[
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the Cartesian coordinate system, the side OB of $\\triangle OAB$ lies on the y-axis with $\\angle ABO=90^\\circ$, $AB=4$. Point C is on line AB with $BC=\\frac{1}{4}AB$, and $\\angle BOC=\\angle A$. If the hyperbola $y=\\frac{k}{x}$ passes through point C, what is the value of $k$?\n",
    "completion": "\\textbf{Solution:} Given BC $= \\frac{1}{4}$ and AB = 4, the x-coordinate of point C is 1.\\\\\nSince in the right triangles $\\triangle ABO$ and $\\triangle OBC$, $\\angle ABO = \\angle OBC$ and $\\angle BOC = \\angle A$,\\\\\nit follows that $\\triangle ABO \\sim \\triangle OBC$,\\\\\nwhich implies that $ \\frac{AB}{OB} = \\frac{OB}{BC}$,\\\\\nleading to $OB^2 = AB \\cdot BC = 4 \\cdot 1 = 4$,\\\\\nthus, OB = 2.\\\\\nThe coordinates of point C are (1, 2). Substituting into $y = \\frac{k}{x}$, we get: $k = \\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51446382.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, the square $ABCD$ and the equilateral triangle $\\triangle CDE$ share the common side $CD$. Connect $AE$ and $AC$. What is the measure of $\\angle EAC$ in degrees?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral $ABCD$ is a square, \\\\\nit follows that $DA=DC$, $\\angle ADC=90^\\circ$, and $\\angle CAD=45^\\circ$, \\\\\nsince $\\triangle CDE$ is an equilateral triangle, \\\\\nit follows that $DE=DC$, $\\angle CDE=60^\\circ$, \\\\\ntherefore, $DA=DE$, $\\angle ADE=90^\\circ+60^\\circ=150^\\circ$, \\\\\nthus, $\\triangle ADE$ is an isosceles triangle, \\\\\ntherefore, $\\angle DAE=\\angle DEA$, \\\\\nhence, $\\angle DAE = \\frac{1}{2}(180^\\circ-150^\\circ)=15^\\circ$, \\\\\ntherefore, $\\angle EAC=\\angle CAD - \\angle DAE = 45^\\circ - 15^\\circ = \\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53182928.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</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 $A$ 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 value of the tangent of $\\angle ACB$?\n",
    "completion": "\\textbf{Solution:} Draw $BE\\perp AC$ at point E,\\\\\nLet $y=-x^2-2x+3=0$\\\\\nThen $x^2+2x-3=0$\\\\\n$\\therefore x_1=-3$, $x_2=1$\\\\\nHence $A(-3,0)$, $B(1,0)$\\\\\n$\\therefore AB=1-(-3)=4$, $OC=3$\\\\\n$\\therefore S_{\\triangle ABC}=\\frac{1}{2}\\cdot AB\\cdot OC=\\frac{1}{2}\\times 4\\times 3=6$\\\\\n$\\because A(-3, 0)$, $C(0, 3)$\\\\\n$\\therefore AC=\\sqrt{(-3-0)^2+(0-3)^2}=3\\sqrt{2}$\\\\\n$\\because S_{\\triangle ABC}=\\frac{1}{2}\\cdot AC\\cdot BE=\\frac{1}{2}\\times 3\\sqrt{2}\\times BE$\\\\\n$\\therefore \\frac{1}{2}\\times 3\\sqrt{2}\\times BE=6$\\\\\n$\\therefore BE=2\\sqrt{2}$\\\\\n$\\because BC^2=OC^2+OB^2=3^2+1^2=10$\\\\\n$\\therefore EC=\\sqrt{BC^2-BE^2}=\\sqrt{10-8}=\\sqrt{2}$\\\\\n$\\therefore \\tan\\angle ACB=\\frac{BE}{EC}=\\frac{2\\sqrt{2}}{\\sqrt{2}}=\\boxed{2}$",
    "image_path": "dataset/mm_math/images/MM_Math/51319005_105.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</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 the extension lines of $CF$ and $DE$ intersect at point $F$. If $AB=4$ and $CF=3$, what is the length of $BD$?\n",
    "completion": "\\textbf{Solution:} Since $CF\\parallel AB$,\\\\\nit follows that $\\angle ADF=\\angle CFD$ and $\\angle DAC=\\angle ACF$,\\\\\nsince $E$ is the midpoint of side $AC$,\\\\\nit follows that $AE=EC$,\\\\\nin $\\triangle ADE$ and $\\triangle CFE$,\\\\\nwe have $\\left\\{\\begin{array}{l}\n\\angle ADF=\\angle CFD \\\\\n\\angle DAC=\\angle ACF \\\\\nAE=EC\n\\end{array}\\right.$,\\\\\ntherefore, $\\triangle ADE\\cong \\triangle CFE$,\\\\\ntherefore, $AD=CF=3$,\\\\\ntherefore, $BD=AB−AD= \\boxed{1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53010085.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{3}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in rectangle $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$, $\\angle ABD = 60^\\circ$, $BD = 4$. What is the area of rectangle $ABCD$?\n",
    "completion": "\\textbf{Solution:} In rectangle $ABCD$, $\\angle BAD=90^\\circ$ and $AC=BD=4$,\\\\\n$OA=\\frac{1}{2}AC$, $OB=\\frac{1}{2}BD=2$,\\\\\nthus $OA=OB$,\\\\\nand since $\\angle ABD=60^\\circ$,\\\\\ntherefore $\\triangle AOB$ is an equilateral triangle,\\\\\nthus $AB=OB=2$,\\\\\nsince in $\\triangle ABD$ with $BD=4$,\\\\\ntherefore $AD=\\sqrt{4^2-2^2}=2\\sqrt{3}$,\\\\\nhence, the area of rectangle $ABCD$ is $2\\times 2\\sqrt{3}=\\boxed{4\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52387027.png"
  },
  {
    "solution": "\\boxed{140^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, points $I$ and $O$ are the incenter and circumcenter of $\\triangle ABC$, respectively. If $\\angle AIB = 125^\\circ$, then what is the degree measure of $\\angle AOB$?\n",
    "completion": "\\textbf{Solution:} Construct the circumscribed circle $\\odot O$ of $\\triangle ABC$, and the inscribed circle $\\odot I$ of $\\triangle ABC$,\\\\\nSince point I is the incenter of $\\triangle ABC$,\\\\\nTherefore, AI bisects $\\angle CAB$, BI bisects $\\angle ABC$,\\\\\nTherefore, $\\angle IAB=\\frac{1}{2}\\angle CAB$, $\\angle IBA=\\frac{1}{2}\\angle CBA$,\\\\\nSince $\\angle AIB=125^\\circ$,\\\\\nTherefore, $\\angle IAB+\\angle IBA=180^\\circ-\\angle AIB=55^\\circ$,\\\\\nTherefore, $\\angle CAB+\\angle CBA=2(\\angle IAB+\\angle IBA)=110^\\circ$,\\\\\nTherefore, $\\angle ACB=180^\\circ-(\\angle CAB+\\angle CBA)=70^\\circ$,\\\\\nSince point O is the circumcenter of $\\triangle ACB$,\\\\\nTherefore, $\\angle AOB=2\\angle ACB=140^\\circ$,\\\\\nHence, the answer is: $\\boxed{140^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52537017.png"
  },
  {
    "solution": "\\boxed{\\frac{3}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, it is known that $AB$ is the diameter of the circle $\\odot O$, $PD$ is tangent to $\\odot O$ at point C, and intersects the extension line of $AB$ at point D, and $\\angle D = 2\\angle CAD$. If $CD=2$, find the length of the arc $\\overset{\\frown}{AC}$.\n",
    "completion": "Solution: Since $\\angle COB=\\angle D$, and $CD=2$,\nthus $CO=CD=2$.\nSince $\\angle COB=45^\\circ$,\nthus $\\angle AOC=135^\\circ$.\nTherefore, the length of the arc $\\overset{\\frown}{AC}$, $l=\\frac{n\\pi R}{180}=\\frac{135\\times 2\\times \\pi }{180}=\\boxed{\\frac{3}{2}\\pi}$",
    "image_path": "dataset/mm_math/images/MM_Math/51403374_131.png"
  },
  {
    "solution": "\\boxed{y=x^2-2x-3}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</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 $C(2,n)$. Find the analytical expression of the parabola.\n",
    "completion": "Solution: Substituting A ($-1$, $0$) and B ($3$, $0$) into $y=ax^2+bx-3$, we get\n\\[\n\\left\\{\n\\begin{array}{l}\na-b-3=0 \\\\\n9a+3b-3=0\n\\end{array}\n\\right.\n\\]\nSolving this, we obtain\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/51446075_226.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, point $P$ is a point on the angle bisector $AD$ of $\\angle BAC$, where $AC=9$, $AB=5$, and $PB=3$. What could be the possible length of $PC$?\n",
    "completion": "\\textbf{Solution:} Let's intercept AE=AB=5 on line AC, and connect PE. \\\\\nSince AC=9, \\\\\nthus CE=AC-AE=9-5=4, \\\\\nSince point P is on the angle bisector AD of $\\angle$BAC, \\\\\nthus $\\angle$CAD=$\\angle$BAD, \\\\\nIn $\\triangle APE$ and $\\triangle APB$, $\\left\\{\\begin{aligned} AE&=AB \\\\ \\angle CAP&=\\angle BAD \\\\ AP&=AP \\end{aligned}\\right.$, \\\\\nthus $\\triangle APE\\cong \\triangle APB$ (SAS), \\\\\nthus PE=PB=3, \\\\\nSince 4-3＜PC＜4+3, \\\\\nWe find that $1＜PC＜7$, \\\\\nUpon examining the four options, the length of PC can be $\\boxed{6}$, \\\\",
    "image_path": "dataset/mm_math/images/MM_Math/52653758.png"
  },
  {
    "solution": "\\boxed{A_1}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As illustrated in the Cartesian coordinate system, the line \\( l: y=-\\frac{\\sqrt{3}}{3}x+4 \\) intersects the x-axis and y-axis at points M and N, respectively. An equilateral triangle \\(\\triangle ABC\\) with a height of 3 units, and side \\(BC\\) lying on the x-axis, is translated in the positive direction of the x-axis. During this translation, \\(\\triangle A_1B_1C_1\\) is obtained. When point \\(B_1\\) coincides with the origin, determine the coordinates of point \\(A_1\\) and ascertain whether \\(A_1\\) lies on line \\(l\\)?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $A_1D\\perp OM$ through point $A_1$, with D being the foot of the perpendicular.\\\\\nSince $\\triangle A_1B_1C_1$ is an equilateral triangle, and $A_1D\\perp OM$,\\\\\ntherefore, $\\angle B_1A_1D=30^\\circ$,\\\\\ntherefore, in $Rt\\triangle A_1DB_1$, $B_1D=\\frac{1}{2}A_1B_1$.\\\\\nSince $A_1D=3$,\\\\\ntherefore, $A_1B_1^2-B_1D^2=3B_1D^2=A_1D^2=9$,\\\\\ntherefore, $B_1D=\\sqrt{3}, A_1B_1=2\\sqrt{3}$.\\\\\ntherefore, the coordinates of point $A_1$ are \\(\\boxed{(\\sqrt{3}, 3)}\\).\\\\\nFrom the equation of line l, we get\\\\\nwhen $x=\\sqrt{3}$, $y=-\\frac{\\sqrt{3}}{3}\\times \\sqrt{3}+4=3$\\\\\ntherefore, point \\(\\boxed{A_1}\\) lies on the line l.",
    "image_path": "dataset/mm_math/images/MM_Math/52251020_95.png"
  },
  {
    "solution": "\\boxed{y=3x-1}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, 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. If the $x$-coordinate of point $D$ is 1, determine the analytical expression of the linear function $y=kx+b$.\n",
    "completion": "\\textbf{Solution}: From the problem, we know that the x-coordinate of point $D$ is 1, and point $D$ lies on the line, thus $y=x+1$.\n\n$\\therefore$ Substituting $x=1$ into $y=x+1$, we get $y=2$.\n\n$\\therefore D(1, 2)$.\n\nSubstituting points $B(0, -1)$ and $D(1, 2)$ into $y=kx+b$, we obtain $\\begin{cases}b=-1 \\\\ k+b=2\\end{cases}$. Solving these equations, we find $\\begin{cases}k=3 \\\\ b=-1\\end{cases}$.\n\n$\\therefore$ The equation of line $BD$ is $\\boxed{y=3x-1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50482144_76.png"
  },
  {
    "solution": "\\boxed{y=2x-4}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in the Cartesian coordinate system, the rectangle $OABC$ has two adjacent sides on the coordinate axes, with vertex $B(6, 4)$. A line passing through point $P(4, m)$ on side $BC$ bisects the area of the rectangle. What is the equation of this line?\n",
    "completion": "\\textbf{Solution:} \n\nSince the adjacent sides of rectangle OABC are on the coordinate axes, and the vertex B is at (6, 4),\\\\\ntherefore, the intersection point of the diagonal lines of the rectangle is at (3, 2),\\\\\nsince point P(4, m) is a point on side BC,\\\\\ntherefore, P(4, 4),\\\\\nsince a line passing through the intersection point of the diagonal lines bisects the rectangle,\\\\\ntherefore, let the line passing through P(4, m) and bisecting the rectangle be $y=kx+b$,\\\\\nsubstituting points (4, 4) and (3, 2) yields $\\begin{cases}4k+b=4\\\\ 3k+b=2\\end{cases}$,\\\\\nsolving, we get $\\begin{cases}k=2\\\\ b=-4\\end{cases}$,\\\\\ntherefore, the equation of this line is $y=2x-4$.\\\\\nHence, the answer is: $\\boxed{y=2x-4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50682792.png"
  },
  {
    "solution": "\\boxed{\\frac{25}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: In figure, in $ \\triangle ABC$, $AB=AC$, $AD$ is the median to side $BC$, $DE\\perp AB$ at point $E$. Triangles $ \\triangle BDE \\sim \\triangle CAD$. If $AC=13$, $BC=10$, find the length of the line segment $BE$.\n",
    "completion": "\\textbf{Solution:} Since $AD$ is the median line on side $BC$ and $BC=10$,\\\\\nsince $BD=CD=\\frac{1}{2}BC=5$,\\\\\nsince $\\triangle BDE\\sim \\triangle CAD$ and $AC=13$,\\\\\ntherefore $\\frac{BE}{CD}=\\frac{BD}{AC}$,\\\\\nwhich means $\\frac{BE}{5}=\\frac{5}{13}$,\\\\\ntherefore $BE=\\boxed{\\frac{25}{13}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53080826_102.png"
  },
  {
    "solution": "\\boxed{\\sqrt{10}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, the side length of square $ABCD$ is $2$. Diagonals $AC$ and $BD$ intersect at point $O$, and $E$ is a point on the extension line of $AC$, with $OE = 2CO$. What is the length of $BE$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral $ABCD$ is a square, \\\\\n$\\therefore AC\\perp BD$, $AO=BO=CO=DO$, \\\\\nSince the side length of square $ABCD$ is 4, \\\\\n$\\therefore BC=2$, \\\\\nIn $\\triangle BOC$ (right-angled triangle), \\\\\n$BO^{2}+CO^{2}=BC^{2}$, \\\\\nwhich means $2CO^{2}=2^{2}$, \\\\\nsolving this gives $CO=\\sqrt{2}$, \\\\\nSince $OE=2CO$, \\\\\n$\\therefore OE=2\\sqrt{2}$, \\\\\nIn $\\triangle BOE$ (right-angled triangle), \\\\\n$BE=\\sqrt{OE^{2}+BO^{2}}=\\sqrt{(2\\sqrt{2})^{2}+(\\sqrt{2})^{2}}=\\boxed{\\sqrt{10}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53313397.png"
  },
  {
    "solution": "\\boxed{70^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in $\\triangle ABC$, $AB=CB$, and $\\angle ABC=90^\\circ$. Point F lies on the extension of side AB, and point E is on side BC, with $AE=CF$. $Rt\\triangle ABE \\cong Rt\\triangle CBF$. If $\\angle CAE=25^\\circ$, what is the measure of $\\angle CFA$?\n",
    "completion": "\\textbf{Solution:} Since $AB=CB$ and $\\angle ABC=90^\\circ$,\n\nit follows that $\\angle BAC=\\angle BCA=45^\\circ$.\n\nGiven $\\angle CAE=25^\\circ$,\n\nthen $\\angle BAE=45^\\circ-25^\\circ=20^\\circ$.\n\nSince Rt$\\triangle ABE \\cong$ Rt$\\triangle CBF$,\n\nit implies that $\\angle BCF=\\angle BAE=20^\\circ$,\n\ntherefore, $\\angle CFA=90^\\circ-20^\\circ=\\boxed{70^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51369714_96.png"
  },
  {
    "solution": "\\boxed{y=-x+1}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</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 $B$. The line segment $BC$ is perpendicular to the $x$-axis, with the foot of the perpendicular being $C$, and the area of $\\triangle AOC$ is 1. Find the equation of line $AC$.\n",
    "completion": "\\textbf{Solution:} Given that the line $y=mx$ intersects the hyperbola $y=\\frac{n}{x}$ at points $A(-1, a)$ and $B$,\\\\\nthe x-coordinate of point $B$ is 1,\\\\\nsince $BC$ is perpendicular to the x-axis with the foot at $C$,\\\\\nwe have $C(1, 0)$, and the distance $OC=1$,\\\\\nconsidering the area of $\\triangle AOC$ is 1,\\\\\nwe get $\\frac{1}{2}\\times 1\\cdot a=1$,\\\\\nsolving this equation yields $a=2$,\\\\\nthus, $A(-1, 2)$,\\\\\nLet the equation of line $AC$ be $y=kx+b$,\\\\\nsubstituting points $A(-1, 2), C(1, 0)$ into the equation yields: $\\left\\{\\begin{array}{l}-k+b=2 \\\\ k+b=0 \\end{array}\\right.$, solving these equations gives $\\left\\{\\begin{array}{l}k=-1 \\\\ b=1 \\end{array}\\right.$,\\\\\ntherefore, the equation of line $AC$ is $\\boxed{y=-x+1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53498218_194.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC$, $\\angle BAC=90^\\circ$, point $D$ is on segment $BC$, $\\angle EDB=\\frac{1}{2}\\angle ACB$, $BE\\perp DE$, and $DE$ intersects $AB$ at point $F$. If $BE=4$, what is the length of $DF$?\n",
    "completion": "\\textbf{Solution:} Construct $DH\\parallel AC$ through $D$ and meet $AB$ at $H$. Extend $BE$ to intersect with the extension of $DH$ at point $G$, as shown in the figure:\\\\\nIn $\\triangle ABC$, $AB=AC$, $\\angle BAC=90^\\circ$,\\\\\n$\\therefore \\angle ABC=\\angle C=45^\\circ$,\\\\\n$\\because DH\\parallel AC$,\\\\\n$\\therefore \\angle BDH=\\angle C=45^\\circ$, $\\angle BHD=\\angle A=90^\\circ$,\\\\\n$\\therefore \\triangle HBD$ is an isosceles right triangle,\\\\\n$\\therefore HB=HD$,\\\\\n$\\because \\angle EDB=\\frac{1}{2}\\angle ACB=22.5^\\circ$,\\\\\n$\\therefore \\angle GDE=\\angle EDB=22.5^\\circ$,\\\\\n$\\therefore DE$ bisects $\\angle BDG$,\\\\\n$\\because DE\\perp BG$,\\\\\n$\\therefore \\angle GED=\\angle BED=90^\\circ$,\\\\\nIn $\\triangle BDE$ and $\\triangle GDE$,\\\\\n$\\left\\{\\begin{array}{l}\n\\angle GED=\\angle BED=90^\\circ \\\\\nED=ED \\\\\n\\angle GDE=\\angle BDE=22.5^\\circ\n\\end{array}\\right.$,\\\\\n$\\therefore \\triangle BDE\\cong \\triangle GDE\\left(AAS\\right)$,\\\\\n$\\therefore BE=GE$, that is $BG=2BE$,\\\\\nIn $\\triangle BGH$ and $\\triangle DFH$,\\\\\n$\\left\\{\\begin{array}{l}\n\\angle GBH=\\angle FDH=22.5^\\circ \\\\\nBH=DH \\\\\n\\angle GHB=\\angle FHD\n\\end{array}\\right.$,\\\\\n$\\therefore \\triangle BGH\\cong \\triangle DFH(ASA)$,\\\\\n$\\therefore BG=DF$,\\\\\n$\\therefore DF=2BE=\\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52715507.png"
  },
  {
    "solution": "\\boxed{9}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, it is known that in quadrilateral $ABCD$, $AC$ and $BD$ intersect at point $O$, $AB\\perp AC$, and $CD\\perp BD$. $\\triangle AOD\\sim \\triangle BOC$; if $\\sin\\angle ABO=\\frac{2}{3}$ and the area of $\\triangle AOD=4$, what is the value of the area of $\\triangle BOC$?\n",
    "completion": "\\textbf{Solution:} In the right triangle $\\triangle AOB$, we have $\\sin\\angle ABO=\\frac{2}{3}$,\\\\\nhence $\\frac{AO}{BO}=\\frac{2}{3}$,\\\\\nsince $\\triangle AOD\\sim \\triangle BOC$,\\\\\nit follows that $\\frac{{S}_{\\triangle AOD}}{{S}_{\\triangle BOC}}=\\left(\\frac{AO}{BO}\\right)^{2}=\\left(\\frac{2}{3}\\right)^{2}=\\frac{4}{9}$,\\\\\ngiven that $S_{\\triangle AOD}=4$,\\\\\ntherefore $S_{\\triangle BOC}=\\boxed{9}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52917610_91.png"
  },
  {
    "solution": "\\boxed{x\\le -3}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: The graph of the line $y_1=k_1x+b$ and the line $y_2=k_2x$ in the same Cartesian coordinate system is shown in the figure. What is the solution set of the inequality $k_1x+b\\le k_2x$ with respect to $x$?\n",
    "completion": "\\textbf{Solution:} Given that ${k}_{1}x+b\\le {k}_{2}x$ graphically represents that the line ${y}_{1}={k}_{1}x+b$ is below the line ${y}_{2}={k}_{2}x$. From the graph, it is known that the solution set is $\\boxed{x\\le -3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52251808.png"
  },
  {
    "solution": "\\boxed{45^\\circ + \\frac{1}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, point $O$ lies on line $AB$, $\\angle BOD$ and $\\angle COD$ are complementary, $\\angle BOC = n\\angle EOC$. If $n=4$, let $\\angle AOD = \\alpha$, find the degree measure of $\\angle DOE$ (express the degree measure of $\\angle DOE$ in terms of $\\alpha$).\n",
    "completion": "\\textbf{Solution:} Given the problem statement,\\\\\n\\(\\because \\angle BOD\\) and \\(\\angle COD\\) are complementary, \\\\\n\\(\\therefore \\angle BOD + \\angle AOD = 180^\\circ\\),\\\\\n\\(\\therefore \\angle COD = \\angle AOD = \\alpha\\),\\\\\nLet \\(\\angle COE = y\\),\\\\\nand \\(n = 4\\),\\\\\nthen \\(\\angle BOC = 4y\\), \\(\\angle BOE = 3y\\),\\\\\n\\(\\therefore 4y + 2\\alpha = 180^\\circ\\),\\\\\n\\(\\therefore y = 45^\\circ - \\frac{1}{2}\\alpha\\),\\\\\n\\(\\therefore \\angle DOE = \\alpha + 45^\\circ - \\frac{1}{2}\\alpha = 45^\\circ + \\frac{1}{2}\\alpha\\).\\\\\nHence, the measure of \\(\\angle DOE\\) is \\(\\boxed{45^\\circ + \\frac{1}{2}\\alpha}\\).",
    "image_path": "dataset/mm_math/images/MM_Math/52774176_156.png"
  },
  {
    "solution": "\\boxed{80^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACD=110^\\circ$, $\\angle B=30^\\circ$, then what is the measure of $\\angle A$ in degrees?\n",
    "completion": "\\textbf{Solution:} Since $\\angle ACD$ is an exterior angle of $\\triangle ABC$,\\\\\ntherefore $\\angle ACD = \\angle A + \\angle B$,\\\\\nthus $\\angle A = 110^\\circ - 30^\\circ = \\boxed{80^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52755870.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, it is known that $AB \\parallel CD$, $AE$ and $CE$ bisect $\\angle BAC$ and $\\angle ACD$ respectively, $AE$ and $CE$ intersect at point $E$, draw line $ED \\perp CD$ with the foot of the perpendicular being $D$, intersecting $AB$ at point $B$. If $AC=8, BD=6$, what is the area of $\\triangle ACE$?\n",
    "completion": "\\textbf{Solution:} Draw $EF\\perp AC$ at point F from point E,\\\\\nsince $AB\\parallel CD$ and $ED\\perp CD$,\\\\\ntherefore $EB\\perp AB$,\\\\\nsince $AE$ and $CE$ bisect $\\angle BAC$ and $\\angle ACD$ respectively,\\\\\ntherefore $BE=EF=ED$,\\\\\nthus point E is the midpoint of $BD$,\\\\\nsince $AC=8$ and $BD=6$,\\\\\ntherefore $EF=\\frac{1}{2}BD=3$,\\\\\nthus, the area of $\\triangle ACE$ is $\\frac{1}{2}\\times AC\\times EF=\\frac{1}{2}\\times 8\\times 3=\\boxed{12}$",
    "image_path": "dataset/mm_math/images/MM_Math/52827122.png"
  },
  {
    "solution": "\\boxed{130^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</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 $\\angle BAC=40^\\circ$, then what is the degree measure of $\\angle ADC$?\n",
    "completion": "\\textbf{Solution:} Connect $CB$, \\\\\n$\\because$ $AB$ is the diameter of $\\odot O$, $\\angle BAC = 40^\\circ$\\\\\n$\\therefore$ $\\angle ACB = 90^\\circ$, \\\\\n$\\because$ $\\angle BAC = 40^\\circ$, \\\\\n$\\therefore$ $\\angle CBA = 50^\\circ$, \\\\\n$\\because$ quadrilateral $ABCD$ is inscribed in $\\odot O$, \\\\\n$\\therefore$ $\\angle ADC + \\angle CBA = 180^\\circ$, \\\\\n$\\therefore$ $\\angle ADC = \\boxed{130^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53102184.png"
  },
  {
    "solution": "\\boxed{1\\colon 2}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in $ \\triangle ABC$, D and E are the midpoints of sides $AB$ and $AC$, respectively. What is the ratio of the perimeters of $ \\triangle ADE$ and $ \\triangle ABC$?\n",
    "completion": "\\textbf{Solution:} Given that point D and point E are the midpoints of side AB and AC, respectively, \\\\\nit follows that DE is the mid-segment of $\\triangle ABC$, \\\\\nthus, DE$\\parallel$BC, and the ratio DE$\\colon$BC = 1$\\colon$2, \\\\\ntherefore, $\\triangle ADE\\sim \\triangle ABC$, \\\\\nhence, the ratio of the perimeters of $\\triangle ADE$ to $\\triangle ABC$ is $\\boxed{1\\colon 2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51379157.png"
  },
  {
    "solution": "\\boxed{(-2,0)}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, point $N(0, 6)$ lies on the plane, and point $M$ is on the negative $x$-axis with $ON = 3OM$. $A$ is a point on line segment $MN$, $AB \\perp x$-axis with $B$ as the foot of the perpendicular, and $AC \\perp y$-axis with $C$ as the foot of the perpendicular. Write down the coordinates of point $M$.\n",
    "completion": "\\textbf{Solution:} Given, $\\because N(0,6)$,\\\\\n$\\therefore ON=6$,\\\\\n$\\because ON=3OM$,\\\\\n$\\therefore OM=2$,\\\\\n$\\therefore$ the coordinates of point M are $\\boxed{(-2,0)}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52225566_109.png"
  },
  {
    "solution": "\\boxed{7}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in $\\triangle ABC$, the line $DE$ is the perpendicular bisector of the side $AC$, and $AE$ is connected. If $AB=3$ and $BC=5$, what is the perimeter of $\\triangle ABE$?\n",
    "completion": "\\textbf{Solution:} Since $DE$ is the perpendicular bisector of side $AC$,\\\\\n$\\therefore AE=EC$,\\\\\n$\\therefore AB+BE+AE=AB+BE+EC=AB+BC=2+5=\\boxed{7}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52838015.png"
  },
  {
    "solution": "\\boxed{P(-4,5)}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, the parabola $y=ax^2+bx+c$ ($a\\ne 0$) has its axis of symmetry as the line $x=-1$, intersects with the $x$-axis at points $A$ and $B$, and intersects with the $y$-axis at point $C$. It is given that $OA=OC$, and the coordinates of point $A$ are $(-3,0)$. If point $P$ lies on the parabola, and the area of $\\triangle POC$ is 4 times the area of $\\triangle BOC$, what are the coordinates of point $P$?\n",
    "completion": "\\textbf{Solution:} When $x=0$, $y=-3$,\\\\\n$\\therefore$ point $C(0, -3)$, i.e., $OC=3$,\\\\\n$\\because$ A and B are symmetric about the axis of symmetry,\\\\\n$\\therefore$ B$(1,0)$, i.e., $OB=1$,\\\\\n$\\therefore$ $S_{\\triangle BOC}=\\frac{1}{2}OB\\times OC=\\frac{3}{2}$,\\\\\nLet $P(x,{x}^{2}+2x-3)$,\\\\\n$\\therefore$ $S_{\\triangle POC}=\\frac{1}{2}\\times3\\times|x|$,\\\\\n$\\because$ $S_{\\triangle POC}=4S_{\\triangle BOC}$,\\\\\n$\\therefore$ $\\frac{3}{2}|x|=4\\times \\frac{3}{2}$,\\\\\n$\\therefore$ $x=\\pm 4$,\\\\\n$\\therefore$ the coordinates of point $P$ are $\\boxed{P(4,21)}$, $\\boxed{P(-4,5)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52765835_126.png"
  },
  {
    "solution": "\\boxed{-8}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, there is a point $A(3, 0)$ on the x-axis. Point $D$ is the symmetric point of point $A$ with respect to the y-axis. Point $B$ is on the graph of the inverse proportional function $y=\\frac{k}{x}$ $(x<0)$. When connecting $BD$ and intersecting the graph of the inverse proportional function at point $C$, if $OC\\parallel AB$, and the area of $\\triangle ABD$ is $24$, what is the value of $k$?\n",
    "completion": "\\textbf{Solution:} Let $BE \\perp x$-axis at point E, $CF \\perp x$-axis at point F, \\\\\nsince point D is the symmetric point of point A with respect to the y-axis, \\\\\nthus the coordinates of D are $(-3, 0)$, \\\\\nthus $AD=6$, \\\\\nsince the area of $\\triangle ABD = \\frac{1}{2}BE \\cdot AD = 24$, \\\\\nthus $BE=8$, \\\\\nsince $OC \\parallel AB$, \\\\\nthus $\\triangle DCO \\sim \\triangle DBA$, \\\\\nthus $\\frac{OC}{AB} = \\frac{DO}{AD} = \\frac{1}{2}$, \\\\\nsince $\\triangle OCF \\sim \\triangle ABE$, \\\\\nthus $\\frac{CF}{BE} = \\frac{OC}{AB} = \\frac{OF}{AE} = \\frac{1}{2}$, \\\\\nthus $CF = \\frac{1}{2}BE = 4$, \\\\\nsince B and C are on the graph, \\\\\nthus $B\\left(\\frac{k}{8},8\\right)$, $C\\left(\\frac{k}{4},4\\right)$, \\\\\nsince $AE = 3 - \\frac{k}{8}$, $OF = -\\frac{k}{4}$, \\\\\nthus $\\frac{-\\frac{k}{4}}{3-\\frac{k}{8}} = \\frac{1}{2}$, \\\\\nsolving gives $k = \\boxed{-8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52537112.png"
  },
  {
    "solution": "\\boxed{7}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC$, $\\angle BAC=90^\\circ$. Line $l$ passes through point $A$, and from point $B$, $BE\\perp l$ at point $E$, and from point $C$, $CF\\perp l$ at point $F$. If $BE=2$, $CF=5$, what is the length of $EF$?\n",
    "completion": "\\textbf{Solution:}\\\\\nSince $BE\\perp l$ and $CF\\perp l$,\\\\\nit follows that $\\angle AEB=\\angle CFA=90^\\circ$,\\\\\nthus, $\\angle EAB+\\angle ABE=90^\\circ$,\\\\\nsince $\\angle BAC=90^\\circ$,\\\\\nit follows that $\\angle EAB+\\angle FAC=90^\\circ$,\\\\\nthus, $\\angle ABE=\\angle FAC$,\\\\\nsince $AB=AC$,\\\\\nit follows that $\\triangle AEB\\cong \\triangle CFA$ (AAS),\\\\\ntherefore, $AF=BE=2$, $AE=CF=5$,\\\\\nthus, $EF=AE+AF=5+2=\\boxed{7}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52661292.png"
  },
  {
    "solution": "\\boxed{(12,−8) \\text { or}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in the Cartesian coordinate system, $O$ is the origin. The coordinate of point $A$ is $(0, 8)$, and the coordinate of point $B$ is $(-4, 0)$. Point $P$ is a moving point on the line $l: x+y=4$. If $\\angle PAB = \\angle ABO$, what are the coordinates of point $P$?\n",
    "completion": "\\textbf{Solution:} As illustrated:\\\\\n(1) When point P is on the left side of the y-axis, connect AP, $\\because$ $\\angle PAB = \\angle ABO$, $\\therefore$ AP$\\parallel$OB, $\\therefore$ the y-coordinate of point P is 8. When y=8, x+8=4, solving this gives x=−4, $\\therefore$ P(−4, 8);\\\\\n(2) When point P is on the right side of the y-axis, connect AP intersecting the x-axis at point C, $\\because$ $\\angle PAB = \\angle ABO$, $\\therefore$ AC=BC, let OC=a, then BC=AC=4+a. In $\\triangle AOC$, by Pythagoras' theorem, we get $(4+a)^{2}=a^{2}+8^{2}$, solving for a gives a=6, $\\therefore$ C(6,0). Let the linear equation of line AC be y=kx+b. Substituting A(0,8) and C(6,0) into y=kx+b gives $\\left\\{\\begin{array}{l}b=8\\\\ 6k+b=0\\end{array}\\right.$, solving this system gives $\\left\\{\\begin{array}{l}k=-\\frac{4}{3}\\\\ b=8\\end{array}\\right.$, $\\therefore$ $y=-\\frac{4}{3}x+8$. $\\because$ Point P is the intersection of line l and line AC, $\\therefore$ $\\left\\{\\begin{array}{l}y=-\\frac{4}{3}x+8\\\\ x+y=4\\end{array}\\right.$, solving this system gives $\\left\\{\\begin{array}{l}x=12\\\\ y=-8\\end{array}\\right.$, $\\therefore$ P(12,−8), hence the answer is: $\\boxed{(12,−8) \\text { or } (−4, 8)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50431224.png"
  },
  {
    "solution": "\\boxed{125^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the diagram, lines AB and CD intersect at point O, and OA bisects $\\angle EOC$ where $\\angle EOC=110^\\circ$. What is the measure of $\\angle BOC$?\n",
    "completion": "\\textbf{Solution:} Since OA bisects $\\angle EOC$ and $\\angle EOC=110^\\circ$,\\\\\nit follows that $\\angle AOC= \\frac{1}{2}\\angle COE=55^\\circ$\\\\\nGiven $\\angle AOC+\\angle BOC$ form a straight angle,\\\\\nit implies $\\angle BOC=180^\\circ-55^\\circ=\\boxed{125^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/53404466.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: The \"Zhao Shuang Chord Diagram\" is acclaimed as the totem of Chinese mathematics. It consists of four congruent right-angled triangles and a smaller square within a larger square, as shown in the diagram. By connecting $AE$, if the area of the large square $ABCD$ is $25$ and the area of $\\triangle ABE$ is $8$, what is the area of the smaller square $EFGH$?\n",
    "completion": "\\textbf{Solution:} Let $AG=BH=b$ and $AH=BE=a$,\n\nsince the area of the large square ABCD is 25,\n\nthus $AB=5$,\n\ntherefore $a^{2}+b^{2}=25$,\n\nsince the area of $\\triangle ABE$ is 8,\n\nthus $\\frac{1}{2}a^{2}=8$,\n\ntherefore $a=4$ (negative value discarded),\n\nthus $b=\\sqrt{25-16}=3$,\n\ntherefore, the area of the small square EFGH $=\\left(a-b\\right)^{2}=\\left(4-3\\right)^{2}=\\boxed{1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52828910.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, $\\angle C=90^\\circ$, $AD$ bisects $\\angle BAC$ and intersects $BC$ at $D$. If $BC=7\\text{cm}$ and $BD=4\\text{cm}$, what is the distance from point $D$ to $AB$ in $\\text{cm}$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw DH $\\perp$ AB at point H, \\\\\n$\\because$ BC = 7cm, BD = 4cm, \\\\\n$\\therefore$ CD = 7 - 4 = 3cm, \\\\\nAlso, $\\because$ AD is the bisector of $\\angle BAC$, $\\angle C=90^\\circ$, and DH $\\perp$ AB, \\\\\n$\\therefore$ DH = CD = 3cm, \\\\\n$\\therefore$ the distance from point D to AB is \\boxed{3} cm.",
    "image_path": "dataset/mm_math/images/MM_Math/52807552.png"
  },
  {
    "solution": "\\boxed{\\frac{13}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, $O$ is the midpoint of the diagonal $BD$ of rectangle $ABCD$, and $E$ is the midpoint of side $AB$. Given that $AB = 12$ and $OE = \\frac{5}{2}$, what is the length of segment $OC$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rectangle, \\\\\n$\\therefore$$\\angle$A=$\\angle$BCD=$90^\\circ$, DO=OB,\\\\\nsince E is the midpoint of side AB,\\\\\n$\\therefore$OE is the median of $\\triangle$ADB, OE=$\\frac{5}{2}$,\\\\\n$\\therefore$AD=$2$OE=$5$,\\\\\nsince AB=$12$,\\\\\n$\\therefore$BD=$\\sqrt{AD^{2}+AB^{2}}=\\sqrt{12^{2}+5^{2}}=13$,\\\\\nsince O is the midpoint of the diagonal BD of rectangle ABCD,\\\\\n$\\therefore$OC=$\\frac{1}{2}$BD=$\\boxed{\\frac{13}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52250596.png"
  },
  {
    "solution": "\\boxed{20\\sqrt{3}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, the perimeter of the rhombus flower bed ABCD is 80m, with $\\angle ABC=120^\\circ$. If two paths AC and BD are built along the diagonals of the rhombus, what is the length of path AC?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, let the diagonals AC and BD intersect at point O, \\\\\nsince quadrilateral ABCD is a rhombus, and $\\angle ABC=120^\\circ$, \\\\\ntherefore $\\angle ADB=\\angle CDB=60^\\circ$, and AC$\\perp$BD, \\\\\nthus $\\triangle AOD$ is a right triangle, with $\\angle DAO=30^\\circ$, \\\\\nsince the perimeter of the rhombus is 80, \\\\\ntherefore AD=80÷4=20, \\\\\nhence OD=10, \\\\\nusing the Pythagorean theorem, we get: $OA=\\sqrt{AD^{2}-OD^{2}}=10\\sqrt{3}$, \\\\\nbased on the properties of a rhombus, we can derive: AC=2OA=$\\boxed{20\\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52159827.png"
  },
  {
    "solution": "\\boxed{C(1, 5.5)}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, points $A$ and $B$ are two points in the Cartesian coordinate system. Connect $OA$ and $OB$, where $OA=5$, $OB=10$, and $OA \\perp OB$. If the abscissa of point $A$ is $-4$, and the graph of the inverse proportion function $y=\\frac{k_2}{x}$ passes through point $B$, while the graph of the inverse proportion function $y=\\frac{k_1}{x}$ passes through point $A$. If point $C$ is on the line segment $AB$, and the area of $\\triangle OBC$ is $\\frac{1}{2}$ the area of $\\triangle OAB$, find the coordinates of point $C$.\n",
    "completion": "\\textbf{Solution:} Since $S_{\\triangle OBC} = \\frac{1}{2}S_{\\triangle OAB}$,\\\\\nit follows that point C is the midpoint of AB.\\\\\nDraw AD$\\perp$BN at point D, and draw CE$\\parallel$y-axis intersecting AD at point E, as shown in the figure.\\\\\nSince BN$\\perp$x-axis,\\\\\nit follows that BN$\\parallel$CE,\\\\\nthus $\\triangle ACE \\sim \\triangle ABD$,\\\\\nwhich implies $\\frac{AE}{AD} = \\frac{CE}{BD} = \\frac{AC}{AB} = \\frac{1}{2}$,\\\\\ntherefore, AD=2AE, BD=2CE,\\\\\nmeaning point E is the midpoint of AD.\\\\\nSince quadrilateral AMND is a rectangle,\\\\\nit follows that DN=AM=3, AD=MN=10,\\\\\nhence BD=BN−DN=5,\\\\\nLet the coordinates of point C be $(m, n)$,\\\\\nthen $6−m=\\frac{1}{2}\\times 10$, $n−3=\\frac{1}{2}\\times 5$,\\\\\nthus $m=1$, $n=5.5$,\\\\\nhence $\\boxed{C(1, 5.5)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52601676_119.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{5}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</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$). When is $AD$ such that $\\triangle ACD$ is an isosceles triangle?\n",
    "completion": "\\textbf{Solution:} \n(1) When $AC=AD=8$, $\\triangle ACD$ is an isosceles triangle,\\\\\n(2) When $AC=CD=8$, $\\triangle ACD$ is an isosceles triangle,\\\\\nAs shown in the figure, draw $CF\\perp AD$ at $F$,\\\\\n$\\therefore AD=2DF$, $\\angle CFD=90^\\circ$,\\\\\n$\\because \\angle B=\\angle D$, $\\angle CFD=\\angle ACB=90^\\circ$,\\\\\n$\\therefore \\triangle CDF\\sim \\triangle ABC$,\\\\\n$\\therefore \\frac{CD}{AB}=\\frac{DF}{BC}$,\\\\\n$\\therefore \\frac{8}{10}=\\frac{DF}{6}$,\\\\\n$\\therefore DF=\\frac{24}{5}$,\\\\\n$\\therefore AD=\\frac{48}{5}$;\\\\\n(3) When $AD=CD$, $\\triangle ACD$ is an isosceles triangle,\\\\\nAs shown in the figure, when point $D$ is on the major arc $ADC$, connect $DO$ and extend it to meet $AC$ at $M$,\\\\\nthen $AM=CM=\\frac{1}{2}AC=4$, $DM\\perp AC$,\\\\\n$\\because AO=BO$,\\\\\n$\\therefore OM=\\frac{1}{2}BC=3$,\\\\\n$\\therefore DM=OM+OD=8$,\\\\\n$\\therefore AD=\\sqrt{AM^2+DM^2}=\\sqrt{4^2+8^2}=4\\sqrt{5}$,\\\\\nWhen point $D$ is on the minor arc $AD$, by similar reasoning, we get $AD=2\\sqrt{5}$,\\\\\nIn summary, when $AD$ is $8$ or $\\boxed{\\frac{48}{5}}$ or $\\boxed{4\\sqrt{5}}$ or $\\boxed{2\\sqrt{5}}$, $\\triangle ACD$ is an isosceles triangle.",
