\\frac{5}{2}, \\frac{7}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, within the Cartesian coordinate system $xOy$, it is known that the graph of the linear function $y=x+m$ intersects the x-axis and y-axis at points A and B, respectively. It intersects the graph of the linear function $y=-3x+n\\ (n>m)$ at point P. Additionally, the graph of the linear function $y=-3x+n$ intersects the x-axis and y-axis at points C and D, respectively. Given the conditions in (2), there exists a point M on line AB and a point N in the plane such that the quadrilateral formed by vertices D, P, M, N is a rhombus. Please directly provide the coordinates of point N.\n",
    "completion": "\\textbf{Solution:} Given that $m=2, n=6$, we have $D(0,6)$ and $P(1,3)$, thus $OD=6$ and $DP= \\sqrt{(0-1)^{2}+(6-3)^{2}}=\\sqrt{10}$. Since $\\angle PAC=45^\\circ$ and $\\angle AOB=90^\\circ$, it follows that $\\angle ABO=45^\\circ$. We discuss in three scenarios:\n\n(1) Draw a line from point N such that $NE\\perp y$-axis at point E. Since quadrilateral DPMN is a rhombus, it follows that $DN\\parallel PM$, and hence $\\angle EDN=\\angle ABO=45^\\circ$, with $DN=DP= \\sqrt{10}$. Therefore, $EN=DE=\\sqrt{5}$, and thus $OE=6+\\sqrt{5}$, which gives $N(\\sqrt{5},6+\\sqrt{5})$;\n\n(2) Draw a line from point N such that $NF\\perp y$-axis at point F. Since quadrilateral DPMN is a rhombus, it follows that $DN\\parallel PM$, and hence $\\angle FDN=\\angle ABO=45^\\circ$, with $DN=DP= \\sqrt{10}$. Therefore, $EF=DF=\\sqrt{5}$, and thus $OE=6-\\sqrt{5}$, which gives $N(-\\sqrt{5},6-\\sqrt{5})$;\n\n(3) Connect MN to intersect DP at point G. Since quadrilateral DMPN is a rhombus, it follows that $DN\\parallel PM$, $DG=PG$, and $DP\\perp MN$. Given the equation of line AB as $y=x+2$, with $D(0,6)$, the equation for DN becomes $y=x+6$. Suppose $N(n,n+6)$, and since $D(0,6)$, $P(1,3)$, and $DG=PG$, we have $G\\left(\\frac{1}{2},\\frac{9}{2}\\right)$. Considering $DN^{2}=DG^{2}+NG^{2}$, we find $(n-0)^{2}+(n+6-6)^{2}=\\left(0-\\frac{1}{2}\\right)^{2}+\\left(6-\\frac{9}{2}\\right)^{2}+\\left(n-\\frac{1}{2}\\right)^{2}+\\left(n+6-\\frac{9}{2}\\right)^{2}$, which gives $n=-\\frac{5}{2}$. Therefore, $N\\left(-\\frac{5}{2},\\frac{7}{2}\\right)$, leading to $N_1\\left(\\sqrt{5}, \\sqrt{5}+6\\right)$, $N_2\\left(-\\sqrt{5}, 6-\\sqrt{5}\\right)$, and $N_3\\left(-\\frac{5}{2}, \\frac{7}{2}\\right)$.\n\nTherefore, the answer is $N_1\\left(\\boxed{\\sqrt{5}, \\sqrt{5}+6}\\right)$, $N_2\\left(\\boxed{-\\sqrt{5}, 6-\\sqrt{5}}\\right)$, $N_3\\left(\\boxed{-\\frac{5}{2}, \\frac{7}{2}}\\right)$.",
    "image_path": "dataset/mm_math/images/MM_Math/52205271_98.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, $CD \\perp AB$ at $D$, $AD=9$, $CD=6$, then what is the length of $BD$?\n",
    "completion": "\\textbf{Solution:} Given that $CD\\perp AB$,\\\\\nit follows that $\\angle ADC=\\angle CDB=90^\\circ$, $\\angle A+\\angle ACD=90^\\circ$,\\\\\nsince $\\angle ACB=90^\\circ$,\\\\\nit follows that $\\angle ACD+\\angle BCD=90^\\circ$,\\\\\ntherefore $\\angle A=\\angle BCD$,\\\\\ntherefore $\\triangle ADC\\sim \\triangle CDB$,\\\\\ntherefore $\\frac{AD}{CD}=\\frac{CD}{BD}$,\\\\\nsince $AD=9, CD=6$,\\\\\nit follows that $\\frac{9}{6}=\\frac{6}{BD}$,\\\\\ntherefore $BD=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53065888.png"
  },
  {
    "solution": "\\boxed{108^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, $AB$ and $CD$ are diameters of $\\odot O$, with chords $DE$ and $BF$ intersecting the radii $AO$ and $CO$ at points $G$ and $H$, respectively, and $\\angle FBA=\\angle EDC$. Given $DE=BF$, and if $\\overset{\\frown}{AE}=\\overset{\\frown}{EF}=\\overset{\\frown}{FC}$, and $\\angle DOB=\\angle EGO$, what is the degree measure of $\\overset{\\frown}{AC}$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, since $OB = OD$, it follows that $\\angle 1 = \\angle 2$, therefore $\\angle DOB = 180^\\circ - 2\\angle 1$. Since $\\angle EGO = \\angle EDB + \\angle ABD = \\angle 3 + \\angle 1 + \\angle 2 = \\angle 3 + 2\\angle 1$, and $\\angle DOB = \\angle EGO$, it follows that $180^\\circ - 2\\angle 1 = \\angle 3 + 2\\angle 1$, therefore $\\angle 3 = 180^\\circ - 4\\angle 1$. Since $\\overset{\\frown}{AE} = \\overset{\\frown}{EF} = \\overset{\\frown}{FC}$, it follows that $\\angle 3 = 2\\angle ADE$, therefore $\\angle ADE = \\frac{1}{2}\\angle 3$. Since $CD$ is the diameter of $\\odot O$, it follows that $\\overset{\\frown}{AD} + \\overset{\\frown}{AE} + \\overset{\\frown}{CE} = 180^\\circ$, therefore $\\angle 2 + \\angle ADE + \\angle 3 = 90^\\circ$, hence $\\angle 1 +  \\frac{1}{2} \\times (180^\\circ - 4\\angle 1) + (180^\\circ - 4\\angle 1) = 90^\\circ$, therefore $5\\angle 1 = 180^\\circ$, thus $\\angle 1 = 36^\\circ$, hence $\\angle DOB = 180^\\circ - 36^\\circ \\times 2 = 108^\\circ$, therefore $\\angle AOC = 108^\\circ$, hence the degree of $\\overset{\\frown}{AC}$ is $\\boxed{108^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52755194_131.png"
  },
  {
    "solution": "\\boxed{y=x+2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, line $l_1$ passes through points A$(-2, 0)$ and B$(0, 2)$, and line $l_2: y = -4x + 14$ intersects $l_1$ at point C. Find the analytical formula of line $l_1$.\n",
    "completion": "\\textbf{Solution:} Let the equation of line $ {l}_{1}$ be $y=ax+b$.\\\\\nSince line $ {l}_{1}$ passes through points A($-2,0$) and B($0,2$),\\\\\nwe have $\\begin{cases} -2a+b=0\\\\ b=2 \\end{cases}$, solving these gives us $\\begin{cases} a=1\\\\ b=2 \\end{cases}$.\\\\\nThus, the equation of line $ {l}_{1}$ is $\\boxed{y=x+2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52252074_71.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, rectangle OABC is divided into six smaller rectangles by three lines. If D and E are the trisection points on the side CO, and the inverse proportion function $y=\\frac{k}{x}\\left(k\\ne 0\\right)$ just passes through the vertices F, G of the smaller rectangles. If the area of the shaded rectangles in the figure $S_{1}+S_{2}=5$, then what is the value of the inverse proportionality constant $k$?\n",
    "completion": "\\textit{Solution}: Since D and E are the trisection points on side CO, \\\\\nit follows that $S_{1}+S_{2}=\\frac{1}{3}S_{\\text{rectangle} OABC}$, \\\\\nhence $S_{\\text{rectangle} OAGD}=\\frac{2}{3}S_{\\text{rectangle} OABC}$, \\\\\ntherefore $S_{\\text{rectangle} OAGD}=2(S_{1}+S_{2})=10$, \\\\\nthus, $k=\\boxed{10}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52076721.png"
  },
  {
    "solution": "\\boxed{\\frac{4}{13}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, it is known that in the Cartesian coordinate system $xOy$, an isosceles right-angled triangle $ABC$ has its hypotenuse $AB \\perp x$-axis at point $A$. The graph of the inverse proportion function $y = \\frac{k}{x}$ ($k>0$, $x>0$) passing through point $B$ intersects side $AC$ at point $D$. Connect $OB$ and $OC$. If point $D$ is the midpoint of $AC$ and the area of $\\triangle OBC$ is $1$, then what is the value of $k$?\n",
    "completion": "\\textbf{Solution:}\n\nDraw line $EF \\perp y$-axis at point F through point B. Draw line $CE \\perp EF$ at point E through point C. Draw line $CM \\perp AB$ at point M through point C.\\\\\nTherefore, $\\angle ABE = \\angle E = \\angle BMC = 90^\\circ$,\\\\\nThus, quadrilateral $BMCE$ is a rectangle,\\\\\nSince isosceles right triangle $\\triangle ABC$ has hypotenuse $AB \\perp x$-axis at point A,\\\\\nTherefore, $\\angle ABC = 45^\\circ$,\\\\\nTherefore, $\\angle EBC = 45^\\circ$,\\\\\nTherefore, $\\triangle EBC$ is an isosceles right triangle,\\\\\nTherefore, $BE = CE$,\\\\\nTherefore, quadrilateral $BMCE$ is a square,\\\\\nTherefore, $BE = CE = CM = \\frac{1}{2}AB$,\\\\\nSince the graph of the inverse proportion function $y = \\frac{k}{x}$ ($k>0$, $x>0$) passing through point B intersects side $AC$ at point D,\\\\\nLet point A be $(x,0)$, then $B\\left(x, \\frac{k}{x}\\right)$\\\\\nTherefore, $BE = CE = CM = \\frac{k}{2x}$, $AB = \\frac{k}{x}$,\\\\\nTherefore, point $C\\left(x+\\frac{k}{2x}, \\frac{k}{2x}\\right)$\\\\\nSince point D is the midpoint of $AC$,\\\\\nTherefore, $D\\left(x+\\frac{k}{4x}, \\frac{k}{4x}\\right)$,\\\\\nSince $S_{\\triangle OBC} = S_{\\text{trapezoid} EFOC} - S_{\\triangle BFO} - S_{\\triangle BEC} = 1$\\\\\nTherefore, $\\frac{\\left(\\frac{k}{2x}+\\frac{k}{x}\\right)\\left(x+\\frac{k}{2x}\\right)}{2}-\\frac{1}{2}x\\cdot\\frac{k}{x}-\\frac{1}{2}\\cdot\\frac{k}{2x}\\cdot\\frac{k}{2x}=1$\\\\\nRearranging gives $k^2 = (4-k)x^2$;\\\\\nSince points B and D are on the graph of the inverse proportion function,\\\\\nTherefore, $k = \\left(x+\\frac{k}{4x}\\right)\\cdot\\frac{k}{4x}$\\\\\nRearranging gives $x^2 = \\frac{k}{12}$\\\\\nSubstituting the above equation yields $k^2 = (4-k) \\times \\frac{k}{12}$\\\\\nSince $k \\neq 0$\\\\\nSolving yields: $k = \\boxed{\\frac{4}{13}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52204871.png"
  },
  {
    "solution": "\\boxed{13}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: \\textbf{Definition:} A convex quadrilateral in which one pair of opposite sides is equal and the other pair is unequal is called an \"equilateral quadrilateral\". As shown in the figure, in $Rt\\triangle PBC$, $\\angle PCB=90^\\circ$, point A is on side BP, and point D is on side CP. If $BC=11$, $\\tan\\angle PBC=\\frac{12}{5}$, $AB=13$, and quadrilateral ABCD is an \"equilateral quadrilateral\", then what is the length of CD?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, the position of point D is illustrated as follows: \\\\\n(1) If $CD=AB$, at this time point D is at the position $D_{1}$, $CD_{1}=AB=13$; \\\\\n(2) If $AD=BC=11$, at this time point D is at the positions $D_{2}$, $D_{3}$, $AD_{2}=AD_{3}=BC=11$, \\\\\nThrough point A, draw $AE\\perp BC$, $AF\\perp PC$, with E and F as the feet of the perpendiculars, respectively, \\\\\nLet $BE=x$, \\\\\n$\\because$ $\\tan \\angle PBC = \\frac{12}{5}$, \\\\\n$\\therefore$ $AE=\\frac{12}{5}x$, \\\\\nIn $\\triangle ABE$, $AE^{2}+BE^{2}=AB^{2}$, \\\\\ni.e., $x^{2}+\\left(\\frac{12}{5}x\\right)^{2}=13^{2}$, \\\\\nSolving yields: $x_{1}=5$, $x_{2}=-5$ (discard), \\\\\n$\\therefore$ $BE=5$, $AE=12$, \\\\\n$\\therefore$ $CE=BC-BE=6$, \\\\\nSince quadrilateral $AECF$ is a rectangle, we have $AF=CE=6$, $CF=AE=12$, \\\\\nIn $\\triangle AFD_{2}$, $FD_{2}=\\sqrt{AD_{2}^{2}-AF^{2}}=\\sqrt{85}$, \\\\\n$\\therefore$ $CD_{2}=CF-FD_{2}=12-\\sqrt{85}$, \\\\\n$CD_{3}=CF+FD_{2}=12+\\sqrt{85}$, \\\\\nIn summary, the length of CD can be $13$, $12-\\sqrt{85}$, or $12+\\sqrt{85}$. \\\\\nHence, the answer is: $\\boxed{13}$, $12-\\sqrt{85}$, or $12+\\sqrt{85}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51403273.png"
  },
  {
    "solution": "\\boxed{9\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Fold rectangle $ABCD$ as shown in the figure, with $BE$, $EG$, $FG$ being the creases. If vertices $A$, $C$, $D$ all coincide at point $O$, and points $B$, $O$, $G$ lie on the same line, while points $E$, $O$, $F$ lie on another line. If $AD=4\\sqrt{2}$, what is the area of quadrilateral $BEGF$?\n",
    "completion": "\\textbf{Solution:} Hence, the area of the quadrilateral $BEGF$ is $\\boxed{9\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51212662_29.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the line $y=\\frac{1}{2}x+b$ intersects the x-axis at point A, and the line $y=-x+2$ intersects the x-axis at point B. The two lines intersect at point $C(-2, a)$. What are the values of $a$ and $b$?\n",
    "completion": "\\textbf{Solution:} Since point $C$ lies on the line $y=-x+2$, it follows that $a=2+2=\\boxed{4}$. Also, since point $C(-2, 4)$ lies on the line $y=\\frac{1}{2}x+b$, it follows that $\\frac{1}{2}\\times(-2)+b=4$, solving for $b$ gives $b=\\boxed{5}$; therefore, $a=4$, $b=5$.",
    "image_path": "dataset/mm_math/images/MM_Math/52423819_216.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the right triangle $ABC$, with $\\angle ACB=90^\\circ$, $AC=BC=2$, point $P$ is a moving point on segment $AB$. Connect $CP$, and rotate segment $CP$ $90^\\circ$ clockwise around point $C$ to obtain segment $CQ$. Connect $PQ$ and $AQ$. What is the maximum area of $\\triangle PAQ$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, rotate the line segment CP clockwise around point C to obtain line segment CQ, \\\\\n$\\therefore \\angle PCQ=90^\\circ$, CP=CQ,\\\\\n$\\therefore \\angle ACP+\\angle ACQ=90^\\circ$,\\\\\nAlso, $\\because \\angle ACB=90^\\circ$,\\\\\n$\\therefore \\angle BCP+\\angle ACP=90^\\circ$,\\\\\n$\\therefore \\angle BCP=\\angle ACP$,\\\\\n$\\because AC=BC$,\\\\\n$\\therefore \\triangle BPC\\cong \\triangle AQC$ (SAS),\\\\\n$\\therefore \\angle B=\\angle CAQ$, BP=AQ,\\\\\n$\\because BC=AC=2$,\\\\\n$\\therefore \\angle B=\\angle CAQ=\\angle BAC=45^\\circ$,\\\\\n$\\therefore \\angle PAQ=\\angle BAC+\\angle CAQ=90^\\circ$,\\\\\nIn $\\text{Rt}\\triangle ABC$, by the Pythagorean theorem AB= $\\sqrt{BC^2+AC^2}=\\sqrt{2^2+2^2}=2\\sqrt{2}$,\\\\\nLet BP=AQ=x, then $AP=2\\sqrt{2}-x$,\\\\\n$\\therefore S_{\\triangle PAQ}=\\frac{1}{2}x(2\\sqrt{2}-x)=-\\frac{1}{2}(x^2-2\\sqrt{2}x)=-\\frac{1}{2}(x-\\sqrt{2})^2+1$,\\\\\n$\\because a=-\\frac{1}{2}<0$, the function opens downward, and the function has a maximum value,\\\\\nWhen $x=\\sqrt{2}$, $S_{\\triangle PAQ_{\\max}}=\\boxed{1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51285790.png"
  },
  {
    "solution": "\\boxed{9+3\\sqrt{7}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $O$ is the midpoint of the diagonal $AC$ of the rectangle $ABCD$, and point $E$ is the midpoint of $AD$. If $AB=6, \\angle ACB=30^\\circ$, then what is the perimeter of $\\triangle BOE$?\n",
    "completion": "\\textbf{Solution:} Given in rectangle $ABCD$, $AB=6$,\n\nthus $CD=AB=6$, $AD=BC$, $\\angle ABC=\\angle BAD=90^\\circ$,\n\nsince $\\angle ACB=30^\\circ$,\n\nthus $AC=2AB=12$, $AD=BC=\\sqrt{AC^{2}-AB^{2}}=6\\sqrt{3}$,\n\nsince point $O$ is the midpoint of the diagonal $AC$ of rectangle $ABCD$,\n\nthus $OB=\\frac{1}{2}AC=6$,\n\nsince point $E$ is the midpoint of $AD$,\n\nthus $OE=\\frac{1}{2}CD=3$, $AE=\\frac{1}{2}AD=3\\sqrt{3}$,\n\nthus $BE=\\sqrt{AB^{2}+AE^{2}}=3\\sqrt{7}$,\n\nthen the perimeter of $\\triangle BOE$ is $OB+OE+BE=6+3+3\\sqrt{7}=\\boxed{9+3\\sqrt{7}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52386463.png"
  },
  {
    "solution": "\\boxed{5\\times\\left(\\frac{3}{2}\\right)^{4040}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the Cartesian coordinate system, the position of the square $ABCD$ is shown in the diagram, with the coordinates of point $A$ being $\\left(1, 0\\right)$ and the coordinates of point $D$ being $\\left(0, 2\\right)$. Extend $CB$ to meet the $x$-axis at point ${A}_{1}$, and construct the square ${A}_{1}{B}_{1}{C}_{1}C$; extend ${C}_{1}{B}_{1}$ to meet the $x$-axis at point ${A}_{2}$, and construct the square ${A}_{2}{B}_{2}{C}_{2}{C}_{1}$, and so on according to this pattern, what is the area of the 2021st square?\n",
    "completion": "\\textbf{Solution:} \\\\\nLet point A be (1,0), and point D be (0,2),\\\\\ntherefore, OA=1, OD=2, BC=AB=AD=$\\sqrt{5}$,\\\\\nconsidering squares ABCD and A$_{1}$B$_{1}$C$_{1}$C,\\\\\ntherefore, $\\angle$OAD+$\\angle$A$_{1}$AB=$90^\\circ$, $\\angle$ADO+$\\angle$OAD=$90^\\circ$,\\\\\nthus, $\\angle$A$_{1}$AB=$\\angle$ADO,\\\\\nsince $\\angle$AOD=$\\angle$A$_{1}$BA=$90^\\circ$,\\\\\nit follows that $\\triangle$AOD $\\sim$ $\\triangle$A$_{1}$BA,\\\\\nthus, $\\frac{AO}{{A}_{1}B}=\\frac{OD}{AB}$,\\\\\nthus, $\\frac{1}{{A}_{1}B}=\\frac{2}{\\sqrt{5}}$,\\\\\ntherefore, ${A}_{1}B=\\frac{\\sqrt{5}}{2}$,\\\\\nhence, ${A}_{1}{B}_{1}={A}_{1}C={A}_{1}B+BC=\\frac{3}{2}\\sqrt{5}$,\\\\\nSimilarly, it can be derived that, ${A}_{2}{B}_{2}=\\frac{9}{4}\\sqrt{5}=\\left(\\frac{3}{2}\\right)^{2}\\sqrt{5}$,\\\\\nSimilarly, it can be derived that, ${A}_{3}{B}_{3}=\\left(\\frac{3}{2}\\right)^{3}\\sqrt{5}$,\\\\\nSimilarly, it can be derived that, ${A}_{2020}{B}_{2020}=\\left(\\frac{3}{2}\\right)^{2020}\\sqrt{5}$,\\\\\nTherefore, the area of the 2021st square is $\\left[\\left(\\frac{3}{2}\\right)^{2021−1}\\times\\sqrt{5}\\right]^{2}=5\\times\\left(\\frac{3}{2}\\right)^{4040}$.\\\\\nHence, the answer is: $\\boxed{5\\times\\left(\\frac{3}{2}\\right)^{4040}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52537043.png"
  },
  {
    "solution": "\\boxed{\\frac{125}{24}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle B=90^\\circ$, $AB=8$, $BC=6$, and D, E, F are points on sides AB, BC, and AC, respectively. Given that $\\angle BED + \\angle C = 90^\\circ$ and $\\triangle BED$ is symmetrical to $\\triangle FED$ with respect to DE, what is the length of DE?\n",
    "completion": "\\textbf{Solution:} Draw $FN\\perp AB$, with the foot at N, through point E draw $EM\\perp NF$, intersecting the extended line of NF at point M,\\\\\n$\\therefore$ $\\angle FND=\\angle FME=90^\\circ$,\\\\\n$\\because$ $\\angle B=90^\\circ$,\\\\\n$\\therefore$ Quadrilateral NBEM is a rectangle,\\\\\n$\\therefore$ $NB=ME$,\\\\\n$\\because$ $\\angle BED+\\angle C=90^\\circ$, $\\angle C+\\angle A=90^\\circ$,\\\\\n$\\therefore$ $\\angle BED=\\angle A$,\\\\\n$\\because$ $\\angle B=\\angle B$,\\\\\n$\\therefore$ $\\triangle BED\\sim \\triangle BAC$,\\\\\n$\\therefore$ $\\frac{BD}{BE}=\\frac{BC}{BA}=\\frac{6}{8}=\\frac{3}{4}$,\\\\\n$\\because$ Triangle $BED$ and $\\triangle FED$ are symmetric with respect to DE,\\\\\n$\\therefore$ $\\triangle BED\\cong \\triangle FED$,\\\\\n$\\therefore$ $\\angle DFE=\\angle B=90^\\circ$, $DF=BD$, $EF=BE$,\\\\\n$\\therefore$ $\\frac{DF}{EF}=\\frac{BD}{BE}=\\frac{3}{4}$,\\\\\n$\\because$ $\\angle FME=90^\\circ$,\\\\\n$\\therefore$ $\\angle MEF+\\angle MFE=90^\\circ$,\\\\\n$\\because$ $\\angle MFE+\\angle NFD=90^\\circ$,\\\\\n$\\therefore$ $\\angle MEF=\\angle NFD$,\\\\\n$\\therefore$ $\\triangle NDF\\sim \\triangle MFE$,\\\\\n$\\therefore$ $\\frac{NF}{ME}=\\frac{DF}{FE}=\\frac{3}{4}$,\\\\\n$\\therefore$ Let $NF=3x$, $ME=4x$,\\\\\n$\\because$ $\\angle ANF=\\angle B=90^\\circ$, $\\angle A=\\angle A$,\\\\\n$\\therefore$ $\\triangle ANF\\sim \\triangle ABC$,\\\\\n$\\therefore$ $\\frac{NF}{BC}=\\frac{AN}{AB}$,\\\\\n$\\therefore$ $\\frac{3x}{6}=\\frac{8-4x}{8}$,\\\\\n$\\therefore$ $x=1$,\\\\\n$\\therefore$ $NF=3$, $ME=NB=4$,\\\\\nLet $BD=DF=y$,\\\\\nthen $ND=NB-BD=4-y$,\\\\\nIn $\\triangle NDF$, $NF^{2}+ND^{2}=DF^{2}$,\\\\\n$\\therefore$ $3^{2}+(4-y)^{2}=y^{2}$,\\\\\n$\\therefore$ $y=\\frac{25}{8}$,\\\\\n$\\therefore$ $BD=\\frac{25}{8}$,\\\\\n$\\because$ $\\angle B=90^\\circ$, $AB=8$, $BC=6$,\\\\\n$\\therefore$ $AC=\\sqrt{AB^{2}+BC^{2}}=\\sqrt{8^{2}+6^{2}}=10$,\\\\\n$\\because$ $\\triangle BED\\sim \\triangle BAC$,\\\\\n$\\therefore$ $\\frac{DE}{AC}=\\frac{BD}{BC}$,\\\\\n$\\therefore$ $DE=\\boxed{\\frac{125}{24}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52872379.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ is inscribed in $\\odot O$, where $CD$ is the diameter of $\\odot O$. Connect $AD$. If $CD = 2AD$ and $AB = BC = 6$, then what is the radius of $\\odot O$?\n",
    "completion": "Solution: Since CD is a diameter, \\\\\nit follows that $\\angle DAC=90^\\circ$, \\\\\nsince CD=2AD, \\\\\nit follows that $\\angle ACD=30^\\circ$, $\\angle D=60^\\circ$, \\\\\nsince $\\overset{\\frown}{AC} = \\overset{\\frown}{AC}$, \\\\\nit follows that $\\angle D=\\angle B=60^\\circ$, \\\\\nsince AB=BC, \\\\\nit follows that $\\triangle ABC$ is an equilateral triangle, \\\\\ntherefore BC=AC=6; \\\\\nthus $AD^{2}+AC^{2}=CD^{2}$, that is, $AD^{2}+36=4AD^{2}$ \\\\\nSolving this, we find AD=$2\\sqrt{3}$. \\\\\nTherefore, the radius of the circle is $\\boxed{2\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52774800.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the quadrilateral $ABCD$ is inscribed within the semicircle $O$, with $AB$ being the diameter and $C$ is the midpoint of arc $\\overset{\\frown}{BD}$. Extending $AD$ and $BC$ to meet at point $E$, we have $CE = CD$. Given $AB = 5$ and $BC = \\sqrt{5}$, find the length of $AD$.\n",
    "completion": "\\textbf{Solution:} From (1) we have: $AE=AB=5$, $CD=CE=BC=\\sqrt{5}$, $\\triangle CED \\sim \\triangle AEB$,\\\\\n$\\therefore \\frac{CE}{AE}=\\frac{DE}{BE}$, $\\therefore DE=\\frac{\\sqrt{5}\\times 2\\sqrt{5}}{5}=2.$\\\\\n$\\therefore AD=AE-DE=\\boxed{3}.$",
    "image_path": "dataset/mm_math/images/MM_Math/51385629_70.png"
  },
  {
    "solution": "\\boxed{40^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, semicircle $O$ with diameter $AB$ of $\\triangle ABC$ intersects sides $AC$ and $BC$ at points $E$ and $D$, respectively, and $D$ is the midpoint of arc $\\overset{\\frown}{BE}$. If $\\angle A=80^\\circ$, what is the measure of $\\angle DBE$?\n",
    "completion": "\\textbf{Solution:} Draw $OD$, intersect $BE$ at point $H$, connect $ED$,\n\n$\\because D$ is the midpoint of $\\overset{\\frown}{BE}$,\n\n$\\therefore \\overset{\\frown}{DE}=\\overset{\\frown}{DB}$,\n\n$OD\\perp BE$, $BH=EH$,\n\n$\\therefore DE=DB$, $\\angle DEB=\\angle DBE$,\n\n$\\because AB$ is the diameter, then $\\angle AEB=90^\\circ$,\n\n$\\therefore \\angle BEC=90^\\circ$,\n\n$\\therefore \\angle CBE+\\angle BCE=90^\\circ$,\n\n$\\angle BED+\\angle CED=90^\\circ$,\n\n$\\therefore \\angle CBE+\\angle CED=90^\\circ$,\n\n$\\therefore \\angle C=\\angle CED$,\n\n$\\therefore DC=DE$,\n\n$\\therefore DC=DB$,\n\n$\\because OA=OB$,\n\n$\\therefore OD$ is the median of $\\triangle ABC$,\n\n$\\therefore OD \\parallel AC$,\n\n$\\therefore \\angle DOB=\\angle CAB$,\n\n$\\angle ODB=\\angle C$,\n\n$\\because \\angle CAB=80^\\circ$, then $\\angle DOB=80^\\circ$,\n\n$\\therefore \\angle ODB=\\angle OBD=\\frac{1}{2}(180^\\circ-80^\\circ)=50^\\circ$,\n\nAlso, $\\angle OBE=90^\\circ-80^\\circ=10^\\circ$,\n\nThen, $\\angle DBE=\\angle OBD-\\angle OBE=50^\\circ-10^\\circ=\\boxed{40^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51265419_141.png"
  },
  {
    "solution": "\\boxed{y=x^{2}-2x-3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the quadratic function $y=ax^{2}+bx-3$ intersects the $x$-axis at points $A(-1,0)$ and $B(3,0)$, and intersects the $y$-axis at point $C$. The vertex of the parabola is point $M$. What is the analytical expression of the quadratic function?\n",
    "completion": "\\textbf{Solution:} Substitute point A ($-1$, $0$) and point B ($3$, $0$) into $y=ax^{2}+bx-3$,\\\\\nwe get $\\begin{cases}a-b-3=0\\\\ 9a+3b-3=0\\end{cases}$.\\\\\nSolving this, we find $\\begin{cases}a=1\\\\ b=-2\\end{cases}$.\\\\\nTherefore, the equation of the quadratic function is $\\boxed{y=x^{2}-2x-3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52536969_108.png"
  },
  {
    "solution": "\\boxed{118^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\triangle ABC$, it is known that $ \\angle A=58^\\circ, \\angle B=60^\\circ$. What is the degree measure of $ \\angle ADC$?\n",
    "completion": "\\textbf{Solution:} Given: $\\because$ $\\angle 1=58^\\circ$, $\\angle B=60^\\circ$\\\\\n$\\therefore$ $\\angle ADC=\\angle 1+\\angle B=58^\\circ+60^\\circ=\\boxed{118^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/52655364.png"
  },
  {
    "solution": "\\boxed{x<4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given the function $y=2x+b$ and the function $y=kx-3$ intersect at point $P(4, -6)$, what is the solution set of the inequality $kx-3>2x+b$?\n",
    "completion": "\\[ \\text{{\\bf Solution:}} \\, \\text{Given that the function} \\, y=2x+b \\, \\text{and the function} \\, y=kx-3 \\, \\text{intersect at point} \\, P(4,-6), \\]\n\\[ \\text{therefore, the solution set of the inequality} \\, kx-3>2x+b \\, \\text{is} \\, \\boxed{x<4}. \\]",
    "image_path": "dataset/mm_math/images/MM_Math/52381567.png"
  },
  {
    "solution": "\\boxed{70^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB \\parallel CD$, $\\angle 1 = 30^\\circ$, $\\angle 2 = 40^\\circ$, then what is the degree measure of $\\angle 3$?\n",
    "completion": "\\textbf{Solution:} Given $\\because$ AB$\\parallel$CD, $\\angle 1=30^\\circ$,\\\\\n$\\therefore$ $\\angle A=\\angle 1=30^\\circ$,\\\\\nAlso, $\\because$ $\\angle 2=40^\\circ$,\\\\\n$\\therefore$ $\\angle 3=\\angle A+\\angle 2=70^\\circ$,\\\\\nTherefore, $\\angle 3=\\boxed{70^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/53033276.png"
  },
  {
    "solution": "\\boxed{80^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC$. Circle $\\odot O$ is drawn with $AB$ as the diameter, intersecting $BC$ at point $D$ and $AC$ at point $E$. The arc $\\overset{\\frown}{BD}=\\overset{\\frown}{DE}$. If $\\angle BAC=50^\\circ$, find the degree measure of the arc $\\overset{\\frown}{AE}$.\n",
    "completion": "\\textbf{Solution:} Connect $OE$, as shown in Figure 2:\\\\\n$\\because$ $AB$ is the diameter of $\\odot O$,\\\\\n$\\therefore$ $OA$ is the radius,\\\\\n$\\therefore$ $OA=OE$,\\\\\n$\\therefore$ $\\angle OEA=\\angle BAC=50^\\circ$,\\\\\n$\\therefore$ $\\angle AOE=180^\\circ-50^\\circ-50^\\circ=80^\\circ$,\\\\\n$\\therefore$ The degree measure of $\\overset{\\frown}{AE}$ is $\\boxed{80^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53080790_88.png"
  },
  {
    "solution": "\\boxed{\\frac{7}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the right-angled triangle ABC, $\\angle ACB=90^\\circ$, $AB=7$, the point D is the midpoint of $AB$, the point P is a moving point on the hypotenuse $AB$, $FG$ is the perpendicular bisector of the line segment $CP$, Q is a moving point on $FG$. What is the minimum value of $PQ+QD$?\n",
    "completion": "\\textbf{Solution:} Let's connect $CQ$, $CD$ as shown in the figure,\\\\\nsince point P is a moving point on the hypotenuse $AB$, $FG$ is the perpendicular bisector of the line segment $CP$, Q is a moving point on $FG$,\\\\\ntherefore $CQ=PQ$,\\\\\nthus $PQ+QD=CQ+DQ\\ge CD$,\\\\\nsince $\\angle ACB=90^\\circ$, $AB=7$, and point D is the midpoint of $AB$,\\\\\ntherefore $CD=\\frac{7}{2}$.\\\\\nHence, the minimum value of $PQ+QD$ is $\\boxed{\\frac{7}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52933308.png"
  },
  {
    "solution": "\\boxed{\\frac{5}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in rectangle ABCD, \\(AB = 4\\), \\(BC = 6\\). Fold \\(\\triangle ABC\\) along \\(AC\\), making point \\(B\\) coincide with point \\(E\\), and the intersection of \\(CE\\) and \\(AD\\) is point \\(F\\). What is the length of \\(DF\\)?\n",
    "completion": "\\textbf{Solution:} \nLet rectangle ABCD be folded along the diagonal AC so that $\\triangle ABC$ falls into the position of $\\triangle ACE$,\\\\\ntherefore, AE=AB and $\\angle E=\\angle B=90^\\circ$,\\\\\nFurthermore, since ABCD is a rectangle,\\\\\nit follows that AB=CD,\\\\\nthus, AE=DC,\\\\\nand $\\angle AFE=\\angle DFC$,\\\\\nin triangles $\\triangle AEF$ and $\\triangle CDF$,\\\\\n\\[\n\\left\\{\\begin{array}{l}\n\\angle AFE=\\angle CFD \\\\\n\\angle E=\\angle D \\\\\nAE=CD\n\\end{array}\\right.\n\\]\nthus, $\\triangle AEF\\cong \\triangle CDF$ (AAS),\\\\\nhence, EF=DF;\\\\\nSince ABCD is a rectangle,\\\\\nit follows that AD=BC=6, and CD=AB=4,\\\\\nSince $\\triangle AEF\\cong \\triangle CDF$,\\\\\nit follows that FC=FA,\\\\\nLet FA=x, then FC=x, FD=6−x,\\\\\nin $\\triangle CDF$, CF$^{2}$=CD$^{2}$+DF$^{2}$,\\\\\nthat is, $x^{2}=4^{2}+(6−x)^{2}$,\\\\\nsolving leads to $x=\\frac{13}{3}$,\\\\\ntherefore, $FD=6−x=\\boxed{\\frac{5}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53498363.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, $AC=BC=6$. Points D and E are located on AC and BC, respectively, $CD=2$. If the circle $\\odot O$ with diameter DE intersects the midpoint F of AB, then what is the diameter of $\\odot O$?\n",
    "completion": "\\textbf{Solution:} Construct $FG\\perp AC$ and $FH\\perp CB$, with the feet of the perpendiculars being G and H, respectively, as shown in the figure. \\\\\nThen, quadrilateral BCGF is a rectangle, with $AC\\parallel FH$ and $CB\\parallel FG$, \\\\\nsince $AC=BC=6$ and point F is the midpoint of AB, \\\\\nthus, $CG=CH=GF=HF=\\frac{1}{2} \\times 6=3$, \\\\\ntherefore, quadrilateral BCGF is a square, \\\\\ntherefore, $\\angle GFH=90^\\circ$, \\\\\nsince DE is a diameter, then $\\angle DFE=90^\\circ$, \\\\\ntherefore, $\\angle DFG+\\angle DFH=\\angle DFH+\\angle EFH=90^\\circ$, \\\\\ntherefore, $\\angle DFG=\\angle EFH$, \\\\\nsince $\\angle DGF=\\angle EHF=90^\\circ$ and $GF=HF=3$, \\\\\ntherefore, $\\triangle DFG \\cong \\triangle EFH$, \\\\\ntherefore, DF=EF, \\\\\nsince, in the right-angled $\\triangle DFG$, $DG=3-2=1$ and $GF=3$, \\\\\ntherefore, $DF=\\sqrt{1^2+3^2}=\\sqrt{10}=EF$, \\\\\nIn the right-angled $\\triangle DEF$, \\\\\n$DE=\\sqrt{(\\sqrt{10})^2+(\\sqrt{10})^2}=2\\sqrt{5}$; \\\\\nThus, the answer is: $\\boxed{2\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52731636.png"
  },
  {
    "solution": "\\boxed{D(2,1)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the parabola passes through points A$(4, 0)$, B$(1, 0)$, and C$(0, -2)$. Is there a point D on the parabola above line AC for which the area of $\\triangle DCA$ is maximized? If such a point exists, find the coordinates of point D and the maximum area of $\\triangle DCA$; if it does not exist, please explain why.\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect AC and DC. Assume there exists a point D on the parabola, and connect AD and CD. Draw $DE \\perp x$ axis from point D and let it intersect line AC at point E. Assume the expression for line AC is $y=kx+b$ ($k\\neq 0$), substituting A(4, 0) and C(0, -2) into this expression, we get:\n\\[\n\\begin{cases}\n0=4k+b\\\\\n-2=b\n\\end{cases}\n\\]\nFrom which, we solve:\n\\[\n\\begin{cases}\nk=\\frac{1}{2}\\\\\nb=-2\n\\end{cases}\n\\]\n$\\therefore$ The equation of line AC is $y=\\frac{1}{2}x-2$. Assume point D has coordinates $\\left(x, -\\frac{1}{2}x^{2}+ \\frac{5}{2}x-2\\right)$, then coordinates of point E are $\\left(x, \\frac{1}{2}x-2\\right)$,\\\\\n$\\therefore$ The area of $\\triangle DCA$ equals to $S_{\\triangle DCA}=S_{\\triangle DCE}+S_{\\triangle DAE}=\\frac{1}{2}\\cdot DE\\cdot x_{E}+\\frac{1}{2}\\cdot DE\\cdot(x_{A}-x_{E})=\\frac{1}{2}\\cdot DE\\cdot x_{A}=\\frac{1}{2}\\cdot DE\\cdot 4=2DE$,\\\\\n$\\because$ $DE=\\left(-\\frac{1}{2}x^{2}+\\frac{5}{2}x-2\\right)-\\left(\\frac{1}{2}x-2\\right)=-\\frac{1}{2}x^{2}+2x$,\\\\\n$\\therefore$ The area of $\\triangle DCA$ equals $2DE=2\\left(-\\frac{1}{2}x^{2}+2x\\right)=-x^{2}+4x=-\\left(x-2\\right)^{2}+4$,\\\\\n$\\therefore$ When $x=2$, the area of $\\triangle DCA$ reaches its maximum value, which is $\\boxed{4}$,\\\\\nSubstituting $x=2$ into the equation $y=-\\frac{1}{2}x^{2}+\\frac{5}{2}x-2$, we solve and get $y=1$,\\\\\n$\\therefore$ Point D is at \\boxed{D(2,1)}.",
    "image_path": "dataset/mm_math/images/MM_Math/52874656_287.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, $P$ is a point on line segment $MN$, and $Q$ is the midpoint of line segment $PN$. If $MN = 10$ and $MP = 6$, what is the length of $MQ$?\n",
    "completion": "\\textbf{Solution:} Since $MN = 10$ and $MP = 6$, \\\\\nit follows that $NP = MN - MP = 4$, \\\\\nbecause point Q is the midpoint of NP, \\\\\nit follows that $PQ = QN = \\frac{1}{2} NP = 2$, \\\\\nthus $MQ = MP + PQ = 6 + 2 = \\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52837558.png"
  },
  {
    "solution": "\\boxed{130^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $a \\parallel b$, if $\\angle 1=50^\\circ$, then what is the measure of $\\angle 2$ in degrees?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, \\\\\n$\\because a \\parallel b$, \\\\\n$\\therefore \\angle 3 = \\angle 1$, \\\\\n$\\because \\angle 1 = 50^\\circ$, \\\\\n$\\therefore \\angle 3 = 50^\\circ$ \\\\\n$\\because \\angle 2 + \\angle 3 = 180^\\circ$ \\\\\n$\\therefore \\angle 2 = 180^\\circ - \\angle 3 = 180^\\circ - 50^\\circ = \\boxed{130^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/51212472.png"
  },
  {
    "solution": "\\boxed{-x^{2}+2x+3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, the parabola $y = ax^2 + bx + 3$ intersects the x-axis at points $A(3,0)$ and $B(-1,0)$, and intersects the y-axis at point $C$. Point $P$ is a moving point on the parabola in the first quadrant, and point $F$ is a moving point on the y-axis. Connect $PA$, $PF$, $AF$. What is the function expression corresponding to this parabola?\n",
    "completion": "\\textbf{Solution:} \\\\\nSince the parabola $y=ax^{2}+bx+3$ intersects the $x$-axis at points $A(3,0)$ and $B(-1,0)$, \\\\\nwe have $\\begin{cases}9a+3b+3=0 \\\\ a-b+3=0 \\end{cases}$ \\\\\nSolving these, we get $\\begin{cases}a=-1 \\\\ b=2 \\end{cases}$ \\\\\nTherefore, the corresponding function expression for the parabola is $y=-x^{2}+2x+3=\\boxed{-x^{2}+2x+3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52660930_104.png"
  },
  {
    "solution": "\\boxed{\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line of symmetry of rectangle ABCD intersects AB at point E and CD at point F respectively. If rectangle AEFD is similar to rectangle ABCD, what is the value of AB$\\colon$ BC?\n",
    "completion": "\\textbf{Solution:} Since $ABCD$ is a rectangle,\\\\\nit follows that $AD=BC$,\\\\\nsince the axes of symmetry of rectangle $ABCD$ intersect $AB$ at point $E$, and $CD$ at point $F$,\\\\\nit follows that $AE=\\frac{1}{2}AB$,\\\\\nsince rectangle $AEFD$ is similar to rectangle $ABCD$,\\\\\nit follows that $\\frac{AB}{BC}=\\frac{AD}{AE}$,\\\\\nthus $\\frac{AB}{BC}=\\frac{BC}{\\frac{1}{2}AB}$,\\\\\nhence $\\frac{1}{2}AB^2=BC^2$,\\\\\ntherefore $AB^2=2BC^2$,\\\\\nthus $AB=\\sqrt{2}BC$,\\\\\nhence $AB\\colon BC=\\boxed{\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51403191.png"
  },
  {
    "solution": "\\boxed{(-2,-1)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the $xy$-plane coordinate system, point $A(0, 3)$, point $B(2, 1)$, and point $C(2, -3)$. Through drawing operations, it can be known: what are the coordinates of the center of the circumscribed circle of $\\triangle ABC$?\n",
    "completion": "\\textbf{Solution:} Since the circumcenter of $\\triangle ABC$ is the intersection point of the perpendicular bisectors of the triangle's sides, as shown in the figure: the intersection point $O^{\\prime}$ of EF and MN is the desired circumcenter of $\\triangle ABC$, therefore, the coordinates of the circumcenter of $\\triangle ABC$ are $\\boxed{(-2,-1)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51351832.png"
  },
  {
    "solution": "\\boxed{(-3, 1)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, given the quadratic function $y=ax^2+bx+c$ with part of its graph as illustrated, if $y>0$, what is the range of $x$?\n",
    "completion": "\\textbf{Solution:} From the figure, we know that the axis of symmetry of the parabola is the line $x=-1$, and one intersection point with the $x$-axis has an abscissa of $-3$,\\\\\n$\\therefore$ the abscissa of the other intersection point of the parabola with the $x$-axis is $1$,\\\\\n$\\because y>0$,\\\\\n$\\therefore -3<x<1$,\\\\\nTherefore, the answer is: $x \\in \\boxed{(-3, 1)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52793773.png"
  },
  {
    "solution": "\\boxed{(-2,0)\\cup(4,+\\infty)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y_{1}=kx+2$ intersects with the graph of the inverse proportion function $y_{2}=-\\frac{8}{x}$ at points $A(a，-2a)$ and $B(4，-2)$. Considering the graphs, directly write down the solution set of the inequality $kx+2+\\frac{8}{x}<0$:\n",
    "completion": "\\textbf{Solution:} From (1) it is known that: Points A(-2, 4) and B(4, -2),\\\\\nsince $kx+2+\\frac{8}{x}<0$,\\\\\ntherefore $kx+2<-\\frac{8}{x}$, which means when $y_1<y_2$, the graph of the linear function is below that of the hyperbolic function,\\\\\nthus $-2<x<0$ or $x>4$. Hence, the solution set is $x\\in\\boxed{(-2,0)\\cup(4,+\\infty)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52050093_181.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the line $l$ passes through points $A(0,1)$ and $B(-2,0)$. If $y=-x+b$ passes through point $A$ and intersects the $x$-axis at point $C$, what is the area of $\\triangle ABC$?\n",
    "completion": "Solution: The line $y = -x + b$ passes through point $A(0,1)$, thus $b = 1$. Therefore, the line $y = -x + 1$ intersects the $x$-axis at point $C(1,0)$.\\\\\nTherefore, the area of $\\triangle ABC$ is $= \\frac{1}{2} \\times (2+1) \\times 1 = \\boxed{\\frac{3}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51662963_204.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, \\(l_1, l_2, l_3\\) are a set of parallel lines, and the lines AC, DF intersect this set of parallel lines successively at points A, B, C, and D, E, F, respectively. If \\(\\frac{AB}{BC}=\\frac{2}{3}\\), then what is the value of \\(\\frac{EF}{DF}\\)?\n",
    "completion": "\\textbf{Solution:} Since $l_1 \\parallel l_2 \\parallel l_3$, we have $\\frac{EF}{DF}=\\frac{BC}{AC}$. Furthermore, since $\\frac{AB}{BC}=\\frac{2}{3}$, it follows that $\\frac{BC}{AC}=\\frac{3}{5}$. Hence, $\\frac{EF}{DF}=\\boxed{\\frac{3}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51385606.png"
  },
  {
    "solution": "\\boxed{8\\,\\text{cm}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $\\triangle ABC$ is an equilateral triangle with the height $AH=8\\,\\text{cm}$. $P$ is a moving point on $AH$, and $D$ is the midpoint of $AB$. What is the minimum value of $PD+PB$ in $\\,\\text{cm}$?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABC$ is an equilateral triangle and $AH$ is the height,\\\\\n$\\therefore$ point $H$ is the midpoint of $BC$,\\\\\npoint $B$ is symmetric to point $C$ with respect to $AH$, connecting $DC$ intersects $AH$ at a point $P$, at this moment there is a minimum value, the minimum value is the length of $DC$,\\\\\nsince $D$ is the midpoint of $AB$,\\\\\n$\\therefore DC$ is the height from $AB$,\\\\\n$\\therefore DC=AH=8$,\\\\\n$\\therefore PD+PB=PD+PC=DC=\\boxed{8\\,\\text{cm}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52720328.png"
  },
  {
    "solution": "\\boxed{125^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure below, $\\angle A=70^\\circ$, BP and CP bisect $\\angle ABC$ and $\\angle ACB$ respectively. What is the measure of $\\angle P$?\n",
    "completion": "\\textbf{Solution:} In $\\triangle ABC$, $\\angle A = 70^\\circ$,\\\\\n$\\therefore \\angle ABC + \\angle ACB = 180^\\circ - \\angle A = 180^\\circ - 70^\\circ = 110^\\circ$,\\\\\n$\\because BP$ and $CP$ bisect $\\angle ABC$ and $\\angle ACB$ respectively,\\\\\n$\\therefore \\angle PBC = \\frac{1}{2} \\angle ABC$, $\\angle PCB = \\frac{1}{2} \\angle ACB$,\\\\\nIn $\\triangle PBC$,\\\\\n$\\angle P + \\angle PBC + \\angle PCB = 180^\\circ$,\\\\\n$\\therefore \\angle P = 180^\\circ - (\\angle PBC + \\angle PCB) = 180^\\circ - \\frac{1}{2} (\\angle ABC + \\angle ACB) = 180^\\circ - \\frac{1}{2} \\times 110^\\circ = \\boxed{125^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52837420.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ABC=60^\\circ$, $\\angle ACB=40^\\circ$, $BD$ bisects $\\angle ABC$, $CD$ bisects $\\angle ACB$, connect $AD$, construct $DE\\perp AB$, $DE=2$, and $AC=4$. What is the area of $\\triangle ADC$?\n",
    "completion": "\\textbf{Solution:} Let $DF\\perp AC$ at $F$, and $DH\\perp BC$ at $H$, as shown in Figure 2,\\\\\nsince $BD$ bisects $\\angle ABC$, $DE\\perp AB$, and $DH\\perp BC$,\\\\\ntherefore $DH=DE=2$,\\\\\nsince $CD$ bisects $\\angle ACB$, $DF\\perp AC$, and $DH\\perp BC$,\\\\\ntherefore $DF=DH=2$,\\\\\nthus, the area of $\\triangle ADC$ is $\\frac{1}{2}DF\\cdot AC=\\frac{1}{2}\\times 2\\times 4=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52978993_121.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that point $E$ is a point on side $AD$ of rectangle $ABCD$. Quadrilateral $BCDE$ is folded along the line $BE$. After folding, the corresponding point of point $C$ is $C'$, and the corresponding point of point $D$ is $D'$. If point $A$ is on $C'D'$, and $AB=10$, $BC=8$, what is the length of $AE$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rectangle,\\\\\n$\\therefore$ $\\angle D=\\angle C=\\angle DAB=90^\\circ$, AB=DC=10, AD=BC=8,\\\\\nAccording to the properties of folding, we get: $\\angle D'=\\angle D=90^\\circ$, $\\angle C'=\\angle C=90^\\circ$, BC'=BC=8, D'C'=DC=10,\\\\\n$\\therefore$ $AC'=\\sqrt{AB^{2}-BC'^{2}}=6$,\\\\\n$\\therefore$ AD'=D'C'-AC'=10-6=4,\\\\\nLet DE=D'E=x, then AE=8-x,\\\\\n$\\therefore$ $(8-x)^{2}=4^{2}+x^{2}$;\\\\\n$\\therefore$ x=3,\\\\\n$\\therefore$ AE=$\\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52933867.png"
  },
  {
    "solution": "\\boxed{\\frac{120}{13}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the rhombus $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$. A perpendicular is drawn from point $A$ to $BC$ at right foot $E$, and $BC$ is extended to point $F$ such that $CF=BE$. Segment $DF$ is drawn. Segment $OE$ is drawn. If $AB=13$ and $OE=5$, what is the length of $AE$?\n",
    "completion": "\\textbf{Solution:} Since point O is the midpoint of AC, and triangle AEC is a right-angled triangle,\\\\\nwe have $AC=2OE=10$.\\\\\nLet $EC=x$, then $BE=BC-EC=13-x$.\\\\\nHence, $AB^2-BE^2=AC^2-EC^2$, that is $13^2-(13-x)^2=10^2-x^2$.\\\\\nSolving this, we find $x=\\frac{50}{13}$.\\\\\nTherefore, $AE=\\sqrt{AC^2-EC^2}=\\boxed{\\frac{120}{13}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52344804_172.png"
  },
  {
    "solution": "\\boxed{x<1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, given the linear function $y=kx+b(k\\ne 0)$ whose graph is illustrated, what is the solution set of the inequality $kx+b>0$?\n",
    "completion": "\\textbf{Solution:} When $x<1$, we have $y>0$,\\\\\nthus the solution set for the inequality $kx+b>0$ is $x<1$.\\\\\nTherefore, the answer is $\\boxed{x<1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50446738.png"
  },
  {
    "solution": "\\boxed{16}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle BAC=90^\\circ$, $AB=5\\text{cm}$, $BC=13\\text{cm}$. Point $D$ is on segment $AC$, and $CD=7\\text{cm}$. A moving point $P$ starts from point $E$, which is $15\\text{cm}$ away from point $B$, and moves in the direction of ray $EA$ at a speed of $2\\text{cm}$ per second, for a time of $t$ seconds. During the motion, is there a moment when $\\triangle ABC$ and $\\triangle ADP$ are congruent? If yes, please calculate the value of $t$; if no, please explain the reason.\n",
    "completion": "\\textbf{Solution:} There exists a moment such that $\\triangle ABC$ is congruent to $\\triangle ADP$, for the following reasons:\\\\\n$\\because$ $\\angle DAP=\\angle BAC=90^\\circ$, and $DA=BA$,\\\\\n$\\therefore$ when $AP=AC$, $\\triangle ABC\\cong \\triangle ADP$,\\\\\n$\\therefore$ $20-2t=12$ or $2t-20=12$,\\\\\nSolving this, we get $t=\\boxed{4}$ or $t=16$,\\\\\n$\\therefore$ the values of $t$ that satisfy the conditions are \\boxed{4} or \\boxed{16}.",
    "image_path": "dataset/mm_math/images/MM_Math/55588005_82.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\angle ACD = \\angle BCE = 90^\\circ$. If $\\angle ACB = 150^\\circ$, can you calculate the measure of $\\angle DCE$? Please write down the solution process.\n",
    "completion": "\\textbf{Solution:} Since $\\angle ACD = \\angle BCE = 90^\\circ$,\\\\\nit follows that $\\angle ACD + \\angle ECB = 180^\\circ$\\\\\nAlso, since $\\angle ACB + \\angle ECD = \\angle ACD + \\angle ECB = 180^\\circ$, and $\\angle ACB = 150^\\circ$,\\\\\nit follows that $\\angle ECD = 180^\\circ - \\angle ACB = \\boxed{30^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/52580479.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{2k+1}}{2k+1}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $D$ is a point on $AC$ and $AC=kAD$. $E$ is a point on $BC$, and $BD$ intersects $AE$ at point $O$. If $\\angle AOB + \\angle BAC = 180^\\circ$. In figure 2, if $BE=2CE$, find the value of $\\frac{AO}{AB}$.\n",
    "completion": "\\textbf{Solution:} As shown in Figure 2, construct $EF \\parallel BD$ intersecting $AC$ at $F$, let $AD=a$, let $OD=x$, thus $AC=k\\cdot a$, $CD=(k-1)\\cdot a$,\n\n$\\therefore \\triangle EFC\\sim \\triangle BDC$, $\\triangle AOD\\sim \\triangle AEF$\n\n$\\therefore \\frac{EF}{BD}=\\frac{CF}{CD}=\\frac{CE}{BC}=\\frac{1}{3}$, $\\frac{OD}{EF}=\\frac{AD}{AF}$,\n\n$\\therefore BD=3EF$, $CF=\\frac{1}{3}CD=\\frac{1}{3}(k-1)\\cdot a$,\n\n$\\therefore \\frac{x}{EF}=\\frac{a}{ka-\\frac{1}{3}(k-1)a}$,\n\n$\\therefore EF=\\frac{2k+1}{3}x$,\n\n$\\therefore BD=(2k+1)\\cdot x$,\n\n$\\because \\angle CAE=\\angle ACD$, $\\angle ADO=\\angle ADB$,\n\n$\\therefore \\triangle AOD\\sim \\triangle BAD$,\n\n$\\therefore \\frac{AD}{BD}=\\frac{OD}{AD}=\\frac{AO}{AB}$,\n\n$\\therefore \\frac{a}{(2k+1)\\cdot x}=\\frac{x}{a}$,\n\n$\\therefore x=\\sqrt{\\frac{1}{2k+1}}\\cdot a$\n\n$\\therefore \\frac{AO}{AB}=\\frac{\\sqrt{\\frac{1}{2k+1}}}{a}\\cdot a=\\boxed{\\frac{\\sqrt{2k+1}}{2k+1}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51402867_153.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, AB is the diameter of semicircle O with AB = 4. Points C and D are on the semicircle, and $OC \\perp AB$. It is given that $\\overset{\\frown}{BD} = 2\\overset{\\frown}{CD}$. Point P is a moving point on OC. What is the minimum value of $BP + DP$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect AD, PA, PD, OD. \\\\\n$\\because$ OC $\\perp$ AB, OA = OB, \\\\\n$\\therefore$ PA = PB, $\\angle$ COB = $90^\\circ$, \\\\\n$\\because$ $\\overset{\\frown}{BD} = 2\\overset{\\frown}{CD}$, \\\\\n$\\therefore$ $\\angle$ DOB = $\\frac{2}{3} \\times 90^\\circ = 60^\\circ$, \\\\\n$\\because$ OD = OB, \\\\\n$\\therefore$ $\\triangle$ OBD is an equilateral triangle, \\\\\n$\\therefore$ $\\angle$ ABD = $60^\\circ$, \\\\\n$\\because$ AB is the diameter, \\\\\n$\\therefore$ $\\angle$ ADB = $90^\\circ$, \\\\\n$\\therefore$ AD = AB $\\cdot \\sin\\angle$ ABD = $2\\sqrt{3}$, \\\\\n$\\because$ PB + PD = PA + PD $\\geq$ AD, \\\\\n$\\therefore$ PD + PB $\\geq 2\\sqrt{3}$, \\\\ \n$\\therefore$ The minimum value of PD + PB is $\\boxed{2\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51318402.png"
  },
  {
    "solution": "\\boxed{18}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AD$ is the median of $\\triangle ABC$. It is known that the perimeter of $\\triangle ABD$ is $25\\,\\text{cm}$, and $AB$ is $7\\,\\text{cm}$ longer than $AC$. What is the perimeter of $\\triangle ACD$?\n",
    "completion": "\\textbf{Solution:} Since $AD$ is the median of $\\triangle ABC$,\\\\\nit follows that $BD=CD$,\\\\\ntherefore, the difference in the perimeters of $\\triangle ABD$ and $\\triangle ACD$ is $(AB+BD+AD)-(AC+CD+AD)=AB-AC$.\\\\\nSince the perimeter of $\\triangle ABD$ is $25\\,cm$ and $AB$ is $7\\,cm$ longer than $AC$,\\\\\nthe perimeter of $\\triangle ACD$ is: $25-7=\\boxed{18}\\,(cm)$.",
    "image_path": "dataset/mm_math/images/MM_Math/53102440.png"
  },
  {
    "solution": "\\boxed{-\\frac{1}{2}x^2-\\frac{3}{2}x+2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=\\frac{1}{2}x+2$ intersects with the $x$-axis and the $y$-axis at points $A$ and $C$, respectively. The parabola $y=-\\frac{1}{2}x^2+bx+c$ passes through points $A$ and $C$, and intersects with the $x$-axis at another point $B$. Point $D$ is a moving point on the parabola. What is the analytical expression of the parabola?\n",
    "completion": "\\textbf{Solution:} Let $y=0$, we get, $\\frac{1}{2}x+2=0$, which gives $x=-4$,\\\\\nLet $x=0$, we get, $y=2$,\\\\\n$\\therefore A(-4, 0)$, $C(0, 2)$\\\\\n$\\because$ the parabola $y=-\\frac{1}{2}x^2+bx+c$ passes through points $A$ and $C$,\\\\\n$\\therefore \\left\\{\\begin{array}{l}\n-\\frac{1}{2}\\times (-4)^2-4b+c=0 \\\\\nc=2\n\\end{array}\\right.$,\\\\\nUpon solving, we get $\\left\\{\\begin{array}{l}\nb=-\\frac{3}{2} \\\\\nc=2\n\\end{array}\\right.$,\\\\\n$\\therefore y=-\\frac{1}{2}x^2-\\frac{3}{2}x+2=\\boxed{-\\frac{1}{2}x^2-\\frac{3}{2}x+2}$",
    "image_path": "dataset/mm_math/images/MM_Math/52622506_110.png"
  },
  {
    "solution": "\\boxed{7}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB = 90^\\circ$, $AC = BC$. A straight line DE is drawn outside $\\triangle ABC$ through point C, and perpendicular lines are drawn from points A and B to line DE, with the feet of the perpendiculars being E and D, respectively. If $AE=4$ and $BD=3$, then what is the length of $DE$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle ACB=90^\\circ$, \\\\\nit follows that $\\angle ACE + \\angle BCD = 90^\\circ$.\\\\\nSince $\\angle ACE + \\angle CAE = 90^\\circ$,\\\\\nit implies that $\\angle BCD = \\angle CAE$.\\\\\nIn $\\triangle ACE$ and $\\triangle CBD$, \\\\\n$\\begin{array}{c}\n\\angle CAE = \\angle BCD,\\\\\n\\angle E = \\angle D = 90^\\circ,\\\\\nAC = BC,\n\\end{array}$\\\\\nhence, $\\triangle ACE \\cong \\triangle CBD$ by AAS,\\\\\nthus, CE = BD = 3, CD = AE = 4,\\\\\ntherefore, DE = CE + CD = \\boxed{7}.",
    "image_path": "dataset/mm_math/images/MM_Math/51435468.png"
  },
  {
    "solution": "\\boxed{\\left(\\frac{17}{3},\\frac{44}{9}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that a parabola passing through point $B(4, 1)$, given by $y=\\frac{1}{2}x^{2}-\\frac{5}{2}x+c$, intersects with the coordinate axes at points A and C as shown. Lines AC, BC, and AB are connected. Within the first quadrant, a moving point M is on the parabola. A perpendicular line $AM$ is drawn from point M to meet the y-axis at point P. When point P is above point A, and $\\triangle AMP$ is similar to $\\triangle ABC$, what are the coordinates of point M?\n",
    "completion": "\\textbf{Solution:} Substituting point B $(4,1)$ into $y=\\frac{1}{2}x^{2}-\\frac{5}{2}x+c$, we get:\\\\\n$\\frac{1}{2}\\times 1^{2}-\\frac{5}{2}\\times 1+c=4$\\\\\n$\\therefore c=3$\\\\\nHence, the equation of the parabola is $y=\\frac{1}{2}x^{2}-\\frac{5}{2}x+3$\\\\\nLetting x=0, we find y=3,\\\\\n$\\therefore$A$(0,3)$\\\\\nLetting y=0, we get $\\frac{1}{2}x^{2}-\\frac{5}{2}x+3=0$\\\\\nSolving this yields $x_{1}=2$, $x_{2}=3$\\\\\n$\\therefore$C$(3,0)$\\\\\n$\\therefore$AC$=\\sqrt{(3-0)^{2}+(0-3)^{2}}=3\\sqrt{2}$\\\\\nBecause B$(4,1)$\\\\\n$\\therefore$BC$=\\sqrt{(4-3)^{2}+(1-0)^{2}}=\\sqrt{2}$, AB$=\\sqrt{(0-4)^{2}+(3-1)^{2}}=2\\sqrt{5}$\\\\\n$\\therefore AC^{2}+BC^{2}=AB^{2}$\\\\\n$\\therefore \\triangle ABC$ is a right triangle, and $\\frac{BC}{AC}=\\frac{1}{3}$,\\\\\nDraw MG perpendicular to the y-axis at G from the point M, then $\\angle$MGA=90$^\\circ$,\\\\\nLet the x-coordinate of point M be x, and since M is to the right of the y-axis, we have x>0, thus MG=x,\\\\\nBecause PM$\\perp$MA, $\\angle$ACB=90$^\\circ$,\\\\\n$\\therefore \\angle AMP=\\angle ACB=90^\\circ$,\\\\\n(1) As shown in the figure, when $\\angle MAP=\\angle CBA$, then $\\triangle MAP\\sim \\triangle CBA$,\\\\\n$\\therefore \\frac{AM}{MP}=\\frac{BC}{AC}=\\frac{1}{3}$\\\\\nSimilarly, $\\triangle AGM\\sim \\triangle AMP$\\\\\n$\\therefore \\frac{AG}{MG}=\\frac{AM}{MP}=\\frac{1}{3}$\\\\\n$\\therefore$AG$=\\frac{1}{3}$MG$=\\frac{1}{3}$x, therefore M$(x,3+\\frac{1}{3}x)$,\\\\\nSubstituting M$(x,3+\\frac{1}{3}x)$ into $y=\\frac{1}{2}x^{2}-\\frac{5}{2}x+3$, we obtain\\\\\n$\\frac{1}{2}x^{2}-\\frac{5}{2}x+3=3+\\frac{1}{3}x$,\\\\\nSolving this, $x_{1}=0$ (discard), $x_{2}=\\frac{17}{3}$,\\\\\n$\\therefore$3+$\\frac{1}{3}$x=3+$\\frac{17}{9}=\\frac{44}{9}$\\\\\n$\\therefore$M$\\left(\\frac{17}{3},\\frac{44}{9}\\right)$;\\\\\nIn summary, the coordinates of point M are $\\boxed{\\left(\\frac{17}{3},\\frac{44}{9}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52872299.png"
  },
  {
    "solution": "\\boxed{3:4:5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the square $ABCD$, with point $E$ on side $CD$, connect $AE$. Construct $AF\\perp AE$, intersecting the extension of $CB$ at point $F$, and connect $EF$. Let $AG$ bisect $\\angle FAE$, with $AG$ intersecting $BC$ and $EF$ at points $G$ and $H$, respectively. Connect $EG$ and $DH$. If $DE=CE$, find the value of $CE:CG:EG$?\n",
    "completion": "\\textbf{Solution:}\nGiven (1), we have $\\triangle ADE \\cong \\triangle ABF$,\\\\\n$\\therefore BF=DE$,\\\\\nFrom (2), we have $\\triangle AGF \\cong \\triangle AGE$,\\\\\n$\\therefore GE=GF$,\\\\\nLet $DE=EC=a$, $BG=x$, thus $BF=a$, $BC=CD=2a$,\\\\\n$\\therefore FG= a+x$,\\\\\n$\\therefore EG=a+x$, $GC=2a-x$,\\\\\nIn $\\triangle ECG$, $EG^2=EC^2+CG^2$,\\\\\n$\\therefore (x+a)^2=a^2+(2a-x)^2$, solving gives $x=\\frac{2}{3}a$,\\\\\n$\\therefore CG=\\frac{4}{3}a$, $EG=\\frac{5}{3}a$,\\\\\n$\\therefore CE:CG:EG=a:\\frac{4}{3}a:\\frac{5}{3}a=3:4:5$.\\\\\n\nTherefore, the answer is $CE:CG:EG=\\boxed{3:4:5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52188888_130.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, point $D$ is on side $BC$, $DB=AB=\\frac{3}{5}BC$, $BE \\perp AD$ at point $E$. If the area of $\\triangle ABE$ is 6, what is the area of $\\triangle EBC$?\n",
    "completion": "\\textbf{Solution:} Since $DB=AB$ and $BE \\perp AD$, \\\\\nit follows that $AE=DE$, \\\\\ntherefore $S_{\\triangle BDE}=S_{\\triangle ABE}=6$, \\\\\nsince $DB=\\frac{3}{5}BC$, \\\\\nit therefore follows that $S_{\\triangle BDE} : S_{\\triangle EDC}=3:2$, \\\\\nhence $S_{\\triangle EDC}=4$, \\\\\nthus $S_{\\triangle BCE}=S_{\\triangle BDE}+S_{\\triangle EDC}=\\boxed{10}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53010087.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, AB is the diameter of $\\odot O$, and chord CD $\\perp$ AB at point E. If AB $=$ 10, BE $=$ 2, what is the length of chord CD?\n",
    "completion": "\\textbf{Solution:} Draw line $OC$ as shown:\\\\\nSince $AB$ is the diameter of $\\odot O$ and $CD\\perp AB$,\\\\\nit follows that $CE=DE=\\frac{1}{2}CD$, and $OC=OA=OB=5$,\\\\\nthus $OE=OB-EB=5-2=3$,\\\\\nin the right triangle $\\triangle OCE$, by the Pythagorean theorem, we have: $CE=\\sqrt{OC^{2}-OE^{2}}=\\sqrt{5^{2}-3^{2}}=4$,\\\\\ntherefore, $CD=2CE=\\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51445744.png"
  },
  {
    "solution": "\\boxed{90^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, lines AB, CD, and EF intersect at point O. If $\\angle AOC=70^\\circ$ and $\\angle BOF=20^\\circ$, what are the measures of $\\angle BOC$ and $\\angle DOE$?\n",
    "completion": "\\textbf{Solution}: Given the problem,\\\\\nsince $\\angle AOC=70^\\circ$, and $\\angle AOC + \\angle BOC= 180^\\circ$,\\\\\nthus $\\angle BOC= \\boxed{110^\\circ}$.\\\\\nSince $\\angle BOF= 20^\\circ$,\\\\\nthus $\\angle COF=\\angle BOC-\\angle BOF= 90^\\circ$,\\\\\ntherefore $\\angle DOE=\\boxed{90^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/54927579_137.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, rectangle $OABC$ has vertices $A$ and $C$ located on the positive semiaxes of the $x$-axis and $y$-axis, respectively, with the coordinate of point $B$ being $(2\\sqrt{3}, 4)$. The graph of the linear function $y=-\\frac{\\sqrt{3}}{3}x+b$ intersects sides $OC$, $AB$, and the $x$-axis at points $D$, $E$, and $F$, respectively, with $\\angle DFO=30^\\circ$, and it satisfies $OD=BE$, where point $M$ is a moving point on line segment $DF$. What is the value of $b$?\n",
    "completion": "\\textbf{Solution:} Given $y=-\\frac{\\sqrt{3}}{3}x+b$, let $x=0$, then $y=b$,\\\\\n$\\therefore$ the coordinates of point $D$ are $(0, b)$,\\\\\n$\\because OD=BE$ and $B(2\\sqrt{3}, 4)$,\\\\\n$\\therefore E(2\\sqrt{3}, 4-b)$,\\\\\nSubstituting $E(2\\sqrt{3}, 4-b)$ into $y=-\\frac{\\sqrt{3}}{3}x+b$ we get:\\\\\n$4-b=-\\frac{\\sqrt{3}}{3}\\times 2\\sqrt{3}+b$,\\\\\nSolving for $b$ yields: $b=\\boxed{3}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51152292_112.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, points $D$, $E$, $F$ are the midpoints of $AB$, $BC$, $AC$, respectively. If the area of $\\triangle ABC$ is $1$, what is the area of $\\triangle DEF$?\n",
    "completion": "\\textbf{Solution:} Given points $D$, $E$, $F$ are the midpoints of $AB$, $BC$, $AC$ respectively,\\\\\nthus $EF=\\frac{1}{2}BA$, $DF=\\frac{1}{2}BC$, $DE=\\frac{1}{2}AC$,\\\\\nthus $\\frac{EF}{BA}=\\frac{DF}{BC}=\\frac{DE}{AC}=\\frac{1}{2}$,\\\\\nthus $\\triangle ABC\\sim \\triangle EDF$,\\\\\nthus $\\frac{{S}_{\\triangle EDF}}{{S}_{\\triangle ABC}}={\\left(\\frac{1}{2}\\right)}^{2}=\\frac{1}{4}$,\\\\\ngiven ${S}_{\\triangle ABC}=1$,\\\\\nthus ${S}_{\\triangle EDF}=\\boxed{\\frac{1}{4}}$,",
    "image_path": "dataset/mm_math/images/MM_Math/53058709.png"
  },
  {
    "solution": "\\boxed{20}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that points $B$ and $C$ divide the line segment $AD$ into three parts in the ratio $2:5:3$. Let $M$ be the midpoint of $AD$, and $BM=6$. Find the length of $AD$.\n",
    "completion": "\\textbf{Solution:} Since points B and C divide line segment AD into three parts in the ratio 2:5:3,\\\\\nit follows that AB=$\\frac{1}{5}$AD, and CD=$\\frac{3}{10}$AD.\\\\\nSince M is the midpoint of AD,\\\\\nwe have AM=$\\frac{1}{2}$AD.\\\\\nSince BM=AM−AB,\\\\\nwe find that $\\frac{1}{2}$AD−$\\frac{1}{5}$AD=6.\\\\\nSolving this gives: AD=\\boxed{20}.",
    "image_path": "dataset/mm_math/images/MM_Math/53253574.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{18}{x}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the line $y=-\\frac{1}{2}x$ intersects the graph of the inverse proportion function $y=\\frac{k}{x}$ at two points $A$ and $B$ that are symmetric about the origin. Given that the y-coordinate of point $A$ is 3, what is the expression for the inverse proportion function?\n",
    "completion": "\\textbf{Solution:} Let the linear function be $y=-\\frac{1}{2}x$ and set $y=3$, then $3=-\\frac{1}{2}x$. Solving this, we get: $x=-6$, thus the coordinates of point $A$ are $\\left(-6, 3\\right)$.\\\\\nSince point $A\\left(-6, 3\\right)$ lies on the graph of the inverse proportion function $y=\\frac{k}{x}$,\\\\\nit follows that $k=-6\\times 3=-18$,\\\\\nhence, the expression for the inverse proportion function is $\\boxed{y=-\\frac{18}{x}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50496356_64.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{241}}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, the vertices $A$ and $C$ of rectangle $OABC$ are located on the positive half-axes of the $x$-axis and $y$-axis, respectively, with point $B(4,3)$. The graph of the inverse proportion function $y=\\frac{k}{x}$ ($x>0$) intersects $AB$ and $BC$ at points $D$ and $E$ respectively, with $BD=1$. Point $P$ is a moving point on segment $OA$. As shown in Figure 2, when $PE$ and $PD$ are connected, what is the minimum value of $PD+PE$?\n",
    "completion": "\\textbf{Solution:} In Figure 2, construct the symmetric point $D'$ of point D about the x-axis. Draw a line from $D'$ to E, intersecting the x-axis at point P. Connect PD. At this point, $PD+PE$ reaches its minimum value, which is equal to $D'E$. \\\\\n$\\because$ The coordinates of point D are (4, 2), \\\\\n$\\therefore$ The coordinates of point $D'$ are (4, −2). \\\\\nAlso, $\\because$ The coordinates of point E are $\\left(\\frac{8}{3}, 3\\right)$, \\\\\n$\\therefore D'E=\\sqrt{\\left(4-\\frac{8}{3}\\right)^{2}+\\left(-2-3\\right)^{2}}=\\frac{\\sqrt{241}}{3}$. \\\\\n$\\therefore$ The minimum value of $PD+PE$ is $\\boxed{\\frac{\\sqrt{241}}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51446330_268.png"
  },
  {
    "solution": "\\boxed{\\sqrt{3} + \\frac{1}{3}\\pi}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a sector BAC with a radius of 2 and a central angle of $90^\\circ$ is rotated counterclockwise around point A. During the rotation, point B falls on point B' on the arc of sector BAC, and the corresponding point for point C is point C'. What is the area of the shaded region?\n",
    "completion": "\\textbf{Solution:} Connect $BB'$, and draw $AF \\perp BB'$ at $F$, then $\\angle AFB = 90^\\circ$, as shown in the figure,\\\\\n$\\because$ when the sector $BAC$ with radius $2$ and a central angle of $90^\\circ$ is rotated counterclockwise around point $A$, during the rotation, point $B$ falls on the point $B''$ on the arc of sector $BAC$, and the corresponding point for point $C$ is point $C'$,\\\\\n$\\therefore$ the area of sector $ABC$ is equal to the area of sector $AB'C'$, and $AB = AB' = BC = BB' = 2$,\\\\\n$\\therefore$ $\\triangle ABB'$ is an equilateral triangle,\\\\\n$\\therefore$ $\\angle ABF = 60^\\circ$,\\\\\n$\\therefore$ $\\angle BAF = 30^\\circ$,\\\\\n$\\therefore$ $BF = \\frac{1}{2} AB = \\frac{1}{2} \\times 2 = 1$, by Pythagoras' theorem, we have: $AF = \\sqrt{2^2 - 1^2} = \\sqrt{3}$,\\\\\n$\\therefore$ the area of the shaded part $S = S_{\\text{sector } ABC} - (S_{\\text{sector } ABB'} - S_{\\triangle ABB'})$\\\\\n$= \\frac{90^\\circ\\pi \\times 2^2}{360} - \\left(\\frac{60^\\circ\\pi \\times 2^2}{360} - \\frac{1}{2} \\times 2 \\times \\sqrt{3}\\right)$\\\\\n$= \\boxed{\\sqrt{3} + \\frac{1}{3}\\pi}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51368374.png"
  },
  {
    "solution": "\\boxed{2^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle BAC=128^\\circ$, $P_{1}$ is the intersection point of the bisector of the interior angle $\\angle ABC$ of $\\triangle ABC$, which is $BP_{1}$, and the bisector of the exterior angle $\\angle ACE$, which is $CP_{1}$. $P_{2}$ is the intersection point of the bisector of the interior angle $\\angle P_{1}BC$ of $\\triangle BP_{1}C$, which is $BP_{2}$, and the bisector of the exterior angle $\\angle P_{1}CE$, which is $CP_{2}$. $P_{3}$ is the intersection point of the bisector of the interior angle $\\angle P_{2}BC$ of $\\triangle BP_{2}C$, which is $BP_{3}$, and the bisector of the exterior angle $\\angle P_{2}CE$, which is $CP_{3}$. Continuing in this manner, what is the measure of $\\angle P_{6}$?\n",
    "completion": "\\textbf{Solution:} Given that the internal angle bisector BP and the external angle bisector CP1 of $\\triangle ABC$ intersect at $P_{1}$,\\\\\nhence, $\\angle P_{1}BC = \\frac{1}{2}\\angle ABC$, $\\angle P_{1}CE = \\frac{1}{2}\\angle ACE$,\\\\\nsince $\\angle ACE = \\angle A + \\angle ABC$, $\\angle P_{1}CE = \\angle P_{1}BC + \\angle P_{1}$,\\\\\nthus, $\\frac{1}{2}(\\angle A + \\angle ABC) = \\angle P_{1}BC + \\angle P_{1} = \\frac{1}{2}\\angle ABC + \\angle P_{1}$,\\\\\ntherefore, $\\angle P_{1} = \\frac{1}{2}\\angle A = \\frac{1}{2} \\times 128^\\circ = 64^\\circ$,\\\\\nsimilarly, $\\angle P_{2} = \\frac{1}{2}\\angle P_{1} = 32^\\circ$,\\\\\nhence, $\\angle P_{6} = \\boxed{2^\\circ}$,",
    "image_path": "dataset/mm_math/images/MM_Math/52655333.png"
  },
  {
    "solution": "\\boxed{\\frac{12\\sqrt{3}}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, in quadrilateral $ABCD$, $\\angle ABD = \\angle BCD = 90^\\circ$, $DB$ bisects $\\angle ADC$, given that $CD = 6$, $AD = 8$. As shown in Figure 2, draw $BM \\parallel CD$ through point $B$ to intersect $AD$ at $M$, and connect $CM$ to intersect $DB$ at $N$. Find the length of $DN$.\n",
    "completion": "\\textbf{Solution:} Given that $BM\\parallel CD$ and $DB$ bisects $\\angle ADC$,\\\\\n$\\therefore \\angle MBD=\\angle BDC$, $\\angle BDC=\\angle BDM$,\\\\\n$\\therefore \\angle MBD=\\angle BDM$,\\\\\n$\\therefore MB=MD$,\\\\\nGiven $\\angle MBD+\\angle MBA=\\angle ABD=90^\\circ$ and $\\angle BDM+\\angle A=90^\\circ$,\\\\\n$\\therefore \\angle MBA=\\angle A$,\\\\\n$\\therefore MB=MA$,\\\\\n$\\therefore MB=\\frac{1}{2}AD=4$,\\\\\nGiven $BM\\parallel CD$,\\\\\n$\\therefore \\triangle MNB\\sim \\triangle CND$,\\\\\n$\\therefore \\frac{BM}{DC}=\\frac{BN}{DN}$,\\\\\n$\\therefore \\frac{BN}{DN}=\\frac{4}{6}=\\frac{2}{3}$,\\\\\n$\\therefore \\frac{DN}{DB}=\\frac{3}{5}$,\\\\\nGiven $BD=4\\sqrt{3}$,\\\\\n$\\therefore DN=\\frac{3}{5}DB=\\frac{3}{5}\\times 4\\sqrt{3}=\\boxed{\\frac{12\\sqrt{3}}{5}}$,\\\\\ni.e., the length of $DN$ is $\\frac{12\\sqrt{3}}{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51402165_150.png"
  },
  {
    "solution": "\\boxed{y=-x^{2}-2x+8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y = -x^2 + bx + c$ intersects the $x$-axis at points $A(2, 0)$ and $B(-4, 0)$. What is the equation of this parabola?\n",
    "completion": "\\textbf{Solution:} Substituting A(2,0) and B(-4,0) into $y=-x^{2}+bx+c$, we get:\n\\[\n\\begin{cases}\n-4+2b+c=0\\\\\n-16-4b+c=0\n\\end{cases}\n\\]\nSolving this, we get:\n\\[\n\\begin{cases}\nb=-2\\\\\nc=8\n\\end{cases}\n\\]\n$\\therefore$ The equation of the parabola is $y=-x^{2}-2x+8$. Therefore, the answer is $\\boxed{y=-x^{2}-2x+8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52602758_129.png"
  },
  {
    "solution": "\\boxed{\\frac{5}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the linear function $y_{1}=x+b$ intersects with the hyperbolic function $y_{2}=\\frac{k}{x}$ at points $A(1, a)$ and $D(-4, -1)$, and intersects with the $y$-axis and $x$-axis at points $B$ and $C$, respectively. Draw $AE\\perp y$-axis at point $E$, and connect $DE$. What is the area of $\\triangle ADE$?\n",
    "completion": "\\textbf{Solution:} Since point $A(1, 4)$ lies on the graph of the inverse proportion function $y=\\frac{4}{x}$,\\\\\nit follows that $a=4$,\\\\\nhence the coordinates of point $A$ are $(1, 4)$,\\\\\nsince $AE \\perp y$-axis,\\\\\nit follows that $AE=1, OE=4$,\\\\\nDraw $DM\\perp AE$ intersecting the extension of $AE$ at point $M$, and intersecting the $x$-axis at point $N$.\\\\\nHence, $\\angle NME=90^\\circ$,\\\\\nsince $AE \\perp y$-axis,\\\\\nit follows that $\\angle MEO=90^\\circ$,\\\\\nsince $\\angle EON=90^\\circ$\\\\\nit follows that $\\angle NME=\\angle MEO=\\angle EON=90^\\circ$,\\\\\nhence, quadrilateral EONM is a rectangle,\\\\\ntherefore $MN=OE=4, ND=1$,\\\\\nhence, $MD=5$,\\\\\ntherefore, ${S}_{\\triangle ADE}=\\frac{1}{2}\\times AE\\times DM=\\frac{1}{2}\\times 1\\times 5=\\boxed{\\frac{5}{2}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51379335_197.png"
  },
  {
    "solution": "\\boxed{\\left(\\frac{7}{2}, 3\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, side BC of the parallelogram ABCD is on the x-axis, with point A at $(0, 3)$ and point B at $(-1, 0)$. If the line $y=-2x+4$ perfectly bisects the area of the parallelogram ABCD, what are the coordinates of point D?\n",
    "completion": "\\textbf{Solution:} Let's consider the diagram, where we assume that the line $y=-2x+4$ intersects with AD and BC at points H and G, respectively.\\\\\nSince ABCD is a parallelogram with A at (0, 3),\\\\\nit follows that H is at $\\left(\\frac{1}{2}, 3\\right)$, and G is at (2, 0), with AD $\\parallel$ BC and AD=BC,\\\\\nLet's assume D is at (d, 3),\\\\\nthus DH=AD−AH=d−$\\frac{1}{2}$, and BG=2−(−1)=3,\\\\\nSince the line $y=-2x+4$ precisely bisects the area of the parallelogram ABCD into two congruent trapezoids,\\\\\nit follows that DH=BG,\\\\\nthus d−$\\frac{1}{2}$=3,\\\\\nwhich implies d=$\\frac{7}{2}$,\\\\\nTherefore, the coordinates of D are $\\boxed{\\left(\\frac{7}{2}, 3\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51825222.png"
  },
  {
    "solution": "\\boxed{y=(1-\\sqrt{2})x+8-4\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, the coordinates of point C are $(4\\sqrt{2}, 4\\sqrt{2})$. Perpendiculars are drawn from point C to the x-axis and y-axis, meeting at points B and D, respectively. Point E is on the line segment OD (not coinciding with points O or D). Connect BE, construct the symmetric point $O'$ of point O with respect to line BE, connect $CO'$, point P is the midpoint of $CO'$, connect BP, extend $CO'$ to intersect the extended line of BE at point F, connect DF. $\\angle PBF=45^\\circ$; As shown in Figure 2, connect BD, when point $O'$ just falls on the line segment BD, what is the equation of line BF?\n",
    "completion": "\\textbf{Solution:}\n\nConnect $EO'$, as shown in the figure: \\\\\nIn $\\triangle BOD$, we have $OB=OD=4\\sqrt{2}$, \\\\\n$\\therefore BD=\\sqrt{OB^2+OD^2}=8$, \\\\\n$\\because$ point O is the symmetric point of line BE with respect to O', \\\\\n$\\therefore OE=O'E$, $O'B=OB=4\\sqrt{2}$, $\\angle EO'B=\\angle EOB=90^\\circ$, \\\\\n$\\therefore \\angle DOE=90^\\circ$, $DO'=BD-O'B=8-4\\sqrt{2}$, \\\\\nLet $OE=O'E=x$, then $DE=4\\sqrt{2}-x$, \\\\\nIn $\\triangle DOE$, we have $DO'^2+O'E^2=DE^2$, \\\\\n$\\therefore (8-4\\sqrt{2})^2+x^2=(4\\sqrt{2}-x)^2$, \\\\\nSolving for $x$ gives $x=8-4\\sqrt{2}$, \\\\\n$\\therefore E(0,8-4\\sqrt{2})$, \\\\\nLet the equation of line BF be $y=kx+b$. Substituting $B(4\\sqrt{2},0)$ and $E(0,8-4\\sqrt{2})$ into this gives: \\\\\n$\\begin{cases} 0=4\\sqrt{2}k+b \\\\ 8-4\\sqrt{2}=b \\end{cases}$, solving yields $\\begin{cases} k=1-\\sqrt{2} \\\\ b=8-4\\sqrt{2} \\end{cases}$, \\\\\n$\\therefore$ The equation of line BF is $\\boxed{y=(1-\\sqrt{2})x+8-4\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52019276_101.png"
  },
  {
    "solution": "\\boxed{4:25}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, $E$ is a point on $CD$. Connect $BE$ and extend it to meet the extended line of $AD$ at point $F$. If $DE:EC=2:3$, then what is the value of $\\frac{S_{\\triangle DEF}}{S_{\\triangle ABF}}$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a parallelogram,\\\\\nit follows that AB$\\parallel$CD and AB＝CD,\\\\\nthus $\\triangle DEF\\sim \\triangle ABF$,\\\\\nhence $\\frac{S_{\\triangle DEF}}{S_{\\triangle ABF}} = \\left(\\frac{DE}{AB}\\right)^2$,\\\\\nsince DE:EC＝2:3,\\\\\nit follows that DE:CD＝DE:AB＝2:5,\\\\\ntherefore, S$_{\\triangle DEF}$:S$_{\\triangle ABF}$＝4:25.\\\\\nThus the answer is: $\\boxed{4:25}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51317421.png"
  },
  {
    "solution": "\\boxed{8\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the isosceles right $\\triangle ABC$, where $\\angle C=90^\\circ$, point D lies on AC. If $CD=6$ and $\\sin\\angle CBD=\\frac{3}{5}$, then what is the length of AB?\n",
    "completion": "\\textbf{Solution:} Given: $\\because$ $\\angle C=90^\\circ$,\\\\\n$\\therefore$ $\\sin\\angle CBD=\\frac{CD}{BD}=\\frac{3}{5}$,\\\\\n$\\because$ $CD=6$,\\\\\n$\\therefore$ $\\frac{6}{BD}=\\frac{3}{5}$, solving this we get: $BD=10$,\\\\\n$\\therefore$ $BC=\\sqrt{BD^{2}-CD^{2}}=\\sqrt{10^{2}-6^{2}}=8$,\\\\\n$\\because$ isosceles right $\\triangle ABC$,\\\\\n$\\therefore$ $AC=BC=8$,\\\\\n$\\therefore$ $AB=\\sqrt{AC^{2}+BC^{2}}=\\sqrt{8^{2}+8^{2}}=\\boxed{8\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53110760.png"
  },
  {
    "solution": "\\boxed{\\frac{5}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ and $\\triangle ADE$ are both equilateral triangles, with point D on side BC, and DE intersecting AC at point F. If AB=6, AD=5, and CD=4, what is the length of EF?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABC$ and $\\triangle ADE$ are both equilateral triangles,\\\\\nit follows that $\\angle B=\\angle C=\\angle ADE=60^\\circ$, and $DE=AD=5$,\\\\\nSince $\\angle ADC=\\angle ADE+\\angle CDE=\\angle B+\\angle BAD$,\\\\\nit follows that $\\angle CDE=\\angle BAD$,\\\\\ntherefore, $\\triangle ABD\\sim \\triangle DCF$,\\\\\nthus, $\\frac{AB}{DC}=\\frac{AD}{DF}$,\\\\\nwhich means $\\frac{6}{4}=\\frac{5}{DF}$,\\\\\ntherefore, $DF=\\frac{10}{3}$,\\\\\ntherefore, $EF=DE-DF=\\boxed{\\frac{5}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51155345.png"
  },
  {
    "solution": "\\boxed{20^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $OB$ bisects $\\angle AOD$, $\\angle BOC = \\frac{2}{3} \\angle AOC$, and if $\\angle AOD=100^\\circ$, find the degree measure of $\\angle BOC$.\n",
    "completion": "\\textbf{Solution:} Given that OB bisects $\\angle AOD$, and $\\angle AOD = 100^\\circ$, \\\\\n$\\therefore \\angle AOB = \\frac{1}{2} \\angle AOD = 50^\\circ$, \\\\\nGiven that $\\angle BOC + \\angle AOC = \\angle AOB$, and $\\angle BOC = \\frac{2}{3} \\angle AOC$, \\\\\n$\\therefore \\frac{2}{3} \\angle AOC + \\angle AOC = 50^\\circ$, \\\\\n$\\therefore \\angle AOC = 30^\\circ$, \\\\\n$\\therefore \\angle BOC = \\frac{2}{3} \\angle AOC = \\boxed{20^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51482322_178.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, $\\triangle OAB$ is an isosceles right-angled triangle with $\\angle OAB=90^\\circ$. The coordinate of point B is $(0, 2\\sqrt{2})$, and point A is in the first quadrant. When the triangle is translated to the right along the $x$ axis to get $\\triangle O'B'C$, the coordinate of point B' is $(2\\sqrt{2}, 2\\sqrt{2})$. What is the area of the region swept by $\\triangle OAB$ during the translation process?\n",
    "completion": "\\textbf{Solution:} Given that the coordinates of point B' are $(2\\sqrt{2}, 2\\sqrt{2})$, and after shifting the triangle along the x-axis to the right, we obtain the right-angled triangle $\\triangle O'B'C$, \\\\\n$\\therefore$AC=BB'=$2\\sqrt{2}$,\\\\\n$\\because$ $\\triangle OAB$ is an isosceles right-angled triangle,\\\\\n$\\therefore$ the coordinates of A are $(\\sqrt{2}, \\sqrt{2})$, and OA=AB=O'C=CB'=2,\\\\\n$\\therefore$ the area of the shape swept by $\\triangle OAB$ during the shifting process is: $2\\sqrt{2}\\times 2\\sqrt{2}+\\frac{1}{2}\\times 2\\times 2=\\boxed{10}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53839196.png"
  },
  {
    "solution": "\\boxed{36^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown, the regular pentagon $ABCDE$ is inscribed in circle $\\odot O$. What is the measure of $\\angle DAE$ in degrees?\n",
    "completion": "\\textbf{Solution:} Since the regular pentagon $ABCDE$ is inscribed in circle $\\odot O$, \\\\\ntherefore, $\\angle AED=\\frac{(5-2)\\times 180^\\circ}{5}=108^\\circ$, and $AE=DE$,\\\\\nhence, $\\angle DAE=\\angle ADE=\\frac{1}{2}\\times (180^\\circ-108^\\circ)=\\boxed{36^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53078987.png"
  },
  {
    "solution": "\\boxed{\\frac{55}{16}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in rectangle $ABCD$, with $AB=4$ and $AD=3$, point $E$ is the midpoint of side $BC$. Line $AE$ is drawn, and $\\triangle ABE$ is folded along $AE$ to obtain $\\triangle AFE$. When $AF$ is extended to intersect $CD$ at point $G$, what is the length of $DG$?\n",
    "completion": "\\textbf{Solution:} Based on the properties of folding, we know that: $EB=EF$, $AB=AF$, and $\\angle EBA=\\angle EFA=90^\\circ$,\\\\\nsince point $E$ is the midpoint of side $BC$,\\\\\nthus $CE=BE$,\\\\\ntherefore $CE=BE=EF$,\\\\\nAs shown in the figure: connect $EG$,\\\\\nsince $DC\\perp CE$, $EF\\perp AC$,\\\\\ntherefore $\\angle ECG=\\angle EFG=90^\\circ$,\\\\\nIn $\\triangle ECG$ and $\\triangle EFG$,\\\\\n$\\left\\{\\begin{array}{l}EC=EF\\\\ EG=EG\\end{array}\\right.$\\\\\ntherefore $\\triangle ECG\\cong \\triangle EFG\\left(\\text{HL}\\right)$,\\\\\nthus $GC=GF$,\\\\\nLet $GC=GF=x$, then $DG=4-x$, $AG=4+x$,\\\\\nIn $\\triangle DAG$, $AD^{2}+DG^{2}=AG^{2}$,\\\\\nwhich means $3^{2}+(4-x)^{2}=(4+x)^{2}$,\\\\\nSolving this, we get: $x=\\frac{9}{16}$,\\\\\nThus, the answer is: $DG=4-x=\\boxed{\\frac{55}{16}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52424728.png"
  },
  {
    "solution": "\\boxed{62^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $ \\triangle ABC\\cong \\triangle AED$, with point E on line segment BC, and $ \\angle 1=56^\\circ$, what is the measure of $ \\angle AED$?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABC\\cong \\triangle AED$,\\\\\nthen $\\angle BAC=\\angle EAD$, and $AB=AE$,\\\\\nthus $\\angle BAE=\\angle 1=56^\\circ$,\\\\\ntherefore $\\angle B=\\angle AEB=\\frac{1}{2}(180^\\circ-56^\\circ)=62^\\circ$,\\\\\nhence $\\angle AED=\\angle B=\\boxed{62^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52837423.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, it demonstrates the principle of pinhole imaging, where $OA=30\\,cm$, $OC=10\\,cm$, and $AB\\parallel CD$. If the height of the object $AB$ is $15\\,cm$, what is the height of the image $CD$ in $cm$?\n",
    "completion": "\\textbf{Solution:} Since AB $\\parallel$ CD,\\\\\ntherefore $\\frac{OA}{OC}=\\frac{AB}{CD}$,\\\\\nthus $\\frac{30}{10}=\\frac{15}{CD}$,\\\\\ntherefore CD = \\boxed{5} cm.",
    "image_path": "dataset/mm_math/images/MM_Math/52933996.png"
  },
  {
    "solution": "\\boxed{\\frac{9}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=ax^2+bx+c$ ($a\\neq 0$) has its axis of symmetry as the line $x=-1$, intersects the $x$-axis at points $A$ and $B$, and intersects the $y$-axis at point $C$, with $OA=OC$. The coordinate of point $A$ is $(-3,0)$. Suppose point $Q$ is a moving point on segment $AC$, and a line is drawn from $Q$ perpendicular to the $x$-axis, intersecting the parabola at point $D$. What is the maximum length of segment $QD$?\n",
    "completion": "\\textbf{Solution:} Given point A ($-3$, $0$) and point C ($0$, $-3$),\\\\\nhence, the equation of line AC is $y=-x-3$,\\\\\nLet point Q be ($m$, $-m-3$) where $-3\\leq m\\leq 0$,\\\\\nthen point $D(m, m^2+2m-3)$,\\\\\nthus $QD=-m-3-(m^2+2m-3)=-\\left(m+\\frac{3}{2}\\right)^2+\\frac{9}{4}$,\\\\\nTherefore, when $m=-\\frac{3}{2}$, the maximum value of $QD$ is $\\boxed{\\frac{9}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52765835_127.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the trapezoid ABCD, AD $\\parallel$ BC, $\\angle BAD = 90^\\circ$, and the diagonal BD $\\perp$ DC. If AD = 4 and BC = 9, what is the length of BD?\n",
    "completion": "\\textbf{Solution:} Given: $\\because$ AD$\\parallel$BC $\\therefore$ $\\angle$ADB=$\\angle$CBD (When two lines are parallel, the corresponding alternate interior angles are equal)\\\\\n$\\because$ $\\angle$BAD=$90^\\circ$, BD$\\perp$DC, $\\therefore$ $\\angle$BAD=$\\angle$BDC=$90^\\circ$,\\\\\n$\\because$ $\\angle$ADB=$\\angle$CBD, $\\angle$BAD=$\\angle$BDC, $\\therefore$ $\\triangle$ABD$\\sim$$\\triangle$DCB (Two triangles are similar if two angles are equal respectively)\\\\\n$\\therefore$ $\\frac{BD}{BC}=\\frac{AD}{BD}$ (Corresponding sides of similar triangles are proportional)\\\\\n$\\therefore$ $BD^2=AD\\times BC$, $\\because$AD=4, BC=9\\\\\n$\\therefore$BD=\\boxed{6}",
    "image_path": "dataset/mm_math/images/MM_Math/51155313.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $ \\triangle ABC$ is inscribed in $ \\odot O$, with chords AE and BC intersecting at point D. Line BO is drawn such that $ \\angle CAE = \\angle ABO$ and $ AE \\perp BC$. If $ ED = \\sqrt{3} CD$, find the degree measure of $ \\angle ABC$.\n",
    "completion": "Solution: As shown in the figure, connect CE,\\\\\n$\\therefore \\angle ABC=\\angle AEC$,\\\\\n$\\because ED=\\sqrt{3}CD$, AE$\\perp$BC,\\\\\n$\\therefore \\tan\\angle AEC=\\frac{CD}{DE}=\\frac{\\sqrt{3}}{3}$,\\\\\n$\\therefore \\angle AEC=30^\\circ$,\\\\\n$\\therefore \\angle ABC=\\boxed{30^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51318321_117.png"
  },
  {
    "solution": "\\boxed{120^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rhombus $ABCD$, $E$ is the midpoint of $AB$, and $DE\\perp AB$, with $AE=2$. What is the measure of $\\angle ABC$ in degrees?\n",
    "completion": "\\textbf{Solution:} Given that $E$ is the midpoint of $AB$, and $DE \\perp AB$,\\\\\n$\\therefore AD = BD$.\\\\\nAlso, in rhombus $ABCD$, $AD = AB$,\\\\\n$\\therefore AD = AB = BD$,\\\\\n$\\therefore \\triangle ABD$ is an equilateral triangle,\\\\\n$\\therefore \\angle ABD = 60^\\circ$,\\\\\n$\\therefore \\angle ABC = \\boxed{120^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51614472_112.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\triangle ABC$, $DE$ is the median line of $ \\triangle ABC$. $DC$ and $BE$ intersect at point $F$. If $ {S}_{\\triangle DEF} = 1$, then what is $ {S}_{\\triangle ADE}$?\n",
    "completion": "\\textbf{Solution:} Since $DE$ is the median of $\\triangle ABC$, \\\\\nit follows that $DE \\parallel BC$ and $BC = 2DE$, \\\\\nhence $\\triangle DEF \\sim \\triangle CBF$, \\\\\nthus $\\frac{{S}_{\\triangle CBF}}{{S}_{\\triangle DEF}} = \\left(\\frac{BC}{DE}\\right)^2 = 2^2$, \\\\\ntherefore ${S}_{\\triangle CBF} = 4{S}_{\\triangle DEF}$, \\\\\nsince ${S}_{\\triangle DEF} = 1$, \\\\\nthen ${S}_{\\triangle CBF} = 4$, \\\\\nsince $BE$ is the median, \\\\\nit follows that ${S}_{\\triangle ABE} = {S}_{\\triangle CBE}$, \\\\\nsince $DE$ is the median of $\\triangle ABC$, \\\\\nit follows that $DE \\parallel BC$, \\\\\nthus ${S}_{\\triangle BDE} = {S}_{\\triangle CDE}$, \\\\\nhence ${S}_{\\triangle BDF} = {S}_{\\triangle CFE}$, \\\\\ntherefore ${S}_{\\triangle BDF} + {S}_{\\triangle ADE} + {S}_{\\triangle DEF} = {S}_{\\triangle CFE} + {S}_{\\triangle CBF}$, \\\\\nthus ${S}_{\\triangle ADE} + {S}_{\\triangle DEF} = {S}_{\\triangle CBF}$, \\\\\ntherefore ${S}_{\\triangle ADE} = \\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51433924.png"
  },
  {
    "solution": "\\boxed{110^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ABC=82^\\circ$, $\\angle C=58^\\circ$, $BD\\perp AC$ at $D$, $AE$ bisects $\\angle CAB$, and $BD$ intersects $AE$ at point $F$. What is $\\angle AFB$?\n",