    "image_path": "dataset/mm_math/images/MM_Math/52793804_150.png"
  },
  {
    "solution": "\\boxed{\\frac{3\\sqrt{5}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, points $D$, $E$, and $F$ are the midpoints of the sides $AB$, $BC$, and $AC$ of $\\triangle ABC$, respectively. Extend $DE$ to point $G$ such that $DE=EG$, and connect $AE$, $FG$. Quadrilateral $AEGF$ is a parallelogram. If $\\angle BAC=90^\\circ$ and $AD=AC=3$, find the length of $FG$.\n",
    "completion": "\\textbf{Solution:} From (1), we know: $DE=\\frac{1}{2}AC$, $DE\\parallel AC$, since $\\angle BAC=90^\\circ$, therefore $\\angle EAD=90^\\circ$. Also, since $AC=AD=3$, therefore $DE=\\frac{3}{2}$, hence $AE=\\sqrt{DE^2+AD^2}=\\sqrt{\\left(\\frac{3}{2}\\right)^2+3^2}=\\frac{3\\sqrt{5}}{2}$. From (1), it is known that quadrilateral AEGF is a parallelogram, therefore $FG=\\boxed{\\frac{3\\sqrt{5}}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51825337_74.png"
  },
  {
    "solution": "\\boxed{y=\\frac{1}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, 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)$, and intersects the graph of the inverse proportion function $y=\\frac{6}{x}$ within the first quadrant at point $A$. Line $AD$ is drawn perpendicular to the x-axis at point $D$, with $OD=2$. Find the analytic equation of line $AB$?\n",
    "completion": "Solution:\n\nSince AD is perpendicular to the x-axis, and OD = 2, the x-coordinate of point D is 2. By substituting x = 2 into $y= \\frac{6}{x}$, we get y = 3. Therefore, A(2, 3). Let the equation of line AB be $y=kx+b$ (where $k\\neq 0$). Substituting points C(0, 2) and A(2, 3) into $y=kx+b$ gives $\\begin{cases}b=2\\\\ 2k+b=3\\end{cases}$. Therefore, $\\begin{cases}b=2\\\\ k=\\frac{1}{2}\\end{cases}$. Hence, the equation of line AB is $y=\\frac{1}{2}x+2$; so, the answer is $\\boxed{y=\\frac{1}{2}x+2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51458203_264.png"
  },
  {
    "solution": "\\boxed{0<x<5}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</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 proportional 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. Directly write the range of values of $x$ when $y_1<y_2$.\n",
    "completion": "\\textbf{Solution:} When $y_1<y_2$, the range of $x$ is \\boxed{x<-2} or \\boxed{0<x<5}.",
    "image_path": "dataset/mm_math/images/MM_Math/52601939_123.png"
  },
  {
    "solution": "\\boxed{rhombus}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, place two strips of paper with parallel opposite sides and equal widths crosswise at will. By rotating one of them, the overlapping part forms a quadrilateral. What can this quadrilateral be?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw heights AE and AF from point D to sides AB and BC, respectively,\\\\\nsince quadrilateral ABCD is formed by overlapping two strips of equal width,\\\\\ntherefore AB$\\parallel$CD, and AD$\\parallel$BC,\\\\\ntherefore, quadrilateral ABCD is a parallelogram,\\\\\nsince DE$\\perp$AB, and DF$\\perp$BC,\\\\\ntherefore DE=DF,\\\\\nsince the area of parallelogram ABCD, $S_{\\text{parallelogram } ABCD}$=AB$\\cdot$DE=BC$\\cdot$DF,\\\\\ntherefore AB=BC,\\\\\ntherefore, quadrilateral ABCD is a rhombus.\\\\\nHence, the answer is: Quadrilateral ABCD is a \\boxed{rhombus}.",
    "image_path": "dataset/mm_math/images/MM_Math/51614526.png"
  },
  {
    "solution": "\\boxed{x < 1}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, the line $y = x + 1$ and $y = ax + 4$ intersect at point $A(1, m)$. Write the solution set for the inequality $x + 1 < ax + 4$.\n",
    "completion": "\\textbf{Solution:} From the figure, we can see that the solution set of the inequality $x + 1 < ax + 4$ is $\\boxed{x < 1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52284181_69.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, $AB$ is the diameter of the semicircle $O$, and $C$, $D$ are two points on the semicircle $O$, with $OD\\parallel BC$, and $OD$ intersects $AC$ at point $E$. If $AB=13$ and $AC=12$, find the length of $DE$.\n",
    "completion": "\\textbf{Solution:} Given that $AB=13$ and $AC=12$,\\\\\nit follows that $BC=\\sqrt{AB^{2}-AC^{2}}=5$.\\\\\nSince $OD\\parallel BC$ and $\\angle C=90^\\circ$,\\\\\nit implies $OD\\perp AC$,\\\\\nwhich means $AE=CE$.\\\\\nGiven $OA=OB$,\\\\\nit can be deduced that $OE=\\frac{1}{2}BC=2.5$.\\\\\nSince $OD=OA=\\frac{1}{2}AB=6.5$,\\\\\nit follows that $DE=OD-OE=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53009507_139.png"
  },
  {
    "solution": "\\boxed{1\\colon 9}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the diagram, in the Cartesian coordinate system, $\\triangle OAB$ is scaled up with the origin $O$ as the center of similarity to obtain $\\triangle OCD$. If $B(0, 1)$ and $D(0, 3)$, what is the ratio of the areas of $\\triangle OAB$ to $\\triangle OCD$?\n",
    "completion": "\\textbf{Solution:} Let $\\triangle OAB$ be enlarged from the origin $O$ to form $\\triangle OCD$, where B(0,1), D(0,3),\\\\\n$\\therefore OB\\colon OD=1\\colon 3$\\\\\n$\\therefore$ The ratio of the areas of $\\triangle OAB$ and $\\triangle OCD$ is $\\boxed{1\\colon 9}$",
    "image_path": "dataset/mm_math/images/MM_Math/51406931.png"
  },
  {
    "solution": "\\boxed{41}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in rectangle $ABCD$, $AB=9$ and $AD=14$. Points $E$ and $F$ are located on $AD$ and $BC$ respectively, with $AE=CF=1$. Point $G$ is a moving point on side $DC$, and point $H$ is a moving point on side $AB$. What is the minimum value of $EG+HG+HF$?\n",
    "completion": "\\textbf{Solution:} Let point $E'$ be the symmetric point of $E$ with respect to line $CD$, and point $F'$ be the symmetric point of $F$ with respect to line $AB$,\\\\ \nthus, $EG= E'G$ and $HF=HF'$, \\\\ \n$\\therefore EG+HG+HF=E'G+HG+HF'$, \\\\ \nConnect points $E'$ and $F'$, intersecting lines $AB$ and $CD$ at points $H$ and $G$ respectively, the minimum value of $EG+HG+HF$ is exactly the length of $E'F'$, \\\\ \nDraw a line from $E'$ such that $E'H\\perp BC$ and intersects the extension line of $BC$ at point $H$, \\\\\nthus, $F'H=14+2\\times 13=40$, $E'H =9$, \\\\ \nIn $\\triangle E'HF'$, by the Pythagorean theorem, we obtain $E'F'=\\sqrt{9^2+40^2}=\\boxed{41}$, \\\\",
    "image_path": "dataset/mm_math/images/MM_Math/52539248.png"
  },
  {
    "solution": "\\boxed{10+5\\sqrt{5}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, a rectangular piece of paper ABCD is folded along EF and MN, resulting in a pentagon GEFNM. Herein, the vertices A and D coincide at point G, and the overlapping section GHIJ forms a square, with vertex I located on EM. If FN equals $4\\sqrt{5}$ cm and EM equals 10 cm, what is the length of side BC in centimeters?\n",
    "completion": "\\textbf{Solution:} Construct IR$\\perp$BC at R, as shown in the figure:\\\\\n$\\because$ A rectangle paper ABCD is folded along EF, MN, resulting in a pentagon GEFNM, where, vertex A coincides with D at point G, FN=$4\\sqrt{5}$cm, EM=10cm,\\\\\n$\\therefore$ $\\angle$IEF=$\\angle$BFE=$\\angle$EFJ, $\\angle$IMN=$\\angle$MNC=$\\angle$MNH, AB=GJ=CD=HG,\\\\\n$\\therefore$ EI=FI, MI=NI,\\\\\nLet EI=FI=xcm, then MI=NI=EM-EI=(10-x)cm,\\\\\n$\\because$ Quadrilateral GHIJ is a square,\\\\\n$\\therefore$ GH=HI=IJ=GJ, $\\angle$HIJ=$90^\\circ$=$\\angle$FIN,\\\\\nIn $\\triangle$FIN, $EI^{2}+NI^{2}=FN^{2}$,\\\\\n$\\therefore$ $x^{2}+(10-x)^{2}=(4\\sqrt{5})^{2}$,\\\\\nSolving this, we get $x=5-\\sqrt{15}$ or $x=5+\\sqrt{15}$,\\\\\nAssume $x=5-\\sqrt{15}$ (The result is the same for $x=5+\\sqrt{15}$),\\\\\nThen EI=FI=($5-\\sqrt{15}$)cm, MI=NI=($5+\\sqrt{15}$)cm,\\\\\n$\\because$2S$\\triangle$FIN=FN$\\cdot$IR=FI$\\cdot$NI,\\\\\n$\\therefore$ $IR=\\frac{FI\\cdot NI}{FN}=\\frac{(5-\\sqrt{15})(5+\\sqrt{15})}{4\\sqrt{5}}=\\frac{\\sqrt{5}}{2}$cm,\\\\\n$\\therefore$ AB=CD=$\\frac{\\sqrt{5}}{2}$cm,\\\\\n$\\therefore$ GJ=HG=$\\frac{\\sqrt{5}}{2}$cm,\\\\\n$\\because$ Quadrilateral GHIJ is a square,\\\\\n$\\therefore$ HI=JI=$\\frac{\\sqrt{5}}{2}$cm,\\\\\n$\\therefore$ HN=HI+NI=$\\left(\\frac{\\sqrt{5}}{2}+5+\\sqrt{15}\\right)$cm, $FJ=JI+FI=\\left(\\frac{\\sqrt{5}}{2}+5-\\sqrt{15}\\right)$cm,\\\\\n$\\therefore$ CN=HN=$\\left(\\frac{\\sqrt{5}}{2}+5+\\sqrt{15}\\right)$cm, BF=FJ=$\\left(\\frac{\\sqrt{5}}{2}+5-\\sqrt{15}\\right)$cm,\\\\\n$\\therefore$ BC=CN+FN+BF=$\\left(\\frac{\\sqrt{5}}{2}+5+\\sqrt{15}+4\\sqrt{5}+\\frac{\\sqrt{5}}{2}+5-\\sqrt{15}\\right)=\\boxed{10+5\\sqrt{5}}$cm.",
    "image_path": "dataset/mm_math/images/MM_Math/53498365.png"
  },
  {
    "solution": "\\boxed{75^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: Given that $O$ is a point on line $AB$, and $OC \\perp OE$. Under these conditions, as shown in the figure, draw ray $OM$ inside $\\angle BOE$, such that $\\angle COM + \\frac{17}{10}\\angle AOE = 2\\angle BOM + \\angle FOM$, what is the degree measure of $\\angle BOM$?\n",
    "completion": "\\textbf{Solution:} Given $\\because$OC$\\perp$OE,\\\\\n$\\therefore \\angle AOC=90^\\circ-\\angle AOE=40^\\circ$\\\\\nLet the measure of $\\angle BOM$ be $x$\\\\\n$\\therefore \\angle COM=\\angle AOC+\\angle AOM=40^\\circ+180^\\circ-x=220^\\circ-x$, and $\\angle FOM=\\angle AOM-\\angle AOF=180^\\circ-x-25^\\circ=155^\\circ-x$\\\\\n$\\because \\angle COM+ \\frac{17}{10}\\angle AOE=2\\angle BOM+\\angle FOM$,\\\\\n$\\therefore 220^\\circ-x+ \\frac{17}{10}\\times 50^\\circ=2x+155^\\circ-x$\\\\\nSolving gives $x=\\boxed{75^\\circ}$\\\\\n$\\therefore$ The measure of $\\angle BOM$ is $\\boxed{75^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51438726_166.png"
  },
  {
    "solution": "\\boxed{30}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in $\\triangle ABC$, $BO$ bisects $\\angle ABC$, $CO$ bisects $\\angle ACB$, $MN$ passes through point $O$ and is parallel to $BC$, intersecting $AB$ and $AC$ at $M$ and $N$, respectively. Given $AB=12$, $AC=18$, and $MN=BM+CN$, what is the perimeter of $\\triangle AMN$?\n",
    "completion": "\\textbf{Solution:} From (1) we know that $MN = BM + CN$,\\\\\n$\\therefore$ the perimeter of $\\triangle AMN = AM + AN + MN = AM + AN + BM + CN = AB + AC$,\\\\\n$\\because$ $AB = 12$, $AC = 18$,\\\\\n$\\therefore$ the perimeter of $\\triangle AMN = AB + AC = \\boxed{30}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52793539_112.png"
  },
  {
    "solution": "\\boxed{100^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in $\\triangle ABC$, $DM$ and $EN$ are the perpendicular bisectors of $AB$ and $AC$, respectively, with footpoints $M$ and $N$. They intersect $BC$ at points $D$ and $E$, respectively. If $\\angle DAE=20^\\circ$, what is the measure of $\\angle BAC$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle$DAE=$20^\\circ$,\\\\\nthen $\\angle$ADE+$\\angle$DEA=$160^\\circ$.\\\\\nSince DM and EN are the perpendicular bisectors of AB and AC respectively,\\\\\nthen AD=BD, AE=EC,\\\\\nthen $\\angle$B=$\\angle$BAD, $\\angle$C=$\\angle$CAE,\\\\\nMoreover, since $\\angle$ADE, $\\angle$DEA are the external angles of $\\triangle$ABD and $\\triangle$AEC respectively,\\\\\nthen $\\angle$ADE=$\\angle$B+$\\angle$BAD=$2\\angle$B, $\\angle$DEA=$\\angle$C+$\\angle$CAE=$2\\angle$C,\\\\\nthen $\\angle$B+$\\angle$C=$80^\\circ$,\\\\\nthen $\\angle$BAC=$180^\\circ$−($\\angle$B+$\\angle$C)=$\\boxed{100^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52096771.png"
  },
  {
    "solution": "\\boxed{P_3\\left(\\frac{-3+\\sqrt{41}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: Given 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 being to the left of point B. The coordinates of point B are $(1, 0)$, and $OC=3OB$. If point E lies on the x-axis and point P lies on the parabola, is there a parallelogram with vertices A, C, E, and P that has AC as one of its sides? If such a parallelogram exists, provide the coordinates of point P; if it does not exist, explain why.\n",
    "completion": "\\textbf{Solution:} There exist $\\boxed{P_1(-3, -3)}$, $\\boxed{P_2\\left(\\frac{-3-\\sqrt{41}}{2}, 3\\right)}$, $\\boxed{P_3\\left(\\frac{-3+\\sqrt{41}}{2}, 3\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53080621_39.png"
  },
  {
    "solution": "\\boxed{20^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, it is known that $O$ is a point on the line $AB$, $\\angle COD=90^\\circ$, $OE$ bisects $\\angle BOC$. If $\\angle AOC=40^\\circ$, what is the degree measure of $\\angle DOE$?\n",
    "completion": "\\textbf{Solution}: Thus, we have $\\angle DOE = \\boxed{20^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53119355_127.png"
  },
  {
    "solution": "\\boxed{15}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, $A$, $B$, $C$, and $D$ are four consecutive vertices of a regular polygon, and $O$ is the center of the regular polygon. If $\\angle ADB = 12^\\circ$, how many sides does this regular polygon have?\n",
    "completion": "Solution: As shown in the figure, let the circumcircle of the polygon be $\\odot O$, and connect OA and OB,\\\\\n$\\because$ $\\angle ADB = 12^\\circ$,\\\\\n$\\therefore$ $\\angle AOB = 2\\angle ADB = 24^\\circ$,\\\\\nand $360^\\circ \\div 24^\\circ = 15$,\\\\\n$\\therefore$ the polygon is a regular pentadecagon,\\\\\nthus, the answer is: $\\boxed{15}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53009784.png"
  },
  {
    "solution": "\\boxed{(2, -3)}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, the coordinate of point \\(A\\) is \\( \\left(2, 3\\right) \\), the coordinate of point \\(B\\) is \\( \\left(4, 0\\right) \\), and point \\(C\\) is located in the fourth quadrant. If the quadrilateral formed by the vertices \\(O\\), \\(A\\), \\(B\\), and \\(C\\) is a rhombus, what are the coordinates of point \\(C\\)?\n",
    "completion": "\\textbf{Solution:} Given the property of a diamond that its diagonals are perpendicular and bisect each other, it is known that points $A$ and $C$ are symmetrical about the $x$-axis,\\\\\nsince $A(2, 3)$,\\\\\nthe symmetry about the $x$-axis keeps the $x$-coordinate the same and changes the sign of the $y$-coordinate,\\\\\ntherefore $C(2, -3)$,\\\\\nthus, the answer is: $C=\\boxed{(2, -3)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50553838.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in the Cartesian coordinate system, the sides of the rhombus OABC are located with OC on the x-axis, and point 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 with OB at point E. Find the value of OE: EB.\n",
    "completion": "\\textbf{Solution:} Set $E(a, \\frac{48}{a})$. Draw line $EG$ perpendicular to the x-axis at point $G$. According to the problem, it is known that $OG=a$ and $EG=\\frac{48}{a}$. From the drawing, we know that $EG \\parallel BF$,\\\\\n$\\therefore \\triangle OGE \\sim \\triangle OBF$,\\\\\n$\\therefore \\frac{\\frac{48}{a}}{6}=\\frac{a}{18}$. Solving this gives $a=12$,\\\\\n$\\therefore \\frac{OE}{OB}=\\frac{OG}{OF}=\\frac{12}{18}=\\frac{2}{3}$,\\\\\n$\\therefore \\frac{OE}{EB}=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51433106_157.png"
  },
  {
    "solution": "\\boxed{x\\ge \\frac{3}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</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)$. It is given that $OA=5$, $\\tan\\angle AOC=\\frac{3}{4}$, and the $y$-coordinate of point $B$ is $-8$. Directly write the solution set for the inequality $k_1x+b\\le \\frac{k_2}{x}$.\n",
    "completion": "\\textbf{Solution:} The solution set to the inequality ${k}_{1}x+b\\le \\frac{{k}_{2}}{x}$ is: $-4\\le x<0$ or $x\\ge \\frac{3}{2}$.\n\nTherefore, the solution set to the inequality is: $-4\\le x<0$ or $x\\ge \\frac{3}{2}$.\n\nHence, the answer is: \\boxed{-4\\le x<0} or \\boxed{x\\ge \\frac{3}{2}}.",
    "image_path": "dataset/mm_math/images/MM_Math/51446358_249.png"
  },
  {
    "solution": "\\boxed{\\frac{4}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in the Cartesian coordinate system, the coordinates of points $P$ and $A$ are (1, 0) and (2, 4), respectively. Point $B$ is a moving point on the $y$-axis. A line passing through point $A$ is perpendicular to $AB$ and intersects the $x$-axis at point $C$. Point $M$ is the midpoint of segment $BC$. What is the minimum value of $PM$?\n",
    "completion": "\\textbf{Solution:} Draw $AF\\perp y$-axis at point $F$, connect $AM$, $OM$\\\\\nSince $\\angle BAC=\\angle BOC=90^\\circ$, and $M$ is the midpoint of $BC$,\\\\\nthus $AM=OM$\\\\\nTherefore, point $M$ is on the perpendicular bisector of segment $AO$\\\\\nConstruct the perpendicular bisector of segment $AO$ to intersect the $y$-axis and $x$-axis at points $D$, $E$. When $PM\\perp DE$, $PM$ is minimized\\\\\nConnect $AD$, then $AD=OD$\\\\\nGiven $A(2,4)$\\\\\nTherefore $AF=2$, $OF=4$\\\\\nLet $OD=AD=t$, then $FD=4-t$,\\\\\nSince $FD^2+AF^2=AD^2$\\\\\nThus $(4-t)^2+2^2=t^2$\\\\\nTherefore $t=\\frac{5}{2}$\\\\\nTherefore $OD=\\frac{5}{2}$\\\\\nSince $\\angle FOA+\\angle AOE=90^\\circ$, $\\angle AOE+\\angle OED=90^\\circ$\\\\\nTherefore $\\angle FOA=\\angle OED$\\\\\nSince $\\angle AFO=\\angle DOE=90^\\circ$\\\\\nTherefore $\\triangle FAO\\sim \\triangle ODE$\\\\\nTherefore $\\frac{AF}{OD}=\\frac{OF}{OE}$, i.e., $AF\\cdot OE=OD\\cdot OF$,\\\\\nTherefore $OE=5$\\\\\nGiven $P(1,0)$\\\\\nTherefore $PE=4$\\\\\nSince in $\\triangle AFO$, $OA=\\sqrt{OF^2+AF^2}=\\sqrt{4^2+2^2}=2\\sqrt{5}$\\\\\nWhen $PM\\perp DE$, $PM$ is minimized\\\\\nTherefore $\\angle PME=\\angle AFO=90^\\circ$\\\\\n$\\triangle PME\\sim \\triangle AFO$\\\\\nTherefore $\\frac{PM}{AF}=\\frac{PE}{OA}$\\\\\nTherefore $\\frac{PM}{2}=\\frac{4}{2\\sqrt{5}}$\\\\\nTherefore $PM=\\boxed{\\frac{4}{5}\\sqrt{5}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51284430.png"
  },
  {
    "solution": "\\boxed{155^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, $O$ is a point on the line $AB$, $\\angle AOC=50^\\circ$, $OD$ bisects $\\angle AOC$, and $\\angle DOE=90^\\circ$. What is the measure of $\\angle BOD$ in degrees?\n",
    "completion": "\\textbf{Solution:} Since $\\angle AOD + \\angle BOD = 180^\\circ$,\\\\\nit follows that $\\angle BOD = 180^\\circ - \\angle AOD = \\boxed{155^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52916280_88.png"
  },
  {
    "solution": "\\boxed{\\frac{1}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, it is known that in rectangle $ABCD$, the diagonals $AC$ and $BD$ intersect at point $O$. Point $E$ is a moving point on side $AB$, and $CE$ intersects $BD$ at point $F$. Connect $OE$. If point $E$ is the midpoint of $AB$, what is the value of $\\frac{OF}{FB}$?\n",
    "completion": "Solution: $\\frac{OF}{FB} = \\boxed{\\frac{1}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52977191_45.png"
  },
  {
    "solution": "\\boxed{\\frac{2}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, a regular hexagon (the shaded area in the figure) is drawn on a regular triangular dartboard. Assuming that the probability of the dart hitting any point on the board is equal (if the dart hits the boundary or misses the board, then rethrow once), what is the probability that a dart thrown at random hits the shaded area?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $AN\\perp BC$ passing through point $A$ and intersect $DI$ at point $M$. Let the side length of the regular hexagon be $a$,\n\nsince hexagon $DEFGHI$ is a regular hexagon,\n\ntherefore $DE=EF=FG=GH=HI=DI=a$,\n\nthe degree of each external angle is: $360^\\circ\\div6=60^\\circ$,\n\ntherefore $\\angle ADI=\\angle AID=\\angle BEF=\\angle BFE=\\angle CGH=\\angle CHG=60^\\circ$,\n\ntherefore $\\triangle ADI$, $\\triangle BFE$, and $\\triangle CHG$ are all equilateral triangles, and their side lengths are all equal to $a$,\n\nsince $\\triangle ABC$ is an equilateral triangle,\n\ntherefore $BC=CA=AB=AD+DE+BE=a+a+a=3a$,\n\n$\\angle BAC=\\angle B=60^\\circ$\n\nIn $\\triangle ADI$, $\\triangle BFE$, and $\\triangle CHG$\n\n$\\begin{cases}\n\\angle ADI=\\angle BFE=\\angle CHG\\\\\nDI=FE=HG\\\\\n\\angle AID=\\angle BEF=\\angle CGH\n\\end{cases}$\n\ntherefore $\\triangle ADI\\cong \\triangle BFE\\cong \\triangle CHG\\left(ASA\\right)$,\n\ntherefore ${S}_{\\triangle ADI}={S}_{\\triangle BFE}={S}_{\\triangle CHG}$,\n\nsince $\\angle ADI=\\angle B=60^\\circ$,\n\ntherefore $DI\\parallel BC$,\n\nsince $AN\\perp BC$,\n\ntherefore $\\angle ANB=90^\\circ$,\n\ntherefore $\\angle AMD=\\angle ANB=90^\\circ$,\n\nsince $\\triangle ADI$ and $\\triangle ABC$ are both equilateral triangles and their side lengths are $a$ and $3a$ respectively,\n\ntherefore $DM=\\frac{1}{2}a$, $BN=\\frac{3}{2}a$,\n\ntherefore $AM=\\sqrt{AD^{2}-DM^{2}}=\\sqrt{a^{2}-\\left(\\frac{1}{2}a\\right)^{2}}=\\frac{\\sqrt{3}}{2}a$,\n\n$AN=\\sqrt{AB^{2}-BN^{2}}=\\sqrt{\\left(3a\\right)^{2}-\\left(\\frac{3}{2}a\\right)^{2}}=\\frac{3\\sqrt{3}}{2}a$,\n\ntherefore ${S}_{\\triangle ADI}=\\frac{1}{2}DI\\cdot AM=\\frac{1}{2}a\\times \\frac{\\sqrt{3}}{2}a=\\frac{\\sqrt{3}}{4}a^{2}$,\n\n${S}_{\\triangle ABC}=\\frac{1}{2}BC\\cdot AN=\\frac{1}{2}\\times 3a\\times \\frac{3\\sqrt{3}}{2}a=\\frac{9\\sqrt{3}}{4}a^{2}$,\n\ntherefore ${S}_{\\triangle BFE}={S}_{\\triangle CHG}={S}_{\\triangle ADI}=\\frac{\\sqrt{3}}{4}a^{2}$,\n\ntherefore ${S}_{\\text{regular hexagon }DEFGHI}={S}_{\\triangle ABC}-3{S}_{\\triangle ADI}=\\frac{9\\sqrt{3}}{4}a^{2}-3\\times \\frac{\\sqrt{3}}{4}a^{2}=\\frac{3\\sqrt{3}}{2}a^{2}$,\n\ntherefore the probability of the dart hitting the shaded area is $\\frac{{S}_{\\text{regular hexagon }DEFGHI}}{{S}_{\\triangle ABC}}=\\frac{\\frac{3\\sqrt{3}}{2}a^{2}}{\\frac{9\\sqrt{3}}{4}a^{2}}=\\boxed{\\frac{2}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53080524.png"
  },
  {
    "solution": "\\boxed{1\\text{cm}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, five squares with a side length of 1cm are arranged as illustrated. The points $A_1$, $A_2$, ..., $A_n$ represent the centers of the squares, respectively. What is the total area of the overlapping regions formed by the overlap of the five squares?\n",
    "completion": "As shown in the diagram, through the center O of the square ABCD, draw OM$\\perp$CD at M, and draw ON$\\perp$BC at N,\\\\\nthen $\\angle EOM=\\angle FON$, and OM=ON,\\\\\nin $\\triangle OEM$ and $\\triangle OFN$,\\\\\n$\\left\\{\\begin{array}{l}\n\\angle OME=\\angle ONF \\\\\nOM=ON \\\\\n\\angle EOM=\\angle FON\n\\end{array}\\right.$,\\\\\n$\\therefore \\triangle OEM\\cong \\triangle OFN$ (ASA),\\\\\nthus, the area of the quadrilateral OECF is equal to the area of the square OMCN,\\\\\nif the side length of the square ABCD is 1, then the area of OMCN is $\\frac{1}{4}\\text{cm}^2$,\\\\\n$\\therefore$ the area of the shaded part is equal to $\\frac{1}{4}$ of the square's area,\\\\\n$\\therefore$ the total area of the shaded parts (overlaps) in 5 such squares is $\\frac{1}{4} \\times 4 = \\boxed{1\\text{cm}^2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53320014.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the diagram, the square $MNPQ$ is inscribed in $\\triangle ABC$, with points M, N on side $BC$, and points P, Q on sides $AC$ and $AB$ respectively. Given that the height $AD$ from side $BC$ is $6\\,cm$ and $BC=12\\,cm$, what is the side length of the square $MNPQ$ in centimeters?\n",
    "completion": "\\textbf{Solution:} Given that: square $MNPQ$ is inscribed in $\\triangle ABC$, with the height from $BC$ being $AD=6\\,cm$, \\\\\n$\\therefore$ $\\angle ADC=\\angle ADB=\\angle CNP=\\angle BMQ=90^\\circ$, \\\\\nSince $\\angle B=\\angle B$, $\\angle C=\\angle C$, \\\\\n$\\therefore$ $\\triangle BMQ\\sim \\triangle BDA$, $\\triangle CNP\\sim \\triangle CDA$, \\\\\n$\\therefore$ $\\frac{BM}{BD}=\\frac{QM}{AD}$, $\\frac{NP}{AD}=\\frac{CN}{CD}$, \\\\\nSince $QM=NP$, let the side length of the square be $x$, \\\\\n$\\therefore$ $BM=\\frac{x}{AD}\\cdot BD$, $CN=\\frac{x}{AD}\\cdot CD$ \\\\\n$\\therefore$ $BM+CN=\\frac{x}{AD}\\cdot (BD+CD)=\\frac{x}{AD}\\cdot BC=BC-x$ \\\\\nSince $AD=6\\,cm$, $BC=12\\,cm$, \\\\\n$\\therefore$ $\\frac{x}{6}\\cdot 12=12-x$, \\\\\nSolving gives: $x=\\boxed{4}$ \\\\\n$\\therefore$ The side length of square $MNPQ$ is $4\\,cm$.",
    "image_path": "dataset/mm_math/images/MM_Math/53101929.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, the graph of the linear function $y=kx+b$ passes through the points $A(1, 6)$ and $B(-3, -2)$. What is the area of $\\triangle AOB$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, the line intersects the y-axis at point D,\\\\\nfor $y=2x+4$, by setting $x=0$, we get $y=4$,\\\\\n$\\therefore$ the coordinates of point D are: $(0,4)$,\\\\\n$\\therefore$ OD=4,\\\\\n$\\therefore$ ${S}_{\\triangle AOB}={S}_{\\triangle AOD}+{S}_{\\triangle BOD}$,\\\\\n$\\therefore$ ${S}_{\\triangle AOB}=\\frac{1}{2}\\times 4\\times 1+\\frac{1}{2}\\times 4\\times \\left|-3\\right|=2+6=\\boxed{8}$;\\\\\n$\\therefore$ the area of $\\triangle AOB$ is $\\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52055836_198.png"
  },
  {
    "solution": "\\boxed{triangular prism}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the diagram, what is the solid represented by its net?\n",
    "completion": "\\textbf{Solution:} According to the unfolded figure, there are a total of 5 faces, with two triangles serving as the base and three rectangles as the lateral faces,\\\\\n$\\therefore$ this solid is a \\boxed{triangular prism}.",
    "image_path": "dataset/mm_math/images/MM_Math/55025245.png"
  },
  {
    "solution": "\\boxed{y = -4x - 8}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</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)$. What is the analytical expression of line ${l}_{2}$?\n",
    "completion": "\\textbf{Solution:} Let's substitute point C $(-1, a)$ into $y = x - 3$, yielding $a = -4$,\\\\\n$\\therefore$ the coordinates of point C are $(-1, -4)$,\\\\\nLet the equation of line $l_2$ be $y = kx + b$,\\\\\n$\\therefore$ $\\begin{cases} -2k+b=0 \\\\ -k+b=-4 \\end{cases}$,\\\\\nSolving these, we get $\\begin{cases} k=-4 \\\\ b=-8 \\end{cases}$,\\\\\n$\\therefore$ the equation of line $l_2$ is $\\boxed{y = -4x - 8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52395325_102.png"
  },
  {
    "solution": "\\boxed{\\frac{5}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, the four vertices of rectangle $EFGH$ are respectively on the four sides of rhombus $ABCD$, with $BE=BF$. $\\triangle AEH$ and $\\triangle CFG$ are folded along edges $EH$ and $FG$ respectively. When the overlapping section forms a rhombus whose area is $\\frac{1}{16}$ of the area of rhombus $ABCD$, then what is the value of $\\frac{AE}{EB}$?\n",
    "completion": "\\textbf{Solution:} Let the overlapping rhombus have a side length of $x$, and $BE=BF=y$,\\\\\ndue to the symmetry of the rectangle and the rhombus as well as the property of folding, the quadrilateral $AHME$ and the quadrilateral $BENF$ are rhombuses,\\\\\n$\\therefore AE=EM$, $EN=BE=y$, $EM=x+y$,\\\\\n$\\because$ when the overlapping part is a rhombus and its area is $\\frac{1}{16}$ of the area of the rhombus $ABCD$, and the two rhombuses are similar,\\\\\n$\\therefore AB=4MN=4x$,\\\\\n$\\therefore AE=AB-BE=4x-y$,\\\\\n$\\therefore 4x-y=x+y$,\\\\\nSolving gives: $x=\\frac{2}{3}y$,\\\\\n$\\therefore AE=\\frac{5}{3}y$,\\\\\n$\\therefore \\frac{AE}{EB}=\\frac{\\frac{5}{3}y}{y}=\\boxed{\\frac{5}{3}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/50584152.png"
  },
  {
    "solution": "\\boxed{y=-2x+6}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the diagram, the graph of the linear function $y=x+3$, denoted by ${l}_{1}$, intersects the $x$-axis at point $B$, and intersects the graph of another linear function passing through point $A(3, 0)$, denoted by ${l}_{2}$, at point $C(1, m)$. What is the equation of the linear function represented by the graph ${l}_{2}$?\n",
    "completion": "\\textbf{Solution:} Since $C(1, m)$ is on the graph of $l_1\\colon y=x+3$,\\\\\n$\\therefore m=1+3=4$;\\\\\nAssume the expression for the linear function graph $l_2$ is $y=kx+b\\ (k\\ne 0)$,\\\\\nSubstituting the points $A(3, 0)$ and $C(1, 4)$, we get: $\\left\\{\\begin{array}{l}3k+b=0 \\\\ k+b=4\\end{array}\\right.$,\\\\\nSolving this, we find $\\left\\{\\begin{array}{l}k=-2 \\\\ b=6\\end{array}\\right.$,\\\\\n$\\therefore$ The equation of $l_2$ is: \\boxed{y=-2x+6}.",
    "image_path": "dataset/mm_math/images/MM_Math/52250956_85.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, a rectangular piece of paper $ABCD$ with sides of length 4 and 8 is folded such that point $C$ coincides with point $A$. What is the length of $BF$?\n",
    "completion": "\\textbf{Solution:} Given the property of the fold, we know: $AF=CF$.\\\\\nLet $BF=m$, then $AF=CF=8-m$,\\\\\nIn $\\triangle ABF$, $\\angle ABF=90^\\circ$, $AB=4$, $BF=m$, $AF=8-m$,\\\\\n$\\therefore AF^2=AB^2+BF^2$, meaning $(8-m)^2=4^2+m^2$,\\\\\n$\\therefore m=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52189194.png"
  },
  {
    "solution": "\\boxed{-2k+3}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: In the Cartesian coordinate system, the coordinates of point A are $(6, 4)$. Through point A, lines $AB \\perp x$-axis at point B and $AC \\perp y$-axis at point C are drawn, respectively. The graph of the linear function $y = kx + b$ passes through point $P(2, 3)$. Express $b$ in an algebraic formula involving $k$.\n",
    "completion": "\\textbf{Solution}: Substitute point P$(2,3)$ into $y=kx+b$, we get $3=2k+b$, \\\\\n$\\therefore b=\\boxed{-2k+3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52342947_114.png"
  },
  {
    "solution": "\\boxed{\\left(2, \\frac{3}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in the Cartesian coordinate system, the vertices of rectangle $OABC$ are such that vertex $O$ coincides with the origin of the coordinates, vertices $A$ and $C$ are respectively on the x-axis and y-axis, and the coordinates of point $B$ are $\\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$. Connect $DE$. Find the value of $k$ and the coordinates of point $E$.\n",
    "completion": "\\textbf{Solution:} In rectangle $OABC$,\\\\\nsince the coordinates of point $B$ are $(2, 3)$,\\\\\nthen the coordinates of midpoint $D$ on side $BC$ are $(1, 3)$,\\\\\nand since the inverse proportion function $y=\\frac{k}{x}$ passes through point $D(1, 3)$,\\\\\nthen $3=\\frac{k}{1}$,\\\\\ntherefore $k=\\boxed{3}$,\\\\\nsince point $E$ is on $AB$,\\\\\nthen the x-coordinate of $E$ is 2,\\\\\nand since $y=\\frac{3}{x}$ passes through point $E$,\\\\\nthen the y-coordinate of $E$ is $\\frac{3}{2}$,\\\\\ntherefore, the coordinates of point $E$ are $\\boxed{\\left(2, \\frac{3}{2}\\right)}$,",
    "image_path": "dataset/mm_math/images/MM_Math/52827689_140.png"
  },
  {
    "solution": "\\boxed{\\pi -2}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: There is a rectangular piece of paper ABCD, with the known dimensions $AB=2\\sqrt{2}$ and $AD=4$. On it, there is a semicircle with $AD$ as the diameter (as shown in Figure 1). Let $E$ be a point on side $AB$. When the paper is folded along $DE$ such that point $A$ exactly falls on $BC$, what is the area of the part of the semicircle that is still exposed (as shown in Figure 2, the shaded area)?\n",
    "completion": "\\textbf{Solution:} Let the center of the circle containing the shaded region be $O$, and let $AD$ intersect the semicircle at point $F$, as shown in the figure. Connect $OF$,\n\nBy the property of folds, we know: $AD=BC=4$,\n\nBy the property of rectangles, we know: $CD=AB=2\\sqrt{2}$, $\\angle DCA=90^\\circ$,\n\n$\\therefore AC=\\sqrt{AD^{2}-CD^{2}}=\\sqrt{4^{2}-(2\\sqrt{2})^{2}}=2\\sqrt{2}=CD$,\n\n$\\therefore \\triangle ADC$ is an isosceles right triangle,\n\n$\\therefore \\angle DAC=45^\\circ$,\n\nBecause $OD\\parallel BC$,\n\n$\\therefore \\angle ODF=\\angle DAC=45^\\circ$,\n\nBecause $OD=OF=2$,\n\n$\\therefore \\angle OFD=\\angle ODF=45^\\circ$,\n\n$\\therefore \\angle DOF=180^\\circ-45^\\circ-45^\\circ=90^\\circ$,\n\n$\\therefore S_{\\text{shaded}}=S_{\\text{sector} DOF}-S_{\\triangle DOF}=\\frac{90^\\circ \\pi \\times 2^{2}}{360}-\\frac{1}{2}\\times 2\\times 2=\\boxed{\\pi -2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51918252.png"
  },
  {
    "solution": "\\boxed{2 + \\sqrt{3}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, fold the equilateral triangle $\\triangle ABC$ such that point $C$ falls on point $D$ on side $AB$, where $EF$ is the crease. If $\\angle ADE = 90^\\circ$ and $AD = 1$, what is the length of $AC$?\n",
    "completion": "\\textbf{Solution:} Given an equilateral $\\triangle ABC$ is folded such that point C falls on the point D on side AB,\\\\\ntherefore $DE = CE$, and $\\angle A = 60^\\circ$,\\\\\nsince $\\angle ADE = 90^\\circ$,\\\\\nit follows that $\\angle AED = 30^\\circ$,\\\\\nalso, given $AD = 1$,\\\\\nthus $AE = 2AD = 2$,\\\\\ntherefore $DE = \\sqrt{AE^2 - AD^2} = \\sqrt{3}$,\\\\\nhence $AC = AE + CE = AE + DE = 2 + \\sqrt{3}$,\\\\\nthus, the answer is: $AC = \\boxed{2 + \\sqrt{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52914126.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in quadrilateral $ABCD$, $BD=AD$. Extend $CB$ to point $E$ such that $BE=BD$, and connect $AE$. Quadrilateral $AEBD$ is a rhombus; connect $DE$ and intersect $AB$ at point $F$. If $DC=\\sqrt{10}$ and the ratio $DC:DE=1:3$, what is the length of $AD$?\n",
    "completion": "\\textbf{Solution}: As shown in the diagram, extend DE to meet AB at F, \\\\\n$\\because$ Quadrilateral AEBD is a rhombus,\\\\\n$\\therefore$ AB$\\perp$DE,\\\\\n$\\therefore$ $\\angle$EFB$=90^\\circ$. \\\\\n$\\because$ Quadrilateral ABCD is a parallelogram,\\\\\n$\\therefore$ AB$ \\parallel $DC.\\\\\n$\\therefore$ $\\angle$EDC $=\\angle$EFB$=90^\\circ$.\\\\\n$\\because$ DC$= \\sqrt{10}$ and the ratio DC$\\colon$ DE$=1\\colon 3$,\\\\\n$\\therefore$ DE$= 3\\sqrt{10}$.\\\\\nIn $\\triangle$EDC, by the Pythagorean theorem, we have $EC=\\sqrt{ED^{2}+DC^{2}}=10$\\\\\n$\\therefore$ AD$=\\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53333420_83.png"
  },
  {
    "solution": "\\boxed{24}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in rectangle $ABCD$, where $AB=3$ and $AD=4$, two lines passing through points $A$ and $C$ intersect at point $E$. If $AE=13$ and $CE=12$, find the area of the shaded region in the figure.\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral $ABCD$ is a rectangle, with $AB=3$ and $AD=4$,\\\\\nit follows that $CD=AB=3$ and $\\angle D=90^\\circ$,\\\\\nthus, in right triangle $ACD$, $AC=\\sqrt{AD^2+CD^2}=\\sqrt{4^2+3^2}=5$,\\\\\nin triangle $ACE$, where $AE=13$, $CE=12$, and $AC=5$,\\\\\nsince $AC^2+CE^2=5^2+12^2=169$ and $AE^2=13^2=169$,\\\\\nit follows that $AC^2+CE^2=AE^2$,\\\\\nthus, $\\angle ACE=90^\\circ$,\\\\\nhence, the area of the shaded region, ${S}_{\\text{shaded}}={S}_{\\triangle ACE}-{S}_{\\triangle ACD}$\\\\\n$=\\frac{1}{2}AC\\cdot CE-\\frac{1}{2}AD\\cdot CD$\\\\\n$=\\frac{1}{2}\\times 5\\times 12-\\frac{1}{2}\\times 4\\times 3$\\\\\n$=30-6$\\\\\n$=\\boxed{24}$.\\\\\nTherefore, the area of the shaded region in the figure is $24$.",