    "completion": "\\textbf{Solution:} Given: $\\because \\angle CAB = 180^\\circ - \\angle ABC - \\angle C$,\\\\\nand $\\angle ABC = 82^\\circ, \\angle C = 58^\\circ$,\\\\\n$\\therefore \\angle CAB = 40^\\circ$,\\\\\n$\\because$ AE bisects $\\angle CAB$,\\\\\n$\\therefore \\angle DAF = 20^\\circ$,\\\\\n$\\because$ BD $\\perp$ AC at D,\\\\\n$\\therefore \\angle ADB = 90^\\circ$,\\\\\n$\\therefore \\angle AFB = \\angle ADB + \\angle DAF = 90^\\circ + 20^\\circ = \\boxed{110^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52699728.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, the graph of the inverse proportion function $y=\\frac{k}{x}$ passes through points $A(1, m)$ and $B(2, 1)$. The coordinates of point $P$ are $(6, 1)$. As shown in Figure 2, the coordinates of points $C$ and $D$ are $(a, 0)$ and $(0, a)$ ($0<a\\leq 6$), respectively. When the area of $\\triangle PCD$ is $3$, what is the value of $a$?\n",
    "completion": "\\textbf{Solution:} Construct a perpendicular $PN$ from point P to the x-axis, with N being the foot of the perpendicular, \\\\\n$\\because S_{\\triangle PCD}=S_{\\text{trapezoid}} PNOD-S_{\\triangle COD}-S_{\\triangle PCN}$, \\\\\n$\\therefore \\frac{1}{2}(1+a)\\times 6-\\frac{1}{2}a^{2}-\\frac{1}{2}(6-a)\\times 1=3$, \\\\\nwe obtain $a_{1}=\\boxed{1}$, $a_{2}=6$.",
    "image_path": "dataset/mm_math/images/MM_Math/51445690_118.png"
  },
  {
    "solution": "\\boxed{\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in square $ABCD$, $AB=3$, $E$ is a point on $AD$, $AE=2$. Connect $BE$ and draw $AF\\perp BE$ to meet $DC$ at $F$. What is the length of $EF$?\n",
    "completion": "\\textbf{Solution:}\n\nSince quadrilateral ABCD is a square, \\\\\n$\\therefore AB=3=AD$, $\\angle BAE=90^\\circ=\\angle ADF$,\\\\\n$\\therefore \\angle DAF+\\angle AFD=90^\\circ$,\\\\\nSince $AF\\perp BE$,\\\\\n$\\therefore \\angle AEB+\\angle DAF=90^\\circ$,\\\\\n$\\therefore \\angle AFD=\\angle AEB$,\\\\\nIn $\\triangle ABE$ and $\\triangle DAF$,\\\\\n$\\left\\{\\begin{array}{l}\n\\angle AEB=\\angle AFD \\\\\n\\angle BAE=\\angle ADF \\\\\nAB=AD\n\\end{array}\\right.$,\\\\\n$\\therefore \\triangle ABE \\cong \\triangle DAF \\left(AAS\\right)$,\\\\\n$\\therefore AE=DF=2$,\\\\\nSince $AD=3$, $AE=2$,\\\\\n$\\therefore DE=AD-AE=3-2=1$,\\\\\nIn $Rt\\triangle DEF$, $EF=\\sqrt{DE^2+DF^2}=\\sqrt{1^2+2^2}=\\boxed{\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52160361.png"
  },
  {
    "solution": "\\boxed{y=\\frac{3}{4}x^{2}+\\frac{9}{4}x-3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=ax^2+3ax+c$ ($a>0$) intersects the y-axis at point C and the x-axis at points A and B, with point A to the left of point B. The coordinate of point B is $(1, 0)$, and $OC=3OB$. Find the equation of the parabola.\n",
    "completion": "\\textbf{Solution:} Because the coordinates of B are $(1,0)$,\\\\\ntherefore $OB = 1$.\\\\\nBecause $OC = 3OB = 3$, and point C is below the x-axis,\\\\\ntherefore C is $(0, -3)$.\\\\\nBecause substituting B $(1,0)$, C $(0, -3)$ into the quadratic equation yields $\\begin{cases}4a+c=0\\\\ c=-3\\end{cases}$, solving gives: $\\begin{cases}a=\\frac{3}{4}\\\\ c=-3\\end{cases}$,\\\\\ntherefore the equation of the parabola is $\\boxed{y=\\frac{3}{4}x^{2}+\\frac{9}{4}x-3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53080621_37.png"
  },
  {
    "solution": "\\boxed{50^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of $\\odot O$, $AP$ is a tangent to $\\odot O$, and $PB$ intersects $\\odot O$ at point $C$. If point $D$ is on $\\odot O$ and $\\angle ADC=40^\\circ$, what is the degree measure of $\\angle P$?\n",
    "completion": "\\textbf{Solution:} Given $\\angle ADC=40^\\circ$,\\\\\nit follows that $\\angle ABC=40^\\circ$,\\\\\nsince $AB$ is the tangent of $\\odot O$ and point $A$ is the point of tangency,\\\\\nit follows that $\\angle OAB=90^\\circ$,\\\\\nthus, $\\angle P=90^\\circ-\\angle ABC=90^\\circ-40^\\circ=\\boxed{50^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51446085.png"
  },
  {
    "solution": "\\boxed{(0,\\frac{3}{2})}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $l\\colon y = \\frac{4}{3}x+b$ passes through the point $A(-3,0)$ and intersects with the $y$-axis at point $B$. The bisector of $\\angle OAB$ intersects the $y$-axis at point $C$. A perpendicular is drawn from point $C$ to line $AB$, intersecting the $x$-axis at point $E$, with the foot of the perpendicular being point $D$. What are the coordinates of points $B$ and $C$?\n",
    "completion": "\\textbf{Solution}: Substituting point $A(-3,0)$ into $y=\\frac{4}{3}x+b$, we find $b=4$,\\\\\n$\\therefore B(0,4)$,\\\\\n$\\therefore OB=4$,\\\\\n$\\because A(-3,0)$,\\\\\n$\\therefore OA=3$,\\\\\nIn $\\triangle AOB$, $\\angle AOB=90^\\circ$,\\\\\n$\\therefore AB=\\sqrt{OA^2+OB^2}=5$.\\\\\n$\\because AC$ bisects $\\angle OAB$, $CD\\perp AB$, $CO\\perp OA$,\\\\\n$\\therefore CD=CO$, $\\angle ACD=\\angle ACO$,\\\\\n$\\because AC=AC$,\\\\\n$\\therefore \\triangle ACD\\cong \\triangle ACO$ (SAS),\\\\\n$\\therefore AD=AO=3$, $BD=AB-AD=2$.\\\\\nLet $CD=CO=m$, then $BC=4-m$,\\\\\nIn $\\triangle BDC$,\\\\\nby the Pythagorean theorem, $CD^2+BD^2=BC^2$,\\\\\n$\\therefore m^2+2^2=(4-m)^2$,\\\\\nSolving, we get $m=\\frac{3}{2}$,\\\\\n$\\therefore C(0,\\frac{3}{2})$;\\\\\n$\\therefore$ The coordinates of $B$ and $C$ are $B(0,4)$ and $C(0,\\frac{3}{2})$ respectively.\\\\\nTherefore, the coordinates of points $B$ and $C$ are $B=\\boxed{(0,4)}$ and $C=\\boxed{(0,\\frac{3}{2})}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52189067_133.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points C, M, and N are on line segment AB, with $AC=12$, $BC=6$, $AM=\\frac{1}{3}AC$, and $BN=\\frac{1}{3}BC$. What is the length of line segment MN?\n",
    "completion": "\\textbf{Solution:}\\\\\nSince $AC=12$,\\\\\nit follows that $AM=\\frac{1}{3}AC=4$,\\\\\nSince $BC=6$,\\\\\nit follows that $BN=\\frac{1}{3}BC=2$,\\\\\nTherefore, $MN=AB-AM-BN=AC+BC-AM-BN=12+6-4-2=\\boxed{12}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53404179.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, $\\angle D=90^\\circ$, $\\angle B=60^\\circ$, $AD=6$, $AB^2=\\frac{100}{3}$, $AB\\perp AC$. A point $E$ is chosen on $CD$, and line $AE$ is drawn. Triangle $ADE$ is folded along $AE$ so that point $D$ falls on point $F$ on $AC$. Find: What is the length of $CD$?\n",
    "completion": "\\textbf{Solution:} Since $AB \\perp AC$ and $\\angle B = 60^\\circ$,\\\\\nit follows that $\\angle BCA = 30^\\circ$,\\\\\nwhich implies $BC = 2AB$,\\\\\nhence $BC^2 = 4AB^2 = \\frac{400}{3}$,\\\\\ntherefore $AC^2 = BC^2 - AB^2 = 100$,\\\\\nleading to $AC = 10$.\\\\\nSince $\\angle D = 90^\\circ$,\\\\\nin $\\triangle ADC$, where $AD = 6, AC = 10$,\\\\\nby the Pythagorean theorem, it is found that $CD = \\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52731085_118.png"
  },
  {
    "solution": "\\boxed{7}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In quadrilateral $ABCD$, $AD \\parallel BC$, $AB= \\sqrt{5}$, $AD=2$, $DC= 2\\sqrt{5}$, $\\tan \\angle ABC=2$ (as shown in the figure). Point $E$ is on ray $AD$, and point $F$ is on side $BC$, with $BE$ and $EF$ connected, and $\\angle BEF=\\angle DCB$. What is the length of segment $BC$?\n",
    "completion": "\\textbf{Solution:} Draw $AH\\perp BC$ and $DG\\perp BC$ through points A and D respectively, with H and G as the foot of the perpendiculars.\\\\ \nWe can get: $AD=HG=2$, $AH=DG$,\\\\ \n$\\because \\tan\\angle ABC=2$, $AB= \\sqrt{5}$,\\\\ \n$\\therefore AH=2$, $BH=1$,\\\\ \n$\\therefore DG=2$,\\\\ \n$\\because DC= 2\\sqrt{5}$,\\\\ \n$\\therefore CG= \\sqrt{DC^{2}-DG^{2}}=4$,\\\\ \n$\\therefore BC=BH+HG+GC=1+2+4=\\boxed{7}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51379498_103.png"
  },
  {
    "solution": "\\boxed{\\sqrt{13}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $O$ is a point inside the equilateral triangle $ABC$, with $\\angle BOC = 150^\\circ$. The triangle $BOC$ is rotated counterclockwise around point $C$ to obtain triangle $ADC$, and then lines $OD$ and $OA$ are drawn. If $OB=2$ and $OC=3$, what is the length of $AO$?\n",
    "completion": "\\textbf{Solution}: From the property of rotation, we have $AD=OB=2$,\\\\\nsince $\\triangle OCD$ is an equilateral triangle,\\\\\ntherefore $OD=OC=3$,\\\\\nsince $\\angle BOC=150^\\circ$, $\\angle ODC=60^\\circ$,\\\\\ntherefore $\\angle ADO=90^\\circ$,\\\\\nin $\\triangle AOD$, by Pythagoras' theorem, we get $AO=\\sqrt{AD^2+OD^2}=\\sqrt{2^2+3^2}=\\boxed{\\sqrt{13}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51212821_101.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the rectangle $ABCD$, where $BC=4$, the diagonals $AC$ and $BD$ intersect at point $O$, $AN \\perp BD$ with the foot being $N$, $BN = 3DN$, what is the length of $AN$?\n",
    "completion": "\\textbf{Solution:} Since $BN=3DN$,\\\\\nit follows that $BD=4DN$,\\\\\nSince quadrilateral $ABCD$ is a rectangle and $BC=4$,\\\\\nit follows that $AD=BC=4$ and $BO=DO$,\\\\\nthus $DO=\\frac{1}{2}BD=2DN$,\\\\\nhence $DN=ON$,\\\\\nSince $AN\\perp BD$,\\\\\nit follows that $\\angle ANO=\\angle AND=90^\\circ$,\\\\\ntherefore, $\\triangle ANO$ and $\\triangle AND$ are right-angled triangles,\\\\\nIn $\\triangle ANO$ and $\\triangle AND$,\\\\\n$\\left\\{\\begin{array}{l}\nAN=AN\\\\\n\\angle ANO=\\angle AND\\\\\nNO=ND\n\\end{array}\\right.$\\\\\nhence $\\triangle ANO\\cong \\triangle AND$ (SAS),\\\\\nthus $AO=AD$, $\\angle NAO=\\angle NAD$,\\\\\ntherefore, $AO=AD=OD$,\\\\\nhence $\\triangle AOD$ is an equilateral triangle,\\\\\nthus $\\angle OAD=60^\\circ$,\\\\\ntherefore, $\\angle NAD=\\frac{1}{2}\\angle OAD=\\frac{1}{2}\\times 60^\\circ=30^\\circ$,\\\\\nthus $ND=\\frac{1}{2}AD=\\frac{1}{2}\\times 4=2$,\\\\\nthus $AN=\\sqrt{AD^2-ND^2}=\\sqrt{4^2-2^2}=\\boxed{2\\sqrt{3}}$",
    "image_path": "dataset/mm_math/images/MM_Math/52019249.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB=12\\,cm$, $C$ is the midpoint of $AB$, and point $D$ is on segment $AC$, with $AD\\colon CB=1\\colon 3$, what is the length of $DB$ in $cm$?\n",
    "completion": "\\textbf{Solution:} Given $\\because$AB=12cm, and C is the midpoint of AB,\\\\\n$\\therefore$AC=BC=$\\frac{1}{2}$AB=6(cm),\\\\\n$\\because$AD$\\colon$ CB=1$\\colon$ 3,\\\\\n$\\therefore$AD=2cm,\\\\\n$\\therefore$DC=AC−AD=4(cm),\\\\\n$\\therefore$DB=DC+BC=\\boxed{10}(cm).",
    "image_path": "dataset/mm_math/images/MM_Math/53119292.png"
  },
  {
    "solution": "\\boxed{\\frac{4}{25}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in parallelogram $ABCD$, $E$ is a point on $CD$. $DE:CE=2:3$. Connect $AE$ and $BD$ which intersect at point $F$. What is the value of $\\frac{S_{\\triangle DEF}}{S_{\\triangle ABF}}$?\n",
    "completion": "\\textbf{Solution:} In parallelogram $ABCD$, we have $DC=AB$ and $DC\\parallel AB$,\\\\\n$\\therefore$ $\\triangle DEF\\sim \\triangle BAF$,\\\\\n$\\because$ $DE\\colon CE=2\\colon 3$,\\\\\n$\\therefore$ $DE\\colon DC=2\\colon 5$,\\\\\n$\\therefore$ $\\frac{DE}{AB}=\\frac{2}{5}$,\\\\\n$\\therefore$ $\\frac{S_{\\triangle DEF}}{S_{\\triangle ABF}}=\\left(\\frac{2}{5}\\right)^2=\\boxed{\\frac{4}{25}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53078877.png"
  },
  {
    "solution": "\\boxed{40^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $O$ lies on line $AB$, $OE$ bisects $\\angle AOC$, $\\angle DOC=90^\\circ$, and $\\angle EOC=65^\\circ$. Find the measure of $\\angle BOD$.\n",
    "completion": "\\textbf{Solution:} Since $OE$ bisects $\\angle AOC$,\\\\\nwe have $\\angle AOC=2\\angle EOC=2\\times 65^\\circ=130^\\circ$,\\\\\nthus $\\angle BOC=180^\\circ-\\angle AOC=180^\\circ-130^\\circ=50^\\circ$,\\\\\ngiven that $\\angle DOC=90^\\circ$,\\\\\nit follows $\\angle BOD=\\angle DOC-\\angle BOC=90^\\circ-50^\\circ=\\boxed{40^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53253420.png"
  },
  {
    "solution": "\\boxed{2\\colon1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, parallelogram $ABCD$ has diagonals $AC$ and $BD$ intersecting at point $O$. Point $F$ is the midpoint of $BC$, and connecting $DF$ intersects $AC$ at point $E$. What is the ratio of $DE:EF$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect OF, \\\\\n$\\because$ Quadrilateral ABCD is a parallelogram,\\\\\n$\\therefore$ O is the midpoint of BD,\\\\\n$\\because$ F is the midpoint of BC,\\\\\n$\\therefore$ OF is the median of $\\triangle$BCD,\\\\\n$\\therefore$ CD $=$ 2OF, and OF $\\parallel$ CD,\\\\\n$\\therefore$ $\\triangle$DEC $\\sim$ $\\triangle$FEO,\\\\\n$\\therefore$ DE$\\colon$EF $=$ CD$\\colon$OF $=$ \\boxed{2\\colon1}.",
    "image_path": "dataset/mm_math/images/MM_Math/52745993.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AD \\perp BC$ at point $D$, $\\cos B = \\frac{4}{5}$, $\\tan C = \\sqrt{3}$, $AB = 5$. Find the length of $AC$.\n",
    "completion": "\\textbf{Solution:}\\\\\nSince $AD\\perp BC$,\\\\\nit follows that $\\angle ADB=\\angle ADC=90^\\circ$,\\\\\nSince $\\cos B=\\frac{4}{5}$ and $AB=5$,\\\\\nit follows that $BD=AB\\cdot\\cos B=5\\times \\frac{4}{5}=4$,\\\\\nTherefore, $AD=\\sqrt{AB^{2}-BD^{2}}=\\sqrt{5^{2}-4^{2}}=3$,\\\\\nSince $\\tan C=\\sqrt{3}$,\\\\\nit follows that $CD=\\frac{AD}{\\tan C}=\\frac{3}{\\sqrt{3}}=\\sqrt{3}$,\\\\\nTherefore, $AC=\\sqrt{AD^{2}+CD^{2}}=\\sqrt{3^{2}+(\\sqrt{3})^{2}}=\\boxed{2\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53019671.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=x+1$ and $y=ax+4$ intersect at point $A(1, m)$. Find the value of $m$.\n",
    "completion": "\\textbf{Solution:} Substitute $A\\left(1, m\\right)$ into $y=x+1$,\\\\\nyielding $m=1+1=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52284181_68.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, $D$ is the midpoint of $AB$, $DE\\perp AB$, intersecting $AC$ at $E$. If $\\frac{AE}{EC}=\\frac{5}{3}$, then what is the value of $\\tan\\angle A$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect BE. Since $D$ is the midpoint of $AB$ and $DE \\perp AB$, it follows that AE = BE. Since $\\frac{AE}{EC} = \\frac{5}{3}$, we can set AE = BE = 5k and CE = 3k. By the Pythagorean theorem, we get BC = 4k. Therefore, AC = AE + CE = 8k. Therefore, $\\tan A = \\frac{BC}{AC} = \\frac{4k}{8k} = \\boxed{\\frac{1}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55533861.png"
  },
  {
    "solution": "\\boxed{x<-2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, given that the graph of function $y=k_{1}x+b_{1}$ intersects with the graph of function $y=k_{2}x+b_{2}$ at point $A(-2,1)$, what is the solution set of the inequality $k_{1}x+b_{1}>k_{2}x+b_{2}$ with respect to $x$?\n",
    "completion": "\\textbf{Solution:} Since from the graph of the function, it is known that when $x<-2$, the line $y=k_{1}x+b_{1}$ is above the line $y=k_{2}x+b_{2}$,\\\\\ntherefore, the solution set for the inequality $k_{1}x+b_{1}>k_{2}x+b_{2}$ with respect to $x$ is $x<-2$.\\\\\nThus, the answer is: $\\boxed{x<-2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50589675.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $C$ is the midpoint of line segment $AB$, and point $D$ is on $AB$, with $AD=\\frac{1}{3}AB$. If $AD=6$, find the length of line segment $CD$.\n",
    "completion": "\\textbf{Solution:} Given $\\because AD= \\frac{1}{3} AB$ and $AD=6$,\\\\\n$\\therefore AB=18$,\\\\\n$\\because C$ is the midpoint of $AB$,\\\\\n$\\therefore AC= \\frac{1}{2} AB=9$,\\\\\n$\\therefore CD=AC-AD=9-6=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51419997_163.png"
  },
  {
    "solution": "\\boxed{\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given a square $ABCD$ with side length 1, let $E$ and $F$ be two movable points on sides $BC$ and $CD$, respectively, satisfying $BE=CF$. Connect $AE$ and $AF$. What is the minimum value of $AE+AF$?\n",
    "completion": "\\textbf{Solution:} Draw line DE, \\\\\nsince $BE=CF$ and the quadrilateral ABCD is a square, \\\\\nthus $CD-CF=BC-BE$, i.e., $DF=CE$, in $\\triangle ADF$ and $\\triangle DCE$, \\\\\n\\[\\begin{cases}\nAD=DC, \\\\\n\\angle ADF=\\angle DCE, \\\\\nDF=CE,\n\\end{cases}\\]\nthus $\\triangle ADF \\cong \\triangle DCE$, \\\\\ntherefore, $AF=DE$; $AE+AF=AE+DE$, \\\\\nby taking BC as the axis of symmetry, construct the symmetrical point $A'$ of point A with respect to BC and connect $DA'$, intersecting BC at point E, \\\\\nsince point A and point $A'$ are symmetric with respect to BC, \\\\\nthus AE=$A'E$, \\\\\n$AE+DE=A'E+DE=A'D$, \\\\\nby the Pythagorean theorem, we have: $A'D=\\sqrt{AD^{2}+AA'^{2}}=\\sqrt{2^{2}+1^{2}}=\\sqrt{5}$, \\\\\ntherefore, the minimum value of $AE+AF$ is $\\boxed{\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52211561.png"
  },
  {
    "solution": "\\boxed{100^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within the circle $ \\odot O$, $AB = AC$. If $\\angle ABC = 65^\\circ$, what is the degree measure of $\\angle BOC$?\n",
    "completion": "\\textbf{Solution:} Given that $AB=AC$, \\\\\nhence $\\angle ABC=\\angle ACB=65^\\circ$, \\\\\ntherefore $\\angle A=180^\\circ-2\\angle ABC=180^\\circ-2\\times65^\\circ=50^\\circ$, \\\\\nsince $\\overset{\\frown}{BC}=\\overset{\\frown}{BC}$, \\\\\nthus $\\angle BOC=2\\angle A=2\\times50^\\circ=\\boxed{100^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53081440.png"
  },
  {
    "solution": "\\boxed{0<x<3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=kx+b$ intersects the parabola $y=-x^2+2x+3$ at points $A$ and $B$, where point $A$ is at $(0, 3)$ and point $B$ is at $(3, 0)$. The parabola intersects the x-axis at another point $C(-1, 0)$. What is the solution set of the inequality $-x^2+2x+3>kx+b$?\n",
    "completion": "\\textbf{Solution:} From the graph and the problem statement, it is given that:\\\\\nThe solution set for the inequality $-x^{2}+2x+3>kx+b$ is $0<x<3$;\\\\\nTherefore, the answer is $\\boxed{0<x<3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53042850.png"
  },
  {
    "solution": "\\boxed{(4,-5)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the quadrilateral $ABCD$ is a rhombus, where the coordinates of points $A$ and $B$ are $\\left(0,3\\right)$ and $\\left(4,0\\right)$, respectively, and point $D$ is on the $y$-axis. What are the coordinates of point $C$?\n",
    "completion": "\\textbf{Solution:} Given that the coordinates of points A and B are respectively $(0,3)$ and $(4,0)$,\\\\\nthus OA=3, OB=4,\\\\\nhence AB$=\\sqrt{AO^{2}+BO^{2}}=5$,\\\\\nsince quadrilateral ABCD is a rhombus,\\\\\nthus BC=AB=5, and AD$\\parallel$BC,\\\\\nsince A and D are on the y-axis,\\\\\nthus BC$\\parallel$y-axis,\\\\\nhence the coordinates of C are $\\boxed{(4,-5)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52381590.png"
  },
  {
    "solution": "\\boxed{70^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the direction of ray $OA$ is $15^\\circ$ east of north, and the direction of ray $OB$ is $40^\\circ$ west of north. Ray $OA$ is the angle bisector of $\\angle BOC$. $OF$ is the extension line in the opposite direction of $OB$. Find: the direction of ray $OC$.\n",
    "completion": "\\textbf{Solution:} From the diagram, it is known that: $\\angle AOB=15^\\circ+40^\\circ=55^\\circ$,\\\\\nsince $OA$ is the angle bisector of $\\angle BOC$,\\\\\nthus $\\angle AOC=55^\\circ$,\\\\\ntherefore $\\angle NOC=\\angle NOA+\\angle AOC$\\\\\n$=15^\\circ+55^\\circ=70^\\circ$,\\\\\nhence, the ray $OC$ is in the direction of \\boxed{70^\\circ} East of North.",
    "image_path": "dataset/mm_math/images/MM_Math/53404231_75.png"
  },
  {
    "solution": "\\boxed{20^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, E is the midpoint of the hypotenuse AB of $\\triangle ABC$ and $\\triangle ABD$, with $AC = BC$ and $\\angle DBA = 25^\\circ$. What is the measure of $\\angle DCE$?\n",
    "completion": "\\textbf{Solution:} Given that point $E$ is the midpoint of the hypotenuse $AB$ of the right triangle $ABD$,\\\\\nhence, $ED=EB=\\frac{1}{2}AB$,\\\\\ntherefore, $\\angle EDB=\\angle DBA=25^\\circ$,\\\\\nthus, $\\angle DEA=\\angle EDB+\\angle DBA=50^\\circ$,\\\\\nsince point $E$ is also the midpoint of the hypotenuse $AB$ of the right triangle $ABC$, and $AC=BC$,\\\\\nhence, $EC=\\frac{1}{2}AB$, and $CE\\perp AB$,\\\\\nthus, $\\angle AEC=90^\\circ$,\\\\\ntherefore, $\\angle DEC=\\angle DEA+\\angle AEC=140^\\circ$, and since $ED=EC$,\\\\\nhence, $\\angle DCE=\\frac{1}{2}\\times \\left(180^\\circ-140^\\circ\\right)=\\boxed{20^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52991477.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in square $ABCD$, $AB = 4$, point $M$ is on side $CD$, and $DM = 1$. Triangles $AEM$ and $ADM$ are symmetric about the line containing $AM$. Rotating $\\triangle ADM$ $90^\\circ$ clockwise around point $A$ results in $\\triangle ABF$. When connecting $EF$, what is the length of segment $EF$?\n",
    "completion": "\\textbf{Solution:} \\\\\nSince $\\triangle AEM$ and $\\triangle ADM$ are symmetrical about the line containing AM,\\\\\ntherefore $\\triangle AEM \\cong \\triangle ADM$\\\\\nthus $AE=AD=AB=4$\\\\\nConnect BM, as shown in the figure,\\\\\nsince $\\triangle ADM$ is rotated clockwise by $90^\\circ$ around point A to get $\\triangle ABF$,\\\\\ntherefore $\\triangle ABF \\cong \\triangle ADM$\\\\\nthus $AF=AM$, $\\angle FAB=\\angle MAD$\\\\\nthus $\\angle FAB=\\angle MAE$\\\\\ntherefore $\\angle FAB+\\angle BAE=\\angle MAE + \\angle BAE$\\\\\ntherefore $\\angle FAE=\\angle MAB$\\\\\nthus $\\triangle FAE\\cong \\triangle MAB$ (SAS)\\\\\nthus $EF=BM$\\\\\nIn the square ABCD,\\\\\ntherefore $BC=CD=AB=4$\\\\\nsince $DM=1$\\\\\ntherefore $CM=3$\\\\\nIn $\\triangle BCM$,\\\\\n$BM=\\sqrt{3^2+4^2}=5$\\\\\ntherefore $EF=\\boxed{5}$",
    "image_path": "dataset/mm_math/images/MM_Math/51578581.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, take the hypotenuse $BC$ of the right triangle $\\triangle ABC$ as a side to construct a square $BCEF$ on the same side of $\\triangle ABC$. Let the center of the square be $O$, and connect $AO$. If $AB=2$ and $AO=3\\sqrt{2}$, what is the length of $AC$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, draw line OD $\\perp$ OA, intersecting AC at D, \\\\\n$\\because$ Quadrilateral BCEF is a square, \\\\\n$\\therefore$ $\\angle$BOC=90$^\\circ$, OB=OC, \\\\\n$\\therefore$ $\\angle$AOB+$\\angle$BOD=90$^\\circ$, $\\angle$DOC+$\\angle$BOD=90$^\\circ$, \\\\\n$\\therefore$ $\\angle$AOB=$\\angle$DOC, \\\\\n$\\because$ $\\angle$ODC=$\\angle$OAC+$\\angle$AOD, $\\angle$OAB=$\\angle$OAC+$\\angle$BAC, $\\angle$AOD=$\\angle$BAC=90$^\\circ$, \\\\\n$\\therefore$ $\\angle$OAB=$\\angle$ODC, \\\\\nIn $\\triangle$ABO and $\\triangle$DCO, $ \\{\\begin{array}{l}\\angle OAB=\\angle ODC\\hfill \\\\ \\angle AOB=\\angle DOC\\hfill \\\\ OB=OC\\hfill \\end{array}$, \\\\\n$\\therefore$ $\\triangle$ABO$\\cong$$\\triangle$DCO, \\\\\n$\\therefore$ OA=OD= $3\\sqrt{2}$, CD=AB=2, \\\\\n$\\therefore$ $\\triangle$OAD is an isosceles right triangle, \\\\\n$\\therefore$ AD= $ \\sqrt{2}OA$ =6, \\\\\n$\\therefore$ AC=AD+CD=\\boxed{8}.",
    "image_path": "dataset/mm_math/images/MM_Math/50496622.png"
  },
  {
    "solution": "\\boxed{\\frac{9}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the figure, within $\\triangle ABC$, points $D$ and $E$ lie on sides $AB$ and $AC$ respectively, with $DE\\parallel BC$. Given that $AB=3$, $AC=4$, and $EC=1$, determine the length of $AD$.\n",
    "completion": "\\textbf{Solution:} \\\\\nSince $DE\\parallel BC$, \\\\\nit follows that $\\frac{DB}{AB}=\\frac{EC}{AC}$, that is $\\frac{DB}{3}=\\frac{1}{4}$, \\\\\ntherefore, $DB=\\frac{3}{4}$, \\\\\nthus, $AD=AB−DB=3−\\frac{3}{4}=\\boxed{\\frac{9}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51446425.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{2}-2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within $\\triangle ABC$, $\\angle ACB=90^\\circ$, $AC=BC$, points $D$ and $E$ are on $AB$ and $BC$ respectively, connecting $CD$ and $DE$. If $BC=BD$, $AC=2$, and $\\angle CDE=45^\\circ$, what is the length of $BE$?\n",
    "completion": "\\textbf{Solution:} Given $\\because \\angle ACB=90^\\circ$ and AC=BC,\\\\\n$\\therefore \\angle A=\\angle B=45^\\circ$,\\\\\n$\\because \\angle BDC=\\angle CDE+\\angle BDE=\\angle A+\\angle ACD$ and $\\angle CDE=45^\\circ$,\\\\\n$\\therefore \\angle BDE=\\angle ACD$,\\\\\n$\\because BC=BD$,\\\\\n$\\therefore BD=AC$,\\\\\nIn $\\triangle BDE$ and $\\triangle ACD$,\\\\\n$\\left\\{\\begin{array}{l}\n\\angle B=\\angle A \\\\\nBD=AC \\\\\n\\angle BDE=\\angle ACD\n\\end{array}\\right.$,\\\\\n$\\therefore \\triangle BDE \\cong \\triangle ACD$ (ASA),\\\\\n$\\therefore BE=AD$,\\\\\n$\\because BC=AC=2$ and $\\angle ACB=90^\\circ$,\\\\\n$\\therefore AB =\\sqrt{AC^2+BC^2}=\\sqrt{2^2+2^2}=2\\sqrt{2}$,\\\\\n$\\because BD=BC=2$,\\\\\n$\\therefore AD=AB-BD=2\\sqrt{2}-2$,\\\\\n$\\therefore BE=\\boxed{2\\sqrt{2}-2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52699755.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, within triangle paper $ABC$, $\\angle C=90^\\circ$, $\\angle A=30^\\circ$, and $AC=6$. When folding the paper so that point $C$ lands on point $D$ on side $AB$, and the fold line $BE$ intersects $AC$ at point $E$, what is the length of the fold line $BE$?\n",
    "completion": "\\textbf{Solution:} Given, in $\\triangle ABC$, $\\angle C=90^\\circ$, $\\angle A=30^\\circ$,\\\\\nhence $\\angle ABC=60^\\circ$,\\\\\nby the property of folding, we get: $\\angle ABE=\\frac{1}{2}\\angle ABC=30^\\circ$, $\\angle BDE=\\angle C=90^\\circ$, $DE=CE$,\\\\\nthus $\\angle ABE=\\angle A$,\\\\\nthus $BE=AE$,\\\\\ngiven, in $Rt\\triangle ADE$, $\\angle ADE=90^\\circ$, $\\angle A=30^\\circ$,\\\\\nthus $DE=\\frac{1}{2}AE$,\\\\\nsince $DE+AE=CE+AE=AC=6$,\\\\\ntherefore $\\frac{1}{2}AE+AE=6$, solving gives $AE=4$,\\\\\nhence $BE=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51212567.png"
  },
  {
    "solution": "\\boxed{\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a straight line intersects the positive semi-axes of two coordinate axes at points $A$ and $B$, respectively. Let $P$ be any point on the line segment $AB$ (excluding the endpoints). Perpendicular lines are drawn from point $P$ to the two coordinate axes, forming a rectangle with a perimeter of 4. What is the minimum value of $OP$?\n",
    "completion": "\\textbf{Solution:} Let the coordinates of point P be $(x,y)$,\\\\\nthen: $2(x+y)=4$,\\\\\nafter rearranging: $y=-x+2$,\\\\\nsince P is any point on the line segment AB,\\\\\ntherefore the equation of the line containing AB is: $y=-x+2$,\\\\\nwhen $x=0$, $y=2$,\\\\\nwhen $y=0$, $x=2$,\\\\\ntherefore, OA=OB=2,\\\\\nthus, in the $\\triangle OAB$, according to the Pythagorean theorem: $AB=\\sqrt{OA^2+OB^2}=\\sqrt{2^2+2^2}=2\\sqrt{2}$,\\\\\ntherefore, when OP $\\perp$ AB, OP is the shortest,\\\\\nat this time, $S_{\\triangle OAB}=\\frac{1}{2}OA\\cdot OB=\\frac{1}{2}AB\\cdot OP$,\\\\\nwhich gives: $4=2\\sqrt{2}OP$,\\\\\nsolving this gives: $OP=\\boxed{\\sqrt{2}}$,\\\\\ntherefore, the minimum value of the line segment $OP$ is: $\\sqrt{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52250453.png"
  },
  {
    "solution": "\\boxed{6\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In $\\triangle ABC$, $AB=6$, $\\angle B$ is an acute angle and $\\cos B=\\frac{1}{2}$, $\\tan C=3\\sqrt{3}$. Find the area of $\\triangle ABC$.\n",
    "completion": "\\textbf{Solution:} Let $S_{\\triangle ABC}=\\frac{1}{2}BC\\cdot AH=\\frac{1}{2}\\times 4\\times 3\\sqrt{3}=\\boxed{6\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53080305_100.png"
  },
  {
    "solution": "\\boxed{110^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, AB is the diameter of $\\odot O$, and D, E are two points on $\\odot O$ located on opposite sides of AB. Extend BD to point C such that CD=BD, connect AC which intersects $\\odot O$ at point F, and connect AE, DE, DF. $\\angle E=\\angle C$; If $\\angle E=55^\\circ$, what is the degree measure of $\\angle BDF$?\n",
    "completion": "\\textbf{Solution:}\n\nSince the quadrilateral AEDF is inscribed in circle O,\n\\[\\therefore \\angle AFD=180^\\circ-\\angle E\\]\nSince \\(\\angle CFD=180^\\circ-\\angle AFD\\),\n\\[\\therefore \\angle CFD=\\angle E=55^\\circ\\]\nFrom (1), we have \\(\\angle E=\\angle C=55^\\circ\\),\n\\[\\therefore \\angle BDF=\\angle C+\\angle CFD=55^\\circ+55^\\circ=\\boxed{110^\\circ}\\]",
    "image_path": "dataset/mm_math/images/MM_Math/53059881_30.png"
  },
  {
    "solution": "\\boxed{\\frac{2\\sqrt{5}}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, if the vertices of $\\triangle ABC$ are lattice points on a square grid, what is the value of $\\cos\\angle C$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect BE in the diagram, \\\\\nby the Pythagorean theorem, we find: $CE=\\sqrt{4^2+2^2}=2\\sqrt{5}$, $BE=\\sqrt{1^2+2^2}=\\sqrt{5}$, $BC=\\sqrt{3^2+4^2}=5$, \\\\\n$\\therefore CE^2+BE^2=(2\\sqrt{5})^2+(\\sqrt{5})^2=25=BC^2$, \\\\\n$\\therefore \\triangle CEB$ is a right triangle, $\\angle CEB=90^\\circ$, \\\\\n$\\therefore \\cos C=\\frac{CE}{CB}=\\frac{2\\sqrt{5}}{5}=\\boxed{\\frac{2\\sqrt{5}}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51298870.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point E is on the diagonal BD of the rhombus ABCD, and DE = DA. Line AE is drawn. If $\\angle CAE = 15^\\circ$, what is the degree measure of $\\angle ABC$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rhombus, \\\\\nit follows that AC $\\perp$ BD, \\\\\nsince $\\angle$CAE = 15$^\\circ$, \\\\\nthus, $\\angle$AED = 90$^\\circ$ - $\\angle$EAC = 90$^\\circ$ - 15$^\\circ$ = 75$^\\circ$, \\\\\nsince DE = DA, \\\\\nit follows that $\\angle$EAD = $\\angle$AED = 75$^\\circ$, \\\\\ntherefore, $\\angle$ADB = $\\angle$ABD = 30$^\\circ$, \\\\\ntherefore, $\\angle$ABC = 2$\\angle$ADB = $2 \\times 30^\\circ$ = \\boxed{60^\\circ},",
    "image_path": "dataset/mm_math/images/MM_Math/52344116.png"
  },
  {
    "solution": "\\boxed{ellipse}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Intersecting a cylinder as shown in the figure with a plane, it is impossible for the cross-section to be what?\n",
    "completion": "\\textbf{Solution:} In this problem, using a plane to cut through a cylinder, a horizontal cut results in a circle, a vertical cut results in a rectangle, and an oblique cut results in an \\boxed{ellipse}. The only impossibility is a trapezoid.",
    "image_path": "dataset/mm_math/images/MM_Math/53124523.png"
  },
  {
    "solution": "\\boxed{(30^\\circ, 2\\sqrt{3})}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, select a fixed point O in the plane and draw a directed ray $Ox$. By choosing a unit length, the position of any point M on the plane can be determined by the degree $\\theta$ of $\\angle MOx$ and the length m of $OM$, forming an ordered pair $\\left(\\theta, m\\right)$, which is referred to as the \"polar coordinates\" of point M. The coordinate system established in this manner is called the \"polar coordinate system\".\n\nApplication: Given a polar coordinate system as in Figure 2, if a regular hexagon has a side length of 2 and one side OA lies on the ray $Ox$, what would be the polar coordinates of the vertex C of the hexagon?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw CB$\\perp$OA at point B,\\\\\n$\\therefore$ $\\angle$OBC=$90^\\circ$,\\\\\n$\\because$ the side length of the regular hexagon is 2,\\\\\n$\\therefore$ OA=CA=2, $\\angle$OAC=$120^\\circ$,\\\\\n$\\therefore$ $\\angle$COA=$\\angle$OCA=$\\angle$ACB=$30^\\circ$,\\\\\n$\\therefore$ AB=$\\frac{1}{2}$AC=1, BC=$\\sqrt{3}$,\\\\\n$\\therefore$ OC=2BC=2$\\sqrt{3}$,\\\\\n$\\therefore$ the polar coordinates of vertex C of the regular hexagon should be $\\boxed{(30^\\circ, 2\\sqrt{3})}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52936145.png"
  },
  {
    "solution": "\\boxed{\\frac{10}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, $\\angle B = 90^\\circ$, $AB=8$, and $BC=6$. Point $M$ is the midpoint of diagonal $AC$, and point $N$ is the midpoint of side $AD$. Connecting $BM$, $MN$, if $BM = 3MN$, what is the length of the line segment $CD$?\n",
    "completion": "\\textbf{Solution:} Since $AC= \\sqrt{AB^{2}+BC^{2}}=\\sqrt{8^{2}+6^{2}}=10$,\\\\\nand in $\\triangle ABC$ with right angle at $C$, $M$ is the midpoint of $AC$,\\\\\nthus $BM= \\frac{1}{2}AC=5$,\\\\\nand $MN= \\frac{1}{3}BM=\\frac{5}{3}$,\\\\\nsince $M$ and $N$ are midpoints of $AC$ and $AD$ respectively,\\\\\nit follows that $CD=2MN=2\\times \\frac{5}{3}=\\boxed{\\frac{10}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50475208.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, fold the equilateral triangle $\\triangle ABC$ such that point $B$ falls exactly on point $D$ on side $AC$. The crease is $EF$, and $O$ is a moving point on crease $EF$. If $AD=1$ and $AC=3$, then what is the minimum perimeter of $\\triangle OCD$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect BD, OB, \\\\\n$\\because$ by folding the equilateral $\\triangle ABC$ so that point B falls exactly on point D on side AC, \\\\\n$\\therefore$ EF is the axis of symmetry of BD, \\\\\n$\\therefore$ OB=OD, \\\\\n$\\because$ AD=1, AC=3, \\\\\n$\\therefore$ CD=2, \\\\\n$\\because$ the perimeter of $\\triangle OCD$ = CD+OD+OC = 2+BO+OC, \\\\\n$\\therefore$ when points B, O, and C are collinear, the minimum perimeter of $\\triangle OCD$ = 2+BC = \\boxed{5},",
    "image_path": "dataset/mm_math/images/MM_Math/52603721.png"
  },
  {
    "solution": "\\boxed{-5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graphs of the hyperbolas $C_1\\colon y=\\frac{k}{x} (k\\neq 0)$ and $C_2\\colon y=-\\frac{1}{x}$ are illustrated. Point A lies on $C_1$. Perpendiculars are drawn from point A to the x-axis and the y-axis, meeting them at points B and C, respectively. The line segment $AB$ intersects $C_2$ at point D. If the area of the triangle $\\triangle AOD$ is 2, what is the value of $k$?\n",
    "completion": "Solution: Since $S_{\\triangle AOD} = S_{\\triangle AOB} - S_{\\triangle DOB}$,\n\\[\n\\therefore \\frac{|k|}{2}-\\frac{1}{2}=2,\n\\]\n\\[\n\\therefore |k|=5,\n\\]\nSince the inverse proportion function is located in the third quadrant,\n\\[\n\\therefore k=\\boxed{-5}.\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/52601721.png"
  },
  {
    "solution": "\\boxed{65^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, $\\angle A=40^\\circ$. The bisector of the exterior angle $\\angle CBD$ of $\\triangle ABC$ intersects the extension of $AC$ at point $E$. What is the measure of $\\angle CBE$ in degrees?\n",
    "completion": "\\textbf{Solution:}\nGiven $\\triangle ABC$ where $\\angle ACB=90^\\circ$ and $\\angle A=40^\\circ$,\\\\\nthus $\\angle ABC=90^\\circ-\\angle A=50^\\circ$.\\\\\nThus, $\\angle CBD=130^\\circ$.\\\\\nSince $BE$ bisects $\\angle CBD$,\\\\\nthus, $\\angle CBE= \\frac{1}{2} \\angle CBD=\\boxed{65^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50710244_209.png"
  },
  {
    "solution": "\\boxed{\\left(-\\frac{1}{3},0\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the parabola $y=x^{2}+bx+c$ intersects the x-axis at points A ($-1$, $0$) and B ($3$, $0$), and intersects the y-axis at point C, with the vertex being point D. If point P is a point on the x-axis, when $\\triangle BDP$ is similar to $\\triangle ABC$, determine the coordinates of point P.\n",
    "completion": "\\textbf{Solution:} We have $\\because \\tan\\angle ACO=\\frac{AO}{OC}=\\frac{1}{3}$,\n$\\therefore \\angle ACO=\\angle CBD$.\n$\\because OC=OB$,\n$\\therefore \\angle OCB=\\angle OBC=45^\\circ$.\n$\\therefore \\angle ACO+\\angle OCB=\\angle CBD+\\angle OBC$.\nThat is: $\\angle ACB=\\angle DBO$.\n$\\therefore$, when $\\triangle BDP$ is similar to $\\triangle ABC$, point P is to the left of point B.\n(i) When $\\frac{AC}{CB}=\\frac{DB}{BP}$,\n$\\therefore \\frac{\\sqrt{10}}{3\\sqrt{2}}=\\frac{2\\sqrt{5}}{BP}$.\n$\\therefore BP=6$.\n$\\therefore P(-3,0)$.\n(ii) When $\\frac{AC}{CB}=\\frac{BP}{DB}$,\n$\\therefore \\frac{\\sqrt{10}}{3\\sqrt{2}}=\\frac{BP}{2\\sqrt{5}}$.\n$\\therefore BP=\\frac{10}{3}$.\n$\\therefore P\\left(-\\frac{1}{3},0\\right)$.\nIn conclusion, the coordinates of point P are $\\boxed{(-3,0)}$ or $\\boxed{\\left(-\\frac{1}{3},0\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51379494_85.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $P$ is a point on the line segment $AB$, and points $M$ and $N$ are the midpoints of line segments $AP$ and $PB$, respectively. If point $P$ is any point on the line segment $AB$, and $AB=12\\, \\text{cm}$, what is the length of the line segment $MN$ in $\\text{cm}$?\n",
    "completion": "\\textbf{Solution:} Since points M and N are the midpoints of the line segments AP and PB, respectively,\\\\\n$\\therefore AP=2MP$, $BP=2PN$.\\\\\n$\\therefore AP+BP=2MP+2PN=2MN$, that is $AB=2MN$.\\\\\n$\\because AB=12\\ \\text{cm}$,\\\\\n$\\therefore MN=\\boxed{6}\\ \\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55431962_67.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{2}-2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within the Cartesian coordinate system, the quadrilateral \\(OABC\\) is a square. Point \\(B\\) has the coordinates \\((2\\sqrt{2}, 0)\\). Point \\(D\\) is a moving point on ray \\(OB\\) that does not coincide with point \\(O\\). Segment \\(CD\\) is rotated \\(90^\\circ\\) clockwise around point \\(D\\) to obtain \\(ED\\), and segments \\(AD\\) and \\(AE\\) are drawn. \\(DA=DE\\); As in figure 2, by connecting \\(AC\\) and \\(BE\\), when triangles \\(CDA\\) and \\(DBE\\) are similar, what is the length of \\(BD\\)?\n",
    "completion": "\\textbf{Solution:} As shown in Figure 2, let AC intersect OB at point J, and draw DK $\\perp$ BC at point K. \\\\\n$\\because$ DA=DC=DE, \\\\\n$\\therefore$ $\\angle$DAC=$\\angle$DCA, $\\angle$DAE=$\\angle$DEA, \\\\\n$\\because$ $\\triangle ADC$ is similar to $\\triangle EBD$, \\\\\n$\\therefore$ $\\angle$ACD=$\\angle$DEB, \\\\\n$\\therefore$ $\\angle$CAD=$\\angle$DAE, \\\\\n$\\because$ $\\angle$CAB=$\\angle$ACB=$45^\\circ$, \\\\\n$\\therefore$ $\\angle$CAD=$\\angle$DAB=$\\angle$ACD=$\\angle$BCD=$22.5^\\circ$, \\\\\n$\\because$ DJ$\\perp$CJ, DK$\\perp$CB, \\\\\n$\\therefore$ DJ=DK, \\\\\n$\\because$ B$(2\\sqrt{2},0)$, \\\\\n$\\therefore$ OJ=JB=$\\sqrt{2}$, \\\\\n$\\therefore$ DJ+$\\sqrt{2}$DJ=$\\sqrt{2}$, \\\\\n$\\therefore$ DJ=$2-\\sqrt{2}$, \\\\\n$\\therefore$ OD=2, \\\\\n$\\therefore$ BD=$2\\sqrt{2}-2$; \\\\\nTherefore, the length of BD is $BD=\\boxed{2\\sqrt{2}-2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52861918_131.png"
  },
  {
    "solution": "\\boxed{26^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within circle $\\odot O$, $OC \\perp AB$, and $\\angle ADC=32^\\circ$, what is the degree measure of $\\angle OBA$?\n",
    "completion": "\\textbf{Solution:} Since in circle $O$, $OC\\perp AB$,\\\\\nit follows that $\\overset{\\frown}{AC}=\\overset{\\frown}{BC}$, and $\\angle BOC+\\angle OBA=90^\\circ$,\\\\\nthus $\\angle BOC=2\\angle ADC=64^\\circ$,\\\\\nhence $\\angle OBA=90^\\circ-\\angle BOC=90^\\circ-64^\\circ=\\boxed{26^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51402725.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$ and $\\triangle ADE$, $\\triangle ABD \\sim \\triangle ACE$; $\\angle BAC = \\angle DAE = 90^\\circ$, $\\angle ABC = \\angle ADE = 30^\\circ$, $AC$ and $DE$ intersect at point F. Point D is on side $BC$, $\\frac{AD}{BD} = \\sqrt{3}$. Find the value of $\\frac{DF}{CF}$.\n",
    "completion": "\\textbf{Solution:} As shown in Figure 1, draw line EC,\\\\\n$\\because \\angle BAC = \\angle DAE = 90^\\circ$, $\\angle ABC = \\angle ADE = 30^\\circ$,\\\\\n$\\therefore \\triangle ABC \\sim \\triangle ADE$,\\\\\nFrom (1), we know $\\triangle ABD \\sim \\triangle ACE$,\\\\\n$\\therefore \\frac{AE}{CE} = \\frac{AD}{BD} = \\sqrt{3}$, $\\angle ACE = \\angle ABD = \\angle ADE$,\\\\\nIn $\\triangle ADE$, $\\angle ADE = 30^\\circ$,\\\\\n$\\therefore \\frac{AD}{AE} = \\sqrt{3}$,\\\\\n$\\therefore \\frac{AD}{EC} = \\frac{AD}{AE} \\times \\frac{AE}{CE} = \\sqrt{3} \\times \\sqrt{3} = 3$.\\\\\n$\\because \\angle ADF = \\angle ECF$, $\\angle AFD = \\angle EFC$,\\\\\n$\\therefore \\triangle ADF \\sim \\triangle ECF$,\\\\\n$\\therefore \\frac{DF}{CF} = \\frac{AD}{CE} = \\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51403000_134.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle APB$, $C$ and $D$ are respectively points on $AP$ and $BP$. Moreover, $\\angle CDP = \\angle A$, $AB = 8\\text{cm}$, $\\frac{BP}{CP} = \\frac{4}{3}$. Find the length of $CD$.\n",
    "completion": "\\textbf{Solution:} Since $\\left\\{\\begin{array}{l}\n\\angle CDP=\\angle A \\\\\n\\angle P=\\angle P\n\\end{array}\\right.$,\\\\\nit follows that $\\triangle PCD\\sim \\triangle PBA$,\\\\\nhence $\\frac{BP}{CP}=\\frac{AB}{CD}$,\\\\\ngiven that $AB=8\\text{cm}$ and $\\frac{BP}{CP}=\\frac{4}{3}$,\\\\\nwe have $\\frac{4}{3}=\\frac{8}{CD}$,\\\\\ntherefore, $CD=\\boxed{6}\\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53080403.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{1}{2}x+1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the Cartesian coordinate system, the graph of the linear function $y=ax+b$ ($a\\neq 0$) intersects with the graph of the inverse proportion function $y=\\frac{k}{x}$ ($k\\neq 0$) at points $A$ and $B$ within the second and fourth quadrants, respectively, and with the $y$-axis at point $C$. Drawing a line perpendicular to the $y$-axis from point $A$ to point $H$ such that $AH \\perp y$-axis and the foot of the perpendicular is $H$, with $OH=3$ and $\\tan\\angle AOH =\\frac{4}{3}$. The coordinates of point $B$ are ($m$, $-2$). Determine the analytical expressions of the inverse proportion function and the linear function.\n",
    "completion": "\\textbf{Solution:} Let $\\because$OH$=3$, $AH=4$, and point A is in the second quadrant\\\\\n$\\therefore$ $A\\left(-4, 3\\right)$ \\\\\n$\\because$ the graph of the inverse proportion function $y=\\frac{k}{x}$ (where $k\\neq 0$) passes through point A\\\\\n$\\therefore$ $3=\\frac{k}{-4}$\\\\\n$\\therefore$ $k=-12$ \\\\\n$\\therefore$ the equation of the inverse proportion function is: $y=-\\frac{12}{x}$;\\\\\n$\\because$ the graph of the inverse proportion function passes through point B, and the coordinates of point B are $(6, -2)$\\\\\n$\\therefore$ $-2=-\\frac{12}{6}$\\\\\n$\\therefore$ $m=6$ \\\\\nUpon verification, $m=6$ is a solution to $-2=-\\frac{12}{m}$;\\\\\n$\\therefore$ $B\\left(6, -2\\right)$ \\\\\n$\\because$ the graph of the linear function $y=ax+b$ (where $a\\neq 0$) passes through points A and B\\\\\n$\\therefore$ $\\left\\{\\begin{array}{l}-4a+b=3 \\\\ 6a+b=-2\\end{array}\\right.$ \\\\\n$\\therefore$ $\\left\\{\\begin{array}{l}a=-\\frac{1}{2} \\\\ b=1\\end{array}\\right.$ \\\\\n$\\therefore$ the equation of the linear function is: $\\boxed{y=-\\frac{1}{2}x+1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51367530_123.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, it is known that point $B$ is on line segment $AC$, and point $D$ is on line segment $AB$. If $AB=10\\, \\text{cm}$, $BC=6\\, \\text{cm}$, and $D$ is the midpoint of line segment $AC$, what is the length of line segment $DB$ in $\\text{cm}$?\n",
    "completion": "\\textbf{Solution:} As shown in Figure 1 in the problem, \\\\\nsince AB=10cm, and BC=6cm, \\\\\ntherefore AC=AB+BC=10+6=16cm. \\\\\nAlso, since D is the midpoint of segment AC, \\\\\ntherefore DC=$\\frac{1}{2}AC=\\frac{1}{2}\\times 16=8$cm, \\\\\ntherefore DB=DC−BC=8−6=$\\boxed{2}$cm.",
    "image_path": "dataset/mm_math/images/MM_Math/54927244_8.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{2}{x}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the graph of the linear function $y = -2x$ intersects with the graph of the inverse proportion function $y = \\frac{k}{x}$ at a point $A(-1, n)$. What is the equation of the inverse proportion function $y = \\frac{k}{x}$?\n",
    "completion": "\\textbf{Solution:} Since point A $(-1, 2)$ lies on the graph of the linear function $y=-2x$,\\\\\n$\\therefore n=-2\\times(-1)=2$\\\\\n$\\therefore$ The coordinates of point A are $(-1, 2)$\\\\\nSince point A lies on the graph of the inverse proportion function,\\\\\n$\\therefore k=-2$\\\\\n$\\therefore$ The equation of the inverse proportion function is $\\boxed{y=-\\frac{2}{x}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51402287_57.png"
  },
  {
    "solution": "\\boxed{\\frac{5\\sqrt{13}}{6}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $BD$ is the diagonal. The rectangle $ABCD$ is folded along the lines containing $BE$ and $BF$, making point $A$ coincide with point $M$ on $BD$, and point $C$ with point $N$ on $BD$, then $EF$ is drawn. Given that $AB=3$ and $BC=4$, what is the length of $EF$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rectangle, \\\\\n$\\therefore$AB=CD=3, AD=BC=4, $\\angle$A=$\\angle$C=$\\angle$EDF=90$^\\circ$,\\\\\n$\\therefore$BD=$\\sqrt{AB^2+AD^2}=\\sqrt{3^2+4^2}=5$,\\\\\nSince rectangle ABCD is folded along the straight line containing BE, making point A fall on point M on BD,\\\\\n$\\therefore$AE=EM, $\\angle$A=$\\angle$BME=90$^\\circ$,\\\\\n$\\therefore$$\\angle$EMD=90$^\\circ$,\\\\\nSince $\\angle$EDM=$\\angle$ADB,\\\\\n$\\therefore$$\\triangle EDM\\sim \\triangle BDA$,\\\\\n$\\therefore$$\\frac{ED}{BD}=\\frac{EM}{AB}$,\\\\\nLet DE=x, then AE=EM=4−x,\\\\\n$\\therefore$$\\frac{x}{5}=\\frac{4−x}{3}$,\\\\\nSolving for x, we get x=$\\frac{5}{2}$,\\\\\n$\\therefore$DE=$\\frac{5}{2}$,\\\\\nSimilarly, $\\triangle DNF\\sim \\triangle DCB$,\\\\\n$\\therefore$$\\frac{DF}{BD}=\\frac{NF}{BC}$,\\\\\nLet DF=y, then CF=NF=3−y,\\\\\n$\\therefore$$\\frac{y}{5}=\\frac{3−y}{4}$,\\\\\nSolving for y, we get y=$\\frac{5}{3}$,\\\\\n$\\therefore$DF=$\\frac{5}{3}$,\\\\\n$\\therefore$EF=$\\sqrt{DE^2+DF^2}=\\sqrt{(\\frac{5}{2})^2+(\\frac{5}{3})^2}=\\boxed{\\frac{5\\sqrt{13}}{6}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51212855.png"
  },
  {
    "solution": "\\boxed{y = -\\frac{5}{3}x + \\frac{34}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In parallelogram $ABCD$, diagonal $AC$ is perpendicular to side $CD$, $\\frac{AB}{AC}=\\frac{3}{4}$, and the perimeter of quadrilateral $ABCD$ is $16$. Point $E$ is on the extension of line $AD$, and point $F$ is on the ray $AB$, $\\angle CED=\\angle CDF$. As shown in Figure 2, point $F$ is a point on side $AB$. Let $AE=x$, $BF=y$, find the functional relationship between $y$ and $x$?\n",
    "completion": "\\textbf{Solution:} From the given information, \\\\\n$\\because CD \\parallel AB$, \\\\\n$\\therefore \\angle EDC = \\angle FAD$, $\\angle CDF = \\angle AFD$, \\\\\n$\\because \\angle CED = \\angle CDF$, \\\\\n$\\therefore \\angle CED = \\angle AFD$, \\\\\n$\\therefore \\triangle CDE \\sim \\triangle DAF$, \\\\\n$\\therefore \\frac{DE}{AF} = \\frac{DC}{AD}$, \\\\\n$\\because AE = x$, $BF = y$, $AD = BC = 5$, \\\\\n$\\therefore DE = x - 5$, $DC = AB = 3$, $AF = 3 - y$, \\\\\n$\\therefore \\frac{x - 5}{3 - y} = \\frac{3}{5}$, \\\\\n$\\therefore y = -\\frac{5}{3}x + \\frac{34}{3}$, \\\\\nThe domain is: $5 < x \\le \\frac{34}{5}$. \\\\\n$\\therefore \\boxed{y = -\\frac{5}{3}x + \\frac{34}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51379461_94.png"
  },
  {
    "solution": "\\boxed{1+\\sqrt{10}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system, the graph of the inverse proportion function $y=\\frac{k}{x}$ ($x>0$) intersects with side $AB$ of rectangle $OABC$ at point $M(1,3)$, and intersects with side $BC$ at point $N$. If the symmetric point $B'$ of point $B$ with respect to line $MN$ lies exactly on the $x$-axis, what is the length of $OC$?\n",
    "completion": "Solution:It is known from the problem that the point $M(1,3)$ lies on the graph of the inverse proportion function $y=\\frac{3}{x}(x>0)$,\n\n$\\therefore$ $3=\\frac{k}{1}$, solving for $k$, we get $k=3$,\n\n$\\therefore$ the equation of this inverse proportion function is $y=\\frac{3}{x}(x>0)$.\n\nLet $N(a,\\frac{3}{a})$, then $B(a,3)$, $C(a,0)$,\n\n$\\therefore$ $CN=\\frac{3}{a}$\n\n$\\therefore$ $BN=BC-CN=3-\\frac{3}{a}$, $BM=AB-AM=a-1$.\n\nAs shown in the figure, connect $MB',NB'$, and draw $MD\\perp x$ axis at point D through point M.\n\n$\\therefore$ $DM=3$, $OD=1$.\n\nSince the symmetric point $B'$ of point B about line $MN$ exactly lies on the x-axis,\n\n$\\therefore$ $B'N=BN=3-\\frac{3}{a}$, $B'M=BM=a-1$.\n\nSince $\\angle DMB'+\\angle DB'M=\\angle CB'N+\\angle DB'M=90^\\circ$,\n\n$\\therefore$ $\\angle DMB'=\\angle CB'N$.\n\nAlso, since $\\angle MDB'=\\angle B'CN=90^\\circ$,\n\n$\\therefore$ $\\triangle DMB'\\sim \\triangle CB'N$,\n\n$\\therefore$ $\\frac{B'M}{B'N}=\\frac{DM}{B'C}=\\frac{B'D}{CN}$, that is, $\\frac{a-1}{3-\\frac{3}{a}}=\\frac{3}{B'C}=\\frac{B'D}{\\frac{3}{a}}$,\n\nSolving this, we find $B'C=\\frac{9}{a}$, $B'D=1$.\n\nSince $B'M^2=MD^2+B'D^2$, i.e., $(a-1)^2=3^2+1^2$,\n\n$\\therefore$ $a_1=1+\\sqrt{10}$, $a_2=1-\\sqrt{10}$ (discard),\n\n$\\therefore$ $OC=1+\\sqrt{10}$.\n\nTherefore, the answer is: $OC=\\boxed{1+\\sqrt{10}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53080525.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{2}}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, circle $ \\odot A$ with a radius of 3 passes through the origin $O$ and point $C(0, 2)$. Point $B$ is a point on the major arc of $\\odot A$ on the left side of the y-axis. What is the value of $\\tan\\angle OBC$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, connect CD,\\\\\n$\\because$ $\\angle$DOC=$90^\\circ$,\\\\\n$\\therefore$ CD is the diameter of circle A,\\\\\n$\\because$ the circle $A$ with radius $3$ passes through the origin $O$ and the point $C(0,2)$,\\\\\n$\\therefore$ CD=6, OC=2,\\\\\n$\\therefore$ DO=$\\sqrt{CD^{2}-OC^{2}}=\\sqrt{6^{2}-2^{2}}=4\\sqrt{2}$,\\\\\n$\\therefore$ $\\tan\\angle$ODC=$\\frac{OC}{OD}=\\frac{2}{4\\sqrt{2}}=\\frac{\\sqrt{2}}{4}$,\\\\\n$\\because$ $\\angle$ODC=$\\angle$OBC,\\\\\n$\\therefore$ $\\tan\\angle$OBC=$\\boxed{\\frac{\\sqrt{2}}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52601801.png"
  },
  {
    "solution": "\\boxed{35}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC=20\\,cm$, $DE$ is perpendicular bisector of $AB$ with the foot of the perpendicular as $E$, and intersects $AC$ at $D$. If $BC=15\\,cm$, then what is the perimeter of $\\triangle DBC$?\n",
    "completion": "\\textbf{Solution:} Since $DE$ bisects $AB$ perpendicularly,\\\\\nit follows that $AD=BD$,\\\\\nhence the perimeter of $\\triangle DBC = BD+DC+BC = AD+DC+BC = AC+BC$,\\\\\nsince $AB=AC=20\\,cm$, $BC=15\\,cm$,\\\\\ntherefore, the perimeter of $\\triangle DBC =AC+BC=20+15=\\boxed{35}\\,cm$.",
    "image_path": "dataset/mm_math/images/MM_Math/52828221.png"
  },
  {
    "solution": "\\boxed{x>4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: If the graph of the linear function \\(y=kx+b(k\\ne 0)\\) is shown as in the figure, what is the solution set of the inequality \\(kx+b<0\\)?\n",
    "completion": "\\textbf{Solution}: Given $y=kx+b\\ (k \\neq 0)$ intersects the $x$-axis at $(4,0)$,\\\\\n$\\therefore kx+b<0$ has the solution set: $x>4$,\\\\\nthus, the answer is: $\\boxed{x>4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53182636.png"
  },
  {
    "solution": "\\boxed{165^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Place a right-angle triangle set as shown in the diagram, with the short right-angle side of the triangle containing the $30^\\circ$ angle coinciding with one of the right-angle sides of the triangle containing the $45^\\circ$ angle. What is the measure of $\\angle \\alpha$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, $ \\angle ABC=45^\\circ$, $ \\angle D=30^\\circ$, \\\\\n$ \\because \\angle ABC+\\angle DBC=180^\\circ$, \\\\\n$ \\therefore \\angle DBC=135^\\circ$, \\\\\n$ \\because \\angle DBC+\\angle D+\\angle BFD=180^\\circ$, \\\\\n$ \\therefore \\angle BFD=15^\\circ$, \\\\\n$ \\therefore \\angle \\alpha =180^\\circ−15^\\circ=\\boxed{165^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52458505.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, point O is a point on side BC, a circle with center O and radius OB intersects side AB at point D, connect DC, and $DC=AC$. $DC$ is the tangent to $\\odot O$; If the radius of $\\odot O$ is 3, and $CD=4$, find the length of BC.\n",
    "completion": "Solution: In $\\triangle \\mathrm{ODC}$, we have $OD=3$ and $CD=4$, \\\\\n$\\therefore OC=\\sqrt{OD^2+CD^2}=5$ \\\\\n$\\therefore BC=OB+OC=3+5=\\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52602982_154.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the direct proportion function $y=kx$ ($k \\neq 0$) intersects with the graph of the linear function $y=ax+b$ ($a \\neq 0$) at point A(1, 1). What is the solution set of the inequality $kx \\ge ax+b$?\n",
    "completion": "\\textbf{Solution:} When $x\\ge 1$, the graph of $y=kx$ is above the graph of $y=ax+b$,\\\\\n$\\therefore kx\\ge ax+b$,\\\\\nwhich means the solution set for $kx\\ge ax+b$ is: $x\\ge \\boxed{1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50481639.png"
  },
  {
    "solution": "\\boxed{35^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points A, B, C, and D are on the circle $\\odot O$, with $\\angle AOC=140^\\circ$. Point B is the midpoint of the arc $\\overset{\\frown}{AC}$. What is the degree measure of $\\angle D$?\n",
    "completion": "\\textbf{Solution:} Draw line OB as shown in the figure,\\\\\nsince point B is the midpoint of $\\widehat{AC}$, and $\\angle$AOC=$140^\\circ$,\\\\\ntherefore, $\\angle$AOB=$\\frac{1}{2}\\angle$AOC=$70^\\circ$,\\\\\nby the Inscribed Angle Theorem, we have $\\angle$D=$\\frac{1}{2}\\angle$AOB=$\\boxed{35^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52623030.png"
  },
  {
    "solution": "\\boxed{120^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point O is a point on the line AB, $\\angle COD$ is a right angle, and $OE$ bisects $\\angle BOC$. As shown in figure (2), if $\\angle COE = \\angle DOB$, what is the degree measure of $\\angle AOC$?\n",
    "completion": "\\textbf{Solution:} \\\\\n$\\because$ OE bisects $\\angle BOC$, \\\\\n$\\therefore \\angle COE = \\angle BOE$, \\\\\n$\\because \\angle COE = \\angle BOD$, \\\\\n$\\therefore \\angle COE = \\angle BOE = \\angle DOB$, \\\\\n$\\because \\angle COD = 90^\\circ$, \\\\\n$\\therefore \\angle COE = \\angle BOE = \\frac{1}{3} \\times 90^\\circ = 30^\\circ$, \\\\\n$\\therefore \\angle AOC = 180^\\circ - 30^\\circ - 30^\\circ = \\boxed{120^\\circ}.$",
    "image_path": "dataset/mm_math/images/MM_Math/51438807_183.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the diagonals AC and BD of the rhombus ABCD intersect at point O, and point E is the midpoint of side AB. If $OE = 6$, what is the length of BC?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rhombus, \\\\\nit follows that $OA=OC$, \\\\\nsince point E is the midpoint of side AB, \\\\\nit follows that $OE$ is the median of $\\triangle ABC$, \\\\\nhence $BC=2OE=2\\times 6=\\boxed{12}$",
    "image_path": "dataset/mm_math/images/MM_Math/52355086.png"
  },
  {
    "solution": "\\boxed{(2\\sqrt{2}-2,0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the plane Cartesian coordinate system, the quadrilateral OABC is a square. The coordinates of point B are $(2\\sqrt{2}, 0)$. Point D is a moving point on ray OB that does not coincide with point O. The line segment CD is rotated clockwise by $90^\\circ$ around point D to obtain ED. Connect AD and AE. When the symmetric point $A'$ of point A with respect to line ED falls on the edge of the square, find the coordinates of point D.\n",
    "completion": "\\textbf{Solution:} As shown in Figure 3-1, when point $A^{\\prime}$ falls on BC,\\\\\ndue to symmetry, it can be known that $\\angle CDE=\\angle ADA^{\\prime}=90^\\circ$, $\\angle EDA=\\angle EDA^{\\prime}=45^\\circ$,\\\\\n$\\therefore \\angle ADC=135^\\circ$,\\\\\n$\\therefore \\angle BDC=\\angle BDA=67.5^\\circ$,\\\\\n$\\because \\angle DBC=45^\\circ$,\\\\\n$\\therefore \\angle BCD=\\angle BDC=67.5^\\circ$,\\\\\n$\\therefore BD=BC=2$,\\\\\n$\\therefore OD=2\\sqrt{2}-2$,\\\\\n$\\therefore D(2\\sqrt{2}-2,0)$;\\\\\nWhen point D is on AC, $A^{\\prime}$ coincides with point C, satisfying the condition, at this time $D(\\sqrt{2},0)$;\\\\\nAs shown in Figure 3-2, when point $A^{\\prime}$ falls on AC, by the same method, it can be proved that $OD=OC=2$, so we get $D(2,0)$.\\\\\nWhen point D coincides with B, A coincides with $A^{\\prime}$, satisfying the condition, at this time $D(2\\sqrt{2},0)$.\\\\\nIn summary, the coordinates of point D that meet the condition are $\\boxed{(2\\sqrt{2}-2,0)}$ or $\\boxed{(\\sqrt{2},0)}$ or $\\boxed{(2,0)}$ or $\\boxed{(2\\sqrt{2},0)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52861918_132.png"
  },
  {
    "solution": "\\boxed{54^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points $A$, $B$, and $C$ are all on the circle $\\odot O$. If $\\angle ACB = 36^\\circ$, what is the measure of $\\angle OAB$?\n",
    "completion": "\\textbf{Solution:} Given that $\\because \\angle ACB=\\frac{1}{2}\\angle AOB$ and $\\angle ACB=36^\\circ$, \\\\\n$\\therefore \\angle AOB=2\\times \\angle ACB=72^\\circ$. \\\\\n$\\because OA=OB$, \\\\\n$\\therefore \\triangle OAB$ is an isosceles triangle, \\\\\n$\\because \\angle AOB+\\angle OAB+\\angle OBA=180^\\circ$, \\\\\n$\\therefore \\angle OAB=\\frac{1}{2}\\left(180^\\circ−\\angle AOB\\right)=54^\\circ$, \\\\\nThus, the answer is: $\\boxed{54^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52457089.png"
  },
  {
    "solution": "\\boxed{90^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated in the diagram, Island C is located at $50^\\circ$ east of north from Island A, and Island C is at $40^\\circ$ west of north from Island B. What is the angle $\\angle ACB$ between Islands A and B as seen from Island C?\n",
    "completion": "\\textbf{Solution:} Draw line CD through point C such that CD$\\parallel$AE, \\\\\nsince AE$\\parallel$BF, \\\\\ntherefore CD$\\parallel$AE$\\parallel$BF, \\\\\ntherefore $\\angle$ACD＝$\\angle$EAC＝50°, $\\angle$BCD＝$\\angle$CBF＝40°, \\\\\ntherefore $\\angle$ACB＝$\\angle$ACD＋$\\angle$BCD＝$50^\\circ+40^\\circ$＝\\boxed{90^\\circ}.",
    "image_path": "dataset/mm_math/images/MM_Math/42461641.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{2}{3}x^{2}+\\frac{4}{3}x+2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the line $y=-\\frac{2}{3}x+2$ intersects the x-axis at point A and the y-axis at point B. The parabola $y=-\\frac{2}{3}x^2+bx+c$ passes through points A and B. What is the expression of this parabola?\n",
    "completion": "\\textbf{Solution:} From $y=-\\frac{2}{3}x+2$, we obtain:\\\\\nWhen $x=0$, $y=2$; when $y=0$, $x=3$,\\\\\n$\\therefore \\boxed{A(3,0)}, \\boxed{B(0,2)}$,\\\\\nSubstituting the coordinates of $A$, $B$ into $y=-\\frac{2}{3}x^{2}+bx+c$ yields:\\\\\n$\\left\\{\\begin{array}{l}\n-\\frac{2}{3}\\times 9+3b+c=0 \\\\\nc=2\n\\end{array}\\right.$,\\\\\nSolving this gives: $\\left\\{\\begin{array}{l}\nb=\\frac{4}{3} \\\\\nc=2\n\\end{array}\\right.$,\\\\\n$\\therefore$ The equation of the parabola is: \\boxed{y=-\\frac{2}{3}x^{2}+\\frac{4}{3}x+2}.",
    "image_path": "dataset/mm_math/images/MM_Math/51379530_200.png"
  },
  {
    "solution": "\\boxed{y=x-1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the rectangle $ABCD$ has its side $BC$ on the $x$-axis, point $E$ is the midpoint of diagonal $BD$, and the graph of the function $y=\\frac{2}{x}$ passes through points $A$ and $E$. If $\\angle ABD=45^\\circ$, what is the function expression corresponding to line $BD$?\n",
    "completion": "\\textbf{Solution:} Solve: When $ \\angle ABD=45^\\circ$, then $AB=AD$, \\\\\n$\\therefore$ the rectangle $ABCD$ is a square, \\\\\nLet the coordinates of point $A$ be $(a, \\frac{2}{a})$, then $B(a, 0)$, $D\\left(a+\\frac{2}{a}, \\frac{2}{a}\\right)$, \\\\\n$\\because$ $E$ is the midpoint of the diagonal $BD$, \\\\\n$\\therefore$ point $E\\left(a+\\frac{1}{a}, \\frac{1}{a}\\right)$, \\\\\n$\\because$ the graph of the function $y=\\frac{2}{x}$ passes through points $A$ and $E$, \\\\\n$\\therefore \\left(a+\\frac{1}{a}\\right)\\cdot \\frac{1}{a}=2$,  \\\\\nSolving, we get $a=1$ (negatives are discarded), \\\\\n$\\therefore B(1, 0)$, $D(3, 2)$, \\\\\nLet the equation of line $BD$ be $y=kx+b$, \\\\\n$\\therefore$ $\\begin{cases}k+b=0\\\\ 3k+b=2\\end{cases}$, solving we get $\\begin{cases}k=1\\\\ b=-1\\end{cases}$, \\\\\n$\\therefore$ the expression for line $BD$ is $y=x-1$, \\\\\nHence, the answer is $\\boxed{y=x-1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50594939.png"
  },
  {
    "solution": "\\boxed{\\frac{2}{1}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point E is the midpoint on side CD of square $ABCD$. The diagonals intersect at point O. If BE intersects AC at point F, what is the ratio of $CF:OF$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral $ABCD$ is a square,\\\\\nit follows that $AB=CD$ and $AB\\parallel CD$,\\\\\nsince $E$ is the midpoint of $CD$,\\\\\nit follows that $CE=\\frac{1}{2}CD=\\frac{1}{2}AB$,\\\\\nsince the diagonals of square $ABCD$ intersect at $O$,\\\\\nit follows that $AO=CO=OF+CF$,\\\\\nsince $AB\\parallel CD$,\\\\\nit follows that $\\triangle ABF\\sim \\triangle CEF$,\\\\\nhence, $\\frac{CE}{AB}=\\frac{CF}{AF}=\\frac{1}{2}$,\\\\\nsince $AF=AO+OF$ and $AO=OF+CF$,\\\\\nit follows that $\\frac{CF}{AF}=\\frac{CF}{AO+OF}=\\frac{CF}{CF+2OF}=\\frac{1}{2}$,\\\\\nthus, $2CF=CF+2OF$,\\\\\ntherefore, $CF=2OF$,\\\\\nhence, $\\frac{CF}{OF}=\\boxed{\\frac{2}{1}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52866265.png"
  },
  {
    "solution": "\\boxed{20\\sin70^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $A_1B_1$ is the orthogonal projection of the line segment $AB$ on the projection plane $P$, $AB=20\\,\\text{cm}$, $\\angle ABB_1=70^\\circ$. What is the length of the projection $A_1B_1$?\n",
    "completion": "\\textbf{Solution:} Construct $AC \\perp BB_1$ at point C, \\\\\n$\\therefore$ Quadrilateral $AA_1B_1C$ is a rectangle, \\\\\n$\\therefore AC=A_1B_1$, \\\\\nIn $\\triangle ABC$, $AC=AB\\cdot \\sin\\angle ABB_1=20\\sin70^\\circ$, \\\\\n$\\therefore A_1B_1=20\\sin70^\\circ$, \\\\\nTherefore, the answer is: $A_1B_1=\\boxed{20\\sin70^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52796953.png"
  },
  {
    "solution": "\\boxed{8\\text{cm}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, AB is the diameter of $\\odot O$, and CD is a chord with AB $\\perp$ CD at point E. Connect AC, OC, and BC. If AE = 2cm and BE = 8cm, what is the length of chord CD in cm?\n",
    "completion": "\\textbf{Solution:} Given $AE=2\\text{cm}$ and $BE=8\\text{cm}$,\\\\\nit follows that $OA=5\\text{cm}$ and $OE=3\\text{cm}$,\\\\\nIn $\\triangle OCE$, $CE= \\sqrt{5^2-3^2}=4\\text{cm}$,\\\\\nSince $AB\\perp CD$,\\\\\nit implies $CE=DE$,\\\\\nTherefore, $CD=2CE=\\boxed{8\\text{cm}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52874751_92.png"
  },
  {
    "solution": "\\boxed{50^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, line $AB$ intersects line $CD$ at point $O$, $OE \\perp CD$ and $OE$ bisects $\\angle BOF$. If $\\angle BOD$ is $10^\\circ$ larger than $\\angle BOE$, what is the degree measure of $\\angle COF$?\n",
    "completion": "\\textbf{Solution:}\nSince $OE \\perp CD$,\\\\\nit follows that $\\angle DOE = 90^\\circ$,\\\\\nwhich means $\\angle DOB + \\angle EOB = 90^\\circ$,\\\\\nSince $\\angle DOB = \\angle EOB + 10^\\circ$,\\\\\nit follows that $\\angle EOB = 40^\\circ$,\\\\\nSince $OE$ bisects $\\angle BOF$,\\\\\nit follows that $\\angle EOF = \\angle BOE = 40^\\circ$,\\\\\nTherefore, $\\angle COF = 90^\\circ - \\angle EOF = \\boxed{50^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52792852_148.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{7}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, line AB $ \\parallel $ CD $ \\parallel $ EF. If AC = 3 and CE = 4, what is the value of $\\frac{BD}{BF}$?\n",
    "completion": "\\textbf{Solution:} Given that $\\because$ AB $\\parallel$ CD $\\parallel$ EF\\\\\n$\\therefore$ $\\frac{BD}{BF}=\\frac{AC}{AE}$\\\\\n$\\because$ AC=3, CE=4\\\\\n$\\therefore$ $\\frac{BD}{BF}=\\boxed{\\frac{3}{7}}$",
    "image_path": "dataset/mm_math/images/MM_Math/52536480.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, AB is the diameter of $\\odot$O. If AC=4 and $\\angle D=60^\\circ$, what is the length of BC?\n",
    "completion": "\\textbf{Solution:} Since AB is the diameter of the circle $O$,\\\\\nit follows that $\\angle ACB=90^\\circ$,\\\\\nsince $\\angle A=\\angle D=60^\\circ$,\\\\\nthus, $\\angle ABC=90^\\circ-\\angle A=30^\\circ$,\\\\\nsince AC=4,\\\\\nit follows that AB=2AC=8.\\\\\nTherefore, $BC= \\sqrt{AB^{2}-AC^{2}}=\\sqrt{8^{2}-4^{2}}=\\boxed{4\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52622671.png"
  },
  {
    "solution": "\\boxed{\\frac{4}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, $AC=3$, $BC=4$, $BE$ bisects $\\angle ABC$, $CD\\perp AB$ at $D$, and $BE$ intersects $CD$ at $F$. What is the length of $CF$?\n",
    "completion": "\\textbf{Solution:} Construct line EG such that EG $\\perp$ AB at point G. Since CD $\\perp$ AB, it follows that EG $\\parallel$ CD, thus $\\angle$BEG = $\\angle$CFE. Since BE bisects $\\angle$ABC and EG $\\perp$ AB, with $\\angle$ACB = 90$^\\circ$ implying AC $\\perp$ BC, it follows that EG = CE. In right triangles $\\triangle$BCE and $\\triangle$BGE, we have \n\\[\\left\\{\\begin{array}{l}\nBE = BE\\\\ \nCE = EG\n\\end{array}\\right.\\]\nTherefore, right $\\triangle$BCE $\\cong$ right $\\triangle$BGE (HL), leading to BC = BG = 4, and $\\angle$BEG = $\\angle$BEC = $\\angle$CFE, thus CE = CF = EG. In right $\\triangle$ABC, we have \n\\[AB = \\sqrt{AC^{2} + BC^{2}} = \\sqrt{3^{2} + 4^{2}} = 5,\\]\ntherefore AG = AB - BG = 5 - 1 = 4. Let CF = x, then AE = 3 - x. In right $\\triangle$AEG, we have \n\\[AG^{2} + EG^{2} = AE^{2}\\]\nwhich means \n\\[1^{2} + x^{2} = (3 - x)^{2}.\\]\nSolving this equation, we find \n\\[x = \\frac{4}{3},\\]\ntherefore CF = \\boxed{\\frac{4}{3}}.",
    "image_path": "dataset/mm_math/images/MM_Math/53059799.png"
  },
  {
    "solution": "\\boxed{-2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y_1=kx+2$ intersects with the graph of the inverse proportion function $y_2=-\\frac{8}{x}$ at points $A(a, -2a)$ and $B(4, -2)$. What are the values of $a$ and $k$?\n",
    "completion": "\\textbf{Solution:} Since the graph of the linear function $y_{1}=kx+2$ intersects with the graph of the reciprocal function $y_{2}=-\\frac{8}{x}$ at point A$(a, -2a)$ and B$(4, -2)$,\\\\\nit follows that $-2=4k+2$,\\\\\nwhich implies $k=-1$,\\\\\nthus $y_{1}=-x+2$,\\\\\nand therefore $-2a=-a+2$,\\\\\nleading to $a=\\boxed{-2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52050093_180.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of $\\odot O$, the chord $CD \\perp AB$ at point $E$, and $AD$ is connected. If $AB=10$ and $CD=8$, what is the length of $AD$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, connect OD.\n\n\\[\n\\because AB \\perp CD,\n\\]\n\\[\n\\therefore CE = ED = 4,\n\\]\n\\[\n\\because \\angle OED = 90^\\circ, \\text{ and } OD = 5,\n\\]\n\\[\n\\therefore OE = \\sqrt{CD^{2} - ED^{2}} = \\sqrt{5^{2} - 4^{2}} = 3,\n\\]\n\\[\n\\therefore AE = OA + OE = 8,\n\\]\n\\[\n\\therefore AD = \\sqrt{AE^{2} + DE^{2}} = \\sqrt{8^{2} + 4^{2}} = \\boxed{4\\sqrt{5}}.\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/51318907.png"
  },
  {
    "solution": "\\boxed{26}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points E and F are located on sides BC and CD of rectangle ABCD, respectively, and $CE = \\frac{2}{3}BC$, $CF = \\frac{2}{3}CD$. If the perimeter of $\\triangle ABC$ is 39, what is the perimeter of $\\triangle CEF$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral ABCD is a rectangle,\\\\\n$\\therefore \\angle B=\\angle ECF=90^\\circ$, $AB=CD$\\\\\n$\\because CE=\\frac{2}{3}BC$, $CF=\\frac{2}{3}CD$\\\\\n$\\therefore \\frac{CE}{BC}=\\frac{CF}{CD}=\\frac{CF}{AB}=\\frac{2}{3}$\\\\\n$\\therefore \\triangle CEF \\sim \\triangle BCA$, with a similarity ratio of $\\frac{2}{3}$.\\\\\n$\\because$ The perimeter of $\\triangle ABC$ is 39, $\\therefore$ the perimeter of $\\triangle CEF$ is $39\\times \\frac{2}{3}=\\boxed{26}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53059521.png"
  },
  {
    "solution": "\\boxed{x\\ge 4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $l_1: y=kx+4$ intersects with the line $l_2: y=ax+b$ at point $P(4, m)$. What is the solution set of the inequality $kx+4 \\ge ax+b$ with respect to $x$?\n",
    "completion": "\\textbf{Solution:} To solve: for the line $l_{1}: y=kx+4$ and the line $l_{2}: y=ax+b$ intersect at point $P(4, m)$,\\\\\nto satisfy the inequality $kx+4\\ge ax+b$,\\\\\nthe graph of $l_{1}: y=kx+4$ must be above that of $l_{2}: y=ax+b$,\\\\\nfrom the graph, it is known to intersect at point $P(4, m)$,\\\\\nwith the x-coordinate being 4,\\\\\nhence, the solution set for the inequality $kx+4\\ge ax+b$ is: $\\boxed{x\\ge 4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52342160.png"
  },
  {
    "solution": "\\boxed{x \\geq 8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given line AB: $y = \\frac{1}{2}x + 1$ intersects the x-axis and y-axis at points A and B, respectively, and line CD: $y = x + b$ intersects the x-axis and y-axis at points C and D, respectively. Line AB and line CD intersect at point P, and the area of $\\triangle ABD = 4$. Please directly write the solution set for the inequality $\\frac{1}{2}x + 1 \\leq x + b$.\n",
    "completion": "\\textbf{Solution:} By the given condition, the solution set of the inequality $\\frac{1}{2}x + 1 \\leq x + b$ is $\\boxed{x \\geq 8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52251756_79.png"
  },
  {
    "solution": "\\boxed{\\sqrt{22}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point A is inside circle $O$ with a radius of 5, and $OA = \\sqrt{3}$. Let P be a moving point on circle $O$. When $\\angle OPA$ reaches its maximum value, what is the length of PA?\n",
    "completion": "\\textbf{Solution:} As shown in the figure:\\\\\n$\\because$ OA and OP are constants,\\\\\n$\\therefore$ when PA $\\perp$ OA, $\\angle$OPA is the maximum,\\\\\nIn the right-angled triangle OPA, OA $= \\sqrt{3}$, OP $= 5$,\\\\\n$\\therefore$ PA $= \\sqrt{OP^2 - OA^2}$ $= \\boxed{\\sqrt{22}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52872298.png"
  },
  {
    "solution": "\\boxed{x<1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the graph, in the Cartesian coordinate system, the graph of the linear function $y=kx+b$ passes through point $A(-2,6)$, and intersects with the $x$-axis at point $B$, and intersects with the graph of the direct proportion function $y=3x$ at point $C$, where the abscissa of point $C$ is $1$. Please directly write down the solution set of the inequality $kx+b-3x>0$.\n",
    "completion": "\\textbf{Solution:} From (1), we know that the linear function $y=kx+b$ is $y=-x+4$, i.e., the inequality $kx+b-3x>0$ is transformed into: the inequality $-x+4>3x$. According to the graph of the function, the solution set of the inequality $-x+4>3x$ is $\\boxed{x<1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52381468_175.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, D is a point on AC, if $DA=DB=5$, the area of $\\triangle ABD$ is $10$, what is the length of CD?\n",
    "completion": "\\textbf{Solution:} Given that the area of $\\triangle ABD$ is 10,\\\\\nit follows that $\\frac{1}{2}AD\\cdot BC=10$, thus $\\frac{1}{2}\\times 5BC=10$. Solving this gives $BC=4$,\\\\\nin the right-angled $\\triangle BCD$, we have $DC=\\sqrt{BD^{2}-CB^{2}}=\\sqrt{5^{2}-4^{2}}=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52861117.png"
  },
  {
    "solution": "\\boxed{25^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, AB is the diameter of $\\odot O$. A tangent to $\\odot O$ is drawn through the point C on $\\odot O$, intersecting the extension of AB at point D. If $\\angle D=40^\\circ$, what is the magnitude of $\\angle A$?\n",
    "completion": "\\textbf{Solution:} Let:\\\\\nSince $CD$ is the tangent to $\\odot O$,\\\\\n$\\therefore$ $\\angle OCD = 90^\\circ$,\\\\\nSince $\\angle D = 40^\\circ$,\\\\\n$\\therefore$ $\\angle COD = 50^\\circ$,\\\\\nSince $AB$ is the diameter of $\\odot O$,\\\\\n$\\therefore$ $\\angle A$ and $\\angle COD$ are respectively the inscribed angle and the central angle that subtend the same arc $\\overset{\\frown}{BC}$,\\\\\n$\\therefore$ $\\angle A = \\frac{1}{2} \\angle COD = \\boxed{25^\\circ}$,",
    "image_path": "dataset/mm_math/images/MM_Math/51458132.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB = AC$, $EF$ is the perpendicular bisector of $AC$, intersecting $AC$ at point $E$ and $AB$ at point $F$. $D$ is the midpoint of side $BC$, and $M$ is a moving point on segment $EF$. If $BC = 4$ and the area of $\\triangle ABC$ is 16, what is the minimum perimeter of $\\triangle CDM$?\n",
    "completion": "\\textbf{Solution:}\n\nSince the perimeter of $\\triangle CDM$ is equal to $CM+MD+CD$, \\\\\nthen, when $CM+MD$ is at its minimum, the perimeter of $\\triangle CDM$ is also at its minimum. \\\\\nSince EF is the perpendicular bisector of AC, \\\\\nthen points A and C are symmetric with respect to EF, \\\\\nhence $CM+MD=AM+MD\\ge AD$. \\\\\nBy connecting AD, and intersecting EF at point M, at this point, the perimeter of $\\triangle CDM$ is minimized. \\\\\nSince AB=AC and D is the midpoint of side BC, \\\\\nthen $AD\\perp BC$, \\\\\nthus, the area of $\\triangle ABC$ is $\\frac{1}{2}BC\\cdot AD=\\frac{1}{2}\\times 4AD=16$, \\\\\nhence, $AD=8$, \\\\\ntherefore, the perimeter of $\\triangle CDM$ is $AD+CD=8+2=\\boxed{10}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52827564.png"
  },
  {
    "solution": "\\boxed{\\left(5-\\sqrt{2}, 6\\sqrt{2}-2\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=-x^2+bx+c$ intersects the x-axis at points $A(-1,0)$ and $B(5,0)$, and intersects the y-axis at point $C$. The line $y=-x+n$ passes through point $B$. Point $P$ is a moving point on the parabola above line $BC$. Through point $P$, draw $PF\\perp x$ axis intersecting line $BC$ at point $E$. Suppose the x-coordinate of point $P$ is $m$. If point $E'$ is the symmetric point of $E$ with respect to line $PC$, is there any point $P$ such that point $E'$ lies on the y-axis? If yes, find the coordinates of point $P$; if no, explain why.\n",
    "completion": "\\textbf{Solution:} Construct ME $\\perp$ y-axis at point M,\\\\\nsince points E$^{\\prime}$ and E are symmetric about PC,\\\\\ntherefore $\\angle$E$^{\\prime}$CP$=\\angle$PCE, CE=CE$^{\\prime}$, PE=PE$^{\\prime}$,\\\\\nsince PE$\\parallel$y-axis,\\\\\ntherefore $\\angle$E$^{\\prime}$CP$=\\angle$CPE$=\\angle$PCE,\\\\\ntherefore CE=PE=EC,\\\\\ntherefore quadrilateral E$^{\\prime}$PEC is a rhombus,\\\\\nlet point P $(m, -m^{2}+4m+5)$, E$(m, -m+5)$,\\\\\ntherefore PE$=-m^{2}+4m+5+m-5=-m^{2}+5m$;\\\\\nsince OB=OC=5,\\\\\ntherefore $\\triangle$OBC is an isosceles right triangle,\\\\\ntherefore $\\angle$MCE$=45^\\circ$,\\\\\ntherefore $\\triangle$MCE is an isosceles right triangle,\\\\\ntherefore CM=ME=m,\\\\\ntherefore CE$= \\sqrt{2}m$,\\\\\ntherefore solving $-m^{2}+5m=\\sqrt{2}m$: \\\\\n$m = 5-\\sqrt{2}$, m=0 (discard),\\\\\ntherefore the y-coordinate of point P is $-\\left(5-\\sqrt{2}\\right)^{2}+4\\left(5-\\sqrt{2}\\right)+5=6\\sqrt{2}-2$,\\\\\ntherefore the coordinates of point P are $\\boxed{\\left(5-\\sqrt{2}, 6\\sqrt{2}-2\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52756020_84.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in an acute-angled $\\triangle ABC$, point E is a point on side AB, with $BE=CE$, $AD\\perp BC$ at point D, and $AD$ intersects $EC$ at point G. $\\triangle AEG$ is an isosceles triangle. If $BE=10$, $CD=3$, and $G$ is the midpoint of $CE$, what is the length of $AG$?\n",
    "completion": "\\textbf{Solution:} Draw EF$\\perp$AG at point E with F as the foot, \\\\\n$\\therefore \\angle EFG=90^\\circ$, \\\\\n$\\because EA=EG$ and EF$\\perp$AG, \\\\\n$\\therefore AG=2FG$, \\\\\n$\\because G$ is the midpoint of $CE$, \\\\\n$\\therefore EG=GC=\\frac{1}{2}EC$, \\\\\n$\\because EB=EC=10$, \\\\\n$\\therefore GC=\\frac{1}{2}EC=5$, \\\\\n$\\because \\angle EFG=\\angle CDG=90^\\circ$ and $\\angle EGF=\\angle CGD$, \\\\\n$\\therefore \\triangle EFG \\cong \\triangle CDG$ (AAS), \\\\\n$\\therefore FG=DG$, \\\\\nIn $\\triangle CDG$, $CD=3$, \\\\\n$\\therefore DG=\\sqrt{CG^2-CD^2}=\\sqrt{5^2-3^2}=4$, \\\\\n$\\therefore FG=DG=4$, \\\\\n$\\therefore AG=2FG=8$, \\\\\n$\\therefore$ The length of $AG$ is $\\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52913842_126.png"
  },
  {
    "solution": "\\boxed{x>-2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the line $l_{1}: y= \\frac{3}{2}x+6$ and the line $l_{2}: y= -\\frac{5}{2}x-2$ intersect at point $P(-2,3)$. What is the solution set of the inequality $\\frac{3}{2}x+6 > -\\frac{5}{2}x-2$?\n",
    "completion": "\\textbf{Solution:} Given that the line $l_{1}\\colon y= \\frac{3}{2}x+6$ intersects with the line $l_{2}\\colon y= -\\frac{5}{2}x-2$ at the point (-2, 3), and when $x>-2$, the line $l_{1}$ is above line $l_{2}$,\\\\\nthe solution set for the inequality $\\frac{3}{2}x+6 > -\\frac{5}{2}x-2$ is $x>-2$.\\\\\nThus, the answer is: $\\boxed{x>-2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52019512.png"
  },
  {
    "solution": "\\boxed{when\\ -1<x<3,\\ y<0}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given the parabola $y=x^{2}-2x-3$ as shown in the graph. Based on the graph, answer the following: For what values of $x$ is $y>0$? For what values of $x$ is $y<0$?\n",
    "completion": "\\textbf{Solution:} From the graph and the coordinates of the intersection points of the parabola with the x-axis, we know that\\\\\nwhen $x>3$ or $x<-1$, $y>0$;\\\\\nwhen $-1<x<3$, $y<0$.\\\\\n$\\therefore \\boxed{when\\ x>3\\ or\\ x<-1,\\ y>0};\\ \\boxed{when\\ -1<x<3,\\ y<0}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52808108_37.png"
  },
  {
    "solution": "\\boxed{2-\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in quadrilateral $ABCD$, we have $A(-10, 0)$, $B(-2, 0)$, and $C(0, 4)$. Points $P$ and $Q$ are located on line segments $OA$ and $CD$, respectively, with $OP=2DQ$. By connecting $PQ$ and $PC$, and denoting $DQ$ as $x$, construct the symmetric point $Q'$ of point $Q$ with respect to line $PC$. When $Q'$ lies on an axis, determine the value of $x$ that satisfies this condition.\n",
    "completion": "\\textbf{Solution:} When point $Q'$ is on the y-axis, as shown in the diagram:\\\\\n\\[\n\\because \\text{Point Q and point $Q'$ are symmetric with respect to PC,}\n\\]\n\\[\n\\therefore QQ' \\perp PC, \\text{ CQ=C$Q'$},\n\\]\n\\[\n\\therefore \\text{PC bisects } \\angle DCO,\n\\]\n\\[\n\\therefore \\angle PCO=45^\\circ,\n\\]\n\\[\n\\because \\text{PO}\\perp\\text{CQ},\n\\]\n\\[\n\\therefore \\text{PO=CO},\n\\]\n\\[\n\\therefore 2x=4,\n\\]\n\\[\n\\therefore x=\\boxed{2};\n\\]\nWhen point $Q'$ is on the x-axis, as shown in the diagram:\\\\\nConnect C$Q'$, C$Q'$=CQ=8−x,\\\\\n\\[\n\\because \\text{DQ=x, CD=8, OP=2DQ},\n\\]\n\\[\n\\therefore \\text{CQ=CD−DQ}=8−x, \\text{OP=2x},\n\\]\n\\[\n\\because \\text{Point Q's symmetric point with respect to line PC is $Q'$},\n\\]\n\\[\n\\therefore \\text{PC}\\perp QQ', \\text{ QE=$Q'E$}, \\text{C$Q'$=CQ=8−x},\n\\]\n\\[\n\\because \\text{AB}\\parallel\\text{CD},\n\\]\n\\[\n\\therefore \\angle QCE=\\angle Q'PE,\n\\]\nIn $\\triangle QCE$ and $\\triangle Q'PE$,\n\\[\n\\left\\{ \\begin{array}{l}\n\\angle QCE=\\angle Q'PE \\\\\n\\angle CEQ=\\angle Q'EP \\\\\nQE=Q'E\n\\end{array} \\right.,\n\\]\n\\[\n\\therefore \\triangle QCE \\cong \\triangle Q'PE,\n\\]\n\\[\n\\therefore \\text{CQ=PQ'}=8−x,\n\\]\n\\[\n\\therefore \\text{OQ'}=\\left|OP-PQ'\\right|=\\left|3x−8\\right|,\n\\]\nIn $\\triangle OCQ'$,\n\\[\nOQ'^{2}+OC^{2}=CQ'^{2},\n\\]\n\\[\n\\text{i.e., } \\left(3x−8\\right)^{2}+4^{2}=\\left(8−x\\right)^{2},\n\\]\nSolving this, we find $x=2+\\sqrt{2}$ or $x=2-\\sqrt{2}$,\\\\\nIn conclusion, when $Q'$ lies on the coordinate axes, the values of $x$ that satisfy the conditions are \\boxed{2} or \\boxed{2+\\sqrt{2}} or \\boxed{2-\\sqrt{2}}.",
    "image_path": "dataset/mm_math/images/MM_Math/53498608_81.png"
  },
  {
    "solution": "\\boxed{M(0, -1)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system $xOy$, it is known that $A(0, 4)$ and $B(4, 0)$. The graph of the linear function $y=-2x$ intersects with line $AB$ at point $P$. If point $M$ is located on the $y$ axis, and the area of $\\triangle PMA$ is equal to $10$, what are the coordinates of point $M$?\n",
    "completion": "\\textbf{Solution:} From the problem, we have\n\\[\n\\because S_{\\triangle PMA} = \\frac{1}{2}AM \\times |x_{P}|,\n\\]\n\\[\n\\therefore \\frac{1}{2} \\times AM \\times 4 = 10,\n\\]\n\\[\n\\therefore AM = 5,\n\\]\n\\[\n\\because A(0, 4),\n\\]\n\\[\n\\therefore M(0, 9) \\text{ or } M(0, -1).\n\\]\n\nThus, the coordinates of point $M$ are $\\boxed{M(0, 9)}$ or $\\boxed{M(0, -1)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50560455_70.png"
  },
  {
    "solution": "\\boxed{\\frac{16}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB = 90^\\circ$, $CD \\perp AB$ with the foot of the perpendicular being $D$. $\\triangle ABC \\sim \\triangle ACD$. Given that $AC = 4$ and $BC = 3$, what is the length of $AD$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle ACB=90^\\circ$, $AC=4$, and $BC=3$, it follows that $AB=5$. Since $\\triangle ABC \\sim \\triangle ACD$, we have $\\frac{AC}{AD}=\\frac{AB}{AC}$. Therefore, $\\frac{4}{AD}=\\frac{5}{4}$, which implies that $AD=\\boxed{\\frac{16}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52934010_175.png"
  },
  {
    "solution": "\\boxed{-2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, the line $y=-\\frac{1}{2}x+2$ intersects the x-axis and the y-axis at points $A$ and $B$ respectively, and point $E$ is a point on the negative half-axis of the y-axis, with the area of $\\triangle ABE$ being $12$. As shown in Figure 2, the line $y=mx$ intersects line $AB$ at point $M$ and intersects line $AE$ at point $N$. When the area of $\\triangle OEN$ is twice the area of $\\triangle OBM$, what is the value of $m$?\n",
    "completion": "\\textbf{Solution:} From equation (1), we know that $OE=4$,\\\\\nsince $S_{\\triangle OEN}=2S_{\\triangle OBM}$, i.e., $\\frac{1}{2}\\times OB\\times |x_{M}|=\\frac{1}{2}\\times \\frac{1}{2}\\times OE\\times x_{N}$,\\\\\nwhich implies $\\frac{1}{2}\\times 2\\times |x_{M}|=\\frac{1}{4}\\times 4\\times x_{N}$, i.e., $x_{M}=-x_{N}$, then $y_{M}=-y_{N}$,\\\\\nlet the coordinates of point $N$ be $(n, n-4)$, then the coordinates of point $M$ would be $(-n, -n+4)$,\\\\\nSubstituting the coordinates of point $M$ into $y=-\\frac{1}{2}x+2$ yields:\\\\\n$-n+4=-\\frac{1}{2}(-n)+2$,\\\\\nsolving this gives $n=\\frac{4}{3}$,\\\\\ntherefore, the coordinates of point $N$ are $\\left(\\frac{4}{3}, -\\frac{8}{3}\\right)$,\\\\\nSubstituting the coordinates of point $N$ into $y=mx$ yields:\\\\\n$-\\frac{8}{3}=\\frac{4}{3}m$,\\\\\nsolving this gives $m=\\boxed{-2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50560383_74.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, AD is the diameter of the circle $O$, with $AD=8$ and $\\angle DAC = \\angle ABC$. What is the length of AC?\n",
    "completion": "\\textbf{Solution:} Let's connect CD\\\\\n$\\because \\angle DAC=\\angle ABC$\\\\\n$\\therefore AC=DC$\\\\\nAlso, $\\because AD$ is the diameter of the circle $O$\\\\\n$\\therefore \\angle ACD=90^\\circ$\\\\\n$\\therefore AC^{2}+DC^{2}=AD^{2}$\\\\\n$\\therefore 2AC^{2}=AD^{2}$\\\\\n$\\therefore AC=\\frac{\\sqrt{2}}{2}AD=\\frac{\\sqrt{2}}{2}\\times 8=4\\sqrt{2}$\\\\\nTherefore, the answer is: $AC=\\boxed{4\\sqrt{2}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51402723.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the line $l_1: y=kx+b$ ($k\\ne 0$) passing through point $B(1, 0)$ intersects with line $l_2: y=2x+4$ at point $P(-1, a)$. What is the area of $\\triangle ABP$?\n",
    "completion": "Solution: Since the line $l_2\\colon y=2x+4$ intersects the x-axis at point A, let $y=0$, then $2x+4=0$. Solving this gives: $x=-2$, therefore, the coordinates of point A are $(-2,0)$. Given that $AB=3$, and since $P(-1, 2)$, the area of triangle $ABP$ is $\\frac{1}{2}\\times3\\times2=\\boxed{3}",
    "image_path": "dataset/mm_math/images/MM_Math/52386847_117.png"
  },
  {
    "solution": "\\boxed{3\\sqrt{2}-4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the square ABCD, points E and F are located on BC and CD, respectively. AE intersects BF at point G, and AC intersects BF at point H. If CE = DF, BG = GH, and AB = 2, what is the area of $\\triangle CFH$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $FM\\perp CH$ at point $M$. \\\\\n$\\because$ Quadrilateral $ABCD$ is a square,\\\\\n$\\therefore \\angle ABC=\\angle BCD=\\angle BAD=90^\\circ$, $AB=BC=2$, $\\angle BAC=\\frac{1}{2}\\angle BAD$, $\\angle ACD=\\frac{1}{2}\\angle BCD$.\\\\\n$\\therefore AC=\\sqrt{AB^{2}+BC^{2}}=\\sqrt{2^{2}+2^{2}}=2\\sqrt{2}$, $\\angle BAC=45^\\circ$, $\\angle ACD=45^\\circ$.\\\\\nMoreover, $\\because CE=DF$,\\\\\n$\\therefore BC-CE=CD-DF$.\\\\\n$\\therefore BE=CF$.\\\\\nIn $\\triangle ABE$ and $\\triangle BCF$,\\\\\n$\\begin{cases} AB=BC, \\\\ \\angle ABE=\\angle BCF, \\\\ BE=CF. \\end{cases}$\\\\\n$\\therefore \\triangle ABE\\cong \\triangle BCF$ (SAS).\\\\\n$\\therefore \\angle 1=\\angle 2$.\\\\\nMoreover, $\\because \\angle ABE=\\angle ABG+\\angle 2=90^\\circ$,\\\\\n$\\therefore \\angle ABG+\\angle 1=90^\\circ$.\\\\\n$\\therefore \\angle AGB=180^\\circ-(\\angle ABG+\\angle 1)=90^\\circ$.\\\\\nMoreover, $\\because BG=GH$,\\\\\n$\\therefore AH=AB=2$.\\\\\n$\\therefore CH=AC-AH=2\\sqrt{2}-2$.\\\\\nIn $\\triangle ABG$ and $\\triangle AHG$,\\\\\n$\\begin{cases} AB=AH, \\\\ AG=AG, \\\\ BG=HG. \\end{cases}$\\\\\n$\\therefore \\triangle ABG\\cong \\triangle AHG$ (SSS).\\\\\n$\\therefore \\angle 1=\\angle HAG$.\\\\\n$\\therefore \\angle BAC=\\angle 1+\\angle HAG=2\\angle 1=45^\\circ$.\\\\\n$\\therefore \\angle 1=22.5^\\circ$.\\\\\n$\\therefore \\angle 2=\\angle 1=22.5^\\circ$.\\\\\n$\\therefore \\angle BFC=180^\\circ-\\angle 2-\\angle BCF=180^\\circ-22.5^\\circ-90^\\circ=67.5^\\circ$.\\\\\n$\\therefore \\angle CHF=180^\\circ-\\angle HCF-\\angle HFC=180^\\circ-45^\\circ-67.5^\\circ=67.5^\\circ$.\\\\\n$\\therefore \\angle CHF=\\angle CFH$.\\\\\n$\\therefore CH=CF=2\\sqrt{2}-2$.\\\\\n$\\because MF\\perp HC$, $\\angle ACD=45^\\circ$,\\\\\n$\\therefore \\angle MFC=180^\\circ-\\angle FMC-\\angle MCF=45^\\circ$\\\\\n$\\therefore FM=MC$.\\\\\nIn $\\triangle FMC$, $FC^{2}=MC^{2}+MF^{2}$.\\\\\n$\\therefore {(2\\sqrt{2}-2)}^{2}=2MF^{2}$.\\\\\n$\\therefore MF=2-\\sqrt{2}$.\\\\\n$\\therefore S_{\\triangle FHC}=\\frac{1}{2}MF\\cdot HC=\\frac{1}{2}(2\\sqrt{2}-2)(\\sqrt{2}-2)=\\boxed{3\\sqrt{2}-4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50496693.png"
  },
  {
    "solution": "\\boxed{y=-x^{2}+2x+3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given the diagram, it is known that the line $AB$ intersects the parabola $C: y=ax^2+2x+c$ at points $A(-1, 0)$ and $B(2, 3)$. What is the function expression of the parabola $C$?\n",
    "completion": "\\textbf{Solution:} By substituting the points $(-1,0)$ and $(2,3)$ into $y=ax^{2}+2x+c$, we obtain\n\\[\n\\begin{cases}\na-2+c=0\\\\\n4a+4+c=3\n\\end{cases},\n\\]\nfrom which we get\n\\[\n\\begin{cases}\na=-1\\\\\nc=3\n\\end{cases}.\n\\]\nThus, the equation of parabola $C$ is: \\boxed{y=-x^{2}+2x+3}.",
    "image_path": "dataset/mm_math/images/MM_Math/53101984_148.png"
  },
  {
    "solution": "\\boxed{24}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the line $y=-\\frac{4}{3}x+8$ intersects the x-axis at point A and the y-axis at point B. A moving point C starts from point B and moves towards point A at a uniform speed of 1 unit per second along direction $BA$, while another moving point D starts from point A and moves towards point O at a uniform speed of 2 units per second along direction $AO$. When one point stops moving, the other point stops as well. Let the time of movement be t seconds ($t>0$). What is the area of $\\triangle AOB$?\n",
    "completion": "\\textbf{Solution:} Solve: In the equation $y=-\\frac{4}{3}x+8$,\\\\\nwhen $x=0$, $y=8$,\\\\\nwhen $y=0$, $-\\frac{4}{3}x+8=0$,\\\\\nwe get: $x=6$,\\\\\n$\\therefore A(6,0), B(0,8)$,\\\\\nthus $OA=6, OB=8$,\\\\\n$\\therefore {S}_{\\triangle AOB}=\\frac{1}{2}OA\\cdot OB=\\frac{1}{2}\\times 6\\times 8=\\boxed{24}$,\\\\\nmeaning the area of $\\triangle AOB$ is $\\boxed{24}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52917352_78.png"
  },
  {
    "solution": "\\boxed{\\frac{9}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated in the figure, within $ \\triangle ABC$, D is the midpoint of AB. A line passing through D intersects AC at E and intersects the extension of BC at F. When $ BF=9$ and $ CF=4$, what is the value of $ \\frac{AE}{EC}$?\n",
    "completion": "\\textbf{Solution:} Construct $CM\\parallel AB$ through $C$, intersecting $DF$ at point $M$,\\\\\n$\\therefore$ $\\triangle CME\\sim \\triangle ADE$, $\\triangle FMC\\sim \\triangle FDB$,\\\\\n$\\therefore$ $\\frac{AE}{EC}=\\frac{AD}{CM}$, $\\frac{BD}{CM}=\\frac{BF}{CF}$,\\\\\nMoreover, $\\because$ $AD=BD$,\\\\\n$\\therefore$ $\\frac{AE}{EC}=\\frac{BF}{CF}$,\\\\\n$\\because$ $BF=9$, $CF=4$,\\\\\n$\\therefore$ $\\frac{AE}{EC}=\\boxed{\\frac{9}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52977968.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC=\\sqrt{5}, BC=2$, and a line is drawn from point $B$ perpendicular to $AC$, intersecting $AC$ at point $D$. Find the value of $\\cot\\angle ACB$.\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $AG \\perp BC$ at point $G$,\\\\\nsince $AB=AC$,\\\\\nthen $CG= \\frac{1}{2}BC=1$, and $AG \\perp BC$,\\\\\nin $\\triangle AGC$, by the Pythagorean theorem, we have $AG=\\sqrt{AC^{2}-CG^{2}}=\\sqrt{5-1}=2$,\\\\\nthus, $\\cot\\angle ACB=\\frac{CG}{AG}=\\boxed{\\frac{1}{2}}$,",
    "image_path": "dataset/mm_math/images/MM_Math/52977907_122.png"
  },
  {
    "solution": "\\boxed{105^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in an equilateral $\\triangle ABC$, points $E$ and $D$ are located on sides $AC$ and $BC$, respectively, with $AE=CD$. Lines $AD$ and $BE$ intersect at point $O$. $\\triangle ABE \\cong \\triangle CAD$; If $\\angle OBD=45^\\circ$, find the measure of $\\angle ADC$.\n",
    "completion": "\\textbf{Solution:} From (1) we have $\\triangle ABE \\cong \\triangle CAD$,\\\\\n$\\therefore \\angle ABE = \\angle CAD$,\\\\\n$\\therefore \\angle BOD = \\angle ABE + \\angle BAO = \\angle BAC = 60^\\circ$,\\\\\n$\\therefore \\angle ADC = \\angle OBD + \\angle BOD = 45^\\circ + 60^\\circ = \\boxed{105^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51419677_117.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $PA$ is the tangent line to $\\odot O$ at point $A$, and $PO$ intersects $\\odot O$ at point $B$. Given $\\tan P = \\frac{3}{4}$ and $OB = 6$, what is the length of $PB$?\n",
    "completion": "\\textbf{Solution:} Given that $PA$ is the tangent of circle $O$ with A as the point of tangency,\\\\\nthus $OA\\perp AP$,\\\\\nhence $\\angle OAP=90^\\circ$\\\\\nIn $\\triangle OAP$,\\\\\nsince $\\tan P=\\frac{OA}{AP}=\\frac{3}{4}$ and $OA=OB=6$,\\\\\nwe have $\\frac{6}{AP}=\\frac{3}{4}$,\\\\\ntherefore $AP=8$,\\\\\nthus $OP=\\sqrt{OA^{2}+AP^{2}}=10$,\\\\\nhence $PB=OP-OB=10-6=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53019842.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $BD$ bisects $\\angle ABC$, $CD \\perp BD$ with the foot of the perpendicular being D, and E is the midpoint of $AC$. If $AB=30$, $BC=18$, what is the length of $DE$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure:\\\\\nExtend CD to intersect AB at F.\\\\\nIn $\\triangle BDC$ and $\\triangle BDF$, we have\\\\\n$\\left\\{\\begin{array}{l}\n\\angle DBC=\\angle DBF\\\\\nBD=BD\\\\\n\\angle BDC=\\angle BDF=90^\\circ\n\\end{array}\\right.$\\\\\n$\\therefore \\triangle BDC \\cong \\triangle BDF$ (ASA)\\\\\n$\\therefore BF=BC=18$, CD=DF\\\\\n$\\therefore AF=AB-BF=12$,\\\\\n$\\because E$ is the midpoint of AC,\\\\\n$\\therefore CE=EA$\\\\\n$\\because CD=DF$,\\\\\n$\\therefore DE=\\frac{1}{2}AF=\\boxed{6}.$",
    "image_path": "dataset/mm_math/images/MM_Math/52395607.png"
  },
  {
    "solution": "\\boxed{20^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, within quadrilateral $ABCD$, $AC$ and $BD$ intersect at point $F$. Point $E$ is on $BD$, and $\\angle BAE=\\angle CAD$, $\\frac{AB}{AE}=\\frac{AC}{AD}$. If $\\angle BAE=20^\\circ$, find the measure of $\\angle CBD$.\n",
    "completion": "Solution: Since $\\triangle ABC\\sim \\triangle AED$, \\\\\nit follows that $\\angle ADB=\\angle ACB$, \\\\\nAlso, since $\\angle AFD=\\angle BFC$, by virtue of vertical angles being equal, \\\\\nit implies $\\triangle AFD\\sim \\triangle BFC$, \\\\\nthus $\\angle CBD=\\angle CAD$, \\\\\nGiven $\\angle BAE=\\angle CAD$, and $\\angle BAE=20^\\circ$, \\\\\nit results in $\\angle CAD=20^\\circ$, \\\\\nTherefore, the answer is: $\\boxed{20^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52872215_90.png"
  },
  {
    "solution": "\\boxed{15^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $\\angle AOB=2\\angle BOC$, and $OD$ is the bisector of $\\angle AOC$. If $\\angle AOB=60^\\circ$, what is the degree measure of $\\angle BOD$?\n",
    "completion": "\\textbf{Solution:} Given $\\because \\angle AOB=2\\angle BOC$ and $\\angle AOB=60^\\circ$,\\\\\n$\\therefore \\angle BOC= \\frac{1}{2}\\angle AOB=30^\\circ$,\\\\\n$\\therefore \\angle AOC=\\angle AOB+\\angle BOC=90^\\circ$,\\\\\n$\\because$ OD is the bisector of $\\angle AOC$,\\\\\n$\\therefore \\angle COD= \\frac{1}{2}\\angle AOC=45^\\circ$,\\\\\n$\\therefore \\angle BOD=\\angle COD-\\angle BOC=45^\\circ-30^\\circ=\\boxed{15^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55253062_92.png"
  },
  {
    "solution": "\\boxed{18.0}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in order to measure the heights of an ancient tree BH and a teaching building CG, a math interest group first takes a measurement at point A with a theodolite that is 1.5 meters tall. They measure the angle of elevation to the top of the ancient tree H to be $\\angle HDE=45^\\circ$. At this moment, the top of the teaching building G is exactly in line with sight DH. Then, moving forward 7 meters to point B, they measure the angle of elevation to the top of the teaching building G to be $\\angle GEF=60^\\circ$. Points A, B, and C are on the same horizontal line. Calculate the height of the teaching building CG. (Reference data: $\\sqrt{2} \\approx 1.4$, $\\sqrt{3} \\approx 1.7$)\n",
    "completion": "\\textbf{Solution:} Let $HJ \\perp CG$ at $G$. Then $\\triangle HJG$ is an isosceles triangle, and quadrilateral $BCJH$ is a rectangle. Let $HJ=GJ=BC=x$. In $\\triangle EFG$, $\\tan 60^\\circ = \\frac{GF}{EF}$,\n\n\\[\n\\therefore \\sqrt{3} = \\frac{7+x}{x},\n\\]\n\n\\[\n\\therefore x = \\frac{7}{2}(\\sqrt{3}+1),\n\\]\n\n\\[\n\\therefore GF = \\sqrt{3}x \\approx 16.45,\n\\]\n\n\\[\n\\therefore CG = CF + FG = 1.5 + 16.45 \\approx \\boxed{18.0} \\text{ meters.}\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/51103651_87.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a rectangular window frame in the shape of the character \"目\" is made from 48 meters of wood (the crossbars EF and GH are also made of wood). Where AB $\\parallel$ EF $\\parallel$ GH $\\parallel$ CD, to maximize the area of the window frame ABCD, what should be the length of AB?\n",
    "completion": "\\textbf{Solution:} Let the length of AB be $x$ meters, then the length of AD is $\\frac{48-4x}{2}$ meters.\\\\\nUsing the formula for the area of a rectangle, we get: $S_{\\text{rectangle}} = \\text{AD} \\cdot \\text{AB} = x \\cdot \\frac{48-4x}{2} = -2x^2 + 24x = -2(x-6)^2 + 72$,\\\\\n$\\because 48-4x > 0$,\\\\\n$\\therefore x < 12$,\\\\\n$\\therefore 0 < x < 12$,\\\\\n$\\because -2 < 0$,\\\\\n$\\therefore$ when $x = \\boxed{6}$, the area of the rectangle has its maximum value.",
    "image_path": "dataset/mm_math/images/MM_Math/52514047.png"
  },
  {
    "solution": "\\boxed{\\left(3, −1\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given: As shown in the figure, the linear function $y_{1}=kx-2$ intersects the $x$-axis at point $B(-2, 0)$, and $y_{2}=x+b$ intersects the $x$-axis at point $C(4, 0)$. The graphs of these two functions intersect at point $A$. Connect $OA$. Is there a point $P$ on the line $y_{2}=x+b$ such that $S_{\\triangle OCP}=\\frac{1}{3}S_{\\triangle OAC}$? If so, find the coordinates of point $P$.\n",
    "completion": "\\textbf{Solution:} Given that OC=4, and A(1, −3),\n\n\\[S_{\\triangle OAC}=\\frac{1}{2}\\times 3\\times 4=6,\\]\n\n\\[S_{\\triangle OCP}=\\frac{1}{3}S_{\\triangle OAC}=2,\\]\n\nLet the coordinates of point P be (x, y),\n\n\\[S_{\\triangle OCP}=\\frac{1}{2}OC\\times |y|,\\]\n\n\\[2=\\frac{1}{2}\\times 4\\times |y|,\\]\n\n\\[|y|=1,\\]\n\nWhen y=1, $1=x+b$,\n\nx=5, the coordinates of point P are $\\boxed{\\left(5, 1\\right)}$;\n\nWhen y=−1, $−1=x+b$,\n\nx=3, the coordinates of point P are $\\boxed{\\left(3, −1\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52381120_149.png"
  },
  {
    "solution": "\\boxed{6\\sqrt{2}-4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, with the hypotenuse $AB$ of the right-angled $\\triangle ABC$ as one side, construct a square $ABED$ on the right side of $AB$, with the diagonals of the square intersecting at point $O$. Connect $CO$. If $AC=4$ and $CO=6$, what is the length of $BC$?\n",
    "completion": "\\textbf{Solution:} As illustrated, extend $CB$ to the point $G$, such that $BG=AC=4$,\\\\\n$\\because$ according to the problem, quadrilateral $ABED$ is a square,\\\\\n$\\therefore \\angle 4=\\angle 5=45^\\circ$, $\\angle$EBA$=90^\\circ$,\\\\\n$\\therefore \\angle 1+\\angle 2=90^\\circ$,\\\\\nFurthermore, $\\because \\triangle ABC$ is a right-angled triangle with $AB$ being the hypotenuse,\\\\\n$\\therefore \\angle 2+\\angle 3=90^\\circ$,\\\\\n$\\therefore \\angle 1=\\angle 3$,\\\\\n$\\because \\angle 1+\\angle 5=\\angle 3+\\angle 4$,\\\\\n$\\therefore \\angle CAO=\\angle GBO$,\\\\\nIn $\\triangle CAO$ and $\\triangle GBO$,\\\\\n$\\left\\{\\begin{array}{l}CA=GB \\\\ \\angle CAO=\\angle GBO \\\\ AO=BO \\end{array}\\right.$,\\\\\n$\\therefore \\triangle CAO \\cong \\triangle GBO$ (SAS),\\\\\n$\\therefore CO=GO=6$, $\\angle 6=\\angle 8$,\\\\\n$\\because \\angle 7+\\angle 8=90^\\circ$,\\\\\n$\\therefore \\angle 6+\\angle 7=90^\\circ$,\\\\\n$\\therefore \\angle COG=90^\\circ$,\\\\\n$\\therefore CG=\\sqrt{6^2+6^2}=6\\sqrt{2}$,\\\\\n$\\therefore BC=CG-BG=6\\sqrt{2}-4$.\\\\\nThus, the answer is: $BC=\\boxed{6\\sqrt{2}-4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52189624.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\odot O$ is the circumcircle of quadrilateral $ABCD$, with $AD$ being the diameter of $\\odot O$. Connect $BD$. If $\\overset{\\frown}{AC} = \\overset{\\frown}{BD}$ and $\\angle 1 = \\angle 2$, when $AD = 4\\sqrt{2}$ and $BC = 4$, what is the area of $\\triangle ABD$?\n",
    "completion": "\\textbf{Solution:} Draw $OE\\perp BC$ at point E. \\\\\n$\\therefore BE=CE=\\frac{1}{2}BC=2$, \\\\\n$\\because AD$ is the diameter of $\\odot O$, \\\\\n$\\therefore OB=\\frac{1}{2}AD=2\\sqrt{2}$, \\\\\n$\\therefore OE=\\sqrt{OB^2-BE^2}=\\sqrt{(2\\sqrt{2})^2-2^2}=2$, \\\\\n$\\therefore {S}_{\\triangle ABD}=\\frac{1}{2}AD\\cdot OE=\\frac{1}{2}\\times 4\\sqrt{2}\\times 2=\\boxed{4\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51209230_104.png"
  },
  {
    "solution": "\\boxed{(0, \\frac{4\\sqrt{3}}{3})}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, circle $ \\odot A$ passes through the points O, D, C, and B. Line segments CO and CD are drawn. Given that $\\angle DCO = 30^\\circ$ and $B(4, 0)$, what are the coordinates of point D?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect DB,\\\\\nsince $\\angle DCO=30^\\circ$ and $\\angle DOB=90^\\circ$,\\\\\ntherefore, $\\angle OBD=30^\\circ$, DB is the diameter,\\\\\nsince B(4,0),\\\\\ntherefore, $OB=4$,\\\\\nin $\\triangle DOB$, $\\tan\\angle OBD= \\frac{OD}{OB}$,\\\\\nthus $\\frac{OD}{4}=\\frac{\\sqrt{3}}{3}$,\\\\\ntherefore, $OD=\\frac{4\\sqrt{3}}{3}$,\\\\\ntherefore, the coordinates of point D are $\\boxed{(0, \\frac{4\\sqrt{3}}{3})}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52532703.png"
  },
  {
    "solution": "\\boxed{80}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: A car passes through a certain section of the road at a constant speed. The required time $t$ (h) and the driving speed $v$ (km/h) meet the functional relationship: $t=\\frac{k}{v}$, whose graph is a curve as shown, with endpoints $A(40, 1)$ and $B(m, 0.5)$. What are the values of $k$ and $m$?\n",
    "completion": "\\textbf{Solution:} From the given, the function $t=\\frac{k}{v}$ passes through the point $(40,1)$, \\\\\n$1=\\frac{k}{40}$, then we get $k=\\boxed{40}$, \\\\\nHence, the function equation is: $t=\\frac{40}{v}$ \\\\\nSubstituting $(m,0.5)$ into $t=\\frac{40}{v}$, we find $m=\\boxed{80}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51446664_251.png"
  },
  {
    "solution": "\\boxed{\\frac{9}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, within $ \\square ABCD$, where E is on DC, if $ DE: EC=1: 2$, what is the value of $ \\frac{{S}_{\\triangle ABF}}{{S}_{\\triangle CEF}}$?\n",
    "completion": "\\textbf{Solution:} Given $\\because DE:EC=1:2$,\\\\\n$\\therefore \\frac{CD}{CE}=\\frac{3}{2}$,\\\\\n$\\because$ quadrilateral ABCD is a parallelogram,\\\\\n$\\therefore$ AB $\\parallel$ CD, AB=CD,\\\\\n$\\therefore \\triangle ABF\\sim \\triangle CEF$,\\\\\n$\\therefore \\frac{S_{\\triangle ABF}}{S_{\\triangle CEF}}=\\left(\\frac{AB}{CE}\\right)^2=\\left(\\frac{3}{2}\\right)^2=\\boxed{\\frac{9}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52623490.png"
  },
  {
    "solution": "\\boxed{2:5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in square $ABCD$, $E$ is a point on side $BC$. If $BE: EC=2:3$ and $AE$ intersects $BD$ at $F$, what is the ratio of $BF: FD$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a parallelogram, \\\\\n$\\therefore$ AD $\\parallel$ BC and AD $=$ BC, \\\\\nSince BE: EC $=$ 2: 3, \\\\\n$\\therefore$ BE: AD $=$ 2: 5, \\\\\nSince AD $\\parallel$ BC, \\\\\n$\\therefore$ $\\triangle$ADF $\\sim$ $\\triangle$EBF, \\\\\n$\\therefore$ BF: FD $=$ BE: AD $=$ 2: 5, \\\\\nHence, the answer is: \\boxed{2:5}.",
    "image_path": "dataset/mm_math/images/MM_Math/53010266.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, AB is a generatrix of the cone, and BC is the diameter of the base. Given that BC = 6 cm and the lateral surface area of the cone is $15\\pi$, what is the value of $\\cos\\angle ABO$?\n",
    "completion": "\\textbf{Solution:} From the given information, we have $\\frac{1}{2}\\pi \\times 6\\times AB=15\\pi$, from which we find $AB=5\\,\\text{cm}$. Since $BO=\\frac{1}{2}$BC$=3\\,\\text{cm}$, it follows that $\\cos\\angle ABO=\\frac{BO}{AB}=\\boxed{\\frac{3}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55498457.png"
  },
  {
    "solution": "\\boxed{P_{2}\\left(3, \\frac{8}{3}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line passing through the origin $y=2x$ intersects the inverse proportional function $y_{1}=\\frac{2}{x}$ at point $B$, and intersects another inverse proportional function $y_{2}=\\frac{k}{x}$ at point $C$, with $OB=BC$. The point $A$ has an abscissa of $4$ and lies on the curve $y_{1}=\\frac{2}{x}$. A perpendicular line $BD \\perp x$-axis is drawn from point $B$, with $D$ being the foot of the perpendicular. Is there a point $P$ on the graph of the inverse proportional function $y_{2}=\\frac{k}{x}$ within the first quadrant such that $S_{\\triangle ABP}=\\frac{5}{11}S_{\\text{quadrilateral} ADBP}$? If such a point exists, find the coordinates of point $P$; if not, please state the reason.\n",
    "completion": "\\textbf{Solution:} From the given problem, we have $A(4, \\frac{1}{2}), B(1, 2), D(1, 0)$,\\\\\n$\\therefore S_{\\triangle ABD}=\\frac{1}{2}BD\\cdot |x_{A}-x_{B}|$, $AB=\\sqrt{(4-1)^{2}+(\\frac{1}{2}-2)^{2}}=\\frac{3\\sqrt{5}}{2}$\\\\\nLet the equation of line AB be $y=mx+n$. Substituting $A(4, \\frac{1}{2}), B(1, 2)$, we get $\\left\\{\\begin{array}{l}m+n=2 \\\\ 4m+n=\\frac{1}{2}\\end{array}\\right.$, solving this gives, $\\left\\{\\begin{array}{l}m=-\\frac{1}{2} \\\\ n=\\frac{5}{2}\\end{array}\\right.$\\\\\n$\\therefore$ The equation of line AB is $y=-\\frac{1}{2}x+\\frac{5}{2}$\\\\\nConstruct PR$\\perp$x-axis from P, meeting AB at point F,\\\\\n$\\therefore S_{\\triangle ABP}=S_{\\triangle AFP}+S_{\\triangle BFP}$\\\\\n$=\\frac{1}{2}PF\\cdot |x_{P}-x_{B}|+\\frac{1}{2}PF\\cdot |x_{A}-x_{P}|$\\\\\n$=\\frac{1}{2}PF\\cdot |x_{A}-x_{B}|$\\\\\n$\\because S_{\\triangle ABP}=\\frac{5}{11}S_{\\text{quadrilateral}ADBP}$,\\\\\n$\\therefore S_{\\triangle ABP}=\\frac{5}{6}S_{\\triangle ABD}$\\\\\n$\\therefore PF=\\frac{5}{6}BD=\\frac{5}{6}\\times 2=\\frac{5}{3}$\\\\\nLet $P(x, \\frac{8}{x}), F(x, -\\frac{1}{2}x+\\frac{5}{2})$,\\\\\n$\\therefore PF=\\frac{8}{x}-\\left(-\\frac{1}{2}x+\\frac{5}{2}\\right)=\\frac{8}{x}+\\frac{1}{2}x-\\frac{5}{2}=\\frac{5}{3}$\\\\\nRearranging gives: $3x^{2}-25x+48=0$\\\\\nSolving, we find $x_{1}=\\frac{16}{3}, x_{2}=3$.\\\\\nUpon verification, $x_{1}=\\frac{16}{3}, x_{2}=3$ are the roots of the original equation\\\\\n$\\therefore \\boxed{P_{1}\\left(\\frac{16}{3}, \\frac{3}{2}\\right)}, \\boxed{P_{2}\\left(3, \\frac{8}{3}\\right)}$",
    "image_path": "dataset/mm_math/images/MM_Math/51322681_143.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ is an equilateral triangle, point D is a point on side AB, an equilateral $\\triangle CDE$ is constructed upwards with CD as its side, and AE is connected. $\\triangle BCD \\cong \\triangle ACE$. If $AE=1$, $AB=3$, what is the length of $AD$?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle BCD \\cong \\triangle ACE$, $AE=1$, $AB=3$,\\\\\n$\\therefore$ $BD=AE=1$,\\\\\n$\\therefore$ $AD=AB-BD=3-1=\\boxed{2}$,\\\\\n$\\therefore$ the length of $AD$ is \\boxed{2}.",
    "image_path": "dataset/mm_math/images/MM_Math/52914139_113.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In rectangle $ABCD$, $AB=4$, $BC=5$, P is a moving point on ray $BC$, draw $PE\\perp AP$, $PE$ intersects ray $DC$ at point E, ray $AE$ intersects ray $BC$ at point F, let $BP=x$, $CF=y$. When $\\sin\\angle APB=\\frac{4}{5}$, what is the length of $CE$?\n",
    "completion": "\\textbf{Solution:} In the right triangle $Rt\\triangle APB$, where $AB=4$ and $\\sin\\angle APB=\\frac{4}{5}$,\n\\[\n\\therefore \\sin\\angle APB=\\frac{AB}{AP}=\\frac{4}{5}, \\text{ thus } \\frac{4}{AP}=\\frac{4}{5},\n\\]\n\\[\n\\therefore AP=5,\n\\]\n\\[\n\\therefore PB=\\sqrt{AP^2-AB^2}=3,\n\\]\n\\[\n\\therefore PC=BC-PB=5-3=2,\n\\]\n\\[\n\\because PE\\perp AP,\n\\]\n\\[\n\\therefore \\angle CPE+\\angle APB=90^\\circ,\n\\]\n\\[\n\\because \\angle BAP+\\angle APB=90^\\circ,\n\\]\n\\[\n\\therefore \\angle CPE=\\angle BAP,\n\\]\n\\[\n\\because \\angle ECP=\\angle B=90^\\circ,\n\\]\n\\[\n\\therefore \\triangle PCE\\sim \\triangle ABP,\n\\]\n\\[\n\\therefore \\frac{CE}{BP}=\\frac{PC}{AB},\n\\]\n\\[\n\\therefore CE=\\frac{BP\\cdot PC}{AB}=\\frac{3\\times 2}{4}=\\boxed{\\frac{3}{2}}\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/52977987_134.png"
  },
  {
    "solution": "\\boxed{\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of $\\odot O$, point $C$ is a point on $\\odot O$, the bisector of $\\angle CAB$ intersects $\\overset{\\frown}{BC}$ at point $D$, a line is drawn through point $D$ such that $DE\\parallel BC$ and it intersects the extension of $AC$ at point $E$. $DE$ is a tangent line of $\\odot O$; if $DE=2$ and $CE=1$, find the length of $BD$.\n",
    "completion": "\\textbf{Solution}: Connect CD,\\\\\n$\\because$ $\\angle OAD = \\angle CAD$,\\\\\n$\\therefore$ $\\overset{\\frown}{CD} = \\overset{\\frown}{BD}$,\\\\\n$\\therefore$ CD = BC,\\\\\nIn $\\triangle CED$, $\\angle E = 90^\\circ$, DE = 2, CE = 1,\\\\\n$\\therefore$ $CD = \\sqrt{DE^2 + CE^2} = \\sqrt{2^2 + 1^2} = \\sqrt{5}$,\\\\\n$\\therefore$ BD = $\\boxed{\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52719941_172.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $OD$ bisects $\\angle AOB$, $OE$ bisects $\\angle BOC$, $\\angle COD=20^\\circ$, $\\angle AOB=140^\\circ$. What is the measure of $\\angle DOE$ in degrees?\n",
    "completion": "\\textbf{Solution:} Since $OE$ bisects $\\angle BOC$,\\\\\nit follows that $\\angle COE = \\frac{1}{2}\\angle BOC = \\frac{1}{2} \\times 50^\\circ = 25^\\circ$,\\\\\ntherefore, $\\angle DOE = \\angle COE + \\angle COD = 25^\\circ + 20^\\circ = \\boxed{45^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53119651_131.png"
  },
  {
    "solution": "\\boxed{48^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a regular hexagon and a regular pentagon are placed in the manner shown, with a common vertex O, and side AB of the hexagon and side DE of the pentagon lying on the same straight line. What is the measure of $\\angle BOE$?\n",
    "completion": "\\textbf{Solution:} Similarly, we can obtain: $\\angle DEO=108^\\circ$,\\\\\n$\\therefore \\angle OEB=180^\\circ-108^\\circ=72^\\circ$, $\\angle OBE=180^\\circ-120^\\circ=60^\\circ$,\\\\\n$\\therefore \\angle BOE=180^\\circ-72^\\circ-60^\\circ=\\boxed{48^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52905816_133.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, on a square sheet of paper ABCD, there is a point P, with $PA=1$, $PD=2$, and $PC=3$. Now, cut out $\\triangle PCD$ and place it as shown in the figure (where C coincides with A, P coincides with G, and D coincides with D). Find the length of the segment PG.\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a square, \\\\\nit follows that AD=CD, and $\\angle ADC=90^\\circ$, \\\\\nsince PA=1, PD=2, PC=3, \\\\\nby cutting $\\triangle PCD$ and pasting it as shown in the figure (C coincides with A, P with G, and D with D), \\\\\ntherefore PD=GD=2, $\\angle CDP=\\angle ADG$, AG=PC=3, \\\\\ntherefore $\\angle PDG=\\angle ADC=90^\\circ$, \\\\\nthus, $\\triangle PDG$ is an isosceles right triangle, \\\\\nhence $\\angle GPD=45^\\circ$, PG=$\\sqrt{2}$PD=$\\boxed{2\\sqrt{2}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52188913_141.png"
  },
  {
    "solution": "\\boxed{(4, -2)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the vertices A and C of the rhombus ABCD are located at (3,3) and (-1,-1), respectively. The diagonal BD intersects AC at point M and intersects the x-axis at point N. If BN=2ND, what are the coordinates of point D?\n",
    "completion": "\\textbf{Solution:} As illustrated in the figure, draw AE $\\perp$ x-axis at point E, draw DF $\\perp$ x-axis at point F, and draw MP $\\perp$ x-axis at point P,\\\\\n$\\therefore$ MP $\\parallel$ DF, \\\\\n$\\because$ Quadrilateral ABCD is a rhombus, \\\\\n$\\therefore$ MA=MC, MB=MD, AC$\\perp$BD, \\\\\n$\\because$ Point A is (3, 3), and point C is (-1, -1), \\\\\n$\\therefore$ M is (1, 1), \\\\\n$\\therefore$ OP=MP=1, \\\\\n$\\therefore$ $\\triangle$OPM is an isosceles right triangle, \\\\\n$\\therefore$ $\\angle$MOP = 45$^\\circ$, \\\\\n$\\because$ AC$\\perp$BD, $\\angle$OMN = 90$^\\circ$, \\\\\n$\\therefore$ $\\triangle$OMN is an isosceles right triangle, \\\\\n$\\therefore$ MN=MO, \\\\\n$\\therefore$ NP=MP=OP=1, \\\\\n$\\because$ BN=2ND, \\\\\n$\\therefore$ BD=3DN, \\\\\n$\\because$ BD=2DM, \\\\\n$\\therefore$ $DM=\\frac{3}{2}DN$, \\\\\n$\\therefore$ DN=2MN, \\\\\n$\\because$ MP$\\parallel$DF, \\\\\n$\\therefore$ $\\triangle$PMN$\\sim$$\\triangle$FDN, \\\\\n$\\therefore$ $\\frac{DF}{MP}=\\frac{FN}{PN}=\\frac{DN}{MN}=2$, \\\\\n$\\therefore$ DF=2, FN=2, \\\\\n$\\therefore$ OF=4, \\\\\n$\\therefore$ The coordinates of point D are \\boxed{(4, -2)}.",
    "image_path": "dataset/mm_math/images/MM_Math/52189536.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$. Draw a line through point $B$ parallel to $AC$ and a line through point $C$ parallel to $BD$, and these two parallel lines intersect at point $E$. Given that $AB=2$ and $AD=2\\sqrt{3}$, and quadrilateral $OBEC$ is a rhombus; calculate the area of rhombus $OBEC$.\n",
    "completion": "\\textbf{Solution:} Let quadrilateral $ABCD$ be a rectangle,\\\\\n$\\therefore \\angle ABC=90^\\circ$, $BC=AD=2\\sqrt{3}$, $OA=OC$,\\\\\n$\\therefore S_{\\triangle ABC}=2S_{\\triangle OBC}$,\\\\\n$\\because S_{\\text{rhombus }OBEC}=2S_{\\triangle OBC}$,\\\\\n$\\therefore S_{\\text{rhombus }OBEC}=S_{\\triangle ABC}=\\frac{1}{2}AB\\cdot BC=\\frac{1}{2}\\times 2\\times 2\\sqrt{3}=\\boxed{2\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52284147_107.png"
  },
  {
    "solution": "\\boxed{24}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, $AB\\parallel CD$, $AD\\perp CD$, $\\angle B=45^\\circ$. Extend $CD$ to point $E$ such that $DE=DA$, and connect $AE$. Quadrilateral $ABCE$ is a parallelogram. If $AB=6$, $CD=2$, find the area of quadrilateral $ABCE$.\n",
    "completion": "\\textbf{Solution:} Given from (1) $AB=CE$,\\\\\nsince $CD=2$, $AB=6$,\\\\\nthus $DE=4$,\\\\\nsince $AD=DE$,\\\\\nthus $AD=4$,\\\\\ntherefore, the area of quadrilateral $ABCE=AB\\times AD=6\\times 4=\\boxed{24}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51920372_130.png"
  },
  {
    "solution": "\\boxed{14^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle B=\\angle C$, $D$ is a point on side $BC$, and $E$ is a point on side $AC$, with $\\angle ADE=\\angle AED$. If $\\angle BAD=28^\\circ$, what is the measure of $\\angle CDE$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle ADC$ is an exterior angle of $\\triangle ABD$, \\\\\nit follows that $\\angle ADC = \\angle B + \\angle BAD = \\angle B + 28^\\circ$, \\\\\nand since $\\angle AED$ is an exterior angle of $\\triangle CDE$, \\\\\nwe have $\\angle AED = \\angle C + \\angle EDC$, \\\\\nconsidering $\\angle B = \\angle C$ and $\\angle ADE = \\angle AED$, \\\\\nthus $\\angle C + \\angle EDC = \\angle ADC - \\angle EDC = \\angle B + 28^\\circ - \\angle EDC$, \\\\\nleading to the solution $\\angle EDC = \\boxed{14^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52836308.png"
  },
  {
    "solution": "\\boxed{64^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, two triangular plates of the same size containing $45^\\circ$, triangular plate ACF and triangular plate CFB are placed as shown. Point D is on side AC, and point E is on side BC, with $\\angle CFE=13^\\circ$ and $\\angle CFD=32^\\circ$. What is the measure of $\\angle DEC$ in degrees?\n",
    "completion": "\\textbf{Solution:}\n\nConstruct $FH \\perp FE$, intersecting $AC$ at point $H$.\\\\\n$\\because$ $\\angle AFC=\\angle EFH=90^\\circ$\\\\\nAlso, $\\because$ $\\angle AFC=\\angle AFH+\\angle CFH$ and $\\angle HFE=\\angle CFE+\\angle CFH$\\\\\n$\\therefore$ $\\angle AFH=\\angle CFE=13^\\circ$\\\\\n$\\because$ $\\angle A=\\angle FCE=45^\\circ$ and $FA=CF$\\\\\n$\\therefore$ $\\triangle FAH\\cong \\triangle FCE$ (ASA)\\\\\n$\\therefore$ $FH=FE$\\\\\n$\\because$ $\\angle DFE=\\angle DFC+\\angle EFC=32^\\circ+13^\\circ=45^\\circ$\\\\\n$\\because$ $\\angle DFH=\\angle HFE-\\angle DFE=90^\\circ-45^\\circ=45^\\circ$\\\\\n$\\therefore$ $\\angle DFE=\\angle DFH$\\\\\nAlso, $\\because$ $DF=DF$\\\\\n$\\therefore$ $\\triangle HDF\\cong \\triangle EDF$ (SAS)\\\\\n$\\therefore$ $\\angle DHF=\\angle DEF$\\\\\n$\\because$ $\\angle DHF=\\angle A+\\angle HFA=45^\\circ+13^\\circ=58^\\circ$\\\\\n$\\therefore$ $\\angle DEF=58^\\circ$\\\\\n$\\because$ $\\angle CFE+\\angle CEF+\\angle FCE=180^\\circ$\\\\\n$\\therefore$ $\\angle CEF=180^\\circ-\\angle CFE-\\angle FCE=180^\\circ-13^\\circ-45^\\circ=122^\\circ$\\\\\n$\\therefore$ $\\angle DEC=\\angle CEF-\\angle DEF=122^\\circ-58^\\circ=\\boxed{64^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/53686577.png"
  },
  {
    "solution": "\\boxed{y=x^{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y = -x + 2$ passes through point $A(2, 0)$ on the $x$-axis and intersects the parabola $y = ax^2$ at points $B$ and $C$, where the coordinates of point $B$ are $(1, 1)$. What is the equation of the parabola?\n",
    "completion": "\\textbf{Solution:} Since point B$(1,1)$ lies on the parabola $y=ax^{2}$,\\\\\nit follows that $1=a$,\\\\\ntherefore, the equation of the parabola is $\\boxed{y=x^{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52550623_138.png"
  },
  {
    "solution": "\\boxed{x>-2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, the line $y=kx+b \\left(k \\ne 0\\right)$ passes through the point $\\left(-2, 0\\right)$. What is the solution set of the inequality $kx+b>0$?\n",
    "completion": "\\textbf{Solution:} From the graphic, we have: when $x>-2$, $kx+b>0$, \\\\\nTherefore, the answer is: $\\boxed{x>-2}$",
    "image_path": "dataset/mm_math/images/MM_Math/52381269.png"
  },
  {
    "solution": "\\boxed{2.5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the diagram, within the quadrilateral $ABCD$, points $G$ and $H$ are respectively the midpoints of $AB$ and $CD$, points $E$ and $F$ lie on the diagonal $AC$, and $AE=CF$. Quadrilateral $EGFH$ is a parallelogram; connecting $BD$ intersects with $AC$ at point $O$, if $BD=10$, $AE+CF=EF$, what is the length of $EG$?\n",
    "completion": "\\textbf{Solution:} Draw line $BD$ intersecting $AC$ at point $O$, as shown in the figure:\\\\\nSince quadrilateral $ABCD$ is a parallelogram,\\\\\n$\\therefore OA=OC$, $OB=OD$,\\\\\nSince $BD=10$,\\\\\n$\\therefore OB=OD=5$,\\\\\nSince $AE=CF$, $OA=OC$,\\\\\n$\\therefore OE=OF$,\\\\\nSince $AE+CF=EF$,\\\\\n$\\therefore 2AE=EF=2OE$,\\\\\n$\\therefore AE=OE$,\\\\\nMoreover, since point $G$ is the midpoint of $AB$,\\\\\n$\\therefore EG$ is the median of $\\triangle ABO$,\\\\\n$\\therefore EG=\\frac{1}{2}OB=2.5$. \\\\\nTherefore, the length of $EG$ is $\\boxed{2.5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51865207_98.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the hyperbola $y=\\frac{k}{x}\\ (x>0)$ intersects the side BC of the square ABCD at point E, and intersects the side CD at point F. Given $BE=3CE$ and $A(4, 0)$, $B(8, 0)$, what is the length of $CF$?\n",
    "completion": "\\textbf{Solution:} Given $\\because BE=3CE$,\n\n$\\therefore CE=a$, $BE=3a$,\n\n$\\therefore E(8,3a)$, $C(8,4a)$,\n\n$\\because$ point E lies on $y=\\frac{k}{x}$,\n\n$\\therefore k=24a$,\n\n$\\therefore$ the equation of the hyperbola is: $y=\\frac{24a}{x}$,\n\nFrom the coordinates of point C, we find that the y-coordinate of point F is $4a$,\n\n$\\because$ point F also lies on $y=\\frac{k}{x}$, by substituting the y-coordinate, we find,\n\n$\\therefore x=6$,\n\n$\\therefore CF=8-6=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52865974.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, $AC=3$, $BC=4$, then what is the value of $\\cos A$?\n",
    "completion": "\\textbf{Solution:} In the right triangle $\\triangle ABC$, with $\\angle C=90^\\circ$, $AC=3$, and $BC=4$,\\\\\nby the Pythagorean theorem, we obtain\\\\\n$AB= \\sqrt{AC^{2}+BC^{2}}= 5$.\\\\\n$\\cos A= \\frac{AC}{AB} = \\boxed{\\frac{3}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51174753.png"
  },
  {
    "solution": "\\boxed{\\frac{75}{8}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=ax^2+bx+3$ intersects the x-axis at points $A$ and $B$ (with $A$ to the left of $B$), and intersects the y-axis at point $C$, where $\\tan\\angle CAB=3$. The hyperbola $y=\\frac{k}{x}$ (with $k\\ne 0$) passes through the vertex $D$ of the parabola $y=ax^2+bx+3$, with the x-coordinate of point $D$ being 1. Let point $P$ be a moving point on the parabola, located in the first quadrant. Connect points $BP$ and $CP$. Find the coordinates of point $P$ and the maximum value of the quadrilateral $ABPC$ when it reaches its maximum value.\n",
    "completion": "\\textbf{Solution:} As shown in Figure 1: Connect $BC$, and draw $PE\\perp AB$ through point P, intersecting $BC$ at point E.\\\\\n$\\because AB=4, OC=3$,\\\\\n$\\therefore {S}_{\\triangle ABC}=\\frac{1}{2}AB\\times OC=\\frac{1}{2}\\times 4\\times 3=6$.\\\\\nLet the equation of line BC be $y=kx+b$, substituting $(3, 0)$ and $(0, 3)$ into it gives: $3k+b=0$, $b=3$,\\\\\nsolving this yields $b=3, k=-1$,\\\\\n$\\therefore$ the equation of line BC is $y=-x+3$.\\\\\nLet the coordinates of point P be $\\left(x, -{x}^{2}+2x+3\\right)$, then the coordinates of point E are $\\left(x, -x+3\\right)$.\\\\\n$\\therefore PE=-{x}^{2}+3x$,\\\\\n$\\therefore {S}_{\\triangle PBC}=\\frac{1}{2}PE\\cdot OB=\\frac{1}{2}\\times 3\\times \\left(-{x}^{2}+3x\\right)=-\\frac{3}{2}{x}^{2}+\\frac{9}{2}x$.\\\\\n$\\therefore {S}_{\\text{quadrilateral} ABPC}=-\\frac{3}{2}{x}^{2}+\\frac{9}{2}x+6=-\\frac{3}{2}{\\left(x-\\frac{3}{2}\\right)}^{2}+\\frac{75}{8}$,\\\\\nSubstituting $x=\\frac{3}{2}$ into the equation of the parabola yields: $y=\\frac{15}{4}$,\\\\\n$\\therefore$ The coordinates of point P are $\\left(\\frac{3}{2}, \\frac{15}{4}\\right)$, ${S}_{\\text{quadrilateral ABPC}\\text{ maximum}}=\\boxed{\\frac{75}{8}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53043696_145.png"
  },
  {
    "solution": "\\boxed{\\frac{9}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, $AC=12$, $BC=5$. Point $D$ is outside $\\triangle ABC$, and segments $AD$ and $BD$ are drawn. Point $E$ is the midpoint of segment $BD$, $AD=4$, $\\angle CAD=\\angle CAB$. What is the length of segment $CE$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, extend AD and BC to meet at point F\\\\\n$\\because$ $\\angle ACB=90^\\circ$,\\\\\n$\\therefore$ $\\angle ACF=90^\\circ=\\angle ACB$,\\\\\nAlso, $\\because$ $\\angle CAD=\\angle CAB$ and AC=AC,\\\\\n$\\therefore$ $\\triangle ABC \\cong \\triangle AFC$,\\\\\n$\\therefore$ AF=AB and BC=CF\\\\\nIn $\\triangle ABC$, AC=12 and BC=5,\\\\\n$\\therefore$ $AB=\\sqrt{AC^2+BC^2}=\\sqrt{12^2+5^2}=13$\\\\\n$\\therefore$ AF=13\\\\\n$\\because$ AD=4,\\\\\n$\\therefore$ DF=AF−AD=13−4=9\\\\\n$\\because$ E is the midpoint of BD,\\\\\n$\\therefore$ CE is the midline of $\\triangle BDF$\\\\\n$\\therefore$ $CE=\\frac{1}{2}DF=\\boxed{\\frac{9}{2}}$",
    "image_path": "dataset/mm_math/images/MM_Math/52189286.png"
  },
  {
    "solution": "\\boxed{45}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, building AD is 30m high. There is a distant tower BC. Someone at the bottom of the building A measures the elevation angle to the top of the tower to be $60^\\circ$. After climbing to the top of the building D, the elevation angle to the top of the tower is measured to be $30^\\circ$. What is the height of the tower BC in meters?\n",
    "completion": "\\textbf{Solution:} According to the problem, we have: $BC=\\frac{AC}{\\tan 30^\\circ}=\\sqrt{3}AC$,\\\\\nsince $BE=DE\\tan 30^\\circ=AC\\tan 30^\\circ=\\frac{\\sqrt{3}}{3}AC$,\\\\\ntherefore, the height of the building $AD=BC-BE=\\left(\\sqrt{3}-\\frac{\\sqrt{3}}{3}\\right)AC=30$.\\\\\nThus, we find: $AC=15\\sqrt{3}$.\\\\\nHence, $BC=\\sqrt{3}AC=\\boxed{45}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51403020.png"
  },
  {
    "solution": "\\boxed{16}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the vertices $E$, $G$ of the rectangle $EFGH$ are located on the sides $AD$, $BC$ of the rhombus $ABCD$ respectively, and the vertices $F$, $H$ are on the diagonal $BD$ of the rhombus $ABCD$. Given that $BG=DE$ and $E$ is the midpoint of $AD$, with $FH=4$ cm, what is the perimeter of the rhombus $ABCD$?\n",
    "completion": "\\textbf{Solution:} Connect $EG$. In the rhombus $ABCD$, $AD\\parallel BC$, $AD=BC$.\\\\\nSince $E$ is the midpoint of $AD$,\\\\\nit follows that $AE=ED$,\\\\\nsince $BG=DE$,\\\\\nit follows that $AE\\parallel BG$ and $AE=BG$.\\\\\nTherefore, in quadrilateral $EFGH$, $EG=FH=4$.\\\\\nTherefore, $AB=4$,\\\\\nhence, the perimeter of the rhombus is $\\boxed{16}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50496876_86.png"
  },
  {
    "solution": "\\boxed{\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated in the Cartesian coordinate system, the vertices of $\\triangle ABC$ are respectively at $A(2, 2)$, $B(4, 2)$, and $C(4, 4)$. By taking the origin as the center of similitude, and drawing $\\triangle DEF$ on the opposite side of the origin such that $\\triangle DEF$ is similar to $\\triangle ABC$ with a similarity ratio of $1:2$, what is the length of the line segment DF?\n",
    "completion": "\\textbf{Solution:}\n\nGiven points A(2,2), B(4,2), and C(4,4),\\\\\nhence, AB = 2, and BC = 2,\\\\\nby the Pythagorean theorem, we find that AC = $\\sqrt{AB^{2} + BC^{2}} = 2\\sqrt{2}$,\\\\\nsince a triangle DEF is drawn on the opposite side of the origin with the origin as the center of dilation, such that triangle DEF and triangle ABC are similar with a scale factor of 1:2,\\\\\ntherefore, the length of segment DF is $\\frac{1}{2}AC = \\boxed{\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51445039.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of the circle $\\odot O$. The chord $CD \\perp AB$ at point $E$, $\\angle CDB = 30^\\circ$, and $CD = 2\\sqrt{3}$. Find the length of the diameter $AB$.\n",
    "completion": "\\textbf{Solution:} Since $CD\\perp AB$ and AB is a diameter,\\\\\nwe have $CE=DE=\\sqrt{3}$.\\\\\nSince $\\angle COB=2\\angle CDB=2\\times 30^\\circ=60^\\circ$,\\\\\nwe find $OC=\\frac{\\sqrt{3}}{\\sin 60^\\circ}=2$.\\\\\nTherefore, the diameter $AB=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52872308_103.png"
  },
  {
    "solution": "\\boxed{9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, AB is the diameter of $\\odot O$, C and D are two points on $\\odot O$, $\\angle CDB=30^\\circ$, and BC=4.5, then what is the length of AB?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect $AC$.\\\\\nSince $AB$ is the diameter of circle $O$,\\\\\nit follows that $\\angle ACB=90^\\circ$,\\\\\nSince $\\angle CAB=\\angle CDB=30^\\circ$, and $BC=4.5$,\\\\\nit follows that $AB=2BC=\\boxed{9}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51379599.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the points A, B, C, D, and E are on the same straight line, with point D being the midpoint of segment AB, and point E being the midpoint of segment BC. If the length of segment AC is 12, what is the length of segment DE?\n",
    "completion": "\\textbf{Solution:} Given that $D$ is the midpoint of the line segment $AB$, it follows that $AD=BD$,\\\\\nSince $E$ is the midpoint of the line segment $BC$, it follows that $BE=CE$,\\\\\nGiven that $AC=12$, it follows that $AD+CD=12$, hence $BD+CD=12$,\\\\\nAdditionally, since $BD=2CE+CD$, it follows that $2CE+CD+CD=12$,\\\\\nwhich simplifies to $2(CE+CD)=12$, hence $CE+CD=6$,\\\\\nmeaning the length of the segment $DE$ equals $\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51212520.png"
  },
  {
    "solution": "\\boxed{14}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $AE=BE$, $DE$ is the perpendicular bisector of $AB$, and $F$ is a point on $DE$. If $BF=11\\text{cm}$ and $CF=3\\text{cm}$, what is the length of $AC$ in cm?\n",
    "completion": "\\textbf{Solution:} Since $AE=BE$, and $DE$ is the perpendicular bisector of $AB$,\\\\\nhence, $DE$ is the median of $AB$,\\\\\ntherefore, $DE$ is the perpendicular bisector of $AB$,\\\\\nsince $F$ is a point on $DE$,\\\\\ntherefore, $AF=BF$,\\\\\nthus, $AC = AF + CF = BF + CF$,\\\\\nsince $BF=11\\,cm$, and $CF=3\\,cm$,\\\\\ntherefore, $AC = \\boxed{14}\\,cm$",
    "image_path": "dataset/mm_math/images/MM_Math/52603481.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the diagonals $AC$ and $BD$ of rectangle $ABCD$ intersect at point $O$. A line passing through point $O$ intersects $AD$ and $BC$ at points $E$ and $F$, respectively. Given $AB=2$ and $BC=3$, what is the area of the shaded region in the figure?\n",
    "completion": "\\textbf{Solution:} Since the diagonals AC and BD of rectangle ABCD intersect at point O, \\\\\nit follows that the area of the blank triangle within quadrilateral ABFE is equal to the area of the shaded triangle within quadrilateral EDCF.\\\\\nTherefore, finding the area of the shaded part can be considered as finding the area of quadrilateral ABFE.\\\\\nHence, the area of the shaded part is: $\\frac{2\\times 3}{2}=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53359697.png"
  },
  {
    "solution": "\\boxed{22}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, the perpendicular bisector of AC intersects AC at E and BC at D. The perimeter of $\\triangle ABD$ is $13\\text{cm}$, and $AE=4.5\\text{cm}$. What is the perimeter of $\\triangle ABC$?\n",
    "completion": "\\textbf{Solution:} Since DE is the perpendicular bisector of AC,\\\\\nit follows that AD = CD,\\\\\ntherefore, the perimeter of $\\triangle ABD$ equals to AB + BD + AD = AB + BD + CD = AB + BC,\\\\\nalso, since AE = 4.5cm,\\\\\nit follows that AC = 2AE = 2×4.5 = 9cm,\\\\\ntherefore, the perimeter of $\\triangle ABC$ = 13 + 9 = \\boxed{22}cm.",
    "image_path": "dataset/mm_math/images/MM_Math/52603599.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{7}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that point $E$ is a moving point on the diagonal $AC$ of rectangle $ABCD$, and the vertices $G$ and $H$ of square $EFGH$ are both on side $AD$. If $AB=3$ and $BC=4$, then what is the value of $\\tan\\angle AFE$?\n",
    "completion": "\\textbf{Solution:} Given that $EH\\parallel CD$,\\\\\n$\\triangle AEH \\sim \\triangle ACD$,\\\\\nhence $\\frac{EH}{AH} = \\frac{CD}{AD} = \\frac{3}{4}$,\\\\\nlet $EH = 3a$, then $AH = 4a$,\\\\\nthus, $HG = GF = EH = 3a$,\\\\\nsince $EF\\parallel AD$,\\\\\nthus $\\angle AFE = \\angle FAG$,\\\\\nhence $\\tan\\angle AFE = \\tan\\angle FAG = \\frac{GF}{AG} = \\frac{3a}{4a+3a} = \\boxed{\\frac{3}{7}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51212701.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As the diagram shows, points $A$ and $B$ are separated by a pond and their distance cannot be measured directly. Therefore, Xiao Ming selects a point $C$ on the shore, connects $CA$ and $CB$, and extends them to points $M$ and $N$ respectively, such that $AM=AC$ and $BN=BC$. Given that the length of $MN$ is measured to be $20m$, what is the distance between points $A$ and $B$ in meters?\n",
    "completion": "\\textbf{Solution:} Since $AM=AC$, and $BN=BC$\\\\\ntherefore, A and B are respectively the midpoints of CM and CN,\\\\\ntherefore, AB is the mid-segment of $\\triangle CMN$,\\\\\ntherefore, $MN=2AB$,\\\\\ntherefore, AB= $\\frac{1}{2}MN=\\boxed{10}$ m",
    "image_path": "dataset/mm_math/images/MM_Math/50496567.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that part of the graph of the parabola $y = -x^2 + bx - c$. Find the axis of symmetry and the maximum value of $y$ for the parabola, respectively.\n",
    "completion": "\\textbf{Solution:} Given the function $y = -x^{2} - 2x + 3$,\\\\\nit can be rewritten as $y = -(x + 1)^{2} + 4$,\\\\\ntherefore, the axis of symmetry of the parabola is the line $\\boxed{x = -1}$,\\\\\nthus, the maximum value of $y$ is $\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52693909_98.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, $\\Delta DBE$ is inscribed in $\\odot O$, with $BD$ as the diameter and $DE=EB$. Point $C$ is on $\\odot O$ (not coincident with $D$, $B$, or $E$), $\\angle A=45^\\circ$, and point $A$ is on line $CD$. Connect $AB$. If line $BC$ intersects line $DE$ at point $F$, when $\\frac{DC}{CB}=\\frac{1}{3}$, find the value of $\\frac{BF}{DF}$.\n",
    "completion": "\\textbf{Solution:} Join $DG$, construct $EH \\perp BC$ at point $H$,\\\\\nsince $\\frac{DC}{CB} = \\frac{1}{3}$, let $DC = k$, $CB = 3k$,\\\\\nas $\\triangle ABC$ is an isosceles right-angled triangle, then $BC = AC = 3k$, $AD = 2k$,\\\\\nsince $BD$ is the diameter,\\\\\ntherefore $\\angle DGB = \\angle DGA = 90^\\circ$,\\\\\nsince $\\angle A = 45^\\circ$,\\\\\ntherefore $DG = AD\\sin\\angle A = \\sqrt{2}k$,\\\\\nsince $\\angle ABC = \\angle DBE$, that is $\\angle DBG + \\angle CBD = \\angle CBH + \\angle CBD$,\\\\\ntherefore $\\angle DBG = \\angle EBH$,\\\\\ntherefore $\\triangle DBG \\sim \\triangle EHB$,\\\\\ntherefore $\\frac{DB}{EB} = \\frac{DG}{EH} = \\sqrt{2}$,\\\\\ntherefore $EH = k$,\\\\\nsince $DC \\parallel EH$,\\\\\ntherefore $\\triangle DCF \\sim \\triangle EHF$,\\\\\ntherefore $\\frac{EF}{DF} = \\frac{EH}{DC} = 1$;\\\\\nAs shown in figure 2, join $DG$, construct $EH \\perp BC$ at point $H$,\\\\\nsince $\\frac{DC}{CB} = \\frac{1}{3}$, let $DC = k$, $CB = 3k$,\\\\\nas $\\triangle ABC$ is an isosceles right-angled triangle, then $BC = AC = 3k$, $AD = 4k$,\\\\\nsince $BD$ is the diameter,\\\\\ntherefore $\\angle DGB = \\angle DGA = 90^\\circ$,\\\\\nsince $\\angle A = 45^\\circ$,\\\\\ntherefore $DG = \\sqrt{2}k$,\\\\\nsince $\\angle ABC = \\angle DBE$,\\\\\ntherefore $\\angle DBG = \\angle EBH$,\\\\\ntherefore $\\triangle DBG \\sim \\triangle EHB$,\\\\\ntherefore $\\frac{DB}{EB} = \\frac{DG}{EH} = \\sqrt{2}$,\\\\\ntherefore $EH = 2k$,\\\\\nsince $DC \\parallel EH$,\\\\\ntherefore $\\triangle DCF \\sim \\triangle EHF$,\\\\\ntherefore $\\frac{EF}{DF} = \\frac{EH}{DC} = 2$;\\\\\nIn conclusion, the value of $\\frac{BF}{DF}$ is $\\boxed{1}$ or $\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51369960_131.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{10}}{10}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure (1), $BC$ is the diameter of $\\odot O$, and points $D, F$ are on $\\odot O$. Connect $CD, BF$ and extend them to intersect at point $A$, with $AD=2CD=8$, $AF=3BF$. $\\triangle AFD\\sim \\triangle ACB$. Find: $\\cos\\angle ADF$?\n",
    "completion": "\\textbf{Solution:} Connect $CF$,\\\\\nsince $BC$ is the diameter of $\\odot O$,\\\\\ntherefore $\\angle BFC=90^\\circ$,\\\\\nfrom (1) we know $\\triangle AFD\\sim \\triangle ACB$,\\\\\ntherefore $\\frac{AF}{AC}=\\frac{AD}{AB}$,\\\\\nsince $AD=2CD=8$,\\\\\ntherefore $CD=4$,\\\\\nsince $AF=3BF$,\\\\\ntherefore $AB=AF+BF=3BF+BF=4BF$,\\\\\nthus $\\frac{3BF}{12}=\\frac{8}{4BF}$,\\\\\ntherefore $BF=2\\sqrt{2}$,\\\\\ntherefore $AF=6\\sqrt{2}$,\\\\\nin $Rt\\triangle ACF$, $CF=\\sqrt{AC^{2}-AF^{2}}=\\sqrt{12^{2}-(6\\sqrt{2})^{2}}=6\\sqrt{2}$,\\\\\nin $Rt\\triangle BCF$, $BC=\\sqrt{BF^{2}+CF^{2}}=\\sqrt{(2\\sqrt{2})^{2}+(6\\sqrt{2})^{2}}=4\\sqrt{5}$,\\\\\ntherefore $\\cos\\angle ADF=\\cos\\angle B=\\frac{BF}{BC}=\\frac{2\\sqrt{2}}{4\\sqrt{5}}=\\boxed{\\frac{\\sqrt{10}}{10}}$",
    "image_path": "dataset/mm_math/images/MM_Math/53148615_112.png"
  },
  {
    "solution": "\\boxed{20}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle B= 90^\\circ$, $AB=2$, and $BC=4$, the quadrilateral $ADEC$ is a square. What is the area of the square $ADEC$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle B= 90^\\circ$, AB=2, BC=4,\\\\\nthus $AC=\\sqrt{AB^{2}+BC^{2}}$\\\\\n$=\\sqrt{2^{2}+4^{2}}$\\\\\n$=2\\sqrt{5}$\\\\\nTherefore, the area of the square ADEC is $\\boxed{20}$",
    "image_path": "dataset/mm_math/images/MM_Math/53728740.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ and $AC$ are the sides of an inscribed square and an inscribed equilateral triangle in the circle $\\odot O$, respectively, and $BC$ is one side of an inscribed regular $n$-sided polygon in the circle. What is the value of $n$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw lines AO, BO, and CO.\\\\\nSince AB and AC are respectively one side of a square and an equilateral triangle inscribed in $\\odot$O,\\\\\nit follows that $ \\angle AOB=\\frac{360^\\circ}{4}=90^\\circ$ and $ \\angle AOC=\\frac{360^\\circ}{3}=120^\\circ$,\\\\\ntherefore $ \\angle BOC=30^\\circ$,\\\\\nhence $ n=\\frac{360^\\circ}{30^\\circ}=\\boxed{12}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53066236.png"
  },
  {
    "solution": "\\boxed{square}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $M$ is the midpoint of side $AD$, and $P$ is a moving point on side $BC$, such that $PE\\perp MC$, $PF\\perp BM$, with the perpendicular feet being $E$ and $F$, respectively. If quadrilateral $PEMF$ is a rectangle, when does point $P$ move to a position that makes quadrilateral $PEMF$ a square? Prove your conclusion.\n",
    "completion": "\\textbf{Solution:} When point $P$ moves to the midpoint of side $BC$, the rectangle $PEMF$ becomes a square, at which point $BP=CP$. \\\\\n\\textit{Proof:} Since quadrilateral $PEMF$ is a rectangle, \\\\\n$\\therefore$ $\\angle PFM=\\angle PFB=\\angle PEC=90^\\circ$. \\\\\nIn $\\triangle BFP$ and $\\triangle CEP$, $\\left\\{\\begin{array}{l}\\angle FBP=\\angle ECP=45^\\circ \\\\ \\angle PFB=\\angle PEC \\\\ BP=CP \\end{array}\\right.$, \\\\\n$\\therefore$ $\\triangle BFP\\cong \\triangle CEP$ (AAS), \\\\\n$\\therefore$ $PF=PE$. \\\\\nMoreover, since quadrilateral $PEMF$ is a rectangle, \\\\\n$\\therefore$ rectangular $PEMF$ is a square, \\\\\nthat is, when point $P$ moves to the midpoint of $BC$, quadrilateral $PEMF$ is a \\boxed{square}.",
    "image_path": "dataset/mm_math/images/MM_Math/52076067_94.png"
  },
  {
    "solution": "\\boxed{20}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB=24$, $C$ is the midpoint of $AB$, point $D$ is on the line segment $AC$, and $AD:CB=1:3$, what is the length of $DB$?\n",
    "completion": "\\textbf{Solution:} Given that $AB=24$ and point C is the midpoint of AB,\\\\\nit follows that $BC= \\frac{1}{2} AB= \\frac{1}{2} \\times 24=12$,\\\\\nsince $AD\\colon CB=1\\colon 3$,\\\\\nthus $AD= \\frac{1}{3} \\times 12=4$,\\\\\ntherefore $DB=AB−AD=24−4=\\boxed{20}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51434642.png"
  },
  {
    "solution": "\\boxed{F\\left(6, \\frac{3}{2}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, the equation of line $AB$ is $y=kx+6$, and the coordinate of point $D$ is $(8, 0)$. The symmetric point of point $O$ with respect to line $AB$ is point $C$, which lies on line $AD$. As shown in Figure 2, if $OC$ intersects $AB$ at point $E$, is there a point $F$ on segment $AD$ such that the area of $\\triangle ABC$ is equal to the area of $\\triangle AEF$? Determine the coordinates of point $F$.\n",
    "completion": "Solution: From (1), we have $BC=OB=3$, $CD=4$, and $BD=5$. Thus, $\\frac{1}{2}\\times 3\\times 4=\\frac{1}{2}\\times 5\\times y_c$. Solving this gives: $y_c=\\frac{12}{5}$. Substituting $y_c=\\frac{12}{5}$ into $y=-\\frac{3}{4}x+6$ yields: $x_c=\\frac{24}{5}$. Therefore, $C\\left(\\frac{24}{5}, \\frac{12}{5}\\right)$. The equation of line $AD$ is: $y=-\\frac{3}{4}x+6$. $\\therefore$ The equation of line $OC$ is: $y=\\frac{1}{2}x$. $\\because S_{\\triangle ABC}=S_{\\triangle AEF}$, $\\therefore S_{\\triangle BEC}=S_{\\triangle ECF}$, $\\therefore BF\\parallel OC$. Let the equation of line $BF$ be: $y=\\frac{1}{2}x+b$. Point $B(3, 0)$ lies on line $BF$, therefore $\\frac{1}{2}\\times 3+b=0$, hence $b=-\\frac{3}{2}$. Thus, the equation of line $BF$ is: $y=\\frac{1}{2}x-\\frac{3}{2}$. Solving the system of equations $\\begin{cases}y=\\frac{1}{2}x-\\frac{3}{2}\\\\ y=-\\frac{3}{4}x+6\\end{cases}$, we get $\\begin{cases}x=6\\\\ y=\\frac{3}{2}\\end{cases}$. Therefore, point $\\boxed{F\\left(6, \\frac{3}{2}\\right)}$ exists.",
    "image_path": "dataset/mm_math/images/MM_Math/50560467_66.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ABC=90^\\circ$, $CD$ bisects $\\angle ACB$, through point B draw $BD\\perp CD$ with the foot of the perpendicular at point D, connect AD. If $AB=3$ and $BC=4$, what is the area of $\\triangle ABD$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, extend $BD$ to intersect $AC$ at point $E$. In the right triangle $\\triangle ABC$, $AB=3$, $BC=4$,\\\\\n$\\therefore AC=\\sqrt{AB^{2}+BC^{2}}=\\sqrt{3^{2}+4^{2}}=5$,\\\\\n$\\because CD$ bisects $\\angle ACB$, and $BD\\perp CD$,\\\\\n$\\therefore BC=CE=4$, $AE=AC-CE=1$, $BD=ED$,\\\\\n$\\therefore \\frac{S_{\\triangle ABE}}{S_{\\triangle ABC}}=\\frac{AE}{AC}=\\frac{1}{5}$,\\\\\n$\\therefore S_{\\triangle ABE}=\\frac{1}{5}S_{\\triangle ABC}=\\frac{1}{5}\\times\\frac{1}{2}\\times3\\times4=\\frac{6}{5}$,\\\\\n$\\therefore S_{\\triangle ABD}=\\frac{1}{2}S_{\\triangle ABE}=\\frac{1}{2}\\times\\frac{6}{5}=\\boxed{\\frac{3}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52807549.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the isosceles right-angled $\\triangle ABC$, where $AC=BC$ and $\\angle ACB=90^\\circ$, point D is a moving point on side $AC$ (not coinciding with A or C). A line is drawn from point A perpendicular to $BD$ at point E, and $AE$ is extended to intersect the extension of $BC$ at point F. $CE$ is connected. Then, what is the measure of $\\angle BEC$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, draw $CH \\perp AF$ at $H$, and $CG \\perp BE$ at $G$,\\\\\n$\\therefore \\angle AHC=\\angle BGC=90^\\circ$,\\\\\n$\\because \\angle ACB=90^\\circ$, and $AF\\perp BE$,\\\\\n$\\therefore \\angle AEB=\\angle BCD=\\angle BEF=90^\\circ$,\\\\\nAlso, since $\\angle ADE=\\angle BDC$,\\\\\n$\\therefore \\angle CAH=\\angle CBG$,\\\\\nAlso, since $AC=BC$,\\\\\n$\\therefore \\triangle AHC\\cong \\triangle BCG$ (AAS),\\\\\n$\\therefore CH=CG$,\\\\\nSince $CH\\perp EF$, and $CG\\perp BE$,\\\\\n$\\therefore CE$ bisects $\\angle BEF$,\\\\\n$\\therefore \\angle BEC= \\frac{1}{2}\\angle BEF=\\boxed{45^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52653732.png"
  },
  {
    "solution": "\\boxed{y=-x^{2}+3x+4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, the parabola $y=-x^{2}+bx+c$ passes through points $A(-1,0)$ and $B(4,0)$ and intersects the $y$-axis at point $C$. Connect points $B$ and $C$. Point $P$ is a moving point on the parabola. A perpendicular line $l$ to the $x$-axis is drawn from point $P$, intersects line $BC$ at point $G$, and intersects the $x$-axis at point $E$. What is the expression of the parabola?\n",
    "completion": "\\textbf{Solution:} To solve, substitute the coordinates of points A (-1,0) and B (4,0) into the expression of the function, we obtain:\n\\[\n\\begin{cases}\n-1-b+c=0, \\\\\n-16+4b+c=0.\n\\end{cases}\n\\]\nRearranging gives:\n\\[\n\\begin{cases}\n-b+c=1, \\\\\n4b+c=16,\n\\end{cases}\n\\]\nSolving for the variables, we get:\n\\[\n\\begin{cases}\nb=3, \\\\\nc=4,\n\\end{cases}\n\\]\n$\\therefore$The equation of the parabola is $y=-x^{2}+3x+4$; thus, the answer is $\\boxed{y=-x^{2}+3x+4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51458091_166.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ is an isosceles right-angled triangle, with $\\angle ACB=90^\\circ$, $AC=BC=6$. Point $D$ is on segment $BC$, and $E$ is a point on segment $AD$. Now taking $CE$ as the right angle side, with $C$ as the right angle vertex, an isosceles right-angled triangle $\\triangle ECF$ is constructed below $CE$, and $BF$ is connected. $\\triangle AEC \\cong \\triangle BFC$. When points $A$, $E$, and $F$ are collinear, as shown in figure 2, if $AF=8$, what is the length of $BF$?\n",
    "completion": "Solution: As shown in Figure 2,\\\\\n$\\because CA=CB=6$, $\\angle ACB=90^\\circ$,\\\\\n$\\therefore AB=6\\sqrt{2}$,\\\\\n$\\because \\triangle ACE \\cong \\triangle BCF$,\\\\\n$\\therefore \\angle CAD=\\angle DBF$,\\\\\n$\\because \\angle ADC=\\angle BDF$,\\\\\n$\\therefore \\angle ACD=\\angle DFB=90^\\circ$,\\\\\n$\\therefore BF=\\sqrt{AB^{2}-AF^{2}}=\\sqrt{(6\\sqrt{2})^{2}-8^{2}}=\\boxed{2\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55403719_77.png"
  },
  {
    "solution": "\\boxed{\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the rectangle $ABCD$ is divided into 5 congruent rectangles. If all these 5 rectangles are similar to rectangle $ABCD$, what is the value of $AD\\colon AB$?\n",
    "completion": "\\textbf{Solution:} Let AE = a, \\\\\n$\\because$ the five small rectangles are congruent, \\\\\n$\\therefore$ AD = 5AE = 5a, \\\\\n$\\because$ each small rectangle is similar to rectangle ABCD, \\\\\n$\\therefore$  $\\frac{AD}{AB}$ =  $\\frac{AB}{AE}$, \\\\\n$\\therefore$AB$^{2}$=AD$\\cdot$AE=5AE$^{2}$=5a$^{2}$, \\\\\nAB =  $\\sqrt{5}$ a, \\\\\n$\\therefore$AD : AB = 5a : $\\sqrt{5}$ a = $\\boxed{\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51458314.png"
  },
  {
    "solution": "\\boxed{\\sqrt{13}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In rectangle $ABCD$, $AB = 3$, $BC = 2\\sqrt{10}$. Point $E$ is on side $BC$. Connect $DE$. Fold $\\triangle DEC$ along $DE$ to get $\\triangle DEC'$. $C'E$ intersects $AD$ at point $F$. Connect $AC'$. If point $F$ is the midpoint of $AD$, what is the length of $AC'$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $C'H \\perp AD$ at point H,\n\nsince point F is the midpoint of AD, and AD = BC = $2\\sqrt{10}$,\n\nthus AF = DF = $\\sqrt{10}$,\n\nsince $\\triangle DEC$ is folded along DE,\n\nthus CD = C'D = 3, $\\angle C = \\angle EC'D = 90^\\circ$,\n\nin $\\triangle DC'F$, C'F = $\\sqrt{DF^{2} - C'D^{2}} = \\boxed{1}$,\n\nsince $S_{\\triangle C'F} = \\frac{1}{2} \\times DF \\times C'H = \\frac{1}{2} \\times C'F \\times C'D$,\n\nthus $\\sqrt{10} \\times C'H = 1 \\times 3$,\n\nthus C'H = $\\frac{3\\sqrt{10}}{10}$,\n\nthus FH = $\\sqrt{C'F^{2} - C'H^{2}} = \\frac{\\sqrt{10}}{10}$,\n\nthus AH = AF + FH = $\\frac{11\\sqrt{10}}{10}$,\n\nin $\\triangle AC'H$, AC'$ = \\sqrt{AH^{2} + C'H^{2}} = \\boxed{\\sqrt{13}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51540105.png"
  },
  {
    "solution": "\\boxed{\\frac{25}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the rectangle \\(ABCD\\), where \\(AB=6\\) and \\(BC=8\\). The rectangle is folded along \\(BD\\) so that point \\(A\\) coincides with point \\(E\\) (as shown in Figure 1). \\(DE\\) and \\(BC\\) intersect at point \\(F\\), and \\(ABDF\\) forms an isosceles triangle. Determine the length of \\(BF\\).\n",
    "completion": "\\textbf{Solution:} Given that when the rectangle is folded along BD, making point A coincide with point E,\\\\\nthus AD$\\parallel$BC and $\\angle$ADB = $\\angle$EDB,\\\\\nhence $\\angle$ADB = $\\angle$DBF = $\\angle$EDB,\\\\\ntherefore BF=DF,\\\\\nimplying that $\\triangle$BDF is an isosceles triangle;\\\\\nLet BF=DF=x, then CF=8−x,\\\\\nin $\\triangle$DCF, we have CD$^{2}$+CF$^{2}$=DF$^{2}$, that is $6^{2}$+(8−x)$^{2}$=x$^{2}$,\\\\\nsolving this, we get: x= \\boxed{\\frac{25}{4}},\\\\\ntherefore, the length of BF is \\boxed{\\frac{25}{4}}.",
    "image_path": "dataset/mm_math/images/MM_Math/51614698_97.png"
  },
  {
    "solution": "\\boxed{\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in square $ABCD$, where $AB=\\sqrt{5}$, $E$ is a point inside square $ABCD$, $DE=AB$, and $\\angle EDC=\\alpha$ $\\left(0^\\circ<\\alpha<90^\\circ\\right)$. Connect $CE$ and $AE$. Through point $D$, draw $DF\\perp AE$, where $F$ is the foot of the perpendicular, and it intersects the extension of $CE$ at point $G$. Connect $AG$. When $GF=1$, what is the length of $CE$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, connect $AC$,\n\nsince the quadrilateral $ABCD$ is a square,\n\ntherefore $AC=\\sqrt{2}AB=\\sqrt{10}$,\n\nsince $\\triangle AEG$ is an isosceles right triangle and $GF\\perp AE$,\n\ntherefore $GF=AF=EF=1$,\n\ntherefore $AG=GE=\\sqrt{2}$,\n\nsince $AC^{2}=AG^{2}+GC^{2}$,\n\ntherefore $10=2+(EC+\\sqrt{2})^{2}$,\n\ntherefore $EC=\\boxed{\\sqrt{2}}$ (negative values have been disregarded).",
    "image_path": "dataset/mm_math/images/MM_Math/50481469_139.png"
  },
  {
    "solution": "\\boxed{45^\\circ+\\frac{1}{4}x^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=BC$, $AD \\perp BC$ at point $D$, $CE$ bisects $\\angle ACB$, intersects $AB$ at point $E$, and intersects $AD$ at point $P$. If $\\angle B=x^\\circ$, what is the measure of $\\angle APE$?\n",
    "completion": "\\textbf{Solution:} Since $AB=BC$ and $\\angle B=x^\\circ$,\\\\\nit follows that $\\angle BAC=\\angle ACB$,\\\\\nthus $\\angle ACB= \\frac{1}{2}(180^\\circ-\\angle B)= \\frac{1}{2}(180^\\circ-x^\\circ)=90^\\circ- \\frac{1}{2}x^\\circ$;\\\\\nSince $CE$ bisects $\\angle ACB$,\\\\\nthen $\\angle DCP= \\frac{1}{2}\\angle ACB= \\frac{1}{2}(90^\\circ- \\frac{1}{2}x^\\circ)=45^\\circ- \\frac{1}{4}x^\\circ$;\\\\\nSince $AD \\perp BC$,\\\\\nthen $\\angle ADC=90^\\circ$,\\\\\nthus $\\angle APE=\\angle DPC=90^\\circ-\\angle DCP=90^\\circ-(45^\\circ- \\frac{1}{4}x^\\circ)=45^\\circ+ \\frac{1}{4}x^\\circ.$\\\\\nTherefore, the answer is: $\\boxed{45^\\circ+\\frac{1}{4}x^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51419892.png"
  },
  {
    "solution": "\\boxed{3\\sqrt{5}-3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In $\\triangle ABC$, $AB=AC=6$, $\\angle BAC=108^\\circ$, point $D$ is on side $BC$, $\\angle BAD=36^\\circ$. $\\triangle BAD\\sim \\triangle BCA$; what is the length of $AD$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle B=\\angle BAD$, it follows that $BD=AD=x$. Because $\\angle ADC=\\angle B+\\angle BAD=72^\\circ$ and $\\angle DAC=108^\\circ-36^\\circ=72^\\circ$, we have $\\angle ADC=\\angle DAC$, thus $DC=AC=6$. Since $\\triangle BAD\\sim \\triangle BCA$, we have $\\frac{AD}{AB}=\\frac{AB}{AC}$, that is $\\frac{x}{6}=\\frac{6}{x+6}$. Solving this, we get: $x=3\\sqrt{5}-3$. Therefore, $AD=\\boxed{3\\sqrt{5}-3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53059675_156.png"
  },
  {
    "solution": "\\boxed{\\left(-1, -3\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system, the line $l_1\\colon y=-2x+1$ intersects with the y-axis at point A, and the line $l_2$ intersects with the y-axis at point B(0, -2) and intersects with line $l_1$ at point C, where the y-coordinate of point C is -1. Point D is any point on line $l_2$. A perpendicular line is drawn from point D to the x-axis, intersecting line $l_1$ at point E. When $DE = 2AB$, find the coordinates of point D?\n",
    "completion": "\\textbf{Solution:} Let $x=0$, we get $y=-2\\times 0+1=1$, \\\\\n$\\therefore$ the coordinates of point $A$ are $\\left(0, 1\\right)$, \\\\\n$\\therefore$ $AB=3$, let the abscissa of point $D$ be $m$, then the coordinates of point $D$ are $\\left(m, m-2\\right)$, \\\\\n$\\because$ $DE$ is parallel to the y-axis, \\\\\n$\\therefore$ the coordinates of point $E$ are $\\left(m, -2m+1\\right)$, \\\\\n$\\therefore$ $DE=\\left|\\left(m-2\\right)-\\left(-2m+1\\right)\\right|=\\left|3m-3\\right|$, \\\\\n$\\because$ $DE=2AB=6$, \\\\\n$\\therefore$ $\\left|3m-3\\right|=6$, solving gives $m=3$ or $m=-1$, \\\\\nWhen $m=3$, the coordinates of point $D$ are $\\left(3, 1\\right)$, \\\\\nWhen $m=-1$, the coordinates of point $D$ are $\\left(-1, -3\\right)$. \\\\\nTo summarize: The coordinates of point $D$ are $\\boxed{\\left(3, 1\\right)}$ or $\\boxed{\\left(-1, -3\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50555108_102.png"
  },
  {
    "solution": "\\boxed{\\frac{13}{6}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, $AC=12$, and $BC=5$, what is the distance between the centroid $P$ and the circumcentre $Q$ of $\\triangle ABC$?\n",
    "completion": "Solution: According to the given information, points C, P, and Q are collinear.\n\nIn $\\triangle ABC$, $\\angle C=90^\\circ$, $AC=12$, $BC=5$,\n\n$\\therefore AB=\\sqrt{AC^2+BC^2}=\\sqrt{12^2+5^2}=13$,\n\n$\\because$ the circumcenter of $\\triangle ABC$ is $Q$,\n\n$\\therefore Q$ is the midpoint of the hypotenuse $AB$,\n\n$\\therefore CQ=\\frac{1}{2}AB=\\frac{13}{2}$,\n\n$\\because$ the centroid of $\\triangle ABC$ is $P$,\n\n$\\therefore PQ=\\frac{1}{3}CQ=\\boxed{\\frac{13}{6}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53102366.png"
  },
  {
    "solution": "\\boxed{20}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the diagonals of the rhombus $ABCD$ are $AC=6$ and $BD=8$. What is the perimeter of the rhombus $ABCD$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure:\\\\\nLet the intersection point of AC and BD be O,\\\\\nsince in rhombus ABCD, AC=6, BD=8,\\\\\nthus OA= $\\frac{1}{2}$ AC=3, OB= $\\frac{1}{2}$ BD=4, and AC$\\perp$BD,\\\\\ntherefore, $AB=\\sqrt{OA^{2}+OB^{2}}=5$,\\\\\nthus, the perimeter of rhombus ABCD=4$\\times$5=\\boxed{20},",
    "image_path": "dataset/mm_math/images/MM_Math/52050219.png"
  },
  {
    "solution": "\\boxed{15^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, an equilateral triangle $ADE$ is constructed on the exterior side of square $ABCD$. What is the measure of $\\angle AEB$?\n",
    "completion": "\\textbf{Solution}: Let an equilateral triangle $\\triangle ADE$ be constructed on the exterior of square ABCD.\\\\\nTherefore, $AB=AD$, $AD=AE$, $\\angle BAD=90^\\circ$, $\\angle DAE=60^\\circ$,\\\\\nThus, $AB=AE$, $\\angle BAE=150^\\circ$,\\\\\nHence, $\\angle ABE=\\angle AEB=\\boxed{15^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52386497.png"
  },
  {
    "solution": "\\boxed{25^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $E$ is a point on the diagonal $AC$ of the square $ABCD$. Line $DE$ is drawn, and through point $E$, line $EF\\perp DE$ is drawn, which intersects the ray $BC$ at point $F$. A rectangle $DEFG$ is constructed with $DE$ and $EF$ as adjacent sides, and $CG$ is connected. When the angle between line segment $DE$ and one side of square $ABCD$ is $25^\\circ$, directly write down the degree measure of $\\angle EFC$.\n",
    "completion": "\\textbf{Solution:} We have $\\angle EFC = \\boxed{115^\\circ}$ or $\\boxed{25^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52354937_89.png"
  },
  {
    "solution": "\\boxed{(1, -2)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=ax^{2}+bx-3\\ (a\\ne 0)$ intersects the x-axis at points $A(-1, 0)$ and $B(3, 0)$, and intersects the y-axis at point C. Find a point $Q$ on the axis of symmetry such that the perimeter of $\\triangle ACQ$ is minimized. Determine the coordinates of point $Q$.\n",
    "completion": "\\textbf{Solution:}\n\nLet the line $BC$ intersect the axis of symmetry at point $Q$,\\\\\nsince $y=x^{2}-2x-3=(x-1)^{2}-4$,\\\\\nthus, the axis of symmetry of the parabola is the line $x=1$,\\\\\nsince points A and B are symmetric about the axis $x=1$,\\\\\nthus $AQ=BQ$,\\\\\nthus, $AC+AQ+CQ=AC+CQ+BQ\\ge AC+BC$,\\\\\nwhen points C, B, and Q are collinear, the perimeter of $\\triangle ACQ$ is minimized,\\\\\nsince $C(0, -3)$, $B(3, 0)$,\\\\\nlet the equation of line $BC$ be $y=kx+b$,\\\\\nthus $\\begin{cases}b=-3 \\\\ 3k+b=0\\end{cases}$, solving gives $\\begin{cases}k=1 \\\\ b=-3\\end{cases}$,\\\\\nthus, the equation of line $BC$ is $y=x-3$,\\\\\nthus, the coordinates of $Q\\left(1, -2\\right)$ are $\\boxed{(1, -2)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52796987_46.png"
  },
  {
    "solution": "\\boxed{\\frac{6}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, within $\\triangle ABC$, $AB=6$, $AC=5$, $BC=3$. Point $M$ is on $AB$ with $AM=2$. A line $MN$ intersects $\\triangle ABC$ at point $M$, and it satisfies $\\angle AMN = \\angle C$. What is the length of $MN$?\n",
    "completion": "\\textit{Solution}: In triangles $\\triangle AMN$ and $\\triangle ACB$,\\\\\n$\\because$ $\\left\\{\\begin{array}{l}\n\\angle AMN=\\angle C \\\\\n\\angle MAN=\\angle CAB\n\\end{array}\\right.$ \\\\\n$\\therefore$ $\\triangle AMN\\sim \\triangle ACB$ \\\\\n$\\therefore$ $\\frac{MN}{BC}=\\frac{AM}{AC}$ \\\\\n$\\therefore$ $\\frac{MN}{3}=\\frac{2}{5}$ \\\\\n$\\therefore$ $MN=\\boxed{\\frac{6}{5}}$",
    "image_path": "dataset/mm_math/images/MM_Math/53065525.png"
  },
  {
    "solution": "\\boxed{3\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, it is known that $AE\\perp BD$ at $E$, $\\angle DBC=30^\\circ$, and $BE=3\\text{cm}$. What is the length of $AE$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral ABCD is a rectangle,\\\\\n$\\therefore$ $\\angle ABC = \\angle BAD = 90^\\circ$,\\\\\nSince $\\angle DBC = 30^\\circ$,\\\\\n$\\therefore$ $\\angle ABE = 60^\\circ$,\\\\\nGiven that AE$\\perp$BD,\\\\\n$\\therefore$ $\\angle BAE = 30^\\circ$,\\\\\n$\\therefore$ $AE = \\sqrt{3}BE = \\boxed{3\\sqrt{3}}\\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53333441.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\angle ACB = 90^\\circ$, D is the midpoint of AB, and point E is the midpoint of AF, such that E, C, D are collinear, and $CE = \\frac{1}{4}CD$. If $BF = 10$, what is the length of AB?\n",
    "completion": "\\textbf{Solution:} Since point D is the midpoint of AB and point E is the midpoint of AF, and BF = 10,\\\\\nit follows that $DE = \\frac{1}{2}BF = 5$,\\\\\nSince $CE = \\frac{1}{4}CD$,\\\\\nit then follows that $CD = \\frac{4}{5}DE = 4$,\\\\\nSince $\\angle ACB = 90^\\circ$ and point D is the midpoint of AB,\\\\\nthus $AB = 2CD = \\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51863890.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{2}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, E is a point on the diagonal BD of a square ABCD with a side length of 1, and $BE=BC$. Let P be any point on CE, and let PQ be perpendicular to BC at point Q, and PR be perpendicular to BE at point R. What is the value of $PQ+PR$?\n",
    "completion": "\\textbf{Solution:} Draw BP and construct CM $\\perp$ BD through C, \\\\ \n$\\because$BC=BE, \\\\ \n$\\therefore$Area of $\\triangle BCE$ = Area of $\\triangle BPE$ + Area of $\\triangle BPC$ \\\\ \n= BC $\\times$ PQ $\\times \\frac{1}{2}$ + BE $\\times$ PR $\\times \\frac{1}{2}$ \\\\ \n= BE $\\times$ (PQ+PR) $\\times \\frac{1}{2}$ \\\\ \n= BE $\\times$ CM $\\times \\frac{1}{2}$, \\\\ \n$\\therefore$ PQ+PR=CM, \\\\ \n$\\because$ BE=BC=1, and BD is the diagonal of the square ABCD, \\\\ \n$\\therefore$ BD=$\\sqrt{2}$BC=$\\sqrt{2}$, \\\\ \n$\\because$ BC=CD, and CM$\\perp$ BD, \\\\ \n$\\therefore$ M is the midpoint of BD, \\\\ \n$\\because$ $\\triangle$BDC is a right triangle, \\\\ \n$\\therefore$ CM=$\\frac{1}{2}$BD=$\\frac{\\sqrt{2}}{2}$, \\\\ \nthus, the value of PQ+PR is $\\boxed{\\frac{\\sqrt{2}}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52344746.png"
  },
  {
    "solution": "\\boxed{2\\text{cm}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in triangle \\(ABC\\), \\(AB=10\\text{cm}\\), \\(AC=6\\text{cm}\\), \\(D\\) is the midpoint of \\(BC\\), and \\(E\\) is a point on side \\(AB\\). If the perimeter of triangle \\(BDE\\) is equal to the perimeter of quadrilateral \\(ACDE\\), find the length of segment \\(AE\\).\n",
    "completion": "\\textbf{Solution:} Given the diagram, we have:\\\\\nThe perimeter of triangle $BDE = BE + BD + DE$, the perimeter of quadrilateral $ACDE = AE + AC + DC + DE$,\\\\\nAnd since the perimeter of triangle $BDE$ equals the perimeter of quadrilateral $ACDE$, and $D$ is the midpoint of $BC$,\\\\\nTherefore, $BD = DC$, $BE + BD + DE = AE + AC + DC + DE$,\\\\\nTherefore, $BE = AE + AC$,\\\\\nAnd since $AB= 10\\text{cm}$, $AC = 6\\text{cm}$, $BE = AB - AE$,\\\\\nTherefore, $AE + AC = AB - AE$,\\\\\nTherefore, $10 - AE = AE + 6$,\\\\\nTherefore, $AE = \\boxed{2\\text{cm}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53066781.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, in the quadrilateral $ABCD$, $\\angle ABD = \\angle BCD = 90^\\circ$, and $DB$ bisects $\\angle ADC$. If $CD = 6$ and $AD = 8$, what is the length of $BD$?\n",
    "completion": "\\textbf{Solution:} Given that $DB$ bisects $\\angle ADC$,\\\\\nthus $\\angle ADB=\\angle BDC$,\\\\\nsince $\\angle ABD=\\angle BCD=90^\\circ$,\\\\\nit follows that $\\triangle ADB\\sim \\triangle BDC$,\\\\\nhence $\\frac{AD}{BD}=\\frac{BD}{CD}$,\\\\\ngiven that $CD=6$, $AD=8$,\\\\\nthus $\\frac{8}{BD}=\\frac{BD}{6}$,\\\\\nsolving yields: $BD=\\boxed{4\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51402165_149.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $l_1$ passing through point $B(2, 0)$ intersects with the line $l_2: y=2x+8$ at point $P(-1, n)$. What is the value of $n$?\n",
    "completion": "\\textbf{Solution:} Substitute $P(-1, n)$ into $y=2x+8$ to get $n=2\\times (-1)+8$. Solving this, we find $n=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50553735_104.png"
  },
  {
    "solution": "\\boxed{(-2, 3)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $A(-1, 6)$ is a point on the hyperbola $y=\\frac{k}{x}$ $(x<0)$. Let $P$ be a point on the positive half of the $y$-axis. Rotating point $A$ counterclockwise by $90^\\circ$ around point $P$ precisely lands it on another point $B$ on the hyperbola. What are the coordinates of point $B$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $AM \\perp OM$ at point M through point A, and draw $BN \\perp OM$ at point N through point B, \\\\\n$\\therefore \\angle PMA = \\angle BNP = 90^\\circ$, \\\\\n$\\because$ Rotating point A $90^\\circ$ counterclockwise around point P precisely lands on another point B on the hyperbola, \\\\\n$\\therefore PA=PB$, $\\angle APB=90^\\circ$, \\\\\n$\\therefore \\angle APM + \\angle BPN = \\angle BPN + \\angle PBN = 90^\\circ$, \\\\\n$\\therefore \\angle APM = \\angle PBN$, \\\\\n$\\therefore \\triangle APM \\cong \\triangle BNP$ (AAS), \\\\\n$\\therefore AM=PN$, $PM=BN$, \\\\\n$\\because$ Point A $(-1,6)$ lies on the hyperbola $y=\\frac{k}{x}$ ($x>0$), \\\\\n$\\therefore k = -1 \\times 6 = -6$, $AM=PN=1$, \\\\\nLet point P be $(m,0)$, then $OP=m$, \\\\\n$\\therefore PM=BN=6-m$, $ON=OP-PN=m-1$, \\\\\n$\\therefore$ The coordinates of point B are $(m-6, m-1)$, \\\\\n$\\therefore k = (m-6) \\cdot (m-1) = -6$, solving for $m$ gives $m=3$ or $m=4$, \\\\\n$\\therefore$ The coordinates of point B are $\\boxed{(-3, 2)}$ or $\\boxed{(-2, 3)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51445135.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{10}}{10}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AC=6$, $BC=8$, and $D$, $E$ are on $CA$, $CB$ respectively, with point $F$ inside $\\triangle ABC$. If quadrilateral $CDFE$ is a square with a side length of $2$, then what is $\\sin\\angle FBA$?\n",