    "image_path": "dataset/mm_math/images/MM_Math/53360000.png"
  },
  {
    "solution": "\\boxed{5\\sqrt{10}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, AB is the diameter of $\\odot O$, point C is the midpoint of $\\overset{\\frown}{BD}$, CF is a chord of $\\odot O$, and CF $\\perp$ AB with the foot of the perpendicular at point E. Connect BD and CF at point G, and connect CD, AD, BF. $\\triangle BFG \\cong \\triangle CDG$; if AD=10 and EF=15, find the length of BE.\n",
    "completion": "\\textbf{Solution:} Draw line OF,\\\\\n$\\because$ arc CD = arc CB = arc BF,\\\\\n$\\therefore$ arc CF = arc DB,\\\\\n$\\therefore$ BD = CF = 2, EF = 30,\\\\\n$\\because$ AD = 10, $\\angle ADB = 90^\\circ$,\\\\\n$\\therefore$ AB = $10\\sqrt{10}$,\\\\\n$\\therefore$ OF = $5\\sqrt{10}$,\\\\\n$\\because$ CF $\\perp$ AB,\\\\\n$\\therefore$ OE = 5, BE = $5\\sqrt{10} - 5 = \\boxed{5\\sqrt{10} - 5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52861206_119.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, the line $l_{1}\\colon y=-2x+4$ intersects with the x-axis at point D and with the y-axis at point A, the line $l_{2}\\colon y=x+1$ intersects with the x-axis at point C, the two lines $l_{1}$ and $l_{2}$ intersect at point B, and AC is connected. What is the area of $\\triangle ABC$?\n",
    "completion": "\\textbf{Solution:} Let the line $l_{2}\\colon y=x+1$ intersect the y-axis at point E. By setting $x=0$, $y=1$,\\\\\n$\\therefore$ point E$(0,1)$,\\\\\n$\\therefore AE=4-1=3$,\\\\\n$\\therefore S_{\\triangle ABC}= S_{\\triangle AEC}+S_{\\triangle ABE}=\\frac{1}{2}AE\\cdot OC+\\frac{1}{2}AE\\cdot x_{B}=\\frac{1}{2}\\times 3\\times 1+\\frac{1}{2}\\times 3\\times 1=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51435342_111.png"
  },
  {
    "solution": "\\boxed{24}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, within the quadrilateral $ABCD$, points E and F are located on sides AD and BC, respectively, with $DE=BF$. Here, line segments $CE$ and $AF$ are drawn. Quadrilateral $AECF$ is a parallelogram; Given that E is the midpoint of AD and $CE \\perp AD$, when $CE=4$ and $AB=5$, what is the area of quadrilateral $ABCD$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a parallelogram,\\\\\nit follows that CD = AB = 5,\\\\\nsince CE $\\perp$ AD,\\\\\nit follows that $\\angle$DEC = $90^\\circ$,\\\\\nthus DE = $\\sqrt{CD^{2} - CE^{2}} = \\sqrt{5^{2} - 4^{2}} = 3$,\\\\\nsince E is the midpoint of AD,\\\\\nit follows that AD = 2DE = 6,\\\\\ntherefore, the area of quadrilateral ABCD = AD × CE = 6 × 4 = $\\boxed{24}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50496670_101.png"
  },
  {
    "solution": "\\boxed{\\frac{5}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</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 is the area of $\\triangle AOB$?\n",
    "completion": "\\textbf{Solution:} Since the coordinates of point D are $ (0, \\frac{5}{3})$.\n\nHence, $S_{\\triangle AOB}=S_{\\triangle AOD}+S_{\\triangle BOD}$\n\n$=\\frac{1}{2}\\times \\frac{5}{3}\\times 2+\\frac{1}{2}\\times \\frac{5}{3}\\times 1$\n\n$=\\boxed{\\frac{5}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51699361_173.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the diagram, within circle $O$, chord $AB=8$, and point $C$ is on circle $O$ (and is distinct from $A$ and $B$). Connect $CA$ and $CB$, and draw $OD \\perp AC$ and $OE \\perp BC$ with perpendicular foot at points $D$ and $E$, respectively. The distance from point $O$ to $AB$ is 3. What is the radius of circle $O$?\n",
    "completion": "\\textbf{Solution:} Draw $OH\\perp AB$ through point $O$, with the foot as point $H$, where $OH=3$, and connect $OA$.\\\\\nSince $OH$ passes through the center $O$,\\\\\nit follows that $AH=BH=\\frac{1}{2}AB$,\\\\\nSince $AB=8$,\\\\\nit follows that $AH=4$,\\\\\nIn $\\triangle AHO$, we have $AH^2+OH^2=AO^2$,\\\\\ntherefore $AO=\\boxed{5}$,\\\\\nwhich means the radius of circle $O$ is $\\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53080238_91.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{6}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the diagram, in rectangle $ABCD$, we have $AB=4$ and $AE=\\frac{1}{3}AD$. After folding $\\triangle ABE$ along $BE$, $\\triangle GBE$ is obtained. Extending $BG$ to intersect $CD$ at point $F$, if $F$ is the midpoint of $CD$, what is the length of $BC$?\n",
    "completion": "\\textbf{Solution:} Extend BF to meet the extension of AD at point H, \\\\\n$\\because$ Quadrilateral ABCD is a rectangle, \\\\\n$\\therefore$ AD=BC, AD$\\parallel$BC, $\\angle$A=$\\angle$BCF=90$^\\circ$, \\\\\n$\\therefore$ $\\angle$H=$\\angle$CBF, \\\\\nIn $\\triangle BCF$ and $\\triangle HDF$, \\\\\n$ \\left\\{\\begin{array}{l}\\angle CBF=\\angle H \\\\ \\angle BCF=\\angle DFH \\\\ CF=DF \\end{array}\\right.$  , \\\\\n$\\therefore$ $\\triangle BCF \\cong \\triangle HDF$ (AAS), \\\\\n$\\therefore$ BC=DH, \\\\\n$\\because$ Folding $\\triangle ABE$ along BE yields $\\triangle GBE$, \\\\\n$\\therefore$ $\\angle$A=$\\angle$BGE=90$^\\circ$, AE=EG, \\\\\n$\\therefore$ $\\angle$EGH=90$^\\circ$, \\\\\n$\\because$ AE=$ \\frac{1}{3}$AD, \\\\\n$\\therefore$ Let AE=EG=x, then AD=BC=DH=3x, \\\\\n$\\therefore$ ED=2x, \\\\\n$\\therefore$ EH=ED+DH=5x, \\\\\nIn right $\\triangle EGH$, $\\sin\\angle H=\\frac{EG}{EH}=\\frac{x}{5x}=\\frac{1}{5}$  , \\\\\n$\\therefore$ $\\sin\\angle$CBF=$ \\frac{CF}{BF}=\\frac{1}{5}$  , \\\\\n$\\because$ AB=CD=4, and F is the midpoint of CD, \\\\\n$\\therefore$ CF=2, \\\\\n$\\therefore$ $\\frac{2}{BF}=\\frac{1}{5}$ , \\\\\n$\\therefore$ BF=10, upon verification, this is consistent with the problem statement, \\\\\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/51262289.png"
  },
  {
    "solution": "\\boxed{\\left(-2, 3\\right)}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</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$. Line $BC$ is drawn and it intersects the axis of symmetry of the parabola at point $E$, with the vertex being point $D$. Point $P$ is a moving point on the left side of the parabola's axis of symmetry, and point $Q$ lies on the ray $ED$. If the triangle formed by points $P$, $Q$, and $E$ is similar to $\\triangle BOC$, please write down the coordinates of point $P$ directly.\n",
    "completion": "\\textbf{Solution:} The coordinates of point $P$ are $\\boxed{\\left(-1-\\sqrt{2}, 2\\right)}$ or $\\boxed{\\left(-2, 3\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52602919_160.png"
  },
  {
    "solution": "\\boxed{108^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the diagram, it is known that $ \\angle AOC : \\angle BOC = 1 : 5$, $OD$ bisects $\\angle AOB$, and $\\angle COD = 36^\\circ$. What is the degree measure of $\\angle AOB$?\n",
    "completion": "\\textbf{Solution:} Let $\\angle AOC=x$, then $\\angle BOC=5x$,\\\\\nsince $OD$ bisects $\\angle AOB$,\\\\\nthus $\\angle AOD=\\frac{1}{2}\\angle AOB=\\frac{1}{2}(x+5x)=3x$,\\\\\nalso, since $\\angle COD=\\angle AOD-\\angle AOC$,\\\\\nthus $3x-x=36^\\circ$,\\\\\nsolving this, we get $x=18^\\circ$,\\\\\ntherefore $\\angle AOB=\\angle AOC+\\angle BOC=x+5x=\\boxed{108^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53106950.png"
  },
  {
    "solution": "\\boxed{\\frac{27\\sqrt{3}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, AB is the diameter of $\\odot O$, point D and E are on $\\odot O$, and quadrilateral BDEO is a parallelogram. A line through point D, $DC\\perp AE$, intersects the extension of AE at point C. CD is the tangent of $\\odot O$. If $AC=9$, what is the area of the shaded region?\n",
    "completion": "\\textbf{Solution:}\n\nGiven (1) we have $\\angle EAO = \\angle AOE = \\angle OBD = \\angle BOD = 60^\\circ$, and $ED \\parallel AB$,\n\n$\\therefore \\angle EAO = \\angle CED = 60^\\circ$,\n\n$\\because \\angle AOE + \\angle EOD + \\angle BOD = 180^\\circ$,\n\n$\\therefore \\angle EOD = 60^\\circ$,\n\n$\\therefore \\triangle DEO$ is an equilateral triangle,\n\n$\\therefore ED = OE = AE$,\n\n$\\because CD \\perp AE$, $\\angle CED = 60^\\circ$,\n\n$\\therefore \\angle CDE = 30^\\circ$,\n\n$\\therefore ED = 2CE = AE$,\n\n$\\because AC = 9$,\n\n$\\therefore CE = 3$, $AE = OE = ED = 6$,\n\n$\\therefore CD = \\sqrt{ED^2 - CE^2} = 3\\sqrt{3}$,\n\nLet the height of $\\triangle OED$ be $h$,\n\n$\\therefore h = OE \\cdot \\sin 60^\\circ = 3\\sqrt{3}$,\n\n$\\therefore S_{\\text{sector} ED} = S_{\\text{sector} OED} - S_{\\triangle OED} = \\frac{\\pi r^{2}n}{360}-\\frac{1}{2}ED\\cdot h = 6\\pi - 9\\sqrt{3}$,\n\n$\\therefore S_{\\text{shaded area}} = S_{\\triangle CED} - S_{\\text{arc} ED} = \\frac{1}{2}CE \\cdot CD - (6\\pi - 9\\sqrt{3}) = \\boxed{\\frac{27\\sqrt{3}}{2} - 6\\pi}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51433735_109.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in quadrilateral $ABCD$, the diagonals $AC \\perp BD$ intersect at point $O$, and points $E$, $F$, $G$, $H$ are the midpoints of sides $AD$, $AB$, $BC$, $CD$, respectively. If $AC=8$ and $BD=6$, what is the area of quadrilateral $EFGH$?\n",
    "completion": "\\textbf{Solution}: Since points $E$, $F$ are the midpoints of the sides $AD$, $AB$ of the quadrilateral $ABCD$, respectively,\\\\\nit follows that $EF \\parallel BD$ and $EF = \\frac{1}{2}BD = 3.$\\\\\nSimilarly, it is found that $EH \\parallel AC \\parallel GF$, and $EH = GF = \\frac{1}{2}AC = 4$,\\\\\nMoreover, since $AC \\perp BD$,\\\\\nit follows that $EF \\parallel GH$, $FG \\parallel HE$ and $EF \\perp FG$.\\\\\nThe quadrilateral $EFGH$ is a rectangle.\\\\\nTherefore, the area of quadrilateral $EFGH$ is $EF \\cdot EH = 3 \\times 4 = \\boxed{12}$, which means the area of quadrilateral $EFGH$ is 12.",
    "image_path": "dataset/mm_math/images/MM_Math/53320141.png"
  },
  {
    "solution": "\\boxed{(3,0)}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the diagram, 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 downwards by $m$ units to obtain the line $l_{2}$. It is known that the line $l_{2}$ passes through point $(-1, -2)$ and intersects the x-axis at point C. Calculate the value of $m$ and the coordinates of point C.\n",
    "completion": "\\textbf{Solution:} Let's move the line $l_{1}$ downwards by $m$ units to get the line $l_{2}$, thus the equation of line $l_{2}$ is $y= \\frac{1}{2}x+1-m$, \\\\\n$\\because$ The line $l_{2}$ passes through the point $(-1,-2)$, \\\\\n$\\therefore -2= \\frac{1}{2} \\times (-1)+1-m$, solving this gives $m= \\boxed{\\frac{5}{2}}$, \\\\\n$\\therefore$ The equation of line $l_{2}$ is $y= \\frac{1}{2}x- \\frac{3}{2}$, \\\\\nLet $y=0$, then solving gives $x=3$, \\\\\n$\\therefore$ The coordinates of point C are $\\boxed{(3,0)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50446864_230.png"
  },
  {
    "solution": "\\boxed{50^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, $\\angle ACD$ is an exterior angle of $\\triangle ABC$, $CE$ bisects $\\angle ACD$. If $\\angle A=60^\\circ$ and $\\angle B=40^\\circ$, then what is the measure of $\\angle DCE$ in degrees?\n",
    "completion": "\\textbf{Solution}: Since $CE$ bisects $\\angle ACD$, \\\\\nit follows that $\\angle ACD=2\\angle ECD$, \\\\\nsince $\\angle ACD$ is an exterior angle of $\\triangle ABC$, \\\\\nthus $\\angle ACD=2\\angle ECD=\\angle A+\\angle B$, \\\\\ntherefore, $2\\angle DCE=60^\\circ+40^\\circ=100^\\circ$, \\\\\nhence, $\\angle DCE=\\boxed{50^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52756083.png"
  },
  {
    "solution": "\\boxed{\\left( \\frac{3}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: In the figure, the line $y=-\\frac{1}{2}x+2$ intersects the $x$ axis and the $y$ axis at points $A$ and $B$, respectively. Point $E$ is on the negative half-axis of the $y$ axis, and the area of $\\triangle ABE=12$. In the figure, point $P$ is on the line $y=-x-1$. If $\\angle ABP=45^\\circ$, please directly write down the coordinates of point $P$.\n",
    "completion": "\\textbf{Solution:} The coordinates of point $P$ are $\\left( \\frac{3}{2}, -\\frac{5}{2} \\right)=\\boxed{\\left( \\frac{3}{2}, -\\frac{5}{2} \\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50560383_75.png"
  },
  {
    "solution": "\\boxed{x<0$ or $x>3}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, the parabola \\(y_{1}=ax^{2}+bx+c\\) intersects the line \\(y_{2}=mx+n\\) at points \\((3,0)\\) and \\((0,3)\\). If \\(ax^{2}+bx+c>mx+n\\), what is the range of values for \\(x\\)?\n",
    "completion": "\\textbf{Solution:} According to the graph of the functions, \\\\\nwhen $x<0$ or $x>3$, $y_1>y_2$, \\\\\nthus the solution set for $ax^2+bx+c>mx+n$ is $x<0$ or $x>3$. \\\\\nTherefore, the answer is: $\\boxed{x<0$ or $x>3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52605419.png"
  },
  {
    "solution": "\\boxed{70^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in $\\triangle ABC$, $D$ and $E$ are the midpoints of sides $AB$ and $AC$ respectively. If $\\angle A=75^\\circ$ and $\\angle ADE=35^\\circ$, what is the degree measure of $\\angle C$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle A=75^\\circ$ and $\\angle ADE=35^\\circ$,\\\\\nit follows that $\\angle AED=180^\\circ-\\angle A-\\angle ADE=70^\\circ$,\\\\\nsince $D$ and $E$ are the midpoints of sides $AB$ and $AC$ respectively, implying that $DE$ is the midline of $\\triangle ABC$,\\\\\nthus $DE\\parallel BC$,\\\\\ntherefore, $\\angle C=\\angle AED=\\boxed{70^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53599250.png"
  },
  {
    "solution": "\\boxed{\\frac{1}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</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 point A ($-1$, $0$) and point B ($3$, $0$), and intersects the y-axis at point C. The vertex is point D. Connecting BC and BD, what is the value of the tangent of $\\angle CBD$?\n",
    "completion": "\\textbf{Solution:} Let's connect CD, and draw DE $\\perp$ y-axis at point E,\\\\\n$\\because y={x}^{2}-2x-3={(x-1)}^{2}-4$,\\\\\n$\\therefore$ the coordinates of point D are (1, -4).\\\\\n$\\because$ B(3, 0), C(0, -3), D(1, -4), E(0, -4),\\\\\n$\\therefore$ OB=OC=3, CE=DE=1,\\\\\n$\\therefore$ BC=$3\\sqrt{2}$, DC=$\\sqrt{2}$, BD=$2\\sqrt{5}$.\\\\\n$\\therefore BC^{2}+DC^{2}=18+2=20=DB^{2}$.\\\\\n$\\therefore\\angle BCD=90^\\circ$.\\\\\n$\\therefore\\tan\\angle CBD=\\frac{DC}{BC}=\\frac{\\sqrt{2}}{3\\sqrt{2}}=\\boxed{\\frac{1}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51379494_84.png"
  },
  {
    "solution": "\\boxed{20^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, within square $ABCD$, point $E$ is on side $CD$. Connect $AE$, and draw $AF\\perp AE$ intersecting the extension of $CB$ at point $F$. Connect $EF$, and $AG$ bisects $\\angle FAE$, with $AG$ intersecting $BC$ and $EF$ at points $G$ and $H$, respectively. Connect $EG$ and $DH$. Given $AF=AE$; if $\\angle BAG=10^\\circ$, what is the degree measure of $\\angle EGC$?\n",
    "completion": "\\textbf{Solution:} Given that $AG$ bisects $\\angle EAF$,\\\\\nit follows that $\\angle GAF = \\angle GAE$,\\\\\ngiven that $AF=AE$, and $AG=AG$,\\\\\nit follows that $\\triangle AGF \\cong \\triangle AGE$ (SAS),\\\\\nthus $\\angle AGF = \\angle AGE = 90^\\circ - \\angle BAG$,\\\\\nsince $\\angle BAG = 10^\\circ$,\\\\\nit follows that $\\angle AGF = \\angle AGE = 80^\\circ$,\\\\\nthus $\\angle EGC = 180^\\circ - \\angle AGF - \\angle AGE = \\boxed{20^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52188888_129.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{1}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</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})$. Find the expressions of the linear function and the inverse proportion function?\n",
    "completion": "\\textbf{Solution:} Given the inverse proportion function $y_{2}=\\frac{m}{x}$ (where $m\\neq 0$) passes through point A ($-3$, $2$),\n\n$\\therefore m=-3\\times 2=-6$,\n\n$\\therefore$ the expression of the inverse proportion function is $y=-\\frac{6}{x}$.\n\nGiven point B ($n$, $-\\frac{3}{2}$) lies on the graph of the inverse proportion function $y=-\\frac{6}{x}$,\n\n$\\therefore n=4$.\n\n$\\therefore$ the coordinates of point B are ($4$, $-\\frac{3}{2}$).\n\nGiven the linear function $y_{1}=kx+b$ (where $k\\neq 0$) passes through points A ($-3$, $2$) and B ($4$, $-\\frac{3}{2}$),\n\n$\\therefore \\left\\{\\begin{array}{l}-3k+b=2 \\\\ 4k+b=-\\frac{3}{2}\\end{array}\\right.$,\n\nsolving this gives $\\left\\{\\begin{array}{l}k=-\\frac{1}{2} \\\\ b=\\frac{1}{2}\\end{array}\\right.$,\n\n$\\therefore$ the linear function is $\\boxed{y=-\\frac{1}{2}x+\\frac{1}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52076225_177.png"
  },
  {
    "solution": "\\boxed{\\frac{13\\sqrt{2}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, points E and F are inside the square ABCD, and AE $\\perp$ EF, CF $\\perp$ EF. Given AE = 9, EF = 5, FC = 3, what is the side length of the square ABCD?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect AC, and draw CG $\\perp$ AE extended to intersect at point G. \\\\\nSince AE $\\perp$ EF and CF $\\perp$ EF, \\\\\nthen the quadrilateral EFCG is a rectangle. \\\\\nIn $\\triangle ACG$, $AG=9+3=12$, $CG=EF=5$, \\\\\nthus, $AC=\\sqrt{AG^{2}+GC^{2}}=13$, \\\\\ntherefore, $AD=\\frac{13\\sqrt{2}}{2}$. \\\\\nHence, the answer is: $AD=\\boxed{\\frac{13\\sqrt{2}}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52302577.png"
  },
  {
    "solution": "\\boxed{(30\\sqrt{3}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, Alpha and Beta are two buildings, with a horizontal distance $BC = 30m$ between them. The elevation angle $\\angle EAD$ measured from point A to point D is $45^\\circ$, and the elevation angle $\\angle CBD$ measured from point B to point D is $60^\\circ$. What is the height of building Beta?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $AF\\perp CD$ at point F. In $Rt\\triangle BCD$,\\\\\n$\\angle DBC=60^\\circ$, $BC=30m$, $\\tan\\angle DBC=\\frac{CD}{BC}$\\\\\n$\\therefore CD=BC\\cdot \\tan60^\\circ=30\\sqrt{3}m$,\\\\\nIn $Rt\\triangle ADF$, $\\angle DAF=45^\\circ$,\\\\\n$\\therefore DF=AF=BC=30m$,\\\\\n$\\therefore AB=CF=CD-DF=(30\\sqrt{3}-30)m$,\\\\\nthis means the height of building B is $\\boxed{(30\\sqrt{3}-30)m}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53043705.png"
  },
  {
    "solution": "\\boxed{\\frac{1}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in diamond $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$, and $E$ is the midpoint of $BC$. If we connect $OE$, what must the length of $OE$ be?\n",
    "completion": "\\textbf{Solution}: Since ABCD is a rhombus, \\\\\ntherefore, AC $\\perp$ BD, i.e., $\\angle$BOC = $90^\\circ$, \\\\\nAlso, since E is the midpoint of BC, \\\\\ntherefore OE = BE = EC = $\\frac{1}{2}$BC. \\\\\nHence, $OE = \\boxed{\\frac{1}{2}BC}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51435944.png"
  },
  {
    "solution": "\\boxed{1.2}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</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 along side $AB$ towards point B at a speed of $1\\,cm/s$, and point Q starts from point B and moves along side $BC$ towards point C at a speed of $2\\,cm/s$. If P and Q start simultaneously from A and B respectively, and the moving time is denoted as $t\\,(s)$. When $t$ is how much, will $\\triangle PBQ$ be similar to $\\triangle ABC$?\n",
    "completion": "\\textbf{Solution:} Given the motion, we have $BQ=2t$ and $PB=6-t$.\\\\\nSince $\\triangle PQB$ and $\\triangle ABC$ are right-angled triangles,\\\\\ntherefore, when $\\triangle PQB$ is similar to $\\triangle ABC$, there are two cases,\\\\\nwhich are $\\frac{BP}{AB}=\\frac{BQ}{BC}$ or $\\frac{BP}{BC}=\\frac{BQ}{AB}$.\\\\\nWhen $\\frac{BP}{AB}=\\frac{BQ}{BC}$, we get $\\frac{6-t}{6}=\\frac{2t}{12}$, solving for $t$ yields $t=\\boxed{3}$;\\\\\nWhen $\\frac{BP}{BC}=\\frac{BQ}{AB}$, we get $\\frac{6-t}{12}=\\frac{2t}{6}$, solving for $t$ yields $t=\\boxed{1.2}$;\\\\\nTherefore, when $t=1.2$ or $3$, $\\triangle PQB$ is similar to $\\triangle ABC$.",
    "image_path": "dataset/mm_math/images/MM_Math/53106523_97.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, the quadrilateral $ABCD$ is inscribed in the circle $\\odot O$. Given that $\\angle ADC = 150^\\circ$, what is the measure of $\\angle AOC$ in degrees?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is inscribed in circle O, \\\\\nit follows that $\\angle B + \\angle ADC = 180^\\circ$, \\\\\nGiven that $\\angle ADC = 150^\\circ$, \\\\\nit follows that $\\angle B = 180^\\circ - 150^\\circ = 30^\\circ$. \\\\\nTherefore, $\\angle AOC = 2\\angle B = \\boxed{60^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52866046.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, AB is a chord of circle $O$, radius OC is perpendicular to AB at point D, and AB = 6, OD = 4. What is the length of DC?\n",
    "completion": "\\textbf{Solution:} Extend line OB. \\\\\nSince AB is a chord of circle O, and radius OC is perpendicular to AB at point D, and AB=6, \\\\\ntherefore BD=3, let CD=x, then OC=OB=x+4, \\\\\nin $\\triangle BDO$, $OD^{2}+BD^{2}=OB^{2}$, \\\\\nthus $4^{2}+3^{2}=(x+4)^{2}$, \\\\\nsolving this, we find $x=\\boxed{1}$, \\\\\ntherefore, CD=1.",
    "image_path": "dataset/mm_math/images/MM_Math/52874564.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, the cross-section of a horizontally placed cylindrical drainage pipe is $\\odot O$, with the height of the water-containing segment being 2, and the chord $AB=4\\sqrt{3}$. What is the radius of the cross-section?\n",
    "completion": "\\textbf{Solution:} Draw OD$\\perp$AB through point O, with the foot at point C, intersecting $\\odot O$ at point D,\\\\\nsince $AB=4\\sqrt{3}$ and $OD\\perp AB$,\\\\\ntherefore $AC=\\frac{1}{2}AB=2\\sqrt{3}$,\\\\\nLet the radius be $r$,\\\\\nsince $CD=2$,\\\\\ntherefore $OC=r-2$,\\\\\nIn $\\triangle OAC$, by the Pythagorean theorem, we obtain:\\\\\n$OC^2+AC^2=OA^2$, which means ${(r-2)}^2+{(2\\sqrt{3})}^2=r^2$,\\\\\nSolving yields: $r=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52871726.png"
  },
  {
    "solution": "\\boxed{\\sqrt{13}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in the quadrilateral $ABCD$, $E$ is the midpoint of $AD$, and $F$ is a point on the extension of $BC$, with $CF = \\frac{1}{2}BC$. Connect $CE$ and $DF$. Quadrilateral $CEDF$ is a parallelogram; if $AB=4$, $AD=6$, and $\\angle B=60^\\circ$, find the length of $DF$.\n",
    "completion": "\\textbf{Solution:} Let the line through $DH$ perpendicular to $BF$ at $H$, \\\\\nsince quadrilateral $ABCD$ is a parallelogram, it follows that $AB \\parallel CD$ and $CD=AB=4$. Hence, $\\angle DCH=\\angle B=60^\\circ$, thus $\\angle CDH=30^\\circ$. Therefore, $CH=\\frac{1}{2}CD=2$, and thus $DH=\\sqrt{CD^{2}-CH^{2}}=\\sqrt{4^{2}-2^{2}}=2\\sqrt{3}$. Since $CF=\\frac{1}{2}AD=\\frac{1}{2}BC=3$, it follows that $HF=CF-CH=3-2=1$. Therefore, $DF=\\sqrt{DH^{2}+HF^{2}}=\\sqrt{12+1}=\\boxed{\\sqrt{13}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52345537_113.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, the diagonals AC and BD of the rhombus ABCD intersect at point O, and $DE\\parallel OC$, $CE\\parallel OD$. The quadrilateral OCED is a rectangle; Given that $AC=4\\sqrt{5}$ and $BD=2\\sqrt{5}$, find the perimeter and area of the rectangle OCED.\n",
    "completion": "\\textbf{Solution:} \\[\\quad \\text{Since in the rhombus } ABCD, \\; OC=\\frac{1}{2}AC=2\\sqrt{5}, \\; OD=\\frac{1}{2}BD=\\sqrt{5}, \\\\\n\\therefore \\text{the perimeter of the rectangle } OCED \\text{ is: } (2\\sqrt{5}+\\sqrt{5})\\times 2=\\boxed{6\\sqrt{5}}, \\text{ and the area of } OCED \\text{ is: } 2\\sqrt{5}\\times \\sqrt{5}=\\boxed{10}\\].",
    "image_path": "dataset/mm_math/images/MM_Math/53313215_101.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{2}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in rectangle $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$, with $DE \\parallel AC$ and $CE \\parallel BD$. Quadrilateral $ODEC$ is a rhombus; connect $OE$. If $BC = 2\\sqrt{2}$, find the length of $OE$.\n",
    "completion": "\\textbf{Solution}: As shown in the figure, connect OE, intersecting CD at point F,\\\\\n$\\because$ from (1) we know that quadrilateral OCED is a rhombus,\\\\\n$\\therefore$ OE $\\perp$ CD,\\\\\n$\\therefore \\angle ADC=\\angle OFC=90^\\circ$,\\\\\n$\\therefore$ AD $\\parallel$ OE,\\\\\n$\\because$ DE $\\parallel$ AC,\\\\\n$\\therefore$ quadrilateral AOED is a parallelogram,\\\\\n$\\therefore$ OE=AD=BC=$\\boxed{2\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52387374_118.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the diagram, it is known that $AB$ is the diameter of $\\odot O$, and $C$ is a point on $\\odot O$. The bisector of $\\angle OCB$ intersects $\\odot O$ at point $D$. A tangent to $\\odot O$ at point $D$ intersects the extension of line $CB$ at point $E$. $CE \\perp DE$; if $AB=10$, and $\\tan A= \\frac{1}{3}$, find the length of $DE$.\n",
    "completion": "\\textbf{Solution:} Given $OD\\parallel CE$,\\\\\nit follows that $\\angle ODC=\\angle DCE$.\\\\\nGiven $OD=OA$ and $OC=OD$,\\\\\nit implies $\\angle A=\\angle ADO$ and $\\angle ODC=\\angle OCD$.\\\\\nSince $\\angle A=\\angle DCE$,\\\\\nwe have $\\angle A=\\angle OCD=\\angle ADO=\\angle CDO$.\\\\\nGiven $OA=OC$,\\\\\nit results in $\\triangle ADO\\cong \\triangle CDO$ (AAS),\\\\\nthus, $CD=AD$.\\\\\nSince $AB$ is the diameter, we have $\\angle ADB=90^\\circ$,\\\\\nand given $\\tan A=\\frac{DB}{AD}=\\frac{1}{3}$ and $AB=10$,\\\\\nfrom $AD^2+DB^2=AB^2=100$\\\\\nwe find $BD=\\sqrt{10}$, $AD=3\\sqrt{10}$,\\\\\nthus, $CD=AD=3\\sqrt{10}$.\\\\\nSince $\\angle A=\\angle DCE$ and $\\angle ADB=\\angle E=90^\\circ$,\\\\\n$\\triangle ADB\\sim \\triangle CED$,\\\\\nit follows that $\\frac{AB}{CD}=\\frac{BD}{DE}$,\\\\\nthus, $\\frac{10}{3\\sqrt{10}}=\\frac{\\sqrt{10}}{DE}$,\\\\\nresulting in $DE=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52601859_148.png"
  },
  {
    "solution": "\\boxed{\\frac{1}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, D and E are the midpoints of AB and AC, respectively. What is the value of $\\frac{S_{\\triangle ADE}}{S_{\\triangle ABC}}$?\n",
    "completion": "\\textbf{Solution:} Since D and E are the midpoints of AB and AC respectively,\\\\\nit follows that DE is the midline of the triangle,\\\\\ntherefore, the ratio DE: BC = 1: 2,\\\\\ntherefore, the area ratio $S_{\\triangle ADE} : S_{\\triangle ABC}$ = 1: 4.\\\\\nHence, the answer is: $S_{\\triangle ADE} : S_{\\triangle ABC} = \\boxed{\\frac{1}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51155485.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</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$. A perpendicular line is drawn from point $P$, denoted as $PF\\perp x$-axis, meeting point $F$, and intersects line $BC$ at point $E$. Assume the x-coordinate of point $P$ is $m$. If $PF=3EF$, what is the value of $m$?\n",
    "completion": "\\textbf{Solution:} Let point C be $(0,5)$, then the slope of BC is $k$, with $y=kx+5$.\\\\\n$\\therefore 5k+5=0$, solving for $k$ gives $k=-1$.\\\\\n$\\therefore$ The equation for BC is $y=-x+5$.\\\\\nSince $PF\\perp x$-axis and passes through point F, intersecting line BC at point E,\\\\\n$\\therefore$ Let point P be $(m, -m^2+4m+5)$, and point E be $\\left(m, -m+5\\right)$.\\\\\n$PF=-m^2+4m+5$, $EF=-m+5$.\\\\\nWe obtain $-m^2+4m+5=3(-m+5)$, solving for $m$ yields $m=\\boxed{2}$, $m=5$ (discard).",
    "image_path": "dataset/mm_math/images/MM_Math/52756020_83.png"
  },
  {
    "solution": "\\boxed{(3+3\\sqrt{2}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in the Cartesian coordinate system, the point $A(3, 3)$ forms a rhombus $AOBC$ with $OA$ as one of its sides. What are the coordinates of point $C$?\n",
    "completion": "\\textbf{Solution:} Let's draw AE$\\perp$x-axis at E, and CF$\\perp$x-axis at F,\\\\\nthen quadrilateral AEFC is a rectangle,\\\\\n$\\therefore$AE=CF, and AC=EF,\\\\\n$\\because$ point A is (3, 3),\\\\\n$\\therefore$AE=OE=3,\\\\\n$\\therefore$OA=$3\\sqrt{2}$,\\\\\n$\\because$ quadrilateral AOBC is a rhombus,\\\\\n$\\therefore$AC=OA=$3\\sqrt{2}$,\\\\\n$\\therefore$CF=AE=3, and OF=3+$3\\sqrt{2}$,\\\\\n$\\therefore$ the coordinates of point C are (3+$3\\sqrt{2}$, 3),\\\\\nso, the answer is: The coordinates of point C are $\\boxed{(3+3\\sqrt{2}, 3)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53728510.png"
  },
  {
    "solution": "\\boxed{80^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC=4$, $\\angle B=\\angle C=40^\\circ$. Point D moves along the segment $BC$ (D does not coincide with B or C). Connect $AD$ and construct $\\angle ADE=40^\\circ$, where $DE$ intersects segment $AC$ at $E$. During the movement of point D, is $\\triangle ADE$ an isosceles triangle? Calculate the degree measure of $\\angle BDA$.\n",
    "completion": "\\textbf{Solution:} Solve: When the degree of $\\angle BDA$ is $110^\\circ$ or $80^\\circ$, the shape of $\\triangle ADE$ is an isosceles triangle,\\\\\n$\\because$ when $\\angle BDA = 110^\\circ$,\\\\\n$\\therefore$ $\\angle ADC = 70^\\circ$,\\\\\n$\\because$ $\\angle C = 40^\\circ$,\\\\\n$\\therefore$ $\\angle DAE = 70^\\circ$,\\\\\n$\\therefore$ $\\angle AED = 180^\\circ - 70^\\circ - 40^\\circ=70^\\circ$\\\\\n$\\therefore$ the shape of $\\triangle ADE$ is an isosceles triangle;\\\\\n$\\because$ when the degree of $\\angle BDA$ is $80^\\circ$,\\\\\n$\\therefore$ $\\angle ADC = 100^\\circ$,\\\\\n$\\because$ $\\angle C = 40^\\circ$,\\\\\n$\\therefore$ $\\angle DAE = 40^\\circ$,\\\\\n$\\therefore$ $\\angle DAE = \\angle ADE$,\\\\\n$\\therefore$ the shape of $\\triangle ADE$ is an isosceles triangle.\\\\\nTherefore, the degree of $\\angle BDA$ is $\\boxed{110^\\circ}$ or $\\boxed{80^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52913913_104.png"
  },
  {
    "solution": "\\boxed{\\frac{2\\sqrt{5}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in the given grid, the side length of the small square is 1, and the points $A$, $B$, and $O$ are all on the lattice points. Then, what is the value of $\\cos \\angle A$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, according to the problem we have: $OC=2$, $AC=4$, \\\\\nBased on the Pythagorean theorem, we get $AO=\\sqrt{AC^2+OC^2}=\\sqrt{4^2+2^2}=2\\sqrt{5}$, \\\\\n$\\therefore$ $\\cos \\angle A=\\frac{AC}{AO}=\\frac{4}{2\\sqrt{5}}=\\boxed{\\frac{2\\sqrt{5}}{5}}$",
    "image_path": "dataset/mm_math/images/MM_Math/50934083.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, D, E, and F are 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:} Since D, E, F are the midpoints of the sides of $\\triangle ABC$ respectively,\\\\\nit follows that DE, DF, and EF are the midlines of $\\triangle ABC$,\\\\\ntherefore, BC = 2DF, AC = 2DE, AB = 2EF,\\\\\nthus, the perimeter of $\\triangle ABC$ = AB + BC + AC = 2(DF + EF + DE) = 12,\\\\\nhence, DF + EF + DE = $12 \\times \\frac{1}{2} = \\boxed{6}$, which means the perimeter of $\\triangle DEF$ is 6.",
    "image_path": "dataset/mm_math/images/MM_Math/52076214.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{4}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, the rectangle $ABCD$ has two sides $AD$, $AB$ with lengths $3$, $8$ respectively, and $E$ is the midpoint of $DC$. The graph of the inverse proportion function $y=\\frac{m}{x}$ (where $x<0$) passes through point $E$, and intersects $AB$ at point $F$. Given that the coordinates of point $B$ are $(-6, 0)$, find the value of $m$ and the expression of the linear function that passes through points $A$ and $E$.\n",
    "completion": "\\textbf{Solution:} Given that the coordinates of point B are $(-6,0)$, $AD=3$, $AB=8$, and $E$ is the midpoint of $CD$, thus $A(-6,8)$ and $E(-3,4)$. Since the graph of the inverse proportion function passes through point $E$, it follows that $m=-3\\times 4=\\boxed{-12}$. Suppose the expression of the linear function passing through points $A$ and $E$ is $y=kx+b(k\\ne 0)$. Then,\n\\[\n\\left\\{\n\\begin{array}{l}\n-6k+b=8, \\\\\n-3k+b=4,\n\\end{array}\n\\right.\n\\]\nfrom which we can solve\n\\[\n\\left\\{\n\\begin{array}{l}\nk=-\\frac{4}{3}, \\\\\nb=0,\n\\end{array}\n\\right.\n\\]\ntherefore, the expression of the linear function is $\\boxed{y=-\\frac{4}{3}x}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51615141_75.png"
  },
  {
    "solution": "\\boxed{5^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in $\\triangle ABC$, AE is the altitude on side BC. If AD is the bisector of $\\angle BAC$, $\\angle B=40^\\circ$, and $\\angle C=50^\\circ$, what is the measure of $\\angle DAE$?\n",
    "completion": "\\textbf{Solution}:\\\\\nSince $\\angle B=40^\\circ$ and $\\angle C=50^\\circ$,\\\\\nit follows that $\\angle BAC=90^\\circ$,\\\\\nSince $AD$ is the bisector of $\\angle BAC$,\\\\\nit results in $\\angle BAD=45^\\circ$,\\\\\nSince $\\angle ADE$ is an exterior angle of $\\triangle ABD$,\\\\\nit follows that $\\angle ADE=\\angle B+\\angle BAD=40^\\circ+45^\\circ=85^\\circ$,\\\\\nTherefore, $\\angle DAE=90^\\circ-85^\\circ=\\boxed{5^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52693985_125.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, the line $y=-2x+2$ intersects the $x$-axis and the $y$-axis at points $A$ and $B$, respectively, and the ray $AP\\perp AB$ at point $A$. If point $C$ is a moving point on ray $AP$, and point $D$ is a moving point on the $x$-axis, and the triangle with vertices $C$, $D$, $A$ is congruent to $\\triangle AOB$, then what is the length of $OD$?\n",
    "completion": "\\textbf{Solution:} In $y = -2x + 2$, \\\\\nwhen $x = 0$, then $y = 2$; when $y = 0$, then $x = 1$,\\\\\n$\\therefore$ $OA = 1$, $OB = 2$, by the Pythagorean theorem we get $AB = \\sqrt{OA^2 + OB^2} = \\sqrt{5}$,\\\\\n(1) When $\\angle ACD = 90^\\circ$, as in figure 1,\\\\\n$\\because \\triangle AOB \\cong \\triangle DCA$,\\\\\n$\\therefore AD = AB = \\sqrt{5}$,\\\\\n$\\therefore OD = 1 + \\sqrt{5}$;\\\\\n(2) When $\\angle ADC = 90^\\circ$, as in figure 2,\\\\\n$\\because AP \\perp AB$,\\\\\n$\\therefore \\angle BAP = \\angle AOB = 90^\\circ$,\\\\\n$\\therefore \\angle ABO + \\angle BAO = \\angle CAD + \\angle BAO = 90^\\circ$,\\\\\n$\\therefore \\angle ABO = \\angle CAD$,\\\\\n$\\because \\triangle AOB \\cong \\triangle CDA$,\\\\\n$\\therefore AD = OB = 2$,\\\\\n$\\therefore OA + AD = 3$,\\\\\nIn conclusion: the length of $OD$ is $1 + \\sqrt{5}$ or $\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52096265.png"
  },
  {
    "solution": "\\boxed{\\frac{6}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, square $ABCD$ has a side length of 10, and contains 6 identical smaller squares within. The vertices $E$, $F$, $G$, and $H$ of a small square are located on sides $AD$, $AB$, $BC$, and $CD$ respectively. What is the length of $DH$?\n",