    "completion": "\\textbf{Solution:} Construct $FG \\perp AB$ at G, and connect AF. \\\\\nSince quadrilateral CDFE is a square with side length 2, \\\\\nthus $CD = CE = DF = EF = 2$, $\\angle C = \\angle ADF = 90^\\circ$, \\\\\nsince $AC = 6$, $BC = 8$, \\\\\nthus $AD = 4$, $BE = 6$, \\\\\ntherefore $AB = \\sqrt{AC^{2} + BC^{2}} = 10$, $AF = \\sqrt{AD^{2} + DF^{2}} = 2\\sqrt{5}$, $BF = \\sqrt{BE^{2} + EF^{2}} = 2\\sqrt{10}$, \\\\\nlet $BG = x$, \\\\\nsince $FG^{2} = AF^{2} - AG^{2} = BF^{2} - BG^{2}$, \\\\\nthus $(2\\sqrt{5})^{2} - (10 - x)^{2} = (2\\sqrt{10})^{2} - x^{2}$, solving gives: $x = 6$, \\\\\ntherefore $FG = \\sqrt{BF^{2} - BG^{2}} = 2$, \\\\\ntherefore $\\sin\\angle FBA = \\frac{FG}{BF} = \\boxed{\\frac{\\sqrt{10}}{10}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51445412.png"
  },
  {
    "solution": "\\boxed{13}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the cross-section of a certain dam of a reservoir is a trapezoid $ABCD$, with the height of the dam $DE = 5m$, and the slope ratio of slope $BC$ is $5\\colon 12$. What is the length of slope $BC$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, draw line CF$\\perp$AB through point C, with F being the foot of the perpendicular. Thus,\\\\\n$\\because$ Dam height $DE=5m$, and CF$\\perp$AB,\\\\\n$\\therefore$DE=CF=5cm\\\\\nMoreover, the slope ratio of slope $BC$ is $5:12$\\\\\n$\\therefore$BF=12cm,\\\\\nIn $\\triangle BCF$\\\\\n$BC= \\sqrt{CF^{2}+BF^{2}}$\\\\\n$= \\sqrt{5^{2}+12^{2}}$\\\\\n$=\\boxed{13}$cm",
    "image_path": "dataset/mm_math/images/MM_Math/51446343.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the figure, within circle $O$, $AB$ is the diameter of circle $O$, and chord $CD$ is perpendicular to $AB$ at point H. If $AH=5$ and $HB=1$, then what is the length of $CD$?\n",
    "completion": "\\textbf{Solution:} Link $OD$, \\\\\n$\\because$ $AB$ is the diameter of the circle $\\odot O$, and the chord $CD\\perp AB$, \\\\\n$\\therefore$ $CD=2HD$, \\\\\n$\\because$ $AH=5$, $HB=1$, \\\\\n$\\therefore$ $AB=AH+HB=6$, \\\\\n$\\therefore$ $OD=OA=3$, \\\\\n$\\therefore$ $OH=AH−OA=2$, \\\\\nIn $\\triangle ODH$, $DH=\\sqrt{OD^{2}−OH^{2}}=\\sqrt{5}$, \\\\\n$\\therefore$ $CD=2DH=2\\sqrt{5}$, \\\\\nThus, the answer is: $CD=\\boxed{2\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52991253.png"
  },
  {
    "solution": "\\boxed{(-1,-\\sqrt{3})}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system, a regular hexagon $ABCDEF$ with side length 2 has its center coinciding with the origin $O$, and $AB$ is parallel to the x-axis, intersecting the y-axis at point $P$. If $\\triangle OAP$ is rotated clockwise around point $O$, rotating $90^\\circ$ each time, what are the coordinates of point $A$ after the 2022nd rotation?\n",
    "completion": "\\textbf{Solution:} Given a regular hexagon ABCDEF with side length 2, and its center coincides with the origin O,\\\\\n\\[ \\because OA=AB=2, \\angle BAO=60^\\circ, \\]\n\\[ \\because AB\\parallel x \\text{ axis}, \\]\n\\[ \\therefore \\angle APO=90^\\circ, \\]\n\\[ \\therefore \\angle AOP=30^\\circ, \\]\n\\[ \\therefore AP=1, OP=\\sqrt{3}, \\]\n\\[ \\therefore A(1,\\sqrt{3}), \\]\n\\[ \\because \\text{rotating } \\triangle OAP \\text { clockwise around point O by } 90^\\circ\\text{ each time, it is known that point A}_2 \\text{ coincides with } D, \\]\n\\[ \\text{Since } 360^\\circ \\div 90^\\circ=4, \\]\n\\[ \\text{indicating each cycle consists of 4 rotations,} \\]\n\\[ \\therefore 2022 \\div 4=505\\ldots 2, \\]\n\\[ \\therefore \\text{point A}_{2022} \\text{ coincides with point A}_2, \\]\n\\[ \\because \\text{point A}_2 \\text{ is symmetric to point A concerning the origin } O, \\]\n\\[ \\therefore A_2(-1,-\\sqrt{3}), \\]\n\\[ \\therefore \\text{after the 2022nd rotation, the coordinates of point A are} \\boxed{(-1,-\\sqrt{3})}\\].",
    "image_path": "dataset/mm_math/images/MM_Math/52634229.png"
  },
  {
    "solution": "\\boxed{2, 3.5, or 4.5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in $ \\triangle ABC$, $ \\angle ACB=90^\\circ$, $ \\angle ABC=60^\\circ$, and $BC=2\\,cm$. Point D is the midpoint of BC. If a moving point E starts from point A and moves in the direction of A→B→A at a speed of $1\\,cm/s$, let the movement time of point E be $t\\,s$ ($0 \\leq t < 6$). Connect DE. When $ \\triangle BDE$ is similar to $ \\triangle ABC$, what is the value of t?\n",
    "completion": "\\textbf{Solution:} Given that in $\\triangle ABC$, $\\angle ACB=90^\\circ$, $\\angle ABC=60^\\circ$, and $BC=2cm$,\\\\\n$\\therefore$AB=2BC=4(cm),\\\\\nSince $BC=2cm$, and $D$ is the midpoint of $BC$, and a moving point $E$ starts from $A$ at a speed of 1cm/s,\\\\\n$\\therefore$BD=$\\frac{1}{2}$BC=1(cm), BE=AB−AE=4−t(cm),\\\\\nWhen $\\triangle BDE$ is similar to $\\triangle ABC$, $\\triangle BDE$ is a right triangle,\\\\\nIf $\\angle BED=90^\\circ$,\\\\\nAs A moves towards B, since $\\angle ABC=60^\\circ$,\\\\\n$\\therefore$$\\angle BDE=30^\\circ$,\\\\\n$\\therefore$BE=$\\frac{1}{2}$BD=$\\frac{1}{2}$(cm),\\\\\n$\\therefore$t=3.5,\\\\\nAs B moves towards A, t=4+0.5=4.5.\\\\\nIf $\\angle BDE=90^\\circ$,\\\\\nAs A moves towards B, since $\\angle ABC=60^\\circ$,\\\\\n$\\therefore$$\\angle BED=30^\\circ$,\\\\\n$\\therefore$BE=2BD=2(cm),\\\\\n$\\therefore$t=4−2=$2$,\\\\\nAs B moves towards A, t=4+2=6 (discard).\\\\\nIn summary, the value of $t$ is either \\boxed{2, 3.5, or 4.5}.",
    "image_path": "dataset/mm_math/images/MM_Math/52623528.png"
  },
  {
    "solution": "\\boxed{y=-2x+6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=-\\frac{1}{2}x+3$ intersects with the coordinate axes at points A and B, respectively, and intersects with the line $y=x$ at point C. A point Q on line segment OA moves from point O to point A at a uniform speed of 1 unit per second, with the time of motion denoted as $t(s)$. If line segment CQ bisects the area of $\\triangle OAC$, what is the functional expression of line CQ?\n",
    "completion": "\\textbf{Solution:} Let $x=6$,\\\\\nhence A$(6,0)$,\\\\\n$\\because$ CQ bisects the area of $\\triangle OAC$,\\\\\n$\\therefore$ Q$(3,0)$,\\\\\nAssume the equation of line CQ is $y=kx+b$,\\\\\nSubstituting C$(2,2)$ and Q$(3,0)$ into it, we get:\\\\\n$\\begin{cases} 3k+b=0 \\\\ 2k+b=2 \\end{cases}$,\\\\\nSolving the system, we find:\\\\\n$\\begin{cases} k=-2 \\\\ b=6 \\end{cases}$,\\\\\n$\\therefore$ the function corresponding to line CQ is: \\boxed{y=-2x+6}.",
    "image_path": "dataset/mm_math/images/MM_Math/51663132_208.png"
  },
  {
    "solution": "\\boxed{95^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, quadrilateral ABCD is a rhombus. Points E and F are located on sides BC and CD, respectively, with BE = DF and AB = AE. If $\\angle EAF=75^\\circ$, what is the degree measure of $\\angle C$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral ABCD is a rhombus, \\\\\nit follows that AB = AD, $\\angle B$ = $\\angle D$, and $\\angle C$ = $\\angle BAD$. \\\\\nIn $\\triangle ABE$ and $\\triangle ADF$, \\\\\nsince $\\left\\{\\begin{array}{l} AB = AD \\\\ \\angle B = \\angle D \\\\ BE = DF \\end{array}\\right.$, \\\\\nit implies that $\\triangle ABE \\cong \\triangle ADF$ (SAS), \\\\\nthus, $\\angle DAF = \\angle BAE$. \\\\\nLet $\\angle BAE = \\angle DAF = x$, \\\\\nthen $\\angle DAE = 75^\\circ + x$, \\\\\nand since AD $\\parallel$ BC, \\\\\n$\\angle AEB = 75^\\circ + x$. \\\\\nGiven that AB = AE, \\\\\n$\\angle B = \\angle AEB = 75^\\circ + x$. \\\\\nSince $\\angle BAE + \\angle ABE + \\angle AEB = 180^\\circ$, \\\\\n$x + 75^\\circ + x + 75^\\circ + x = 180^\\circ$, \\\\\nthus $x = 10^\\circ$, \\\\\ntherefore $\\angle BAD = 95^\\circ$, \\\\\nwhich means $\\angle C = \\boxed{95^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50496730.png"
  },
  {
    "solution": "\\boxed{6cm}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points C and D are the trisection points of line segment $AB$, and point E is the midpoint of line segment $DB$. Given $AB=12\\,\\text{cm}$, find the length of line segment $CE$.\n",
    "completion": "\\textbf{Solution:} Since points C and D are the trisection points of $AB$ and $AB=12cm$,\\\\\nit follows that $CD=DB=\\frac{1}{3}AB=4cm$,\\\\\nSince point E is the midpoint of the line segment $DB$,\\\\\nit follows that $DE=\\frac{1}{2}DB=2cm$,\\\\\ntherefore, $CE=CD+DE=\\boxed{6cm}$",
    "image_path": "dataset/mm_math/images/MM_Math/53206226.png"
  },
  {
    "solution": "\\boxed{4\\,\\text{cm}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, the perpendicular bisector of side $AB$, $OM$, and the perpendicular bisector of side $AC$, $ON$, intersect at point $O$. These two perpendicular bisectors intersect $BC$ at points $D$ and $E$, respectively. It is known that the perimeter of $\\triangle ADE$ is $7\\,\\text{cm}$. The lines $OA$, $OB$, and $OC$ are drawn. If the perimeter of $\\triangle OBC$ is $15\\,\\text{cm}$, find the length of $OA$.\n",
    "completion": "\\textbf{Solution:} Connect OA, OB, OC,\\\\\n$\\because$ the perimeter of $\\triangle ADE$ is $7\\,\\text{cm}$,\\\\\n$\\therefore$ $AD+DE+EA=7\\,\\text{cm}$,\\\\\n$\\therefore$ $BC=DB+DE+EC=AD+DE+EA=7\\,\\text{cm}$;\\\\\n$\\because$ the perimeter of $\\triangle OBC$ is $15$,\\\\\n$\\therefore$ $OB+OC+BC=15$,\\\\\n$\\because$ $BC=7$,\\\\\n$\\therefore$ $OB+OC=8$,\\\\\n$\\because$ $OM$ is the perpendicular bisector of $AB$,\\\\\n$\\therefore$ $OA=OB$,\\\\\nSimilarly, $OA=OC$,\\\\\n$\\therefore$ $OA=OB=OC=\\boxed{4\\,\\text{cm}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53147720_119.png"
  },
  {
    "solution": "\\boxed{3\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the square $ABCD$, where $AB=6$, $E$ is the midpoint of $BC$, and $F$ is a point on side $CD$. The triangle $CEF$ is folded along $EF$ to obtain the triangle $GEF$. Connect $CG$ and extend it to intersect $EF$ and $AB$ at points $O$ and $H$, respectively. If $G$ is the midpoint of $OH$, what is the length of $CF$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral $ABCD$ is a square,\\\\\nit follows that $\\angle B=90^\\circ$,\\\\\nsince $AB=6$ and E is the midpoint of $BC$,\\\\\nthen $CE=BE=\\frac{1}{2}AB=3$,\\\\\nsince $\\triangle CEF$ is folded along $EF$ to obtain $\\triangle GEF$,\\\\\nit follows that $EF$ bisects $CG$ perpendicularly,\\\\\nthus $OG=OC$, $\\angle COE=90^\\circ$\\\\\nGiven G is the midpoint of $OH$,\\\\\nthen $OG=GH$,\\\\\ntherefore $OG=GH=OC=\\frac{1}{3}CH$,\\\\\nlet $OC=x$, then $CH=3x$,\\\\\nsince $\\angle COE=\\angle B=90^\\circ$, $\\angle OCE=\\angle BCH$,\\\\\nit follows that $\\triangle COE\\sim \\triangle CBH$,\\\\\nwhich means $\\frac{CO}{BC}=\\frac{CE}{CH}$,\\\\\nthus $\\frac{x}{6}=\\frac{3}{3x}$,\\\\\nsolving gives $x=\\sqrt{6}$,\\\\\ntherefore $CO=\\sqrt{6}$,\\\\\nhence $OE=\\sqrt{CE^2-CO^2}=\\sqrt{3}$,\\\\\ntherefore $\\sin\\angle OCE=\\frac{OE}{CE}=\\frac{\\sqrt{3}}{3}$,\\\\\nsince $\\angle CFO+\\angle OCF=\\angle OCE+\\angle OCF=90^\\circ$,\\\\\nit follows that $\\angle CFO=\\angle OCE$,\\\\\nthus $\\sin\\angle CFO=\\sin\\angle OCE=\\frac{OC}{CF}=\\frac{\\sqrt{3}}{3}$,\\\\\nhence $CF=\\sqrt{3}OC=\\sqrt{3}\\times \\sqrt{6}=\\boxed{3\\sqrt{2}}$",
    "image_path": "dataset/mm_math/images/MM_Math/53110868.png"
  },
  {
    "solution": "\\boxed{40^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the rhombus $ABCD$, $E$ is a point on $CD$. The line $AE$ intersects the diagonal $BD$ at point $F$, and $CF$ is connected. If $\\angle AED = 40^\\circ$, what is the degree measure of $\\angle BCF$?\n",
    "completion": "\\textbf{Solution:} Given: Quadrilateral ABCD is a rhombus,\\\\\n$\\therefore$ AB = CB, AB $\\parallel$ DC, $\\angle$ABF = $\\angle$CBF,\\\\\nSince AB = CB, $\\angle$ABF = $\\angle$CBF, BF = BF,\\\\\n$\\therefore$ $\\triangle$ABF $\\cong$ $\\triangle$CBF (by SAS),\\\\\n$\\therefore$ $\\angle$BAF = $\\angle$BCF,\\\\\nSince $\\angle$AED = $40^\\circ$, AD $\\parallel$ BC,\\\\\n$\\therefore$ $\\angle$AED = $\\angle$BAF,\\\\\n$\\therefore$ $\\angle$BCF = \\boxed{40^\\circ},",
    "image_path": "dataset/mm_math/images/MM_Math/52423430.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{13}\\pi }{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in a grid diagram where each small square has a side length of 1, a circular arc passes through lattice points $A$, $B$, $C$. The line connecting lattice points $C$ and $D$ intersects $\\arc{BC}$ at point $E$. What is the length of $\\arc{EC}$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect AE, AC, and AD, \\\\\n$\\because \\angle ABC=90^\\circ$, \\\\\n$\\therefore AC$ is the diameter, \\\\\n$\\therefore \\angle ABC=\\angle AEC=90^\\circ$, \\\\\nAccording to the grid figure, we know: $AC=AD=\\sqrt{13}$, $CD=\\sqrt{26}$, \\\\\n$\\therefore AC^{2}+AD^{2}=CD^{2}=26$, \\\\\n$\\therefore \\triangle ACD$ is an isosceles right-angled triangle, \\\\\n$\\therefore \\angle CAD=90^\\circ$, $\\angle ACE=45^\\circ$, \\\\\n$\\therefore \\angle EAC=45^\\circ$, \\\\\n$\\therefore$ the central angle corresponding to $\\overset{\\frown}{EC}$ is $90^\\circ$, \\\\\n$\\therefore$ the length of $\\overset{\\frown}{EC}$ is $\\frac{1}{4}$ of the circumference of the circle with diameter AC, \\\\\nThat is, the length of $\\overset{\\frown}{EC}$ is: $\\frac{1}{4}\\times \\pi \\times \\sqrt{13}=\\boxed{\\frac{\\sqrt{13}\\pi }{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52457425.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Point \\(C\\) lies on segment \\(AB\\), and point \\(M\\) is the midpoint of \\(AC\\), with \\(AM=1\\), \\(BC=4\\). If point \\(N\\) is the midpoint of \\(BC\\), what is the length of \\(MN\\)?\n",
    "completion": "\\textbf{Solution:} Given that, \\\\\n$ \\because $ point $ M$ is the midpoint of $ AC$, and $ AM=1$, \\\\\n$ \\therefore CM=AM=1$, \\\\\n$ \\because $ point $ N$ is the midpoint of $ BC$, and $ BC=4$, \\\\\n$ \\therefore CN=\\frac{1}{2}BC=2$, \\\\\n$ \\therefore MN=CM+CN=1+2=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52794258_175.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, two squares are constructed externally on the sides AC and AB of $\\triangle ABC$, denoted as $S_1$ and $S_2$ representing the areas of these two squares respectively. If $S_1=8$ and $S_2=17$, what is the length of BC?\n",
    "completion": "\\textbf{Solution:} We have $BC = \\sqrt{S_2 - S_1} = \\sqrt{17 - 8} = \\boxed{3}$",
    "image_path": "dataset/mm_math/images/MM_Math/52251811.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, lines $l_1 \\parallel l_2 \\parallel l_3$, and lines AC, DF intersect $l_1$, $l_2$, $l_3$ at points A, B, C and D, E, F, respectively. Connect AF. Let $BG \\parallel AF$. If $\\frac{DE}{EF} = \\frac{2}{3}$ and $BG = 6$, then what is the length of AF?\n",
    "completion": "\\textbf{Solution:} Since $l_1 \\parallel l_2 \\parallel l_3$, $\\frac{DE}{EF}=\\frac{2}{3}$\\\\\nTherefore, $\\frac{AB}{BC}=\\frac{DE}{EF}=\\frac{2}{3}$\\\\\nTherefore, $\\frac{BC}{AC}=\\frac{3}{5}$\\\\\nSince $BG \\parallel AF$,\\\\\nTherefore, $\\triangle CBG \\sim \\triangle CAF$,\\\\\nTherefore, $\\frac{CB}{AC}=\\frac{BG}{AF}$\\\\\nTherefore, $\\frac{3}{5}=\\frac{6}{AF}$\\\\\nTherefore, $AF=\\boxed{10}$",
    "image_path": "dataset/mm_math/images/MM_Math/52872371.png"
  },
  {
    "solution": "\\boxed{x > -3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=ax+b\\ (a\\neq 0)$ passes through points A and B. What is the solution set of the inequality $ax+b>0$?\n",
    "completion": "\\textbf{Solution:} According to the function graph, it is known that when $x > -3$, $y > 0$, \\\\\ntherefore, the solution set of the inequality $ax+b>0$ is $x > -3$. \\\\\nHence, the answer is: $\\boxed{x > -3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52387412.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{7}+4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point C is any point on the semicircle with AB as the diameter, where $AB=8$. After connecting AC, the line segment AC is rotated $120^\\circ$ counterclockwise around point A to obtain the line segment $AC'$. What is the maximum value of $BC'$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, find the midpoint $O$ of the line segment $AB$, connect $OC$, and rotating the line segment $AC$ around point $A$ by $120^\\circ$ can be considered as rotating $\\triangle AOC$ around point $A$ by $120^\\circ$ to obtain $\\triangle AO'C'$,\\\\\n$\\therefore \\triangle AOC \\cong \\triangle AO'C'$,\\\\\n$\\therefore OC=O'C'=4$,\\\\\nTaking point $O'$ as the center and $O'C'$ as the radius to draw a circle, when points $O'$, $C'$, and $B$ are collinear, $BC'$ takes the maximum value,\\\\\nAs shown in the figure: $\\angle DAO' =60^\\circ$, $O'A=4$, $AB=8$,\\\\\n$AD=O'A\\cdot \\cos 60^\\circ=2$, $O'D=2\\sqrt{3}$,\\\\\n$\\therefore BO'=\\sqrt{DB^2+O'B^2}=\\sqrt{10^2+(2\\sqrt{3})^2}=4\\sqrt{7}$,\\\\\n$\\therefore BC' =BO'+C'O' =4\\sqrt{7}+4$.\\\\\nTherefore, the answer is: $BC'=\\boxed{4\\sqrt{7}+4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51299079.png"
  },
  {
    "solution": "\\boxed{15\\text{cm}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, on the number line, points $C$ and $D$ divide the line segment $AB$ into three parts in the ratio $2:5:3$, with $E$ being the midpoint of $AB$. If $ED=3\\,cm$, what is the length of the line segment $AB$?\n",
    "completion": "\\textbf{Solution:} From the given information, \\\\\nsince $E$ is the midpoint of $AB$, \\\\\nthus $EB=\\frac{1}{2}AB$, \\\\\nsince points $C$, $D$ on the number line divide the segment $AB$ into three parts in the ratio $2:5:3$, \\\\\nthus $BD=\\frac{3}{2+5+3}AB=\\frac{3}{10}AB$, \\\\\nthus $ED=EB-BD=\\frac{1}{2}AB-\\frac{3}{10}AB=\\frac{1}{5}AB$, \\\\\nalso, since $ED=3\\text{cm}$, \\\\\nthus $\\frac{1}{5}AB=3\\text{cm}$, \\\\\nthus $AB=\\boxed{15\\text{cm}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52792740_146.png"
  },
  {
    "solution": "\\boxed{1\\colon\\sqrt{5}\\colon2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$ and rectangle $AEGH$, the ratios $AD:AB=AH:AE=1:2$. What are the ratios of $DH:CG:BE$?\n",
    "completion": "\\textbf{Solution:} Connect AC, AG,\\\\\n$\\because$ in rectangles ABCD and AEGH,\\\\\n$\\therefore$ $\\angle$DAH+$\\angle$HAB=90$^\\circ$, $\\angle$BAE+$\\angle$HAB=90$^\\circ$, HG=AE,\\\\\n$\\therefore$ $\\angle$DAH=$\\angle$BAE,\\\\\n$\\because$ AD$\\colon$AB=AH$\\colon$AE=1$\\colon$2,\\\\\n$\\therefore$ $\\triangle$ADH$\\sim$ $\\triangle$ABE,\\\\\n$\\therefore$ DH$\\colon$BE=AD$\\colon$AB=1$\\colon$2,\\\\\nIn right $\\triangle$AHG, HG=2AH,\\\\\n$\\therefore$ AG=$\\sqrt{AH^2+HG^2}=\\sqrt{5}AH$,\\\\\n$\\therefore$ AH$\\colon$AG$\\colon$AE=1$\\colon\\sqrt{5}\\colon$2,\\\\\nWhen AH overlaps with DA, rotate rectangle AHGE around point A to rectangle AHGE,\\\\\n$\\therefore$ $\\triangle$ACG$\\sim$ $\\triangle$ABE$\\sim$ $\\triangle$ADH,\\\\\n$\\therefore$ $\\frac{HD}{CG}=\\frac{AH}{AG}=\\frac{1}{\\sqrt{5}}$, $\\frac{CG}{BE}=\\frac{AG}{AE}=\\frac{\\sqrt{5}}{2}$\\\\\n$\\therefore$ DH$\\colon$CG$\\colon$BE=1$\\colon\\sqrt{5}\\colon$2.\\\\\nThus, the answer is: DH$\\colon$CG$\\colon$BE=$\\boxed{1\\colon\\sqrt{5}\\colon2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52963745.png"
  },
  {
    "solution": "\\boxed{\\frac{2}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle BCA=90^\\circ$, $BC=8$, and $AC=6$. Point $D$ is the midpoint on side $AB$, and point $E$ is a moving point on side $BC$. Connect $DE$, and fold $\\triangle BDE$ along $DE$ to get $\\triangle FDE$. As shown in figure (1), segment $DF$ intersects segment $BC$ at point $G$. When $BE=2$, what is the value of $\\frac{GE}{GD}$?\n",
    "completion": "\\textbf{Solution}: Solve: $\\frac{GE}{GD}=\\boxed{\\frac{2}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53151227_93.png"
  },
  {
    "solution": "\\boxed{(-1, \\frac{3}{2})}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the parabola $y = -x^2 + bx + c$ passes through the points A, B ($-3, 0$), and C ($0, 3$). Find a point M on the axis of symmetry of the parabola such that the sum of the distances from point M to points O and C is minimized. Determine the coordinates of point M at this time.\n",
    "completion": "\\textbf{Solution:} \\\\\nSince the axis of symmetry of the parabola is the line $x=-1$, construct point $C(0,3)$'s symmetric point $D(-2,3)$ about the line $x=-1$.\\\\\nFind a point $M$ on the axis of symmetry of the parabola such that the sum of distances from point $M$ to points $O$ and $C$ is minimized, that is $OM+CM=OM+MD\\leq OD$,\\\\\nThe sum is shortest when points $D$, $M$, and $O$ are collinear,\\\\\nLet the equation of line $OD$ be: $y=kx$, passing through point $D$,\\\\\nTherefore, $3=-2k, k=-\\frac{3}{2}$,\\\\\nTherefore, the equation of line $OD$ is: $y=-\\frac{3}{2}x$.\\\\\nWhen $x=-1$, $y=\\frac{3}{2}$,\\\\\nTherefore, the coordinates of $M$ are $\\boxed{(-1, \\frac{3}{2})}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52602225_173.png"
  },
  {
    "solution": "\\boxed{y=-3x-3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the graph of the inverse proportion function $y= \\frac{k}{x}$ intersects with the line $y=ax+b$ at point $A\\left( \\frac{1}{3}, m+9\\right)$ and point $B(1, m)$. Determine the analytical expressions for the inverse proportion function $y= \\frac{k}{x}$ and the line $y=ax+b$.\n",
    "completion": "\\textbf{Solution:} Let's solve this. Since the graph of the inverse proportion function $y= \\frac{k}{x}$ intersects with the line $y=ax+b$ at points A$\\left(\\frac{1}{3}m, m+9\\right)$ and B$(1, m)$.\n\nTherefore, substituting point A into $y= \\frac{k}{x}$ gives: $k=\\frac{1}{3}m(m+9)$, and substituting point B gives: $k=m$,\n\nTherefore, $\\frac{1}{3}m(m+9)=m$, solving this gives: $m_1=0$ (discard), $m_2=-6$,\n\nTherefore, A$(-2, 3)$, B$(1, -6)$.\n\nTherefore, substituting point A$(-2, 3)$ into $y= \\frac{k}{x}$ gives: $k=-6$,\n\nTherefore, the equation of the inverse proportion function $y= \\frac{k}{x}$ is $\\boxed{y=-\\frac{6}{x}}$,\n\nSubstituting points A$(-2, 3)$, B$(1, -6)$ into the line $y=ax+b$, gives: $\\begin{cases}3=-2a+b\\\\ -6=a+b\\end{cases}$, solving this gives: $\\begin{cases}a=-3\\\\ b=-3\\end{cases}$,\n\nTherefore, the equation of the line $y=ax+b$ is $\\boxed{y=-3x-3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51368308_106.png"
  },
  {
    "solution": "\\boxed{3\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points $A\\left(5a-1, 2\\right)$ and $B\\left(8, 1\\right)$ both lie on the graph of the inverse proportion function $y=\\frac{k}{x}\\left(k\\ne 0\\right)$. Point $P$ is a moving point on the line $y=x$. What is the minimum value of $PA+PB$?\n",
    "completion": "\\textbf{Solution:} Since points A ($5a-1$, 2) and B (8, 1) are both on the graph of the inverse proportion function $y=\\frac{k}{x}$,\\\\\nwe have $(5a-1) \\times 2 = 8$,\\\\\nthus, $a = 1$,\\\\\ntherefore, A (4, 2) and B (8, 1),\\\\\ntherefore, the symmetrical point of A about the line $y=x$ is A' (2, 4),\\\\\nthus, $AB = \\sqrt{(8-2)^{2} + (1-4)^{2}} = 3\\sqrt{5}$,\\\\\ntherefore, the minimum value of $PA + PB$ is $\\boxed{3\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51446351.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given: $\\angle AOB = 90^\\circ$, draw rays $OC$, $OD$ is the bisector of $\\angle AOC$, $OE$ is the bisector of $\\angle BOC$. As shown in figure (2), if ray $OC$ rotates around point $O$ within $\\angle AOB$, when $\\angle BOC = a$, what is the degree measure of $\\angle DOE$?\n",
    "completion": "\\textbf{Solution:} The size of $\\angle DOE$ remains constant. The reason is as follows:\n\nSince $\\angle AOB=90^\\circ$, $\\angle BOC=\\alpha$,\n\nTherefore, $\\angle AOD=90^\\circ-\\alpha$.\n\nAlso, since $OE$, $OD$ are the angle bisectors of $\\angle BOC$ and $\\angle AOC$ respectively,\n\nTherefore, $\\angle EOC=\\frac{1}{2}\\alpha$, $\\angle COD=\\frac{1}{2}\\left(90^\\circ-\\alpha \\right)$.\n\nTherefore, $\\angle DOE=\\angle EOC+\\angle COD$\n\n$=\\frac{1}{2}\\alpha +\\frac{1}{2}\\left(90^\\circ-\\alpha \\right)=\\boxed{45^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/47364521_203.png"
  },
  {
    "solution": "\\boxed{10^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AE$ and $AD$ are respectively the altitude and the angle bisector of $\\triangle ABC$, and $\\angle B=40^\\circ$, $\\angle C=60^\\circ$, find the degree measure of $\\angle DAE$.\n",
    "completion": "\\textbf{Solution:} Since $\\angle B = 40^\\circ$ and $\\angle C = 60^\\circ$, \\\\\nit follows that $\\angle BAC = 180^\\circ - \\angle B - \\angle C = 80^\\circ$.\\\\\nSince $AD$ is the angle bisector of $\\triangle ABC$, \\\\\nit follows that $\\angle BAD = \\angle DAC = \\frac{1}{2} \\angle BAC = 40^\\circ$, \\\\\ntherefore $\\angle ADE = \\angle B + \\angle BAD = 80^\\circ$, \\\\\ntherefore $\\angle DAE = 90^\\circ - \\angle ADE = 90^\\circ - 80^\\circ = \\boxed{10^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52914135.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC \\cong \\triangle DEF$, where $BC = 7$ and $EC = 4$. What is the length of $CF$?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABC \\cong \\triangle DEF$,\\\\\nit follows that $BC=EF=7$,\\\\\nthus, $CF=EF-EC=7-4=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52733484.png"
  },
  {
    "solution": "\\boxed{100^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $O$ is a point on the line $CE$. With $O$ as the vertex, we make $ \\angle AOB=90^\\circ$, and $OA$, $OB$ are on opposite sides of the line $CE$, $ \\angle AOC=50^\\circ$, and $OB$ bisects $ \\angle COD$. Find the degree measure of $ \\angle DOE$?\n",
    "completion": "\\textbf{Solution:} \n\nGiven that, $\\angle BOC = \\angle AOB - \\angle AOC = 90^\\circ - 50^\\circ = 40^\\circ$,\\\\\nsince OB bisects $\\angle COD$,\\\\\nit follows that $\\angle COD = 2\\angle BOC = 80^\\circ$,\\\\\nthus, $\\angle DOE = 180^\\circ - \\angle COD = \\boxed{100^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52251792.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: It is known that the parabola $y=\\frac{1}{4}x^{2}+1$ has the following property: for any point on the parabola, its distance to the fixed point $F(0, 2)$ is equal to its distance to the x-axis, as shown in the figure. The coordinate of point $M$ is $(\\sqrt{3}, 3)$, and $P$ is a moving point on the parabola $y=\\frac{1}{4}x^{2}+1$. What is the minimum perimeter of $\\triangle PMF$?\n",
    "completion": "\\textbf{Solution:} Draw $ME\\perp x$ axis at point $E$, and $ME$ intersects the parabola at point ${P}^{\\prime}$.\\\\\nSince the distance from any point on the parabola to the fixed point $F(0, 2)$ is equal to its distance to the x-axis,\\\\\nit follows that ${P}^{\\prime}F={P}^{\\prime}E$,\\\\\nand since $MF$ is constant,\\\\\nit follows that when point $P$ moves to point ${P}^{\\prime}$, the perimeter of $\\triangle PMF$ takes its minimum value,\\\\\nsince $F(0, 2)$, $M(\\sqrt{3}, 3)$,\\\\\nit follows that $ME=3$, $FM=\\sqrt{(\\sqrt{3}-0)^{2}+(3-2)^{2}}=2$,\\\\\nthus $MP^{\\prime}+P^{\\prime}F+MF=MP^{\\prime}+P^{\\prime}E+MF=ME+MF=3+2=\\boxed{5}$,\\\\\ntherefore, the minimum perimeter of $\\triangle PMF$ is $\\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52978121_182.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: The cross-section of a drainpipe is shown in the figure below. Given that the radius of the drainpipe is $OB=10$, and the width of the water surface $AB=12$. If more water is added, such that the width of the water surface $AB$ changes to $16$, then by how much does the height of the water surface $AB$ increase?\n",
    "completion": "\\textbf{Solution:} Given $EF \\parallel AB$, $EF=16$. Draw $OC \\perp AB$ at point $C$, intersecting $EF$ at point $G$. Connect $OE$.\n\n$\\therefore OC \\perp EF$,\n\n$\\therefore \\angle OGE=\\angle OCB=90^\\circ$, $BC=\\frac{1}{2}AB=6$, $EG=\\frac{1}{2}EF=8$,\n\nIn $\\triangle OCB$, $OC=\\sqrt{OB^{2}-CB^{2}}=\\sqrt{10^{2}-6^{2}}=8$;\n\nIn $\\triangle EOG$, $OG=\\sqrt{OE^{2}-EG^{2}}=\\sqrt{10^{2}-8^{2}}=6$\n\nWhen $EF$ and $AB$ are on the same side of the center $O$, $CG=OC-OG=8-6=2$;\n\nWhen $EF$ and $AB$ are on opposite sides of the center $O$, $CG=OC+OG=8+6=14$;\n\n$\\therefore$ When the width of the water surface $AB$ changes to 16, the height by which the water surface $AB$ rises is $\\boxed{2}$ or 14.",
    "image_path": "dataset/mm_math/images/MM_Math/53059867.png"
  },
  {
    "solution": "\\boxed{4.8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the diagram, within $\\triangle ABC$, $\\angle A=90^\\circ$, $AB=8$, $AC=6$, point $D$ is a moving point on $BC$. Lines are drawn from $D$ such that $DE\\perp AB$ and $DF\\perp AC$, with $E$ and $F$ being the feet of the perpendiculars, respectively. What is the minimum value of $EF$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect CD,\n\\[\n\\because \\angle BAC = 90^\\circ, AB = 8, AC = 6,\n\\]\n\\[\n\\therefore BC = \\sqrt{AB^2 + AC^2} = 10,\n\\]\n\\[\n\\because DE \\perp AB, DF \\perp CA,\n\\]\n\\[\n\\therefore \\angle DEA = \\angle DFA = 90^\\circ,\n\\]\n\\[\n\\because \\angle BAC = 90^\\circ,\n\\]\n\\[\n\\therefore Quadrilateral FDEA is a rectangle,\n\\]\n\\[\n\\therefore EF = DA,\n\\]\nGiven that a perpendicular segment is the shortest, when DA $\\perp$ BC, the length of segment DA is the minimum, and hence the length of segment EF is also the minimum,\nAt this point, $S_{\\triangle ABC} = \\frac{1}{2}AB \\cdot AC = \\frac{1}{2}BC \\cdot AD$,\n\\[\n\\text{indicating } \\frac{1}{2} \\times 8 \\times 6 = \\frac{1}{2} \\times 10 \\cdot AD\n\\]\nSolving gives: DA $= \\frac{24}{5} = 4.8$\n\\[\n\\therefore \\text{The minimum value of } EF \\text{ is } \\boxed{4.8}\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/53359994.png"
  },
  {
    "solution": "\\boxed{\\frac{25}{9}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that in rhombus $ABCD$, $AB=5$ and $\\cos B=\\frac{3}{5}$. Points E and F are located on sides $BC$ and $CD$ respectively, and the extension line of $AF$ intersects the extension line of $BC$ at point G, with $\\angle EAF=\\frac{1}{2}\\angle BAD$. The extensions of $AE$ and $DC$ intersect at point H. Lines $GH$ and $AC$ are connected. If $\\triangle AGH$ is similar to $\\triangle ABC$, what is the length of segment $BE$?\n",
    "completion": "\\textbf{Solution:} Let us denote:\n\nSince $\\angle BAC = \\angle EAF$,\n\nTherefore, when $\\frac{AH}{AG}=\\frac{AB}{AC}$ or $\\frac{AH}{AG}=\\frac{AC}{AB}$, $\\triangle AGH$ is similar to $\\triangle ABC$.\n\n(i) When $\\frac{AH}{AG}=\\frac{AB}{AC}$,\n\nSince $\\angle BAC=\\frac{1}{2}\\angle BAD$, and $\\angle EAF=\\frac{1}{2}\\angle BAD$,\n\nTherefore, $\\angle BAC=\\angle EAF$. Therefore, $\\angle BAE=\\angle CAG$.\n\nSince $AB\\parallel CD$, therefore $\\angle BAE=\\angle AHC$. Therefore, $\\angle CAG=\\angle AHC$.\n\nAlso, since $\\angle EAC=\\angle AGC$, therefore, $\\triangle AHC \\sim \\triangle GAC$. Therefore, $\\frac{AH}{AG}=\\frac{CH}{AC}$.\n\nAlso, $\\frac{AH}{AG}=\\frac{AB}{AC}$, therefore, $CH=AB$.\n\nSince $AB\\parallel CD$, therefore $\\frac{BE}{EC}=\\frac{AB}{CH}$. Therefore, $BE=EC=\\frac{5}{2}$.\n\n(ii) When $\\frac{AH}{AG}=\\frac{AC}{AB}$, similarly, it can be derived that $\\triangle AHC \\sim \\triangle GAC$.\n\nTherefore, $\\frac{AH}{AG}=\\frac{AC}{CG}$. Also, since $\\frac{AH}{AG}=\\frac{AC}{AB}$, therefore, $CG=AB$. Therefore, $BE=\\boxed{\\frac{25}{9}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52942187_156.png"
  },
  {
    "solution": "\\boxed{\\frac{16}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $D$ is a point on the circle $\\odot O$, and point $C$ is on the extension line of diameter $BA$, with $CD^{2}=CA\\cdot CB$; $CD$ is the tangent of $\\odot O$; a tangent to $\\odot O$ is drawn through point $B$ and intersects the extension line of $CD$ at point $E$. If $BC=10$ and $\\tan\\angle CDA=\\frac{3}{5}$, what is the length of $BE$?\n",
    "completion": "\\textbf{Solution:} From the given information, $BE$ and $CE$ are tangent to the circle $O$, \\\\\n$\\therefore ED=EB$, \\\\\n$\\because \\triangle DCA\\sim \\triangle BCD$, \\\\\n$\\therefore \\angle DBA=\\angle CDA$, \\\\\n$\\therefore \\frac{DC}{BC}=\\frac{DA}{BD}=\\tan\\angle DBA=\\tan\\angle CDA=\\frac{3}{5}$, \\\\\n$\\therefore CD=\\frac{3}{5}BC=6$, \\\\\nLet $BE=x$, then $DE=x$, $CE=x+6$, \\\\\nIn $\\triangle CBE$, $(x+6)^{2}=x^{2}+10^{2}$, \\\\\nSolving this gives: $x=\\frac{16}{3}$, \\\\\n$\\therefore BE=\\boxed{\\frac{16}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51368194_107.png"
  },
  {
    "solution": "\\boxed{\\frac{4\\sqrt{5}}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the rectangle $ABCD$, where $AB=2$ and $AD=4$, point $P$ is distinct from points $A$ and $D$. Perpendiculars to $AC$ and $BD$ are drawn from point $P$, meeting them at points $E$ and $F$, respectively. What is the value of $PE+PF$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, connect OP, \\\\\n$\\because$ AB = 2, AD = 4, \\\\\nby the Pythagorean theorem, we have BD = $\\sqrt{2^2+4^2}=2\\sqrt{5}$, \\\\\n$S_{\\triangle ABD}=\\frac{1}{2}\\times AB \\times AD=\\frac{1}{2}\\times 2\\times 4=4$, \\\\\nin rectangle ABCD, OA = OD = OB = $\\frac{1}{2}$BD = $\\sqrt{5}$, \\\\\n$\\because S_{\\triangle AOD}=S_{\\triangle AOP}+S_{\\triangle DOP}=\\frac{1}{2}S_{\\triangle ABD}$, \\\\\n$\\therefore \\frac{\\sqrt{5}}{2}PE+\\frac{\\sqrt{5}}{2}PF=2$, \\\\\n$\\therefore PE+PF=\\boxed{\\frac{4\\sqrt{5}}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52387026.png"
  },
  {
    "solution": "\\boxed{y=\\frac{4}{x}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, $\\triangle AOB$ is an isosceles right triangle with the hypotenuse $OB$ on the x-axis. The graph of the linear function $y=3x-4$ passes through point A and intersects the y-axis at point C. The graph of the inverse proportion function $y=\\frac{k}{x} (x>0)$ also passes through point A. What is the analytical expression of the inverse proportion function?\n",
    "completion": "\\textbf{Solution}: Construct $AM\\perp y$-axis at M and $AN\\perp x$-axis at N through point A, as shown in the figure,\\\\\n$\\therefore$ Quadrilateral $AMON$ is a rectangle,\\\\\n$\\because$ $\\triangle AOB$ is an isosceles right triangle,\\\\\n$\\therefore$ $\\angle AON=45^\\circ$,\\\\\n$\\therefore$ $ON=NA$,\\\\\n$\\therefore$ Quadrilateral $AMON$ is a square,\\\\\n$\\therefore$ $AM=AN$,\\\\\nLet $A\\left(a, a\\right)$,\\\\\n$\\because$ Point $A$ lies on the line $y=3x-4$,\\\\\n$\\therefore$ $a=3a-4$,\\\\\nSolving gives $a=2$,\\\\\n$\\therefore$ $A\\left(2, 2\\right)$,\\\\\n$\\because$ The graph of the inverse proportion function $y=\\frac{k}{x}\\left(x>0\\right)$ passes through point A,\\\\\n$\\therefore$ $2=\\frac{k}{2}$,\\\\\n$\\therefore$ $k=4$,\\\\\n$\\therefore$ The analytical expression of the inverse proportion function is $\\boxed{y=\\frac{4}{x}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52602253_186.png"
  },
  {
    "solution": "\\boxed{20}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within circle $O$, there is a broken line $OABC$, where $OA=8$, $AB=12$, and $\\angle A=\\angle B=60^\\circ$. What is the length of $BC$?\n",
    "completion": "\\textbf{Solution:} Extend $AO$ to meet $BC$ at point $D$, and draw $OE\\perp BC$ at point $E$.\\\\\n$\\because \\angle A=\\angle B=60^\\circ$,\\\\\n$\\therefore \\angle ADB=60^\\circ$,\\\\\n$\\therefore \\triangle ADB$ is an equilateral triangle,\\\\\n$\\therefore BD=AD=AB=12$,\\\\\n$\\therefore OD=4$,\\\\\nFurthermore, $\\because \\angle ADB=60^\\circ$,\\\\\n$\\therefore DE=OD\\cdot \\cos 60^\\circ=\\frac{1}{2}OD=2$,\\\\\n$\\therefore BE=10$,\\\\\n$\\because OE\\perp BC$,\\\\\n$\\therefore BC=2BE=\\boxed{20}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52871876.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, we have $AB=AC=2\\sqrt{2}$ and $\\angle BAC=90^\\circ$. There is a moving point $D$ on the line segment $BC$. We connect $AD$. Through point $A$, we draw $AD\\perp AE$ where $AD=AE$. We then connect $DE$. What is the length of $BC$?\n",
    "completion": "\\textbf{Solution:} According to the problem, we have $BC=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55502350_43.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, and $CD$ is the altitude to side $AB$. Fold side $AC$ such that the crease is $EF$, and connect $CE$. If $CD$ bisects $\\angle BCE$, find the measure of $\\angle A$.\n",
    "completion": "\\textbf{Solution.} From the given problem, $EF$ is the axis of symmetry of $AC$,\\\\\nhence $FA=FC$, $EA=EC$,\\\\\nhence $\\angle ECA=\\angle A$.\\\\\nMoreover, since $CD$ is the height from $AB$,\\\\\nhence $\\angle CDE=\\angle CDB=90^\\circ$.\\\\\nAlso, since $CD$ bisects $\\angle BCE$,\\\\\nhence $\\angle DCE=\\angle DCB$.\\\\\nAlso, since $CD=CD$,\\\\\nhence $\\triangle CDE\\cong \\triangle CDB$ (ASA),\\\\\nhence $\\angle CED=\\angle CBD$.\\\\\nMoreover, since $\\angle CED=\\angle A+\\angle ECA$, $\\angle ECA=\\angle A$,\\\\\nhence $\\angle CBD=2\\angle A$.\\\\\nAlso, since $\\angle ACB=90^\\circ$,\\\\\nhence $\\angle A+\\angle B=90^\\circ$,\\\\\nhence $3\\angle A=90^\\circ$,\\\\\nhence $\\angle A=\\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51369213_98.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in rectangle $ABCD$, $BC>CD$ with $BC$ and $CD$ being the two roots of the quadratic equation $x^2-14x+48=0$. Connect $BD$. A moving point $P$ starts from $B$ and moves towards $D$ along $BD$ at a speed of 1 unit per second, while another moving point $Q$ starts from $D$ and moves along ray $DA$ at the same speed. When point $P$ reaches point $D$, point $Q$ stops moving. Let the motion time be $t$ seconds. Construct triangle $\\triangle PQM$ with $PQ$ as the hypotenuse such that point $M$ is on the line segment $BD$. When triangle $\\triangle PQM$ is similar to $\\triangle BCD$, what is the value of $t$?\n",
    "completion": "\\textbf{Solution:} From the given information, $\\because \\angle QMD = \\angle QMP = \\angle C = 90^\\circ$, $AD \\parallel BC$, $\\therefore \\angle DBC = \\angle QDM$, $\\therefore \\triangle DQM \\sim \\triangle BDC$, $\\therefore \\frac{QM}{DC} = \\frac{QD}{BD} = \\frac{DM}{CB}$, that is, $\\frac{QM}{6} = \\frac{t}{10} = \\frac{DM}{8}$. Solving these equations gives: $QM = \\frac{3}{5}t$, $DM = \\frac{4}{5}t$.\n\n$\\because \\triangle PQM \\sim \\triangle BCD$, $\\therefore \\frac{QM}{DC} = \\frac{PM}{BC}$ or $\\frac{QM}{BC} = \\frac{PM}{DC}$, $\\therefore QM = \\frac{3}{4}PM$ or $QM = \\frac{4}{3}PM$.\n\nWhen point M is on segment PD, $PM = 10 - t - \\frac{4}{5}t = 10 - \\frac{9}{5}t$, $\\therefore QM = \\frac{3}{4}PM$ that is $\\frac{3}{5}t = \\frac{3}{4}\\left(10 - \\frac{9}{5}t\\right)$. Solving this equation yields: $t = \\boxed{\\frac{50}{13}}$.\n\nWhen $QM = \\frac{4}{3}PM$, $\\frac{3}{5}t = \\frac{4}{3}\\left(10 - \\frac{9}{5}t\\right)$. Solving this equation yields: $t = \\boxed{\\frac{40}{9}}$.\n\nWhen point M is on PB, $PM = t + \\frac{4}{5}t - 10 = \\frac{9}{5}t - 10$, $QM = \\frac{4}{3}PM$ that is $\\frac{3}{5}t = \\frac{4}{3}\\left(\\frac{9}{5}t - 10\\right)$. Solving this equation yields: $t = \\boxed{\\frac{200}{27}}$.\n\nWhen $QM = \\frac{3}{4}PM$, $\\frac{3}{5}t = \\frac{3}{4}\\left(\\frac{9}{5}t - 10\\right)$. Solving this equation yields: $t = \\boxed{10}$.\n\nTherefore, the possible values for $t$ are \\boxed{\\frac{50}{13}}, \\boxed{\\frac{40}{9}}, \\boxed{\\frac{200}{27}}, or \\boxed{10}.",
    "image_path": "dataset/mm_math/images/MM_Math/52774682_59.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{10}}{10}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the vertices of $\\triangle ABC$ are on the grid points of a square grid. What is the value of $\\cos\\angle ACB$?\n",
    "completion": "\\textbf{Solution:} Draw line $AD \\perp BC$ at $D$,\\\\\n$\\therefore$ $DC = 1$, $AD = 3$,\\\\\n$\\therefore$ $AC =\\sqrt{AD^{2}+DC^{2}}=\\sqrt{10}$,\\\\\n$\\therefore$ $\\cos\\angle ACB =\\frac{DC}{AC}=\\frac{1}{\\sqrt{10}}=\\frac{\\sqrt{10}}{10}=\\boxed{\\frac{\\sqrt{10}}{10}}$",
    "image_path": "dataset/mm_math/images/MM_Math/52601690.png"
  },
  {
    "solution": "\\boxed{9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, segment $AB=12$, point $C$ is the midpoint of segment $AB$, and point $D$ is the midpoint of segment $BC$. What is the length of segment $AD$?\n",
    "completion": "\\textbf{Solution:} Given $\\because AB=12$ and C is the midpoint of AB,\\\\\n$\\therefore AC=BC=6$,\\\\\n$\\because D$ is the midpoint of $BC$,\\\\\n$\\therefore CD=\\frac{1}{2}BC=3$,\\\\\n$\\therefore AD=AC+CD=\\boxed{9}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52732986_65.png"
  },
  {
    "solution": "\\boxed{25\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rhombus $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$, $DE \\parallel AC$, and $AE \\parallel BD$. Quadrilateral $AODE$ is a rectangle; if $AB=10$ and $\\angle ABC=60^\\circ$, what is the area of quadrilateral $AODE$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral $ABCD$ is a rhombus,\\\\\n$\\therefore$ $AD=AB=BC=10$, $OA=OC$, $AC\\perp BD$,\\\\\nSince $\\angle ABC=60^\\circ$,\\\\\n$\\therefore$ $\\triangle ABC$ is an equilateral triangle,\\\\\n$\\therefore$ $AC=AB=10$,\\\\\n$\\therefore$ $OA=\\frac{1}{2}AC=5$,\\\\\nIn $Rt\\triangle AOD$, by the Pythagorean theorem, $OD=\\sqrt{AD^{2}-OA^{2}}=\\sqrt{10^{2}-5^{2}}=5\\sqrt{3}$,\\\\\nFrom (1), quadrilateral $AODE$ is a rectangle,\\\\\n$\\therefore$ the area of quadrilateral $AODE$ $=OA\\cdot OD=5\\times 5\\sqrt{3}=\\boxed{25\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53333469_86.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{2}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $DE\\parallel BC$, $S_{\\triangle ADE} = S_{\\text{trapezoid } DBCE}$, then what is the ratio $DE: BC$?\n",
    "completion": "\\textbf{Solution:} Given that, $S_{\\triangle ADE}=S_{\\text{trapezoid} DBCE}$,\\\\\nthen $S_{\\triangle ADE} : S_{\\triangle ABC} = 1 : 2$,\\\\\nsince $DE \\parallel BC$,\\\\\nit follows that $\\triangle ADE \\sim \\triangle ABC$,\\\\\nlet the similarity ratio be $k$,\\\\\nthen the ratio of the areas is $k^2 = \\frac{1}{2}$,\\\\\nthus the similarity ratio is $\\frac{\\sqrt{2}}{2}$,\\\\\ntherefore $DE : BC = \\boxed{\\frac{\\sqrt{2}}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51284349.png"
  },
  {
    "solution": "\\boxed{7}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, where $AB=24$ and $BC=25$, an arc is drawn with point $B$ as the center and the length of $BC$ as the radius, intersecting side $AD$ at point $E$. What is the length of $AE$?\n",
    "completion": "\\textbf{Solution:}\nConnect BE as shown in the figure,\\\\\nas given, BE=BC=25,\\\\\nsince quadrilateral ABCD is a rectangle,\\\\\ntherefore, $\\angle A=\\angle D=90^\\circ$, AB=DC=24, AD=BC=25,\\\\\nin $\\triangle ABE$,\\\\\n$AE=\\sqrt{BE^{2}-AB^{2}}=\\sqrt{25^{2}-24^{2}}=\\boxed{7}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52355149.png"
  },
  {
    "solution": "\\boxed{\\frac{7}{25}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\triangle ABC$, $ AB=AC=5$, $ BC=6$, find the value of $\\cos A$.\n",
    "completion": "\\textbf{Solution:} Construct $AE\\perp BC$ at $E$, and construct $BD\\perp AC$ at $D$,\\\\\n$\\therefore$ $AB=AC=5$, $BC=6$,\\\\\n$\\therefore$ $BE=\\frac{1}{2}BC=3$,\\\\\n$\\therefore$ $AE=\\sqrt{AB^{2}-BE^{2}}=4$,\\\\\n$\\therefore$ the area of triangle $ABC =\\frac{1}{2}AC\\cdot BD=\\frac{1}{2}BC\\cdot AE$,\\\\\n$\\therefore$ $BD=\\frac{BC\\cdot AE}{AC}=\\frac{6\\times 4}{5}=\\frac{24}{5}$,\\\\\n$\\therefore$ $AD=\\sqrt{AB^{2}-BD^{2}}=\\sqrt{5^{2}-\\left(\\frac{24}{5}\\right)^{2}}=1.4$,\\\\\n$\\therefore$ $\\cos A=\\frac{AD}{AB}=\\frac{1.4}{5}=\\boxed{\\frac{7}{25}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51175037.png"
  },
  {
    "solution": "\\boxed{103^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, quadrilateral $ABCD$ is inscribed in circle $O$, with $\\overset{\\frown}{AD} = \\overset{\\frown}{CD}$, $\\angle ABD = 33^\\circ$, and $\\angle ACB = 44^\\circ$. What is the measure of $\\angle BAD$?\n",
    "completion": "Solution: Since $\\overset{\\frown}{CD}=\\overset{\\frown}{CD}$,\\\\\n$\\therefore \\angle DAC=\\angle DBC=33^\\circ$,\\\\\n$\\therefore \\angle BAD=\\angle DAC+\\angle CAB=33^\\circ+70^\\circ=\\boxed{103^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52774665_128.png"
  },
  {
    "solution": "\\boxed{65^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, point $P$ is the intersection of the perpendicular bisectors of $AB$ and $BC$, and point $Q$ is symmetric to point $P$ with respect to $AC$. Connect $PC$, $PQ$, and $CQ$. If in $\\triangle PCQ$ there is an angle of $50^\\circ$, then what is the measure of $\\angle B$?\n",
    "completion": "\\textbf{Solution:} Connect AP and BP as shown in the diagram: \\\\\n$\\because$ Point P is the intersection of the perpendicular bisectors of AB and BC,\\\\\n$\\therefore$ PA = PB = PC,\\\\\n$\\therefore$ $\\angle$PAB = $\\angle$PBA, $\\angle$PBC = $\\angle$PCB, $\\angle$PAC = $\\angle$PCA,\\\\\n$\\because$ Point Q is symmetric to point P with respect to AC,\\\\\n$\\therefore$ PC = QC, $\\angle$PCA = $\\angle$QCA,\\\\\n$\\therefore$ $\\angle$CPQ = $\\angle$CQP,\\\\\n(1) When $\\angle$CPQ = $\\angle$CQP = 50$^\\circ$, $\\angle$PCQ = 80$^\\circ$,\\\\\n$\\therefore$ $\\angle$PCA = 40$^\\circ$,\\\\\n$\\therefore$ $\\angle$PAC = 40$^\\circ$,\\\\\n$\\therefore$ $\\angle$PAB + $\\angle$PBA + $\\angle$PBC + $\\angle$PCB = 180$^\\circ$ - $\\angle$PAC - $\\angle$PCA = 100$^\\circ$,\\\\\n$\\therefore$ 2$\\angle$ABP + 2$\\angle$PBC = 100$^\\circ$,\\\\\n$\\therefore$ $\\angle$ABP + $\\angle$PBC = 50$^\\circ$, i.e., $\\angle$ABC = 50$^\\circ$,\\\\\n(2) When $\\angle$PCQ = 50$^\\circ$, $\\angle$PCA = 25$^\\circ$,\\\\\n$\\therefore$ $\\angle$PAC = 25$^\\circ$,\\\\\n$\\therefore$ $\\angle$PAB + $\\angle$PBA + $\\angle$PBC + $\\angle$PCB = 180$^\\circ$ - $\\angle$PAC - $\\angle$PCA = 130$^\\circ$,\\\\\n$\\therefore$ 2$\\angle$ABP + 2$\\angle$PBC = 130$^\\circ$,\\\\\n$\\therefore$ $\\angle$ABP + $\\angle$PBC = 65$^\\circ$, i.e., $\\angle$ABC = 65$^\\circ$,\\\\\nIn conclusion, $\\angle$ABC is \\boxed{50^\\circ} or \\boxed{65^\\circ}.",
    "image_path": "dataset/mm_math/images/MM_Math/52836749.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, line AB intersects with line CD at point O, $\\angle BOE = 90^\\circ$, $\\angle AOD = 30^\\circ$, and OF is the angle bisector of $\\angle BOD$. What is the degree measure of $\\angle EOC$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle BOC = \\angle AOD = 30^\\circ$ and $\\angle BOE = 90^\\circ$,\n\\[\n\\therefore \\angle COE = 90^\\circ - 30^\\circ = \\boxed{60^\\circ}.\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/46785011_222.png"
  },
  {
    "solution": "\\boxed{\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in triangles \\( \\triangle ABC \\) and \\( \\triangle ADE \\), we have \\( \\angle BAC = \\angle DAE = 90^\\circ \\), \\( AB = 3\\sqrt{3} \\), \\( AD = \\sqrt{3} \\), \\( BC = 6 \\), \\( DE = 2 \\). \\( \\triangle ABC \\sim \\triangle ADE \\); find the value of \\( \\frac{BD}{CE} \\).\n",
    "completion": "\\textbf{Solution:} From (1) we have $\\frac{AB}{AC} = \\frac{AD}{AE}$, \\\\\n$\\therefore \\frac{AB}{AD} = \\frac{AC}{AE}$, \\\\\n$\\because \\angle BAD=\\angle CAE=90^\\circ-\\angle CAD$, \\\\\n$\\therefore \\triangle ABD\\sim \\triangle ACE$, \\\\\n$\\therefore \\frac{BD}{CE} = \\frac{AB}{AC} = \\sqrt{3}$, \\\\\n$\\therefore$ the value of $\\frac{BD}{CE}$ is $\\boxed{\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52866587_119.png"
  },
  {
    "solution": "\\boxed{\\frac{29}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=20$, $AC=9$. The point $M$ is the midpoint of $BC$. The line $AD$ bisects the exterior angle $\\angle CAE$ of $\\triangle ABC$, intersecting the extended line of $BC$ at point $D$. Through point $M$, draw $MN \\parallel to AD$, intersecting $AB$ at point $N$. What is the length of $AN$?\n",
    "completion": "\\textbf{Solution:} Let NA intersect NF at BN, and connect CF, as shown in the diagram.\\\\\n$\\because$BM=MC, NF=BN,\\\\\n$\\therefore$MN$\\parallel$CF,\\\\\n$\\because$MN$\\parallel$AD,\\\\\n$\\therefore$CF$\\parallel$AD,\\\\\nThus, $\\angle$AFC=$\\angle$EAD, $\\angle$ACF=$\\angle$DAC,\\\\\n$\\because$AD bisects $\\angle$CAE,\\\\\n$\\therefore \\angle$DAC=$\\angle$EAD,\\\\\n$\\therefore \\angle$ACF=$\\angle$AFC,\\\\\n$\\therefore$AF=AC=9,\\\\\n$\\therefore$BF=AB−AF=11,\\\\\n$\\because$MN is the midline of $\\triangle$BCF,\\\\\n$\\therefore$BN=NF=$\\frac{11}{2}$,\\\\\n$\\therefore$AN=NF+AF=$\\boxed{\\frac{29}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52354580.png"
  },
  {
    "solution": "\\boxed{y = -x + 4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y=kx+b$ intersects the $x$ axis at point $A$, with $OA=4$, and intersects the graph of the direct proportion function $y=3x$ at point $B$, where the $x$-coordinate of point $B$ is 1. Find the equation of the linear function $y=kx+b$.\n",
    "completion": "\\textbf{Solution:} Given that $OA=4$, we have $A(4,0)$. Substituting $x=1$ into $y=3x$, we get $y=3$, thus $B(1,3)$. Substituting $A(4,0)$ and $B(1,3)$ into $y=kx+b$, we have the system of equations\n\\[\\left\\{\\begin{array}{l}\n4k + b = 0 \\\\\nk + b = 3\n\\end{array}\\right.\\]\nThus, we find that\n\\[\\left\\{\\begin{array}{l}\nk = -1 \\\\\nb = 4\n\\end{array}\\right.\\]\nTherefore, the linear equation is $\\boxed{y = -x + 4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52381215_152.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, points D and E are the midpoints of AB and AC, respectively. If the area of $\\triangle ADE$ is 1, what is the area of quadrilateral BDEC?\n",
    "completion": "Solution: Since points D and E are the midpoints of AB and AC respectively,\n\nit follows that DE is the midline of $\\triangle ABC$,\n\ntherefore DE$\\parallel$BC, and DE$=\\frac{1}{2}$BC,\n\nthus, $\\triangle ADE$ is similar to $\\triangle ABC$,\n\nwhich implies $\\frac{S_{\\triangle ADE}}{S_{\\triangle ABC}}=\\left(\\frac{DE}{BC}\\right)^{2}=\\frac{1}{4}$,\n\ngiven that the area of $\\triangle ADE$ is 1,\n\nit follows that $S_{\\triangle ABC}=4$,\n\nhence, the area of the quadrilateral BCED equals $S_{\\triangle ABC}-S_{\\triangle ADE}=3$,\n\nTherefore, the answer is: $S_{\\text{quadrilateral } BCED}=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51322485.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the circle $\\odot O$, $AB$ is the diameter of $\\odot O$, and the chord $CD\\perp AB$ at point $H$. If $AH=5$ and $HB=1$, then what is the length of $CD$?\n",
    "completion": "\\textbf{Solution:} Connect OD, \\\\\n$\\because$ AB is the diameter of $\\odot$O, and chord CD $\\perp$ AB,\\\\\n$\\therefore$ CD = 2DH,\\\\\n$\\because$ AH = 5, HB = 1,\\\\ \n$\\therefore$ AB = AH + HB = 6,\\\\\n$\\therefore$ OD = OA = 3,\\\\\n$\\therefore$ OH = AH - OA = 2,\\\\\nIn $\\triangle ODH$, DH = $\\sqrt{OD^{2}-OH^{2}} = \\sqrt{3^{2}-2^{2}} = \\sqrt{5}$,\\\\\n$\\therefore$ CD = 2DH = $2\\sqrt{5}$,\\\\\nTherefore, $CD=\\boxed{2\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52755172.png"
  },
  {
    "solution": "\\boxed{-2<x<0}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y_1=x+1$ intersects with the graph of the inverse proportion function $y_2= \\frac{2}{x}$ at points $A$ and $B$. A line is drawn from point $A$ perpendicular to the $x$-axis, forming a perpendicular foot at point $H$, and then connecting $BH$. When $y_1>y_2$, please directly state the range of the independent variable $x$ using the graph.\n",
    "completion": "Solution: As can be seen from the figure, when $y_1>y_2$, we have \\boxed{x>1} or \\boxed{-2<x<0}.",
    "image_path": "dataset/mm_math/images/MM_Math/50475225_91.png"
  },
  {
    "solution": "\\boxed{-x^{2}-2x+3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the parabola $y=-x^2+bx+c$ passes through points $A(-3,0)$ and $B(1,0)$ and intersects the $y$-axis at point $C$. Line segment $BC$ is drawn, and point $P$ is a moving point on the parabola, which is above the $x$-axis. A perpendicular line $PE$ is drawn from $P$ to the $x$-axis, with $E$ being the foot of the perpendicular. What is the equation of the parabola?\n",
    "completion": "\\textbf{Solution:} By substituting A ($-3$, $0$) and B ($1$, $0$) into $y=-x^{2}+bx+c$, we get:\n\\[\n\\left\\{\n\\begin{array}{l}\n0=-9+3b+c \\\\\n0=-1+b+c\n\\end{array}\n\\right.\n\\]\nSolving this, we find:\n\\[\n\\left\\{\n\\begin{array}{l}\nb=-2 \\\\\nc=3\n\\end{array}\n\\right.\n\\]\n$\\therefore$ The equation of the parabola is $y=-x^{2}-2x+3=\\boxed{-x^{2}-2x+3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51403139_38.png"
  },
  {
    "solution": "\\boxed{4.8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the quadrilateral ABCD is a rhombus, with the diagonal AC = 8 and BD = 6. If DH $\\perp$ AB at point H, what is the length of DH?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rhombus, AC=8, and BD=6, \\\\\ntherefore AC$\\perp$BD, OA=OC=4, and OB=OD=3, \\\\\nthus AB=5cm, \\\\\ntherefore the area of rhombus ABCD, $S_{\\text{rhombus} ABCD} = \\frac{1}{2} AC \\cdot BD = AB \\cdot DH$, \\\\\nthus $DH = \\frac{AC \\cdot BD}{2AB} = \\frac{6 \\times 8}{2 \\times 5} = \\boxed{4.8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53360040.png"
  },
  {
    "solution": "\\boxed{18\\frac{2}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system, the vertices of the rectangle $OABC$ with vertex O coinciding with the origin, and vertices A and C located on the coordinate axes respectively, whereas the coordinates of vertex B are $\\left(6, 4\\right)$. E is the midpoint of $AB$, and the line passing through points $D\\left(8, 0\\right)$ and E intersects with $BC$ and the y-axis at points F and G, respectively. Calculate the area of the quadrilateral $OHFG$.\n",
    "completion": "\\textbf{Solution:}\n\nThe equation of line $FH$ is: $y=\\frac{3}{2}x-2$,\n\nWhen $y=0$, we have $\\frac{3}{2}x-2=0$, thus $x=\\frac{4}{3}$,\n\nTherefore, point $H\\left(\\frac{4}{3}, 0\\right)$,\n\nSince $G$ is the intersection of line $DE$ with the y-axis, thus point $G\\left(0, 8\\right)$,\n\nTherefore, $OH=\\frac{4}{3}$, $CF=4$, $OC=4$, $CG=OG-OC=4$,\n\nTherefore, the area of quadrilateral $OHFG$ is the sum of the area of trapezoid $OHFC$ and the area of triangle $CFG$: ${S}_{\\text{quadrilateral} OHFG}={S}_{\\text{trapezoid} OHFC}+{S}_{\\triangle CFG}=\\frac{1}{2}\\times \\left(\\frac{4}{3}+4\\right)\\times 4+\\frac{1}{2}\\times 4\\times 4=\\boxed{18\\frac{2}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51663234_194.png"
  },
  {
    "solution": "\\boxed{\\frac{3\\sqrt{3}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In rhombus $ABCD$, $\\angle BAD=60^\\circ$, $AB=3$, point $E$ is on side $AB$, and point $F$ is a moving point on diagonal $AC$. Lines $BF$ and $EF$ are connected. What is the minimum value of $BF+EF$?\n",
    "completion": "Solution: Draw lines BD and DF, and construct DH$\\perp$AB at H.\\\\\nSince quadrilateral ABCD is a rhombus,\\\\\nit follows that $AD=AB$,\\\\\nSince $\\angle BAD=60^\\circ$,\\\\\n$\\triangle ADB$ is an equilateral triangle,\\\\\nSince B and D are symmetric about AC,\\\\\nit follows that $BF=DF$,\\\\\nThus, $BF+FE=DF+FE$,\\\\\nAccording to the principle that vertical segments are the shortest, when D, F, and E are collinear and coincide with DH, the value of $BF+EF$ is minimized, and the minimum value is the length of DH,\\\\\nIn $\\triangle ADH$, $\\angle BAD=60^\\circ$, $AD=AB=3$,\\\\\nthus $\\angle ADH=30^\\circ$\\\\\ntherefore $AH=\\frac{1}{2}AD=\\frac{3}{2}$,\\\\\ntherefore $DH=\\sqrt{AD^{2}-AH^{2}}=\\frac{3\\sqrt{3}}{2}$,\\\\\nHence, the minimum value of $BF+EF$ is $\\boxed{\\frac{3\\sqrt{3}}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53572645.png"
  },
  {
    "solution": "\\boxed{30\\text{cm}^2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the rectangle $ABCD$ is folded along the line $AE$, with the vertex $D$ precisely landing on point $F$ on side $BC$. Given that $CE=3\\,\\text{cm}$ and $AB=8\\,\\text{cm}$, what is the area of the shaded section?\n",
    "completion": "\\textbf{Solution:} Given the properties of folding, $EF=DE=CD-CE=5$, $AD=AF=BC$,\\\\\nby the Pythagorean theorem, $CF=4$, $AF^2=AB^2+BF^2$,\\\\\nthus $AD^2=8^2+(AD-4)^2$,\\\\\nSolving gives, $AD=10$,\\\\\n$\\therefore BF=6$,\\\\\nThe area of the shaded part in the diagram is $S_{\\triangle ABF}+S_{\\triangle CEF}=\\boxed{30\\text{cm}^2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53639871.png"
  },
  {
    "solution": "\\boxed{7}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in $\\triangle ABC$, $\\angle ACB=90^\\circ$. Squares are constructed externally on the three sides of $\\triangle ABC$, with their areas being $S_1$, $S_2$, and $S_3$, respectively, and it is given that $S_1=9$ and $S_3=16$. What is the value of $S_2$?\n",
    "completion": "\\textbf{Solution:} Given that ${S}_{1}=9$ and ${S}_{3}=16$,\\\\\nit follows that $BC^{2}=9$ and $AB^{2}=16$,\\\\\nsince $\\angle ACB=90^\\circ$,\\\\\nin $\\triangle ABC$,\\\\\nit implies that $BC^{2}+AC^{2}=AB^{2}$,\\\\\ntherefore $9+AC^{2}=16$,\\\\\nhence $AC^{2}=7$,\\\\\nthus, ${S}_{2}=\\boxed{7}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52699615.png"
  },
  {
    "solution": "\\boxed{y=\\frac{4}{x}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, if the graph of the linear function $y=-x+5$ intersects with the graph of the reciprocal function $y=\\frac{k}{x}$ ($k\\ne 0$) at points $A(1, n)$ and $B$ in the first quadrant, what is the equation of the reciprocal function?\n",
    "completion": "\\textbf{Solution:} Since $A(1, 4)$ lies on the line $y=-x+5$,\\\\\nit follows that $n=-1+5=4$,\\\\\ntherefore $A(1, 4)$,\\\\\nsubstituting $A(1, 4)$ into $y=\\frac{k}{x}$ yields $k=4$,\\\\\ntherefore, the equation of the inverse proportion function is $\\boxed{y=\\frac{4}{x}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52551188_145.png"
  },
  {
    "solution": "\\boxed{y=x+1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given that in the Cartesian coordinate system, point $A(1, 2)$ lies on the graph of the inverse proportion function $y=\\frac{k_1}{x}$, and the line passing through point $A$ intersects the other branch of the hyperbola at point $B(-2, m)$. What is the analytical equation of line $AB$?\n",
    "completion": "\\textbf{Solution:} To solve, substitute point A$(1,2)$ into $y=\\frac{2}{x}$, solving for $k_1=2$,\\\\\n$\\therefore$ the inverse proportion function is $y=\\frac{2}{x}$.\\\\\nSubstituting point B$(-2,m)$ into $y=\\frac{2}{x}$ yields $m=-1$,\\\\\n$\\therefore$ the coordinates of point B are $(-2,-1)$.\\\\\nLet the equation for line AB be $y=kx+b$,\\\\\nSubstituting A$(1,2)$ and B$(-2,-1)$ into the equation results in\n\\[\n\\left\\{\n\\begin{array}{l}\nk+b=2 \\\\\n-2k+b=-1\n\\end{array}\n\\right.\n\\]\nSolving this system gives\n\\[\n\\left\\{\n\\begin{array}{l}\nk=1 \\\\\nb=1\n\\end{array}\n\\right.\n\\]\n$\\therefore$ the functional expression for line AB is $\\boxed{y=x+1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50555566_84.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the square grid, the position of $\\triangle ABC$ is shown in the figure, what is the value of $\\sin\\angle BAC$?\n",
    "completion": "\\textbf{Solution:} Connect the lattice points DC, BD.\\\\\nIn $\\triangle ABD$,\\\\\n$\\because CD=3$, $AD=4$,\\\\\n$\\therefore AC= \\sqrt{CD^2+AD^2}=5$.\\\\\n$\\therefore \\sin\\angle BAC= \\frac{CD}{AC}=\\boxed{\\frac{3}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51446308.png"
  },
  {
    "solution": "\\boxed{\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $E$ is a point on the diagonal $AC$ of the square $ABCD$. Connect $DE$ and draw $EF\\perp DE$ to intersect ray $BC$ at point $F$. Construct rectangle $DEFG$ with $DE$ and $EF$ as adjacent sides, and connect $CG$. Rectangle $DEFG$ is a square; if $AB=4$ and $CE=3\\sqrt{2}$, what is the length of $CG$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw EH $\\perp$ AB at point H, \\\\\nthen $\\angle EHB=90^\\circ$, since $\\angle ENB=90^\\circ$, and $\\angle B=90^\\circ$, thus quadrilateral BNEH is a rectangle, and therefore $EN=BH$. Within square ABCD, $AD=CD$, and $\\angle ADC=90^\\circ$. In square DEFG, $DE=DG$, and $\\angle EDG=90^\\circ$, thus $\\angle ADE=\\angle CDG$. In $\\triangle ADE$ and $\\triangle CDG$, we have $\\left\\{ \\begin{array}{l} AD=CD \\\\ \\angle ADE=\\angle CDG \\\\ DE=DG \\end{array} \\right.$, therefore $\\triangle ADE \\cong \\triangle CDG$ (SAS), hence $CG=AE$. Since $CE=3\\sqrt{2}$, $\\angle ECN=45^\\circ$, and $\\angle ENC=90^\\circ$, therefore $\\angle NEC=45^\\circ$, thus $NC=NE=3$. Given that $AB=4$, thus $AH=4-3=1$. As $\\angle HAE=45^\\circ$, therefore $\\angle HEA=45^\\circ$, thus $AH=HE=1$, hence $AE=\\sqrt{1^2+1^2}=\\sqrt{2}$, thus $CG=\\boxed{\\sqrt{2}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52354937_88.png"
  },
  {
    "solution": "\\boxed{18}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $l_{1}$, $l_{2}$, $l_{3}$, $l_{4}$ are a set of parallel lines, $l_{5}$, $l_{6}$ intersect this set of parallel lines successively at points A, B, C, D, and E, F, G, H, respectively. If the ratio of $AB:BC:CD$ is $2:3:4$, and $EG=10$, then what is the length of $EH$?\n",
    "completion": "\\textbf{Solution:}\\\\\nSince $AB:BC:CD=2:3:4$,\\\\\nit follows that $AC:CD=5:4$,\\\\\nSince $l_{1}\\parallel l_{3}\\parallel l_{4}$,\\\\\nit results in $AC:CD=EG:GH$, i.e., $5:4=10:GH$,\\\\\nthus $GH=8$,\\\\\nUpon verification, $GH=8$ is a solution to the given equation and is consistent with the problem statement,\\\\\ntherefore $EH=EG+GH=10+8=\\boxed{18}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52935299.png"
  },
  {
    "solution": "\\boxed{12\\,cm}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the length of line segment $AD=30\\,cm$, points C and B are both on line segment $AD$, and point E is the midpoint of $AB$. If $BD=6\\,cm$, what is the length of line segment $AE$ in $cm$?\n",
    "completion": "\\textbf{Solution:} Given $\\because AD=30\\,cm$, $BD=6\\,cm$,\\\\\n$\\therefore AB=AD-BD=24\\,cm$,\\\\\n$\\because E$ is the midpoint of $AB$,\\\\\n$\\therefore AE= \\boxed{12\\,cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51482139_104.png"
  },
  {
    "solution": "\\boxed{\\left(\\frac{25}{6},0\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the quadrilateral $OABC$ is a square piece of paper placed in the Cartesian coordinate system, where point $O$ coincides with the origin, point $A$ is on the x-axis, and point $C$ is on the y-axis, with $OC=5$. Point $E$ is on side $BC$, and the coordinates of point $N$ are $\\left(3,0\\right)$. A line through point $N$ and parallel to the y-axis, $MN$, intersects $EB$ at point $M$. Now, the paper is folded so that vertex $C$ lands on point $G$ on $MN$, with the fold line being $OE$. On the positive x-axis, there exists a point $P$ such that the triangle formed by $P$, $O$, and $G$ is an isosceles triangle. What are the coordinates of point $P$?\n",
    "completion": "\\textbf{Solution:} Given that the coordinates of point N are $(3,0)$, \\\\\nhence $ON=3$, \\\\\nfrom the folding, we have $OG=OC=5$, \\\\\nsince $\\angle ONG=90^\\circ$, \\\\\nthus $NG=\\sqrt{OG^{2}-ON^{2}}=\\sqrt{5^{2}-3^{2}}=4$, \\\\\ntherefore, $G(3,4)$, \\\\\nas shown in Figure 1, when $OP=OG=5$, \\\\\nwe get $P(5,0)$; \\\\\nas shown in Figure 2, when $PG=OG$, \\\\\nsince $GN\\perp PO$, \\\\\nthus $PN=ON=3$, \\\\\ntherefore $OP=6$, \\\\\nhence, $P(6,0)$; \\\\\nas shown in Figure 3, when $PO=PG$, let $P(x,0)$, \\\\\nfrom $PO=PG$ we have $PO^{2}=PG^{2}$, \\\\\nthus $x^{2}=(x-3)^{2}+(0-4)^{2}$, \\\\\nsolving this gives $x=\\frac{25}{6}$, \\\\\ntherefore, $P\\left(\\frac{25}{6},0\\right)$; \\\\\nIn summary, the coordinates of point P are $(5,0)$ or $(6,0)$ or $\\left(\\frac{25}{6},0\\right)$, \\\\\nthus, the answer is: $\\boxed{(5,0)}$ or $\\boxed{(6,0)}$ or $\\boxed{\\left(\\frac{25}{6},0\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52891079.png"
  },
  {
    "solution": "\\boxed{\\frac{30}{13}\\le AM<6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle BAC=90^\\circ$, $AB=5$, $AC=12$. $P$ is a moving point on side $BC$, $PE\\perp AB$ at $E$, $PF\\perp AC$ at $F$, and $M$ is the midpoint of $EF$. What is the range of values for $AM$?\n",
    "completion": "\\textbf{Solution:} Let's consider that PE$\\perp$AB and PF$\\perp$AC, and $\\angle$BAC$=90^\\circ$, \\\\\ntherefore, quadrilateral AEPF is a rectangle, \\\\\nsince M is the midpoint of EF, \\\\\nthus AM$= \\frac{1}{2}$AP, and when AP$\\perp$BC, the value of AP is minimized, and when point P coincides with point C, the value of AC is maximized, \\\\\nthus AP$= \\frac{AB\\times AC}{BC}=\\frac{60}{13}$, and the maximum value of AP is 12, \\\\\ntherefore, $ \\frac{30}{13}\\le AM<6$. \\\\\nHence, the answer is: $ \\boxed{\\frac{30}{13}\\le AM<6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52386890.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, AB is the diameter of $\\odot O$, PD is tangent to $\\odot O$ at point C, and intersects the extension line of AB at point D, and $\\angle D = 2\\angle CAD$. What is the degree measure of $\\angle D$?\n",
    "completion": "\\textbf{Solution:} Since $OA=OC$,\\\\\nit follows that $\\angle A=\\angle ACO$,\\\\\nthus $\\angle COD=\\angle A+\\angle ACO=2\\angle A$,\\\\\ngiven $\\angle D=2\\angle A$,\\\\\nit implies $\\angle D=\\angle COD$,\\\\\nsince line $PD$ is tangent to circle $O$ at point $C$,\\\\\ntherefore $\\angle OCD=90^\\circ$,\\\\\nhence $\\angle D=\\angle COD=\\boxed{45^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51446430_115.png"
  },
  {
    "solution": "\\boxed{20}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in right triangle $ABC$ with $\\angle ACB=90^\\circ$, we have $AB=25$ and $AC=15$. Point $P$ starts from point $B$ and moves towards $A$ at a speed of $2$ units per second. Simultaneously, point $Q$ starts from $C$ and moves towards $B$ at a speed of $1$ unit per second. When one point reaches its endpoint, the other point stops moving. Connect $PQ$ and on the ray $PB$, take $PN=PQ$ to form quadrilateral $PQMN$ with $PN$ and $PQ$ as sides. Let the movement time be $t$ seconds ($t>0$). What is the length of $BC$?\n",
    "completion": "\\textbf{Solution:} From the given conditions, we have $BC=\\boxed{20}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51445941_162.png"
  },
  {
    "solution": "\\boxed{2 + 2\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, quadrilateral ABCD is a parallelogram. The circle $\\odot O$ with diameter AD intersects AB at point E. Line DE is drawn, DA $= 2\\sqrt{2}$, DE $= \\sqrt{7}$, and DC $= 5$. A line $l$ is drawn through point E. Through point C, draw CH $\\perp l$, with the foot of the perpendicular being H. Line BH is connected. As line $l$ rotates around point E, what is the maximum value of BH?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect CE and take the midpoint K of CE. Draw KM $\\perp$ BE at M,\\\\\n$\\because$ CH $\\perp$ EH,\\\\\n$\\therefore \\angle CHE = 90^\\circ$,\\\\\n$\\therefore$ H lies on the circle centered at K with CE as diameter,\\\\\n$\\because BH \\le HK + BK$,\\\\\n$\\therefore$ as shown in the figure, when H moves to the position of H, i.e., H, B, K are collinear at this moment, BH reaches its maximum value $BH$,\\\\\n$\\because$ quadrilateral ABCD is a parallelogram,\\\\\n$\\therefore$ AB = CD = 5, AB $\\parallel$ CD,\\\\\n$\\therefore$ BE = AB - AE = 4, $\\angle CDE = \\angle AED = 90^\\circ$, $\\angle DCE = \\angle MEK$,\\\\\n$\\therefore$ CE = KE = $\\sqrt{DE^2 + CD^2} = 4\\sqrt{2}$,\\\\\n$\\therefore$ KH = $\\frac{1}{2}CE = 2\\sqrt{2}$,\\\\\n$\\because \\angle CDE = \\angle EMK = 90^\\circ$,\\\\\n$\\therefore \\triangle CDE \\sim \\triangle EMK$,\\\\\n$\\therefore \\frac{KM}{DE} = \\frac{EK}{CE} = \\frac{EM}{CD} = \\frac{1}{2}$,\\\\\n$\\therefore$ KM = $\\frac{1}{2}DE = \\frac{\\sqrt{7}}{2}$, $EM = \\frac{1}{2}CD = \\frac{5}{2}$,\\\\\n$\\therefore$ BM = AB - AE - EM = $\\frac{3}{2}$,\\\\\n$\\therefore$ BK = $\\sqrt{KM^2 + BM^2} = 2$,\\\\\n$\\therefore$ BH = $2 + 2\\sqrt{2}$,\\\\\n$\\therefore$ The maximum value of BH is $\\boxed{2 + 2\\sqrt{2}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51351656_170.png"
  },
  {
    "solution": "\\boxed{90^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In $\\triangle ABC$, $\\angle B=40^\\circ$, $D$ is a point on side $BC$. The triangle is folded along $AD$ such that $AC$ falls on $AB$ and point $C$ coincides with point $E$. If $\\triangle BDE$ is a right-angled triangle, what is the measure of $\\angle C$?\n",
    "completion": "\\textbf{Solution:} When $\\angle BED=90^\\circ$, $\\angle C=\\angle AED=90^\\circ$.\\\\\nWhen $\\angle BDE=90^\\circ$, $\\angle C=\\angle AED=\\angle B+\\angle BDE=130^\\circ$.\\\\\nIn summary, the measure of $\\angle C$ is either $90^\\circ$ or $130^\\circ$.\\\\\nTherefore, the answer is: $\\boxed{90^\\circ}$ or $130^\\circ$.",
    "image_path": "dataset/mm_math/images/MM_Math/53009754.png"
  },
  {
    "solution": "\\boxed{115^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: During our study in the latter half of seventh grade, we delved into the problems concerning the angles between bisectors of interior angles and the angles between external and internal angle bisectors. Throughout this process, Kun found a new category of problems during his independent exploration. His exploration process is as follows:\n\n\\textbf{Exploratory Research}: As shown in Figure 1, in $\\triangle ABC$, $BM$ bisects $\\angle ABC$ and $CM$ bisects $\\angle ACB$, with $BM$ and $CM$ intersecting at point $M$. If $\\angle A=50^\\circ$, then what is the measure of $\\angle BMC$?\n",
    "completion": "\\textbf{Solution:} We have $\\angle BMC = \\boxed{115^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/54661756_91.png"
  },
  {
    "solution": "\\boxed{Q\\left(3, 2\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, the parabola $y=ax^2+bx+\\frac{5}{2}$ intersects line $AB$ at points $A(-1, 0)$ and $B(4, \\frac{5}{2})$. Point $D$ is a moving point on the parabola between points $A$ and $B$ (not coinciding with points $A$ and $B$), and line $CD$ is parallel to the y-axis, intersecting line $AB$ at point $C$. Lines $AD$ and $BD$ are drawn. Point $D$ is the vertex of the parabola, point $P$ is a moving point on the parabola, and point $Q$ is a moving point on line $AB$. When the quadrilateral with vertices at points $P$, $Q$, $C$, and $D$ forms a parallelogram, determine the coordinates of point $Q$.\n",
    "completion": "Solution: Suppose there exist points P and Q such that the quadrilateral formed by the vertices P, Q, C, and D is a parallelogram.\\\\\n$\\because y=-\\frac{1}{2}x^2+2x+\\frac{5}{2}=-\\frac{1}{2}(x-2)^2+\\frac{9}{2}$, point D is the vertex of the parabola, $CD\\parallel y$-axis,\\\\\n$\\therefore D(2, \\frac{9}{2})$, $C(2, \\frac{3}{2})$,\\\\\n(1) When $CD$ is a side of the parallelogram, $PQ\\parallel CD, PQ=CD$ \\\\\\\\\nLet $P(x, -\\frac{1}{2}x^2+2x+\\frac{5}{2})$, then: $Q(x, \\frac{1}{2}x+\\frac{1}{2})$\\\\\nAs shown in the figure, when P is above Q: $PQ=-\\frac{1}{2}x^2+2x+\\frac{5}{2}-\\frac{1}{2}x-\\frac{1}{2}=3$\\\\\nSolving, $x=1$ or $x=2$ (discard),\\\\\n$\\therefore \\boxed{Q\\left(1, 1\\right)}$;\\\\\nWhen P is below Q: $PQ=\\frac{1}{2}x+\\frac{1}{2}+\\frac{1}{2}x^2-2x-\\frac{5}{2}=3$,\\\\\nSolving: $x=5$ or $x=-2$,\\\\\n$\\therefore \\boxed{Q\\left(5, 3\\right)}$ or \\boxed{Q\\left(-2, -\\frac{1}{2}\\right)};\\\\\n(2) When $CD$ is a diagonal of the parallelogram,\\\\\nDraw $PE\\perp CD$ at point E from point P, and draw $QF\\perp CD$ at point F from point Q, then $PE=QF, DE=FC$.\\\\\nLet $P(x, -\\frac{1}{2}x^2+2x+\\frac{5}{2})$, then $E(2, -\\frac{1}{2}x^2+2x+\\frac{5}{2})$,\\\\\n$\\therefore Q(4-x, \\frac{5}{2}-\\frac{1}{2}x)$, $F(2, \\frac{5}{2}-\\frac{1}{2}x)$,\\\\\nFrom $DE=CF$, we get $\\frac{9}{2}-(\\frac{5}{2}-\\frac{1}{2}x^2+2x+\\frac{5}{2})=\\frac{5}{2}-\\frac{1}{2}x-\\frac{3}{2}$,\\\\\nSolving, $x=1$ or $x=2$ (discard),\\\\\n$\\therefore \\boxed{Q\\left(3, 2\\right)}$\\\\\nIn summary: when \\boxed{Q\\left(1, 1\\right)} or \\boxed{Q\\left(5, 3\\right)} or \\boxed{Q\\left(-2, -\\frac{1}{2}\\right)} or \\boxed{Q\\left(3, 2\\right)}, the quadrilateral formed by the vertices P, Q, C, and D is a parallelogram.",
    "image_path": "dataset/mm_math/images/MM_Math/53043850_142.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the line $y_1=k_1x+b$ intersects with the hyperbola $y_2=\\frac{k_2}{x}$ at points $A(4, 2)$ and $B(m, -4)$. A line is drawn from point B perpendicular to the x-axis intersecting it at point P. What is the area of $\\triangle ABP$?\n",
    "completion": "\\textbf{Solution:} Given $\\because A(4, 2)$, $B(-2, -4)$, $\\therefore BP=4$, $\\therefore {S}_{\\triangle ABP}=\\frac{1}{2}\\times 4\\times (4+2)=\\boxed{12}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52381008_157.png"
  },
  {
    "solution": "\\boxed{\\sqrt{5}-2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the area of square $ABCD$ is $5$ and the area of square $BEFG$ is $4$. What is the area of $\\triangle GCE$?\n",
    "completion": "\\textbf{Solution:} Given that the area of square ABCD is 5, and the area of square BEFG is 4,\\\\\nit follows that the side length of square ABCD is $\\sqrt{5}$, and the side length of square BEFG is 2,\\\\\ntherefore, CE＝$\\sqrt{5}-2$,\\\\\nthe area of $\\triangle GCE$＝$\\frac{1}{2}$CE·BG＝$\\frac{1}{2}\\times(\\sqrt{5}-2)\\times2=\\boxed{\\sqrt{5}-2}$",
    "image_path": "dataset/mm_math/images/MM_Math/51510280.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, D and E are points on sides AB and AC of $\\triangle ABC$ respectively, with DE $\\parallel$ BC, and the area ratio $S_{\\triangle ADE} : S_{\\triangle ABC} = 1 : 9$. What is the value of the ratio AD : BD?\n",
    "completion": "\\textbf{Solution:} Since $DE\\parallel BC$,\\\\\nit follows that $\\triangle ADE\\sim \\triangle ABC$,\\\\\nwhich implies $\\frac{S_{\\triangle ADE}}{S_{\\triangle ABC}}=\\frac{AD^2}{AB^2}=\\frac{1}{9}$\\\\\nTherefore, $\\frac{AD}{AB}=\\frac{1}{3}$\\\\\nConsequently, $\\frac{AD}{BD}=\\boxed{\\frac{1}{2}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51458070.png"
  },
  {
    "solution": "\\boxed{\\frac{5}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the figure, within $\\triangle ABC$, $\\angle ACB=90^\\circ$, CD is the median. Extend CB to point E such that BE=BC; connect DE, let F be the midpoint of DE, and connect BF. If AC=8, BC=6, what is the length of BF?\n",
    "completion": "\\textbf{Solution:} In right-angled triangle $\\triangle ABC$, $\\angle ACB=90^\\circ$, $AC=8$, and $BC=6$,\\\\\n$\\therefore AB=\\sqrt{AC^{2}+BC^{2}}=10$,\\\\\n$\\because CD$ is the median to the hypotenuse $AB$,\\\\\n$\\therefore CD=\\frac{1}{2}AB=5$,\\\\\n$\\because BE=BC$, and $F$ is the midpoint of $DE$,\\\\\n$\\therefore BF=\\frac{1}{2}CD=\\boxed{\\frac{5}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52189596.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{2}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the square $ABCD$ with a side length of 2, points $E$ and $F$ are the midpoints of sides $AB$ and $BC$, respectively. Connect $EC$ and $FD$. Points $G$ and $H$ are the midpoints of $EC$ and $FD$, respectively. Connect $GH$. What is the length of $GH$?\n",
    "completion": "\\textbf{Solution:} As illustrated, let $DF$ and $CE$ intersect at $O$,\\\\\n$\\because$ Quadrilateral $ABCD$ is a square,\\\\\n$\\therefore \\angle B=\\angle DCF=90^\\circ$, $BC=CD=AB=2$,\\\\\n$\\because$ Points $E$ and $F$ are the midpoints of sides $AB$ and $BC$ respectively,\\\\\n$\\therefore$ $BE=CF=1$,\\\\\n$\\therefore \\triangle CBE \\cong \\triangle DCF$ (SAS),\\\\\n$\\therefore CE=DF$, $\\angle BCE=\\angle CDF$,\\\\\n$\\because \\angle CDF+\\angle CFD=90^\\circ$,\\\\\n$\\therefore \\angle BCE+\\angle CFD=90^\\circ$,\\\\\n$\\therefore \\angle COF=90^\\circ$,\\\\\n$\\therefore$ $DF \\perp CE$,\\\\\n$\\therefore CE=DF=\\sqrt{2^2+1^2}=\\sqrt{5}$,\\\\\n$\\because$ Points $G$ and $H$ are the midpoints of $EC$ and $FD$ respectively,\\\\\n$\\therefore$ $CG=FH=\\frac{\\sqrt{5}}{2}$,\\\\\n$\\because \\angle DCF=90^\\circ$, $CO \\perp DF$,\\\\\n$\\therefore CF^2=OF\\cdot DF$,\\\\\n$\\therefore OF=\\frac{CF^2}{DF}=\\frac{1}{\\sqrt{5}}=\\frac{\\sqrt{5}}{5}$,\\\\\n$\\therefore$ $OH=\\frac{3\\sqrt{5}}{10}$, $OD=\\frac{4\\sqrt{5}}{5}$,\\\\\n$\\because OC^2=OF\\cdot OD$,\\\\\n$\\therefore OC=\\sqrt{\\frac{\\sqrt{5}}{5}\\times \\frac{4\\sqrt{5}}{5}}=\\frac{2\\sqrt{5}}{5}$,\\\\\n$\\therefore$ $OG=CG-OC=\\frac{\\sqrt{5}}{2}-\\frac{2\\sqrt{5}}{5}=\\frac{\\sqrt{5}}{10}$,\\\\\n$\\therefore$ $HG=\\sqrt{OG^2+OH^2}=\\sqrt{\\left(\\frac{\\sqrt{5}}{10}\\right)^2+\\left(\\frac{3\\sqrt{5}}{10}\\right)^2}=\\boxed{\\frac{\\sqrt{2}}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52688312.png"
  },
  {
    "solution": "\\boxed{65^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $OC$ is a ray inside $\\angle AOB$, $OD$ bisects $\\angle BOC$, and $OE$ bisects $\\angle AOC$. If $\\angle AOB=130^\\circ$, find the degree measure of $\\angle DOE$.\n",
    "completion": "\\textbf{Solution:} When $\\angle AOB = 130^\\circ$,\\\\\nwe have $\\angle DOE = \\frac{1}{2}\\angle AOB$\\\\\n$= \\frac{1}{2}\\times 130^\\circ$\\\\\n$= 65^\\circ$.\\\\\nTherefore, the measure of $\\angle DOE$ is $\\boxed{65^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52949136_165.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the diamond $ABCD$, $AB=BD=5$, what is the degree measure of $\\angle BAC$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rhombus, \\\\\n$\\therefore$ AB=AD, and AC bisects $\\angle$BAD, \\\\\nSince AB=BD=5, \\\\\n$\\therefore$ AB=AD=BD, \\\\\n$\\therefore$ $\\triangle$ABD is an equilateral triangle, \\\\\n$\\therefore$ $\\angle$BAD$=60^\\circ$, \\\\\n$\\therefore$ $\\angle$BAC$=\\boxed{30^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53359704_116.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is a chord of the circle $\\odot O$, $OD\\perp AB$, the foot of the perpendicular is $E$, intersecting $\\odot O$ at points $C$ and $D$. Given that $AB=2\\sqrt{5}$ and $ED=1$, what is the length of the radius of $\\odot O$?\n",
    "completion": "\\textbf{Solution:} Let the radius be $r$. Since $OD\\perp AB$ and $AB=2\\sqrt{5}$, it follows that $AE=BE=\\frac{1}{2}AB=\\sqrt{5}$. In $\\triangle AOE$, we have $OE^2+AE^2=OA^2$. Given $ED=1$ and the radius is $r$, we have $OE=r-1$. Thus, $(r-1)^2+(\\sqrt{5})^2=r^2$, which solves to $r=\\boxed{3}$. Therefore, $OA=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53081489_82.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the graph of the line $y=2x+3$ and the graph of the line $y=-2x-1$ are as shown in the figure. Find the area of $\\triangle ABC$.\n",
    "completion": "\\textbf{Solution:} Draw $CD\\perp AB$ through point C, intersecting the y-axis at point D;\\\\\n$\\therefore CD=1$;\\\\\n$\\because AB=3-(-1)=4$;\\\\\n$\\therefore S_{\\triangle ABC}=\\frac{1}{2}AB\\cdot CD=\\frac{1}{2}\\times 4\\times 1=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51699291_163.png"
  },
  {
    "solution": "\\boxed{130^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\angle A=70^\\circ$, $\\angle ABD=\\angle BCE=30^\\circ$, and $CE$ bisects $\\angle ACB$. What is the measure of $\\angle BEC$ in degrees?\n",
    "completion": "\\textbf{Solution:} Given $CE$ bisects $\\angle ACB$,\\\\\nhence $\\angle ACE = \\angle BCE = 30^\\circ$,\\\\\nsince $\\angle A = 70^\\circ$,\\\\\nthus $\\angle ABC + \\angle ACB = 110^\\circ$,\\\\\ngiven $\\angle ABE = \\angle ACE = 30^\\circ$,\\\\\nthus $\\angle EBC + \\angle ECB = 110^\\circ - 60^\\circ = 50^\\circ$,\\\\\ntherefore $\\angle BEC = 180^\\circ - 50^\\circ = \\boxed{130^\\circ}$,",
    "image_path": "dataset/mm_math/images/MM_Math/52096232.png"
  },
  {
    "solution": "\\boxed{-7}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the inverse proportion function $y=\\frac{k}{x}$ passes through the vertex $P$ of the rectangle $AOBP$, and the two sides $AO$, $BO$ of the rectangle are respectively on the x-axis and y-axis. If the area of the rectangle is $7$, what is the value of $k$?\n",
    "completion": "\\textbf{Solution}: Since the area of rectangle $OAPB$ is 7\\\\\nit follows that $\\left|k\\right|=7$\\\\\nGiven that the graph is in the second quadrant, it implies $ k=-7$\\\\\nTherefore, the answer is: $k=\\boxed{-7}$",
    "image_path": "dataset/mm_math/images/MM_Math/52602480.png"
  },
  {
    "solution": "\\boxed{-2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y=-2x$ intersects with the graph of $y=kx+b$ at point $A(-2, 4)$. What is the solution of the equation $kx+b+2x=0$ with respect to $x$?\n",
    "completion": "\\textbf{Solution:} Since the graph of the linear function $y=-2x$ intersects with the graph of $y=kx+b$ at point A ($-2$, $4$), \\\\\ntherefore, the solution to the equation in $x$, $-2x=kx+b$, for which $x=-2$, i.e., the solution to $kx+b+2x=0$ is $x=\\boxed{-2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52251666.png"