    "completion": "\\textbf{Solution:} As illustrated in the figure:\\\\\n$\\because$ The side length of square ABCD is 10,\\\\\n$\\therefore \\angle A=\\angle B=90^\\circ$, $AB=10$,\\\\\nDraw GP$\\perp$AD at point G, with P being the foot of the perpendicular, thus $\\angle 4=\\angle 5=90^\\circ$,\\\\\n$\\therefore$ Quadrilateral APGB is a rectangle,\\\\\n$\\therefore \\angle 2+\\angle 3=90^\\circ$, $PG=AB=10$,\\\\\n$\\because$ Six squares of identical size are placed within the large square as shown in the figure,\\\\\n$\\therefore \\angle 1+\\angle 2=90^\\circ$,\\\\\n$\\therefore \\angle 1=\\angle 3$,\\\\\n$\\therefore \\triangle BGF \\sim \\triangle PGE$,\\\\\n$\\therefore \\frac{BG}{PG}=\\frac{FG}{EG}$,\\\\\n$\\therefore \\frac{BG}{10}=\\frac{1}{5}$,\\\\\n$\\therefore GB=2$,\\\\\n$\\therefore AP=2$,\\\\\nSimilarly, $DE=2$,\\\\\n$\\therefore PE=AD-AP-DE=6$,\\\\\n$\\therefore EG=\\sqrt{10^2+6^2}=2\\sqrt{34}$,\\\\\n$\\therefore$ The side length of the small square is $\\frac{2\\sqrt{34}}{5}$,\\\\\n$\\therefore DH=\\sqrt{EH^2-DE^2}=\\sqrt{\\left(\\frac{2\\sqrt{34}}{5}\\right)^2-2^2}=\\boxed{\\frac{6}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52513833.png"
  },
  {
    "solution": "\\boxed{\\frac{n-1}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, $n$ squares, each with a side length of 1, are arranged as depicted. The points $A_{1}$, $A_{2}$, ..., $A_{n}$ represent the centers of these squares, respectively. What is the total area of the overlapping parts formed by these $n$ squares?\n",
    "completion": "\\textbf{Solution:} Given the problem statement, the area of one shaded part is equal to $\\frac{1}{4}$ of the area of a square, which is $\\frac{1}{4}$.\\\\\nThe total area of the shaded parts (overlap area) for $n$ such squares is: $\\frac{1}{4}\\times (n-1) = \\boxed{\\frac{n-1}{4}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51212188.png"
  },
  {
    "solution": "\\boxed{\\left(-\\frac{3}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: Synthesis and Inquiry: As shown in the diagram, 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. If point D is a moving point on the parabola in the third quadrant, and lines $AD$ and $CD$ are drawn, find the maximum area of the quadrilateral $AOCD$ and the coordinates of point D at that time.\n",
    "completion": "\\textbf{Solution:} Let's draw a figure, passing through point D draw $DF\\perp x$-axis at point F, let $D\\left(m, m^{2}+2m-3\\right)$  $\\left(-3<m<0\\right)$.\\\\\n$\\therefore DF=-m^{2}-2m+3$, $AF=m+3$, $OF=-m$.\\\\\nWhen $x=0$, $y=-3$.\\\\\n$\\therefore$ The coordinates of point C are $\\left(0, -3\\right)$.\\\\\n$\\therefore OC=3$.\\\\\n$\\therefore S_{\\text{quadrilateral}AOCD}=\\frac{1}{2}AF\\cdot DF+\\frac{1}{2}\\left(OC+DF\\right)\\cdot OF$,\\\\\n$=\\frac{1}{2}\\left(m+3\\right)\\left(-m^{2}-2m+3\\right)+\\frac{1}{2}\\left(3-m^{2}-2m+3\\right)\\left(-m\\right)$,\\\\\n$=-\\frac{3}{2}m^{2}-\\frac{9}{2}m+\\frac{9}{2}$,\\\\\n$=-\\frac{3}{2}\\left(m+\\frac{3}{2}\\right)^{2}+\\frac{63}{8}$.\\\\\n$\\therefore$ When $m=-\\frac{3}{2}$, $S_{\\text{quadrilateral}AOCD}$ is maximized, and the maximum value is $\\boxed{\\frac{63}{8}}$.\\\\\nAt this time, the coordinates of point D are $\\boxed{\\left(-\\frac{3}{2}, -\\frac{15}{4}\\right)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52720059_108.png"
  },
  {
    "solution": "\\boxed{40^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, point D is on the circumcircle of the equilateral triangle $\\triangle ABC$, and $\\angle DAC = 20^\\circ$. What is the measure of $\\angle ACD$?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABC$ is an equilateral triangle,\\\\\nit follows that $\\angle BAC = \\angle ACB = 60^\\circ$,\\\\\nthus $\\angle BAD = \\angle BAC + \\angle DAC = 60^\\circ + 20^\\circ = 80^\\circ$,\\\\\nsince quadrilateral ABCD is a cyclic quadrilateral,\\\\\nit follows that $\\angle BCD = 180^\\circ - \\angle BAD = 180^\\circ - 80^\\circ = 100^\\circ$,\\\\\ntherefore $\\angle ACD = \\angle BCD - \\angle ACB = 100^\\circ - 60^\\circ = \\boxed{40^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52861180.png"
  },
  {
    "solution": "\\boxed{-1,0}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, the quadrilateral $ABCD$ is a rhombus, with known coordinates $A(4, 0)$ and $B(0, 3)$. Directly write down the coordinates of points $C$ and $D$.\n",
    "completion": "\\textbf{Solution:} Since $A(4,0)$ and $B(0,3)$, it follows that $OA=4, OB=3$. Therefore, $AB=\\sqrt{AO^{2}+BO^{2}}=\\sqrt{4^{2}+3^{2}}=5$. Since quadrilateral ABCD is a rhombus, it follows that $BC=AD=AB=5$. Consequently, $C(\\boxed{-5,3})$ and $D(\\boxed{-1,0})$.",
    "image_path": "dataset/mm_math/images/MM_Math/52423104_78.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, the bisector of $\\angle CAB$ intersects $BC$ at point $D$, and $DE$ is the perpendicular bisector of $AB$, with $E$ being the foot of the perpendicular. If $BC=3$, what is the length of $DE$?\n",
    "completion": "\\textbf{Solution:} Since $AD$ is the bisector of $\\angle CAB$, and $DC \\perp AC$, $DE \\perp AB$,\\\\\nit follows that $DC=DE$,\\\\\nLet $DC=y$, then $DE=y$, and $BD=3-y$,\\\\\nFurthermore, since $\\angle B=30^\\circ$,\\\\\nwe have $y=\\frac{3-y}{2}$,\\\\\nSolving this, we get $\\boxed{y=1}$,\\\\\nthus $DE=\\boxed{1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52095924_111.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle A=40^\\circ$, $AB=AC$, and the perpendicular bisector $MN$ of $AB$ intersects $AC$ at point D and $AB$ at point E. The perimeter of $\\triangle ABC$ is $13cm$, $AB=5cm$. What is the degree measure of $\\angle DBC$?\n",
    "completion": "\\textbf{Solution:} \\\\\nSince $AB=AC$, \\\\\nit follows that $\\triangle ABC$ is an isosceles triangle, \\\\\nbecause $\\angle A=40^\\circ$, \\\\\nthus $\\angle ABC=\\angle ACB=\\frac{1}{2}\\left(180^\\circ-40^\\circ\\right)=70^\\circ$, \\\\\nsince $MN$ is the perpendicular bisector of $AB$, \\\\\nit follows that $\\triangle ABD$ is an isosceles triangle, \\\\\ntherefore $\\angle ABD=\\angle A=40^\\circ$, \\\\\nhence $\\angle DBC=\\angle ABC-\\angle ABD=70^\\circ-40^\\circ=\\boxed{30^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52793526_108.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ABC=90^\\circ$, $BC=3$. Point D is on the extension of AC, such that $AC = 3CD$. Through point D, draw $DH \\parallel AB$, which intersects the extension of BC at point H. Find the value of $BD \\cdot \\cos\\angle HBD$?\n",
    "completion": "\\textbf{Solution:} From the given information, \\\\\nsince $DH\\parallel AB$, \\\\\nthen $\\angle BHD=\\angle ABC =90^\\circ$, \\\\\nthus $\\triangle ABC\\sim \\triangle DHC$, \\\\\nresulting in $\\frac{AC}{CD}=\\frac{BC}{CH}$, \\\\\nsince $AC=3CD$ and $BC=3$, \\\\\nit follows that $CH=1$, \\\\\n$BH=BC+CH=4$, \\\\\nIn $\\triangle BHD$, $\\cos\\angle HBD= \\frac{BH}{BD}$, \\\\\nthus $BD\\cdot\\cos\\angle HBD=BH=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52623501_107.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, for a regular hexagon $ABCDEF$ inscribed in a circle, with the radius $OA=4$, what is the length of the side of the hexagon?\n",
    "completion": "\\textbf{Solution:} Draw line OB, as shown in the diagram,\\\\\n$\\because$ Hexagon ABCDEF is inscribed in circle $O$,\\\\\n$\\therefore \\angle$AOB=$ \\frac{360^\\circ}{6}=60^\\circ$, OA=OB,\\\\\n$\\therefore \\triangle$OAB is an equilateral triangle,\\\\\n$\\therefore$AB=OA= \\boxed{4},\\\\\ni.e., the side length of this regular hexagon is 4.",
    "image_path": "dataset/mm_math/images/MM_Math/51446421.png"
  },
  {
    "solution": "\\boxed{20}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the rhombus $ABCD$ has its edge $AB$ on the x-axis and vertex $C$ on the y-axis. It is known that the coordinates of points $A$ and $C$ are $\\left(-2, 0\\right)$ and $\\left(0, 4\\right)$, respectively. What is the area of the rhombus $ABCD$?\n",
    "completion": "\\textbf{Solution:} From the problem, we have the area of rhombus $ABCD$, $S_{\\text{rhombus} ABCD} = AB \\times OC = 5 \\times 4 = \\boxed{20}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53182915_205.png"
  },
  {
    "solution": "\\boxed{\\sqrt{2}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC=2\\sqrt{10}$, and $AD$ is the altitude on side $BC$, with $DE\\parallel AC$ intersecting $AB$ at point $E$. $\\triangle ADE$ is an isosceles triangle; connect $CE$ and intersect $AD$ at point $H$. If $\\angle DCE=45^\\circ$, find the length of $EH$.\n",
    "completion": "\\textbf{Solution:} Draw $EG\\parallel BC$, intersecting $AD$ at $G$,\\\\\n$\\because AE=BE$,\\\\\n$\\therefore AG=DG$,\\\\\n$\\therefore EG= \\frac{1}{2} BD= \\frac{1}{2} CD$,\\\\\n$\\because EG\\parallel BC$,\\\\\n$\\therefore \\frac{GH}{DH}=\\frac{EH}{CH}=\\frac{EG}{CD} = \\frac{1}{2}$,\\\\\n$\\therefore GH= \\frac{1}{2} DH$, $EH= \\frac{1}{2} CH$,\\\\\n$\\because AD\\perp BC$, $\\angle DCE=45^\\circ$,\\\\\n$\\therefore \\triangle CDH$ is an isosceles right triangle,\\\\\n$\\therefore DH=DC$,\\\\\n$\\therefore AD=3DC$,\\\\\n$\\because AB=AC=2\\sqrt{10}$, $AC^2=AD^2+DC^2$,\\\\\n$\\therefore 40=9DC^2+DC^2$,\\\\\n$\\therefore DC=2$,\\\\\n$\\therefore DH=DC=2$,\\\\\n$\\therefore CH= \\sqrt{2^2+2^2} =2\\sqrt{2}$,\\\\\n$\\therefore EH= \\frac{1}{2} CH= \\boxed{\\sqrt{2}}$,\\\\",
    "image_path": "dataset/mm_math/images/MM_Math/52861967_90.png"
  },
  {
    "solution": "\\boxed{x > 1}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</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 $kx>mx+n$ in terms of $x$?\n",
    "completion": "\\textbf{Solution:} From the graph, it's known that the intersection point of the two functions is $(1,2)$,\\\\\nthus the solution set of the linear inequality in $x$, $kx - n > mx$, is $x > 1$,\\\\\ntherefore, the answer is: $\\boxed{x > 1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50545552.png"
  },
  {
    "solution": "\\boxed{\\frac{5}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in right-angled triangle $ABC$ with $\\angle C=90^\\circ$, $D$ is the midpoint of $AB$, $AD=5$, $BC=8$, and $E$ is a moving point on line $BC$. After folding $\\triangle BDE$ along line $ED$, point $B$ falls at point $F$. When $FD \\perp BC$, what is the length of segment $BE$?\n",
    "completion": "\\textbf{Solution:} Suppose when point F is below BC, DF intersects BC at point P, as shown in Figure 1:\\\\\n$\\because$D is the midpoint of AB,\\\\\n$\\therefore$AB $=$ 2AD $=$ 10,\\\\\n$\\because$$\\angle$C $= 90^\\circ$,\\\\\n$\\therefore$AC $=$ $\\sqrt{AB^{2}-BC^{2}}$ $=$ $\\sqrt{10^{2}-8^{2}}$ $=$ 6,\\\\\n$\\because$FD$\\perp$BC, $\\angle$C $= 90^\\circ$,\\\\\n$\\therefore$FD$\\parallel$AC, and D is the midpoint of AB,\\\\\n$\\therefore$$\\triangle PBD\\sim \\triangle CBA$\\\\\n$\\therefore$$\\frac{BD}{BA}$ $=$ $\\frac{BP}{BC}$ $=$ $\\frac{DP}{AC}$ $=$ $\\frac{1}{2}$,\\\\\n$\\therefore$BP $=$ PC $=$ $\\frac{1}{2}$BC $=$ 4, DP $=$ $\\frac{1}{2}$AC $=$ 3,\\\\\n$\\because$$\\triangle BDE$ is folded along line ED,\\\\\n$\\therefore$FD $=$ BD $=$ 5, FE $=$ BE,\\\\\n$\\therefore$FP $=$ FD - DP $=$ 5 - 3 $=$ 2,\\\\\nIn $\\triangle FPE$, EF$^{2}$ $=$ FP$^{2}$ + PE$^{2}$,\\\\\n$\\therefore$BE$^{2}$ $=$ $2^{2}$ + $(4-BE)^{2}$,\\\\\nSolving: BE $=$ $\\boxed{\\frac{5}{2}}$;\\\\\nSuppose when point F is above BC, the extension of FD intersects BC at point P, as shown in Figure 2,\\\\\n$\\because$FD = BD = 5, DP $=$ $\\frac{1}{2}$AC $=$ 3,\\\\\n$\\therefore$FP $=$ DP + FD $=$ 3 + 5 $=$ 8,\\\\\nIn $\\triangle EFP$, EF$^{2}$ $=$ FP$^{2}$ + EP$^{2}$,\\\\\n$\\therefore$BE$^{2}$ $=$ 64 + $(BE-4)^{2}$,\\\\\nSolving: BE $=$ 10;",
    "image_path": "dataset/mm_math/images/MM_Math/52745997.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: Given points $B$ and $D$ on line segment $AC$, as shown in the figure, if $AC=20$, $AB=8$, and point $D$ is the midpoint of line segment $AC$, what is the length of line segment $BD$?\n",
    "completion": "\\textbf{Solution:} Given that $AC=20$ and point $D$ is the midpoint of segment $AC$, \\\\\n$\\therefore AD=DC=\\frac{1}{2}AC=\\frac{1}{2}\\times 20=10$, \\\\\nSince $AB=8$, \\\\\n$\\therefore BD=AD-AB=10-8=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50009912_220.png"
  },
  {
    "solution": "\\boxed{33^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, $\\triangle ABC$ and $\\triangle ADE$ are symmetric with respect to line $MN$, and the intersection $F$ of $BC$ and $DE$ lies on line $MN$. If $BC \\parallel AD$, $AE$ bisects $\\angle BAM$, and $\\angle BFE + \\angle C = 81^\\circ$, what is the degree measure of $\\angle EAF$?\n",
    "completion": "\\textbf{Solution:} Given $BC\\parallel AD$,\\\\\nit follows that $\\angle C = \\angle CAD$,\\\\\ngiven $\\angle DAC = \\angle BAE$, and $\\angle EAF = \\angle CAF$,\\\\\nand since $AE$ bisects $\\angle BAM$,\\\\\nit follows $\\angle DAC = \\angle CAF = \\angle EAF = \\angle BAE$,\\\\\ngiven $\\angle BFE + \\angle C = 81^\\circ$,\\\\\nit follows $\\angle D + \\angle DAC = 81^\\circ$,\\\\\ntherefore $\\angle CAF + \\angle EAF + \\angle E = 180^\\circ - 81^\\circ = 99^\\circ$,\\\\\ngiven $\\angle C = \\angle E$,\\\\\nit follows $3\\angle EAF = 99^\\circ$,\\\\\nhence $\\angle EAF = \\boxed{33^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52991684_118.png"
  },
  {
    "solution": "\\boxed{2^{2020}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the diagram, $\\angle MON=30^\\circ$, points $A_1, A_2, A_3, \\ldots$ are on ray ON, points $B_1, B_2, B_3, \\ldots$ are on ray OM, $\\triangle A_1B_1A_2$, $\\triangle A_2B_2A_3$, $\\triangle A_3B_3A_4,\\ldots$ are all equilateral triangles. The side length of the first equilateral triangle from the left is denoted as $a_1$, the side length of the second equilateral triangle is denoted as $a_2$, and so on. If $OA_1=1$, then what is the value of $a_{2021}$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure,\\\\\nsince $\\triangle A_1B_1A_2$ is an equilateral triangle,\\\\\nthus $A_1B_1=A_2B_1$, $\\angle 3=\\angle 4=\\angle 12=60^\\circ$,\\\\\ntherefore $\\angle 2=120^\\circ$,\\\\\nsince $\\angle MON=30^\\circ$,\\\\\nthus $\\angle 1=180^\\circ-120^\\circ-30^\\circ=30^\\circ$,\\\\\nalso since $\\angle 3=60^\\circ$,\\\\\nthus $\\angle 5=180^\\circ-60^\\circ-30^\\circ=90^\\circ$,\\\\\nsince $\\angle MON=\\angle 1=30^\\circ$,\\\\\nthus $OA_1=A_1B_1=1$, that is, the side length of $\\triangle A_1B_1A_2$ is $a_1=1=2^0$;\\\\\ntherefore $A_2B_1=1$,\\\\\nsince $\\triangle A_2B_2A_3$, $\\triangle A_3B_3A_4$ are equilateral triangles,\\\\\nthus $\\angle 11=\\angle 10=60^\\circ$, $\\angle 13=60^\\circ$,\\\\\nsince $\\angle 4=\\angle 10=\\angle 11=60^\\circ$, $\\angle 12=\\angle 13=60^\\circ$,\\\\\nthus $A_1B_1\\parallel A_2B_2\\parallel A_3B_3$, $B_1A_2\\parallel B_2A_3$,\\\\\nthus $\\angle 1=\\angle 6=\\angle 7=30^\\circ$, $\\angle 5=\\angle 8=90^\\circ$,\\\\\nthus $A_2B_2=2B_1A_2=2=2^1$, that is, the side length of $\\triangle A_2B_2A_3$ is $a_2=2^1$\\\\\nSimilarly, $B_3A_3=2B_2A_3=4=2^2$, that is, the side length of $\\triangle A_3B_3A_4$ is $a_3=2^2$,\\\\\n... ,\\\\\nthus, the side length of $\\triangle A_nB_nA_{n+1}$ is $a_n=2^{n-1}$,\\\\\nthus, the side length of $\\triangle A_{2021}B_{2021}A_{2022}$ is $a_{2021}=\\boxed{2^{2020}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52603194.png"
  },
  {
    "solution": "\\boxed{80^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, points A, B, and C are three points on the circle $ \\odot O$. If $ \\angle BAC=40^\\circ$, what is the degree 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=2\\times 40^\\circ=\\boxed{80^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53043215.png"
  },
  {
    "solution": "\\boxed{x<0}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, the graph of the linear function $y=kx+b$ (where $k$ and $b$ are constants) is depicted as illustrated. What is the solution set of the inequality $kx+b<1$?\n",
    "completion": "\\textbf{Solution:} From the graph, we know that the linear function $y=kx+b$ ($k,b$ are constants, $k\\neq 0$) passes through the point $(0,1)$, and the value of $y$ increases as $x$ increases. Therefore, the solution set of the inequality $kx+b<1$ is $\\boxed{x<0}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52425090.png"
  },
  {
    "solution": "\\boxed{24}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in rhombus $ABCD$, the lengths of the two diagonals are $AC=6$ and $BD=8$. What is the area of this rhombus?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rhombus, \\\\\nthe area of the rhombus = $\\frac{1}{2}$AC$\\cdot$BD = $\\frac{1}{2} \\times 6 \\times 8 = \\boxed{24}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52211187.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, it is known that in quadrilateral $ABCD$, $BD$ bisects $\\angle ABC$, $AB=4$, and $BC=9$. If $\\triangle ABD$ is similar to $\\triangle DBC$, what is the length of $BD$?\n",
    "completion": "\\textbf{Solution:} Given that $\\triangle ABD \\sim \\triangle DBC$,\\\\\nit follows that $\\frac{BD}{BC}=\\frac{AB}{BD}$,\\\\\nwhich implies $BD^{2}=AB \\cdot BC=4 \\times 9=36$,\\\\\ntherefore, $BD=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52942171.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in parallelogram $ABCD$, points $E$ and $F$ are on sides $BC$ and $AD$ respectively, with $BE=DF$, $\\angle AEC=90^\\circ$. Quadrilateral $AECF$ is a rectangle; $BF$ is drawn, and if $AB=4$, $\\angle ABC=60^\\circ$, and $BF$ bisects $\\angle ABC$, what is the length of $AD$?\n",
    "completion": "\\textbf{Solution:} Given the construction, $BF$ bisects $\\angle ABC$,\\\\\n$\\therefore \\angle ABF = \\angle FBC$,\\\\\n$\\because BC \\parallel AD$,\\\\\n$\\therefore \\angle AFB = \\angle FBC$,\\\\\n$\\therefore \\angle AFB = \\angle ABF$,\\\\\n$\\therefore AF = AB = 4$,\\\\\nIn $\\triangle ABE$,\\\\\n$\\angle AEB = 90^\\circ$, $\\angle ABE = 60^\\circ$, $AB = 4$,\\\\\n$\\therefore \\angle BAE = 30^\\circ$,\\\\\n$\\therefore BE = 2$,\\\\\n$\\therefore FD = BE = 2$,\\\\\n$\\therefore AD = AF + FD = \\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53183067_158.png"
  },
  {
    "solution": "\\boxed{-2}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</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)$. What are the values of $a$ and $k$?\n",
    "completion": "\\textbf{Solution:} Substituting $A\\left(1, 3\\right)$ into $y=3x$, we have $3a=3$. Solving this gives $a=\\boxed{1}$. Therefore, the coordinates of point A are $\\left(1, 3\\right)$. Since the graph of the function $y=kx+5$ passes through point A, we have $k+5=3$. Solving this yields $k=\\boxed{-2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52416343_86.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, the side length of square $ABCD$ is 8, $M$ is on $DC$, and $DM=2$, $N$ is a moving point on $AC$, then what is the minimum value of $DN+MN$?\n",
    "completion": "\\textbf{Solution:} Let's connect BN, BD, and BM, where BM intersects AC at point E, \\\\\ngiven that ABCD is a square, then AC and BD are perpendicular bisectors of each other, \\\\\n$\\therefore ND=NB$, \\\\\nwhen point N and E are not coincident, in $\\triangle NBM$ we have $NB+NM>BM$, \\\\\nwhen point N and E coincide, $NB+NM=BM$, \\\\\n$\\therefore NB+NM\\geq BM$, hence the minimum value of $DN+MN$ is $BM$, \\\\\nsince ABCD is a square, we have $BC=CD=8$, $\\angle BCD=90^\\circ$, \\\\\n$\\therefore CM=CD-DM=8-2=6$, \\\\\n$\\therefore BM= \\sqrt{BC^{2}+CM^{2}}=\\boxed{10}$, \\\\\nthus, the minimum value of $DN+MN$ is $10$.",
    "image_path": "dataset/mm_math/images/MM_Math/53183057.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{10}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, $A$, $B$, and $D$ are consecutively on the same straight line, $\\angle A=\\angle D=\\angle CBE=90^\\circ$, $AC=BD$, $CB=BE$. If $AC=2$ and $AD=6$, what is the length of $CE$?\n",
    "completion": "Solution: Since $AC=BD=2$ and $AD=6$,\\\\\nthen $AB=AD-BD=6-2=4$,\\\\\nthus $CB^{2}=AC^{2}+AB^{2}=2^{2}+4^{2}=20$,\\\\\nwhich means $CB^{2}=BE^{2}=20$,\\\\\ntherefore $CE= \\sqrt{CB^{2}+BE^{2}}=\\sqrt{20+20}=2\\sqrt{10}$,\\\\\nhence the length of $CE$ is $CE=\\boxed{2\\sqrt{10}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52873822_151.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{2}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, a regular octagon is assembled with four quadrilaterals as depicted in Figure 1 and four diamonds as illustrated in Figure 2 (as in Figure 3). What is the ratio of the area of the shaded part to the area of the blank part in Figure 3?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $BE\\perp AD$ at E through the vertex B of the rhombus; Let the center of the regular octagon be point O, with one side being MN. Connect OM and ON, and draw $MP\\perp ON$ at P through point M. Suppose the side length of the regular octagon is $a$, then we have $AB=AD=MN=a$,\\\\\n$\\therefore \\angle ABC=180^\\circ-360^\\circ\\div 8=135^\\circ$, $\\angle MON=360^\\circ\\div 8=45^\\circ$,\\\\\n$\\because AD\\parallel BC$,\\\\\n$\\therefore \\angle BAE=180^\\circ-\\angle ABC=45^\\circ$,\\\\\n$\\therefore BE=\\frac{\\sqrt{2}}{2}AB=\\frac{\\sqrt{2}}{2}a$,\\\\\n$\\therefore S_{\\text{rhombus}ABCD}=AD\\cdot BE=\\frac{\\sqrt{2}}{2}a^{2}$,\\\\\n$\\therefore$ the area of the blank part $=4\\times \\frac{\\sqrt{2}}{2}a^{2}=2\\sqrt{2}a^{2}$,\\\\\n$\\because \\angle MON=45^\\circ$,\\\\\n$\\therefore OP=PM$, let $OP=PM=b$, then $OM=ON=\\sqrt{2}b$,\\\\\n$\\therefore PN=(\\sqrt{2}-1)b$,\\\\\nIn $\\triangle MPN$, $PM^{2}+PN^{2}=MN^{2}$,\\\\\n$\\therefore b^{2}+(\\sqrt{2}-1)^{2}b^{2}=a^{2}$,\\\\\nUpon rearranging: $b^{2}=\\frac{2+\\sqrt{2}}{4}a^{2}$,\\\\\n$\\therefore S_{\\triangle OMN}=\\frac{1}{2}ON\\cdot PM=\\frac{1}{2}\\times \\sqrt{2}b\\cdot b=\\frac{\\sqrt{2}}{2}b^{2}=\\frac{1+\\sqrt{2}}{4}a^{2}$,\\\\\n$\\therefore$ the area of the regular octagon is: $8\\times \\frac{1+\\sqrt{2}}{4}a^{2}=2(1+\\sqrt{2})a^{2}$,\\\\\n$\\therefore$ the area of the shaded part is: $2(1+\\sqrt{2})a^{2}-2\\sqrt{2}a^{2}=2a^{2}$,\\\\\n$\\therefore$ the ratio of the area of the shaded part to the area of the blank part $=\\frac{2a^{2}}{2\\sqrt{2}a^{2}}=\\boxed{\\frac{\\sqrt{2}}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52874671.png"
  },
  {
    "solution": "\\boxed{y<-2}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: It is known that the graph of the linear function $y=kx+b$ is as shown in the figure. What is the range of $y$ when $x<1$?\n",
    "completion": "\\textbf{Solution:} \\\\\nConsider the linear function $y=kx+b$. Its graph intersects the $y$-axis at $(0, -4)$, \\\\\n$\\therefore b=-4$, and intersects the $x$-axis at $(2, 0)$, \\\\\n$\\therefore 0=2k-4$, \\\\\n$\\therefore k=2$, \\\\\n$\\therefore y=kx+b=2x-4$, \\\\\n$\\therefore x=(y+4)/2<1$, \\\\\n$\\therefore \\boxed{y<-2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50447025.png"
  },
  {
    "solution": "\\boxed{\\frac{6}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: Given: As shown in the figure, in the right trapezoid $ABCD$, $AB\\parallel CD$, $AB\\perp BC$, $AD=CD$, $E$ is the midpoint of $AC$, extend line $BE$ to intersect line segment $AD$ at point $F$. $\\triangle AEF\\sim \\triangle BAF$, if $CD=3$, $AB=5$, find the length of $DF$.\n",
    "completion": "\\textbf{Solution:} Let $CD=3$ and $AB=5$,\\\\\nHence $AD=CD=3$,\\\\\nExtend $BF$ to meet the extension of $CD$ at point G, as shown in the diagram:\\\\\nIn $\\triangle ECG$ and $\\triangle EAB$, we have $\\left\\{\\begin{array}{l}\n\\angle ECG=\\angle EAB \\\\\nCE=AE \\\\\n\\angle CEG=\\angle AEB\n\\end{array}\\right.$,\\\\\nTherefore, $\\triangle ECG\\cong \\triangle EAB$ (ASA),\\\\\nHence $CG=AB=5$,\\\\\nThus $DG=CG-CD=2$,\\\\\nSince $AB\\parallel CD$,\\\\\n$\\triangle DFG\\sim \\triangle AFB$,\\\\\nHence $\\frac{DF}{AF}=\\frac{DG}{AB}$,\\\\\nThus $AB\\cdot DF=DG\\cdot AF$,\\\\\nHence $5DF=2\\times (3-DF)$,\\\\\nTherefore $DF=\\boxed{\\frac{6}{7}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52977910_145.png"
  },
  {
    "solution": "\\boxed{24}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in rectangle $ABCD$, the perpendicular bisector of the diagonal $AC$, $EF$, intersects $BC$ and $AD$ at points $E$ and $F$ respectively. If $BE=3$ and $AF=5$, what is the perimeter of rectangle $ABCD$?\n",
    "completion": "\\textbf{Solution:} Connect AE, \\\\\n$\\because$ EF is the perpendicular bisector of AC, \\\\\n$\\therefore$ $\\angle$AOF = $\\angle$COE = 90$^\\circ$, AO = CO, AE = CE, \\\\\n$\\because$ Quadrilateral ABCD is a rectangle, \\\\\n$\\therefore$ AD$\\parallel$BC, $\\angle$B = 90$^\\circ$, AD = BC, \\\\\n$\\therefore$ $\\angle$FAO = $\\angle$ECO, \\\\\nIn $\\triangle$AOF and $\\triangle$COE, \\\\\n$\\left\\{\\begin{array}{l} \\angle FAO = \\angle ECO\\hfill \\\\ AO = CO\\hfill \\\\ \\angle AOF = \\angle COE\\hfill \\end{array}\\right.$ , \\\\\n$\\therefore$ $\\triangle$AOF $\\cong$ $\\triangle$COE (ASA), \\\\\n$\\therefore$ AF = CE, \\\\\n$\\therefore$ AE = CE, \\\\\n$\\because$ BE = 3, AF = 5, \\\\\n$\\therefore$ CE = 5, \\\\\n$\\therefore$ AE = 5, BC = BE + CE = 8, \\\\\n$\\therefore$ AB = $ \\sqrt{AE^{2}-BE^{2}}$ = 4, \\\\\n$\\therefore$ The perimeter of rectangle ABCD is 2(4 + 8) = $\\boxed{24}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51968496.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, $AD = \\frac{1}{2}DB$, and $E$ is the midpoint of $BC$, with $BE = \\frac{1}{5}AC = 2\\, \\text{cm}$. How long is segment $DE$ in centimeters?\n",
    "completion": "\\textbf{Solution:} By the problem statement, we know $DB=\\frac{2}{3}AB$, $BE=EC=2\\text{cm}$, and $AC=10\\text{cm}$. \\\\\n$\\therefore$ $AB=AC-BC=6\\text{cm}$. \\\\\n$\\therefore$ $DB=\\frac{2}{3}AB=4\\text{cm}$. \\\\\n$\\therefore$ $DE=DB+BE=\\boxed{6}\\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52916301.png"
  },
  {
    "solution": "\\boxed{24}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, after pouring some water into a cylindrical container with a diameter of 78 cm, the cross-section is as shown in the figure. If the width of the water surface $AB = 72cm$, what is the maximum depth of the water?\n",
    "completion": "\\textbf{Solution:} Connect $OB$, and through point O draw $OC\\perp AB$ at point D, intersecting $\\odot O$ at point C, as shown in the diagram:\\\\\nThen $BD=\\frac{1}{2}AB=36\\,\\text{cm}$,\\\\\nSince the diameter of $\\odot O$ is $78\\,\\text{cm}$,\\\\\nTherefore $OB=OC=39\\,\\text{cm}$,\\\\\nIn $\\triangle OBD$ right-angled at D, $OD=\\sqrt{OB^{2}-BD^{2}}=\\sqrt{39^{2}-36^{2}}=15\\,\\text{cm}$,\\\\\nTherefore, $CD=OC-OD=39-15=\\boxed{24}\\,\\text{cm}$,\\\\\nwhich means the maximum depth of water is $24\\,\\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51350809.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in the Cartesian coordinate system $xoy$, the line $y=k_1x+4$ intersects the $y$-axis at point C and intersects the graph of the inverse proportion function $y=\\frac{k_2}{x}$ at point B in the first quadrant. Connect BO. If the area of $\\triangle OBC$ is $2$ and $\\tan\\angle BOC=\\frac{1}{5}$, what is the value of $k_2$?\n",
    "completion": "\\textbf{Solution:} Given that the line $y=k_1x+4$ intersects with the x-axis at point A, and with the y-axis at point C,\\\\\nthus, the coordinates of point C are (0, 4),\\\\\nhence, OC=4,\\\\\ndraw BD $\\perp$ y-axis at D,\\\\\ngiven that $S_{\\triangle OBC}=2$,\\\\\nthus, $\\frac{1}{2}OC\\cdot BD=\\frac{1}{2}\\times 4\\cdot BD=2$,\\\\\nhence, BD=1,\\\\\ngiven that $\\tan \\angle BOC=\\frac{1}{5}$,\\\\\nthus, $\\frac{BD}{OD}=\\frac{1}{5}$,\\\\\nhence, OD=5,\\\\\nthus, the coordinates of point B are (1, 5),\\\\\ngiven that the graph of the inverse proportion function $y=\\frac{k_2}{x}$ intersects at point B within the first quadrant,\\\\\nthus, $k_{2}=1\\times 5=\\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51322318.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</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 segment $AC$, and point $E$ is the midpoint of segment $BC$. As shown in Figure 1, point $C$ is on the segment $AB$; if $AC=6$ and $BC=4$, how long is segment $DE$?\n",
    "completion": "\\textbf{Solution:} Since $D$ is the midpoint of line segment $AC$, and $E$ is the midpoint of line segment $BC$,\\\\\ntherefore $DC=\\frac{1}{2}AC=3$, and $CE=\\frac{1}{2}BC=2$,\\\\\nthus $DE=DC+CE=3+2=\\boxed{5}$;",
    "image_path": "dataset/mm_math/images/MM_Math/55393350_77.png"
  },
  {
    "solution": "\\boxed{\\frac{4}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</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}$. What is the value of $\\frac{AD}{CD}$?\n",
    "completion": "\\textbf{Solution:}\n\nGiven that $BD$ bisects $\\angle ABC$, \\\\\nit follows that $\\angle ABD = \\angle CBD$, \\\\\nsince $\\angle BDC = \\angle A = 90^\\circ$, \\\\\nit implies that $\\triangle ABD \\sim \\triangle DBC$. \\\\\nTherefore, $\\frac{AD}{CD} = \\frac{AB}{BD}$. \\\\\nIn $\\triangle ABD$, \\\\\nsince $\\cos\\angle ABD = \\frac{AB}{BD} = \\frac{4}{5}$, \\\\\nit results in $\\frac{AD}{CD} = \\boxed{\\frac{4}{5}}$;",
    "image_path": "dataset/mm_math/images/MM_Math/52977904_141.png"
  },
  {
    "solution": "\\boxed{\\frac{1}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in $\\triangle ABC$, points D and E are on sides AB and AC respectively, $AD:DB = 3:2$, $AE:EC = 1:2$. The line ED and the extension of CB intersect at point F. What is the value of $\\frac{FB}{FC}$?\n",
    "completion": "\\textbf{Solution:} Draw BG$\\parallel$AC through point B, intersecting EF at point G,\\\\\nsince AD$\\colon$DB = 3$\\colon$2, AE$\\colon$EC = 1$\\colon$2,\\\\\nthus AE = $\\frac{1}{2}$EC,\\\\\nthus $\\triangle$BFG$\\sim$$\\triangle$CEF, $\\triangle$BGD$\\sim$$\\triangle$AED,\\\\\nthus $\\frac{BG}{AE}=\\frac{BD}{AD}=\\frac{2}{3}$,\\\\\n$\\frac{BG}{EC}=\\frac{FB}{FC}$,\\\\\nthus $\\frac{BG}{EC}=\\frac{1}{3}$\\\\\nthus $\\frac{FB}{FC}=\\boxed{\\frac{1}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53081375.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{2}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the diagram, in quadrilateral $ABCD$, diagonal $BD$ bisects $\\angle ABC$. Quadrilateral $ABCD$ is a rhombus. A line through point $A$ is drawn such that $AE \\parallel BD$, intersecting the extension of line $CD$ at point $E$. A line through point $E$ is drawn such that $EF \\perp BC$, intersecting the extension of line $BC$ at point $F$. If $\\angle ABC=45^\\circ$ and $BC=2$, what is the length of $EF$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rhombus, \\\\\nit follows that AB$\\parallel$DE, \\\\\nsince AE$\\parallel$BD, \\\\\nit follows that quadrilateral ABDE is a parallelogram, \\\\\nhence, AB=DE, \\\\\nsince quadrilateral ABCD is a rhombus, \\\\\nit follows that CD=AB=BC=2, \\\\\nthus, DE=AB=2, \\\\\ntherefore, CE=CD+DE=4, \\\\\nsince $\\triangle$CFE is an isosceles right triangle, \\\\\nit follows that EF=$\\frac{\\sqrt{2}}{2}$CE=$\\boxed{2\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51614495_110.png"
  },
  {
    "solution": "\\boxed{55^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, the quadrilateral $ABCD$ is inscribed in a semicircle, where $AB$ is the diameter, and the arc $DC$ is equal to the arc $CB$. If $\\angle C = 110^\\circ$, then what is the degree measure of $\\angle ABC$?\n",
    "completion": "\\textbf{Solution:} Given that quadrilateral ABCD is inscribed in a semicircle, and $\\angle C=110^\\circ$,\n\\[\n\\therefore \\angle DAB=180^\\circ-\\angle C=70^\\circ,\n\\]\nsince arc DC equals arc CB,\n\\[\n\\therefore CD=CB,\n\\]\n\\[\n\\therefore \\angle CDB=\\angle CBD=35^\\circ,\n\\]\nsince AB is the diameter,\n\\[\n\\therefore \\angle ADB=90^\\circ,\n\\]\n\\[\n\\therefore \\angle ABD=90^\\circ-\\angle ADB=90^\\circ-70^\\circ=20^\\circ,\n\\]\n\\[\n\\therefore \\angle ABC=\\angle ABD+\\angle CBD=20^\\circ+35^\\circ=\\boxed{55^\\circ}.\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/52808102.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, the graph of the linear function $y=(m-2)x-4$ intersects the negative x-axis at point A and intersects the y-axis at point B, and the area of $\\triangle OAB$ is 8. Find the value of $m$ and the coordinates of point A.\n",
    "completion": "\\textbf{Solution:} When $x=0$, we have $y=(m-2)x-4=-4$, thus $B(0, -4)$,\\\\\n$\\therefore OB=4$,\\\\\n$\\because$ The area of $\\triangle OAB$ is 8,\\\\\n$\\therefore \\frac{1}{2}\\times OA\\times 4=8$, solving this gives: $OA=4$,\\\\\n$\\therefore \\boxed{A(-4,0)}$.\\\\\nSubstituting point $A(-4,0)$ into $y=(m-2)x-4$, gives: $-4m+8-4=0$,\\\\\n$\\therefore m=\\boxed{1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52095492_167.png"
  },
  {
    "solution": "\\boxed{16}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle BAC=90^\\circ$, points $D$ and $E$ are the midpoints of $AB$ and $BC$ respectively, point $F$ is on the extension of $CA$, $\\angle FDA=\\angle B$, $AC=6$, $AB=8$. What is the perimeter of quadrilateral $AEDF$?\n",
    "completion": "Solution: In right triangle $ABC$, \\\\\nbecause $AC=6$ and $AB=8$, \\\\\ntherefore $BC=10$, \\\\\nbecause $E$ is the midpoint of $BC$, \\\\\ntherefore $AE=BE=5$, \\\\\ntherefore $\\angle BAE=\\angle B$, \\\\\nbecause $\\angle FDA=\\angle B$, \\\\\ntherefore $\\angle FDA=\\angle BAE$, \\\\\ntherefore $DF\\parallel AE$, \\\\\nbecause $D$ and $E$ are the midpoints of $AB$ and $BC$ respectively, \\\\\ntherefore $DE\\parallel AC$, and $DE= \\frac{1}{2}AC=3$ \\\\\ntherefore the quadrilateral $AEDF$ is a parallelogram \\\\\ntherefore the perimeter of the quadrilateral $AEDF$ is $2\\times(3+5)=\\boxed{16}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52250664.png"
  },
  {
    "solution": "\\boxed{\\sqrt{13}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the diagram, in rectangle $ABCD$, where $AB=4$ and $AD=2$, a circle is drawn with center B and radius $BC$, intersecting side $AB$ at point E. Point P is a moving point on the arc $CE$. Connecting $PD$ and $PA$, what is the minimum value of $\\frac{1}{2}AP+DP$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect BP and take the midpoint G of BE, then connect PG,\\\\\n$\\because AD=BC=BP=2$, $AB=4$,\\\\\n$\\therefore \\frac{BP}{BA}=\\frac{2}{4}=\\frac{1}{2}$,\\\\\n$\\because$ G is the midpoint of BE,\\\\\n$\\therefore \\frac{BG}{BP}=\\frac{1}{2}$,\\\\\n$\\therefore \\frac{BP}{BA}=\\frac{BG}{BP}$,\\\\\n$\\because \\angle PBG=\\angle ABP$,\\\\\n$\\therefore \\triangle BPG\\sim \\triangle BAP$,\\\\\n$\\therefore \\frac{PG}{AP}=\\frac{BP}{BA}=\\frac{1}{2}$,\\\\\n$\\therefore PG=\\frac{1}{2}AP$,\\\\\nthus $\\frac{1}{2}AP+DP=PG+DP$, when points P, D, G are collinear, it takes the minimum value, that is, the length of DG,\\\\\n$DG=\\sqrt{AD^2+AG^2}=\\sqrt{4+9}=\\boxed{\\sqrt{13}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51299523.png"