  },
  {
    "solution": "\\boxed{(-1, 2)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=kx+b$ passes through points $A(2,1)$ and $B(-1,-2)$. What is the solution set for the inequality $-2<kx+b<1$?\n",
    "completion": "\\textbf{Solution:} Because the line $y=kx+b$ passes through points $A(2, 1)$ and $B(-1, -2)$, \\\\\ntherefore $\\begin{cases}2k+b=1\\\\ -k+b=-2\\end{cases}$, solving this yields $\\begin{cases}k=1\\\\ b=-1\\end{cases}$,\\\\\nBecause $-2<kx+b<1$,\\\\\ntherefore $-2<x-1<1$,\\\\\ntherefore $\\begin{cases}x-1<1\\\\ x-1>-2\\end{cases}$, solving this gives $-1<x<2$,\\\\\nhence, the answer is $x \\in \\boxed{(-1, 2)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50555959.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{61}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in $\\square ABCD$, $AB=13$, $AD=5$, and $AC\\perp BC$, what is the length of $BD$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, draw DE $\\perp$ BC, intersecting the extended line of BC at E,\\\\\nsince quadrilateral ABCD is a parallelogram,\\\\\ntherefore BC=AD=5, AD$\\parallel$BC,\\\\\nsince $AC\\perp BC$,\\\\\ntherefore $\\angle ACB=90^\\circ$,\\\\\nby the Pythagorean theorem, we have $AC= \\sqrt{AB^{2}-BC^{2}}=\\sqrt{13^{2}-5^{2}}=12$,\\\\\nsince DE$\\perp$BC,\\\\\ntherefore AC$\\parallel$DE,\\\\\ntherefore quadrilateral ACED is a parallelogram,\\\\\ntherefore CE=AD=5, DE=AC=12,\\\\\ntherefore BE=BC+CE=10,\\\\\nsince DE$\\perp$BC,\\\\\ntherefore, by the Pythagorean theorem, we have $BD= \\sqrt{BE^{2}+DE^{2}}=\\sqrt{12^{2}+10^{2}}=\\boxed{2\\sqrt{61}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52250417.png"
  },
  {
    "solution": "\\boxed{35^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in square $ABCD$, $E$ is a point on side $CD$, and $F$ is a point on the extension of side $BC$, with $CE=CF$. If $\\angle BEC=80^\\circ$, what is the degree measure of $\\angle EFD$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a square,\\\\\nit follows that BC=CD, and $\\angle$BCE=$\\angle$DCF=90$^\\circ$,\\\\\nsince CE=CF,\\\\\nit follows that $\\triangle$BCE$\\cong$$\\triangle$DCF, $\\angle$CFE=$\\angle$CEF=45$^\\circ$,\\\\\ntherefore $\\angle$CFD=$\\angle$BEC=80$^\\circ$,\\\\\ntherefore $\\angle$EFD=$\\angle$CFD−$\\angle$CFE=80$^\\circ$−45$^\\circ=\\boxed{35^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51614616.png"
  },
  {
    "solution": "\\boxed{50^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given in the figure, points B, D, and C lie on a straight line, with AB = AD, AC = AE, and $\\angle BAD = \\angle CAE$. If $\\angle B = 65^\\circ$ and BC = DE, what is the measure of $\\angle CDE$?\n",
    "completion": "\\textbf{Solution:} Given that $AB=AD$ and $\\angle B=65^\\circ$, \\\\\nit follows that $\\angle ADB=\\angle B=65^\\circ$, \\\\\nsince $\\triangle BAC \\cong \\triangle DAE$, \\\\\nit implies that $\\angle ADE=\\angle B=65^\\circ$, \\\\\ntherefore, $\\angle CDE=180^\\circ-\\angle ADE-\\angle ADB \\\\\n=180^\\circ-2\\times65^\\circ \\\\\n=\\boxed{50^\\circ}.$",
    "image_path": "dataset/mm_math/images/MM_Math/52861964_99.png"
  },
  {
    "solution": "\\boxed{25^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC$, and D is the midpoint of BC. If $\\angle BAC=50^\\circ$, what is the measure of $\\angle BAD$?\n",
    "completion": "\\textbf{Solution:} Since $AB=AC$, and point $D$ is the midpoint of $BC$, $\\angle BAC=50^\\circ$,\n\n$\\therefore AD$ is the angle bisector of $\\angle BAC$,\n\n$\\therefore \\angle BAD=\\angle CAD=\\frac{1}{2}\\angle BAC=\\boxed{25^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/51545511.png"
  },
  {
    "solution": "\\boxed{(3,2)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the line $y=kx+5$ intersects the $x$-axis at point $A$ and the $y$-axis at point $B$, and the coordinates of point $A$ are $(5,0)$. The line $y=2x-4$ intersects the $x$-axis at point $D$ and intersects the line $AB$ at point $C$. What are the coordinates of point $C$?\n",
    "completion": "\\textbf{Solution:} Given that the line $y=kx+5$ passes through point $A(5,0)$,\\\\\nthus $5k+5=0$ and we find $k=-1$.\\\\\nTherefore, the equation of line $AB$ is: $y=-x+5$;\\\\\nBy solving the system of equations for lines $AB$ and $CD$ simultaneously $\\left\\{\\begin{array}{l}y=-x+5 \\\\ y=2x-4\\end{array}\\right.$, we find: $\\left\\{\\begin{array}{l}x=3 \\\\ y=2\\end{array}\\right.$,\\\\\nhence, the coordinates of point $C$ are $\\boxed{(3,2)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52354168_186.png"
  },
  {
    "solution": "\\boxed{(2, 2)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given a parabola $y= \\frac{1}{4}x^{2}+bx+c$ passing through the points $A(0, 2)$ and $C(4, 0)$, and intersecting the $x$-axis at another point $B$. There is a point $M$ on the parabola above the line $AC$. What is the maximum area of the triangle $ACM$ and the coordinates of the point $M$ at that moment?\n",
    "completion": "\\textbf{Solution:} Through point $M$, construct $MN\\perp x$ axis, intersecting $AC$ at point $N$, as shown in Figure 1,\\\\\nLet $M(a, -\\frac{1}{4}a^2+ \\frac{1}{2}a+2)$, then $N(a, -\\frac{1}{2}a+2)$,\\\\\n$S_{\\triangle ACM} = \\frac{1}{2}MN \\cdot OC = \\frac{1}{2}(-\\frac{1}{4}a^2+a) \\times 4 = -\\frac{1}{2}a^2+2a$,\\\\\n$\\therefore S_{\\triangle ACM} = -\\frac{1}{2}a^2+2a$,\\\\\n$\\therefore$ when $a=2$, the area of triangle $ACM$ is maximized, with its maximum value being $\\boxed{2}$,\\\\\nat this moment, the coordinates of $M$ are $\\boxed{(2, 2)}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51628092_132.png"
  },
  {
    "solution": "\\boxed{y=-x^2+3x+4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown, the parabola $y=ax^2 + bx - 4a$ ($a\\neq 0$) passes through points $A(-1, 0)$ and $C(0, 4)$, and intersects the $x$-axis at another point $B$. Lines $AC$ and $BC$ are drawn. What is the equation of the parabola?\n",
    "completion": "\\textbf{Solution:} Given the parabola $y=ax^2+bx-4a$ passes through two points,\\\\\n$\\therefore \\begin{cases} a-b-4a=0 \\\\ -4a=4 \\end{cases}$,\\\\\n$\\therefore \\begin{cases} a=-1 \\\\ b=3 \\end{cases}$,\\\\\n$\\therefore$ the equation of the parabola is $\\boxed{y=-x^2+3x+4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52536660_152.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points $D$ and $E$ are on $AB$ and $AC$ respectively, with $DE \\parallel BC$. Given that $AD = 6$, $AE = 4$, and $CE = 2$, what is the length of $BD$?\n",
    "completion": "\\textbf{Solution:} Since DE$\\parallel$BC, \\\\\nit follows that $\\frac{AD}{BD}=\\frac{AE}{CE}$, \\\\\ngiven AD=6, AE=4, CE=2, \\\\\nthus $\\frac{6}{BD}=\\frac{4}{2}$, \\\\\nsolving gives $BD=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51299716.png"
  },
  {
    "solution": "\\boxed{(0, \\frac{60}{13})}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, place rectangle OABC in the Cartesian coordinate system with points OA and OC along the positive half-axis of x and y respectively. Here AB=15, and the equation of the line containing diagonal AC is $y=- \\frac{5}{3}x+b$. Fold rectangle OABC along BE, such that point A falls on point D located on side OC. If point P is a moving point on the y-axis, there exists a point P for which the perimeter of $\\triangle PBE$ is minimized. What are the coordinates of point P?\n",
    "completion": "\\textbf{Solution:} Construct point $E'$, which is symmetric to point E with respect to the y-axis, and draw line $BE'$ intersecting the y-axis at point P. Under these conditions, the perimeter of $\\triangle BPE$ is minimized.\\\\\nSince $E(4,0)$,\\\\\nit follows that $E'(-4,0)$.\\\\\nLet the equation of the line $BE'$ be $y=kx+b$, then we have\n\\[\n\\begin{cases}\n9k+b=15,\\\\\n4k+b=0.\n\\end{cases}\n\\]\nSolving these equations yields\n\\[\n\\begin{cases}\nk=\\frac{15}{13},\\\\\nb=\\frac{60}{13}.\n\\end{cases}\n\\]\nHence, the equation of the line $BE'$ is $y=\\frac{15}{13}x+\\frac{60}{13}$,\\\\\ntherefore, the coordinates of P are $\\boxed{(0, \\frac{60}{13})}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50675997_173.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the square $ABCD$, $AB=4$. Let $E$ be a point on side $AB$, and let $F$ be a point on side $BC$ such that $BF=1$. Rotate point $E$ clockwise around point $F$ by $90^\\circ$ to obtain point $G$. Connect $DG$. What is the minimum length of $DG$?\n",
    "completion": "\\textbf{Solution:} Draw line $GP\\perp BC$ at point P, extend PG to intersect AD at point H,\\\\\nthen $\\angle GPF=90^\\circ$,\\\\\nsince quadrilateral ABCD is a square,\\\\\n$\\therefore \\angle ADC=\\angle C=\\angle B=90^\\circ$,\\\\\n$\\therefore$ quadrilateral CDHP is a rectangle,\\\\\n$\\therefore CD=PH=AB=4$, $PC=DH$,\\\\\nsince $\\angle EFG=90^\\circ$,\\\\\n$\\therefore \\angle BFE+\\angle PFG=90^\\circ$,\\\\\nalso, $\\angle BFE+\\angle BEF=90^\\circ$,\\\\\n$\\therefore \\angle PFG=\\angle BEF$,\\\\\nsince $FE=FG$, $\\angle B=\\angle GPF=90^\\circ$,\\\\\n$\\therefore \\triangle BEF\\cong \\triangle PFG\\left(AAS\\right)$,\\\\\n$\\therefore BE=PF$, $PG=BF=1$,\\\\\n$\\therefore GH=PH-PG=4-1=3$,\\\\\nlet $BE=PF=x$, then $PC=DH=4-1-x=3-x$,\\\\\nin $\\triangle DGH$, by the Pythagorean theorem,\\\\\n$DG^2=DH^2+GH^2=(3-x)^2+3^2=(3-x)^2+9$,\\\\\nwhen $x=3$, $DG^2$ has its minimum value of $9$,\\\\\n$\\therefore$ the minimum value of $DG$ is $\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52049885.png"
  },
  {
    "solution": "\\boxed{(2.5, 0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the edges AO and CO of the rectangle ABCO lie exactly on the coordinate axes, with AB = 4 and OA = 2. Point D is a point on line segment OC, and point E is a point on line segment AB. When folded along DE, such that point B coincides with point O and point C moves to point $C'$, what are the coordinates of point D at that moment?\n",
    "completion": "\\textbf{Solution:} Since AB$\\parallel$OC,\\\\\nit follows that $\\angle$BED=$\\angle$EDO,\\\\\nby the property of folding we get BE=OE and $\\angle$BED=$\\angle$OED,\\\\\nthus $\\angle$EDO=$\\angle$OED,\\\\\nwhich implies OE=OD,\\\\\nLet BE=OE=x, then AE=4-x,\\\\\nhence in the Rt$\\triangle$AEO, by Pythagorean theorem, we have $2^2+(4-x)^2=x^2$,\\\\\nsolving for $x$ gives $x=2.5$,\\\\\ntherefore, OE=OD=2.5,\\\\\nthus the coordinates of point D are $\\boxed{(2.5, 0)}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52720394.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, it is known that points $A(8, 1)$ and $B(0, -3)$. The graph of the inverse proportion function $y=\\frac{k}{x}$ $(x>0)$ passes through point A. The moving line $x=t$ $(0<t<8)$ intersects with the graph of the inverse proportion function at point M, and intersects with line $AB$ at point N. When $t=4$, what is the area of $\\triangle BMA$?\n",
    "completion": "\\textbf{Solution:} Since $k=8$\\\\\nthus, the equation of the inverse proportion function is: $y=\\frac{8}{x}(x>0)$.\\\\\nLet the equation of line $AB$ be $y=k'x+b$,\\\\\nsubstituting points $A\\left(8, 1\\right)$ and $B\\left(0, -3\\right)$, we get $\\left\\{\\begin{array}{l}8k'+b=1\\\\ b=-3\\end{array}\\right.$,\\\\\nsolving this yields $\\left\\{\\begin{array}{l}k'=\\frac{1}{2}\\\\ b=-3\\end{array}\\right.$,\\\\\nthus, the equation of line $AB$ is $y=\\frac{1}{2}x-3$,\\\\\nthus, in $y=\\frac{1}{2}x-3$, when $x=t=4$, we find $y=-1$,\\\\\nthus, point $N\\left(4, -1\\right)$,\\\\\nin $y=\\frac{8}{x}(x>0)$, when $x=t=4$, we find $y=2$,\\\\\nthus, point $M\\left(4, 2\\right)$,\\\\\nthus, the distance $MN=2-\\left(-1\\right)=3$,\\\\\nsince points $A\\left(8, 1\\right)$ and $B\\left(0, -3\\right)$,\\\\\nthus, the area of $\\triangle BMA=\\frac{1}{2}MN\\times \\left({x}_{A}-{x}_{B}\\right)=\\frac{1}{2}\\times 3\\times \\left(8-0\\right)=\\boxed{12}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53490748_198.png"
  },
  {
    "solution": "\\boxed{16}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system $xOy$, where $O$ is the origin, point $B$ is on the positive half-axis of the x-axis, and quadrilateral $OBCA$ is a parallelogram. It is given that $\\sin\\angle AOB=\\frac{4}{5}$. The inverse proportion function $y=\\frac{m}{x}$ ($m>0$) passes through point $A$ in the first quadrant and intersects $BC$ at point $F$. If point $F$ is the midpoint of $BC$, and the area of $\\triangle AOF$ is 12, then what is the value of $m$?\n",
    "completion": "\\textbf{Solution:} Draw $AM\\perp OB$ at $M$, and $FN\\perp OB$ at $N$, where $OA=5k$,\\\\\nsince $\\sin\\angle AOB=\\frac{4}{5}$,\\\\\ntherefore $AM=4k$, $OM=3k$, $m=12k^{2}$,\\\\\nsince quadrilateral OACB is a parallelogram and F is the midpoint of $BC$,\\\\\ntherefore $FN=2k$, $ON=6k$,\\\\\nsince ${S}_{\\triangle AOM}={S}_{\\triangle OFN}$,\\\\\n${S}_{\\text{trapezium} OAFN}={S}_{\\text{trapezium} AMNF}+{S}_{\\triangle AOM}={S}_{\\triangle AOF}+{S}_{\\triangle OFN}$,\\\\\ntherefore ${S}_{\\text{trapezium} AMNF}={S}_{\\triangle AOF}=12$,\\\\\ntherefore $\\frac{1}{2}(4k+2k)\\cdot 3k=12$,\\\\\ntherefore $k^{2}=\\frac{4}{3}$,\\\\\ntherefore $m=12k^{2}=\\boxed{16}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53043823.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, point $E$ lies on $AD$, and $EC$ bisects $\\angle BED$, $AB=\\sqrt{3}$, $\\angle ABE=30^\\circ$, what is the length of $DE$?\n",
    "completion": "Solution: Since quadrilateral $ABCD$ is a rectangle,\\\\\nit follows that $AB=CD=\\sqrt{3}$, and $\\angle A=\\angle D=90^\\circ$,\\\\\nSince $\\angle ABE=30^\\circ$,\\\\\nit follows that $\\angle AEB=60^\\circ$,\\\\\nhence $\\angle BED=120^\\circ$,\\\\\nSince $EC$ bisects $\\angle BED$,\\\\\nit follows that $\\angle DEC=60^\\circ$,\\\\\nhence $\\angle DCE=30^\\circ$,\\\\\nthus $CD=\\sqrt{3}DE=\\sqrt{3}$,\\\\\ntherefore, $DE=\\boxed{1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52551361.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{12}{x}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y = -x + b$ intersects with the graph of the inverse proportion function $y = \\frac{k}{x}$ ($x < 0$) at point $A(-6, m)$, and intersects with the x-axis at point $B(-4, 0)$. What are the expressions for the linear function and the inverse proportion function?\n",
    "completion": "\\textbf{Solution:} Given $B(-4, 0)$ lies on $y=-x+b$, we find that $b=-4$,\\\\\nthus the expression of the linear function is $y=-x-4$.\\\\\nGiven $A(-6, m)$ lies on $y=-x-6$, we find that $m=-2$,\\\\\ntherefore $A(-6, -2)$.\\\\\nSubstituting $A(-6, -2)$ into $y=\\frac{k}{x}$ yields $k=-12$,\\\\\ntherefore the expression of the inverse proportionality function is $\\boxed{y=-\\frac{12}{x}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52602187_169.png"
  },
  {
    "solution": "\\boxed{-\\frac{1}{2}m + 5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the sides $OA$ and $OC$ of rectangle $OABC$ lie on the positive x-axis and the positive y-axis, respectively, and the lengths of $OA$ and $OC$ are the two roots of the equation ${x}^{2}-20x+100=0$. In figure 2, point D is on $CB$, point E is on the extension line of $BA$, and $CD=AE$. Connect $AD$ and $CE$. Draw $OF\\perp AD$ through point O and intersect $CE$ at point G, with the foot of the perpendicular being point F. Let the length of $CD$ be $m$, and let the x-coordinate of point G be $n$. What is the functional relationship between $n$ and $m$?\n",
    "completion": "\\textbf{Solution:} Extend $OF$ to intersect $AB$ at $H$,\\\\\n\\[\n\\because \\angle AFO = \\angle B = \\angle OAB = 90^\\circ,\n\\]\n\\[\n\\therefore \\angle OAF + \\angle BAD = \\angle AOH + \\angle OAF = 90^\\circ,\n\\]\n\\[\n\\therefore \\angle AOB = \\angle BAD,\n\\]\n\\[\n\\because OA = AB,\n\\]\n\\[\n\\therefore \\triangle OAH \\cong \\triangle ABD,\n\\]\n\\[\n\\therefore AH = BD = 10 - m, \\text{ and } AE = CD = m,\n\\]\n\\[\n\\therefore EH = 10 = OC,\n\\]\n\\[\n\\therefore \\triangle EGH \\cong \\triangle CGO,\n\\]\n\\[\n\\therefore G \\text{ is the midpoint of } EC,\n\\]\nDraw $EM \\perp OC$ intersecting the extended line of $CO$ at $M$, take the midpoint $N$ of $MC$, connect $GN$,\\\\\n\\[\n\\therefore GN \\text{ is the midline of } \\triangle EMC,\n\\]\n\\[\n\\therefore GN \\parallel EM, \\text{ i.e., } GN \\perp OC,\n\\]\n\\[\n\\because ON = OC - NC = 10 - \\frac{10 + m}{2} = -\\frac{m}{2} + 5,\n\\]\n\\[\n\\therefore n = \\boxed{-\\frac{1}{2}m + 5}.\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/53728996_96.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, DE is the median line of $\\triangle ABC$, with point F on DE, and $\\angle AFB=90^\\circ$. If $AB=8$ and $BC=10$, what is the length of EF?\n",
    "completion": "\\textbf{Solution:} Given that DE is the median of $\\triangle ABC$, with BC=10, \\\\\nthus DE=$\\frac{1}{2}$BC=5, \\\\\nGiven that $\\angle AFB=90^\\circ$, and D is the midpoint of AB, with AB=8, \\\\\nthus DF=$\\frac{1}{2}$AB=4, \\\\\ntherefore EF=DE−DF=5−4=\\boxed{1}.",
    "image_path": "dataset/mm_math/images/MM_Math/52065494.png"
  },
  {
    "solution": "\\boxed{72^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AC=BC$, circle O is the circumcircle of $\\triangle ABC$, and the extension of $BO$ intersects side $AC$ at point D. When $\\triangle ABD$ is an isosceles triangle, what is the measure of $\\angle A$?\n",
    "completion": "\\textbf{Solution:} Connect OC, \\\\\n$\\because$ $CA=CB$, \\\\\n$\\therefore$ $\\overset{\\frown}{CA}=\\overset{\\frown}{CB}$, $\\angle CAB=\\angle CBA$, \\\\\n$\\therefore$ $OC\\perp AB$, \\\\\n$\\therefore$ $\\angle DCO=\\angle BCO$, \\\\\n$\\because$ $OC=OB$, \\\\\n$\\therefore$ $\\angle OBC=\\angle BCO$, \\\\\n$\\therefore$ $\\angle ADB=3\\angle OBC$, \\\\\nWhen $BA=BD$, $\\angle A=\\angle BDA=3\\angle OBC$, \\\\\nThen $8\\angle OBC=180^\\circ$, \\\\\nSolving for: $\\angle OBC=22.5^\\circ$, \\\\\n$\\therefore$ $\\angle A=67.5^\\circ$; \\\\\nWhen $AB=AD$, $\\angle ABD=\\angle BDA=3\\angle OBC$, \\\\\nThen $10\\angle OBC=180^\\circ$, \\\\\nSolving for: $\\angle OBC=18^\\circ$, \\\\\n$\\therefore$ $\\angle A=\\angle ABD+\\angle OBC=4\\angle OBC=72^\\circ$, \\\\\nThe case for DA=DB does not exist, \\\\\nIn summary, when $\\triangle ABD$ is an isosceles triangle, the degree of $\\angle A$ is $67.5^\\circ$ or $72^\\circ$, \\\\\nHence the answer is: $\\boxed{67.5^\\circ}$ or $\\boxed{72^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52991440.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the quadrilateral $ABCD$ is inscribed in circle $\\odot O$, with $\\angle ADC=2\\angle B$, and point $D$ is the midpoint of $\\overset{\\frown}{AC}$. What is the measure of $\\angle B$ in degrees?\n",
    "completion": "\\textbf{Solution:} Since $\\angle ADC=2\\angle B$\\\\\nAnd since $\\angle ADC+\\angle B=180^\\circ$\\\\\nTherefore, $\\angle B=\\boxed{60^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/53080829_102.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the equilateral triangle $ABC$, $BD$ is the median to the side $AC$, and through point $D$, $DE \\perp BC$ at point $E$. If $AB=2$, then what is the length of $CE$?\n",
    "completion": "\\textbf{Solution:} Given: $\\because$ $\\triangle ABC$ is an equilateral triangle, $AB=2$,\\\\\n$\\therefore$ $\\angle ABC = \\angle C = 60^\\circ$, $AB = BC = AC = 2$,\\\\\nAlso $\\because$ $BD$ is the median to side $AC$,\\\\\n$\\therefore$ $\\angle DBC = 30^\\circ$, $\\angle BDC = 90^\\circ$, $DC = 1$,\\\\\nAlso $\\because$ $DE \\perp BC$,\\\\\n$\\therefore$ $\\angle EDC = 30^\\circ$,\\\\\n$\\therefore$ $EC = \\frac{1}{2} DC = \\boxed{\\frac{1}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51385696.png"
  },
  {
    "solution": "\\boxed{D(3, 1)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the equation of line $AB$ is $y=x+2$, and the equation of line $AC$ is $y=-x+4$. The two lines intersect at point $A$, and they intersect the $x$-axis at points $B$ and $C$, respectively. Point $D$ is a point on line segment $AC$, and line $BD$ is drawn. If $BD=\\sqrt{26}$, what are the coordinates of point $D$?\n",
    "completion": "\\textbf{Solution:} Draw $DF\\perp x$-axis at point $F$ through point $D$. Suppose the x-coordinate of point $D$ is $m$. Since point $D$ is on line $AC$, we have $y=-m+4$. Thus, $D(m, -m+4)$. Therefore, $DF=-m+4$ and $BF=m+2$. In right triangle $\\triangle DBF$, with $\\angle DFB=90^\\circ$, according to the Pythagorean theorem, we get $BD^{2}=DF^{2}+BF^{2}$. Hence, $\\left(\\sqrt{26}\\right)^{2}=\\left(-m+4\\right)^{2}+\\left(m+2\\right)^{2}$. After simplification, we obtain $m^{2}-2m-3=0$. Solving this yields $m_{1}=3$, $m_{2}=-1$ (this solution is rejected as it is not consistent with the conditions of the problem). Therefore, $D(3, 1)$. Thus, the coordinates of point $D$ are $\\boxed{D(3, 1)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50590416_150.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, the line $y=ax+6$ intersects the line $y=2x$ at point A ($m, 4$), and it intersects the x-axis at point B, and the y-axis at point C. Calculate the area of $\\triangle AOC$.\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, draw a perpendicular line from A to the y-axis with the foot at D, thus $AD=2$,\\\\\nwhen $x=0$, we have $y=-x+6=6$,\\\\\n$\\therefore$ the coordinates of point C are $(0,6)$\\\\\n$\\therefore$ the area of $\\triangle AOC$ is $\\frac{1}{2}\\times 6\\times 2=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52189216_155.png"
  },
  {
    "solution": "\\boxed{75}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated, there are two islands, C and D, in the sea. A fishing boat at point A found island D to the northeast, at a distance of $30\\sqrt{2}$ nautical miles. The boat then traveled from west to east for a period of time to reach point B, where it measured that island C was exactly to the north of point B, at a distance of $75$ nautical miles, and also measured that point B was $30\\sqrt{5}$ nautical miles away from island D. What is the distance between islands C and D in nautical miles?\n",
    "completion": "\\textbf{Solution:} Construct line DF such that DF $\\perp$ BC at F. In the right triangle $\\triangle BED$, where DE = 30 and BD = $30\\sqrt{5}$, \\\\\nthus, BE = $\\sqrt{BD^{2}-DE^{2}}=60$, \\\\\nsince quadrilateral BFDE is a rectangle, \\\\\nthus, DF = EB = 60 and BF = DE = 30, \\\\\ntherefore, CF = BC−BF = 45, \\\\\nin the right triangle $\\triangle CDF$, CD = $\\sqrt{DF^{2}+CF^{2}}=\\boxed{75}$, \\\\\ntherefore, the distance between islands C and D is \\boxed{75} nautical miles.",
    "image_path": "dataset/mm_math/images/MM_Math/53080937_134.png"
  },
  {
    "solution": "\\boxed{57^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, lines $AB$ and $CD$ intersect at point $O$. $OE$ bisects $\\angle BOC$, $OF\\perp OE$, and $\\angle AOD=66^\\circ$. What is the degree measure of $\\angle BOF$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle AOD = 66^\\circ$,\\\\\nhence $\\angle BOC = \\angle AOD = 66^\\circ$,\\\\\nsince OE bisects $\\angle BOC$,\\\\\ntherefore $\\angle BOE = \\frac{1}{2} \\angle BOC = 33^\\circ$,\\\\\nsince OF is perpendicular to OE,\\\\\ntherefore $\\angle EOF = 90^\\circ$,\\\\\nthus $\\angle BOF = 90^\\circ - 33^\\circ = \\boxed{57^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55191222.png"
  },
  {
    "solution": "\\boxed{165^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that: $OA\\perp OB$, $\\angle BOC=50^\\circ$, and $\\angle AOD : \\angle COD=4 : 7$. Draw the angle bisector OE of $\\angle BOC$, and find the degree measure of $\\angle DOE$.\n",
    "completion": "\\textbf{Solution:} As shown in the figure:\\\\\nSince OA$\\perp$OB,\\\\\nIt follows that $\\angle$AOB=$90^\\circ$,\\\\\nSince $\\angle$AOD$\\colon$ $\\angle$COD=4$\\colon$ 7,\\\\\nLet $\\angle$AOD=$4x^\\circ$ and $\\angle$COD=$7x^\\circ$,\\\\\nSince $\\angle$AOB+$\\angle$AOD+$\\angle$COD+$\\angle$BOC=$360^\\circ$, and $\\angle$BOC=$50^\\circ$,\\\\\nIt follows that $90+7x+4x+50=360$,\\\\\nThus, $x=20$,\\\\\nTherefore, $\\angle$COD=$140^\\circ$.\\\\\nSince OE is the bisector of $\\angle$BOC,\\\\\nIt follows that $\\angle COE=\\frac{1}{2}\\angle$BOC=$25^\\circ$,\\\\\nTherefore, $\\angle$DOE=$\\angle$COD+$\\angle$COE=\\boxed{165^\\circ}.",
    "image_path": "dataset/mm_math/images/MM_Math/6972741.png"
  },
  {
    "solution": "\\boxed{\\frac{378}{23}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points B and C are on line segment $AD$, and $AB:BC:CD=2:3:4$. Point M is the midpoint of line segment $AC$, and point N is a point on line segment $CD$, with $MN=9$. When $CN=\\frac{1}{3}CD$, find the length of $BD$.\n",
    "completion": "\\textbf{Solution:}\\\\\nSince CN = $\\frac{1}{3}$CD,\\\\\nand AB:BC:CD = 2:3:4,\\\\\nlet AB = 2x, BC = 3x, CD = 4x.\\\\\nThus, AC = 5x,\\\\\nand since point M is the midpoint of segment AC,\\\\\nCM = $\\frac{1}{2}$AC = 2.5x,\\\\\nand since CN = $\\frac{1}{3}$CD = $\\frac{4}{3}$x,\\\\\nCM+CN = $\\frac{5}{2}$x+$\\frac{4}{3}$x = MN = 9,\\\\\nthus, x = $\\frac{54}{23}$,\\\\\nHence, BD = 7x = $\\boxed{\\frac{378}{23}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/47162039_158.png"
  },
  {
    "solution": "\\boxed{x>\\frac{2}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the line $y=4x-6$ intersects the x-axis at point $A$, and the line $y=-2x-2$ intersects the x-axis at point $B$, and these two lines intersect at point $C$. Write the range of the independent variable $x$ for which the value of the function $y=4x-6$ is greater than the value of $y=-2x-2$.\n",
    "completion": "\\textbf{Solution:} From the graph, it is known that the range of the independent variable $x$ that makes the value of the function $y=4x-6$ greater than the value of the function $y=-2x-2$ is $\\boxed{x>\\frac{2}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52423437_63.png"
  },
  {
    "solution": "\\boxed{AP=\\left(2t-20\\right)\\text{cm} \\text{ for } t>10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle BAC=90^\\circ$, $AB=5\\,cm$, and $BC=13\\,cm$. Point $D$ is on line segment $AC$, and $CD=7\\,cm$. Moving point $P$ starts from point $E$, which is $15\\,cm$ away from point $B$, and moves in the direction of ray $EA$ at a speed of $2\\,cm$ per second, for a time of $t$ seconds. Express the length of $AP$ in centimeters as an algebraic expression containing $t$.\n",
    "completion": "\\textbf{Solution:} Given that $BE=15$ and $AB=5$,\\\\\ntherefore $AE=20$,\\\\\nthus when $P$, $A$ coincide, $t=\\frac{20}{2}=10$,\\\\\ntherefore for $0\\le t\\le 10$, $AP=\\left(20-2t\\right)\\text{cm}$,\\\\\nwhen $t>10$, $AP=\\left(2t-20\\right)\\text{cm}$.\\\\\n$\\therefore \\boxed{AP=\\left(20-2t\\right)\\text{cm} \\text{ for } 0\\le t\\le 10}$,\\\\\n$\\boxed{AP=\\left(2t-20\\right)\\text{cm} \\text{ for } t>10}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55588005_81.png"
  },
  {
    "solution": "\\boxed{7}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point B is a point on line segment AC, with $AB=21$ and $BC= \\frac{1}{3}AB$. If point O is the midpoint of line segment AC, what is the length of line segment QB?\n",
    "completion": "\\textbf{Solution:} Given that point O is the midpoint of line segment AC, \\\\\nthus $OA = \\frac{1}{2}AB = \\frac{1}{2} \\times 28 = 14$, \\\\\n$OB = AB - OA = 21 - 14 = \\boxed{7}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55307630_4.png"
  },
  {
    "solution": "\\boxed{30cm^{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC$, $M$, $N$ are the midpoints of $AB$, $AC$ respectively, and $D$, $E$ are points on $BC$, connecting $DN$, $EM$. If $AB=13$ cm, $BC=10$ cm, and $DE=5$ cm, what is the area of the shaded region in the figure?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, \\\\\nconnect MN, then MN is the median of $\\triangle ABC$, \\\\\n$\\therefore$ MN=$\\frac{1}{2}$BC=5cm, \\\\\nDraw AF$\\perp$BC at point F through point A, \\\\\n$\\because$ AB=AC, \\\\\n$\\therefore$ BF=$\\frac{1}{2}$BC=5cm, \\\\\n$\\therefore$ AF=$\\sqrt{AB^{2}-BF^{2}}=\\sqrt{13^{2}-5^{2}}=12cm$, \\\\\n$\\because$ the shaded part of the diagram consists of three triangles each with a base of 5cm and the sum of their heights is 12cm, \\\\\n$\\therefore$ $S_{\\text{shaded}}=\\frac{1}{2}\\times 5\\times 12=\\boxed{30cm^{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52304101.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AC$ bisects $\\angle BAD$, and $\\angle BAD = \\angle BCD$. What is the measure of $\\angle DBC$?\n",
    "completion": "\\textbf{Solution:} We have $\\angle DBC = \\boxed{45^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53102367_61.png"
  },
  {
    "solution": "\\boxed{\\frac{30+3\\sqrt{34}}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the quadrilateral $AOBC$ is a rectangle with point $O(0,0)$, point $A(5,0)$, and point $B(0,3)$. The rectangle $AOBC$ is rotated clockwise around the point $A$ to get the rectangle $ADEF$, where $O$, $B$, $C$ correspond to $D$, $E$, $F$, respectively. Let $K$ be the intersection point of the diagonals of rectangle $AOBC$, what is the maximum area of $\\triangle KDE$?\n",
    "completion": "\\textbf{Solution:} Let $A(5, 0)$, and $B(0, 3)$,\\\\\nthus $OA=5$, $OB=3$,\\\\\nsince quadrilateral $AOBC$ is a rectangle,\\\\\nthus $AC=OB=3$, $OA=BC=5$, $\\angle OBC=\\angle C=90^\\circ$,\\\\\nthus $AB=\\sqrt{OA^{2}+OB^{2}}=\\sqrt{25+9}=\\sqrt{34}$\\\\\nthus $BK=AK=\\frac{1}{2}AB=\\frac{\\sqrt{34}}{2}$\\\\\nSince rectangle $ADEF$ is obtained by rotating rectangle $AOBC$,\\\\\nthus $AD=AO=5$,\\\\\nAs shown in the figure, when point $D$ is on segment $BK$, the area of $\\triangle DEK$ is the smallest,\\\\\nWhen point $D$ is on the extension of $BA$, the area of $\\triangle D'E'K$ is the largest,\\\\\nThe maximum area $=\\frac{1}{2}\\times D'E' \\times KD'=\\frac{1}{2}\\times 3\\times \\left(5+\\frac{\\sqrt{34}}{2}\\right)=\\boxed{\\frac{30+3\\sqrt{34}}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53498213.png"
  },
  {
    "solution": "\\boxed{F(8,3)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, pass point C(8, 6) to draw $CB\\perp x$-axis and $CA\\perp y$-axis respectively, with perpendicular feet at points B and A. Point F is a mobile point on segment BC but does not coincide with points B and C. The graph of the inverse proportion function $y=\\frac{k}{x}$ ($k>0$) passes through point F and intersects segment AC at point E. Connect EF. When point E is the midpoint of segment AC, what are the coordinates of point F?\n",
    "completion": "\\textbf{Solution:} Given that point E is the midpoint of line segment AC,\\\\\nthus E(4, 6),\\\\\nsince point E lies on the graph of the inverse proportion function $y=\\frac{24}{x}$,\\\\\nthus $k=xy=4\\times 6=24$,\\\\\ntherefore $y=\\frac{24}{x}$,\\\\\nwhen $x=8$, $y=3$,\\\\\nthus \\boxed{F(8,3)}.",
    "image_path": "dataset/mm_math/images/MM_Math/51445937_234.png"
  },
  {
    "solution": "\\boxed{y=\\frac{2}{x}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the graph of the linear function $y=-x+m$ intersects with the graph of the inverse proportionality function $y=\\frac{k}{x}$ (where $x>0$) at points $A$ and $B$. Given $A(1,2)$, what are the equations of the linear and the inverse proportionality functions?\n",
    "completion": "\\textbf{Solution:} Substitute point A$(1,2)$ into $y=-x+m$, to obtain $-1+m=2$, $\\therefore m=3$, $\\therefore$ the linear function can be expressed as $y=-x+3$; Substitute point A$(1,2)$ into $y=\\frac{k}{x}$, $\\therefore k=1\\times 2=2$, $\\therefore$ the inverse proportional function can be expressed as $\\boxed{y=-x+3}$ and $\\boxed{y=\\frac{2}{x}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52694023_94.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the parabola $y=-x^2+bx+c$ passes through points $A(-1,0)$ and $B(3,0)$. When $0<x<3$, directly write out the range of $y$.\n",
    "completion": "\\textbf{Solution:} When $0<x<3$, the range of $y$ is $0<y\\le \\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52977350_188.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a circle with center $O$ on side $BC$ of $\\triangle ABC$ passes through points $A$ and $B$, intersecting side $BC$ at point $E$, with $DO\\perp BE$ at point $O$. Line $AD$ is drawn to intersect $BC$ at point $F$, given that $AC=FC$. $AC$ is the tangent of circle $O$: Given $BF=8$ and $DF= 2\\sqrt{10}$, what is the radius of circle $O$?\n",
    "completion": "\\textbf{Solution:} Let the radius of $\\odot O$ be $r$, \\\\\n$\\therefore BO=DO=r$, \\\\\n$\\because BF=8$, \\\\\n$\\therefore OF=8-r$, \\\\\n$\\because \\angle DOF=90^\\circ$, \\\\\n$\\therefore$ in $\\triangle ODF$ right, by the Pythagorean theorem, we have $OF^2+OD^2=DF^2$, \\\\\n$\\because DF=2\\sqrt{10}$, \\\\\n$\\therefore (8-r)^2+r^2=(2\\sqrt{10})^2$, \\\\\nSolving, we get: $r=6$ or $r=2$ (discarded as it does not meet the problem statement), \\\\\nHence, the radius of $\\odot O$ is $\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52623066_100.png"
  },
  {
    "solution": "\\boxed{125^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, \\(PA\\) and \\(PB\\) are tangents to circle \\(\\odot O\\) at points \\(A\\) and \\(B\\), respectively, and \\(\\angle P=70^\\circ\\). If point \\(C\\) is on circle \\(\\odot O\\) and does not coincide with \\(A\\) or \\(B\\), what is the measure of \\(\\angle ACB\\)?\n",
    "completion": "\\textbf{Solution:} Draw OA, OB, \\\\\n$ \\because $ PA, PB are tangents to circle $O$ at points A, B, and $ \\angle P=70^\\circ$, \\\\\n$ \\therefore \\angle AOB=360^\\circ-90^\\circ-90^\\circ-\\angle P=110^\\circ$\\\\\nWhen C is on the major arc $ \\overset{\\frown}{ACB}$,\\\\\n$ \\angle ACB=\\frac{1}{2}\\angle AOB=55^\\circ$\\\\\nWhen point C is on the minor arc $ \\overset{\\frown}{AB}$, quadrilateral ACBC$_{1}$ is a circumscribed quadrilateral\\\\\nThen $ \\angle A{C}_{1}B+\\angle C=180^\\circ$\\\\\n$ \\therefore \\angle A{C}_{1}B=125^\\circ$\\\\\nHence, the answer is: $\\boxed{55^\\circ}$ or $\\boxed{125^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51445528.png"
  },
  {
    "solution": "\\boxed{\\frac{17}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the parabola $y=ax^2+bx-1$ ($a\\neq 0$) intersects the x-axis at points $A(-2, 0)$ and $B(2, 0)$, and intersects the y-axis at point D. The line through point $C(0, 1)$, line AC, intersects the parabola at points A and F. Point P is a movable point on the parabola below line AF. A line parallel to the y-axis passing through point P meets line AC at point Q, and a line parallel to the x-axis passing through point P meets the y-axis at point E. Determine the maximum value of $PQ+PE$ and the corresponding coordinates of point P.\n",
    "completion": "Solution: Let the equation of line AC be $y=mx+n$, \\\\\n$\\left\\{\\begin{array}{l}\n-2m+n=0\\\\\nn=1\n\\end{array}\\right.$, solving this system gives $\\left\\{\\begin{array}{l}\nm=\\frac{1}{2}\\\\\nn=1\n\\end{array}\\right.$, \\\\\n$\\therefore$ the equation of line AC is $y=\\frac{1}{2}x+1$, \\\\\nLet P$(x, \\frac{1}{4}x^{2}-1)$, then Q$(x, \\frac{1}{2}x+1)$, E$(0, \\frac{1}{4}x^{2}-1)$, \\\\\nWhen point P is to the right of the y-axis, \\\\\n$\\therefore PQ+PE=(\\frac{1}{2}x+1)-(\\frac{1}{4}x^{2}-1)+x=-\\frac{1}{4}(x-3)^{2}+\\frac{17}{4}$, \\\\\nWhen $x=3$, $PQ+PE$ reaches its maximum value $\\boxed{\\frac{17}{4}}$, \\\\\nAt this time, the coordinates of point P are $(3, \\frac{5}{4})$; \\\\\nWhen point P is to the left of the y-axis, \\\\\n$\\therefore PQ+PE=(\\frac{1}{2}x+1)-(\\frac{1}{4}x^{2}-1)+(0-x)=-\\frac{1}{4}(x+1)^{2}+\\frac{9}{4}$, \\\\\nWhen $x=-1$, $PQ+PE$ reaches its maximum value $\\frac{9}{4}$, \\\\\nAt this time, the coordinates of point P are $(-1, -\\frac{3}{4})$; \\\\\nIn summary, $PQ+PE$ reaches its maximum value $\\boxed{\\frac{17}{4}}$, the coordinates of point P are $(3, \\frac{5}{4})$;",
    "image_path": "dataset/mm_math/images/MM_Math/52620662_109.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $Rt\\triangle AOB \\cong Rt\\triangle CDA$, and the coordinates of points A and B are $(-1, 0)$ and $B(0, 2)$, respectively. What is the length of $OD$?\n",
    "completion": "\\textbf{Solution:} Since $A(-1,0)$ and $B(0,2)$, \\\\\nhence $OA=1$, $OB=2$, \\\\\nsince $\\triangle AOB \\cong \\triangle CDA$, \\\\\nhence $OB=AD=2$, \\\\\ntherefore $OD=AD+AO=2+1=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51435251.png"
  },
  {
    "solution": "\\boxed{110^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $OB$ is a ray inside $\\angle AOC$, $OM$ is a ray inside $\\angle AOB$, and $ON$ is a ray inside $\\angle BOC$. As shown in Figure 1, $OM$ and $ON$ are the angle bisectors of $\\angle AOB$ and $\\angle BOC$, respectively. Given that $\\angle AOB=30^\\circ$ and $\\angle MON=70^\\circ$, what is the degree measure of $\\angle BOC$?\n",
    "completion": "\\textbf{Solution:} Given that $OM$ is the bisector of $\\angle AOB$ and $\\angle AOB=30^\\circ$,\\\\\nit follows that $\\angle BOM=\\frac{1}{2}\\angle AOB=15^\\circ$,\\\\\nGiven that $\\angle MON=70^\\circ$,\\\\\nit follows that $\\angle BON=\\angle MON-\\angle BOM=55^\\circ$,\\\\\nSince $ON$ is the bisector of $\\angle BOC$,\\\\\nit follows that $\\angle BOC=2\\angle BON=\\boxed{110^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53403960_72.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{4}{3}x-4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=-\\frac{4}{3}x+4$ intersects the x-axis and y-axis at points A and B, respectively. A new line is obtained by shifting the line $y=-\\frac{4}{3}x+4$ to the left, which intersects the x-axis and y-axis at points C and D, respectively. If $BD-AB=3$, then the equation of line CD is $y=$?\n",
    "completion": "\\textbf{Solution:} Hence, the equation of the line CD is $y=-\\frac{4}{3}x-4$. Therefore, $\\boxed{y=-\\frac{4}{3}x-4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52416336_33.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, points $E$ and $F$ are respectively on sides $AD$ and $DC$, $\\triangle ABE\\sim \\triangle DEF$, $DF>DE$, $AB=9$, $AE=12$, $DE=3$, what is the length of $FC$?\n",
    "completion": "\\textbf{Solution:} Given that $\\triangle ABE\\sim \\triangle DEF$, \\\\\ntherefore $\\frac{AB}{DE}=\\frac{AE}{DF}$, \\\\\nsince $AB=9$, $AE=12$, and $DE=3$, \\\\\nthen $\\frac{9}{3}=\\frac{12}{DF}$, \\\\\nsolving this gives $DF=4$, \\\\\nsince quadrilateral $ABCD$ is a rectangle, \\\\\ntherefore $DC=AB=9$, \\\\\nthus $FC=DC−DF=9−4=\\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52933571.png"
  },
  {
    "solution": "\\boxed{$\\triangle AEF \\sim \\triangle ACD$}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within $\\triangle ABC$, points E and F are located on side AD and diagonal AC respectively. Line segment EF is drawn, and $\\angle AEF = \\angle CAB$. Prove that: $\\triangle AEF \\sim \\triangle ABC$.\n",
    "completion": "\\textbf{Solution:} Proof: \\\\\n$\\because$ Quadrilateral ABCD is a parallelogram, \\\\\n$\\therefore$ AB $\\parallel$ CD, \\\\\n$\\therefore$ $\\angle$CAB $=$ $\\angle$ACD, \\\\\n$\\because$ $\\angle$AEF $=$ $\\angle$CAB, \\\\\n$\\therefore$ $\\angle$AEF $=$ $\\angle$ACD, \\\\\n$\\because$ $\\angle$EAF $=$ $\\angle$CAD, \\\\\n$\\therefore$ \\boxed{$\\triangle AEF \\sim \\triangle ACD$}.",
    "image_path": "dataset/mm_math/images/MM_Math/51212753_73.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, within the Cartesian coordinate system, the rectangle $OABC$ has its vertex O coincide with the coordinate origin, and vertices A and C respectively lie on the coordinate axes, with vertex B at the coordinates $\\left(6, 4\\right)$. E is the midpoint of $AB$, and the line passing through points $D\\left(8, 0\\right)$ and E intersects $BC$ and the y-axis at points F and G, respectively. The graph of the function $y=mx-2$ goes through point F and intersects the x-axis at point H. Determine the coordinates of point F and the value of $m$?\n",
    "completion": "\\textbf{Solution:} When $y=4$, $-x+8=4$, hence $x=4$,\\\\\ntherefore, the coordinates of point F are $\\boxed{\\left(4, 4\\right)}$,\\\\\nsince the graph of the function $y=mx-2$ passes through point F,\\\\\ntherefore $4m-2=4$, yielding $m=\\boxed{\\frac{3}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51663234_193.png"
  },
  {
    "solution": "\\boxed{y=\\frac{1}{2}x+3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $l_{1}\\colon y=-x-3$ intersects with the line $l_{2}$ passing through the point $A(0,3)$ at point $C(m,1)$, and intersects with the y-axis at point $B$. Find the analytical expression of the line $l_{2}$.\n",
    "completion": "\\textbf{Solution:} Substitute C($m$,1) into $y=-x-3$ to get $-m-3=1$,\\\\\n$\\therefore m=-4$,\\\\\n$\\therefore$ C($-4$,1). Let the equation of line $l_{2}$ be $y=kx+b$. Substituting A(0,3) and C($-4$,1) into it, we get\n\\[\n\\begin{cases}\nb=3, \\\\\n-4k+b=1,\n\\end{cases}\n\\]\nwhich, upon solving, gives\n\\[\n\\begin{cases}\nk=\\frac{1}{2}, \\\\\nb=3,\n\\end{cases}\n\\]\n$\\therefore$ The equation of line $l_{2}$ is $\\boxed{y=\\frac{1}{2}x+3}$;",
    "image_path": "dataset/mm_math/images/MM_Math/50682805_118.png"
  },
  {
    "solution": "\\boxed{\\sqrt{2}-1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, square $ABCD$ has a side length of 1, and point $E$ is on the extension line of $BC$. If $BE=BD$, then what is the length of $CE$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a square, \\\\\nit follows that BC=CD=1, \\\\\nBy the Pythagorean theorem, we have $BD=\\sqrt{BC^{2}+CD^{2}}=\\sqrt{2}$, \\\\\ntherefore, $CE=BE-BC=BD-BC= \\boxed{\\sqrt{2}-1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50496611.png"
  },
  {
    "solution": "\\boxed{B(0, 1)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the Cartesian coordinate system, the line $y=\\frac{1}{2}x+m$ intersects the $x$ axis at point $A$ and the $y$ axis at point $B$, where the coordinates of point $A$ are $\\left(-2, 0\\right)$, and the coordinates of point $C$ are $\\left(3, 0\\right)$. What are the coordinates of point $B$?\n",
    "completion": "\\textbf{Solution:} Substitute $A(-2, 0)$ into $y=\\frac{1}{2}x+m$, yielding $\\frac{1}{2}\\times (-2)+m=0$. Solving for $m$ gives $m=1$. Therefore, the equation of line $AB$ is $y=\\frac{1}{2}x+1$. Letting $x=0$, we find $y=1$, thus $B(0, 1)$. Hence, the coordinates of point $B$ are $\\boxed{B(0, 1)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50553714_107.png"
  },
  {
    "solution": "\\boxed{90^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AE \\perp BD$, $CD \\perp BD$, $AB = BC$, $BE = CD$. What is the measure of $\\angle ABC$ in degrees?\n",
    "completion": "\\textbf{Solution:} Given $\\because AE\\perp BD$, $CD\\perp BD$,\\\\\n$\\therefore \\angle AEB=\\angle BDC=90^\\circ$,\\\\\nIn $Rt\\triangle AEB$ and $Rt\\triangle BDC$\\\\\n$\\because \\left\\{\\begin{array}{l}AB=BC\\\\ BE=CD\\end{array}\\right.$,\\\\\n$\\therefore Rt\\triangle AEB\\cong Rt\\triangle BDC$,\\\\\n$\\therefore \\angle A=\\angle CBD$,\\\\\n$\\because \\angle AEB=90^\\circ$,\\\\\n$\\therefore \\angle A+\\angle ABE=90^\\circ$,\\\\\n$\\therefore \\angle ABC=\\angle ABE+\\angle CBD=\\angle ABE+\\angle A=90^\\circ$.\\\\\nHence, the measure of $\\angle ABC$ is $\\boxed{90^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53033295.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the perimeter of the rhombus ABCD is 16, and $\\angle DAB=60^\\circ$. The point E is the midpoint of the side BC, and the point P is a moving point on the diagonal AC. When connecting BP and EP, what is the minimum value of $PB+PE$?\n",
    "completion": "\\textbf{Solution:} Let's connect BD, which intersects AC at O, and connect DE which intersects AC at P, as shown in the diagram.\\\\\nSince the diagonals of a rhombus are perpendicular bisectors of each other, it can be inferred that B and D are symmetrical about the line AC, hence PD=PB.\\\\\nConsidering that the shortest distance between two points is a straight line,\\\\\n$\\therefore PE+PB=PE+PD=DE$\\\\\nmeaning DE is the minimum value of $PB+PE$.\\\\\n$\\because$ Quadrilateral ABCD is a rhombus with a perimeter of 16, and $\\angle DAB=60^\\circ$,\\\\\n$\\therefore \\angle DCB=\\angle DAB=60^\\circ$, and $BC=DC=4$.\\\\\n$\\therefore \\triangle BDC$ is an equilateral triangle.\\\\\n$\\because$ Point E is the midpoint of side BC,\\\\\n$\\therefore CE=\\frac{1}{2}BC=2$, and $DE\\perp BC$.\\\\\n$\\therefore \\angle DEC=90^\\circ$.\\\\\nIn $\\triangle DCE$, by the Pythagorean theorem, we find that $DE=\\sqrt{BD^2-BE^2}=2\\sqrt{3}$.\\\\\nThus, the minimum value of $PB+PE$ is $\\boxed{2\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51863891.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{1}{2}{x}^{2}+x}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, point $P$ is an arbitrary point on side $BC$ (point $P$ does not coincide with points B and C), and the vertices $F$ and $E$ of rectangle $AFPE$ are on $AB$ and $AC$, respectively. The seventh grade corresponds to the area of rectangle $AFPE$, while the eighth and ninth grades correspond to the areas of $\\triangle PEC$ and $\\triangle BPF$, respectively. Given that $BC=2$ and $S_{\\triangle ABC}=1$. Let $BP=x$, and the area of rectangle $AFPE$ be $y$, find the function relationship between $y$ and $x$.\n",
    "completion": "\\textbf{Solution}: From (1) we have: $\\triangle BFP \\sim \\triangle BAC$, \\\\\n$\\because BP=x, BC=2$, \\\\\n$\\therefore \\frac{{S}_{\\triangle BFP}}{{S}_{\\triangle BAC}}=\\left(\\frac{x}{2}\\right)^{2}$, \\\\\n$\\because {S}_{\\triangle ABC}=1$, \\\\\n$\\therefore {S}_{\\triangle BFP}=\\frac{{x}^{2}}{4}$, \\\\\nSimilarly, ${S}_{\\triangle PEC}=\\left(\\frac{2-x}{2}\\right)^{2}=\\frac{4-4x+{x}^{2}}{4}$, \\\\\n$\\therefore y=1-\\frac{{x}^{2}}{4}-\\frac{4-4x+{x}^{2}}{4}=-\\frac{1}{2}{x}^{2}+x$, \\\\\n$\\therefore$ the function relationship between y and x is $\\boxed{y=-\\frac{1}{2}{x}^{2}+x}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53065579_142.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB\\parallel CD\\parallel EF$. If $AC: CE = 2: 3$ and $BF = 10$, what is the length of the line segment $DF$?\n",
    "completion": "Solution: Since $AB\\parallel CD\\parallel EF$,\\\\\nhence $\\frac{BD}{DF}=\\frac{AC}{CE} = \\frac{2}{3}$,\\\\\nSince $BF=10$,\\\\\nthus $DF=10 \\times \\frac{3}{5} = \\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52514049.png"
  },
  {
    "solution": "\\boxed{\\frac{6}{7}\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the square $ABCD$ with side length 2, point $E$ is the midpoint of $AD$. Connect $BE$ and fold $\\triangle ABE$ along $BE$ to obtain $\\triangle FBE$. $BF$ intersects $AC$ at point $G$. What is the length of $CG$?\n",
    "completion": "\\textbf{Solution:} Extend $BF$ to meet $CD$ at $H$, and connect $EH$,\\\\\nsince $\\triangle FBE$ is obtained by folding $\\triangle ABE$ along $BE$,\\\\\ntherefore $EA=EF$, $\\angle EFB=\\angle EAB=90^\\circ$,\\\\\nsince $E$ is the midpoint of $AD$, and the side length of square $ABCD$ is $2$,\\\\\ntherefore $EA=ED=1$,\\\\\ntherefore $ED=EF=1$,\\\\\nsince quadrilateral $ABCD$ is a square,\\\\\ntherefore $\\angle D=\\angle EFB=\\angle EFH=90^\\circ$,\\\\\nin $Rt\\triangle EDH$ and $Rt\\triangle EFH$,\\\\\n$\\left\\{\\begin{array}{l}ED=EF\\\\ EH=EH\\end{array}\\right.$,\\\\\ntherefore $Rt\\triangle EDH \\cong Rt\\triangle EFH\\ (HL)$,\\\\\ntherefore $\\angle DEH=\\angle FEH$,\\\\\nalso since $\\angle AEB=\\angle FEB$,\\\\\ntherefore $\\angle DEH+\\angle AEB=90^\\circ$,\\\\\nsince $\\angle ABE+\\angle AEB=90^\\circ$,\\\\\ntherefore $\\angle ABE=\\angle DEH$,\\\\\ntherefore $\\triangle DHE \\sim \\triangle AEB$,\\\\\ntherefore $\\frac{DH}{DE}=\\frac{AE}{AB}=\\frac{1}{2}$,\\\\\ntherefore $DH=\\frac{1}{2}$,\\\\\ntherefore $CH=CD−DH=2−\\frac{1}{2}=\\frac{3}{2}$,\\\\\nsince $CH\\parallel AB$,\\\\\ntherefore $\\triangle HGC \\sim \\triangle BGA$,\\\\\ntherefore $\\frac{CG}{AG}=\\frac{CH}{AB}=\\frac{3}{4}$,\\\\\ntherefore $CG=\\frac{3}{4}AG=\\frac{3}{4}(AC−CG)$,\\\\\nsince $AB=2$, $CB=2$, $\\angle CBA=90^\\circ$,\\\\\ntherefore $AC=2\\sqrt{2}$,\\\\\ntherefore $CG=\\frac{3}{4}(2\\sqrt{2}−CG)$,\\\\\ntherefore $CG=\\boxed{\\frac{6}{7}\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53151242.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, fold the square paper $ABCD$ so that $AB$ coincides with $CD$, with the crease being $EF$. As shown in Figure 2, unfold it and then fold it once more, making point C coincide with point E, with the crease being $GH$. The corresponding point of point B is point M, and $EM$ intersects $AB$ at N. What is the ratio of $AE$ to $AN$?\n",
    "completion": "\\textbf{Solution:} Let the side length of the square be $2a$ and $DH=b$. Therefore, $CH=2a−b$. Because when the square paper $ABCD$ is folded to make $AB$ coincide with $CD$, the crease is $EF$. As shown in Figure 2, after unfolding and then folding again to make point C coincide with point E, the crease is $GH$, and the corresponding point to point B is point M. Therefore, $EH=CH=2a−b$, $DE=AE=\\frac{1}{2}\\times2a=a$, $\\angle BEH=\\angle C=90^\\circ$. In $\\triangle DEH$, $DE^{2}+DH^{2}=EH^{2}$, i.e., $a^{2}+b^{2}=(2a−b)^{2}$. Solving gives $b=\\frac{3}{4}a$. Because $\\angle DEH+\\angle AEN=90^\\circ$ and $\\angle AEN+\\angle DEH=90^\\circ$, therefore $\\angle ANE=\\angle DEH$. Therefore, $\\tan\\angle ANE=\\tan\\angle DEH=\\frac{AE}{AN}=\\frac{DH}{DE}=\\frac{b}{a}=\\frac{\\frac{3}{4}a}{a}=\\boxed{\\frac{3}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53081476.png"
  },
  {
    "solution": "\\boxed{-\\frac{1}{2}x^{2}+2x}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, two balls are thrown from point A on the slope and pass through the basketball hoop at point B, landing at point O at the bottom of the slope. A rectangular coordinate system is established with O as the origin, and the coordinates of B are $\\left(\\frac{1}{2}, \\frac{7}{8}\\right)$. The slope of the slope is $1\\colon 2$, and the height of point A from the ground is 1.5 meters. The trajectory of the balls can be described by a quadratic function. What is the expression of the quadratic function?\n",
    "completion": "Solution: Draw $AD\\perp x$-axis through point A, with D as the foot of the perpendicular.\\\\\nAccording to the problem, $AD:OD=1:2$ and $AD=1.5$,\\\\\n$\\therefore OD=2AD=3$,\\\\\n$\\therefore A(3, 1.5)$.\\\\\n$\\because$ The graph of the quadratic function passes through the origin O,\\\\\n$\\therefore$ let the expression of the quadratic function be $y=ax^{2}+bx$.\\\\\nSubstituting the coordinates of points A and B, we get:\\\\\n$\\begin{cases}9a+3b=1.5\\\\ \\frac{1}{4}a+\\frac{1}{2}b=\\frac{7}{8}\\end{cases}$,\\\\\nSolving this, we find $\\begin{cases}a=-\\frac{1}{2}\\\\ b=2\\end{cases}$,\\\\\n$\\therefore$ The expression of the quadratic function is $y=-\\frac{1}{2}x^{2}+2x=\\boxed{-\\frac{1}{2}x^{2}+2x}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52796984_39.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in parallelogram $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$. $E$ is the midpoint of side $CD$, and the length of $OE$ is $3$ cm. What is the length of $BC$ in centimeters?\n",
    "completion": "\\textbf{Solution:} Since ABCD is a parallelogram, \\\\\ntherefore $OB=OD$, \\\\\nand since point E is the midpoint of side CD, \\\\\ntherefore $BC=2OE=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52049813.png"
  },
  {
    "solution": "\\boxed{9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point O is the midpoint of the diagonal $AC$ of rectangle $ABCD$, and point M is the midpoint of $AD$. If $AB=3$ and $BC=4$, what is the perimeter of quadrilateral $ABOM$?\n",
    "completion": "\\textbf{Solution:} Since $AB=3$ and $BC=4$,\\\\\nit follows that $AC=\\sqrt{3^2+4^2}=5$. Since point O is the midpoint of AC, then $BO=\\frac{1}{2}AC=2.5$.\\\\\nAlso, M is the midpoint of AD, thus MO is the midline of $\\triangle ACD$, hence $OM=\\frac{1}{2}CD=1.5$.\\\\\nTherefore, the perimeter of the quadrilateral ABOM is $AB+BO+MO+AM=3+2.5+2+1.5=\\boxed{9}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52423898.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in figure, within the rectangle $ABCD$, the lengths of sides $AB$ and $BC$ $(AB<BC)$ are the two roots of the equation ${x}^{2}-7x+12=0$. Point $P$ starts from point $A$ and moves along the perimeter of $\\triangle ABC$ in the direction of $A\\to B\\to C\\to A$ at a speed of $1$ unit per second, and the movement time is $t$ seconds. When point $P$ moves onto side $BC$, try to find the value of the movement time $t$, so that the length of $AP$ is $\\sqrt{10}$.\n",
    "completion": "\\textbf{Solution:} By the given, we have $3^2+(t-3)^2=10$,\\\\\n$\\therefore t_1=4$, $t_2=2$ (discarded),\\\\\nthus, for $t=\\boxed{4}$, we have $AP=\\sqrt{10}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51722482_164.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=-\\frac{1}{2}x+3$ intersects the coordinate axes at points A and B, and intersects the line $y=x$ at point C. A point Q on the line segment OA moves from point O to point A at a uniform speed of 1 unit per second. Let the motion time be $t(s)$. Connect C and Q. If $\\triangle OQC$ is an isosceles right triangle, find the value of $t$.\n",
    "completion": "\\textbf{Solution:}\\\\\n(1) As shown in Figure 1, when $\\angle CQO=90^\\circ$, CQ=OQ,\\\\\n$\\because C(2,2)$,\\\\\n$\\therefore OQ=CQ=2$,\\\\\n$\\therefore t=2$;\\\\\n(2) As shown in Figure 2, when $\\angle OCQ=90^\\circ$, OC=CQ, draw CM$\\perp$OA at M through point C,\\\\\n$\\because C(2,2)$,\\\\\n$\\therefore CM=OM=2$,\\\\\n$\\therefore QM=OM=2$,\\\\\n$\\therefore t=2+2=4$,\\\\\nthus the value of t is 2 or 4;\\\\\nTherefore, the value of $t$ is $\\boxed{2}$ or $\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51663132_207.png"
  },
  {
    "solution": "\\boxed{9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that lines $a\\parallel b\\parallel c$, and lines $m, n$ intersect with $a, b, c$ at points $A, C, E, B, D, F$, respectively. If $AC=8$, $CE=12$, and $BD=6$, what is the value of $DF$?\n",
    "completion": "\\textbf{Solution:} Since $a\\parallel b\\parallel c$,\\\\\nit follows that $\\frac{AC}{CE}=\\frac{BD}{DF}$,\\\\\nGiven $AC=8$, $CE=12$, and $BD=6$,\\\\\nwe have $\\frac{8}{12}=\\frac{6}{DF}$. Solving this yields: $DF=\\boxed{9}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51457856.png"
  },
  {
    "solution": "\\boxed{40}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the distance from point $M$, the midpoint of one side of rhombus $ABCD$, to point $O$, the intersection of the diagonals, is $5\\text{cm}$. What is the perimeter of rhombus $ABCD$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rhombus,\\\\\nit follows that $\\angle$AOB=$90^\\circ$,\\\\\nsince M is the midpoint of side AB,\\\\\nit follows that AB=2OM=10,\\\\\nthus, the perimeter of the rhombus ABCD is $10\\times 4=\\boxed{40}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52416363.png"
  },
  {
    "solution": "\\boxed{\\sqrt{13}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rhombus $ABCD$, $AC=6$, and $BD=4$. What is the length of $AB$?\n",
    "completion": "\\textbf{Solution:} Since ABCD is a rhombus,\\\\\nit follows that AC $\\perp$ BD, AO = OC, BO = DO,\\\\\nsince AC = 6, BD = 4,\\\\\nit follows that AO = 3, BO = 2,\\\\\ntherefore AB = $\\sqrt{AO^{2}+BO^{2}}$ = $\\sqrt{3^{2}+2^{2}}$ = $\\boxed{\\sqrt{13}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52082489.png"
  },
  {
    "solution": "\\boxed{120^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, line $AB$ intersects with line $CD$ at point $O$, with $OE \\perp AB$. Given that $\\angle BOD = 30^\\circ$, what is the value of $\\angle COE$?\n",
    "completion": "\\textbf{Solution:} Given that $OE\\perp AB$\\\\\nHence, $\\angle AOE=90^\\circ$\\\\\nGiven that $\\angle BOD=30^\\circ$\\\\\nTherefore, $\\angle AOC=\\angle BOD=30^\\circ$\\\\\nThus, $\\angle COE=\\angle AOC+\\angle AOE=\\boxed{120^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/51370077.png"
  },
  {
    "solution": "\\boxed{\\frac{13}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $E$, $F$, and $G$ are on $AB$, $BC$, and $CD$ respectively, $DE \\perp EF$, $EF \\perp FG$, $BE=3$, $BF=2$, and $FC=6$. What is the length of $DG$?\n",
    "completion": "\\textbf{Solution:} Let us consider right-angled triangle $BEF$, where $BF=2$ and $BE=3$\\\\\nThus, $EF= \\sqrt{BE^{2}+BF^{2}}=\\sqrt{3^{2}+2^{2}}=\\sqrt{13}$\\\\\nDraw a line from $G$ such that $GH \\perp DE$, with $H$ as the foot of the perpendicular,\\\\\nSince $DE \\perp EF$ and $EF \\perp FG$\\\\\nHence, quadrilateral $EFGH$ is a rectangle\\\\\nTherefore, $HG=EF= \\sqrt{13}$\\\\\nGiven rectangle $ABCD$\\\\\nTherefore, $\\angle A=\\angle B=90^\\circ$\\\\\nThus, $\\angle AED+\\angle ADE=90^\\circ$\\\\\nSince $DE \\perp EF$\\\\\nTherefore, $\\angle AED+\\angle BEF=90^\\circ$\\\\\nThus, $\\angle BEF=\\angle ADE$\\\\\nAgain, since $\\angle A=\\angle B=90^\\circ$\\\\\nTherefore, $\\triangle EBF\\sim \\triangle DAE$\\\\\nSimilarly, $\\triangle DAE\\sim \\triangle GHD$\\\\\nThus, $\\triangle EBF\\sim \\triangle GHD$\\\\\nTherefore, $\\frac{DG}{EF}=\\frac{HG}{BE}$, that is $\\frac{DG}{\\sqrt{13}}=\\frac{\\sqrt{13}}{3}$, solving this gives $DG= \\boxed{\\frac{13}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51433270.png"
  },
  {
    "solution": "\\boxed{C(10, 0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the Cartesian coordinate system, the sides $OA$ and $OC$ of rectangle $OABC$ are located on the positive $x$-axis and the positive $y$-axis, respectively. The lengths of $OA$ and $OC$ are the two roots of the equation ${x}^{2}-20x+100=0$. Find the coordinates of point $C$.\n",
    "completion": "\\textbf{Solution:} From the given, we get $(x-10)^{2}=0$, \\\\\n$\\therefore x_{1}=x_{2}=10$, \\\\\n$\\therefore OA=OC=10$, \\\\\n$\\therefore \\boxed{C(10, 0)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53728996_95.png"
  },
  {
    "solution": "\\boxed{\\frac{12}{13}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: If the point $P(12, a)$ lies on the graph of the inverse proportion function $y=\\frac{60}{x}$, then what is the value of $\\cos\\angle POH$?\n",
    "completion": "\\textbf{Solution:} Given that point P lies on the graph of the inverse proportion function $y=\\frac{60}{x}$,\\\\\nthus $a=\\frac{60}{12}=5$,\\\\\nhence, the coordinates of point P are (12, 5),\\\\\ntherefore, $OH=12$ and $PH=5$,\\\\\nin $\\triangle OHP$, by the Pythagorean theorem, $OP=\\sqrt{OH^2+PH^2}$,\\\\\nthus, $OP=\\sqrt{12^2+5^2}=13$,\\\\\ntherefore, $\\cos\\angle POH=\\frac{OH}{OP}=\\boxed{\\frac{12}{13}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52623491.png"
  },
  {
    "solution": "\\boxed{\\frac{15}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, three squares with side lengths of 2, 3, and 5, respectively, are arranged as shown in the figure. What is the area of the shaded region in the figure?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, \\\\\n$\\because$VB$\\parallel$ED, and the side lengths of the three squares are 2, 3, and 5 respectively, \\\\\n$\\therefore$VB$\\colon$ DE=AB$\\colon$ AD, that is, VB$\\colon$ 5=2$\\colon$ (2+3+5)=1$\\colon$ 5, \\\\\n$\\therefore$VB=1, \\\\\n$\\because$CF$\\parallel$ED, \\\\\n$\\therefore$CF$\\colon$ DE=AC$\\colon$ AD, that is, CF$\\colon$ 5=5$\\colon$ 10, \\\\\n$\\therefore$CF=2.5, \\\\\n$\\because$Area of trapezoid $VBFC$= $\\frac{1}{2}$(BV+CF)$\\cdot$BC= $\\frac{21}{4}$, \\\\\n$\\therefore$ the area of the shaded part=Area of square $BCQW$−Area of trapezoid $VBCF$= $\\boxed{\\frac{15}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51457951.png"
  },
  {
    "solution": "\\boxed{16}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, the net of a cube with edge length of 1 has two edges respectively on $AC$, $BC$, and two vertices on the hypotenuse $AB$. What is the area of $\\triangle ABC$?\n",
    "completion": "\\textbf{Solution:}\\\\\nFrom the given information, we have: $DE \\parallel MF$,\\\\\n$\\therefore \\triangle BDE \\sim \\triangle BMF$,\\\\\n$\\therefore \\frac{BD}{BM} = \\frac{DE}{MF}$, which gives $\\frac{BD}{BD+1} = \\frac{2}{4}$,\\\\\nSolving gives $BD=1$,\\\\\nSimilarly, we find that $AN=6$;\\\\\nFurthermore, since quadrilateral $DENC$ is a rectangle,\\\\\n$\\therefore DE=CN=2$, $DC=EN=3$,\\\\\n$\\therefore BC=BD+DC=4$, $AC=CN+AN=8$,\\\\\n$\\therefore$ The area of $\\triangle ABC$ $= BC \\times AC \\div 2 = 4 \\times 8 \\div 2 = \\boxed{16}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52977308.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, extend the line segment $AB$ to $C$ so that $BC=4AB$. Let $D$ be the midpoint of line segment $BC$. If $CD=4\\text{cm}$, and let $E$ be the midpoint of line segment $AC$, what is the length of $ED$ in cm?\n",
    "completion": "\\textbf{Solution:} Given that point $E$ is the midpoint of segment $AC$,\\\\\nit follows that $AE=\\frac{1}{2}AC=5\\text{cm}$,\\\\\nsince $BD=4\\text{cm}$ and $AB=2\\text{cm}$,\\\\\nwe have $AD=AB+BD=6\\text{cm}$,\\\\\ntherefore $ED=AD-AE=6-5=\\boxed{1}\\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53106951_63.png"
  },
  {
    "solution": "\\boxed{625}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ABC=90^\\circ$, squares are constructed externally on the sides AB, BC, and AC, with areas of 225, 400, and $S$ respectively. What is the value of $S$?\n",
    "completion": "\\textbf{Solution:} By the Pythagorean theorem, we have $AB^{2}+BC^{2}=AC^{2}$,\\\\\nthus $S=AC^{2}=225+400=625$,\\\\\ntherefore, $S=\\boxed{625}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52604265.png"
  },
  {
    "solution": "\\boxed{y=\\frac{1}{2}x-1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, two lines $l_{1}$ and $l_{2}$ pass through point A(2,0) and intersect the y-axis at points B and C, respectively, where point B is above the origin and point C is below the origin. It is known that $AB=\\sqrt{13}$. If $BC=4$, find the equation of line $l_{2}$.\n",
    "completion": "\\textbf{Solution:} Since $BC=4$,\\\\\nthe coordinates of point C are (0, -1),\\\\\nlet the equation of $l_{2}$ be $y=kx+b$,\\\\\nsubstituting A(2, 0), C(0, -1) into it, we get:\\\\\n$\\therefore \\left\\{\\begin{array}{l}2k+b=0 \\\\ b=-1 \\end{array}\\right.$\\\\\nSolving, we find: $\\left\\{\\begin{array}{l}k=\\frac{1}{2} \\\\ b=-1 \\end{array}\\right.$\\\\\nHence, the equation of $l_{2}$ is $y=\\frac{1}{2}x-1=\\boxed{y=\\frac{1}{2}x-1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52056156_205.png"
  },
  {
    "solution": "\\boxed{21}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the quadratic function $y=-x^2+bx+c$ intersects the $x$-axis at points $A$ and $B$, and intersects the $y$-axis at point $C(0, 6)$, and passes through the point $(1, 10)$. Find: What is the area of $\\triangle ABC$?\n",
    "completion": "\\textbf{Solution:} We have the quadratic function $y=-x^2+5x+6$ intersects the $x$-axis at points A and B,\\\\\ntherefore $-x^2+5x+6=0$, solving this gives $x_1=-1$, $x_2=6$,\\\\\nthus, the coordinates of the two intersection points of the quadratic function $y=-x^2+5x+6$ with the $x$-axis are $A(-1, 0)$, $B(6, 0)$.\\\\\nTherefore, the area of $\\triangle ABC$ is $\\frac{1}{2}AB\\times OC=\\frac{1}{2}\\times |6-(-1)|\\times 6=\\boxed{21}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53065902_78.png"
  },
  {
    "solution": "\\boxed{60^{\\circ}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, PA and PB are tangent to circle \\(O\\) at points A and B, respectively. Point E is a point on circle \\(O\\), and \\(\\angle AEB = 60^\\circ\\). What is the measure of \\(\\angle P\\)?\n",
    "completion": "\\textbf{Solution:} Draw OA, BO;\\\\\nSince $\\angle AOB = 2\\angle E = 120^{\\circ}$,\\\\\nthus $\\angle OAP = \\angle OBP = 90^{\\circ}$,\\\\\ntherefore $\\angle P = 180^{\\circ} - \\angle AOB = \\boxed{60^{\\circ}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51262022.png"
  },
  {
    "solution": "\\boxed{35^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ and $\\triangle ADE$ are symmetric with respect to line $MN$, and the intersection point $F$ of $BC$ and $DE$ lies on line $MN$. If $\\angle BAC=100^\\circ$ and $\\angle CAD=30^\\circ$, what is the degree measure of $\\angle EAF$?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABC$ and $\\triangle ADE$ are symmetrical with respect to line MN, \\\\\n$\\therefore \\triangle ABC \\cong \\triangle ADE$, $\\angle CAF = \\angle EAF$,\\\\\n$\\therefore \\angle DAE = \\angle BAC = 100^\\circ$,\\\\\n$\\because \\angle CAD = 30^\\circ$,\\\\\n$\\therefore \\angle CAE = 100^\\circ - 30^\\circ = 70^\\circ$,\\\\\n$\\therefore \\angle EAF = \\frac{70^\\circ}{2} = \\boxed{35^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52991684_117.png"
  },
  {
    "solution": "\\boxed{155^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $AB\\parallel CD$, $AD\\parallel BC$, $\\angle DCE=90^\\circ$, point E is on segment AB, $\\angle FCG=90^\\circ$, point F is on line AD, $\\angle AHG=90^\\circ$. If $\\angle ECF=25^\\circ$, what is the degree measure of $\\angle BCD$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle ECF=25^\\circ$ and $\\angle DCE=90^\\circ$,\n\n$\\therefore \\angle FCD=65^\\circ$. Also, since $\\angle BCF=90^\\circ$,\n\n$\\therefore \\angle BCD=65^\\circ+90^\\circ=\\boxed{155^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51434356_68.png"
  },
  {
    "solution": "\\boxed{\\frac{12}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the rhombus ABCD, $AC=8$, $BD=6$. Let $E$ be a moving point on side $CD$. Draw $EF\\perp OC$ at point $F$ and $EG\\perp OD$ at point $G$, and connect $FG$. What is the minimum value of $FG$?\n",
    "completion": "\\textbf{Solution:}\\\\\nSince quadrilateral ABCD is a rhombus,\\\\\nit follows that AC$\\perp$BD and AD=CD,\\\\\nGiven that EF$\\perp$OC at point F, and EG$\\perp$OD at point G,\\\\\nit follows that quadrilateral OGEF is a rectangle,\\\\\nBy connecting OE, then OE=GF,\\\\\nWhen OE$\\perp$DC, the value of GF is minimal,\\\\\nSince BD=6 and AC=8,\\\\\nit follows that OD=3 and OC=4,\\\\\nTherefore, CD= $\\sqrt{OC^{2}+OD^{2}}=5$,\\\\\nSince the area of $\\triangle$COD is $\\frac{1}{2}OC\\cdot OD=\\frac{1}{2}CD\\cdot OE$,\\\\\nit follows that $OE=\\frac{OC\\cdot OD}{CD}=\\frac{4\\times 3}{5}=\\frac{12}{5}$,\\\\\nTherefore, the minimum value of FG is $\\boxed{\\frac{12}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52303783.png"
  },
  {
    "solution": "\\boxed{y= -\\frac{1}{4}x^{2}+ \\frac{3}{2}x+2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the parabola $y= \\frac{1}{4}x^{2}+bx+c$ passes through point $A(0, 2)$ and point $C(4, 0)$, and intersects the x-axis at another point $B$. What is the analytic equation of the parabola?\n",
    "completion": "\\textbf{Solution:} Substituting points A and C into $y= -\\frac{1}{4}x^2+bx+c$ yields,\n\\[\n\\left\\{\\begin{array}{l}\nc=2\\\\\n4+4b+c=0\n\\end{array}\\right.\n\\] Solving this gives \\[\n\\left\\{\\begin{array}{l}\nb=\\frac{3}{2}\\\\\nc=2\n\\end{array}\\right.\n\\]\n$\\therefore$ The equation of the parabola is $\\boxed{y= -\\frac{1}{4}x^{2}+ \\frac{3}{2}x+2}$",
    "image_path": "dataset/mm_math/images/MM_Math/51628092_131.png"
  },
  {
    "solution": "\\boxed{(2,0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the vertex $C$ of the parabola passing through points $A\\left(0, 4\\right)$ and $B\\left(5, 9\\right)$ is on the positive half-axis of the $x$-axis. Find the coordinates of point $C$.\n",
    "completion": "\\textbf{Solution:}\nSince the equation of the parabola is $y=(x-2)^2$, \\\\\ntherefore, the coordinates of point $C$ are $\\boxed{(2,0)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51369908_193.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, where $AC=3$, $BC=4\\sqrt{2}$, and $\\angle ACB=45^\\circ$, let $D$ be a moving point within $\\triangle ABC$. Let $\\odot O$ be the circumcircle of $\\triangle ACD$. The line $BD$ intersects $\\odot O$ at point $P$ and intersects $BC$ at point $E$. If the arc $AE=CP$, what is the minimum value of $AD$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure,\\\\\n$\\because$ arc AE = arc CP,\\\\\n$\\therefore$ $\\angle$CDP=$\\angle$ACB=$45^\\circ$,\\\\\n$\\therefore$ $\\angle$BDC=$135^\\circ$,\\\\\n$\\therefore$ when AD passes through $O_{1}$, AD has its minimum value,\\\\\n$\\because$ $\\angle$BDC=$135^\\circ$,\\\\\n$\\therefore$ $\\angle$BO$_{1}$C=$90^\\circ$,\\\\\n$\\because$ BC=$4\\sqrt{2}$,\\\\\n$\\therefore$ O$_{1}$C=O$_{1}$B=4=O$_{1}$D,\\\\\n$\\therefore$ $\\angle$O$_{1}$CB=$45^\\circ$,\\\\\n$\\therefore$ $\\angle$AO$_{1}$C=$45^\\circ+45^\\circ=90^\\circ$,\\\\\n$\\because$ AC=3,\\\\\n$\\therefore$ AO$_{1}$=5,\\\\\n$\\therefore$ AD=AO$_{1}$−O$_{1}$D=5−4=\\boxed{1}.",
    "image_path": "dataset/mm_math/images/MM_Math/52874523.png"
  },
  {
    "solution": "\\boxed{51^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the perpendicular bisectors $m$ and $n$ of sides $AB$ and $AC$ of $\\triangle ABC$ intersect at point $D$. Connect $CD$. If $\\angle 1=39^\\circ$, then what is the measure of $\\angle BCD$ in degrees?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect AD, BD,\\\\\nsince the perpendicular bisectors m and n of sides AB and AC of $\\triangle ABC$ intersect at point D,\\\\\ntherefore, AD=BD=CD,\\\\\nsince $\\angle 1=39^\\circ$,\\\\\ntherefore, $\\angle EDF=180^\\circ-\\angle 1=141^\\circ$,\\\\\nthus, $\\angle EDA+\\angle ADF=141^\\circ$,\\\\\nsince DA=DC, and DE$\\perp$AC,\\\\\ntherefore, $\\angle ADE=\\angle CDE$,\\\\\nsimilarly, it can be derived that $\\angle ADF=\\angle BDF$,\\\\\ntherefore, $\\angle CDE+\\angle BDF=141^\\circ$,\\\\\ntherefore, $\\angle CDB=360^\\circ-\\angle EDA-\\angle ADF-\\angle CDE-\\angle BDF=78^\\circ$,\\\\\nsince DC=DB,\\\\\ntherefore, $\\angle DCB=\\frac{180^\\circ-\\angle CDB}{2}=\\boxed{51^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/52603574.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in $\\triangle ABC$, $\\angle BAC=95^\\circ$, $\\angle B=25^\\circ$, $\\angle CAD=75^\\circ$, determine the measure of $\\angle ADC$.\n",
    "completion": "\\textbf{Solution:} Since $\\angle BAC=95^\\circ$ and $\\angle CAD=75^\\circ$,\\\\\nit follows that $\\angle BAD=\\angle BAC-\\angle CAD=20^\\circ$,\\\\\nSince $\\angle B=25^\\circ$,\\\\\nit follows that $\\angle ADC=\\angle ABC+\\angle BAD=25^\\circ+20^\\circ=\\boxed{45^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52826811.png"
  },
  {
    "solution": "\\boxed{\\sqrt{10}-\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\triangle ABC$, $\\angle ACB=90^\\circ$, $AC=BC$, point D is a moving point on side $AC$, connecting $BD$, a circle with $CD$ as its diameter intersects $BD$ at point E. If the length of $AB$ is 4, what is the minimum length of segment $AE$?\n",
    "completion": "\\textbf{Solution:} Join CE. Since CD is the diameter,\\\\\n$\\therefore \\angle CED=90^\\circ=\\angle BEC$,\\\\\n$\\therefore$ point E moves on the circle $\\odot O$ with the midpoint O of BC as its center and BC as its diameter,\\\\\nJoin AO, intersecting $\\odot O$ at point E, then at this moment AE=AO−OE is the minimum,\\\\\n$\\because \\angle ACB=90^\\circ$, $AC=BC$, $AB=4$,\\\\\n$\\therefore \\angle ABC=\\angle BAC=45^\\circ$,\\\\\n$\\therefore AC=BC=AB\\cdot\\sin45^\\circ=2\\sqrt{2}$, $OB=OC=OE=\\sqrt{2}$,\\\\\n$\\therefore AO=\\sqrt{(2\\sqrt{2})^2+(\\sqrt{2})^2}=\\sqrt{10}$,\\\\\n$\\therefore AE=\\sqrt{10}-\\sqrt{2}=\\boxed{\\sqrt{10}-\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51323911.png"
  },
  {
    "solution": "\\boxed{(3, 0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the vertices of rectangle $OABC$ are such that vertex $O$ is at the origin, side $OC$ is on the positive half-axis of the $x$-axis, and side $OA$ is on the positive half-axis of the $y$-axis, with $OA=3$, $AB=4$. The graph of the inverse proportion function $y=\\frac{k}{x}$ (where $k>0$) intersects the sides $AB$ and $BC$ of the rectangle at points $D$ and $E$, respectively, and it is given that $BD=2AD$. For a point $P$ on the segment $OC$, does there exist a point $P$ such that $\\angle APE=90^\\circ$? If such a point exists, find the coordinates of point $P$; if it does not exist, please explain why.\n",
    "completion": "\\textbf{Solution:} Exist. Assume there exists the required point $P$ with coordinates $(m,0)$,\\\\\n$\\therefore OP=m$, $CP=4-m$.\\\\\n$\\because \\angle APE=90^\\circ$,\\\\\n$\\therefore \\angle APO+\\angle EPC=90^\\circ$,\\\\\nAlso $\\because \\angle APO+\\angle OAP=90^\\circ$,\\\\\n$\\therefore \\angle EPC=\\angle OAP$,\\\\\nAlso $\\because \\angle AOP=\\angle PCE=90^\\circ$,\\\\\n$\\therefore \\triangle AOP\\sim \\triangle PCE$,\\\\\n$\\therefore \\frac{OA}{CP}=\\frac{OP}{CE}$,\\\\\n$\\because OA\\cdot CE=OP\\cdot CP$.\\\\\n$\\therefore m(4-m)=3$,\\\\\nSolving: $m=1$ or $m=3$.\\\\\n$\\therefore$ There exists the required point $P$, with coordinates $\\boxed{(1, 0)}$ or $\\boxed{(3, 0)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51402967_36.png"
  },
  {
    "solution": "\\boxed{9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in quadrilateral $ABCD$, diagonals $AC$ and $BD$ intersect at point $E$. Extend $CD$ to point $F$ such that $DF=CD$, and connect $AF$. Quadrilateral $ABDF$ is a parallelogram; if $AB\\perp AC$ and $AB=2$, $AF=5$, what is the area of quadrilateral $ABCF$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral $ABCD$ is a parallelogram and $AB\\perp AC$, it follows that $CD=AB=2$ and $\\angle ACD=\\angle BAC=90^\\circ$. Since $CD=DF$, we have $CF=2CD=4$. Given that $AF=5$, in $\\triangle ACF$, we have $AC=\\sqrt{AF^{2}-CF^{2}}=\\sqrt{5^{2}-4^{2}}=3$. Therefore, the area of quadrilateral $ABCF$ is $S_{\\text{quadrilateral } ABCF}=S_{\\triangle BAC}+S_{\\triangle ACF}=\\frac{1}{2}AB\\cdot AC+\\frac{1}{2}AC\\cdot CF=\\frac{1}{2}\\times 2\\times 3+\\frac{1}{2}\\times 3\\times 4=\\boxed{9}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52251359_132.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, $AB=10$, $BC=6$, E is a point on side $AC$, $AE=5$, $ED\\perp AB$, and the foot of the perpendicular is point D. Then, what is the value of $DE$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle C=90^\\circ$ and $AB=10$, $BC=6$, we have $\\sin A=\\frac{BC}{AB}=\\frac{6}{10}=\\frac{3}{5}$. Given that $ED \\perp AB$, it follows that $\\sin A=\\frac{DE}{AE}=\\frac{3}{5}$. Hence, $AE=5$ and so $DE=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51384325.png"
  },
  {
    "solution": "\\boxed{(-2, 6)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system, the parabola $y = -\\frac{1}{4}x^2 + bx + c$ intersects the line $KB$ at point $B(2, 0)$ and point $K(-12, -14)$, and intersects the y-axis at point $C$. Connect $BC$. Let point $P$ be a moving point on the parabola above line segment $BK$. Draw $PZ \\parallel x$-axis intersecting $CB$ at point $Z$, and draw $PQ \\parallel CB$ intersecting line $KB$ at point $Q$. Find the maximum value of $\\frac{6}{5}\\sqrt{5}PQ + PZ$ and the coordinates of point $P$ at this time?\n",
    "completion": "textbf{Solution:} Let $P\\left(t, -\\frac{1}{4}t^{2}-\\frac{3}{2}t+4\\right)\\left(-12<t<2\\right)$,\n\ndraw $PM\\parallel y$ axis crossing $BK$ at $M$, and through $Q$ draw $QN\\parallel x$ axis crossing $PM$ at $N$,\n\nsince $B(2, 0)$, $K(-12, -14)$,\n\ntherefore, the equation of line $BK$ is $y=x-2$,\n\nsince $PM\\parallel y$ axis,\n\ntherefore, $P$ and $M$ have the same $x$-coordinate,\n\ntherefore, $M\\left(t, t-2\\right)$,\n\nthus, $PM=-\\frac{1}{4}t^{2}-\\frac{3}{2}t+4-(t-2)=-\\frac{1}{4}t^{2}-\\frac{5}{2}t+6$,\n\nit's easy to find $C\\left(0, 4\\right)$,\n\nsince $B(2, 0)$,\n\ntherefore, the equation of line $BC$ is $y=-2x+4$,\n\nsince $QN\\parallel x$ axis,\n\ntherefore, $P$ and $Z$ have the same $y$-coordinate,\n\ntherefore, the $y$-coordinate of $Z$ is $-\\frac{1}{4}t^{2}-\\frac{3}{2}t+4$,\n\nthus, $-\\frac{1}{4}t^{2}-\\frac{3}{2}t+4=-2x+4$,\n\nsolving for $x$ yields $x=\\frac{1}{8}t^{2}+\\frac{3}{4}t$,\n\nthus, $Z\\left(\\frac{1}{8}t^{2}+\\frac{3}{4}t, -\\frac{1}{4}t^{2}-\\frac{3}{2}t+4\\right)$,\n\ntherefore, $PZ=\\frac{1}{8}t^{2}+\\frac{3}{4}t-t=\\frac{1}{8}t^{2}-\\frac{1}{4}t$,\n\nbecause $PZ\\parallel x$ axis, $BC\\parallel PQ$, $PZ\\parallel x$ axis $\\parallel NQ$,\n\ntherefore, $\\angle OBC=\\angle PZC$, $\\angle PZC=\\angle ZPQ$, $\\angle ZPQ=\\angle PQN$,\n\ntherefore, $\\angle OBC=\\angle PQN$,\n\nalso $\\angle BOC=\\angle QNP=90^\\circ$,\n\ntherefore, $\\triangle BOC\\sim \\triangle QNP$,\n\ntherefore, $\\frac{BO}{NQ}=\\frac{BC}{PQ}=\\frac{OC}{PN}$,\n\nsince $BO=2$, $CO=4$,\n\ntherefore, $BC=\\sqrt{BO^{2}+CO^{2}}=2\\sqrt{5}$,\n\ntherefore, $\\frac{2}{NQ}=\\frac{2\\sqrt{5}}{PQ}=\\frac{4}{PN}$,\n\ntherefore, $NQ=\\frac{\\sqrt{5}}{5}PQ$, $PN=2QN$,\n\nsince $QN\\parallel x$ axis, the acute angle formed by $y=x-2$ and the $x$ axis is $45^\\circ$\n\ntherefore, $\\angle NQM=45^\\circ$,\n\ntherefore, $\\triangle MNQ$ is an isosceles right triangle,\n\ntherefore, $MN=NQ$,\n\ntherefore, $NQ=\\frac{1}{3}PM$,\n\ntherefore, $\\frac{1}{3}PM=\\frac{\\sqrt{5}}{5}PQ$,\n\ntherefore, $PQ=\\frac{\\sqrt{5}}{3}PM$,\n\ntherefore, $\\frac{6}{5}\\sqrt{5}PQ+PZ=\\frac{6}{5}\\sqrt{5} \\times \\boxed{\\frac{\\sqrt{5}}{3}PM} + \\boxed{\\frac{1}{8}t^{2}-\\frac{1}{4}t}$,\n\ntherefore, $\\frac{6}{5}\\sqrt{5}PQ+PZ=\\boxed{2PM} + \\boxed{\\frac{1}{8}t^{2}-\\frac{1}{4}t}$,\n\ntherefore, $\\frac{6}{5}\\sqrt{5}PQ+PZ=\\boxed{2\\left(-\\frac{1}{4}t^{2}-\\frac{5}{2}t+6\\right)} + \\boxed{\\frac{1}{8}t^{2}-\\frac{1}{4}t}$,\n\ntherefore, $\\frac{6}{5}\\sqrt{5}PQ+PZ=\\boxed{-\\frac{1}{8}t^{2}-\\frac{1}{4}t+12}$,\n\ntherefore, $\\frac{6}{5}\\sqrt{5}PQ+PZ$ reaches its maximum value when $t=-2$,\n\ntherefore, the maximum value of $\\frac{6}{5}\\sqrt{5}PQ+PZ$ is $\\boxed{13}$,\n\ntherefore, the coordinates of point $P$ at this time are $\\boxed{(-2, 6)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53080553_30.png"
  },
  {
    "solution": "\\boxed{24-\\sqrt{511}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, rectangle $ABCD$ is inscribed in circle $O$, with $AB=40$ and $BC=55$. Let $E$ be a point on arc $BC$, and draw lines $AE$ and $CE$, where $CE=20$. What is the length of $AE$? Let $F$ be a moving point on ray $CE$ such that $\\angle BFE>90^\\circ$. When $BF=AB$, what is the length of $EF$?\n",
    "completion": "\\textbf{Solution:} Draw $BM\\perp CF$ at point $M$, connect $AC$, let $AE$ intersect $BC$ at point $G$,\\\\\nsince quadrilateral ABCD is a rectangle,\\\\\ntherefore $AB=CD=40$, $AD=BC=55$, $\\angle ABC=\\angle ADC=90^\\circ$,\\\\\ntherefore $AC$ is the diameter,\\\\\ntherefore $\\angle AEC=90^\\circ$,\\\\\nin $\\triangle ADC$, we have $AC^{2}=AD^{2}+DC^{2}=4625$,\\\\\nsince $CE=20$,\\\\\ntherefore $AE=\\sqrt{AC^{2}-CE^{2}}=\\boxed{65}$,\\\\\nin $\\triangle ABG$ and $\\triangle CEG$,\\\\\n$\\left\\{\\begin{array}{l}\n\\angle ABG=\\angle CEG=90^\\circ \\\\\n\\angle AGB=\\angle CGE\n\\end{array}\\right.$,\\\\\ntherefore $\\triangle ABG\\sim \\triangle CEG$,\\\\\ntherefore $\\frac{BG}{EG}=\\frac{AG}{CG}=\\frac{AB}{CE}=\\frac{40}{20}=2$,\\\\\ntherefore $BG=2EG$, $AG=2CG$,\\\\\nsince $AG+EG=AE$,\\\\\ntherefore $2CG+EG=65$ (1),\\\\\nsince $BG+CG=BC$,\\\\\ntherefore $2EG+CG=55$ (2),\\\\\nfrom (1) and (2), we get: $CG=25$, $EG=15$,\\\\\ntherefore $BG=2EG=30$, $AG=2CG=50$,\\\\\nin $\\triangle CBM$ and $\\triangle AGB$,\\\\\n$\\left\\{\\begin{array}{l}\n\\angle BCM=\\angle BAG \\\\\n\\angle CMB=\\angle ABG=90^\\circ\n\\end{array}\\right.$,\\\\\ntherefore $\\triangle CBM\\sim \\triangle AGB$,\\\\\ntherefore $\\frac{BM}{BG}=\\frac{CM}{AB}=\\frac{BC}{AG}$,\\\\\ntherefore $\\frac{BM}{30}=\\frac{CM}{40}=\\frac{55}{50}$,\\\\\ntherefore $BM=33$, $CM=44$,\\\\\nsince $BM\\perp CF$, $BF=AB=40$,\\\\\ntherefore $FM=\\sqrt{BF^{2}-BM^{2}}=\\sqrt{40^{2}-33^{2}}=\\sqrt{511}$,\\\\\ntherefore $EF=CM-CE-FM=44-20-\\sqrt{511}=\\boxed{24-\\sqrt{511}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53080819.png"
  },
  {
    "solution": "\\boxed{\\frac{11}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=6$, $AC=BC=5$. $\\triangle ABC$ is rotated clockwise around point $A$ to obtain $\\triangle ADE$, with the angle of rotation being $\\alpha$ ($0^\\circ<\\alpha<180^\\circ$), where point $B$ corresponds to point $D$, and point $C$ corresponds to point $E$. When the angle between the lines $AB$ and $AD$ equals $\\angle ACB$, what is the length of $BE$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, connect BD,\n\n$\\because$ $AC=BC$\n\n$\\therefore$ $\\angle CAB=\\angle CBA$\n\nBased on the property of rotation,\n\n$\\angle CBA=\\angle ADE$,\n\n$AD=AB$\n\n$\\therefore$ $\\angle ADB=\\angle ABD$\n\n$\\because$ $\\angle ACB=\\angle DAB$\n\n$\\therefore$ $\\angle CAB=\\angle CBA=\\angle ADE=\\angle ADB=\\angle ABD$\n\n$\\therefore$ $D$, $E$, $B$ are collinear\n\n$\\because$ $\\frac{AC}{AD}=\\frac{BC}{AB}$,\n\n$\\angle ACB=\\angle DAB$\n\n$\\therefore$ $\\triangle ACB\\sim \\triangle DAB$\n\n$\\therefore$ $\\frac{AC}{AD}=\\frac{AB}{DB}$, thus\n\n$\\frac{5}{6}=\\frac{6}{DB}$\n\nSolving gives\n\n$DB=\\frac{36}{5}$\n\n$\\therefore$ $BE=DB-DE=\\frac{36}{5}-5=\\boxed{\\frac{11}{5}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51655022.png"