  },
  {
    "solution": "\\boxed{\\frac{7}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, D is a point on side AB of the equilateral $\\triangle ABC$, and $AD: DB = 2: 3$. Now, fold $\\triangle ABC$ such that point C coincides with D, with the fold line being EF, where points E and F lie on AC and BC, respectively. What is the ratio $CE: CF$?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABC$ is an equilateral triangle,\\\\\nit follows that $\\angle A = \\angle B = \\angle C = 60^\\circ$.\\\\\nFrom the properties of folding, we know that $\\angle EDF = \\angle A = 60^\\circ$, $DE = CE$, and $DF = CF$,\\\\\nhence, $\\angle ADE + \\angle BDF = 180^\\circ - \\angle EDF = 120^\\circ$, and $CE:CF = DE:DF$.\\\\\nSince $\\angle AED + \\angle ADE = 180^\\circ - \\angle A = 120^\\circ$,\\\\\nit follows that $\\angle AED = \\angle BDF$,\\\\\nhence, $\\triangle AED \\sim \\triangle BDF$,\\\\\nthus, $\\frac{DE}{DF} = \\frac{7x}{8x} = \\frac{7}{8}$,\\\\\ntherefore, $CE:CF = \\boxed{\\frac{7}{8}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53065559.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{13}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, the radius $OD$ of circle $O$ is perpendicular to chord $AB$ at point $C$. Extend $AO$ to meet circle $O$ at point $E$, and connect $EC$. If $AB=8$ and $OC=3$, what is the length of $EC$?\n",
    "completion": "\\textbf{Solution:} Connect BE, \\\\\n$\\because$AE is the diameter of $\\odot$O, \\\\\n$\\therefore$$\\angle$ABE$=90^\\circ$, \\\\\n$\\because$OD$\\perp$AB and OD passes through O, \\\\\n$\\therefore$AC$=$BC$=\\frac{1}{2}$AB$=\\frac{1}{2}\\times 8=4$, \\\\\n$\\because$AO$=$OE, \\\\\n$\\therefore$BE$=2$OC, \\\\\n$\\because$OC$=3$, \\\\\n$\\therefore$BE$=6$, \\\\\nIn the right triangle $\\triangle$CBE, EC$=\\sqrt{BE^{2}+CB^{2}}=\\sqrt{4^{2}+6^{2}}=\\boxed{2\\sqrt{13}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51403360.png"
  },
  {
    "solution": "\\boxed{144^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: It is known that $O$ is a point on line $AB$, and the right-angle vertex of a right-angled triangle $OMN$ is placed at point $O$. Ray $OC$ bisects $\\angle MOB$. Rotate the right-angled triangle $OMN$ in Figure 1 clockwise about vertex $O$ to the position in Figure 2. When $\\angle AOC = 3\\angle BON$, what is the measure of $\\angle AOM$?\n",
    "completion": "\\textbf{Solution:} Let $\\angle AOM = x$, then $\\angle BOM = 180^\\circ - x$,\n\n$\\because$ OC bisects $\\angle BOM$,\n\n$\\therefore \\angle MOC = \\frac{1}{2}\\angle BOM = \\frac{1}{2}(180^\\circ - x) = 90^\\circ - \\frac{1}{2}x$,\n\n$\\therefore \\angle AOC = \\angle AOM + \\angle MOC = x + 90^\\circ - \\frac{1}{2}x = 90^\\circ + \\frac{1}{2}x$,\n\n$\\because \\angle MON = 90^\\circ$,\n\n$\\therefore \\angle BON = \\angle MON - \\angle BOM = 90^\\circ - (180^\\circ - x) = x - 90^\\circ$,\n\n$\\because \\angle AOC = 3\\angle BON$,\n\n$\\therefore 90^\\circ + \\frac{1}{2}x = 3(x - 90^\\circ)$,\n\nSolving gives $x = \\boxed{144^\\circ}$,\n\n$\\therefore \\angle AOM = 144^\\circ$.",
    "image_path": "dataset/mm_math/images/MM_Math/52169848_190.png"
  },
  {
    "solution": "\\boxed{(0,3)}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</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 a point on OB. If \\(\\triangle ABM\\) is folded along AM, point B exactly falls on point \\(B^{\\prime}\\) on the x-axis. Find the coordinates of point M.\n",
    "completion": "\\textbf{Solution:} Solve: When $x=0$, $y=8$, that is, $B(0,8)$. When $y=0$, $0=-\\frac{4}{3}x+8$, solving this, we get $x=6$, that is, $A(6,0)$;\\\\\n$\\therefore$ $OA=6$, $OB=8$,\\\\\n$\\because$ $\\angle AOB=90^\\circ$,\\\\\n$\\therefore$ $AB= \\sqrt{OA^{2}+OB^{2}} =10$,\\\\\nBy the property of folding, we get: $AB=AB^{\\prime}=10$,\\\\\n$\\therefore$ $OB^{\\prime}=AB^{\\prime}-OA=10-6=4$,\\\\\nLet $MO=x$, then $MB=MB^{\\prime}=8-x$,\\\\\nIn $\\triangle OMB^{\\prime}$, $OM^{2}+OB^{\\prime 2}=B^{\\prime}M^{2}$,\\\\\nthat is, $x^{2}+4^{2}=(8-x)^{2}$,\\\\\nSolving for $x$ yields: $x=3$,\\\\\n$\\therefore$ The coordinates of point M are $\\boxed{(0,3)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50590744_146.png"
  },
  {
    "solution": "\\boxed{1+\\sqrt{2}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, for the square $ABCD$, with points $E$ and $F$ located on the diagonal $AC$, $\\angle EBF=45^\\circ$. The lines $BE$ and $BF$ are drawn. The triangles $ABE$ and $GBE$ are symmetrical about the line $BE$. Point $G$ is on $BD$, and the line $FG$ is drawn. As shown in the auxiliary figure, extend $BF$ to intersect $CD$ at point $H$. Draw the line $HG$. Find the value of $\\frac{CD}{CH}$.\n",
    "completion": "\\textbf{Solution:}\n\nGiven that\\\\\n\\begin{equation}\n    \\because \\triangle ABE \\text{ and } \\triangle GBE \\text{ are symmetrical about the line } BE\n\\end{equation}\n\\begin{equation}\n    \\therefore AB=BG=BC\n\\end{equation}\n\\begin{equation}\n    \\therefore \\left\\{\\begin{array}{l}BG=BC \\\\ \\angle DBF=\\angle FBC \\\\ BF=BF \\end{array}\\right.\n\\end{equation}\n\\begin{equation}\n    \\therefore \\triangle GBF \\cong \\triangle CBF(SAS)\n\\end{equation}\n\\begin{equation}\n    \\therefore GF=FC, \\angle BGF=\\angle BCF=45^\\circ\n\\end{equation}\n\\begin{equation}\n    \\because \\left\\{\\begin{array}{l}BG=BC \\\\ \\angle DBF=\\angle FBC \\\\ BH=BH \\end{array}\\right.\n\\end{equation}\n\\begin{equation}\n    \\therefore \\triangle HGB \\cong \\triangle HCB(SAS)\n\\end{equation}\n\\begin{equation}\n    \\therefore \\angle BGH=\\angle BCH=90^\\circ\n\\end{equation}\n\\begin{equation}\n    \\because \\text{Square } ABCD\n\\end{equation}\n\\begin{equation}\n    \\therefore AC \\perp BD\n\\end{equation}\n\\begin{equation}\n    \\therefore \\angle BOC=90^\\circ\n\\end{equation}\n\\begin{equation}\n    \\therefore \\angle BOC=\\angle BGH=90^\\circ\n\\end{equation}\n\\begin{equation}\n    \\therefore OF \\parallel GH\n\\end{equation}\n\\begin{equation}\n    \\therefore OC \\parallel GH\n\\end{equation}\n\\begin{equation}\n    \\therefore CF \\parallel GH\n\\end{equation}\n\\begin{equation}\n    \\because \\angle BGF=45^\\circ, \\angle BOC=90^\\circ\n\\end{equation}\n\\begin{equation}\n    \\therefore \\angle OFG=45^\\circ\n\\end{equation}\n\\begin{equation}\n    \\therefore \\angle OFG=\\angle ACD=45^\\circ\n\\end{equation}\n\\begin{equation}\n    \\therefore GF \\parallel CD\n\\end{equation}\n\\begin{equation}\n    \\therefore GF \\parallel CH\n\\end{equation}\n\\begin{equation}\n    \\because CF \\parallel GH\n\\end{equation}\n\\begin{equation}\n    \\therefore \\text{Quadrilateral } GHCF \\text{ is a parallelogram}\n\\end{equation}\n\\begin{equation}\n    \\because GF=FC\n\\end{equation}\n\\begin{equation}\n    \\therefore \\text{Quadrilateral } GHCF \\text{ is a rhombus;}\n\\end{equation}\n(2) Given that $\\angle BGH=90^\\circ$\n\\begin{equation}\n    \\therefore \\angle HGD=90^\\circ\n\\end{equation}\n\\begin{equation}\n    \\because \\angle BDC=45^\\circ\n\\end{equation}\n\\begin{equation}\n    \\therefore \\angle GHD=45^\\circ\n\\end{equation}\n\\begin{equation}\n    \\therefore \\triangle DGH \\text{ is an isosceles right triangle}\n\\end{equation}\n\\begin{equation}\n    \\therefore DG=GH\n\\end{equation}\n\\begin{equation}\n    \\therefore DH=\\sqrt{2}GH\n\\end{equation}\n\\begin{equation}\n    \\because \\text{Quadrilateral } GHCF \\text{ is a rhombus}\n\\end{equation}\n\\begin{equation}\n    \\therefore GH=CH\n\\end{equation}\n\\begin{equation}\n    \\therefore CD=CH+DH=GH+DH=GH+\\sqrt{2}GH=\\left(1+\\sqrt{2}\\right)GH\n\\end{equation}\n\\begin{equation}\n    \\therefore \\frac{CD}{CH}=\\frac{\\left(1+\\sqrt{2}\\right)CH}{CH}=1+\\sqrt{2}\n\\end{equation}\nThus, $\\frac{CD}{CH}=\\boxed{1+\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52065370_100.png"
  },
  {
    "solution": "\\boxed{\\sqrt{2}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, AB is the diameter of the circle $\\odot O$, PD is tangent to $\\odot O$ at point C, and intersects the extension of AB at point D, and $\\angle D=2\\angle CAD$. If CD=1, what is the length of BD?\n",
    "completion": "\\textbf{Solution}: Since $\\angle D = \\angle COD$ and $CD=1$,\\\\\nit follows that $OC=OB=CD=1$,\\\\\nIn $\\triangle OCD$, by the Pythagorean theorem, we have $OD=\\sqrt{1^2+1^2}=\\sqrt{2}$,\\\\\ntherefore, $BD=OD-OB=\\boxed{\\sqrt{2}-1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51446430_116.png"
  },
  {
    "solution": "\\boxed{50^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, triangle $ABC$ is rotated $22^\\circ$ clockwise around point B to obtain triangle $DBE$. If $\\angle C = 28^\\circ$ and side $DE$ intersects side $BC$ at point F, then what is the measure of $\\angle CFE$ in degrees?\n",
    "completion": "\\textbf{Solution:} Rotate $\\triangle ABC$ clockwise around point B by $22^\\circ$ to get $\\triangle DBE$,\\\\\n$\\therefore$ $\\angle CBE=22^\\circ$.\\\\\n$\\because$ $\\angle C=28^\\circ$,\\\\\n$\\therefore$ $\\angle E=\\angle C=28^\\circ$.\\\\\n$\\because$ $\\angle CFE$ is the external angle of $\\triangle BEF$,\\\\\n$\\therefore$ $\\angle CFE=\\angle CBE+\\angle E=22^\\circ+28^\\circ=\\boxed{50^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53078697.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, points A and B lie on the graph of the linear function $y=kx+1$, with point A in the first quadrant and point B on the x-axis. Point D is located on the positive half of the x-axis, and point C has coordinates $(1, -2)$. The quadrilateral OADC is a rhombus. What is the area of $\\triangle ABD$?\n",
    "completion": "\\textbf{Solution:} From (1), the linear equation can be expressed as $y=x+1$,\\\\\nwhen $y=0$, then $x=-1$,\\\\\n$\\therefore$ point B$(-1,0)$,\\\\\nwhich means $OB=1$,\\\\\n$\\therefore$ $BD=1+2=3$,\\\\\n$\\therefore$ the area of $\\triangle ABD$ is $\\frac{1}{2} \\times 3 \\times 2=\\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52211814_81.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</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. The vertex is point E. Connect AC, CE, AE. What is the area of $\\triangle ACE$?\n",
    "completion": "\\textbf{Solution:} As illustrated in figure 1, let the point of intersection between line AE and the y-axis be denoted as H,\\\\\ngiven the equation of the parabola as $y=-x^2+2x+3$,\\\\\nsetting $y=0$, yields $-x^2+2x+3=0$,\\\\\n$\\therefore x=-1$ or $x=3$,\\\\\n$\\therefore A(-1,0)$,\\\\\nwhen $x=1$, $y=-1+2+3=4$,\\\\\n$\\therefore E(1,4)$,\\\\\n$\\therefore$ the equation of line AE is $y=2x+2$,\\\\\n$\\therefore H(0,2)$,\\\\\n$\\therefore CH=3-2=1$,\\\\\n$\\therefore S_{\\triangle ACE}=\\frac{1}{2}CH\\cdot|x_E-x_A|=\\frac{1}{2}\\times 1\\times 2=\\boxed{1};$",
    "image_path": "dataset/mm_math/images/MM_Math/53080571_42.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, within $\\triangle ABC$, G is any point on side BC, and D, E, and F are the midpoints of AG, BD, and CE respectively. If the area of $\\triangle ABC$ is $48$, then what is the value of the area of $\\triangle DEF$?\n",
    "completion": "Solution: Draw the line CD as shown in the figure: \\\\\n$\\because$ Point D is the midpoint of AG,\\\\\n$\\therefore S_{\\triangle ABD}=\\frac{1}{2}S_{\\triangle ABG}$, $S_{\\triangle ACD}=\\frac{1}{2}S_{\\triangle AGC}$,\\\\\n$\\therefore S_{\\triangle ABD}+S_{\\triangle ACD}=\\frac{1}{2}S_{\\triangle ABC}=24$,\\\\\n$\\therefore S_{\\triangle BCD}=\\frac{1}{2}S_{\\triangle ABC}=24$,\\\\\n$\\because$ Point E is the midpoint of BD,\\\\\n$\\therefore S_{\\triangle CDE}=\\frac{1}{2}S_{\\triangle BCD}=12$,\\\\\n$\\because$ Point F is the midpoint of CE,\\\\\n$\\therefore S_{\\triangle DEF}=\\frac{1}{2}S_{\\triangle CDE}=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53107080.png"
  },
  {
    "solution": "\\boxed{95^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, $CE$ is the angle bisector of the exterior angle $\\angle ACD$ of $\\triangle ABC$, and $CE$ intersects the extension of $BA$ at point $E$. If $\\angle B=35^\\circ$ and $\\angle E=25^\\circ$, what is the measure of $\\angle CAE$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle DCE$ is an exterior angle of $\\triangle BCE$ and $\\angle B=35^\\circ$, $\\angle E=25^\\circ$,\n\\[\\therefore \\angle DCE=\\angle B+\\angle E=60^\\circ,\\]\nSince $CE$ bisects $\\angle ACD$,\n\\[\\therefore \\angle ACE=\\angle DCE=60^\\circ,\\]\n\\[\\therefore \\angle CAE=180^\\circ-\\angle ACE-\\angle E=\\boxed{95^\\circ}.\\]",
    "image_path": "dataset/mm_math/images/MM_Math/53059214_137.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</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. What is the area of the square ABCD?\n",
    "completion": "\\textbf{Solution:} For the line $y=\\frac{1}{2}x+1$, by setting $x=0$, we find $y=1$; by setting $y=0$, we find $x=-2$,\\\\\n$\\therefore$A$(-2,0)$, B$(0,1)$,\\\\\n$\\therefore$in $\\triangle AOB$, $OA=2$, $OB=1$,\\\\\n$\\therefore$according to the Pythagorean theorem: $AB=\\sqrt{2^2+1^2}=\\sqrt{5}$,\\\\\n$\\therefore$the area of the square $ABCD$ is $\\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53706609_105.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, point $O$ is on line $AB$, $\\angle BOD$ and $\\angle COD$ are complementary, and $\\angle BOC = n\\angle EOC$. If $DO \\perp OE$, what is the value of $n$?\n",
    "completion": "\\textbf{Solution:} Given that, since $\\angle BOD$ and $\\angle COD$ are complementary,\\\\\nwe have $\\angle BOD + \\angle AOD = 180^\\circ$,\\\\\nthus $\\angle COD = \\angle AOD$,\\\\\nLet $\\angle COE = x$,\\\\\nthen $\\angle BOE = nx - x$,\\\\\nsince $OD \\perp OE$,\\\\\nit follows that $\\angle COD + \\angle COE = 90^\\circ = \\angle AOD + \\angle BOE$,\\\\\nthus $\\angle COE = \\angle BOE$,\\\\\nhence $nx - x = x$,\\\\\nand given $x \\ne 0$,\\\\\nwe find $n = \\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52774176_155.png"
  },
  {
    "solution": "\\boxed{24}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, the diameter \\(CD=30\\) of the circle \\(\\odot O\\), and \\(AB\\) is a chord of \\(\\odot O\\) such that \\(AB \\perp CD\\) with the foot of the perpendicular being \\(M\\), and \\(OM:OC=3:5\\). What is the length of \\(AB\\)?\n",
    "completion": "\\textbf{Solution:} Since the diameter $CD$ of the circle $\\odot O$ is 30, and $AB\\perp CD$, with $OM\\colon OC=3\\colon 5$,\\\\\nit follows that $AO=OC=15$ and $OM=\\frac{3}{5}OC=\\frac{3}{5}\\times 15=9$. Given that $AM=MB$,\\\\\nthen $AM=\\sqrt{AO^{2}-MO^{2}}=\\sqrt{15^{2}-9^{2}}=12$,\\\\\nthus, $AB=2AM=\\boxed{24}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52905957.png"
  },
  {
    "solution": "\\boxed{\\frac{4}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in a grid where the side length of each small square is 1, the vertices A, B, and C of $\\triangle ABC$ all fall on the grid points. What is the value of $\\cos \\angle ABC$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $AD\\perp BC$ with the foot of the perpendicular at D,\\\\\n$AD=3$,\\\\\n$BD=4$,\\\\\n$\\therefore AB=\\sqrt{AD^{2}+BD^{2}}=5$,\\\\\n$\\therefore \\cos\\angle ABC=\\frac{BD}{AB}=\\boxed{\\frac{4}{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51407004.png"
  },
  {
    "solution": "\\boxed{68^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, quadrilateral $ABCD$ is inscribed in circle $O$. Point $I$ is the intersection point of the angle bisectors of $\\triangle ABC$, and $\\angle AIC = 124^\\circ$. Point $E$ is on the extension line of side $AD$. What is the measure of $\\angle CDE$?\n",
    "completion": "\\textbf{Solution:} Given that point I is the intersection of the angle bisectors of $\\triangle ABC$, \\\\\n$\\therefore \\angle BAC = 2\\angle IAC$, $\\angle ACB = 2\\angle ICA$, \\\\\nGiven $\\angle AIC = 124^\\circ$, \\\\\n$\\therefore \\angle IAC + \\angle ICA = 180^\\circ - \\angle AIC = 56^\\circ$, \\\\\n$\\therefore \\angle B = 180^\\circ - (\\angle BAC + \\angle ACB) = 180^\\circ - 2(\\angle IAC + \\angle ICA) = 180^\\circ - 2 \\times 56^\\circ = 68^\\circ$, \\\\\nFurthermore, since quadrilateral ABCD is inscribed in circle O, \\\\\n$\\therefore \\angle CDE = \\angle B = \\boxed{68^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52874635.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, square $ABCD$ has a side length of $2$, and points $E$, $F$ are movable points on diagonal $AC$, with the length of $EF$ being $1$. Connecting $BE$ and $BF$, what is the minimum perimeter of $\\triangle BEF$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, draw $BH \\parallel AC$ such that $BH = EF = 1$, and connect $DH$ intersecting $AC$ at $E$, thus the perimeter of $\\triangle BEF$ is minimized,\\\\\n$\\because BH = EF$, $BH \\parallel EF$,\\\\\n$\\therefore$ the quadrilateral $EFBH$ is a parallelogram,\\\\\n$\\therefore$ $BF = EH$,\\\\\n$\\because$ the quadrilateral $ABCD$ is a square,\\\\\n$\\therefore$ $AC$ bisects $BD$ perpendicularly,\\\\\n$\\therefore$ $EB = ED$,\\\\\n$\\therefore$ $BE + BF = EH + ED = DH$,\\\\\n$\\because$ the quadrilateral $ABCD$ is a square,\\\\\n$\\therefore$ $AC \\perp BD$,\\\\\n$\\because$ $BH \\parallel AC$,\\\\\n$\\therefore$ $BD \\perp BH$, that is, $\\angle DBH = 90^\\circ$,\\\\\nIn $\\triangle DBH$,\\\\\n$DH = \\sqrt{BD^2 + BH^2} = \\sqrt{(2\\sqrt{2})^2 + 1^2} = 3$,\\\\\n$\\therefore$ the minimum possible perimeter of $\\triangle BEF$ is $3 + 1 = \\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52082450.png"
  },
  {
    "solution": "\\boxed{\\frac{12}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in $ \\triangle ABC$, $ \\angle C=90^\\circ$, O is a point on side AB, circle $O$ is tangent to both AC and BC. If $ BC=3$ and $ AC=4$, what is the radius of circle $O$?\n",
    "completion": "\\textbf{Solution:} Construct OD$\\perp$AC at D, OE$\\perp$BC at E, as shown in the figure, let the radius of $\\odot O$ be $r$,\\\\\nsince $\\odot O$ is tangent to both AC and BC,\\\\\nthus OD=OE=r,\\\\\nand $\\angle C=90^\\circ$,\\\\\nthus quadrilateral ODCE is a square,\\\\\nthus CD=OD=r,\\\\\nsince OD$\\parallel$BC,\\\\\nthus $\\triangle ADO\\sim \\triangle ACB$,\\\\\nthus $\\frac{AF}{AC}=\\frac{OF}{BC}$ \\\\\nsince AF=AC−r, BC=3, AC=4,\\\\\nsubstituting in, we get $\\frac{4−r}{4}=\\frac{r}{3}$\\\\\nthus $r=\\boxed{\\frac{12}{7}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52536784.png"
  },
  {
    "solution": "\\boxed{(3,0)}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in the Cartesian coordinate system xOy, OA = OB, $\\angle AOB = 90^{\\circ}$. A line parallel to the x-axis is drawn through point B, which intersects with the line passing through point A, $y = -x + b$, at point C. Given that the coordinates of point A are (2, 1), what are the coordinates of points C and D?\n",
    "completion": "\\textbf{Solution:} Let the line $y=-x+b$ pass through point A, with the coordinates of point A being $(2,1)$.\\\\\nTherefore, we have $1=-2+b$,\\\\\nwhich implies $b=3$,\\\\\nthus, the equation of the line is: $y=-x+3$,\\\\\nGiven that $\\angle AOB=90^\\circ$,\\\\\nwe find $B(-1,2)$,\\\\\nConstructing a line parallel to the x-axis through point B,\\\\\nimplies that the y-coordinates of points B and C are the same, both are 2,\\\\\nGiven that C lies on the line,\\\\\nthe coordinates of point C are $\\boxed{(1,2)}$,\\\\\nThe line $y=-x+3$ intersects the x-axis at point D,\\\\\nthus, the coordinates of point D are $\\boxed{(3,0)}$;",
    "image_path": "dataset/mm_math/images/MM_Math/50677077_157.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{6}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: In $\\triangle ABC$, $\\angle BAC=60^\\circ$, $\\angle ABC=45^\\circ$, $AB=2$, and $D$ is a moving point on line segment $BC$. With $AD$ as the diameter, draw $\\odot O$ intersecting $AB$, $AC$ at $E$, $F$ respectively. Connect $EF$. What is the minimum length of line segment $EF$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect OE, OF, and draw OH$\\perp$EF through point O with H as the foot,\\\\\n$\\therefore$ $EH=\\frac{1}{2}EF$,\\\\\n$\\because$ OE=OF, OH$\\perp$EF, $\\angle$BAC$=60^\\circ$,\\\\\n$\\therefore$ $\\angle EOH=\\angle FOH=\\frac{1}{2}\\angle EOF=\\angle BAC=60^\\circ$,\\\\\n$\\therefore$ $\\angle$OEH$=30^\\circ$,\\\\\n$\\therefore$ OH$=\\frac{1}{2}OE$,\\\\\n$\\therefore$ $EH=\\sqrt{OE^2-OH^2}=\\frac{\\sqrt{3}}{2}OE$,\\\\\n$\\therefore$ $EF=\\sqrt{3}OE$,\\\\\n$\\therefore$ To minimize EF, that is to minimize the radius OE, which means to minimize the diameter AD,\\\\\n$\\therefore$ According to the property of the perpendicular segment, when AD is the height from side BC of $\\triangle ABC$, the diameter AD is the shortest,\\\\\n$\\because$ In $\\triangle ADB$, $\\angle ABC=45^\\circ$, AB$=2$,\\\\\n$\\therefore$ AD$=$BD, $BD^2+AD^2=AB^2$,\\\\\n$\\therefore$ $2AD^2=4$,\\\\\n$\\therefore$ AD$=$BD$=\\sqrt{2}$,\\\\\n$\\therefore$ $EF=\\frac{\\sqrt{3}}{2}AD=\\boxed{\\frac{\\sqrt{6}}{2}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51318916.png"
  },
  {
    "solution": "\\boxed{\\frac{5}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in quadrilateral $ABCD$, a circle $\\odot O$ with diameter $AB$ exactly passes through point $C$, and $AC$ and $DO$ intersect at point $E$. It is known that $AC$ bisects $\\angle BAD$, $\\angle ADC=90^\\circ$, and $CD:BC=2:\\sqrt{5}$. What is the value of $CE:AE$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect OC\\\\\n$\\because$ AB is the diameter of the circle,\\\\\n$\\therefore$ $\\angle$ACB=$\\angle$ADC=90$^\\circ$,\\\\\n$\\because$ AC bisects $\\angle$BAD,\\\\\n$\\therefore$ $\\angle$DAC=$\\angle$CAB, $\\angle$DAB=2$\\angle$CAB,\\\\\n$\\therefore$ $\\triangle ADC\\sim \\triangle ACB$,\\\\\n$\\therefore$  $\\frac{CD}{BC}=\\frac{AC}{AB}=\\frac{2}{\\sqrt{5}}=\\frac{AD}{AC}$,\\\\\n$\\therefore$  $AC=\\frac{2\\sqrt{5}}{5}AB$,\\\\\n$\\therefore$  $BC=\\sqrt{AB^{2}-AC^{2}}=\\frac{\\sqrt{5}}{5}AB$,  $AD=\\frac{2}{\\sqrt{5}}AC=\\frac{4}{5}AB$\\\\\n$\\therefore$  $CD=\\frac{2}{\\sqrt{5}}BC=\\frac{2}{5}AB$,\\\\\nAlso $\\because$ $\\angle$BOC=2$\\angle$CAB,\\\\\n$\\therefore$ $\\angle$BOC=$\\angle$DAB,\\\\\n$\\therefore$ AD$\\parallel$OC,\\\\\n$\\therefore$ $\\triangle OCE\\sim \\triangle DAE$,\\\\\n$\\therefore$  $\\frac{CE}{AE}=\\frac{OC}{DA}=\\frac{\\frac{1}{2}AB}{\\frac{4}{5}AB}=\\boxed{\\frac{5}{8}}$,\\\\",
    "image_path": "dataset/mm_math/images/MM_Math/51179012.png"
  },
  {
    "solution": "\\boxed{(4,0)}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown 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. Find the coordinates of point C and the equation of line BC.\n",
    "completion": "\\textbf{Solution:} Solve: The line $y=\\frac{5}{2}x+5$ intersects the x-axis at point A and the y-axis at point B, \\\\\n$\\therefore A(-2,0), B(0,5)$, \\\\\nthat is $OA=2, OB=5$, \\\\\nsince the area of $\\triangle ABC$ is $15$, \\\\\n$\\therefore \\frac{1}{2}(OA+OC) \\cdot OB=15$, \\\\\n$\\therefore OC=4$, \\\\\n$\\therefore C(4,0)$, \\\\\nLet the expression of line BC be $y=kx+b$, \\\\\nSubstituting the coordinates of points B and C into the linear equation yields: $\\left\\{\\begin{array}{l}4k+b=0 \\\\ b=5\\end{array}\\right.$ \\\\\nSolving this, we get: $\\left\\{\\begin{array}{l}k=-\\frac{5}{4} \\\\ b=5\\end{array}\\right.$ \\\\\n$\\therefore$ The expression for line BC is: $y=-\\frac{5}{4}x+5$; \\\\\nHence, the coordinates of point C are $C=\\boxed{(4,0)}$, and the expression for line BC is $y=-\\frac{5}{4}x+5$.",
    "image_path": "dataset/mm_math/images/MM_Math/52019535_93.png"
  },
  {
    "solution": "\\boxed{y_{2}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in the Cartesian coordinate system, 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 $A$ and $B$ within the second and fourth quadrants, and intersects with the $x$-axis at point $C$. The coordinates of point $B$ are $(n, -2)$. The length of segment $OA$ is 5. Point $E$ is located on the negative half of the x-axis, and $\\sin\\angle AOE = \\frac{4}{5}$. Determine the analytic expressions of the inverse proportion function and the linear function.\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $AD\\perp x$-axis at point D through point A,\\\\\n$\\because \\sin\\angle AOE=\\frac{4}{5}$ and OA=5,\\\\\n$\\therefore \\frac{AD}{OA}=\\frac{4}{5}$,\\\\\n$\\therefore AD=4$,\\\\\nBy the Pythagorean theorem, we have $OD=\\sqrt{OA^{2}-AD^{2}}=\\sqrt{5^{2}-4^{2}}=3$,\\\\\n$\\therefore OD=3$,\\\\\n$\\therefore$The coordinates of point A are (-3, 4),\\\\\nSubstituting point A (-3, 4) into $y_{2}=\\frac{m}{x}$,\\\\\nWe get $4=\\frac{m}{-3}$, solving for $m$ yields $m=-12$,\\\\\n$\\therefore$The equation of the inverse proportion function is $y_{2}=-\\frac{12}{x}$,\\\\\nSubstituting B (n, -2) into $y_{2}=-\\frac{12}{x}$ gives $n=6$,\\\\\n$\\therefore$The coordinates of point B are (6, -2),\\\\\nSubstituting points A (-3, 4) and B (6, -2) into $y_{1}=kx+b$,\\\\\nWe get $\\begin{cases}4=-3k+b \\\\ -2=6k+b\\end{cases}$, solving for $\\begin{cases}k=-\\frac{2}{3} \\\\ b=2\\end{cases}$,\\\\\n$\\therefore$The equation of the linear function is $y_{1}=-\\frac{2}{3}x+2$,\\\\\n$\\therefore$The equation for the linear function is $\\boxed{y_{1}=-\\frac{2}{3}x+2}$, and the equation for the inverse proportion function is $\\boxed{y_{2}=-\\frac{12}{x}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52765948_99.png"
  },
  {
    "solution": "\\boxed{80^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the diagram, $\\triangle ABC$ is rotated clockwise by $50^\\circ$ around point $C$ to obtain $\\triangle A'B'C'$. If $\\angle A=40^\\circ$ and $\\angle B=110^\\circ$, what is the degree measure of $\\angle BCA'$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle A=40^\\circ$ and $\\angle B=110^\\circ$, \\\\\nit follows that $\\angle ACB=30^\\circ$, \\\\\nsince $\\triangle ABC$ is rotated $50^\\circ$ clockwise around point C to obtain $\\triangle A'B'C$, \\\\\nthus $\\angle ACA'=50^\\circ$, \\\\\ntherefore, $\\angle BCA'=\\angle BCA+\\angle ACA'=30^\\circ+50^\\circ=\\boxed{80^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51654883.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure above, in $\\triangle ABC$, $\\angle A=90^\\circ$, $AB=AC$, $BD$ bisects $\\angle ABC$, $CE \\perp BD$ at $E$. If $BD=12$, what is the length of $CE$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, extend BA and CE to intersect at point F, \\\\\n$\\because$ $\\angle$ABD + $\\angle$ADB = $90^\\circ$, $\\angle$CDE + $\\angle$ACF = $90^\\circ$, \\\\\n$\\therefore$ $\\angle$ABD = $\\angle$ACF, \\\\\n$\\because$ $\\angle$BAC = $90^\\circ$, \\\\\n$\\therefore$ $\\angle$FAC = $\\angle$BAC = $90^\\circ$, \\\\\nIn $\\triangle ABD$ and $\\triangle ACF$, \\\\\n$\\left\\{\\begin{array}{l}\n\\angle ABD = \\angle ACF\\\\\nAB = AC\\\\\n\\angle BAC = \\angle FAC\n\\end{array}\\right.$, \\\\\n$\\therefore$ $\\triangle ABD \\cong \\triangle ACF$ (ASA), \\\\\n$\\therefore$ BD = CF, \\\\\n$\\because$ BD bisects $\\angle$ABC, \\\\\n$\\therefore$ $\\angle$ABE = $\\angle$CBE, \\\\\n$\\because$ CE $\\perp$ BD, \\\\\n$\\therefore$ $\\angle$BEF = $\\angle$BEC = $90^\\circ$, \\\\\nIn $\\triangle BEF$ and $\\triangle BEC$, \\\\\n$\\left\\{\\begin{array}{l}\n\\angle ABE = \\angle CBE\\\\\nBE = BE\\\\\n\\angle BEF = \\angle BEC\n\\end{array}\\right.$, \\\\\n$\\therefore$ $\\triangle BEF \\cong \\triangle BEC$, \\\\\n$\\therefore$ EF = EC, \\\\\n$\\therefore$ EC = $\\frac{1}{2}$CF, \\\\\n$\\therefore$ CE = $\\frac{1}{2}$BD, \\\\\n$\\because$ BD = 12, \\\\\n$\\therefore$ CE = \\boxed{6}.",
    "image_path": "dataset/mm_math/images/MM_Math/52907833.png"
  },
  {
    "solution": "\\boxed{\\sqrt{6}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</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 $A$, $B$). Connect $PD$ and rotate the line segment $PD$ clockwise by $90^\\circ$ around point $P$ to obtain the line segment $PE$, where $PE$ intersects side $BC$ at point $F$. Connect $BE$ and $DF$. If $AP=\\sqrt{3}$, what is the length of $BE$?\n",
    "completion": "\\textbf{Solution:} Draw $EQ \\perp AB$ through point E to intersect the extended line of AB at point Q, hence $\\angle EQP = \\angle A = 90^\\circ$,\\\\\nIn $\\triangle PAD$ and $\\triangle EQP$,\\\\\n$\\left\\{\\begin{array}{l}\n\\angle A = \\angle EQP \\\\\n\\angle ADP = \\angle EPB \\\\\nPD = PE\n\\end{array}\\right.$,\\\\\n$\\therefore \\triangle PAD \\cong \\triangle EQP$ (AAS),\\\\\n$\\therefore EQ = AP = \\sqrt{3}$, AD = AB = PQ,\\\\\n$\\therefore AP = EQ = BQ$,\\\\\n$\\therefore \\angle CBE = \\angle EBQ = 45^\\circ$;\\\\\n$\\therefore BE = \\sqrt{2}EQ = \\boxed{\\sqrt{6}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51322751_128.png"
  },
  {
    "solution": "\\boxed{\\sqrt{26}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, point $P$ lies on the extension line of the diagonal $BD$ of square $ABCD$, point $M$ is on side $AD$. Lines $CP$, $BM$, and $MP$ are connected. Given that $AB=4$, $AM=1$, and $BM=PM$, what is the length of $CP$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw ME$\\perp$BP at E through point M and draw PF$\\perp$BC intersecting the extended line of BC at F, \\\\\n$\\because$ Quadrilateral ABCD is a square, \\\\\n$\\therefore$ AD=AB=4, $\\angle$MDE=45$^\\circ$, $\\angle$A=90$^\\circ$ \\\\\n$\\therefore$ MD=AD−AM=3, $\\angle$DME=$\\angle$DBC=45$^\\circ$, \\\\\n$\\therefore$ ME=DE, \\\\\n$\\because$ $M{D}^{2}=M{E}^{2}+D{E}^{2}$, \\\\\n$\\therefore$ ${3}^{2}=2M{E}^{2}$, \\\\\n$\\therefore$ ME=DE=$\\frac{3\\sqrt{2}}{2}$, \\\\\n$\\because$ $B{D}^{2}=A{B}^{2}+A{D}^{2}$, \\\\\n$\\therefore$ BD=$\\sqrt{A{B}^{2}+A{D}^{2}}=4\\sqrt{2}$, \\\\\n$\\therefore$ BE=BD−DE=$\\frac{5\\sqrt{2}}{2}$, \\\\\n$\\because$BM=PM, \\\\\n$\\therefore$ BP=2BE=$5\\sqrt{2}$, \\\\\n$\\because$$\\angle$PBC=45$^\\circ$, $\\angle$PFB=90$^\\circ$, \\\\\n$\\therefore$$\\angle$BPF=45$^\\circ$, \\\\\n$\\therefore$BF=PF, $ B{P}^{2}=P{F}^{2}+B{F}^{2}$, \\\\\n$\\therefore$ ${\\left(5\\sqrt{2}\\right)}^{2}=2P{F}^{2}$, \\\\\n$\\therefore$PF=BF=5, \\\\\n$\\therefore$CF=BF−BC=1, \\\\\n$\\therefore$ $PC=\\sqrt{P{F}^{2}+C{F}^{2}}=\\boxed{\\sqrt{26}}$,",
    "image_path": "dataset/mm_math/images/MM_Math/52019443.png"
  },
  {
    "solution": "\\boxed{-16}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in the Cartesian coordinate system, A(8, 0), point B is a moving point on the graph of the linear function $y=x$. Taking OB as a side, a square OBCD is constructed. When AB is at its minimum length, point D just falls on the graph of the inverse proportion function $y=\\frac{k}{x}$. Then, $k=$ (\\quad)\n",
    "completion": "\\textbf{Solution:} According to the principle that a perpendicular line segment is the shortest, when AB is perpendicular to the line $y=x$, AB is at its shortest,\\\\\n$\\because$ $\\angle AOB=45^\\circ$\\\\\n$\\therefore$ $\\angle BAO=45^\\circ$\\\\\n$\\therefore$ $\\triangle AOB$ is an isosceles right triangle,\\\\\n$\\because$ the coordinates of point A are (8, 0)\\\\\n$\\therefore$ OA=8\\\\\n$\\therefore$ $BO=BA=4\\sqrt{2}$\\\\\n$\\because$ quadrilateral OBCD is a square,\\\\\n$\\therefore$ $DO=BO=4\\sqrt{2}$, $\\angle DOB=90^\\circ$\\\\\n$\\therefore$ $\\angle DOC=\\angle BOC=45^\\circ$\\\\\nDraw DE$\\perp$x-axis through point D at point E,\\\\\n$\\therefore$ $\\angle ODE=\\angle DOE=45^\\circ$\\\\\n$\\therefore$ $\\triangle DEO$ is an isosceles right triangle,\\\\\n$\\therefore$ $DE=OE=\\frac{DO}{\\sqrt{2}}=\\frac{4\\sqrt{2}}{\\sqrt{2}}=4$\\\\\n$\\because$ point D is in the second quadrant,\\\\\n$\\therefore$ D(−4, 4)\\\\\nAlso, point D is on the graph of the reciprocal function $y=\\frac{k}{x}$\\\\\n$\\therefore$ $k=(−4)\\times 4=\\boxed{-16}$",
    "image_path": "dataset/mm_math/images/MM_Math/50496566.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{7}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, the side length of rhombus $ABCD$ is $2$, $\\angle A=60^\\circ$, point $G$ is the midpoint of $AB$, a rhombus $BEFG$ is constructed with $BG$ as its side, where point $E$ is on the extension line of $CB$, and point $P$ is the midpoint of $FD$. What is the length of $PB$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect BF and BD, \\\\\n$\\because$ the side length of the rhombus ABCD is 2,\\\\\n$\\therefore$ AB = BC = CD = 2,\\\\\n$\\because$ $\\angle$A = $60^\\circ$,\\\\\n$\\therefore$ $\\triangle$BCD is an equilateral triangle,\\\\\n$\\therefore$ BD = BC = 2, $\\angle$DBC = $60^\\circ$,\\\\\n$\\therefore$ $\\angle$DBA = $60^\\circ$,\\\\\n$\\because$ point G is the midpoint of AB,\\\\\n$\\therefore$ the side length of rhombus BEFG is 1,\\\\\nthat is, BE = EF = BG = 1,\\\\\n$\\because$ point E is on the extension line of CB, $\\angle$GBE = $60^\\circ$,\\\\\n$\\therefore$ $\\angle$FBG = $30^\\circ$,\\\\\nconnect EG,\\\\\n$\\therefore$ EG $\\perp$ FB at point O,\\\\\n$\\therefore$ OB = $\\frac{\\sqrt{3}}{2}$,\\\\\n$\\therefore$ FB = $\\sqrt{3}$,\\\\\n$\\because$ $\\angle$DBF = $\\angle$DBA + $\\angle$FBG = $90^\\circ$,\\\\\naccording to the Pythagorean theorem, we get\\\\\nDF = $\\sqrt{DB^2 + BF^2} = \\sqrt{7}$,\\\\\n$\\because$ point P is the midpoint of FD,\\\\\n$\\therefore$ PB = $\\frac{1}{2}$DF = $\\boxed{\\frac{\\sqrt{7}}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52210861.png"
  },
  {
    "solution": "\\boxed{y=2x}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</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-axis and y-axis at points A and B, respectively. The graph of the direct proportionality function, denoted as $l_{2}$, intersects $l_{1}$ at point C $(m, 4)$. What is the value of $m$ and the equation of $l_{2}$?\n",
    "completion": "\\textbf{Solution:} Substitute $C(m,4)$ into the linear function $y=-\\frac{1}{2}x+5$, we get: $4=-\\frac{1}{2}m+5$, solving for $m$ gives: $m=\\boxed{2}$.\\\\\nTherefore, for $C(2,4)$, let the equation of line $l_{2}$ be $y=ax$, then we have $4=2a$, solving for $a$ gives $a=2$,\\\\\ntherefore, the equation of line $l_{2}$ is $\\boxed{y=2x}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50676403_167.png"
  },
  {
    "solution": "\\boxed{E(1,\\sqrt{3}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, the vertices of rectangle $OABC$ are $O(0, 0)$ and $B(-2, 2\\sqrt{3})$. If the rectangle rotates counterclockwise about point $O$ at a rate of $60^\\circ$ per second, what are the coordinates of the diagonal intersection point $D$ at the 2021st second?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, according to the problem, we know that $D(-1,\\sqrt{3})$, and the trajectory of point $D$ is a circle centered at $O$ with $OD$ as the radius. Construct $DM\\perp x$-axis, where $M$ is the foot of the perpendicular.\\\\\n$\\therefore$ $\\tan\\angle DOM=\\frac{DM}{OM}=\\sqrt{3}$\\\\\n$\\therefore$ $\\angle DOM=60^\\circ$\\\\\n$\\because$ $2021\\times 60^\\circ=336\\times 360^\\circ+300^\\circ$\\\\\n$\\therefore$ it is known that at the 2021st second, the intersection point of the diagonals of the rectangle, $D$, is at point $E$, with $\\angle DOE=60^\\circ$\\\\\n$\\therefore$ $D$ and $E$ are symmetric with respect to the $y$-axis\\\\\n$\\therefore$ $E(1,\\sqrt{3})$\\\\\nTherefore, the answer is: $E(1,\\sqrt{3})$, which means the coordinates of $D$ are $\\boxed{E(1,\\sqrt{3})}$",