  },
  {
    "solution": "\\boxed{x \\leq -2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the graph, there is a linear function $y_1=x+a$ and $y_2=kx+b$ where $k\\neq 0$, what is the solution set of the inequality $kx+b \\geq x+a$ with respect to $x$?\n",
    "completion": "\\textbf{Solution:} Since $\\because kx+b \\geq x+a$,\\\\\n$\\therefore y_2 > y_1$,\\\\\n$\\therefore$ combining with the graph, it is known that when $x \\leq -2$, $y_2 > y_1$.\\\\\nThus, the answer is: $\\boxed{x \\leq -2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52251344.png"
  },
  {
    "solution": "\\boxed{3\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, parallelogram $ABCD$ has its diagonal $BD$ bisecting $\\angle ABC$. Given $BC = 6$, $\\angle ABC = 45^\\circ$, and a moving point $P$ on diagonal $BD$ and another moving point $Q$ on side $BC$ such that $PQ + PC$ reaches its minimum, what is this minimum value?\n",
    "completion": "\\textbf{Solution:} Let's start. Since quadrilateral ABCD is a parallelogram, \\\\\nit follows that AD $\\parallel$ BC, \\\\\nthus $\\angle ADB = \\angle CBD$, \\\\\nsince BD bisects $\\angle ABC$, \\\\\nit follows that $\\angle ABD = \\angle CBD$, \\\\\ntherefore, $\\angle ABD = \\angle ADB$, \\\\\nthus AB=AD, \\\\\nhence, parallelogram ABCD is a rhombus, \\\\\ntherefore, points A and C are symmetric with respect to BD, \\\\\nDraw AQ $\\perp$ BC at Q, intersecting BD at P, \\\\\nthen the minimum value of PQ+PC is AQ, \\\\\nsince $\\angle ABC = 45^\\circ$, \\\\\nit follows that $\\triangle ABQ$ is an isosceles right triangle, \\\\\nsince AB=BC=6, \\\\\nit follows that AQ=$\\frac{\\sqrt{2}}{2}$AB=$3\\sqrt{2}$, \\\\\nthus, this minimal value is $\\boxed{3\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51798867.png"
  },
  {
    "solution": "\\boxed{16}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: The point $P$ is a point on the line segment $AB$, and the points $M$ and $N$ are the midpoints of the line segments $AP$ and $PB$, respectively. As shown in the figure, if the point $P$ is the midpoint of the line segment $AB$, and $MP=4\\, \\text{cm}$, what is the length of the line segment $AB$ in cm?\n",
    "completion": "\\textbf{Solution:} Given that point $M$ and $N$ are the midpoints of line segments $AP$ and $PB$ respectively,\\\\\n$\\therefore AP=2MP$, $BP=2PN$.\\\\\n$\\because MP=4\\, \\text{cm}$, $\\therefore AP=8\\, \\text{cm}$.\\\\\n$\\because P$ is the midpoint of $AB$, $\\therefore AB=2AP=\\boxed{16}\\, \\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55431962_66.png"
  },
  {
    "solution": "\\boxed{(0,3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, two lines $l_{1}$ and $l_{2}$ passing through point A (1, 0) intersect the y-axis at points B and C, respectively, where point B is above the origin and point C is below the origin. It is known that $AB = \\sqrt{10}$. Find the coordinates of point B.\n",
    "completion": "\\textbf{Solution:} Since the coordinates of point A are $(1,0)$, it follows that $AO= 1$. Also, since $AB= \\sqrt{10}$, it follows that $BO= \\sqrt{AB^{2}-AO^{2}}=3$. Since point B is above the origin, the coordinates of point B are $\\boxed{(0,3})$;",
    "image_path": "dataset/mm_math/images/MM_Math/52690999_57.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown, the moving point P starts from vertex A of the rectangle ABCD, and moves uniformly along the direction A→B→C on the sides AB, BC at a speed of $1\\, \\text{cm/s}$ to point C. The area $S$ ($\\text{cm}^{2}$) of $\\triangle APC$ as a function of the movement time $t$ (s) is illustrated in the figure. What is the length of AB?\n",
    "completion": "\\textbf{Solution:} Given that point P moves from point A to B, the distance S increases with time t. When point P moves from point B to C, S decreases as t increases.\\\\\nTherefore, based on Graph 2, the duration of point P's movement from B to C is 4s,\\\\\nSince the speed of point P is 1cm/s,\\\\\nThus, $BC=1\\times4=4$ (cm),\\\\\nGiven that when point P moves along the line AB reaching point B, the area of $\\triangle APC$ is at its maximum,\\\\\nHence, from Graph 2: the area of $\\triangle APC$ is $6\\text{cm}^{2}$,\\\\\nTherefore, the area of $\\triangle ABC$, ${S}_{\\triangle ABC}=\\frac{1}{2}AB\\cdot BC=6$,\\\\\nThus, $AB=\\boxed{3}$cm.",
    "image_path": "dataset/mm_math/images/MM_Math/52386065.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, point $D$ is on side $AB$, and point $E$ is on side $AC$. The angle $\\angle ADE$ is equal to $\\angle C$, and the area of quadrilateral $DBCE$ is three times the area of $\\triangle ADE$. If $DE = 3$, what is the length of $BC$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle A = \\angle A$ and $\\angle ADE = \\angle C$,\\\\\nit follows that $\\triangle ADE \\sim \\triangle ACB$,\\\\\nthus, $\\frac{S_{\\triangle ADE}}{S_{\\triangle ABC}} = \\left(\\frac{DE}{BC}\\right)^2$.\\\\\nGiven that the area of quadrilateral DBCE is three times the area of $\\triangle ADE$,\\\\\nwe get $\\frac{S_{\\triangle ADE}}{S_{\\triangle ABC}} = \\frac{1}{4}$,\\\\\nhence $\\left(\\frac{DE}{BC}\\right)^2 = \\frac{1}{4}$.\\\\\nSince $DE = 3$,\\\\\nit follows that $BC = \\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52623554.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C = 90^\\circ$, if $\\sin B = \\frac{4}{5}$, what is the value of $\\sin A$?\n",
    "completion": "\\textbf{Solution:} In $\\triangle ABC$, where $\\angle C=90^\\circ$ and $\\sin B=\\frac{AC}{AB}=\\frac{4}{5}$,\\\\\nlet $AC=4a$ and $AB=5a$,\\\\\nthus $BC=\\sqrt{AB^{2}-AC^{2}}=3a$,\\\\\ntherefore $\\sin A=\\frac{BC}{AB}=\\frac{3a}{5a}=\\boxed{\\frac{3}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52601954.png"
  },
  {
    "solution": "\\boxed{\\frac{27}{16}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system $xOy$, A and B are two points on the x-axis, C and D are two points on the y-axis. A segment of a parabola passing through points A, C, and B, denoted as $C_1$, and a segment of a parabola passing through points A, D, and B, denoted as $C_2$, combine to form a closed curve, which we call the \"egg curve\". It is known that the coordinate of point C is (0, ), and point M is the vertex of the parabola $C_2$: $y=mx^2-2mx-3m$ ($m<0$). Is there a point P on the \"egg curve\" in the fourth quadrant such that the area of $\\triangle PBC$ is maximized? If such a point exists, find the maximum area of $\\triangle PBC$; if not, provide a reason.\n",
    "completion": "\\textbf{Solution:} There exists a solution. The reason is as follows:\\\\\n$\\because$ Let the expression of parabola $C_1$ be $y=\\frac{1}{2}(x+1)(x-3)$. Substituting $C(0, -\\frac{3}{2})$ into it, we get $a=\\frac{1}{2}$.\\\\\n$\\therefore$ The expression of $C_1$ is: $y=\\frac{1}{2}x^2-x-\\frac{3}{2}$.\\\\\nLet $P(p, \\frac{1}{2}p^2-p-\\frac{3}{2})$,\\\\\n$\\therefore S_{\\triangle PBC}=S_{\\triangle POC}+S_{\\triangle BOP}-S_{\\triangle BOC}=-\\frac{3}{4}\\left(p-\\frac{3}{2}\\right)^2+\\frac{27}{16}$.\\\\\n$\\because a=-\\frac{3}{4}<0$, $\\therefore$ when $p=\\frac{3}{2}$, the maximum value of $S_{\\triangle PBC}$ is $\\boxed{\\frac{27}{16}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52977501_202.png"
  },
  {
    "solution": "\\boxed{58^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the hexagon $ABCDEF$, each interior angle is equal. If $\\angle 1=58^\\circ$, what is the degree measure of $\\angle 2$?\n",
    "completion": "\\textbf{Solution:} Since all internal angles of the hexagon ABCDEF are equal, \\\\\n$\\therefore \\angle B = \\angle C = \\angle CDE = (6-2) \\times 180^\\circ \\div 6 = 120^\\circ$, \\\\\n$\\therefore \\angle$CDA $= 360^\\circ - \\angle1 - \\angle B - \\angle C = 62^\\circ$, \\\\\n$\\therefore \\angle2 = \\angle CDE - \\angle CDA = \\boxed{58^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52189417.png"
  },
  {
    "solution": "\\boxed{}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in square $ABCD$, point $E$ is on side $BC$, and point $F$ is on side $AD$, with $AF=CE$, and $EF$ intersects the diagonal $BD$ at point $O$. $O$ is the midpoint of $BD$. If $EF\\perp BD$ and the perimeter of square $ABCD$ is 24, upon connecting $BF$, what is the perimeter of $\\triangle ABF$?\n",
    "completion": "\\textbf{Solution:} Since there's insufficient information provided by the question, specific steps and results cannot be provided. Please provide the complete question information for an accurate solution.",
    "image_path": "dataset/mm_math/images/MM_Math/53498505_77.png"
  },
  {
    "solution": "\\boxed{(2,2)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=-\\frac{1}{2}x+3$ intersects the coordinate axes at points A and B, respectively, and intersects the line $y=x$ at point C. A point Q on the line segment OA moves from point O towards point A at a constant speed of 1 unit per second. Let the movement time be $t(s)$. Connect C and Q. Find the coordinates of point C.\n",
    "completion": "\\textbf{Solution:} Since the line $y=-\\frac{1}{2}x+3$ intersects the line $y=x$ at point C,\\\\\nit follows that $\\begin{cases} y=-\\frac{1}{2}x+3 \\\\ y=x \\end{cases}$,\\\\\nsolving these equations yields $\\begin{cases} x=2 \\\\ y=2 \\end{cases}$,\\\\\ntherefore, the coordinates of point C are $\\boxed{(2,2)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51663132_206.png"
  },
  {
    "solution": "\\boxed{90^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points $E$ and $F$ are the midpoints of sides $AB$ and $AC$ of $\\triangle ABC$, respectively. $BD$ bisects $\\angle ABC$ and intersects $EF$ at point $D$. Connect $AD$. Given $AB=6$ and $BC=8$, what is the length of $DF$? What is the measure of $\\angle ADB$ in degrees?\n",
    "completion": "\\textbf{Solution:} Given that points $E$ and $F$ are the midpoints of sides $AB$ and $AC$ of $\\triangle ABC$ respectively, and $BC=8$, $AB=6$, \\\\\nit follows that $EF\\parallel BC$ and $EF=\\frac{1}{2}BC=4$, and $AE=BE=3$, \\\\\ntherefore, $\\angle EDB=\\angle DBC$, \\\\\nsince $BD$ bisects $\\angle ABC$, \\\\\nit follows that $\\angle EBD=\\angle DBC$, \\\\\nhence, $\\angle EBD=\\angle EDB$, \\\\\nthus, $DE=BE=3$, \\\\\ntherefore, $DF=EF-DE=1$, \\\\\nsince $AE=DE=BE=3$, \\\\\nit follows that $\\angle EAD=\\angle EDA$, and since $\\angle EBD=\\angle EDB$, \\\\\ntherefore, $\\angle BAD+\\angle ABD+\\angle ADB=2(\\angle ADE+\\angle BDE)=180^\\circ$, \\\\\nthus, $\\angle ADE+\\angle BDE=90^\\circ=\\angle ADB.$\\\\\nConsequently, the answer is: $DF=\\boxed{1}$, $\\angle ADB=\\boxed{90^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52423088.png"
  },
  {
    "solution": "\\boxed{12\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, within rectangle $ABCD$, point $E$ is on $BC$, and $AE$ bisects $\\angle BAC$, $AE=CE$, $BE=2$. What is the area of rectangle $ABCD$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rectangle,\\\\\n$\\therefore$ $\\angle B=90^\\circ$,\\\\\n$\\therefore$ $\\angle BAC + \\angle BCA = 90^\\circ$,\\\\\nSince AE bisects $\\angle BAC$ and AE=CE,\\\\\n$\\therefore$ $\\angle BAE = \\angle EAC = \\angle ECA$,\\\\\n$\\therefore$ $\\angle BAE + \\angle EAC + \\angle ECA = 90^\\circ$,\\\\\n$\\therefore$ $\\angle BAE = \\angle EAC = \\angle ECA = 30^\\circ$,\\\\\n$\\therefore$ AE=CE=2BE=4, AB=$\\sqrt{4^2 - 2^2} = 2\\sqrt{3}$,\\\\\n$\\therefore$ BC=BE+CE=6,\\\\\n$\\therefore$ The area of rectangle ABCD=AB×BC=$2\\sqrt{3} \\times 6 = \\boxed{12\\sqrt{3}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53359742.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given: As shown in the diagram, in $\\triangle ABC$, $\\angle ACB = 90^\\circ$, and $BC = BD$, $AC = AE$, then what is the degree measure of $\\angle DCE$?\n",
    "completion": "\\textbf{Solution:} Given that AC=AE and BC=BD,\\\\\nlet $\\angle$AEC=$\\angle$ACE=x and $\\angle$BDC=$\\angle$BCD=y,\\\\\nthen $\\angle$A=180$^\\circ$-2x and $\\angle$B=180$^\\circ$-2y,\\\\\nsince $\\angle$ACB+$\\angle$A+$\\angle$B=$180^\\circ$,\\\\\nthus 90$^\\circ$+(180$^\\circ$-2x)+(180$^\\circ$-2y)=180$^\\circ$,\\\\\nhence x+y=135$^\\circ$,\\\\\ntherefore, $\\angle$DCE=180$^\\circ$-($\\angle$AEC+$\\angle$BDC)=180$^\\circ$-(x+y)=\\boxed{45^\\circ}.",
    "image_path": "dataset/mm_math/images/MM_Math/52653337.png"
  },
  {
    "solution": "\\boxed{\\frac{4\\sqrt{3}}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of a semicircle. The bisector of $\\angle ABC$ intersects the chord $AC$ and the semicircle at points $E$ and $D$, respectively. If $BE=2DE$ and $AB=4$, what is the length of $AE$?\n",
    "completion": "\\textbf{Solution:}\n\nSince $\\angle CDB = \\angle CAB$ and $\\angle DCA = \\angle DBA$,\\\\\nit follows that $\\triangle CDE \\sim \\triangle ABE$,\\\\\nhence $\\frac{CD}{AB} = \\frac{DE}{BE} = \\frac{1}{2}$,\\\\\nthus $CD = \\frac{1}{2}AB = 2$.\\\\\nSince $BD$ bisects $\\angle ABC$,\\\\\nit implies that $\\angle CBD = \\angle ABD$,\\\\\nand so $\\angle DAC = \\angle CBD = \\angle ABD = \\angle ACD$,\\\\\ntherefore $AD = CD = 2$.\\\\\nSince $AB$ is the diameter of a semicircle,\\\\\nit results in $\\angle ADB = 90^\\circ$,\\\\\nthus $\\sin\\angle ABD = \\frac{AD}{AB} = \\frac{2}{4} = \\frac{1}{2}$,\\\\\ntherefore $\\angle ABD = 30^\\circ$,\\\\\nhence $\\angle DAE = 30^\\circ$,\\\\\ntherefore $AE = \\frac{AD}{\\cos30^\\circ} = \\boxed{\\frac{4\\sqrt{3}}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52991509.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, AB is the diameter of $\\odot O$, and AC intersects $\\odot O$ at point C. The bisector of $\\angle BAC$ intersects $\\odot O$ at point D, DE $\\perp$ AC at point E. DE is the tangent to $\\odot O$; if the diameter AB $= 10$, and the chord AC $= 6$, what is the length of DE?\n",
    "completion": "\\textbf{Solution:} Draw OF$\\perp$AC, with F being the foot of the perpendicular.\\\\\n$\\therefore AF=\\frac{1}{2}AC=3$,\\\\\nIn $\\triangle AFO$, $AF^{2}+OF^{2}=AO^{2}$, $AO=\\frac{1}{2}AB=5$,\\\\\n$\\therefore 3^{2}+OF^{2}=5^{2}$,\\\\\n$\\therefore OF=4$,\\\\\n$\\because \\angle AED=\\angle ODE=\\angle OFE=90^\\circ$,\\\\\n$\\therefore$ Quadrilateral ODEF is a rectangle,\\\\\n$\\therefore DE=OF=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52720273_184.png"
  },
  {
    "solution": "\\boxed{24}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $O$ is the midpoint of line segment $AB$, with $OB=14\\,cm$. Point $P$ divides line segment $AB$ into two parts, with $AP:PB=5:2$. Point $M$ is on line segment $AB$. If the distance from point $M$ to point $P$ is $4\\,cm$, what is the length of line segment $AM$ in $cm$?\n",
    "completion": "Solution: As shown in Figure 1, when point M is to the left of point P,\\\\\n$AM=AB-(PM+BP)=28-(4+8)=\\boxed{16}\\,(\\text{cm})$,\\\\\nAs shown in Figure 2, when point M is to the right of point P,\\\\\n$AM=AB-BM=AB-(BP-PM)=28-(8-4)=\\boxed{24}\\,(\\text{cm})$.\\\\\nIn summary, $AM=16\\,\\text{cm}$ or $24\\,\\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/42460301_218.png"
  },
  {
    "solution": "\\boxed{\\frac{4}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a rectangle I, a small square, and an L-shaped figure composed of four squares of the same size are placed inside rectangle ABCD. One vertex of rectangle I is located at the vertex E of a square within the L-shaped figure, with its other vertices on the sides of rectangle ABCD. The squares within the L-shape have three vertices precisely on the sides of rectangle ABCD, with another vertex coinciding with a vertex of the small square. If rectangle I is similar to rectangle ABCD, what is the value of \\(AB:BC\\)?\n",
    "completion": "\\textbf{Solution:} Let the side length of the four identical squares be $a$, and $DF=x$, that is, $GH=IJ=KJ=a$, $IH=EG=2a$,\n\nIn $\\triangle IJH$, we have $JH= \\sqrt{IJ^2+IH^2}=\\sqrt{5}a$,\n\nSince quadrilateral ABCD and quadrilateral MEFD are rectangles, and quadrilateral NBLK is a rectangle,\n\nThe 4 identical small square templates are placed as shown in the figure,\n\nTherefore, $\\angle B=\\angle C=\\angle KLJ=\\angle EFG=\\angle JIH=90^\\circ$,\n\nTherefore, $\\angle LKJ+\\angle LJK=\\angle IJH+\\angle IHJ=\\angle GHC+\\angle HGC=\\angle EGF+\\angle GEF=90^\\circ$,\n\n$\\angle KJL+\\angle IJH=\\angle IHJ+\\angle GHC=\\angle EGF+\\angle HGC=90^\\circ$,\n\nTherefore, $\\angle KJL=\\angle IHJ=\\angle HGC=\\angle GEF$,\n\nTherefore, $\\triangle KJL\\cong \\triangle HGC$, $\\triangle JHI\\sim \\triangle HGC\\sim \\triangle GEF$,\n\nTherefore, $BL=KL=HC$, $LJ=GC$,\n\n$\\frac{IJ}{FG}=\\frac{IH}{EF}=\\frac{JH}{EG}$,\n\n$\\frac{IJ}{HC}=\\frac{IH}{GC}=\\frac{JH}{GH}$, that is,\n\n$\\frac{a}{FG}=\\frac{2a}{EF}=\\frac{\\sqrt{5}a}{2a}$,\n\n$\\frac{a}{HC}=\\frac{2a}{GC}=\\frac{\\sqrt{5}a}{a}$,\n\nTherefore, $FG= \\frac{2\\sqrt{5}a}{5}$, $EF= \\frac{4\\sqrt{5}a}{5}$, $BL=KL=HC= \\frac{\\sqrt{5}a}{5}$, $LJ=GC= \\frac{2\\sqrt{5}a}{5}$,\n\nTherefore, $CD=DF+FG+GC=x+ \\frac{2\\sqrt{5}a}{5}+\\frac{2\\sqrt{5}a}{5} = \\frac{4\\sqrt{5}a}{5}+x$,\n\n$BC=BL+LJ+JH+HC= \\frac{\\sqrt{5}a}{5} + \\frac{2\\sqrt{5}a}{5}+\\sqrt{5}a+\\frac{\\sqrt{5}a}{5} = \\frac{9\\sqrt{5}a}{5}$,\n\nWhen rectangle MEFD $\\sim$ rectangle ABCD,\n\n$\\frac{CD}{DF}=\\frac{BC}{EF}$, that is,\n\n$\\frac{\\frac{4\\sqrt{5}a}{5}+x}{x}=\\frac{\\frac{9\\sqrt{5}a}{5}}{\\frac{4\\sqrt{5}a}{5}}$,\n\nSolving gives: $x= \\frac{16\\sqrt{5}a}{25}$,\n\nThe ratio of AB:BC is\n\n$\\frac{\\frac{16\\sqrt{5}a}{25}+\\frac{4\\sqrt{5}a}{5}}{\\frac{9\\sqrt{5}a}{5}}=\\boxed{\\frac{4}{5}}$;\n\nWhen rectangle MDFE $\\sim$ rectangle ABCD,\n\n$\\frac{CD}{EF}=\\frac{BC}{DF}$, that is,\n\n$\\frac{\\frac{4\\sqrt{5}a}{5}+x}{\\frac{4\\sqrt{5}a}{5}}=\\frac{\\frac{9\\sqrt{5}a}{5}}{x}$,\n\nSolving gives: $x= \\frac{10\\sqrt{2}-2\\sqrt{5}}{5}a$ (negative values discarded),\n\nThe ratio of AB:BC is\n\n$\\frac{\\frac{10\\sqrt{2}-2\\sqrt{5}}{5}a+\\frac{4\\sqrt{5}a}{5}}{\\frac{9\\sqrt{5}a}{5}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51458355.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the vertices B and E of squares ABCO and CDEF, respectively, lie on the hyperbola $y= \\frac{4}{x}$ ($x>0$). When OB, OE, and BE are connected, what is the value of $S_{\\triangle OBE}$?\n",
    "completion": "\\textbf{Solution:} Connect CE. \\\\\nSince quadrilateral ABCO and quadrilateral DEFC are both squares, \\\\\nthus, $\\angle$ECF=$\\angle$BOC=$45^\\circ$, \\\\\nhence, CE$\\parallel$OB, \\\\\ntherefore, $S_{\\triangle OBE}$=$S_{\\triangle OBC}$, \\\\\nsince $S_{\\triangle OBC}$=$\\frac{1}{2} \\times 4=2$, \\\\\nthus, $S_{\\triangle OBE}$= $\\frac{1}{2} \\times 2 \\times 2=\\boxed{2}$,",
    "image_path": "dataset/mm_math/images/MM_Math/52303921.png"
  },
  {
    "solution": "\\boxed{-x^2+5x+6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the quadratic function $y=-x^2+bx+c$ intersects the $x$ axis at points $A$ and $B$, and intersects the $y$ axis at point $C(0, 6)$, and also passes through the point $(1, 10)$. Determine: What is the expression of this parabola?\n",
    "completion": "Solution: Since the graph of the quadratic function $y=-x^2+bx+c$ passes through the points $C(0, 6)$ and $(1, 10)$,\n\\[\n\\begin{cases}\nc=6\\\\\n-1+b+c=10\n\\end{cases}\n\\]\nSolving this system of equations gives\n\\[\n\\begin{cases}\nb=5\\\\\nc=6\n\\end{cases}\n\\]\nTherefore, the expression for the quadratic function is $y=-x^2+5x+6=\\boxed{-x^2+5x+6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53065902_77.png"
  },
  {
    "solution": "\\boxed{\\left(\\frac{3}{2},-\\frac{5}{4}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system, the parabola passes through points A $(-1, 0)$, B $(4, 0)$, and C $(0, -2)$. Is there a point P on the axis of symmetry of this parabola such that the perimeter of $\\triangle PAC$ is minimized? If such a point exists, find the coordinates of point P; if not, explain why.\n",
    "completion": "\\textbf{Solution}: Exist, the reason is as follows:\\\\\nConnect PB\\\\\nDue to the symmetry of the parabola: $PA=PB$\\\\\n$\\therefore$ the perimeter of $\\triangle PAC$ $PA+PC+AC=PB+PC+AC$,\\\\\n$\\therefore$ when points B, P, C are collinear, $PB+PC$ is minimized,\\\\\nthat is, when points B, P, C are collinear, the perimeter of $\\triangle PAC$ is minimized,\\\\\nSuppose the equation of line BC is $y=kx+m$,\\\\\nSubstitute points B$(4,0)$ and C$(0,-2)$ into it,\\\\\nwe get $\\begin{cases}0=4k+m\\\\ -2=m\\end{cases}$, and solve to get: $\\begin{cases}k=\\frac{1}{2}\\\\ m=-2\\end{cases}$,\\\\\nthus, the equation of line BC is $y=\\frac{1}{2}x-2$.\\\\\nLet $x=\\frac{3}{2}$, then $y=\\frac{1}{2}\\times \\frac{3}{2}-2=-\\frac{5}{4}$,\\\\\nthus, the coordinates of point P are $\\left(\\frac{3}{2},-\\frac{5}{4}\\right)$.\\\\\n$\\therefore$ There exists a point P on the axis of symmetry of this parabola that minimizes the perimeter of $\\triangle PAC$, at which the coordinates of point P are $\\boxed{\\left(\\frac{3}{2},-\\frac{5}{4}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52766565_146.png"
  },
  {
    "solution": "\\boxed{110^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: The protractor with center $O$ and the right-angled triangle $ABC$ are placed as shown in the figure, with the 0-degree line of the protractor coinciding with the hypotenuse $AB$. Point $D$ is a point on the hypotenuse $AB$. A ray $CD$ is drawn intersecting the arc $AB$ at point $E$. If the reading corresponding to point $E$ is $50^\\circ$, what is the measure of $\\angle BDE$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect $OE$,\\\\\nthe reading corresponding to point $E$ is $50^\\circ$,\\\\\n$\\therefore \\angle AOE=50^\\circ$,\\\\\n$\\because AB$ is the diameter, $\\angle ACB=90^\\circ$,\\\\\n$\\therefore$ point $C$ lies on $\\odot O$,\\\\\n$\\therefore \\angle ACE=\\frac{1}{2}\\angle AOE=\\frac{1}{2}\\times 50^\\circ=25^\\circ$,\\\\\n$\\therefore \\angle BCE=90^\\circ-25^\\circ=65^\\circ$,\\\\\n$\\because \\angle BDE$ is the exterior angle of $\\triangle BDC$,\\\\\n$\\therefore \\angle BDE=\\angle BCE+\\angle DBC=65^\\circ+45^\\circ=\\boxed{110^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51402150.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ and $\\triangle ADE$ are equilateral triangles. Connect $BE$, $BD$, $CD$, $EC$. As shown in Figure 1, if $\\angle ADC=30^\\circ$ and $AE=7$, $EC=9$, what is the length of $EB$?\n",
    "completion": "\\textbf{Solution:} Given that, $\\triangle ABC$ and $\\triangle ADE$ are equilateral triangles,\\\\\n$\\therefore AC=AB, AE=AD$, and $\\angle ADE=60^\\circ$, $\\angle EAD=\\angle CAB=60^\\circ$\\\\\n$\\therefore \\angle EDC=90^\\circ$\\\\\nSince $\\angle ADC=30^\\circ$, $AE=7$, and $EC=9$,\\\\\n$\\therefore ED=AE=7$\\\\\nIn $\\triangle CDE$, $CD=\\sqrt{EC^{2}-ED^{2}}=\\sqrt{9^{2}-7^{2}}=4\\sqrt{2}$,\\\\\nSince $\\angle EAD=\\angle CAB=60^\\circ$\\\\\n$\\therefore \\angle EAD+\\angle DAB=\\angle DAB+\\angle AC$\\\\\nTherefore, $\\angle CAD=\\angle BAE$\\\\\n$\\therefore \\triangle EAB\\cong \\triangle DAC$\\\\\n$\\therefore EB=CD=4\\sqrt{2}$, hence the length of $EB$ is $\\boxed{4\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51918275_105.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In parallelogram $ABCD$, an arc is drawn with A as the center and the length of $AB$ as the radius, intersecting $AD$ at F. Then, with B and F as centers, arcs longer than $\\frac{1}{2}BF$ are drawn, intersecting at point G. If $BF=6$ and $AB=5$, what is the length of $AE$?\n",
    "completion": "\\textbf{Solution:} Connect $EF$, as shown in the diagram:\\\\\nBased on the problem statement, we have: $AE$ bisects $\\angle DAB$, $AF=AB=5$,\\\\\n$\\therefore \\angle AOF=90^\\circ$, $OB=OF=3$, $\\angle BAE=\\angle FAE$,\\\\\n$\\therefore OA=\\sqrt{AF^{2}-OF^{2}}=4$,\\\\\n$\\because$ Quadrilateral $ABCD$ is a parallelogram,\\\\\n$\\therefore AD\\parallel BC$,\\\\\n$\\therefore \\angle FAE=\\angle AEB$,\\\\\n$\\therefore \\angle BAE=\\angle AEB$,\\\\\n$\\therefore BE=AB=AF$,\\\\\n$\\therefore$ Quadrilateral $ABEF$ is a parallelogram,\\\\\n$\\therefore OA=OE=\\frac{1}{2}AE$,\\\\\n$\\therefore AE=2OA=\\boxed{8}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52424336.png"
  },
  {
    "solution": "\\boxed{\\sqrt{7}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $AD=2\\sqrt{3}$ and $AB=5$. Let $M$ be a point on $CD$. $\\triangle ADM$ is folded along the straight line $AM$ to obtain $\\triangle ANM$. If $AN$ bisects $\\angle MAB$, what is the length of $CN$?\n",
    "completion": "\\textbf{Solution}: As shown in the diagram, draw a perpendicular line from point N to CD, intersecting $CD$ at point E. \\\\\nFrom the fold, we know: $DM=MN$, \\\\\n$DA=AN=2\\sqrt{3}$, \\\\\n$\\angle DAM=\\angle NAM$ \\\\\nBecause AN bisects $\\angle MAB$, \\\\\nTherefore, $\\angle NAM=\\angle BAN$ \\\\\nTherefore, $\\angle DAM=\\angle NAM=\\angle BAN$ \\\\\nBecause $\\angle DAB=90^\\circ$, \\\\\nTherefore, $\\angle DAM=30^\\circ$, \\\\\n$\\angle DAN=60^\\circ$ \\\\\nTherefore, $DM=AD\\cdot\\tan30^\\circ=2\\sqrt{3}\\times \\frac{\\sqrt{3}}{3}=2$ \\\\\nTherefore, $MN=DM=2$ \\\\\nBecause $\\angle DAN+\\angle DMN=180^\\circ$, \\\\\n$\\angle CMN+\\angle DMN=180^\\circ$ \\\\\nTherefore, $\\angle CMN=\\angle DAN=60^\\circ$ \\\\\nTherefore, $NE=MN\\cdot\\sin60^\\circ=2\\times \\frac{\\sqrt{3}}{2}=\\sqrt{3}$, \\\\\n$ME=1$ \\\\\nTherefore, $CE=5-2-1=2$ \\\\\nTherefore, in right triangle $\\triangle ECN$, by the Pythagorean theorem, we get: \\\\\n$CN=\\sqrt{{(\\sqrt{3})}^{2}+{2}^{2}}=\\sqrt{3+4}=\\boxed{\\sqrt{7}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51654777.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, line $l_{1}: y=kx (k\\neq 0)$ and line $l_{2}: y=ax-2 (a\\neq 0)$ intersect at point B. Line $l_{2}$ intersects the x-axis at point A ($-4$, $0$). If the x-coordinate of point B is $-2$, what is the value of $k$?\n",
    "completion": "\\textbf{Solution:} Since line $l_{2}\\colon y=ax-2\\left(a\\ne 0\\right)$ intersects the x-axis at point A ($-4$, $0$),\\\\\n$\\therefore -4a-2=0$,\\\\\nSolving for $a$, we get: $a=-\\frac{1}{2}$,\\\\\n$\\therefore$ the equation for line $l_{2}$ is $y=-\\frac{1}{2}x-2$,\\\\\nSince line $l_{1}$ intersects line $l_{2}$ at point B, and the x-coordinate of point B is $-2$,\\\\\n$\\therefore$ for line $l_{2}\\colon y=-\\frac{1}{2}x-2$,\\\\\nwhen $x=-2$, $y=1-2=-1$,\\\\\n$\\therefore$ the coordinates of point B are ($-2$, $-1$),\\\\\nSubstituting point ($-2$, $-1$) into $y=kx$,\\\\\nwe get: $-2k=-1$,\\\\\nSolving for $k$, we find: $k=\\boxed{\\frac{1}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52380814.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in square $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$. $E$ is the midpoint of $BC$, and $DE$ intersects $AC$ at $F$. If $DE=12$, what is the length of $EF$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral $ABCD$ is a square, \\\\\nhence $AD=BC$, $AD\\parallel BC$, \\\\\nsince $E$ is the midpoint of $BC$, \\\\\nhence $CE=\\frac{1}{2}BC=\\frac{1}{2}AD$, \\\\\nsince $AD\\parallel BC$, \\\\\nhence $\\triangle ADF\\sim \\triangle CEF$, \\\\\nhence $\\frac{EF}{DF}=\\frac{CE}{AD}=\\frac{1}{2}$, \\\\\nsince $DE=12$, \\\\\nhence $EF=\\frac{1}{3}DE=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53106789.png"
  },
  {
    "solution": "\\boxed{-3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y=mx+2$ passes through point $A(2, 4)$ and $B(n, -1)$. Find the values of $m$ and $n$.\n",
    "completion": "\\textbf{Solution:} Since the graph of the linear function $y=mx+2$ passes through point $A(2, 4)$,\\\\\nthus $2m+2=4$, we find $m=\\boxed{1}$.\\\\\nTherefore, the expression for the linear function is $y=x+2$.\\\\\nSince the graph of the linear function $y=x+2$ passes through point $B(n, -1)$,\\\\\nthus $n+2=-1$, we find $n=\\boxed{-3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50590307_152.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{10}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $A$, $B$, $C$, and $D$ are four points on the circle $\\odot O$. $AB=AC$, $AD$ intersects $BC$ at point $E$, $AE=4$, and $ED=6$. What is the length of $AC$?\n",
    "completion": "\\[ \\textbf{Solution:} \\text{Since } AB=AC, \\]\n\\[ \\therefore \\angle ABE=\\angle ACE, \\]\n\\[ \\text{Since } \\angle ACE=\\angle ADB, \\]\n\\[ \\therefore \\angle ABE=\\angle ADB, \\]\n\\[ \\text{Since } \\angle BAE=\\angle DAB, \\]\n\\[ \\therefore \\triangle ABE\\sim \\triangle ADB, \\]\n\\[ \\therefore \\frac{AB}{AD}=\\frac{AE}{AB}, \\]\n\\[ \\text{Thus, } AB^2=AD\\cdot AE, \\]\n\\[ \\text{Since } AE=4, ED=6, \\]\n\\[ \\therefore AD=10, \\]\n\\[ \\therefore AB=\\sqrt{AD\\cdot AE}=\\sqrt{10\\times 4}=2\\sqrt{10}, \\]\n\\[ \\therefore AC=AB=2\\sqrt{10}, \\]\n\\[ \\text{Hence, the answer is: } AC=\\boxed{2\\sqrt{10}}. \\]",
    "image_path": "dataset/mm_math/images/MM_Math/51368291.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, triangle $ABC$ is inscribed in circle $O$, with $AB$ as the diameter and point $C$ located above $AB$, such that $\\angle CAB > \\angle B$. Connect $OC$. $CG$ is the altitude from $C$ to side $AB$. A line through point $O$ and perpendicular to $OC$ intersects the extension of $CG$ at point $D$ and intersects $BC$ at point $E$. It is given that $\\angle ACG = \\angle BCO$. When $OE = OD$, find the value of $\\frac{OG}{OC}$.\n",
    "completion": "\\textbf{Solution:} Since $DE \\perp OC$, it follows that $\\angle BCO + \\angle CEO = 90^\\circ$. Since $\\angle ACG = \\angle BCO$, we have $\\angle CEO = \\angle A$.\\\\\nSince $\\angle A + \\angle ACG = 90^\\circ$ and also $\\angle BCG + \\angle ACG = 90^\\circ$, it follows that $\\angle BCG = \\angle A$, hence $\\angle BCG = \\angle CEO$,\\\\\ntherefore $DC = DE$. Since $OE = OD$ and $CO \\perp DE$, we have $DC = CE$, thus $DC = DE = CE$,\\\\\ntherefore, $\\triangle CDE$ is an equilateral triangle, hence $\\angle DCE = 60^\\circ$, and thus $\\angle OCG = 30^\\circ$, therefore $\\frac{OG}{OC} = \\sin 30^\\circ = \\boxed{\\frac{1}{2}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/55497848_71.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the isosceles triangle $ABC$, $\\angle C=90^\\circ$, $AC=BC=6$, and point $D$ is a point on $AC$, $\\tan\\angle DBA=\\frac{1}{\\sqrt{5}}$. What is the length of $AD$?\n",
    "completion": "\\textbf{Solution:} Draw line $DE \\perp AB$ at point E, as shown in the diagram,\\\\\n$\\therefore$ $\\angle AED=\\angle DEB=90^\\circ$,\\\\\n$\\because$ $\\angle C=90^\\circ$, $AC=BC=6$,\\\\\n$\\therefore$ $AB=\\sqrt{AC^2+BC^2}=\\sqrt{2}AC=6\\sqrt{2}$, $\\angle A=45^\\circ$,\\\\\n$\\therefore$ $\\triangle ADE$ is an isosceles right triangle,\\\\\n$\\therefore$ $AE=DE$,\\\\\nIn $\\triangle ADE$, let $AE=x$, then $DE=x$,\\\\\n$\\therefore$ $AD=\\sqrt{AE^2+DE^2}=\\sqrt{2}AE=\\sqrt{2}x$\\\\\nIn $\\triangle BED$, $\\tan\\angle DBE=\\frac{DE}{BE}=\\frac{1}{5}$,\\\\\n$\\therefore$ $BE=5x$,\\\\\n$\\therefore$ $AB=AE+BE=x+5x=6\\sqrt{2}$,\\\\\n$\\therefore$ $x=\\sqrt{2}$\\\\\n$\\therefore$ $AD=\\sqrt{2}x=\\boxed{2}$",
    "image_path": "dataset/mm_math/images/MM_Math/52977737.png"
  },
  {
    "solution": "\\boxed{18^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\angle AOB$ is a straight angle, $\\angle BOC=36^\\circ$, $OD$ bisects $\\angle AOC$, $\\angle DOE=90^\\circ$. Find the measure of $\\angle AOE$.\n",
    "completion": "\\textbf{Solution:} Since $\\angle AOB$ is a straight angle and $\\angle BOC=36^\\circ$,\\\\\nit follows that $\\angle AOC=180^\\circ-\\angle BOC=180^\\circ-36^\\circ=144^\\circ$.\\\\\nSince $OD$ bisects $\\angle AOC$,\\\\\nthen $\\angle AOD= \\frac{1}{2}\\angle AOC=\\frac{1}{2}\\times144^\\circ=72^\\circ$.\\\\\nSince $\\angle DOE=90^\\circ$,\\\\\nthus $\\angle AOE=\\angle DOE-\\angle AOD=90^\\circ-72^\\circ=\\boxed{18^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55431957.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the hyperbola $y=\\frac{k}{x} \\ (k \\ne 0)$ and the line $y=mx+n$ intersect at points $A$ and $B$, with the coordinates of point $B$ being $(2, -3)$. The line $AC$ is perpendicular to the $y$-axis at point $C$, with $AC = \\frac{3}{2}$. If the area of $\\triangle AOB$ is $S_{1}$ and the area of $\\triangle ACB$ is $S_{2}$, what is the value of $\\frac{S_{1}}{S_{2}}$?\n",
    "completion": "\\textbf{Solution:} In the equation $y=-2x+1$, let $y=0$, then we solve for $x=\\frac{1}{2}$,\\\\\n$\\therefore$ the coordinates of the intersection point of line $AB$ and the x-axis are $\\left(\\frac{1}{2}, 0\\right)$,\\\\\n$\\therefore$ ${S}_{1}=\\frac{1}{2}\\times \\frac{1}{2}\\times 4+\\frac{1}{2}\\times \\frac{1}{2}\\times 3=\\frac{7}{4}$,\\\\\n$\\because$ ${S}_{2}=\\frac{1}{2}\\times \\frac{3}{2}\\times (4+3)=\\frac{21}{4}$\\\\\n$\\therefore$ $\\frac{{S}_{1}}{{S}_{2}}=\\boxed{\\frac{1}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52622054_225.png"
  },
  {
    "solution": "\\boxed{3\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, let $\\odot O$ have a diameter $AB=6$, with $CD$ being a chord of $\\odot O$ and $CD\\perp AB$ at the perpendicular foot $P$. Connect $AC$ and $AD$. If $\\triangle ACD$ is an equilateral triangle, what is the length of $CD$?\n",
    "completion": "\\textbf{Solution:} Connect $OC$, as shown in the diagram:\\\\\n$\\because$ $CD\\perp AB$,\\\\\n$\\therefore$ $PC=PD=\\frac{1}{2}CD$, $\\angle OPC=90^\\circ$,\\\\\n$\\because$ the diameter $AB$ of $\\odot O$ is $6$,\\\\\n$\\therefore$ $OC=3$,\\\\\n$\\because$ $\\triangle ACD$ is an equilateral triangle,\\\\\n$\\therefore$ $\\angle COP=60^\\circ$, $\\angle OCP=30^\\circ$,\\\\\n$\\therefore$ $OP=\\frac{3}{2}$,\\\\\nBy the Pythagorean theorem we get:\\\\\n$PC=\\sqrt{OC^{2}-OP^{2}}=\\sqrt{3^{2}-\\left(\\frac{3}{2}\\right)^{2}}=\\sqrt{\\frac{27}{4}}=\\frac{3\\sqrt{3}}{2}$,\\\\\n$\\therefore$ $CD=2PC=\\boxed{3\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53066061.png"
  },
  {
    "solution": "\\boxed{\\text{Yes}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a small ball M is thrown from point O on the slope $OA$. The trajectory of the ball is part of a parabola. A Cartesian coordinate system is established as shown in the figure, and the slope can be described by the linear function $y=\\frac{1}{2}x$. If the highest point that the ball reaches has the coordinates $(4, 8)$, answer the following question: There is a tree at point B on slope $OA$, where the x-coordinate of B is 2 and the height of the tree is 3.5. Can ball M fly over this tree? Explain your reasoning through calculation.\n",
    "completion": "\\textbf{Solution:} By substituting $x_B=2$ into $y=\\frac{1}{2}x$, we get: $y_B=\\frac{1}{2}\\times 2=1$,\\\\\n$\\therefore$ B(2, 1).\\\\\n$\\because$ The height of the tree is 3.5,\\\\\n$\\therefore$ the coordinates of the top of the tree are $(2, 4.5)$.\\\\\nBy substituting $x=2$ into $y=-\\frac{1}{2}x^2+4x$, we get: $y=-\\frac{1}{2}\\times 2^2+4\\times 2=6$,\\\\\n$\\therefore$ at this moment $M(2, 6)$,\\\\\n$\\therefore$ $y_M>y_{\\text{top of the tree}}$,\\\\\n$\\therefore$ the ball M can fly over the tree, hence the answer is $\\boxed{\\text{Yes}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52934216_235.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the quadrilateral $ABCD$ is inscribed in circle $\\odot O$, with $\\angle C=2\\angle A$. $DE$ is the diameter of circle $\\odot O$, and $BD$ is connected. If the diameter of $\\odot O$ is 4, what is the length of $BD$?\n",
    "completion": "\\textbf{Solution:} Connect $BE$,\\\\\nsince by the Inscribed Angle Theorem we have $\\angle BED=\\angle A=60^\\circ$,\\\\\nand since $DE$ is the diameter of circle $\\odot O$,\\\\\nit follows that $\\angle EBD=90^\\circ$,\\\\\nthus $\\angle BDE=90^\\circ-60^\\circ=30^\\circ$,\\\\\ntherefore $BE=\\frac{1}{2}DE=\\frac{1}{2}\\times 4=2$,\\\\\ntherefore $BD=\\sqrt{DE^{2}-BE^{2}}=2\\sqrt{3}$.\\\\\nHence, the length of $BD$ is $\\boxed{2\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53019741_127.png"
  },
  {
    "solution": "\\boxed{(1, 0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system, the line $y=\\frac{1}{2}x+2$ intersects with the x-axis at point A, and intersects with the y-axis at point C. The axis of symmetry of the parabola $y=ax^{2}+bx+c$ is $x=-\\frac{3}{2}$, and it passes through points A and C, with another intersection point on the x-axis being point B. What are the coordinates of points A and B?\n",
    "completion": "Solution: From the equation of the line, we get $y=\\frac{1}{2}x+2$. When $x=0$, $y=2$; when $y=0$, $x=-4$.\\\\\n$\\therefore$ $C(0, 2)$ and $A(-4, 0)$. Due to the symmetry of the parabola, point A and point B are symmetric about the line $x=-1.5$,\\\\\n$\\therefore$ $B\\left(1, 0\\right)$.\\\\\nTherefore, the coordinates of points A and B are respectively $A=\\boxed{(-4, 0)}$, $B=\\boxed{(1, 0)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51457966_268.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the radius of the circle $\\odot O$ is 2, and the chord $AB=2\\sqrt{3}$. Point C is a moving point on chord AB, and the length of OC is an integer. What is the length of OC?\n",
    "completion": "\\textbf{Solution:} Draw $OH \\perp AB$ at $H$, as shown in the figure, \\\\\nsince $OH \\perp AB$, \\\\\ntherefore $AH=BH$, \\\\\nthus $AH=BH=\\frac{1}{2}AB=\\frac{1}{2} \\times 2\\sqrt{3} = \\sqrt{3}$, \\\\\nin $\\triangle BOH$, where $OB=2$, and $BH=\\sqrt{3}$, \\\\\ntherefore $OH=\\sqrt{OB^{2}-BH^{2}}=\\sqrt{2^{2}-\\left(\\sqrt{3}\\right)^{2}}=1$, \\\\\nsince point $C$ is a moving point on chord $AB$, \\\\\ntherefore $1 \\le OC \\le 2$, \\\\\nsince the length of $OC$ is an integer, \\\\\nthus $OC=\\boxed{1}$ or $\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52623271.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graphs of the functions $y=2x$ and $y=nx+6$ intersect at point $A(m, 4)$. Then, how many integer solutions does the system of inequalities $0<nx+6<2x$ have?\n",
    "completion": "\\textbf{Solution:} Since point A($m$, 4) lies on the graph of the function $y=2x$, \\\\\nit follows that $4=2m$, solving this gives $m=2$, \\\\\ntherefore, point A is (2, 4), \\\\\nsubstituting point A(2, 4) into $y=nx+6$, we get: $4=2n+6$, solving for $n$ gives: $n=-1$, \\\\\nthus, the equation is: $y=-x+6$, \\\\\nsubstituting $y=0$ into $y=-x+6$, we find: $x=6$, \\\\\ntherefore, point B is (6, 0), \\\\\nfrom the graph of the function, it is known that when $2<x<6$, both functions $y=2x$ and $y=nx+6$ are above the $x$-axis, and the graph of $y=2x$ is above that of $y=nx+6$, \\\\\ntherefore, the solution set of the system of inequalities $0<nx+6<2x$ is: $2<x<6$, \\\\\ntherefore, there are 3 integer solutions: $3, 4, 5$, for a total of $\\boxed{3}$ solutions.",
    "image_path": "dataset/mm_math/images/MM_Math/50561168.png"
  },
  {
    "solution": "\\boxed{100^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, $AB \\perp BC$ at point $B$, $AE \\perp DE$ at point $E$, $AB = AE$, $\\angle ACB = \\angle ADE$, $\\angle ACD = 65^\\circ$, $\\angle BAD = 75^\\circ$. What is the measure of $\\angle BAE$?\n",
    "completion": "\\textbf{Solution:} Given that $AB\\perp BC$ and $AE\\perp DE$,\\\\\nthus $\\angle B=\\angle E=90^\\circ$,\\\\\nin $\\triangle ABC$ and $\\triangle AED$,\\\\\n\\[\n\\left\\{\\begin{array}{l}\n\\angle B=\\angle E\\\\\n\\angle ACB=\\angle ADE\\\\\nAB=AE\n\\end{array}\\right.,\n\\]\ntherefore, $\\triangle ABC \\cong \\triangle AED$ (by AAS),\\\\\nhence $\\angle BAC=\\angle EAD$, $AC=AD$\\\\\nGiven $\\angle ACD=65^\\circ$,\\\\\nit follows that $\\angle ACD=\\angle ADC=65^\\circ$\\\\\nHence, $\\angle CAD=180^\\circ-\\angle ACD-\\angle ADC=50^\\circ$\\\\\nGiven $\\angle BAD=75^\\circ$,\\\\\nit implies that $\\angle BAC=\\angle EAD=\\angle BAD-\\angle CAD=75^\\circ-50^\\circ=25^\\circ$,\\\\\nTherefore, $\\angle BAE=\\angle BAD+\\angle DAE=\\angle BAD+\\angle BAC=75^\\circ+25^\\circ=\\boxed{100^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52978712.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $DE \\parallel BC$, $AD=2$, $BD=3$, $AC=10$, what is the length of $AE$?\n",
    "completion": "\\textbf{Solution:} Since $DE\\parallel BC$, \\\\\nit follows that $\\frac{AD}{BD}=\\frac{AE}{EC}=\\frac{2}{3}$, \\\\\nthus, $\\frac{AE}{AC}=\\frac{2}{5}$, \\\\\ntherefore, $AE=\\frac{2}{5}AC=\\frac{2}{5}\\times 10=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53066266.png"
  },
  {
    "solution": "\\boxed{-\\frac{45}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points A and B are the intersections of the inverse proportion function $y_{1} = \\frac{10}{x}$ and the linear function $y=2x$. Point C is on the inverse proportion function $y_{2} = \\frac{k}{x}$. Connect OC, draw $AD \\perp x$-axis through point A to intersect OC at point D, and connect BD. If $AD=BD$ and $OC=3OD$, then what is the value of $k$?\n",
    "completion": "\\textbf{Solution:} Solve the system of equations $\\begin{cases}y=2x\\\\ y=\\frac{10}{x}\\end{cases}$,\\\\\nwe get $\\begin{cases}x_{1}=-\\sqrt{5}\\\\ y_{1}=-2\\sqrt{5}\\end{cases}$, $\\begin{cases}x_{2}=\\sqrt{5}\\\\ y_{2}=2\\sqrt{5}\\end{cases}$,\\\\\n$\\therefore$ the coordinates of point A are ($-\\sqrt{5}$, $-2\\sqrt{5}$), and the coordinates of point B are ($\\sqrt{5}$, $2\\sqrt{5}$),\\\\\n$\\because$ A and B are symmetric about the origin,\\\\\n$\\therefore$ O is the midpoint of AB,\\\\\nalso, $\\because$ AD = BD,\\\\\n$\\therefore$ point D lies on the perpendicular bisector of segment AB,\\\\\n$\\therefore$ CO$\\perp$AB,\\\\\nalso, $\\because$ AH$\\perp$x-axis,\\\\\n$\\therefore$ $\\angle$AOH + $\\angle$OAH = $\\angle$AOH + $\\angle$COH = $90^\\circ$,\\\\\n$\\therefore$ $\\angle$OAH = $\\angle$COH,\\\\\ndraw CE$\\perp$x-axis at point E,\\\\\n$\\because$ OC = 3OD, and the x-coordinate of point D is $-\\sqrt{5}$,\\\\\n$\\therefore$ the x-coordinate of point C is $-3\\sqrt{5}$,\\\\\n$\\because  \\tan \\angle$OAH = $\\tan \\angle$COH = $\\frac{OH}{AH}$ = $\\frac{CE}{OE}$ = $\\frac{1}{2}$,\\\\\n$\\therefore$ CE = $\\frac{1}{2}$OE = $\\frac{3\\sqrt{5}}{2}$,\\\\\n$\\therefore$ the coordinates of point C are ($-3\\sqrt{5}$, $\\frac{3\\sqrt{5}}{2}$),\\\\\n$\\therefore$ k = $-3\\sqrt{5} \\times \\frac{3\\sqrt{5}}{2}$ = $\\boxed{-\\frac{45}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50987126.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated, the graph of the inverse proportion function $y=\\frac{k}{x}$ (where $k>0$) intercepts in the first quadrant at points A(1, 6) and B(3, b). The line $AB$ intersects the x-axis at point C, and D is a point on the line segment $OA$. If $CA\\cdot AD=AB\\cdot AO$, and by connecting $CD$, let the areas of $\\triangle ADC$ and $\\triangle DOC$ be $S_{1}$ and $S_{2}$ respectively, then what is the value of $S_{1}-S_{2}$?\n",
    "completion": "\\textbf{Solution:} Let $\\because$ A(1, 6) is on the graph of the inverse proportion function $y=\\frac{6}{x}$,\\\\\n$\\therefore$ k=6, the analytical expression of the inverse proportion function is: $y=\\frac{6}{x}$,\\\\\n$\\because$ B(3, b) is on the graph of the inverse proportion function $y=\\frac{6}{x}$,\\\\\n$\\therefore$ b=2, that is, B(3, 2).\\\\\nLet the line AB be: $y=mx+n(m\\ne 0)$,\\\\\n$\\therefore$\n$\\begin{cases}\nm+n=6\\\\\n3m+n=2\n\\end{cases}$, solving we get:\n$\\begin{cases}\nm=-2\\\\\nn=8\n\\end{cases}$,\\\\\n$\\therefore$ the analytical expression for line AB is: $y=-2x+8$.\\\\\n$\\therefore$ For $y=-2x+8$, when $y=0$, i.e., $-2x+8=0$,\\\\\nsolving we get: $x=4$,\\\\\n$\\therefore$ C(4, 0),\\\\\n$\\therefore$ ${S}_{\\triangle AOC}=\\frac{1}{2}OC\\cdot y_{A}=\\frac{1}{2}\\times 4\\times 6=12$,\\\\\n$\\because$ $CA\\cdot AD=AB\\cdot AO$,\\\\\n$\\therefore$ $\\frac{CA}{AB}=\\frac{AO}{AD}$,\\\\\nFurthermore, $\\because$ $\\angle DAB=\\angle OAC$,\\\\\n$\\therefore$ $\\triangle DAB\\sim \\triangle OAC$,\\\\\n$\\therefore$ $\\angle ADB=\\angle AOC$,\\\\\n$\\therefore$ $DB\\parallel OC$,\\\\\n$\\therefore$ $y_{D}=y_{B}=2$,\\\\\n$\\therefore$ ${S}_{\\triangle DOC}=\\frac{1}{2}OC\\cdot y_{D}=\\frac{1}{2}\\times 4\\times 2=4$,\\\\\n$\\therefore$ ${S}_{\\triangle ADC}={S}_{\\triangle AOC}-{S}_{\\triangle DOC}=12-4=8$,\\\\\n$\\therefore$ ${S}_{1}-{S}_{2}={S}_{\\triangle ADC}-{S}_{\\triangle DOC}=8-4=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52745996.png"
  },
  {
    "solution": "\\boxed{\\frac{4}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: A car passes a certain section of the road at a constant speed. The time $t$ (h) required and the driving speed $v$ (km/h) satisfy the function relationship: $t=\\frac{k}{v}$, and its graph is a curve as shown, with endpoints at $A(40, 1)$ and $B(m, 0.5)$. If the driving speed cannot exceed $50\\text{km/h}$, what is the minimum amount of time needed for the car to pass this section of the road?\n",
    "completion": "\\textbf{Solution:} Substituting $v=50$ into $t=\\frac{40}{v}$, we get $t=\\frac{4}{5}$,\\\\\nsince $t$ decreases as $v$ increases,\\\\\ntherefore, the minimum time required for the car to pass through the section is $\\boxed{\\frac{4}{5}}$ hours.",
    "image_path": "dataset/mm_math/images/MM_Math/51446664_252.png"
  },
  {
    "solution": "\\boxed{y= \\frac{\\sqrt{3}}{9}x^{2}+ \\frac{2\\sqrt{3}}{9}x- \\frac{8\\sqrt{3}}{9}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the parabola \\(y=\\frac{k}{8}(x-2)(x+4)\\) (where \\(k\\) is a constant and \\(k>0\\)) intersects the x-axis at points A and B from left to right, and intersects the y-axis at point C. Point D is on the graph of the parabola in the first quadrant, and \\(\\angle BAD=30^\\circ\\). If the x-coordinate of point D is 5, what is the equation of the parabola?\n",
    "completion": "\\textbf{Solution:} Let the parabola be $y=\\frac{k}{8}(x-2)(x+4)$. Setting $y=0$, we obtain $(x-2)(x+4)=0$, which solves to $x=2$ or $x=-4$. Thus, $A(-4,0)$ and $B(2,0)$. Therefore, $AO=4$. Construct $DE\\perp$ x-axis at point $E$, hence $OE=5$, $AE=AO+OE=4+5=9$. Since $\\angle BAD=30^\\circ$, it follows that $DE=AE\\cdot\\tan{30^\\circ}=9\\times \\frac{\\sqrt{3}}{3}=3\\sqrt{3}$. Hence, point $D$ is at $(5,3\\sqrt{3})$. Given that point $D$ lies on the parabola $y=\\frac{k}{8}(5-2)(5+4)=3\\sqrt{3}$, it follows that $k= \\frac{8\\sqrt{3}}{9}$. Thus, the expression for the parabola is: $y= \\frac{\\sqrt{3}}{9}(x-2)(x+4)$. In other words, $\\boxed{y= \\frac{\\sqrt{3}}{9}x^{2}+ \\frac{2\\sqrt{3}}{9}x- \\frac{8\\sqrt{3}}{9}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50890989_139.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $CM$ bisects $\\angle ACB$ and intersects $AB$ at point $M$. A line through point $M$ is drawn such that $MN \\parallel BC$ and intersects $AC$ at point $N$, and $MN$ bisects $\\angle AMC$. If $AN=1$, what is the length of $BC$?\n",
    "completion": "Solution:In a right-angled triangle $ABC$, let $CM$ bisect $\\angle ACB$ and intersect $AB$ at point $M$. Construct $MN \\parallel BC$ through point $M$ to intersect $AC$ at point $N$, and let $MN$ bisect $\\angle AMC$,\\\\\n$\\therefore \\angle AMN = \\angle NMC = \\angle B$, $\\angle NCM = \\angle BCM = \\angle NMC$,\\\\\n$\\therefore \\angle ACB = 2\\angle B$, $NM = NC$,\\\\\n$\\therefore \\angle B = 30^\\circ$,\\\\\nSince $AN = 1$,\\\\\n$\\therefore MN = 2$,\\\\\n$\\therefore AC = AN + NC = 3$,\\\\\n$\\therefore BC = \\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51212564.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the Cartesian coordinate system $xOy$, the graph of the linear function $y=x$ intersects with the graph of the quadratic function $y=-x^2+bx$ (where $b$ is a constant) at points $O$ and $A$. The coordinates of point $A$ are $(3, m)$. If point $P$ is the vertex of the parabola, and lines $OP$ and $AP$ are drawn, what is the area of $\\triangle POA$?\n",
    "completion": "\\textbf{Solution:} Construct $PC\\perp x$-axis through point P, with the foot at C, and intersecting OA at point D. Construct $AE\\perp PC$ through point A, with the foot at E,\\\\\n$\\because y=-x^{2}+4x=-(x-2)^{2}+4$,\\\\\n$\\therefore$ vertex P$(2,4)$,\\\\\nSubstituting $x=2$ into $y=x$ yields:\\\\\n$y=2$,\\\\\n$\\therefore D(2,2)$,\\\\\n$\\therefore PD=4-2=2$,\\\\\n$\\because$ the area of $\\triangle POA$ equals the sum of the areas of $\\triangle OPD$ and $\\triangle APD$,\\\\\n$\\therefore$ the area of $\\triangle POA$ $=\\frac{1}{2}PD\\cdot OC+\\frac{1}{2}PD\\cdot AE$\\\\\n$=\\frac{1}{2}PD(OC+AE)$\\\\\n$=\\frac{1}{2}\\times 2\\times 3$\\\\\n$=\\boxed{3}$,\\\\",
    "image_path": "dataset/mm_math/images/MM_Math/52872343_272.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points D and E are located on the sides $AB$ and $AC$ of $\\triangle ABC$, respectively, with $DE \\parallel BC$. If ${S}_{\\triangle ADE} = 2$ and ${S}_{\\triangle BCE} = 7.5$, what is the area of ${S}_{\\triangle BDE}$?\n",
    "completion": "\\textbf{Solution:} Given $DE\\parallel BC$, it follows that $\\frac{AD}{DB}=\\frac{AE}{EC}$. \\\\\nAlso, since $\\frac{S_{\\triangle ADE}}{S_{\\triangle BDE}}=\\frac{AD}{BD}$, it implies $\\frac{S_{\\triangle ABE}}{S_{\\triangle BCE}}=\\frac{AE}{EC}$, hence $\\frac{S_{\\triangle ADE}}{S_{\\triangle BDE}}=\\frac{S_{\\triangle ABE}}{S_{\\triangle BCE}}$. \\\\\nLet $S_{\\triangle BDE}=x$, with $S_{\\triangle ADE}=2$, and $S_{\\triangle BCE}=7.5$, therefore $\\frac{2}{x}=\\frac{2+x}{7.5}$. Consequently, $x_1=3$, $x_2=-5$ (discard). Therefore, $S_{\\triangle DBE}=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52942179.png"
  },
  {
    "solution": "\\boxed{100^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, the perpendicular bisectors of sides $AB$ and $AC$ intersect at point $P$. Connect $BP$ and $CP$. If $\\angle A = 50^\\circ$, what is the measure of $\\angle BPC$?\n",
    "completion": "\\textbf{Solution:} As the diagram shows, connect AP and extend BP to intersect AC at D,\\\\\n$\\because$ the perpendicular bisectors of AB and AC intersect at point P,\\\\\n$\\therefore$ point P is the circumcenter of $\\triangle ABC$,\\\\\n$\\therefore$ $\\angle BPC = 2\\angle BAC$\\\\\n$\\because$ $\\angle BAC = 50^\\circ$,\\\\\n$\\therefore$ $\\angle BPC = \\boxed{100^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52808095.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $AB=10$, and $BC=5$. If points $M$ and $N$ are two movable points on line segments $AC$ and $AB$ respectively, what is the perimeter of $\\triangle BMN$ when $BM+MN$ is minimized?\n",
    "completion": "\\textbf{Solution:}\n\nAs shown in the figure, construct the point $B'$ as the symmetric point of $B$ with respect to $AC$, and draw line $B'A$ intersecting $DC$ at point $E$. Therefore, the minimum value of $BM+MN$ is equal to the minimum value of $B'M+MN$,\\\\\n$\\therefore$ when $B'N \\perp AB$, the minimum value of $BM+MN$ is achieved,\\\\\n$\\therefore$ construct $B'N' \\perp AB$ intersecting $AC$ at $M'$, then $B'N'$ is the minimum value of $BM+MN$;\\\\\n$\\because$ Quadrilateral $ABCD$ is a rectangle,\\\\\n$\\therefore$ $\\angle D=90^\\circ$, $DC \\parallel AB$,\\\\\n$\\therefore$ $\\angle DCA=\\angle BAC$,\\\\\nAlso, $\\because$ $\\angle B'AC=\\angle BAC$,\\\\\n$\\therefore$ $\\angle B'AC=\\angle DCA$,\\\\\n$\\therefore$ $AE=CE$, let $EC=AE=x$,\\\\\n$\\therefore$ in $\\triangle AED$, $DE^2+AD^2=AE^2$, i.e., $(10-x)^2+5^2=x^2$, solving we get: $x=\\frac{25}{4}$,\\\\\n$\\therefore B'E=AB'−AE=10−\\frac{25}{4}=\\frac{15}{4}$,\\\\\nLet the height from $E$ to side $EC$ in $\\triangle B'EC$ be $h$, by symmetry we have $B'C=BC=5$,\\\\\n$\\angle AB'C=\\angle ABC=90^\\circ$,\\\\\n$\\therefore$ $S_{\\triangle B'CE}=\\frac{1}{2}\\times \\frac{15}{4}\\times 5=\\frac{1}{2}\\cdot \\frac{25}{4}\\cdot h$, solving we get: $h=3$,\\\\\n$B'N'=h+5=8$, that is, the minimum value of $BM+MN$ is 8,\\\\\n$\\therefore$ in $\\triangle AB'N'$, $AN'=\\sqrt{AB'^{2}-B'N'^{2}}=\\sqrt{10^{2}-8^{2}}=6$,\\\\\n$\\therefore$ $BN'=AB-AN'=10-6=4$,\\\\\n$\\therefore$ the perimeter of $\\triangle BMN=BN'+B'M'+M'N'=BN'+B'N'=4+8=\\boxed{12}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51368293.png"
  },
  {
    "solution": "\\boxed{160^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, quadrilateral ABCD is inscribed in circle $O$. If $\\angle C = 100^\\circ$, what is the measure of $\\angle BOD$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is inscribed in circle $O$, and $\\angle BCD=100^\\circ$, \\\\\nit follows that $\\angle A=180^\\circ-\\angle BCD=80^\\circ$, \\\\\nBy the inscribed angle theorem, we have $\\angle BOD=2\\angle A= \\boxed{160^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52828738.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, fold the triangular paper ABC along AD so that point C falls on point E on side BD. If $\\angle C = 45^\\circ$, $\\angle B = 30^\\circ$, and AD = 2, what is the value of $AB^2 - AC^2$?\n",
    "completion": "\\textbf{Solution:} Given that the triangle paper ABC is folded along AD, making point C fall on point E on side BD,\\\\\nthus, $\\angle ADC=\\angle ADE=90^\\circ$,\\\\\nsince $\\angle C=45^\\circ$, AD=2,\\\\\nthus, AC= $\\sqrt{2}$ AD= $2\\sqrt{2}$,\\\\\nsince $\\angle B=30^\\circ$,\\\\\nthus, AB=2AD=$2\\times2$=4,\\\\\ntherefore, AB$^{2}$−AC$^{2}$=4$^{2}$−($2\\sqrt{2}$)$^{2}$=\\boxed{8},",
    "image_path": "dataset/mm_math/images/MM_Math/52837759.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the diagonals AC and BD of the parallelogram ABCD intersect at point O, and E and F are the midpoints of AO and BO, respectively. If $AC+BD=24$ cm, and the perimeter of $\\triangle OAB$ is 18 cm, then how long is $EF$ in cm?\n",
    "completion": "Solution: Since $ABCD$ is a parallelogram, \\\\\nit follows that $OB=OD$ and $OA=OC$, \\\\\nsince $AC+BD=24cm$, it follows that $OA+OB=12cm$, \\\\\nsince the perimeter of $\\triangle OAB$ is $18cm$, it follows that $AB=18-12=6cm$, \\\\\nsince $E$ and $F$ are the midpoints of $AO$ and $BO$ respectively, it follows that $EF$ is the midline of $\\triangle OAB$, \\\\\ntherefore $EF=\\frac{1}{2}AB=3cm$, \\\\\nthus the answer is: $\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52049891.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, on the graph of the inverse proportion function $y=\\frac{2}{x}$ ($x>0$), there are points $P_1$, $P_2$, $P_3$, $P_4$ whose abscissae are 1, 2, 3, 4, respectively. Perpendiculars are drawn from these points to the $x$-axis and the $y$-axis. The areas of the shaded regions formed in the diagram from left to right are $S_1$, $S_2$, and $S_3$, respectively. What is the value of $S_1+S_2+S_3$?\n",
    "completion": "\\textbf{Solution:} After translation, as shown in the figure, \\\\\nwhen $x=4$, $y=\\frac{1}{2}$, \\\\\n$\\therefore$ the area of rectangle AOCB is $1\\times\\frac{1}{2}=\\frac{1}{2}$; \\\\\nwhen $x=1$, $y=2$, \\\\\n$\\therefore S_{1}+S_{2}+S_{3}+\\text{Area of rectangle AOCB}=2$ \\\\\n$\\therefore S_{1}+S_{2}+S_{3}=2-\\frac{1}{2}=\\boxed{\\frac{3}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52694241.png"
  },
  {
    "solution": "\\boxed{0<x<10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y=mx+n$ intersects with the graph of the inverse proportion function $y=\\frac{k}{x}$ at points $A\\left(-2,2\\right)$ and $B\\left(1, a\\right)$. What is the solution set for the inequality $mx+n>\\frac{k}{x}$?\n",
    "completion": "\\textbf{Solution:} The solution set of the inequality $mx+n>\\frac{k}{x}$ is $x<-2$ or $0<x<10$.\n\nTherefore, the answer is $x<-2$ or $\\boxed{0<x<10}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53065859_81.png"
  },
  {
    "solution": "\\boxed{\\sqrt{3}-1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the rhombus $ABCD$ as shown in the figure, $AB=2$ and $\\angle BAD=60^\\circ$. The rhombus $ABCD$ is rotated counterclockwise around point $A$ to get the rhombus $AEFG$, where point $E$ is on $AC$, and $EF$ intersects $CD$ at point $P$. What is the length of $DP$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect BD intersecting AC at point O, \\\\\n$\\because$ Quadrilateral ABCD is a rhombus, \\\\\n$\\therefore$ CD=AB=2, $\\angle$BCD=$\\angle$BAD=$60^\\circ$, $\\angle$ACD=$\\angle$BAC=$\\frac{1}{2}\\angle$BAD=$30^\\circ$, OA=OC, AC$\\perp$BD, \\\\\n$\\therefore$ OB=$\\frac{1}{2}$AB=1, \\\\\n$\\therefore$ OA=$\\sqrt{3}$OB=$\\sqrt{3}$, \\\\\n$\\therefore$ AC=2OA=2$\\sqrt{3}$, \\\\\nBy the property of rotation: AE=AB=2, $\\angle$EAG=$\\angle$BAD=$60^\\circ$, \\\\\n$\\therefore$ CE=AC−AE=2$\\sqrt{3}$−2, \\\\\n$\\because$ Quadrilateral AEFG is a rhombus, \\\\\n$\\therefore$ EF$\\parallel$AG, \\\\\n$\\therefore$ $\\angle$CEP=$\\angle$EAG=$60^\\circ$, \\\\\n$\\therefore$ $\\angle$CEP+$\\angle$ACD=$90^\\circ$, \\\\\n$\\therefore$ $\\angle$CPE=$90^\\circ$, \\\\\n$\\therefore$ PE=$\\frac{1}{2}$CE=$\\sqrt{3}$−1, \\\\\n$\\therefore$ PC=$\\sqrt{3}$PE=3−$\\sqrt{3}$, \\\\\n$\\therefore$ DP=CD−PC=2−(3−$\\sqrt{3}$)=$\\boxed{\\sqrt{3}-1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51672730.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in quadrilateral $ABCD$, $AD \\parallel BC$, $\\angle B=90^\\circ$, $AB=8\\, \\text{cm}$, $AD=12\\, \\text{cm}$, $BC=18\\, \\text{cm}$. Point $P$ starts moving from point $A$ towards point $D$ at a speed of $1\\, \\text{cm/s}$; simultaneously, point $Q$ starts moving from point $C$ towards point $B$ at a speed of $2\\, \\text{cm/s}$. The movement of point $P$ stops when point $Q$ reaches point $B$. Let the time during which points $P$ and $Q$ are moving be $t\\, \\text{(s)}$. At what value of $t$ will quadrilateral $PQBA$ be a rectangle?\n",
    "completion": "\\paragraph{Solution:} Since point P starts from point A and moves toward point D at a speed of $1\\,\\text{cm/s}$, and point Q starts from point C at the same time, moving toward point B at a speed of $2\\,\\text{cm/s}$,\\\\\nit follows that $AP=t$, $CQ=2t$, $BQ=18-2t$,\\\\\nSince rectangle $PQBA$,\\\\\nit follows that $AP=BQ$,\\\\\nThus, $t=18-2t$,\\\\\nSolving this gives $t=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51614702_94.png"
  },
  {
    "solution": "\\boxed{(5\\sqrt{3}, -3)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system, the line $l\\colon y=-\\frac{\\sqrt{3}}{3}x+4$ intersects the x-axis and y-axis at points M and N, respectively. An equilateral triangle ABC with a height of 3, where side BC is on the x-axis, is translated in the positive direction of the x-axis. During the translation process, $\\triangle A_1B_1C_1$ is obtained. When point $B_1$ coincides with the origin, find a point P on the right side of line $A_1C_1$ such that the quadrilateral formed by vertices P, $A_1$, $C_1$, and M is a parallelogram. Find the coordinates of point P.\n",
    "completion": "\\textbf{Solution:}\n\n(1) If we consider $\\angle A_1C_1M$ as an internal angle of the parallelogram, then the desired parallelogram is parallelogram $A_1C_1MP$.\n\nAs shown in Figure (1), draw $A_1E\\perp ON$ from point $A_1$, with E being the foot of the perpendicular.\n\nFrom the analytic equation of line $l$, we get\n\nwhen $y=0$, $-\\frac{\\sqrt{3}}{3}x+4=0$,\n\ntherefore $x=4\\sqrt{3}$.\n\nTherefore, the coordinates of point M are $(4\\sqrt{3}, 0)$\n\nTherefore, $OM=4\\sqrt{3}$.\n\nBecause $B_1C_1=A_1B_1=2\\sqrt{3}$.\n\nTherefore, $C_1M=OM-B_1C_1=4\\sqrt{3}-2\\sqrt{3}=2\\sqrt{3}$\n\nTherefore, $A_1P=C_1M=2\\sqrt{3}$\n\nTherefore, $PE=A_1E+A_1P=\\sqrt{3}+2\\sqrt{3}=3\\sqrt{3}$.\n\nTherefore, the coordinates of point P are $\\boxed{(3\\sqrt{3}, 3)}$;\n\n(2) If we consider $\\angle C_1A_1M$ as an internal angle of the parallelogram, then the desired parallelogram is $A_1C_1PM$. As shown in Figure (2), draw $PG\\perp OM$ from point P, with G being the foot of the perpendicular.\n\nBecause $\\triangle A_1B_1C_1$ is an equilateral triangle,\n\ntherefore $\\angle A_1C_1B_1=60^\\circ$,\n\ntherefore $\\angle AC_1G=120^\\circ$,\n\nbecause parallelogram $A_1C_1PM$ is a parallelogram,\n\ntherefore $A_1C_1\\parallel PM$, $PM=A_1C_1=A_1B_1=2\\sqrt{3}$,\n\ntherefore $\\angle PMC_1=120^\\circ$,\n\ntherefore $\\angle PMG=60^\\circ$,\n\ntherefore, in $\\triangle PMG$, $\\angle MPG=90^\\circ-\\angle PMG=90^\\circ-60^\\circ=30^\\circ$.\n\nTherefore, $MG=\\frac{1}{2}PM=\\frac{1}{2}\\times 2\\sqrt{3}=\\sqrt{3}$,\n\n$PG=\\sqrt{PM^2-MG^2}=\\sqrt{(2\\sqrt{3})^2-(\\sqrt{3})^2}=3$,\n\ntherefore $OG=OM+MG=4\\sqrt{3}+\\sqrt{3}=5\\sqrt{3}$.\n\nTherefore, the coordinates of point P are $\\boxed{(5\\sqrt{3}, -3)}$\n\nIn conclusion, the coordinates of point P are either $\\boxed{(3\\sqrt{3}, 3)}$ or $\\boxed{(5\\sqrt{3}, -3)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52251020_96.png"
  },
  {
    "solution": "\\boxed{30}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, if $D$, $E$, and $F$ are the midpoints of the sides of $\\triangle ABC$, with $EF=6\\,\\text{cm}$, $DE=4\\,\\text{cm}$, and $DF=5\\,\\text{cm}$, then what is the perimeter of $\\triangle ABC$?\n",
    "completion": "\\textbf{Solution:} In $\\triangle ABC$,\\\\\nsince points D, E, F are the midpoints of the sides of $\\triangle ABC$,\\\\\ntherefore $EF = \\frac{1}{2}AB$, $DE = \\frac{1}{2}BC$, $DF = \\frac{1}{2}AC$,\\\\\nthus, the perimeter of $\\triangle ABC$ is: $AB+BC+AC=2EF+2DE+2DF$\\\\\n$=2(EF+DE+DF)=2\\times (6+4+5)$\\\\\n$=\\boxed{30}$ (cm)",
    "image_path": "dataset/mm_math/images/MM_Math/53333921.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $AB$ is the diameter of $\\odot O$, $CD$ is a chord, and $AB\\perp CD$ at point $E$. $G$ is any point on the arc $AC$ (not coinciding with $A$ or $C$), and $AD$ and $GD$ are connected. If $EB=4\\,\\text{cm}$ and $CD=16\\,\\text{cm}$, what is the radius of $\\odot O$?\n",
    "completion": "\\textbf{Solution:} Connect $OD$, as shown in the figure:\\\\\nSince $EB=4\\,\\text{cm}$ and $CD=16\\,\\text{cm}$, and $AB$ is the diameter of $\\odot O$ with $AB\\perp CD$,\\\\\nTherefore, $DE=\\frac{1}{2}CD=8\\,\\text{cm}$,\\\\\nLet $OD=OB=r\\,\\text{cm}$, then $OE=(r-4)\\,\\text{cm}$,\\\\\nTherefore, in $\\triangle OED$, we have $r^2=(r-4)^2+8^2$,\\\\\nSolving this, we find: $r=\\boxed{10}\\,\\text{cm}$,\\\\\nTherefore, the radius of $\\odot O$ is \\boxed{10}\\,\\text{cm}.",
    "image_path": "dataset/mm_math/images/MM_Math/53149307_137.png"
  },
  {
    "solution": "\\boxed{\\left(6, 6\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the vertices A and C of the square $OABC$ are located on the positive semiaxes of the coordinate axes, respectively, and point B is a point in the first quadrant on the line $y=\\frac{1}{2}x+3$. What are the coordinates of point B?\n",
    "completion": "\\textbf{Solution:} From the problem, we know that point B lies on the line $y=\\frac{1}{2}x+3$ and point B is a vertex of the square ABCD,\\\\\nLet $B\\left(x, \\frac{1}{2}x+3\\right)$,\\\\\n$\\therefore x=\\frac{1}{2}x+3$,\\\\\nSolving yields: $x=6$,\\\\\n$\\therefore \\frac{1}{2}x+3=6$,\\\\\n$\\therefore B\\left(6, 6\\right)$;\\\\\nThus, the answer is $B\\boxed{\\left(6, 6\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52342684.png"
  },
  {
    "solution": "\\boxed{y=-2x+4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, in the Cartesian coordinate system, the line $l_{1}$: $y=kx+b\\ (k \\neq 0)$ intersects the x-axis at point $A(a, 0)$ and the y-axis at point $B(0, b)$. The line $l_{2}$, which passes through the origin, intersects the line $l_{1}$ at point C. If $a=2$ and $b=4$, what is the equation of the line $l_{1}$?\n",
    "completion": "\\textbf{Solution:} Let's suppose $a=2$, $b=4$, then point $A(2, 0)$ and point $B(0, 4)$ are given. Substituting into the equation $y=kx+b$, we get:\n\\begin{align*}\n\\begin{cases}\n2k+b=0\\\\\nb=4\n\\end{cases}\n\\end{align*},\nwhich solves to\n\\begin{align*}\n\\begin{cases}\nk=-2\\\\\nb=4\n\\end{cases}\n\\end{align*},\n$\\therefore$ the equation for the line ${l}_{1}$ is $\\boxed{y=-2x+4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52386704_119.png"
  },
  {
    "solution": "\\boxed{\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the diagram, within circle $\\odot O$, radius $OC$ is perpendicular to chord $AB$ at $D$. Point $E$ is on circle $\\odot O$, with $\\angle E=22.5^\\circ$, and $AB=2$. What is the length of the radius $OB$?\n",
    "completion": "\\textbf{Solution:} Given that $\\because$ radius $OC\\perp$ chord AB at point $D$,\\\\\n$\\therefore \\overset{\\frown}{AC}=\\overset{\\frown}{BC}$,\\\\\n$\\therefore \\angle E=\\frac{1}{2}\\angle BOC=22.5^\\circ$,\\\\\n$\\therefore \\angle BOD=45^\\circ$,\\\\\n$\\therefore \\triangle ODB$ is an isosceles right-angled triangle,\\\\\n$\\because AB=2$,\\\\\n$\\therefore DB=OD=1$,\\\\\n$\\therefore OB=\\sqrt{1^2+1^2}=\\boxed{\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51445862.png"
  },
  {
    "solution": "\\boxed{2 \\sqrt{6}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, for the rhombus $ABCD$ with side length $AB=2\\sqrt{3}+2$ and $\\angle ABC=60^\\circ$, two points $E$ and $H$ are taken on the diagonal $BD$ such that $BE=DH$ and $AE\\parallel FH$. When $\\angle AED=75^\\circ$, what is the sum $AE+FH$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, construct FM$\\perp$BD at M, EN$\\perp$AB at N. \\\\\nBecause quadrilateral ABCD is a rhombus and $\\angle$ABC=$60^\\circ$, \\\\\nit follows that $\\angle$ABD=$\\angle$ADB=$30^\\circ$, \\\\\nSince $\\angle$AED=$75^\\circ$=$\\angle$ABE+$\\angle$BAE, \\\\\nit results in $\\angle$BAE=$45^\\circ$, \\\\\nGiven $\\angle$ENA=$90^\\circ$, \\\\\nit leads to $\\angle$NAE=$\\angle$NEA=$45^\\circ$, \\\\\nhence NA=NE, \\\\\nLet NA=NE=x, then BE=2x, and BN=$ \\sqrt{3}$x, \\\\\nSince AB=2$ \\sqrt{3}$+2, \\\\\nit implies that $ \\sqrt{3}$x+x=2$ \\sqrt{3}$+2, \\\\\nthus x=2, \\\\\nTherefore, BE=4, AE=2$ \\sqrt{2}$, \\\\\nSince $\\angle$HFD=$180^\\circ$−$\\angle$FHD−$\\angle$FDH=$75^\\circ$ and $\\angle$FHD=$75^\\circ$, \\\\\nit concludes that $\\angle$DFH=$\\angle$DHF, \\\\\nTherefore, DF=DH=BE=2, \\\\\nIn the right triangle $\\triangle$FMD, \\\\\nsince $\\angle$FMD=$90^\\circ$ and $\\angle$FDM=$30^\\circ$ with DF=4, \\\\\nit follows that FM=2, DM=2$ \\sqrt{3}$, and HM=4−2$ \\sqrt{3}$, \\\\\ntherefore, FH=$ \\sqrt{FM^{2}+HM^{2}}$ \\\\\n=$ \\sqrt{2^{2}+(4−2\\sqrt{3})^{2}}$ \\\\\n=2$ \\sqrt{6}$−2$ \\sqrt{2}$, \\\\\nThus, AE+FH=2$ \\sqrt{6}$−2$ \\sqrt{2}+2$ $\\sqrt{2}$ \\\\\n=$\\boxed{2 \\sqrt{6}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53538188.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\angle AOB$ is a right angle, $\\angle AOC = 30^\\circ$, $ON$ is the bisector of $\\angle AOC$, and $OM$ is the bisector of $\\angle BOC$. What is the measure of $\\angle MON$ in degrees?\n",
    "completion": "\\textbf{Solution:} Given that $ON$ is the bisector of $\\angle AOC$ and $OM$ is the bisector of $\\angle BOC$, \\\\\nit follows that $\\angle NOC= \\frac{1}{2} \\angle AOC$ and $\\angle MOC= \\frac{1}{2} \\angle BOC$\\\\\nTherefore, $\\angle MON=\\angle MOC-\\angle NOC$\\\\\n$= \\frac{1}{2} \\angle BOC- \\frac{1}{2} \\angle AOC$\\\\\n$= \\frac{1}{2} (\\angle BOC-\\angle AOC)$\\\\\n$= \\frac{1}{2} \\angle AOB$\\\\\n$= \\frac{1}{2} \\times 90^\\circ$\\\\\n$= \\boxed{45^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/50056777.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of the circle $\\odot O$, $C$ is a point on $\\odot O$, line $AD$ is perpendicular to the line passing through point $C$ with the foot of the perpendicular being $D$, and $AC$ bisects $\\angle DAB$. $DC$ is a tangent line of $\\odot O$; If the radius of $\\odot O$ is 2, and $AC=2\\sqrt{3}$, what is the length of the line segment $AD$?\n",
    "completion": "\\textbf{Solution:} Connect $BC$,\n\nSince $AB$ is the diameter of $\\odot O$,\n\nTherefore, $\\angle ACB=90^\\circ$,\n\nSince $\\angle DAC=\\angle OAC$, $\\angle ADC=\\angle ACB=90^\\circ$,\n\nTherefore, $\\triangle ADC\\sim \\triangle ACB$,\n\nTherefore, $\\frac{AD}{AC}=\\frac{AC}{AB}$,\n\nSince the radius of $\\odot O$ is 2, $AC=2\\sqrt{3}$,\n\nTherefore, $\\frac{AD}{2\\sqrt{3}}=\\frac{2\\sqrt{3}}{4}$,\n\nTherefore, $AD=\\boxed{3}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53149410_116.png"
  },
  {
    "solution": "\\boxed{0<x<4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the graph of the linear function $y=kx+b$ intersects with the graph of the inverse proportion function $y=-\\frac{8}{x}$ at points $A$ and $B$, and the $x$-coordinate of point $A$ and the $y$-coordinate of point $B$ are both $-2$. Find: For what value of $x$ is the function value of the linear function greater than that of the inverse proportion function?\n",
    "completion": "\\textbf{Solution:} Observe the graph of the function and find: \\\\\nWhen $x<-2$ or $0<x<4$, the graph of the linear function is above the graph of the inverse proportion function, \\\\\n$\\therefore$ The range of $x$ for which the function value of the linear function is greater than the function value of the inverse proportion function is $\\boxed{x<-2}$ or $\\boxed{0<x<4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52765753_97.png"
  },
  {
    "solution": "\\boxed{C\\left(\\frac{1}{2}, 0\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system $xOy$, it is known that the parabola $y=x^{2}+bx$ passes through point A $(2,0)$ and point $B(-1, m)$, with its vertex at point D. Suppose the line segment BD intersects the $x$-axis at point P. If point C is on the $x$-axis, and $\\triangle ABC$ is similar to $\\triangle ABP$, find the coordinates of point C.\n",
    "completion": "Solution: Let the equation of line BD be $y=k_1x+b_1\\left(k_1\\ne 0\\right)$.\\\\\nSubstituting points $D\\left(1, -1\\right)$ and $B\\left(-1, 3\\right)$ yields:\\\\\n$\\begin{cases}k_1+b_1=-1\\\\ -k_1+b_1=3\\end{cases}$, solving this gives: $\\begin{cases}k_1=-2\\\\ b_1=1\\end{cases}$,\\\\\n$\\therefore$ the equation of line BD is $y=-2x+1$,\\\\\nWhen $y=0$, $x=\\frac{1}{2}$,\\\\\n$\\therefore$ the coordinates of point P are $P\\left(\\frac{1}{2}, 0\\right)$,\\\\\nWhen $\\triangle ABP\\sim \\triangle ABC$, $\\angle ABC=\\angle APB$,\\\\\nAs illustrated, draw BQ⊥x-axis at point Q from point B, then BQ=3, OQ=1,\\\\\n$\\because \\triangle ABP\\sim \\triangle ABC$,\\\\\n$\\therefore \\angle ABD=\\angle BCQ$,\\\\\nFrom (2), we know $\\tan\\angle ABD=\\frac{1}{3}$,\\\\\n$\\therefore \\tan\\angle BCQ=\\frac{1}{3}$,\\\\\n$\\therefore \\frac{BQ}{CQ}=\\frac{1}{3}$,\\\\\n$\\therefore CQ=9$,\\\\\n$\\therefore OC=OQ+CQ=10$,\\\\\n$\\therefore$ the coordinates of point C are $C\\left(-10, 0\\right)$;\\\\\nWhen $\\triangle ABP\\sim \\triangle ABC$, $\\angle APB=\\angle ACB$, in this case, point C coincides with point P,\\\\\n$\\therefore$ the coordinates of point C are $C\\left(\\frac{1}{2}, 0\\right)$,\\\\\nIn conclusion, the coordinates of point C are either $C\\left(-10, 0\\right)$ or $\\boxed{C\\left(\\frac{1}{2}, 0\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51403345_147.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB=AC=AD=2$, $\\angle BCD=120^\\circ$, then what is the value of $BD$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, with A as the center and AB as the radius, draw $\\odot A$, extend BA to intersect $\\odot A$ at T, and connect DT, \\\\\n$\\because$ $AB=AC=AD$,\\\\\n$\\therefore$ point A is the center of the circumscribed circle of $\\triangle BCD$,\\\\\n$\\because$ $\\angle BCD+\\angle T=180^\\circ$, $\\angle BCD=120^\\circ$,\\\\\n$\\therefore$ $\\angle T=60^\\circ$,\\\\\n$\\because$ BT is the diameter,\\\\\n$\\therefore$ $BT=2AB=4$, $\\angle BDT=90^\\circ$,\\\\\n$\\therefore$ $\\angle DBT=90^\\circ-60^\\circ=30^\\circ$,\\\\\n$\\therefore$ $DT=\\frac{1}{2}BT=2$,\\\\\n$\\therefore$ $BD=\\sqrt{BT^2-TD^2}=\\sqrt{4^2-2^2}=\\boxed{2\\sqrt{3}}$,\\\\",
    "image_path": "dataset/mm_math/images/MM_Math/51213351.png"
  },
  {
    "solution": "\\boxed{\\frac{b^4}{a^3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC=a$, $BC=b$ ($a>b$). Within $\\triangle ABC$, construct $\\angle CBD=\\angle A$, $\\angle DCE=\\angle CBD$, $\\angle EDF=\\angle DCE$, then what is the length of $EF$?\n",
    "completion": "\\textbf{Solution:} \\\\\n\\[\n\\because \\angle CBD = \\angle A, \\angle BCD = \\angle ACB,\n\\]\n\\[\n\\therefore \\triangle BCD \\sim \\triangle ABC,\n\\]\n\\[\n\\therefore \\frac{BC}{CD} = \\frac{AB}{BC},\n\\]\n\\[\n\\therefore \\frac{b}{CD} = \\frac{a}{b},\n\\]\n\\[\n\\therefore CD = \\frac{b^2}{a},\n\\]\n\\[\n\\because \\triangle BCD \\sim \\triangle ABC,\n\\]\n\\[\n\\therefore \\frac{AB}{AC} = \\frac{BC}{BD},\n\\]\n\\[\n\\because AB = AC,\n\\]\n\\[\n\\therefore BC = BD = b,\n\\]\n\\[\n\\because \\angle DCE = \\angle CBD, \\angle CDE = \\angle BDC,\n\\]\n\\[\n\\therefore \\triangle CDE \\sim \\triangle BDC,\n\\]\n\\[\n\\therefore \\frac{DE}{CD} = \\frac{CD}{BC},\n\\]\n\\[\n\\therefore DE = \\frac{CD^2}{BC} = \\frac{b^4}{a^2}{b} = \\frac{b^3}{a^2},\n\\]\n\\[\n\\because \\triangle CDE \\sim \\triangle BDC,\n\\]\n\\[\n\\therefore \\frac{BD}{BC} = \\frac{CD}{CE},\n\\]\n\\[\n\\because BD = BC,\n\\]\n\\[\n\\therefore CD = CE = \\frac{b^2}{a},\n\\]\n\\[\n\\because \\angle DEC = \\angle CED, \\angle EDF = \\angle DCE,\n\\]\n\\[\n\\therefore \\triangle DEF \\sim \\triangle CDE,\n\\]\n\\[\n\\therefore \\frac{DE}{EF} = \\frac{CD}{DE},\n\\]\n\\[\n\\therefore EF = \\frac{DE^2}{CD} = \\frac{\\left(\\frac{b^6}{a^4}\\right)}{\\left(\\frac{b^2}{a}\\right)} = \\boxed{\\frac{b^4}{a^3}}.\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/52513827.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that in $\\triangle ABC$, $D$ and $E$ are the midpoints of $AB$ and $AC$, respectively. Connect $DE$ and extend it to $F$ such that $EF=DE$. Connect $CF$. If $\\angle B=45^\\circ$, what is the measure of $\\angle F$?\n",
    "completion": "\\textbf{Solution:} Since point E is the midpoint of AC, and point D is the midpoint of AB, \\\\\ntherefore $DE \\parallel BC$ and $BC=2DE$. \\\\\nSince EF=DE, \\\\\ntherefore $DF=2DE$, \\\\\ntherefore $DF=BC$, \\\\\ntherefore quadrilateral BCFD is a parallelogram, \\\\\ntherefore $\\angle F=\\angle B=45^\\circ$. \\\\\nHence, the answer is: $\\boxed{45^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52386366.png"
  },
  {
    "solution": "\\boxed{(1,6)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ACB$, $\\angle ACB = 90^\\circ$, $AC = BC$, the coordinates of point $C$ are $(-2, 0)$, and the coordinates of point $A$ are $(-8, 3)$. What are the coordinates of point $B$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, extend $AD \\perp OC$ at $D$, and $BE \\perp OC$ at $E$, such that $\\angle 1=\\angle 2=\\angle 3=90^\\circ$,\\\\\n$\\therefore$ $\\angle$DAC+$\\angle$ACD=$\\angle$ACD+$\\angle$ECB=$90^\\circ$,\\\\\n$\\therefore$ $\\angle DAC=\\angle ECB$,\\\\\n$\\therefore$ $\\angle ACD=\\angle CBE$,\\\\\nIn $\\triangle ACD$ and $\\triangle CEB$,\\\\\n$\\because$ $\\left\\{\\begin{array}{l}\\angle DAC=\\angle ECB\\hfill \\\\ AC=CB\\hfill \\\\ \\angle ACD=\\angle CBE\\hfill \\end{array}\\right.$,\\\\\n$\\therefore$ $\\triangle ADC\\cong \\triangle CEB$ (ASA),\\\\\n$\\therefore$ $DC=BE$, $AD=CE$,\\\\\n$\\because$ the coordinates of point $C$ are $(-2,0)$, and the coordinates of point $A$ are $(-8,3)$,\\\\\n$\\therefore$ $OC=2$, $AD=CE=3$, $OD=8$,\\\\\n$\\therefore$ $CD=OD-OC=6$, $OE=CE-OC=1$,\\\\\n$\\therefore$ $BE=6$,\\\\\n$\\therefore$ the coordinates of point $B$ are $\\boxed{(1,6)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53044181.png"
  },
  {
    "solution": "\\boxed{\\frac{17}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ is inscribed in $\\odot O(AC>BC)$, with $AB$ being the diameter of $\\odot O$. Points E, C, D lie on $\\odot O$, and the arcs $\\overset{\\frown}{EC}=\\overset{\\frown}{CB}=\\overset{\\frown}{BD}$. Lines $ED$ intersect $AC, AB$ at points F and G, respectively. $\\triangle EFA\\sim \\triangle BCA$. If $BC=5, BG=4$, calculate the length of $AE$.\n",
    "completion": "\\textbf{Solution:} Given that $AB$ is the diameter of $\\odot O$,\\\\\n$\\therefore \\angle ACB=90^\\circ$,\\\\\n$\\because \\triangle EFA\\sim \\triangle BCA$,\\\\\n$\\therefore \\angle EFA=\\angle C=90^\\circ$, $\\frac{EF}{AE}=\\frac{BC}{AB}$,\\\\\nAlso $\\because \\angle CAE=\\angle CAB$, $AF=AF$,\\\\\n$\\therefore \\triangle AEF\\cong \\triangle AFG$ (ASA),\\\\\n$\\therefore AE=AG$, $EF=FG$,\\\\\n$\\because \\angle AEG=\\angle ABD$, $\\angle AGE=\\angle BGD$,\\\\\n$\\therefore \\triangle AEG\\sim \\triangle DBG$,\\\\\n$\\therefore \\frac{BD}{BG}=\\frac{AE}{EG}=\\frac{5}{4}$,\\\\\nLet $EF=FG=2x, AE=AG=5x$,\\\\\n$\\therefore AB=5x+4$,\\\\\n$\\therefore \\frac{BC}{AB}=\\frac{EF}{AE}=\\frac{5}{5x+4}=\\frac{2}{5}$,\\\\\n$\\therefore x=\\frac{17}{10}$,\\\\\n$\\therefore AE=5x=\\boxed{\\frac{17}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52731766_143.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the equilateral $\\triangle ABC$, $AD$ is the height from $A$ to side $BC$, points $M$ and $N$ are respectively on $AD$ and $AC$, and $AM=CN$. Connect $BM$ and $BN$. When $BM+BN$ is at its minimum, what is the measure of $\\angle MBN$?\n",
    "completion": "\\textbf{Solution:} As shown in Figure 1, construct $CH\\perp BC$ such that $CH=BC$, and connect $NH$, $BH$.\\\\\nSince $\\triangle ABC$ is an equilateral triangle, $AD\\perp BC$, $CH\\perp BC$,\\\\\nit follows that $\\angle DAC=\\angle DAB=30^\\circ$, $AD\\parallel CH$,\\\\\nthus $\\angle HCN=\\angle CAD=\\angle BAM=30^\\circ$,\\\\\nsince $AM=CN$, $AB=BC=CH$,\\\\\nit follows that $\\triangle ABM\\cong \\triangle CHN$ (SAS),\\\\\nhence, $BM=HN$,\\\\\nsince $BN+HN\\geq BH$,\\\\\nit follows that when $B$, $N$, and $H$ are collinear, the value of $BM+BN=NH+BN$ is minimized,\\\\\nAs shown in Figure 2, when $B$, $N$, and $H$ are collinear,\\\\\nsince $\\triangle ABM\\cong \\triangle CHN$,\\\\\nit follows that $\\angle ABM=\\angle CHB=\\angle CBH=45^\\circ$,\\\\\nsince $\\angle ABD=60^\\circ$,\\\\\nit follows that $\\angle DBM=15^\\circ$,\\\\\nhence, $\\angle MBN=45^\\circ-15^\\circ=\\boxed{30^\\circ}$,\\\\\nthus, when the value of $BM+BN$ is minimized, $\\angle MBN=30^\\circ$.",
    "image_path": "dataset/mm_math/images/MM_Math/52720610.png"
  },
  {
    "solution": "\\boxed{x>1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y=x+b$ intersects with the line $y=kx+6$ at the point $P(1,3)$. What is the solution set for the inequality $x+b>kx+6$ with respect to $x$?\n",
    "completion": "\\textbf{Solution:} When $x>1$, we have $x+b>kx+6$,\\\\\nwhich means the solution set of the inequality $x+b>kx+6$ is $x>1$,\\\\\ntherefore, the answer is $x>1$.\\\\\nHence, the answer is: $\\boxed{x>1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50561251.png"
  },
  {
    "solution": "\\boxed{\\left(-\\frac{7}{3},-1\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $y_1=3x+6$ intersects the x-axis and y-axis at points A and B, respectively. The line $y_2=kx+b$ passes through point D(3, 0) and intersects line $y_1=3x+6$ at point C(m, 3). There exists a point P on line AB, and through point P, a line PQ is drawn parallel to the y-axis, intersecting line CD at point Q. If the length of segment PQ is 5, what are the coordinates of point P?\n",
    "completion": "\\textbf{Solution:} Given that point $P$ lies on line $AB$ and point $Q$ lies on line $CD$, and $PQ \\parallel y$-axis, \\\\\nlet the coordinates of point $P$ be $(n, 3n+6)$, then the coordinates of point $Q$ are $\\left(n, -\\frac{3}{4}n+\\frac{9}{4}\\right)$, \\\\\nsince the length of segment $PQ$ is $3$, \\\\\n$\\therefore \\left|3n+6+\\frac{3}{4}n-\\frac{9}{4}\\right|=5$, \\\\\nsolving this gives $n=\\frac{1}{3}$ or $n=-\\frac{7}{3}$, \\\\\nwhen $n=\\frac{1}{3}$, $3n+6=7$; \\\\\nwhen $n=-\\frac{7}{3}$, $3n+6=-1$, \\\\\n$\\therefore$ the coordinates of point P are $\\boxed{\\left(\\frac{1}{3},7\\right)}$ or $\\boxed{\\left(-\\frac{7}{3},-1\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52159946_165.png"
  },
  {
    "solution": "\\boxed{\\frac{2}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that in $\\triangle ABC$, $\\angle ABC=90^\\circ$, point D is the midpoint of side AC, and BD is connected. $\\triangle ABC$ is rotated counterclockwise around point A, such that point B precisely falls on point E on ray BD, and point C falls on point F, and then FD and FC are connected. If $AB=1$ and $BC=2$, what is the value of the tangent of $\\angle CFD$?\n",
    "completion": "\\textbf{Solution:} After triangle $\\triangle ABC$ is rotated as shown in the figure, draw $AG\\perp BE$ at $G$ through $A$, draw $CN\\perp AF$ at $N$ through $C$, draw $FH\\perp BE$ through $F$ intersecting the extended line of $BE$ at $H$, draw $DM\\perp CF$ at $M$ through $D$,\\\\\nsince $\\angle ABC=90^\\circ$, $D$ is the midpoint of $AC$, $AB=1$, $BC=2$,\\\\\ntherefore $DB=DA=DC$, $AC=\\sqrt{5}$,\\\\\nthus $\\angle DAB=\\angle ABD$,\\\\\nBy rotation, it follows that $AE=AB=1$, $\\angle FAE=\\angle BAD$, $EF=BC=2$, $AC=AF=\\sqrt{5}$,\\\\\nthus $\\angle ABE=\\angle AEB=\\angle FAE$,\\\\\ntherefore $AF\\parallel BD$, $\\angle GAF=90^\\circ$,\\\\\nthus, quadrilateral $AGHF$ is a rectangle,\\\\\nsince $\\angle AEF=90^\\circ$,\\\\\nthus $\\angle AEG+\\angle FEH=90^\\circ=\\angle AEG+\\angle GAE$,\\\\\ntherefore $\\angle GAE=\\angle FEH$, similarly $\\angle AFE=\\angle HEF$,\\\\\nthus $\\angle GAE=\\angle FEH=\\angle AFE$,\\\\\nthus $\\tan\\angle GAE=\\tan\\angle FEH=\\tan\\angle AFE$,\\\\\nthus $\\frac{EG}{AG}=\\frac{FH}{EH}=\\frac{AE}{EF}=\\frac{1}{2}$,\\\\\nLet $GE=x$, then $FH=AG=2x$, $EH=4x$,\\\\\nthus $GH=AF=AC=5x$, then $5x=\\sqrt{5}$, so $x=\\frac{\\sqrt{5}}{5}$,\\\\\nthus $AG=\\frac{2\\sqrt{5}}{5}=FH$, $EH=\\frac{4\\sqrt{5}}{5}$,\\\\\nthus $BG=\\sqrt{AB^{2}-AG^{2}}=\\frac{\\sqrt{5}}{5}$, and $BD=AD=CD=\\frac{\\sqrt{5}}{2}$,\\\\\nthus $GD=\\sqrt{AD^{2}-AG^{2}}=\\frac{3\\sqrt{5}}{10}$,\\\\\nsince $AF\\parallel BD$, thus $\\angle NAC=\\angle ADG$,\\\\\nthus $\\cos\\angle ADG=\\frac{DG}{AD}=\\frac{3}{5}=\\cos\\angle NAC=\\frac{AN}{AC}$,\\\\\nthus $AN=\\frac{3\\sqrt{5}}{5}$, $NF=\\frac{2\\sqrt{5}}{5}$,\\\\\nthus $NC=\\sqrt{AC^{2}-AN^{2}}=\\frac{4\\sqrt{5}}{5}$, $CF=\\sqrt{NF^{2}+CN^{2}}=2$,\\\\\nwhile $GH=AF=\\sqrt{5}$, $DG=\\frac{3\\sqrt{5}}{10}$,\\\\\nthus $DH=GH-DG=\\frac{7\\sqrt{5}}{10}$,\\\\\nConnect $FD$,\\\\\nthus $DF^{2}=FH^{2}+DH^{2}=\\frac{13}{4}$, $DC^{2}=\\frac{5}{4}$,\\\\\nLet $CM=y$, then $FM=2-y$,\\\\\nfrom $DC^{2}-CM^{2}=DF^{2}-FM^{2}$,\\\\\nsolving for $y$, yields $y=\\frac{1}{2}$, then $FM=2-\\frac{1}{2}=\\frac{3}{2}$,\\\\\nthus $DM=\\sqrt{DC^{2}-CM^{2}}=1$,\\\\\nthus $\\tan\\angle CFD=\\frac{DM}{FM}=\\boxed{\\frac{2}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51051524.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\triangle ABC$, $ \\angle A=55^\\circ$, $ \\angle B=45^\\circ$, points $ D$ and $ E$ are on the sides $ AB$ and $ AC$ of $ \\triangle ABC$, respectively, and $ \\angle ADE=80^\\circ$. $ \\triangle AED\\sim \\triangle ABC$. If $ AD=4$, $ BD=6$, and $ AE=5$, find the length of $ CE$.\n",
    "completion": "\\textbf{Solution:} From (1) we have $\\triangle AED\\sim \\triangle ABC$,\\\\\n$\\therefore \\frac{AD}{AC}=\\frac{AE}{AB}$,\\\\\n$\\because AD=4$, $BD=6$,\\\\\n$\\therefore AB=10$,\\\\\n$\\because AD=4$, $AB=10$, $AE=5$,\\\\\n$\\therefore AC=8$,\\\\\n$\\therefore CE=AC-AE=8-5=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53065996_149.png"
  },
  {
    "solution": "\\boxed{-3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the parabola $y=x^2+bx+c$ passes through points A($-3$, $0$) and C($0$, $-3$). What are the values of $b$ and $c$?\n",
    "completion": "\\textbf{Solution}: Substitute $(-3, 0)$ and $(0, -3)$ into $y=x^{2}+bx+c$ to obtain:\\\\\n\\[\n\\begin{cases}\n9-3b+c=0\\\\\nc=-3\n\\end{cases},\n\\]\nSolving this gives:\n\\[\n\\begin{cases}\nb=\\boxed{2}\\\\\nc=\\boxed{-3}\n\\end{cases},\n\\]\nTherefore, the values of $b$ and $c$ are $2$ and $-3$, respectively.",
    "image_path": "dataset/mm_math/images/MM_Math/52622522_218.png"
  },
  {
    "solution": "\\boxed{\\frac{63}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AD$ is the altitude from $A$ to the side $BC$ of $\\triangle ABC$, with $E$ being a point on $AD$, and it's given that $BD=AD$ and $DE=DC$. If $AE=5$ and $CD=2$, what is the area of $\\triangle ABC$?\n",
    "completion": "\\textbf{Solution:} Given that $Rt\\triangle BDE\\cong Rt\\triangle ADC$ and $CD=2$,\nit follows that $DE=CD=2$ and $BD=AD$.\nSince $AE=5$,\nwe have $AD=AE+DE=5+2=7$.\nThus, $BC=BD+CD=7+2=9$.\nTherefore, the area of $\\triangle ABC$ is $\\frac{1}{2}BC\\cdot AD=\\frac{1}{2}\\times 9\\times 7=\\boxed{\\frac{63}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53044193_89.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB \\parallel EF$, $\\angle ABC = 75^\\circ$, $\\angle CDF = 135^\\circ$. What is the measure of $\\angle BCD$ in degrees?\n",
    "completion": "\\textbf{Solution:} Let point $G$ be the intersection of $BC$ and $EF$, as shown in the diagram below: \\\\\n$\\because$ $AB\\parallel EF$, $\\angle ABC=75^\\circ$, $\\angle CDF=135^\\circ$,\\\\\n$\\therefore$ $\\angle EGC=\\angle ABC=75^\\circ$, $\\angle EDC=180^\\circ-\\angle CDF=180^\\circ-135^\\circ=45^\\circ$,\\\\\nFurthermore, $\\because$ $\\angle EGC=\\angle BCD+\\angle EDC$,\\\\\n$\\therefore$ $\\angle BCD=75^\\circ-45^\\circ=\\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50982437.png"
  },
  {
    "solution": "\\boxed{\\sqrt{2}+1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the diameter $AB=2$ of the circle $\\odot O$, and $C$ is a moving point on $\\odot O$. Connect $CB$ and rotate $CB$ $90^\\circ$ counterclockwise around point $C$ to get $CD$. Connect $OD$. What is the maximum value of $OD$?\n",
    "completion": "\\textbf{Solution:} Construct line EO$\\perp$AB intersecting circle $\\odot$O at point E, with O being the perpendicular foot. Connect BE, ED, OC, and BD. \\\\\nIn circle $\\odot$O, OB=OE, $\\angle BOE=90^\\circ$, \\\\\n$\\therefore$ $BE^2=OB^2+OE^2=2OB^2$, \\\\\n$\\therefore$ $BE=\\sqrt{2}OB$, \\\\\nIn $\\triangle BCD$, BC=CD, $\\angle BCD=90^\\circ$, \\\\\n$\\therefore$ $BD=\\sqrt{2}BC$, \\\\\n$\\therefore$ $\\frac{BE}{OB}=\\frac{BD}{BC}=\\sqrt{2}$, \\\\\n$\\because$ $\\angle OBE= \\angle CBD= 45^\\circ$, \\\\\n$\\therefore$ $\\angle OBE+\\angle EBC = \\angle EBC +\\angle CBD$, that is, $\\angle OBC= \\angle EBD$, \\\\\n$\\therefore$ $\\triangle OBC\\sim \\triangle EBD$, \\\\\n$\\therefore$ $\\frac{DE}{OC}=\\frac{BE}{OB}=\\sqrt{2}$, \\\\\n$\\because$ $OC=\\frac{1}{2}AB=1$, \\\\\n$\\therefore$ $DE=\\sqrt{2}$, \\\\\nHence, point D traces a circle centered at E with radius $\\sqrt{2}$. The maximum length of OD, when points O, E, and D are collinear, is $OD=DE+OE=\\sqrt{2}+1$. \\\\\nTherefore, the answer is: $\\boxed{\\sqrt{2}+1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52872179.png"
  },
  {
    "solution": "\\boxed{13}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, a goalkeeper kicks a high ball from point $O$ on the soccer field. The ball takes off from point $A$, which is 1 meter above the ground and located on the $y$-axis, and reaches its highest point $M$ directly above player B, who is 6 meters away from point $O$, at approximately 4 meters above the ground. After the ball lands, it bounces back into the air. Experimental measurements have shown that the trajectory of the ball after bouncing forms a parabola that is identical in shape to its initial trajectory, but its maximum height is reduced to half of the original maximum height. How many meters away from the goalkeeper is the first landing point $C$ of the ball?\n",
    "completion": "Solution: Let $y=0$, then $-\\frac{1}{12}(x-6)^2+4=0$. \\\\\n$\\therefore (x-6)^2=48$. The solutions are ${x}_1=4\\sqrt{3}+6\\approx 13$ and ${x}_2=-4\\sqrt{3}+6<0$ (discard the latter). \\\\\n$\\therefore$ The first landing point $C$ of the football is approximately $\\boxed{13}$ meters away from the goalkeeper.",
    "image_path": "dataset/mm_math/images/MM_Math/52977497_175.png"
  },
  {
    "solution": "\\boxed{\\frac{-1+\\sqrt{5}}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the side length of the square $ABCD$ is $1$, the area of the square $CEFG$ is $S_1$, point $E$ is on side $CD$, and point $G$ is on the extension of side $BC$. Suppose the area of the rectangle with sides $AD$ and $DE$ as adjacent edges is $S_2$ and $S_1=S_2$. What is the length of the line segment $CE$?\n",
    "completion": "\\textbf{Solution:} Let the side length of the square $CEFG$ be $a$,\\\\\nsince the side length of the square $ABCD$ is $1$, therefore $DE=1-a$, \\\\\nsince $S_1=S_2$, therefore $a^2=1\\times(1-a)$, \\\\\ntherefore $a=\\frac{-1+\\sqrt{5}}{2}$ or $a=\\frac{-1-\\sqrt{5}}{2}$ (discard), \\\\\ntherefore $CE=\\boxed{\\frac{-1+\\sqrt{5}}{2}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51614632_103.png"
  },
  {
    "solution": "\\boxed{\\frac{37}{17}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given the lines $y_{1}=x$, $y_{2}=\\frac{1}{3}x+1$, $y_{3}=-\\frac{4}{5}x+5$ as shown in the figure, if for any value of $x$, $y$ always takes the minimum value among $y_{1}$, $y_{2}$, and $y_{3}$, what is the maximum value of $y$?\n",
    "completion": "\\textbf{Solution:}\n\nAs shown in the figure, we find the coordinates of the intersection points of $y_{1}$, $y_{2}$, and $y_{3}$ as $A\\left(\\frac{3}{2},\\frac{3}{2}\\right)$; $B\\left(\\frac{25}{9},\\frac{25}{9}\\right)$; $C\\left(\\frac{60}{17},\\frac{37}{17}\\right)$\n\nWhen $x<\\frac{3}{2}$, $y=y_{1}$\n\nWhen $\\frac{3}{2}\\le x<\\frac{25}{9}$, $y=y_{2}$\n\nWhen $\\frac{25}{9}\\le x<\\frac{60}{17}$, $y=y_{2}$\n\nWhen $x\\ge \\frac{60}{17}$, $y=y_{3}$\n\nSince $y$ always takes the minimum value among $y_{1}$, $y_{2}$, and $y_{3}$,\n\nTherefore, the maximum value of $y$ is $\\boxed{\\frac{37}{17}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51662879.png"
  },
  {
    "solution": "\\boxed{37.5^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: When two acute angle vertices of a set of triangular boards coincide, $\\angle AOB=45^\\circ$, $\\angle COD=30^\\circ$, $OM$ and $ON$ are the angle bisectors of $\\angle AOC$ and $\\angle BOD$ respectively. As shown in Figure 1, when $OB$ coincides with $OC$, what is the magnitude of $\\angle MON$?\n",
    "completion": "\\textbf{Solution:} We have $\\angle MON = \\boxed{37.5^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53195491_128.png"
  },
  {
    "solution": "\\boxed{80^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $\\triangle ABC$ with $AB=AC$ and $\\angle A=70^\\circ$. Points $O$ and $D$ are the midpoints of $BC$ and $AB$, respectively. A circle is drawn with $O$ as the center and $OD$ as the radius, intersecting $AB$ at another point $E$ and intersecting $AC$ at points $G$ and $F$. What is the degree measure of $\\angle DOE + \\angle FOG$?\n",
    "completion": "\\textbf{Solution:} Since $O$ and $D$ are the midpoints of $BC$ and $AB$ respectively, and $AB=AC$,\\\\\nit follows that $OD\\parallel AC$, and $OG=OD= \\frac{1}{2}AC= \\frac{1}{2}AB$,\\\\\ntherefore, $\\angle ODE=\\angle A=70^\\circ$, and $OG\\parallel AB$,\\\\\nthus, $\\angle OGF=\\angle A=70^\\circ$,\\\\\nsince $OE=OD=OG=OF$,\\\\\nit implies that $\\angle ODE=\\angle OED=70^\\circ$, $\\angle OFG=\\angle OGF=70^\\circ$,\\\\\nthus, $\\angle DOD=\\angle GOF=40^\\circ$,\\\\\nhence, $\\angle DOE+\\angle FOG=\\boxed{80^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51209219.png"
  },
  {
    "solution": "\\boxed{\\frac{54}{5}\\pi}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the regular pentagon ABCDE has a side length of 6. A circle is drawn with A as the center and AB as the radius. What is the area of the shaded portion in the figure?\n",
    "completion": "\\textbf{Solution:} Given that pentagon ABCDE is a regular pentagon with side length 6,\\\\\nit follows that $AB=AE=6$ and $\\angle BAE=\\frac{180^\\circ \\times (5−2)}{5}=108^\\circ$,\\\\\nthus, the area of the shaded region in the diagram is $\\frac{108\\pi \\times {6}^{2}}{360}=\\boxed{\\frac{54}{5}\\pi}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51445522.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\triangle ABC$, $AD$ is the median to side $BC$, and point $E$ is on $AD$ with $ \\frac{DE}{AE}=\\frac{1}{3}$, ray $CE$ intersects $AB$ at point $F$. Find the value of $ \\frac{AF}{FB}$.\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $DM\\parallel AB$ through $D$, intersecting $CF$ at $M$,\\\\\n$\\therefore \\triangle DME\\sim \\triangle AFE$, $\\triangle CDM\\sim \\triangle CBF$,\\\\\n$\\therefore \\frac{DE}{AE}=\\frac{DM}{AF}=\\frac{1}{3}$, $\\frac{CD}{CB}=\\frac{DM}{BF}$,\\\\\n$\\because D$ is the midpoint of $BC$, $\\therefore \\frac{CD}{CB}=\\frac{DM}{BF}=\\frac{1}{2}$,\\\\\n$\\therefore \\frac{AF}{BF}=\\boxed{\\frac{3}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51213332.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the graph of the linear function $y=kx+k+1$ intersects with the graph of the linear function $y=-x+4$ at point $A(1, a)$. Considering the graph, when $x>2$, find the range of values of $y$ for the linear function $y=-x+4$.\n",
    "completion": "\\textbf{Solution:} When $x=2$, $y=-x+4=-2+4=2$,\\\\\nsince $k=-1<0$,\\\\\nthus, according to this function, as $x$ increases, it decreases,\\\\\ntherefore, when $x>2$, $y<2$.\\\\\nHence, the range of the function value $y$ is $y\\in(-\\infty, \\boxed{2})$.",
    "image_path": "dataset/mm_math/images/MM_Math/52345530_208.png"
  },
  {
    "solution": "\\boxed{\\frac{8\\sqrt{2}}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $AB$ is the diameter of the circle $O$, with $AB=6$, and $\\sin C= \\frac{1}{3}$. A line is drawn through point $D$ such that $DE \\perp AB$, with $E$ as the foot of the perpendicular, and the extension of $DE$ intersects circle $O$ at point $F$. Find the length of $DF$.\n",
    "completion": "\\textbf{Solution:} Given that $AB$ is the diameter and $DE \\perp AB$, it follows that $DF = 2DE$. Connect $DO$ and let $OE = x$. Thus, we have $3^{2} - x^{2} = 2^{2} - (3 - x)^{2}$. This implies $x = \\frac{7}{3}$. By the Pythagorean theorem, we obtain $DE = \\frac{4\\sqrt{2}}{3}$. Therefore, $DF = \\boxed{\\frac{8\\sqrt{2}}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51299813_124.png"
  },
  {
    "solution": "\\boxed{4 \\colon 9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, quadrilateral $ABCD$ and quadrilateral $AEFG$ are similar figures with point A as the center of similarity, and $AC: AF = 2:3$, then what is the ratio of the area of quadrilateral $ABCD$ to the area of quadrilateral $AEFG$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral $ABCD$ and quadrilateral $AEFG$ are similar figures with point $A$ as the center of similitude and the ratio of $AC \\colon AF = 2 \\colon 3$, \\\\\ntherefore, the similarity ratio between quadrilateral $ABCD$ and quadrilateral $AEFG$ is $2 \\colon 3$, \\\\\ntherefore, the ratio of the areas of quadrilateral $ABCD$ to quadrilateral $AEFG$ is $\\boxed{4 \\colon 9}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51458235.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $ \\angle ACD=90^\\circ$, $ \\angle D=15^\\circ$, point B is on the perpendicular bisector of $ AD$, if $ AC=4$, then what is the length of $ AB$?\n",
    "completion": "\\textbf{Solution:} Since point B is on the perpendicular bisector of $AD$,\\\\\nit follows that $AB=BD$,\\\\\nthus, $\\angle BAD=\\angle D=15^\\circ$,\\\\\ntherefore, $\\angle ABC=\\angle BAD+\\angle D=30^\\circ$,\\\\\nsince $\\angle ACD=90^\\circ$ and $AC=4$,\\\\\nit follows that $AB=2AC=\\boxed{8}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52978596.png"
  },
  {
    "solution": "\\boxed{\\frac{15}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the linear function $y=kx+b$ intersects with the graph of the inverse proportion function $y=\\frac{m}{x}\\ (x>0)$ at points $A(1, 4)$ and $B(4, n)$. Find the area of $\\triangle AOB$.\n",
    "completion": "Solution: As shown in the figure, let the graph of the linear function intersect the coordinate axis at points C and D, and draw AE perpendicular to the y-axis at E and BF perpendicular to the x-axis at F, \\\\\n$\\because A\\left(1, 4\\right)$ , $B(4, 1)$,\\\\\n$\\therefore$AE=BF=1,\\\\\n$\\because$ the equation of the linear function is: $y=-x+5$,\\\\\n$\\therefore$OC=OD=5,\\\\\n$\\therefore$ $ {S}_{\\triangle AOC}={S}_{\\triangle BOD}=\\frac{1}{2}OA\\times AE=\\frac{5}{2}$,\\\\\n$ {S}_{\\triangle COD}=\\frac{1}{2}OC\\times OD=\\frac{25}{2}$,\\\\\n$\\therefore$ $ {S}_{\\triangle AOB}={S}_{\\triangle COD}-{S}_{\\triangle AOC}-{S}_{\\triangle BOD}=\\frac{25}{2}-\\frac{5}{2}-\\frac{5}{2}=\\boxed{\\frac{15}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52205237_121.png"
  },
  {
    "solution": "\\boxed{\\frac{20}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, $AC=5$, $D$ is the midpoint of $AB$, $BE\\perp CD$ with foot $E$. If $BE=4$, then what is the length of $BC$?\n",
    "completion": "\\textbf{Solution:} Let CF$\\perp$AB at F, \\\\\nsince $\\angle$ACB=$90^\\circ$ and D is the midpoint of AB, \\\\\ntherefore CD=$\\frac{1}{2}$AB=BD, \\\\\nin $\\triangle CDF$ and $\\triangle BDE$, \\\\\n$\\left\\{\\begin{array}{l}\n\\angle CDF=\\angle BDE \\\\\n\\angle CFD=\\angle BED=90^\\circ \\\\\nCD=BD\n\\end{array}\\right.$, \\\\\ntherefore $\\triangle CDF \\cong \\triangle BDE$ (AAS), \\\\\ntherefore CF=BE=4, \\\\\nin $\\triangle ACF$, AF=$\\sqrt{AC^{2}-CF^{2}}=\\sqrt{5^{2}-4^{2}}=3$, \\\\\nlet BF=x, then AB=x+3, \\\\\nby Pythagoras' theorem, BC$^{2}$=BF$^{2}$+CF$^{2}$, \\\\\nthus, x$^{2}$+4$^{2}$=$(x+3)^{2}-5^{2}$, \\\\\nsolving this, x=$\\frac{16}{3}$, \\\\\ntherefore BC$=\\sqrt{CF^{2}+BF^{2}}=\\boxed{\\frac{20}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52933309.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given: As shown in the figure, in the circle $\\odot O$ with a radius of 4, $AB$ and $CD$ are two diameters, $M$ is the midpoint of $OB$, the extension line of $CM$ intersects $\\odot O$ at point $E$, and $EM>MC$. Connect $DE$, where $DE=\\sqrt{15}$. It is known that $AM\\cdot MB=EM\\cdot MC$; find the length of $EM$.\n",
    "completion": "\\textbf{Solution}: From the given problem, we know that\\\\\nsince $DC$ is the diameter of the circle $O$,\\\\\nthus $\\angle DEC = 90^\\circ$,\\\\\ntherefore $DE^2 + EC^2 = DC^2$,\\\\\nsince $DE = \\sqrt{15}$ and $CD = 8$, and $EC$ is a positive number,\\\\\nthus $EC = 7$,\\\\\nsince $M$ is the midpoint of $OB$,\\\\\nthus $BM = 2$, $AM = 6$,\\\\\nsince $AM \\cdot BM = EM \\cdot CM = EM(EC - EM) = EM(7 - EM) = 12$, and $EM > MC$,\\\\\ntherefore $EM = \\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53066251_85.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given that $AB$ is the diameter of circle $O$, and $\\angle ACD$ is the angle subtended by the arc $\\overset{\\frown}{AD}$ on the circumference of the circle with $\\angle ACD=30^\\circ$. What is the measure of $\\angle DAB$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, connect BD,\\\\\n$\\because \\angle ACD=30^\\circ$,\\\\\n$\\therefore \\angle B=\\angle ACD=30^\\circ$,\\\\\n$\\because AB$ is the diameter of $\\odot O$,\\\\\n$\\therefore \\angle ADB=90^\\circ$,\\\\\n$\\therefore \\angle DAB=90^\\circ-\\angle B= \\boxed{60^\\circ};$",
    "image_path": "dataset/mm_math/images/MM_Math/51073825_133.png"
  },
  {
    "solution": "\\boxed{32}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the diagonals AC and BD of the rhombus $ABCD$ intersect at point $O$, and $E$ is the midpoint of $AB$. Line $OE$ is drawn. If $OE = 4$, what is the perimeter of the rhombus $ABCD$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rhombus,\\\\\nit follows that OA=OC, and AB=BC=CD=AD,\\\\\nSince point E is the midpoint of AB,\\\\\nit follows that OE is the median of $\\triangle ABC$,\\\\\nhence BC=2OE=8,\\\\\nTherefore, the perimeter of the rhombus ABCD is $8\\times 4=\\boxed{32}$",
    "image_path": "dataset/mm_math/images/MM_Math/52303773.png"
  },
  {
    "solution": "\\boxed{y=3x+3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the quadrilateral $ABCD$ is a rhombus, with $A(4, 0)$ and $B(0, 3)$. What is the analytical expression of the line passing through points $B$ and $D$?\n",
    "completion": "\\textbf{Solution:} Let the equation of the line passing through points $B$ and $D$ be $y=mx+n$. Substituting $B(0,3)$, $D(-1,0)$ into it, we get: \n\\[\n\\left\\{\n\\begin{array}{l}\nn=3 \\\\\n-m+n=0\n\\end{array}\n\\right.\n\\]\nSolving this, we find:\n\\[\n\\left\\{\n\\begin{array}{l}\nm=3 \\\\\nn=3\n\\end{array}\n\\right.\n\\]\n$\\therefore$ The equation of the line passing through points $B$ and $D$ is \\boxed{y=3x+3}.",
    "image_path": "dataset/mm_math/images/MM_Math/52423104_79.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, in $\\triangle ABC$, $BO\\perp AC$ at point $O$, $AO=BO=3$, $OC=1$. A line is drawn through point $A$ such that $AH\\perp BC$ at point $H$, and it intersects $BO$ at point $P$. Connect $OH$. What is the degree measure of $\\angle AHO$?\n",
    "completion": "\\textbf{Solution:} Draw $OM \\perp CB$ at point M and $ON \\perp HA$ at point N, as shown in Figure 1:\\\\\nIn quadrilateral $OMHN$, $\\angle MON = 360^\\circ - 3 \\times 90^\\circ = 90^\\circ$,\\\\\n$\\therefore \\angle COM = \\angle PON = 90^\\circ - \\angle MOP$.\\\\\nIn $\\triangle COM$ and $\\triangle PON$,\\\\\n$\\left\\{\\begin{array}{l}\n\\angle COM = \\angle PON \\\\\n\\angle OMC = \\angle ONP = 90^\\circ \\\\\nOC = OP\n\\end{array}\\right.$,\\\\\n$\\therefore \\triangle COM \\cong \\triangle PON$ (AAS),\\\\\n$\\therefore OM = ON$.\\\\\nBecause $OM \\perp CB$ and $ON \\perp HA$,\\\\\n$\\therefore HO$ bisects $\\angle CHA$,\\\\\n$\\therefore \\angle AHO = \\frac{1}{2} \\angle AHC = \\boxed{45^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52913848_122.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point C is the midpoint of line segment AB, point D is a point on line segment CB, and point E is the midpoint of line segment DB. Given $AB=20$ and $EB=3$, find the length of line segment DB.\n",
    "completion": "\\textbf{Solution:} Since point E is the midpoint of segment DB, \\\\\ntherefore $DB = 2EB = 2 \\times 3 = \\boxed{6}$;",
    "image_path": "dataset/mm_math/images/MM_Math/46945019_172.png"
  },
  {
    "solution": "\\boxed{72^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the rhombus $ABCD$, $\\angle C = 108^\\circ$, the perpendicular bisector of $AD$ intersects the diagonal $BD$ at point $P$, and the foot of the perpendicular is $E$. Connecting $AP$, what is the measure of $\\angle APB$?\n",
    "completion": "\\textbf{Solution:} Given: Since $ABCD$ is a rhombus,\\\\\ntherefore $\\angle C = \\angle DAB = 108^\\circ$, and $AD = AB$\\\\\ntherefore $\\angle ADP = (180^\\circ - 108^\\circ) \\div 2 = 36^\\circ$,\\\\\nsince the perpendicular bisector of $AD$ intersects diagonal $BD$ at point $P$,\\\\\ntherefore $AP = DP$,\\\\\ntherefore $\\angle ADP = \\angle DAP = 36^\\circ$,\\\\\ntherefore $\\angle APB = \\angle ADP + \\angle DAP = 36^\\circ + 36^\\circ = \\boxed{72^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51614856.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in parallelogram $ABCD$, points $E$ and $F$ are located on sides $AB$ and $BC$ respectively, with $DE=AF$ and $DE \\perp AF$ at point $G$. If $AB=AD$, $\\angle AED=60^\\circ$, $AE=6$, and $BF=2$, what is the length of $DE$?\n",
    "completion": "\\textbf{Solution:} Extend $CB$ to point $H$, such that $BH=AE$, and connect $AH$, as shown in the figure:\\\\\n$\\because$ Quadrilateral $ABCD$ is a parallelogram, and $AB=AD$,\\\\\n$\\therefore$ Quadrilateral $ABCD$ is a rhombus,\\\\\n$\\therefore$ $AD\\parallel BC$,\\\\\n$\\therefore$ $\\angle ABH=\\angle BAD$.\\\\\n$\\because$ $BH=AE$,\\\\\n$\\therefore$ $\\triangle DAE\\cong \\triangle ABH$ (SAS),\\\\\n$\\therefore$ $DE=AH$, $\\angle AHB=\\angle DEA=60^\\circ$,\\\\\n$\\because$ $DE=AF$,\\\\\n$\\therefore$ $AH=AF$,\\\\\n$\\therefore$ $\\triangle AHF$ is an equilateral triangle,\\\\\n$\\therefore$ $AH=HF=BH+BF=AE+BF=6+2=8$,\\\\\n$\\therefore$ $DE=AH=\\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52344940_166.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=ax^2+bx-3a$ intersects the negative half of the x-axis at point $A(-1, 0)$, intersects the x-axis at another point $B$, and intersects the positive half of the y-axis at point $C(0, 3)$. The axis of symmetry of the parabola intersects line $BC$ at point $M$, and intersects the x-axis at point $G$. Find the analytical expression of the parabola and the axis of symmetry?\n",
    "completion": "Solution: Substituting $A(-1, 0)$ and $C(0, 3)$ into $y=ax^{2}+bx-3a$ yields:\n\\[\n\\left\\{\n\\begin{array}{l}\na-b-3a=0 \\\\\n-3a=3\n\\end{array}\n\\right.\n\\]\nSolving this, we get:\n\\[\n\\left\\{\n\\begin{array}{l}\na=-1 \\\\\nb=2\n\\end{array}\n\\right.\n\\]\nThus, the equation of the parabola is $y=-x^{2}+2x+3$, and its axis of symmetry is $x=-\\frac{b}{2a}=\\boxed{1}$",
    "image_path": "dataset/mm_math/images/MM_Math/52622585_238.png"
  },
  {
    "solution": "\\boxed{\\sqrt{15}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, quadrilateral $ABCD$ is a rectangle, point $E$ is on the extension of segment $CB$, connecting $DE$ intersects $AB$ at point $F$, $\\angle AED = 2\\angle DEC$, $G$ is the midpoint of $DF$, if $BE=1, DF=8$, then what is the length of $AB$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral $ABCD$ is a rectangle,\\\\\nit follows that $\\angle ABC=\\angle BAD=90^\\circ$, $AD\\parallel BC$,\\\\\ntherefore, $\\angle ADG=\\angle DEC$,\\\\\nsince G is the midpoint of $DF$ and $DF=8$,\\\\\nit follows that AG=FG=DG=4,\\\\\nthus, $\\angle GAD=\\angle GDA=\\angle DEC$,\\\\\nsince $\\angle AGE=\\angle GAD+\\angle GDA=2\\angle DEC$ and $\\angle AED=2\\angle DEC$,\\\\\nit follows that $\\angle AEG=\\angle AGE$,\\\\\ntherefore, AE=AG=4,\\\\\nin right triangle $ABE$,\\\\\n$AB=\\sqrt{AE^{2}-EB^{2}}=\\sqrt{4^{2}-1^{2}}=\\sqrt{15}$.\\\\\nTherefore, $AB=\\boxed{\\sqrt{15}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53359559.png"
  },
  {
    "solution": "\\boxed{100^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, line AB and line CD intersect at point O, with OM $\\perp$ AB. If $\\angle 1=40^\\circ$ and $\\angle 2= 30^\\circ$, what is the degree measure of $\\angle$NOD?\n",
    "completion": "\\textbf{Solution:} Since OM$\\perp$AB,\\\\\nit follows that $\\angle$AOM= $90^\\circ$.\\\\\nSince $\\angle$1=$40^\\circ$,\\\\\nit follows that $\\angle$AOC=$\\angle$AOM−$\\angle$1= $90^\\circ−40^\\circ= 50^\\circ$,\\\\\nSince $\\angle$2=$30^\\circ$,\\\\\nit follows that $\\angle$NOD=$\\angle$DOC−$\\angle$AOC−$\\angle$2= $\\boxed{100^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/54927620_133.png"
  },
  {
    "solution": "\\boxed{y=-x^{2}-2x+3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y=ax^2+bx+3$ ($a\\neq 0$) intersects the x-axis at points $A(1,0)$ and $B(-3,0)$ and intersects the y-axis at point $C$. Connect $BC$ and intersect the axis of symmetry of the parabola at point $E$, with the vertex being point $D$. What is the equation of the parabola?\n",
    "completion": "\\textbf{Solution:} Since the parabola $y=ax^{2}+bx+3$ ($a\\neq 0$) intersects the x-axis at points A $(1,0)$ and B $(-3,0)$,\\\\\nit follows that $\\begin{cases} a+b+3=0 \\\\ 9a-3b+3=0 \\end{cases}$,\\\\\nsolving these equations yields $\\begin{cases} a=-1 \\\\ b=-2 \\end{cases}$,\\\\\ntherefore, the equation of the parabola is $\\boxed{y=-x^{2}-2x+3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53059187_104.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $I$ is the centroid of $\\triangle ABC$, with $S_{\\triangle ABC}=9$. If $\\angle BAC$ is translated so that its vertex coincides with point $I$, what is the area of the shaded region in the diagram?\n",
    "completion": "\\textbf{Solution:} Extend AI to meet BC at D, as shown in the figure, \\\\\n$\\because$ point I is the centroid of $\\triangle ABC$, \\\\\n$\\therefore$ AI=2ID, \\\\\n$\\therefore$ ID=$\\frac{1}{3}AD$, \\\\\n$\\because$ angle BAC is translated so that its vertex coincides with point I, \\\\\n$\\therefore$ $\\angle$EIF=$\\angle$BAC, $\\angle$IEF=$\\angle$B, \\\\\n$\\therefore$ $\\triangle IEF\\sim \\triangle ABC$, \\\\\n$\\therefore$ $\\frac{S_{\\triangle IEF}}{S_{\\triangle ABC}}=\\left(\\frac{DI}{DA}\\right)^2=\\left(\\frac{1}{3}\\right)^2=\\frac{1}{9}$, \\\\\n$\\therefore$ $S_{\\triangle IEF}=\\frac{1}{9}S_{\\triangle ABC}=1$, \\\\\n$\\therefore$ the area of the shaded part in the figure $=9-1=\\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52872330.png"
  },
  {
    "solution": "\\boxed{105^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, if a set of triangular boards are stacked as illustrated, what is the degree measure of $\\angle 1$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure,\\\\\n$\\because$ $\\angle ABD=90^\\circ$, $\\angle ABC=45^\\circ$, $\\angle D=60^\\circ$,\\\\\n$\\therefore$ $\\angle CDB=\\angle ABD-\\angle ABC=90^\\circ-45^\\circ=45^\\circ$,\\\\\n$\\because$ $\\angle 1=\\angle D+\\angle CBD$,\\\\\n$\\therefore$ $\\angle 1=60^\\circ+45^\\circ=\\boxed{105^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52861340.png"
  },
  {
    "solution": "\\boxed{4.8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within rectangle $ABCD$, $EF$ is the perpendicular bisector of the diagonal $BD$, intersecting $AD$ and $BC$ at points $E$ and $F$, respectively. Connect $AO$. If $ AO=4$ and $EF=6$, then what is the length of $AB$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect BE, \\\\\nsince EF is the perpendicular bisector of diagonal BD of rectangle ABCD, and AO = 4, \\\\\nthus BD = 2DO = 2AO = 8, BE = DE, $\\angle$DOE = $90^\\circ$, \\\\\ntherefore, DO = 4, \\\\\nsince quadrilateral ABCD is a rectangle, \\\\\nthus AD $\\parallel$ BC, \\\\\nthus $\\angle$EDO = $\\angle$FBO, \\\\\nsince OB = OD, $\\angle$EOD = $\\angle$FOB, \\\\\ntherefore, $\\triangle EOD \\cong \\triangle FOB$ (ASA), \\\\\nthus EO = OF, \\\\\nsince EF = 6, \\\\\nthus EO = 3, \\\\\nlet AE = x, \\\\\nin $Rt\\triangle OBE$, $BE = DE = \\sqrt{3^2 + 4^2} = 5$, \\\\\nin $Rt\\triangle ABD$ and $Rt\\triangle ABE$, $AB^2 = BD^2 - AD^2 = BE^2 - AE^2$, \\\\\nthus $8^2 - (5 + x)^2 = 5^2 - x^2$, \\\\\nsolving this, we get: $x = \\frac{7}{5}$, \\\\\ntherefore, $AB = \\sqrt{5^2 - \\left(\\frac{7}{5}\\right)^2} = \\boxed{4.8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52424415.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, it is known that points D, E, and F are the midpoints of sides BC, AD, and CE, respectively, and the area of $\\triangle ABC$ is $8\\,\\text{cm}^2$. What is the area of the shaded region $\\triangle BEF$ in $\\text{cm}^2$?\n",
    "completion": "\\textbf{Solution:}\n\nSince point D is the midpoint of side BC,\\\\\nit follows that AD is the median of $\\triangle ABC$,\\\\\nhence, the area of $\\triangle ABD$ equals the area of $\\triangle ADC$ equals $\\frac{1}{2}$ the area of $\\triangle ABC$ equals $\\frac{1}{2} \\times 8 = 4$;\\\\\nSince E is the midpoint of side AD,\\\\\nthen BE and CE are the medians,\\\\\nhence, the area of $\\triangle EBD$ equals the area of $\\triangle EDC$ equals $\\frac{1}{2}$ the area of $\\triangle ABD$ equals $\\frac{1}{2} \\times 4 = 2$;\\\\\ntherefore, the area of $\\triangle BEC$ equals $2 + 2 = 4$,\\\\\nSimilarly, it is known that BF is the median of $\\triangle BEC$,\\\\\ntherefore, the area of $\\triangle BEF$ equals $\\frac{1}{2} \\times 4 = \\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52861344.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\triangle ABC$ is an equilateral triangle with a side length of 4. $AD$ is the median on side $BC$, and $F$ is a moving point on segment $AD$. $E$ is a point on side $AC$. If $AE = 2$, when the sum of $EF+CF$ attains its minimum value, what is the measure of $\\angle ECF$?\n",
    "completion": "\\textbf{Solution:} Draw $EM \\parallel BC$ passing through $E$ and intersecting $AD$ at $N$,\\\\\nsince $AC=4$ and $AE=2$,\\\\\nthus $EC=2=AE$,\\\\\nthus $AM=BM=2$,\\\\\nthus $AM=AE$,\\\\\nsince $AD$ is the median of side $BC$, and $\\triangle ABC$ is an equilateral triangle,\\\\\nthus $AD \\perp BC$,\\\\\nsince $EM \\parallel BC$,\\\\\nthus $AD \\perp EM$,\\\\\nsince $AM=AE$,\\\\\nthus $E$ and $M$ are symmetric with respect to $AD$,\\\\\ndraw $CM$ intersecting $AD$ at $F$, and connect $EF$,\\\\\nat this moment, the sum of $EF+CF$ is minimized,\\\\\nsince $\\triangle ABC$ is an equilateral triangle,\\\\\nthus $\\angle ACB=60^\\circ$, and $AC=BC$,\\\\\nsince $AM=BM$,\\\\\nthus $\\angle ECF=\\frac{1}{2}\\angle ACB=30^\\circ$,\\\\\ntherefore, the answer is: $\\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53009847.png"
  },
  {
    "solution": "\\boxed{\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $l\\colon y = \\sqrt{3}x + \\sqrt{3}$ intersects the two coordinate axes at points $A$ and $B$ respectively. Point $M$ is the midpoint of line segment $AB$. If line $l'$ intersects the $x$-axis at point $C$, and the line $MC$ intersects the $y$-axis at point $N$, connect $OM$. Find $\\frac{OM}{ON}$.\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, draw $MD\\perp x$-axis at D. In right triangle $MDO$, by the Pythagorean theorem we have: \n\\[OM=\\sqrt{OD^{2}+DM^{2}}=\\sqrt{\\left(\\frac{1}{2}\\right)^{2}+\\left(\\frac{\\sqrt{3}}{2}\\right)^{2}}=1,\\] \nAlso, let the equation of line $MC$ be $y=ax+b$, \nthen \\[\\left\\{\\begin{array}{l}\n-\\frac{1}{2}a+b=\\frac{\\sqrt{3}}{2}, \\\\\na+b=0.\n\\end{array}\\right.\\] Solving the system of equations, we get \\[\\left\\{\\begin{array}{l}\na=-\\frac{\\sqrt{3}}{3} \\\\\nb=\\frac{\\sqrt{3}}{3}\n\\end{array}\\right.,\\] \n\\(\\therefore\\) The equation of line $MC$ is $y=-\\frac{\\sqrt{3}}{3}x+\\frac{\\sqrt{3}}{3}$, \n\\(\\therefore\\) The intersection of line $MC$ with the y-axis is $N\\left(0, \\frac{\\sqrt{3}}{3}\\right)$, \n\\(\\therefore \\frac{OM}{ON}=1\\div\\frac{\\sqrt{3}}{3}=\\boxed{\\sqrt{3}}.\\)",
    "image_path": "dataset/mm_math/images/MM_Math/52690944_62.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle CAB=90^\\circ$. Three squares are constructed on the sides of $\\triangle ABC$. A line is drawn from point H perpendicular to $ED$ at point J, and $HE$ is connected. Extend $HE$ to intersect $FC$ at point K. If $K$ is the midpoint of $FC$, and the area of $\\triangle ABC$ minus the area of $\\triangle EHJ$ equals 16, then what is the length of $AB$?\n",
    "completion": "\\textbf{Solution:} Extend $HJ$ and $FC$ to intersect at point N.\\\\\n$\\because$ Quadrilateral $EABD$ is a square and $HJ\\perp ED$,\\\\\n$\\therefore$ Quadrilateral $ECNJ$ is a rectangle.\\\\\n$\\therefore$ $CE=NJ$, $EJ=CN$.\\\\\n$CE\\parallel HN$, $ED\\parallel CN$.\\\\\n$\\therefore$ $\\angle ECH=\\angle MHN$.\\\\\n$\\because$ $\\angle ECH+\\angle BCA=90^\\circ$ and $\\angle CBA+\\angle BCA=90^\\circ$,\\\\\n$\\therefore$ $\\angle ECH=\\angle CBA$.\\\\\n$\\therefore$ $\\angle CBA=\\angle MHN$.\\\\\n$\\therefore$ $\\triangle HCN\\cong \\triangle BCA$.\\\\\n$\\therefore$ $CN=CA$, $HN=AB$.\\\\\nLet $CN=CA=a$, $HN=AB=b$.\\\\\n$\\therefore$ $CE=b-a$, $HJ=HN-JN=b-(b-a)=a$.\\\\\n$\\therefore$ $HJ=EJ=a$.\\\\\n$\\therefore$ $\\angle HEJ=45^\\circ$.\\\\\nAlso, $\\because$ $ED\\parallel KN$,\\\\\n$\\therefore$ $\\angle EKC=45^\\circ$.\\\\\n$\\therefore$ $\\triangle EKC$ is an isosceles right triangle.\\\\\n$\\therefore$ $CK=CE=b-a$.\\\\\n$\\because$ K is the midpoint of $CF$, we have $CK=\\frac{1}{2}a$.\\\\\n$\\therefore$ $\\frac{1}{2}a=b-a$.\\\\\n$\\therefore$ $b=\\frac{3}{2}a$.\\\\\n$\\because$ ${S}_{\\triangle ABC}-{S}_{\\triangle EHJ}=16$,\\\\\n$\\therefore$ $\\frac{1}{2}\\times a\\times b-\\frac{1}{2}a^2=16$.\\\\\n$\\therefore$ Solving gives $a=8$.\\\\\n$\\therefore$ $b=\\frac{3}{2}\\times 8=\\boxed{12}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53490767.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB\\parallel CD\\parallel EF$, if $BF=3DF$, then what is the value of $\\frac{AC}{CE}$?\n",
    "completion": "\\textbf{Solution:} Since $BF=3DF$,\\\\\nit follows that $BD=2DF$,\\\\\ndue to $AB\\parallel CD\\parallel EF$,\\\\\nit follows that $\\frac{AC}{CE}=\\frac{BD}{DF}$,\\\\\nthus, $\\frac{AC}{CE}=\\frac{2DF}{DF}=\\boxed{2}$",
    "image_path": "dataset/mm_math/images/MM_Math/51445633.png"
  },
  {
    "solution": "\\boxed{\\frac{10}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC=6$, $BC=4$, point $D$ is on side $AC$, $BD=BC$, then what is the length of $AD$?\n",
    "completion": "\\textbf{Solution}: Given that $AB=AC$, and $BD=BC$,\\\\\nit follows that $\\angle ABC = \\angle C$, and $\\angle C = \\angle BDC$,\\\\\ntherefore $\\triangle ABC \\sim \\triangle BDC$,\\\\\nwhich implies $ \\frac{AB}{BD}=\\frac{BC}{CD}$, \\\\\ngiven $AB=AC=6$, $BC=4$, and $BD=BC$,\\\\\nit follows $ \\frac{6}{4}=\\frac{4}{CD}$, \\\\\nthus $ CD=\\frac{8}{3}$, \\\\\ntherefore $AD=AC-CD=6- \\frac{8}{3}=\\boxed{\\frac{10}{3}}$,",
    "image_path": "dataset/mm_math/images/MM_Math/51403331.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the quadrilateral $ABCD$ is a rhombus with the coordinates of points $A$ and $B$ being $(\\sqrt{3}, 0)$ and $(0, 1)$, respectively. If points $C$ and $D$ are on the coordinate axes, what is the perimeter of the rhombus $ABCD$?\n",
    "completion": "Solution: Since the coordinates of points $A$ and $B$ are $(\\sqrt{3}, 0)$ and $(0,1)$ respectively,\\\\\n$\\therefore OB=1$, $OA=\\sqrt{3}$,\\\\\n$\\because \\angle BOA=90^\\circ$,\\\\\n$\\therefore AB=\\sqrt{OB^2+OA^2}=\\sqrt{1+3}=2$,\\\\\n$\\because$ quadrilateral $ABCD$ is a rhombus,\\\\\n$\\therefore AB=BC=CD=DA=2$,\\\\\n$\\therefore$ the perimeter of rhombus $ABCD$ is equal to $4\\times 2=\\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53183060.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in parallelogram $ABCD$, $AB=4$, $BC=2$, point $F$ is on the extension of $BC$, $AF$ intersects $CD$ at point $E$, and $\\angle 1=\\angle F$, find the lengths of $CF$ and $DE$.\n",
    "completion": "\\textbf{Solution:} Since $\\angle B = \\angle B$ and $\\angle 1 = \\angle F$,\\\\\nit follows that $\\triangle ABC \\sim \\triangle FBA$,\\\\\ntherefore, $\\frac{AB}{BF} = \\frac{BC}{AB}$,\\\\\nthus, BF = $\\frac{4 \\times 4}{2} = 8$,\\\\\nhence, CF = BF - BC = 8 - 2 = \\boxed{6},\\\\\nIn the parallelogram ABCD,\\\\\nthus, AD $\\parallel$ BF, CD = AB = 4, AD = BC = 2,\\\\\ntherefore, $\\triangle ADE \\sim \\triangle FCE$,\\\\\nthus, $\\frac{DE}{CE} = \\frac{AD}{CF}$,\\\\\ntherefore, $\\frac{DE}{4 - DE} = \\frac{2}{6}$,\\\\\nthus, DE = \\boxed{1}.",
    "image_path": "dataset/mm_math/images/MM_Math/52765253.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AC$ is the diameter of $\\odot O$, $PA$ is the tangent of $\\odot O$ with $A$ being the point of tangency, point $B$ is on $\\odot O$, $PA=PB$, chord $AB$ intersects $PC$ at point $M$, if $AP=AC$, $PO$ intersects $AB$ at point $D$, $PC$ intersects $\\odot O$ at point $N$, draw $DN$, then what is the value of $\\frac{DP}{DN}$?\n",
    "completion": "\\textbf{Solution:} We have $\\frac{DP}{DN}=\\boxed{2\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51299739_171.png"
  },
  {
    "solution": "\\boxed{D\\left(-\\frac{3}{2}, \\frac{3}{2}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the parabola $y = ax^2 + bx + 3$ intersects the x-axis at points A and $B(1,0)$, and intersects the y-axis at point C. Line segment $AC$ is drawn, and a moving point D moves on segment $AC$. A perpendicular line to the x-axis is drawn from point D, intersects the parabola at point E, and intersects the x-axis at point F. It is given that $AB = 4$. Let the x-coordinate of point D be m. Lines $AE$ and $CE$ are drawn. When the area of $\\triangle ACE$ is maximized, find the maximum area of $\\triangle ACE$ and the coordinates of point D.\n",
    "completion": "\\textbf{Solution:} Suppose the equation of line AC is $y=k'x+b'$,\n\n$\\therefore \\left\\{\\begin{array}{l}\n-3k'+b'=0 \\\\\nb'=3\n\\end{array}\\right.$, solving this we get: $\\left\\{\\begin{array}{l}\nk'=1 \\\\\nb'=3\n\\end{array}\\right.$,\n\n$\\therefore$ the equation of line AC is $y=x+3$,\n\n$\\therefore D(m, m+3), E(m, -m^2-2m+3)$,\n\n$\\therefore DE=-m^2-3m$,\n\n$\\therefore {S}_{\\triangle ACE}=\\frac{1}{2}\\times 3\\times (-m^2-3m)=-\\frac{3}{2}(m+\\frac{3}{2})^2+\\frac{27}{8}$,\n\n$\\therefore$ when $m=-\\frac{3}{2}$, the maximum value of ${S}_{\\triangle ACE}$ is $\\boxed{\\frac{27}{8}}$,\n\n$\\therefore \\boxed{D\\left(-\\frac{3}{2}, \\frac{3}{2}\\right)}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53107133_272.png"
  },
  {
    "solution": "\\boxed{x<-3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the line $l_1\\colon y=kx+b$ intersects with the line $l_2\\colon y=mx+n$ at point $P(-3, -2)$, what is the solution set of the inequality $mx+n>kx+b$ with respect to $x$?\n",
    "completion": "\\textbf{Solution:} Given that line $l_1\\colon y=kx+b$ intersects with line $l_2\\colon y=mx+n$ at point $P(-3, -2)$.\\\\\nTherefore, the solution set for $mx+n>kx+b$ is $\\boxed{x<-3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50496203.png"
  },
  {
    "solution": "\\boxed{-2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In a Cartesian coordinate system, point \\(O\\) is the origin. The parabola \\(y=ax^2+bx+3\\) passes through point \\(B(-3, 0)\\), point \\(C(1, 0)\\), and intersects the y-axis at point \\(A\\). What are the values of \\(a\\) and \\(b\\)?\n",
    "completion": "Solution: Since the parabola $y=ax^2+bx+3$ passes through points $B(-3, 0)$ and $C(1, 0)$,\n\\[\n\\therefore \n\\begin{cases}\n9a-3b+3=0 \\\\\na+b+3=0\n\\end{cases}\n\\]\n\\[\n\\therefore \n\\begin{cases}\na=-1 \\\\\nb=-2\n\\end{cases}\n\\]\nHence, $a=\\boxed{-1}$, $b=\\boxed{-2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53019865_160.png"
  },
  {
    "solution": "\\boxed{58^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of $\\odot O$, and $P$ is a point on $\\odot O$. If $\\angle PAB=32^\\circ$, what is the degree measure of $\\angle PBA$?\n",
    "completion": "\\textbf{Solution:} Since $AB$ is the diameter of $\\odot O$, \\\\\n$\\therefore \\angle APB=90^\\circ$ \\\\\nSince $\\angle PAB=32^\\circ$, \\\\\n$\\therefore \\angle PBA=\\boxed{58^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53149073.png"
  },
  {
    "solution": "\\boxed{\\frac{20}{\\tan 36^\\circ}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the roof of the workshop is an isosceles triangular (in the shape of a chevron) steel frame with the central column $AD$ (where $D$ is the midpoint of the base) measuring 10 meters in length, and $\\angle B = 36^\\circ$. What is the length of the span $BC$?\n",
    "completion": "\\textbf{Solution:} Given that: $AB=AC$, $BD=CD$, $AD=10$ meters, \\\\\n$\\therefore$ $AD\\perp BC$, \\\\\n$\\because$ $\\angle B=36^\\circ$, \\\\\n$\\therefore$ $BD=\\frac{AD}{\\tan B}=\\frac{10}{\\tan 36^\\circ}$ meters, \\\\\n$\\therefore$ $BC=2BD=\\frac{20}{\\tan 36^\\circ}$ meters. \\\\\nHence, the answer is: $BC=\\boxed{\\frac{20}{\\tan 36^\\circ}}$ meters.",
    "image_path": "dataset/mm_math/images/MM_Math/53043745.png"
  },
  {
    "solution": "\\boxed{1536}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, it is known that $\\angle MON=30^\\circ$, points $A_1$, $A_2$, $A_3$, ... are on ray $ON$, and points $B_1$, $B_2$, $B_3$, ... are on ray $OM$. The triangles $\\triangle A_1B_1A_2$, $\\triangle A_2B_2A_3$, $\\triangle A_3B_3A_4$, ... are all equilateral triangles. If $OA_1=3$, what is the side length of $\\triangle A_{10}B_{10}A_{11}$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle MON=30^\\circ$ and $\\angle B_{1}A_{1}A_{2}=60^\\circ$,\\\\\nit follows that $\\angle OB_{1}A_{1}=\\angle B_{1}A_{1}A_{2}-\\angle MON=30^\\circ$.\\\\\nTherefore, $\\angle MON=\\angle OB_{1}A_{1}$.\\\\\nThus, we have $A_{1}O=B_{1}A_{1}=3$.\\\\\nConsequently, $A_{2}O=A_{1}O+A_{1}A_{2}=3\\times 2$.\\\\\nSimilarly, we can deduce $AO_{3}=A_{3}B_{3}=3\\times 2^{2}$.\\\\\nFollowing the same pattern $A_{4}O=A_{4}B_{4}=3\\times 2^{3}$.\\\\\n$A_{5}O=A_{5}B_{5}=3\\times 2^{4}$.\\\\\n$\\cdots$\\\\\n$A_{n}O=A_{n}B_{n}=3\\times 2^{n-1}$.\\\\\nTherefore, the side length of $\\triangle A_{10}B_{10}A_{11}$ is $3\\times 2^{9}=\\boxed{1536}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52905809.png"
  },
  {
    "solution": "\\boxed{50^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated, it is known that the line $AB$ intersects with $CD$ at point $O$, $OE\\perp CD$, and $\\angle AOC=40^\\circ$, with $OF$ being the angle bisector of $\\angle AOD$. What is the degree measure of $\\angle EOB$?\n",
    "completion": "\\textbf{Solution:} Since $OE \\perp CD$,\\\\\n$\\therefore \\angle EOD=90^\\circ$,\\\\\nand since $\\angle BOD=\\angle AOC=40^\\circ$,\\\\\n$\\therefore \\angle EOB=\\angle EOD-\\angle BOD=50^\\circ$.\\\\\nTherefore, the measure of $\\angle EOB$ is $\\boxed{50^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/46869571_216.png"
  },
  {
    "solution": "\\boxed{\\frac{8}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in a right-angled triangle paper, where $\\angle C=90^\\circ$, $AB=10$, $BC=8$, and $AC=6$, fold the triangle along the line passing through point $B$ such that point $C$ falls on point $E$ on side $AB$, with the crease being $BD$. Find the length of $DE$.\n",
    "completion": "\\textbf{Solution:} Let $CD=DE=x$, then $AD=AC-CD=6-x$. By the property of folding, we know that $\\angle DEB=\\angle C=90^\\circ$, $\\therefore \\angle DEA=90^\\circ$, $\\therefore AD^2=AE^2+DE^2$, $\\therefore (6-x)^2=2^2+x^2$, solving this, we find $x=\\frac{8}{3}$, \\\\\n$\\therefore DE=\\boxed{\\frac{8}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51368158_114.png"
  },
  {
    "solution": "\\boxed{\\sqrt{34}-3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the right triangle $\\triangle ABC$, $\\angle ABC=90^\\circ$, $AB=6$, $BC=5$. $P$ is a moving point inside $\\triangle ABC$, and it satisfies $\\angle PAB=\\angle PBC$. What is the minimum value of the length of the segment $CP$?\n",
    "completion": "\\textbf{Solution:} Given that $\\because$ $\\angle ABC=90^\\circ$ and $\\angle PAB=\\angle PBC$, \\\\\n$\\therefore$ $\\angle PBA+\\angle PBC=90^\\circ$ and $\\angle PBA+\\angle PAB=90^\\circ$, \\\\\n$\\therefore$ $\\angle BPA=90^\\circ$, \\\\\n$\\therefore$ point P lies on the circle with diameter AB, as shown in the figure, with O as the center of the circle. Connect OC, with P being the intersection point of OC and circle O, thus CP represents the minimum value. \\\\\n$\\because$ AB=6, \\\\\n$\\therefore$ OB=OP=3, \\\\\n$\\because$ BC=5, \\\\\n$\\therefore$ OC=$\\sqrt{OB^2+BC^2}=\\sqrt{3^2+5^2}=\\sqrt{34}$, \\\\\n$\\therefore$ CP=$\\sqrt{34}-3$, \\\\\nTherefore, the answer is: $CP=\\boxed{\\sqrt{34}-3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52622379.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the line $y_{1}=2x-2$ intersects the y-axis at point A, and the graph of the line $y_{2}=-2x+6$ intersects the y-axis at point B. The two lines intersect at point C. What is the area of $\\triangle ABC$?\n",
    "completion": "\\textbf{Solution:} Let $x=0$, then $y_1=-2$, and $y_2=6$,\\\\\nTherefore $A(0, -2)$ and $B(0, 6)$.\\\\\nTherefore $AB=8$.\\\\\nTherefore, the area of $\\triangle ABC$ is $\\frac{1}{2} \\times 8 \\times 2 = \\boxed{8}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52423206_71.png"
  },
  {
    "solution": "\\boxed{134}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: According to legend, the ancient Greek mathematician and astronomer Thales once used the principle of similar triangles by erecting a wooden pole at the top of the shadow of a pyramid. With the help of sunlight, he constructed two similar triangles to measure the height of the pyramid. Given that the wooden pole $EF$ is $2m$ long and its shadow $FD$ is $3m$ long, and $OA$ is measured to be $201m$, what is the height of the pyramid $BO$?\n",
    "completion": "\\textbf{Solution:} Let the height of the pyramid be $x$ meters, \\\\\nAccording to the problem, we have: $\\frac{OB}{OA}=\\frac{EF}{FD}$, \\\\\n$\\frac{x}{201}=\\frac{2}{3}$, \\\\\nSolving this equation, we get: $x=\\boxed{134}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52719959.png"
  },
  {
    "solution": "\\boxed{\\frac{48}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the quadrilateral ABCD is a rhombus with $AC=16$ and $DB=12$. If $DH \\perp AB$ at point H, what is the length of $DH$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral ABCD is a rhombus,\\\\\nthus $\\angle AOB = 90^\\circ$, OA = OC = 8, OB = OD = 6,\\\\\nthus AB = $\\sqrt{OA^{2} + OB^{2}}$ = $\\sqrt{8^{2} + 6^{2}}$ = 10,\\\\\nthus, since the area of rhombus $ABCD = AB \\cdot DH = \\frac{1}{2}AC \\times BD$,\\\\\nthus DH = $\\frac{AC \\cdot BD}{2AB}$ = $\\frac{16 \\times 12}{2 \\times 10}$ = $\\boxed{\\frac{48}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51770528.png"
  },
  {
    "solution": "\\boxed{\\frac{2}{7}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Folding the parallelogram-shaped paper ABCD as shown in Figure 1, so that the vertices A, B both fall on point M on the segment HF, and the vertices C, D both fall on point N on the segment HF, where $AD=6$, $AE=x$. If $AE=2$ and $AF$ is connected such that $AF\\perp BC$ (as shown in Figure 2), find the value of $\\frac{BF}{FC}$.\n",
    "completion": "\\textbf{Solution:} Solution Method (1): Given $AE=EM=BE$, \\\\\n$\\therefore AB=2AE=4$, \\\\\n$\\because AF\\perp BC$, $\\therefore EF=\\frac{1}{2}AB=2$, $HG=EF=2$, \\\\\nSimilarly, $CG=DG=2$, \\\\\n$\\therefore CH\\perp AD$, $CH\\perp BC$, $CH=AF$, \\\\\nLet $BF=\\alpha$, then $CF=6-\\alpha$, \\\\\n$\\therefore AF^{2}=AB^{2}-BF^{2}=16-\\alpha^{2}$, $CH^{2}=FH^{2}-FC^{2}=36-(6-\\alpha)^{2}$, \\\\\n$\\therefore 16-\\alpha^{2}=36-(6-\\alpha)^{2}$, \\\\\n$\\therefore \\alpha=\\frac{4}{3}$, $6-\\alpha=\\frac{14}{3}$, $\\therefore \\frac{BF}{CF}=\\boxed{\\frac{2}{7}}$; \\\\\nMethod (2): As shown in the diagram, connect EG intersecting AF at point P, \\\\\nGiven: $AE=EM=BE$, \\\\\n$\\therefore AB=2AE=4$, \\\\\n$\\because AP\\perp BC$, $\\therefore EF=\\frac{1}{2}AB=2$, $EG=FH=6$, \\\\\n$\\therefore FG=4\\sqrt{2}$, \\\\\n$\\because EG\\parallel BC$, $AF\\perp BC$, $\\therefore EG\\perp AF$, \\\\\n$\\because AE=EF$, $\\therefore AP=PF=\\frac{EF\\cdot FG}{EG}=\\frac{4\\sqrt{2}}{3}$, $AF=\\frac{8\\sqrt{2}}{3}$, \\\\\nIn $\\triangle ABF$, $AB^{2}=AF^{2}+BF^{2}$, \\\\\n$\\therefore BF=\\frac{4}{3}$, $FC=\\frac{14}{3}$, $\\therefore \\frac{BF}{FC}=\\boxed{\\frac{2}{7}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52077012_88.png"
  },
  {
    "solution": "\\boxed{(-16,-\\frac{1}{2})}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given: As shown in the figure, the graph of the linear function $y=-2x+10$ intersects with the graph of the hyperbolic function $y=\\frac{k}{x}$ at points $A$ and $B$ (with $A$ to the right of $B$), and the x-coordinate of point $A$ is $4$. On the other branch of the hyperbolic function graph, is there a point $P$ such that $\\triangle PAB$ is a right triangle with $AB$ as one of its legs? If such point $P$ exists, find the coordinates of all points $P$ that meet the conditions; if it does not exist, please explain why.\n",
    "completion": "\\textbf{Solution:} The point exists,\\\\\nReason:\\\\\n(1) If $\\angle BAP=90^\\circ$,\\\\\ndraw $AH\\perp OE$ at H from point A, let the intersection of AP and the x-axis be M, as shown in figure 1,\\\\\nfor $y=-2x+10$,\\\\\nwhen $y=0$, $-2x+10=0$, solving gives $x=5$,\\\\\n$\\therefore$ point E(5,0), $OE=5$.\\\\\n$\\because A(4,2)$,\\\\\n$\\therefore OH=4$, $AH=2$,\\\\\n$\\therefore HE=5-4=1$.\\\\\n$\\because AH\\perp OE$,\\\\\n$\\therefore \\angle AHM=\\angle AHE=90^\\circ$.\\\\\nFurthermore, $\\because \\angle BAP=90^\\circ$,\\\\\n$\\therefore \\angle AME+\\angle AEM=90^\\circ$, $\\angle AME+\\angle MAH=90^\\circ$,\\\\\n$\\therefore \\angle MAH=\\angle AEM$,\\\\\n$\\therefore \\triangle AHM\\sim \\triangle EHA$,\\\\\n$\\therefore \\frac{AH}{EH}=\\frac{MH}{AH}$, i.e., $\\frac{2}{1}=\\frac{MH}{2}$\\\\\n$\\therefore MH=4$,\\\\\n$\\therefore M(0,0)$,\\\\\nwe can set the equation of the line AP as $y=mx$,\\\\\nthen $4m=2$, solving gives $m=\\frac{1}{2}$,\\\\\n$\\therefore$ the equation of line AP is $y=\\frac{1}{2}x$,\\\\\nSolve the system of equations $\\begin{cases}y=\\frac{1}{2}x\\\\ y=\\frac{8}{x}\\end{cases}$ to obtain $\\begin{cases}x=4\\\\ y=2\\end{cases}$, $\\begin{cases}x=-4\\\\ y=-2\\end{cases}$\\\\\n$\\therefore$ the coordinates of point P are $(-4,-2)$.\\\\\n(2) If $\\angle ABP=90^\\circ$,\\\\\nSimilarly, we get: the coordinates of point P are $(-16,-\\frac{1}{2})$.\\\\\nIn summary: The coordinates of point P that meet the conditions are $\\boxed{(-4,-2)}$, $\\boxed{(-16,-\\frac{1}{2})}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53080947_110.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, D and E are respectively the midpoints of sides AB and AC of $\\triangle ABC$. If $DE = \\frac{3}{2}$, then what is the length of BC?\n",
    "completion": "\\textbf{Solution:} Since D and E are the midpoints of sides AB and AC of $\\triangle ABC$, \\\\\nthen DE is the midline of $\\triangle ABC$, \\\\\nhence, BC = 2DE = 2 $\\times$ $\\frac{3}{2} = \\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52251738.png"
  },
  {
    "solution": "\\boxed{28}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $D$ is the midpoint of line segment $AC$, and point $E$ is the midpoint of line segment $BC$, with $BC=\\frac{4}{7}AC$. If $DE=6$, what is the length of $AC$?\n",
    "completion": "\\textbf{Solution:} Given from the problem, $BC=\\frac{4}{7}AC$,\n\n$\\therefore AC=AB+BC=\\frac{7}{4}BC$,\n\n$\\because$ point $D$ is the midpoint of line segment $AC$, and point $E$ is the midpoint of line segment $BC$,\n\n$\\therefore DC=DA=\\frac{1}{2}AC=\\frac{7}{8}BC$, $EC=BE=\\frac{1}{2}BC$,\n\n$\\therefore DE=DC-EC=\\frac{7}{8}BC-\\frac{1}{2}BC=6$,\n\nSolving this, we find $BC=16$,\n\n$\\therefore AC=\\frac{7}{4}\\times 16=\\boxed{28}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52531561_132.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within circle $\\odot O$, chord $AB$ perpendicularly bisects radius $OC$. Find the degree measure of $\\angle C$?\n",
    "completion": "\\textbf{Solution:} Since chord $AB$ perpendicularly bisects radius $OC$,\n\nit follows that $OB=BC$,\n\nand since $OB=OC$,\n\nit follows that $\\triangle OCB$ is an equilateral triangle,\n\nhence, $\\angle C = \\boxed{60^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52977640_147.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $AB$ is the diameter of $\\odot O$, and $BD$ is the tangent of $\\odot O$ with the point of tangency at $B$. Through the point $C$ on $\\odot O$, draw $CD\\parallel AB$, intersecting $BD$ at point $D$. Connect $AC$ and $BC$. If $DC$ is the tangent of $\\odot O$ with the point of tangency at $C$. What are the measures of $\\angle BCD$ and $\\angle DBC$?\n",
    "completion": "\\textbf{Solution:} Given that AB is the diameter of circle O, and BD is a tangent to circle O at point B,\\\\\n\\[\\therefore \\text{DB} \\perp \\text{AB},\\]\n\\[\\therefore \\angle \\text{DBA} = 90^\\circ,\\]\nGiven that DC is a tangent to circle O at point C,\\\\\n\\[\\therefore \\text{DC} = \\text{DB},\\]\nGiven that CD is parallel to AB,\\\\\n\\[\\therefore \\angle D + \\angle \\text{DBA} = 180^\\circ,\\]\n\\[\\therefore \\angle D = 90^\\circ,\\]\n\\[\\therefore \\angle \\text{BCD} = \\angle \\text{DBC} = \\boxed{45^\\circ};\\]",
    "image_path": "dataset/mm_math/images/MM_Math/52720307_180.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $O$ is a point inside the equilateral triangle $ABC$, with $\\angle BOC = 150^\\circ$. Triangle $BOC$ is rotated counterclockwise around point $C$ to obtain triangle $ADC$. Connect $OD$ and $OA$. What is the degree measure of $\\angle ODC$?\n",
    "completion": "\\textbf{Solution:} From the property of rotation, we have $CD=CO$, $\\angle ACD=\\angle BCO$,\\\\\nsince $\\angle ACB=60^\\circ$,\\\\\nhence $\\angle DCO=60^\\circ$,\\\\\ntherefore, $\\triangle OCD$ is an equilateral triangle,\\\\\nthus $\\angle ODC=\\boxed{60^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51212821_100.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, square $ABCD$ has a side length of $4$, point $M$ is on $AB$ with $AM=1$, and $N$ is a moving point on $BD$. What is the minimum value of $AN+MN$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, we connect MC and CN,\n\nsince the quadrilateral ABCD is a square,\n\nhence A and C are symmetric about BD,\n\ntherefore AN = CN,\n\ntherefore AN + MN = CN + NM $\\geq$ CM,\n\nin triangle BMC,\n\nsince $\\angle$BCM $= 90^\\circ$, BC = 4, BM = AB - AM = 4 - 1 = 3,\n\ntherefore CM = $\\sqrt{MB^{2}+BC^{2}}$ = 5.\n\ntherefore AN + MN $\\geq$ 5,\n\nthus, the minimum value of $AN+MN$ is $\\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52395463.png"
  },
  {
    "solution": "\\boxed{(4,3)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system $xOy$, points A and B are respectively the intersections of the line $y=\\frac{3}{4}x+3$ with the x-axis and the y-axis. Given that point C is in the first quadrant, if quadrilateral $ABCO$ is a parallelogram, what are the coordinates of point C?\n",
    "completion": "\\textbf{Solution:} Let $x=0$, then $y=3$; let $y=0$, then $\\frac{3}{4}x+3=0$, \\\\\n$\\therefore x=-4$, \\\\\n$\\because$ Points A and B are the intersections of the line $y=\\frac{3}{4}x+3$ with the $x$ and $y$ axes, respectively, \\\\\n$\\therefore$ A$(-4,0)$, B$(0,3)$, \\\\\n$\\because$ Point C is in the first quadrant, if quadrilateral ABCO is a parallelogram, \\\\\n$\\therefore$ AB$\\parallel$CD, and AB=CD, \\\\\nBased on coordinates and translations, we can derive that C$(4,3)$. \\\\\nHence, the coordinates of point C is $\\boxed{(4,3)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50446991_226.png"
  },
  {
    "solution": "\\boxed{70^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $AB=AC$, $AD=AE$, $\\angle BAC=\\angle DAE$, $\\angle 1=30^\\circ$, $\\angle 2=40^\\circ$. What is the degree measure of $\\angle 3$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle BAC=\\angle DAE$,\\\\\nit follows that $\\angle BAC-\\angle DAC=\\angle DAE-\\angle DAC$, i.e., $\\angle BAD=\\angle CAE$,\\\\\nthus $\\triangle ABD\\cong \\triangle ACE\\ (SAS)$,\\\\\nhence $\\angle ABD=\\angle 2=40^\\circ$,\\\\\nsince $\\angle 1=30^\\circ$,\\\\\ntherefore $\\angle 3=\\boxed{70^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55502380.png"
  },
  {
    "solution": "\\boxed{\\left(-\\frac{7}{2}, -6\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $AB: y=2x-k$ passes through point $M(k, 2)$ and intersects the $x$-axis and $y$-axis at points $A$ and $B$, respectively. The line $AB$ is translated upward by 3 units to obtain line $l$. If $C$ is a point on line $l$, and the area of $\\triangle AOC=3$, find the coordinates of point $C$.\n",
    "completion": "\\textbf{Solution:} Since the equation of line AB is $y=2x-2$, when $y=0$, $2x-2=0$. Solving this gives $x=1$, hence point A is $(1,0)$, and thus OA$=1$. Translating line $AB$ upwards by 3 units gives line $l$, with the equation $y=2x+1$. Given that $C$ is a point on line $l$, let $C(m,2m+1)$. Since the area of $\\triangle AOC$ is $3$, we have $\\frac{1}{2}\\times 1\\times |2m+1|=3$. Solving this yields $m_1=\\frac{5}{2}$ and $m_2=-\\frac{7}{2}$. When $m_1=\\frac{5}{2}$, $2m+1=6$; and when $m_1=-\\frac{7}{2}$, $2m+1=-6$; hence the coordinates of point $C$ are $\\boxed{\\left(\\frac{5}{2}, 6\\right)}$ or $\\boxed{\\left(-\\frac{7}{2}, -6\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52345458_211.png"
  },
  {
    "solution": "\\boxed{7}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated, let $\\triangle ABC$ be an equilateral triangle. Draw a perpendicular from point $D$ on side $AB$ to $BC$ at point $E$. Construct $EF \\perp DE$, intersecting $AC$ at point $F$. Draw $FG \\perp AC$, meeting $BC$ at point $G$, where $FG$ and $DE$ intersect at point $M$. If $DM = ME$ and $GE = 3$, what is the length of $AB$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw MN $\\perp$ BC at point N, \\\\\n$\\because \\triangle ABC$ is an equilateral triangle, \\\\\n$\\therefore \\angle B=\\angle C=60^\\circ$, \\\\\n$\\because DE\\perp AB$, $EF\\perp DE$, $FG\\perp AC$, \\\\\n$\\therefore \\angle BDE=90^\\circ$, $\\angle DEF=90^\\circ$, $\\angle GFC=90^\\circ$, \\\\\n$\\therefore \\angle DEB=180^\\circ-90^\\circ-60^\\circ=30^\\circ$, $\\angle FGC=180^\\circ-90^\\circ-60^\\circ=30^\\circ$, \\\\\n$\\therefore \\angle FEC=180^\\circ-90^\\circ-30^\\circ=60^\\circ$, $\\angle DEB=\\angle FGC$, \\\\\n$\\therefore GN=EN=\\frac{1}{2}GE=\\frac{3}{2}$, \\\\\nIn $Rt\\triangle MNE$, $\\angle MEN=30^\\circ$, $EN=\\frac{3}{2}$, \\\\\n$\\therefore \\frac{NE}{ME}=\\cos30^\\circ=\\frac{\\sqrt{3}}{2}$, \\\\\n$\\therefore ME=\\frac{\\frac{3}{2}}{\\frac{\\sqrt{3}}{2}}=\\sqrt{3}$, \\\\\n$\\because DM=ME$, \\\\\n$\\therefore DE=2ME=2\\sqrt{3}$, \\\\\nIn $Rt\\triangle BDE$, $\\angle DEB=30^\\circ$, \\\\\n$\\therefore BE=\\frac{DE}{\\frac{\\sqrt{3}}{2}}=\\frac{2\\sqrt{3}}{\\frac{\\sqrt{3}}{2}}=4$, \\\\\n$\\because \\angle C=60^\\circ$, $\\angle FEC=60^\\circ$, \\\\\n$\\therefore \\triangle EFC$ is an equilateral triangle, \\\\\n$\\therefore \\angle EFC=60^\\circ$, $EF=CE$, \\\\\n$\\therefore \\angle GFE=90^\\circ-60^\\circ=30^\\circ=\\angle FGC$, \\\\\n$\\therefore EF=GE=3$, \\\\\n$\\therefore CE=3$, \\\\\n$\\therefore BC=BE+CE=4+3=7$, \\\\\n$\\therefore AB=BC=\\boxed{7}$, \\\\",
    "image_path": "dataset/mm_math/images/MM_Math/52914190.png"
  },
  {
    "solution": "\\boxed{60}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, there is an observation tower $EF$ at the top $E$ of a mountain. It is known that the angle $\\angle EAB$ between the mountain slope and the horizontal plane is $30^\\circ$, the height of the mountain $EB$ is 120 meters, point $C$ is 180 meters away from the base of the mountain $A$, $CD\\parallel AB$, and intersects $EB$ at point $D$. The elevation angle $\\angle FCD$ of the top end $F$ of the observation tower measured at point $C$ is $60^\\circ$. What is the height of the observation tower $EF$ in meters?\n",
    "completion": "\\textbf{Solution:} According to the question, we have: $\\angle ABE=90^\\circ$, \\\\\n$\\because$ $\\angle EAB=30^\\circ$, \\\\\n$\\therefore$ $\\angle AEB=60^\\circ$, $AE=2EB$, \\\\\n$\\because$ $EB=120$, \\\\\n$\\therefore$ $AE=2EB=240$, \\\\\n$\\because$ $AC=180$, \\\\\n$\\therefore$ $CE=AE-AC=60$, \\\\\n$\\because$ $CD\\parallel AB$, \\\\\n$\\therefore$ $\\angle CDE=\\angle ABE=90^\\circ$, $\\angle ECD=\\angle EAB=30^\\circ$, \\\\\n$\\because$ $\\angle FCD=60^\\circ$, $\\angle FCE=\\angle FCD-\\angle ECD=30^\\circ$, \\\\\n$\\therefore$ $\\angle FCE=\\angle CFE=30^\\circ$, \\\\\n$\\therefore$ $EF=CE=60$, \\\\\nthus, the height of the tower is \\boxed{60} meters.",
    "image_path": "dataset/mm_math/images/MM_Math/53066576.png"
  },
  {
    "solution": "\\boxed{\\frac{3\\sqrt{5}}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the right triangle $ABC$, $\\angle ACB=90^\\circ$. A square $ACED$ is constructed on the side $AC$ to the left of $AC$. A line is drawn through point $E$ such that $EF\\perp AB$, with the foot of the perpendicular at point $F$, intersecting $AC$ at point $G$. It is given that $CG=CB$; if point $G$ is the midpoint of $AC$, and the length of $EC$ is 2, find the length of $BF$.\n",
    "completion": "\\textbf{Solution:} Given that point $G$ is the midpoint of $AC$ and the length of $EC$ is $2$,\\\\\n$\\therefore CG=\\frac{1}{2}AC=\\frac{1}{2}EC=1=AG$,\\\\\nFrom (1) we know $CB=CG$,\\\\\n$\\therefore CB=1$,\\\\\n$\\therefore EG=\\sqrt{CE^2+CG^2}=\\sqrt{5}=AB$,\\\\\nLet $BF=x$,\\\\\nIn $\\triangle BFE$, by the Pythagorean theorem: $BF^2+EF^2=BE^2$,\\\\\n$\\therefore x^2+(\\sqrt{5}+GF)^2=3^2$, rearranging: $x^2+5+2\\sqrt{5}GF+GF^2=9$ (1),\\\\\n$\\therefore GF^2=4-x^2-2\\sqrt{5}GF$,\\\\\nIn $\\triangle AGF$, by the Pythagorean theorem: $GF^2+AF^2=AG^2$,\\\\\n$\\therefore GF^2+(\\sqrt{5}-x)^2=1^2$, rearranging: $GF^2+5-2\\sqrt{5}x+x^2=1$ (2),\\\\\n$\\therefore 4-x^2-2\\sqrt{5}GF+5-2\\sqrt{5}x+x^2=1$,\\\\\n$\\therefore GF+x=\\frac{4\\sqrt{5}}{5}$ (3),\\\\\nDraw line $BG$,\\\\\nIn $\\triangle BGF$, by the Pythagorean theorem: $GF^2+BF^2=BG^2$,\\\\\n$\\because BG^2=1^2+1^2=2$,\\\\\n$\\therefore GF^2+x^2=2$,\\\\\n(1) + (2) gives: $x^2+5+2\\sqrt{5}GF+GF^2+GF^2+5-2\\sqrt{5}x+x^2=10$,\\\\\nRearranging gives: $x-GF=\\frac{2\\sqrt{5}}{5}$④,\\\\\n(3) + ④ gives: $2x=\\frac{6\\sqrt{5}}{5}$,\\\\\n$\\therefore x=\\boxed{\\frac{3\\sqrt{5}}{5}}$,\\\\\nTherefore, the length of $BF$ is $\\boxed{\\frac{3\\sqrt{5}}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53359714_120.png"
  },
  {
    "solution": "\\boxed{40^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure of rhombus $ABCD$, if $\\angle ABC=80^\\circ$, then what is the measure of $\\angle BDA$?\n",
    "completion": "\\textbf{Solution:} Since ABCD is a rhombus,\\\\\n$\\therefore$ $\\angle ADC = \\angle ABC = 80^\\circ$, and DB bisects $\\angle ADC$ and $\\angle ABC$,\\\\\n$\\therefore$ $\\angle BDA = \\angle BDC = 40^\\circ$,\\\\\nTherefore, $\\angle BDA = \\boxed{40^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/53359987.png"
  },
  {
    "solution": "\\boxed{M\\left(\\frac{2\\sqrt{3}}{3}, \\frac{7}{3}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, the vertices $A$ and $C$ of the rectangle $OABC$ are on the positive half-axes of $x$ and $y$, respectively, and the coordinates of point $B$ are $(2\\sqrt{3}, 4)$. The graph of the linear function $y=-\\frac{\\sqrt{3}}{3}x+b$ intersects with sides $OC$, $AB$, and the $x$ axis at points $D$, $E$, and $F$, respectively, with $\\angle DFO=30^\\circ$, and satisfying $OD=BE$. Point $M$ is a moving point on the segment $DF$. Connect $OM$. If the ratio of the area of $\\triangle ODM$ to the area of the quadrilateral $OAEM$ is $1:3$, what are the coordinates of point $M$?\n",
    "completion": "\\textbf{Solution:} From (1), we obtain the linear function $y=-\\frac{\\sqrt{3}}{3}x+3$, with points $D(0, 3)$ and $E(2\\sqrt{3}, 1)$,\\\\\n$\\therefore OD=3$, $AE=1$, $OA=2\\sqrt{3}$,\\\\\n$\\therefore S_{\\text{quadrilateral }OADE}=\\frac{1}{2}(OD+AE)\\cdot OA=\\frac{1}{2}\\times (3+1)\\times 2\\sqrt{3}=4\\sqrt{3}$,\\\\\n$\\because$ the area ratio of $\\triangle ODM$ to the quadrilateral $OAEM$ is $1:3$,\\\\\n$\\therefore$ the area ratio of $\\triangle ODM$ to the quadrilateral $OADE$ is $1:4$,\\\\\n$\\therefore S_{\\triangle ODM}=\\frac{1}{4}S_{\\text{quadrilateral }OADE}=\\sqrt{3}$,\\\\\nLet the x-coordinate of point $M$ be $a$, hence $\\frac{1}{2}\\times 3a=\\sqrt{3}$,\\\\\nSolving this yields $a=\\frac{2\\sqrt{3}}{3}$,\\\\\nSubstituting $x=\\frac{2\\sqrt{3}}{3}$ into $y=-\\frac{\\sqrt{3}}{3}x+3$ gives $y=\\frac{7}{3}$,\\\\\n$\\therefore \\boxed{M\\left(\\frac{2\\sqrt{3}}{3}, \\frac{7}{3}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51152292_113.png"
  },
  {
    "solution": "\\boxed{5\\text{cm}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, construct squares outwardly using the two right-angle sides of the right-angled triangle $ABC$, with their areas being $S_1$ and $S_2$ respectively. If $S_1=8\\text{cm}^2$ and $S_2=17\\text{cm}^2$, what is the length of the hypotenuse $AB$?\n",
    "completion": "\\textbf{Solution:} Let $S_{1}=8\\text{cm}^{2}$, and $S_{2}=17\\text{cm}^{2}$,\\\\\nTherefore, $BC^{2}=8$, and $AC^{2}=17$,\\\\\nBecause $\\angle ACB=90^\\circ$,\\\\\nThus $AB^{2}=BC^{2}+AC^{2}$, that is, $AB^{2}=8+17=25$,\\\\\nTherefore, $AB=\\boxed{5\\text{cm}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52189159.png"
  },
  {
    "solution": "\\boxed{-2, -6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the graph of the line $y_1=2x-2$ intersects the y-axis at point A, and the graph of the line $y_2=-2x+6$ intersects the y-axis at point B. The two lines intersect at point C. There exists another point $P$ on the graph of the line $y_1=2x-2$, different from point $C$, such that the area of $\\triangle ABC$ is equal to the area of $\\triangle ABP$. Find the coordinates of point $P$.\n",
    "completion": "\\[ \\textbf{Solution:} \\text{ Let } P(x_0, 2x_0-2), \\text{ then } S_{\\triangle ABP}=\\frac{1}{2}\\times 8\\times |x_0|=8, \\]\n\\[ \\therefore x_0=\\pm 2. \\]\n\\[ \\because \\text{ Point } P \\text{ is different from point } C, \\]\n\\[ \\therefore x_0=-2, 2x_0-2=-6. \\]\n\\[ \\therefore P(\\boxed{-2, -6}). \\]",
    "image_path": "dataset/mm_math/images/MM_Math/52423206_72.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, in the rhombus ABCD, AC is a diagonal, and AB = AC = 6. Points E and F are moving points on sides AB and BC, respectively, satisfying AE = BF. Lines AF and CE intersect at point G. Find the measure of $\\angle CGF$.\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral ABCD is a rhombus, and AB=AC, \\\\\n$\\therefore$ AB=AC=BC=AD=CD, \\\\\n$\\therefore$ $\\triangle ABC$, $\\triangle ACD$ are equilateral triangles. \\\\\n$\\therefore$ $\\angle ABC=\\angle CAE$, in $\\triangle ABF$ and $\\triangle CAE$, \\\\\n$\\left\\{\\begin{array}{l}\nAB=CA\\\\\n\\angle ABF=\\angle CAE\\\\\nBF=AE\n\\end{array}\\right.$ \\\\\n$\\therefore$ $\\triangle ABF\\cong \\triangle CAE$ (SAS), \\\\\n$\\therefore$ $\\angle BAF=\\angle ACE$, \\\\\n$\\therefore$ $\\angle CGF=\\angle GAC+\\angle ACG=\\angle GAC+\\angle BAF=\\angle BAC=60^\\circ$; \\\\\nHence, the measure of $\\angle CGF$ is $\\boxed{60^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51152250_119.png"
  },
  {
    "solution": "\\boxed{\\frac{6}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points B and C are on $y=2x$ and $y=kx-2a$, respectively. Points A and D are two points on the x-axis, and the y-coordinate of point B is $a$. If quadrilateral ABCD is a rectangle and $AB=\\frac{1}{2}AD$, then what is the value of $k$?\n",
    "completion": "\\textbf{Solution:}\n\nGiven that point B is on the line $y = 2x$ and the y-coordinate of point B is $a$, \n\nit follows that $2x = a$ and $AB = a$,\n\nfrom which we find $x = \\frac{a}{2}$,\n\nand since $AB = \\frac{1}{2}AD$,\n\nit follows that $AD = 2a$,\n\nthus $OD = OA + AD = \\frac{a}{2} + 2a = \\frac{5a}{2}$,\n\ntherefore $C\\left(\\frac{5a}{2}, a\\right)$,\n\nhence $\\frac{5a}{2}k - 2a = a$,\n\nsolving this gives $k = \\boxed{\\frac{6}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52354235.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in rectangle $ABCD$, $AB=4$, $AD=10$, point $E$ is on side $AD$, it is known that points $B$ and $E$ are symmetrical with respect to line $l$, line $l$ intersects sides $AD$ and $BC$ at points $M$ and $N$, respectively. Connect $BM$ and $NE$. Quadrilateral $BMEN$ is a rhombus. If $DE=2$, what is the length of $NC$?\n",
    "completion": "\\textbf{Solution:} Let the side length of the rhombus be $x$, then $AM=8-x$. In $\\triangle ABM$, we have $4^2+(8-x)^2=x^2$. Solving for $x$, we obtain $x=5$. $\\therefore NC=\\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51614835_88.png"
  },
  {
    "solution": "\\boxed{180^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the quadrilateral $ABCD$, $\\angle BAD=121^\\circ$, $\\angle B=\\angle D=90^\\circ$, and points $M,N$ are on $BC,CD$ respectively. When the perimeter of $\\triangle AMN$ is minimized, what is the sum of $\\angle AMN+\\angle ANM$ in degrees?\n",
    "completion": "\\textbf{Solution:} It is known from the problem statement that when the perimeter of $\\triangle AMN$ is minimized, $\\angle AMN + \\angle ANM = 180^\\circ - \\angle MNA$.\\\\\nSince $\\angle MNA$ is the external angle of $\\triangle AMN$,\\\\\n$\\therefore \\angle AMN + \\angle ANM = 180^\\circ - \\angle MNA = \\boxed{180^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52653516_89.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AD$ is tangent to the circle $\\odot O$ at point $D$, and point $A$ is on the extension of the diameter $CB$. $\\angle DCB = \\angle ADB$; If $\\angle DCB = 30^\\circ$ and $AC = 3\\sqrt{3}$, find the length of $AD$.\n",
    "completion": "\\textbf{Solution:} Since $\\because \\angle DCB = \\angle ADB$ and $\\angle DAC = \\angle CAD$,\\\\\n$\\therefore \\triangle ADB \\sim \\triangle ACD$,\\\\\n$\\therefore \\frac{DB}{CD} = \\frac{AD}{AC}$,\\\\\nSince $CB$ is the diameter,\\\\\n$\\therefore \\angle CDB = 90^\\circ$, $\\angle DCB = 30^\\circ$,\\\\\n$\\therefore \\tan \\angle DCB = \\frac{DB}{CD} = \\frac{\\sqrt{3}}{3}$,\\\\\n$\\therefore \\frac{AD}{AC} = \\frac{\\sqrt{3}}{3}$,\\\\\nSince $AC = 3\\sqrt{3}$,\\\\\n$\\therefore AD = \\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52536497_100.png"
  },
  {
    "solution": "\\boxed{y=-2x+6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system, the line $y=kx+b(k\\neq 0)$ intersects the line $y=x$ at point $A(2, a)$ and intersects the y-axis at point $B(0, 6)$. Find the analytical expression of the function corresponding to the line $y=kx+b$.\n",
    "completion": "\\textbf{Solution:} Substituting point A (2, a) into $y=x$, \\\\\nwe get $a=2$, \\\\\n$\\therefore$ A (2, 2), \\\\\nSubstituting points A (2, 2), B (0, 6) into $y=kx+b$, \\\\\nwe obtain $\\begin{cases}2k+b=2\\\\ b=6\\end{cases}$, \\\\\nSolving this, we find $\\begin{cases}k=-2\\\\ b=6\\end{cases}$, \\\\\n$\\therefore$ the equation of the line is $\\boxed{y=-2x+6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53182647.png"
  },
  {
    "solution": "\\boxed{x>3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given the graphs of the linear functions $y_1=kx+b$ and $y_2=x+a$ as shown in the figure, what is the solution set of the inequality $kx+b<x+a$?\n",
    "completion": "Since when $x>3$, the graph of the linear function $y_{1}=kx+b$ is below the graph of the linear function $y_{2}=x+a$, the solution set of the inequality $kx+b<x+a$ is $x>3$. \\\\\nTherefore, the solution set is $\\boxed{x>3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50520447.png"
  },
  {
    "solution": "\\boxed{(-3,-2)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated in the Cartesian coordinate system, the line $l_{1}$ intersects with the x-axis at point A$(-3, 0)$ and with the y-axis at point B, while its intersection with line $l_{2}\\colon y=\\frac{4}{3}x$ is point C$(a, 4)$. If the quadrilateral formed by points O, D, B, C is a parallelogram, write down the coordinates of point D directly.\n",
    "completion": "\\textbf{Solution:} Given that $B(0,2)$ and $C(3,4)$. When $OB$ is a side, by drawing a parallel line to $OB$ through point $C$ and setting $CD=OB$, point $C$ can be translated up or down by 2 units. Thus, the positions of point $D$ are $D_1(3,2)$ and $D_2(3,6)$. When $OB$ is a diagonal, by constructing point $D_1$'s symmetric point $D_3$ with respect to point $O$ and given that $D_1(3,2)$, we find that $D_3(-3,-2)$. Therefore, the coordinates of point $D$ are $\\boxed{(3,2)}$, $\\boxed{(3,6)}$, or $\\boxed{(-3,-2)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52226206_106.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $\\triangle ABC$ is an equilateral triangle with a side length of 4, and $\\triangle DBC$ is an isosceles triangle with a vertex angle of $120^\\circ$. The moving points $E$ and $F$ are on the sides $AB$ and $AC$, respectively, and $\\angle EDF=60^\\circ$. What is the perimeter of $\\triangle AEF$?\n",
    "completion": "\\textbf{Solution:} Extend $EB$ to $G$, such that $BG=FC$, and connect $DG$,\\\\\nsince $\\triangle ABC$ is an equilateral triangle,\\\\\nthus $\\angle ABC = \\angle ACB = 60^\\circ$,\\\\\nfurthermore, since $BD=CD$,\\\\\nthus $\\angle DCB = \\angle DBC = \\frac{180^\\circ - \\angle BDC}{2} = 30^\\circ$,\\\\\ntherefore $\\angle DCF = \\angle DBE = 90^\\circ$,\\\\\nin right-angled $\\triangle DCF$ and $\\triangle DBG$,\\\\\n$\\left\\{\\begin{array}{l}DC=DB \\\\ \\angle DCF=\\angle DBG \\\\ CF=BG \\end{array}\\right.$,\\\\\ntherefore $\\triangle DCF \\cong \\triangle DBG$,\\\\\nthus $DG=DF$, $\\angle FDC = \\angle GDB$,\\\\\ntherefore $\\angle GDF = \\angle BDC = 120^\\circ$,\\\\\nfurthermore, since $\\angle EDF = 60^\\circ$,\\\\\nthus $\\angle EDG = 60^\\circ$,\\\\\nin $\\triangle EDG$ and $\\triangle EDF$,\\\\\n$\\left\\{\\begin{array}{l}DG=DF \\\\ \\angle EDG=\\angle EDF \\\\ DE=DE \\end{array}\\right.$,\\\\\nthus $\\triangle EDG \\cong \\triangle EDF$,\\\\\nthus $EF=EG=EB+GB=EB+FC$,\\\\\ntherefore, the perimeter of $\\triangle AEF$ is: $AE+AF+EF=AE+AF+BE+FC=AB+AC=\\boxed{8}$,\\\\",
    "image_path": "dataset/mm_math/images/MM_Math/53010081.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{2}a}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given that point $C$ is on line $AB$, and points $M$ and $N$ are the midpoints of $AC$ and $BC$, respectively. If point $C$ is on the extension of segment $AB$, and it satisfies $AC-BC=a$ cm, please draw a diagram according to the given conditions, and calculate the length of $MN$ (express the result in terms of $a$).\n",
    "completion": "\\textbf{Solution:} According to the problem, the position of point $M$ can be in two situations:\n\n(1) The position of point $M$ is to the left of point $B$, as shown in the diagram:\\\\\nSince $M$ is the midpoint of $AC$,\\\\\nTherefore, $CM=\\frac{1}{2}AC$,\\\\\nSince $N$ is the midpoint of $BC$,\\\\\nTherefore, $CN=\\frac{1}{2}BC$,\\\\\nTherefore, $MN=CM-CN=\\frac{1}{2}AC-\\frac{1}{2}BC=\\frac{1}{2}(AC-BC)=\\boxed{\\frac{1}{2}a}\\,cm$;\\\\\n(2) The position of point $M$ is to the right of point $B$, as shown in the diagram:\\\\\nSince $M$ is the midpoint of $AC$,\\\\\nTherefore, $CM=\\frac{1}{2}AC$,\\\\\nSince $N$ is the midpoint of $BC$,\\\\\nTherefore, $CN=\\frac{1}{2}BC$,\\\\\nTherefore, $MN=CM-CN=\\frac{1}{2}AC-\\frac{1}{2}BC=\\frac{1}{2}(AC-BC)=\\boxed{\\frac{1}{2}a}\\,cm$.",
    "image_path": "dataset/mm_math/images/MM_Math/52872897_140.png"
  },
  {
    "solution": "\\boxed{15^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the isosceles triangle $ABC$, where $AB=AC$, $DE$ is the perpendicular bisector of $AB$, intersecting $AB$ at point $E$ and $AC$ at point $D$. If $\\angle ADE=40^\\circ$, what is the measure of $\\angle CBD$?\n",
    "completion": "\\textbf{Solution:} Given $DE$ bisects $AB$ perpendicularly,\\\\\nhence $AD=BD$, $\\angle AED=90^\\circ$,\\\\\ngiven $\\angle ADE=40^\\circ$,\\\\\nthus $\\angle ABD=\\angle A=50^\\circ$,\\\\\ntherefore, $\\angle ABC=\\angle C=\\frac{1}{2}\\times(180^\\circ-50^\\circ)=65^\\circ$,\\\\\nthus, $\\angle CBD=65^\\circ-50^\\circ=\\boxed{15^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53059213.png"
  },
  {
    "solution": "\\boxed{\\frac{4\\sqrt{3}}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, a moving point $P$ starts from vertex $A$ of the rhombus $ABCD$, moves along the path $A \\to C \\to D$ at a speed of $1cm/s$, and stops at point $D$. Let the movement time of point $P$ be $x(s)$, and the area of $\\triangle PAB$ be $y(cm^2)$. The graph showing the functional relationship between $y$ and $x$ is shown in Figure 2. What is the value of $a$?\n",
    "completion": "\\textbf{Solution:} Given the graph of the function, we can conclude that:\\\\\nWhen $x=4s$, points P and C coincide, and area $S\\triangle ABP=a$, which is the maximum area,\\\\\n$\\therefore AC=1\\times 4=4$,\\\\\nWhen $x=a+4$, points P and D coincide,\\\\\n$\\therefore AB=CD=1\\times (a+4)-4=a$,\\\\\nAs shown in the diagram, draw CK$\\perp$AB at K through C,\\\\\n$\\therefore \\frac{1}{2}a\\cdot CK=a$,\\\\\n$\\therefore CK=2$,\\\\\n$\\therefore \\sin\\angle CAK=\\frac{CK}{CA}=\\frac{1}{2}$,\\\\\n$\\therefore \\angle CAK=30^\\circ$,\\\\\n$\\therefore \\angle ACK=60^\\circ$,\\\\\nGiven that rhombus ABCD,\\\\\n$\\therefore AB=BC=a$, $\\angle BCA=\\angle BAC=30^\\circ$,\\\\\n$\\therefore \\angle BCK=60^\\circ-30^\\circ=30^\\circ$,\\\\\nSince $\\cos\\angle BCK=\\frac{CK}{BC}$,\\\\\n$\\therefore \\frac{2}{a}=\\cos30^\\circ=\\frac{\\sqrt{3}}{2}$,\\\\\n$\\therefore \\sqrt{3}a=4$,\\\\\n$\\therefore a=\\boxed{\\frac{4\\sqrt{3}}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51284464.png"
  },
  {
    "solution": "\\boxed{90^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, quadrilaterals $ABCD$, $CDEF$, and $EFGH$ are all squares. What is the sum of $\\angle ACB+\\angle AFB+\\angle AGB$ in degrees?\n",
    "completion": "\\textbf{Solution:} Let the side length of the square be $a$, \\\\\n$AC=\\sqrt{a^2+a^2}=\\sqrt{2}a$, \\\\\n$\\because \\frac{CA}{CF}=\\frac{\\sqrt{2}a}{a}=\\sqrt{2}$, $\\frac{CG}{AC}=\\frac{2a}{\\sqrt{2}a}=\\sqrt{2}$, \\\\\n$\\therefore \\frac{AC}{CF}=\\frac{CG}{AC}$, \\\\\n$\\because \\angle ACF=\\angle ACF$, \\\\\n$\\therefore \\triangle ACF\\sim \\triangle GCA$; \\\\\n$\\therefore \\angle AGB=\\angle CAF$, \\\\\n$\\because \\angle AGB+\\angle AFB=\\angle CAF+\\angle AFB=\\angle ACB=45^\\circ$, \\\\\n$\\therefore \\angle ACB + \\angle AFB + \\angle AGB = 45^\\circ + 45^\\circ = \\boxed{90^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52917673.png"
  },
  {
    "solution": "\\boxed{y=\\frac{2}{3}x+\\frac{5}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the rectangle $OABC$ has its vertex O coinciding with the origin, vertices A and C respectively on the x-axis and y-axis, and the coordinate of point B is $\\left(2, 3\\right)$. The graph of the inverse proportion function $y=\\frac{k}{x}$ passes through the midpoint D of BC, and intersects AB at point E, with DE being joined. If point F is a point on the side OC, and $\\triangle FBC\\sim \\triangle DEB$, find the equation of the straight line FB.\n",