    "image_path": "dataset/mm_math/images/MM_Math/53498209.png"
  },
  {
    "solution": "\\boxed{7}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, a parabola (with a fixed shape) intersects the x-axis at points E and F (point E is to the left of point F), and its vertex P moves along the segment $AB$. If the coordinates of points A and B are respectively $(-2, -3)$ and $(4, -3)$, and the minimum value of the x-coordinate of point E is $-5$, what is the maximum value of the x-coordinate of point F?\n",
    "completion": "\\textbf{Solution:} Let the axis of symmetry be the line $x=a$, and E, F are symmetric about $x=a$,\\\\\nthen $a-x_{E}=x_{F}-a$, $2a=x_{F}+x_{E}$,\\\\\nsince the shape of the parabola is fixed,\\\\\nthus, the opening size of the parabola remains unchanged,\\\\\nby translation, it is known that, when P is at point A, the x-coordinate of E is the minimum,\\\\\nwhen P is at point A, we have $2 \\times (-2) = -5 + x_{F}$, solving gives $x_{F}=1$,\\\\\nthus $x_{F}-x_{E}$ is a constant value $1-(-5)=6$,\\\\\nwhen P moves to B, the maximum of $x_{F}$ is $4+\\frac{6}{2}=\\boxed{7}$,",
    "image_path": "dataset/mm_math/images/MM_Math/52602069.png"
  },
  {
    "solution": "\\boxed{\\sqrt{97}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in quadrilateral ABCD, AB $\\parallel$ CD, $\\angle B = \\angle D$. Point E is on the extension of BC. Line segment AE is drawn and intersects CD at point H. The bisector of $\\angle DCE$ intersects AE at point G. If AB $= 2$AD $= 10$, point H is the midpoint of CD, and HE $= 6$, then what is the value of AC?\n",
    "completion": "\\textbf{Solution:} Given that $\\because$ AB＝2AD＝10,\\\\\n$\\therefore$AD=5,\\\\\n$\\because$AB$\\parallel$CD,\\\\\n$\\therefore$$\\angle$B=$\\angle$DCE,\\\\\n$\\because$$\\angle$B=$\\angle$D,\\\\\n$\\therefore$$\\angle$B=$\\angle$DCE,\\\\\n$\\therefore$AD$\\parallel$BC,\\\\\n$\\therefore$Quadrilateral ABCD is a parallelogram, $\\angle$DAH=$\\angle$E,\\\\\n$\\therefore$CD=AB=10,\\\\\n$\\because$ point H is the midpoint of CD,\\\\\n$\\therefore$CH=DH=5,\\\\\n$\\therefore$$\\triangle $AHD$\\cong $$\\triangle $ECH (AAS),\\\\\n$\\therefore$CE=AD=5=CH, AH=EH=6,\\\\\n$\\because$ CG bisects $\\angle$DCE,\\\\\n$\\therefore$CG$\\perp$EH, HG=EG=3,\\\\\n$\\therefore$CG=$ \\sqrt{CH^{2}-HG^{2}}=\\sqrt{5^{2}-3^{2}}$=4, AG=AH+HG=9,\\\\\n$\\therefore$AC=$ \\sqrt{AG^{2}+CG^{2}}=\\sqrt{9^{2}+4^{2}}=\\boxed{\\sqrt{97}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51827661.png"
  },
  {
    "solution": "\\boxed{\\frac{8}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the diagram, $AB$ is the diameter of circle $O$, and points $C$ and $D$ are on circle $O$ with the condition that point $C$ is the midpoint of arc $AD$, and $AD$ intersects $OC$ at point $E$. Given $AD=8$, $EC=2$. A line parallel to $AB$ passing through point $C$ intersects chord $AD$ at point $F$. What is the length of segment $EF$?\n",
    "completion": "\\textbf{Solution}: Since $CF\\parallel AB$,\\\\\nit follows that $\\frac{CE}{OE}=\\frac{EF}{AE}$,\\\\\nthus $\\frac{2}{3}=\\frac{EF}{4}$,\\\\\nsolving this gives $EF=\\boxed{\\frac{8}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53148601_134.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, $l_{1}\\parallel l_{2}\\parallel l_{3}$, and lines a, b intersect $l_{1}$, $l_{2}$, $l_{3}$ at points A, B, C, and D, E, F, respectively. If $\\frac{AB}{BC}=\\frac{2}{3}$ and $DE=4$, what is the length of $DF$?\n",
    "completion": "\\textbf{Solution:} Since $l_1 \\parallel l_2 \\parallel l_3$,\\\\\nit follows that $\\frac{DE}{EF} = \\frac{AB}{BC} = \\frac{2}{3}$. Given that $DE=4$,\\\\\nit implies that $EF=6$,\\\\\nhence $DF=DE+EF=\\boxed{10}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52536877.png"
  },
  {
    "solution": "\\boxed{\\frac{50}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, AB represents a ski jump track. The angle of elevation of point B from point A is measured to be $35^\\circ$. The distance between the bottom point C and the top point B is 50 meters. What is the length of the track AB? Meters.\n",
    "completion": "\\textbf{Solution:} In the right triangle $ABC$,\\\\\nsince $\\angle A=35^\\circ$ and $BC=50$ meters,\\\\\nit follows that $\\sin 35^\\circ = \\frac{BC}{AB}$,\\\\\ntherefore, $AB = \\frac{50}{\\sin 35^\\circ}$ (meters).\\\\\nThus, $AB = \\boxed{\\frac{50}{\\sin 35^\\circ}}$ (meters).",
    "image_path": "dataset/mm_math/images/MM_Math/53042711.png"
  },
  {
    "solution": "\\boxed{\\frac{16}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: In quadrilateral $ABCD$, $AB\\parallel CD$, $\\angle BCD=90^\\circ$, $AB=AD=10\\text{cm}$, $BC=8\\text{cm}$. Point $P$ starts moving from point $A$ at a constant speed of $3\\text{cm/s}$ along the polyline $ABCD$ direction; point $Q$ starts moving from point $D$ at a constant speed of $2\\text{cm/s}$ along the line segment $DC$ direction. It is known that both points start moving at the same time, and when one point reaches its destination, the other point stops moving, let the motion time be $t(s)$. During the movement of points $P$ and $Q$, there exists a moment when the area of $\\triangle BPQ$ is $16\\text{cm}^2$. Find all possible values of $t$ that satisfy this condition?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, we connect $PQ$, $BQ$, and let the time of motion be $t(s)$,\\\\\n$\\because AB=10$, $BC=8$, the speed of point P is $3cm/s$, $DC=16$, the speed of point Q is $2cm/s$,\\\\\n$\\therefore$ the time for point P to reach point B is $\\frac{10}{3}$ seconds, the time for point P to reach point C is $6$ seconds, and the time for point Q to reach point C is $8$ seconds,\\\\\n(1) When $0<t<\\frac{10}{3}$, $BP=10-3t$, the height of $\\triangle BPQ$ is $BC=8$,\\\\\n$\\therefore \\frac{1}{2}BP\\cdot BC=\\frac{1}{2}\\times (10-3t)\\times 8=16$, solving this equation, we get $t=\\boxed{2}$,\\\\\nVerification: When $t=2$, $BP=10-3t=10-3\\times 2=4$, $BC=8$, the area of $\\triangle BPQ$ is $16cm^2$, and $0<2<\\frac{10}{3}$,\\\\\n$\\therefore t=2$, satisfies the condition;\\\\\n(2) When $t=\\frac{10}{3}$, point P coincides with point B, $\\triangle BPQ$ does not exist, which does not meet the requirements, hence $t\\ne \\frac{10}{3}$;\\\\\n(3) When $\\frac{10}{3}<t\\le 6$, as shown in the figure,\\\\\n$\\therefore BP=3t-10$, $DQ=2t$, $CQ=16-2t$,\\\\\n$\\therefore$ the base of $\\triangle BPQ$ is $BP=3t-10$, and the height is $CQ=16-2t$,\\\\\n$\\therefore$ the area of $\\triangle BPQ$ is $\\frac{1}{2}BP\\cdot CQ=\\frac{1}{2}\\times (3t-10)\\times (16-2t)=16$, after simplification, $3t^2-34t+96=0$, solving this equation, we get $t_1=\\frac{16}{3}$, $t_2=6$,\\\\\n$\\therefore$when $t=6$ or $t=\\boxed{\\frac{16}{3}}$, the condition is met;\\\\\n④ When $6<t<8$, as shown in the figure,\\\\\n$\\therefore CP=3t-18$, $CQ=16-2t$,\\\\\n$\\therefore PQ=CQ-CP=16-2t-(3t-18)=34-5t$, the height of $\\triangle BPQ$ is $BC=8$,\\\\\n$\\therefore$ the area of $\\triangle BPQ$ is $\\frac{1}{2}PQ\\cdot BC=\\frac{1}{2}\\times (34-5t)\\times 8=16$, solving this equation, we get $t=6$,\\\\\n$\\therefore$ when $6<t<8$, the condition is not met;\\\\\n⑤ When $t=8$, point Q coincides with point C, motion stops, as shown in the figure,\\\\\n$\\therefore$ the position of point P is $CP=3t-18=3\\times 8-18=6$, the height of $\\triangle BPQ$ is $BC=8$,\\\\\n$\\therefore$ the area of $\\triangle BPQ$ is $\\frac{1}{2}CP\\cdot BC=\\frac{1}{2}\\times 6\\times 8=24$, which does not meet the condition.\\\\\nIn summary, when the value of $t$ is $2$ or $\\frac{16}{3}$ or $6$, the area of $\\triangle BPQ$ is $16cm^2$.",
    "image_path": "dataset/mm_math/images/MM_Math/53490651_88.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, point O is on line AB, $\\angle COE$ is a right angle, OF bisects $\\angle AOE$, $\\angle COF=30^\\circ$, what is the measure of $\\angle EOB$?\n",
    "completion": "\\textbf{Solution:} Given, $\\because$ $\\angle COE$ is a right angle,\\\\\n$\\therefore$ $\\angle COE=90^\\circ$,\\\\\n$\\because$ $\\angle COE=\\angle COF+\\angle FOE$, $\\angle COF=30^\\circ$,\\\\\n$\\therefore$ $\\angle FOE=\\angle COE−\\angle COF=90^\\circ−30^\\circ=\\boxed{60^\\circ}$,\\\\\n$\\because$ OF bisects $\\angle AOE$,\\\\\n$\\therefore$ $\\angle FOE=\\angle AOF=\\frac{1}{2}\\angle AOE$,\\\\\n$\\therefore$ $\\angle AOE=2\\angle FOE=120^\\circ$,\\\\\n$\\because$ $\\angle AOE+\\angle BOE=180^\\circ$,\\\\\n$\\therefore$ $\\angle BOE=180^\\circ−120^\\circ=60^\\circ$",
    "image_path": "dataset/mm_math/images/MM_Math/52915719.png"
  },
  {
    "solution": "\\boxed{118^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the diagram, $\\angle AOB$ is a right angle, $OB$ bisects $\\angle COD$, $\\angle AOC=62^\\circ$, find the degree measure of $\\angle AOD$.\n",
    "completion": "\\textbf{Solution:} Since $\\angle AOB$ is a right angle,\\\\\nit follows that $\\angle AOB=90^\\circ$,\\\\\ntherefore, $\\angle BOC=\\angle AOB-\\angle AOC=90^\\circ-62^\\circ=28^\\circ$.\\\\\nSince $OB$ bisects $\\angle COD$,\\\\\nit follows that $\\angle BOD=\\angle BOC=28^\\circ$,\\\\\ntherefore, $\\angle AOD=\\angle AOB+\\angle BOD=90^\\circ+28^\\circ=\\boxed{118^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51208120.png"
  },
  {
    "solution": "\\boxed{\\sqrt{2}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, quadrilateral $OABC$ is a rectangle, with point $A$ in the third quadrant. The point symmetrical to point $A$ with respect to $OB$ is point $D$. Points $B$ and $D$ are both on the graph of the function $y=-\\frac{3\\sqrt{2}}{x}$ $(x<0)$. $BE\\perp y$-axis at point $E$. If the extension line of $DC$ intersects the $y$-axis at point $F$, and when the area of rectangle $OABC$ is $6$, what is the value of $\\frac{OF}{OE}$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect BF and DO,\\\\\ndue to symmetry and the properties of rectangles, it's easy to see that $\\triangle OAB \\cong \\triangle BCO \\cong \\triangle BDO$,\\\\\n$\\therefore$BO$\\parallel$DF,\\\\\n$\\therefore S_{\\triangle BFE} = S_{\\triangle BDO} = \\frac{1}{2}S_{\\text{rectangle} OABC} = 3$,\\\\\nMoreover, since point B lies on the graph of the inverse proportional function $y = -\\frac{3\\sqrt{2}}{x}$,\\\\\n$\\therefore S_{\\triangle BEO} = \\frac{3\\sqrt{2}}{2}$,\\\\\n$\\therefore \\frac{OF}{OE} = \\frac{S_{\\triangle BFE}}{S_{\\triangle BEO}} = \\frac{3}{\\frac{3\\sqrt{2}}{2}} = \\boxed{\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52586399.png"
  },
  {
    "solution": "\\boxed{0}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, 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 line segment $AC$ such that $CE=BC$. What is the length of $AE$?\n",
    "completion": "\\textbf{Solution:} Since $CE=BC$ and $BC=4$, \\\\\nit follows that $CE=4$, \\\\\nthus, $AE=AC-CE=4-4=\\boxed{0}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52579842_72.png"
  },
  {
    "solution": "\\boxed{\\frac{15}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the diagram, a right-angled triangle is placed with point C on the extension of FD, AB $\\parallel$ CF, $\\angle F=\\angle ACB=90^\\circ$, $\\angle E=45^\\circ$, $\\angle A=60^\\circ$, and AC=5. Determine the length of CD.\n",
    "completion": "\\textbf{Solution:} Draw BM$\\perp$FD at point M,\\\\\nIn $\\triangle ACB$, $\\angle ACB=90^\\circ$, $\\angle A=60^\\circ$, AC=5,\\\\\n$\\therefore \\angle ABC=30^\\circ$, BC=AC$\\cdot \\tan 60^\\circ=5\\sqrt{3}$,\\\\\n$\\because AB\\parallel CF$,\\\\\n$\\therefore \\angle BCM=30^\\circ$,\\\\\n$\\therefore$BM=BC$\\cdot \\sin 30^\\circ=\\frac{5\\sqrt{3}}{2}$, CM=BC$\\cdot \\cos 30^\\circ=\\frac{15}{2}$,\\\\\nIn $\\triangle EFD$, $\\angle F=90^\\circ$, $\\angle E=45^\\circ$,\\\\\n$\\therefore \\angle EDF=45^\\circ$,\\\\\n$\\therefore \\angle EDF=\\angle DBM=45^\\circ$,\\\\\n$\\therefore$MD=BM=$\\frac{5\\sqrt{3}}{2}$,\\\\\n$\\therefore$CD=$CM-MD=\\boxed{\\frac{15}{2}-\\frac{5\\sqrt{3}}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51298675.png"
  },
  {
    "solution": "\\boxed{\\sqrt{37}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the diagram, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, $BC=4$, $CA=6$. Circle $C$ has a radius of 2, and $P$ is a moving point on the circle. $AP$ and $BP$ are connected. What is the minimum value of $AP+\\frac{1}{2}BP$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure,\\\\\ntake a point $D$ on $CB$, such that $CD=1$, then we have $\\frac{CD}{CP}=\\frac{CP}{CB}=\\frac{1}{2}$,\\\\\nand since $\\angle PCD=\\angle BCP$,\\\\\ntherefore, $\\triangle PCD\\sim \\triangle BCP$,\\\\\nthus, $\\frac{PD}{BP}=\\frac{1}{2}$,\\\\\nhence, $PD=\\frac{1}{2}BP$,\\\\\ntherefore, $AP+\\frac{1}{2}BP=AP+PD$.\\\\\nTo minimize $AP+\\frac{1}{2}BP$, $AP+PD$ must be minimized,\\\\\nwhen points $A$, $P$, $D$ are on the same line, $AP+PD$ is minimized,\\\\\nthat is, the minimum value of $AP+\\frac{1}{2}BP$ is $AD$,\\\\\nin $\\triangle ACD$, where $CD=1$, $AC=6$,\\\\\ntherefore, $AD=\\sqrt{AC^2+CD^2}=\\boxed{\\sqrt{37}}$,\\\\\nthe minimum value of $AP+\\frac{1}{2}BP$ is $\\sqrt{37}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52872012.png"
  },
  {
    "solution": "\\boxed{13.5}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, the mathematics interest group measures the height of the flagpole $CD$ and the academic building $AB$. First, at point E, they use a theodolite $EF$ that is 1.5 meters high to measure the elevation angle $\\angle AFH$ of the top of the academic building A 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 $CD$), they measure the elevation angle $\\angle AGH$ of the top of the academic building A again as $60^\\circ$. Points B, C, and E are on the same horizontal line. Calculate the height of the flagpole $CD$ in meters.\n",
    "completion": "\\textbf{Solution:} Given the problem, we have: $\\angle FDG = \\angle AFH = 45^\\circ$, $EF = CG = BH = 1.5$ meters, $GF = CE = 12$ meters,\\\\\nin $\\triangle AFH$, since $\\angle AFH = 45^\\circ$, $\\therefore \\triangle DFG$ is an isosceles right triangle,\\\\\n$\\therefore DG = FG = 12$ meters, $\\therefore CD = CG + DG = 1.5 + 12 = \\boxed{13.5}$ (meters).",
    "image_path": "dataset/mm_math/images/MM_Math/52602002_144.png"
  },
  {
    "solution": "\\boxed{\\frac{3}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in rectangle $OABC$, where $OA=4$ and $AB=3$, point $D$ is on side $BC$, and $CD=3DB$. Point $E$ is a point on side $OA$, and $DE$ is connected. When quadrilateral $ABDE$ is folded along $DE$, and the symmetric point $A'$ of point $A$ exactly falls on side $OC$, then what is the length of $OE$?\n",
    "completion": "\\textbf{Solution:} Connect $A^{\\prime}D$, $AD$,\\\\\nsince quadrilateral $OABC$ is a rectangle,\\\\\nthus BC=OA=4, OC=AB=3, $\\angle C=\\angle B=\\angle O=90^\\circ$,\\\\\nsince $CD=3DB$,\\\\\nthus $CD=3$, $BD=1$,\\\\\nthus $CD=AB$,\\\\\nsince quadrilateral $ABDE$ is folded along $DE$, if the symmetric point of $A^{\\prime}$ exactly falls on side $OC$,\\\\\nthus $A^{\\prime}D=AD$, $A^{\\prime}E=AE$,\\\\\nin $\\triangle A^{\\prime}CD$ and $\\triangle DBA$, $\\left\\{\\begin{array}{l}CD=AB \\\\ A^{\\prime}D=AD\\end{array}\\right.$,\\\\\nthus $\\triangle A^{\\prime}CD\\cong \\triangle DBA$ (HL),\\\\\nthus $A^{\\prime}C=BD=1$,\\\\\nthus $A^{\\prime}O=2$,\\\\\nsince $A^{\\prime}O^{2}+OE^{2}=A^{\\prime}E^{2}$,\\\\\nthus $2^{2}+OE^{2}=(4-OE)^{2}$,\\\\\nthus $OE=\\boxed{\\frac{3}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52387487.png"
  },
  {
    "solution": "\\boxed{\\frac{25}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: In rectangle $ABCD$, $AB=6$, and $BC=8$. The rectangle is folded so that point $B$ coincides with point $D$ (as shown in Figure 2). What is the length of $DG$?\n",
    "completion": "\\textbf{Solution:}\n\\[DG = \\boxed{\\frac{25}{4}}.\\]",
    "image_path": "dataset/mm_math/images/MM_Math/51614698_98.png"
  },
  {
    "solution": "\\boxed{\\frac{6}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in the acute-angled triangle $ABC$, where $AB=AC$, the circle $\\odot O$ with $BC$ as the diameter intersects $AB$ and $AC$ at points $D$ and $E$ respectively. Lines $OD$, $OE$, and $DE$ are connected. $DE\\parallel BC$. If the radius of $\\odot O$ is $m$, and $\\tan\\angle ABC=2$, what is the area of the quadrilateral $ADOE$ (expressed in an algebraic expression involving $m$)?\n",
    "completion": "\\textbf{Solution:} Given that $AB=AC$ and $BO=OC$,\\\\\nwe have $AO\\perp BC$,\\\\\nthus $\\tan\\angle ABC=\\frac{AO}{OC}=2$,\\\\\nsince $OB=m$,\\\\\nit follows that $OA=2m$,\\\\\nsince $\\tan\\angle C=\\frac{BE}{EC}=2$,\\\\\nwe find $EC=\\frac{2\\sqrt{5}}{5}m$,\\\\\ngiven $AB=AC=\\sqrt{5}m$,\\\\\nthus $AE=AC-EC=\\sqrt{5}m-\\frac{2\\sqrt{5}}{5}m=\\frac{3\\sqrt{5}}{5}m$,\\\\\nsince $DE\\parallel CB$,\\\\\nit follows that $\\frac{DE}{BC}=\\frac{AE}{AC}$,\\\\\nthus $\\frac{DE}{2m}=\\frac{\\frac{3\\sqrt{5}}{5}m}{\\sqrt{5}m}$,\\\\\nconsequently, $DE=\\frac{6}{5}m$,\\\\\nsince $AO\\perp BC$ and $DE\\parallel BC$,\\\\\nit follows that $AO\\perp DE$,\\\\\nhence the area of quadrilateral $ADOE$ is $\\frac{1}{2}\\cdot AO\\cdot DE=\\frac{1}{2}\\times 2m\\times \\frac{6}{5}m=\\boxed{\\frac{6}{5}m^2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52872091_114.png"
  },
  {
    "solution": "\\boxed{66^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, $AD$ is the altitude of $\\triangle ABC$, $BE$ bisects $\\angle ABC$ and intersects $AD$ at $E$. If $\\angle C=70^\\circ$ and $\\angle BED=68^\\circ$, what is the measure of $\\angle BAC$?\n",
    "completion": "\\textbf{Solution:} Given that $AD$ is the height of $\\triangle ABC$, and $\\angle BED=68^\\circ$,\n\\[\n\\therefore \\angle EBD=90^\\circ-\\angle BED=90^\\circ-68^\\circ=22^\\circ,\n\\]\nSince $BE$ bisects $\\angle ABC$ and intersects $AD$ at E,\n\\[\n\\therefore \\angle ABC=2\\angle EBD=2\\times 22^\\circ=44^\\circ,\n\\]\nGiven that $\\angle C=70^\\circ$,\n\\[\n\\therefore \\angle BAC=180^\\circ-\\angle ABC-\\angle C=180^\\circ-44^\\circ-70^\\circ=\\boxed{66^\\circ}.\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/52793336.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in quadrilateral $ABCD$, $CD$ bisects the angle between the extensions of sides $AC$ and $BC$, $AD \\perp DC$, and point $E$ is the midpoint of $AB$. If $AC=4$ and $BC=6$, what is the length of segment $DE$?\n",
    "completion": "\\textbf{Solution:} Extend $BC$ and $AD$ respectively to meet at point $G$, \\\\\n$\\because$ $CD$ bisects the angle between the extended lines of $AC$ and $BC$\\\\\n$\\therefore \\angle 1=\\angle 2$\\\\\n$\\because$ $AD\\perp DC$\\\\\n$\\therefore \\angle CDA=\\angle CDG=90^\\circ$\\\\\n$\\because CD=CD$\\\\\n$\\therefore \\triangle CDG\\cong \\triangle CDA$ (ASA)\\\\\n$\\therefore CG=AC$, $AD=GD$\\\\\n$\\because$ $AC=4$, $BC=6$\\\\\n$\\therefore BG=BC+CG=6+4=10$\\\\\nAlso, $\\because$ point $E$ is the midpoint of $AB$\\\\\n$\\therefore DE=\\frac{1}{2}BG=\\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50482456.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{2}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, points A, B, and C are all located at the vertices of the small square, and the side length of each small square is 1. What is the value of $\\cos\\angle BAC$?\n",
    "completion": "Solution: Connect BC. Since the side length of each small square is 1, we have AB=$\\sqrt{5}$, BC=$\\sqrt{5}$, AC=$\\sqrt{10}$. Since $(\\sqrt{5})^{2}+(\\sqrt{5})^{2}=(\\sqrt{10})^{2}$, triangle ABC is a right-angled triangle. Therefore, $\\cos\\angle BAC=\\frac{AB}{AC}=\\frac{\\sqrt{5}}{\\sqrt{10}}=\\frac{\\sqrt{2}}{2}$.\n\nHence, $\\cos\\angle BAC = \\boxed{\\frac{\\sqrt{2}}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50982861.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, point C is a point on the line segment AB, where M and N are the midpoints of AB and CB, respectively. Given that $AC=10\\,cm$ and $NB=6\\,cm$, find the length of CM.\n",
    "completion": "\\textbf{Solution:} \\\\\n$\\because$ N is the midpoint of BC, and NB = 6cm, \\\\\n$\\therefore$ BC = 2NB = $2 \\times 6$ = 12cm, \\\\\n$\\because$ AC = 10cm, \\\\\n$\\therefore$ AB = AC + BC = 10 + 12 = 22cm, \\\\\n$\\because$ M is the midpoint of AB, \\\\\n$\\therefore$ BM = AM = $\\frac{1}{2}$AB = $\\frac{1}{2} \\times 22$ = 11cm, \\\\\n$\\therefore$ CM = AM - AC = 11 - 10 = \\boxed{1}cm.",
    "image_path": "dataset/mm_math/images/MM_Math/51494975_205.png"
  },
  {
    "solution": "\\boxed{110^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle B=40^\\circ$, $\\angle C=30^\\circ$, and point D is a point on side BC. After folding $\\triangle ADC$ along line AD, point C falls onto point E. If $DE\\parallel AB$, then what is the measure of $\\angle ADE$?\n",
    "completion": "\\textbf{Solution:} In $\\triangle ABC$, $\\angle B=40^\\circ$, $\\angle C=30^\\circ$,\\\\\nthus $\\angle BAC=180^\\circ-\\angle B-\\angle C=180^\\circ-40^\\circ-30^\\circ=110^\\circ$;\\\\\nBecause after folding $\\triangle ADC$ along the line AD, point C falls on point E,\\\\\nthus $\\angle C=\\angle E=30^\\circ$, $\\angle EAD=\\angle CAD$, $\\angle ADE=\\angle ADC$,\\\\\nBecause DE$\\parallel$AB,\\\\\nthus $\\angle BAE=\\angle E=30^\\circ$,\\\\\nthus $\\angle EAD=\\angle CAD= \\frac{1}{2}(\\angle BAC-\\angle BAE)= \\frac{1}{2}(110^\\circ-30^\\circ)=40^\\circ$;\\\\\nthus $\\angle ADE=\\angle ADC=180^\\circ-\\angle C-\\angle CAD=180^\\circ-30^\\circ-40^\\circ=\\boxed{110^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52661185.png"
  },
  {
    "solution": "\\boxed{\\frac{120}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in quadrilateral $ABCD$, $AB \\parallel DC$, $AB=AD$, and the diagonals $AC$ and $BD$ intersect at point $O$, with $AC$ bisecting $\\angle BAD$. Quadrilateral $ABCD$ is a rhombus; through point $C$, draw $CE \\perp AB$ to the extended line of $AB$ at point $E$, given that $AB=13$ and $BD=10$, what is the length of $CE$?\n",
    "completion": "\\textbf{Solution:} In the rhombus ABCD, we have OA = OC, OB = OD, and AC $\\perp$ BD, \\\\\n$\\therefore$ OB = $\\frac{1}{2}$BD = 5, $\\angle$AOB = 90$^\\circ$, \\\\\nIn $\\triangle AOB$, AO = $\\sqrt{AB^2−AO^2}$ = $\\sqrt{13^2−5^2}$ = 12, \\\\\n$\\therefore$ AC = 2AO = 24; \\\\\n$\\because$ The area of the rhombus ABCD = $\\frac{1}{2} \\cdot AC \\cdot BD = AB \\cdot CE$, \\\\\n$\\therefore$ $\\frac{1}{2} \\times 24 \\times 10 = 13 \\times CE$, \\\\\n$\\therefore$ CE = $\\boxed{\\frac{120}{13}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52189638_119.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{10}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As illustrated, in the isosceles triangle $ABC$ where $AB=AC$, $AD$ and $BE$ are the altitudes of the isosceles triangle $ABC$. When $DE$ is connected and given $AE=4$ and $CE=1$, what is the length of $DE$?\n",
    "completion": "\\textbf{Solution:} \nGiven that $AE=4$ and $CE=1$, hence $AC=AB=AE+CE=4+1=5$. Since $AD$ and $BE$ are the altitudes of isosceles triangle $ABC$, it follows that $\\angle AEB=\\angle BEC=90^\\circ$ and $BD=CD$. In right triangle $ABE$, we have $BE=\\sqrt{AB^{2}-AE^{2}}=\\sqrt{5^{2}-4^{2}}=3$; In right triangle $BEC$, $BC=\\sqrt{BE^{2}+CE^{2}}=\\sqrt{3^{2}+1^{2}}=\\sqrt{10}$. In right triangle $BEC$, with $DE$ being the median, therefore $DE=\\frac{1}{2}BC=\\frac{\\sqrt{10}}{2}$. Hence, the answer is: $DE=\\boxed{\\frac{\\sqrt{10}}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52756087.png"
  },
  {
    "solution": "\\boxed{\\left(\\frac{93}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the diagram, point A is located at (0, 2), B at (3, 3), and $C(7, -1)$. The quadrilateral OBCD is a rectangle, and BD intersects with the x-axis at point E. If point P is a moving point on line AB, with the area of $\\triangle POB$ denoted as $S_{1}$ and the area of $\\triangle OBE$ denoted as $S_{2}$, and given that $S_{1} = 2S_{2}$, find the coordinates of point P.\n",
    "completion": "Solution: By the problem statement, we have\n\n\\[\\because \\text{the area of } \\triangle OBE \\text{ is } S_{2} = \\frac{1}{2}OE \\cdot yB = \\frac{1}{2} \\times \\frac{24}{7} \\times 3 = \\frac{36}{7},\\]\n\n\\[\\therefore S_{1} = 2S_{2} = \\frac{72}{7}, \\text{draw } BM \\perp y \\text{ axis at point } M,\\]\n\n(1) When point $P$ is on the left side of the y-axis, connect $OP$, construct $PN \\perp y$ axis at point $N$,\n\n\\[\\because S_{\\triangle POB} = S_{\\triangle POA} + S_{\\triangle AOB},\\]\n\n\\[\\therefore S_{\\triangle POA} = S_{\\triangle POB} - S_{\\triangle AOB} = S_{1} - \\frac{1}{2} \\cdot OA \\cdot xB = \\frac{72}{7} - \\frac{1}{2} \\times 2 \\times 3 = \\frac{51}{7},\\]\n\n\\[\\therefore \\frac{1}{2}OA \\cdot |xP| = \\frac{1}{2} \\times 2 \\cdot (-xP) = \\frac{51}{7},\\] solving for $xP = -\\frac{51}{7}$, substitute $x = -\\frac{51}{7}$ into $y = \\frac{1}{3}x + 2$ to get $y = -\\frac{51}{7} \\times \\frac{1}{3} + 2 = -\\frac{3}{7}$,\n\n\\[\\therefore \\text{the coordinates of point } P \\text{ are } \\boxed{\\left(-\\frac{51}{7}, -\\frac{3}{7}\\right)}.\\]\n\n(2) When point $P$ is on the right side of the y-axis, connect $OP$, construct $PN \\perp y$ axis at point $N$,\n\n\\[\\because S_{\\triangle POB} = S_{\\triangle POA} - S_{\\triangle AOB} = \\frac{72}{7},\\]\n\n\\[\\therefore S_{\\triangle POA} = S_{\\triangle POB} + S_{\\triangle AOB} = \\frac{72}{7} + \\frac{1}{2} \\times 2 \\times 3 = \\frac{93}{7},\\]\n\n\\[\\therefore \\frac{1}{2}OA \\cdot xP = \\frac{93}{7},\\] solving for $xP = \\frac{93}{7}$, substitute $x = \\frac{93}{7}$ into $y = \\frac{1}{3}x + 2$ to get $y = \\frac{45}{7}$,\n\n\\[\\therefore \\text{the coordinates of point } P \\text{ are } \\boxed{\\left(\\frac{93}{7}, \\frac{45}{7}\\right)}.\\]\n\nIn summary, the coordinates of point $P$ are \\boxed{\\left(-\\frac{51}{7}, -\\frac{3}{7}\\right)} or \\boxed{\\left(\\frac{93}{7}, \\frac{45}{7}\\right)}.",
    "image_path": "dataset/mm_math/images/MM_Math/52691041_55.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in rhombus ABCD, the diagonals AC and BD intersect at point O, DE $\\perp$ AB at point E, and OE is connected. If DE $= \\sqrt{3}$, and BE $= 1$, what is the degree measure of $\\angle ABD$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rhombus, \\\\\n$\\therefore$ AC $\\perp$ BD, DO=BO, \\\\\nSince DE $\\perp$ AB, DE=$\\sqrt{3}$, BE=1, \\\\\n$\\therefore$ BD=$\\sqrt{DE^2+BE^2}=2$, \\\\\n$\\therefore$ DO=BO=1, \\\\\nBecause DE $\\perp$ BA, DO=BO, \\\\\n$\\therefore$ EO=DO=BO=1, \\\\\n$\\therefore$ BE=BO=EO=1, \\\\\n$\\therefore$ $\\triangle$BEO is an equilateral triangle, \\\\\n$\\therefore$ $\\angle$ABD=$\\boxed{60^\\circ}$,",
    "image_path": "dataset/mm_math/images/MM_Math/53436916.png"
  },
  {
    "solution": "\\boxed{y=4x+4}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, the line $l_{1}: y=-2x+4$ intersects the x-axis at point D and the y-axis at point A, while the line $l_{2}: y=x+1$ intersects the x-axis at point C. The two lines $l_{1}$ and $l_{2}$ intersect at point B. Connect A to C. Find the coordinates of point B and the equation of the line AC.\n",
    "completion": "\\textbf{Solution:} Since lines $l_1$ and $l_2$ intersect at point B, by solving the system of equations $\\begin{cases}y=-2x+4\\\\ y=x+1\\end{cases}$, we obtain $\\begin{cases}x=1\\\\ y=2\\end{cases}$. Therefore, the coordinates of point B are $\\boxed{(1,2)}$. Since line $l_1: y=-2x+4$ intersects the y-axis at point A, by setting $x=0$, we get $y=4$. Therefore, the coordinates of point A are $\\boxed{(0,4)}$. Since line $l_2: y=x+1$ intersects the x-axis at point C, by setting $y=0$, we solve $x+1=0$ to get $x=-1$. Therefore, the coordinates of point C are $\\boxed{(-1,0)}$. Let the equation of line AC be $y=kx+b$. Substituting the coordinates, we have $\\begin{cases}0=-k+b\\\\ b=4\\end{cases}$, which gives us $\\begin{cases}k=4\\\\ b=4\\end{cases}$. Therefore, the equation of line AC is $\\boxed{y=4x+4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51435342_110.png"
  },
  {
    "solution": "\\boxed{\\frac{1 + \\sqrt{5}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, it is known that quadrilateral $ABCD$ is a rectangle, point $E$ is on the extension line of $BA$, and $AE=AD$, $EC$ intersects $AD$, $BD$ at points $F$, $G$ respectively. If $AF=AB$, what is the value of $AD\\colon AB$?\n",
    "completion": "\\textbf{Solution:} Let quadrilateral ABCD be a rectangle.\\\\\nTherefore, DC $\\parallel$ AB,\\\\\nthus $\\angle$DCE = $\\angle$AEC, $\\angle$CDA = $\\angle$EAD\\\\\ntherefore, $\\triangle EAF \\sim \\triangle CDF$\\\\\nhence, $\\frac{AE}{AF} = \\frac{CD}{DF}$\\\\\nLet AB = AF = CD = x, AE = AD = y,\\\\\nthen we have $x^2 - y^2 + xy = 0$\\\\\nDivide both sides of the equation by $x^2$, we have $1 - \\left(\\frac{y}{x}\\right)^2 + \\frac{y}{x} = 0$\\\\\nLet $\\frac{y}{x}$ be t, then we have $t^2 - t - 1 = 0$\\\\\nSolving this, we get $t_1 = \\frac{1 + \\sqrt{5}}{2}$, $t_2 = \\frac{1 - \\sqrt{5}}{2}$ (discard this root)\\\\\nthus, $t = \\frac{y}{x} = \\frac{1 + \\sqrt{5}}{2}$\\\\\nTherefore, $\\frac{AD}{AB} = \\boxed{\\frac{1 + \\sqrt{5}}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51284459.png"
  },
  {
    "solution": "\\boxed{x<2}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, it is known that the line $y=kx$ intersects the line $y=ax+4$ at the point $A(2,1)$. What is the solution set of the inequality $ax+4>kx$?\n",
    "completion": "\\textbf{Solution:} From the graph, we can see that the coordinates of $A$ are $(2,1)$,\\\\\nwhen $x<2$, the line $y=ax+4$ is above the line $y=kx$,\\\\\nmeaning the solution set for the inequality $ax+4>kx$ with respect to $x$ is: $x<2$,\\\\\nhence, the answer is: $\\boxed{x<2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52423346.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in $\\triangle ABC$, $AC = 10$, the perpendicular bisector of $AB$ intersects $AB$ at point $M$ and intersects $AC$ at point $D$. The perimeter of $\\triangle BDC$ is $18$. What is the length of $BC$?\n",
    "completion": "\\textbf{Solution:} Since $MN$ is the perpendicular bisector of $AB$,\\\\\nthen $BD = AD$,\\\\\nthus the perimeter of $\\triangle BDC = BD + CD + BC = AD + CD + BC = AC + BC = 18$,\\\\\nsince $AC = 10$,\\\\\ntherefore $BC = \\boxed{8}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55587989.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, 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 $\\angle AFC=120^\\circ$. Calculate how many degrees is $\\angle BCD$?\n",
    "completion": "\\textbf{Solution}: $\\angle BCD = \\boxed{30^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53066385_61.png"
  },
  {
    "solution": "\\boxed{y=-x+3}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</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, where the coordinate of point B is $(0, 3)$. The line $l_{2}\\colon y=2x$ intersects the line $l_{1}$ at point C, where the x-coordinate of point C is 1. What is the equation of line $l_{1}$?\n",
    "completion": "\\[ \\textbf{Solution:} \\]\nSince point C is on the line \\( l_{2} \\) with an x-coordinate of 1,\\\\\nthus substituting \\( x=1 \\) into \\( y=2x \\) gives \\( y=2\\times 1=2 \\),\\\\\nhence, the coordinates of point C are \\( (1, 2) \\).\\\\\nAssume the equation of line \\( l_{1} \\) is \\( y=kx+b \\). Substituting the coordinates of points B and C, we get \n\\[ \n\\left\\{ \\begin{array}{l}\n3=k\\times 0+b, \\\\ \n2=k\\times 1+b, \n\\end{array} \\right.\n\\]\nfrom which we derive \n\\[\n\\left\\{ \\begin{array}{l}\nk=-1, \\\\ \nb=3.\n\\end{array} \\right.\n\\]\nTherefore, the equation of line \\( l_{1} \\) is \\(\\boxed{y=-x+3}\\).",
    "image_path": "dataset/mm_math/images/MM_Math/51662414_210.png"
  },
  {
    "solution": "\\boxed{\\frac{29}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, a truck is traveling from Place A to Place B, and a sedan is traveling from Place B to Place A. Both vehicles set off from their respective locations along the same road at the same time, moving at constant speeds. It is known that the sedan travels $20\\,km$ per hour faster than the truck. After the vehicles meet, they rest for a while and then continue to travel at the same time. The relationship between the distance $y(km)$ separating the two vehicles and the travel time $x(h)$ of the truck is represented by the line segments $AB-BC-CD-DE$ in the figure. Based on the figure, answer the following questions: What is the function equation for segment $CD$? After how many hours from the truck's departure, is the distance between the truck and the sedan $20\\,km$?\n",
    "completion": "\\textbf{Solution:} Let the $x$-coordinate of point $D$ be $x$. Then we have:\n\\[80(x-1.5)+100(x-1.5)=144\\]\nSolving this gives $x=2.3$. Hence, the coordinates of point D are $(2.3, 144)$. Let the function of line segment $CD$ be $y=kx+b (k\\ne 0)$. Then:\n\\[\n\\left\\{\n\\begin{array}{l}\n1.5k+b=0\\\\\n2.3k+b=144\n\\end{array}\n\\right.\n\\]\nSolving these equations gives:\n\\[\n\\left\\{\n\\begin{array}{l}\nk=180\\\\\nb=-270\n\\end{array}\n\\right.\n\\]\nTherefore, $y=180x-270$. When $180x-270=20$, solving for $x$ gives $x=\\frac{29}{18}$. Let the equation for $AB$ be $y=mx+n (m\\ne 0)$. Then:\n\\[\n\\left\\{\n\\begin{array}{l}\nn=180\\\\\nm+n=0\n\\end{array}\n\\right.\n\\]\nSolving this yields:\n\\[\n\\left\\{\n\\begin{array}{l}\nm=-180\\\\\nn=180\n\\end{array}\n\\right.\n\\]\nTherefore, the equation for line segment $AB$ is: $y=-180x+180$. When $-180x+180=20$, solving for $x$ gives $x=\\frac{8}{9}$. Hence, the truck departs after $\\boxed{\\frac{8}{9}}$ hours or $\\boxed{\\frac{29}{18}}$ hours and will be $20km$ away from the car.",
    "image_path": "dataset/mm_math/images/MM_Math/52381910_139.png"
  },
  {
    "solution": "\\boxed{25}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, $AB$ is the diameter of circle $O$, and $D,E$ are two points on circle $O$ on opposite sides of $AB$. Connect $BD$ and extend it to point $C$ such that $CD=BD$. Connect $AC$ and let it intersect circle $O$ at point $F$. Connect $AE$, $DE$, $DF$. Suppose $DE$ intersects $AB$ at point $G$, and if $AB=10$ and $E$ is the midpoint of arc $AEB$, what is the value of $EG\\cdot ED$?\n",
    "completion": "\\textbf{Solution:} Since $E$ is the midpoint of $\\overset{\\frown}{AEB}$, it follows that $AE=BE$\\\\\nTherefore, $\\angle EAB=\\angle EBA=45^\\circ$\\\\\nAlso, since $\\angle EAB$ and $\\angle EDB$ are the angles subtended by arc $\\widehat{BE}$ on the circle, it follows that $\\angle EAB=\\angle EDB=45^\\circ$\\\\\nThus, $\\angle EDB=\\angle EBA=45^\\circ$\\\\\nAdditionally, since $\\angle DEB=\\angle BEG$\\\\\nIt follows that $\\triangle DEB \\sim \\triangle BEG$\\\\\nTherefore, $\\frac{EG}{EB}=\\frac{EB}{ED}$\\\\\n$EG\\cdot ED=EB^2$\\\\\nThat is, in $\\triangle AEB$, $BE=AB\\cos45^\\circ=5\\sqrt{2}$\\\\\nThus, $EG\\cdot ED=\\boxed{25}$",
    "image_path": "dataset/mm_math/images/MM_Math/53059881_31.png"
  },
  {
    "solution": "\\boxed{y=\\frac{4}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in the rectangle $AOBC$, using the lines through $OB$ and $OA$ as the x-axis and y-axis respectively, a Cartesian coordinate system is established as shown in the figure. The coordinates of point A are $(0, 3)$, and the coordinates of point B are $(4, 0)$. F is a moving point on line segment $BC$ (not coinciding with B or C). The graph of the inverse proportion function $y=\\frac{k}{x} (x > 0)$ passing through point F intersects the edge $AC$ at point E. Lines $OE$ and $OF$ are then drawn, and line EF is constructed. If $CF=2$，find the new equation of the inverse proportion function.\n",
    "completion": "\\textbf{Solution:} Since quadrilateral AOBC is a rectangle, and the coordinates of point A are (0, 3), and the coordinates of point B are (4, 0), \\\\\nthus $AC=OB=4$, and $BC=OA=3$, \\\\\nsince $CF=2$, \\\\\ntherefore, the coordinates of F are (4, 1), \\\\\nSubstituting the coordinates of F (4, 1) into $y=\\frac{k}{x}$ gives $1=\\frac{k}{4}$, \\\\\nSolving this 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/51446287_230.png"
  },
  {
    "solution": "\\boxed{15}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: Given: Line segment $AB = 6$, point $C$ is the midpoint of line segment $AB$, extend line segment $AB$ to $D$, such that $BD = 3BC$. Find the length of line segment $AD$?\n",
    "completion": "\\textbf{Solution:} Since point C is the midpoint of line segment AB,\\\\\ntherefore $BC=\\frac{1}{2}AB$,\\\\\nsince AB = 6,\\\\\ntherefore BC = 3,\\\\\nsince BD= 3BC,\\\\\ntherefore BD= 9,\\\\\ntherefore AD=AB+BD=6+9=\\boxed{15}",
    "image_path": "dataset/mm_math/images/MM_Math/51379764.png"
  },
  {
    "solution": "\\boxed{71^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, point $D$ is on $AB$. When folded along $CD$, point $A$ falls on point $E$ on side $BC$. If $\\angle B=26^\\circ$, what is the measure of $\\angle CDE$?\n",
    "completion": "\\textbf{Solution:} Given that $\\angle ACB=90^\\circ$ and $\\angle B=26^\\circ$, \\\\\nthus $\\angle A=90^\\circ-26^\\circ=64^\\circ$. \\\\\nBy the fold, $\\angle CDE=\\angle ADC$ and $\\angle ACD=\\angle BCD=45^\\circ$, \\\\\ntherefore $\\angle ADC=180^\\circ-45^\\circ-64^\\circ=71^\\circ$, \\\\\nthus $\\angle CDE=\\angle ADC=\\boxed{71^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52634675.png"
  },
  {
    "solution": "\\boxed{62^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, point $A$, $O$, and $B$ are on the same line, and $OC$ bisects $\\angle AOB$. If $\\angle COD = 28^\\circ$, what is the measure of $\\angle BOD$?\n",
    "completion": "\\textbf{Solution:} Since points A, O, and B are on the same line, and OC bisects $\\angle AOB$,\\\\\ntherefore $\\angle AOC = \\angle BOC = 90^\\circ$,\\\\\nthus $\\angle BOD = \\angle BOC - \\angle COD = 90^\\circ - 28^\\circ = \\boxed{62^\\circ}$;",
    "image_path": "dataset/mm_math/images/MM_Math/53224379_69.png"
  },
  {
    "solution": "\\boxed{25}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in the Cartesian coordinate system, the vertices $A$ and $B$ of the isosceles $\\triangle ABC$ are located on the $x$-axis and $y$-axis, respectively. The graph of the inverse proportion function $y= \\frac{k}{x}$ ($k>0$, $x>0$) passes through the right-angle vertex $C$. If $OA=7$ and $OB=3$, what is the value of $k$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, construct $CE\\perp y$-axis, $CD\\perp x$-axis. Since $\\angle O=90^\\circ$, it follows that quadrilateral $ODCE$ is a rectangle. Therefore, $\\angle ECD=90^\\circ$. Since $\\angle BCA=90^\\circ$, it follows that $\\angle ECB=\\angle DCA$. Therefore, in $\\triangle CEB$ and $\\triangle CDA$, we have $\\left\\{\\begin{array}{l} \\angle CEB=\\angle CDA \\\\ \\angle ECB=\\angle DCA \\\\ CB=CA \\end{array}\\right.$ Therefore, $\\triangle CEB\\cong \\triangle CDA\\left(AAS\\right)$, which implies $CE=CD$, $EB=DA$. Consequently, quadrilateral $ODCE$ is a square. Let $EB=DA=x$, so $OD=OA-DA=7-x$, $OE=OB+BE=3+x$. Since $OE=OD$, we get $7-x=3+x$, which is solved to find $x=2$. Therefore, $OD=7-2=5$, $OE=3+2=5$. Hence, the coordinates of point C are $(5,5)$. Substituting $C(5,5)$ into $y=\\frac{k}{x}$ gives $5=\\frac{k}{5}$. Solving this yields $k=\\boxed{25}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50497050.png"
  },
  {
    "solution": "\\boxed{y=\\frac{1}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</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 the y-axis at points A and B, respectively. Point C is on the positive half of the x-axis, with $AB=AC$. Construct line $BC$. $AD$ is the median of $\\triangle ABC$, intersecting the y-axis at point F. What is the equation of line $AD$?\n",
    "completion": "\\textbf{Solution:} Given that $AB=AC=10$, $OA=6$,\\\\\nhence $OC=AC-OA=4$,\\\\\nsince $AD$ is the median of $\\triangle ABC$,\\\\\nit follows that $BD=CD$, and $AD\\perp BC$,\\\\\nLet the midpoint of $OC$ be $Q$, and connect $DQ$, hence $DQ$ is the median of $\\triangle OBC$,\\\\\nthus $DQ=\\frac{1}{2}OB=4$, $OQ=\\frac{1}{2}OC=2$,\\\\\nhence $D(2, 4)$, $Q(2, 0)$,\\\\\nLet the equation of line $AD$ be $y=kx+b$, substituting $A(-6, 0), D(2, 4)$ into it yields:\\\\\n$\\begin{cases}\n-6k+b=0\\\\\n2k+b=4\n\\end{cases}$, solving gives $\\begin{cases}\nk=\\frac{1}{2}\\\\\nb=3\n\\end{cases}$,\\\\\nthus $\\boxed{y=\\frac{1}{2}x+3}$",
    "image_path": "dataset/mm_math/images/MM_Math/53728859_92.png"
  },
  {
    "solution": "\\boxed{50^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, $AB$ is the diameter of $\\odot O$, point $D$ is on the extension of $AB$, and $C$ is a point on $\\odot O$. Given that $\\angle A=25^\\circ$ and $\\angle D=40^\\circ$, what is the degree measure of $\\angle DOC$?\n",
    "completion": "\\textbf{Solution:} Given $OA = OC$, $\\angle A = 25^\\circ$, \\\\\n$\\therefore \\angle A = \\angle OCA = 25^\\circ$, \\\\\n$\\therefore \\angle DOC = 2\\angle A = \\boxed{50^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53110813_69.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{2}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, $\\triangle ADE\\sim \\triangle ABC$, and $\\frac{AC}{AB}=\\frac{2}{3}$. Point D is inside $\\triangle ABC$, and lines $BD$, $CD$, $CE$ are connected. $\\triangle ABD\\sim \\triangle ACE$. If $CD=CE$, $BD=3$, and $\\angle ABD + \\angle ACD = 90^\\circ$, what is the length of $DE$?\n",
    "completion": "\\textbf{Solution:} Given $\\triangle ABD\\sim \\triangle ACE$, \\\\\n$\\therefore$ $\\frac{AC}{AB}=\\frac{CE}{BD}=\\frac{2}{3}$, $\\angle ABD=\\angle ACE$, \\\\\nAlso given $BD=3$, \\\\\n$\\therefore$ $CE=2$, \\\\\n$\\therefore$ $CD=CE=2$, \\\\\nGiven $\\angle ABD+\\angle ACD=90^\\circ$, \\\\\n$\\therefore$ $\\angle ACD+\\angle ACE=90^\\circ$, \\\\\n$\\therefore$ $\\angle DCE=90^\\circ$, \\\\\n$\\therefore$ $DE=\\sqrt{2}CD=2\\sqrt{2}$. \\\\\n\nHence, the length of $DE$ is $\\boxed{2\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52917339_76.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, $\\angle AOE = 100^\\circ$, $\\angle BOF = 80^\\circ$, OE bisects $\\angle BOC$, and OF bisects $\\angle AOC$. What is the measure of $\\angle EOF$?\n",
    "completion": "\\textbf{Solution:} \n\nSince $OE$ bisects $\\angle BOC$, and $OF$ bisects $\\angle AOC$\\\\\nTherefore, $\\angle AOB=2\\angle EOF$\\\\\nAlso, since $\\angle AOB=\\angle BOF+\\angle AOE-\\angle EOF=80^\\circ+100^\\circ-\\angle EOF=180^\\circ-\\angle EOF$\\\\\nTherefore, $2\\angle EOF=180^\\circ-\\angle EOF$\\\\\nSolving this gives $\\angle EOF=\\boxed{60^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/51495482.png"
  },
  {
    "solution": "\\boxed{\\frac{10}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</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. As illustrated in figure 2, when line AB is translated m units to the right along the x-axis, it intersects the graph of the inverse proportion function $y=\\frac{2}{x}$ $(x>0)$ at points P and Q. By connecting AP and BQ, if the area of the quadrilateral ABQP is exactly equal to $m^{2}$, find the value of m.\n",
    "completion": "\\textbf{Solution:} Let the equation of the line after translation be $y_{PQ}=-x+3+m$. Draw $DF\\perp PQ$ at point F from D, thus $ED=m$ and $DF= \\frac{\\sqrt{2}m}{2}$.\\\\\nThe area of $\\triangle ABPQ$ is $\\frac{(AB+PQ)\\cdot \\frac{\\sqrt{2}m}{2}}{2}=\\frac{\\sqrt{2}m(\\sqrt{2}+PQ)}{4}=m^{2}$.\\\\\nTherefore, $\\sqrt{2}+PQ=2\\sqrt{2}m$,\\\\\nHence, $PQ= 2\\sqrt{2}m - \\sqrt{2}$.\\\\\nAccording to the problem, $\\begin{cases}y=-x+3+m\\\\ y=\\frac{x}{2}\\end{cases}$.\\\\\nSolving, we get $x_{1}=\\frac{m+3+\\sqrt{m^{2}+6m+1}}{2}$ and $x_{2}=\\frac{m+3-\\sqrt{m^{2}+6m+1}}{2}$,\\\\\nThus, $QH=x_{1}-x_{2}= \\sqrt{m^{2}+6m+1}$,\\\\\nTherefore, $PQ=\\sqrt{2}\\sqrt{m^{2}+6m+1}$,\\\\\nTherefore, $\\sqrt{2}\\sqrt{m^{2}+6m+1} = 2\\sqrt{2}m - \\sqrt{2}$.\\\\\nTherefore, $m^{2}+6m+1=4m^{2}-4m+1$,\\\\\nSolving we get $m_{1}=0$ (discarded), $m_{2}=\\frac{10}{3}$,\\\\\nthus $m=\\boxed{\\frac{10}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51407111_117.png"
  },
  {
    "solution": "\\boxed{(4, 5)}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</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 $B(3,0)$. Point $P$ is a point on the parabola. If the area of $\\triangle PAB$ is $10$, find the coordinates of point $P$.\n",
    "completion": "\\textbf{Solution:} Given $A(-1,0)$ and $B(3,0)$,\\\\\nthus $AB=4$.\\\\\nLet $P(x,y)$, then the area $S_{\\triangle PAB}=\\frac{1}{2}AB|y|=10$,\\\\\nthus $|y|=5$,\\\\\ntherefore $y=\\pm 5$.\\\\\n(1) When $y=5$,\\\\\n$x^{2}-2x-3=5$,\\\\\nsolving this, we find: $x_{1}=-2$, $x_{2}=4$,\\\\\nthus, the coordinates of point $P$ are $(-2,5)$ or $(4, 5)$;\\\\\n(2) When $y=-5$,\\\\\n$x^{2}-2x-3=-5$,\\\\\nthis equation has no solution;\\\\\nIn conclusion, the coordinates of point $P$ are $\\boxed{(-2,5)}$ or $\\boxed{(4, 5)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52978190_215.png"
  },
  {
    "solution": "\\boxed{70^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the diagram, point $O$ is on line $AB$, and ray $OD$ is the bisector of $\\angle AOC$. If $\\angle COB = 40^\\circ$, what is the degree measure of $\\angle DOC$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle COB=40^\\circ$, \\\\\nit follows that $\\angle AOC=180^\\circ-\\angle COB=140^\\circ$, \\\\\nbecause OD is the angle bisector of $\\angle AOC$, \\\\\ntherefore $\\angle DOC= \\frac{1}{2}\\angle AOC= \\frac{1}{2}\\times140^\\circ=\\boxed{70^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53078554.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, the line segment $AB=8$, point C is the midpoint of line segment AB, and point D is the midpoint of line segment BC. Find the length of line segment AD.\n",
    "completion": "\\textbf{Solution:} Since $AB=8$ and C is the midpoint of AB,\\\\\nit follows that $AC=BC=4$.\\\\\nSince D is the midpoint of BC,\\\\\nit follows that $CD=DB=\\frac{1}{2}BC=2$.\\\\\nTherefore, $AD=AC+CD=4+2=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50733799_145.png"
  },
  {
    "solution": "\\boxed{8}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in $ \\triangle ABC$, $ DE\\parallel BC$, $ AD=9$, $ DB=3$, $ CE=2$, what is the length of $ AC$?\n",
    "completion": "\\textbf{Solution:} Since DE $\\parallel$ BC, with AD = 9, DB = 3, CE = 2,\\\\\nit follows that $\\frac{AD}{DB} = \\frac{AE}{EC}$, i.e., $\\frac{9}{3} = \\frac{AE}{2}$,\\\\\nSolving this, we find AE = 6,\\\\\nTherefore, AC = AE + EC = \\boxed{8}",
    "image_path": "dataset/mm_math/images/MM_Math/51458167.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, point C is a point on line segment AB, M and N are the midpoints of AB and CB respectively, with $AC=10\\,cm$ and $NB=6\\,cm$. Find the length of MN.\n",
    "completion": "\\textbf{Solution:} From (1) we have $BM=11\\,\\text{cm}$,\n\n$\\because NB=6\\,\\text{cm}$,\n\n$\\therefore MN=BM-NB=11-6=\\boxed{5}\\,\\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51494975_206.png"
  },
  {
    "solution": "\\boxed{1}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, the diagonals $AC$ and $BD$ of rectangle $ABCD$ intersect at point $O$, and $\\angle AOB = 60^\\circ$, $AC = 2$. What is the length of side $AB$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rectangle and $AC=2$, \\\\\nit follows that $AC=BD=2$, and $OA=\\frac{1}{2}AC=\\frac{1}{2}BD=OB=1$, \\\\\nbecause $\\angle AOB = 60^\\circ$, \\\\\ntherefore $\\triangle AOB$ is an equilateral triangle, \\\\\nhence $AB=OA=OB=\\boxed{1}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53599249.png"
  },
  {
    "solution": "\\boxed{118^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, $\\angle BAC=33^\\circ$, points $D$ and $E$ are on sides $AB$ and $AC$ respectively, connect $DE$. When $\\angle A$ is folded along $DE$, the corresponding point of point $A$ is $A'$. If $\\angle 1 + \\angle 2 = 170^\\circ$, what is the measure of $\\angle 2$?\n",
    "completion": "\\textbf{Solution:} Let point O be the intersection of AB and $AE$,\\\\\nfrom the property of folding, we have: $\\angle A = \\angle BAC = 33^\\circ$,\\\\\nsince $\\angle 2 = \\angle BAC + \\angle AOE$, and $\\angle AOE = \\angle 1 + \\angle A$,\\\\\nthus $\\angle 2 = \\angle BAC + \\angle 1 + \\angle A = \\angle 1 + 66^\\circ$, that is $\\angle 1 = \\angle 2 - 66^\\circ$,\\\\\nsince $\\angle 1 + \\angle 2 = 170^\\circ$,\\\\\ntherefore $\\angle 2 = \\boxed{118^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51378984.png"
  },
  {
    "solution": "\\boxed{54^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, line 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 $\\angle AOD=2\\angle BOD$ and $\\angle BOC=3\\angle AOC$, what is the measure of $\\angle BON$?\n",
    "completion": "\\textbf{Solution:} Let $\\angle AOC=x^\\circ$, then $\\angle BOC=3x^\\circ$,\\\\\nsince $\\angle COD=\\angle MON=70^\\circ$,\\\\\nthus $\\angle BOD=\\angle BOC-\\angle COD= 3x^\\circ- 70^\\circ$.\\\\\nTherefore, $\\angle AOD=\\angle AOC+\\angle COD=x^\\circ+70^\\circ$.\\\\\nSince $\\angle AOD=2\\angle BOD$,\\\\\ntherefore $x+70= 2(3x-70)$,\\\\\nsolving this, we get $x=42$,\\\\\nthus $\\angle BOD=3x^\\circ- 70^\\circ=3\\times42^\\circ- 70^\\circ=56^\\circ$,\\\\\ntherefore, $\\angle BON= 180^\\circ-\\angle MON-\\angle BOD= 180^\\circ-70^\\circ-56^\\circ=\\boxed{54^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/54927588_140.png"
  },
  {
    "solution": "\\boxed{65^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, the lines $AB$ and $CD$ intersect at point $O$, with $OE$ bisecting $\\angle AOD$ and $\\angle FOC = 90^\\circ$, $\\angle 1 = 40^\\circ$. Find the measure of $\\angle 2$.\n",
    "completion": "\\textbf{Solution:} Given that $ \\because \\angle FOC=90^\\circ$ and $ \\angle 1=40^\\circ$,\\\\\n$ \\therefore \\angle AOC=180^\\circ-\\angle FOC-\\angle 1=180^\\circ-90^\\circ-40^\\circ=50^\\circ$,\\\\\n$ \\therefore \\angle AOD=180^\\circ-\\angle AOC=180^\\circ-50^\\circ=130^\\circ$,\\\\\n$ \\because OE$ bisects $\\angle AOD$,\\\\\n$ \\therefore \\angle 2=\\frac{1}{2}\\angle AOD=\\frac{1}{2}\\times 130^\\circ=\\boxed{65^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53119496.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, it is known that the area of the rhombus $ABCD$ is 24 and the area of the square $AECF$ is 18. What is the side length of the rhombus?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, connect AC and BD, intersecting at point O, \\\\\n$\\because$ the area of square AECF is 18, \\\\\n$\\therefore$ AC $= \\sqrt{18} \\times \\sqrt{2} = 6$, \\\\\n$\\therefore$ AO $= 3$, \\\\\n$\\because$ the area of diamond ABCD is 24, \\\\\n$\\therefore$ BD $= \\frac{24 \\times 2}{6} = 8$, \\\\\n$\\therefore$ BO $= 4$, \\\\\n$\\therefore$ in $\\triangle$AOB, $AB = \\sqrt{AO^{2} + BO^{2}} = \\sqrt{3^{2} + 4^{2}} = \\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52691062.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, $\\triangle ABC \\cong \\triangle EDC$, where point D is on $AB$, and $\\angle B=60^\\circ$, then what is the degree measure of $\\angle EDA$?\n",
    "completion": "\\textbf{Solution:} Since $\\triangle ABC \\cong \\triangle EDC$,\\\\\nit follows that $\\angle B = \\angle EDC = 60^\\circ$, $CD = CB$,\\\\\nhence $\\angle CDB = \\angle B = 60^\\circ$,\\\\\ntherefore $\\angle EDA = 180^\\circ - \\angle EDC - \\angle CDB = \\boxed{60^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52905808.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in $\\triangle ABC$ and $\\triangle ADC$, $\\angle B=\\angle D=90^\\circ$, $CB=CD$, $\\angle 1=30^\\circ$, what is the value of $\\angle 2$?\n",
    "completion": "\\textbf{Solution}: Since $\\angle B=\\angle D=90^\\circ$\\\\\nTherefore, both $\\triangle ABC$ and $\\triangle ADC$ are right-angled triangles. \\\\\nIn $\\triangle ABC$ and $\\triangle ADC$: \\\\\nSince $\\left\\{\\begin{array}{l}CB=CD\\\\ AC=AC\\end{array}\\right.$\\\\\nTherefore, $\\triangle ABC\\cong \\triangle ADC$ (HL criterion)\\\\\nTherefore, $\\angle 1 = \\angle CAD$\\\\\nSince $\\angle 2 + \\angle CAD + \\angle D = 180^\\circ$\\\\\nTherefore, $\\angle 2 = 180^\\circ - 90^\\circ - 30^\\circ = \\boxed{60^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/52096450.png"
  },
  {
    "solution": "\\boxed{4\\sqrt{5}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</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 $2\\sqrt{3}$. Find the perimeter of the rhombus ABCD.\n",
    "completion": "\\textbf{Solution:} \\\\\nSince quadrilateral ABCD is a rhombus, \\\\\nit follows that $AC\\perp BD$, $AO=CO$, $BO=DO$\\\\\nSince $AC=2\\sqrt{3}$, $BD=2\\sqrt{2}$ \\\\\nit follows that $AO=\\sqrt{3}$, $BO=\\sqrt{2}$\\\\\nTherefore, $AB=\\sqrt{BO^2+AO^2}=\\sqrt{5}$\\\\\nThus, the perimeter of the rhombus ABCD is $4\\sqrt{5}=\\boxed{4\\sqrt{5}}$",
    "image_path": "dataset/mm_math/images/MM_Math/51510287_138.png"
  },
  {
    "solution": "\\boxed{\\frac{1}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</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 $D$. What is the value of $\\tan(\\angle ABD)$?\n",
    "completion": "\\textbf{Solution:} Connect BD and AD, \n\n\\[\n\\because y=x^{2}-2x=(x-1)^{2}-1,\n\\]\n\n\\[\n\\therefore \\text{the coordinates of point } D \\text{ are } D(1, -1),\n\\]\n\n\\[\n\\because A(2,0), B(-1, 3),\n\\]\n\n\\[\n\\therefore AB^{2}=(-1-2)^{2}+3^{2}=18, AD^{2}=(2-1)^{2}+(-1)^{2}=2, BD^{2}=(-1-1)^{2}+(-1-3)^{2}=20,\n\\]\n\n\\[\n\\therefore AB^{2}+AD^{2}=BD^{2},\n\\]\n\n\\[\n\\therefore \\triangle ABD \\text{ is a right-angled triangle},\n\\]\n\n\\[\n\\therefore \\tan\\angle ABD=\\frac{AD}{AB}=\\frac{\\sqrt{2}}{\\sqrt{18}}=\\boxed{\\frac{1}{3}}\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/51403345_146.png"
  },
  {
    "solution": "\\boxed{x>-3}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, it is known that the graphs of the functions $y=ax+b$ and $y=kx$ intersect at point $P$. According to the graph, what is the solution set of the linear inequality in one variable $ax+b>kx$ with respect to $x$?\n",
    "completion": "\\textbf{Solution:} Since the graph of the function indicates that when $x>-3$, the linear function $y=ax+b$ is above the graph of the linear function $y=kx$, \\\\\ntherefore, the solution to the inequality in $x$, $ax+b>kx$, is $\\boxed{x>-3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51663047.png"
  },
  {
    "solution": "\\boxed{\\frac{5}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, point D is on side $AC$ and $AD=BD$, $M$ is the midpoint of $BD$. If $AC=8$ and $BC=4$, what is the length of $CM$?\n",
    "completion": "\\textbf{Solution:} Since $\\angle ACB=90^\\circ$ and M is the midpoint of BD,\\\\\n$\\therefore CM=\\frac{1}{2}BD$,\\\\\nLet $CM=x$, then $BD=AD=2x$,\\\\\nSince $AC=8$,\\\\\n$\\therefore CD=AC-AD=8-2x$,\\\\\nIn $\\triangle BCD$, according to the Pythagorean theorem,\\\\\n$BC^2+CD^2=BD^2$,\\\\\ni.e., $4^2+(8-2x)^2=(2x)^2$,\\\\\nSolving for $x$, we get $x=\\boxed{\\frac{5}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52699750.png"
  },
  {
    "solution": "\\boxed{(2,0)}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the Cartesian coordinate system, a circular arc passes through the lattice points A, B, and C. What are the coordinates of the center of the circle?\n",
    "completion": "\\textbf{Solution:} Based on the corollary of the Perpendicular Diameter Theorem: the perpendicular bisector of a chord always passes through the center of the circle,\\\\\none can construct the perpendicular bisectors of chords AB and BC, and their intersection point will be the center of the circle.\\\\\nAs shown in the figure, thus the center of the circle is $\\boxed{(2,0)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52536782.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: Given: In $\\triangle ABC$, $CD \\perp AB$, $\\sin A = \\frac{4}{5}$, $CD = 4$, $AB = 5$. Find the length of $AD$ and the value of $\\tan B$.\n",
    "completion": "\\textbf{Solution:}\n\nSince $CD \\perp AB$,\\\\\nit follows that $\\angle CDA = 90^\\circ$\\\\\nSince $\\sin A = \\frac{CD}{AC} = \\frac{4}{5}$, and $CD = 4$,\\\\\nit follows that $AC = 5$ \\\\\nBy the Pythagorean theorem,\\\\\n$AD = \\sqrt{AC^{2} - CD^{2}} = \\boxed{3}$\\\\\n$BD = AB - AD = 4$\\\\\nTherefore, $\\tan B = \\frac{CD}{BD} = \\boxed{2}$",
    "image_path": "dataset/mm_math/images/MM_Math/53019955.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, $AB$ is the diameter of $\\odot O$, with chord $CD\\perp AB$ at $H$, $\\angle A=30^\\circ$, $OH=1$, then what is the radius of $\\odot O$?\n",
    "completion": "\\textbf{Solution:} Draw $OC$, \\\\\n$\\because$ $\\angle A=30^\\circ$, \\\\\n$\\therefore$ $\\angle COH=2\\angle A=60^\\circ$, \\\\\n$\\because$ chord $CD \\perp AB$ at $H$, \\\\\n$\\therefore$ $\\angle OHC=90^\\circ$, \\\\\n$\\therefore$ $\\angle OCH=30^\\circ$, \\\\\n$\\because$ $OH=1$, \\\\\n$\\therefore$ $OC=2OH=\\boxed{2}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51433156.png"
  },
  {
    "solution": "\\boxed{\\frac{1}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</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, and $DE \\parallel BC$, $EF \\parallel AB$, and $AD = 2BD$. What is the value of $\\frac{CF}{CB}$?\n",
    "completion": "\\textbf{Solution:} Given that $AD=2BD$,\\\\\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/52977388.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{19}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, $\\triangle ABC$ is inscribed in circle $\\odot O$, with chords AE and BC intersecting at point D. BO is connected, and $\\angle CAE = \\angle ABO$. Through point O, construct $OH \\perp BC$ at point H, and extend HO to intersect AB at point P. If $HO = 1$ and $AP = 6$, find the length of the radius of circle $\\odot O$.\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw $OF$ perpendicular to $AB$ at $F$, \\\\\n$\\therefore BF = AF$, \\\\\nLet $FP = x$, \\\\\n$\\therefore BF = AF = AP + PF = 6 + x$, \\\\\n$\\therefore BP = BF + PF = 6 + 2x$, \\\\\nSince $\\angle ABC = 30^\\circ$, and $PH \\perp BC$, \\\\\n$\\therefore \\angle BPH = 60^\\circ$, and $BP = 2PH$, \\\\\nAlso, since $\\angle OF \\perp AB$, \\\\\n$\\therefore \\angle OFP = 90^\\circ$, \\\\\n$\\therefore \\angle POF = 30^\\circ$, \\\\\n$\\therefore OP = 2FP = 2x$, \\\\\n$\\therefore PH = OP + OH = 1 + 2x$, \\\\\n$\\therefore BP = 2 + 4x$, \\\\\n$\\therefore 2 + 4x = 6 + 2x$, \\\\\nSolving gives $x = 2$, \\\\\n$\\therefore PF = 2$, $BF = 8$, $PO = 4$, \\\\\n$\\therefore OF = \\sqrt{PO^{2} - PF^{2}} = 2\\sqrt{3}$, \\\\\n$\\therefore OB = \\sqrt{OF^{2} + BF^{2}} = \\boxed{2\\sqrt{19}}$, \\\\\n$\\therefore$ The radius of circle $O$ is $2\\sqrt{19}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51318321_118.png"
  },
  {
    "solution": "\\boxed{24\\, \\text{cm}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: Given that point \\(D\\) is the midpoint of line segment \\(AB\\), and point \\(C\\) lies on line segment \\(AB\\). As shown in Figure 2, if \\(BC = 2CD\\), and point \\(E\\) is the midpoint of \\(BD\\), with \\(AE = 18cm\\), find the length of line segment \\(AB\\).\n",
    "completion": "\\textbf{Solution:} Let $CD=x\\, \\text{cm}$.\\\\\nSince $BC=2CD$,\\\\\nit follows that $BC=2x\\, \\text{cm}$.\\\\\nSince $BD=CD+BC$,\\\\\nit follows that $BD=x+2x=3x\\, \\text{cm}$.\\\\\nSince $E$ is the midpoint of $BD$,\\\\\nit follows that $DE=\\frac{1}{2}BD=\\frac{3}{2}x\\, \\text{cm}$.\\\\\nMoreover, since $D$ is the midpoint of $AB$,\\\\\nit follows that $AD=BD=3x\\, \\text{cm}$.\\\\\nSince $AE=AD+DE$,\\\\\nit follows that $AE=3x+\\frac{3}{2}x=\\frac{9}{2}x\\, \\text{cm}$.\\\\\nSince $AE=18\\, \\text{cm}$,\\\\\nit follows that $\\frac{9}{2}x=18$, $x=4$,\\\\\nthus $AB=2BD=6x=24\\, \\text{cm}$,\\\\\ntherefore, the length of segment $AB$ is $\\boxed{24\\, \\text{cm}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51000645_152.png"
  },
  {
    "solution": "\\boxed{2\\sqrt{13}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, the diagonals $AC$ and $BD$ of the quadrilateral $\\square ABCD$ intersect at point $Q$, with $E$ and $F$ being the midpoints of $OB$ and $OD$, respectively. Lines $AE$, $AF$, $CE$, and $CF$ are connected. Quadrilateral $AECF$ is a parallelogram. If $AB \\perp AC$ and $AB=3$, $BC=5$, what is the length of $BD$?\n",
    "completion": "\\textbf{Solution:} Given that $AB \\perp AC$, it follows that $\\angle BAC = 90^\\circ$. In right triangle $\\triangle ABC$, we have $AC = \\sqrt{BC^2 - AB^2} = \\sqrt{5^2 - 3^2} = 4$. Thus, $OA = \\frac{1}{2}AC = 2$. In right triangle $\\triangle AOB$, $BO = \\sqrt{AB^2 + OA^2} = \\sqrt{3^2 + 2^2} = \\sqrt{13}$, hence $BD = 2BO = 2\\sqrt{13}$. Therefore, the length of $BD$ is $\\boxed{2\\sqrt{13}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51825026_80.png"
  },
  {
    "solution": "\\boxed{y=-\\frac{6}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, the line $y = -x + 1$ intersects with the graph of the inverse proportion function $y = \\frac{k}{x}$, where the $x$-coordinate of one of the intersection points is $-2$. What is the expression of the inverse proportion function?\n",
    "completion": "\\textbf{Solution:} Substituting $x=-2$ into $y=-x+1=3$ gives us one of the points of intersection as $\\left(-2,3\\right)$.\\\\\nSubstituting $\\left(-2,3\\right)$ into the expression for the inverse proportionality function and solving yields: $k=-2\\times 3=-6$,\\\\\n$\\therefore$ the expression for the inverse proportionality function is: $\\boxed{y=-\\frac{6}{x}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52091246_97.png"
  },
  {
    "solution": "\\boxed{(-1, 2)}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure is the graph of the function $y_{1}=|x|$. It is known that the graph of the function $y_{2}=\\frac{1}{3}x+\\frac{4}{3}$ intersects with the graph of $y_{1}=|x|$ at points A and B, and $A(-1, 1)$, then what is the range of values of $x$ for which $y_{2}>y_{1}$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure,\\\\\nby solving the system of equations $\\begin{cases}y=x \\\\ y=\\frac{1}{3}x+\\frac{4}{3}\\end{cases}$, we get: $\\begin{cases}x=2 \\\\ y=2\\end{cases}$,\\\\\n$\\therefore$ Point B is (2, 2),\\\\\nFrom the graph, it is found that when $y_{2}>y_{1}$, $-1<x<2$,\\\\\nthus, the answer is: $x \\in \\boxed{(-1, 2)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52381528.png"
  },
  {
    "solution": "\\boxed{5\\sqrt{2}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: In $\\triangle ABC$, where $BC=2$ and $AC=5$, a perpendicular line is drawn from side $AB$ to meet $AC$ at point $F$ and the extended line of $BC$ at point $E$. Connect $AE$. If it happens that $EF=AB$, what is the length of $AE$?\n",
    "completion": "\\textbf{Solution:}\n\nGiven that $AB\\perp DE$,\n\nIt follows that $\\angle BDE=\\angle ACB=90^\\circ$,\n\nThus $\\angle B+\\angle BAC=\\angle B+\\angle BED=90^\\circ$,\n\nTherefore, $\\angle BAC=\\angle CEF$,\n\nSince $\\angle ACB=\\angle ECF=90^\\circ$ and $AB=EF$,\n\nIt means that $\\triangle ABC\\cong \\triangle EFC$ (AAS),\n\nHence, $CE=AC=5$,\n\nIn $\\triangle ACE$,\n\nIt results in $AE=\\sqrt{AC^2+CE^2}=\\sqrt{5^2+5^2}=\\boxed{5\\sqrt{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52457337.png"
  },
  {
    "solution": "\\boxed{y=-2x+4}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, the graph of the linear function $y=ax+b$ passes through the points $(1,2)$ and $(-1,6)$. Question: What is the analytic expression of this linear function?\n",
    "completion": "\\textbf{Solution:} According to the problem, \\\\\nwhen $x=1$, $y=2$; \\\\\nwhen $x=-1$, $y=6$. \\\\\nTherefore, $\\begin{cases}2=a+b\\\\ 6=-a+b\\end{cases}$ \\\\\nSolving this system, we get $\\begin{cases}a=-2\\\\ b=4\\end{cases}$ \\\\\n$\\therefore$ the linear equation is: $y=-2x+4$. \\\\\nHence, the answer is $\\boxed{y=-2x+4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52344480_213.png"
  },
  {
    "solution": "\\boxed{-2}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, the line $y_1 = x + 2$ intersects the x and y axes at points A and B, respectively. The line $y_2 = kx + 4$ intersects the x and y axes at points C and D, respectively. The two lines intersect at point $E\\left(\\frac{2}{3}, n\\right)$. What is the value of $k$?\n",
    "completion": "\\textbf{Solution:} Because the line $y_{1}=x+2$ intersects with the line $y_{2}=kx+4$ at point $E\\left(\\frac{2}{3}, n\\right)$,\\\\\ntherefore, when $x= \\frac{2}{3}$, $n= \\frac{2}{3} +2= \\frac{8}{3}$,\\\\\nthus, the coordinates of point E are $\\left( \\frac{2}{3} , \\frac{8}{3}\\right)$. Substituting point E$\\left( \\frac{2}{3} , \\frac{8}{3}\\right)$ into $y_{2}=kx+4$, we get $ \\frac{8}{3} = \\frac{2}{3}k+4$. Solving this, we find $k=\\boxed{-2}$,\\\\\ntherefore, $y_{2}=-2x+4$, which means the value of $k$ is $-2$.",
    "image_path": "dataset/mm_math/images/MM_Math/52381906_141.png"
  },
  {
    "solution": "\\boxed{}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AD=DC$. If $\\angle BAC=78^\\circ$, what is the degree measure of $\\angle C$?\n",
    "completion": "\\textbf{Solution:} Since the given information is insufficient, it is impossible to provide a solution.",
    "image_path": "dataset/mm_math/images/MM_Math/53019706_106.png"
  },
  {
    "solution": "\\boxed{m\\le 0}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: Given that 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 $y=ax^2-4ax+2\\ (a\\neq 0)$. Point $B(x_2, y_2)$ is the intersection of the parabola and line $l$ in the first quadrant. If $y_1 \\leq y_2$, directly write out the range of values for $m$.\n",
    "completion": "\\textbf{Solution}: We have $\\boxed{m\\le 0}$ or $m\\ge 4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53147494_61.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle ACB=90^\\circ$, $\\angle B=15^\\circ$. Point D is the midpoint of AB, and $DE \\perp AB$ intersects BC at point E. Given $BE=8\\text{cm}$, what is the length of $AC$ in cm?\n",
    "completion": "\\textbf{Solution:} Since point D is the midpoint of AB, and DE is perpendicular to AB, intersecting BC at point E, \\\\\ntherefore DE bisects AB perpendicularly, \\\\\nthus AE=BE=8, \\\\\ntherefore $\\angle B=\\angle BAE=15^\\circ$, \\\\\nsince $\\angle AEC=\\angle B+\\angle BAE$, \\\\\ntherefore $\\angle AEC=30^\\circ$, \\\\\nsince $\\angle ACB=90^\\circ$, and AE=8, \\\\\ntherefore AC=\\boxed{4}.",
    "image_path": "dataset/mm_math/images/MM_Math/52653587.png"
  },
  {
    "solution": "\\boxed{5}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in rhombus $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$. $H$ is the midpoint of side $AD$. If the perimeter of rhombus $ABCD$ is $40$, what is the length of $OH$?\n",
    "completion": "Solution: Since the perimeter of the rhombus ABCD is 40,\\\\\ntherefore $CD=10$,\\\\\nand since H is the midpoint of side AD and point O is the midpoint of AC,\\\\\ntherefore $OH$ is the median of $\\triangle ACD$,\\\\\ntherefore $OH=\\frac{1}{2}CD=\\frac{1}{2} \\times 10=\\boxed{5}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51510278.png"
  },
  {
    "solution": "\\boxed{\\frac{5}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in right triangle $ABC$, the right-angle sides $AC=3\\text{cm}$ and $BC=4\\text{cm}$. Let $P$ and $Q$ be points moving along $AB$ and $BC$, respectively. Point $P$ moves from point $A$ towards point $B$ at a constant speed, while point $Q$ moves from point $B$ towards point $C$ at the same constant speed of $1\\text{cm}$ per second. The movement of $P$ stops as soon as $Q$ reaches point $C$. Let the movement time of $P$ and $Q$ be $t$ seconds. For what value of $t$ is $\\triangle PBQ$ an isosceles triangle with $\\angle B$ as the vertex angle?\n",
    "completion": "\\textbf{Solution:} Given that the leg $AC=3\\text{cm}$ and $BC=4\\text{cm}$,\\\\\nthus, by the Pythagorean theorem, we have $AB=\\sqrt{AC^2+BC^2}=\\sqrt{3^2+4^2}=5$,\\\\\ntherefore, $AP=t$ and $BP=5-t$, $BQ=t$,\\\\\nsince $\\triangle PBQ$ is an isosceles triangle with $\\angle B$ as the vertex angle,\\\\\nthus, $BP=BQ$, that is, $5-t=t$, solving this gives $t=\\boxed{\\frac{5}{2}}$ seconds,\\\\\ntherefore, when $t=\\boxed{\\frac{5}{2}}$ seconds, $\\triangle PBQ$ forms an isosceles triangle with $\\angle B$ as the vertex angle.",
    "image_path": "dataset/mm_math/images/MM_Math/53043122_179.png"
  },
  {
    "solution": "\\boxed{16}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in rhombus $ABCD$, $E$ and $F$ are the midpoints of $AB$ and $AC$, respectively. If $EF=2$, what is the perimeter of rhombus $ABCD$?\n",
    "completion": "\\textbf{Solution:} Given that in rhombus $ABCD$, $E$ and $F$ are the midpoints of $AB$ and $AC$ respectively, and $EF=2$,\\\\\nit follows that $BC=2EF=2\\times 2=4$. Therefore, $AB=BC=CD=AD=4$.\\\\\nHence, the perimeter of the rhombus is $4BC=4\\times 4=\\boxed{16}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51212848.png"
  },
  {
    "solution": "\\boxed{A(10,0)}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: Given that 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); on the x-axis, there is a moving point P. A perpendicular line to the x-axis is drawn from point P, intersecting the graphs of the functions $y=-\\frac{1}{2}x+b$ and $y=2x$ at points C and D, respectively. Find the equation of line AB and the coordinates of point A.\n",
    "completion": "\\textbf{Solution:} Substituting point M$(2,4)$ into $y=-\\frac{1}{2}x+b$ yields: $4=-\\frac{1}{2}\\times 2+b$. Solving for $b$ gives $b=5$. Thus, the equation of line AB is $y=-\\frac{1}{2}x+5$.\\\\\nSubstituting $y=0$ into $y=-\\frac{1}{2}x+5$ yields $x=10$, meaning the coordinates of point \\boxed{A(10,0)}.",