    "completion": "\\textbf{Solution:} From (1) we have $BD=1$, $BE=\\frac{3}{2}$, $CB=2$,\n\nsince $\\triangle FBC\\sim \\triangle DEB$,\n\nthen $\\frac{BD}{CF}=\\frac{BE}{CB}$, which is $\\frac{1}{CF}=\\frac{\\frac{3}{2}}{2}$,\n\ntherefore $CF=\\frac{4}{3}$,\n\nthus $OF=\\frac{5}{3}$, meaning the coordinates of point F are $\\left(0, \\frac{5}{3}\\right)$,\n\nAssume the equation of line $FB$ is $y=k_1x+b\\ (k_1\\ne 0)$, and line $FB$ passes through $B(2, 3)$, $F\\left(0, \\frac{5}{3}\\right)$,\n\ntherefore $\\begin{cases}3=2k_1+b \\\\ \\frac{5}{3}=b \\end{cases}$,\n\nhence $k_1=\\frac{2}{3}, b=\\frac{5}{3}$,\n\ntherefore, the equation of line $FB$ is $\\boxed{y=\\frac{2}{3}x+\\frac{5}{3}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52827689_141.png"
  },
  {
    "solution": "\\boxed{80^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, PA and PB respectively tangent to the circle O at points A and B, point E is a point on the circle O, and $ \\angle E=50^\\circ$, then what is the degree measure of $ \\angle P$?\n",
    "completion": "\\textbf{Solution:} Connect AO, BO,\\\\\n$ \\because $ PA and PB are tangents to circle O at points A and B, respectively,\\\\\n$ \\therefore \\angle PAO=\\angle PBO=90^\\circ$\\\\\n$ \\because \\angle E=50^\\circ$\\\\\n$ \\therefore \\angle AOB=2\\angle E=100^\\circ$\\\\\n$ \\therefore \\angle P=360^\\circ−\\angle PAO−\\angle PBO−\\angle AOB=360^\\circ−90^\\circ−90^\\circ−100^\\circ=\\boxed{80^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/52536856.png"
  },
  {
    "solution": "\\boxed{-3 < x < -2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the intersection point of the lines $y=-x+a$ and $y=x+b$ has an abscissa of $-2$, and the abscissas of the points where the two lines intersect with the $x$ axis are $-1$ and $-3$, respectively. What is the solution set of the inequality $-x+a>x+b>0$ with respect to $x$?\n",
    "completion": "\\textbf{Solution:} Given $\\because$ the intersection point of the line $y = -x + a$ and $y = x + b$ has an x-coordinate of $-2$, and the x-intercepts of the two lines are $-1$ and $-3$ respectively, \\\\\n$\\therefore$ the solution set for the inequality in $x$, $-x + a > x + b > 0$, is the range of $x$ values for which the line $y = -x + a$ is above the line $y = x + b$, that is, $-3 < x < -2$.\\\\\nThus, the answer is: $\\boxed{-3 < x < -2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50613655.png"
  },
  {
    "solution": "\\boxed{15^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in an acute-angled $\\triangle ABC$, where $\\angle A=75^\\circ$, $DE$ and $DF$ are the perpendicular bisectors of sides $AB$ and $AC$, respectively. What is the degree measure of $\\angle DBC$?\n",
    "completion": "\\textbf{Solution:} Draw DA, DC, \\\\\nsince $\\angle BAC=75^\\circ$, \\\\\nthus $\\angle ABC + \\angle ACB = 180^\\circ - 75^\\circ = 105^\\circ$, \\\\\nsince DE and DF are the perpendicular bisectors of AB and AC respectively, \\\\\nthus DA=DB, DA=DC, \\\\\nthus DB=DC, $\\angle DBA=\\angle DAB$, $\\angle DAC=\\angle DCA$, \\\\\nthus $\\angle DBA+\\angle DCA=\\angle DAB+\\angle DAC=75^\\circ$, \\\\\nthus $\\angle DBC=\\angle DBC= \\frac{1}{2} \\times (105^\\circ - 75^\\circ) = \\boxed{15^\\circ}$, \\\\",
    "image_path": "dataset/mm_math/images/MM_Math/53066350.png"
  },
  {
    "solution": "\\boxed{\\left(2\\sqrt{2}, 4-2\\sqrt{2}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, it is known that the coordinates of point C are $(4\\sqrt{2}, 4\\sqrt{2})$. Perpendiculars are drawn from point C to the x-axis and y-axis, with the feet of the perpendiculars being points B and D, respectively. Point E is a point on line segment OD (not coinciding with points O or D). Line segment BE is connected. The symmetric point of O with respect to line BE is constructed as \\(O'\\). Line segment \\(CO'\\) is drawn, and point P is the midpoint of \\(CO'\\). Line segment BP is connected. Line \\(CO'\\) is extended to intersect the extended line of BE at point F. Line segment DF is connected. Is there a point M in the plane such that the quadrilateral formed by vertices M, O, \\(O'\\), and F is a parallelogram? If it exists, please write down the coordinates of point M directly; if it does not exist, please explain the reason.\n",
    "completion": "\\textbf{Solution:} Yes, it exists. The coordinates of $M$ are \\(\\boxed{\\left(6\\sqrt{2}-8, 4+2\\sqrt{2}\\right)}\\), \\(\\boxed{\\left(-2\\sqrt{2}, 2\\sqrt{2}-4\\right)}\\), or \\(\\boxed{\\left(2\\sqrt{2}, 4-2\\sqrt{2}\\right)}\\).",
    "image_path": "dataset/mm_math/images/MM_Math/52019276_102.png"
  },
  {
    "solution": "\\boxed{3\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the square $ABCD$, $AB=6$, point $Q$ is a moving point on side $AB$ (point $Q$ does not coincide with point $B$), and points $M$, $N$ are the midpoints of $DQ$ and $BQ$, respectively. What is the length of segment $MN$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, connect BD,\\\\\n$\\because$ M and N are the midpoints of BQ and DQ, respectively,\\\\\n$\\therefore$ MN is the midline of $\\triangle QBD$,\\\\\n$\\because$ MN=$\\frac{1}{2}$BD,\\\\\n$\\because$ Quadrilateral ABCD is a square,\\\\\n$\\therefore$ $\\angle$A=$90^\\circ$, AB=AD=6,\\\\\n$\\therefore$ BD=$\\sqrt{AB^{2}+AD^{2}}=6\\sqrt{2}$,\\\\\n$\\therefore$ MN= $\\boxed{3\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52225948.png"
  },
  {
    "solution": "\\boxed{15}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, the perpendicular bisector of AB intersects AC at point D. Given that $BC=10$ and the perimeter of $\\triangle BDC$ is $25$, what is the length of AC?\n",
    "completion": "\\textbf{Solution:} Since $DE$ is the perpendicular bisector of $AB$,\\\\\nit follows that $DB=DA$,\\\\\nSince the perimeter of $\\triangle BDC$ is 25,\\\\\nit follows that $BC+BD+CD=25$,\\\\\nhence $BC+CD+DA=25$,\\\\\nthus $BC+CA=25$,\\\\\nsince $BC=10$,\\\\\nit follows that $AC=25-10=\\boxed{15}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53009851.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $AB \\parallel CD \\parallel EF$, which intersect lines $l_1, l_2$ at points $A, D, F$ and $B, C, E$ respectively. If $\\frac{AD}{DF} = \\frac{2}{3}$ and $BE=20$, what is the length of segment $BC$?\n",
    "completion": "\\textbf{Solution:} Since $AB\\parallel CD\\parallel EF$,\\\\\nit follows that $\\frac{BC}{CE}=\\frac{AD}{DF}=\\frac{2}{3}$,\\\\\nthus $CE=\\frac{3}{2}BC$,\\\\\nGiven $BC+CE=BE=20$,\\\\\nwe have $\\frac{3}{2}BC+BC=20$,\\\\\nsolving this gives $BC=\\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51379396.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\triangle ABC$, $ \\angle C=90^\\circ$ , $AC=12$, and the perpendicular bisector $MN$ of $AB$ intersects $AC$ at $D$, and $BD$ is connected. If $\\cos\\angle CDB=\\frac{3}{5}$, then how much is $BC$?\n",
    "completion": "\\textbf{Solution:} Given $\\cos\\angle CDB=\\frac{3}{5}$,\\\\\n$\\therefore$ $\\frac{CD}{BD}=\\frac{3}{5}$,\\\\\nLet $CD=3x$ and $BD=5x$,\\\\\n$\\therefore$ $BC=\\sqrt{{(5x)}^{2}-{(3x)}^{2}}=4x$,\\\\\nSince MN is the perpendicular bisector of AB,\\\\\n$\\therefore$ AD=BD=5x,\\\\\n$\\therefore$ AC=CD+AD=3x+5x=8x,\\\\\nGiven that AC=12,\\\\\n$\\therefore$ 8x=12,\\\\\n$\\therefore$ x= $\\frac{3}{2}$,\\\\\n$\\therefore$ BC=4x=\\boxed{6}.",
    "image_path": "dataset/mm_math/images/MM_Math/51083570.png"
  },
  {
    "solution": "\\boxed{\\frac{4}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, $AC=3$, $BC=4$. Points $E$ and $F$ are on the hypotenuse $AB$. The side $AC$ is folded along $CE$ making point $A$ fall on point $D$ on $AB$. Then, side $BC$ is folded along $CF$ making point $B$ fall on point $B'$ on the extension line of $CD$. What is the length of the segment $B'F$?\n",
    "completion": "\\textbf{Solution:}\n\nSince $\\angle ACB=90^\\circ, AC=3, BC=4$,\n\nHence, $AB=\\sqrt{AC^2+BC^2}=5$,\n\nBy the property of folding, we have $B'F=BF, \\angle AEC=\\angle DEC=90^\\circ, \\angle ACE=\\angle DCE, \\angle B'CF=\\angle BCF$,\n\nTherefore, $S_{\\triangle ABC}=\\frac{1}{2}AB\\cdot CE=\\frac{1}{2}AC\\cdot BC$, that is, $\\frac{1}{2}\\times 5CE=\\frac{1}{2}\\times 3\\times 4$,\n\nSolving this, we find $CE=\\frac{12}{5}$,\n\nTherefore, $AE=\\sqrt{AC^2-CE^2}=\\frac{9}{5}$,\n\nAlso, since $\\angle ACE=\\angle DCE, \\angle B'CF=\\angle BCF, \\angle ACB=90^\\circ$,\n\nTherefore, $\\angle DCE+\\angle B'CF=\\frac{1}{2}\\angle ACB=45^\\circ$, which means $\\angle ECF=45^\\circ$,\n\nTherefore, right-angled $\\triangle CEF$ is an isosceles right triangle, $EF=CE=\\frac{12}{5}$,\n\nTherefore, $BF=AB-AE-EF=\\frac{4}{5}$,\n\nTherefore, $B'F=BF=\\boxed{\\frac{4}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52836392.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, it is known that the equilateral triangle $\\triangle ABC$ has a side length of $12$, $D$ is a moving point on $AB$, a line is drawn through $D$ such that $DE \\perp BC$ at point $E$, a line is drawn through $E$ such that $EF \\perp AC$ at point $F$, and a line is drawn through $F$ such that $FG \\perp AB$ at point $G$. When $G$ coincides with $D$, what is the length of $AD$?\n",
    "completion": "\\textbf{Solution:} Let AD=x, \\\\\nsince $\\triangle ABC$ is an equilateral triangle, \\\\\ntherefore $\\angle A = \\angle B = \\angle C = 60^\\circ$, \\\\\nsince DE$\\perp$BC at point E, EF$\\perp$AC at point F, FG$\\perp$AB at point G, \\\\\ntherefore $\\angle BDF = \\angle DEB = \\angle EFC = 90^\\circ$, \\\\\ntherefore AF=2x, \\\\\ntherefore CF=12-2x, \\\\\ntherefore CE=2CF=24-4x, \\\\\ntherefore BE=12-CE=4x-12, \\\\\ntherefore BD=2BE=8x-24, \\\\\nsince AD+BD=AB, \\\\\ntherefore 8x-24+x=12, \\\\\ntherefore x=\\boxed{4}, \\\\\ntherefore AD=4.",
    "image_path": "dataset/mm_math/images/MM_Math/52836211.png"
  },
  {
    "solution": "\\boxed{17.5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: It is known that point $C$ is a point on segment $AB$, point $M$ is the midpoint of segment $AC$, and point $N$ is the midpoint of segment $BC$. If $AC:CB=3:2$ and $NB=3.5\\,cm$, then how many $cm$ does $AB$ equal?\n",
    "completion": "\\textbf{Solution:} Given that $\\because NB=3.5\\,cm$,\\\\\n$\\therefore BC=7\\,cm$,\\\\\n$\\therefore AB=7\\div \\frac{2}{5}=17.5\\,cm$.\\\\\nHence, $AB=\\boxed{17.5}\\,cm$.",
    "image_path": "dataset/mm_math/images/MM_Math/43067255_149.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $OE$ is the angle bisector of $\\angle AOC$, and $OD$ is the angle bisector of $\\angle BOC$. If $\\angle AOB=90^\\circ$ and $\\angle BOC=\\alpha$, what is the degree measure of $\\angle DOE$?\n",
    "completion": "\\textbf{Solution:} Given $\\because$ $\\angle AOB=90^\\circ$, $\\angle BOC=\\alpha$ \\\\\n$\\therefore$ $\\angle AOC = 90^\\circ+\\alpha$ \\\\\n$\\because$ OE is the bisector of $\\angle AOC$, OD is the bisector of $\\angle BOC$. \\\\\n$\\therefore$ $\\angle AOE= \\frac{1}{2} \\angle AOC$, $\\angle COD= \\frac{1}{2}\\angle BOC$ \\\\\n$\\angle AOE= \\frac{1}{2}(90^\\circ+\\alpha)$, $\\angle COD= \\frac{1}{2}\\alpha$ \\\\\n$\\angle DOE=\\angle AOC - \\angle AOE-\\angle COD$ \\\\\n$= (90^\\circ+\\alpha) - \\frac{1}{2}(90^\\circ+\\alpha) - \\frac{1}{2}\\alpha = \\boxed{45^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/51355508_173.png"
  },
  {
    "solution": "\\boxed{y=\\frac{8}{x}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, the graph of the linear function $y=kx-3$ ($k\\neq 0$) intersects the y-axis at point B and intersects the graph of the inverse proportion function $y=\\frac{m}{x}$ ($x>0$) at point A (8,1). What are the analytical expressions of the linear function and the inverse proportion function?\n",
    "completion": "\\textbf{Solution:} Given that point A$(8,1)$ lies on the line $y=kx-3$,\\\\\ntherefore $1=8k-3$,\\\\\nsolving for $k$, we find $k=\\frac{1}{2}$,\\\\\nthus, the linear function can be expressed as $\\boxed{y=\\frac{1}{2}x-3}$,\\\\\nsince A$(8,1)$ also lies on the graph of $y=\\frac{m}{x}$ (for $x>0$),\\\\\ntherefore $1=\\frac{m}{8}$,\\\\\nsolving for $m$, we obtain $m=8$,\\\\\nhence, the inverse proportionality function is expressed as $\\boxed{y=\\frac{8}{x}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51445655_274.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that $AB \\perp AC$, $AB=AC$, $DE$ passes through point $A$, and $CD \\perp DE$, $BE \\perp DE$, with the feet of the perpendiculars being points $D$ and $E$, respectively, $CD=5$, $BE=3$. What is the length of $DE$?\n",
    "completion": "\\textbf{Solution:} Given $\\because \\angle CAB=90^\\circ$,\\\\\n$\\therefore \\angle DAC+\\angle BAE=90^\\circ$,\\\\\n$\\because CD\\perp DE, BE\\perp DE$,\\\\\n$\\therefore \\angle D=\\angle E=90^\\circ$,\\\\\n$\\therefore \\angle DAC+\\angle DCA=90^\\circ$,\\\\\n$\\therefore \\angle BAE=\\angle DCA$,\\\\\n$\\because AB=AC$,\\\\\n$\\therefore \\triangle ADC\\cong \\triangle BEA(AAS)$,\\\\\n$\\therefore AD=BE, CD=AE$,\\\\\n$\\because CD=5, BE=3$,\\\\\n$\\therefore AD=3, AE=5$,\\\\\n$\\therefore DE=\\boxed{8}$",
    "image_path": "dataset/mm_math/images/MM_Math/53044345.png"
  },
  {
    "solution": "\\boxed{6+2m}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\triangle MNP$, $ \\angle P=60^\\circ$, $ MN=NP$, $ MQ\\perp PN$ with the foot being Q. Extend MN to G, taking $ NG=NQ$. If the perimeter of $ \\triangle MNP$ is 12 and $ MQ=m$, what is the perimeter of $ \\triangle MGQ$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle P=60^\\circ$,\\\\\n$MN=NP$,\\\\\nthus, $\\triangle PMN$ is an equilateral triangle,\\\\\ngiven that $MQ\\perp PN$,\\\\\nthus, $QN=PQ= \\frac{1}{2}MN$, $\\angle QMN=30^\\circ$, $\\angle QNM=60^\\circ$,\\\\\ngiven that $NG=NQ$, therefore $\\angle GQN=\\angle G=30^\\circ$, $QN=NG= \\frac{1}{2}MN$,\\\\\nthus, $\\angle QMN=\\angle G=30^\\circ$,\\\\\nthus, $QM=QG$,\\\\\ngiven that the perimeter of $\\triangle MNP$ is 12,\\\\\n$MQ=m$,\\\\\nthus, MN=4, QN=NC=2, QM=QG=m,\\\\\nthus, the perimeter of $\\triangle MGQ$ is QM+QG+MN+NG=$\\boxed{6+2m}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51435441.png"
  },
  {
    "solution": "\\boxed{y=2x+4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: The graph of the linear function $y=kx+b$ passes through points $A(1, 6)$ and $B(-3, -2)$. Find the equation of this linear function.\n",
    "completion": "\\textbf{Solution:} Let's substitute points A$(1,6)$ and B$(-3,-2)$ into $y=kx+b$ to get:\\\\\n\\[\n\\begin{cases}\nk+b=6\\\\\n-3k+b=-2\n\\end{cases},\n\\]\nfrom which we find:\n\\[\n\\begin{cases}\nk=2\\\\\nb=4\n\\end{cases},\n\\]\ntherefore the expression of the linear function is: $\\boxed{y=2x+4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52055836_197.png"
  },
  {
    "solution": "\\boxed{B(21,12)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated in the Cartesian coordinate system, $AB\\parallel OC$, with $A(0, 12)$, $B(a, c)$, $C(b, 0)$, and $a, b$ satisfying $\\left|a-21\\right|+\\left(b-16\\right)^2=0$. A moving point $P$ departs from point $A$ and travels towards point $B$ along line segment $AB$ at a speed of 2 units per second; another moving point $Q$ starts from point $O$ and moves towards point $C$ along line segment $OC$ at a speed of 1 unit per second. Points $P$ and $Q$ commence their movements from points $A$ and $O$ respectively, simultaneously. When point $P$ reaches point $B$, it stops, and point $Q$ stops accordingly. Let the duration of the movement be $t$ seconds. What are the coordinates of points $B$ and $C$?\n",
    "completion": "\\textbf{Solution:} We have $\\because$ $\\left|a-21\\right|+\\left(b-16\\right)^{2}=0$,\n\n$\\therefore$ $\\left\\{\\begin{array}{l}\na-21=0\\\\ \nb-16=0\n\\end{array}\\right.$, solving this gives us: $a=21$, $b=16$,\n\n$\\therefore$ $\\boxed{C(16,0)}$, $\\because$ $AB\\parallel OC$, $OA=12$,\n\n$\\therefore$ $\\boxed{B(21,12)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52345544_115.png"
  },
  {
    "solution": "\\boxed{90^\\circ-\\frac{\\alpha}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given: As shown in the figure, $OC$ is the bisector of $\\angle AOB$. When $\\angle AOB = \\alpha$, $\\angle EOC = 90^\\circ$, write down the degree measure of $\\angle AOE$ directly. (Express in terms of $\\alpha$)\n",
    "completion": "\\textbf{Solution:} We have $\\angle AOE=90^\\circ+\\frac{\\alpha}{2}$ or $\\angle AOE=90^\\circ-\\frac{\\alpha}{2}$. Therefore, the degree measure of $\\angle AOE$ is \\boxed{90^\\circ+\\frac{\\alpha}{2}} or \\boxed{90^\\circ-\\frac{\\alpha}{2}}.",
    "image_path": "dataset/mm_math/images/MM_Math/52795995_115.png"
  },
  {
    "solution": "\\boxed{\\frac{2}{7}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points B and C are located on the lines $y=2x$ and $y=kx$, respectively. Points A and D are two points on the x-axis. If quadrilateral ABCD is a rectangle, and the ratio of $AB$ to $AD$ is $1:3$, what is the value of $k$?\n",
    "completion": "\\textbf{Solution:} Let point A be (a, 0),\\\\\nsince point B lies on y=2x,\\\\\nthus point B is (a, 2a),\\\\\ntherefore AB=2a,\\\\\nsince the ratio AB : AD = 1 : 3,\\\\\nthus AD=3AB=$3\\times2a$=6a,\\\\\ntherefore point D is (7a, 0),\\\\\nsince quadrilateral ABCD is a rectangle,\\\\\nthus point C is (7a, 2a),\\\\\nSubstituting point C (7a, 2a) into y=kx gives 7ak=2a,\\\\\ntherefore k=$\\boxed{\\frac{2}{7}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51708499.png"
  },
  {
    "solution": "\\boxed{50\\sqrt{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in rhombus $ABCD$, $\\angle B=60^\\circ$, and the diagonal $AC=10$. What is the perimeter and the area of the rhombus?\n",
    "completion": "\\textbf{Solution:} Draw line BD intersecting AC at O, as shown in the figure:\\\\\n$\\because$ Quadrilateral $ABCD$ is a rhombus,\\\\\n$\\therefore$ AB=AC, $AC\\perp BD$,\\\\\n$\\because$ $\\angle ABC=60^\\circ$,\\\\\n$\\therefore$ $\\triangle ABC$ is an equilateral triangle,\\\\\n$\\therefore$ AC=AB=10, AO=5,\\\\\nBy Pythagoras' theorem, BO=$5\\sqrt{3}$,\\\\\n$\\therefore$ The perimeter of the rhombus is $4\\times 10=\\boxed{40}$, the area of the rhombus is $2AO\\times BO=10\\times 5\\sqrt{3}=\\boxed{50\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52422318.png"
  },
  {
    "solution": "\\boxed{BP=BM}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Consider the diagram, where we have an equilateral triangle $\\triangle ABC$ with point $M$ on side $BC$, and $MN\\perp AC$ at point $N$, intersecting $AB$ at point $P$. Prove that: $BP=BM$.\n",
    "completion": "\\textbf{Solution:} Proof:\n\nSince $\\triangle ABC$ is an equilateral triangle,\n\n$\\therefore \\angle ABC=\\angle C=60^\\circ$,\n\nSince $MN \\perp AC$ at $N$,\n\n$\\therefore \\angle BMP=\\angle CMN=90^\\circ-60^\\circ=30^\\circ$,\n\n$\\therefore \\angle P+\\angle BMP=\\angle ABC=60^\\circ$,\n\n$\\therefore \\angle P=30^\\circ$, that is, $\\angle P=\\angle BMP=30^\\circ$,\n\n$\\therefore \\boxed{BP=BM}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51368161_116.png"
  },
  {
    "solution": "\\boxed{85^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACD=125^\\circ$, $\\angle B=40^\\circ$, then what is the measure of $\\angle A$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle ACD$ is the exterior angle of $\\triangle ABC$, and $\\angle ACD=125^\\circ$, $\\angle B=40^\\circ$,\n\\[\n\\therefore \\angle A=\\angle ACD-\\angle B=\\boxed{85^\\circ}\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/52837762.png"
  },
  {
    "solution": "\\boxed{13}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $M$ is on the side $BC$ of the equilateral triangle $\\triangle ABC$, with $BM = 8$. The ray $CD$ is perpendicular to $BC$ at point $C$. Point $P$ is a moving point on ray $CD$, and point $N$ is a moving point on line segment $AB$. When the value of $MP+NP$ is minimized, $BN = 9$. What is the length of $AC$?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABC$ is an equilateral triangle, \\\\\nthus $AC=BC$ and $\\angle B=60^\\circ$, \\\\\nLet point $M$ be the symmetric point with respect to line $CD$ and construct $GN\\perp AB$ at $N$ intersecting $CD$ at $P$, \\\\\nthen, at this time, the value of $MP+PN$ is minimized, \\\\\nSince $\\angle B=60^\\circ$ and $\\angle BNG=90^\\circ$, \\\\\nthus $\\angle G=30^\\circ$, \\\\\nSince $BN=9$, \\\\\nthus $BG=2BN=18$, \\\\\nthus $MG=BG-BM=18-8=10$, \\\\\nthus $CM=CG=5$, \\\\\nthus $AC=BC=\\boxed{13}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52699756.png"
  },
  {
    "solution": "\\boxed{(-2, 1)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the parabola $y_1=ax^2\\ (a\\neq 0)$ intersects with the line $y_2=bx+c\\ (b\\neq 0)$ at two points, whose coordinates are $A(-2, 4)$ and $B(1, 1)$, respectively. What is the range of $x$ when $y_1<y_2$?\n",
    "completion": "\\textbf{Solution:} From the graph, it is known that when the line $y_2=bx+c\\ (b\\ne 0)$ is above the parabola $y_1=ax^2\\ (a\\ne 0)$, the range of $x$ is $-2<x<1$,\\\\\n$\\therefore$ when $y_1<y_2$, the range of $x$ is $-2<x<1$.\\\\\nTherefore, the answer is: $x \\in \\boxed{(-2, 1)}$",
    "image_path": "dataset/mm_math/images/MM_Math/52977265.png"
  },
  {
    "solution": "\\boxed{2.5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, and point D is the midpoint of AB. Connect CD. If $CD+AB=7.5$, what is the length of $CD$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle ACB=90^\\circ$ and point D is the midpoint of AB, \\\\\nit follows that CD=$\\frac{1}{2}$AB, \\\\\nalso, since CD+AB=7.5, \\\\\nit follows that 3CD=7.5, \\\\\ntherefore, CD=\\boxed{2.5}.",
    "image_path": "dataset/mm_math/images/MM_Math/52807544.png"
  },
  {
    "solution": "\\boxed{120^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that there is a line $AE$, and $O$ is a point on line $AE$. $OB$ is the bisector of $\\angle AOC$, $OD$ is the bisector of $\\angle COE$. If $\\angle AOB=30^\\circ$, what is the degree measure of $\\angle COE$?\n",
    "completion": "\\textbf{Solution:} Since OB is the bisector of $\\angle AOC$ and $\\angle AOB=30^\\circ$,\\\\\nit follows that $\\angle BOC=\\angle AOB=30^\\circ$,\\\\\nthus $\\angle AOC=2\\angle AOB=60^\\circ$.\\\\\nSince $\\angle AOC+\\angle COE=180^\\circ$,\\\\\nwe have $\\angle COE=180^\\circ-\\angle AOC=180^\\circ-60^\\circ=\\boxed{120^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52579729_196.png"
  },
  {
    "solution": "\\boxed{\\frac{17\\sqrt{13}}{65}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within the square $ABCD$, where $AB=3$, point $M$ is on side $CD$, and $DM=1$. The triangle $AEM$ is symmetrical to triangle $ADM$ about the line on which $AM$ is located. By rotating triangle $ADM$ $90^\\circ$ clockwise around point $A$, we obtain triangle $ABF$. Connecting $EF$, what is the value of $\\cos\\angle EFC$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw HG $\\perp$ AB at point H, intersecting CD at point G, then $\\angle AHE = \\angle EGM = 90^\\circ$, and quadrilateral AHGD is a rectangle. \\\\\nBy symmetry, $\\angle AEM = \\angle D = 90^\\circ$, AE = AD = 3, EM = DM = 1, \\\\\n$\\therefore \\angle AEH + \\angle MEG = 90^\\circ$, \\\\\n$\\because \\angle AEH + \\angle HAE = 90^\\circ$, \\\\\n$\\therefore \\angle MEG = \\angle HAE$, \\\\\n$\\therefore \\triangle AHE \\sim \\triangle EGM$, \\\\\n$\\therefore \\frac{AH}{EG} = \\frac{HE}{GM} = \\frac{AE}{EM}$, \\\\\nLet MG = x, then AH = MG + MD = x + 1, \\\\\n$\\therefore \\frac{x + 1}{EG} = \\frac{HE}{x} = \\frac{3}{1}$, \\\\\n$\\therefore HE = 3x$, EG = $\\frac{x + 1}{3}$, \\\\\nIn $\\triangle AHE$, $AH^2 + HE^2 = AE^2$, \\\\\n$\\therefore (x + 1)^2 + (3x)^2 = 3^2$, \\\\\nSolving this gives: x = $\\frac{4}{5}$ or x = -1 (discard), \\\\\n$\\therefore AH = \\frac{9}{5}$, HE = $\\frac{12}{5}$, EG = $\\frac{3}{5}$, \\\\\nDraw EN $\\perp$ BC at point N through point E, then $\\angle ENF = 90^\\circ$, and quadrilateral ENBH is a rectangle, \\\\\n$\\therefore EN = BH = AB - AH = 3 - \\frac{9}{5} = \\frac{6}{5}$, BN = HE = $\\frac{12}{5}$, \\\\\nBy rotation, BF = DM = 1, \\\\\n$\\therefore FN = FB + BN = 1 + \\frac{12}{5} = \\frac{17}{5}$, \\\\\n$\\therefore EF = \\sqrt{EN^2 + FN^2} = \\sqrt{\\left(\\frac{6}{5}\\right)^2 + \\left(\\frac{17}{5}\\right)^2} = \\sqrt{13}$, \\\\\n$\\therefore \\cos\\angle EFN = \\frac{FN}{EF} = \\frac{\\frac{17}{5}}{\\sqrt{13}} = \\boxed{\\frac{17\\sqrt{13}}{65}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52931044.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated in the rectangle $ABCD$, where $AB=4$ and $BC=5$, $P$ is a moving point on ray $BC$. Construct $PE\\perp AP$ with $PE$ intersecting ray $DC$ at point $E$, and ray $AE$ intersecting ray $BC$ at point $F$. Let $BP=x$ and $CF=y$. When $\\sin\\angle APB=\\frac{4}{5}$, what is the length of $CE$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral $ABCD$ is a rectangle,\\\\\n$\\therefore \\angle B=90^\\circ$,\\\\\nIn $\\triangle APB$, $AB=4$, $\\sin\\angle APB=\\frac{4}{5}$,\\\\\n$\\therefore \\sin\\angle APB=\\frac{AB}{AP}=\\frac{4}{5}$,\\\\\n$\\therefore AP=5$,\\\\\n$\\therefore PB=\\sqrt{AP^2-AB^2}=\\sqrt{5^2-4^2}=3$,\\\\\n$\\therefore PC=BC-PB=5-3=2$,\\\\\nBecause $PE\\perp AP$,\\\\\n$\\therefore \\angle APE=90^\\circ$,\\\\\n$\\therefore \\angle CPE+\\angle APB=90^\\circ$,\\\\\nBecause $\\angle BAP+\\angle APB=90^\\circ$,\\\\\n$\\therefore \\angle CPE=\\angle BAP$,\\\\\nBecause $\\angle ECP=\\angle B=90^\\circ$,\\\\\n$\\therefore \\triangle PCE\\sim \\triangle ABP$,\\\\\n$\\therefore \\frac{CE}{BP}=\\frac{PC}{AB}$,\\\\\ni.e., $\\frac{CE}{3}=\\frac{2}{4}$,\\\\\nSolving for: $CE=\\boxed{\\frac{3}{2}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52917616_86.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{6}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, within the rectangle $ABCD$, $E$ is the midpoint of $AD$. After folding $\\triangle ABE$ along $BE$, we obtain $\\triangle GBE$. Extending $BG$ intersects $CD$ at point $F$. If $CF=2$ and $FD=4$, what is the length of $BC$?\n",
    "completion": "\\textbf{Solution:} Connect EF, \\\\\n$\\because$ $CF=2$, $FD=4$, \\\\\n$\\therefore$ $CD=CF+FD=2+4=6$, \\\\\n$\\because$ the quadrilateral ABCD is a rectangle, \\\\\n$\\therefore$ $AB=CD=6$, \\\\\nAs known from the folding, $EG=EA$, \\\\\nAlso, $\\because$ E is the midpoint of AD, \\\\\n$\\therefore$ $ED=EA$, \\\\\n$\\therefore$ $ED=EG$, \\\\\n$\\because$ $EF=EF$, \\\\\n$\\therefore$ $\\triangle EGF \\cong \\triangle EDF$ (HL), \\\\\n$\\therefore$ $GF=DF=4$, \\\\\nBy the property of folding: $BG=AB=6$, \\\\\n$\\therefore$ $BF=BG+GF=6+4=10$, \\\\\n$\\therefore$ $BC=\\sqrt{BF^{2}-CF^{2}}=\\sqrt{10^{2}-2^{2}}=\\boxed{4\\sqrt{6}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51721868.png"
  },
  {
    "solution": "\\boxed{24}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AD \\parallel BC$, E is the midpoint of $CD$, extend line $AE$ to intersect the extension of line $BC$ at point F. $\\triangle ADE \\cong \\triangle FCE$. If $\\angle BAF = 90^\\circ$, $AB = 8$, and $EF = 3$, what is the area of quadrilateral $ABCD$?\n",
    "completion": "\\textbf{Solution:} Given that $ \\because \\triangle ADE \\cong \\triangle FCE$,\\\\\n$ \\therefore AE=EF=3$, $ {S}_{\\triangle ADE}={S}_{\\triangle FCE}$\\\\\n$ \\therefore AF=AE+EF=6$\\\\\n$ \\because \\angle BAF=90^\\circ$,\\\\\n$ {S}_{\\triangle ABF}=\\frac{1}{2}AB\\cdot AF=\\frac{1}{2}\\times 8\\times 6=24$\\\\\n$ \\therefore {S}_{\\text{quadrilateral }ABCD}={S}_{\\text{quadrilateral }ABCE}+{S}_{\\triangle ADE}={S}_{\\text{quadrilateral }ABCE}+{S}_{\\triangle FCE}={S}_{\\triangle ABF}=\\boxed{24}$",
    "image_path": "dataset/mm_math/images/MM_Math/55502397_95.png"
  },
  {
    "solution": "\\boxed{y=\\frac{1}{2}(1+x)\\sqrt{2+2x^{2}}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given: As shown in the diagram, the side length of the square is 1. Take a point E on ray AB, connect DE, and rotate ADE $90^\\circ$ about point D. Point E falls on point F. Connect EF, intersecting the line where diagonal BD is located at point M, and intersecting ray DC at point N. When point E is on segment AB, if $AE=x$ and $FM=y$, find the analytic expression of $y$ as a function of $x$, and state its domain.\n",
    "completion": "\\textbf{Solution:} Given the problem, we have \\\\\n$\\because$AE=x,\\\\\n$\\therefore$BE=1−x,\\\\\n$\\because$ Rotating $\\triangle ADE$ $90^\\circ$ around point D gives $\\triangle DCF$,\\\\\n$\\therefore$CF=AE=x, ED=FD=$\\sqrt{AD^{2}+AE^{2}}=\\sqrt{1+x^{2}}$,\\\\\n$\\therefore$BF=BC+CF=1+x,\\\\\nIn $\\triangle EBF$, EF=$\\sqrt{BE^{2}+BF^{2}}=\\sqrt{(1−x)^{2}+(1+x)^{2}}=\\sqrt{2+2x^{2}}$,\\\\\n$\\because$$\\angle EDF=90^\\circ$, ED=FD,\\\\\n$\\therefore$$\\triangle DEF$ is an isosceles right triangle,\\\\\n$\\therefore$$\\angle DFE=\\angle DEF=45^\\circ$,\\\\\n$\\therefore$$\\angle EBM=\\angle MFD=45^\\circ$,\\\\\n$\\because$$\\angle EMB=\\angle DMF$,\\\\\n$\\therefore$$\\triangle BEM\\sim \\triangle FDM$,\\\\\n$\\therefore$$\\frac{BE}{DF}=\\frac{BM}{FM}$, that is, $\\frac{1−x}{\\sqrt{1+x^{2}}}=\\frac{BM}{y}$,\\\\\n$\\because$$\\angle DEM=\\angle FBM=45^\\circ$, $\\angle EMD=\\angle BMF$,\\\\\n$\\therefore$$\\triangle EMD\\sim \\triangle BMF$,\\\\\n$\\therefore$$\\frac{ED}{BF}=\\frac{EM}{BM}$, that is, $\\frac{\\sqrt{1+x^{2}}}{1+x}=\\frac{\\sqrt{2+2x^{2}}−y}{BM}$,\\\\\n$\\therefore$$\\frac{1−x}{\\sqrt{1+x^{2}}}\\times \\frac{\\sqrt{1+x^{2}}}{1+x}=\\frac{BM}{y}\\times \\frac{\\sqrt{2+2x^{2}}−y}{BM}$,\\\\\n$\\therefore$$\\frac{1−x}{1+x}=\\frac{\\sqrt{2+2x^{2}}−y}{y}$,\\\\\n$\\therefore$$\\frac{1−x+1+x}{1+x}=\\frac{\\sqrt{2+2x^{2}}−y+y}{y}$, that is, $\\frac{2}{1+x}=\\frac{\\sqrt{2+2x^{2}}}{y}$,\\\\\n$\\therefore$$\\boxed{y=\\frac{1}{2}(1+x)\\sqrt{2+2x^{2}}}$, with the domain $0\\leq x\\leq 1$.",
    "image_path": "dataset/mm_math/images/MM_Math/51403292_159.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $D$ is a point on side $BC$ of $\\triangle ABC$, with $\\angle CAD = \\angle B$. Given $AD=4$, $AB=8$, $AC=6$, what is the length of $DC$?\n",
    "completion": "Solution: In $\\triangle ADC$ and $\\triangle BAC$, we have $\\left\\{\\begin{array}{l}\n\\angle CAD=\\angle B \\\\\n\\angle C=\\angle C\n\\end{array}\\right.$,\\\\\n$\\therefore \\triangle ADC\\sim \\triangle BAC$,\\\\\n$\\therefore \\frac{AD}{AB}=\\frac{DC}{AC}$,\\\\\n$\\because AD=4$, $AB=8$, $AC=6$,\\\\\n$\\therefore \\frac{4}{8}=\\frac{DC}{6}$,\\\\\nSolving for $DC$, we find $DC=\\boxed{3}$,\\\\\nTherefore, the length of $DC$ is 3.",
    "image_path": "dataset/mm_math/images/MM_Math/52793606.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{5}}{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, for square $ABCD$ with a side length of 1, $O$ is the midpoint of diagonal $BD$. Point $M$ is on side $AB$, and $BM=2AM$. Point $N$ is on side $BC$, and $BN=AM$. Lines $AN$ and $MD$ intersect at point $P$. Connect $OP$. What is the length of $OP$?\n",
    "completion": "\\textbf{Solution:} Let AN and BD intersect at point Q,\\\\\nsince the side length of square ABCD is 1,\\\\\ntherefore $BD=\\sqrt{AB^2+AD^2}=\\sqrt{2}$,\\\\\nsince $BM=2AM$,\\\\\ntherefore $BM+AM=AB=1$,\\\\\ntherefore $AM=BN=\\frac{1}{3}$, $BM=\\frac{2}{3}$,\\\\\nsince $AD\\parallel BN$,\\\\\ntherefore $\\triangle NQB\\sim \\triangle AQD$,\\\\\ntherefore $\\frac{BN}{AD}=\\frac{BQ}{QD}=\\frac{NQ}{AQ}=\\frac{1}{3}$,\\\\\ntherefore $DQ=3BQ$,\\\\\ntherefore $DQ=\\frac{3\\sqrt{2}}{4}$, $BQ=\\frac{\\sqrt{2}}{4}$,\\\\\nsince O is the midpoint of BD,\\\\\ntherefore $OD=\\frac{\\sqrt{2}}{2}$,\\\\\ntherefore $OQ=DQ-OD=\\frac{\\sqrt{2}}{4}$,\\\\\nIn $\\triangle ADM$ and $\\triangle BAN$,\\\\\n$\\begin{cases}AD=AB\\\\ \\angle DAM=\\angle ABN\\\\ AM=BN\\end{cases}$,\\\\\ntherefore $\\triangle ADM\\cong \\triangle BAN$ (SAS),\\\\\ntherefore $\\angle ADM=\\angle BAN$,\\\\\nsince $\\angle PAD+\\angle BAN=90^\\circ$,\\\\\ntherefore $\\angle PAD+\\angle ADM=90^\\circ$,\\\\\ntherefore $\\angle APD=90^\\circ$,\\\\\nsince $\\angle DAM=90^\\circ$, $AM=\\frac{1}{3}$, $AD=1$,\\\\\ntherefore $DM= \\sqrt{AM^2+AD^2}=\\frac{\\sqrt{10}}{3}$,\\\\\nsince $S_{\\triangle ADM}=\\frac{1}{2}\\times AD\\cdot AM=\\frac{1}{2}\\times DM\\cdot AP$,\\\\\ntherefore $AP=\\frac{1\\times \\frac{1}{3}}{\\frac{\\sqrt{10}}{3}}=\\frac{\\sqrt{10}}{10}$,\\\\\ntherefore $PM=\\sqrt{AM^2-AP^2}=\\frac{\\sqrt{10}}{30}$,\\\\\ntherefore $PD=DM-PM=\\frac{3\\sqrt{10}}{10}$,\\\\\nsince $\\triangle ADM\\cong \\triangle BAN$,\\\\\ntherefore $AN=DM= \\frac{\\sqrt{10}}{3}$,\\\\\nsince $\\frac{NQ}{AQ}=\\frac{1}{3}$,\\\\\ntherefore $NQ=\\frac{\\sqrt{10}}{12}$, $AQ=\\frac{\\sqrt{10}}{4}$,\\\\\ntherefore $PQ=AQ-AP=\\frac{\\sqrt{10}}{4}-\\frac{\\sqrt{10}}{10}=\\frac{3\\sqrt{10}}{20}$,\\\\\nAs shown in the diagram, draw OG$\\perp$DM at point G through point O,\\\\\nsince $OG\\parallel PQ$,\\\\\ntherefore $\\triangle OGD\\sim \\triangle QPD$,\\\\\ntherefore $\\frac{DO}{DQ}=\\frac{DG}{DP}=\\frac{OG}{QP}=\\frac{2}{3}$,\\\\\ntherefore $DG=\\frac{2}{3}DP=\\frac{2}{3}\\times \\frac{3\\sqrt{10}}{10}=\\frac{\\sqrt{10}}{5}$, $OG=\\frac{2}{3}QP=\\frac{2}{3}\\times \\frac{3\\sqrt{10}}{20}=\\frac{\\sqrt{10}}{10}$,\\\\\ntherefore $PG=DP-DG=\\frac{3\\sqrt{10}}{10}-\\frac{\\sqrt{10}}{5}=\\frac{\\sqrt{10}}{10}$,\\\\\ntherefore $OP=\\sqrt{OG^2+PG^2}=\\sqrt{\\frac{1}{10}+\\frac{1}{10}}=\\boxed{\\frac{\\sqrt{5}}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51403069.png"
  },
  {
    "solution": "\\boxed{-5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the Cartesian coordinate system, the graph of the linear function $y=kx+b$ passes through the vertices A and C of the square OABC. Given that the coordinates of point A are $\\left(1, -3\\right)$, what is the value of $b$?\n",
    "completion": "\\textbf{Solution:} Construct $CM \\perp x$-axis at point M, and construct $AN \\perp y$-axis at point N. \\\\\nHence, $\\angle OMC = \\angle ONA$, \\\\\nSince quadrilateral OABC is a square, \\\\\nThus, $OA=OC$, $\\angle AOC=90^\\circ$, \\\\\nSince $\\angle COM + \\angle MOA = \\angle MOA + \\angle NOA = 90^\\circ$, \\\\\nTherefore, $\\angle NOA = \\angle COM$, \\\\\nAlso, since $OA=OC$, \\\\\nTherefore, $Rt\\triangle OCM \\cong Rt\\triangle OAN$ (AAS), \\\\\nThus, $OM=ON$, $CM=AN$, \\\\\nSince the coordinate of point A is $(1, -3)$, \\\\\nTherefore, $AN=1$, $ON=3$, \\\\\nThus, $OM=3$, $CM=1$, \\\\\nHence, the coordinate of point C is $(3, 1)$, \\\\\nSubstituting the coordinates of points A and C into $y=kx+b$, \\\\\nWe obtain $\\begin{cases}k+b=-3 \\\\ 3k+b=1\\end{cases}$, \\\\\nSolving gives: $\\begin{cases}k=2 \\\\ b=\\boxed{-5}\\end{cases}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52355068.png"
  },
  {
    "solution": "\\boxed{\\frac{5}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In rectangle $ABCD$, where $AB=5$ and $BC=5\\sqrt{3}$, point $P$ is a moving point on $BC$. Connect $AP$ and construct an equilateral triangle $APE$ with $AP$ as one of its sides. Connect $DE$. What is the minimum value of segment $DE$ as point $P$ moves?\n",
    "completion": "\\textbf{Solution:} Construct an equilateral triangle ABF with AB as its side, draw a line connecting FE, and draw line DH perpendicular to the extension of FE at point H, as shown in the figure,\\\\\nsince quadrilateral ABCD is a rectangle, it follows that $\\angle ABP=\\angle BAD=90^\\circ$,\\\\\nsince $\\triangle APE$ and $\\triangle ABF$ are equilateral triangles, it follows that $\\angle BAF=\\angle PAE=60^\\circ$, $BA=FA$, PA=AE, thus $\\angle BAP=\\angle FAE$,\\\\\ntherefore $\\triangle BAP\\cong \\triangle FAE$, hence $\\angle ABP=\\angle AFE=90^\\circ$,\\\\\nsince $\\angle FAG=90^\\circ-60^\\circ=30^\\circ$, it follows that $\\angle AGE=90^\\circ-30^\\circ=60^\\circ$,\\\\\nsince AB=AF=5, it follows that $AG=AF\\div\\cos30^\\circ=\\frac{10\\sqrt{3}}{3}$,\\\\\ntherefore, point E moves on the ray FH, since AD=BC=$5\\sqrt{3}$, it follows that DG=AD−AG=$\\frac{5\\sqrt{3}}{3}$,\\\\\nsince DH is perpendicular to FE, $\\angle AGE=\\angle HGD=60^\\circ$, it follows that $DH=DG\\times \\sin60^\\circ=\\frac{5}{2}$,\\\\\nalso, since the perpendicular segment is the shortest, it follows that when point E coincides with point H, DE has its minimum value, and the minimum value is $\\boxed{\\frac{5}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/54967313.png"
  },
  {
    "solution": "\\boxed{\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of $\\odot O$, and chord $CD\\perp AB$ at point $E$, with $AD, BD$ connected. $\\angle ADC=\\angle ABD$; draw $OF\\perp AD$ at point $F$. If the radius of $\\odot O$ is $5$ and $OE=3$, find the length of $OF$.\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect $OD$.\\\\\nIn $Rt\\triangle OED$, we have $DE=\\sqrt{OD^{2}-OE^{2}}=\\sqrt{5^{2}-3^{2}}=4$,\\\\\nIn $Rt\\triangle ADE$, $AD=\\sqrt{AE^{2}+DE^{2}}=\\sqrt{8^{2}+4^{2}}=4\\sqrt{5}$,\\\\\n$\\because \\sin\\angle A=\\frac{OF}{OA}=\\frac{DE}{AD}$,\\\\\n$\\therefore \\frac{OF}{5}=\\frac{4}{4\\sqrt{5}}$,\\\\\n$\\therefore OF=\\boxed{\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52872119_107.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the line $l_{1}: y=2x+2$ intersects the $x$-axis at point $A$ and the $y$-axis at point $C$; the line $l_{2}: y=kx+b$ intersects the $x$-axis at point $B(3, 0)$ and intersects with the line $l_{1}$ at point $D$, and the $y$-coordinate of point $D$ is 4. Find the equation of line $l_{2}$ and the area of $\\triangle CDE$.\n",
    "completion": "\\textbf{Solution:} Substitute $B(3, 0)$ and $D(1, 4)$ into $y=kx+b$ to obtain: $\\begin{cases}3k+b=0\\\\ k+b=4\\end{cases}$\\\\\n$\\therefore \\begin{cases}k=-2\\\\ b=6\\end{cases}$\\\\\n$\\therefore$ the equation of line $l_{2}$ is: $y=-2x+6$\\\\\nLet $x=0$\\\\\n$\\therefore y=6$\\\\\n$\\therefore E(0, 6)$\\\\\nFrom $y=2x+2$, we know: $C(0, 2)$\\\\\n$\\therefore Area_{\\triangle CDE}=\\frac{1}{2}\\times (6-2)\\times 1=\\boxed{2}$\\\\\n$\\therefore$ The area of $\\triangle CDE$ is $\\boxed{2}$",
    "image_path": "dataset/mm_math/images/MM_Math/50555324_95.png"
  },
  {
    "solution": "\\boxed{8-\\sqrt{39}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given that, in $\\triangle ABC$, $AB=AC=5$, $BC=8$. Point $E$ is a moving point on ray $CA$, and point $O$ is a moving point on side $BC$, with $OC=OE$. Ray $OE$ intersects ray $BA$ at point $D$. When point $E$ is on side $AC$, connecting $BE$ and $CD$, $\\angle DBE = \\angle CDO$, find the length of segment $OC$.\n",
    "completion": "\\textbf{Solution:} From equation (1), we have $\\angle B = \\angle OEC$,\\\\\n$\\because \\angle OEC + \\angle OEA = 180^\\circ$,\\\\\n$\\therefore \\angle B + \\angle OEA = 180^\\circ$,\\\\\n$\\therefore$ Points A, B, O, and E are concyclic,\\\\\n$\\therefore \\angle DBE = \\angle AOD$,\\\\\n$\\because \\angle DBE = \\angle CDO$,\\\\\n$\\therefore \\angle AOD = \\angle CDO$,\\\\\n$\\therefore AO \\parallel DC$,\\\\\n$\\therefore \\triangle AOE \\sim \\triangle CDE, \\triangle ABO \\sim \\triangle DBC$,\\\\\n$\\therefore \\frac{AO}{DC} = \\frac{BO}{CB}$, $\\frac{AO}{DC} = \\frac{AE}{CE}$,\\\\\n$\\therefore \\frac{AE}{EC} = \\frac{BO}{CB}$,\\\\\nLet OC=x, OB=8-x,\\\\\n$\\because \\triangle ABC \\sim \\triangle OEC$,\\\\\n$\\therefore \\frac{AC}{OC} = \\frac{BC}{CE}$,\\\\\n$\\therefore \\frac{5}{x} = \\frac{8}{CE}$,\\\\\nSolving, $CE=1.6x$,\\\\\n$\\therefore AE=5-1.6x$\\\\\n$\\therefore \\frac{5-1.6x}{1.6x} = \\frac{8-x}{8}$,\\\\\nSolving, $x_1=8-\\sqrt{39}$, $x_2=8+\\sqrt{39}$ (discard),\\\\\nHence, the length of OC is $\\boxed{8-\\sqrt{39}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51319009_114.png"
  },
  {
    "solution": "\\boxed{0^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, it is known that point $O$ is on line $AB$, with rays $OD$ and $OC$ located respectively above and below line $AB$, and $\\angle COD = 80^\\circ$. Ray $OE$ is always the bisector of $\\angle AOD$. If $\\angle AOE = 10^\\circ$, what is the degree measure of $\\angle COE$?\n",
    "completion": "\\textbf{Solution:} Since $OE$ bisects $\\angle AOD$,\\\\\nit follows that $\\angle AOE=\\angle DOE=10^\\circ$,\\\\\nhence $\\angle COE-\\angle COD-\\angle DOE=80^\\circ-10^\\circ-70^\\circ=\\boxed{0^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53414133_151.png"
  },
  {
    "solution": "\\boxed{\\frac{3}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point C is a point on segment $AB$ ($AC>BC$), D is on segment $BC$, $BD=2CD$, and point E is the midpoint of $AB$. If $AC=2BC$, find the value of $\\frac{EC}{BD}$.\n",
    "completion": "\\textbf{Solution:} From the given information, we have $BC=CD+BD=3x$, $AC=2BC=6x$,\n\\[\\therefore AB=3x+6x=9x,\\]\n\\[\\because \\text{E is the midpoint of } AB,\\]\n\\[\\therefore EB=\\frac{1}{2}AB=4.5x,\\]\n\\[\\therefore EC=EB-BC=1.5x,\\]\n\\[\\therefore \\frac{EC}{BD}=\\frac{1.5x}{2x}=\\boxed{\\frac{3}{4}}.\\]",
    "image_path": "dataset/mm_math/images/MM_Math/53206007_57.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in right-angled triangle $\\triangle ABC$, $\\angle ACB=90^\\circ$, $AC=6$, and $AB=10$. Point D moves from point B in the direction of ray BC at a speed of $5$ units per second, and point F moves from point A towards point B along edge AB at the same speed. When F reaches point B, point D stops moving. Let the movement time of point D be $t$ seconds. A square $DEFG$ is constructed with $DF$ as its diagonal. During the movement, if one side of square $DEFG$ happens to coincide with one side of the right-angled triangle $ABC$, then what is the value of $t$?\n",
    "completion": "\\textbf{Solution:}\n\nGiven that in the right triangle $ABC$, $AC=6$ and $AB=10$, \\\\\n$\\therefore BC=\\sqrt{AB^2-AC^2}=8$, \\\\\nbecause point $D$ moves from point $B$ in the direction of ray $BC$ at a speed of 5 units per second, and point $F$ moves from point $A$ towards point $B$ along edge $AB$ at the same speed, \\\\\n$\\therefore AF=DB=5t$, \\\\\nAs shown in the figure, taking $B$ as the origin and the line along $BC$ as the x-axis, a Cartesian coordinate system is established, \\\\\n$\\therefore A(8,6)$ \\\\\nLet the equation of line $BA$ be $y=kx$, then $6=8k$ \\\\\nSolving gives $k=\\frac{3}{4}$, \\\\\n$\\therefore$ the equation of line $BA$ is $y=\\frac{3}{4}x$, \\\\\nLet $F(m,\\frac{3}{4}m)$, $0\\le m\\le 8$, \\\\\n(1) As shown in the figure, when $DE$ is on side $BC$, draw $FM\\perp AC$ at $M$. \\\\\n$\\therefore FM=8-m$, $AM=6-\\frac{3}{4}m$, \\\\\n$\\therefore AF^2=AM^2+FM^2=\\left(8-m\\right)^2+\\left(6-\\frac{3}{4}m\\right)^2=\\left(\\frac{5m-40}{4}\\right)^2$, \\\\\nbecause $AF=5t$, \\\\\n$\\therefore 5t=\\left|\\frac{5m-40}{4}\\right|$, \\\\\nSolving gives $t=\\left|\\frac{m-8}{4}\\right|$, \\\\\nbecause $0\\le m\\le 8$, \\\\\n$\\therefore t=\\frac{8-m}{4}$ \\\\\n$\\therefore$ $FM=EC=4t$, $AM=3t$, $CM=EF=DE=6-3t$, \\\\\nbecause $BD+DE+EC=8$, \\\\\n$\\therefore$ $5t+6-3t+4t=8$, \\\\\nSolving gives $t=\\boxed{\\frac{1}{3}}$, \\\\\n(2) As shown in the figure, when $FG$ is on side $AB$, \\\\\nIn right triangle $\\triangle BGD$, with $DB=5t$, similar to (1) we can conclude $DG=FG=3t$, then $BG=4t$, \\\\\nbecause $BG+FG+AF=10$, \\\\\n$\\therefore$ $4t+3t+5t=10$, \\\\\nSolving gives $t=\\frac{5}{6}$, \\\\\n(3) When $DG$ is on side $BC$, \\\\\nthen $FG=DG=6-3t$, $BG=8-4t$, \\\\\nbecause $BD=BG+DG=5t$, \\\\\n$\\therefore$ $8-4t+6-3t=5t$, \\\\\nSolving gives $t=\\frac{7}{6}$; \\\\\n(4) When $EF$ is on side $AB$, \\\\\nSimilar to (1) we can conclude $BE=4t$, $DE=EF=3t$, \\\\\nbecause $BE-EF=BF$, \\\\\n$\\therefore$ $4t-3t=10-5t$, \\\\\nSolving gives $t=\\frac{5}{3}$; \\\\\nIn summary, $t=\\frac{1}{3}$ or $\\frac{5}{6}$ or $\\frac{7}{6}$ or $\\frac{5}{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53490772.png"
  },
  {
    "solution": "\\boxed{(1,1)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated, $\\triangle OAB$ and $\\triangle ODC$ are similar figures with point O as the center of similarity and a similarity ratio of $1\\colon2$. $\\angle OCD=90^\\circ$, and $CO=CD$. If $A(1, 0)$, what are the coordinates of point C?\n",
    "completion": "\\textbf{Solution:} Draw line $AC$, \\\\\nsince $CO=CD$ and $\\angle OCD=90^\\circ$, \\\\\ntherefore $\\triangle OCD$ is an isosceles right triangle, \\\\\nthus $\\angle COD=45^\\circ$, \\\\\nsince $\\triangle OAB$ and $\\triangle ODC$ are similar figures with center of similarity at $O$ and a similarity ratio of $1:2$, \\\\\nthus $2OA=OD$, and $\\triangle OAB$ is an isosceles right triangle, \\\\\nhence $A$ is the midpoint of $OD$, \\\\\ntherefore $AC \\perp OD$ (the three lines of an isosceles triangle coincide), \\\\\nthus $\\triangle OAC$ is an isosceles right triangle, \\\\\nsince $A(1,0)$, \\\\\nthus $OA=AC=1$, \\\\\nhence the coordinates of point $C$ are $\\boxed{(1,1)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51910252.png"
  },
  {
    "solution": "\\boxed{74}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In $\\triangle ACB$, $\\angle ABC=90^\\circ$, $AB=4$, and $BC=3$. Squares are constructed externally on each of the three sides. Connecting the vertices of these squares forms a hexagon DEFGHI. What is the area of hexagon DEFGHI?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, \\\\\nin $\\triangle ABC$, since $AB=4$ and $BC=3$, \\\\\ntherefore $AC=\\sqrt{AB^{2}+BC^{2}}=\\sqrt{4^{2}+3^{2}}=5$. \\\\\nSince quadrilaterals ABDE, BCGF, and ACHM are all squares, \\\\\ntherefore $\\angle ABD=\\angle CBF=\\angle BAE=\\angle CAM=\\angle ACH=\\angle GCH=90^\\circ$, BD=BA, AM=AC, CBN=CG, $S_{\\text{square ABDE}}=4^{2}=16$, $S_{\\text{square ACHM}}=5^{2}=25$, $S_{\\text{square BCGF}}=3^{2}=9$, \\\\\ntherefore, $\\angle DBF+\\angle ABC=180^\\circ$, $\\angle MAE+\\angle BAC=180^\\circ$, $\\angle ACB+\\angle HCG=180^\\circ$, \\\\\ndraw IM$\\perp$DA to intersect the extension line of DA at M, \\\\\ntherefore $\\angle M=\\angle ABC=90^\\circ$, \\\\\nsince $\\angle DAI+\\angle MAI=\\angle DAI+\\angle BAC=180^\\circ$, \\\\\ntherefore $\\angle IAM=\\angle BAC$, \\\\\nin $\\triangle AMI$ and $\\triangle BAC$, \\\\\n$\\left\\{\\begin{array}{l}\\angle M=\\angle ABC \\\\ \\angle BAC=\\angle MAI \\\\ AC=AI \\end{array}\\right.$ \\\\\ntherefore $\\triangle AMI \\cong \\triangle ABC$, \\\\\nhence AB=AM, \\\\\nthus AD=AM, \\\\\nthus $S_{\\triangle AMI}=S_{\\triangle ABC}=S_{\\triangle ADI}$, \\\\\nSimilarly, $S_{\\triangle BEF}=S_{\\triangle ABC}$, $S_{\\triangle CHG}=S_{\\triangle ABC}$, \\\\\nthus $S_{\\triangle DBF}=S_{\\triangle MAE}=S_{\\triangle HCG}=S_{\\triangle ABC}=\\frac{1}{2}\\times 3\\times 4=6$, \\\\\ntherefore the area of the hexagon DEMHGF=$25+16+9+4\\times6=\\boxed{74}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52836896.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the right-angled triangle $ABC$ with $\\angle B=90^\\circ$, $AB=4$, and $BC=3$, the shaded area is a rectangle, and $AE=1$. What is the area of the shaded region?\n",
    "completion": "\\textbf{Solution:} In $\\triangle ABC$, where $\\angle B=90^\\circ$, $AB=4\\text{cm}$, $BC=3\\text{cm}$,\\\\\nby the Pythagorean theorem, we have $AB^2+BC^2=AC^2$,\\\\\nthus $4^2+3^2=AC^2$\\\\\n$\\therefore AC= \\sqrt{3^2+4^2}=5\\text{cm}$,\\\\\n$\\because AE=1\\text{cm}$,\\\\\n$\\therefore$ the area of rectangle $ACDE$ is $5\\times 1=\\boxed{5}\\text{cm}^2$",
    "image_path": "dataset/mm_math/images/MM_Math/52096620.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the straight line AO is perpendicular to OB with the foot of the perpendicular at O. The length of segment AO is 3, and the length of BO is 4. Taking point A as the center and the length of AB as the radius, an arc is drawn to intersect the straight line AO at point C. What is the length of OC?\n",
    "completion": "\\textbf{Solution:} Since $AO \\perp OB$, and the length of segment AO = 3, BO = 4,\\\\\ntherefore, in $\\triangle AOB$, we have $AB = \\sqrt{AO^{2} + OB^{2}} = 5$,\\\\\naccording to the problem, we know: AC = AB = 5,\\\\\ntherefore, $OC = AC - AO = \\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52826614.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC$, the perpendicular bisector of $AC$ intersects $AB$ and $AC$ at $D$ and $E$, respectively. Connect $CD$. If $\\angle B=70^\\circ$, what is the measure of $\\angle DCB$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle B=70^\\circ$ and AB=AC,\\\\\nit follows that $\\angle A=180^\\circ-2\\times70^\\circ=40^\\circ$,\\\\\nfurthermore, since DE is perpendicular bisector of AC,\\\\\nit implies that DA=DC,\\\\\nthus, $\\angle ACD=\\angle A=40^\\circ$,\\\\\ntherefore, $\\angle BCD=\\angle ACB-\\angle ACD=70^\\circ-40^\\circ=\\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52653753.png"
  },
  {
    "solution": "\\boxed{15}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $DE$ is the perpendicular bisector of side $BC$ of $\\triangle ABC$, intersecting sides $AB$ and $BC$ at points $D$ and $E$, respectively, and $AB=9$, $AC=6$. What is the perimeter of $\\triangle ACD$?\n",
    "completion": "\\textbf{Solution:} Since DE is the perpendicular bisector of side BC in $\\triangle ABC$, \\\\\nit follows that BD = CD, \\\\\nand since the perimeter of $\\triangle ACD$ is AD + CD + AC = AD + BD + AC = AB + AC = 9 + 6 = \\boxed{15}.",
    "image_path": "dataset/mm_math/images/MM_Math/52756081.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, points D and E are on sides $AB$ and $AC$ respectively, with $DE\\parallel BC$. Given that $AE=2$ and $\\frac{AD}{AB}=\\frac{1}{3}$, what is the length of $EC$?\n",
    "completion": "\\textbf{Solution:} Given $DE\\parallel BC$, \\\\\nhence $\\angle ADE=\\angle ABC$, $\\angle AED=\\angle ACB$, \\\\\ntherefore $\\triangle ADE\\sim \\triangle ABC$, \\\\\nthus $\\frac{AE}{AC}=\\frac{AD}{AB}$, \\\\\ngiven $\\frac{AD}{AB}=\\frac{1}{3}$, $AE=2$, \\\\\nthus $AC=6$, \\\\\nhence $EC=AC-AE=6-2=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53019770.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given an equilateral triangle $\\triangle ABC$, $M$ is on side $BC$, $MN\\perp AC$ at $N$, and intersects $AB$ at point $P$. If $E$, $F$ are respectively on $AB$, $AC$, and $\\triangle MEF$ is an equilateral triangle, when the value of $\\frac{S_{\\triangle MEF}}{S_{\\triangle ABC}}$ is minimal, what is the value of $\\frac{BM}{BC}$?\n",
    "completion": "\\textbf{Solution:} When the value of $\\frac{S_{\\triangle MEF}}{S_{\\triangle ABC}}$ is minimal, $\\frac{BM}{BC}=\\boxed{\\frac{1}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51368161_118.png"
  },
  {
    "solution": "\\boxed{15}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB$ is the diameter of $\\odot O$, point $D$ is the midpoint of the arc $AC$, draw $DE \\perp AB$ at point $E$, extend $DE$ to intersect $\\odot O$ at point $F$. If $AC = 12$ and $AE = 3$, what is the length of the diameter of $\\odot O$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect OF.\\\\\nSince DE$\\perp$AB, it follows that DE=EF, $\\overset{\\frown}{AD} = \\overset{\\frown}{AF}$. Since point D is the midpoint of arc AC, it follows $\\overset{\\frown}{AD} = \\overset{\\frown}{CD}$, hence $\\overset{\\frown}{AC} = \\overset{\\frown}{DF}$, therefore AC=DF=12, thus EF=$\\frac{1}{2}$DF=6. Let OA=OF=x, in the right triangle $\\triangle$OEF, we have $x^{2}=6^{2}+(x-3)^{2}$. Solving this, we get $x=\\frac{15}{2}$, thus AB=2x=\\boxed{15}.",
    "image_path": "dataset/mm_math/images/MM_Math/52754802.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $BC=8$, the perpendicular bisector of $AB$ intersects $BC$ at $D$, and the perpendicular bisector of $AC$ intersects $BC$ at $E$. Then, what is the perimeter of $\\triangle ADE$?\n",
    "completion": "\\textbf{Solution:} Given that the perpendicular bisector of line segment AB intersects BC at point D,\\\\\n$ \\therefore DB=DA$\\\\\nGiven that the perpendicular bisector of line segment AC intersects BC at point F,\\\\\n$ \\therefore EA=EC$,\\\\\n$ \\therefore $ the perimeter of $\\triangle ADE = AD+DE+EA = DB+DE+EC = BC = \\boxed{8}$",
    "image_path": "dataset/mm_math/images/MM_Math/52603847.png"
  },
  {
    "solution": "\\boxed{18^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In quadrilateral $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$, where $AO=CO=10$, $BO=DO$, and $AB=12$, $BC=16$. Quadrilateral $ABCD$ is a rectangle. If $\\angle ADF : \\angle FDC=3:2$, and $DF\\perp AC$ at point $E$, what is the measure of $\\angle BDF$?\n",
    "completion": "\\textbf{Solution:}\nGiven that $\\angle ADC=90^\\circ$ and $\\angle ADF : \\angle FDC = 3 : 2$,\nit follows that $\\angle FDC=36^\\circ$.\nGiven also that $DF \\perp AC$,\nit implies that $\\angle DCO=90^\\circ - 36^\\circ = 54^\\circ$.\nSince quadrilateral ABCD is a rectangle,\nit follows that $OC=OD$,\nwhich means $\\angle ODC=54^\\circ$.\nTherefore, $\\angle BDF=\\angle ODC - \\angle FDC=\\boxed{18^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51614832_85.png"
  },
  {
    "solution": "\\boxed{y= -\\frac{2}{3}x+ \\frac{8}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the quadrilateral OABC is a rectangle, with points A and C located on the x-axis and y-axis, respectively. Triangle $ODE$ is obtained by rotating triangle $OCB$ $90^\\circ$ clockwise around point O. Point D is on the x-axis, line BD intersects the y-axis at point F, and intersects OE at point H. The coordinates of point B are $(-2, 4)$. What is the equation of line BD?\n",
    "completion": "\\textbf{Solution:} Given: $D(4,0)$ \\\\\nLet the expression of line BD be: $y=kx+b$ \\\\\nThen we have $\\begin{cases} -2k+b=4 \\\\ 4k+b=0 \\end{cases}$ \\\\\nSolving, we get: $\\begin{cases} b=\\frac{8}{3} \\\\ k=-\\frac{2}{3} \\end{cases}$ \\\\\n$\\therefore$The expression of line BD is: $\\boxed{y= -\\frac{2}{3}x+ \\frac{8}{3}}$",
    "image_path": "dataset/mm_math/images/MM_Math/52251770_74.png"
  },
  {
    "solution": "\\boxed{x\\geq 1}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: The coordinates of the intersection point of the lines $y_{1}=k_{1}x+a$ and $y_{2}=k_{2}x+b$ are $(1, 2)$. What is the range of values of $x$ for which $y_{1} \\ge y_{2}$?\n",
    "completion": "\\textbf{Solution:} Given the graph of the function, the inequality $y_{1}\\ge y_{2}$, which translates to $k_{1}x+a\\ge k_{2}x+b$, corresponds to the set of x-values for which the graph of $y_{1}=k_{1}x+a$ is above the graph of $y_{2}=k_{2}x+b$. Therefore, the solution set to the inequality $y_{1}\\ge y_{2}$ is $x\\geq 1$, \\\\ \nhence the answer is: $\\boxed{x\\geq 1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53182680.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, in the Cartesian coordinate system $xOy$, the parabola $y=ax^2+bx$ passes through the point $\\left(-2, 5\\right)$, and intersects with the line $y=-\\frac{1}{2}x$ at point A in the second quadrant. A line perpendicular to the $x$-axis passing through point A meets the $x$-axis at point $B\\left(-4, 0\\right)$. Let P be a moving point on the parabola above line OA, draw $PC\\perp x$-axis at point C, and intersect line OA at point D. Connect $OP$ and $PA$. What is the maximum value of the area $S$ of $\\triangle AOP$?\n",
    "completion": "\\textbf{Solution}: Let point $P\\left(t, -{t}^{2}-\\frac{9}{2}t\\right)$, then point $D\\left(t, -\\frac{1}{2}t\\right)$,\\\\\n$\\therefore PD=\\left(-{t}^{2}-\\frac{9}{2}t\\right)-\\left(-\\frac{1}{2}t\\right)=-{t}^{2}-4t$,\\\\\n$\\therefore S=\\frac{1}{2}\\cdot PD\\cdot 4=\\frac{1}{2}\\cdot \\left(-{t}^{2}-4t\\right)\\cdot 4=-2{\\left(t+2\\right)}^{2}+8$,\\\\\n$\\therefore$ when $t=-2$, $S_{\\max}=\\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51323941_197.png"
  },
  {
    "solution": "\\boxed{120^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in Figure 1, point O is a point on the line AB. A ray OC is drawn through point O such that $\\angle AOC=60^\\circ$. A right-angled triangle plate is placed with its right-angle vertex at point O, one side OM on the ray OB, and the other side ON below the line AB. The triangle plate in Figure 1 is rotated counterclockwise around point O to Figure 2, such that one side OM is inside $\\angle BOC$ and exactly bisects $\\angle BOC$. What are the measures of $\\angle CON$ and $\\angle AOM$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle AOC = 60^\\circ$, it follows that $\\angle BOC = 120^\\circ$. Since OM bisects $\\angle BOC$, we have $\\angle COM = \\frac{1}{2}\\angle BOC = 60^\\circ$. Thus, $\\angle CON = \\angle COM + 90^\\circ = 150^\\circ$, and $\\angle AOM = \\angle AOC + \\angle COM = 60^\\circ + 60^\\circ = 120^\\circ$; therefore, the measure of $\\angle CON$ is $\\boxed{150^\\circ}$, and the measure of $\\angle AOM$ is $\\boxed{120^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52441579_83.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, the altitude $AE$ intersects $BC$ at point $E$. If $\\frac{1}{2}\\angle ABE + \\angle C = 45^\\circ$, $CE = 5\\sqrt{2}$, and the area of $\\triangle ABC$ is 20, what is the length of $AB$?\n",
    "completion": "\\textbf{Solution:} Draw CD $\\perp$ BA from point C, intersecting the extension line of BA at point D,\\\\\nthus $\\angle ADC = 90^\\circ$,\\\\\nsince $\\frac{1}{2}\\angle ABE + \\angle ACB = 45^\\circ$,\\\\\nthus $\\angle ABE + 2\\angle ACB = 90^\\circ$,\\\\\nsince $\\angle ABE + \\angle ACB + \\angle ACD = 90^\\circ$,\\\\\nthus $\\angle ACB = \\angle ACD$,\\\\\nsince AE is the altitude of $\\triangle ABC$,\\\\\nthus $\\angle AEC = \\angle ADC = 90^\\circ$,\\\\\nin $\\triangle AEC$ and $\\triangle ADC$,\\\\\n$\\left\\{\\begin{array}{l}\n\\angle AEC=\\angle ADC \\\\\n\\angle ACE=\\angle ACD \\\\\nAC=AC\n\\end{array}\\right.$\\\\\nthus $\\triangle AEC\\cong \\triangle ADC$ (AAS),\\\\\nthus CE$=CD=5\\sqrt{2}$,\\\\\nsince $S_{\\triangle ABC}=\\frac{1}{2}\\text{AB}\\cdot\\text{CD}$,\\\\\nthus $20=\\frac{1}{2}\\times\\text{AB}\\times 5\\sqrt{2}$,\\\\\nthus AB$=\\boxed{4\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52861346.png"
  },
  {
    "solution": "\\boxed{0.5\\ km/s}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Please provide the text you need translated into English and formatted in standard LaTeX.\n",
    "completion": "\\textbf{Solution:} In $Rt\\triangle ARL$, $AL=AR\\cdot \\sin 39^\\circ \\approx 5\\times 0.63=3.15$ ($km$),\\\\\nIn $Rt\\triangle BRL$, $BL=RL\\approx 3.90$ ($km$),\\\\\n$\\therefore AB=BL-AL=3.90-3.15\\approx 0.75$ ($km$),\\\\\n$\\therefore$ the average velocity of the rocket from A to B is $0.75\\div 1.5=0.5$ ($km/s$),\\\\\n\\textbf{Answer:} The average velocity of the rocket from A to B is $\\boxed{0.5\\ km/s}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53043157_37.png"
  },
  {
    "solution": "\\boxed{25}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, the areas of the two smaller squares are 9 and 16, respectively. What is the area of the square represented by the letter A?\n",
    "completion": "\\textbf{Solution:} The area of the square represented by the letter A is: $16+9= \\boxed{25}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53232740.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AD \\perp BC$ at point $D$, $BE \\perp AC$ at point $E$, and $BE$ intersects $AD$ at point $F$. If $AD=BD=5$ and $CD=3$, what is the length of $AF$?\n",
    "completion": "\\textbf{Solution:} Since $BE\\perp AC$ and $AD\\perp BC$,\\\\\nit follows that $\\angle AEB=\\angle ADC=\\angle BDF=90^\\circ$,\\\\\nGiven $\\angle AFE=\\angle BFD$ and $\\angle FBD+\\angle BDF+\\angle BFD=180^\\circ$, $\\angle AEB+\\angle AFE+\\angle DAC=180^\\circ$,\\\\\nit implies $\\angle DAC=\\angle DBF$,\\\\\nIn $\\triangle BDF$ and $\\triangle ADC$, we have\\\\\n$\\left\\{\\begin{array}{l}\n\\angle DBF=\\angle DAC\\\\\nBD=AD\\\\\n\\angle BDF=\\angle ADC\n\\end{array}\\right.$,\\\\\nwhich leads to $\\triangle BDF \\cong \\triangle ADC\\left(ASA\\right)$,\\\\\nthus $DF=CD=3$,\\\\\nSince $AF+DF=AD=5$,\\\\\nit results in $AF=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53059206.png"
  },
  {
    "solution": "\\boxed{\\frac{1}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that in $\\triangle ABC$, $\\angle ABC=90^\\circ$, the circle $\\odot O$ with diameter $AB$ intersects $AC$ at point $D$, and point $E$ is the midpoint of $BC$, connect $BD$, $DE$. If $\\frac{AD}{AB}=\\frac{1}{3}$, find $\\sin C$.\n",
    "completion": "\\textbf{Solution:} From the given information, we know that $AB$ is a diameter,\\\\\n$\\therefore \\angle ADB=90^\\circ$,\\\\\n$\\therefore \\angle ABD + \\angle BAD=90^\\circ$,\\\\\n$\\because \\angle ABC=90^\\circ$,\\\\\n$\\therefore \\angle C + \\angle BAC=90^\\circ$,\\\\\n$\\therefore \\angle C = \\angle ABD$,\\\\\n$\\because \\frac{AD}{AB}=\\frac{1}{3}$,\\\\\n$\\therefore \\sin \\angle ABD=\\frac{1}{3}$,\\\\\n$\\therefore \\sin C = \\boxed{\\frac{1}{3}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/51322908_131.png"
  },
  {
    "solution": "\\boxed{18\\text{cm}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $ \\triangle ABC$, $ \\angle ABC = 90^\\circ$, $ \\angle A = 60^\\circ$, and $D$ is a point on side $AC$, with $ \\angle DBC = 30^\\circ$ and $ AC = 12\\,cm$. What is the perimeter of $ \\triangle ABD$?\n",
    "completion": "\\textbf{Solution:} Given that in $\\triangle ABC$, $\\angle ABC=90^\\circ$, $\\angle A=60^\\circ$,\\\\\nthus $\\angle C=30^\\circ$,\\\\\nand since $\\angle DBC=30^\\circ$,\\\\\nit follows that $\\angle ABD=\\angle ABC-\\angle DBC=60^\\circ$, and $BD=DC$,\\\\\nthus $\\angle ABD=\\angle A=60^\\circ$,\\\\\ntherefore $\\triangle ABD$ is an equilateral triangle,\\\\\nthus $BD=AD=AB$,\\\\\nsince $BD=DC$ and $AC=12\\text{cm}$,\\\\\nit follows that $AD=DC=\\frac{1}{2}AC=6\\text{cm}$,\\\\\nthus $BD=AD=AB=6\\text{cm}$,\\\\\ntherefore, the perimeter of $\\triangle ABD$ is: $BD+AD+AB=6\\times 3=\\boxed{18\\text{cm}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52978711.png"
  },
  {
    "solution": "\\boxed{24\\,cm}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point $C$ is the midpoint of line segment $AB$, point $D$ is the midpoint of line segment $AC$, and $BE=\\frac{1}{3}BC$. If $DE=14\\,\\text{cm}$, find the length of $AB$.\n",
    "completion": "\\textbf{Solution:} Let $BE = x\\,cm$, hence $BC = 3x\\,cm$,\\\\\nsince point $C$ is the midpoint of segment $AB$,\\\\\ntherefore $AB = 2BC = 6x\\,cm$, $AC = 3x\\,cm$,\\\\\nsince point $D$ is the midpoint of segment $AC$,\\\\\ntherefore $AD = \\frac{1}{2}AC = \\frac{3}{2}x\\,cm$,\\\\\ntherefore $DE = AB - AD - BE = 6x - \\frac{3}{2}x - x = \\frac{7}{2}x\\,cm$,\\\\\nsince $DE = 14\\,cm$,\\\\\ntherefore $\\frac{7}{2}x = 14$, solving for $x$ yields: $x = 4$,\\\\\ntherefore $AB = 6x = \\boxed{24\\,cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53205976_51.png"
  },
  {
    "solution": "\\boxed{9}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that line segment $AB = 23$, $BC = 15$, and point $M$ is the midpoint of $AC$. Take a point $N$ on $CB$ such that the ratio $CN:NB = 1:2$. What is the length of line segment $MN$?\n",
    "completion": "Solution: Since $BC=15$ and $CN:NB=1:2$, \\\\\nit follows that $CN=\\frac{1}{3}BC=\\frac{1}{3}\\times 15=5$, \\\\\nand since point M is the midpoint of AC, and $AC=8$, \\\\\nit follows that $MC=\\frac{1}{2}AC=4$, \\\\\ntherefore $MN=MC+NC=4+5=\\boxed{9}$, meaning the length of MN is 9.",
    "image_path": "dataset/mm_math/images/MM_Math/53210853_93.png"
  },
  {
    "solution": "\\boxed{C}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: In the Cartesian coordinate system, the ray OA in the first quadrant forms an angle $\\alpha$ with the positive x-axis. If $\\sin\\alpha =\\frac{3}{5}$, what could be the coordinates of point P located on ray OA?\n",
    "completion": "\\textbf{Solution:} Draw PB $\\perp$ OB at point B through point P.\\\\\nA, $\\because$ point P (3, 5), $\\therefore$ OB = 3, PB = 5,\\\\\n$\\therefore$ OP = $\\sqrt{34}$ (by the Pythagorean theorem),\\\\\n$\\therefore \\sin\\alpha = \\frac{PB}{OP} = \\frac{5}{\\sqrt{34}} = \\frac{5\\sqrt{34}}{34} \\ne \\frac{3}{5}$, thus this option does not meet the criteria;\\\\\nB, $\\because$ point P (5, 3), $\\therefore$ OB = 5, PB = 3,\\\\\n$\\therefore$ OP = $\\sqrt{34}$ (by the Pythagorean theorem),\\\\\n$\\therefore \\sin\\alpha = \\frac{PB}{OP} = \\frac{3}{\\sqrt{34}} = \\frac{3\\sqrt{34}}{34} \\ne \\frac{3}{5}$, thus this option does not meet the criteria;\\\\\nC, $\\because$ point P (4, 3), $\\therefore$ OB = 4, PB = 3,\\\\\n$\\therefore$ OP = $5$ (by the Pythagorean theorem),\\\\\n$\\therefore \\sin\\alpha = \\frac{PB}{OP} = \\frac{3}{5}$, thus this option meets the criteria;\\\\\nD, $\\because$ point P (3, 4), $\\therefore$ OB = 3, PB = 4,\\\\\n$\\therefore$ OP = $5$ (by the Pythagorean theorem),\\\\\n$\\therefore \\sin\\alpha = \\frac{PB}{OP} = \\frac{4}{5} \\ne \\frac{3}{5}$, thus this option does not meet the criteria;\\\\\nTherefore, the answer is: \\boxed{C}.",
    "image_path": "dataset/mm_math/images/MM_Math/51103664.png"
  },
  {
    "solution": "\\boxed{\\left(\\frac{\\sqrt{2}}{2}, \\frac{\\sqrt{2}}{2}\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in the Cartesian coordinate system, the isosceles right triangle $ABC$ and the isosceles right triangle $DEF$ are similar figures, with their hypotenuses perpendicular to the x-axis. Let $O$ be the center of similarity, $\\angle ABC = \\angle DEF = 90^\\circ$, and the points $O$, $B$, $C$, $E$, $F$ are collinear. If the area ratio ${S}_{\\triangle DEF} \\colon {S}_{\\triangle ABC} = 1 \\colon 2$, and the coordinates of point $D$ are $(-1, 0)$, what are the coordinates of point $B$?\n",
    "completion": "\\textbf{Solution:} Since isosceles right triangle $ABC$ and isosceles right triangle $DEF$ are homothetic figures, it follows that $\\triangle ABC\\sim \\triangle DEF$, $\\frac{OD}{OA}=\\frac{DE}{AB}$. Given that the area ratio $S_{\\triangle DEF} : S_{\\triangle ABC}=1 : 2$, then $\\frac{S_{\\triangle DEF}}{S_{\\triangle ABC}}=\\frac{1}{2}=\\left(\\frac{DE}{AB}\\right)^2$. Hence, $\\frac{OD}{OA}=\\frac{DE}{AB}=\\frac{1}{\\sqrt{2}}$. Since point $D$ is at $(-1,0)$, OD equals 1, thus OA equals $\\sqrt{2}$. Given $\\triangle ABC$ is an isosceles right triangle, we have $\\angle C=\\angle CAB=45^\\circ$, $\\angle ABC=\\angle OBA=90^\\circ$. Therefore, AC is perpendicular to the x-axis, leading to $\\angle CAB+\\angle BAO=90^\\circ$, which means $\\angle BAO=\\angle OBA=45^\\circ$. Consequently, OB=AB. Draw BG perpendicular to OA, thus BG=OG=$\\frac{1}{2}$OA=$\\frac{\\sqrt{2}}{2}$. Therefore, the coordinates of point B are $\\boxed{\\left(\\frac{\\sqrt{2}}{2}, \\frac{\\sqrt{2}}{2}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53081475.png"
  },
  {
    "solution": "\\boxed{\\sqrt{10}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $AB=AC=2\\sqrt{2}$, $\\angle BAC=90^\\circ$. On the line segment $BC$, there is a moving point $D$. Connect $AD$. Through point $A$, construct $AD \\perp AE$ and $AD=AE$. Connect $DE$. When $BD=1$, what is the length of $DE$?\n",
    "completion": "\\textbf{Solution:} When $BD = 1$, the length of $DE$ is $\\boxed{\\sqrt{10}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55502350_44.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given: As shown in the figure, $OC$ is the bisector of $\\angle AOB$. When $\\angle AOB = 60^\\circ$, what is the measure of $\\angle AOC$?\n",
    "completion": "\\textbf{Solution:} Given that OC bisects $\\angle AOB$,\\\\\nit follows that $\\angle AOC= \\frac{1}{2}\\angle AOB$,\\\\\nsince $\\angle AOB=60^\\circ$,\\\\\nwe find $\\angle AOC=\\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52795995_113.png"
  },
  {
    "solution": "\\boxed{\\frac{2}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that in trapezoid $ABCD$, $AB \\parallel CD$, and diagonals $AC$ and $BD$ intersect at point $O$, with $CO = \\frac{2}{5} AC$. What is the value of $\\frac{CD}{AB}$?\n",
    "completion": "\\textbf{Solution:} Because $CO = \\frac{2}{5}AC$, \\\\\nthus $CO: OA = 2:3$, \\\\\nBecause $CD \\parallel AB$, \\\\\nthus $\\frac{CD}{AB} = \\frac{OC}{OA} = \\boxed{\\frac{2}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52917601_83.png"
  },
  {
    "solution": "\\boxed{3.5}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in $\\triangle ABC$, $AB=12$, $AC=16$, $BC=20$. If point $P$ is a point on line segment $AC$, connect $BP$, and $BP=CP$, what is the length of $AP$?\n",
    "completion": "\\textbf{Solution:} Let $AP=x$, then $BP=CP=16-x$.\\\\\nIn $\\triangle ABP$ right-angled at $B$, since $AB^2+AP^2=BP^2$,\\\\\nhence $12^2+x^2=(16-x)^2$,\\\\\nsolving gives $x=\\boxed{3.5}$, therefore the length of $AP$ is \\boxed{3.5}.",
    "image_path": "dataset/mm_math/images/MM_Math/53081172_110.png"
  },
  {
    "solution": "\\boxed{\\frac{2m}{3}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that line segment $AB=m$ (where $m$ is a constant), point $C$ is a point on the line $AB$ (not coinciding with $A$ or $B$), and points $P$, $Q$ are respectively on segments $BC$, $AC$, and satisfy $CQ=2AQ$, $CP=2BP$. As shown in Figure 1, with point $C$ on segment $AB$, what is the length of $PQ$ (expressed in terms of $m$)?\n",
    "completion": "\\textbf{Solution:} Let $AQ=a$ and $BP=b$,\\\\\nthen $CQ=2a$ and $CP=2b$,\\\\\nsince $AB=AQ+CQ+CP+BP=a+2a+2b+b=3a+3b=m$,\\\\\nit follows that $a+b=\\frac{m}{3}$,\\\\\nthus $PQ=CQ+CP=2a+2b=2(a+b)=\\boxed{\\frac{2m}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/54927374_14.png"
  },
  {
    "solution": "\\boxed{72}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the diagram, in $\\triangle ABC$, points $D$ and $E$ are on $AB$ and $AC$ respectively. If $\\frac{AD}{AB}=\\frac{AE}{AC}=\\frac{1}{3}$, and the area of $\\triangle ADE$ is 9, what is the area of quadrilateral $BCED$?\n",
    "completion": "\\textbf{Solution}: Since $\\because \\frac{AD}{AB}=\\frac{AE}{AC}=\\frac{1}{3}$ and $\\angle A=\\angle A$,\\\\\n$\\therefore \\triangle ADE \\sim \\triangle ABC$ with a similarity ratio of $k=\\frac{1}{3}$,\\\\\n$\\therefore S_{\\triangle ADE} : S_{\\triangle ABC} = 1 : 9$,\\\\\n$\\because S_{\\triangle ADE} = 9$,\\\\\n$\\therefore S_{\\triangle ABC} = 81$,\\\\\n$\\therefore$ the area of the quadrilateral $BCED$ is $S_{\\triangle ABC} - S_{\\triangle ADE} = 81 - 9 = \\boxed{72}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52872003.png"
  },
  {
    "solution": "\\boxed{[-4,0)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the parabola $y=x^2+bx+c$ passes through points $A(-1,0)$ and $B(3,0)$. What is the range of $y$ when $0<x<3$?\n",
    "completion": "\\textbf{Solution:} From the graph, we can determine that when $0<x<3$,\\\\\nwe have $-4\\le y<0$.\\\\\nTherefore, the range of $y$ is $y\\in\\boxed{[-4,0)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52978190_214.png"
  },
  {
    "solution": "\\boxed{\\frac{32\\pi }{3}-8\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, $\\odot O$ is a circle centered at the coordinate origin $O$ with a radius of $4\\sqrt{2}$. The coordinate of point $P$ is $(2, 2)$, and the chord $AB$ passes through point $P$. What is the minimum area of the shaded region in the figure?\n",
    "completion": "\\textbf{Solution:} From the given conditions, when OP$\\perp$AB, the area of the shaded region is minimized. \\\\\nSince P(2,2), \\\\\nthus OP=2$\\sqrt{2}$, \\\\\nSince OA=OB=4$\\sqrt{2}$, \\\\\nthus PA=PB=2$\\sqrt{6}$, \\\\\nthus $\\tan\\angle POB=\\frac{PB}{PO}=\\sqrt{3}$, \\\\\nthus $\\angle AOP=\\angle BOP=60^\\circ$, \\\\\nthus $\\angle AOB=120^\\circ$, \\\\\nthus the area of the shaded region $S_{\\text{shaded}}=S_{\\text{sector} AOB}-S_{\\triangle AOB}=\\frac{120\\times \\pi \\times {\\left(4\\sqrt{2}\\right)}^{2}}{360}-\\frac{1}{2}\\times 4\\sqrt{6}\\times 2\\sqrt{2}=\\boxed{\\frac{32\\pi }{3}-8\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51179563.png"
  },
  {
    "solution": "\\boxed{55^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, point \\(O\\) is a point inside \\(\\triangle ABC\\), \\(\\angle ABO=30^\\circ\\), \\(\\angle ACO=15^\\circ\\), \\(\\angle BOC=100^\\circ\\), then what is the degree measure of \\(\\angle A\\)?\n",
    "completion": "\\textbf{Solution:} Since, in $\\triangle OBC$, $\\angle BOC=100^\\circ$,\\\\\nit follows that $\\angle OBC+\\angle OCB=180^\\circ-\\angle BOC=80^\\circ$,\\\\\nIn $\\triangle ABC$,\\\\\n$\\angle A=180^\\circ-(\\angle ABO+\\angle ACO+\\angle OBC+\\angle OCB)=180^\\circ-(30^\\circ+15^\\circ+80^\\circ)=\\boxed{55^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53058542.png"
  },
  {
    "solution": "\\boxed{x<2}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: The line $y_{1} = -x + 3$ and the line $y_{2} = kx - 2$ respectively intersect the $y$-axis at points $A$ and $B$, and intersect each other at point $C(2, m)$. Based on the graph, directly write the range of the independent variable $x$ when $y_{1} > y_{2}$.\n",
    "completion": "\\textbf{Solution:} As shown in the figure, when $\\boxed{x<2}$, $y_{1}>y_{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51968557_186.png"
  },
  {
    "solution": "\\boxed{48}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, in the rhombus $ABCD$, the diagonals $AC$ and $BD$ intersect at point $O$. Given that $AC = 8$ and $BD = 12$, what is the area of the rhombus $ABCD$?\n",
    "completion": "\\textbf{Solution:} In the rhombus $ABCD$, $AC=8$, $BD=12$,\\\\\nsince $AC\\perp BD$,\\\\\ntherefore the area of the rhombus $ABCD$ $=\\frac{1}{2}AC\\cdot BD=\\frac{1}{2}\\times 8\\times 12=\\boxed{48}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53359524.png"
  },
  {
    "solution": "\\boxed{3000\\left(\\sqrt{3}-1\\right)}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As illustrated, one day, a large fishing vessel operating in the South China Sea suddenly encountered a dangerous situation. Upon receiving the distress signal, the relevant departments immediately dispatched a helicopter and a fishery administration ship that was patrolling in the South China Sea for rescue. When the helicopter reached a height of $3000\\,m$ above the sea surface at point $C$, the depression angle to the fishery administration ship at point $A$ was measured to be $45^\\circ$, and the depression angle to the distressed fishing vessel at point $B$ was measured to be $30^\\circ$. What is the distance $AB$ between the fishery administration ship and the fishing vessel at this moment?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, according to the given information, CE$\\parallel$BD,\\\\\n$\\therefore \\angle CBA=30^\\circ$, $\\angle CAD=45^\\circ$, and CD=3000m,\\\\\nIn $\\triangle ACD$, AD=CD=3000m,\\\\\nIn $\\triangle BCD$,\\\\\nBD=$ \\frac{CD}{\\tan \\angle CBA}$=$ \\frac{3000}{\\frac{\\sqrt{3}}{3}}$=$ 3000\\sqrt{3}$m,\\\\\n$\\therefore$AB=BD−AD=$ 3000\\sqrt{3}$−3000=$ 3000\\left(\\sqrt{3}−1\\right)$ (m),\\\\\nTherefore, the answer is: $AB=\\boxed{3000\\left(\\sqrt{3}-1\\right)}$ (m).",
    "image_path": "dataset/mm_math/images/MM_Math/52537144.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{2}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that the area of the rhombus ABCD is $2\\sqrt{6}$, and the length of the diagonal AC is $2\\sqrt{3}$. What is the length of the diagonal BD?\n",
    "completion": "\\textbf{Solution:} Since the area of the rhombus $ABCD$ is $2\\sqrt{6}$ and the length of diagonal $AC$ is $2\\sqrt{3}$, \\\\\n\\therefore $\\frac{1}{2}\\times 2\\sqrt{3}\\times BD=2\\sqrt{6}$, solving this we get $BD=\\boxed{2\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51510287_137.png"
  },
  {
    "solution": "\\boxed{\\frac{5}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, given a rectangular paper $ABCD$, where point $E$ is on side $AB$ and $BE=2, BC=3$, if $\\triangle CBE$ is folded along the line $CE$ such that point $B$ falls on point $G$, and extending $EG$ intersects $CD$ at point $F$, then what is the length of segment $FG$?\n",
    "completion": "\\textbf{Solution:} Given that by folding the triangle $\\triangle CBE$ along the line $CE$, with $BE=2$ and $BC=3$, \\\\\nit follows that $\\angle BEC=\\angle FEC$ and $\\angle CGE=\\angle B=90^\\circ$. Hence, $BE=GE=2$ and $BC=CG=3$. \\\\\nSince we are dealing with the rectangle $ABCD$, \\\\\nit implies that $AB\\parallel CD$, \\\\\nwhich leads to the conclusion that $\\angle BEC=\\angle FCE$, \\\\\ntherefore, $\\angle FEC=\\angle FCE$, \\\\\nindicating that $FE=FC$. \\\\\nLet $FG=x$, then $FC=x+2$, \\\\\nthus, in the right triangle $\\triangle FGC$, we have $(x+2)^2=x^2+3^2$, \\\\\nsolving this, we get $x=\\frac{5}{4}$, which means $FG=\\boxed{\\frac{5}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52828224.png"
  },
  {
    "solution": "\\boxed{11}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, points $C$ and $D$ divide the line segment $AB$ into three parts in the ratio $2\\colon 5\\colon 3$, with $E$ being the midpoint of $AB$. If the numbers represented by points $A$, $B$, and $D$ are $-10$, $+20$, and $x$ respectively, what is the value of $x$?\n",
    "completion": "\\textbf{Solution:} Given the problem, the numbers represented by points $A$, $B$, $D$ are respectively $-10$, $+20$, $x$,\\\\\n$\\therefore AB=20-(-10)=30$, $BD=20-x$,\\\\\n$\\because$ On the number line, points $C$ and $D$ divide segment $AB$ into three parts in the ratio $2\\colon 5\\colon 3$,\\\\\n$\\therefore BD=\\frac{3}{2+5+3}AB=\\frac{3}{10}\\times 30=9$,\\\\\n$\\therefore 20-x=9$,\\\\\n$\\therefore x=\\boxed{11}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52792740_145.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{10}}{4}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given that $AC$ and $EC$ are the diagonals of the rectangles $ABCD$ and $EFCG$ respectively, point $E$ is inside $\\triangle ABC$, and $\\frac{AB}{BC}=\\frac{EF}{FC}=k$, $\\angle CAE+\\angle CBE=90^\\circ$. $\\triangle CAE\\sim \\triangle CBF$; if $BE=2$, $AE=4$, $CE=6$, find the value of $k$.\n",
    "completion": "\\textbf{Solution:}\\\\\nSince $\\triangle CAE\\sim \\triangle CBF$,\\\\\nwe have $\\frac{AC}{BC}=\\frac{EF}{FC}=k$,\\\\\nthus $\\angle CBF=\\angle CAE$,\\\\\nand since $\\angle CAE+\\angle CBE=90^\\circ$,\\\\\nit follows that $\\angle EBF=90^\\circ$,\\\\\nlet $BC=a$, then $AB=ka$, let $FC=b$, then $EF=kb$,\\\\\nthus $AC=\\sqrt{AB^2+BC^2}=a\\sqrt{k^2+1}$,\\\\\nand $EC=\\sqrt{EF^2+FC^2}=b\\sqrt{k^2+1}$,\\\\\nhence $\\frac{AE}{BF}=\\frac{AC}{BC}=\\sqrt{k^2+1}$,\\\\\ngiven that $AE=4$,\\\\\nit follows that $BF=\\frac{4}{\\sqrt{k^2+1}}$,\\\\\nsince $\\angle EBF=90^\\circ$,\\\\\nit follows that $EF^2=BE^2+BF^2=4+\\frac{16}{k^2+1}=k^2b^2$,\\\\\nsince $CE^2=EF^2+CF^2$, i.e., $6^2=(kb)^2+b^2$,\\\\\nwe get $b^2=\\frac{36}{k^2+1}$, hence $\\frac{36k^2}{k^2+1}=4+\\frac{16}{k^2+1}$,\\\\\nsolving for $k$ yields $k=\\boxed{\\frac{\\sqrt{10}}{4}}$ or $k=-\\frac{\\sqrt{10}}{4}$ (discard as irrelevant).\\\\\nTherefore, the value of $k$ is $\\boxed{\\frac{\\sqrt{10}}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51368311_104.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{5}}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: As shown in the figure, it is known that in $\\triangle ABC$, $CD \\perp AB$ with the foot of the perpendicular at point $D$, $AD=2$, $BD=6$, $\\tan\\angle B=\\frac{2}{3}$. Point $E$ is the midpoint of side $BC$. Find the length of side $AC$.\n",
    "completion": "\\textbf{Solution:} Since $CD \\perp AB$ \\\\\nTherefore, both $\\triangle ACD$ and $\\triangle BCD$ are right-angled triangles, \\\\\nSince $\\tan B = \\frac{2}{3}$ \\\\\nTherefore $\\frac{CD}{BD} = \\frac{2}{3}$ \\\\\nSince $BD = 6,$ \\\\\nTherefore $CD = \\frac{2}{3}BD = \\frac{2}{3} \\times 6 = 4$ \\\\\nSince $AD = 2$ \\\\\nBy the Pythagorean theorem, we have $AC = \\sqrt{CD^{2}+AD^{2}} = \\sqrt{4^{2}+2^{2}} = \\boxed{2\\sqrt{5}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51379561_97.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of \boxed{your answer}.\\n Question: Given point $A$ is on the graph of $y= \\frac{k}{x}$ ($x>0$), and point $B$ is on the negative half of the x-axis. Connect $AB$, intersecting the y-axis at point $C$. If $AC=BC$ and $S_{\\triangle BOC}=1$, what is the value of $k$?\n",
    "completion": "\\textbf{Solution:}\\\\\nConstruct AD $\\perp$ y-axis through point A, with D being the foot of the perpendicular,\\\\\n$\\therefore \\angle ADC=\\angle BOC=90^\\circ$,\\\\\nAlso, $\\because \\angle ACD=\\angle BCO$ and AC=BC,\\\\\n$\\therefore \\triangle BOC \\cong \\triangle ADC$,\\\\\n$\\therefore S_{\\triangle BOC}=S_{\\triangle ADC}$\\\\\nConnect OA, then $S_{\\triangle ACO}= S_{\\triangle BOC}=1$,\\\\\n$\\therefore S_{\\triangle ADO}=2$,\\\\\n$\\therefore k=2\\times 2=\\boxed{4}$",
    "image_path": "dataset/mm_math/images/MM_Math/52601845.png"
  }
]