    "image_path": "dataset/mm_math/images/MM_Math/52251693_81.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{2}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, the vertices of $\\triangle ABC$ are on the grid points. What is the value of $\\cos B$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, we select lattice point $D$ such that $\\angle ADB=90^\\circ$, and connect $AD$ and $BD$.\\\\\nSince $AD=3$ and $BD=3$\\\\\nTherefore, $AB=\\sqrt{AD^{2}+BD^{2}}=\\sqrt{3^{2}+3^{2}}=3\\sqrt{2}$\\\\\nTherefore, $\\cos B=\\frac{BD}{AB}=\\frac{3}{3\\sqrt{2}}=\\boxed{\\frac{\\sqrt{2}}{2}}$",
    "image_path": "dataset/mm_math/images/MM_Math/53019660.png"
  },
  {
    "solution": "\\boxed{16}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, the angle bisectors of $\\angle ABC$ and $\\angle ACB$ intersect at point $F$. Through point $F$, draw $EG \\parallel BC$ intersecting $AB$ and $AC$ at points $E$ and $G$, respectively. If $BE=6$ and $CG=10$, what is the length of segment $EG$?\n",
    "completion": "\\textbf{Solution:} Since $EG\\parallel BC$,\n\nit follows that $\\angle EFB=\\angle FBC$,\n\nsince $BF$ bisects $\\angle ABC$,\n\nit follows that $\\angle EBF=\\angle FBC$,\n\ntherefore, $\\angle EBF=\\angle EFB$,\n\nhence $EF=BE$,\n\nsimilarly, $FG=GC$,\n\ntherefore, $EG=EF+FG=BE+GC=\\boxed{16}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53009908.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</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 with the $x$-axis at points $A$ and $B$ (with point $A$ on the left side of point $B$), and intersects with the $y$-axis at point $C$. Point $D$ is the symmetric point of point $C$ with respect to the axis of symmetry of the parabola. Connect $CD$, draw $DH \\perp x$-axis at point $H$ through point $D$, and draw $AE \\perp AC$ through point $A$, intersecting the extension line of $DH$ at point $E$. Under the conditions of question (2), the obtained $\\triangle CFP$ is translated along the line $AE$ to get $\\triangle C'F'P'$, and $\\triangle C'F'P'$ is folded along $C'P'$ to get $\\triangle C'P'F\"$, denoting that during the translation process, the line $F'P'$ intersects the $x$-axis at point $K$. When $\\triangle NF'F\"K$ is an isosceles triangle, directly write out the value of $OK$.\n",
    "completion": "\\textbf{Solution:} We have $OK = \\boxed{3}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51708944_114.png"
  },
  {
    "solution": "\\boxed{60^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in rectangle $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$. If $\\angle ADB=30^\\circ$, what is the measure of $\\angle AOB$?\n",
    "completion": "\\textbf{Solution:} Since ABCD is a rectangle,\\\\\nit follows that $\\angle$DAB = 90$^\\circ$.\\\\\nSince $\\angle$ADB = 30$^\\circ$,\\\\\nit follows that $\\angle$ABD = 90$^\\circ - \\angle$ADB = 60$^\\circ$.\\\\\nSince OA = OB,\\\\\nit follows that $\\triangle$ABO is an equilateral triangle,\\\\\ntherefore, $\\angle$AOB = \\boxed{60^\\circ}.",
    "image_path": "dataset/mm_math/images/MM_Math/53333671.png"
  },
  {
    "solution": "\\boxed{16}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, two congruent rectangular sheets of paper, ABCD and AECF, are overlapped as illustrated, with AB=AF, AE=BC. AE and BC intersect at point G, AD and CF intersect at point H, and $\\angle AGB=30^\\circ$, AB=2. What is the perimeter of the quadrilateral AGCH?\n",
    "completion": "\\textbf{Solution:} Given two congruent rectangular paper pieces ABCD and AECF overlapped as shown in the diagram, with AB=AF and AE=BC, and $\\angle AGB=30^\\circ$. \\\\\nTherefore, $AD\\parallel BC$ and $FC\\parallel AE$, with $\\angle B=\\angle F=90^\\circ$. \\\\\nThus, $\\angle HAG=\\angle AGB=30^\\circ$ and $\\angle FHA=\\angle HAG=30^\\circ$. \\\\\nHence, $AG=2AB$ and $AH=2AF$. \\\\\nGiven that $AB=2$, \\\\\nit follows that $AG=AH=4$. \\\\\nSince $AG\\parallel HC$ and $AH\\parallel GC$, \\\\\ntherefore, quadrilateral $AGCH$ is a parallelogram. \\\\\nBecause $AG=AH$, \\\\\nthus, quadrilateral $AGCH$ is a rhombus. \\\\\nTherefore, the perimeter of quadrilateral $AGCH$ is $\\boxed{16}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53436625.png"
  },
  {
    "solution": "\\boxed{10}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in Figure 1, it is known that point P lies on segment \\(AC\\), \\(BP = DP\\), \\(AB = 3\\), and \\(CD = 7\\). If \\(\\angle A = \\angle C = \\angle BPD = 90^\\circ\\), what is the length of \\(AC\\)?\n",
    "completion": "\\textbf{Solution:} Since $\\angle A=\\angle C=\\angle BPD=90^\\circ$\\\\\nTherefore, $\\angle APB=\\angle CDP=90^\\circ-\\angle CPD$ \\\\\nSince $BP=DP$\\\\\nTherefore, $\\triangle APB\\cong \\triangle CDP$ \\\\\nTherefore, $AB=PC=3$ and $AP=CD=7$ \\\\\nTherefore, $AC=\\boxed{10}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52827553_120.png"
  },
  {
    "solution": "\\boxed{x>20}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, the line $y=x+8$ intersects with the line $y=ax+b$ at the point $P$. Based on the graph, what is the solution set of the inequality $x+8>ax+b$?\n",
    "completion": "\\textbf{Solution:} From the given information, we can determine that the line $y=x+8$ intersects with the line $y=ax+b$ at point P(20, 25).\\\\\nBy examining the graphical representation of the functions, it is evident that for $x>20$,\\\\\nthe graph of the line $y=x+8$ is located above the graph of the line $y=ax+b$,\\\\\nmeaning the solution set for the inequality $x+8>ax+b$ regarding $x$ is: $x>20$,\\\\\ntherefore, the answer is: $\\boxed{x>20}$.",
    "image_path": "dataset/mm_math/images/MM_Math/50555839.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{5}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, it is known that $\\odot O$ is the circumcircle of $\\triangle ABC$, $AD$ is the diameter of $\\odot O$, and $CD$ is connected. If $AD=3$, $AC=2$, then what is the value of $\\cos B$?\n",
    "completion": "\\textbf{Solution:} Given that $AD$ is the diameter of the circle $\\odot O$,\\\\\nthus $\\angle ACD=90^\\circ$,\\\\\ntherefore, by the Pythagorean theorem, $CD=\\sqrt{AD^{2}-AC^{2}}=\\sqrt{3^{2}-2^{2}}=\\sqrt{5}$,\\\\\nsince $\\angle B=\\angle D$,\\\\\nthus, $\\cos B=\\cos D=\\frac{CD}{AD}=\\frac{\\sqrt{5}}{3}=\\boxed{\\frac{\\sqrt{5}}{3}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51265245.png"
  },
  {
    "solution": "\\boxed{3}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</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 $PQ$, $AQ$, and $CP$. Let the time for points $P$ and $Q$ to move be $t$ seconds. For what value of $t$ is quadrilateral $ABQP$ a rectangle?\n",
    "completion": "\\textbf{Solution:} From the given information, we have $BQ=DP=t, \\, AP=CQ=6-t$,\\\\\nIn rectangle $ABCD$, $\\angle B=90^\\circ$, $AD\\parallel BC$,\\\\\nWhen $BQ=AP$, quadrilateral $ABQP$ forms a rectangle,\\\\\n$\\therefore t=6-t$, solving for $t$ yields $t=\\boxed{3}$,\\\\\nTherefore, when $t=3$, quadrilateral $ABQP$ is a rectangle.",
    "image_path": "dataset/mm_math/images/MM_Math/53313476_110.png"
  },
  {
    "solution": "\\boxed{15-5\\sqrt{5}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC$, $\\angle A=36^\\circ$, $BD$ bisects $\\angle ABC$ and intersects $AC$ at point $D$, $CE$ bisects $\\angle ACB$ and intersects $BD$ at point $E$. If $AD= 5\\sqrt{5}-5$, what is the length of $BE$?\n",
    "completion": "\\textbf{Solution:} Given $\\because$ $\\angle A=36^\\circ$ and $AB=AC$,\\\\\n$\\therefore$ $\\angle ABC=\\angle ACB=72^\\circ$,\\\\\n$\\because$ BD bisects $\\angle ABC$ and CE bisects $\\angle ACB$,\\\\\n$\\therefore$ $\\angle ABD=\\angle DBC=36^\\circ$ and $\\angle ACE=\\angle BCE=36^\\circ$,\\\\\n$\\therefore$ $BD=AD=5\\sqrt{5}-5$ and $BE=CE=CD$,\\\\\n$\\because$ $\\angle BDC=\\angle CDE$\\\\\nand $\\angle DBC=\\angle DCE$,\\\\\n$\\therefore$ $\\triangle BDC \\sim \\triangle CDE$,\\\\\n$\\therefore$ $\\frac{DE}{CD}=\\frac{CD}{BD}$,\\\\\n$\\therefore$ $\\frac{BD-BE}{BE}=\\frac{BE}{BD}$,\\\\\nSolving this gives: $BE^2+(5\\sqrt{5}-5)BE-(5\\sqrt{5}-5)^2=0$,\\\\\nSolutions are: $BE=15-5\\sqrt{5}$ or $BE=-10$ (discard the negative value),\\\\\n$\\therefore$ $BE=\\boxed{15-5\\sqrt{5}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51155343.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, it is known that segment $AB=20$, $C$ is a point on segment $AB$, $D$ is a point on $CB$, $E$ is the midpoint of $DB$, and $DE=3$. If $C$ is the midpoint of $AB$, what is the length of $CD$?\n",
    "completion": "\\textbf{Solution:} Since $C$ is the midpoint of $AB$, and $AB=20$,\n\n$\\therefore$ $AC=CB=10$. Given also that $DB=6$,\n\n$\\therefore$ $CD=CB-DB=10-6=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/42461901_215.png"
  },
  {
    "solution": "\\boxed{2}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, within circle $O$, chord $AB=4$, point $C$ moves along $AB$. Connect $OC$, and draw $CD\\perp OC$ intersecting circle $O$ at point $D$. What is the maximum value of $CD$?\n",
    "completion": "\\textbf{Solution:} As shown in the diagram, connect $OD$,\\\\\nsince $CD \\perp OC$,\\\\\nit follows that $\\angle DCO = 90^\\circ$,\\\\\nthus $CD = \\sqrt{OD^{2} - OC^{2}} = \\sqrt{r^{2} - OC^{2}}$,\\\\\nwhen the value of $OC$ is minimal, the value of $CD$ is maximal,\\\\\n$OC \\perp AB$ when $OC$ is minimal, at this time points $D$ and $B$ coincide,\\\\\nthus $CD = CB = \\frac{1}{2}AB = 2$,\\\\\nthat is, the maximal value of $CD$ is $\\boxed{2}$,",
    "image_path": "dataset/mm_math/images/MM_Math/53009880.png"
  },
  {
    "solution": "\\boxed{45^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in rectangle ABCD, point E is the midpoint of CD, point F is a point on BC, and FC = 2BF. Connect AE, EF, and AF. If AB = 2 and AD = 3, what is the measure of $\\angle AEF$?\n",
    "completion": "\\textbf{Solution:} Since quadrilateral ABCD is a rectangle, and AD = 3, AB = 2, \\\\\nthus $\\angle B = \\angle C = 90^\\circ$, CD = AB = 2, BC = AD = 3.\\\\\nSince point E is the midpoint of CD, and FC = 2BF, \\\\\nthus CE = DE = 1, BF = 1, CF = 2.\\\\\nThus AB = CF = 2, CE = BF = 1.\\\\\nIn $\\triangle ABF$ and $\\triangle FCE$,\\\\\n$\\begin{cases} AB=CF\\\\ \\angle B = \\angle C\\\\ BF=CE \\end{cases}$, \\\\\nthus $\\triangle ABF \\cong \\triangle FCE$ (by SAS).\\\\\nThus AF = EF, $\\angle BAF = \\angle CFE$.\\\\\nSince $\\angle B = 90^\\circ$, \\\\\nthus $\\angle BAF + \\angle AFB = 90^\\circ$.\\\\\nThus $\\angle CFE + \\angle AFB = 90^\\circ$.\\\\\nThus $\\angle AFE = 180^\\circ - (\\angle CFE + \\angle AFB) = 180^\\circ - 90^\\circ = 90^\\circ$.\\\\\nThus $\\triangle AFE$ is an isosceles right triangle.\\\\\nThus $\\angle AEF = \\boxed{45^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51919593.png"
  },
  {
    "solution": "\\boxed{-4}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in Figure 1, the graph of the quadratic function $y=-\\frac{1}{2}x^{2}+bx+c$ intersects the x-axis at points $A$ and $B$, with the coordinates of point $B$ being $(1,0)$, and intersects the y-axis at point $C(0,2)$. Connect $AC$ and $BC$. What is the expression of this quadratic function and the coordinates of point $A$?\n",
    "completion": "\\textbf{Solution:} Solve by substituting B(1, 0) and C(0, 2) into $y=-\\frac{1}{2}x^{2}+bx+c$,\\\\\n$\\therefore \\left\\{\\begin{array}{l}\n-\\frac{1}{2}+b+c=0\\\\\nc=2\n\\end{array}\\right.$,\\\\\nwe get $\\left\\{\\begin{array}{l}\nb=-\\frac{3}{2}\\\\\nc=2\n\\end{array}\\right.$,\\\\\n$\\therefore y=-\\frac{1}{2}x^{2}-\\frac{3}{2}x+2$,\\\\\nLet $y=0$, then $-\\frac{1}{2}x^{2}-\\frac{3}{2}x+2=0$,\\\\\nwe find $x=-4$ or $x=1$,\\\\\n$\\therefore A(\\boxed{-4},0)$;",
    "image_path": "dataset/mm_math/images/MM_Math/52550318_135.png"
  },
  {
    "solution": "\\boxed{15^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in the circle $\\odot O$, the diameter $CD$ is perpendicular to the chord $AB$ at point $E$, and $OB$ and $BC$ are connected. Given that the radius of $\\odot O$ is $2$ and $AB=2$, what is the measure of $\\angle BCD$?\n",
    "completion": "\\textbf{Solution:} Given that diameter $CD$ is perpendicular to chord $AB$ at point $E$, and $AB=2$,\\\\\nit follows that $\\angle OEB=90^\\circ$, and $BE=1$,\\\\\nsince $OB=2$,\\\\\nthus $\\angle BOE=30^\\circ$,\\\\\ntherefore $\\angle BCD=\\boxed{15^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52874515.png"
  },
  {
    "solution": "\\boxed{6\\sqrt{3}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the diagram, it is known that points $A$, $B$, $C$ are on the circle $\\odot O$, with $\\angle ABC = 30^\\circ$. The minor arc $\\overset{\\frown}{BC}$ is folded along the straight line $CB$ to intersect chord $AB$ at point $D$. If $DB = 7$ and $AD = 4$, what is the length of $BC$?\n",
    "completion": "\\textbf{Solution:} Let point D be on the circle $\\odot O$ and its corresponding point be E. Connect CE, BE, CD, AC, and draw CF $\\perp$ AD at point F, as shown in the figure,\\\\\n$\\because$ quadrilateral ABEC is inscribed in $\\odot O$,\\\\\n$\\therefore$ $\\angle A+\\angle E=180^\\circ$,\\\\\n$\\because$ the corresponding point of D on $\\odot O$ is E,\\\\\n$\\therefore$ according to the property of folding: $\\angle BEC=\\angle BDC$,\\\\\n$\\because$ $\\angle BDC+\\angle CDA=180^\\circ$,\\\\\n$\\therefore$ $\\angle E+\\angle CDA=180^\\circ$,\\\\\n$\\because$ $\\angle A+\\angle E=180^\\circ$,\\\\\n$\\therefore$ $\\angle A=\\angle ADC$,\\\\\n$\\therefore$ $\\triangle ACD$ is an isosceles triangle,\\\\\n$\\because$ $CF\\perp AD$, $AD=4$,\\\\\n$\\therefore$ $AF=FD=\\frac{1}{2}AD=2$,\\\\\n$\\because$ $BD=7$,\\\\\n$\\therefore$ $BF=BD+DF=9$,\\\\\n$\\because$ $CF\\perp AD$,\\\\\n$\\therefore$ $\\triangle CFB$ is a right triangle,\\\\\n$\\because$ $\\angle ABC=30^\\circ$,\\\\\n$\\therefore$ in $Rt\\triangle CFB$, $CF=\\frac{1}{2}BC$,\\\\\n$\\because$ in $Rt\\triangle CFB$, $CF^2+BF^2=BC^2$,\\\\\n$\\therefore$ $\\left(\\frac{1}{2}BC\\right)^2+9^2=BC^2$,\\\\\n$\\therefore$ $BC=\\boxed{6\\sqrt{3}}$, (negative values are discarded).",
    "image_path": "dataset/mm_math/images/MM_Math/52933561.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle BAC=90^\\circ$, $\\angle B=45^\\circ$. A line through point A such that $AD\\parallel BC$ is drawn, and point D is to the right of point A. Points P and Q are located on the rays $AD$ and $CB$ respectively. Point E is a point on the line segment $CQ$, and $CQ=2AP$. Let $AP=x$, $CE=y$, then $y=2x-2$. When point Q is the midpoint of $BC$, $y=3$. There exists a value of x such that the quadrilateral with vertices A, B, E, and P is a parallelogram. Find the value of x.\n",
    "completion": "\\textbf{Solution:} Existence is proven as follows:\n\n(1) As shown in the figure below, when points Q, E are on the line segment BC, if the quadrilateral formed by vertices A, B, E, P is a parallelogram, then $AP=BE=x$,\\\\\nsince $CE=y=2x-2$,\\\\\nthus $BE=BC-CE=10-(2x-2)=12-2x$,\\\\\ntherefore $x=12-2x$,\\\\\nsolving this gives $x=\\boxed{4}$;\\\\\n(2) As shown in the figure below, when points Q, E are on the extension of the line segment CB, if the quadrilateral formed by vertices A, B, E, P is a parallelogram, then $AP=BE=x$,\\\\\nsince $CE=y=2x-2$,\\\\\nthus $BE=CE-BC=2x-2-10=2x-12$,\\\\\ntherefore $x=2x-12$,\\\\\nsolving this gives $x=\\boxed{12}$.\n\nIn conclusion, when the quadrilateral formed by vertices A, B, E, P is a parallelogram, $x=\\boxed{4}$ or \\boxed{12}.",
    "image_path": "dataset/mm_math/images/MM_Math/53498546_74.png"
  },
  {
    "solution": "\\boxed{\\sqrt{5}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: Quadrilateral $ABCD$ is a rectangle, and quadrilateral $AEFG$ is a square, with $AD > AE$, point $E$ is on the left side of segment $AD$. Connect $DE$ and $BG$. As shown in Figure 1, if point $F$ is on side $AD$ and $AD=3$, $AE= \\sqrt{2}$, find the length of $DE$.\n",
    "completion": "\\textbf{Solution:} Let the intersection of EG and AF be point H, as shown in the figure,\\\\\n$\\because$ Quadrilateral AEFG is a square\\\\\n$\\therefore EG\\perp AF$, $EH=HG$, $FH=AF$, $AF=EG$\\\\\n$\\therefore FH=EH=HG=AH=\\frac{1}{2}EG=\\frac{1}{2}FA$\\\\\nIn $\\triangle AEF$,\\\\\n$AF^{2}=EF^{2}+AE^{2}$\\\\\n$\\therefore AF=\\sqrt{2^{2}+2^{2}}=2$\\\\\n$\\therefore EH=AH=1$\\\\\nIn $\\triangle DHE$,\\\\\n$DE^{2}=EH^{2}+DH^{2}$\\\\\nThat is, $DE^{2}=EH^{2}+(AD-AH)^{2}$\\\\\n$\\therefore DE=\\sqrt{1^{2}+(3-1)^{2}}=\\sqrt{5}=\\boxed{\\sqrt{5}}$",
    "image_path": "dataset/mm_math/images/MM_Math/52050232_102.png"
  },
  {
    "solution": "\\boxed{20^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in $ \\triangle ABC$, D is a point on $BC$, $AD=BD$, $\\angle B=40^\\circ$, and $\\triangle ADE$ is symmetrical to $\\triangle ADB$ with respect to $AD$. What is the measure of $\\angle CDE$?\n",
    "completion": "\\textbf{Solution:} Since $AD=BD$ and $\\angle B=40^\\circ$, \\\\\nit follows that $\\angle B=40^\\circ=\\angle BAD$, \\\\\ntherefore, $\\angle ADB=180^\\circ-\\left(\\angle BAD+\\angle B\\right)=100^\\circ$ and $\\angle ADC=\\angle B+\\angle BAD=80^\\circ$, \\\\\nsince $\\triangle ADE$ is symmetric to $\\triangle ADB$ with respect to $AD$, \\\\\nit follows that $\\angle ADE=\\angle ADB=100^\\circ$, \\\\\nthus, $\\angle CDE=\\angle ADE-\\angle ADC=100^\\circ-80^\\circ=\\boxed{20^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52978717.png"
  },
  {
    "solution": "\\boxed{30^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in $ \\triangle ABC$, $D$ is a point on side $BC$, with $AB=AD=DC$. If $ \\angle B=60^\\circ$, find the measure of $ \\angle C$.\n",
    "completion": "\\textbf{Solution:} Since $AB=AD, \\angle B=60^\\circ$,\\\\\nit follows that $\\triangle ABD$ is an equilateral triangle,\\\\\ntherefore $\\angle ADB=60^\\circ$,\\\\\nsince $AD=DC$,\\\\\nit implies that $\\angle C=\\angle DAC=\\frac{1}{2}\\angle ADB=\\boxed{30^\\circ}$",
    "image_path": "dataset/mm_math/images/MM_Math/53044190_143.png"
  },
  {
    "solution": "\\boxed{7}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</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. It is known that $AM = \\frac{5}{4}BC = 5\\,\\text{cm}$. What is the length of MN in centimeters?\n",
    "completion": "\\textbf{Solution:} Since $M$ is the midpoint of segment $AC$,\\\\\n$\\therefore AM=CM=5\\text{cm}$, $BC=4\\text{cm}$.\\\\\nAlso, since $N$ is the midpoint of segment $BC$,\\\\\n$\\therefore CN=\\frac{1}{2}BC=2\\text{cm}$,\\\\\n$\\therefore MN=MC+CN=\\boxed{7}\\text{cm}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53365756.png"
  },
  {
    "solution": "\\boxed{\\frac{\\sqrt{409}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, there is a crescent-shaped pond in a park, with five decorative lamps labeled A, B, C, D, and E along the edge of the pond. To estimate the size of the pond, observers at points A and D noticed that points A, E, C and D, E, B are on the same straight line. Walking in the direction of AD to point F, it was found that $\\angle AFC=90^\\circ$. It was measured that $AD=9.6$ meters, $AE=DE=8$ meters, $DF=2.4$ meters. What is the radius of the circle containing $\\overset{\\frown}{AED}$? What is the radius of the circle containing $\\overset{\\frown}{ACD}$?\n",
    "completion": "\\textbf{Solution:} Draw line EG $\\perp$ AD at point G through point E. Thus, the center of the circle passing through points A, E, D must lie on EG. As shown in the figure, take the center as O, and connect AO. Since $AE=DE=8$ meters, AD=9.6 meters, therefore $AG=$ \\(\\frac{1}{2}AD=4.8\\) meters, $EG=$ $\\sqrt{AE^{2}-AG^{2}}=\\sqrt{8^{2}-4.8^{2}}=6.4$ meters. Let the radius of the circle containing $\\overset{\\frown}{AED}$ be r meters, then $AO=EO=r$ meters, $OG=6.4−r$ (meters). In $\\triangle AOG$, $AG^{2}$+$OG^{2}=AO^{2}$, that is $4.8^{2}+(6.4−r)^{2}=r^{2}$. Solving this yields: $r=\\boxed{5}$ meters, that is, the radius is 5 meters. Extending GE to intersect the larger circle at point H, the center of the containing circle must lie on HG. As shown in the figure, let the center of the large circle be $O_{1}$, make $O_{1}M\\perp AC$ at point M, and connect $O_{1}C$. Since AD=9.6 meters, DF=2.4 meters, therefore $AF=9.6+2.4=12$ meters. Since $ACF=90^\\circ$, therefore $EG\\parallel CF$. Therefore $\\frac{AG}{AF}=\\frac{AE}{AC}$, that is, $\\frac{4.8}{12}=\\frac{8}{AC}$. Therefore $AC=20$ meters. Therefore $AM=CM=10$ meters, $EM=10−8=2$ meters. Since $\\triangle AEG\\sim \\triangle O_{1}EM$, therefore $\\frac{EM}{O_{1}M}=\\frac{EG}{AG}$, that is, $\\frac{2}{O_{1}M}=\\frac{6.4}{4.8}$. Therefore $O_{1}M=\\frac{3}{2}$. Therefore $O_{1}C=\\sqrt{O_{1}M^{2}+MC^{2}}=\\sqrt{\\left(\\frac{3}{2}\\right)^{2}+10^{2}}=\\boxed{\\frac{\\sqrt{409}}{2}}$.",
    "image_path": "dataset/mm_math/images/MM_Math/55497792.png"
  },
  {
    "solution": "\\boxed{12cm}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, the two diagonals of rectangle ABCD intersect at point O, where $ \\angle AOB=60^\\circ$ and $ AB=6$. What is the length of the diagonals of the rectangle?\n",
    "completion": "\\textbf{Solution:} Given rectangle ABCD, \\\\\nit follows that AC = BD, OA = OC, OD = OB, \\\\\ntherefore, OA = OB, \\\\\nsince $\\angle AOB = 60^\\circ$, \\\\\nhence $\\triangle AOB$ is an equilateral triangle, \\\\\ntherefore OA = OB = AB = 6cm, \\\\\nhence AC = BD = 2$\\times$6cm = \\boxed{12cm}, \\\\",
    "image_path": "dataset/mm_math/images/MM_Math/51863571.png"
  },
  {
    "solution": "\\boxed{50^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, it is known that $\\angle AOB = 80^\\circ$, $OC$ is the bisector of $\\angle AOB$, $OD$ is the bisector of $\\angle BOC$. If $\\angle COE = \\frac{1}{4}\\angle COB$, what is the degree measure of $\\angle AOE$?\n",
    "completion": "\\textbf{Solution:} Given from (1) that $\\angle COB=40^\\circ$,\\\\\n$\\therefore \\angle COE=10^\\circ$,\\\\\nwhen OE is above OC, then $\\angle AOE=\\angle AOC-\\angle COE=40^\\circ-10^\\circ=\\boxed{30^\\circ}$;\\\\\nwhen OE is below OC, then $\\angle AOE=\\angle AOC+\\angle COE=40^\\circ+10^\\circ=\\boxed{50^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52796437_118.png"
  },
  {
    "solution": "\\boxed{125^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in $\\triangle ABC$, $AB=AC$, $\\angle BAC=70^\\circ$, $\\odot O$ is the circumcircle of $\\triangle ABC$, point D is on the minor arc $\\overset{\\frown}{AC}$. What is the degree measure of $\\angle D$?\n",
    "completion": "\\textbf{Solution:} Given that AB=AC, and $\\angle$BAC=$70^\\circ$, \\\\\nthus $\\angle$B=$\\angle$ACB=$55^\\circ$, \\\\\nsince quadrilateral ABCD is inscribed in circle O, \\\\\ntherefore $\\angle$D=$180^\\circ$−$\\angle$B=$\\boxed{125^\\circ}$,",
    "image_path": "dataset/mm_math/images/MM_Math/51368322.png"
  },
  {
    "solution": "\\boxed{y=\\frac{5}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</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 with the hyperbola $y=\\frac{m}{x}$ (where $x<0$) at point $A(-1,n)$. The line segment $OA$ is drawn. Find the analytic expression of the line and the hyperbola?\n",
    "completion": "\\textbf{Solution:} By substituting point $C$ into $y=x+b$, we get $b=-4$,\\\\\ntherefore $y=x-4$;\\\\\nsubstituting point $A$ into $y=x-4$, we find $n=-5$,\\\\\ntherefore $A(-1, -5)$,\\\\\nthus $m=-1\\times (-5)=5$,\\\\\ntherefore $y=\\frac{5}{x}$\\\\\ntherefore, the equations of the line and the hyperbola are \\boxed{y=x-4} and \\boxed{y=\\frac{5}{x}}, respectively.",
    "image_path": "dataset/mm_math/images/MM_Math/51379219_203.png"
  },
  {
    "solution": "\\boxed{23^\\circ}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, it is known that CD is the diameter of the circle $\\odot O$, and the chord DE passing through point D is parallel to the radius OA. If the degree of arc CE is $92^\\circ$, what is the degree of $\\angle C$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, connect OE, \\\\\n$\\because$ the degree of arc CE is $92^\\circ$, \\\\\n$\\therefore$ $\\angle COE = 92^\\circ$, \\\\\n$\\therefore$ $\\angle CDE = \\frac{1}{2}\\angle COE = 46^\\circ$, \\\\\n$\\because$ OA $\\parallel$ DE, \\\\\n$\\therefore$ $\\angle AOD = \\angle CDE = 46^\\circ$, \\\\\n$\\therefore$ $\\angle C = \\frac{1}{2}\\angle AOD = \\boxed{23^\\circ}$.",
    "image_path": "dataset/mm_math/images/MM_Math/51446225.png"
  },
  {
    "solution": "\\boxed{\\frac{4}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the graph of the linear function $y = -x + 6$ is denoted as ${l}_{1}$, which intersects the $x$ and $y$ axes at points A and B, respectively. The graph of the direct proportion function, denoted ${l}_{2}$, intersects ${l}_{1}$ at point C$(m,5)$. A line perpendicular to the $x$ axis, $x = a$, intersects lines ${l}_{1}$ and ${l}_{2}$ at points P and Q, respectively. If the length of segment PQ is 2, find the value of $a$.\n",
    "completion": "\\textbf{Solution:} When $x=a$, it intersects with line $l_1$ at point $P(a, 6-a)$ and with line $l_2$ at point $Q(a, 5a)$. The distance PQ= $\\left|5a-(6-a)\\right|=6\\left|a-1\\right|=2$, \\\\\n$\\therefore \\left|a-1\\right|=\\frac{1}{3}$, \\\\\n$\\therefore a-1=\\pm \\frac{1}{3}$, \\\\\n$a=\\boxed{\\frac{4}{3}}$ or $a=\\frac{2}{3}$",
    "image_path": "dataset/mm_math/images/MM_Math/50555762_82.png"
  },
  {
    "solution": "\\boxed{8cm}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, it is known that in $\\triangle ABC$, $AC=6\\,cm$, and points D and E are the midpoints of AC and BC, respectively. DE is connected, and there is a point F on DE such that $EF=1\\,cm$. AF and CF are connected. If $AF \\perp CF$, then what is the length of $AB$ in $cm$?\n",
    "completion": "Solution: Since $AF\\perp CF$ and point D is the midpoint of AC,\\\\\n$\\therefore DF=\\frac{1}{2}AC$\\\\\nGiven that $AC=6cm$\\\\\n$\\therefore DF=3cm$\\\\\nSince points D and E are the midpoints of AC and BC, respectively,\\\\\n$\\therefore DE=\\frac{1}{2}AB$\\\\\nGiven that $EF=1cm$\\\\\n$\\therefore DE=EF+FD=4cm$\\\\\n$\\therefore AB=2DE=\\boxed{8cm}$",
    "image_path": "dataset/mm_math/images/MM_Math/52422259.png"
  },
  {
    "solution": "\\boxed{4\\pi}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown, in $\\triangle ABC$, $AB=6$. After rotating $\\triangle ABC$ $40^\\circ$ counterclockwise around point $A$, we get $\\triangle ADE$. The path traced by point $B$ is $\\overset{\\frown}{BD}$. What is the area of the shaded region in the figure?\n",
    "completion": "\\textbf{Solution:} From the property of rotation, we know that $S_{\\triangle ADE} = S_{\\triangle ABC}$,\\\\\nthus, the area of the shaded part $= S_{\\triangle ADE} + S_{\\text{sector} DAB} - S_{\\triangle ABC}$\\\\\n$= S_{\\text{sector} DAB}$\\\\\n$= \\frac{40\\pi \\times 6^2}{360}$\\\\\n$= \\boxed{4\\pi}$,",
    "image_path": "dataset/mm_math/images/MM_Math/51684405.png"
  },
  {
    "solution": "\\boxed{14}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, a drone interest group is carrying out activities on the playground. At that moment, the drone is at point D, $20\\sqrt{3}$ meters above the ground. The drone measures the depression angle from the controller at point A to be $30^\\circ$, and the depression angle from the top of the teaching building at point C to be $45^\\circ$. Additionally, through manual measurement, the horizontal distance between the controller A and the teaching building along $BC$ is found to be 80 meters. What is the height of the teaching building $BC$? (Note: Points A, B, C, and D are all in the same plane, reference data $\\sqrt{3} \\approx 1.7$)\n",
    "completion": "\\textbf{Solution:} Construct $DE\\perp AB$ at point E, and $CF\\perp DE$ at point F, \\\\\naccording to the problem, $DE=20\\sqrt{3}$, $\\angle A=30^\\circ$, $\\angle DCF=45^\\circ$, \\\\\nin $Rt\\triangle ADE$, $\\angle AED=90^\\circ$, \\\\\ntherefore, $\\tan30^\\circ=\\frac{DE}{AE}=\\frac{\\sqrt{3}}{3}$, \\\\\ntherefore, $AE=\\sqrt{3}DE=60$ (meters), \\\\\nsince $AB=80$ meters, \\\\\ntherefore, $BE=80-60=20$ (meters), \\\\\nsince $\\angle FEB=\\angle CBE=\\angle CFE=90^\\circ$, \\\\\nsince quadrilateral $BCFE$ is a rectangle, \\\\\ntherefore, $CF=BE=20$ meters, \\\\\nin $Rt\\triangle DCF$, $\\angle DFC=90^\\circ$, \\\\\ntherefore, $\\angle CDF=\\angle DCF=45^\\circ$, \\\\\ntherefore, $CF=DF=20$, \\\\\ntherefore, $BC=EF=DE-DF=20\\sqrt{3}-20\\approx 20\\times 1.7-20=\\boxed{14}$ (meters).",
    "image_path": "dataset/mm_math/images/MM_Math/53019950.png"
  },
  {
    "solution": "\\boxed{12}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, points C and D are on line segment AB, where D is the midpoint of line segment AB, $AC = \\frac{1}{3}AD$, and $CD = 4\\, \\text{cm}$. What is the length of line segment AB?\n",
    "completion": "\\textbf{Solution:} Since $AC = \\frac{1}{3}AD$ and $CD = 4cm$,\\\\\nit follows that $CD = AD - AC = AD - \\frac{1}{3}AD = \\frac{2}{3}AD = 4$,\\\\\nhence $AD = 6$,\\\\\ngiven that $D$ is the midpoint of segment $AB$,\\\\\nit follows that $AB = 2AD = 12$,\\\\\nthus $AB = \\boxed{12}$cm.",
    "image_path": "dataset/mm_math/images/MM_Math/52580147.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in $\\triangle ABC$, $\\angle C=90^\\circ$, the angle bisector of $\\angle BAC$ intersects $BC$ at point D. Point O is on $AB$, and the circle with point O as the center and $OA$ as the radius just passes through point D and intersects $AC$, $AB$ at points E, F respectively. If $BD=2\\sqrt{5}$ and $BF=2$, what is the radius of circle $O$?\n",
    "completion": "\\textbf{Solution:} In the right-angled triangle $\\triangle OBD$, let the radius be $r$. According to the Pythagorean theorem, we have\n\\[r^{2}+BD^{2}=(r+BF)^{2},\\]\n\\[\\because BD=2\\sqrt{5}, BF=2,\\]\n\\[\\therefore r^{2}+(2\\sqrt{5})^{2}=(r+2)^{2},\\]\nsolving this gives $r=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53066245_64.png"
  },
  {
    "solution": "\\boxed{6}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, $AB$ is the diameter of $\\odot O$, $BC$ is the chord of $\\odot O$, the radius $OD \\perp BC$ with the foot of the perpendicular being $E$. Given that $BC=6\\sqrt{3}$ and $DE=3$, find the radius of $\\odot O$?\n",
    "completion": "\\textbf{Solution:} Let the radius of $\\odot O$ be $r$,\\\\\nsince radius $OD\\perp BC$,\\\\\nthen it follows that $CE=\\frac{1}{2}BC=3\\sqrt{3}$,\\\\\nsince $OC^{2}=CE^{2}+OE^{2}$,\\\\\nthen $r^{2}=(r-3)^{2}+(3\\sqrt{3})^{2}$,\\\\\nthus $r=\\boxed{6}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53009657_141.png"
  },
  {
    "solution": "\\boxed{4}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, in the Cartesian coordinate system, the vertex $A$ of $\\square OABC$ lies on the inverse proportional function $y=\\frac{1}{x}$, the vertex $B$ lies on the inverse proportional function $y=\\frac{5}{x}$, and point $C$ is located on the positive half of the $x$-axis. What is the area of $\\square OABC$?\n",
    "completion": "\\textbf{Solution:} As shown in the figure, draw perpendiculars from points A and B to the x-axis, with the feet of the perpendiculars being M and N, respectively, and connect OB. \\\\\nSince quadrilateral ABCO is a parallelogram, \\\\\nit follows that $OA=BC$ and $AB\\parallel OC$, \\\\\nwhich means $AM=BN$, \\\\\nIn $\\triangle AMO$ and $\\triangle BNC$ right-angled triangles, \\\\\nsince $OA=CB$ and $AM=BN$, \\\\\nit leads to $\\triangle AMO \\cong \\triangle BNC$ (HL criterion), \\\\\nAlso, since point A lies on the inverse proportion function $y=\\frac{1}{x}$, \\\\\nit implies $Area_{\\triangle AOM}=\\frac{1}{2}\\times 1=\\frac{1}{2}=Area_{\\triangle BNC}$, \\\\\nSince point B lies on the inverse proportion function $y=\\frac{5}{x}$, \\\\\nit implies $Area_{\\triangle BON}=\\frac{1}{2}\\times 5=\\frac{5}{2}$, \\\\\nTherefore, $Area_{\\triangle OBC}=Area_{\\triangle BON}-Area_{\\triangle BNC}=\\frac{5}{2}-\\frac{1}{2}=2=\\frac{1}{2}Area_{\\square ABCO}$, \\\\\nThus, $Area_{\\square ABCO}=\\boxed{4}$.",
    "image_path": "dataset/mm_math/images/MM_Math/52551555.png"
  },
  {
    "solution": "\\boxed{(2, 2)}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, the side length of the square paper ABCD is 3, with points E and F located on BC and CD, respectively. AB and AD are folded along AE and AF, respectively, so that points B and D just coincide at point G. It is known that $3BE=BC$. If points P and Q are moving points on segments AG and AF, respectively, what is the minimum value of $EP+PQ$, and what are the coordinates of point Q at this time?\n",
    "completion": "\\textbf{Solution}: As shown in the figure, connect EQ,\n\n$\\because EP+PQ \\ge EQ$,\n\n$\\therefore$ when points Q, P, and E are collinear, $EP+PQ$ reaches its minimum value EQ,\n\nFurthermore, since the perpendicular from a point to a line is the shortest segment, when $EQ \\perp AF$, EQ reaches its minimum value, that is, $EP+PQ$ reaches its minimum value,\n\nFrom (1) we have $EF=EG+GF=\\frac{3}{2}+1=\\frac{5}{2}$, and by the property of folding, we get AG=AB=3, $\\angle AGE=\\angle AGF=\\angle ABC=90^\\circ$,\n\n$\\therefore AF=\\sqrt{AG^2+GF^2}=\\frac{3\\sqrt{5}}{2}$,\n\n$\\because S_{\\triangle AEF}=\\frac{1}{2}AG\\cdot EF=\\frac{1}{2}AF\\cdot QE$,\n\n$\\therefore QE=\\frac{AG\\cdot EF}{AF}=\\sqrt{5}$,\n\nSuppose the equation of line AF is $y=mx+n$,\n\n$\\therefore \\left\\{\\begin{array}{l}3k+b=\\frac{3}{2} \\\\ b=3 \\end{array}\\right.$,\n\n$\\therefore \\left\\{\\begin{array}{l}k=-\\frac{1}{2} \\\\ b=3 \\end{array}\\right.$,\n\n$\\therefore$ the equation of line AF is $y=-\\frac{1}{2}x+3$;\n\nSuppose the coordinates of point Q are $(t, -\\frac{1}{2}t+3)$,\n\n$\\therefore QE^2=(t-1)^2+\\left(-\\frac{1}{2}t+3\\right)^2=5$,\n\nSolving this yields $t=2$,\n\n$\\therefore$ the coordinates of point Q are $(2, 2)$,\n\n$\\therefore$ the minimum value of $EP+PQ$ is $\\boxed{\\sqrt{5}}$, at this moment, the coordinates of point Q are $\\boxed{(2, 2)}$.",
    "image_path": "dataset/mm_math/images/MM_Math/53182535_152.png"
  },
  {
    "solution": "\\boxed{\\frac{1}",
    "prompt": "Solve the problem and output the answer in the format of <answer>Your answer here</answer>. \\n Question: As shown in the figure, $AB$ is the diameter of the semicircle $O$, and $C,D$ are two points on the semicircle, satisfying $\\angle ADC=120^\\circ$ and $BC=1$. What is the length of $BC$?\n",
    "completion": "\\[\n\\textbf{Solution:} \\text{As shown in the figure, connect OC}\\\\\n\\because \\angle ADC=120^\\circ, \\\\\n\\therefore \\angle ABC=60^\\circ, \\\\\n\\because \\text{OB}=\\text{OC}, \\\\\n\\therefore \\angle OCB=\\angle OBC=\\angle B=60^\\circ, \\\\\n\\text{OB}=\\text{OC}=\\text{BC}=1, \\\\\n\\therefore \\text{the length of } \\overset{\\frown}{BC} \\text{ is } \\frac{60^\\circ \\pi \\times 1}{180}=\\boxed{\\frac{1}{3}\\pi}\n\\]",
    "image_path": "dataset/mm_math/images/MM_Math/52872329.png"
  }
]